Theoretical Fluid Mechanics

Richard Fitzpatrick

Theoretical Fluid Mechanics

Contents

1 Mathematical Models of Fluid Motion 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 What is a Fluid? . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Volume and Surface Forces . . . . . . . . . . . . . . . . . . . . . 21.4 General Properties of Stress Tensor . . . . . . . . . . . . . . . . . 41.5 Stress Tensor in a Static Fluid . . . . . . . . . . . . . . . . . . . . 61.6 Stress Tensor in a Moving Fluid . . . . . . . . . . . . . . . . . . 71.7 Viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.8 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . 101.9 Mass Conservation . . . . . . . . . . . . . . . . . . . . . . . . . 101.10 Convective Time Derivative . . . . . . . . . . . . . . . . . . . . . 111.11 Momentum Conservation . . . . . . . . . . . . . . . . . . . . . . 121.12 Navier-Stokes Equation . . . . . . . . . . . . . . . . . . . . . . . 141.13 Energy Conservation . . . . . . . . . . . . . . . . . . . . . . . . 141.14 Equations of Incompressible Fluid Flow . . . . . . . . . . . . . . 171.15 Equations of Compressible Fluid Flow . . . . . . . . . . . . . . . 181.16 Dimensionless Numbers in Incompressible Flow . . . . . . . . . . 191.17 Dimensionless Numbers in Compressible Flow . . . . . . . . . . 201.18 Fluid Equations in Cartesian Coordinates . . . . . . . . . . . . . 241.19 Fluid Equations in Cylindrical Coordinates . . . . . . . . . . . . . 251.20 Fluid Equations in Spherical Coordinates . . . . . . . . . . . . . 271.21 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2 Hydrostatics 332.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.2 Hydrostatic Pressure . . . . . . . . . . . . . . . . . . . . . . . . 332.3 Buoyancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.4 Equilibrium of Floating Bodies . . . . . . . . . . . . . . . . . . . 352.5 Vertical Stability of Floating Bodies . . . . . . . . . . . . . . . . 372.6 Angular Stability of Floating Bodies . . . . . . . . . . . . . . . . 372.7 Determination of Metacentric Height . . . . . . . . . . . . . . . . 392.8 Energy of a Floating Body . . . . . . . . . . . . . . . . . . . . . 432.9 Curve of Buoyancy . . . . . . . . . . . . . . . . . . . . . . . . . 442.10 Rotational Hydrostatics . . . . . . . . . . . . . . . . . . . . . . . 49

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2.11 Equilibrium of a Rotating Liquid Body . . . . . . . . . . . . . . . 512.12 Maclaurin Spheroids . . . . . . . . . . . . . . . . . . . . . . . . 542.13 Jacobi Ellipsoids . . . . . . . . . . . . . . . . . . . . . . . . . . 582.14 Roche Ellipsoids . . . . . . . . . . . . . . . . . . . . . . . . . . 622.15 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

3 Surface Tension 733.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733.2 Young-Laplace Equation . . . . . . . . . . . . . . . . . . . . . . 743.3 Spherical Interfaces . . . . . . . . . . . . . . . . . . . . . . . . . 763.4 Capillary Length . . . . . . . . . . . . . . . . . . . . . . . . . . 773.5 Angle of Contact . . . . . . . . . . . . . . . . . . . . . . . . . . 783.6 Jurin’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803.7 Capillary Curves . . . . . . . . . . . . . . . . . . . . . . . . . . 813.8 Axisymmetric Soap-Bubbles . . . . . . . . . . . . . . . . . . . . 863.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

4 Incompressible Inviscid Flow 954.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 954.2 Streamlines, Stream Tubes, and Stream Filaments . . . . . . . . . 954.3 Bernoulli’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 964.4 Euler Momentum Theorem . . . . . . . . . . . . . . . . . . . . . 984.5 d’Alembert’s Paradox . . . . . . . . . . . . . . . . . . . . . . . . 984.6 Flow Through an Orifice . . . . . . . . . . . . . . . . . . . . . . 1004.7 Sub-Critical and Super-Critical Flow . . . . . . . . . . . . . . . . 1024.8 Flow over Shallow Bump . . . . . . . . . . . . . . . . . . . . . . 1034.9 Stationary Hydraulic Jumps . . . . . . . . . . . . . . . . . . . . . 1054.10 Tidal Bores . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1084.11 Flow over a Broad-Crested Weir . . . . . . . . . . . . . . . . . . 1104.12 Vortex Lines, Vortex Tubes, and Vortex Filaments . . . . . . . . . 1114.13 Circulation and Vorticity . . . . . . . . . . . . . . . . . . . . . . 1124.14 Kelvin Circulation Theorem . . . . . . . . . . . . . . . . . . . . 1134.15 Irrotational Flow . . . . . . . . . . . . . . . . . . . . . . . . . . 1144.16 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

5 Two-Dimensional Incompressible Inviscid Flow 1195.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1195.2 Two-Dimensional Flow . . . . . . . . . . . . . . . . . . . . . . . 1195.3 Velocity Potentials and Stream Functions . . . . . . . . . . . . . 1225.4 Two-Dimensional Uniform Flow . . . . . . . . . . . . . . . . . . 1235.5 Two-Dimensional Sources and Sinks . . . . . . . . . . . . . . . . 1245.6 Two-Dimensional Vortex Filaments . . . . . . . . . . . . . . . . 1265.7 Two-Dimensional Irrotational Flow in Cylindrical Coordinates . . 1305.8 Flow Past a Cylindrical Obstacle . . . . . . . . . . . . . . . . . . 1325.9 Motion of a Submerged Cylinder . . . . . . . . . . . . . . . . . . 136

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5.10 Inviscid Flow Past a Semi-Infinite Wedge . . . . . . . . . . . . . 1415.11 Inviscid Flow Over a Semi-Infinite Wedge . . . . . . . . . . . . . 1435.12 Two-Dimensional Jets . . . . . . . . . . . . . . . . . . . . . . . . 1445.13 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

6 Two-Dimensional Potential Flow 1496.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1496.2 Complex Functions . . . . . . . . . . . . . . . . . . . . . . . . . 1496.3 Cauchy-Riemann Relations . . . . . . . . . . . . . . . . . . . . . 1506.4 Complex Velocity Potential . . . . . . . . . . . . . . . . . . . . . 1526.5 Complex Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . 1536.6 Method of Images . . . . . . . . . . . . . . . . . . . . . . . . . . 1546.7 Conformal Maps . . . . . . . . . . . . . . . . . . . . . . . . . . 1606.8 Schwarz–Christoffel Theorem . . . . . . . . . . . . . . . . . . . 1656.9 Free Streamline Theory . . . . . . . . . . . . . . . . . . . . . . . 1686.10 Complex Line Integrals . . . . . . . . . . . . . . . . . . . . . . . 1796.11 Blasius Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 1816.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

7 Axisymmetric Incompressible Inviscid Flow 1937.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1937.2 Axisymmetric Flow . . . . . . . . . . . . . . . . . . . . . . . . . 1937.3 Stokes Stream Function . . . . . . . . . . . . . . . . . . . . . . . 1937.4 Axisymmetric Velocity Fields . . . . . . . . . . . . . . . . . . . 1957.5 Axisymmetric Irrotational Flow in Spherical Coordinates . . . . . 1967.6 Uniform Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1977.7 Point Sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1987.8 Dipole Point Sources . . . . . . . . . . . . . . . . . . . . . . . . 1997.9 Flow Past a Spherical Obstacle . . . . . . . . . . . . . . . . . . . 2007.10 Motion of a Submerged Sphere . . . . . . . . . . . . . . . . . . . 2037.11 Conformal Maps . . . . . . . . . . . . . . . . . . . . . . . . . . 2057.12 Flow Around a Submerged Oblate Spheroid . . . . . . . . . . . . 2107.13 Flow Around a Submerged Prolate Spheroid . . . . . . . . . . . . 2137.14 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

8 Incompressible Boundary Layers 2198.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2198.2 No Slip Condition . . . . . . . . . . . . . . . . . . . . . . . . . . 2198.3 Boundary Layer Equations . . . . . . . . . . . . . . . . . . . . . 2208.4 Self-Similar Boundary Layers . . . . . . . . . . . . . . . . . . . 2248.5 Boundary Layer on a Flat Plate . . . . . . . . . . . . . . . . . . . 2298.6 Wake Downstream of a Flat Plate . . . . . . . . . . . . . . . . . . 2348.7 Von Karman Momentum Integral . . . . . . . . . . . . . . . . . . 2398.8 Boundary Layer Separation . . . . . . . . . . . . . . . . . . . . . 2408.9 Criterion for Boundary Layer Separation . . . . . . . . . . . . . . 244

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8.10 Approximate Solutions of Boundary Layer Equations . . . . . . . 2478.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

9 Incompressible Aerodynamics 2579.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2579.2 Theorem of Kutta and Zhukovskii . . . . . . . . . . . . . . . . . 2589.3 Cylindrical Airfoils . . . . . . . . . . . . . . . . . . . . . . . . . 2609.4 Zhukovskii’s Hypothesis . . . . . . . . . . . . . . . . . . . . . . 2639.5 Vortex Sheets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2699.6 Induced Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2709.7 Three-Dimensional Airfoils . . . . . . . . . . . . . . . . . . . . . 2719.8 Aerodynamic Forces . . . . . . . . . . . . . . . . . . . . . . . . 2759.9 Ellipsoidal Airfoils . . . . . . . . . . . . . . . . . . . . . . . . . 2789.10 Simple Flight Problems . . . . . . . . . . . . . . . . . . . . . . . 2819.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283

10 Incompressible Viscous Flow 28710.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28710.2 Flow Between Parallel Plates . . . . . . . . . . . . . . . . . . . . 28710.3 Flow Down an Inclined Plane . . . . . . . . . . . . . . . . . . . . 28910.4 Poiseuille Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . 29110.5 Taylor-Couette Flow . . . . . . . . . . . . . . . . . . . . . . . . 29210.6 Flow in Slowly-Varying Channels . . . . . . . . . . . . . . . . . 29310.7 Lubrication Theory . . . . . . . . . . . . . . . . . . . . . . . . . 29610.8 Stokes Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29910.9 Axisymmetric Stokes Flow . . . . . . . . . . . . . . . . . . . . . 29910.10 Axisymmetric Stokes Flow Around a Solid Sphere . . . . . . . . 30010.11 Axisymmetric Stokes Flow In and Around a Fluid Sphere . . . . . 30610.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309

11 Waves in Incompressible Fluids 31311.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31311.2 Gravity Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31311.3 Gravity Waves in Deep Water . . . . . . . . . . . . . . . . . . . . 31511.4 Gravity Waves in Shallow Water . . . . . . . . . . . . . . . . . . 31711.5 Energy of Gravity Waves . . . . . . . . . . . . . . . . . . . . . . 31811.6 Wave Drag on Ships . . . . . . . . . . . . . . . . . . . . . . . . . 32011.7 Ship Wakes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32311.8 Gravity Waves in a Flowing Fluid . . . . . . . . . . . . . . . . . 32811.9 Gravity Waves at an Interface . . . . . . . . . . . . . . . . . . . . 32911.10 Steady Flow over a Corrugated Bottom . . . . . . . . . . . . . . . 33111.11 Surface Tension . . . . . . . . . . . . . . . . . . . . . . . . . . . 33211.12 Capillary Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . 33411.13 Capillary Waves at an Interface . . . . . . . . . . . . . . . . . . . 33511.14 Wind Driven Waves in Deep Water . . . . . . . . . . . . . . . . . 336

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11.15 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338

12 Terrestrial Ocean Tides 34112.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34112.2 Tide Generating Potential . . . . . . . . . . . . . . . . . . . . . . 34112.3 Decomposition of Tide Generating Potential . . . . . . . . . . . . 34312.4 Expansion of Tide Generating Potential . . . . . . . . . . . . . . 34412.5 Surface Harmonics and Solid Harmonics . . . . . . . . . . . . . . 34512.6 Planetary Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . 34612.7 Total Gravitational Potential . . . . . . . . . . . . . . . . . . . . 34712.8 Planetary Response . . . . . . . . . . . . . . . . . . . . . . . . . 34812.9 Laplace Tidal Equations . . . . . . . . . . . . . . . . . . . . . . . 35312.10 Harmonics of Forcing Term in Laplace Tidal Equations . . . . . . 35712.11 Response to Equilibrium Harmonic . . . . . . . . . . . . . . . . . 35912.12 Global Ocean Tides . . . . . . . . . . . . . . . . . . . . . . . . . 36012.13 Non-Global Ocean Tides . . . . . . . . . . . . . . . . . . . . . . 36612.14 Useful Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36712.15 Transformation of Laplace Tidal Equations . . . . . . . . . . . . 36912.16 Another Useful Lemma . . . . . . . . . . . . . . . . . . . . . . . 37112.17 Basis Eigenfunctions . . . . . . . . . . . . . . . . . . . . . . . . 37112.18 Auxilliary Eigenfunctions . . . . . . . . . . . . . . . . . . . . . . 37312.19 Gyroscopic Coefficients . . . . . . . . . . . . . . . . . . . . . . . 37412.20 Proudman Equations . . . . . . . . . . . . . . . . . . . . . . . . 37612.21 Hemispherical Ocean Tides . . . . . . . . . . . . . . . . . . . . . 38012.22 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385

13 Equilibrium of Compressible Fluids 39313.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39313.2 Isothermal Atmosphere . . . . . . . . . . . . . . . . . . . . . . . 39413.3 Adiabatic Atmosphere . . . . . . . . . . . . . . . . . . . . . . . 39413.4 Atmospheric Stability . . . . . . . . . . . . . . . . . . . . . . . . 39613.5 Eddington Solar Model . . . . . . . . . . . . . . . . . . . . . . . 39713.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404

14 One-Dimensional Compressible Inviscid Flow 40714.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40714.2 Thermodynamic Considerations . . . . . . . . . . . . . . . . . . 40714.3 Isentropic Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . 41114.4 Sound Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41114.5 Bernoulli’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 41414.6 Mach Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41514.7 Sonic Flow through a Nozzle . . . . . . . . . . . . . . . . . . . . 41614.8 Normal Shocks . . . . . . . . . . . . . . . . . . . . . . . . . . . 42014.9 Piston-Generated Shock Wave . . . . . . . . . . . . . . . . . . . 42514.10 Piston-Generated Expansion Wave . . . . . . . . . . . . . . . . . 427

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14.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430

15 Two-Dimensional Compressible Inviscid Flow 43715.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43715.2 Oblique Shocks . . . . . . . . . . . . . . . . . . . . . . . . . . . 43715.3 Supersonic Flow in Corner or over Wedge . . . . . . . . . . . . . 44115.4 Weak Oblique Shocks . . . . . . . . . . . . . . . . . . . . . . . . 44315.5 Supersonic Compression by Turning . . . . . . . . . . . . . . . . 44515.6 Supersonic Expansion by Turning . . . . . . . . . . . . . . . . . 44715.7 Detached Shocks . . . . . . . . . . . . . . . . . . . . . . . . . . 45015.8 Shock-Expansion Theory . . . . . . . . . . . . . . . . . . . . . . 45115.9 Thin-Airfoil Theory . . . . . . . . . . . . . . . . . . . . . . . . . 45315.10 Crocco’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 45815.11 Homenergic Homentropic Flow . . . . . . . . . . . . . . . . . . . 45915.12 Small-Perturbation Theory . . . . . . . . . . . . . . . . . . . . . 46015.13 Subsonic Flow Past a Wave-Shaped Wall . . . . . . . . . . . . . . 46415.14 Supersonic Flow Past a Wave-Shaped Wall . . . . . . . . . . . . . 46615.15 Linearized Subsonic Flow . . . . . . . . . . . . . . . . . . . . . . 46815.16 Linearized Supersonic Flow . . . . . . . . . . . . . . . . . . . . 47015.17 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472

A Vectors and Vector Fields 483A.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483A.2 Scalars and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . 483A.3 Vector Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . 484A.4 Cartesian Components of a Vector . . . . . . . . . . . . . . . . . 486A.5 Coordinate Transformations . . . . . . . . . . . . . . . . . . . . 487A.6 Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489A.7 Vector Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491A.8 Vector Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492A.9 Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495A.10 Scalar Triple Product . . . . . . . . . . . . . . . . . . . . . . . . 496A.11 Vector Triple Product . . . . . . . . . . . . . . . . . . . . . . . . 498A.12 Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . 498A.13 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499A.14 Vector Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . 502A.15 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 502A.16 Vector Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . 504A.17 Volume Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 505A.18 Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506A.19 Grad Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510A.20 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511A.21 Laplacian Operator . . . . . . . . . . . . . . . . . . . . . . . . . 514A.22 Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 516A.23 Useful Vector Identities . . . . . . . . . . . . . . . . . . . . . . . 519

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A.24 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 520

B Cartesian Tensors 525B.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525B.2 Tensors and Tensor Notation . . . . . . . . . . . . . . . . . . . . 525B.3 Tensor Transformation . . . . . . . . . . . . . . . . . . . . . . . 528B.4 Tensor Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531B.5 Isotropic Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . 533B.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536

C Non-Cartesian Coordinates 539C.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539C.2 Orthogonal Curvilinear Coordinates . . . . . . . . . . . . . . . . 539C.3 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . 543C.4 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 546C.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548

D Ellipsoidal Potential Theory 551D.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 551D.2 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 551D.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555

E Calculus of Variations 557E.1 Indroduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557E.2 Euler-Lagrange Equation . . . . . . . . . . . . . . . . . . . . . . 557E.3 Conditional Variation . . . . . . . . . . . . . . . . . . . . . . . . 559E.4 Multi-Function Variation . . . . . . . . . . . . . . . . . . . . . . 562E.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562

Bibliography 565

1Mathematical Models of Fluid Motion

1.1 IntroductionThis chapter sets forth the mathematical models commonly used to describe theequilibrium and dynamics of fluids. Unless stated otherwise, all of the analysis isperformed using a standard right-handed Cartesian coordinate system: (x1, x2, x3).Moreover, the Einstein summation convention is employed (which means that re-peated roman subscripts are assumed to be summed from 1 to 3). (See Appendix B.)More information on the fundamental assumptions that underlie conventional fluidmechanics, as well as the basic equations that describe fluid motion, can be found inBatchelor 2000.

1.2 What is a Fluid?By definition, a solid material is rigid. Although a rigid material tends to shatterwhen subjected to very large stresses, it can withstand a moderate shear stress (i.e.,a stress that tends to deform the material by changing its shape, without necessarilychanging its volume) for an indefinite period. To be more exact, when a shear stressis first applied to a rigid material it deforms slightly, but then springs back to itsoriginal shape when the stress is relieved.

A plastic material, such as clay, also possess some degree of rigidity. However,the critical shear stress above which it yields is relatively small, and once this stressis exceeded the material deforms continuously and irreversibly, and does not recoverits original shape when the stress is relieved.

By definition, a fluid material possesses no rigidity at all. In other words, asmall fluid element is unable to withstand any tendency of an applied shear stressto change its shape. This does not preclude the possibility that such an elementmay offer resistance to shear stress. However, any resistance must be incapable ofpreventing the change in shape from eventually occurring, which implies that theforce of resistance vanishes with the rate of deformation. An obvious corollary isthat the shear stress must be zero everywhere inside a fluid that is in mechanicalequilibrium.

Fluids are conventionally classified as either liquids or gases. The most important

1

2 Theoretical Fluid Mechanics

difference between these two types of fluid lies in their relative compressibility. To bemore exact, gases can be compressed much more easily than liquids. Consequently,any motion that involves significant pressure variations is generally accompanied bymuch larger changes in mass density in the case of a gas than in the case of a liquid.

A macroscopic fluid ultimately consists of a huge number of individual molecules.However, most practical applications of fluid mechanics are concerned with behav-ior on lengthscales that are far longer than the typical intermolecular spacing. Underthese circumstances, it is reasonable to suppose that the bulk properties of a givenfluid are the same as if it were completely continuous in structure. A corollary of thisassumption is that when, in the following, we talk about infinitesimal volume ele-ments, we really mean elements that are sufficiently small that the bulk fluid proper-ties (such as mass density, pressure, and velocity) are approximately constant acrossthem, but are still sufficiently large that they contain a very great number of molecules(which implies that we can safely neglect any statistical variations in the bulk prop-erties). The continuum hypothesis also requires infinitesimal volume elements to bemuch larger than the molecular mean-free-path between collisions.

In addition to the continuum hypothesis, our study of fluid mechanics is premisedon three major assumptions:

1. Fluids are isotropic media. In other words, there is no preferred direction in afluid.

2. Fluids are Newtonian. In other words, there is a linear relationship between thelocal shear stress and the local rate of strain, as first postulated by Isaac Newton(1642-1727). It is also assumed that there is a linear relationship between thelocal heat flux density and the local temperature gradient.

3. Fluids are classical. In other words, the macroscopic motion of ordinary fluids iswell described by Newtonian dynamics, and both quantum and relativistic effectscan be safely ignored.

It should be noted that the previous assumptions are not valid for all fluid types (e.g.,certain liquid polymers, which are non-isotropic; thixotropic fluids, such as jelly orpaint, which are non-Newtonian; and quantum fluids, such as liquid helium, whichexhibit non-classical effects on macroscopic lengthscales). However, most practicalapplications of fluid mechanics involve the equilibrium and motion of bodies of wa-ter or air, extending over macroscopic lengthscales, and situated relatively close tothe Earth’s surface. Such bodies are very well described as isotropic, Newtonian,classical fluids.

1.3 Volume and Surface ForcesGenerally speaking, fluids are acted upon by two distinct types of force. The first typeis long range in nature—that is, such that it decreases relatively slowly with increas-

Mathematical Models of Fluid Motion 3

ing distance between interacting elements—and is capable of completely penetratinginto the interior of a fluid. Gravity is an obvious example of a long-range force. Oneconsequence of the relatively slow variation of long-range forces with position isthat they act equally on all of the fluid contained within a sufficiently small volumeelement. In this situation, the net force acting on the element becomes directly pro-portional to its volume. For this reason, long-range forces are often called volumeforces. In the following, we shall write the total volume force acting at time t on thefluid contained within a small volume element of magnitude dV , centered on a fixedpoint whose position vector is r, as

F(r, t) dV. (1.1)

The second type of force is short range in nature, and is most conveniently mod-eled as momentum transport within the fluid. Such transport is generally due to acombination of the mutual forces exerted by contiguous molecules, and momentumfluxes caused by relative molecular motion. Suppose that πx(r, t) is the net flux den-sity of x-directed fluid momentum due to short-range forces at position r and timet. In other words, suppose that, at position r and time t, as a direct consequenceof short-range forces, x-momentum is flowing at the rate of |πx| newton-seconds permeter squared per second in the direction of vector πx. Consider an infinitesimalplane surface element, dS = n dS , located at point r. Here, dS is the area of theelement, and n its unit normal. (See Section A.7.) The fluid which lies on that sideof the element toward which n points is said to lie on its positive side, and vice versa.The net flux of x-momentum across the element (in the direction of n) is πx · dSnewtons, which implies (from Newton’s second law of motion) that the fluid on thepositive side of the surface element experiences a force πx · dS in the x-direction dueto short-range interaction with the fluid on the negative side. According to Newton’sthird law of motion, the fluid on the negative side of the surface experiences a force−πx ·dS in the x-direction due to interaction with the fluid on the positive side. Short-range forces are often called surface forces, because they are directly proportional tothe area of the surface element across which they act. Let πy(r, t) and πz(r, t) be thenet flux density of y- and z-momentum, respectively, at position r and time t. Bya straightforward extension of previous argument, the net surface force exerted bythe fluid on the positive side of some planar surface element, dS, on the fluid on itsnegative side is

f = (−πx · dS, −πy · dS, −πz · dS). (1.2)

In tensor notation (see Appendix B), the previous equation can be written

fi = σi j dS j, (1.3)

where σ11 = −(πx)x, σ12 = −(πx)y, σ21 = −(πy)x, et cetera. (Note that, because thesubscript j is repeated in the previous equation, it is assumed to be summed from 1 to3. Hence, σi j dS j is shorthand for

∑j=1,3 σi j dS j.) Here, the σi j(r, t) are termed the

local stresses in the fluid at position r and time t, and have units of force per unit area.Moreover, the σi j are the components of a second-order tensor (see Appendix B)known as the stress tensor. [This follows because the fi are the components of a


first-order tensor (as all forces are proper vectors), and the dS i are the componentsof an arbitrary first-order tensor (as surface elements are also proper vectors—seeSection A.7—and Equation (1.3) holds for surface elements whose normals point inany direction), so application of the quotient rule (see Section B.3) to Equation (1.3)reveals that the σi j transform under rotation of the coordinate axes as the componentsof a second-order tensor.] We can interpret σi j(r, t) as the i-component of the forceper unit area exerted, at position r and time t, across a plane surface element normalto the j-direction. The three diagonal components of σi j are termed normal stresses,because each of them gives the normal component of the force per unit area actingacross a plane surface element parallel to one of the Cartesian coordinate planes. Thesix non-diagonal components are termed shear stresses, because they drive shearingmotion in which parallel layers of fluid slide relative to one another.

1.4 General Properties of Stress TensorThe i-component of the total force acting on a fluid element consisting of a fixedvolume V enclosed by a surface S is written

fi =∫

VFi dV +

∮Sσi j dS j, (1.4)

where the first term on the right-hand side is the integrated volume force actingthroughout V , whereas the second term is the net surface force acting across S . Mak-ing use of the tensor divergence theorem (see Section B.4), the previous expressionbecomes

fi =∫

VFi dV +

∫V

∂σi j

∂x jdV. (1.5)

In the limit V → 0, it is reasonable to suppose that the Fi and ∂σi j/∂x j are ap-proximately constant across the element. In this situation, both contributions on theright-hand side of the previous equation scale as V . According to Newtonian dy-namics, the i-component of the net force acting on the volume element is equal tothe i-component of the rate of change of its linear momentum. However, in the limitV → 0, the linear acceleration and mass density of the fluid are both approximatelyconstant across the element. In this case, the rate of change of the element’s linearmomentum also scales as V . In other words, the net volume force, surface force, andrate of change of linear momentum of an infinitesimal fluid element all scale as thevolume of the element, and consequently remain approximately the same order ofmagnitude as the volume shrinks to zero. We conclude that the linear equation ofmotion of an infinitesimal fluid element places no particular restrictions on the stresstensor.

The i-component of the total torque, taken about the origin O of the coordinatesystem, acting on a fluid element that consists of a fixed volume V enclosed by a


surface S is written [see Equations (A.46) and (B.6)]

τi =

∫Vεi jk x j Fk dV +

∮Sεi jk x jσkl dS l, (1.6)

where the first and second terms on the right-hand side are due to volume and sur-face forces, respectively. [Here, εi jk is the third-order permutation tensor. See Equa-tion (B.7).] Making use of the tensor divergence theorem (see Section B.4), theprevious expression becomes

τi =


∫Vεi jk∂(x jσkl)∂xl

dV, (1.7)

which reduces to

τi =


∫Vεi jk σk j dV +

∫Vεi jk x j

∂σkl

∂xldV, (1.8)

because ∂xi/∂x j = δi j. [Here, δi j is the second-order identity tensor. See Equa-tion (B.9).] Assuming that point O lies within the fluid element, and taking the limitV → 0 in which the Fi, σi j, and ∂σi j/∂x j are all approximately constant across theelement, we deduce that the first, second, and third terms on the right-hand side ofthe previous equation scale as V 4/3, V , and V 4/3, respectively (because x ∼ V 1/3).According to Newtonian dynamics, the i-component of the total torque acting onthe fluid element is equal to the i-component of the rate of change of its net angularmomentum about O. Assuming that the linear acceleration and density of the fluidare approximately constant across the element, we deduce that the rate of change ofits angular momentum scales as V 4/3 (because the net rate of change of the linearmomentum scales as V , so the net rate of change of the angular momentum scalesas x V , and x ∼ V 1/3). Hence, it is clear that the rotational equation of motion ofa fluid element, surrounding a general point O, becomes completely dominated bythe second term on the right-hand side of Equation (1.8) in the limit that the volumeof the element approaches zero (because this term is a factor V −1/3 larger than theother terms). It follows that the second term must be identically zero (otherwise aninfinitesimal fluid element would acquire an absurdly large angular velocity). This isonly possible, for all choices of the position of point O, and the shape of the element,if

εi jk σk j = 0 (1.9)

throughout the fluid. The previous relation shows that the stress tensor must be sym-metric: that is,

σ ji = σi j. (1.10)

It immediately follows that the stress tensor only has six independent components(i.e., σ11, σ22, σ33, σ12, σ13, and σ23).

It is always possible to choose the orientation of a set of Cartesian axes in such amanner that the non-diagonal components of a given symmetric second-order tensorfield are all set to zero at a given point in space. (See Exercise B.6.) With refer-ence to such principal axes, the diagonal components of the stress tensor σi j become


so-called principal stresses—σ′11, σ′22, σ′33, say. In general, the orientation of theprincipal axes varies with position. The normal stress σ′11 acting across a surfaceelement perpendicular to the first principal axis corresponds to a tension (or a com-pression if σ′11 is negative) in the direction of that axis. Likewise, for σ′22 and σ′33.Thus, the general state of the fluid, at a particular point in space, can be regarded asa superposition of tensions, or compressions, in three orthogonal directions.

The trace of the stress tensor, σii = σ11 + σ22 + σ33, is a scalar, and, therefore,independent of the orientation of the coordinate axes. (See Appendix B.) Thus, itfollows that, irrespective of the orientation of the principal axes, the trace of the stresstensor at a given point is always equal to the sum of the principal stresses: that is,

σii = σ′11 + σ

′22 + σ

′33. (1.11)

1.5 Stress Tensor in a Static FluidConsider the surface forces exerted on some infinitesimal cubic volume element ofa static fluid. Suppose that the components of the stress tensor are approximatelyconstant across the element. Suppose, further, that the sides of the cube are alignedparallel to the principal axes of the local stress tensor. This tensor, which now haszero non-diagonal components, can be regarded as the sum of two tensors: that is,

13σii, 0, 00, 1

3σii, 00, 0, 1

3σii

, (1.12)

and σ′11 −

13σii, 0, 0

0, σ′22 −13σii, 0

0, 0, σ′33 −13σii

. (1.13)

The tensor (1.12) is isotropic (see Section B.5), and corresponds to the samenormal force per unit area acting inward (because the sign of σii/3 is invariablynegative) on each face of the volume element. This uniform compression acts tochange the element’s volume, but not its shape, and can easily be withstood by thefluid within the element.

The tensor (1.13) represents the departure of the stress tensor from an isotropicform. The diagonal components of this tensor have zero sum, in view of Equa-tion (1.11), and thus represent equal and opposite forces per unit area, acting onopposing faces of the volume element, which are such that the forces on at leastone pair of opposing faces constitute a tension, and the forces on at least one pairconstitute a compression. Such forces necessarily act to change the shape of the vol-ume element, either elongating or compressing it along one of its symmetry axes.Moreover, this tendency cannot be offset by any volume force acting on the element,


because such forces become arbitrarily small compared to surface forces in the limitthat the element’s volume tends to zero (because the ratio of the net volume force tothe net surface force scales as the volume to the surface area of the element, whichtends to zero in the limit that the volume tends to zero—see Section 1.4). We pre-viously defined a fluid as a material that is incapable of withstanding any tendencyof applied forces to change its shape. (See Section 1.2.) It follows that if the diag-onal components of the tensor (1.13) are non-zero anywhere inside the fluid then itis impossible for the fluid at that point to be at rest. Hence, we conclude that theprincipal stresses, σ′11, σ′22, and σ′33, must be equal to one another at all points in astatic fluid. This implies that the stress tensor takes the isotropic form (1.12) every-where in a stationary fluid. Furthermore, this is true irrespective of the orientation ofthe coordinate axes, because the components of an isotropic tensor are rotationallyinvariant. (See Section B.5.)

Fluids at rest are generally in a state of compression, so it is convenient to writethe stress tensor of a static fluid in the form

σi j = −p δi j, (1.14)

where p = −σii/3 is termed the static fluid pressure, and is generally a function ofr and t. It follows that, in a stationary fluid, the force per unit area exerted across aplane surface element with unit normal n is −p n. [See Equation (1.3).] Moreover,this normal force has the same value for all possible orientations of n. This well-known result—namely, that the pressure is the same in all directions at a given pointin a static fluid—is known as Pascal’s law, after its discoverer Blaise Pascal (1623-1662), and is a direct consequence of the fact that a fluid element cannot withstandshear stresses, or, alternatively, any tendency of applied forces to change its shape.

1.6 Stress Tensor in a Moving FluidWe have seen that in a static fluid the stress tensor takes the form

σi j = −p δi j, (1.15)

where p = −σii/3 is the static pressure: that is, minus the normal stress acting inany direction. The normal stress at a given point in a moving fluid generally varieswith direction. In other words, the principal stresses are not equal to one another.However, we can still define the mean principal stress as (σ′11 + σ

′22 + σ

′33)/3 =

σii/3. Moreover, given that the principal stresses are actually normal stresses (in acoordinate frame aligned with the principal axes), we can also regard σii/3 as themean normal stress. It is convenient to define pressure in a moving fluid as minus themean normal stress: that is,

p = −13σii. (1.16)


Thus, we can write the stress tensor in a moving fluid as the sum of an isotropicpart, −p δi j, which has the same form as the stress tensor in a static fluid, and aremaining non-isotropic part, di j, which includes any shear stresses, and also hasdiagonal components whose sum is zero. In other words,

σi j = −p δi j + di j, (1.17)

wheredii = 0. (1.18)

Moreover, because σi j and δi j are both symmetric tensors, it follows that di j is alsosymmetric: that is,

d ji = di j. (1.19)

It is clear that the so-called deviatoric stress tensor, di j, is a consequence of fluidmotion, because it is zero in a static fluid. Suppose, however, that we were to view astatic fluid both in its rest frame and in a frame of reference moving at some constantvelocity relative to the rest frame. We would expect the force distribution within thefluid to be the same in both frames of reference, because the fluid does not acceleratein either. However, in the first frame, the fluid appears stationary and the deviatoricstress tensor is therefore zero, while in the second it has a spatially uniform velocityfield and the deviatoric stress tensor is also zero (because it is the same as in therest frame). We, thus, conclude that the deviatoric stress tensor is zero both in astationary fluid and in a moving fluid possessing no spatial velocity gradients. Thissuggests that the deviatoric stress tensor is driven by velocity gradients within thefluid. Moreover, the tensor must vanish as these gradients vanish.

Let the vi(r, t) be the Cartesian components of the fluid velocity at point r andtime t. The various velocity gradients within the fluid then take the form ∂vi/∂x j.The simplest possible assumption, which is consistent with the previous discussion,is that the components of the deviatoric stress tensor are linear functions of thesevelocity gradients: that is,

di j = Ai jkl∂vk∂xl. (1.20)

Here, Ai jkl is a fourth-order tensor (this follows from the quotient rule because di j

and ∂vi/∂x j are both proper second-order tensors). Any fluid in which the deviatoricstress tensor takes the previous form is termed a Newtonian fluid, because Newtonwas the first to postulate a linear relationship between shear stresses and velocitygradients.

In an isotropic fluid—that is, a fluid in which there is no preferred direction—wewould expect the fourth-order tensor Ai jkl to be isotropic—that is, to have a form inwhich all physical distinction between different directions is absent. As demonstratedin Section B.5, the most general expression for an isotropic fourth-order tensor is

Ai jkl = α δi j δkl + β δik δ jl + γ δil δ jk, (1.21)

where α, β, and γ are arbitrary scalars (which can be functions of position and time).


Thus, it follows from Equations (1.20) and (1.21) that

di j = α∂vk∂xk

δi j + β∂vi∂x j+ γ

∂v j

∂xi. (1.22)

However, according to Equation (1.19), di j is a symmetric tensor, which implies thatβ = γ, and

di j = α ekk δi j + 2 β ei j, (1.23)

where

ei j =12

(∂vi∂x j+∂v j

∂xi

)(1.24)

is called the rate of strain tensor. Finally, according to Equation (1.18), di j is atraceless tensor, which yields 3α = −2 β, and

di j = 2 µ(ei j −

13

ekk δi j

), (1.25)

where µ = β. We, thus, conclude that the most general expression for the stresstensor in an isotropic Newtonian fluid is

σi j = −p δi j + 2 µ(ei j −

13

ekk δi j

), (1.26)

where p(r, t) and µ(r, t) are arbitrary scalars.

1.7 ViscosityThe significance of the parameter µ, appearing in the previous expression for thestress tensor, can be seen from the form taken by the relation (1.25) in the special caseof simple shearing motion. With ∂v1/∂x2 as the only non-zero velocity derivative, allof the components of di j are zero apart from the shear stresses

d12 = d21 = µ∂v1

∂x2. (1.27)

Thus, µ is the constant of proportionality between the rate of shear and the tangentialforce per unit area when parallel plane layers of fluid slide over one another. Thisconstant of proportionality is generally referred to as viscosity. It is a matter ofexperience that the force acting between layers of fluid undergoing relative slidingmotion always tends to oppose the motion, which implies that µ > 0.

The viscosities of dry air and pure water at 20 C and atmospheric pressure areabout 1.8 × 10−5 kg/(m s) and 1.0 × 10−3 kg/(m s), respectively. In neither case doesthe viscosity exhibit much variation with pressure. However, the viscosity of airincreases by about 0.3 percent, and that of water decreases by about 3 percent, perdegree Centigrade rise in temperature (Batchelor 2000).


1.8 Conservation LawsSuppose that θ(r, t) is the density of some bulk fluid property (e.g., mass, momen-tum, or energy) at position r and time t. In other words, suppose that, at time t, aninfinitesimal fluid element of volume dV , located at position r, contains an amountθ(r, t) dV of the property in question. Note, incidentally, that θ can be either a scalar,a component of a vector, or even a component of a tensor. The total amount of theproperty contained within some fixed volume V is

Θ =

∫Vθ dV, (1.28)

where the integral is taken over all elements of V . Let dS be an outward directedelement of the bounding surface of V . Suppose that this element is located at pointr. The volume of fluid that flows per second across the element, and so out of V ,is v(r, t) · dS. Thus, the amount of the fluid property under consideration that isconvected across the element per second is θ(r, t) v(r, t) · dS. It follows that the netamount of the property that is convected out of volume V by fluid flow across itsbounding surface S is

ΦΘ =

∮Sθ v · dS, (1.29)

where the integral is taken over all outward directed elements of S . Suppose, finally,that the property in question is created within the volume V at the rate SΘ per second.The conservation equation for the fluid property takes the form

dΘdt= SΘ −ΦΘ. (1.30)

In other words, the rate of increase in the amount of the property contained withinV is the difference between the creation rate of the property inside V , and the rate atwhich the property is convected out of V by fluid flow. The previous conservationlaw can also be written

dΘdt+ΦΘ = SΘ. (1.31)

Here, ΦΘ is termed the flux of the property out of V , whereas SΘ is called the netgeneration rate of the property within V .

1.9 Mass ConservationLet ρ(r, t) and v(r, t) be the mass density and velocity of a given fluid at point rand time t. Consider a fixed volume V , surrounded by a surface S . The net masscontained within V is

M =∫

Vρ dV, (1.32)


where dV is an element of V . Furthermore, the mass flux across S , and out of V , is[see Equation (1.29)]

ΦM =

∮Sρ v · dS, (1.33)

where dS is an outward directed element of S . Mass conservation requires that therate of increase of the mass contained within V , plus the net mass flux out of V ,should equal zero: that is,

dMdt+ΦM = 0 (1.34)

[cf., Equation (1.31)]. Here, we are assuming that there is no mass generation (ordestruction) within V (because individual molecules are effectively indestructible).It follows that ∫

V

∂ρ

∂tdV +

∮Sρ v · dS = 0, (1.35)

because V is non-time-varying. Making use of the divergence theorem (see Sec-tion A.20), the previous equation becomes∫

V

[∂ρ

∂t+ ∇ · (ρ v)

]dV = 0. (1.36)

However, this result is true irrespective of the size, shape, or location of volume V ,which is only possible if

∂ρ

∂t+ ∇ · (ρ v) = 0 (1.37)

throughout the fluid. The previous expression is known as the equation of fluid con-tinuity, and is a direct consequence of mass conservation.

1.10 Convective Time DerivativeThe quantity ∂ρ(r, t)/∂t, appearing in Equation (1.37), represents the time derivativeof the fluid mass density at the fixed point r. Suppose that v(r, t) is the instantaneousfluid velocity at the same point. It follows that the time derivative of the density, asseen in a frame of reference which is instantaneously co-moving with the fluid atpoint r, is

limδt→0

ρ(r + v δt, t + δt) − ρ(r, t)δt

=∂ρ

∂t+ v · ∇ρ = Dρ

Dt, (1.38)

where we have Taylor expanded ρ(r + v δt, t + δt) up to first order in δt, and where

DDt=∂

∂t+ v · ∇ = ∂

∂t+ vi

∂

∂xi. (1.39)

Clearly, the so-called convective time derivative, D/Dt, represents the time derivativeseen in the local rest frame of the fluid.


The continuity equation (1.37) can be rewritten in the form

1ρ

DρDt=

D ln ρDt

= −∇ · v, (1.40)

because ∇ · (ρ v) = v · ∇ρ + ρ∇ · v. [See Equation (A.174).] Consider a volumeelement V that is co-moving with the fluid. In general, as the element is convectedby the fluid its volume changes. In fact, it is easily seen that

DVDt=

∮S

v · dS =∮

Svi dS i =

∫V

∂vi∂xi

dV =∫

V∇·v dV, (1.41)

where S is the bounding surface of the element, and use has been made of the diver-gence theorem. In the limit that V → 0, and ∇ · v is approximately constant acrossthe element, we obtain

1V

DVDt=

D ln VDt

= ∇ · v. (1.42)

Hence, we conclude that the divergence of the fluid velocity at a given point in spacespecifies the fractional rate of increase in the volume of an infinitesimal co-movingfluid element at that point.

1.11 Momentum ConservationConsider a fixed volume V surrounded by a surface S . The i-component of the totallinear momentum contained within V is

Pi =

∫Vρ vi dV. (1.43)

Moreover, the flux of i-momentum across S , and out of V , is [see Equation (1.29)]

Φi =

∮Sρ vi v j dS j. (1.44)

Finally, the i-component of the net force acting on the fluid within V is

fi =∫

VFi dV +

∮Sσi j dS j, (1.45)

where the first and second terms on the right-hand side are the contributions fromvolume and surface forces, respectively.

Momentum conservation requires that the rate of increase of the net i-momentumof the fluid contained within V , plus the flux of i-momentum out of V , is equal to therate of i-momentum generation within V . Of course, from Newton’s second law of


motion, the latter quantity is equal to the i-component of the net force acting on thefluid contained within V . Thus, we obtain [cf., Equation (1.31)]

dPi

dt+ Φi = fi, (1.46)

which can be written∫V

∂(ρ vi)∂t

dV +∮

Sρ vi v j dS j =

∫V

Fi dV +∮

Sσi j dS j, (1.47)

because the volume V is non-time-varying. Making use of the tensor divergencetheorem, this becomes∫

V

[∂(ρ vi)∂t+∂(ρ vi v j)∂x j

]dV =

∫V

(Fi +

∂σi j

∂x j

)dV. (1.48)

However, the previous result is valid irrespective of the size, shape, or location ofvolume V , which is only possible if

∂(ρ vi)∂t+∂(ρ vi v j)∂x j

= Fi +∂σi j

∂x j(1.49)

everywhere inside the fluid. Expanding the derivatives, and rearranging, we obtain(∂ρ

∂t+ v j

∂ρ

∂x j+ ρ

∂v j

∂x j

)vi + ρ

(∂vi∂t+ v j

∂vi∂x j

)= Fi +

∂σi j

∂x j. (1.50)

In tensor notation, the continuity equation (1.37) is written

∂ρ

∂t+ v j

∂ρ

∂x j+ ρ

∂v j

∂x j= 0. (1.51)

So, combining Equations (1.50) and (1.51), we obtain the following fluid equation ofmotion,

ρ

(∂vi∂t+ v j

∂vi∂x j

)= Fi +

∂σi j

∂x j. (1.52)

An alternative form of this equation is

DviDt=

Fi

ρ+

1ρ

∂σi j

∂x j. (1.53)

The previous equation describes how the net volume and surface forces per unit massacting on a co-moving fluid element determine its acceleration.


1.12 Navier-Stokes EquationEquations (1.24), (1.26), and (1.53) can be combined to give the equation of motionof an isotropic, Newtonian, classical fluid:

ρDviDt= Fi −

∂p∂xi+∂

∂x j

[µ

(∂vi∂x j+∂v j

∂xi

)]− ∂

∂xi

(23µ∂v j

∂x j

). (1.54)

This equation is generally known as the Navier-Stokes equation, and is named afterClaude-Louis Navier (1785-1836) and George Gabriel Stokes (1819-1903). In sit-uations in which there are no strong temperature gradients in the fluid, it is a goodapproximation to treat viscosity as a spatially uniform quantity, in which case theNavier-Stokes equation simplifies somewhat to give

ρDviDt= Fi −

∂p∂xi+ µ

(∂ 2vi∂x j ∂x j

+13∂ 2v j

∂xi ∂x j

). (1.55)

When expressed in vector form, the previous expression becomes

ρDvDt≡ ρ

[∂v∂t+ (v · ∇) v

]= F − ∇p + µ

[∇ 2v +

13∇(∇ · v)

], (1.56)

where use has been made of Equation (1.39). Here,

[(a · ∇)b]i = a j∂bi

∂x j, (1.57)

(∇ 2v)i = ∇ 2vi. (1.58)

Note, however, that the previous identities are only valid in Cartesian coordinates.(See Appendix C.)

1.13 Energy ConservationConsider a fixed volume V surrounded by a surface S . The total energy content ofthe fluid contained within V is

E =∫

VρE dV +

∫V

12ρ vi vi dV, (1.59)

where the first and second terms on the right-hand side are the net internal and kineticenergies, respectively. Here, E(r, t) is the internal (i.e., thermal) energy per unit massof the fluid. The energy flux across S , and out of V , is [cf., Equation (1.29)]

ΦE =

∮Sρ

(E + 1

2vi vi

)v j dS j =

∫V

∂

∂x j

[ρ

(E + 1

2vi vi

)v j

]dV, (1.60)


where use has been made of the tensor divergence theorem. According to the firstlaw of thermodynamics, the rate of increase of the energy contained within V , plusthe net energy flux out of V , is equal to the net rate of work done on the fluid withinV , minus the net heat flux out of V: that is,

dEdt+ΦE =

.W −

.Q, (1.61)

where.

W is the net rate of work, and.

Q the net heat flux. It can be seen that.

W −.

Q isthe effective energy generation rate within V [cf., Equation (1.31)].

The net rate at which volume and surface forces do work on the fluid within V is

.W =

∫Vvi Fi dV +

∮Svi σi j dS j =

∫V

[vi Fi +

∂(viσi j)∂x j

]dV, (1.62)

where use has been made of the tensor divergence theorem.Generally speaking, heat flow in fluids is driven by temperature gradients. Let

the qi(r, t) be the Cartesian components of the heat flux density at position r andtime t. It follows that the heat flux across a surface element dS, located at point r, isq · dS = qi dS i. Let T (r, t) be the temperature of the fluid at position r and time t.Thus, a general temperature gradient takes the form ∂T/∂xi. Let us assume that thereis a linear relationship between the components of the local heat flux density and thelocal temperature gradient: that is,

qi = Ai j∂T∂x j

, (1.63)

where the Ai j are the components of a second-rank tensor (which can be functionsof position and time). In an isotropic fluid we would expect Ai j to be an isotropictensor. (See Section B.5.) However, the most general second-order isotropic tensoris simply a multiple of δi j. Hence, we can write

Ai j = −κ δi j, (1.64)

where κ(r, t) is termed the thermal conductivity of the fluid. It follows that the mostgeneral expression for the heat flux density in an isotropic fluid is

qi = −κ∂T∂xi, (1.65)

or, equivalently,q = −κ∇T. (1.66)

Moreover, it is a matter of experience that heat flows down temperature gradients:that is, κ > 0. We conclude that the net heat flux out of volume V is

.Q = −

∮Sκ∂T∂xi

dS i = −∫

V

∂

∂xi

(κ∂T∂xi

)dV, (1.67)

where use has been made of the tensor divergence theorem.


Equations (1.59)–(1.62) and (1.67) can be combined to give the following energyconservation equation:∫

V

∂

∂t

[ρ

(E + 1

2vi vi

)]+∂

∂x j

[ρ

(E + 1

2vi vi

)v j

]dV

=

∫V

[vi Fi +

∂

∂x j

(vi σi j + κ

∂T∂x j

)]dV. (1.68)

However, this result is valid irrespective of the size, shape, or location of volume V ,which is only possible if

∂

∂t

[ρ

(E + 1

2vi vi

)]+∂

∂x j

[ρ

(E + 1

2vi vi

)v j

]= vi Fi +

∂

∂x j

(viσi j + κ

∂T∂x j

)(1.69)

everywhere inside the fluid. Expanding some of the derivatives, and rearranging, weobtain

ρDDt

(E + 1

2vi vi

)= vi Fi +

∂

∂x j

(vi σi j + κ

∂T∂x j

), (1.70)

where use has been made of the continuity equation, (1.40). The scalar product of vwith the fluid equation of motion, (1.53), yields

ρ viDviDt= ρ

DDt

(12vi vi

)= vi Fi + vi

∂σi j

∂x j. (1.71)

Combining the previous two equations, we get

ρDEDt=∂vi∂x j

σi j +∂

∂x j

(κ∂T∂x j

). (1.72)

Finally, making use of Equation (1.26), we deduce that the energy conservation equa-tion for an isotropic Newtonian fluid takes the general form

DEDt= −

pρ

∂vi

∂xi+

1ρ

[χ +

∂

∂x j

(κ∂T∂x j

)]. (1.73)

Here,

χ =∂vi∂x j

di j = 2 µ(ei j ei j −

13

eii e j j

)= µ

(∂vi∂x j

∂vi∂x j+∂vi∂x j

∂v j

∂xi− 2

3∂vi∂xi

∂v j

∂x j

)(1.74)

is the rate of heat generation per unit volume due to viscosity. When written in vectorform, Equation (1.73) becomes

DEDt= −

pρ∇ · v +

χ

ρ+∇· (κ∇T )

ρ. (1.75)

According to the previous equation, the internal energy per unit mass of a co-movingfluid element evolves in time as a consequence of work done on the element bypressure as its volume changes, viscous heat generation due to flow shear, and heatconduction.


1.14 Equations of Incompressible Fluid FlowIn most situations of general interest, the flow of a conventional liquid, such as water,is incompressible to a high degree of accuracy. A fluid is said to be incompressiblewhen the mass density of a co-moving volume element does not change appreciablyas the element moves through regions of varying pressure. In other words, for anincompressible fluid, the rate of change of ρ following the motion is zero: that is,

DρDt= 0. (1.76)

In this case, the continuity equation (1.40) reduces to

∇ · v = 0. (1.77)

We conclude that, as a consequence of mass conservation, an incompressible fluidmust have a divergence-free, or solenoidal, velocity field. This immediately implies,from Equation (1.42), that the volume of a co-moving fluid element is a constant ofthe motion. In most practical situations, the initial density distribution in an incom-pressible fluid is uniform in space. Hence, it follows from Equation (1.76) that thedensity distribution remains uniform in space and constant in time. In other words,we can generally treat the density, ρ, as a uniform constant in incompressible fluidflow problems.

Suppose that the volume force acting on the fluid is conservative in nature (seeSection A.18): that is,

F = −ρ∇Ψ, (1.78)

where Ψ (r, t) is the potential energy per unit mass, and ρΨ the potential energy perunit volume. Assuming that the fluid viscosity is a spatially uniform quantity, whichis generally the case (unless there are strong temperature variations within the fluid),the Navier-Stokes equation for an incompressible fluid reduces to

DvDt= −∇pρ− ∇Ψ + ν∇ 2v, (1.79)

whereν =

µ

ρ(1.80)

is termed the kinematic viscosity, and has units of meters squared per second. Roughlyspeaking, momentum diffuses a distance of order

√ν t meters in t seconds as a conse-

quence of viscosity. The kinematic viscosity of water at 20 C is about 1.0×10−6 m2/s(Batchelor 2000). It follows that viscous momentum diffusion in water is a relativelyslow process.

The complete set of equations governing incompressible flow is

∇ · v = 0, (1.81)

DvDt= −∇p

ρ− ∇Ψ + ν∇ 2v. (1.82)


Here, ρ and ν are regarded as known constants, and Ψ (r, t) as a known function.Thus, we have four equations—namely, Equation (1.81), plus the three componentsof Equation (1.82)—for four unknowns—namely, the pressure, p(r, t), plus the threecomponents of the velocity, v(r, t). Note that an energy conservation equation isredundant in the case of incompressible fluid flow.

1.15 Equations of Compressible Fluid Flow

In many situations of general interest, the flow of gases is compressible. In otherwords, there are significant changes in the mass density as the gas flows from placeto place. For the case of compressible flow, the continuity equation (1.40), and theNavier-Stokes equation (1.56), must be augmented by the energy conservation equa-tion (1.75), as well as thermodynamic relations that specify the internal energy perunit mass, and the temperature in terms of the density and pressure. For an ideal gas,these relations take the form (Reif 1965)

E = cV

MT, (1.83)

T =MR

pρ, (1.84)

where cV is the molar specific heat at constant volume, R = 8.3145 J K−1 mol−1

the molar ideal gas constant, M the molar mass (i.e., the mass of 1 mole of gasmolecules), and T the temperature in degrees Kelvin. Incidentally, 1 mole corre-sponds to 6.0221 × 1024 molecules. Here, we have assumed, for the sake of simplic-ity, that cV is a uniform constant. It is also convenient to assume that the thermalconductivity, κ, is a uniform constant. Making use of these approximations, Equa-tions (1.40), (1.75), (1.83), and (1.84) can be combined to give

1γ − 1

(DpDt−γ pρ

DρDt

)= χ +

κMR∇ 2(

pρ

), (1.85)

where

γ =cp

cV=

cV + RcV

(1.86)

is the ratio of the molar specific heat at constant pressure, cp, to that at constantvolume, cV . [Incidentally, the result that cp = cV + R for an ideal gas is a standardtheorem of thermodynamics (Reif 1965).] (See Section 14.2.) The ratio of specificheats of dry air at 20 C is 1.40 (Batchelor 2000).


The complete set of equations governing compressible ideal gas flow are

DρDt= −ρ∇ · v, (1.87)

DvDt= −∇p

ρ− ∇Ψ + µ

ρ

[∇ 2v +

13∇(∇ · v)

], (1.88)

1γ − 1

(DpDt−γ pρ

DρDt

)= χ +

κMR∇ 2(

pρ

), (1.89)

where the dissipation function χ is specified in terms of µ and v in Equation (1.74).Here, µ, γ, κ, M, and R are regarded as known constants, and Ψ (r, t) as a knownfunction. Thus, we have five equations—namely, Equations (1.87) and (1.89), plusthe three components of Equation (1.88)—for five unknowns—namely, the density,ρ(r, t), the pressure, p(r, t), and the three components of the velocity, v(r, t).

1.16 Dimensionless Numbers in Incompressible FlowIt is helpful to normalize the equations of incompressible fluid flow, (1.81)–(1.82),in the following manner: ∇ = L∇, v = v/V0, t = (V0/L) t, Ψ = Ψ/(g L), andp = p/(ρV 2

0 + ρ g L + ρ νV0/L). Here, L is a typical spatial variation lengthscale,V0 a typical fluid velocity, and g a typical gravitational acceleration (assuming thatΨ represents a gravitational potential energy per unit mass). All barred quantitiesare dimensionless, and are designed to be comparable with unity. The normalizedequations of incompressible fluid flow take the form

∇ · v = 0, (1.90)

DvDt= −

(1 +

1Fr 2 +

1Re

)∇p −

∇ΨFr 2 +

∇2v

Re, (1.91)

where D/Dt = ∂/∂t + v · ∇, and

Re =L V0

ν, (1.92)

Fr =V0

(g L)1/2 . (1.93)

Here, the dimensionless quantities Re and Fr are known as the Reynolds numberand the Froude number, respectively. [After Osborne Reynolds (1842–1912) andWilliam Froude (1810–1879), respectively.] The Reynolds number is the typical ratioof inertial to viscous forces within the fluid, whereas the square of the Froude numberis the typical ratio of inertial to gravitational forces. Thus, viscosity is relativelyimportant compared to inertia when Re 1, and vice versa. Likewise, gravity is


relatively important compared to inertia when Fr 1, and vice versa. Note that, inprincipal, Re and Fr are the only quantities in Equations (1.90) and (1.91) that can besignificantly greater or smaller than unity.

For the case of water at 20 C, located on the surface of the Earth (Batchelor2000),

Re 1.0 × 106 L(m) V0(m s−1), (1.94)

Fr 3.2 × 10−1 V0(m s−1)/[L(m)]1/2. (1.95)

Thus, if L ∼ 1 m and V0 ∼ 1 m s−1, as is often the case for terrestrial water dynamics,then the previous expressions suggest that Re 1 and Fr ∼ O(1). In this situa-tion, the viscous term on the right-hand side of Equation (1.91) becomes negligible,and the (unnormalized) incompressible fluid flow equations reduce to the followinginviscid, incompressible, fluid flow equations:

∇ · v = 0, (1.96)

DvDt= −∇p

ρ− ∇Ψ. (1.97)

For the case of lubrication oil at 20C, located on the surface of the Earth,ν 1.0 × 10−4 m2 s−1 (i.e., oil is about 100 times more viscous than water), andso (Batchelor 2000)

Re 1.0 × 104 L(m) V0(m s−1), (1.98)

Fr 3.2 × 10−1 V0(m s−1)]/[L(m)]1/2. (1.99)

Suppose that oil is slowly flowing down a narrow lubrication channel such that L ∼10−3 m and V0 10−1 m s−1. It follows, from the previous expressions, that Re 1and Fr 1. In this situation, the inertial term on the left-hand side of (1.91) becomesnegligible, and the (unnormalized) incompressible fluid flow equations reduce to thefollowing inertia-free, incompressible, fluid flow equations:

∇ · v = 0, (1.100)

0 = −∇pρ− ∇Ψ + ν∇ 2v. (1.101)

1.17 Dimensionless Numbers in Compressible FlowIt is helpful to normalize the equations of compressible ideal gas flow, (1.87)–(1.89),in the following manner: ∇ = L∇, v = v/V0, t = (V0/L) t, ρ = ρ/ρ0, Ψ = Ψ/(g L),χ = (L/V0)2 χ, and p = (p− p0)/(ρ0 V 2

0 +ρ0 g L+ρ0 νV0/L). Here, L is a typical spa-tial variation lengthscale, V0 a typical fluid velocity, ρ0 a typical mass density, and g


a typical gravitational acceleration (assuming thatΨ represents a gravitational poten-tial energy per unit mass). Furthermore, p0 corresponds to atmospheric pressure atground level, and is a uniform constant. It follows that p represents deviations fromatmospheric pressure. All barred quantities are dimensionless, and are designed tobe comparable with unity. The normalized equations of compressible ideal gas flowtake the form

DρDt= −ρ∇ · v, (1.102)

DvDt= −

(1 +

1Fr 2 +

1Re

)∇pρ−∇ΨFr 2 +

1Re ρ

[∇

2v −

13∇(∇ · v)

], (1.103)

1γ − 1

[DpDt−γ(

p0 + pρ

)DρDt

]=

χ

1 + Re (1 + 1/Fr2)+

1Re Pr

∇ 2(

p0 + pρ

), (1.104)

p0 =1

γMa 2 (1 + 1/Fr 2 + 1/Re), (1.105)

where D/Dt ≡ ∂/∂t + v · ∇,

Re =L V0

ν, (1.106)

Fr =V0

(g L)1/2 , (1.107)

Pr =ν

κH, (1.108)

Ma =V0√γ p0/ρ0

, (1.109)

and

ν =µ

ρ0, (1.110)

κH =κMR ρ0

. (1.111)

Here, the dimensionless numbers Re, Fr, Pr, and Ma are known as the Reynoldsnumber, Froude number, Prandtl number, and Mach number, respectively. [Thelatter two numbers are named after Ludwig Prandtl (1875–1953) and Ernst Mach(1838–1916), respectively.] The Reynolds number is the typical ratio of inertial toviscous forces within the gas, the square of the Froude number is the typical ratioof inertial to gravitational forces, the Prandtl number is the typical ratio of the mo-mentum and thermal diffusion rates, and the Mach number is the typical ratio of thegas flow and sound propagation speeds. Thus, thermal diffusion is far faster thanmomentum diffusion when Pr 1, and vice versa. Moreover, the gas flow is termedsubsonic when Ma 1, supersonic when Ma 1, and transonic when Ma ∼ O(1).Note that

√γ p0/ρ0 is the speed of sound in the undisturbed gas (Reif 1965). The


quantity κH is called the thermal diffusivity of the gas, and has units of meters squaredper second. Thus, heat typically diffuses through the gas a distance

√κH t meters in

t seconds. The thermal diffusivity of dry air at atmospheric pressure and 20 C isabout κH = 2.1 × 10−5 m2 s−1 (Batchelor 2000). It follows that heat diffusion in airis a relatively slow process. The kinematic viscosity of dry air at atmospheric pres-sure and 20 C is about ν = 1.5 × 10−5 m2 s−1 (Batchelor 2000). Hence, momentumdiffusion in air is also a relatively slow process.

For the case of dry air at atmospheric pressure and 20 C (Batchelor 2000),

Re 6.7 × 104 L(m) V0(m s−1), (1.112)

Fr 3.2 × 10−1 V0(m s−1)/[L(m)]1/2, (1.113)

Pr 7.2 × 10−1, (1.114)

Ma 2.9 × 10−3 V0(m s−1). (1.115)

Thus, if L ∼ 1 m and V0 ∼ 1 m s−1, as is often the case for subsonic air dynamicsclose to the Earth’s surface, then the previous expressions suggest that Re 1,Ma 1, and Fr, Pr ∼ O(1). It immediately follows from Equation (1.105) thatp0 1. However, in this situation, Equation (1.104) is dominated by the second termin square brackets on its left-hand side. Hence, this equation can only be satisfied ifthe term in question is small, which implies that

DρDt 1. (1.116)

Equation (1.102) then gives∇ · v 1. (1.117)

It is evident that subsonic (i.e., Ma 1) gas flow is essentially incompressible. Thefact that Re 1 implies that such flow is also essentially inviscid. In the incompress-ible inviscid limit (in which ∇ · v = 0 and Re 1), the (unnormalized) compressibleideal gas flow equations reduce to the previously derived, inviscid, incompressible,fluid flow equations:

∇ · v = 0, (1.118)

DvDt= −∇p

ρ− ∇Ψ. (1.119)

It follows that the equations which govern subsonic gas dynamics close to the surfaceof the Earth are essentially the same as those that govern the flow of water.

Suppose that L ∼ 1 m and V0 ∼ 300 m s−1, as is typically the case for transonicair dynamics (e.g., air flow over the wing of a fighter jet). In this situation, Equa-tions (1.105) and (1.112)–(1.115) yield Re, Fr 1 and Ma, Pr, p0 ∼ O(1). It followsthat the final two terms on the right-hand sides of Equations (1.103) and (1.104) canbe neglected. Thus, the (unnormalized) compressible ideal gas flow equations reduce


to the following set of inviscid, adiabatic, ideal gas, flow equations:

DρDt= −ρ∇ · v, (1.120)

DvDt= −∇pρ, (1.121)

DDt

(pργ

)= 0. (1.122)

In particular, if the initial distribution of p/ργ is uniform in space, as is often thecase, then Equation (1.122) ensures that the distribution remains uniform as timeprogresses. In fact, it can be shown that the entropy per unit mass of an ideal gas is(Reif 1965)

S = cV

Mln(

pργ

). (1.123)

Hence, the assumption that p/ργ is uniform in space is equivalent to the assumptionthat the entropy per unit mass of the gas is a spatial constant. A gas for which thisis the case is termed homentropic. Equation (1.122) ensures that the entropy of a co-moving gas element is a constant of the motion in transonic flow. A gas for which thisis the case is termed isentropic. In the homentropic case, the previous compressiblegas flow equations simplify somewhat to give

DρDt= −ρ∇ · v, (1.124)

DvDt= −∇pρ, (1.125)

pp0=

(ρ

ρ0

)γ. (1.126)

Here, p0 is atmospheric pressure, and ρ0 is the density of air at atmospheric pressure.Equation (1.126) is known as the adiabatic gas law, and is a consequence of the factthat transonic gas dynamics takes place far too quickly for thermal heat conduction(which is a relatively slow process) to have any appreciable effect on the temperaturedistribution within the gas. Incidentally, a gas in which thermal diffusion is negligibleis generally termed adiabatic.


1.18 Fluid Equations in Cartesian Coordinates

Let us adopt the conventional Cartesian coordinate system, (x, y, z). According toEquation (1.26), the various components of the stress tensor are

σxx = −p + 2 µ∂vx

∂x, (1.127)

σyy = −p + 2 µ∂vy

∂y, (1.128)

σzz = −p + 2 µ∂vz∂z, (1.129)

σxy = σyx = µ

(∂vx

∂y+∂vy

∂x

), (1.130)

σxz = σzx = µ

(∂vx

∂z+∂vz∂x

), (1.131)

σyz = σzy = µ

(∂vy

∂z+∂vz∂y

), (1.132)

where v is the velocity, p the pressure, and µ the viscosity. The equations of com-pressible fluid flow, (1.87)–(1.89) (from which the equations of incompressible fluidflow can easily be obtained by setting ∆ = 0), become

DρDt= −ρ ∆, (1.133)

Dvx

Dt= −

1ρ

∂p∂x−∂Ψ

∂x+µ

ρ

(∇ 2vx +

13∂∆

∂x

), (1.134)

DvyDt= −1

ρ

∂p∂y− ∂Ψ∂y+µ

ρ

(∇ 2vy +

13∂∆

∂y

), (1.135)

DvzDt= −1

ρ

∂p∂z− ∂Ψ∂z+µ

ρ

(∇ 2vz +

13∂∆

∂z

), (1.136)

1γ − 1

(DρDt− γ pρ

DρDt

)= χ +

κMR∇ 2(

pρ

), (1.137)


where ρ is the mass density, γ the ratio of specific heats, κ the heat conductivity, Mthe molar mass, and R the molar ideal gas constant. Furthermore,

∆ =∂vx

∂x+∂vy

∂y+∂vz∂z, (1.138)

DDt=∂

∂t+ vx

∂

∂x+ vy

∂

∂y+ vz

∂

∂z, (1.139)

∇ 2 =∂ 2

∂x 2 +∂ 2

∂y 2 +∂ 2

∂z 2 , (1.140)

χ = 2 µ

(∂vx

∂x

)2+

(∂vy

∂y

)2+

(∂vz∂z

)2+

12

(∂vx

∂y+∂vy

∂x

)2+

12

(∂vx

∂z+∂vz∂x

)2+

12

(∂vy

∂z+∂vz∂y

)2 . (1.141)

Here, γ, µ, κ, and M are treated as uniform constants.

1.19 Fluid Equations in Cylindrical Coordinates

Let us adopt the cylindrical coordinate system, (r, θ, z). Making use of the resultsquoted in Section C.3, the components of the stress tensor are

σrr = −p + 2 µ∂vr∂r, (1.142)

σθθ = −p + 2 µ(1r∂vθ∂θ+vrr

), (1.143)

σzz = −p + 2 µ∂vz∂z, (1.144)

σrθ = σθr = µ

(1r∂vr∂θ+∂vθ∂r− vθ

r

), (1.145)

σrz = σzr = µ

(∂vr∂z+∂vz∂r

), (1.146)

σθz = σzθ = µ

(1r∂vz∂θ+∂vθ∂z

), (1.147)


whereas the equations of compressible fluid flow become

DρDt= −ρ ∆, (1.148)

DvrDt−v 2θ

r= −1

ρ

∂p∂r− ∂Ψ∂r

+µ

ρ

(∇ 2vr −

vr

r 2 −2r 2

∂vθ∂θ+

13∂∆

∂r

), (1.149)

DvθDt+vr vθ

r= −

1ρ r

∂p∂θ−

1r∂Ψ

∂θ

+µ

ρ

(∇ 2vθ +

2r 2

∂vr∂θ− vθ

r 2 +1

3 r∂∆

∂θ

), (1.150)

DvzDt= −1

ρ

∂p∂z− ∂Ψ∂z+µ

ρ

(∇ 2vz +

13∂∆

∂z

), (1.151)

1γ − 1

(DρDt− γ pρ

DρDt

)= χ +

κMR∇ 2(

pρ

), (1.152)

where

∆ =1r∂(r vr)∂r

+1r∂vθ∂θ+∂vz∂z, (1.153)

DDt=∂

∂t+ vr

∂

∂r+vθr∂

∂θ+ vz

∂

∂z, (1.154)

∇ 2 =1r∂

∂r

(r∂

∂r

)+

1r 2

∂ 2

∂θ 2 +∂ 2

∂z 2 , (1.155)

χ = 2 µ

(∂vr∂r

)2+

(1r∂vθ∂θ+vrr

)2+

(∂vz∂z

)2+

12


r

)2+

12

(∂vr∂z+∂vz∂r

)2+

12

(∂vθ∂z+

1r∂vz∂θ

)2 . (1.156)


1.20 Fluid Equations in Spherical Coordinates

Let us, finally, adopt the spherical coordinate system, (r, θ, ϕ). Making use of theresults quoted in Section C.4, the components of the stress tensor are

σrr = −p + 2 µ∂vr∂r, (1.157)

σθθ = −p + 2 µ(1r∂vθ∂θ+vrr

), (1.158)

σϕϕ = −p + 2 µ(

1r sin θ

∂vϕ

∂ϕ+vrr+

cot θ vθr

), (1.159)

σrθ = σθr = µ


r

), (1.160)

σrϕ = σϕr = µ

(1

r sin θ∂vr

∂ϕ+∂vϕ

∂r−vϕ

r

), (1.161)

σθϕ = σϕθ = µ

(1

r sin θ∂vθ∂ϕ+

1r∂vϕ

∂θ−

cot θ vϕr

), (1.162)

whereas the equations of compressible fluid flow become

DρDt= −ρ ∆, (1.163)

DvrDt−v 2θ + v

2ϕ

r= −1

ρ

∂p∂r− ∂Ψ∂r+µ

ρ

(∇ 2vr −

2 vrr 2 −

2r 2

∂vθ∂θ

−2 cot θ vθ

r 2 −2

r 2 sin θ∂vϕ

∂ϕ+

13∂∆

∂r

), (1.164)

DvθDt+vr vθ − cot θ v 2

ϕ

r= − 1

ρ r∂p∂θ− 1

r∂Ψ

∂θ+µ

ρ

(∇ 2vθ +

2r 2

∂vr∂θ

−vθ

r 2 sin2 θ−

2 cot θr 2 sin θ

∂vϕ

∂ϕ+

13 r

∂∆

∂θ

), (1.165)

DvϕDt+vr vϕ + cot θ vθ vϕ

r= − 1

ρ r sin θ∂p∂ϕ− 1

r sin θ∂Ψ

∂ϕ+µ

ρ

(∇ 2vϕ −

vϕ

r 2 sin2 θ

+2

r 2 sin2 θ

∂vr∂ϕ+

2 cot θr 2 sin θ

∂vθ∂ϕ+

13 r sin θ

∂∆

∂ϕ

), (1.166)

1γ − 1

(DρDt− γ pρ

DρDt

)= χ +

κMR∇ 2(

pρ

), (1.167)


where

∆ =1r 2

∂(r 2 vr)∂r

+1

r sin θ∂(sin θ vθ)

∂θ+

1r sin θ

∂vϕ

∂ϕ, (1.168)

DDt=∂

∂t+ vr

∂

∂r+vθ

r∂

∂θ+

vϕ

r sin θ∂

∂ϕ, (1.169)

∇ 2 =1r 2

∂

∂r

(r 2 ∂

∂r

)+

1r 2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1r 2 sin2 θ

∂ 2

∂ϕ 2 , (1.170)

χ = 2 µ

(∂vr∂r

)2+

(1r∂vθ∂θ+vrr

)2+

(1

r sin θ∂vϕ

∂ϕ+vrr+

cot θ vθr

)2+

12


r

)2+

12

(1

r sin θ∂vr∂ϕ+∂vϕ

∂r−vϕ

r

)2+

12

(1

r sin θ∂vθ

∂ϕ+

1r∂vϕ

∂θ−

cot θ vϕr

)2 . (1.171)

1.21 Exercises1.1 Equations (1.66), (1.75), and (1.87) can be combined to give the following

energy conservation equation for a non-ideal compressible fluid:

ρDEDt−

pρ

DρDt= χ − ∇ · q,

where ρ is the mass density, p the pressure, E the internal energy per unitmass, χ the viscous energy dissipation rate per unit volume, and q the heatflux density. We also have

DρDt= −ρ∇ · v,

q = −κ∇T,

where v is the fluid velocity, T the temperature, and κ the thermal conductiv-ity. According to a standard theorem in thermodynamics (Reif 1965),

T dS = dE − pρ 2 dρ,

where S is the entropy per unit mass. Moreover, the entropy flux density at agiven point in the fluid is (Hazeltine and Waelbroeck 2004)

s = ρS v +qT,


where the first term on the right-hand side is due to direct entropy convectionby the fluid, and the second is the entropy flux density associated with heatconduction.

Derive an entropy conservation equation of the form

dSdt+ ΦS = ΘS ,

where S is the net amount of entropy contained in some fixed volume V , ΦS

the entropy flux out of V , and ΘS the net rate of entropy creation within V .Give expressions for S , ΦS , and ΘS . Demonstrate that the entropy creationrate per unit volume is

θ =χ

T+

q · qκ T 2 .

Finally, show that θ ≥ 0, in accordance with the second law of thermodynam-ics.

1.2 The Navier-Stokes equation for an incompressible fluid of uniform mass den-sity ρ takes the form

DvDt= −∇p

ρ− ∇Ψ + ν∇ 2v,

where v is the fluid velocity, p the pressure, Ψ the potential energy per unitmass, and ν the (uniform) kinematic viscosity. The incompressibility con-straint requires that

∇ · v = 0.

Finally, the quantityω ≡ ∇ × v

is generally referred to as the fluid vorticity.

Derive the following vorticity evolution equation from the Navier-Stokesequation:

DωDt= (ω · ∇) v + ν∇ 2ω.

1.3 Consider two-dimensional incompressible fluid flow. Let the velocity fieldtake the form

v = vx(x, y, t) ex + vy(x, y, t) ey.

Demonstrate that the equations of incompressible fluid flow (see Exercise 1.2)can be satisfied by writing

vx = −∂ψ

∂y,

vy =∂ψ

∂x,


where∂ω

∂t+ [ψ, ω] = ν∇ 2ω,

andω = ∇ 2ψ.

Here, [A, B] ≡ ez · ∇A × ∇B, and ∇ 2 = ∂ 2/∂x 2 + ∂ 2/∂y 2. Furthermore, thequantity ψ is termed a stream function, because v · ∇ψ = 0. In other words,the fluid flow is everywhere parallel to contours of ψ.

1.4 Consider incompressible irrotational flow: that is, flow that satisfies

∇ × v = 0,

as well as

DvDt= −∇p

ρ− ∇Ψ + ν∇ 2v,

∇ · v = 0.

Here, v is the fluid velocity, ρ the uniform mass density, p the pressure,Ψ thepotential energy per unit mass, and ν the (uniform) kinematic viscosity.

Demonstrate that the previous equations can be satisfied by writing

v = −∇φ,

where∇ 2φ = 0,

and−∂φ∂t+

12v 2 +

pρ+ Ψ = C(t).

Here, C(t) is a spatial constant. This type of flow is known as potential flow,because the velocity field is derived from a scalar potential.

1.5 The equations of inviscid adiabatic ideal gas flow are


DvDt= −∇p

ρ− ∇Ψ,

DDt

(pργ

)= 0.

Here, ρ is the mass density, v the flow velocity, p the pressure,Ψ the potentialenergy per unit mass, and γ the (uniform) ratio of specific heats. Supposethat the pressure and potential energy are both time independent: that is,∂p/∂t = ∂Ψ/∂t = 0.


Demonstrate thatH =

12v 2 +

γ

γ − 1pρ+ Ψ

is a constant of the motion. In other words, DH/Dt = 0. This result is knownas Bernoulli’s theorem.

1.6 The equations of inviscid adiabatic non-ideal gas flow are


DvDt= −∇p

ρ− ∇Ψ,

DEDt− pρ 2

DρDt= 0.

Here, ρ is the mass density, v the flow velocity, p the pressure,Ψ the potentialenergy per unit mass, and E the internal energy per unit mass. Suppose thatthe pressure and potential energy are both time independent: that is, ∂p/∂t =∂Ψ/∂t = 0. Demonstrate that

H =12v 2 + E + p

ρ+ Ψ

is a constant of the motion. In other words, DH/Dt = 0. This result is a moregeneral form of Bernoulli’s theorem.

1.7 Demonstrate that Bernoulli’s theorem for incompressible, inviscid fluid flowtakes the form DH/Dt = 0, where

H =12v 2 +

pρ+ Ψ.

2Hydrostatics

2.1 IntroductionThis chapter discusses the mechanical equilibrium of incompressible fluids, a topicthat is conventionally termed hydrostatics. More information on the fundamentals ofhydrostatics can be found in Lamb 1928. More information on Maclaurin spheroids,Jacobi ellipsoids, and Roche ellipsoids can be found in Chandrasekhar 1969, andLamb 1993.

2.2 Hydrostatic PressureConsider a body of water which is stationary in a reference frame that is fixed withrespect to the Earth’s surface. In this chapter, such a frame is treated as approximatelyinertial. Let z measure vertical height, and suppose that the region z ≤ 0 is occupiedby water, and the region z > 0 by air. According to Equation (1.79), the air/watersystem remains in mechanical equilibrium (i.e., v = Dv/Dt = 0) provided

0 = ∇p + ρ∇Ψ, (2.1)

where p is the static fluid pressure, ρ the mass density, Ψ = g z the gravitationalpotential energy per unit mass, and g the (approximately uniform) acceleration dueto gravity. Now,

ρ(z) =

0 z > 0ρ0 z ≤ 0

, (2.2)

where ρ0 is the (approximately uniform) mass density of water. Here, the compara-tively small mass density of air has been neglected. Because ρ = ρ(z) and Ψ = Ψ (z),it immediately follows, from Equation (2.1), that p = p(z), where

dpdz= −ρ g. (2.3)

We conclude that constant pressure surfaces in a stationary body of water take theform of horizontal planes. Making use of Equation (2.2), the previous equation can

33


be integrated to give

p(z) =

p0 z > 0p0 − ρ0 g z z ≤ 0

, (2.4)

where p0 105 N m−2 is atmospheric pressure at ground level (Batchelor 2000).According to this expression, pressure in stationary water increases linearly withincreasing depth (i.e., with decreasing z, for z < 0). In fact, given that g 9.8 m s−2

and ρ0 103 kg m−3 (Batchelor 2000), we deduce that hydrostatic pressure in waterrises at the rate of 1 atmosphere (i.e., 105 N m−2) every 10.2 m increase in depthbelow the surface.

2.3 BuoyancyConsider the air/water system described in the previous section. Let V be some vol-ume, bounded by a closed surface S , that straddles the plane z = 0, and is thus par-tially occupied by water, and partially by air. The i-component of the net force actingon the fluid (i.e., either water or air) contained within V is written (see Section 1.3)

fi =∮

Sσi j dS j +

∫V

Fi dV, (2.5)

whereσi j = −p δi j (2.6)

is the stress tensor for a static fluid (see Section 1.5), and

F = −ρ g ez (2.7)

the gravitational force density. (Recall that the indices 1, 2, and 3 refer to the x-, y-,and z-axes, respectively. Thus, f3 ≡ fz, et cetera.) The first term on the right-handside of Equation (2.5) represents the net surface force acting across S , whereas thesecond term represents the net volume force distributed throughout V . Making useof the tensor divergence theorem (see Section B.4), Equations (2.5)–(2.7) yield thefollowing expression for the net force:

f = B +W, (2.8)

whereBi = −

∫V

∂p∂xi

dV, (2.9)

and

Wx = Wy = 0, (2.10)

Wz = −∫

Vρ g dV. (2.11)

Hydrostatics 35

Here, B is the net surface force, and W the net volume force.It follows from Equations (2.4) and (2.9) that

B = M0 g ez, (2.12)

where M0 = ρ0 V0. Here, V0 is the volume of that part of V which lies below thewaterline, and M0 the total mass of water contained within V . Moreover, from Equa-tions (2.2), (2.10), and (2.11),

W = −M0 g ez. (2.13)

It can be seen that the net surface force, B, is directed vertically upward, and ex-actly balances the net volume force, W, which is directed vertically downward. Ofcourse, W is the weight of the water contained within V . On the other hand, B,which is generally known as the buoyancy force, is the resultant pressure of the wa-ter immediately surrounding V . We conclude that, in equilibrium, the net buoyancyforce acting across S exactly balances the weight of the water inside V , so that thetotal force acting on the contents of V is zero, as must be the case for a system inmechanical equilibrium. We can also deduce that the line of action of B (which isvertical) passes through the center of gravity of the water inside V . Otherwise, a nettorque would act on the contents of V , which would contradict our assumption thatthe system is in mechanical equilibrium.

2.4 Equilibrium of Floating BodiesConsider the situation described in the previous section. Suppose that the fluid con-tained within V is replaced by a partially submerged solid body whose outer surfacecorresponds to S . Furthermore, suppose that this body is in mechanical equilibriumwith the surrounding fluid (i.e., it is stationary, and floating on the surface of thewater). It follows that the pressure distribution in the surrounding fluid is unchanged[because the force balance criterion (2.3) can be integrated to give the pressure dis-tribution (2.4) at all contiguous points in the fluid, provided that the fluid remainsin mechanical equilibrium]. We conclude that the net surface force acting across Sis also unchanged (because this is directly related to the pressure distribution in thefluid immediately surrounding V), which implies that the buoyancy force acting onthe floating body is the same as that acting on the displaced water: that is, the waterthat previously occupied V . In other words, from Equation (2.12),

B = W0 ez, (2.14)

where W0 = M0 g and M0 are the weight and mass of the displaced water, respec-tively. The fact that the buoyancy force is unchanged also implies that the verticalline of action of B passes through the center of gravity, H (say), of the displacedwater. Incidentally, H is generally known as the center of buoyancy.


A floating body of weight W is acted upon by two forces: namely, its own weight,

W = −W ez, (2.15)

and the buoyancy force, B = W0 ez, due to the pressure of the surrounding water.Of course, the line of action of W passes through the body’s center of gravity, G(say). So as to remain in equilibrium, the body must be subject to zero net force andzero net torque. The requirement of zero net force yields W0 = W. In other words,in equilibrium, the weight of the water displaced by a floating body is equal to theweight of the body, or, alternatively, in equilibrium, the magnitude of the buoyancyforce acting on a floating body is equal to the weight of the displaced water. Thisfamous result is known as Archimedes’ principle, after Archimedes of Syracuse (c.278–212 BCE). The requirement of zero net torque implies that, in equilibrium, thecenter of gravity, G, and center of buoyancy, H, of a floating body must lie on thesame vertical straight-line.

Consider a floating body of mass M and volume V . Let ρ = M/V be the body’smean mass density. Archimedes’ principle implies that, in equilibrium,

V0

V= s, (2.16)

where

s =ρ

ρ0(2.17)

is termed the body’s specific gravity. (Recall, that V0 is the submerged volume, andρ0 the mass density of water.) We conclude, from Equation (2.16), that the volumefraction of a floating body that is submerged is equal to the body’s specific gravity.Obviously, the specific gravity must be less than unity, because the submerged vol-ume fraction cannot exceed unity. In fact, if the specific gravity exceeds unity thenit is impossible for the buoyancy force to balance the body’s weight, and the bodyconsequently sinks.

Consider a body of volume V and specific gravity s that floats in equilibrium.It follows, from Equation (2.16), that the submerged volume is V0 = s V . Hence,the volume above the waterline is V1 = V − V0 = (1 − s) V . Suppose that the bodyis inverted such that its previously submerged part is raised above the waterline, andvice versa: that is, V0 ↔ V1. According to Equation (2.16), the body can only remainin equilibrium in this configuration if its specific gravity is

s′ =V1

V=

V − V0

V= 1 − s. (2.18)

We conclude that for every equilibrium configuration of a floating body of specificgravity s there exists an inverted equilibrium configuration for a body of the sameshape having the complementary specific gravity 1 − s.

Hydrostatics 37

2.5 Vertical Stability of Floating BodiesConsider a floating body of weight W that, in equilibrium, has a submerged volumeV0. Thus, the body’s downward weight is balanced by the upward buoyancy force,B = ρ0 V0 g: that is, ρ0 V0 g = W. Let A0 be the cross-sectional area of the body atthe waterline (i.e., in the plane z = 0). It is convenient to define the body’s meandraft (or mean submerged depth) as δ0 = V0/A0. Suppose that the body is displacedslightly downward, without rotation, such that its mean draft becomes δ0 + δ1, where|δ1| δ0. Assuming that the cross-sectional area in the vicinity of the waterline isconstant, the new submerged volume is V ′0 = A0 (δ0 + δ1) = V0 + A0 δ1, and the newbuoyancy force becomes B′ = ρ0 V ′0 g = W + ρ0 A0 g δ1 = (1 + δ1/δ0) W. However,the weight of the body is unchanged. Thus, the body’s perturbed vertical equation ofmotion is written

Wg

d 2δ1

dt 2 = W − B′ = −Wδ0δ1, (2.19)

which reduces to the simple harmonic equation

d 2δ1

dt 2 = −g

δ0δ1. (2.20)

We conclude that if a floating body of mean draft δ0 is subject to a small verticaldisplacement then it oscillates about its equilibrium position at the characteristic fre-quency

ω =

√g

δ0. (2.21)

It follows that such a body is unconditionally stable to small vertical displacements.Incidentally, the previous calculation neglects the phenomenon of added mass, bywhich some of the water surrounding the floating body oscillates in sympathy withit, thereby increasing the body’s effective inertia. (See Sections 5.9 and 7.10.)

2.6 Angular Stability of Floating BodiesLet us now investigate the stability of floating bodies to angular displacements. Forthe sake of simplicity, we shall only consider bodies that have two mutually perpen-dicular planes of symmetry. Suppose that when such a body is in an equilibrium statethe two symmetry planes are vertical, and correspond to the x = 0 and y = 0 planes.As before, the z = 0 plane coincides with the surface of the water. It follows, fromsymmetry, that when the body is in an equilibrium state its center of gravity, G, andcenter of buoyancy, H, both lie on the z-axis.

Suppose that the body turns through a small angle θ about some horizontal axis,lying in the plane z = 0, that passes through the origin. Let GH be that, originally


STABLE

θ

H H ′

G

UNSTABLE

M

θ

M

G

H H ′

Figure 2.1Stable and unstable configurations for a floating body.

vertical, straight-line that passes through the body’s center of gravity, G, and originalcenter of buoyancy, H. Owing to the altered shape of the volume of displaced water,the center of buoyancy is shifted to some new position H′. Let the vertical straight-line passing through H′ meet GH at M. (See Figure 2.1.) Point M (in the limitθ → 0) is called the metacenter. In the disturbed state, the body’s weight W actsdownward through G, and the buoyancy ρ0 V0 g acts upward through M. Let usassume that the submerged volume, V0, is unchanged from the equilibrium state(which excludes vertical oscillations from consideration). It follows that the weightand the buoyancy force are equal and opposite, so that there is no net force on thebody. However, as can be seen from Figure 2.1, the weight and the buoyancy forcegenerate a net torque of magnitude τ = W λ sin θ. Here, λ is the length MG: thatis, the distance between the metacenter and the center of gravity. This distance isgenerally known as the metacentric height, and is defined such that it is positivewhen M lies above G, and vice versa. Moreover, as is also clear from Figure 2.1,when M lies above G the torque acts to reduce θ, and vice versa. In the former case,the torque is known as a righting torque. We conclude that a floating body is stableto small angular displacements about some horizontal axis lying in the plane z = 0provided that its metacentric height is positive: that is, provided that the metacenterlies above the center of gravity. Because we have already demonstrated that a floatingbody is unconditionally stable to small vertical displacements (and as it is also fairlyobvious that such a body is neutrally stable to both horizontal displacements andangular displacements about a vertical axis passing through its center of gravity), itfollows that a necessary and sufficient condition for the stability of a floating body toa general small perturbation (made up of arbitrary linear and angular components) is

Hydrostatics 39

that its metacentric height be positive for angular displacements about any horizontalaxis.

2.7 Determination of Metacentric HeightSuppose that the floating body considered in the previous section is in an equilibriumstate. Let A0 be the cross-sectional area at the waterline: that is, in the plane z = 0.Because the body is assumed to be symmetric with respect to the x = 0 and y = 0planes, we have ∫

A0

x dx dy =∫

A0

y dx dy =∫

A0

x y dx dy = 0, (2.22)

where the integrals are taken over the whole cross-section at z = 0. Let δ(x, y) bethe body’s draft: that is, the vertical distance between the surface of the water andthe body’s lower boundary. It follows, from symmetry, that δ(−x, y) = δ(x, y) andδ(x,−y) = δ(x, y). Moreover, the submerged volume is

V0 =

∫A0

∫ δ

0dx dy dz =

∫A0

δ(x, y) dx dy. (2.23)

It also follows from symmetry that∫A0

x δ(x, y) dx dy =∫

A0

y δ(x, y) dx dy = 0. (2.24)

The depth of the unperturbed center of buoyancy below the surface of the water is

h =

∫A0

∫ δ0 z dx dy dz

V0=

12 V0

∫A0

δ 2(x, y) dx dy =δ 2

0 A0

2 V0, (2.25)

where

δ0 =

∫

A0δ 2(x, y) dx dy

A0

1/2

. (2.26)

Finally, from symmetry, the unperturbed center of buoyancy lies at x = y = 0.Suppose that the body now turns through a small angle θ about the x-axis. As is

easily demonstrated, the body’s new draft becomes δ′(x, y) δ(x, y) − θ y. Hence,the new submerged volume is

V ′0 =∫

A0

[δ(x, y) − θ y] dx dy = V0 + θ

∫A0

y dx dy = V0, (2.27)

where use has been made of Equations (2.22) and (2.23). Thus, the submerged vol-ume is unchanged, as should be the case for a purely angular displacement. The new


depth of the center of buoyancy is

h′ =

∫A0

∫ δ′0 z dx dy dz

V0=

12 V0

∫A0

[δ 2(x, y) − 2 θ y δ(x, y) + O(θ 2)

]dx dy = h,

(2.28)where use has been made of Equations (2.24) and (2.25). Thus, the depth of thecenter of buoyancy is also unchanged. Moreover, from symmetry, it is clear that thecenter of buoyancy still lies at x = 0. Finally, the new y-coordinate of the center ofbuoyancy is ∫

A0

∫ δ′0 y dx dy dz

V0=

∫A0y [δ(x, y) − θ y] dx dy

V0= −θ

κ 2x A0

V0, (2.29)

where use has been made of Equation (2.24). Here,

κx =

∫

A0y2 dx dy

A0

1/2

, (2.30)

is the radius of gyration of area A0 about the x-axis.It follows, from the previous analysis, that if the floating body under considera-

tion turns through a small angle θ about the x-axis then its center of buoyancy shiftshorizontally a distance θ κ 2

x A0/V0 in the plane perpendicular to the axis of rotation.In other words, the distance HH′ in Figure 2.1 is θ κ 2

x A0/V0. Simple trigonometryreveals that θ HH′/MH′ (assuming that θ is small). Hence, MH′ = HH′/θ =κ 2

x A0/V0. Now, MH′ is the height of the metacenter relative to the center of buoy-ancy. However, the center of buoyancy lies a depth h below the surface of the water(which corresponds to the plane z = 0). Hence, the z-coordinate of the metacen-ter is zM = κ

2x A0/V0 − h. Finally, if zG and zH = −h are the z-coordinates of the

unperturbed centers of gravity and buoyancy, respectively, then

zM =κ 2

x A0

V0+ zH , (2.31)

and the metacentric height, λ = zM − zG, becomes

λ =κ 2

x A0

V0− zGH , (2.32)

where zGH = zG − zH . Note that, because κ 2x A0/V0 > 0, the metacenter always lies

above the center of buoyancy.A simple extension of the previous argument reveals that if the body turns through

a small angle θ about the y-axis then the metacentric height is

λ =κ 2y A0

V0− zGH , (2.33)

Hydrostatics 41

where

κy =

∫

A0x2 dx dy

A0

1/2

, (2.34)

is radius of gyration of area A0 about the y-axis. Finally, as is easily demonstrated,if the body rotates about a horizontal axis which subtends an angle α with the x-axisthen

λ =κ 2 A0

V0− zGH , (2.35)

whereκ 2 = κ 2

x cos2 α + κ 2y sin2 α. (2.36)

Thus, the minimum value of κ 2 is the lesser of κ 2x and κ 2

y . It follows that the equilib-rium state in question is unconditionally stable provided it is stable to small ampli-tude angular displacements about horizontal axes normal to its two vertical symmetryplanes (i.e., the x = 0 and y = 0 planes).

As an example, consider a uniform rectangular block of specific gravity s floatingsuch that its sides of length a, b, and c are parallel to the x-, y-, and z-axes, respec-tively. Such a block can be thought of as a very crude model of a ship. The volumeof the block is V = a b c. Hence, the submerged volume is V0 = s V = s a b c. Thecross-sectional area of the block at the waterline (z = 0) is A0 = a b. It is easilydemonstrated that δ(x, y) = δ0 = V0/A0 = s c. Thus, the center of buoyancy lies adepth h = δ 2

0 A0/2 V0 = s c/2 below the surface of the water. [See Equation (2.25).]Moreover, by symmetry, the center of gravity is a height c/2 above the bottom sur-face of the block, which is located a depth s c below the surface of the water. Hence,zH = −h = −s c/2, zG = c/2 − s c, and zGH = c (1 − s)/2. Consider the stability ofthe block to small amplitude angular displacements about the x-axis. We have

κ 2x =

∫ a/2−a/2

∫ b/2−b/2 y

2 dx dy

a b=

b 2

12. (2.37)

Hence, from Equation (2.32), the metacentric height is

λ =b 2

12 s c− c

2(1 − s). (2.38)

The stability criterion λ > 0 yields

b 2

6 c 2 − s (1 − s) > 0. (2.39)

Because the maximum value that s (1 − s) can take is 1/4, it follows that the block isstable for all specific gravities when

c < c0 =

√23

b. (2.40)


On the other hand, if c > c0 then the block is unstable for intermediate specificgravities such that s− < s < s+, where

s± =1 ±

√1 − c 2

0 /c2

2, (2.41)

and is stable otherwise. Assuming that the block is stable, its angular equation ofmotion is written

Id 2θ

dt 2 = −W λ sin θ −W λ θ, (2.42)

where

I =WgV

∫ a/2

−a/2

∫ b/2

−b/2

∫ c−s c

−s c(y 2 + z 2) dx dy dz =

W12 g

(b 2 + 4 [(1 − s)3 + s 3] c 2

)(2.43)

is the moment of inertia of the block about the x-axis. Thus, we obtain the the simpleharmonic equation

d 2θ

dt 2 = −ω2 θ, (2.44)

where

ω 2 =W λ

I=g

s cc 2

0 − 4 s (1 − s) c 2

c 20 + (8/3) [(1− s)3 + s 3] c 2

. (2.45)

We conclude that the block executes small amplitude angular oscillations about the x-axis at the angular frequencyω. For the case of rotation about the y-axis, the previousanalysis is unchanged except that a ↔ b. The previous analysis again neglects thephenomenon of added mass, and, therefore, underestimates the effective inertia ofthe block. (See Sections 5.9 and 7.10.)

The metacentric height of a conventional ship whose length greatly exceeds itswidth is typically much less for rolling (i.e., rotation about a horizontal axis runningalong the ship’s length) than for pitching (i.e., rotation about a horizontal axis per-pendicular to the ship’s length), because the radius of gyration for pitching greatlyexceeds that for rolling. As is clear from Equation (2.45), a ship with a relativelysmall metacentric height (for rolling) has a relatively long roll period, and vice versa.An excessively low metacentric height increases the chances of a ship capsizing ifthe weather is rough, if its cargo/ballast shifts, or if the ship is damaged and partiallyflooded. For this reason, maritime regulatory agencies, such as the International Mar-itime Organization, specify minimum metacentric heights for various different typesof sea-going vessel. A relatively large metacentric height, on the other hand, gener-ally renders a ship uncomfortable for passengers and crew, because the ship executesshort period rolls, resulting in large g-forces. Such forces also increase the risk thatcargo may break loose or shift.

We saw earlier, in Section 2.4, that if a body of specific gravity s floats in verti-cal equilibrium in a certain position then a body of the same shape, but of specificgravity 1 − s, can float in vertical equilibrium in the inverted position. We shall nowdemonstrate that these positions are either both stable, or both unstable, provided the

Hydrostatics 43

body is of uniform density. Let V1 and V2 be the volumes that are above and belowthe waterline, respectively, in the first position. Let H1 and H2 be the mean centers ofthese two volumes, and H that of the whole volume. It follows that H2 is the centerof buoyancy in the first position, H1 the center of buoyancy in the second (inverted)position, and H the center of gravity in both positions. Moreover,

V1 H1G = V2 H2G =(

V1 V2

V1 + V2

)H1H2, (2.46)

where H1G is the distance between points H1 and G, et cetera. The metacentricheights in the first and second positions are

λ1 =κ 2 AV1− H1G =

1V1

[A κ 2 −

(V1 V2

V1 + V2

)H1H2

], (2.47)

λ2 =κ 2 AV2− H2G =

1V2

[A κ 2 −

(V1 V2

V1 + V2

)H1H2

], (2.48)

respectively, where A and κ are the area and radius of gyration of the common water-line section, respectively. Thus,

λ1 λ2 =1

V1 V2

[A κ 2 −

(V1 V2

V1 + V2

)H1H2

] 2

≥ 0, (2.49)

which implies that λ1 0 as λ2 0, and vice versa. It follows that the first andsecond positions are either both stable, both marginally stable, or both unstable.

2.8 Energy of a Floating BodyThe conditions governing the equilibrium and stability of a floating body can also bededuced from the principle of energy.

For the sake of simplicity, let us suppose that the water surface area is infinite,so that the immersion of the body does not generate any change in the water level.The potential energy of the body itself is W zG, where W is the body’s weight, andzG the height of its center of gravity, G, relative to the surface of the water. If thebody displaces a volume V0 of water then this effectively means that a weight ρ0 V0 ofwater, whose center of gravity is located at the center of buoyancy, H, is removed, andthen spread as an infinitely thin film over the surface of the water. This involves a gainof potential energy of −ρ0 V0 zH , where zH is the height of H relative to the surface ofthe water. Vertical force balance requires that W = ρ0 V0. Thus, the potential energyof the system is W zGH (modulo an arbitrary additive constant), where zGH = zG − zH

is the height of the center of gravity relative to the center of buoyancy.According to the principles of statics, an equilibrium state corresponds to either

a minimum or a maximum of the potential energy (Fitzpatrick 2012). However, such


r

G

pH0

A

H

B

Figure 2.2Curve of buoyancy for a floating body.

an equilibrium is only stable when the potential energy is minimized. Thus, it followsthat a stable equilibrium configuration of a floating body is such as to minimize theheight of the body’s center of gravity relative to its center of buoyancy.

2.9 Curve of BuoyancyConsider a floating body in vertical force balance that is slowly rotated about a hor-izontal axis normal to one of its vertical symmetry planes. Let us take the center ofgravity, G, which necessarily lies in this plane, as the origin of a coordinate systemthat is fixed with respect to the body. As illustrated in Figure 2.2, as the body rotates,the locus of its center of buoyancy, H, as seen in the fixed reference frame, appearsto traces out a curve, AB, in the plane of symmetry. This curve is known as the curveof buoyancy. Let r represent the radial distance from the origin, G, to some point, H,on the curve of buoyancy. Note that the tangent to the curve of buoyancy is alwaysorientated horizontally. This follows because, as was shown in the previous section,small rotations of a floating body in vertical force balance cause its center of buoy-ancy to shift horizontally, rather than vertically, in the plane perpendicular to the axisof rotation. Thus, the difference in vertical height, zGH , between the center of gravity

Hydrostatics 45

and the center of buoyancy is equal to the perpendicular distance, p, between G andthe tangent to the curve of buoyancy at H. An equilibrium configuration thereforecorresponds to a maximum or a minimum of p as point H moves along the curveof buoyancy. However, the equilibrium is only stable if p is minimized. If R is theradius of curvature of the curve of buoyancy then, according to a standard result indifferential calculus (Lamb 1928),

R = rdrdp. (2.50)

Writing this result in the form

δr =Rrδp, (2.51)

it can be seen that maxima and minima of δp, which are the points on the curve ofbuoyancy where δp = 0, correspond to the points where δr = 0, and are, thus, coin-cident with maxima and minima of r. In other words, an equilibrium configurationcorresponds to a point of maximum or minimum r on the curve of buoyancy: that is,a point at which GH meets the curve at right-angles. At such a point, r = p, and thepotential energy consequently takes the value W r.

Let H0 be a point on the curve of buoyancy, and let r0, p0, and R0 be the corre-sponding values of r, p, and R. For neighboring points on the curve, we can write

r − r0 =drdp

∣∣∣∣∣H0

(p − p0), (2.52)

or

r − r0 =R0

r0(p − p0). (2.53)

It follows that p− p0 has the same sign as r−r0 (because R0 and r0 are both positive).[The fact that R0 is positive (i.e., dr/dp > 0) follows from the previously establishedresult that the metacenter, which is the center of curvature of the curve of buoyancy,always lies above the center of buoyancy, implying that the curve of buoyancy isnecessarily concave upwards.] Hence, the minima and maxima of r occur simultane-ously with those of p. Consequently, a stable equilibrium configuration correspondsto a point of minimum r on the curve of buoyancy: that is, a minimum in the distanceGH between the center of gravity and the center of buoyancy.

We can use the previous result to determine the stable equilibrium configurationsfor a beam of square cross-section, and uniform specific gravity s, that floats with itslength horizontal. In order to achieve this goal, we must calculate the distance GH forall possible configurations of the beam that are in vertical force balance. However,we need only consider cases where s < 1/2, because, according to the analysis ofSection 2.6, for every stable equilibrium configuration with s = s0 < 1/2 there is acorresponding stable inverted configuration with s = 1 − s0 > 1/2, and vice versa.

Let us define fixed rectangular axes, x and y, passing through the center of themiddle section of the beam, and running parallel to its sides. Let us start with the


Q

A B

θP

P ′

O

y

xQ′

Figure 2.3Beam of square cross-section floating with two corners immersed.

case where the waterline PQ is parallel to a side. (See Figure 2.3.) If the length of aside is 2 a then Equation (2.16) yields

AP = BQ = 2 a s. (2.54)

Suppose that the beam is turned through an angle θ > 0 such that the waterlineassumes the position P′Q′ in Figure 2.3, but still intersects two opposite sides. Thelengths AP′ and BQ′ satisfy

BQ′ − AP′ = 2 a tan θ. (2.55)

Moreover, the area of the trapezium P′ABQ′ must match that of the rectangle PABQin order to ensure that the submerged volume remains invariant (otherwise, the beamwould not remain in vertical force balance): that is,

(AP′ + BQ′) a = 4 a 2 s. (2.56)

It follows that

AP′ = a (2 s − tan θ), (2.57)

BQ′ = a (2 s + tan θ). (2.58)

The constraint that the waterline intersect two opposite sides of the beam implies thatAP′ > 0, and, hence, that

tan θ < 2 s. (2.59)

Hydrostatics 47

The coordinates of the center of buoyancy, H, which is the mean center of the trapez-ium P′ABQ′, are

x =

∫ a−a

∫ aa−h(x) x dx dy∫ a

−a

∫ aa−h(x) dx dy

=(2/3) a 3 tan θ

4 a 2 s=

a6 s

tan θ, (2.60)

y =

∫ a−a

∫ aa−h(x) y dx dy∫ a

−a

∫ aa−h(x) dx dy

=4 a 3 s (1 − s) − (1/3) a 3 tan2 θ

4 a 2 s= (1 − s) a − a

12 stan2 θ, (2.61)

whereh(x) = 2 a s + x tan θ. (2.62)

Thus, if u = r 2/a 2 = (x 2 + y 2)/a 2 then

u =t 2

36 s 2 +

[(1 − s) − t 2

12 s

] 2

, (2.63)

where t = tan θ. A stable equilibrium state corresponds to a minimum of r withrespect to θ, and, hence, of u with respect to t. However,

dudt=

t36 s 2

[t 2 − 12 s (1 − s) + 2

], (2.64)

d 2udt 2 =

136 s 2

[3 t 2 − 12 s (1 − s) + 2

]. (2.65)

The minima and maxima of u occur when du/dt = 0, d 2u/dt 2 > 0 and du/dt = 0,d 2u/dt 2 < 0, respectively. It follows that the symmetrical position, t = 0, in whichthe sides of the beam are either parallel or perpendicular to the waterline, is alwaysan equilibrium, but is only stable when

s 2 − s +16> 0 : (2.66)

that is, when s < 1/2 − 1/√

12 = 0.2113. It is also possible to obtain equilibria inasymmetric positions such that t is the root of

t 2 = 12 s (1 − s) − 2. (2.67)

Such equilibria only exist for s > 0.2113, and are stable. Finally, in order to satisfythe constraint (2.59), we must have t < 2 s, which, in combination with the previousequation, implies that

8 s 2 − 6 s + 1 > 0, (2.68)

or s < 0.25.


θ

A B

O

y

x

P ′

Q′

Figure 2.4Beam of square cross-section floating with one corner immersed.

Suppose that the constraint (2.59) is not satisfied, so that the immersed portion ofthe beam’s cross-section is triangular. (See Figure 2.4.) It is clear that

BQ′

BP′= tan θ. (2.69)

Moreover, the area of the triangle P′BQ′ in Figure 2.4 must match that of the rect-angle PABQ in Figure 2.3, in order to ensure that the submerged volume remaininvariant: that is,

12

BP′ BQ′ = 4 a 2 s. (2.70)

It follows that

BP′ = (8 s/ tan θ)1/2 a, (2.71)

BQ′ = (8 s tan θ)1/2 a, (2.72)

or, writing z 2 = tan θ and ω 2 = (8/9) s,

BP′ = 3ω z−1 a, (2.73)

BQ′ = 3ω z a. (2.74)

The coordinates of the center of buoyancy, H, which is the mean center of triangle

Hydrostatics 49

P′BQ′, are

x = a − BP′/3 = a (1 − ω z−1), (2.75)

y = a − BQ′/3 = a (1 − ω z), (2.76)

because the perpendicular distance of the mean center of a triangle from one of itssides is one third of the perpendicular distance from the side to the opposite vertex.Thus, if u = r 2/a 2 = (x 2 + y 2)/a 2 then

u = (1 − ω z−1)2 + (1 − ω z)2, (2.77)

dudz=

2ω 2 (z 2 − 1) (z 2 − ω−1 z + 1)z 3 , (2.78)

d 2udz 2 =

2ω 2 (z 4 − 2ω−1 z + 3)z 4 . (2.79)

Moreover, the constraint (2.59) yields

z >32ω. (2.80)

The stable and unstable equilibria correspond to du/dz = 0, d 2u/dz 2 > 0 and du/dz,d 2u/dz 2 < 0, respectively. It follows that the symmetrical position, z = 1, in whichthe diagonals of the beam are either parallel or perpendicular to the waterline is anequilibrium provided ω < 2/3, or s < 1/2, but is only stable when ω > 1/2, ors > 9/32 = 0.28125. It is also possible to obtain equilibria in asymmetric positionssuch that z is the root of

z 2 − ω−1 z + 1 = 0. (2.81)

Such equilibria only exist for√

2/3 < ω < 1/2, or 1/4 < s < 9/32, and are stable.In summary, the stable equilibrium configurations of a beam of square cross-

section, floating with its length horizontal, are such that the sides are either parallelor perpendicular to the waterline for s < 0.2113, such that two corners are immersedbut the sides and diagonals are neither parallel nor perpendicular to the waterline for0.2113 < s < 0.25, such that only one corner is immersed but the sides and diagonalsare neither parallel nor perpendicular to the waterline for 0.25 < s < 0.28125, andsuch that the diagonals are either parallel or perpendicular to waterline for 0.28125 <s < 0.5. For s > 0.5, the stable configurations are the same as those for a beam withthe complimentary specific gravity 1 − s.

2.10 Rotational HydrostaticsConsider the equilibrium of an incompressible fluid that is uniformly rotating at afixed angular velocity ω in some inertial frame of reference. Of course, such a fluid


appears stationary in a non-inertial co-rotating reference frame. Moreover, accordingto standard Newtonian dynamics (Fitzpatrick 2012), the force balance equation forthe fluid in the co-rotating frame takes the form (cf., Section 2.2)

0 = ∇p + ρ∇Ψ + ρω × (ω × r), (2.82)

where p is the static fluid pressure, ρ the mass density, Ψ the gravitational potentialenergy per unit mass, and r a position vector (measured with respect to an originthat lies on the axis of rotation). The final term on the right-hand side of the previousequation represents the fictitious centrifugal force density. Without loss of generality,we can assume that ω = ω ez. It follows that

0 = ∇p + ρ∇(Ψ + Ψ ′), (2.83)

where

Ψ ′ = −12ω 2 (x 2 + y 2) (2.84)

is the so-called centrifugal potential. Recall, incidentally, that ρ is a uniform constantin an incompressible fluid.

As an example, consider the equilibrium of a body of water, located on the Earth’ssurface, that is uniformly rotating about a vertical axis at the fixed angular velocityω. It is convenient to adopt cylindrical coordinates (see Section C.3), r, θ, z, whosesymmetry axis coincides with the axis of rotation. Let z increase upward. It followsthat Ψ = g z and Ψ ′ = −(1/2)ω 2 r 2. Assuming that the pressure distribution isaxisymmetric, so that p = p(r, z), the force balance equation, (2.83), reduces to

∂p∂r+ ρ

∂(Ψ + Ψ ′)∂r

= 0, (2.85)

∂p∂z+ ρ

∂(Ψ + Ψ ′)∂z

= 0, (2.86)

or

∂p∂r− ρω 2 r = 0, (2.87)

∂p∂z+ ρ g = 0. (2.88)

The previous two equations can be integrated to give

p(r, z) = p0 + ρ

(12ω 2 r 2 − g z

), (2.89)

where p0 is a constant. Thus, constant pressure surfaces in a uniformly rotatingbody of water take the form of paraboloids of revolution about the rotation axis.Suppose that p0 represents atmospheric pressure. In this case, the surface of thewater is the locus of p(r, z) = p0: that is, it is the constant pressure surface whose

Hydrostatics 51

pressure matches that of the atmosphere. It follows that the surface of the water isthe paraboloid of revolution

z =ω 2

2 gr 2, (2.90)

where r is the perpendicular distance from the axis of rotation, and z = 0 the on-axisheight of the surface.

According to the analysis of Section 2.3, the buoyancy force acting on any co-rotating solid body, which is wholly or partially immersed in the water, is the sameas that which would maintain the mass of water displaced by the body in relativeequilibrium. In the case of a floating body, this mass is limited by the continuation ofthe water’s curved surface through the body. Let points G and H represent the centersof gravity and buoyancy, respectively, of the body. Of course, the latter point issimply the center of gravity of the displaced water. Suppose that G and H are locatedperpendicular distances rG and rH from the axis of rotation, respectively. Finally,let M be the mass of the body, and M0 the mass of the displaced water. It followsthat the buoyancy force has an upward vertical component M0 g, and an outwardhorizontal component −M0 ω

2 rH . Thus, according to standard Newtonian dynamics(Fitzpatrick 2012), the equation of horizontal motion of a general co-rotating body is

M (..r − ω 2 rG) = −M0 ω2 rH , (2.91)

where .= d/dt. From Archimedes’ principle, M0 = M for the case of a floating

body that is less dense than water. However, if the body is of uniform density thenrH > rG, as a consequence of the curvature of the water’s surface. Hence, we obtain

..r = −ω 2 (rH − rG) < 0. (2.92)

In other words, a floating body drifts radially inward towards the rotation axis. Onthe other hand, M0 < M for a fully submerged body that is more dense than water.However, if the body is of uniform density then its centers of gravity and buoyancycoincide with one another, so that rH = rG. Hence, we obtain

..r = (M − M0)ω 2 rG > 0. (2.93)

In other words, a fully submerged body drifts radially outward from the rotation axis.The previous analysis accounts for the common observation that objects heavier thanwater, such as grains of sand, tend to collect on the outer side of a bend in a fastflowing river, while floating objects, such as sticks, tend to collect on the inner side.

2.11 Equilibrium of a Rotating Liquid BodyConsider a self-gravitating liquid body in outer space that is rotating uniformly aboutsome fixed axis passing through its center of mass. What is the shape of the body’s


bounding surface? This famous theoretical problem had its origins in investigationsof the figure of a rotating planet, such as the Earth, that were undertaken by New-ton, Maclaurin, Jacobi, Meyer, Liouville, Dirichlet, Dedekind, Riemann, and othercelebrated scientists, in the 17th, 18th, and 19th centuries (Chandrasekhar 1969).Incidentally, it is reasonable to treat the Earth as a liquid, for the purpose of thiscalculation, because the shear strength of the solid rock out of which the terrestrialcrust is composed is nowhere near sufficient to allow the actual shape of the Earth todeviate significantly from that of a hypothetical liquid Earth (Fitzpatrick 2012).

In a co-rotating reference frame, the shape of a self-gravitating, rotating, liquidplanet is determined by a competition between fluid pressure, gravity, and the ficti-tious centrifugal force. The latter force opposes gravity in the plane perpendicularto the axis of rotation. Of course, in the absence of rotation, the planet would bespherical. Thus, we would expect rotation to cause the planet to expand in the planeperpendicular to the rotation axis, and to contract along the rotation axis (in order toconserve volume).

For the sake of simplicity, we shall restrict our investigation to a rotating planetof uniform density whose outer boundary is ellipsoidal. An ellipsoid is the three-dimensional generalization of an ellipse. Let us adopt the right-handed Cartesiancoordinate system x1, x2, x3. An ellipse whose principal axes are aligned along thex1- and x2-axes satisfies

x 21

a 21

+x 2

2

a 22

= 1, (2.94)

where a1 and a2 are the corresponding principal radii. Moreover, as is easily demon-strated,

A =∫

dA = π a1 a2, (2.95)∫x 2

i dA =14

a 2i A, (2.96)∫

x1 x2 dA = 0, (2.97)

where A is the area, dA an element of A, and the integrals are taken over the wholeinterior of the ellipse. Likewise, an ellipsoid whose principal axes are aligned alongthe x1-, x2-, and x3-axes satisfies

x 21

a 21

+x 2

2

a 22

+x 2

3

a 23

= 1, (2.98)

where a1, a2, and a3 are the corresponding principal radii. Moreover, as is easily

Hydrostatics 53

demonstrated,

V =∫

dV =43π a1 a2 a3, (2.99)∫

x 2i dV =

15

a 2i V, (2.100)∫

x1 x2 dV =∫

x2 x3 dV = 0, (2.101)

where V is the volume, dV an element of V , and the integrals are taken over thewhole interior of the ellipsoid.

Suppose that the planet is rotating uniformly about the x3-axis at the fixed angularvelocity ω. The planet’s moment of inertia about this axis is [cf., Equation (2.100)]

I33 =15

M (a 21 + a 2

2 ), (2.102)

where M is its mass. Thus, the planet’s angular momentum is

L = I33 ω =15

M (a 21 + a 2

2 )ω, (2.103)

and its rotational kinetic energy becomes

K =12

I33 ω2 =

110

M (a 21 + a 2

2 )ω 2. (2.104)

According to Equations (2.83) and (2.84), the fluid pressure distribution withinthe planet takes the form

p = p′0 − ρ[Ψ −

12ω 2 (x 2

1 + x 22 )], (2.105)

where Ψ is the gravitational potential (i.e., the gravitational potential energy of aunit test mass) due to the planet, ρ = M/V the uniform planetary mass density, andp′0 a constant. However, it is demonstrated in Appendix D that the gravitational po-tential inside a homogeneous self-gravitating ellipsoidal body can be written (Chan-drasekhar 1969; Lamb 1993)

Ψ = −34

G M

α0 −∑i=1,3

αi x 2i

, (2.106)

where G is the universal gravitational constant (Yoder 1995), and

α0 =

∫ ∞

0

du∆, (2.107)

αi =

∫ ∞

0

du(a 2

i + u)∆, (2.108)

∆ = (a 21 + u)1/2 (a 2

2 + u)1/2 (a 23 + u)1/2. (2.109)


Thus, we obtain

p = p0−12ρ

[(32

G M α1 − ω 2)

x 21 +

(32

G M α2 − ω 2)

x 22 +

32

G M α3 x 23

], (2.110)

where p0 is the central fluid pressure. The pressure at the planet’s outer boundarymust be zero, otherwise there would be a force imbalance across the boundary. Inother words, we require

12ρ

[(32

G M α1 − ω 2)

x 21 +

(32

G M α2 − ω 2)

x 22 +

32

G M α3 x 23

]= p0, (2.111)

wheneverx 2

1

a 21

+x 2

2

a 22

+x 2

3

a 23

= 1. (2.112)

The previous two equations can only be simultaneously satisfied if[α1 −

ω 2

(3/2) G M

]a 2

1 =

[α2 −

ω 2

(3/2) G M

]a 2

2 = α3 a 23 . (2.113)

Rearranging the previous expression, we obtain

ω 2

2πG ρ=

a1 a3

a2(a 2

2 − a 23 )∫ ∞

0

u du(a 2

2 + u) (a 23 + u)∆

, (2.114)

subject to the constraint

(a 21 − a 2

2 )∫ ∞

0

a 21 a 2

2

(a 21 + u) (a 2

2 + u)−

a 23

(a 23 + u)

du∆= 0, (2.115)

where use has been made of Equation (2.99).Finally, according to Appendix D, the net gravitational potential energy of the

planet is

U = − 310

G M 2 α0. (2.116)

Hence, the body’s total mechanical energy becomes

E = K + U =1

10M (a 2

1 + a 22 )ω 2 −

310

G M 2 α0. (2.117)

2.12 Maclaurin SpheroidsOne, fairly obvious, way in which the constraint (2.115) can be satisfied is if a2 = a1.In other words, if the planet is rotationally symmetric about its axis of rotation. Anellipsoid that is rotationally symmetric about a principal axis—or, equivalently, an

Hydrostatics 55

ellipsoid with two equal principal radii—is known as a spheroid. In fact, if a2 = a1

then the cross-section of the planet’s outer boundary in any plane passing though thex3-axis is an ellipse of major radius a1 in the direction perpendicular to the x3-axis,and minor radius a3 in the direction parallel to the x3-axis. Here, we are assumingthat a1 > a3: that is, the planet is flattened along its axis of rotation. The degree offlattening is conveniently measured by the eccentricity,

e13 ≡ (1 − a 23 /a

21 )1/2. (2.118)

Thus, if e13 = 0 then there is no flattening, and the planet is consequently spherical,whereas if e13 → 1 then the flattening is complete, and the planet consequentlycollapses to a disk in the x1-x2 plane.

Let u = a 21 λ and λ = e 2

13/z2 − 1. Setting a2 = a1 in Equation (2.114), we obtain

ω 2

2πG ρ= (1 − e 2

13)1/2 e 213

∫ ∞

0

λ dλ(1 + λ)2 (1 + λ − e 2

13)3/2

=2 (1 − e 2

13)1/2

e 313

[∫ e13

0

z 2 dz(1 − z 2)1/2 − (1 − e 2

13)∫ e13

0

z 2 dz(1 − z 2)3/2

]. (2.119)

Performing the integrals, which are standard (Speigel, Liu, and Lipschutz 1999), wefind that

ω 2

2πG ρ=

3 − 2 e 213

e 313

(1 − e 213)1/2 sin−1 e13 −

3e 2

13

(1 − e 213). (2.120)

This famous result was first obtained by Colin Maclaurin (1698-1746) in 1742. Fi-nally, in order to calculate the potential energy, (2.116), we need to evaluate

α0 =1a1

∫ ∞

0

dλ(1 + λ) (1 + λ − e 2

13)1/2=

2a1 e13

∫ e13

0

dz(1 − z 2)1/2

=2a1

sin−1 e13

e13. (2.121)

Let e13 = sin γ. Thus, γ = 0 corresponds to no rotational flattening, and γ = π/2to complete flattening. Moreover, a1 = a0 (cosγ)−1/3 and a3 = a0 (cos γ)2/3, wherea0 = (a1 a2 a3)1/3 = (3 V/4π)1/3 is the mean radius. It is also helpful to define ω =ω/(2πG ρ)1/2, L = L/(G M 3 a0)1/2, and E = E/(G M 2/a0). The previous analysisleads to the following set of equations that specify the properties of the so-calledMaclaurin spheroids:

ω 2 =cosγsin2 γ

[(1 + 2 cos2 γ)

γ

sin γ− 3 cos γ

], (2.122)

L =

√6

5(cosγ)−1/6

sin γ

[(1 + 2 cos2 γ)

γ

sin γ− 3 cos γ

]1/2, (2.123)

E = − 310

(cosγ)1/3

sin2 γ

[(1 − 4 cos2 γ)

γ

sin γ+ 3 cos γ

]. (2.124)


e13 ω L −E e13 ω L −E0.00 0.00000 0.00000 0.60000 0.60 0.31729 0.18037 0.562330.05 0.02582 0.01266 0.59980 0.65 0.34484 0.20286 0.553200.10 0.05168 0.02540 0.59919 0.70 0.37239 0.22834 0.542000.15 0.07758 0.03830 0.59817 0.75 0.39967 0.25792 0.528000.20 0.10357 0.05144 0.59672 0.80 0.42612 0.29345 0.510010.25 0.12967 0.06491 0.59479 0.85 0.45046 0.33833 0.485870.30 0.15591 0.07882 0.59236 0.90 0.46932 0.39994 0.451070.35 0.18231 0.09329 0.58936 0.95 0.47045 0.50074 0.392720.40 0.20889 0.10846 0.58572 0.96 0.46472 0.53194 0.374850.45 0.23567 0.12450 0.58135 0.97 0.45418 0.57123 0.352730.50 0.26267 0.14163 0.57612 0.98 0.43475 0.62486 0.323510.55 0.28989 0.16013 0.56986 0.99 0.39389 0.71209 0.27916

Table 2.1Properties of the Maclaurin spheroids.

These properties are set out in Table 2.1.In the limit, γ → 0, in which the planet is relatively slowly rotating (i.e., ω 1),

and its degree of flattening consequently slight, Equations (2.122)–(2.124) reduce to

e13 √

152

ω, (2.125)

L √

65ω, (2.126)

E −35. (2.127)

In other words, in the limit of relatively slow rotation, when the planet is almostspherical, its eccentricity becomes directly proportional to its angular velocity. Inthis case, it is more conventional to parameterize angular velocity in terms of

m =ω 2 a0

g0=

32ω 2, (2.128)

where g0 = G M 2/a 20 is the mean surface gravitational acceleration. Furthermore,

the degree of rotational flattening is more conveniently expressed in terms of theellipticity,

ε =a1 − a3

a0

e 213

2. (2.129)

Thus, it follows from (2.125) that

ε 54

m. (2.130)

Hydrostatics 57

0

0.1

0.2

ω 2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1e13

Figure 2.5Normalized angular velocity squared of a Maclaurin spheroid (solid) and a Jacobiellipsoid (dashed) versus the eccentricity e13 in the x1-x3 plane.

For the case of the Earth (ω = 7.27 × 10−5 rad. s−1, a0 = 6.37 × 106 m, g0 =

9.81 m s−1—Yoder 1995), we obtain

m 1291

. (2.131)

Thus, it follows that, were the Earth homogeneous, its figure would be a spheroid,flattened at the poles, of ellipticity

ε 54

1291 1

233. (2.132)

This result was first obtained by Newton. The actual ellipticity of the Earth is about1/294 (Yoder 1995), which is substantially smaller than Newton’s prediction. Thediscrepancy is due to the fact that the Earth is strongly inhomogeneous, being muchdenser at its core than in its outer regions.

Figures 2.5 and 2.6 illustrate the variation of the normalized angular velocity,ω, and angular momentum, L, of a Maclaurin spheroid with its eccentricity, e13,as predicted by Equations (2.122)–(2.124). It can be seen, from Figure 2.5, thatthere is a limit to how large the normalized angular velocity of such a spheroid canbecome. The limiting value corresponds to ω = 0.47399, and occurs when e13 =

0.92995. For values of ω lying below 0.47399 there are two possible Maclaurinspheroids, one with an eccentricity less than 0.92995, and one with an eccentricity


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

L

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1e13

Figure 2.6Normalized angular momentum of a Maclaurin spheroid (solid) and a Jacobi ellip-soid (dashed) versus the eccentricity e13 in the x1-x3 plane.

greater than 0.92995. Note, however, from Figure 2.6, that despite the fact that theangular velocity, ω, of a Maclaurin spheroid varies in a non-monotonic manner withthe eccentricity, e13, the angular momentum, L, increases monotonically with e13,becoming infinite in the limit e13 → 1. It follows that there is no upper limit to theangular momentum of a Maclaurin spheroid.

2.13 Jacobi EllipsoidsIf a2 a1 (i.e., if the outer boundary of the rotating body is ellipsoidal, rather thanspheroidal) then the constraint (2.115) can only be satisfied when∫ ∞

0

a 21 a 2

2

(a 21 + u) (a 2

2 + u)−

a 23

(a 23 + u)

du∆= 0. (2.133)

Without loss of generality, we can assume that a1 ≥ a2 ≥ a3. Let

a2 = a1 cos β, (2.134)

a3 = a1 cos γ, (2.135)

Hydrostatics 59

where γ ≥ β. It follows that the cross-sections of the planet’s outer boundary in thex1-x2 and x1-x3 planes are ellipses of eccentricities

e12 = (1 − a 23 /a

22 )1/2 = sin β, (2.136)

e13 = (1 − a 23 /a

21 )1/2 = sin γ, (2.137)

respectively. It is also helpful to define

α = sin−1(sin β/ sin γ). (2.138)

Let sin2 θ = [a 21 /(a

21 + u)] sin2 γ. Here, u = 0 corresponds to θ = γ, and u = ∞

to θ = 0. Equations (2.115) and (2.114) transform to (Darwin 1886)

E(γ, α) − 2 F(γ, α) +[1 + (sinα tan β cos γ)2

cos2 α

]E(γ, α)

− sinα tan β cosγ (1 + sin2 β)cos2 α

= 0, (2.139)

and

ω 2 = 2[

F(γ, α) − E(γ, α)tan β sin β tan γ

+cos β E(γ, α)tan3 γ cos2 α

− cos2 β

tan2 γ cos2 α

], (2.140)

respectively, where

E(γ, α) =∫ γ

0(1 − sin2 α sin2 θ)1/2 dθ, (2.141)

F(γ, α) =∫ γ

0

dθ

(1 − sin2 α sin2 θ)1/2, (2.142)

are special functions known as incomplete elliptic integrals (Abramowitz and Stegun1965). The integral α0, defined in Equation (2.107), transforms to

α0 =2 (cos β cos γ)1/3

a0 sin γF(γ, α). (2.143)

Finally, making use of some of the analysis in the previous two sections, the nor-malized angular momentum, and normalized mechanical energy, of the planet can bewritten

L =

√6

101 + cos2 β

(cosβ cosγ)2/3 ω, (2.144)

E = −35

(cosβ cosγ)1/3

sin γF(γ, α) +

320

1 + cos2 β

(cosβ cosγ)2/3 ω2, (2.145)

respectively.The constraint (2.139) is obviously satisfied in the limit β → 0, because this


e12 e13 ω L −E e12 e13 ω L −E

0.00 0.81267 0.43257 0.30375 0.50452 0.60 0.85585 0.42827 0.30984 0.501380.05 0.81293 0.43257 0.30375 0.50459 0.65 0.86480 0.42609 0.31296 0.499750.10 0.81372 0.43257 0.30375 0.50459 0.70 0.87510 0.42288 0.31760 0.497340.15 0.81504 0.43256 0.30377 0.50458 0.75 0.88705 0.41807 0.32462 0.493720.20 0.81691 0.43253 0.30380 0.50457 0.80 0.90102 0.41069 0.33562 0.488140.25 0.81934 0.43248 0.30388 0.50453 0.85 0.91761 0.39879 0.35390 0.479080.30 0.82237 0.43237 0.30402 0.50445 0.90 0.93778 0.37787 0.38783 0.462950.35 0.82603 0.43220 0.30427 0.50432 0.95 0.96340 0.33353 0.46860 0.427820.40 0.83037 0.43191 0.30468 0.50410 0.96 0.96950 0.31776 0.50078 0.414990.45 0.83544 0.43146 0.30532 0.50376 0.97 0.97605 0.29691 0.54672 0.397710.50 0.84131 0.43078 0.30628 0.50326 0.98 0.98317 0.26722 0.62003 0.372410.55 0.84808 0.42976 0.30772 0.50250 0.99 0.99101 0.21809 0.76872 0.32842

Table 2.2Properties of the Jacobi ellipsoids.

implies that α → 0 and E(γ, α), F(γ, α) → γ. Of course, this limit corresponds tothe axisymmetric Maclaurin spheroids discussed in the previous section. Carl Jacobi(1804-1851), in 1834, was the first researcher to obtain the very surprising result thatEquation (2.139) also has non-axisymmetric ellipsoidal solutions characterized byβ > 0. These solutions are known as the Jacobi ellipsoids in his honor. The propertiesof the Jacobi ellipsoids, as determined from Equations (2.139), (2.140), (2.144), and(2.145), are set out in Table 2.2, and illustrated in Figures 2.5 and 2.6. It can be seenthat the sequence of Jacobi ellipsoids bifurcates from the sequence of Maclaurinspheroids when e13 = 0.81267. Moreover, there are no Jacobi ellipsoids with e13 <0.81267. However, as e13 increases above this critical value, the eccentricity, e12,of the Jacobi ellipsoids in the x1-x2 plane grows rapidly, approaching unity as e13

approaches unity. Thus, in the limit e13 → 1, in which a Maclaurin spheroid collapsesto a disk in the x1-x2 plane, a Jacobi ellipsoid collapses to a line running along thex1-axis. Note, from Figures 2.5 and 2.6, that, at fixed e13, the Jacobi ellipsoids havelower angular velocity and angular momentum than Maclaurin spheroids (with thesame mass and volume). Furthermore, as is the case for a Maclaurin spheroid, thereis a maximum angular velocity that a Jacobi ellipsoid can have (i.e., ω = 0.43257),but no maximum angular momentum.

Figure 2.7 shows the mechanical energy of the Maclaurin spheroids and Jacobiellipsoids plotted as a function of their angular momentum. It can be seen that theJacobi ellipsoid with a given angular momentum has a lower energy that the corre-sponding Maclaurin spheroid (i.e., the spheroid with the same angular momentum,mass, and volume). This is significant because, in the presence of a small amountof dissipation (i.e., viscosity), we would generally expect an isolated fluid systemto slowly evolve toward the equilibrium state with the lowest energy, subject to anyglobal constraints on the system. For the case of a weakly viscous, isolated, rotat-

Hydrostatics 61

0

0.1

0.2

0.3

0.4

0.5

0.6

−E

0 0.2 0.4 0.6 0.8 1 1.2

L

Figure 2.7Normalized mechanical energy of a Maclaurin spheroid (solid) and a Jacobi ellipsoid(dashed) versus the normalized angular momentum.

ing, liquid planet, the relevant constraints are that the mass, volume, and net angu-lar momentum of the system cannot spontaneously change. Thus, we expect sucha planet to evolve toward the equilibrium state with the lowest energy for a givenmass, volume, and angular momentum. This suggests, from Figure 2.7, that at rela-tively high angular momentum (i.e., L > 0.30375, e13 > 0.81267), when the Jacobiellipsoid solutions exist, they are stable equilibrium states (because there is no lowerenergy state to which the system can evolve), whereas the Maclaurin spheroids areunstable. On the other hand, at relatively low angular momentum (i.e., L < 0.30375,e13 < 0.81267), when there are no Jacobi ellipsoid solutions, the Maclaurin spheroidsare stable equilibrium states (again, because there is no lower energy state to whichthey can evolve). These predictions are borne out by the results of direct stabilityanalysis performed on the Maclaurin spheroids and Jacobi ellipsoids (Chandrasekhar1969). In fact, such stability studies demonstrate that the Maclaurin spheroids areunstable in the presence of weak dissipation for e13 > 0.81267, and unconditionallyunstable for e13 > 0.95289. The Jacobi ellipsoids, on the other hand, are uncondi-tionally stable for e13 < 0.93858, but are unconditionally unstable for e13 > 0.93858,evolving toward lower energy “pear shaped” equilibria (which are, themselves, un-stable in the presence of weak dissipation).


2.14 Roche EllipsoidsConsider a homogeneous liquid moon of mass M that is in a circular orbit of radiusR about a planet of mass M′. Let C, C′, and C′′ be the center of the moon, the centerof the planet, and the center of mass of the moon-planet system, respectively. As iseasily demonstrated, all three points lie on the same straight-line, and the distancesbetween them take the constant values CC′ = R and CC′′ = [M′/(M + M′)] R (Fitz-patrick 2012). Moreover, according to standard Newtonian dynamics, there exists aninertial frame of reference in which C′′ is stationary, and the line CC′ rotates at thefixed angular velocity ω, where (Fitzpatrick 2012)

ω 2 =G (M + M′)

R 3 . (2.146)

In other words, in the inertial frame, the moon and the planet orbit in a fixed planeabout their common center of mass at the angular velocity ω. It is convenient totransform to a non-inertial reference frame that rotates (with respect to the inertialframe), about an axis passing through C′′, at the angular velocity ω. It follows thatpoints C, C′, and C′′ appear stationary in this frame. It is also convenient to adopt thestandard right-handed Cartesian coordinates, x1, x2, x3, and to choose the coordinateaxes such that ω = ω e3, C = (0, 0, 0), C′ = (R, 0, 0), and C′′ = ([M′/(M +M′)] R, 0, 0). Thus, in the non-inertial reference frame, the orbital rotation axis runsparallel to the x3-axis, and the centers of the moon and the planet both lie on thex1-axis.

Suppose that the moon does not rotate (about an axis passing through its centerof mass) in the non-inertial reference frame. This implies that, in the inertial frame,the moon appears to rotate about an axis parallel to the x3-axis, and passing throughC, at the same angular velocity as it orbits about C′′. This type of rotation is termedsynchronous, and ensures that the same hemisphere of the moon is always directedtoward the planet. Such rotation is fairly common in the solar system. For instance,the Moon rotates synchronously in such a manner that the same hemisphere is alwaysvisible from the Earth. Synchronous rotation in the solar system is a consequence ofprocess known as tidal locking (Murray and Dermott 1999).

Because a synchronously rotating moon is completely stationary in the afore-mentioned non-inertial frame, its internal pressure, p, is governed by a force balanceequation of the form [cf., Equation (2.83)]

0 = ∇p + ρ∇(Ψ + Ψ ′ + Ψ ′′), (2.147)

where ρ is the uniform internal mass density, Ψ the gravitational potential due to themoon, Ψ ′ the gravitational potential due to the planet, and

Ψ ′′ = −12ω2

(x1 −M′

M + M′R)2+ x 2

2

(2.148)

the centrifugal potential due to the fact that the non-inertial frame is rotating (about

Hydrostatics 63

an axis parallel to the x3-axis and passing through point C′′) at the angular velocityω [cf., Equation (2.84)]. Suppose that the moon is much less massive that the planet(i.e., M/M′ 1). In this limit, the centrifugal potential (2.148) reduces to

Ψ ′′ −12 − x1

R+

(1/2) x 21 + (1/2) x 2

2

R 2

, (2.149)

where use has been made of Equation (2.146).Suppose that the planet is spherical. It follows that the potential Ψ ′ is the same

as that which would be generated by a point mass M′ located at C′. In other words,

Ψ ′ = −G M′

R

1 − 2x1

R+

x 21 + x 2

2 + x 23

R 2

−1/2

−G M′

R

1 + x1

R+

x 21 − (1/2) x 2

2 − (1/2) x 23

R 2 + · · · , (2.150)

where we have expanded up to second order in x1/R, et cetera.The previous two equations can be combined to give

Ψ ′ + Ψ ′′ −λ(32

x 21 −

12

x 23

), (2.151)

whereλ =

G M′

R 3 , (2.152)

and any constant terms have been neglected. Thus, the net force field experiencedby the moon due to the combined action of the fictitious centrifugal force and thegravitational force field of the planet is

−ρ∇(Ψ ′ + Ψ ′′) = ρ λ (3 x1, 0, −x3). (2.153)

The previous type of force field is known as a tidal force field, and clearly acts toelongate the moon along the axis joining the centers of the moon and planet (i.e.,the x1-axis), and to compress it along the orbital rotation axis (i.e., the x3-axis).Moreover, the magnitude of the tidal force increases linearly with distance from thecenter of the moon. The tidal force field is a consequence of the different spatialvariation of the centrifugal force and the planet’s gravitational force of attraction.This different variation causes these two forces, which balance one another at thecenter of the moon, to not balance away from the center (Fitzpatrick 2012). As aresult of the tidal force field, we expect the shape of the moon to be distorted from asphere. Of course, the moon also generates a tidal force field that acts to distort theshape of the planet. However, we are assuming that the tidal distortion of the planetis much smaller than that of the moon (which justifies our earlier statement thatthe planet is essentially spherical). As will be demonstrated later, this assumptionis reasonable provided the mass of the moon is much less than that of the planet(assuming that the planet and moon have similar densities).


Suppose that the bounding surface of the moon is the ellipsoid

x 21

a 21

+x 2

2

a 22

+x 2

3

a 23

= 1, (2.154)

where a1 ≥ a2 ≥ a3. It follows, from Appendix D, that the gravitational potential ofthe moon at an interior point can be written

Ψ = −34

G M

α0 −∑i=1,3

αi x 2i

, (2.155)

where the integrals αi, for i = 0, 3, are defined in Equations (D.30) and (D.31).Hence, from Equations (2.147) and (2.151), the pressure distribution within the moonis given by

p = p0 −12ρ

[(32

G M α1 − 3 λ)

x 21 +

32

G M α2 x 22 +

(32

G M α3 + λ

)x 2

3

], (2.156)

where p0 is the central pressure. The pressure must be zero on the moon’s boundingsurface, otherwise this surface would not be in equilibrium. Thus, in order to achieveequilibrium, we require

12ρ

[(32

G M α1 − 3 λ)

x 21 +

32

G M α2 x 22 +

(32

G M α3 + λ

)x 2

3

]= p0, (2.157)

wheneverx 2

1

a 21

+x 2

2

a 22

+x 2

3

a 23

= 1. (2.158)

The previous two equations can only be simultaneously satisfied if[α1 −

3 λ(3/2) G M

]a 2

1 = α2 a 22 =

[α3 +

λ

(3/2) G M

]a 2

3 . (2.159)

Let a2 = a1 cos β and a3 = a1 cos γ, where γ ≥ β. It is also helpful to define α =sin−1(sin β/ sin γ). With the help of some of the analysis presented in the previoussection, the integrals αi, for i = 1, 3, can be shown to take the form

α1 =2

a 31 sin3 γ

F(γ, α) − E(γ, α)sin2 α

, (2.160)

α2 =2

a 21 sin3 γ

[E(γ, α)

sin2 α cos2 α− F(γ, α)

sin2 α− cosγ sin γ

cos2 α cos β

], (2.161)

α3 =2

a 21 sin3 γ

[−E(γ, α)

cos2 α+

cos β sin γcos2 α cos γ

], (2.162)

Hydrostatics 65

0

0.01

0.02

0.03

0.04

0.05

e 13−

e 12

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1e13

Figure 2.8Properties of the Roche ellipsoids.

where the incomplete elliptic integrals E(γ, α) and F(γ, α) are defined in Equa-tions (2.141) and (2.142), respectively. Thus, Equation (2.159) yields

µ =1

sin β tan β tan γ

[F(γ, α) (1 + cos2 β) − E(γ, α)

(1 +

cos2 β

cos2 α

)+

sinα sin β cos β cosγcos2 α

], (2.163)

subject to the constraint

0 = cos2 γ

[F(γ, α) − 2 E(γ, α) + E(γ, α)

cos2 β

cos2 α− sinα sin β cos β cosγ

cos2 α

]+ (3 + cos2 γ)

[E(γ, α) + [F(γ, α) cos2 α − 2 E(γ, α)]

cos2 β

cos2 α

+2 sinα sin β cos β cosγ

cos2 α

], (2.164)

where

µ =M′

M

a 30

R 3 , (2.165)

and a0 = (a1 a2 a3)1/3 is the mean radius of the moon. The dimensionless parameter


e12 e13 µ e12 e13 µ

0.00 0.00000 0.00000 0.52 0.56740 0.334400.04 0.04613 0.00213 0.56 0.60632 0.382040.08 0.09223 0.00852 0.60 0.64445 0.430940.12 0.13809 0.01913 0.64 0.68182 0.480270.16 0.18364 0.03392 0.68 0.71848 0.528900.20 0.22879 0.05282 0.72 0.75446 0.575320.24 0.27346 0.07573 0.76 0.78984 0.617290.28 0.31756 0.10253 0.80 0.82472 0.651500.32 0.36104 0.13308 0.84 0.85923 0.672650.36 0.40383 0.16721 0.88 0.89353 0.671510.40 0.44588 0.20470 0.92 0.92793 0.629780.44 0.48718 0.24528 0.96 0.96294 0.501350.48 0.52769 0.28865 1.00 1.00000 0.00000

Table 2.3Properties of the Roche ellipsoids.

µ measures the strength of the tidal distortion field, generated by the planet, that actson the moon. There is an analogous parameter,

µ′ =MM′

a′03

R 3 , (2.166)

where a′0 is the mean radius of the planet, which measures the tidal distortion field,generated by the moon, that acts on the planet. We previously assumed that theformer distortion field is much stronger than the latter, allowing us to neglect thetidal distortion of the planet altogether, and so to treat it as a sphere. This assumptionis only justified if µ µ′, which implies that

MM′ ρ′

ρ, (2.167)

where ρ = M/[(4/3) π a 30 ] and ρ′ = M′/[(4/3) π a′0

3] are the mean densities of themoon and the planet, respectively. Assuming that these densities are similar, theprevious condition reduces to M M′, or, equivalently, a0 < a′0. In other words,neglecting the tidal distortion of the planet, while retaining that of the moon, is gener-ally only reasonable when the mass of the moon is much less than that of the planet,as was previously assumed to be the case.

Equations (2.163) and (2.164), which describe the ellipsoidal equilibria of a syn-chronously rotating, relatively low mass, liquid moon due to the tidal force field ofthe planet about which it orbits, were first obtained by Edouard Roche (1820-1883)in 1850. The properties of the so-called Roche ellipsoids are set out in Table 2.3, andFigures 2.8 and 2.9.

Hydrostatics 67

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

µ

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1e13

Figure 2.9Properties of the Roche ellipsoids.

It can be seen, from Table 2.3 and Figure 2.8, that the eccentricity e12 = sin βof a Roche ellipsoid in the x1-x2 plane is almost equal to its eccentricity e13 = sin γin the x1-x3 plane. In other words, Roche ellipsoids are almost spheroidal in shape,being elongated along the x1-axis (i.e., the axis joining the centers of the moon andthe planet), and compressed by almost equal amounts along the x2- and x3-axes. Inthe limit µ 1, in which the tidal distortion field due to the planet is weak, it iseasily shown that

e 212 e 2

13 152µ. (2.168)

For the case of the tidal distortion field generated by the Earth, and acting on theMoon, which is characterized by M/M′ = 0.01230 and R/a0 = 221.29, we obtainµ = 7.50 × 10−6 (Yoder 1995). It follows that e13 = 7.50 × 10−3, and (a1 − a3)/a1 e 2

13/2 = 2.81× 10−5. In other words, were the Moon a homogeneous liquid body, theelongation generated by the tidal field of the Earth would be about 50 m.

It can be seen, from Table 2.3 and Figure 2.9, that the parameter µ attains amaximum value as the eccentricity of a Roche ellipsoid varies from 0 to 1. In fact,this maximum value, µ = 0.06757, occurs when e12 = 0.8594 and e13 = 0.8759. Itfollows that there is a maximum strength of the tidal distortion field, generated by aplanet, that is consistent with an ellipsoidal equilibrium of a synchronously rotating,homogeneous, liquid moon in a circular orbit about the planet. It is plausible that ifthis maximum strength is exceeded then the moon is tidally disrupted by the planet.


The equilibrium condition µ < 0.06757 is equivalent to

Ra′0> 2.455

(ρ′

ρ

)1/3, (2.169)

where ρ = M/[(4/3) π a 30 ] and ρ′ = M′/[(4/3) π a′0

3] are the mean densities of themoon and the planet, respectively. According to the previous expression, there isa minimum orbital radius of a moon circling a planet. Below this radius, which iscalled the Roche radius, the moon is presumably torn apart by tidal effects. TheRoche radius for a synchronously rotating, self-gravitating, liquid moon in a circularorbit about a spherical planet is about 2.5 times the planet’s radius (assuming thatthe moon and the planet have approximately the same mass density). Of course, rel-atively small objects, such as artificial satellites, which are held together by internaltensile strength, rather than gravity, can orbit inside the Roche radius without beingdisrupted.

2.15 Exercises2.1 A hollow vessel floats in a basin. If, as a consequence of a leak, water flows

slowly into the vessel, how will the level of the water in the basin be affected?(Lamb 1928.)

2.2 A hollow spherical shell made up of material of specific gravity s > 1 hasexternal and internal radii a and b, respectively. Demonstrate that the spherewill only float in water if

ba>

(1 − 1

s

)1/3.

2.3 Show that the equilibrium of a solid of uniform density floating with an edgeor corner just emerging from the water is unstable. (Lamb 1928.)

2.4 Prove that if a solid of uniform density floats with a flat face just above thewaterline then the equilibrium is stable. (Lamb 1928.)

2.5 Demonstrate that a uniform solid cylinder floating with its axis horizontalis in a stable equilibrium provided that its length exceeds the breadth of thewaterline section. [Hint: The cylinder is obviously neutrally stable to rota-tions about its axis, which means that the corresponding metacentric heightis zero.] (Lamb 1928.)

2.6 Show that a uniform solid cylinder of radius a and height h can float in stableequilibrium, with its axis vertical, if h/a <

√2. If the ratio h/a exceeds this

Hydrostatics 69

value, prove that the equilibrium is only stable when the specific gravity ofthe cylinder lies outside the range

12

1 ±√

1 − 2a 2

h 2

.2.7 A uniform, thin, hollow cylinder of radius a and height h is open at both

ends. Assuming that h > 2 a, prove that the cylinder cannot float upright ifits specific gravity lies in the range

12±√

14− a 2

h 2 .

(Lamb 1928.)

2.8 Show that the cylinder of the preceding exercise can float with its axis hori-zontal provided

h2 a

>√

3 sin(s π),

where s is the specific gravity of the cylinder. (Lamb 1928.)

2.9 Prove that any segment of a uniform sphere, made up of a substance lighterthan water, can float in stable equilibrium with its plane surface horizontaland immersed. (Lamb 1928.)

2.10 A vessel carries a tank of oil, of specific gravity s, running along its length.Assuming that the surface of the oil is at sea level, show that the effect ofthe oil’s fluidity on the rolling of the vessel is equivalent to a reduction in themetacentric height by A κ 2 s/V , where V is the displacement of the ship, Athe surface-area of the tank, and κ the radius of gyration of this area. In whatratio is the effect diminished when a longitudinal partition bisects the tank?(Lamb 1928.)

2.11 Find the stable equilibrium configurations of a cylinder of elliptic cross-section, with major and minor radii a and b < a, respectively, made up ofmaterial of specific gravity s, which floats with its axis horizontal.

2.12 A cylindrical tank has a circular cross-section of radius a. Let the center ofgravity of the tank be located a distance c above its base. Suppose that thetank is pivoted about a horizontal axis passing through its center of gravity,and is then filled with fluid up to a depth h above its base. Demonstrate thatthe position in which the tank’s axis is upright is unstable for all filling depthsprovided

c 2 <12

a 2.

Show that if c 2 > (1/2) a 2 then the upright position is stable when h lies inthe range

c ±√

c 2 − a 2/2.


2.13 A thin cylindrical vessel of cross-sectional area A floats upright, being im-mersed to a depth h, and contains water to a depth k. Show that the workrequired to pump out the water is ρ0 A k (h − k) g. (Lamb 1928.)

2.14 A sphere of radius a is just immersed in water that is contained in a cylindricalvessel of radius R whose axis is vertical. Prove that if the sphere is raised justclear of the water then the water’s loss of potential energy is

W a(1 − 2

3a 2

R 2

),

where W is the weight of the water originally displaced by the sphere. (Lamb1928.)

2.15 A sphere of radius a, weight W, and specific gravity s > 1, rests on the bottomof a cylindrical vessel of radius R whose axis is vertical, and which containswater to a depth h > 2 a. Show that the work required to lift the sphere out ofthe vessel is less than if the water had been absent by an amount(

h − a − 23

a 3

R 2

)Ws.

(Lamb 1928.)

2.16 A lead weight is immersed in water that is steadily rotating at an angularvelocity ω about a vertical axis, the weight being suspended from a fixedpoint on this axis by a string of length l. Prove that the position in which theweight hangs vertically downward is stable or unstable depending on whetherl < g/ω 2 or l > g/ω 2, respectively. Also, show that if the vertical positionis unstable then there exists a stable inclined position in which the string isnormal to the surface of equal pressure passing though the weight.

2.17 A thin cylindrical vessel of radius a and height H is orientated such that itsaxis is vertical. Suppose that the vessel is filled with liquid of density ρ tosome height h < H above the base, spun about its axis at a steady angularvelocity ω, and the liquid allowed to attain a steady state. Demonstrate that,provided ω 2 a 2/g < 4 h and ω 2 a 2/g < 4 (H − h), the net radial thrust on thevertical walls of the vessel is

π a h 2 ρ g

(1 +

ω 2 a 2

4 g h

)2.

2.18 A thin cylindrical vessel of radius a with a plane horizontal lid is just filledwith liquid of density ρ, and the whole rotated about a vertical axis at a fixedangular velocity ω. Prove that the net upward thrust of the fluid on the lid is

14π a 4 ρω 2.

Hydrostatics 71

2.19 A liquid-filled thin spherical vessel of radius a spins about a vertical diameterat the fixed angular velocity ω. Assuming that the liquid co-rotates with thevessel, and that ω 2 > g/a, show that the pressure on the wall of the vessel isgreatest a depth g/ω 2 below the center. Also prove that the net normal thrustson the lower and upper hemispheres are

54

M g +3

16Mω 2 a,

and14

M g − 316

Mω 2 a,

respectively, where M is the mass of the liquid.

2.20 A closed cubic vessel filled with water is rotating about a vertical axis passingthrough the centers of two opposite sides. Demonstrate that, as a consequenceof the rotation, the net thrust on a side is increased by

16

a4 ρω 2,

where a is the length of an edge of the cube, and ω the angular velocity ofrotation.

2.21 A closed vessel filled with water is rotating at constant angular velocity ωabout a horizontal axis. Show that, in the state of relative equilibrium, theconstant pressure surfaces in the water are circular cylinders whose commonaxis is a height g/ω 2 above the axis of rotation. (Batchelor 2000.)

2.22 Verify Equations (2.95)–(2.97) and (2.99)–(2.101).

2.23 Consider a homogeneous, rotating, liquid body of mass M, mean radius a0,and angular velocity ω, whose outer boundary is a Maclaurin spheroid ofeccentricity e.

(a) Demonstrate that

e √

52

ω

(G M/a 30 )1/2

in the low rotation limit, ω (G M/a 30 )1/2. Hence, show that e =

0.09262 for the case of a homogeneous body with the same mass andvolume as the Earth, which rotates once every 24 hours.

(b) Show that the critical angular velocity at which the bifurcation to thesequence of Jacobi ellipsoids takes place is

ω = 0.5298 (G M/a 30 )1/2,

and occurs when e = 0.81267. Hence, show that, for the case of ahomogeneous body with the same mass and volume as the Earth, thebifurcation would take place at a critical rotation period of 2 h 39 m.


(c) Demonstrate that the maximum angular velocity consistent with a spheroidalshape is

ω = 0.5805 (G M/a 30 )1/2,

and occurs when e = 0.92995. Hence, show that, for the case of a ho-mogeneous body with the same mass and volume as the Earth, this max-imum velocity corresponds to a minimum rotation period of 2 h 25 m.

3Surface Tension

3.1 Introduction

As is well known, small drops of water in air, and small bubbles of gas in water, tendto adopt spherical shapes. This observation, and a host of other natural phenomena,can only be accounted for on the hypothesis that an interface between two differentmedia is associated with a particular form of energy whose magnitude is directlyproportional to the interfacial area. To be more exact, if S is the interfacial area thenthe contribution of the interface to the Helmholtz free energy of the system takes theform γ S , where γ only depends on the temperature and chemical composition of thetwo media on either side of the interface. It follows, from standard thermodynamics,that γ S is the work that must be performed on the system in order to create theinterface via an isothermal and reversible process (Reif 1965). However, this workis exactly the same as that which we would calculate on the assumption that theinterface is in a state of uniform constant tension per unit length γ. Thus, γ can beinterpreted as both a free energy per unit area of the interface, and a surface tension.This tension is such that a force of magnitude γ per unit length is exerted acrossany line drawn on the interface, in a direction normal to the line, and tangential tothe interface. More information on surface tension can be found in Lamb 1928, andBatchelor 2000.

Surface tension originates from intermolecular cohesive forces. The average freeenergy of a molecule in a given isotropic medium possessing an interface with asecond medium is independent of its position, provided that the molecule does not lietoo close to the interface. However, the free energy is modified when the molecule’sdistance from the interface becomes less than the range of the cohesive forces (whichis typically 10−9 m). Because this range is so small, the number of molecules in amacroscopic system whose free energies are affected by the presence of an interfaceis directly proportional to the interfacial area. Hence, the contribution of the interfaceto the total free energy of the system is also proportional to the interfacial area. Ifonly one of the two media in question is a condensed phase then the parameter γis invariably positive (i.e., such that a reduction in the surface area is energeticallyfavorable). This follows because the molecules of a liquid or a solid are subject toan attractive force from neighboring molecules. However, molecules that are near toan interface with a gas lack neighbors on one side, and so experience an unbalancedcohesive force directed toward the interior of the liquid/solid. The existence of this

73


C t

n

t × n

2

1

S

Figure 3.1Interface between two immiscible fluids.

force makes it energetically favorable for the interface to contract (i.e., γ > 0). Onthe other hand, if the interface separates a liquid and a solid, or a liquid and anotherliquid, then the sign of γ cannot be predicted by this argument. In fact, it is possiblefor both signs of γ to occur at liquid/solid and liquid/liquid interfaces.

The surface tension of a water/air interface at 20 C is γ = 7.28 × 10−2 N m−1

(Batchelor 2000). The surface tension at most oil/air interfaces is much lower—typically, γ 2×10−2 N m−1 (Batchelor 2000). On the other hand, interfaces betweenliquid metals and air generally have very large surface tensions. For instance, thesurface tension of a mercury/air interface at 20 C is 4.87 × 10−1 N m−1 (Batchelor2000).

For some pairs of liquids, such as water and alcohol, an interface cannot generallybe observed because it is in compression (i.e., γ < 0). Such an interface tends tobecome as large as possible, leading to complete mixing of the two liquids. In otherwords, liquids for which γ > 0 are immiscible, whereas those for which γ < 0 aremiscible.

Finally, the surface tension at a liquid/gas or a liquid/liquid interface can be af-fected by the presence of adsorbed impurities at the interface. For instance, thesurface tension at a water/air interface is significantly deceased in the presence ofadsorbed soap molecules. Impurities that tend to reduce surface tension at interfacesare termed surfactants.

3.2 Young-Laplace EquationConsider an interface separating two immiscible fluids that are in equilibrium withone another. Let these two fluids be denoted 1 and 2. Consider an arbitrary segmentS of this interface that is enclosed by some closed curve C. Let t denote a unit tangentto the curve, and let n denote a unit normal to the interface directed from fluid 1 tofluid 2. (Note that C circulates around n in a right-handed manner. See Figure 3.1.)Suppose that p1 and p2 are the pressures of fluids 1 and 2, respectively, on either sideof S . Finally, let γ be the (uniform) surface tension at the interface.

Surface Tension 75

The net force acting on S is

f =∫

S(p1 − p2) n dS + γ

∮C

t × n dr, (3.1)

where dS = n dS is an element of S , and dr = t dr an element of C. Here, thefirst term on the right-hand side is the net normal force due to the pressure differenceacross the interface, whereas the second term is the net surface tension force. Notethat body forces play no role in Equation (3.1), because the interface has zero vol-ume. Furthermore, viscous forces can be neglected, because both fluids are static. Inequilibrium, the net force acting on S must be zero: that is,∫

S(p1 − p2) n dS = −γ

∮C

t × n dr. (3.2)

(In fact, the net force would be zero even in the absence of equilibrium, because theinterface has zero mass.)

Applying the curl theorem (see Section A.22) to the curve C, we find that∮C

F · dr =∫

S∇ × F · dS, (3.3)

where F is a general vector field. This theorem can also be written∮C

F · t dr =∫

S∇ × F · n dS . (3.4)

Suppose that F = g × b, where b is an arbitrary constant vector. We obtain∮C

(g × b) · t dr =∫

S∇ × (g × b) · n dS . (3.5)

However, the vector identity (A.179) yields

∇ × (g × b) = −(∇ · g) b + (b · ∇) g, (3.6)

as b is a constant vector. Hence, we get

b ·∮

Ct × g dr = b ·

∫S

[(∇g) · n − (∇ · g) n] dS , (3.7)

where b · (∇g) · n ≡ bi (∂g j/∂xi) n j. Now, because b is also an arbitrary vector, theprevious equation gives∮

Ct × g dr =

∫S

[(∇g) · n − (∇ · g) n

]dS . (3.8)

Taking g = γ n, we find that

γ

∮C

t × n dr = γ∫

S[(∇n) · n − (∇ · n) n] dS . (3.9)


But, (∇n) · n ≡ (1/2)∇(n2) = 0, because n is a unit vector. Thus, we obtain

γ

∮C

t × n dr = −γ∫

S(∇ · n) n dS , (3.10)

which can be combined with Equation (3.2) to give∫S

[(p1 − p2) − γ (∇ · n)

]n dS = 0. (3.11)

Finally, given that S is arbitrary, the previous expression reduces to the pressurebalance constraint

∆p = γ∇ · n, (3.12)

where ∆p = p1 − p2. The previous relation is generally known as the Young-Laplaceequation, and is named after Thomas Young (1773-1829), who developed the quali-tative theory of surface tension in 1805, and Pierre-Simon Laplace (1749-1827) whocompleted the mathematical description in the following year. The Young-Laplaceequation can also be derived by minimizing the free energy of the interface. (SeeSection 3.8.) Note that ∆p is the jump in pressure seen when crossing the interface inthe opposite direction to n. Of course, a plane interface is characterized by ∇ · n = 0.On the other hand, a curved interface generally has ∇ · n 0. In fact, ∇ · n measuresthe local mean curvature of the interface. Thus, according to the Young-Laplaceequation, there is a pressure jump across a curved interface between two immisciblefluids, the magnitude of the jump being proportional to the surface tension.

3.3 Spherical InterfacesGenerally speaking, the equilibrium shape of an interface between two immisciblefluids is determined by solving the force balance equation (2.1) in each fluid, andthen applying the Young-Laplace equation to the interface. However, in situations inwhich a mass of one fluid is completely immersed in a second fluid—for example,a mist droplet in air, or a gas bubble in water—the shape of the interface is fairlyobvious. Provided that either the size of the droplet or bubble, or the difference indensities on the two sides of the interface, is sufficiently small, we can safely ignorethe effect of gravity. This implies that the pressure is uniform in each fluid, andconsequently that the pressure jump ∆p is constant over the interface. Hence, fromEquation (3.12), the mean curvature ∇ · n of the interface is also constant. Becausea sphere is the only closed surface which possesses a constant mean curvature, weconclude that the interface is spherical. This result also follows from the argumentthat a stable equilibrium state is one which minimizes the free energy of the interface,subject to the constraint that the enclosed volume be constant. Thus, the equilibriumshape of the interface is that which has the least surface area for a given volume: inother words, a sphere.

Surface Tension 77

Suppose that the interface corresponds to the spherical surface r = R, where r isa spherical coordinate. (See Section C.4.) It follows that n = er |r=R. (Note, for futurereference, that n points away from the center of curvature of the interface.) Hence,from Equation (C.65),

∇ · n = 1r 2

∂r 2

∂r

∣∣∣∣∣∣r=R=

2R. (3.13)

The Young-Laplace equation, (3.12), then gives

∆p =2 γR. (3.14)

Thus, given that ∆p is the pressure jump seen crossing the interface in the oppositedirection to n, we conclude that the pressure inside a droplet or bubble exceeds thatoutside by an amount proportional to the surface tension, and inversely proportionalto the droplet or bubble radius. This explains why small bubbles are louder that largeones when they burst at a free surface: for instance, champagne fizzes louder thanbeer. Note that soap bubbles in air have two interfaces defining the inner and outerextents of the soap film. Consequently, the net pressure difference is twice that acrossa single interface.

3.4 Capillary LengthConsider an interface separating the atmosphere from a liquid of uniform density ρthat is at rest on the surface of the Earth. Neglecting the density of air compared tothat of the liquid, the pressure in the atmosphere can be regarded as constant. On theother hand, the pressure in the liquid varies as p = p0 − ρ g z (see Chapter 2), wherep0 is the pressure of the atmosphere, g the acceleration due to gravity, and z measuresvertical height (relative to the equilibrium height of the interface in the absence ofsurface tension). Note that z increases upward. In this situation, the Young-Laplaceequation, (3.12), yields

ρ g z = −γ∇ · n, (3.15)

where n is the normal to the interface directed from liquid to air. If R representsthe typical radius of curvature of the interface then the left-hand side of the previousequation dominates the right-hand side whenever R l, and vice versa. Here,

l =(γ

ρ g

)1/2(3.16)

is known as the capillary length, and takes the value 2.7 × 10−3 m for pure waterat 20 C (Batchelor 2000). We conclude that the effect of surface tension on theshape of an liquid/air interface is likely to dominate the effect of gravity when theinterface’s radius of curvature is much less than the capillary length, and vice versa.


δr

2 1

3

θ

Figure 3.2Interface between a liquid (1), a gas (2), and a solid (3).

3.5 Angle of ContactSuppose that a liquid/air interface is in contact with a solid, as would be the case forwater in a glass tube, or a drop of mercury resting on a table. Figure 3.2 shows asection perpendicular to the edge at which the liquid, 1, the air, 2, and the solid, 3,meet. Suppose that the free energies per unit area at the liquid/air, liquid/solid, andair/solid interfaces are γ12, γ13, and γ23, respectively. If the boundary between thethree media is slightly modified in the neighborhood of the edge, as indicated by thedotted line in the figure, then the area of contact of the air with the solid is increasedby a small amount δr per unit breadth (perpendicular to the figure), whereas that ofthe liquid with the solid is decreased by δr per unit breadth, and that of the liquidwith the air is decreased by δr cos θ per unit breadth. Thus, the net change in freeenergy per unit breadth is

γ23 δr − γ13 δr − γ12 δr cos θ. (3.17)

However, an equilibrium state is one which minimizes the free energy, implying thatthe previous expression is zero for arbitrary (small) δr: that is,

cos θ =γ23 − γ13

γ12. (3.18)

We conclude that, in equilibrium, the angle of contact, θ, between the liquid andthe solid takes a fixed value that depends on the free energies per unit area at theliquid/air, liquid/solid, and air/solid interfaces. Note that the previous formula couldalso be obtained from the requirement that the various surface tension forces actingat the edge balance one another, assuming that it is really appropriate to interpret γ13and γ23 as surface tensions when one of the media making up the interface is a solid.

Surface Tension 79

liquid

a

θ

liquidglass tube

air

R

free surface z = 0

h

air

Figure 3.3Elevation of liquid level in a capillary tube.

As explained in Section 3.1, we would generally expect γ12 and γ23 to be positive.On the other hand, γ13 could be either positive or negative. Now, because | cos θ| ≤ 1,Equation (3.18) can only be solved when γ13 lies in the range

γ23 + γ12 > γ13 > γ23 − γ12. (3.19)

If γ13 > γ23 + γ12 then the angle of contact is 180, which corresponds to the casewhere the free energy at the liquid/solid interface is so large that the liquid does notwet the solid at all, but instead breaks up into beads on its surface. On the other hand,if γ13 < γ23−γ12 then the angle of contact is 0, which corresponds to the case wherethe free energy at the liquid/solid interface is so small that the liquid completely wetsthe solid, spreading out indefinitely until it either covers the whole surface, or itsthickness reaches molecular dimensions.

The angle of contact between water and glass typically lies in the range 25 to29, whereas that between mercury and glass is about 127 (Batchelor 2000).


3.6 Jurin’s LawConsider a situation in which a narrow, cylindrical, glass tube of radius a is dippedvertically into a liquid of density ρ, and the liquid level within the tube rises a height habove the free surface as a consequence of surface tension. (See Figure 3.3.) Supposethat the radius of the tube is much less than the capillary length. A tube for whichthis is the case is generally known as a capillary tube. According to the discussion inSection 3.4, the shape of the internal liquid/air interface within a capillary tube is notsignificantly affected by gravity. Thus, from Section 3.3, the interface is a segmentof a sphere of radius R (say). If θ is the angle of contact of interface with the glassthen simple geometry (see Figure 3.3) reveals that

R =a

cos θ. (3.20)

Hence, from Equation (3.13), the mean curvature of the interface is given by

∇ · n = − 2R= −2 cos θ

a, (3.21)

where γ is the associated surface tension. [The minus sign in the previous expressionarises from the fact that n points towards the center of curvature of the interface,whereas the opposite is true for Equation (3.13).] Finally, from Equation (3.15),application of the Young-Laplace equation to the interface yields

ρ g h =2 γ cos θ

a, (3.22)

which can be rearranged to give

h 2 γ cos θρ g a

. (3.23)

This result, which relates the height, h, to which a liquid rises in a capillary tube ofradius a to the liquid’s surface tension, γ, is known as Jurin’s law, and is named afterits discoverer, James Jurin (1684-1750). The assumption that the radius of the tubeis much less than the capillary length is equivalent to the assumption that the heightof the interface above the free surface of the liquid is much greater than the radius ofthe tube. This follows, from Equations (3.16) and (3.23), because

ha= 2 cos θ

l 2

a 2 . (3.24)

Thus, the ordering a l implies that h a.For the case of water at 20, assuming a contact angle of 25, Jurin’s law yields

h(mm) = 13.5/a(mm) (Batchelor 2000). Thus, water rises a height 13.5 mm in acapillary tube of radius 1 mm, but rises 13.5 cm in a capillary tube of radius 0.1 mm.In the case of a liquid, such a mercury, that has an oblique angle of contact withglass, so that cos θ < 0, the liquid level in a capillary tube is depressed below that ofthe free surface (i.e., h < 0).

Surface Tension 81

3.7 Capillary CurvesLet adopt Cartesian coordinates on the Earth’s surface such that z increases verticallyupward. Suppose that the interface of a liquid of density ρ and surface tension γ withthe atmosphere corresponds to the surface z = f (x), where the liquid occupies theregion z < f (x). The shape of the interface is assumed to be y-independent. The unitnormal to the interface (directed from liquid to air) is thus

n =∇(z − f )|∇(z − f )|

=ez − fx ex

(1 + f 2x )1/2 , (3.25)

where fx = d f /dx. Hence, the mean curvature of the interface is (Riley 1974)

∇ · n = − fxx

(1 + f 2x )3/2 , (3.26)

where fxx = d 2 f /dx 2. According to Equations (3.15) and (3.16), the shape of theinterface is governed by the nonlinear differential equation

f =l 2 fxx

(1 + f 2x )3/2 . (3.27)

where the vertical height, f , of the interface is measured relative to its equilibriumheight in the absence of surface tension. Multiplying the previous equation by fx/l 2,and integrating with respect to x, we obtain

1(1 + f 2

x )1/2 = C − f 2

2 l 2 , (3.28)

where C is a constant. It follows that

C −f 2

2 l 2 ≥ 1, (3.29)

and1fx= ∓

C − f 2/2 l 2

[1 − (C − f 2/2 l 2)2]1/2 . (3.30)

LetC =

2k 2 − 1, (3.31)

where 0 < k < 1, and

f = ±2 lk

(1 − k 2 sin2 φ)1/2. (3.32)

Thus, from Equations (3.31) and (3.32),

C − f 2

2 l 2 = − cos(2 φ), (3.33)


0

1

2

3

z/l

−2 −1 0 1 2x/l

Figure 3.4Capillary curves for π/4 ≤ φ ≤ 3π/4 and (in order from the top to the bottom)k = 0.6, 0.7, 0.8, 0.9, and 0.99.

and so the constraint (3.29) implies that π/4 ≤ φ ≤ 3π/4. Moreover, Equations (3.30)and (3.33) reduce to

1fx=

dxd f= ± 1

tan(2 φ). (3.34)

It follows from Equations (3.32) and (3.34) that

dxdφ=

dxd f

d fdφ= − l k cos(2 φ)

(1 − k 2 sin2 φ)1/2, (3.35)

which can be integrated to give

xl=

∫ π/2

φ

k cos(2 φ)(1 − k 2 sin2 φ)1/2

dφ, (3.36)

assuming that x = 0 when φ = π/2. Thus, we get

xl=

(k −

2k

)F(φ, k) +

2k

E(φ, k), (3.37)

where

E(φ, k) = E(π/2, k) − E(φ, k), (3.38)

F(φ, k) = F(π/2, k) − F(φ, k), (3.39)

Surface Tension 83

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

z/l

−1 −0.5 0 0.5 1x/l

Figure 3.5Liquid/air interface for a liquid trapped between two vertical parallel plates locatedat x = ±l. The contact angle of the interface with the plates is θ = 30.

and

E(φ, k) =∫ φ

0(1 − k 2 sin2 φ)1/2, (3.40)

F(φ, k) =∫ φ

0(1 − k 2 sin2 φ)−1/2, (3.41)

are types of incomplete elliptic integral (Abramowitz and Stegun 1965). In conclu-sion, the interface shape is determined parametrically by

xl=

(k −

2k

)F(φ, k) +

2k

E(φ, k), (3.42)

zl= ±2

k(1 − k 2 sin2 φ)1/2, (3.43)

where π/4 ≤ φ ≤ 3π/4. Here, the parameter k is restricted to lie in the range 0 < k <1.

Figure 3.4 shows the capillary curves predicted by Equations (3.42) and (3.43) forvarious different values of k. Here, we have chosen the plus sign in Equation (3.43).However, if the minus sign is chosen then the curves are simply inverted: that is,x → x and z → −z. In can be seen that all of the curves shown in the figure are


symmetric about x = 0: that is, z → z as x → −x. Consequently, we can use thesecurves to determine the shape of the liquid/air interface which arises when a liquid istrapped between two flat vertical plates (made of the same material) that are parallelto one another. Suppose that the plates in question lie at x = ±d. Furthermore, letthe angle of contact of the interface with the plates be θ, where θ < π/2. Because theangle of contact is acute, we expect the liquid to be drawn upward between the plates,and the interface to be concave (from above). This corresponds to the positive signin Equation (3.43). In order for the interface to meet the plates at the correct angle,we require fx = −1/ tan θ at x = −d and fx = 1/ tan θ at x = +d. However, if one ofthese boundary conditions is satisfied then, by symmetry, the other is automaticallysatisfied. From Equation (3.34) (choosing the positive sign), the latter boundarycondition yields tan(2 φ) = 1/ tan θ at x = +d, which is equivalent to x = +d whenφ = 3π/4−θ/2. Substituting this value of φ into Equation (3.42), we can numericallydetermine the value of k for which x = d. The interface shape is then given byEquations (3.42) and (3.43), using the aforementioned value of k, and φ in the rangeπ/4 + θ/2 to 3π/4 − θ/2. For instance, if d = l and θ = 30 then k = 0.9406, and theassociated interface is shown in Figure 3.5. Furthermore, if we invert this interface(i.e., x → x and z→ −z) then we obtain the interface which corresponds to the sameplate spacing, but an obtuse contact angle of θ = 180 − 30 = 150.

Consider the limit k 1, which is such that the distance between the two platesis much less than the capillary length. It is easily demonstrated that, at small k(Abramowitz and Stegun 1965)

E(φ, k) φ −k 2

4(φ + sin φ cos φ), (3.44)

F(φ, k) φ + k 2

4(φ + sin φ cos φ), (3.45)

where φ = π/2 − φ. Thus, Equations (3.42) and (3.43) reduce to

xl − k

2sin(2 φ), (3.46)

zl 2

k− k

2[1 − cos(2 φ)]. (3.47)

It follows that the interface is a segment of the curved surface of a cylinder whoseaxis runs parallel to the y-axis. If the distance between the plates is 2 d, and thecontact angle is θ, then we require x = d when φ = 3π/4 − θ/2 (which correspondsto φ = −π/4 + θ/2). From Equation (3.46), this constraint yields

dl k

2cos θ. (3.48)

Thus, the height that the liquid rises between the two plates—that is, h = z(x = 0) =z(φ = π/2) 2 l/k—is given by

h γ cos θρ g d

. (3.49)

Surface Tension 85

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

z/l

0 1 2 3 4 5x/l

Figure 3.6Liquid/air interface for a liquid in contact with a vertical plate located at x = 0. Thecontact angle of the interface with the plate is θ = 25.

This result is the form taken by Jurin’s law, (3.23), for a liquid drawn up betweentwo parallel plates of spacing 2 d.

Consider the case k = C = 1, which is such that the distance between the twoplates is infinite. Let the leftmost plate lie at x = 0, and let us completely neglect therightmost plate, because it lies at infinity. Suppose that h = z(x = 0) is the height ofthe interface above the free surface of the liquid at the point where the interface meetsthe leftmost plate. If θ is the angle of contact of the interface with the plate then werequire fx = −1/ tan θ at x = 0. Because C = 1, it follows from Equation (3.28) that

h 2

2 l 2 = 1 − sin θ, (3.50)

orh = 2 l sin(π/4 − θ/2). (3.51)

Furthermore, again recalling that C = 1, Equation (3.30) can be integrated to give

x =∫ h

z

d ffx= l∫ h/2l

z/2l

1 − 2 y 2

y (1 − y 2)1/2 dy, (3.52)

where we have chosen the minus sign, and y = f /(2 l). Making the substitution


y = sin u, this becomes

xl=

∫ sin−1(h/2 l)

sin−1(z/2 l)

(1

sin u− 2 sin u

)du =

[− ln

(1 + cos u

sin u

)+ 2 cos u

]sin−1(h/2 l)

sin−1(z/2 l), (3.53)

which reduces to

xl= cosh−1

(2 lz

)− cosh−1

(2 lh

)+

(4 −

h 2

l 2

)1/2−(4 −

z 2

l 2

)1/2, (3.54)

because cosh−1(z) ≡ ln[z + (z 2 − 1)1/2]. Thus, Equations (3.51) and (3.54) specifythe shape of a liquid/air interface that meets an isolated vertical plate at x = 0. Inparticular, Equation (3.51) gives the height that the interface climbs up the plate(relative to the free surface) due to the action of surface tension. Note that this heightis restricted to lie in the range −2 l ≤ h ≤ 2 l, irrespective of the angle of contact.Figure 3.6 shows an example interface calculated for θ = 25.

3.8 Axisymmetric Soap-BubblesConsider an axisymmetric soap-bubble whose surface takes the form r = f (z) incylindrical coordinates. (See Section C.3.) The unit normal to the surface is

n ≡ ∇(r − f )|∇(r − f )|

=er − fz ez

(1 + f 2z )1/2 , (3.55)

where fz ≡ d f /dz. Hence, from Equation (C.39), the mean curvature of the surfaceis given by

∇ · n = 1f fz

ddz

[f

(1 + f 2z )1/2

]. (3.56)

The Young-Laplace equation, (3.12), then yields

f fza=

ddz

[f

(1 + f 2z )1/2

], (3.57)

wherea =

γ

p0. (3.58)

Here, γ is the net surface tension, including the contributions from the internal andexternal soap/air interfaces. Moreover, p0 = ∆p is the pressure difference betweenthe interior and the exterior of the bubble. Equation (3.57) can be integrated to give

f(1 + f 2

z )1/2 =f 2

2 a+C, (3.59)

Surface Tension 87

where C is a constant.Suppose that the bubble occupies the region z1 ≤ z ≤ z2, where z1 < z2, and

has a fixed radius at its two end-points, z = z1 and z = z2. This could most easilybe achieved by supporting the bubble on two rigid parallel co-axial rings located atz = z1 and z = z2. The net free energy required to create the bubble can be written

E = γ S − p0 V, (3.60)

where S is area of the bubble surface, and V the enclosed volume. The first term onthe right-hand side of the previous expression represents the work needed to over-come surface tension, whereas the second term represents the work required to over-come the pressure difference, −p0, between the exterior and the interior of the bub-ble. From the general principles of statics, we expect a stable equilibrium state ofa mechanical system to be such as to minimize the net free energy, subject to anydynamical constraints (Fitzpatrick 2012). It follows that the equilibrium shape of thebubble is such as to minimize

E = γ∫ z2

z1

2π f (1 + f 2z )1/2 dz − p0

∫ z2

z1

π f 2 dz, (3.61)

subject to the constraint that the bubble radius, f , be fixed at z = z1 and z = z2.Hence, we need to find the function f (z) that minimizes the integral∫ z2

z1

L( f , fz) dz, (3.62)

whereL( f , fz) = 2π γ f (1 + f 2

z )1/2 − π p0 f 2, (3.63)

subject to the constraint that f is fixed at the limits. This is a standard problemin the calculus of variations. (See Appendix E.) In fact, because the functionalL( f , fz) does not depend explicitly on z, the minimizing function is the solution of[see Equation (E.14)]

L − fz∂L∂ fz= C′, (3.64)

where C′ is an arbitrary constant. Thus, we obtain

2π γ[

f(1 + f 2

z )1/2 −f 2

2 a

]= C′, (3.65)

which can be rearranged to give Equation (3.59). Hence, we conclude that applica-tion of the Young-Laplace equation does indeed lead to a bubble shape that mini-mizes the net free energy of the soap/air interfaces.

Consider the case p0 = 0, in which there is no pressure difference across thesurface of the bubble. In this situation, writing C = b > 0, Equation (3.59) reducesto

f = b (1 + f 2z )1/2. (3.66)


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

r/c

−0.6 −0.4 −0.2 0 0.2 0.4 0.6z/c

Figure 3.7Radius versus axial distance for a catenoid soap bubble supported by two parallelco-axial rings of radius c located at z = ±0.65 c.

Moreover, according to the previous discussion, the bubble shape specified by Equa-tion (3.66) is such as to minimize the surface area of the bubble (because the onlycontribution to the free energy of the soap/air interfaces is directly proportional tothe bubble area). The previous equation can be rearranged to give

fz = ±(

f 2

b 2 − 1)1/2

, (3.67)

which leads to

z − z0 =

∫ r

b

d ffz= ±

∫ r

b

d f( f 2/b 2 − 1)1/2 = ±b cosh−1(r/b), (3.68)

orr = b cosh(|z − z0|/b), (3.69)

where z0 is a constant. This expression describes an axisymmetric surface known asa catenoid.

Suppose, for instance, that the soap bubble is supported by identical rings ofradius c that are located a perpendicular distance 2 d apart. Without loss of generality,we can specify that the rings lie at z = ±d. It follows, from Equation (3.69), thatz0 = 0, and

r = b cosh(z/b). (3.70)

Surface Tension 89

Here, the parameter b must be chosen so as to satisfy

c = b cosh(d/b). (3.71)

For example, if d = 0.65 c then b = 0.6416 c, and the resulting bubble shape isillustrated in Figure 3.7.

Let d/c = ζ and d/b = u, in which case the previous equation becomes

G(u) = u − ζ cosh u = 0. (3.72)

Now, the function G(u) attains a maximum value

G(u0) = u0 −1

tanh u0, (3.73)

when u0 = sinh−1(1/ζ). Moreover, if G(u0) > 0 then Equation (3.72) possessestwo roots. It turns out that the root associated with the smaller value of u mini-mizes the interface system energy, whereas the other root maximizes the free energy.Hence, the former root corresponds to a stable equilibrium state, whereas the lattercorresponds to an unstable equilibrium state. On the other hand, if G(u0) < 0 thenEquation (3.72) possesses no roots, implying the absence of any equilibrium state.The critical case G(u0) = 0 corresponds to u = uc and ζ = ζc, where uc tanh uc = 1and ζc = 1/ sinh uc. It is easily demonstrated that uc = 1.1997 and ζc = 0.6627. Weconclude that a stable equilibrium state of a catenoid bubble only exists when ζ ≤ ζc,which corresponds to d ≤ 0.6627 c. If the relative ring spacing d exceeds the criticalvalue 0.6627 c then the bubble presumably bursts.

Consider the case p0 0, in which there is a pressure difference across thesurface of the bubble. In this situation, writing

2 a = α + β, (3.74)

2 a C = α β, (3.75)

Equation (3.59) becomes(α + β) f

(1 + f 2z )1/2 = f 2 + α β, (3.76)


fz = ∓(α 2 − f 2)1/2 ( f 2 − β 2)1/2

f 2 + α β. (3.77)

We can assume, without loss of generality, that |α| > |β|. It follows, from the previousexpression, that |α| ≤ f ≤ |β|. Hence, we can write

f 2 = α 2 cos2 φ + β 2 sin2 φ, (3.78)

k 2 =α 2 − β 2

α 2 , (3.79)


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

r/α

−2 −1 0 1 2z/α

Figure 3.8Radius versus axial distance for an unduloid soap bubble calculated with k = 0.95.

where 0 ≤ φ ≤ π/2 and 0 < k ≤ 1. It follows that

f = |α| (1 − k 2 sin2 φ)1/2, (3.80)

β = sgn(β) |α| (1 − k 2)1/2, (3.81)

anddzdφ=

1fz

d fdφ= ±

(f +

α β

f

), (3.82)


|z| = |α|[E(φ, k) + sgn(α β) (1 − k 2)1/2 F(φ, k)

], (3.83)

where E(φ, k) and F(φ, k) are incomplete elliptic integrals [see Equations (3.40) and(3.41)]. Here, we have assumed that φ = 0 when z = 0. There are three cases ofinterest.

In the first case, α > 0 and β > 0. It follows that (1 − k 2)1/2 ≤ r/α ≤ 1 forπ/2 ≥ φ ≥ 0, and 0.5 ≤ γ/(p0 α) < 1 for 1 ≥ k > 0, where

r = α (1 − k 2 sin2 φ), (3.84)

|z| = α[E(φ, k) + (1 − k 2)1/2 F(φ, k)

]. (3.85)

Surface Tension 91

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

r/α

−0.5 0 0.5z/α

Figure 3.9Radius versus axial distance for a positive pressure nodoid soap bubble calculatedwith k = 0.95.

The axisymmetric curve parameterized by the previous pair of equations is knownas an unduloid. Note that an unduloid bubble always has positive internal pressure(relative to the external pressure): that is, p0 > 0. An example unduloid soap bubbleis illustrated in Figure 3.8

In the second case, α > 0 and β < 0. It follows that (1−k 2)1/4 ≤ r/α ≤ 1 for φ0 ≥φ ≥ 0, and 0 < γ/(p0 α) ≤ 0.5 for 0 < k ≤ 1, where φ0 = sin−1([1− (1−k 2)1/2]1/2/k),and

r = α (1 − k 2 sin2 φ), (3.86)

|z| = α[E(φ, k) − (1 − k 2)1/2 F(φ, k)

]. (3.87)

The axisymmetric curve parameterized by the previous pair of equations is knownas an nodoid. This particular type of nodoid bubble has positive internal pressure:that is, p0 > 0. An example positive pressure nodoid soap bubble is illustrated inFigure 3.9.

In the third case, α < 0 and β > 0. It follows that (1 − k 2)1/2| ≤ r/|α| ≤α| (1 − k 2)1/4 for π/2 ≥ φ ≥ φ0 (or π/2 ≤ φ ≤ π − φ0), and 0 > γ/(p0 |α|) ≥ −0.5 for


0

0.1

0.2

0.3

0.4

0.5

0.6

r/|α|

0.1 0.2 0.3 0.4 0.5z/|α|

Figure 3.10Radius versus axial distance for a negative pressure nodoid soap bubble calculatedwith k = 0.95.

0 < k ≤ 1, where

r = |α| (1 − k 2 sin2 φ), (3.88)

|z| = |α|[E(φ, k) − (1 − k 2)1/2 F(φ, k)

]. (3.89)

The axisymmetric curve parameterized by the previous pair of equations is again anodoid. However, this particular type of nodoid bubble has negative internal pres-sure: that is, p0 < 0. An example negative pressure nodoid soap bubble is illustratedin Figure 3.10.

3.9 Exercises3.1 Show that if N equal spheres of water coalesce so as to form a single spherical

drop then the surface energy is decreased by a factor 1/N 1/3. (Lamb 1928.)

3.2 A circular cylinder of radius a, height h, and specific gravity s floats uprightin water. Show that the depth of the base below the general level of the water

Surface Tension 93

surface is

s h +2 γa

cos θ,

where γ is the surface tension at the air/water interface, and θ the contactangle of the interface with the cylinder. (Lamb 1928.)

3.3 A film of water is held between two parallel plates of glass a small distance2 d apart. Prove that the apparent attraction between the plates is

2 A γ cos θd

+ L γ sin θ,

where γ is the surface tension at the air/water interface, θ the angle of contactof the interface with glass, A the area of the film, and L the circumference ofthe film. (Lamb 1928.)

3.4 Show that if the surface of a sheet of water is slightly corrugated then thesurface energy is increased by

γ

2

∫ (∂ζ

∂x

)2dx

per unit breadth of the corrugations. Here, x is measured horizontally, per-pendicular to the corrugations. Moreover, ζ denotes the elevation of the sur-face above the mean level. Finally, γ is the surface tension at an air/waterinterface. If the corrugations are sinusoidal, such that

ζ = a sin(k x),

show that the average increment of the surface energy per unit area is

(1/4) γ a 2 k 2.

(Lamb 1928.)

3.5 A mass of liquid, which is held together by surface tension alone, revolvesabout a fixed axis at a small angular velocity ω, so as to assume a slightlyspheroidal shape of mean radius a. Prove that the ellipticity of the spheroidis

ε =ρω 2 a 3

8 γ,

where ρ is the uniform mass density, and γ the surface tension. [If r+ is themaximum radius, and r− the minimum radius, then a = (r 2

+ r−)1/3, and theellipticity is defined ε = (r+ − r−)/r+.] (Lamb 1928.)

3.6 A liquid mass rotates, in the form of a circular ring of radius a and smallcross-section, with a constant angular velocity ω, about an axis normal to


the plane of the ring, and passing through its center. The mass is held to-gether by surface tension alone. Show that the section of the ring must beapproximately circular. Demonstrate that

ω =

(2 γρ a c 2

)1/2,

where ρ is the density, γ the surface tension, and c the radius of the cross-section. (Lamb 1928.)

3.7 Two spherical soap bubbles of radii a1 and a2 are made to coalesce. Showthat when the temperature of the gas in the resulting bubble has returned toits initial value the radius a of the bubble satisfies

p0 a 3 + 4 γ a 2 = p0 (a 31 + a 3

2 ) + 4 γ (a 21 + a 2

2 ),

where p0 is the ambient pressure, and γ the surface tension of the soap/airinterfaces. (Batchelor 2000.)

3.8 A rigid sphere of radius a rests on a flat rigid surface, and a small amount ofliquid surrounds the contact point, making a concave-planar lens whose di-ameter is small compared to a. The angle of contact of the liquid/air interfacewith each of the solid surfaces is zero, and the surface tension of the interfaceis γ. Show that there is an adhesive force of magnitude 4π a γ acting on thesphere. (It is interesting to note that the force is independent of the volumeof liquid.) (Batchelor 2000.)

3.9 Two small solid bodies are floating on the surface of a liquid. Show that theeffect of surface tension is to make the objects approach one another if theliquid/air interface has either an acute or an obtuse angle of contact with bothbodies, and to make them move away from one another if the interface has anacute angle of contact with one body, and an obtuse angle of contact with theother. (Batchelor 2000.)

4Incompressible Inviscid Flow

4.1 Introduction

This chapter introduces some of the fundamental concepts that arise in the theory ofincompressible, inviscid (or, to be more exact, high Reynolds number) fluid motion.Further information on these concepts can be found in Faber 1995, Batchelor 2000,and Milne-Thompson 2011.

4.2 Streamlines, Stream Tubes, and Stream Filaments

A line drawn in a fluid such that its tangent at each point is parallel to the localfluid velocity is called a streamline. The aggregate of all the streamlines at a giveninstance in time constitutes the instantaneous flow pattern. The streamlines drawnthrough each point of a closed curve constitute a stream tube. Finally, a streamfilament is defined as a stream tube whose cross-section is a curve of infinitesimaldimensions.

If the flow is unsteady then the configuration of the stream tubes and filamentschanges from time to time. However, if the flow is steady then the stream tubesand filaments are stationary. In the latter case, a stream tube acts like an actual tubethrough which the fluid is flowing. This follows because there can be no flow acrossthe walls, and into the tube, because the flow is, by definition, always tangential tothese walls. Moreover, the walls are fixed in space and time, because the motion issteady. Thus, the motion of the fluid within the tube would be unchanged were thewalls replaced by a rigid frictionless boundary.

Consider a stream filament of an incompressible fluid whose motion is steady.Suppose that the cross-sectional area of the filament is sufficiently small that thefluid velocity is the same at each point on the cross-section. Moreover, let the cross-section be everywhere normal to the direction of this common velocity. Suppose thatv1 and v2 are the flow speeds at two points on the filament where the cross-sectionalareas are S 1 and S 2, respectively. Consider the section of the filament lying betweenthese points. Because the fluid is incompressible, the same volume of fluid must flowinto one end of the section, in a given time interval, as flows out of the other, which

95


A′

B′

CC ′D

D′

B

A

Figure 4.1Bernoulli’s theorem.

implies thatv1 S 1 = v2 S 2. (4.1)

This is the simplest manifestation of the equation of fluid continuity discussed inSection 1.9. The previous result is equivalent to the statement that the product ofthe flow speed and cross-sectional area is constant along any stream filament of anincompressible fluid in steady motion. Thus, a stream filament within such a fluidcannot terminate unless the speed at that point becomes infinite. Leaving this case outof consideration, it follows that stream filaments in steadily flowing incompressiblefluids either form closed loops, or terminate at the boundaries of the fluid. The sameis, of course, true of streamlines.

4.3 Bernoulli’s TheoremIn its most general form, Bernoulli’s theorem—which was discovered by DanielBernoulli (1700–1783)—states that, in the steady flow of an inviscid fluid, the quan-tity

pρ+ T (4.2)

is constant along a streamline, where p is the pressure, ρ the density, and T the totalenergy per unit mass.

The proof is straightforward. Consider the body of fluid bounded by the cross-sectional areas AB and CD of the stream filament pictured in Figure 4.1. Let us

Incompressible Inviscid Flow 97

denote the values of quantities at AB and CD by the suffixes 1 and 2, respectively.Thus, p1, v1, ρ1, S 1, T1 are the pressure, flow speed, mass density, cross-sectionalarea, and total energy per unit mass, respectively, at AB, et cetera. Suppose that, aftera short time interval δt, the body of fluid has moved such that it occupies the sectionof the filament bounded by the cross-sections A′B′ and C′D′, where AA′ = v1 δt andCC′ = v2 δt. Because the motion is steady, the mass m of the fluid between AB andA′B′ is the same as that between CD and C′D′, so that

m = S 1 v1 δt ρ1 = S 2 v2 δt ρ2. (4.3)

Let T denote the total energy of the section of the fluid lying between A′B′ and CD.Thus, the increase in energy of the fluid body in the time interval δt is

(mT2 + T ) − (mT1 + T ) = m (T2 − T1). (4.4)

In the absence of viscous energy dissipation, this energy increase must equal the network done on the fluid by the pressures at AB and CD, which is

p1 S 1 v1 δt − p2 S 2 v2 δt = m(

p1

ρ1− p2

ρ2

). (4.5)

Equating expressions (4.4) and (4.5), we find thatp1

ρ1+ T1 =

p2

ρ2+ T2, (4.6)

which demonstrates that p/ρ + T has the same value at any two points on a givenstream filament, and is therefore constant along the filament. Note that Bernoulli’stheorem has only been proved for the case of the steady motion of an inviscid fluid.However, the fluid in question may either be compressible or incompressible.

For the particular case of an incompressible fluid, moving in a conservative force-field, the total energy per unit mass is the sum of the kinetic energy per unit mass,(1/2) v 2, and the potential energy per unit mass, Ψ , and Bernoulli’s theorem thusbecomes

pρ+

12v 2 + Ψ = constant along a streamline. (4.7)

If we focus on a particular streamline, 1 (say), then Bernoulli’s theorem states that

pρ+

12v 2 + Ψ = C1, (4.8)

where C1 is a constant characterizing that streamline. If we consider a second stream-line, 2 (say), then

pρ+

12v 2 + Ψ = C2, (4.9)

where C2 is another constant. It is not generally the case that C1 = C2. If, however,the fluid motion is irrotational then the constant in Bernoulli’s theorem is the samefor all streamlines (see Section 4.15), so that

pρ+

12v 2 + Ψ = C (4.10)

throughout the fluid.


4.4 Euler Momentum TheoremConsider the stream filament shown in Figure 4.1. Let S 1 and S 2 be the cross-sectional areas at AB and CD, respectively, and let v1 and v2 be the correspondingflow velocities. Assuming the the flow is both steady and incompressible, Euler’smomentum theorem, which is named after Leonhard Euler (1707-1783), states that,neglecting external forces, the resultant force due to the pressure of the surroundingfluid on the walls and ends of the filament is equivalent to forces ρ S 1 v

21 and ρ S 2 v

22

acting normally outward at the ends AB and CD, respectively.The proof is straightforward. According to Newton’s second law of motion, the

resultant force must produce the change in the momentum of the fluid that occu-pies the portion of the filament between AB and CD at any given instant of time, t.Suppose that at time t + δt the fluid in question occupies the portion of the filamentbetween A′B′ and C′D′. The momentum of the fluid in question has then increasedby the momentum of the fluid between CD and C′D′, and decreased by the momen-tum of the fluid between AB and A′B′. Hence, there has been a gain of momentumρ S 2 v2 δt × v2 at CD, and a loss of momentum ρ S 1 v1 δ t × v1 at AB. Thus, the netrate of charge of momentum consists of a gain ρ S 2 v2 v2 at CD, and a loss ρ S 1 v1 v1

at AB. This net rate of change is produced solely by the thrusts acting on the wallsand ends of the filaments. It follows that these thrusts are equivalent to the forcesρ S 1 v

21 and ρ S 2 v

22 acting normally outward at AB and CD, respectively.

If p1 and p2 are the pressures at AB and CB, respectively, then the thrusts act-ing normally inward on the ends of the filament are p1 S 1 at AB and p2 S 2 at CD.According to Euler’s theorem, the thrusts exerted on the walls plus the thrusts actingon the ends are equivalent to the normal outward forces ρ S 1 v

21 at AB and ρ S 2 v

22 at

CD. It follows that the thrusts exerted by the walls on the fluid are equivalent to thenormal outward forces S 1 (p1 + ρ v

21 ) at AB and S 2 (p2 + ρ v

22 ) at CD. Conversely,

the thrusts exerted by the fluid on the walls are equivalent to normal inward forcesS 1 (p1 + ρ v

21 ) at AB and S 2 (p2 + ρ v

22 ) at CD.

Note, finally, that the Euler momentum theorem obviously also applies to astream tube, as long as the flow through the ends of the tube is uniform across thecross-section.

4.5 d’Alembert’s ParadoxConsider a long straight tube through which an inviscid fluid flows at the constantspeed V . If we place a small obstacle, A, in the middle of the tube then the flow in theimmediate neighborhood of A will be modified, but that a great distance upstreamor downstream of A will presumably remain undisturbed. In general, in order tomaintain the obstacle at rest, we need to exert both a force and a couple on it. LetF be the component of the force in the direction of the original flow. Neglecting


S2

VA

V

S1

Figure 4.2The d’Alembert paradox.

external forces, such as gravity, F is the resultant in the direction of the flow of thepressure thrusts acting on the boundary of A.

Consider two cross-sections, S 1 and S 2, a great distance upstream and down-stream of A, respectively. The fluid between these sections can be split into a greatmany stream filaments to each of which Euler’s momentum theorem is applicable.The outer filaments are bounded by the walls of the tube, and so the thrust compo-nents acting on these are directed perpendicular to the flow. The walls of the filamentsin contact with A are acted on by the obstacle, which exerts on them a force whosecomponent in the direction of the flow is −F. By Euler’s theorem, the resultant of allof the thrusts acting on the fluid in the tube is

−ρ S 1 V 2 + ρ S 2 V 2, (4.11)

which vanishes, because S 1 = S 2.By Bernoulli’s theorem, the pressure p1 acting across S 1 is the same as the pres-

sure p2 acting across S 2. Thus, according to Euler’s theorem,

p1 S 1 − F − p2 S 2 = 0, (4.12)

which implies thatF = 0. (4.13)

The surprising result that the net parallel force exerted by an inviscid fluid streamon a stationary obstacle placed in its path is zero is known as d’Alembert’s paradox,after the French scientist Jean-Baptiste d’Alembert (1717–1783).

If we suppose the walls of the tube to recede to infinity then we obtain the case ofan obstacle immersed in a moving stream that is unbounded in every direction. Theprevious proof still shows that F = 0.

Finally, if we impose on the whole system a uniform velocity V in the directionopposite to that of the stream then the fluid at a great distance is reduced to rest, andthe obstacle A moves with the uniform speed V . However, superposing a uniform


v

B

A

Figure 4.3Outflow through an orifice.

flow does not alter the dynamical conditions. Therefore, we conclude that the resis-tance to a solid body moving with uniform velocity through an unbounded inviscidfluid, otherwise at rest, is zero.

4.6 Flow Through an Orifice

Consider the situation, illustrated in Figure 4.3, in which a horizontal jet of fluidemerges from an orifice in the side of a container. As shown in the figure, the jet nar-rows over a short distance beyond the orifice that is comparable with the jet diameterto form what is generally known as a vena contracta—that is, a “contracted vein.”The jet is bound to narrow in this manner because of the curvature of the lines offlow as they pass through the orifice. The narrowing of the jet implies the existenceof a transverse pressure gradient. In other words, the pressure at A, on the axis of thejet, is higher than the atmospheric pressure that acts at B. The pressure excess at Asuggests that the fluid on the axis is still accelerating longitudinally as it leaves theorifice. Only in the vena contracta does the flow velocity becomes uniform, and thepressure atmospheric, all the way across the jet.

Let S be the cross-sectional area of the orifice, and C S that of the vena contracta.Here, C is known as the contraction coefficient. Let us apply Bernoulli’s theorem toa streamline that starts on the surface of the fluid within the container, and ends in thevena contracta. Suppose that the surface of the fluid lies a height h above the orifice.Let us assume that the fluid close to the surface is essentially at rest (which implies


that the outflow through the orifice is not sufficiently strong to cause the surface levelto drop at a significant rate.) Let v be the uniform fluid velocity in the vena contracta.Of course, the pressure is atmospheric both at the surface of the fluid and in the venacontracta. It follows that

g h =12v 2, (4.14)

or

v = (2 g h)1/2. (4.15)

In other words, the efflux velocity of the fluid from the orifice is the same as thatit would have acquired by falling a height h under gravity. This result is known asTorricelli’s law, after Evangelista Torricelli (1608-1647). Finally, the discharge rateof fluid flowing through the orifice is

Q = C S v = C S (2 g h)1/2. (4.16)

Let p = p0 + ρ g h be the hydrostatic pressure at the level of the orifice whenthe orifice is closed. Here, p0 is atmospheric pressure. The fluid experiences athrust S p from the section of the wall directly opposite the orifice, and a thrust−S p from the section of the wall closing the orifice. Let us suppose, as a firstapproximation, that the hydrostatic pressure remains unaltered when the orifice isopened. In this situation, the fluid experiences a thrust S p from the section of thewall directly opposite the orifice, and a thrust −S p0 from the orifice. The net thrust,S (p − p0) = S ρ g h, is responsible for accelerating the jet. Now, the jet’s rate ofmomentum outflow is ρ vC S × v. Momentum conservation yields

S ρ g h = C S ρ v 2 = 2C S ρ g h, (4.17)

where use has been made of Equation (4.15). Thus, we conclude that the contractioncoefficient takes the value 1/2.

In reality, Bernoulli’s theorem suggests that when the orifice is opened the pres-sure on the walls in the neighborhood of the orifice will fall below the hydrostaticvalue, which implies that the accelerating thrust is actually greater than S (p − p0).Consequently, C > 1/2. Obviously, C cannot exceed unity, so we conclude that, ingeneral, 1/2 < C < 1. For instance, if the orifice is a circular hole punched in a thinplate then the contraction coefficient is observed to take the value 0.62 (Batchelor2000).

Suppose, however, that we fit a small cylindrical nozzle projecting inward fromthe orifice, as shown in Figure 4.4. In this case, the original assumption that thepressure on the walls in the neighborhood of the orifice is hydrostatic is essentiallycorrect. This follows because the region where the lines of flow are converging on theorifice is far removed from the walls, and the velocity of the fluid in contact with thewalls is negligible. Thus, the contraction coefficient is exactly 1/2. This arrangementis known as a Borda mouthpiece, after Jean-Charles Borda (1733–1799).


Figure 4.4A Borda mouthpiece.

4.7 Sub-Critical and Super-Critical FlowConsider a shallow stream of depth h, uniform width, and uniform flow velocity v,that is fed from a deep reservoir whose surface lies a height H above the (horizontal)bed of the stream. Here, H is usually referred to as the head height. Assuming thatthe water in the reservoir is effectively stationary, application of Bernoulli’s equationto a streamline lying on the surface of the water (where the pressure is atmospheric)yields

H = h +v 2

2 g. (4.18)

Let Q be the flow rate per unit width of the stream, which is assumed to be fixed. Itfollows that

Q = h v. (4.19)

The previous two equations can be combined to give

H = F(v), (4.20)

where

F(v) =Qv+v 2

2 g. (4.21)

It is easily demonstrated that the function F(v) attains its minimum value,

Fc =32

(Q 2

g

)1/3, (4.22)


when v = vc, wherevc = (Q g)1/3. (4.23)

We conclude that, as long as H > (3/2) (Q 2/g)1/3, Equation (4.20) possesses twopossible solutions that are consistent with a given head height and flow rate. In onesolution, the stream flows at a relatively slow velocity, v−, which is such that v− < vc.In the other, the stream flows at a relatively fast velocity, v+, which is such thatv+ > vc. The corresponding depths are h− = Q/v− and h+ = Q/v+, respectively.

It is helpful to introduce the dimensionless Froude number,

Fr =v√g h. (4.24)

(See Section 1.15.) Note that√g h is the characteristic propagation velocity of a

gravity wave in shallow water of depth h. (See Section 11.4.) Hence, if Fr < 1then the stream’s flow velocity falls below the wave speed—such flow is termedsub-critical. On the other hand, if Fr > 1 then the flow velocity exceeds the wavespeed—such flow is termed super-critical.

We can combine Equations (4.18), (4.19), and (4.24) to give

H = G(Fr), (4.25)

where

G(Fr) =(

Q 2

g

)1/2 ( 1Fr 2/3 +

Fr 4/3

2

). (4.26)

It is easily demonstrated that G(Fr) attains its minimum value

Gc =32

(Q 2

g

)1/3, (4.27)

when Fr = Frc, whereFrc = 1. (4.28)

Hence, we again conclude that, as long as H > (3/2) (Q 2/g)1/3, there are two pos-sible flow velocities of the stream (parameterized by two different Froude numbers)that are consistent with a given head height and flow rate. However, it is now clearthat the smaller velocity is sub-critical (i.e., Fr < 1), whereas the larger velocity issuper-critical (i.e., Fr > 1).

4.8 Flow over Shallow BumpConsider a shallow stream of depth H, uniform width, and uniform flow velocity V .Suppose that there is a very shallow bump of height d H on the (horizontal) bedof the stream, as shown in Figure 4.5. Suppose, further, that, at the point where the


h

d

H

vV

Figure 4.5Flow over a shallow bump.

stream passes over the top of the bump, its velocity is v, and its surface rises a heighth H above the unperturbed surface.

Fluid continuity yieldsH V = (H + h − d) v. (4.29)

Furthermore, application of Bernoulli’s equation to a streamline lying on the surfaceof the water (where the pressure is atmospheric) gives

gH +12

V 2 = g (H + h) +12v 2. (4.30)

The previous equation reduces to

v 2 = V 2 − 2 g h. (4.31)

Eliminating v between Equations (4.29) and (4.31), we obtain

H 2 V 2 = (H + h − d) 2 (V 2 − 2 g h), (4.32)


V 2[(H + h − d)2 − H 2

]= 2 g h (H + h − d)2, (4.33)

or

Fr 2

(1 + h − dH

)2− 1

= 2hH

(1 +

h − dH

) 2

. (4.34)

Here,

Fr =V√gH

(4.35)

is the Froude number of the unperturbed flow. Finally, given that h/H 1 andd/H 1, Equation (4.34) reduces to

h d1 − 1/Fr 2 . (4.36)


v2

zh2

h1 v1

Figure 4.6A hydraulic jump.

It follows, from the previous expression, that if the flow is super-critical, so thatFr > 1, then h is positive. On the other hand, if the flow is sub-critical, so thatFr < 1, then h is negative. Thus, if a super-critical shallow stream passes over a veryshallow bump on its bed then the surface of the stream becomes slightly elevated.On the other hand, if a sub-critical stream passes over such a bump then the surfaceof the stream becomes slightly depressed. A similar effect occurs when there is anarrowing of the channel in the horizontal direction. A more sophisticated versionof the previous calculation, which does not necessarily assume that the stream isshallow, can be found in Section 11.10.

4.9 Stationary Hydraulic Jumps

Under certain circumstances, water flowing within a horizontal open channel (i.e., astream) of constant width is found to have a depth that changes very rapidly over ashort section of the channel. This phenomenon is known as a hydraulic jump, and isillustrated in Figure 4.6.

Suppose that the jump is stationary. Let h1, v1, and p1 be the depth, flow veloc-ity, and pressure, of the water, respectively, upstream of the jump. Similarly let h2,v2, and p2, be the depth, flow velocity and pressure, respectively, downstream of thejump. As before, v1 and v2 are assumed to be uniform across the channel. We are alsoassuming that the flow is laminar (i.e., smooth) both upstream and downstream of thejump. Note that the region of the channel in which the jump occurs is generally asso-ciated with violent mixing that gives rise to significant transfer of mechanical energyfrom the laminar (i.e., smooth) to the turbulent component of the flow. (The lattercomponent is localized in the vicinity of the jump, and is continuously dissipated byviscosity on small-scales.) Consequently, we cannot use Bernoulli’s equation to ana-lyze the jump, because this equation assumes that the laminar component of the flowconserves mechanical energy. However, we can still make use of fluid continuity, as


well as the Euler momentum theorem, neither of which depend on the conservationof mechanical energy.

Fluid continuity yieldsQ = h1 v1 = h2 v2, (4.37)

where Q is the fixed flow rate per unit width. Furthermore, the Euler momentumtheorem (see Section 4.4) implies that

F21 =

∫ h2

0(p2 + ρ v

22 ) dz −

∫ h1

0(p1 + ρ v

21 ) dz. (4.38)

Here, F21 is the horizontal thrust per unit width exerted by the channel bed on thewater lying between points 1 and 2. Moreover, z measured vertical height above thebed. Assuming the usual linear pressure variation with depth, we can write

p1(z) = p0 + ρ g (h1 − z), (4.39)

p2(z) = p0 + ρ g (h2 − z), (4.40)

where p0 is atmospheric pressure. Hence, we deduce that

F21

ρ g=

12

h 22 −

12

h 21 +

Qg

(v2 − v1). (4.41)

The neglect of frictional drag at the channel bed implies that F21 = 0. Thus, weobtain

12

h 22 −

12

h 21 +

Qg

(v2 − v1) = 0. (4.42)

The previous equation possesses the trivial solution h1 = h2 and v1 = v2, whichcorresponds to the absence of a hydraulic jump. Eliminating v1 and v2 between Equa-tions (4.37) and (4.42), and canceling a common factor h2 − h1, we obtain the non-trivial solution

h2 h1 (h2 + h1) =2 Q 2

g. (4.43)

Now, the upstream Froude number is defined

Fr1 =v1√g h1=

Q

g 1/2 h 3/21

. (4.44)

Eliminating Q between the previous two equations, we obtain(h2

h1

)2+

h2

h1− 2 Fr 2

1 = 0, (4.45)

which can be solved to give

h2

h1=

12

(−1 +

√1 + 8 Fr 2

1

). (4.46)


Here, we have neglected an unphysical solution in which h2/h1 is negative. Note that

h2

h1 1 as Fr1 1. (4.47)

The downstream Froude number is defined

Fr2 =v2√g h2=

Q

g 1/2 h 3/22

. (4.48)

Eliminating Q between Equations (4.43) and (4.48), we obtain(h1

h2

)2+

h1

h2− 2 Fr 2

2 = 0, (4.49)

which can be solved to give

h1

h2=

12

(−1 +

√1 + 8 Fr 2

2

). (4.50)

Here, we have again neglected an unphysical solution in which h1/h2 is negative. Wecan combine Equations (4.46) and (4.50) to give

G(Fr1)G(Fr2) = 1, (4.51)

whereG(x) ≡ 1

2

(−1 +

√1 + 8 x 2

). (4.52)

Note that

G(x) 1 as x 1, (4.53)

which implies that

Fr1 1 as Fr2 1. (4.54)

In other words, if the upstream flow is super-critical then the downstream flow issub-critical, and h2 > h1. On the other hand, if the upstream flow is sub-critical thenthe downstream flow is super-critical, and h2 < h1.

By analogy with Equation (4.18), we can define the head heights of the flowsupstream and downstream of the jump as

H1 = h1 +12v 2

1

g, (4.55)

and

H2 = h2 +12v 2

2

g, (4.56)

respectively. Of course, in the absence of any transfer of mechanical energy from the


laminar to the turbulent component of the flow, within the jump, we would expectH1 = H2. (Because this is what Bernoulli’s equation predicts.) In the presence ofthe transfer, we expect H1 > H2. In other words, we expect there to be a head lossacross the jump. Note that it is impossible for H2 to exceed H1, because this wouldimply a transfer of mechanical energy from the turbulent to the laminar componentof the flow, within the jump, which violates the second law of thermodynamics. Let

HL = H1 − H2 = h1 − h2 +v 2

1

2 g−v 2

2

2 g(4.57)

be the positive definite head loss. Making use of some previous definitions, we canwrite

HL

h1= −

(h2

h1− 1)+

12

Fr 21

1 − h 21

h 22

. (4.58)

Now, according to Equation (4.45),

Fr 21 =

12

h2

h1

(h2

h1+ 1). (4.59)

Hence, combining the previous two equations, we obtain

HL =14

h1

h2

(h2

h1− 1) −4

h2

h1+

(h2

h1+ 1)2 , (4.60)

or

HL =(h2/h1 − 1)3

4 h2/h1. (4.61)

Thus, the thermodynamic constraint HL > 0 implies that h2 > h1. However, as wehave already seen, h2 > h1 implies that Fr1 > 1. In other words, a stationary hydraulicjump can only occur when the upstream flow is super-critical, and the downstreamflow sub-critical.

A hydraulic jump often occurs at the base of a spillway from a dam, where theflow is accelerated to super-critical speeds. Because a hydraulic jump always resultsin a loss of mechanical energy from the flow, spillways are sometimes designed topromote jumps, so as to deliberately remove energy from the flow, thereby reducingthe danger from excessive currents in flood control.

4.10 Tidal BoresA tidal bore is a sort of hydraulic jump that propagates up (i.e., upstream) a riverestuary. The upper part of Figure 4.7 shows such a bore in the local rest frame ofthe Earth. The bore is propagating at the velocity V up a river of uniform width,and depth h1, that is flowing downstream at the velocity u1. The flow behind the


h2

Co-moving frame

V + u1

V − u2

u1

u2

Earth frame

V

h1

h1

h2

Figure 4.7A tidal bore.

bore is of depth h2, and is flowing upstream at the velocity u2. The lower part ofthe figure shows the same phenomenon in the rest frame of the bore. In this frame,we observe a stationary hydraulic jump with an upstream depth and flow velocityh1 and v1 = V + u1, respectively, and a downstream depth and flow velocity h2 andv2 = V − u2, respectively. Making use of Equations (4.44) and (4.45), we obtain

(h2

h1

)2+

h2

h1− 2 (u1 + V)2

g h1= 0, (4.62)


V = −u1 +

[g h2 (h1 + h2)

2 h1

]1/2. (4.63)

Thus, we deduce that the speed of the bore, relative to the unperturbed river, is asimple function of the upstream and downstream depths. Note that, in the limith1 h2 h, the previous equation reduces to V −u1 +

√g h. In other words, a

weak bore degenerates into an ordinary shallow water gravity wave propagating atthe characteristic velocity

√g h relative to the stream.

Tidal bores are found in river estuaries where a funneling effect causes the speedof the incoming tide to increase to such a point that the flow becomes super-critical.For example, bores can be observed daily on the River Severn in England.


h(x)

d(x)

v(x)

x

Figure 4.8Flow over a broad-crested weir.

4.11 Flow over a Broad-Crested WeirConsider the situation, illustrated in Figure 4.8, in which a broad-crested weir isplaced in a shallow stream. The purpose of the weir is to impede the flow in such amanner that there is a transition from sub-critical flow, upstream of the weir, to super-critical flow, immediately downstream of the weir. (There is usually a transition backto sub-critical flow, via a hydraulic jump, some way downstream of the weir.)

Let x measure horizontal distance, and let h(x), v(x), and d(x) be the stream depth,stream velocity, and the height of the weir above the stream bed, respectively. (Here,we are assuming that the velocity remains uniform across the stream at any point onthe weir.) A direct generalization of the analysis of Section 4.7 reveals that

H = h + d +v 2

2 g, (4.64)

andQ = h v, (4.65)

where (via continuity) the flow rate, Q, is not a function of x. It follows that

H − d(x) = F(v), (4.66)

where F(v) is defined in Equation (4.21). Suppose that the weir attains its maximumheight, d(0), at x = 0. It follows that H − d(x) passes through a minimum value atx = 0. Hence, we would expect F(v) to also pass through its minimum value at x = 0.It follows that

H − d(0) = Fc =32

(Q 2

g

)1/3, (4.67)

v(0) = vc = (Q g)1/3, (4.68)


where use has been made of Equations (4.22) and (4.23). However,

Q = h(0) v(0), (4.69)

where h(0) is the depth of the stream as its passes over the highest point of the weir.Thus, we deduce that

Q = [g h 3(0)]1/2 =

(23

)3/2 (g [H − d(0)]3

)1/2. (4.70)

In other words, the flow rate (per unit width) is very simply related to the depth ofthe stream as it passes over the highest point of the weir, or, alternatively, the depthof the highest point of the weir below the water surface in the reservoir feeding thestream. For this reason, weirs are commonly used as devices to both measure andcontrol flow rates in streams.

4.12 Vortex Lines, Vortex Tubes, and Vortex FilamentsThe curl of the velocity field of a fluid, which is generally termed vorticity, is usuallyrepresented by the symbol ω, so that

ω = ∇ × v. (4.71)

A vortex line is a line whose tangent is everywhere parallel to the local vorticityvector. The vortex lines drawn through each point of a closed curve constitute thesurface of a vortex tube. Finally, a vortex filament is a vortex tube whose cross-section is of infinitesimal dimensions.

Consider a section AB of a vortex filament. The filament is bounded by thecurved surface that forms the filament wall, as well as two plane surfaces, whosevector areas are S1 and S2 (say), which form the ends of the section at points A and B,respectively. (See Figure 4.9.) Let the plane surfaces have outward pointing normalsthat are parallel (or anti-parallel) to the vorticity vectors, ω1 and ω2, at points A andB, respectively. The divergence theorem (see Section A.20), applied to the section,yields ∮

ω · dS =∫∇ · ω dV, (4.72)

where dS is an outward directed surface element, and dV a volume element. How-ever,

∇ · ω = ∇ · ∇ × v ≡ 0 (4.73)

[see Equation (A.173)], implying that∮ω · dS = 0. (4.74)


S2

A

B

S1

Figure 4.9A vortex filament.

Now, ω · dS = 0 on the curved surface of the filament, because ω is, by definition,tangential to this surface. Thus, the only contributions to the surface integral comefrom the plane areas S1 and S2. It follows that∮

ω · dS = S 2 ω2 − S 1 ω1 = 0. (4.75)

This result is essentially an equation of continuity for vortex filaments. It impliesthat the product of the magnitude of the vorticity and the cross-sectional area, whichis termed the vortex intensity, is constant along the filament. It follows that a vortexfilament cannot terminate in the interior of the fluid. For, if it did, the cross-sectionalarea, S , would have to vanish, and, therefore, the vorticity, ω, would have to be-come infinite. Thus, a vortex filament must either form a closed vortex ring, or mustterminate at the fluid boundary.

Because a vortex tube can be regarded as a bundle of vortex filaments whose netintensity is the sum of the intensities of the constituent filaments, we conclude thatthe intensity of a vortex tube remains constant along the tube.

4.13 Circulation and VorticityConsider a closed curve C situated entirely within a moving fluid. The vector lineintegral (see Section A.14)

ΓC =

∮C

v · dr, (4.76)


where dr is an element of C, and the integral is taken around the whole curve, istermed the circulation of the flow around the curve. The sense of circulation (i.e.,either clockwise or counter-clockwise) is arbitrary.

Let S be a surface having the closed curve C for a boundary, and let dS be anelement of this surface (see Section A.7) with that direction of the normal which isrelated to the chosen sense of circulation around C by the right-hand circulation rule.(See Section A.8.) According to the curl theorem (see Section A.22),

ΓC =

∮C

v · dr =∫

Sω · dS. (4.77)

Thus, we conclude that circulation and vorticity are intimately related to one another.In fact, according to the previous expression, the circulation of the fluid around loopC is equal to the net sum of the intensities of the vortex filaments passing throughthe loop and piercing the surface S (with a filament making a positive, or negative,contribution to the sum depending on whether it pierces the surface in the directiondetermined by the chosen sense of circulation around C and the right-hand circulationrule, or in the opposite direction). One important proviso to Equation (4.77) is thatthe surface S must lie entirely within the fluid.

4.14 Kelvin Circulation TheoremAccording to the Kelvin circulation theorem, which is named after Lord Kelvin(1824–1907), the circulation around any co-moving loop in an inviscid fluid is in-dependent of time. The proof is as follows. The circulation around a given loop C isdefined

ΓC =

∮C

v · dr. (4.78)

However, for a loop that is co-moving with the fluid, we have dv = d(dr/dt) =d(dr)/dt. Thus,

dΓC

dt=

∮C

dvdt· dr +

∮C

v · dv. (4.79)

By definition, dv/dt = Dv/Dt for a co-moving loop. (See Section 1.10.) More-over, the equation of motion of an incompressible inviscid fluid can be written [seeEquation (1.79)]

DvDt= −∇

(pρ+ Ψ

), (4.80)

because ρ is a constant. Hence,

dΓC

dt= −

∮C∇(

pρ− 1

2v 2 + Ψ

)· dr = 0, (4.81)

because v · dv = d(v 2/2) = ∇(v 2/2) · dr (see Section A.18), and p/ρ − v 2/2 + Ψ isobviously a single-valued function.


C ′

C

Figure 4.10A vortex tube.

One corollary of the Kelvin circulation theorem is that the fluid particles thatform the walls of a vortex tube at a given instance in time continue to form thewalls of a vortex tube at all subsequent times. To prove this, imagine a closed loopC that is embedded in the wall of a vortex tube but does not circulate around theinterior of the tube. (See Figure 4.10.) The normal component of the vorticity overthe surface enclosed by C is zero, because all vorticity vectors are tangential to thissurface. Thus, from Equation (4.77), the circulation around the loop is zero. ByKelvin’s circulation theorem, the circulation around the loop remains zero as thetube is convected by the fluid. In other words, although the surface enclosed by Cdeforms, as it is convected by the fluid, it always remains on the tube wall, becauseno vortex filaments can pass through it.

Another corollary of the circulation theorem is that the intensity of a vortex tuberemains constant as it is convected by the fluid. This can be proved by consideringthe circulation around the loop C′ pictured in Figure 4.10.

4.15 Irrotational FlowFlow is said to be irrotational when the vorticity ω has the magnitude zero every-where. It immediately follows, from Equation (4.77), that the circulation around anyarbitrary loop in an irrotational flow pattern is zero (provided that the loop can bespanned by a surface that lies entirely within the fluid). Hence, from Kelvin’s circu-lation theorem, if an inviscid fluid is initially irrotational then it remains irrotationalat all subsequent times. This can be seen more directly from the equation of motionof an inviscid incompressible fluid which, according to Equations (1.39) and (1.79),


takes the form∂v∂t+ (v · ∇) v = −∇

(pρ+ Ψ

), (4.82)

because ρ is a constant. However, from Equation (A.171),

(v · ∇) v = ∇(v 2/2) − v × ω. (4.83)

Thus, we obtain∂v∂t= −∇

(pρ+

12v 2 + Ψ

)+ v ×ω. (4.84)

Taking the curl of this equation, and making use of the vector identities ∇ × ∇φ ≡ 0[see Equation (A.176)], ∇ · ∇ ×A ≡ 0 [see Equation (A.173)], as well as the identity(A.179), and the fact that ∇ · v = 0 in an incompressible fluid, we obtain the vorticityevolution equation

DωDt= (ω · ∇) v. (4.85)

Thus, if ω = 0, initially, then Dω/Dt = 0, and, consequently,ω = 0 at all subsequenttimes.

Suppose that O is a fixed point, and P an arbitrary movable point, in an irrota-tional fluid. Let O and P be joined by two different paths, OAP and OBP (say). Itfollows that OAPBO is a closed curve. Because the circulation around such a curvein an irrotational fluid is zero, we can write∫

OAPv · dr +

∫PBO

v · dr = 0, (4.86)

which implies that ∫OAP

v · dr =∫

OBPv · dr = −φP (4.87)

(say). It is clear that φP is a scalar function whose value depends on the position of P(and the fixed point O), but not on the path taken between O and P. Thus, if O is theorigin of our coordinate system, and P an arbitrary point whose position vector is r,then we have effectively defined a scalar field φ(r) = −

∫ PO v · dr.

Consider a point Q that is sufficiently close to P that the velocity v is constantalong PQ. Let η be the position vector of Q relative to P. It then follows that (seeSection A.18)

−η · ∇φ = −φQ + φP =

∫ Q

Pv · dr v · η. (4.88)

The previous equation becomes exact in the limit that |η| → 0. Because Q is arbi-trary (provided that it is sufficiently close to P), the direction of the vector η is alsoarbitrary, which implies that

v = −∇φ. (4.89)

We, thus, conclude that if the motion of a fluid is irrotational then the associatedvelocity field can always be expressed as minus the gradient of a scalar function of


position, φ(r). This scalar function is called the velocity potential, and flow whichis derived from such a potential is known as potential flow. Note that the velocitypotential is undefined to an arbitrary additive constant.

We have demonstrated that a velocity potential necessarily exists in a fluid whosevelocity field is irrotational. Conversely, when a velocity potential exists the flow isnecessarily irrotational. This follows because [see Equation (A.176)]

ω = ∇ × v = −∇ × ∇φ = 0. (4.90)

Incidentally, the fluid velocity at any given point in an irrotational fluid is normal tothe constant-φ surface that passes through that point.

If a flow pattern is both irrotational and incompressible then we have

v = −∇φ (4.91)

and∇ · v = 0. (4.92)

These two expressions can be combined to give (see Section A.21)

∇ 2φ = 0. (4.93)

In other words, the velocity potential in an incompressible irrotational fluid satisfiesLaplace’s equation.

According to Equation (4.84), if the flow pattern in an incompressible inviscidfluid is also irrotational, so that ω = 0 and v = −∇φ, then we can write

∇(

pρ+

12v 2 + Ψ − ∂φ

∂t

)= 0, (4.94)

which implies thatpρ+

12v 2 + Ψ − ∂φ

∂t= C(t), (4.95)

where C(t) is uniform in space, but can vary in time. In fact, the time variation of C(t)can be eliminated by adding the appropriate function of time (but not of space) to thevelocity potential, φ. Note that such a procedure does not modify the instantaneousvelocity field v derived from φ. Thus, the previous equation can be rewritten

pρ+

12v 2 + Ψ − ∂φ

∂t= C, (4.96)

where C is constant in both space and time. Expression (4.96) is a generalizationof Bernoulli’s theorem (see Section 4.3) that takes non-steady flow into account.However, this generalization is only valid for irrotational flow. For the special caseof steady flow, we get

pρ+

12v 2 + Ψ = C, (4.97)

which demonstrates that for steady irrotational flow the constant in Bernoulli’s theo-rem is the same on all streamlines. (See Section 4.3.)


4.16 Exercises4.1 Liquid is led steadily through a pipeline that passes over a hill of height h

into the valley below, the speed at the crest being v. Show that, by properlyadjusting the ratio of the cross-sectional areas of the pipe at the crest andin the valley, the pressure may be equalized at these two places. (Milne-Thomson 1958.)

4.2 Water of mass density ρ and pressure p flows through a curved pipe of uni-form cross-sectional area S , whose radius of curvature is R, at the uniformspeed v. Demonstrate that there is a net force per unit length S (p−p0+ρ v

2)/Racting on the pipe, and that this force is everywhere directed away from thepipe’s local center of curvature. Here, p0 is atmospheric pressure.

4.3 Water is held in a right circular conical tank whose apex lies vertically belowthe center of its base. The water initially fills the tank to a height h above thevertex. Let a be the initial radius of the surface of the water inside the tank.A small hole of area S (that is much less than π a 2) is made at the bottom ofthe tank. Demonstrate that the time required to empty the tank is at least

52π a 2

S

(h

2 g

)1/2.

4.4 Water is held in a spherical tank of radius a, and initially fills the tank to aheight h < 2 a above its lowest point. A small hole of area S (that is muchless than π a 2) is made at the bottom of the tank. Demonstrate that the timerequired to empty the tank is at least

2πS (2 g)1/2

(23

a h 3/2 − 15

h 5/2).

4.5 Water is held in two contiguous tanks whose cross-sectional areas, A1 andA2, are independent of height. A small hole of area S (where S A1, A2)is made in the wall connecting the tanks. Assuming that the initial differencein water level between the two tanks is h, show that the time required for thewater levels to equilibrate is at least

2S

(A1 A2

A1 + A2

) (h

2 g

)1/2.

4.6 For a channel of width W, having a discharge rate Q, show that there is acritical depth hc, where

hc =

(Q 2

gW 2

)1/3,

which must be exceeded before a hydraulic jump is possible.


4.7 Show that for a stationary hydraulic jump in a rectangular channel, the up-stream Froude number Fr1, and the downstream Froude number Fr2, are re-lated by

Fr 22 =

8 Fr 21

[(1 + 8 Fr 21 )1/2 − 1] 3

.

4.8 Consider a simply-connected volume V whose boundary is the surface S .Suppose that V contains an incompressible fluid whose motion is irrotational.Let the velocity potential φ be constant over S . Prove that φ has the sameconstant value throughout V . [Hint: Consider the identity ∇ · (A∇A) ≡ ∇A ·∇A + A∇ 2A.]

4.9 In Exercise 4.8, suppose that, instead of φ taking a constant value on theboundary, the normal velocity is everywhere zero on the boundary. Showthat φ is constant throughout V .

4.10 An incompressible fluid flows in a simply-connected volume V bounded by asurface S . The normal flow at the boundary is prescribed. Show that the flowpattern with the lowest kinetic energy is irrotational. This result is knownas the Kelvin minimum energy theorem. [Hint: Try writing v = −∇φ + δv,where φ is the velocity potential of the irrotational flow pattern. Let ∇·δv = 0throughout V , and δv · dS = 0 on S . Show that the kinetic energy is lowestwhen δv = 0 throughout V .]

5Two-Dimensional Incompressible InviscidFlow

5.1 IntroductionThis chapter investigates two-dimensional, incompressible, inviscid flow. More in-formation on this subject can be found in Batchelor 2000, and Milne-Thompson2011.

5.2 Two-Dimensional FlowFluid motion is said to be two-dimensional when the velocity at every point is parallelto a fixed plane, and is the same everywhere on a given normal to that plane. Thus,in Cartesian coordinates, if the fixed plane is the x-y plane then we can express ageneral two-dimensional flow pattern in the form

v = vx(x, y, t) ex + vy(x, y, t) ey. (5.1)

Let A be a fixed point in the x-y plane, and let ABP and ACP be two curves,also in the x-y plane, that join A to an arbitrary point P. (See Figure 5.1.) Supposethat fluid is neither created nor destroyed in the region, R (say), bounded by thesecurves. Because the fluid is incompressible, which essentially means that its densityis uniform and constant, fluid continuity requires that the rate at which the fluid flowsinto the region R, from right to left (in Figure 5.1) across the curve ABP, is equal tothe rate at which it flows out the of the region, from right to left across the curveACP. The rate of fluid flow across a surface is generally termed the flux. Thus, theflux (per unit length parallel to the z-axis) from right to left across ABP is equal tothe flux from right to left across ACP. Because ACP is arbitrary, it follows that theflux from right to left across any curve joining points A and P is equal to the fluxfrom right to left across ABP. In fact, once the base point A has been chosen, thisflux only depends on the position of point P, and the time t. In other words, if wedenote the flux by ψ then it is solely a function of the location of P and the time.Thus, if point A lies at the origin, and point P has Cartesian coordinates (x, y), then

119


P

C

B

A

Figure 5.1Two-dimensional flow.

we can writeψ = ψ(x, y, t). (5.2)

The function ψ is known as the stream function. Moreover, the existence of a streamfunction is a direct consequence of the assumed incompressible nature of the flow.

Consider two points, P1 and P2, in addition to the fixed point A. (See Figure 5.2.)Let ψ1 and ψ2 be the fluxes from right to left across curves AP1 and AP2. Usingsimilar arguments to those employed previously, the flux across AP2 is equal to theflux across AP1 plus the flux across P1P2. Thus, the flux across P1P2, from right toleft, is ψ2−ψ1. If P1 and P2 both lie on the same streamline then the flux across P1P2

is zero, because the local fluid velocity is directed everywhere parallel to P1P2. Itfollows that ψ1 = ψ2. Hence, we conclude that the stream function is constant alonga streamline. The equation of a streamline is thus ψ = c, where c is an arbitraryconstant.

Let P1P2 = δs be an infinitesimal arc of a curve that is sufficiently short that itcan be regarded as a straight-line. The fluid velocity in the vicinity of this arc canbe resolved into components parallel and perpendicular to the arc. The componentparallel to δs contributes nothing to the flux across the arc from right to left. Thecomponent perpendicular to δs contributes v⊥ δs to the flux. However, the flux isequal to ψ2 − ψ1. Hence,

v⊥ =ψ2 − ψ1

δs. (5.3)

In the limit δs→ 0, the perpendicular velocity from right to left across ds becomes

v⊥ =dψds. (5.4)

Thus, in Cartesian coordinates, by considering infinitesimal arcs parallel to the x-

Two-Dimensional Incompressible Inviscid Flow 121

P2

A

P1

Figure 5.2Two-dimensional flow.

and y-axes, we deduce that

vx = −∂ψ

∂y, (5.5)

vy =∂ψ

∂x. (5.6)

These expressions can be combined to give

v = ez × ∇ψ = ∇z × ∇ψ. (5.7)

Note that when the fluid velocity is written in this form then it immediately becomesclear that the incompressibility constraint∇·v = 0 is automatically satisfied [because∇·(∇A×∇B) ≡ 0—see Equations (A.175) and (A.176)]. It is also clear that the streamfunction is undefined to an arbitrary additive constant.

The vorticity in two-dimensional flow takes the form

ω = ωz ez, (5.8)

where

ωz =∂vy

∂x− ∂vx

∂y. (5.9)

Thus, it follows from Equations (5.5) and (5.6) that

ωz =∂ 2ψ

∂x 2 +∂ 2ψ

∂y 2 = ∇2ψ. (5.10)


Hence, irrotational two-dimensional flow is characterized by

∇ 2ψ = 0. (5.11)

When expressed in terms of cylindrical coordinates (see Section C.3), Equa-tion (5.7) yields

v = vr(r, θ, t) er + vθ(r, θ, t) eθ, (5.12)

where

vr = −1r∂ψ

∂θ, (5.13)

vθ =∂ψ

∂r. (5.14)

Moreover, the vorticity is ω = ωz ez, where

ωz =1r∂

∂r

(r∂ψ

∂r

)+

1r 2

∂ 2ψ

∂θ 2 . (5.15)

5.3 Velocity Potentials and Stream FunctionsAs we have seen, a two-dimensional velocity field in which the flow is everywhereparallel to the x-y plane, and there is no variation along the z-direction, takes theform

v = vx(x, y, t) ex + vy(x, y, t) ey. (5.16)

Moreover, if the flow is irrotational then∇×v = 0 is automatically satisfied by writingv = −∇φ, where φ(x, y, t) is termed the velocity potential. (See Section 4.15.) Hence,

vx = −∂φ

∂x, (5.17)

vy = −∂φ

∂y. (5.18)

On the other hand, if the flow is incompressible then ∇ · v = 0 is automaticallysatisfied by writing v = ∇z×∇ψ, where ψ(x, y, t) is termed the stream function. (SeeSection 5.2.) Hence,

vx = −∂ψ

∂y, (5.19)

vy =∂ψ

∂x. (5.20)


Finally, if the flow is both irrotational and incompressible then Equations (5.17)–(5.18) and (5.19)–(5.20) hold simultaneously, which implies that

∂φ

∂x=∂ψ

∂y, (5.21)

∂ψ

∂x= −∂φ

∂y. (5.22)

It immediately follows, from the previous two expressions, that

∂ 2φ

∂x 2 =∂ 2ψ

∂x ∂y=∂ 2ψ

∂y ∂x= −∂

2φ

∂y 2 , (5.23)

or∂ 2φ

∂x 2 +∂ 2φ

∂y 2 = 0. (5.24)

Likewise, it can also be shown that

∂ 2ψ

∂x 2 +∂ 2ψ

∂y 2 = 0. (5.25)

We conclude that, for two-dimensional, irrotational, incompressible flow, the veloc-ity potential and the stream function both satisfy Laplace’s equation. Equations (5.21)and (5.22) also imply that

∇φ · ∇ψ = 0. (5.26)

In other words, the contours of the velocity potential and the stream function cross atright-angles.

5.4 Two-Dimensional Uniform FlowConsider a steady two-dimensional flow pattern that is uniform: in other words, apattern which is such that the fluid velocity is the same everywhere in the x-y plane.For instance, suppose that the common fluid velocity is

v = V0 cos θ0 ex + V0 sin θ0 ey, (5.27)

which corresponds to flow at the uniform speed V0 in a fixed direction that subtendsa (counter-clockwise) angle θ0 with the x-axis. It follows, from Equations (5.5) and(5.6), that the stream function for steady uniform flow takes the form

ψ(x, y) = V0 (sin θ0 x − cos θ0 y) . (5.28)

When written in terms of cylindrical coordinates, this becomes

ψ(r, θ) = −V0 r sin(θ − θ0). (5.29)


Note, from Equation (5.28), that ∂ 2ψ/∂x 2 = ∂ 2ψ/∂y 2 = 0. Thus, it follows fromEquation (5.10) that uniform flow is irrotational. Hence, according to Section 4.15,such flow can also be derived from a velocity potential. In fact, it is easily demon-strated that

φ(r, θ) = −V0 r cos(θ − θ0). (5.30)

5.5 Two-Dimensional Sources and SinksConsider a uniform line source, coincident with the z-axis, that emits fluid isotropi-cally at the steady rate of Q unit volumes per unit length per unit time. By symmetry,we expect the associated steady flow pattern to be isotropic, and everywhere directedradially away from the source. (See Figure 5.3.) In other words, we expect

v = vr(r) er, (5.31)

where r = (x 2 + y 2)1/2. Consider a cylindrical surface S of unit height (in the z-direction) and radius r that is co-axial with the source. In a steady state, the rate atwhich fluid crosses this surface must be equal to the rate at which the section of thesource enclosed by the surface emits fluid. Hence,∫

Sv · dS = 2π r vr(r) = Q, (5.32)

which implies that

vr(r) =Q

2π r. (5.33)

According to Equations (5.13) and (5.14), the stream function associated with aline source of strength Q that is coincident with the z-axis is

ψ(r, θ) = − Q2πθ. (5.34)

Note that the streamlines, ψ = c, are directed radially away from the z-axis, as illus-trated in Figure 5.3. Note, also, that the stream function associated with a line sourceis multivalued. However, this does not cause any particular difficulty, because thestream function is continuous, and its gradient is single valued.

It follows from Equation (5.34) that ∂ψ/∂r = ∂ 2ψ/∂θ 2 = 0. Hence, according toEquation (5.15), ωz = −∇ 2ψ = 0. In other words, the steady flow pattern associatedwith a uniform line source is irrotational, and can, thus, be derived from a velocitypotential. In fact, it is easily demonstrated that this potential takes the form

φ(r, θ) = − Q2π

ln r. (5.35)

A uniform line sink, coincident with the z-axis, which absorbs fluid isotropically


y

S

x

Figure 5.3Streamlines of the flow generated by a line source coincident with the z-axis.

at the steady rate of Q unit volumes per unit length per unit time has an associatedsteady flow pattern

v = − Q2π r

er, (5.36)

whose stream function isψ(r, θ) =

Q2πθ. (5.37)

This flow pattern is also irrotational, and can be derived from the velocity potential

φ(r, θ) =Q2π

ln r. (5.38)

Consider a line source and a line sink of equal strength, which both run parallelto the z-axis, and are located a small distance apart in the x-y plane. Such an arrange-ment is known as a doublet or dipole line source. Suppose that the line source, whichis of strength Q, is located at r = d/2 (where r is a position vector in the x-y plane),and that the line sink, which is also of strength Q, is located at r = −d/2. Let thefunction

ψQ(r) = −Q2πθ = −

Q2π

tan−1(y/x) (5.39)

be the stream function associated with a line source of strength Q located at r = 0.Thus, ψQ(r − r0) is the stream function associated with a line source of strength Q


located at r = r0. Furthermore, the stream function associated with a line sink ofstrength Q located at r = r0 is −ψQ(r − r0). We expect the flow pattern associatedwith the combination of a source and a sink to be the vector sum of the flow patternsgenerated by the source and sink taken in isolation. It follows that the overall streamfunction is the sum of the stream functions generated by the source and the sink takenin isolation. In other words,

ψ(r) = ψQ(r − d/2) − ψQ(r + d/2) −d · ∇ψQ(r), (5.40)

to first order in d/r. Hence, if d = d (cos θ0 ex + sin θ0 ey) = d [cos(θ− θ0) er − sin(θ−θ0) eθ], so that the line joining the sink to the source subtends a (counter-clockwise)angle θ0 with the x-axis, then

ψ(r, θ) = − D2π

sin(θ − θ0)r

, (5.41)

where D = Q d is termed the strength of the dipole source. The previous streamfunction is antisymmetric across the line θ = θ0 joining the source to the sink. Itfollows that the associated dipole flow pattern,

vr(r, θ) =D2π

cos(θ − θ0)r 2 , (5.42)

vθ(r, θ) =D2π

sin(θ − θ0)r 2 , (5.43)

is symmetric across this line. Figure 5.4 shows the streamlines associated with adipole flow pattern characterized by D > 0 and θ0 = 0. Note that the flow speed in adipole pattern falls off like 1/r 2.

A dipole flow pattern is necessarily irrotational because it is a linear superpositionof two irrotational flow patterns. The associated velocity potential is

φ(r, θ) =D2π

cos(θ − θ0)r

. (5.44)

5.6 Two-Dimensional Vortex FilamentsConsider a vortex filament of intensity Γ that is coincident with the z-axis. By sym-metry, we expect the associated flow pattern to circulate isotropically around thefilament. (See Figure 5.5.) In other words, we expect

v = vθ(r) eθ. (5.45)

According to Section 4.13, the circulation,∮

v·dr, around any closed curve in the x-yplane is equal to the net intensity of the vortex filaments that pass through the curve.Consider a circular curve of radius r that is concentric with the origin. It follows that

Γr =

∮v · dr = 2π r vθ(r) = Γ, (5.46)


−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

y

−1 −0.5 0 0.5 1x

Figure 5.4Streamlines of the flow generated by a dipole line source (with D > 0) coincidentwith the z-axis, and aligned along the x-axis. The flow is outward along the posi-tive x-axis and inward along the negative x-axis. Positive and negative contours areshown as solid and dashed lines, respectively.

orvθ(r) =

Γ

2π r. (5.47)

According to Equations (5.13) and (5.14), the stream function associated with avortex filament of intensity Γ that is coincident with the z-axis is

ψ(r, θ) =Γ

2πln r. (5.48)

Note that the streamlines, ψ = c, circulate around the z-axis, as illustrated in Fig-ure 5.5.

It can be seen, from Equation (5.48), that (∂/∂r)(r ∂ψ/∂r) = ∂ψ/∂θ = 0. Hence,it follows from Equation (5.15) that ωz = −∇ 2ψ = 0. In other words, the flowpattern associated with a straight vortex filament is irrotational. This is a somewhatsurprising result, because there is a net circulation of the flow around the filament,and, according to Section 4.13, non-zero circulation implies non-zero vorticity. Theparadox can be resolved by supposing that the filament has a small, but finite, radius.In fact, let the filament have the finite radius a, and be such that the vorticity isuniform inside this radius, and zero outside: that is,

ωz =

Γ/(π a 2) r ≤ a0 r > a

. (5.49)


x

y

Figure 5.5Streamlines of the flow generated by a line vortex coincident with the z-axis.

Note that the intensity of the filament (i.e., the product of its vorticity and cross-sectional area) is still Γ. According to Equation (5.15), and assuming that ψ = ψ(r),

1r

d∂r

(r

dψdr

)=

Γ/(π a 2) r ≤ a0 r > a

. (5.50)

The solution that is well behaved at r = 0, and continuous (up to its first derivative)at r = a, is

ψ(r, θ) =

(Γ/4π) (r 2/a 2 − 1) r ≤ a(Γ/2π) ln(r/a) r > a

. (5.51)

This expression is equivalent to Equation (5.48) (apart from an unimportant additiveconstant) outside the filament, but differs inside. The associated circulation velocity,vθ(r) = ∂ψ/∂r, is

vθ(r) =

(Γ/2π) (r/a 2) r ≤ a(Γ/2π) (1/r) r > a

, (5.52)

whereas the circulation, Γr(r) = 2π r vθ(r), is written

Γr(r) =Γ (r/a)2 r ≤ aΓ r > a

. (5.53)

Thus, we conclude that the flow pattern associated with a straight vortex filament is


irrotational outside the filament, but has finite vorticity inside the filament. Moreover,the non-zero internal vorticity generates a constant net circulation of the flow outsidethe filament. In the limit in which the radius of the filament tends to zero, the vorticitywithin the filament tends to infinity (in such a way that the product of the vorticityand the cross-sectional area of the filament remains constant), and the region of thefluid in which the vorticity is non-zero becomes infinitesimal in extent.

Let us determined the pressure profile in the vicinity of a vortex filament of finiteradius. Assuming, from symmetry, that p = p(r), Equation (1.149), yields

dpdr= ρ

v 2θ

r, (5.54)


p = p∞ − ρ∫ ∞

r

v 2θ

rdr, (5.55)

where p∞ is the pressure at infinity. Making use of expression (5.52), we obtain

p(r) =

p∞ − (ρ/2) (Γ/2π a)2 (2 − r 2/a 2) r ≤ ap∞ − (ρ/2) (Γ/2π a)2 (a/r)2 r > a

. (5.56)

It follows that the minimum pressure occurs at the center of the vortex (r = 0), andtakes the value

p0 = p∞ − ρ(Γ

2π a

)2. (5.57)

Under normal circumstances, the pressure in a fluid must remain positive, whichimplies that a vortex filament of intensity Γ, embedded in a fluid of density ρ andbackground pressure p∞, has a minimum radius of order

amin (ρ

p∞

)1/2 (Γ

2π

). (5.58)

Finally, because the flow pattern outside a straight vortex filament is irrotational,it can be derived from a velocity potential. It is easily demonstrated that the appro-priate potential takes the form

φ(r, θ) = − Γ2πθ. (5.59)

Note that the previous potential is multivalued. However, this does not cause anyparticular difficulty, because the potential is continuous, and its gradient is singlevalued.


5.7 Two-Dimensional Irrotational Flow in Cylindrical Coordi-nates

In a two-dimensional flow pattern, we can automatically satisfy the incompressibil-ity constraint, ∇ · v = 0, by expressing the pattern in terms of a stream function.Suppose, however, that, in addition to being incompressible, the flow pattern is alsoirrotational. In this case, Equation (5.10) yields

∇ 2ψ = 0. (5.60)

In cylindrical coordinates, because ψ = ψ(r, θ, t), this expression implies that (seeSection C.3)

1r∂

∂r

(r∂ψ

∂r

)+

1r 2

∂ 2ψ

∂θ 2 = 0. (5.61)

Let us search for a separable solution of Equation (5.61) of the form

ψ(r, θ) = R(r)Θ(θ). (5.62)

It is easily seen thatrR

ddr

(r

dRdr

)= − 1

Θ

d 2Θ

dθ 2 , (5.63)

which can only be satisfied if

rddr

(r

dRdr

)= m 2 R, (5.64)

d 2Θ

dθ 2 = −m 2Θ, (5.65)

where m 2 is an arbitrary (positive) constant. The general solution of Equation (5.65)is a linear combination of exp( i m θ) and exp(−i m θ) factors. However, assumingthat the flow extends over all θ values, the function Θ(θ) must be single-valued inθ, otherwise ∇ψ—and, hence, v—would not be be single-valued (which is unphysi-cal). It follows that m can only take integer values (and that m 2 must be a positive,rather than a negative, constant). The general solution of Equation (5.64) is a linearcombination of r m and r−m factors, except for the special case m = 0, when it is a lin-ear combination of r 0 and ln r factors. Thus, the general stream function for steadytwo-dimensional irrotational flow (that extends over all values of θ) takes the form

ψ(r, θ) = α0 + β0 ln r +∑m>0

(αm r m + βm r−m) sin[m (θ − θm)], (5.66)

where αm, βm, and θm are arbitrary constants. We can recognize the first few termson the right-hand side of the previous expression. The constant term α0 has zerogradient, and, therefore, does not give rise to any flow. The term β0 ln r is the flow


−5

−4

−3

−2

−1

0

1

2

3

4

5

y/a

−5 −4 −3 −2 −1 0 1 2 3 4 5x/a

Figure 5.6Streamlines of the flow generated by a cylindrical obstacle of radius a, whose axisruns along the z-axis, placed in the uniform flow field v = V0 ex. The normalizedcirculation is γ = 0.

pattern generated by a vortex filament of intensity 2π β0, coincident with the z-axis.(See Section 5.6.) The term α1 r sin(θ− θ1) corresponds to uniform flow of speed α1

whose direction subtends a (counter-clockwise) angle θ1 with the minus x-axis. (SeeSection 5.4.) Finally, the term β1 sin(θ − θ1)/r corresponds to a dipole flow pattern.(See Section 5.5.)

The velocity potential associated with the irrotational stream function (5.66) sat-isfies [see Equations (4.89) and (5.7)]

∂φ

∂r=

1r∂ψ

∂θ, (5.67)

1r∂φ

∂θ= −∂ψ

∂r. (5.68)

It follows that

φ(r, θ) = α0 − β0 θ +∑m>0

(αm r m − βm r−m) cos[m (θ − θ0)]. (5.69)


5.8 Flow Past a Cylindrical ObstacleConsider the steady flow pattern produced when an impenetrable rigid cylindricalobstacle is placed in a uniformly flowing, incompressible, inviscid fluid, with thecylinder orientated such that its axis is normal to the flow. For instance, suppose thatthe radius of the cylinder is a, and that its axis corresponds to the line x = y = 0.Furthermore, let the unperturbed fluid velocity be of magnitude V0, and be directedparallel to the x-axis. We expect the flow pattern to remain unperturbed very faraway from the cylinder. In other words, we expect v(r, θ) → V0 ex as r/a → ∞,which corresponds to a boundary condition on the stream function of the form (seeSection 5.4)

ψ(r, θ)→ −V0 r sin θ as r/a→ ∞. (5.70)

Given that the fluid velocity field a large distance upstream of the cylinder is irrota-tional (because we have already seen that the flow pattern associated with uniformflow is irrotational—see Section 5.4), it follows from the Kelvin circulation theorem(see Section 4.14) that the velocity field remains irrotational as it is convected pastthe cylinder. Hence, according to Section 5.2, the stream function of the flow satisfiesLaplace’s equation,

∇ 2ψ = 0. (5.71)

The appropriate boundary condition at the surface of the cylinder is simply that thenormal fluid velocity there be zero, because the fluid must stay in contact with thecylinder, but cannot penetrate its surface. Hence, vr(a, θ) ≡ −(1/a) ∂ψ/∂θ|r=a = 0,which implies that

ψ(a, θ) = 0, (5.72)

because ψ is undetermined to an arbitrary additive constant. It follows that we aresearching for the most general solution of Equation (5.71) that satisfies the bound-ary conditions (5.70) and (5.72). Comparison with Equation (5.66) reveals that thissolution takes the form

ψ(r, θ) = V0 a[−γ ln

( ra

)−( ra− a

r

)sin θ

], (5.73)

whereγ = − Γ

2π a V0, (5.74)

and Γ is the circulation of the flow around the cylinder. (Note that the velocity fieldcan be irrotational, but still possess nonzero circulation around the cylinder, becausea loop that encloses the cylinder cannot be spanned by a surface lying entirely withinthe fluid. Thus, zero fluid vorticity does not necessarily imply zero circulation aroundsuch a loop from the curl theorem.) Let us assume that γ ≥ 0, for the sake ofdefiniteness.

Figure 5.6–5.8 show streamlines of the flow calculated for various different val-ues of the normalized circulation, γ. For γ < 2, there exist a pair of points on the


−5

−4

−3

−2

−1

0

1

2

3

4

5

y/a

−5 −4 −3 −2 −1 0 1 2 3 4 5x/a

Figure 5.7Streamlines of the flow generated by a cylindrical obstacle of radius a, whose axisruns along the z-axis, placed in the uniform flow field v = V0 ex. The normalizedcirculation is γ = 1.

surface of the cylinder at which the flow speed is zero. These are known as stagnationpoints, and can be located in Figures 5.6 and 5.7 as the points at which streamlinesintersect the surface of the cylinder at right-angles. The tangential fluid velocity atthe surface of the cylinder is

vt(θ) = vθ(a, θ) =∂ψ

∂r

∣∣∣∣∣r=a= −V0 (γ + 2 sin θ). (5.75)

The stagnation points correspond to the points at which vt = 0 (because the normalvelocity is automatically zero at the surface of the cylinder). Thus, the stagnationpoints lie at θ = sin−1(−γ/2). When γ > 2, the stagnation points coalesce andmove off the surface of the cylinder, as illustrated in Figure 5.8 (the stagnation pointcorresponds to the point at which two streamlines cross at right-angles).

The irrotational form of Bernoulli’s theorem, (4.97), can be combined with theboundary condition v → V0 as r/a → ∞, as well as the fact that Ψ is constant in thepresent case, to give

p = p0 +12ρ(V 2

0 − v2), (5.76)

where p0 is the constant static fluid pressure a large distance from the cylinder. Inparticular, the fluid pressure on the surface of the cylinder is

P(θ) = p(a, θ) = p0 +12ρ(V 2

0 − v2t

)= p1 + ρV 2

0 (cos 2θ − 2 γ sin θ) , (5.77)


where p1 = p0 − (1/2) ρV 20 (1 + γ 2). The net force per unit length exerted on the

cylinder by the fluid has the Cartesian coordinates

Fx = −∮

P cos θ a dθ, (5.78)

Fy = −∮

P sin θ a dθ. (5.79)

Thus, it follows from Equation (5.77) that

Fx = 0, (5.80)

Fy = 2π γ ρV 20 a = ρV0 (−Γ). (5.81)

The component of the force that a moving fluid exerts on an obstacle, placed in itspath, in a direction parallel to that of the unperturbed flow is usually called drag. Onthe other hand, the component of the force that the fluid exerts in a direction per-pendicular to that of the unperturbed flow is usually called lift. Hence, the previousequations imply that if a cylindrical obstacle is placed in a uniformly flowing inviscidfluid then there is zero drag. On the other hand, as long as there is net circulation ofthe flow around the cylinder, the lift is non-zero. Lift is generated because (negative)circulation tends to increase the fluid speed directly above, and to decrease it directlybelow, the cylinder. Thus, from Bernoulli’s theorem, the fluid pressure is decreasedabove, and increased below, the cylinder, giving rise to a net upward force (i.e., aforce in the +y-direction).

Suppose that the cylinder is placed in a fluid which is initially at rest, and that thefluid’s uniform flow velocity, V0, is then very slowly ramped up (in such a mannerthat no vorticity is induced in the upstream flow at infinity). Because the flow patternis initially irrotational, and because the flow pattern well upstream of the cylinder isassumed to remain irrotational, the Kelvin circulation theorem indicates that the flowpattern around the cylinder also remains irrotational. Consider the time evolution ofthe circulation, Γ =

∮C v · dr, around some fixed curve C that lies entirely within the

fluid, and encloses the cylinder. We have

dΓC

dt=

∮C

∂v∂t· dr =

∮C

[−∇

(pρ+

12v 2)+ v ×ω

]· dr =

∮C

v × ω · dr, (5.82)

where use has been made of Equation (4.84) (with Ψ assumed constant). However,ω = ωz ez in two-dimensional flow, and dr× ez = dS, where dS is an outward surfaceelement of a unit depth (in the z-direction) surface whose normal lies in the x-y plane,and that cuts the x-y plane at C. In other words,

dΓC

dt= −

∮Sωz v · dS. (5.83)

We, thus, conclude that the rate of change of the circulation around C is equal tominus the flux of the vorticity across S [assuming that vorticity is convected by theflow, which follows from Equation (4.85), the fact that ω = ωz ez, and the fact that


−5

−4

−3

−2

−1

0

1

2

3

4

5

y/a

−5 −4 −3 −2 −1 0 1 2 3 4 5x/a

Figure 5.8Streamlines of the flow generated by a cylindrical obstacle of radius a, whose axisruns along the z-axis, placed in the uniform flow field v = V0 ex. The normalizedcirculation is γ = 2.5.

∂/∂z = 0 in two-dimensional flow]. However, we have already seen that the flowfield surrounding the cylinder is irrotational (i.e., such that ωz = 0). It follows thatΓC is constant in time. Moreover, because ΓC = 0 originally, as the fluid surroundingthe cylinder was initially at rest, we deduce that ΓC = 0 at all subsequent times.Hence, we conclude that, in an inviscid fluid, if the circulation of the flow around thecylinder is initially zero then it remains zero. It follows, from the previous analysis,that, in such a fluid, zero drag force and zero lift force are exerted on the cylinderas a consequence of the fluid flow. This result is a manifestation of d’Alembert’sparadox, which was introduced in Section 4.5. d’Alembert’s result is paradoxicalbecause it would seem, at first sight, to be a reasonable approximation to neglectviscosity altogether in high Reynolds number flow. However, if we do this then weend up with the nonsensical prediction that a high Reynolds number fluid is incapableof exerting any force on an obstacle placed in its path.


x

y

Vθ

r

a

Figure 5.9Cylinder moving through an inviscid fluid

5.9 Motion of a Submerged CylinderConsider the situation, illustrated in Figure 5.9, in which an impenetrable rigid cylin-der of radius a and infinite length, whose symmetry axis runs parallel to the z-direction, is moving through an incompressible, inviscid fluid at the time-dependentvelocity V = Vx(t) ex. Assuming that the fluid and cylinder were both initially sta-tionary, it follows that the fluid velocity field was initially irrotational. Thus, ac-cording to the Kelvin circulation theorem, the fluid velocity field remains irrotationalwhen the cylinder starts to move. Thus, we can write

v = −∇φ, (5.84)

where v is the fluid velocity. Moreover, because the fluid is incompressible, we have

∇ · v = ∇ 2φ = 0. (5.85)

Let x, y, z be Cartesian coordinates in the initial rest frame of the fluid, and let r,θ be cylindrical coordinates in a frame of reference that co-moves with the cylinder,as shown in Figure 5.9. In the following, all calculations are performed in the restframe. We expect the fluid a long way from the cylinder to remain stationary. Inother words,

φ(r, θ, t)→ 0 as r → ∞. (5.86)

Moreover, because the cylinder is impenetrable, we require that

V · er = v · er |r=a , (5.87)


or∂φ

∂r

∣∣∣∣∣r=a= −Vx cos θ. (5.88)

It is easily demonstrated that the solution to Equation (5.85), subject to the boundaryconditions (5.86) and (5.88), is

φ(r, θ, t) = Vx(t)a 2

rcos θ. (5.89)

Hence,

vr(r, θ, t) = Vx(t)a 2

r 2 cos θ, (5.90)

vθ(r, θ, t) = Vx(t)a 2

r 2 sin θ. (5.91)

The general form of Bernoulli’s theorem, (4.96), which applies to an irrotationalflow field, yields

pρ+

12v 2 − ∂φ

∂t=

p0

ρ, (5.92)

where ρ is the uniform fluid mass density, and p0 the fluid pressure at infinity. Thus,the pressure distribution at the surface of the cylinder can be written

p(a, θ, t) = p0 −12ρ v 2

∣∣∣∣∣r=a+ ρ

∂φ

∂t

∣∣∣∣∣r=a− ρ V · ∇φ|r=a . (5.93)

The final term on the right-hand side of the previous equation arises because

∂

∂t

∣∣∣∣∣x,y=∂

∂t

∣∣∣∣∣r,θ− V · ∇. (5.94)

Hence, we obtain

p(a, θ, t) = p0 −12ρV 2

x + ρV 2x cos(2 θ) + ρ a

dVx

dtcos θ. (5.95)

The net force per unit length exerted on the cylinder by the fluid has the Cartesiancomponents

Fx(t) = −∮

p(a, θ, t) cos θ a dθ, (5.96)

Fy(t) = −∮

p(a, θ, t) sin θ a dθ. (5.97)

It follows that

Fx(t) = −π a 2 ρdVx

dt, (5.98)

Fy(t) = 0, (5.99)


orF = −m′

dVdt, (5.100)

where m′ = π a 2 ρ is the mass per unit length of the fluid displaced by the cylinder.Suppose that the cylinder is subject to an external (i.e., not due to the fluid) force

per unit length Fext. The equation of motion of the cylinder is thus

mdVdt= F + Fext, (5.101)

where m is the cylinder’s mass per unit length. The previous two equations can becombined to give

(m + m′)dVdt= Fext. (5.102)

In other words, the cylinder moves under the action of the external force, Fext, asif its mass per unit length were m + m′, rather than m. Here, m + m′ is commonlyreferred to as the cylinder’s virtual mass (per unit length), whereas m′ is termed theadded mass (per unit length).

The origin of added mass is easily explained. According to Equations (5.90) and(5.91), the total kinetic energy per unit length of the fluid surrounding the cylinder is

Kfluid =

∫fluid

12ρ v 2 dV = 2π

∫ ∞

a

12ρV 2

x

(ar

)4r dr =

12

m′ V 2x . (5.103)

However, the kinetic energy per unit length of the cylinder is

Kcylinder =12

m V 2x . (5.104)

Thus, the total kinetic energy per unit length is

K = Kfluid + Kcylinder =12

(m + m′) V 2x . (5.105)

In other words, the kinetic energy of the fluid surrounding the cylinder can be ac-counted for by supposing that a mass (per unit length) m′ of the fluid co-moves withthe cylinder, and that the remainder of the fluid remains stationary. This entrainedfluid mass accounts for the added mass of the cylinder. Note that the added massis independent of the speed of the cylinder (i.e., it is the same whether the cylindermoves slowly or rapidly.) In the present case, the added mass is equal to the mass ofthe displaced fluid. However, this is not a general rule. (In general, the added massof a object moving through an inviscid fluid is proportional to the displaced mass,but the constant of proportionality is not necessarily unity, and depends on the shapeof the object.)

Let us generalize the previous calculation to allow the cylinder to move in anydirection in the x-y plane: that is,

V(t) = Vx(t) ex + Vy(t) ey. (5.106)


Furthermore, let the fluid possess the initial circulation Γ in the x-y plane. Accordingto the Kelvin circulation theorem, this circulation remains constant in time. Thus, wemust now solve Equation (5.85) subject to the boundary conditions

φ(r, θ, t)→ − Γ

2π ras r → ∞. (5.107)

and∂φ

∂r

∣∣∣∣∣r=a= −Vx cos θ − Vy sin θ. (5.108)

It is easily demonstrated that the appropriate solution is

φ(r, θ, t) = − Γ

2π r+ Vx(t)

a 2

rcos θ + Vy(t)

a 2

rsin θ. (5.109)

Hence,

vr(r, θ, t) = Vx(t)a 2

r 2 cos θ + Vy(t)a .2

r 2 sin θ, (5.110)

vθ(r, θ, t) =Γ

2π r+ Vx(t)

a 2

r 2 sin θ − Vy(t)a 2

r 2 cos θ. (5.111)

Bernoulli’s theorem yields

pρ+

12v 2 + g r sin θ − ∂φ

∂t=

p0

ρ, (5.112)

where we have assumed that the fluid and cylinder are both situated in a gravitationalfield of uniform acceleration g = −g ey. Thus, the pressure distribution at the surfaceof the cylinder can be written

p(a, θ, t) = p0 −12ρ v 2

∣∣∣∣∣r=a− ρ g a sin θ − ρ ∂φ

∂t

∣∣∣∣∣r=a− ρ V · ∇φ|r=a , (5.113)

which yields

p(a, θ, t) = p0 −12ρ

[V 2 +

(Γ

2π a

)2]+ ρ (V 2

x − V 2y ) cos(2 θ) + 2 ρVx Vy sin(2 θ)

+ ρ adVx

drcos θ + ρ a

dVydr

sin θ − ρ g a sin θ

− ρΓπ a

(Vx sin θ − Vy cos θ

). (5.114)

It follows from Equations (5.96) and (5.97) that the force per unit length exerted onthe cylinder by the fluid has the Cartesian components

Fx(t) = −ρΓ Vy − m′dVx

dt, (5.115)

Fy(t) = ρΓ Vx − m′dVydt+ m′ g. (5.116)


Here, the first terms on the right-hand sides of the previous two equations are thecomponents of the lift (per unit length) acting on the cylinder, due to the fluid cir-culation, whereas the final term on the right-hand side of the second equation is thebuoyancy force (per unit length) acting on the cylinder. The cylinder’s equation ofmotion,

mdVx

dt= Fx, (5.117)

mdVydt= Fy − m g, (5.118)

leads to

d 2Xdt 2 = −ω

2 Y, (5.119)

d 2Ydt 2 = −ω

2 X +(m′ − mm′ + m

)g, (5.120)

where ω = ρΓ/(m + m′), and (X, Y) are the Cartesian components of the cylinder’saxis. Let us assume that X = Y = dX/dt = dY/dt = 0 at t = 0. It follows that

X(t) =(

m − m′

m + m′

)g

ω 2 [sin(ω t) − ω t] , (5.121)

Y(t) = −(

m − m′

m + m′

)g

ω 2 [1 − cos(ω t)] . (5.122)

Consider, first, the case in which there is no circulation of the flow: that is, Γ = 0.In this case, the previous two equations reduce to

X(t) = 0, (5.123)

Y(t) = −(m − m′

m + m′

)g t 2. (5.124)

In other words, the cylinder moves vertically (i.e., in the y-direction) with the con-stant acceleration

ay = −(

s − 1s + 1

)g, (5.125)

where s = m/m′ is the cylinder’s specific gravity. It follows that if the cylinder ismuch denser than the fluid (i.e., s 1) then it accelerates downward at the acceler-ation due to gravity: that is, ay = −g. However, if the cylinder is much less densethan the fluid (i.e., 0 < s 1) then it accelerates upward at the acceleration due togravity: that is ay = +g. Note that, in the latter case, the upward acceleration is lim-ited by the cylinder’s added mass (i.e., in the absence of added mass, the accelerationwould be infinite.)

In the general case, in which the fluid circulation is non-zero, the trajectory ofthe cylinder is a cycloid. In particular, assuming that m > m′, the lift acting on


y

xα π

Figure 5.10Inviscid flow past a wedge.

the cylinder prevents it from falling through the fluid a distance greater than 2 [(m −m′)/(m + m′)] (g/ω 2). Once the cylinder has fallen through this distance, it startsto rise again, until it attains its original height, and the motion then repeats itselfad infinitum. Moreover, the cylinder simultaneously moves horizontally (i.e. in thex-direction) at the mean velocity −[(m − m′)/(m + m′)] (g/ω).

5.10 Inviscid Flow Past a Semi-Infinite WedgeConsider the situation, illustrated in Figure 5.10, in which incompressible irrotationalflow is incident on a impenetrable rigid wedge whose apex subtends an angle απ. Letthe cross-section of the wedge in the x-y plane be both z-independent and symmetricabout the x-axis. Furthermore, let the apex of the wedge lie at x = y = 0. Finally, letthe upstream flow a large distance from the wedge be parallel to the x-axis.

Because the flow is two-dimensional, incompressible, and irrotational, it can berepresented in terms of a stream function that satisfies Laplace’s equation. More-over, in cylindrical coordinates, this equation takes the form (5.61). The boundaryconditions on the stream function are

ψ(r, α π/2) = ψ(r, 2π − απ/2) = ψ(r, π) = 0. (5.126)

The first two boundary conditions ensure that the normal velocity at the surface ofthe wedge is zero. The third boundary condition follows from the observation that,


−3

−2

−1

0

1

2

3

y

−3 −2 −1 0 1 2 3x

Figure 5.11Streamlines of inviscid incompressible irrotational flow past a 90 wedge.

by symmetry, the streamline that meets the apex of the wedge splits in two, and thenflows along its top and bottom boundaries, combined with the well-known result thatψ is constant on a streamline. It is easily demonstrated that

ψ(r, θ) =A

1 + mr 1+m sin [(1 + m) (π − θ)] (5.127)

is a solution of Equation (5.61). Moreover, this solution satisfies the boundary con-ditions provided (1 + m) (1 − α/2) = 1, or

m =α

2 − α. (5.128)

Because, as is well known, the solutions to Laplace’s equation (for problems withwell-posed boundary conditions) are unique (Riley 1974), we can be sure that Equa-tion (5.127) is the correct solution to the problem under investigation. According tothis solution, the tangential velocity on the surface of the wedge is given by

vt(r) = A r m, (5.129)

where m ≥ 0. Note that the tangential velocity is zero at the apex of the wedge.Because the normal velocity is also zero at this point, we conclude that the apex isa stagnation point of the flow. Figure 5.11 shows the streamlines of the flow for thecase α = 1/2.


x

y

α π

α′ π

Figure 5.12Inviscid flow over a wedge.

5.11 Inviscid Flow Over a Semi-Infinite WedgeConsider the situation illustrated in Figure 5.12 in which an incompressible irrota-tional fluid flows over an impenetrable rigid wedge whose apex subtends an angleαπ. Let the cross-section of the wedge in the x-y plane be both z-independent andsymmetric about the y-axis. Furthermore, let the apex of the wedge lie at x = y = 0.Finally, let the upstream flow a large distance from the wedge be parallel to the x-axis.

Because the flow is two-dimensional, incompressible, and irrotational, it canbe represented in terms of a stream function that satisfies Laplace’s equation. Theboundary conditions on the stream function are

ψ (r, [3 − α] π/2) = ψ (r,−[1 − α] π/2) = 0. (5.130)

These boundary conditions ensure that the normal velocity at the surface of thewedge is zero. It is easily demonstrated that

ψ(r, θ) = − A1 − m

r 1−m cos [(1 − m) (θ − π/2)] (5.131)

is a solution of Laplace’s equation, (5.61). Moreover, this solution satisfies theboundary conditions provided that (1 − m) (1 − α/2) = 1/2, or

m =α′

1 + α′, (5.132)


−3

−2

−1

0

1

2

3

y

−3 −2 −1 0 1 2 3x

Figure 5.13Streamlines of inviscid incompressible irrotational flow over a 90 wedge.

where α′ = 1 − α. Because the solutions to Laplace’s equation are unique, wecan again be sure that Equation (5.131) is the correct solution to the problem underinvestigation. According to this solution, the tangential velocity on the surface of thewedge is given by

vt(r) = A r−m, (5.133)

where m ≥ 0. Note that the tangential velocity, and hence the flow speed, is infiniteat the apex of the wedge. However, this singularity in the flow can be eliminated byslightly rounding the apex. Figure 5.13 shows the streamlines of the flow for the caseα = 1/2.

5.12 Two-Dimensional JetsConsider a two-dimensional jet of width h and uniform speed v that impinges on arigid plate, at an angle α to the normal to the plate, and divides into two streams,running along the plate, whose widths are ultimately uniform, and equal to h1 andh2. (See Figure 5.14.) We expect the pressure well upstream of the plate to be atmo-spheric all the way across the jet, because the jet is not yet decelerating. Likewise,we expect the pressure well downstream of the strike point to be atmospheric allthe way across the two streams that run along the plate, because the streams are no


α

h2

h1

h

v

v

v

Figure 5.14A two-dimensional jet impinging on a rigid plate.

longer accelerating (and, hence, their widths are constant). It immediately followsfrom Bernoulli’s theorem that the asymptotic flow speed in the two streams is v, asindicated in Figure 5.14.

Fluid continuity demands that h v = h1 v + h2 v, or

h = h1 + h2. (5.134)

Moreover, because the plate can only exert a normal force on the fluid, it followsthat the parallel (to the plate) momentum flux of the incident jet must equal the netparallel momentum flux of the two streams. In other words,

ρ h v 2 sinα = ρ h1 v2 − ρ h2 v

2, (5.135)

orh sinα = h1 − h2. (5.136)

Equations (5.134) and (5.136) can be combined to give

h1 =h2

(1 + sinα), (5.137)

h2 =h2

(1 − sinα). (5.138)

Let us apply the Euler momentum theorem (see Section 4.4) to the flux tubeshown in Figure 5.14, treating the plate as one of the walls of the tube. By symmetry,


h1 ρ v 2

α

h1/2

h2/2

OF

C

h2 ρ v 2

h ρ v 2

Figure 5.15Determination of the center of pressure.

the net force (per unit width perpendicular to the page) exerted across the walls andends of the tube is the resultant of the three forces shown in Figure 5.15. It is easilydemonstrated that this net force, F, is normal to the plate, as shown in the figure, andof magnitude

F = h ρ v 2 cosα. (5.139)

Obviously, the force in question corresponds to the force exerted by the plate on thefluid. An equal and opposite force is exerted by the fluid on the plate. (Note that thereis no net pressure force on the flux tube because it is entirely surrounded by fluid atthe same pressure: i.e., atmospheric pressure.) The line of action of the resultantforce passes through the so-called center of pressure, C. This is displaced slightlyfrom the point, O, where the middle of the jet strikes the plate. (See Figure 5.15.)Let d be the distance OC. We can calculate d by demanding that the net moment ofthe three forces shown in Figure 5.15 about point C be zero. In other words,

h ρ v 2 cosα d − h1 ρ v2 h1

2+ h2 ρ v

2 h2

2= 0, (5.140)

which yields

d =h2

tanα. (5.141)


5.13 Exercises5.1 For the case of the two-dimensional motion of an incompressible fluid, deter-

mine the condition that the velocity components

vx = a x + b y,

vy = c x + d y

satisfy the equation of continuity. Show that the magnitude of the vorticity isc − b.

5.2 For the case of the two-dimensional motion of an incompressible fluid, showthat

vx = 2 c x y,

vy = c (a 2 + x 2 − y 2)

are the velocity components of a possible flow pattern. Determine the streamfunction and sketch the streamlines. Prove that the motion is irrotational, andfind the velocity potential.

5.3 A cylindrical vortex in an incompressible fluid is co-axial with the z-axis, andsuch that ωz takes the constant value ω for r ≤ a, and is zero for r > a, wherer is a cylindrical coordinate. Show that

1ρ

dpdr=κ 2 ra 4 ,

where p(r) is the pressure at radius r inside the vortex, and the circulation ofthe fluid outside the vortex is 2π κ. Deduce that

p(r) =κ 2 r 2 ρ

2 a 4 + p0,

where p0 is the pressure at the center of the vortex.

5.4 Consider the cylindrical vortex discussed in Exercise 5.3. If p(r) is the pres-sure at radius r external to the vortex, demonstrate that

p(r) = −κ2 ρ

2 r 2 + p∞,

where p∞ is the pressure at infinity.

5.5 Show that the stream function for the cylindrical vortex discussed in Ex-ercises 5.3 and 5.4 is ψ(r) = (1/2)ω a 2 ln(r/a) for r > a, and ψ(r) =(1/4)ω (r 2 − a 2) for r ≤ a.


5.6 Prove that in the two-dimensional motion of a liquid the mean tangential fluidvelocity around any small circle of radius r is ω r, where 2ω is the value of

∂vy

∂x−∂vx

∂y

at the center of the circle. Neglect terms of order r 3.

5.7 Show that the equation of continuity for the two-dimensional motion of anincompressible fluid can be written

∂(r vr)∂r

+∂vθ

∂θ= 0,

where r, θ are cylindrical coordinates. Demonstrate that this equation is sat-isfied when vr = a k r n exp[−k (n + 1) θ] and vθ = a r n exp[−k (n + 1) θ].Determine the stream function, and show that the fluid speed at any point is

(n + 1)ψ√

1 + k 2/r,

where ψ is the stream function at that point (defined such that ψ = 0 at r = 0).

5.8 Demonstrate that streamlines cross at right-angles at a stagnation point intwo-dimensional, incompressible, irrotational flow.

5.9 Consider two-dimensional, incompressible, inviscid flow. Demonstrate thatthe fluid motion is governed by the following equations:

∂ω

∂t+ [ψ, ω] = 0,

∇ 2ψ = ω,

∇ 2χ = ∇ω · ∇ψ + ω 2,

where v = ez × ∇ψ, [A, B] = ez · ∇A × ∇B, and χ = p/ρ + (1/2) v 2 + Ψ .

5.10 For irrotational, incompressible, inviscid motion in two dimensions show that

∇q · ∇q = q∇ 2q,

where q = |v|.

6Two-Dimensional Potential Flow

6.1 IntroductionThis chapter describes the use of complex analysis to facilitate calculations in two-dimensional, incompressible, irrotational fluid dynamics. Incidentally, incompress-ible, irrotational flow is usually referred to as potential flow, because the associ-ated velocity field can be represented in terms of a velocity potential that satisfiesLaplace’s equation. (See Section 4.15.) In the following, all flow patterns are as-sumed to be such that the z-coordinate is ignorable. In other words, the fluid ve-locity is everywhere parallel to the x-y plane, and ∂/∂z = 0. It follows that all linesources and vortex filaments run parallel to the z-axis. Moreover, all solid surfacesare of infinite extent in the z-direction, and have uniform cross-sections. Hence, itis only necessary to specify the locations of line sources, vortex filaments, and solidsurfaces in the x-y plane. More information on the use of complex analysis in two-dimensional fluid mechanics can be found in Batchelor 2000, Milne-Thomson 1958,Milne-Thomson 2011, and Lamb 1993.

6.2 Complex FunctionsThe complex variable is conventionally written

z = x + i y, (6.1)

where i represents the square root of minus one. Here, x and y are both real, and areidentified with the corresponding Cartesian coordinates. (Incidentally, z should notbe confused with a z-coordinate: this is a strictly two-dimensional discussion.) Wecan also write

z = r e i θ, (6.2)

where r =√

x 2 + y 2 and θ = tan−1(y/x) are the modulus and argument of z, re-spectively, but can also be identified with the corresponding plane polar coordinates.Finally, Euler’s theorem (Riley 1974),

e i θ = cos θ + i sin θ, (6.3)

149


implies that

x = r cos θ, (6.4)

y = r sin θ. (6.5)

We can define functions of the complex variable, F(z), in the same way that wedefine functions of a real variable. For instance,

F(z) = z 2, (6.6)

F(z) =1z. (6.7)

For a given function, F(z), we can substitute z = x + i y and write

F(z) = φ(x, y) + iψ(x, y), (6.8)

where φ(x, y) and ψ(x, y) are real two-dimensional functions. Thus, if

F(z) = z 2, (6.9)

thenF(x + i y) = (x + i y)2 = (x 2 − y 2) + 2 i x y, (6.10)

giving

φ(x, y) = x 2 − y 2, (6.11)

ψ(x, y) = 2 x y. (6.12)

6.3 Cauchy-Riemann RelationsWe can define the derivative of a complex function in the same way that we definethe derivative of a real function. In other words,

dFdz= lim |δz|→∞

F(z + δz) − F(z)δz

. (6.13)

However, we now have a problem. If F(z) is a well-behaved function (i.e., finite,single-valued, and differentiable) then it should not matter from which direction inthe complex plane we approach the point z when taking the limit in Equation (6.13).There are, of course, many different possible approach directions, but if we look at aregular complex function, F(z) = z 2 (say), then

dFdz= 2 z (6.14)

Two-Dimensional Potential Flow 151

is perfectly well-defined, and is, therefore, completely independent of the details ofhow the limit is taken in Equation (6.13).

The fact that Equation (6.13) has to give the same result, no matter from whichdirection we approach z, means that there are some restrictions on the forms of thefunctions φ(x, y) and ψ(x, y) in Equation (6.8). Suppose that we approach z along thereal axis, so that δz = δx. We obtain

dFdz= lim |δx|→0

φ(x + δx, y) + iψ(x + δx, y) − φ(x, y) − iψ(x, y)δx

=∂φ

∂x+ i∂ψ

∂x. (6.15)

Suppose that we now approach z along the imaginary axis, so that δz = i δy. We get

dFdz= lim |δy|→0

φ(x, y + δy) + iψ(x, y + δy) − φ(x, y) − iψ(x, y)i δy

= −i∂φ

∂y+∂ψ

∂y. (6.16)

But, if F(z) is a well-behaved function then its derivative must be well-defined, whichimplies that the previous two expressions are equivalent. This requires that

∂φ

∂x=∂ψ

∂y, (6.17)

∂ψ

∂x= −∂φ

∂y. (6.18)

These expressions are called the Cauchy-Riemann relations, and are, in fact, suffi-cient to ensure that all possible ways of taking the limit (6.13) give the same result(Riley 1974).

The Cauchy-Riemann relations can be combined to give

∂ 2φ

∂x 2 +∂ 2φ

∂y 2 =∂ 2ψ

∂x 2 +∂ 2ψ

∂y 2 =∂φ

∂x∂ψ

∂x+∂φ

∂y

∂ψ

∂y= 0. (6.19)

In other words,

∇ 2φ = 0, (6.20)

∇ 2ψ = 0, (6.21)

∇φ · ∇ψ = 0. (6.22)

It follows that the real and imaginary parts of a well-behaved function of the complexvariable both satisfy Laplace’s equation. Furthermore, the contours of these functionscross at right-angles.


6.4 Complex Velocity PotentialEquations (6.17)–(6.18) are identical to Equations (5.21)–(5.22). This suggests thatthe real and imaginary parts of a well-behaved function of the complex variable canbe interpreted as the velocity potential and stream function, respectively, of sometwo-dimensional, irrotational, incompressible flow pattern.

For instance, suppose thatF(z) = −V0 z, (6.23)

where V0 is real. It follows that

φ(r, θ) = −V0 r cos θ, (6.24)

ψ(r, θ) = −V0 r sin θ. (6.25)

It can be seen, by comparison with the analysis of Section 5.4, that the complex ve-locity potential (6.23) corresponds to uniform flow of speed V0 directed along thex-axis. Furthermore, as is easily demonstrated, the complex velocity potential asso-ciated with uniform flow of speed V0 whose direction subtends a (counter-clockwise)angle θ0 with the x-axis is F(z) = −V0 z e−i θ0 .

Suppose that

F(z) = −Q2π

ln z, (6.26)

where Q is real. Because ln z = ln r + i θ (Riley 1974), it follows that

φ(r, θ) = − Q2π

ln r, (6.27)

ψ(r, θ) = − Q2πθ. (6.28)

Thus, according to the analysis of Section 5.5, the complex velocity potential (6.26)corresponds to the flow pattern of a line source, of strength Q, located at the origin.(See Figure 5.3.) As a simple generalization of this result, the complex potential of aline source, of strength Q, located at the point (x0, y0), is F(z) = −(Q/2π) ln(z − z0),where z0 = x0 + i y0. It can be seen, from Equation (6.26), that the complex velocitypotential of a line source is singular at the location of the source.

Suppose that

F(z) = iΓ

2πln z, (6.29)

where Γ is real. It follows that

φ(r, θ) = − Γ2πθ, (6.30)

ψ(r, θ) =Γ

2πln r. (6.31)

Thus, according to the analysis of Section 5.6, the complex velocity potential (6.29)


corresponds to the flow pattern of a vortex filament of intensity Γ located at theorigin. (See Figure 5.5.) As a simple generalization of this result, the complexpotential of a vortex filament, of intensity Γ, located at the point (x0, y0), is F(z) =i (Γ/2π) ln(z − z0), where z0 = x0 + i y0. According to Equation (6.29), the complexvelocity potential of a vortex filament is singular at the location of the filament.

Suppose, finally, that

F(z) = −V0

(z +

a 2

z

)+ i

Γ

2πln( za

), (6.32)

where V0, a, and Γ, are real. It follows that

φ(r, θ) = −V0

(r +

a 2

r

)cos θ − Γ

2πθ, (6.33)

ψ(r, θ) = −V0

(r −

a 2

r

)sin θ +

Γ

2πln( ra

). (6.34)

Thus, according to the analysis of Section 5.8, the complex velocity potential (6.32)corresponds to uniform flow of unperturbed speed V0, running parallel to the x-axis,around an impenetrable circular cylinder of radius a, centered on the origin. (SeeFigures 5.6, 5.7, and 5.8.) Here, Γ is the circulation of the flow about the cylinder.It can be seen that ψ = 0 on the surface of the cylinder (r = a), which ensuresthat the normal velocity is zero on this surface (because the surface corresponds to astreamline), as must be the case if the cylinder is impenetrable.

6.5 Complex VelocityEquations (5.17), (5.20), and (6.15) imply that

dFdz=∂φ

∂x+ i∂ψ

∂x= −vx + i vy. (6.35)

Consequently, dF/dz is termed the complex velocity. It follows that∣∣∣∣∣dFdz

∣∣∣∣∣ 2 = v 2x + v

2y = v

2, (6.36)

where v is the flow speed.A stagnation point is defined as a point in a flow pattern at which the flow speed,

v, falls to zero. (See Section 5.8.) According to the previous expression,

dFdz= 0 (6.37)

at a stagnation point. For instance, the stagnation points of the flow pattern produced


when a cylindrical obstacle of radius a, centered on the origin, is placed in a uniformflow of speed V0, directed parallel to the x-axis, and the circulation of the flow aroundis cylinder is Γ, are found by setting the derivative of the complex potential (6.32) tozero. It follows that the stagnation points satisfy the quadratic equation

dFdz= −V0

(1 − a 2

z 2

)+ i

Γ

2π z= 0. (6.38)

The solutions areza= −i ζ ±

√1 − ζ 2, (6.39)

where ζ = −Γ/(4πV0 a), with the proviso that |z|/a > 1, because the region |z|/a < 1is occupied by the cylinder. Thus, if ζ ≤ 1 then there are two stagnation points onthe surface of the cylinder at x/a = ±

√1 − ζ 2 and y/a = −ζ. On the other hand,

if ζ > 1 then there is a single stagnation point below the cylinder at x/a = 0 andy/a = −ζ −

√ζ 2 − 1.

According to Section 4.15, Bernoulli’s theorem in an steady, irrotational, incom-pressible fluid takes the form

p +12ρ v 2 = p0, (6.40)

where p0 is a uniform constant. Here, gravity (and any other body force) has beenneglected. Thus, the pressure distribution in such a fluid can be written

p = p0 −12ρ

∣∣∣∣∣dFdz

∣∣∣∣∣ 2 . (6.41)

6.6 Method of ImagesLet F1(z) = φ1(x, y) + iψ1(x, y) and F2(z) = φ2(x, y) + iψ2(x, y) be complex velocitypotentials corresponding to distinct, two-dimensional, irrotational, incompressibleflow patterns whose stream functions are ψ1(x, y) and ψ2(x, y), respectively. It fol-lows that both stream functions satisfy Laplace’s equation: that is, ∇ 2ψ1 = ∇ 2ψ2 = 0.Suppose that F3(z) = F1(z) + F2(z). Writing F3(z) = φ3(x, y) + iψ3(x, y), it is clearthat ψ3(x, y) = ψ1(x, y) + ψ2(x, y). Moreover, ∇ 2ψ3 = ∇ 2ψ1 + ∇ 2ψ2 = 0, so ψ3 alsosatisfies Laplace’s equation. We deduce that two complex velocity potentials, cor-responding to distinct, two-dimensional, irrotational, incompressible flow patterns,can be superposed to produce a third velocity potential that corresponds to anothertwo-dimensional, irrotational, incompressible flow pattern. As described in the fol-lowing, this idea can be exploited to determine the flow patterns produced by linesources and vortex filaments in the vicinity of rigid boundaries.

As an example, consider a situation in which there are two line sources of strengthQ located at the points (0, ±a). (See Figure 6.1.) The complex velocity potential of


a

x

Q

Q

y

a

Figure 6.1Two line sources.

the resulting flow pattern is the sum of the complex potentials of each source takenin isolation. Hence, from Section 6.4,

F(z) = −Q2π

ln(z − i a) −Q2π

ln(z + i a) = −Q2π

ln(z 2 + a 2). (6.42)

Thus, the stream function of the flow pattern (which is the imaginary part of thecomplex potential) is

ψ(x, y) = − Q2π

tan−1(

2 x yx 2 − y 2 + a 2

). (6.43)

Note that ψ(x, 0) = 0, which implies that there is zero flow normal to the plane y = 0.Hence, in the region y > 0, we could interpret the previous stream function as thatgenerated by a single line source of strength Q, located at the point (0, a), in thepresence of a planar rigid boundary at y = 0. This follows because the stream func-tion satisfies ∇ 2ψ = 0 everywhere in the region y > 0, has the requisite singularity(corresponding to a line source of strength Q) at (0, a), and satisfies the physicalboundary condition that the normal velocity be zero at the rigid boundary. Moreover,as is well-known, the solutions of Laplace’s equation are unique. The streamlinesof the resulting flow pattern are shown in Figure 6.2. Incidentally, we can think ofthe two sources in Figure 6.1 as the “images” of one another in the boundary plane.Hence, this method of calculation is usually referred to as the method of images.


0

1

2

3

4

5

6

y/a

−3 −2 −1 0 1 2 3x/a

Figure 6.2Stream lines of the two-dimensional flow pattern due to a line source at (0, a) in thepresence of a rigid boundary at y = 0.

The complex velocity associated with the complex velocity potential (6.42) is

dFdz= −Q

π

zz 2 + a 2 . (6.44)

Hence, the flow speed at the boundary is

v(x, 0) =∣∣∣∣∣dF

dz

∣∣∣∣∣ (x, 0) =Qπ a

|x|/a1 + x 2/a 2 . (6.45)

It follows from Equation (6.41) (and the fact that the flow speed at infinity is zero)that the excess pressure on the boundary, due to the presence of the source, is

δp(x, 0) = −ρ2

∣∣∣∣∣dFdz

∣∣∣∣∣ 2y=0= −ρ

2

( Qπ a

)2 x 2/a 2

(1 + x 2/a 2)2 . (6.46)

Thus, the excess force per unit length (in the z-direction) acting on the boundary inthe y-direction is

Fy = −∫ ∞

−∞δp(x, 0) dx =

ρQ 2

2π2 a

∫ ∞

−∞

ζ 2

(1 + ζ 2)2 dζ =ρQ 2

4π a. (6.47)

The fact that the force is positive implies that the boundary is attracted to the source,and vice versa.


−Γ

x

y

a

a

Γ

Figure 6.3Two vortex filaments.

As a second example, consider the situation, illustrated in Figure 6.3, in whichthere are two vortex filaments of intensities Γ and −Γ situated at (0, ±a). As before,the complex velocity potential of the resulting flow pattern is the sum of the complexpotentials of each filament taken in isolation. Hence, from Section 6.4,

F(z) = iΓ

2πln(z − i a) − i

Γ

2πln(z + i a) = i

Γ

2πln(

z − i az + i a

). (6.48)

Thus, the stream function of the flow pattern is

ψ(x, y) =Γ

2πln[

x 2 + (y − a)2

x 2 + (y + a)2

]. (6.49)

As before, ψ(x, 0) = 0, which implies that there is zero flow normal to the plane y = 0(because the plane corresponds to a streamline). Hence, in the region y > 0, we couldinterpret the previous stream function as that generated by a single vortex filamentof intensity Γ, located at the point (0, a), in the presence of a planar rigid boundaryat y = 0. The streamlines of the resulting flow pattern are shown in Figure 6.4. Weconclude that a vortex filament reverses its sense of rotation (i.e., Γ → −Γ) when“reflected” in a boundary plane.

As a final example, consider the situation, illustrated in Figure 6.5, in whichthere is an impenetrable circular cylinder of radius a, centered on the origin, and aline source of strength Q located at (b, 0), where b > a. Consider the so-called analogproblem, also illustrated in Figure 6.5, in which the cylinder is replaced by a source


0

1

2

3

4

5

6

y/a

−3 −2 −1 0 1 2 3x/a

Figure 6.4Stream lines of the two-dimensional flow pattern due to a vortex filament at (0, a) inthe presence of a rigid boundary at y = 0.

of strength Q, located at (c, 0), where c < a, and a source of strength −Q, located atthe origin. We can think of these two sources as the “images” of the external sourcein the cylinder. Moreover, given that the solutions of Laplace’s equation are unique,if the analog problem can be adjusted in such a manner that r = a is a streamline thenthe flow in the region r > a will become identical to that in the actual problem. Thecomplex velocity potential in the analog problem is simply

F(z) = − Q2π

ln(z − b) − Q2π

ln(z − c) +Q2π

ln z = − Q2π

ln[(z − b) (z − c)

z

]. (6.50)

Hence, writing z = r e i θ, the corresponding stream function takes the form

ψ(r, θ) = −Q2π

tan−1[

(r − b c/r) sin θ(r + b c/r) cos θ − (b + c)

]. (6.51)

We require the surface of the cylinder, r = a, to be a streamline: that is, ψ(a, θ) =constant. This is easily achieved by setting c = a 2/b. Thus, the stream functionbecomes

ψ(r, θ) = − Q2π

tan−1[

(r/a − a/r) sin θ(r/a + a/r) cos θ − (b/a + a/b)

]. (6.52)

The corresponding streamlines in the region external to the cylinder are illustrated inFigure 6.6.


ax

b

Q QQx

y y

−Q

c

Figure 6.5A line source in the presence of an impenetrable circular cylinder.

−4

−3

−2

−1

0

1

2

3

4

y/a

−4 −3 −2 −1 0 1 2 3 4

x/a

Figure 6.6Stream lines of the two-dimensional flow pattern due to a line source at (2 a, 0) inthe presence of a rigid circular cylinder of radius a centered on the origin.


6.7 Conformal MapsLet ζ = ξ + i η and z = x + i y, where ξ, η, x, and y are real. Suppose that ζ = f (z),where f is a well-behaved (i.e., single-valued, non-singular, and differentiable) func-tion. We can think of ζ = f (z) as a map from the complex z-plane to the complexζ-plane. In other words, every point (x, y) in the complex z-plane maps to a cor-responding point (ξ, η) in the complex ζ-plane. Moreover, if f (z) is indeed a well-behaved function then this mapping is unique, and also has a unique inverse. Supposethat the point z = z0 in the z-plane maps to the point ζ = ζ0 in the ζ-plane. Let usinvestigate how neighboring points map. We have

ζ0 + dζ′ = f (z0 + dz′), (6.53)

ζ0 + dζ′′ = f (z0 + dz′′). (6.54)

In other words, the points z0 + dz′ and z0 + dz′′ in the complex z-plane map to thepoints ζ0 + dζ′ and ζ0 + dζ′′ in the complex ζ-plane, respectively. If |dz′|, |dz′′| 1then

dζ′ = f ′(z0) dz′, (6.55)

dζ′′ = f ′(z0) dz′′, (6.56)

where f ′(z) = d f /dz. Hence,dζ′′

dζ′=

dz′′

dz′. (6.57)

Thus, it follows that|dζ′′||dζ′|

=|dz′′||dz′|

, (6.58)

andarg(dζ′′) − arg(dζ′) = arg(dz′′) − arg(dz′). (6.59)

We can think of dz′ and dz′′ as infinitesimal vectors connecting neighboring pointsin the complex z-plane to the point z = z0. Likewise, dζ′ and dζ′′ are infinitesimalvectors connecting the corresponding points in the complex ζ-plane. It is clear, fromthe previous two equations, that, in the vicinity of z = z0, the mapping from thecomplex z-plane to the complex ζ-plane is such that the lengths of dz′ and dz′′ expandor contract by the same factor, and the angle subtended between these two vectorsremains the same. (See Figure 6.7.) This type of mapping is termed conformal.

Suppose that F(ζ) = φ(ξ, η) + iψ(ξ, η) is a well-behaved function of the complexvariable ζ. It follows that∇ 2φ = ∇ 2ψ = ∇φ·∇ψ = 0. Hence, the functions φ(ξ, η) andψ(ξ, η) can be interpreted as the velocity potential and stream function, respectively,of some two-dimensional, incompressible, irrotational flow pattern, where ξ and ηare Cartesian coordinates. However, if ζ = f (z), where f (z) is well-behaved, thenF(ζ) = F[ f (z)] = G(z) = φ(x, y) + i ψ(x, y), where G(z) is also well-behaved. Itfollows that ∇ 2φ = ∇ 2ψ = ∇φ · ∇ψ = 0. In other words, the functions φ(x, y) and


dζ ′′

x

y η

ξ

ζ0

dζ ′

dz′′

dz′z0

Figure 6.7A conformal map.

ψ(x, y) can be interpreted as the velocity potential and stream function, respectively,of some new, two-dimensional, incompressible, irrotational flow pattern, where x andy are Cartesian coordinates. In other words, we can use a conformal map to converta given two-dimensional, incompressible, irrotational flow pattern into another, quitedifferent, pattern. Incidentally, a conformal map converts a line source into a linesource of the same strength, and a vortex filament into a vortex filament of the sameintensity. (See Exercise 6.12.)

As an example, consider the conformal map

ζ = i e π z/a. (6.60)

Writing ζ = r e i θ, it is easily demonstrated that x = a ln r/π and y = a (θ/π − 1/2).Hence, the positive ξ-axis (θ = 0) maps to the line y = −a/2, the negative ξ-axis(θ = π) maps to the line y = a/2, and the region η > 0 (0 ≤ θ ≤ π) maps tothe region −a/2 < y < a/2. Moreover, the points ζ = (0, ±1) map to the pointsz = a (0,−1/2±1/2). (See Figure 6.8.) As we saw in Section 6.6, in the region η > 0,the velocity potential

F(ζ) = iΓ

2πln(ζ − iζ + i

)(6.61)

corresponds to the flow pattern generated by a vortex filament of intensity Γ, locatedat the point ζ = (0, 1), in the presence of a rigid plane at η = 0. Hence,

G(z) = F(i e π z/a) = iΓ

2πln tanh

(π z2 a

), (6.62)

corresponds to the flow pattern generated by a vortex filament of intensity Γ, locatedat the origin, in the presence of two rigid planes at y = ±a/2. This follows becausethe line η = 0 is mapped to the lines y = ±a/2, and the point ζ = (0, 1) is mapped


1

y

x

E

η

A′′C BξA′A′ C

A′′ B

D

E

Da

Figure 6.8The conformal map ζ = i e π z/a.

to the origin. Moreover, if the line η = 0 is a streamline in the η-plane then the linesy = ±a/2 are also streamlines in the z-plane. Thus, these lines could all correspond torigid boundaries. The stream function associated with the previous complex velocitypotential,

ψ(x, y) =Γ

πln[

cosh(π x a) − cos(π y/a)cosh(π x/a) + cos(π y/a)

], (6.63)

is shown in Figure 6.9.As a second example, consider the map

ζ = z 2. (6.64)

This maps the positive ξ-axis to the positive x-axis, the negative ξ-axis to the positivey-axis, the region η > 0 to the region x > 0, y > 0, and the point ζ = (0, 2 a 2) to thepoint z = (a, a). As we saw in Section 6.6, in the region η > 0, the velocity potential

F(ζ) =Q2π

ln(ζ 2 + 4 a 4), (6.65)

corresponds to the flow pattern generated by a line source of strength Q, located atthe point ζ = (0, 2 a 2), in the presence of a rigid plane at η = 0. Thus, the complexvelocity potential

G(z) = F(z 2) =Q2π

ln(z 4 + 4 a 4), (6.66)

corresponds to the flow pattern generated by a line source of strength Q, located atthe point z = (a, a), in the presence of two orthogonal rigid planes at y = 0 and x = 0.The stream function associated with the previous complex potential,

ψ(x, y) = −Q2π

tan−1[

4 x y (x 2 − y 2)x 4 − 6 x 2 y 2 + y 4 + 4 a 4

], (6.67)


−0.5

0

0.5y/a

−1 −0.5 0 0.5 1

x/a

Figure 6.9Stream lines of the two-dimensional flow pattern due to a vortex filament at the originin the presence of two rigid planes at y = ±a/2.

0

1

2

3

y/a

0 1 2 3x/a

Figure 6.10Stream lines of the two-dimensional flow pattern due to a line source at (a, a) in thepresence of two rigid planes at x = 0 and y = 0.


is shown in Figure 6.10.As a final example, consider the map

z = ζ +l 2

ζ, (6.68)

where l is real and positive. Writing ζ = r e i θ, we find that

x = 2 l cosh[ln(r/l)] cos θ = (r + l 2/r) cos θ, (6.69)

y = 2 l sinh[ln(r/l)] sin θ = (r − l 2/r) sin θ. (6.70)

Thus, the map converts the circle ξ 2 + η 2 = c 2 in the ζ-plane, where c > l, into theellipse ( x

a

)2+

(y

b

)2= 1 (6.71)

in the z-plane, where

a = 2 l cosh[ln(c/l)] = c + l 2/c, (6.72)

b = 2 l sinh[ln(c/l)] = c − l 2/c. (6.73)

Note that the center of the ellipse lies at the origin, and its major and minor axesrun parallel to the x- and the y-axes, respectively. As we saw in Section 6.4, in theζ-plane, the complex velocity potential

F = −V0

(ζ +

c 2

ζ

), (6.74)

represents uniform flow of unperturbed speed V0, running parallel to the ξ-axis,around a circular cylinder of radius c, centered on the origin. Thus, assuming thatc > l, in the z-plane, the potential represents uniform flow of unperturbed speedV0, running parallel to the x-axis [which follows because at large |z| the map (6.68)reduces to z = ζ, and so the flow at large distances from the origin is the samein the complex z- and ζ-planes], around an elliptical cylinder of major radius a,aligned along the x-axis, and minor radius b, aligned along the y-axis. Note thatl = (a 2 − b 2)1/2/2 and c = (a + b)/2. The corresponding stream function in thez-plane is

ψ(x, y) = −V0

(r − c 2

r

)sin θ, (6.75)

where

r = l exp(cosh−1 p), (6.76)

θ = tan−1(y

xp

[p 2 − 1]1/2

), (6.77)

p =

x 2/l 2 + y 2/l 2 + 4 +([x 2/l 2 + y 2/l 2 + 4]2 − 16 x 2/l 2

)1/28

1/2

. (6.78)

Figure 6.11 shows the streamlines of the flow pattern calculated for c = 1.5 l.


−4

−3

−2

−1

0

1

2

3

4

y/l

−4 −3 −2 −1 0 1 2 3 4x/l

Figure 6.11Stream lines of the two-dimensional flow pattern due to uniform flow parallel to thex-axis around an elliptical cylinder.

6.8 Schwarz–Christoffel TheoremThe Schwarz–Christoffel theorem is an important mathematical result that allowsa polygonal boundary in the z-plane to be mapped conformally onto the real axis,η = 0, in the ζ-plane. It is conventional to map the region inside the polygon in thez-plane onto the upper half, η > 0, of the ζ-plane. If the interior angles of the polygonare α, β, γ, · · · then the appropriate map is

dζdz= K (ζ − a)1−α/π (ζ − b)1−β/π (ζ − c)1−γ/π · · · , (6.79)

where K is a constant, and a, b, c, · · · are the (real) values of ζ that correspond to thevertices of the polygon (Milne-Thompson 2011).

It is often convenient to take the point in the ζ-plane that corresponds to oneof the vertices of the polygon—say, that given by ζ = a—to be at infinity. In thiscase, the factor (ζ − a) in Equation (6.79) becomes effectively constant, and can beabsorbed into a new constant of proportionality, K′.

Consider, for example, a semi-infinite strip in the z-plane, for which α = 0,β = π/2, and γ = π/2. This is mapped onto the upper half of the ζ-plane, with thezero-angle vertex corresponding to a point at infinity in the ζ-plane, by means of the


ξ

x

z-plane

γ

β

y

z0

z0 + i π/K ′

ζ-plane

c b

η

Figure 6.12Conformal transformation of a semi-infinite strip in the z-plane into the upper half ofthe ζ-plane.

transformationdζdz= K′ (ζ − b)1/2 (ζ − c)1/2, (6.80)

where K′ is real and positive. We can integrate the previous expression to give

ζ =12

(b + c) +12

(b − c) cosh[K′ (z − z0)]. (6.81)

Here, the points ζ = b and ζ = c in the ζ-plane correspond to the vertices z = z0 andz = z0 + i π/K′, respectively, in the z-plane. (See Figure 6.12.)

Suppose that b = 0, c = −w, z0 = 0, and π/K′ = w. In this case, the transforma-tion (6.81) becomes

ζ =w

2

[cosh

(π

zw

)− 1], (6.82)

which implies that

ξ =w

2

[cosh

(π

xw

)cos

(πy

w

)− 1], (6.83)

η =w

2sinh

(π

xw

)sin(πy

w

). (6.84)

The transformation (6.82) maps the semi-infinite strip in the z-plane that is boundedby the lines y = 0, x = 0, and y = w into the upper half of the ζ-plane. Thetransformation also maps the origin of the ζ-plane to the origin of the z-plane, andthe ξ-axis to the lines y = w, x = 0, and y = 0. Now, in the ζ-plane,

F = −Qπ

ln ζ (6.85)


0

0.5

1

y/w

0 1 2x/w

Figure 6.13Stream function due to a line source located in the left-hand corner of a semi-infinitestrip bounded by rigid planes at y = 0, x = 0, and y = w.

is the complex potential associated with a line source of strength 2 Q, located at theorigin. By symmetry, the ξ-axis corresponds to a streamline, and can, therefore, bereplaced by a rigid boundary. It follows that, in the z-plane, the same potential isthat due to a line source, of strength Q, located in the lower left-hand corner of asemi-infinite strip bounded by rigid planes at y = 0, x = 0, and y = w. (Incidentally,the source strength in the z-plane is Q, rather than 2 Q, because, by symmetry, halfthe output from the source in the ζ-plane goes below the ξ-axis and, therefore, doesnot map to the semi-infinite strip in the z-plane.) The corresponding stream functionis

ψ = −Qπ

arg(ζ) = −Qπ

tan−1(η

ξ

), (6.86)

and is illustrated in Figure 6.13.An infinite strip in the z-plane is a polygon with two zero-angle vertices, both at

infinity. The required transformation follows from Equation (6.79) by setting α =β = 0, (ζ − a)/a→ −1, and K a→ K′. Thus, we obtain

dζdz= K′ (ζ − b), (6.87)


ζ = b + e K′ (z−z0). (6.88)

This transformation maps the point ζ = b+1 to the point z = z0, the ξ-axis to the twolines y = Im(z0) and y = Im(z0)+π/K′, and the upper half of the ζ-plane to the regionbetween the lines y = Im(z0) and y = Im(z0)+π/K′, as illustrated in Figure 6.14. It isclear that the transformation (6.60), studied in the preceding section, is just a specialcase of the transformation (6.88).


ξ

x

z-planey

b b + 1

z0

ζ-planeη

π/K ′

B

B′A′

A CB

B′ CAA′

Figure 6.14Conformal transformation of an infinite strip in the z-plane into the upper half of theζ-plane.

6.9 Free Streamline TheoryConsider a situation in which steady, incompressible, irrotational flow is partly bound-ed by rigid walls, and partly by free streamlines of unknown shape on which thepressure takes a known constant value. For instance, the free streamlines might cor-respond to the interface of a liquid with the atmosphere, in which case the constantpressure would correspond to that of the atmosphere. According to Bernoulli’s theo-rem, (6.40), (neglecting gravity) the fluid speed is constant on a free streamline.

Let us define the new complex variable

Ω = ln(

dzdF

)= ln

(1

vx − i vy

)= ln

(1v

)+ i θ, (6.89)

where z = x+ i y, F = φ+ i ψ, and v and θ are the magnitude and direction (relative tothe x-axis) of the velocity vector (vx, vy). (See Section 6.5.) Here, F = −F, φ = −φ,and ψ = −ψ. It follows that the real part of Ω is constant on each free streamline,whereas the imaginary part is constant on each straight segment of the rigid boundary.Thus, the whole boundary of the fluid is represented in the Ω-plane by a polygon.We know, from the Schwarz–Christoffel theorem (see Section 6.8), that it is alwayspossible to map the interior (or exterior) of a polygon in one complex plane onto theupper half of another complex plane. Hence, it is possible to find conformal relationsbetween Ω and a new complex variable λ, and between F and λ, such that the flowregion is mapped onto the upper half of the λ-plane in both cases. In this way, wecan find a relation between Ω and F, from which an expression for F in terms of zfollows via integration.


λ-plane

Ω-plane

ψ = −ψ1

ψ = ψ1

b

x

z-plane

y

d C

A

B

B′

A′

θ

π/2

ABCC ′B′A′

C ′

A′

A

B′

C

B

C ′

ψ = ψ1

ψ = −ψ1

ψ = 0

ψ = 0

ψ=

0

C

A′ C ′

A

F -plane

φ

ψB

B′

ln(v/V )

Figure 6.15Conformal transformations required for the determination of the flow from an orificein a plane wall in two dimensions.

As an example of the application of free streamline theory, let us calculate thecontraction ratio of a two-dimensional jet of liquid emerging from an orifice. Sup-pose that the orifice in question is a long thin slot in a plane wall of small thickness,and that the wall forms part of a large vessel containing liquid. The speed of theliquid on the free streamlines that emerge from the edges of the orifice is uniform,and equal to V , say. This is also the speed of the interior of the jet far downstreamof the orifice, where (neglecting the effects of gravity) the streamlines are straightand parallel. (See Figure 6.16.) Let the two streamlines bounding the flow, on whichψ = +ψ1 and ψ = −ψ1, say—be ABC and A′B′C′, respectively, where A, A′, C, andC′ all represent points at infinity. Figure 6.15 shows the corresponding straight-lineboundaries in the F- and Ω-planes, where Ω is now defined in a more convenientmanner as

Ω = ln(V

dzdF

)= ln

(Vv

)+ i θ. (6.90)

We, first, need to map the semi-infinite strip A′B′C′CBA in the Ω-plane to theupper half of the λ-plane. The conformal transformation of a semi-infinite strip ontothe upper half of another complex plane is achieved by the relation (6.81). Adaptingthis relation to our current needs, we take K′ = 1, z0 = −i π/2, and b = −c = 1, sothat

λ = i sinhΩ (6.91)


In particular, this transformation maps the points B′ = i π/2 and B = i π/2 in theΩ-plane to the points B = 1 and B = −1, respectively, in the λ-plane.

Next, we need to map the infinite strip ABCA′B′C′ in the F-plane to the upperhalf of the λ-plane. The conformal transformation of a semi-infinite strip onto theupper half of another complex plane is achieved by the relation (6.88). Adapting thisrelation to our current needs, we take −K′ = π/(2 ψ1), b = 0, z0 = i ψ1, so that

λ = i exp(−π

2Fψ1

). (6.92)

The flow field has now been mapped onto the upper half of the λ-plane in twocoincident ways, giving

λ = i exp(−π

2Fψ1

)= i sinhΩ =

i2

(V

dzdF− 1

VdFdz

). (6.93)

Hence,

VdzdF= −i λ + (1 − λ 2)1/2, (6.94)

where, with cuts in the z-plane at AB and A′B′, the correct branch of (1−λ 2)1/2 is thatwhich is real and positive when ψ = 0. Integration, with the aid of Equation (6.92),yields

π

2Vψ1

(z − z1) = i (λ − 1) − (1 − λ 2)1/2 + tanh−1[(1 − λ 2)1/2], (6.95)

where z1 is a constant. However, λ = 1 at the point B, where z = i d (2 d being thewidth of the orifice), so z1 = i d, and

π

2Vψ1

(z − i d) = i (λ − 1) − (1 − λ 2)1/2 + tanh−1[(1 − λ 2)1/2], (6.96)

Finally, the required relationship between z and F is obtained by eliminating λ be-tween Equations (6.92) and (6.96).

On the free streamline BC, we have

ψ = ψ1, φ = V s, Ω = i θ, (6.97)

where s denotes distance measured along the streamline from B. It follows fromEquations (6.91) and (6.92) that

λ = − sin θ = exp(−π

2Vψ1

s). (6.98)

Thus, making use of Equation (6.96), the equation of streamline BC can be written,in parametric form, as

x =2π

ψ1

V

[tanh−1(cos θ) − cos θ)

], (6.99)

y = d − 2π

ψ1

V(1 + sin θ) . (6.100)


−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

y/d

0 1 2x/d

Figure 6.16Free streamlines of a liquid jet emerging from an orifice in a plane wall in two di-mensions.

Thus, the asymptotic semi-width of the jet is

b = lims→∞

y(s) = d − 2π

ψ1

V. (6.101)

Far from the orifice, which corresponds to λ→ 0, Equation (6.94) yields

dFdz=

1vx − i vy

1V. (6.102)

In other words, as expected, the velocity profile becomes uniform across the jet fardownstream of the orifice, which implies that ψ1 = b V . It follows, from Equa-tion (6.101), that the contraction ratio of a two-dimensional liquid jet emerging froman orifice in a plane wall takes the value

bd=

π

π + 2= 0.61. (6.103)

Somewhat surprisingly, this value is very close to the observed contraction ratio ofa three-dimensional jet emerging from a circular hole in a plane wall. (See Sec-tion 4.6.) According to Equations (6.99), (6.100), and (6.103), the equation of the


0.5

0.6

0.7

0.8

0.9

1

b/d

0 1 2(2/π) α

Figure 6.17Contraction ratio of a liquid jet emerging from a two-dimensional orifice formed bya gap between two semi-infinite plane walls that subtend an angle 2α.

free streamline BC can be written

xd=

2π + 2

[tanh−1(cos ζ) − cos ζ)

], (6.104)

y

d=π + 2 sin ζπ + 2

, (6.105)

for π/2 ≥ ζ ≥ 0. By symmetry, the equation of the free streamline B′C′ is obtainedfrom the previous equation via the transformation x → x, y → −y. The streamlinesBC and B′C′ are plotted in Figure 6.16.

We can easily extend the previous calculation to determine the contraction ratioof a liquid jet emerging from a two-dimensional orifice formed by a gap betweentwo semi-infinite plane walls that subtend an angle 2α. In this case, the free stream-line BC, on which ψ = +ψ1, corresponds to v = V and θ = −α, whereas the freestreamline B′C′, on which ψ = −ψ1, corresponds to v = V and θ = +α. Hence, thetransformation (6.93) generalizes to give

λ = i exp(−π

2Fψ1

)= i sinh

(π

2αΩ), (6.106)

which implies that

VdzdF=[−i λ + (1 − λ 2)1/2

] 2α/π. (6.107)


Far from the orifice, which corresponds to λ → 0, the previous expression yieldsdF/dz 1/V . In other words, a long way downstream of the orifice, the velocityprofile is again uniform across the jet. Hence, we can write ψ1 = b V , where b is thesemi-width of the jet far from the orifice. Combining the previous two equations, weobtain

−π2

dzb=

dλλ

[−i λ + (1 − λ 2)1/2

] 2α/π. (6.108)

Now, λ = 1 corresponds to the point B, at which z = i d, where d is the semi-widthof the orifice. Thus, integration of the previous expression yields

z − i db=

2π

∫ 1

λ

dkk

[−i k + (1 − k 2)1/2

] 2α/π. (6.109)

Making the transformation k = sin β, we get

z − i db=

2π

∫ π/2

sin−1(λ)exp

(−i

2απβ

)dβ

tan β. (6.110)

As before, we have λ = exp[−(π/2) (s/b)] on the free streamline BC, where s denotesdistance measured along the streamline from B. Far downstream of the orifice, λ→ 0and Im(z) → b. Hence, we obtain the following expression for the contraction ratioof the jet:

bd=

(1 +

2π

∫ π/2

0

sin[(2α/π) β]tan β

dβ)−1

. (6.111)

This contraction ratio is plotted as a function of α in Figure 6.17. It can be seen thatthe ratio takes the value 1 when α = 0, which corresponds to a two-dimensional jetemerging from a gap between two parallel planes. Not surprisingly, there is no con-traction in this case, because the streamlines of the flow are already parallel beforethey emerge from the orifice. On the other hand, the ratio takes the value 1/2 whenα = π, which corresponds to the two-dimensional equivalent to the Borda mouth-piece discussed in Section 4.6. In this case, the contraction ratio is exactly the sameas that of a three-dimensional Borda mouthpiece, which is not surprising, becausethere is nothing in the discussion of Section 4.6 that depends crucially on the shapeof the mouthpiece cross-section.

As a second example of the use of free streamline theory, consider the flow pasta flat plate normal to a liquid stream of infinite extent. The pressure in the cavityformed behind the plate is assumed to be equal to that of the undisturbed stream.Thus, the speed of the liquid on the free streamlines bounding the cavity is equal toV , which is the uniform flow speed far upstream of the plate. Let us take ψ = 0 onthe central streamline that divides at the stagnation point O, and later becomes thetwo free streamlines. (See Figure 6.18.)

The correspondence between the various points on the streamline ψ = 0 in thez-, F-, and Ω-planes is specified in Figure 6.18. The flow occupies the whole ofthe F-plane, apart from a thin slit running along the positive section of the real axis.As previously, the procedure is to find transformations that map the flow regions in


B

λ-plane

Ω-plane

y

z-plane

x

C

C ′

Aψ = 0

O

B′

B

b

F -plane

φ

ψ

B

B′

C

C ′A O

AF 1/2-plane

C ′ B′ O B C

ln(V/v)

θ

π/2

O

B′

C ′AC

O

O B C,A, C ′ B′ O

O

Figure 6.18Conformal transformations required for the determination of the flow past a flat platewith a cavity at ambient pressure.

both the F- and Ω-planes coincidentally onto the upper half of the λ-plane. Thesemi-infinite strip in theΩ-plane has the same width, position, and orientation as thatshown in Figure 6.15. Thus, the appropriate relationship between Ω and λ is again

λ = i sinhΩ, (6.112)

with the locations of the points B and B′ again being λ = −1 and λ = +1, respectively.The appropriate relationship between F and λ can be recognized by noting, first, thatthe flow occupies the upper half of the F 1/2-plane, and, second, that an inversion andchange of sign are needed to bring corresponding points on the two real axes intocoincidence. Thus, we obtain

λ = −(

k VF

)1/2, (6.113)

where k is some positive constant.The required relation between F and Ω is given by

λ = −(

k VF

)1/2= i sinhΩ =

i2

(V

dzdF−

1V

dFdz

). (6.114)


Hence,1V

dFdz= −i

(k VF

)1/2+

(1 − k V

F

)1/2, (6.115)

where the relevant branch of (1 − k V/F)1/2 is that which is positive on AO. Integra-tion, making use of the fact that F = 0 when z = 0, yields

zk=

∫ F/k V

0

du−i u−1/2 + (1 − 1/u)1/2 =

∫ F/k V

0

[i u−1/2 + (1 − 1/u)1/2

]du, (6.116)

or

zk=

(F

k V

)1/2 ( Fk V− 1)1/2−ln

( Fk V

)1/2+

(F

k V− 1)1/2+2 i

(F

k V

)1/2+iπ

2. (6.117)

We can now evaluate the constant k making use of the fact that at the point B, whereλ = −1,

z = i b,F

k V= 1. (6.118)

Here, b is the semi-width of the plate. Thus, we obtain

k =2 bπ + 4

. (6.119)

On the free streamline BC,

F = φ = V (k + s), Ω = i θ, (6.120)

where s represents distance along the streamline measured from B. Thus, on thisstreamline,

λ = − sin θ = −(

kk + s

)1/2. (6.121)

Hence, according to Equation (6.117), the equation of the streamline BC can bewritten, in parametric form, as

x = (k + s)1/2 s1/2 − k ln[(

1 +sk

)1/2+

( sk

)1/2], (6.122)

y = 2 (k + s)1/2 k1/2 +π

2k. (6.123)

By symmetry, the equation of the free streamline B′C′ is obtained from the previ-ous equation via the transformation x → x, y → −y. We deduce that the cavitydownstream of the plate extends to infinity, and that its boundary asymptotes to theparabola

y 2 = 4 k x =(

8 bπ + 4

)x. (6.124)


From Bernoulli’s theorem, the net force per unit length exerted by the fluid onthe plate, which is obviously normal to the plate, and, therefore, directed parallel tothe x-axis, is

D =∫ b

−b(p − p0)x=0 dy =

∫ b

−b

(12ρV 2 − 1

2ρ v 2

)dy (6.125)

= ρV 2 b − ρ∫ k V

0

∂φ

∂ydφ. (6.126)

In conventional parlance, this force constitutes a drag, because it is directed parallelto the undisturbed flow. (See Section 5.8.) Now, F/(k V) = φ/(k V) < 1 on OB, so

1V

dFdz= −i γ−1/2 + i

(1γ− 1)1/2

, (6.127)

where γ = φ/(k V). However, dF/dz = vx − i vy = −i ∂φ/∂y. Hence,

D = ρV 2 b − ρV 2 k∫ 1

0

1γ1/2 −

(1γ− 1)1/2 dγ, (6.128)

which yields

D =(

2ππ + 4

)ρV 2 b. (6.129)

Finally, if we define the drag coefficient, in the conventional manner (see Section 8.8),then we obtain

CD =f

ρV 2 b=

2ππ + 4

= 0.88. (6.130)

We can extend the previous calculation to determine the flow past a flat plateinclined at an angle α to a liquid stream of infinite extent. Let the plate lie in theplane x = 0. The flow field then occupies a semi-infinite strip in the Ω-plane thathas the same width, position, and orientation as that in the previous calculation.Consequently, the appropriate relation between Ω and λ is again

λ = i sinhΩ =i2

(V

dzdF−

1V

dFdz

). (6.131)

It follows that1V

dFdz= i λ +

√1 − λ 2. (6.132)

As before, the two edges of the plate, B and B′, correspond to λ = −1 and λ = +1,respectively. Furthermore, the stagnation point, O, on the front surface of the platecorresponds to |λ| → ∞. (See Figure 6.18.) However, the latter point is no longernecessarily located at the plate’s midpoint.

The transformation (6.113) generalizes to give

F =k V

(λ − cosα)2 . (6.133)


At the points A, C, and C′, |F | → ∞ and, hence, λ → cosα. (See Figure 6.18.) Itfollows from Equation (6.132) that dF/dz → V (sinα + i cosα), which implies thatvx → V sinα and vy → −V cosα. In other words, far upstream and downstream ofthe plate, the fluid velocity vector subtends an angle α with the plane x = 0.

On the front surface of the plate, B′OB, |λ| ≥ 1 and dF/dz = −i V ∂φ/∂y. Hence,from Equation (6.132),

1V∂φ

∂y= −λ ±

√λ 2 − 1 =

−1

λ ±√λ 2 − 1

, (6.134)

where the upper/lower signs correspond to λ > 0 and λ < 0, respectively. (The signsare chosen to ensure that ∂φ/∂y → 0 as |λ| → ∞.) Moreover, F = φ on B′OB, soEquation (6.133) implies that

φ =k V

(λ − cosα)2 . (6.135)

It follows thatdydλ=∂y

∂φ

dφ∂λ=

2 k(λ − cosα)3

(λ ±√λ 2 − 1

). (6.136)

Writing

λ =1 − cosα cosωcosα − cosω

, (6.137)

the points B, O, and B′ correspond to ω = 0, ω = α, and ω = π, respectively.Furthermore,

dλ(λ − cosα)3 = −

(cosα − cosω

sin4 α

)sinω dω, (6.138)

±√λ 2 − 1 =

sinα sinωcosα − cosω

. (6.139)

Hence, Equations (6.136)–(6.139) yield

dy = − 2 k

sin4 α(1 − cosα cosω + sinα sinω) sinω dω, (6.140)


y =k

sin4 α

[2 cosω + cosα sin2 ω + sinα sinω cosω + sinα

(π

2− ω

)], (6.141)

where we have chosen the constant of integration so as to ensure that the midpointof the plate lies at y = 0. We expect y = b when ω = 0, where b is the semi-width ofthe plate. We, thus, deduce, from the previous expression, that

k =2 b sin4 α

π sinα + 4. (6.142)


In particular, the stagnation point, O, at which ω = α, lies a distance

y(α) =2 b

π sinα + 4

[2 cosα (1 + sin2 α) + sinα

(π

2− α

)](6.143)

from the plate’s midpoint.Suppose that

λ = cosα + ζ, (6.144)

where |ζ | 1. It follows, from Equation (6.133), that

F =k Vζ 2 , (6.145)

and, from Equation (6.132), that

1V

dFdz′ 1 +

i ζsinα

, (6.146)

where z′ = (sinα + i cosα) z = x′ + i y′. Here, x′ and y′ are Cartesian coordinatesoriented such that the unperturbed flow is parallel to the x′-axis. The previous twoexpressions can be combined to give

z′

k ζ −2 − 2 i

ζ sinα. (6.147)

Now, ζ is real and negative on the free streamline BC, and real and positive on thefree streamline B′C′. (See Figure 6.18.) We conclude that, in the small-|ζ | limit, theequations of these streamlines can be written, in parametric form, as

x′ kζ 2 , (6.148)

y′ − 2 kζ sinα

. (6.149)

In other words, far downstream of the plate, the free streamlines, which form theboundaries of the cavity behind the plate, asymptote to the parabola

y′ 2 =

(4 k

sin2 α

)x′ =

(8 b sin2 α

π sinα + 4

)x′. (6.150)

From Bernoulli’s theorem, the net force per unit length exerted by the fluid onthe plate, which is obviously normal to the plate, and, therefore, directed parallel tothe x-axis, is

f =∫ b

−b(p − p0)x=0 dy =

12ρV 2

∫ b

−b

1 − v 2y

V 2

x=0

dy

=12ρV 2

∫ −1

+1

1 − v 2y

V 2

∂y∂φ dφdλ

dλ

= −ρ k V 2∫ −1

+1

(Vvy−vy

V

)dλ

(λ − cosα)3 , (6.151)


where use has been made of Equation (6.135). However, according to Equations (6.134)and (6.139),

Vvy−vy

V= V

∂y

∂φ− 1

V∂φ

∂y= ∓2

√λ 2 − 1 = − 2 sinα sinω

cosα − cosω. (6.152)

Thus, making use of Equation (6.138), we obtain

f =2 ρ k V 2

sin3 α

∫ π

0sin2 ω dω =

π ρ k V 2

sin3 α. (6.153)

Hence, it follows from Equation (6.142) that

f =(

2π sinαπ sinα + 4

)ρV 2 b. (6.154)

The drag, D, is the component of the force acting on the plate in the direction of theunperturbed flow: that is,

D = f sinα =(

2π sin2 α

π sinα + 4

)ρV 2 b. (6.155)

On the other hand, the lift, L, is the component of the force acting perpendicular tothe direction of the unperturbed flow. (See Section 5.8.) Thus,

L = f cosα =(

2π sinα cosαπ sinα + 4

)ρV 2 b. (6.156)

Finally, it can be seen, from a comparison between Equations (6.124), (6.129),(6.150), and (6.155), that there appears to be a connection between the drag forceexerted on the plate, and the shape of the parabola to which the cavity formed behindthe plate asymptotes. In fact,

DρV 2 = π lim

x→∞

(y 2

4 x

)cavity boundary

, (6.157)

where the unperturbed flow runs parallel to the x-axis. It turns out that this relationis a general one for two-dimensional flow past a body of arbitrary shape with anattached cavity at ambient pressure (Batchelor 2000).

6.10 Complex Line IntegralsConsider the line integral of some function F(z) of the complex variable taken (counter-clockwise) around a closed curve C in the complex plane:

J =∮

CF(z) dz. (6.158)


Because dz = dx + i dy, and writing F(z) = φ(x, y) + iψ(x, y), where φ(x, y) andψ(x, y) are real functions, it follows that J = Jr + i Ji, where

Jr =

∮C

(φ dx − ψ dy), (6.159)

Ji =

∮C

(ψ dx + φ dy). (6.160)

However, we can also write the previous expressions in the two-dimensional vectorform

Jr =

∮C

A · dr, (6.161)

Ji =

∮C

B · dr, (6.162)

where dr = (dx, dy), A = (φ, −ψ), and B = (ψ, φ). According to the curl theorem(see Section A.22), ∮

CA · dr =

∫S

(∇ × A)z dS , (6.163)∮C

B · dr =∫

S(∇ × B)z dS , (6.164)

where S is the region of the x-y plane enclosed by C. Hence, we obtain

Jr = −∫

S

(∂ψ

∂x+∂φ

∂y

)dS , (6.165)

Ji =

∫S

(∂φ

∂x− ∂ψ∂y

)dS . (6.166)

Let

J =∮

CF(z) dz, (6.167)

J′ =∮

C′F(z) dz, (6.168)

where C′ is a closed curve in the complex plane that completely surrounds the smallercurve C. Consider

∆J = J − J′. (6.169)

Writing ∆J = ∆Jr + i∆Ji, a direct generalization of the previous analysis reveals that

∆Jr = −∫

S

(∂ψ

∂x+∂φ

∂y

)dS , (6.170)

∆Ji =

∫S

(∂φ

∂x− ∂ψ∂y

)dS , (6.171)


where S is now the region of the x-y plane lying between the curves C and C′. Sup-pose that F(z) is well-behaved (i.e., finite, single-valued, and differentiable) through-out S . It immediately follows that its real and imaginary components, φ and ψ, re-spectively, satisfy the Cauchy-Riemann relations, (6.17)–(6.18), throughout S . How-ever, if this is the case then it is apparent, from the previous two expressions, that∆Jr = ∆Ji = 0. In other words, if F(z) is well-behaved throughout S then J = J′.

The circulation of the flow about some closed curve C in the x-y plane is defined

Γ =

∮C

(vx dx + vy dy) = −Re∮

C

dFdz

dz, (6.172)

where F(z) is the complex velocity potential of the flow, and use has been made ofEquation (6.35). Thus, the circulation can be evaluated by performing a line integralin the complex z-plane. Moreover, as is clear from the previous discussion, thisintegral can be performed around any loop that can be continuously deformed intothe loop C while still remaining in the fluid, and not passing over a singularity of thecomplex velocity, dF/dz.

6.11 Blasius TheoremConsider some flow pattern in the complex z-plane that is specified by the complexvelocity potential F(z). Let C be some closed curve in the complex z-plane. The fluidpressure on this curve is determined from Equation (6.41), which yields

P = p0 −12ρ

∣∣∣∣∣dFdz

∣∣∣∣∣ 2 . (6.173)

Let us evaluate the resultant force (per unit length), and the resultant moment (perunit length), acting on the fluid within the curve as a consequence of this pressuredistribution.

Consider a small element of the curve C, lying between x, y and x + dx, y + dy,which is sufficiently short that it can be approximated as a straight-line. Let P bethe local fluid pressure on the outer (i.e., exterior to the curve) side of the element.As illustrated in Figure 6.19, the pressure force (per unit length) acting inward (i.e.,toward the inside of the curve) across the element has a component P dy in the minusx-direction, and a component P dx in the plus y-direction. Thus, if X and Y arethe components of the resultant force (per unit length) in the x- and y-directions,respectively, then

dX = −P dy, (6.174)

dY = P dx. (6.175)

The pressure force (per unit length) acting across the element also contributes to a


y

dy

dx

dl

P dx

P dl

P dy

x

Figure 6.19Force acting across a short section of a curve.

moment (per unit length), M, acting about the z-axis, where

dM = x dY − y dX = P (x dx + y dy). (6.176)

Thus, the x- and y-components of the resultant force (per unit length) acting on theof the fluid within the curve, as well as the resultant moment (per unit length) aboutthe z-axis, are given by

X = −∮

CP dy, (6.177)

Y =∮

CP dx, (6.178)

M =∮

CP (x dx + y dy), (6.179)

respectively, where the integrals are taken (counter-clockwise) around the curve C.Finally, given that the pressure distribution on the curve takes the form (6.173), and


that a constant pressure obviously yields zero force and zero moment, we find that

X =12ρ

∮C

∣∣∣∣∣dFdz

∣∣∣∣∣ 2 dy, (6.180)

Y = −12ρ

∮C

∣∣∣∣∣dFdz

∣∣∣∣∣ 2 dx, (6.181)

M = −12ρ

∮C

∣∣∣∣∣dFdz

∣∣∣∣∣ 2 (x dx + y dy). (6.182)

Now, z = x + i y, and z = x − i y, where ¯ indicates a complex conjugate. Hence,dz = dx − i dy, and i dz = dy + i dx. It follows that

X − i Y =12

i ρ∮

C

∣∣∣∣∣dFdz

∣∣∣∣∣ 2 dz. (6.183)

However, ∣∣∣∣∣dFdz

∣∣∣∣∣ 2 dz =dFdz

dFdz

dz =dFdz

dF, (6.184)

where dF = dφ + i dψ and dF = dφ − i dψ. Suppose that the curve C corresponds toa streamline of the flow, in which case ψ = constant on C. Thus, dψ = 0 on C, andso dF = dF. Hence, on C,

dFdz

dF =dFdz

dF =(

dFdz

) 2

dz, (6.185)

which implies that

X − i Y =12

i ρ∮

C

(dFdz

) 2

dz. (6.186)

This result is known as the Blasius theorem, after Paul Blasius (1883–1970).Now, x dx + y dy = Re(z dz). Hence,

M = Re(−1

2ρ

∮C

∣∣∣∣∣dFdz

∣∣∣∣∣ 2 z dz), (6.187)

or, making use of an analogous argument to that employed previously,

M = Re

−12ρ

∮C

(dFdz

) 2

z dz

, (6.188)

In fact, Equations (6.186) and (6.188) hold even when ψ is not constant on thecurve C, as long as C can be continuously deformed into a constant-ψ curve withoutleaving the fluid or crossing over a singularity of (dF/dz) 2.

As an example of the use of the Blasius theorem, consider again the situation,discussed in Section 6.6, in which a line source of strength Q is located at (0, a), and


x

a

Q

C

C ′

y

Figure 6.20Source in the presence of a rigid boundary.

there is a rigid boundary at y = 0. As we have seen, the complex velocity in theregion y > 0 takes the form

dFdz= −Q

π

zz 2 + a 2 . (6.189)

Suppose that we evaluate the Blasius integral, (6.188), about the contour C shown inFigure 6.20. This contour runs along the boundary, and is completed by a semi-circlein the upper half of the z-plane. As is easily demonstrated, in the limit in which theradius of the semi-circle tends to infinity, the contribution of the curved section ofthe contour to the overall integral becomes negligible. In this case, only the straightsection of the contour contributes to the integral. Note that the straight section cor-responds to a streamline (because it is coincident with a rigid boundary). In otherwords, the contour C corresponds to a streamline at all constituent points that make afinite contribution to the Blasius integral, which ensures that C is a valid contour forthe application of the Blasius theorem. In fact, the Blasius integral specifies the netforce (per unit length) exerted on the whole fluid by the boundary. Observe, however,that the contour C can be deformed into the contour C′, which takes the form of asmall circle surrounding the source, without passing over a singularity of (dF/dz) 2.(See Figure 6.20.) Hence, we can evaluate the Blasius integral around C′ withoutchanging its value. Thus,

X − i Y =12

i ρ∮

C′

(dFdz

) 2

dz =12

i ρ(Qπ

)2 ∮C′

z 2

(z 2 + a 2) 2 dz, (6.190)

or

X − i Y =18

i ρ(Qπ

)2 ∮C′

[1

(z − i a)2 +2

(z + i a) (z − i a)+

1(z + i a)

]dz. (6.191)


Writing z = i a + ε e i θ, dz = i ε e i θ dθ, and taking the limit ε → 0, we find that

X − i Y =i ρQ 2

4π a. (6.192)

In other words, the boundary exerts a force (per unit length) F = −(ρQ 2/4π a) ey onthe fluid. Hence, the fluid exerts an equal and opposite force −F = (ρQ 2/4π a) ey onthe boundary. Of course, this result is consistent with Equation (6.47). Incidentally,it is easily demonstrated from Equation (6.188) that there is zero moment (about thez-axis) exerted on the boundary by the fluid, and vice versa.

Consider a line source of strength Q placed (at the origin) in a uniformly flow-ing fluid whose velocity is V = V0 (cos θ0, sin θ0). From Section 6.4, the complexvelocity potential of the net flow is

F(z) = − Q2π

ln z − V0 z e−i θ0 . (6.193)

The net force (per unit length) acting on the source (which is calculated by per-forming the Blasius integral around a large loop that follows streamlines, and thenshrinking the loop to a small circle centered on the source) is (see Exercise 6.1)

F = −ρQ V. (6.194)

This force acts in the opposite direction to the flow. Thus, an external force −F,acting in the same direction as the flow, must be applied to the source in order forit to remain stationary. In fact, the previous result is valid even in a non-uniformlyflowing fluid, as long as V is interpreted as the fluid velocity at the location of thesource (excluding the velocity field of the source itself).

Finally, consider a vortex filament of intensity Γ placed at the origin in a uni-formly flowing fluid whose velocity is V = V0 (cos θ0, sin θ0). From Section 6.4, thecomplex velocity potential of the net flow is

F(z) = iΓ

2πln z − V0 z e−i θ0 . (6.195)

The net force (per unit length) acting on the filament (which is calculated by per-forming the Blasius integral around a small circle centered on the filament) is (seeExercise 6.2)

F = ρΓV × ez. (6.196)

This force is directed at right-angles to the direction of the flow (in the sense obtainedby rotating V through 90 in the opposite direction to the filament’s direction ofrotation). Again, the previous result is valid even in a non-uniformly flowing fluid, aslong as V is interpreted as the fluid velocity at the location of the filament (excludingthe velocity field of the filament itself).


6.12 Exercises6.1 Demonstrate that a line source of strength Q (running along the z-axis) sit-

uated in a uniform flow of (unperturbed) velocity V (lying in the x-y plane)and density ρ experiences a force per unit length

F = −ρQ V.

6.2 Demonstrate that a vortex filament of intensity Γ (running along the z-axis)situated in a uniform flow of (unperturbed) velocity V (lying in the x-y plane)and density ρ experiences a force per unit length

F = −ρΓV × ez.

6.3 Show that two parallel line sources of strengths Q and Q′, located a perpen-dicular distance r apart, exert a radial force per unit length ρQ Q′/(2π r) onone another, the force being attractive if Q Q′ > 0, and repulsive if Q Q′ < 0.

6.4 Show that two parallel vortex filaments of intensities Γ and Γ′, located a per-pendicular distance r apart, exert a radial force per unit length ρΓ Γ′/(2π r)on one another, the force being repulsive if Γ Γ′ > 0, and attractive if Γ Γ′ <0.

6.5 A vortex filament of intensity Γ runs parallel to, and lies a perpendiculardistance a from, a rigid planar boundary. Demonstrate that the boundary ex-periences a net force per unit length ρΓ 2/(4π a) directed toward the filament.

6.6 Two rigid planar boundaries meet at right-angles. A line source of strengthQ runs parallel to the line of intersection of the planes, and is situated a per-pendicular distance a from each. Demonstrate that the source is subject to aforce per unit length

3√

2 ρQ 2

8π a

directed towards the line of intersection of the planes.

6.7 A line source of strength Q is located a distance b from an impenetrablecircular cylinder of radius a < b (the axis of the cylinder being parallel to thesource). Demonstrate that the cylinder experiences a net force per unit length

ρQ 2

2π ba 2

(b 2 − a 2)

directed toward the source.


6.8 A dipole line source consists of a line source of strength Q, running parallel tothe z-axis, and intersecting the x-y plane at (d/2) (cosα, sinα), and a parallelsource of strength −Q that intersects the x-y plane at (d/2)(− cosα, − sinα).Show that, in the limit d → 0, and Q d → D, the complex velocity potentialof the source is

F(z) =D e iα

2π z.

Here, D e iα is termed the complex dipole strength.

6.9 A dipole line source of complex strength D e iα is placed in a uniformlyflowing fluid of speed V0 whose direction of motion subtends a (counter-clockwise) angle θ0 with the x-axis. Show that, while no net force acts onthe source, it is subject to a moment (per unit length) M = ρD V0 sin(α− θ0)about the z-axis.

6.10 Consider a dipole line source of complex strength D1 e iα1 running along thez-axis, and a second parallel source of complex strength D2 e iα2 that inter-sects the x-y plane at (x, 0). Demonstrate that the first source is subject to amoment (per unit length) about the z-axis of

M =ρD1 D2

2π x 2 sin(α1 + α2),

as well as a force (per unit length) whose x- and y-components are

X = −ρD1 D2

π x 3 cos(α1 + α2),

Y =ρD1 D2

π x 3 sin(α1 + α2),

respectively. Show that the second source is subject to the same moment, butan equal and opposite force.

6.11 A dipole line source of complex strength D e iα runs parallel to, and is lo-cated a perpendicular distance a from, a rigid planar boundary. Show that theboundary experiences a force per unit length

ρD 2

8π a 3

acting toward the source.

6.12 Demonstrate that a conformal map converts a line source into a line sourceof the same strength, and a vortex filament into a vortex filament of the sameintensity.

6.13 Consider the conformal map

z = i c cot(ζ/2),


where z = x + i y, ζ = ξ + i η, and c is real and positive. Show that

x =c sinh η

cosh η − cos ξ,

y =c sin ξ

cosh η − cos ξ.

Demonstrate that ξ = ξ0, where 0 ≤ ξ0 ≤ π, maps to a circular arc of center(0, c cot ξ0), and radius c |cosec ξ0|, that connects the points (±c, 0), and liesin the region y > 0. Demonstrate that ξ = ξ0 + π maps to the continuation ofthis arc in the region y < 0. In particular, show that ξ = 0 maps to the region|x| > c on the x-axis, whereas ξ = π maps to the region |x| < c. Finally, showthat η = η0 maps to a circle of center (c coth η0, 0), and radius c |cosech| η0.

6.14 Consider the complex velocity potential

F(z) = −2 c i V0

ncot(ζ/n),

wherez = i c cot(ζ/2).

Here, V0, n, and c are real and positive. Show that

−dFdz=

4V0

n2

sin2(ζ/2)sin2(ζ/n)

.

Hence, deduce that the flow at |z| → ∞ is uniform, parallel to the x-axis, andof speed V0. Demonstrate that

ψ(ξ, η) = −2 V0 cn

sin(2 ξ/n)cosh(2 η/n) − cos(2 ξ/n)

.

Hence, deduce that the streamline ψ = 0 runs along the x-axis for |x| > c, butalong a circular arc connecting the points (±c, 0) for |x| < c. Furthermore,show that if 1 < n < 2 then this arc lies above the x-axis, and is of maximumheight

h = c[cos(π n/2) + 1

sin(π n/2)

],

but if 2 < n < 3 then the arc lies below the x-axis, and is of maximum depth

d = c[cos(π n/2) + 1| sin(π n/2)|

].

Hence, deduce that if 1 < n < 2 then the complex velocity potential underinvestigation corresponds to uniform flow of speed V0, parallel to a planarboundary that possesses a cylindrical bump (whose axis is normal to the flow)of height h and width 2 c, but if 2 < n < 3 then the potential corresponds to


flow parallel to a planar boundary that possesses a cylindrical depression ofdepth d and width 2 c. Show, in particular, that if n = 1 then the bump isa half-cylinder, and if n = 3 then the depression is a half-cylinder. Finally,demonstrate that the flow speed at the top of the bump (in the case 1 < n < 2),or the bottom of the depression (in the case 2 < n < 3) is

v =2 V0

n 2 [1 − cos(π n/2)] .

6.15 Show that z = cosh(π ζ/a) maps the semi-infinite strip 0 ≤ η ≤ a, ξ ≥ 0 inthe ζ-plane onto the upper half (y ≥ 0) of the z-plane. Hence, show that thestream function due to a line source of strength Q placed at ζ = (0, a/2), inthe rectangular region 0 ≤ η ≤ a, ξ ≥ 0 bounded by the rigid planes η = 0,ξ = 0, and η = a, is

ψ(ξ, η) =Q sinh(π ξ/a) sin(π η/a)

2π [sinh2(π ξ/a) + cos2(π η/a)].

6.16 Show that the complex velocity potential

F(z) = − a πV0

tanh(a π/z)

can be interpreted as that due to uniform flow of speed V0 over a cylindricallog of radius a lying on the flat bed of a deep stream (the axis of the log beingnormal to the flow). Demonstrate that the flow speed at the top of the logis (π 2/4) V0. Finally, show that the pressure difference between the top andbottom of the log is π 4 ρV 2

0 /32.

6.17 Show that the complex potential

F(z) = −V(ζ e−iα +

c 2

ζe iα),

where

z = ζ +l 2

ζ.

(l < c) represents uniform flow of unperturbed speed V , whose direction sub-tends a (counter-clockwise) angle α with the x-axis, around an impenetrableelliptic cylinder of major radius a = c + l 2/c, aligned along the x-axis, andminor radius b = c − l 2/c, aligned along the y-axis. Demonstrate that themoment per unit length (about the z-axis) exerted on the cylinder by the flowis

M = −π2ρV 2 (a 2 − b 2) sin(2α).

Hence, deduce that the moment acts to turn the cylinder broadside-on to theflow (i.e., α = π/2 is a dynamically stable equilibrium state), and that theequilibrium state in which the cylinder is aligned with the flow (i.e., α = 0)is dynamically unstable.


6.18 Consider a simply-connected region of a two-dimensional flow pattern boundedon the inside by the closed curve C1 (lying in the x-y plane), and on the out-side by the closed curve C2. Here, C1 and C2 do not necessarily correspondto streamlines of the flow. Demonstrate that the kinetic energy per unit length(in the z-direction) of the fluid lying between the two curves is

K =12ρ

[∫C2

φ dψ −∫

C1

φ dψ −∫

C3

[φ] dψ],

where ρ is the fluid mass density, φ the velocity potential, and ψ the streamfunction. Here, C3 is a curve that runs from C1 to C2, and [φ] denotes theamount by which the velocity potential increases as the argument of x + i yincreases by 2π.

6.19 Show that the complex potential

F(z) = −V(c 2 e iα − l 2 e−iα

)ζ −1

where

z = ζ +l 2

ζ.

(l < c) represents the flow pattern around an impenetrable elliptic cylinderof major radius a = c + l 2/c, aligned along the x-axis, and minor radiusb = c − l 2/c, aligned along the y-axis, moving with speed V , in a directionthat makes a counter-clockwise angle α with the x-axis, through a fluid thatis at rest far from the cylinder. Demonstrate that the kinetic energy per unitlength of the flow pattern is

K =12ρV 2 π

(a 2 sin2 α + b 2 cos2 α

),

where ρ is the fluid mass density. Hence, show that the cylinder’s added massper unit length is

madded = π(a 2 sin2 α + b 2 cos2 α

)ρ.

6.20 Demonstrate from Equation (6.110) that the equation of the free streamlineBC, in the case of a liquid jet emerging from a two-dimensional orifice ofsemi-width d formed by a gap between two semi-infinite plane walls thatsubtend an angle 2α, can be written parametrically as:

xd=

2π

∫ π/2

sin−1(λ)cos

(2απβ

)dβ

tan β

[1 +

2π

∫ π/2

0sin(2απβ

)dβ

tan β

]−1

,

y

d= 1 − 2

π

∫ π/2

sin−1(λ)sin(

2απβ

)dβ

tan β

[1 +

2π

∫ π/2

0sin(

2απβ

)dβ

tan β

]−1

,

where 0 ≤ λ ≤ 1. Here, the orifice corresponds to the plane x = 0, and the


flow a long way from the orifice is in the +y-direction. Show that for thecase of a two-dimensional Borda mouthpiece, α = π, the previous equationsreduce to

xd=

1π

[− ln(λ) − 1 + λ 2

],

y

d=

12+

1π

[sin−1(λ) + λ (1 − λ 2)1/2

].

Finally, show that the previous equations predict that the free streamline isre-entrant, with x/d attaining its minimum value −(1 − ln 2)/(2π) = −0.153when y/d = 3/4 + 1/(2π) = 0.909.

7Axisymmetric Incompressible Inviscid Flow

7.1 IntroductionThis chapter investigates axisymmetric, incompressible, inviscid flow. More infor-mation on this subject can be found in Batchelor 2000, and Milne-Thompson 2011.

7.2 Axisymmetric FlowA flow pattern is said to be axisymmetric when it is identical in every plane thatpasses through a certain straight-line. The straight-line in question is referred toas the symmetry axis. Let us set up a Cartesian coordinate system in which thesymmetry axis corresponds to the z-axis. The flow is most conveniently described interms of the cylindrical coordinates (, ϕ, z), or the spherical coordinates (r, θ, ϕ).Here, = (x 2 + y 2)1/2, ϕ = tan−1(y/x), r = (x 2 + y 2 + z 2)1/2, and θ = cos−1(z/r).(See Appendix C.) In particular, = r sin θ and z = r cos θ.

7.3 Stokes Stream FunctionConsider a fixed point A lying on the symmetry axis, and an arbitrary point P. Letus join A to P via two different curves, AQ1P and AQ2P, that both lie in the sameplane. (See Figure 7.1.) We shall refer to this plane as the meridian plane. Theposition of a given point in the meridian plane can be specified either in terms ofthe cylindrical coordinates (, z), or the spherical coordinates (r, θ). If the meridiancurves AQ1P and AQ2P rotate about the symmetry axis then closed surfaces areformed. Assuming that the flow pattern is incompressible, the flux of fluid from rightto left (in Figure 7.1) across the surface generated by AQ2P must match that in thesame direction across the surface generated by AQ1P. Let us denote the flux acrosseither of these surfaces by 2π ψ. Here, ψ is known as the Stokes stream function. Ifwe keep AQ1P fixed, and replace AQ2P by any other meridian curve joining A to P,then the value of ψ is clearly unaltered. Thus, the stream function ψ depends on the

193


v⊥

z

P

Q1

Q3

Q2

A B

R

R′

δs

Figure 7.1Definition of the Stokes stream function.

position of the arbitrary point P, and, possibly, on that of the fixed point A. In fact,if we take another fixed point B on the symmetry axis, and draw the meridian curveBQ3P, then the flux across the surface generated by BQ3P will be the same as thatacross the surface generated by AQ1P, because, by symmetry, there is no flow acrossAB. (See Figure 7.1.) It follows that the value of ψ does not depend on the particularfixed point that is used in its definition, as long as this point lies on the symmetryaxis. Hence, we conclude that the value of the stream function at P depends solelyon the position of P. Furthermore, if P lies on the axis then ψ = 0.

Consider two neighboring points, R and R′, lying in the meridian plane. (SeeFigure 7.1.) The flux from right to left across the surface generated by revolving anyline joining R to R′ about the symmetry axis is 2π ψR′ − 2π ψR. If the distance RR′

takes the infinitesimal value δs then we can write

2π (ψR′ − ψR) = 2πδs v⊥, (7.1)

where v⊥ is the normal flow velocity (from right to left) across the straight-line RR′

in the meridian plane. (See Figure 7.1.) It follows that

v⊥ =1

∂ψ

∂s. (7.2)

In particular, if we suppose that ds is, in turn, equal to d, dz, dr, and r dθ then weobtain the following expressions for the in-plane velocity components in cylindrical

Axisymmetric Incompressible Inviscid Flow 195

and spherical coordinates:

v =1

∂ψ

∂z, vz = −

1

∂ψ

∂, (7.3)

vr = −1

r sin θ∂ψ

r ∂θ, vθ =

1r sin θ

∂ψ

∂r. (7.4)

7.4 Axisymmetric Velocity Fields

According to the analysis of Appendix C, Equations (7.3) and (7.4) imply that

v = ∇ϕ × ∇ψ. (7.5)

When the fluid velocity is written in this form it becomes obvious that the incom-pressibility constraint ∇ · v = 0 is satisfied [because ∇ · (∇A × ∇B) ≡ 0—see Equa-tions (A.175) and (A.176)]. It is also clear that the Stokes stream function, ψ, isundefined to an arbitrary additive constant.

In fact, the most general expression for an axisymmetric incompressible flowpattern is

v = ∇ϕ × ∇ψ + Ωϕ ∇ϕ, (7.6)

where Ωϕ = Ωϕ(, z) is the angular velocity of flow circulating about the z-axis.(This follows because ∇ · (Ωϕ ∇ϕ) = 0 when ∂Ωϕ/∂ϕ = 0. See Appendix C.) Theprevious expression implies that v · ∇ψ = 0 (because ∇ψ · ∇ϕ = 0 when ∂ψ/∂ϕ =0). In other words, when plotted in the meridian plane, streamlines in a generalaxisymmetric flow pattern correspond to contours of ψ.

Making use of the vector identities (A.176) and (A.178), we can also write Equa-tion (7.6) in the form

v = −∇ × (ψ∇ϕ) + Ωϕ ∇ϕ. (7.7)

It follows from the identity (A.177) that

∇ × v = −∇[∇ · (ψ∇ϕ)] + ∇ 2(ψ∇ϕ) + ∇Ωϕ × ∇ϕ

= ∇ 2(ψ∇ϕ) + ∇Ωϕ × ∇ϕ, (7.8)

because ∇ · (ψ∇ϕ) = 0, by symmetry. Hence, the vorticity of a general axisymmetricflow pattern is written

ω = ∇Ωϕ × ∇ϕ + ωϕ eϕ, (7.9)


where ωϕ = ωϕ(, z), and [see Equations (C.52)–(C.54)]

ωϕ = eϕ · ∇ 2(ψ∇ϕ) = ∇ 2(ψ

)− ψ

3

=∂

∂

(1

∂ψ

∂

)+

1

∂ 2ψ

∂z 2

=1

r sin θ∂ 2ψ

∂r 2 +1r 3

∂

∂θ

(1

sin θ∂ψ

∂θ

). (7.10)

In the following, we shall concentrate on axisymmetric flow patterns in which thereis no circulation about the z-axis (i.e., Ωϕ = vϕ = 0).

7.5 Axisymmetric Irrotational Flow in Spherical CoordinatesIn an irrotational flow pattern, we can automatically satisfy the constraint ∇ × v = 0by writing

v = −∇φ. (7.11)

Suppose, however, that, in addition to being irrotational, the flow pattern is alsoincompressible: that is, ∇ · v = 0. In this case, Equation (7.11) yields

∇ 2φ = 0. (7.12)

In spherical coordinates, assuming that the flow pattern is axisymmetric, so that φ =φ(r, θ), the previous equation leads to (see Section C.4)

1r 2

∂

∂r

(r 2 ∂φ

∂r

)+

1r 2 sin θ

∂

∂θ

(sin θ

∂φ

∂θ

)= 0. (7.13)

Let us search for a separable solution of Equation (7.13) of the form

φ(r, θ) = R(r)Θ(θ). (7.14)

It is easily seen that

1R

ddr

(r 2 dR

dr

)= − 1

Θ sin θddθ

(sin θ

dΘdθ

), (7.15)

which can only be satisfied provided

ddr

(r 2 dR

dr

)− l (l + 1) R = 0, (7.16)

ddµ

[(1 − µ 2)

dΘdµ

]+ l (l + 1)Θ = 0, (7.17)


where µ = cos θ, and l (l + 1) is a constant. The solutions to Equation (7.17) that arewell behaved for µ in the range −1 to +1 are known as the Legendre polynomials,and are denoted the Pl(µ), where l is a non-negative integer (Jackson 1962). (If l isnon-integer then the solutions are singular at µ = ±1.) In fact,

Pl(µ) =(−1) l

2 l l!d l

dµ l

(1 − µ 2

)l. (7.18)

Hence,

P0(µ) = 1, (7.19)

P1(µ) = µ, (7.20)

P2(µ) =12

(3 µ 2 − 1

), (7.21)

P3(µ) =12

(5 µ 3 − 3 µ

), (7.22)

et cetera. The general solution of Equation (7.16) is a linear combination of r l andr−(l+1) factors. Thus, the general axisymmetric solution of Equation (7.12) is written

φ(r, θ) =∑

l=0,∞

[αl r l + βl r−(l+1)

]Pl(cos θ), (7.23)

where the αl and βl are arbitrary coefficients. It follows from Equations (7.4) that thecorresponding expression for the Stokes stream function is

ψ(r, µ) = β0 µ +∑

l=1,∞

(αl

l + 1r l+1 −

βl

lr −l) (

1 − µ 2) dPl

dµ, (7.24)

where µ = cos θ.

7.6 Uniform FlowConsider a uniform steady stream of velocity v = V ez. Consider the flux (in theminus z-direction) across a plane circle of radius that lies in the x-y plane, andwhose center coincides with the z-axis. From the definition of the Stokes streamfunction (see Section 7.3), we have 2π ψ(, z) = −π 2 V , or

ψ(, z) = −12

V 2. (7.25)

When expressed in terms of spherical coordinates, the previous expression yields

ψ(r, θ) = −12

V r 2 sin2 θ. (7.26)


Of course, uniform flow is irrotational [this is clear from a comparison of Equa-tions (7.10) and (7.25)], so we can also represent the flow pattern in terms of a ve-locity potential: that is (see Section 5.4),

φ(, z) = −V z, (7.27)

orφ(r, θ) = −V r cos θ. (7.28)

It follows, from the previous analysis, that the velocity field of a uniform stream,running parallel to the z-axis, can either be written v = ∇ϕ×∇ψ, with ψ specified byEquations (7.25)–(7.26), or v = −∇φ, with φ specified by Equations (7.27)–(7.28).

7.7 Point SourcesConsider a point source, coincident with the origin, that emits fluid isotropically atthe steady rate of Q volumes per unit time. By symmetry, we expect the associatedsteady flow pattern to be isotropic, and everywhere directed radially away from thesource. In other words,

v = v(r) er, (7.29)

where r is a spherical coordinate. Consider a spherical surface S of radius r whosecenter coincides with the source. In a steady state, the rate at which fluid crosses thissurface must be equal to the rate at which the source emits fluid. Hence,∫

Sv · dS = 4π r 2 vr(r) = Q, (7.30)

which implies that

vr(r) =Q

4π r 2 . (7.31)

Of course, vθ = 0.According to Equations (7.4), the Stokes stream function associated with a point

source at the origin is such that ψ = ψ(θ), and is obtained by integrating

vr =Q

4π r 2 = −1

r 2 sin θ∂ψ

∂θ. (7.32)

It follows thatψ =

Q4π

cos θ =Q4π

z( 2 + z 2)1/2 . (7.33)

It is clear, from a comparison of Equations (7.10) and (7.33), that the previouslyspecified flow pattern is irrotational. Hence, this pattern can also be derived from avelocity potential. In fact, by symmetry, we expect that φ = φ(r). The potential itselfis obtained by integrating

vr =Q

4π r 2 = −∂φ

∂r. (7.34)


θ2z

O

θ1

r r2

r1

P

θ

d/2 d/2

Figure 7.2A dipole source.

It follows that

φ =Q

4π r=

Q4π ( 2 + z 2)1/2 . (7.35)

7.8 Dipole Point SourcesConsider the flow pattern generated by point source of strength Q located on thesymmetry axis at z = +d/2, and a point source of strength −Q (i.e., a point sink)located on the symmetry axis at z = −d/2. It follows, by analogy with the analysisof the previous section, that the stream function and velocity potential at a generalpoint, P, lying in the meridian plane, are

ψ =Q4π

(cos θ2 − cos θ1) , (7.36)

and

φ =Q4π

(1r2−

1r1

), (7.37)

respectively. Here, r1, r2, θ1, and θ2 are defined in Figure 7.2.In the limit that the product D = Q d remains constant, while d → 0, we obtain a

so-called dipole point source. According to the sine rule of trigonometry,

r1

sin θ2=

r2

sin θ1=

dsin(θ2 − θ1)

. (7.38)


However, sin(θ2 − θ1) = 2 sin[(θ2 − θ1)/2] cos[(θ2 − θ1)/2], so we obtain

r1 − r2 =d (sin θ2 − sin θ1)

2 sin[(θ2 − θ1)/2] cos[(θ2 − θ1)/2]. (7.39)

In fact, sin θ2 − sin θ1 = 2 cos[(θ2 + θ1)/2] sin[(θ2 − θ1)/2], which leads to

r1 − r2 =d cos[(θ2 + θ1)/2]cos[(θ2 − θ1)/2]

. (7.40)

Thus, in the limit θ2 → θ1 → θ and r2 → r1 → r, we get

r1 − r2 → d cos θ. (7.41)

Hence, according to Equation (7.37),

φ =Q4π

(r1 − r2

r1 r2

)→ D

4πcos θ

r 2 . (7.42)

Equation (7.36) implies that

ψ =Q4π

(z − d/2

r2−

z + d/2r1

)=

Q4π

[z(

1r2−

1r1

)−

d2

(1r2+

1r1

)]. (7.43)

Thus, in the limit θ2 → θ1 → θ and r2 → r1 → r, we obtain

ψ→ Q4π

(d cos2 θ

r− d

r

)= − D

4πsin2 θ

r= − D

4π 2

( 2 + z 2)3/2 , (7.44)

where use has been made of Equation (7.41), as well as the fact that z = r cos θ.Figure 7.3 shows the stream function of a dipole point source located at the origin.

Incidentally, Equations (7.26), (7.28), (7.33), (7.35), (7.42), and (7.44) imply thatthe terms in the expansions (7.23) and (7.24) involving the constants β0, α1, and β1

correspond to a point source at the origin, uniform flow parallel to the z-axis, and adipole point source at the origin, respectively. Of course, the term involving α0 isconstant, and, therefore, gives rise to no flow.

7.9 Flow Past a Spherical ObstacleConsider the steady flow pattern produced when an impenetrable rigid spherical ob-stacle is placed in a uniformly flowing, incompressible, inviscid fluid. For instance,suppose that the radius of the sphere is a, and that its center coincides with the origin.Furthermore, let the unperturbed fluid velocity be of magnitude V , and be directedparallel to the z-axis. We expect the flow pattern to remain unperturbed very far awayfrom the sphere. In other words, we expect v(r, θ, ϕ)→ V ez as r/a→ ∞. Given thatthe fluid velocity field a large distance upstream of the sphere is irrotational (because


−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x

−1 −0.5 0 0.5 1z

Figure 7.3Contours of the stream function of a dipole source located at the origin.

a uniform flow pattern is automatically irrotational), it follows from the Kelvin cir-culation theorem that the velocity field remains irrotational as it is convected past thesphere. (See Section 4.14.) Hence, we can write v = −∇φ, where

∇ 2φ = 0 (7.45)

(because the fluid is incompressible.) The boundary conditions are

φ(r, θ, ϕ) = −V r cos θ as r → ∞, (7.46)

and∂φ(a, θ, ϕ)

∂r= 0. (7.47)

The latter constraint arises because the surface of the sphere is impenetrable, whichimplies that v · er = 0 at r = a.

Let us search for an axisymmetric solution of Equation (7.45) of the form

φ(r, θ) = −V r P1(cos θ) + β1 r −2 P1(cos θ). (7.48)

It can be seen, by comparison with Equation (7.23), that the previous expression def-initely solves Equation (7.45). Moreover, the expression also automatically satisfiesthe boundary condition (7.46) [because P1(cos θ) = cos θ]. The remaining boundarycondition, (7.47), yields β1 = −V/(2 a 3). Hence, we obtain

φ(r, θ) = −V a[

ra+

12

(ar

)2]cos θ, (7.49)


−5

−4

−3

−2

−1

0

1

2

3

4

5

x/a

−5 0 5

z/a

Figure 7.4Contours of the stream function generated by a spherical obstacle of radius a placedin the uniform flow field v = V ez.

or

vr(r, θ) = V(1 − a 3

r 3

)cos θ, (7.50)

vθ(r, θ) = −V(1 +

12

a 3

r 3

)sin θ. (7.51)

Because the solutions of Laplace’s equation, subject to well-posed boundary condi-tions, are unique (Riley 1974), we can be sure that the previous axisymmetric solu-tion is the most general solution to the problem.

It is clear, by comparison with Equations (7.28) and (7.42), that the velocity po-tential (7.49) is the superposition of that associated with uniform flow with velocityV , parallel to the z-axis, and a dipole point source of strength D = −2πV a 3, locatedat the origin. Thus, making use of Equations (7.26) and (7.44), the associated streamfunction takes the form

ψ = −12

V r 2 sin2 θ

(1 −

a 3

r 3

)= −

12

V 2[1 −

a 3

( 2 + z 2)3/2

]. (7.52)

Figure 7.4 show the contours of this stream function.Bernoulli’s theorem yields (see Section 4.3)

pρ+

12v 2 =

p0

ρ+

12ρV 2, (7.53)


where ρ is the uniform fluid mass density, and p0 the fluid pressure at infinity. Thus,making use of Equations (7.50) and (7.51), the pressure distribution on the surfaceof the sphere can be written

p(a, θ, ϕ) = p0 −116ρV 2 +

916ρV 2 cos(2 θ). (7.54)

The net force exerted on the sphere by the fluid has the Cartesian components

Fx = −∮

p(a, θ, ϕ) a 2 sin θ cosϕ dΩ, (7.55)

Fy = −∮

p(a, θ, ϕ) a 2 sin θ sinϕ dΩ, (7.56)

Fz = −∮

p(a, θ, ϕ) a 2 cos θ dΩ, (7.57)

where the integrals are over all solid angle. Thus, it follows that

Fx = Fy = Fz = 0. (7.58)

In other words, the fluid exerts zero net force on the sphere, in accordance withd’Alembert’s paradox. (See Section 4.5.)

7.10 Motion of a Submerged SphereConsider a situation in which an impenetrable rigid sphere of radius a is movingthrough an incompressible, inviscid fluid at the time dependent velocity V = Vz(t) ez.Assuming that the fluid and sphere were both initially stationary, it follows that thefluid velocity field was initially irrotational. Thus, according to the Kelvin circulationtheorem, the fluid velocity field remains irrotational when the sphere starts to move.Hence, we can write v = −∇φ, and

∇ 2φ = 0 (7.59)

(because the fluid is incompressible).Let x, y, z be Cartesian coordinates in the initial rest frame of the fluid, and let r,

θ, ϕ be spherical coordinates in a frame of reference that co-moves with the sphere.In the following, all calculations are performed in the rest frame. We expect the flowpattern set up around the sphere to be axisymmetric (i.e., independent of ϕ.) We alsoexpect the fluid a long way from the sphere to remain stationary. In other words,

φ(r, θ, t)→ 0 as r → ∞. (7.60)

Moreover, because the sphere is impenetrable, we require that

V · er = v · er |r=a , (7.61)


or∂φ

∂r

∣∣∣∣∣r=a= −Vz cos θ. (7.62)

It is easily demonstrated that the solution to Equation (7.59), subject to the boundaryconditions (7.60) and (7.62), is

φ(r, θ, t) =12

Vz(t)a 3

r 2 cos θ. (7.63)

Hence,

vr(r, θ, t) = Vz(t)a 3

r 3 cos θ, (7.64)

vθ(r, θ, t) =12

Vz(t)a 3

r 3 sin θ. (7.65)

The general form of Bernoulli’s theorem, (4.96), which applies to an irrotationalflow field, yields

pρ+

12v 2 + g r cos θ − ∂φ

∂t=

p0

ρ, (7.66)

where ρ is the uniform fluid mass density, p0 the fluid pressure at infinity, and it isassumed that fluid and cylinder are both situated in a gravitational field of uniformacceleration −g ez. Thus, the pressure distribution at the surface of the sphere can bewritten

p(a, θ, t) = p0 −12ρ v 2

∣∣∣∣∣r=a− ρ g a cos θ + ρ

∂φ

∂t

∣∣∣∣∣r=a− ρ V · ∇φ|r=a . (7.67)

The final term on the right-hand side of the previous equation arises because

∂

∂t

∣∣∣∣∣x,y,z=∂

∂t

∣∣∣∣∣r,θ− V · ∇. (7.68)

Hence, we obtain

p(a, θ, ϕ, t) = p0−1

16ρV 2

z +9

16ρV 2

z cos(2 θ)−ρ g a cos θ+12ρ a

dVz

dtcos θ. (7.69)

The net force exerted on the sphere by the fluid is specified by Equations (7.55)–(7.57). It follows that

Fx = Fy = 0, (7.70)

andFz = m′ g − 1

2m′

dVz

dr, (7.71)

where m′ = (4π/3) a 3 ρ is the mass of the fluid displaced by the sphere. The equationof vertical (i.e., in the z-direction) motion of the sphere is thus,

mdVz

dt= Fz − m g, (7.72)


where m is the mass of the sphere. Hence,(m +

12

m′)

dVz

dt= −m g + m′ g. (7.73)

It follows that the sphere moves under the action of its weight, and the buoyancyforce exerted on it by the fluid (i.e., the first and second terms, respectively, on theright-hand side of the previous equation), as if it had the virtual mass m+ (1/2) m′. Inother words, as it moves, the sphere effectively entrains an added mass of fluid equalto half of the displaced fluid mass. The net vertical acceleration of the sphere can bewritten

az = −(

s − 1s + 1/2

)g, (7.74)

where s = m/m′ is the sphere’s specific gravity. For the case of a bubble of gasin a liquid, we expect 0 < s 1, because the gas is inevitably much less densethan the liquid. It follows that the bubble accelerates vertically upward at twice theacceleration due to gravity.

The origin of the sphere’s added mass is easily explained. According to Equa-tions (7.64) and (7.65), the total kinetic energy of the fluid surrounding the sphereis

Kfluid =

∫fluid

12ρ v 2 dV

=12ρV 2

∫ ∞

a

∫ π

0

(ar

)6 (cos2 θ +

14

sin2 θ

)2π r 2 sin θ dr dθ, (7.75)

which reduces toKfluid =

14

m′ V 2. (7.76)

However, the kinetic energy of the sphere is

Ksphere =12

m V 2. (7.77)

Thus, the total kinetic energy is

K = Kfluid + Ksphere =12

(m +

12

m′)

V 2. (7.78)

In other words, the kinetic energy of the fluid surrounding the sphere can be ac-counted for by supposing that a mass m′/2 of the fluid co-moves with the sphere, andthat the remainder of the fluid remains stationary.

7.11 Conformal MapsAs we saw in Section 6.7, conformal maps are extremely useful in the theory oftwo-dimensional, irrotational, incompressible flows. It turns out that such maps also


have applications to the theory of axisymmetric, irrotational, incompressible flows.Consider the general coordinate transformation

z + i = f (ξ + i η), (7.79)

where f is an analytic function. The Cauchy-Riemann relations (see Section 6.3)yield

∂z∂ξ=∂

∂η, (7.80)

∂

∂ξ= −

∂z∂η. (7.81)

It follows, from the previous two expressions, that ∇ξ · ∇η = 0. In other words, ξ andη are orthogonal coordinates in the meridian plane. [Incidentally, we are assumingthat (ξ, η, ) are a right-handed set of coordinates.] Furthermore, it can also beshown from the Cauchy-Riemann relations that

hξ = hη =

( ∂z∂ξ

)2+

(∂

∂ξ

)2 1/2

=

( ∂z∂η

)2+

(∂

∂η

)2 1/2

, (7.82)

where hξ = |∇ξ| −1 and hη = |∇η| −1. Writing the flow velocity in terms of a velocitypotential, so that v = −∇φ, or, alternatively, in terms of a Stokes stream function, sothat v = ∇ϕ × ∇ψ, we get

vξ = −1hξ

∂φ

∂ξ= −

1 hη

∂ψ

∂η, (7.83)

vη = −1hη

∂φ

∂η=

1 hξ

∂ψ

∂ξ. (7.84)

Of course, writing the velocity field in terms of a Stokes stream function ensuresthat the field is incompressible, which also implies that ∇ 2φ = 0. The additionalrequirement that the field be irrotational yields ωϕ = 0. Making use of the analysisof Appendix C, this requirement reduces to

∂(hη vη)∂ξ

−∂(hξ vξ)∂η

= 0, (7.85)

or∂

∂ξ

(1

∂ψ

∂ξ

)+∂

∂η

(1

∂ψ

∂η

)= 0. (7.86)

Let ξ = ξ0 represent the surface of an axisymmetric solid body moving withvelocity V = V ez through an incompressible irrotational fluid that is at rest a longway from the body. Let the fluid occupy the region ξ0 < ξ < ∞, where ξ → ∞ farfrom the body. (See Figure 7.5.) Let η be an angular coordinate such that η = 0 onthe positive z-axis, and η = π on the negative z-axis. The fact that the fluid is at rest


η = 0η = π AF D

ξ = ∞

ξ = ξ0

zC

B

E

V

Figure 7.5An axisymmetric solid body moving through an incompressible irrotational fluid.

at infinity implies that ψ asymptotes to a constant a long way from the body. Withoutloss of generality, we can chose this constant to be zero. Thus, one constraint on thesystem is that

ψ(ξ → ∞, η) = 0. (7.87)

The appropriate constraint at the surface of the body is that

vξ = V · eξ, (7.88)

where eξ = ∇ξ/|∇ξ|. However, we can write V = ∇ϕ × ∇ψ0, where ψ0(, z) =−(1/2) V 2. (See Section 7.6.) Hence, from Equation (7.83), the previous con-straint becomes

∂ψ

∂η=∂ψ0

∂η(7.89)

when ξ = ξ0. Integrating, making use of the constraint (7.87) (which implies thatψ = 0 on the z-axis, where ψ is constant, by symmetry), we obtain

ψ(ξ0, η) = −12

V 2. (7.90)

We can also set the velocity potential, φ, to zero at infinity, and on the z-axis.The total kinetic energy of the fluid surrounding the moving body is

K =12ρ

∫v 2 dV =

12ρ

∫(∇φ)2 dV =

12ρ

∫∇ · (φ∇φ) dV, (7.91)

where we have made use of the fact that ∇ 2φ = 0. Here, ρ is the fluid mass density,


and dV is an element of the volume obtained by rotating the area ABCDEFA, shownin Figure 7.5, about the z-axis. Making use of the divergence theorem, we obtain

K =12ρ

∮φ∇φ · n 2π ds = −1

2

∫ABC

φ1hξ

∂φ

∂ξ2π ds. (7.92)

where ds is an element of the curve ABCDEFA, and n is an outward pointing, unit,normal vector to the area ABCDEFA. Here, we have made use of the fact that thevelocity potential is zero at infinity (i.e., on DEF), and also on the z-axis (i.e., on CDand FA). On the curve ABC, we can write ds = hη dη. Furthermore, it follows fromEquations (7.82) and (7.83) that hξ = hη, and ∂φ/∂ξ = −1 ∂ψ/∂η. Thus,

K = −π ρ∫ π

0φ∂ψ

∂ηdη∣∣∣∣∣ξ=ξ0

, (7.93)

or

K = −π ρ∫ η=π

η=0φ dψ

∣∣∣∣∣∣ξ=ξ0

. (7.94)

As a simple example, consider the conformal map

z + i = c exp (ξ + i η), (7.95)

where c is real and positive. It follows that

z = c e ξ cos η, (7.96)

= c e ξ sin η, (7.97)

which implies thatz 2 + 2 = r 2, (7.98)

wherer(ξ) = c e ξ. (7.99)

Thus, the constant-ξ surfaces are concentric spheres of radius r(ξ). If we set

a = c e ξ0 (7.100)

then the problem reduces to that of a sphere, of radius a, moving through a fluid thatis at rest at infinity. This problem was solved, via different methods, in Section 7.10.The constraints (7.87) and (7.90) yield

ψ(ξ → ∞, η) = 0, (7.101)

ψ(ξ0, η) = −12

V c2 e 2 ξ0 sin2 η, (7.102)

where use has been made of Equation (7.97). This suggests that we can write

ψ(ξ, η) = F(ξ) sin2 η. (7.103)


Substitution into the governing equation, (7.86), gives

ddξ

(e−ξ

dFdξ

)− 2 e−ξ F = 0, (7.104)

whose most general solution is

F(ξ) = A e 2 ξ + B e−ξ. (7.105)

The constraints (7.101) and (7.102) yield

A = 0, (7.106)

B = −12

V c 2 e 3 ξ0 , (7.107)

respectively. Thus, we obtain

ψ(ξ, η) = −12

V a 3 sin2 η

c e ξ. (7.108)

Now, from Equations (7.84) and (7.97),

∂φ

∂η= − 1

∂ψ

∂ξ= −1

2V a 3 sin η

(c e ξ)2 , (7.109)


φ(ξ, η) =12

V a 3 cos η(c e ξ)2 . (7.110)

Note that the previous expression is formally the same as expression (7.63), as longas we make the identifications V → Vz(t), c e ξ → r, and η→ θ.

On the surface of the sphere, ξ = ξ0, we obtain

ψ(ξ0, η) = −12

V a 2 sin2 η, (7.111)

φ(ξ0, η) =12

V a cos η. (7.112)

Thus,

K = −π ρ∫ η=π

η=0φ dψ

∣∣∣∣∣∣ξ=ξ0

=12π a 3 ρV 2

∫ 1

−1µ 2 dµ =

π

3a 3 ρV 2. (7.113)

As is clear from the analysis of Section 7.10, the sphere’s added mass can be written

madded =K

(1/2) V 2 =2π3

a 3 ρ. (7.114)

Hence, we arrive at the standard result that the added mass is half the displaced mass[i.e., half of (4π/3) a 3 ρ].


7.12 Flow Around a Submerged Oblate SpheroidConsider the conformal map

z + i = c sinh(ξ + i η), (7.115)


z = c sinh ξ cos η, (7.116)

= c cosh ξ sin η. (7.117)

Let

a = c cosh ξ0, (7.118)

b = c sinh ξ0. (7.119)

Thus, in the meridian plane, the curve ξ = ξ0 corresponds to the ellipse(

a

)2+

( zb

)2= 1. (7.120)

We conclude that the surface ξ = ξ0 is an oblate spheroid (i.e., the three-dimensionalsurface obtained by rotating an ellipse about a minor axis) of major radius a andminor radius b. The constraints (7.87) and (7.90) yield

ψ(ξ → ∞, η) = 0, (7.121)

ψ(ξ0, η) = −12

V c 2 cosh2 ξ0 sin2 η, (7.122)

respectively. Setting ψ(ξ, η) = F(ξ) sin2 η, and substituting into the governing equa-tion, (7.86), we obtain

ddξ

(1

cosh ξdFdξ

)− 2

cosh ξF = 0, (7.123)


ddξ

(cosh ξ

dFdξ− 2 sinh ξ F

)= 0. (7.124)

On integration, we get

cosh ξdFdξ− 2 sinh ξ F = B, (7.125)


ddξ

(F

cosh2 ξ

)=

B

cosh3 ξ. (7.126)


It follows that

F(ξ) = −B cosh2 ξ

∫ ∞

ξ

dξ′

cosh3 ξ′

= −12

B cosh2 ξ

[sinh ξ′

cosh2 ξ′− tan−1

(1

sinh ξ′

)]∞ξ

=12

B[sinh ξ − cosh2 ξ tan−1

(1

sinh ξ

)], (7.127)

where use has been made of the constraint (7.121). Let e = (1 − b 2/a 2)1/2 be theeccentricity of the spheroid. Thus, cosh ξ0 = 1/e, sinh ξ0 = (1 − e 2)1/2/e, c = e a,and tan−1(1/ sinh ξ0) = sin−1 e. The constraint (7.122) yields

−12

V a 2 = F(ξ0) =12

B[(1 − e 2)1/2

e− sin−1 e

e 2

], (7.128)

or

B = −[

V a 2 e 2

e (1 − e 2)1/2 − sin−1 e

]. (7.129)

Hence,

ψ(ξ, η) = −12

V a 2 e 2[sinh ξ − cosh2 ξ tan−1(1/ sinh ξ)

e (1 − e 2)1/2 − sin−1 e

]sin2 η. (7.130)

Finally, from Equation (7.84),

∂φ

∂η= − 1

c cosh ξ sin η∂ψ

∂ξ, (7.131)


φ(ξ, η) = −V a e[1 − sinh ξ tan−1(1/ sinh ξ)

e (1 − e 2)1/2 − sin−1 e

]cos η. (7.132)

It is easily demonstrated that

ψ(ξ0, η) = −12

V a 2 sin2 η, (7.133)

and

φ(ξ0, η) = V a[e − (1 − e 2)1/2 sin−1 e

sin−1 e − e (1 − e 2)1/2

]cos η. (7.134)

Thus,

K = −π ρ∫ η=π

η=0φ dψ

∣∣∣∣∣∣ξ=ξ0

=23π ρ a 3 V 2

[e − (1 − e 2)1/2 sin−1 e

sin−1 e − e (1 − e 2)1/2

]. (7.135)


−3

−2

−1

0

1

2

3

x/a

−3 −2 −1 0 1 2 3

z/a

Figure 7.6Contours of the Stokes stream function for the case of a disk of radius a, lying in thex-y plane, and placed in a uniform, incompressible, irrotational flow directed parallelto the z-axis.

It follows that the added mass of the spheroid is

madded =43π ρ a 3 G(e), (7.136)

where

G(e) =e − (1 − e 2)1/2 sin−1 e

sin−1 e − e (1 − e 2)1/2(7.137)

is a monotonic function that varies between 1/2 when e = 0 and 2/π when e = 1.In the limit e → 1, our spheroid asymptotes to a thin disk of radius a moving

through the fluid in the direction perpendicular to its plane. Expressions (7.130),(7.132), and (7.136) yield

ψ(ξ, η) = −12

V a 2 sin2 η, (7.138)

φ(ξ, η) =2π

V a cos η, (7.139)

andmadded =

83ρ a 3, (7.140)

respectively. Here, z = sinh ξ cos η and = cosh ξ sin η. It follows that, in the


instantaneous rest frame of the disk,

ψ(, z) = −14

V[a 2 + z 2 + 2 −

√(a 2 + z 2 + 2)2 − 4 a 2 2

]+

12

V 2. (7.141)

This flow pattern, which corresponds to that of a thin disk placed in a uniform flowperpendicular to its plane, is visualized in Figure 7.6.

7.13 Flow Around a Submerged Prolate SpheroidConsider the conformal map

z + i = c cosh(ξ + i η), (7.142)


z = c cosh ξ cos η, (7.143)

= c sinh ξ sin η. (7.144)

Let

a = c cosh ξ0, (7.145)

b = c sinh ξ0. (7.146)

Thus, in the meridian plane, the curve ξ = ξ0 corresponds to the ellipse(

b

)2+

( za

)2= 1. (7.147)

We conclude that the surface ξ = ξ0 is an prolate spheroid (i.e., the three-dimensionalsurface obtained by rotating an ellipse about a major axis) of major radius a andminor radius b. The constraints (7.87) and (7.90) yield

ψ(ξ → ∞, η) = 0, (7.148)

ψ(ξ0, η) = −12

V c 2 sinh2 ξ0 sin2 η, (7.149)

respectively. Setting ψ(ξ, η) = F(ξ) sin2 η, and substituting into the governing equa-tion, (7.86), we obtain

ddξ

(1

sinh ξdFdξ

)− 2

sinh ξF = 0. (7.150)

The solution that satisfied the constraint (7.148) is

F(ξ) = −12

B(cosh ξ + sinh2 ξ ln

[tanh(ξ/2)

]). (7.151)


Let e = (1 − b 2/a 2)1/2 be the eccentricity of the spheroid. Thus, cosh ξ0 = 1/e,sinh ξ0 = (1 − e 2)1/2/e, c = e a, and tanh(ξ0/2) = [(1 − e)/(1 + e)]1/2. The constraint(7.149) yields

B =V a 2 e 2

e (1 − e 2)−1 + (1/2) ln [(1 − e)/(1 + e)]. (7.152)

Hence,

ψ(ξ, η) = −12

V b 2

cosh ξ + sinh2 ξ ln[tanh(ξ/2)

]e−1 + e−2 (1 − e 2) (1/2) ln [(1 − e)/(1 + e)]

sin2 η. (7.153)

Finally, from Equation (7.84),

∂φ

∂η= − 1

c sinh ξ sin η∂ψ

∂ξ, (7.154)


φ(ξ, η) = −V a

1 + cosh ξ ln[tanh(ξ/2)

](1 − e 2)−1 + (1/2 e) ln [(1 − e)/(1 + e)]

cos η. (7.155)


ψ(ξ0, η) = −12

V b 2 sin2 η, (7.156)

and

φ(ξ0, η) = −V a

1 + (1/2 e) ln [(1 − e)/(1 + e)](1 − e 2)−1 + (1/2 e) ln [(1 − e)/(1 + e)]

cos η. (7.157)

Thus,

K = −π ρ∫ η=π

η=0φ dψ

∣∣∣∣∣∣ξ=ξ0

= −23π ρ a b 2 V 2

1 + (1/2 e) ln [(1 − e)/(1 + e)]

(1 − e 2)−1 + (1/2 e) ln [(1 − e)/(1 + e)]

. (7.158)

It follows that the added mass of the spheroid is

madded =43π ρ a b 2 G(e), (7.159)

where

G(e) = −

1 + (1/2 e) ln [(1 − e)/(1 + e)](1 − e 2)−1 + (1/2 e) ln [(1 − e)/(1 + e)]

. (7.160)

is a monotonic function that takes the value 1/2 when e = 0, and asymptotes to(1 − e) ln[1/(1 − e)] as e→ 1.


7.14 Exercises7.1 Demonstrate that the streamlines that pass through a stagnation point in an

incompressible, irrotational, axisymmetric flow pattern cross at right angles.

7.2 A point source of strength Q is placed at the origin in a uniform stream ofincompressible fluid of velocity v = V ez. Show that the stream function ofthe resultant flow pattern is

ψ(r, θ) = −12

V r 2 sin2 θ +Q4π

cos θ, (7.161)

where r, θ, ϕ are spherical coordinates. Demonstrate that the flow patternpossesses a single stagnation point at r = a, θ = π, where

a =( Q4πV

)1/2,

and that the streamline, other than = 0, that passes through this stagnationpoint satisfies

ra=

1| sin(θ/2)|

.

The flow pattern (7.161) can be reinterpreted as that which results whena blunt obstacle lying to the right of the previously specified streamline isplaced in a uniform stream of velocity v = V ez. Show that the obstacle inquestion has the asymptotic (i.e., as z → ∞) thickness 4 a. Demonstrate thatthe pressure distribution over the surface of the obstacle is

p = p0 + ρV 2 sin2(θ/2)[32

sin2(θ/2) − 1],

where ρ is the fluid mass density, and p0 the pressure at infinity. Show that themaximum pressure, p = p0 + (1/2) ρV 2, on the surface occurs at θ = π, andthat the minimum pressure, p = p0−(1/6) ρV 2, occurs at θ = 2 sin−1(1/

√3).

Finally, demonstrate that these are, respectively, the maximum and minimumpressures attained in the whole flow pattern.

7.3 A and B are point sources of strengths Q and −Q′, respectively, in an infi-nite incompressible fluid. Here, Q > Q′ > 0. Show that the equation of astreamline is

Q cos θ − Q′ cos θ′ = constant,

where θ, θ′ are the angles that AP, BP make with AB. P being a general point.In addition, show that the cone defined by the equation

cos θ = 1 −2 Q′

Q

divides the streamlines issuing from A into two sets, one extending to infinity,and the other terminating at B. (Milne-Thompson 2011.)


7.4 A and B are point sources of equal strength Q (where Q > 0) located at(, z) = (0, −a) and (0, +a), respectively in a uniform stream of incom-pressible fluid of velocity v = V ez. Show that the stream function of theresultant flow pattern is

ψ = −12

V 2 +Q4π

(cos θ1 − cos θ2), (7.162)

where θ, θ′ are the angles that AP, BP make with AB. P = (, z) being ageneral point. Demonstrate that the flow pattern has two stagnation points,located at C = (0, −l) and D = (0, +1), where

(l 2 − a 2)2 = 2 a b 2 l,

and b 2 = Q/(2πV). Show that the streamline (other that = 0) that passesthrough these points satisfies

2 = b 2 (cos θ1 − cos θ2),

and also passes through the points E = (h, 0) and F = (−h, 0), where

h 2

b 2 =2 a

(h 2 + a 2)1/2 .

The flow pattern (7.162) can be reinterpreted as that which results when anaxisymmetric solid body of oval cross-section CEDF, lying inside the previ-ously specified streamline, is placed in a uniform stream of velocity v = V ez.Such an obstacle is known as a Rankine solid.

7.5 Verify that

ψ =( Ar 2 cos θ + B r 2

)sin2 θ

is a possible stream function for an axisymmetric, incompressible, irrota-tional flow pattern, and find the corresponding velocity potential. (Milne-Thompson 2011.)

7.6 A sphere, a great depth below the surface of an incompressible fluid, is pro-jected with velocity V at an inclination of 45 to the horizontal. If the densityof the sphere is twice that of the fluid, prove that the greatest height above thepoint of projection attained by the sphere is 5 V 2/(8 g). (Milne-Thompson2011.)

7.7 A sphere of radius a is placed in an incompressible fluid flowing with theuniform velocity V = V ez. Show that the streamlines of the resultant flowpattern satisfy (

a 3 − r 3) sin2 θ

r= constant,

where r, θ are spherical coordinates whose origin coincides with the center


of the sphere. If the sphere is divided into two halves by a diametral planelying in the x-y plane, show that the resultant force between the two parts isless than it would have been if the fluid were at rest, the pressure at infinityremaining the same, by an amount π ρ a 2 V 2/16, where ρ is the fluid density.(Milne-Thompson 2011.)

7.8 A sphere of radius a is moving along the z-axis through an incompressiblefluid with the variable speed V . Show that the pressure on the surface of thesphere is least at the intersection of the surface with the plane

z = − 2 a 2

9 V 2

dVdt,

the center of the sphere being instantaneously at the origin. (Milne-Thompson2011.)

7.9 Consider the conformal map

z + i = c sinh(ξ + i η).

where c is real and positive. Show that the stream function

ψ(ξ, η) =12

c 2 V cos η

can be interpreted as that of incompressible irrotational flow, with mean speedV , through a circular hole of radius c in an infinite plane wall, correspondingto z = 0.

8Incompressible Boundary Layers

8.1 IntroductionPreviously, in Sections 4.5, 5.8 and 7.9, we saw that a uniformly flowing incompress-ible fluid that is modeled as being completely inviscid is incapable of exerting a dragforce on a rigid stationary obstacle placed in its path. This result—which is knownas d’Alembert’s paradox—is surprising because, in practice, a stationary obstacleexperiences a significant drag when situated in such a fluid, even in the limit that theReynolds number tends to infinity (which corresponds to the inviscid limit). In thischapter, we shall attempt to reconcile these two results by introducing the conceptof a boundary layer. This is a comparatively thin layer that covers the surface ofan obstacle placed in a high Reynolds number incompressible fluid. Viscosity is as-sumed to have a significant effect on the flow inside the layer, but a negligible effecton the flow outside. For the sake of simplicity, we shall restrict our discussion tothe two-dimensional boundary layers that form when a high Reynolds number fluidflows transversely around a stationary obstacle of infinite length and uniform cross-section. More information on such boundary layers can be found in Batchelor 2000and Schlichting 1987.

8.2 No Slip ConditionWe saw previously (for instance, in Section 5.8) that when an inviscid fluid flowsaround a rigid stationary obstacle then the normal fluid velocity at the surface ofthe obstacle is required to be zero. However, in general, the tangential velocity isnon-zero. In fact, if the fluid velocity field is both incompressible and irrotationalthen it is derivable from a stream function that satisfies Laplace’s equation. (SeeSection 5.2.) It is a well-known property of Laplace’s equation that we can eitherspecify the solution itself, or its normal derivative, on a bounding surface, but wecannot specify both these quantities simultaneously (Riley 1974). The constraintof zero normal velocity is equivalent to the requirement that the stream functiontake the constant value zero (say) on the surface of the obstacle. Hence, the normalderivative of the stream function, which determines the tangential velocity, cannotalso be specified at this surface, and is, in general, non-zero.

219


In reality, all physical fluids possess finite viscosity. Moreover, when a viscousfluid flows around a rigid stationary obstacle both the normal and the tangential com-ponents of the fluid velocity are found to be zero at the obstacle’s surface. The addi-tional constraint that the tangential fluid velocity be zero at a rigid stationary bound-ary is known as the no slip condition, and is ultimately justified via experimentalobservations.

The concept of a boundary layer was first introduced into fluid mechanics byLudwig Prandtl (1875–1953) in order to account for the modification to the flowpattern of a high Reynolds number irrotational fluid necessitated by the imposition ofthe no slip condition on the surface of an impenetrable stationary obstacle. Accordingto Prandtl, the boundary layer covers the surface of the obstacle, but is relatively thinin the direction normal to this surface. Outside the layer, the flow pattern is thesame as that of an idealized inviscid fluid, and is thus generally irrotational. Thisimplies that the normal fluid velocity is zero on the outer edge of the layer, whereit interfaces with the irrotational flow, but, in general, the tangential velocity is non-zero. However, the no slip condition requires the tangential velocity to be zero onthe inner edge of the layer, where it interfaces with the rigid surface. It follows thatthere is a very large normal gradient of the tangential velocity across the layer, whichimplies the presence of intense internal vortex filaments trapped within the layer.Consequently, the flow within the layer is not irrotational. In the following, we shallattempt to make the concept of a boundary layer more precise.

8.3 Boundary Layer EquationsConsider a rigid stationary obstacle whose surface is (locally) flat, and correspondsto the x-z plane. Let this surface be in contact with a high Reynolds number fluid thatoccupies the region y > 0. (See Figure 8.1.) Let δ be the typical normal thicknessof the boundary layer. The layer thus extends over the region 0 < y <∼ δ. The fluidthat occupies the region δ <∼ y < ∞, and thus lies outside the layer, is assumed tobe both irrotational and (effectively) inviscid. On the other hand, viscosity must beincluded in the equation of motion of the fluid within the layer. The fluid both insideand outside the layer is assumed to be incompressible.

Suppose that the equations of irrotational flow have already been solved to de-termine the fluid velocity outside the boundary layer. This velocity must be suchthat its normal component is zero at the outer edge of the layer (i.e., at y δ). Onthe other hand, the tangential component of the fluid velocity at the outer edge ofthe layer, U(x) (say), is generally non-zero. Here, we are assuming, for the sake ofsimplicity, that there is no spatial variation in the z-direction, so that both the irrota-tional flow and the boundary layer are effectively two-dimensional. Likewise, we arealso assuming that all flows are steady, so that any time variation can be neglected.The motion of the fluid within the boundary layer is governed by the equations ofsteady-state, incompressible, two-dimensional, viscous flow, which take the form

Incompressible Boundary Layers 221

boundary layer

irrotational fluid

solid surface

x

y

δ

U (x)

Figure 8.1A boundary layer.

(see Section 1.14)

∂vx

∂x+∂vy

∂y= 0, (8.1)

vx∂vx

∂x+ vy

∂vx

∂y= −1

ρ

∂p∂x+ ν

(∂ 2vx

∂x 2 +∂ 2vx

∂y 2

), (8.2)

vx∂vy

∂x+ vy

∂vy

∂y= −

1ρ

∂p∂y+ ν

(∂ 2vy

∂x 2 +∂ 2vy

∂y 2

), (8.3)

where ρ is the (constant) density, and ν the kinematic viscosity. Here, Equation (8.1)is the equation of continuity, whereas Equations (8.2) and (8.3) are the x- and y-components of the fluid equation of motion, respectively. The boundary conditionsat the outer edge of the layer, where it interfaces with the irrotational fluid, are

vx(x, y)→ U(x), (8.4)

p(x, y)→ P(x) (8.5)

as y/δ→ ∞. Here, P(x) is the fluid pressure at the outer edge of the layer, and

UdUdx= −1

ρ

dPdx

(8.6)

(because vy = 0, and viscosity is negligible, just outside the layer). The boundaryconditions at the inner edge of the layer, where it interfaces with the impenetrablesurface, are

vx(x, 0) = 0, (8.7)

vy(x, 0) = 0. (8.8)


Of course, the first of these constraints corresponds to the no slip condition.Let U0 be a typical value of the external tangential velocity, U(x), and let L be the

typical variation length-scale of this quantity. It is reasonable to suppose that U0 andL are also the characteristic tangential flow velocity and variation length-scale in thex-direction, respectively, of the boundary layer. Of course, δ is the typical variationlength-scale of the layer in the y-direction. Moreover, δ/L 1, because the layer isassumed to be thin. It is helpful to define the normalized variables

X =xL, (8.9)

Y =y

δ, (8.10)

Vx(X, Y) =vx

U0, (8.11)

Vy(X, Y) =vy

U1, (8.12)

P(X, Y) =pp0, (8.13)

where U1 and p0 are constants. All of these variables are designed to be O(1) insidethe layer. Equation (8.1) yields

U0

L∂Vx

∂X+

U1

δ

∂Vy∂Y= 0. (8.14)

In order for the terms in this equation to balance one another, we need

U1 =δ

LU0. (8.15)

In other words, within the layer, continuity requires the typical flow velocity in they-direction, U1, to be much smaller than that in the x-direction, U0.

Equation (8.2) gives

U 20

L

(Vx∂Vx

∂X+ Vy

∂Vx

∂Y

)= − p0

ρ L∂P∂X+

(νU0

δ 2

) [(δ

L

)2 ∂ 2Vx

∂X 2 +∂ 2Vx

∂Y 2

]. (8.16)

In order for the pressure term on the right-hand side of the previous equation to be ofsimilar magnitude to the advective terms on the left-hand side, we require that

p0 = ρU 20 . (8.17)

Furthermore, in order for the viscous term on the right-hand side to balance the otherterms, we need

δ

L=

U1

U0=

1Re1/2 , (8.18)

whereRe =

U0 Lν

(8.19)


is the Reynolds number of the flow external to the layer. (See Section 1.16.) Theassumption that δ/L 1 can be seen to imply that Re 1. In other words, thenormal thickness of the boundary layer separating an irrotational flow pattern froma rigid surface is only much less than the typical variation length-scale of the patternwhen the Reynolds number of the flow is much greater than unity.

Equation (8.3) yields

1Re

(Vx∂Vy∂X+ Vy

∂Vy∂Y

)= −∂P

∂Y+

1Re

[1

Re∂ 2Vy∂X 2 +

∂ 2Vy∂Y 2

]. (8.20)

In the limit Re 1, this reduces to

∂P∂Y= 0. (8.21)

Hence, P = P(X), wheredPdX= −U

dUdX, (8.22)

U(X) = U/U0, and use has been made of Equation (8.6). In other words, the pressureis uniform across the layer, in the direction normal to the surface of the obstacle, andis thus the same as that on the outer edge of the layer.

Retaining only O(1) terms, our final set of normalized layer equations becomes

∂Vx

∂X+∂Vy∂Y= 0, (8.23)

Vx∂Vx

∂X+ Vy

∂Vy∂Y= U

dU∂X+∂ 2Vy∂Y 2 , (8.24)

subject to the boundary conditions

Vx(X,∞) = U(X), (8.25)

and

Vx(X, 0) = 0, (8.26)

Vy(X, 0) = 0. (8.27)

In unnormalized form, the previous set of layer equations are written

∂vx

∂x+∂vy

∂y= 0, (8.28)

vx∂vx

∂x+ vy

∂vx

∂y= U

dUdx+ ν

∂ 2vx

∂y 2 , (8.29)


vx(x,∞) = U(x) (8.30)


(note that y = ∞ really means y/δ→ ∞), and

vx(x, 0) = 0, (8.31)

vy(x, 0) = 0. (8.32)

Equation (8.28) can be automatically satisfied by expressing the flow velocity interms of a stream function: that is,

vx = −∂ψ

∂y, (8.33)

vy =∂ψ

∂x. (8.34)

In this case, Equation (8.29) reduces to

ν∂ 3ψ

∂y 3 −∂ψ

∂x∂ 2ψ

∂y 2 +∂ψ

∂y

∂ 2ψ

∂x ∂y= U

dUdx, (8.35)


∂ψ(x,∞)∂y

= −U(x), (8.36)

and

ψ(x, 0) = 0, (8.37)

∂ψ(x, 0)∂y

= 0. (8.38)

To lowest order, the vorticity internal to the layer, ω = ω ez, is given by

ω =∂ 2ψ

∂y 2 , (8.39)

whereas the x-component of the viscous force per unit area acting on the surface ofthe obstacle is written (see Section 1.18)

σxy

∣∣∣y=0 = ρ ν

∂vx

∂y

∣∣∣∣∣y=0= −ρ ν

∂ 2ψ

∂y 2

∣∣∣∣∣∣y=0. (8.40)

8.4 Self-Similar Boundary LayersThe boundary layer equation, (8.35), takes the form of a nonlinear partial differ-ential equation that is extremely difficult to solve exactly. However, considerableprogress can be made if this equation is converted into an ordinary differential equa-tion by demanding that its solutions be self-similar. Self-similar solutions are such


that, at a given distance, x, along the layer, the tangential flow profile, vx(x, y),is a scaled version of some common profile: that is, vx(x, y) = U(x) F[y/δ(x)],where δ(x) is a scale-factor, and F(z) a dimensionless function. It follows thatψ(x, y) = −U(x) δ(x) f [y/δ(x)], where f ′(z) = F(z).

Let us search for a self-similar solution to Equation (8.35) of the general form

ψ(x, y) = −[2 νU0 x m+1

m + 1

]1/2f (η) = −U0 x m

[2 ν

(m + 1) U0 x m−1

]1/2f (η), (8.41)

where

η =

[(m + 1) U0 x m−1

2 ν

]1/2y. (8.42)

This implies that δ(x) = [2 ν/(m + 1) U0 x m−1]1/2, and U(x) = U0 x m. Here, U0 andm are constants. Moreover, U0 x m has dimensions of velocity, whereas m, η, and f ,are dimensionless. Transforming variables from x, y to x, η, we find that

∂

∂x

∣∣∣∣∣y=∂

∂x

∣∣∣∣∣η+

m − 12

η

x∂

∂η

∣∣∣∣∣x, (8.43)

∂

∂y

∣∣∣∣∣x=

[(m + 1) U0 x m−1

2 ν

]1/2∂

∂η

∣∣∣∣∣x. (8.44)

Hence,

∂ψ

∂x= −

[νU0 x m−1

2 (m + 1)

]1/2[(m + 1) f + (m − 1) η f ′], (8.45)

∂ψ

∂y= −U0 x m f ′, (8.46)

∂ 2ψ

∂y 2 = − (m + 1) U 3

0 x 3 m−1

2 ν

1/2 f ′′, (8.47)

∂ 2ψ

∂x ∂y= −U0 x m−1

2[2 m f ′ + (m − 1) η f ′′], (8.48)

∂ 3ψ

∂y 3 = −(m + 1) U 2

0 x 2 m−1

2 νf ′′′, (8.49)

where ′ = d/dη. Thus, Equation (8.35) becomes

(m + 1) f ′′′ + (m + 1) f f ′′ − 2m f ′ 2 = − 1U 2

0 x 2 m−1

dU 2

dx. (8.50)

Because the left-hand side of the previous equation is a (non-constant) function of η,while the right-hand side is a function of x (and as η and x are independent variables),the equation can only be satisfied if its right-hand side takes a constant value. In fact,if

1U 2

0 x 2 m−1

dU 2

dx= 2 m (8.51)


thenU(x) = U0 x m (8.52)

(which is consistent with our initial guess), and

f ′′′ + f f ′′ + β (1 − f ′ 2) = 0, (8.53)

whereβ =

2 mm + 1

. (8.54)

Expression (8.53) is known as the Falkner-Skan equation. The solutions to this equa-tion that satisfy the physical boundary conditions (8.36)–(8.38) are such that

f (0) = f ′(0) = 0, (8.55)

and

f ′(∞) = 1, (8.56)

f ′′(∞) = 0. (8.57)

(The final condition corresponds to the requirement that the vorticity tend to zero atthe edge of the layer.) Note, from Equations (8.39), (8.42), (8.47), (8.52), (8.55), and(8.56), that the normally integrated vorticity within the boundary layer is∫ ∞

0ω dy = −U(x). (8.58)

Furthermore, from Equations (8.40), (8.47), and (8.52), the x-component of the vis-cous force per unit area acting on the surface of the obstacle is

σxy

∣∣∣y=0 =

12ρU 2

(ν

U x

)1/2(m + 1)1/2

√2 f ′′(0). (8.59)

It is convenient to parameterize this quantity in terms of a skin friction coefficient,

c f =σxy

∣∣∣y=0

(1/2) ρU 2 . (8.60)

It follows that

c f (x) =(m + 1)1/2

√2 f ′′(0)

[Re(x)]1/2 , (8.61)

whereRe(x) =

U(x) xν

(8.62)

is the effective Reynolds number of the flow on the outer edge of the layer at positionx. Hence, c f (x) ∝ x−(m+1)/2. Finally, according to Equation (8.41), the width of theboundary layer is approximately

δ(x)x 1

[Re(x)] 1/2 , (8.63)


0

0.5

1

1.5

f′′ (

0)

0 1 2β

Figure 8.2f ′′(0) calculated as a function of β for solutions of the Falkner-Skan equation.

which implies that δ(x) ∝ x−(m−1)/2.If m > 0 then the external tangential velocity profile, U(x) = U0 x m, corresponds

to that of irrotational inviscid flow incident, in a symmetric fashion, on a semi-infinitewedge whose apex subtends an angle απ, where α = 2 m/(m+1). (See Section 5.10,and Figure 5.10.) In this case, U(x) can be interpreted as the tangential velocitya distance x along the surface of the wedge from its apex (in the direction of theflow). By analogy, if m = 0 then the external velocity profile corresponds to that ofirrotational inviscid flow parallel to a semi-infinite flat plate (which can be thoughtof as a wedge whose apex subtends zero angle). In this case, U(x) can be interpretedas the tangential velocity a distance x along the surface of the plate from its leadingedge (in the direction of the flow). (See Section 8.5.) Finally, if m < 0 then theexternal velocity profile is that of symmetric irrotational inviscid flow over the backsurface of a semi-infinite wedge whose apex subtends an angle (1 − α′) π, whereα′ = −m/(1 + m). (See Section 5.11, and Figure 5.11.) In this case, U(x) can beinterpreted as the tangential velocity a distance x along the surface of the wedgefrom its apex (in the direction of the flow).

Unfortunately, the Falkner-Skan equation, (8.53), possesses no general analyticsolutions. However, this equation is relatively straightforward to solve via numeri-cal methods. Figure 8.2 shows f ′′(0), calculated numerically as a function of β =2 m/(m+ 1), for the solutions of Equation (8.53) that satisfy the boundary conditions(8.55)–(8.57). In addition, Figure 8.3 shows f ′(η) versus η, calculated numericallyfor various different values of m. Because β→ 2 as m→ ∞, solutions of the Falkner-


−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

f ′

0 1 2 3 4 5 6 7 8 9 10η

Figure 8.3Solutions of the Falkner-Skan equation. In order from the left to the right, the varioussolid curves correspond to forward flow solutions calculated with m = 4, 1, 1/3, 1/9,0, −0.05, and−0.0904, respectively. The dashed curve shows a reversed flow solutioncalculated with m = −0.05.

Skan equation with β > 2 have no physical significance. For 0 < β < 2, it can beseen, from Figures 8.2 and 8.3, that there is a single solution branch characterized byf ′(η) > 0 and f ′′(0) > 0. This branch is termed the forward flow branch, because it issuch that the tangential velocity, vx(η) ∝ f ′(η), is in the same direction as the externaltangential velocity [i.e., vx(∞)] across the whole layer (i.e., 0 < η < ∞). The for-ward flow branch is characterized by a positive skin friction coefficient, c f ∝ f ′′(0).It can also be seen that for β < 0 there exists a second solution branch, which istermed the reversed flow branch, because it is such that the tangential velocity is inthe opposite direction to the external tangential velocity in the region of the layerimmediately adjacent to the surface of the obstacle (which corresponds to η = 0).The reversed flow branch is characterized by a negative skin friction coefficient. Thereversed flow solutions are probably unphysical, because reversed flow close to thewall is generally associated with a phenomenon known as boundary layer separation(see Section 8.10) that invalidates the boundary layer orderings. It can be seen thatthe two solution branches merge together at β = β∗ = −0.1989, which correspondsto m = m∗ = −0.0905. Moreover, there are no solutions to the Falkner-Skan equa-tion with β < β∗ or m < m∗. The disappearance of solutions when m becomes toonegative (i.e., when the deceleration of the external flow becomes too large) is alsorelated to boundary layer separation.


y

x

δ

plateboundary layer

wake

L

U0

Figure 8.4Flow over a flat plate.

8.5 Boundary Layer on a Flat Plate

Consider a flat plate of length L, infinite width, and negligible thickness, that liesin the x-z plane, and whose two edges correspond to x = 0 and x = L. Supposethat the plate is immersed in a low viscosity fluid whose unperturbed velocity field isv = U0 ex. (See Figure 8.4.) In the inviscid limit, the appropriate boundary conditionat the surface of the plate, vy = 0—corresponding to the requirement of zero normalvelocity—is already satisfied by the unperturbed flow. Hence, the original flow is notmodified by the presence of the plate. However, when we take the finite viscosity ofthe fluid into account, an additional boundary condition, vx = 0—corresponding tothe no slip condition—must be satisfied at the plate. The imposition of this additionalconstraint causes thin boundary layers, of thickness δ(x) L, to form above and be-low the plate. The fluid flow outside the boundary layers remains effectively inviscid,whereas that inside the layers is modified by viscosity. It follows that the flow ex-ternal to the layers is unaffected by the presence of the plate. Hence, the tangentialvelocity at the outer edge of the boundary layers is U(x) = U0. This correspondsto the case m = 0 discussed in the previous section. [See Equation (8.52).] (Here,we are assuming that the flow upstream of the trailing edge of the plate, x = L, isunaffected by the edge’s presence, and, is, therefore, the same as if the plate were ofinfinite length. Of course, the flow downstream of the edge is modified as a conse-quence of the finite length of the plate.)

Making use of the analysis contained in the previous section (with m = 0), aswell as the fact that, by symmetry, the lower boundary layer is the mirror image of


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

v x/U

0

−5 −4 −3 −2 −1 0 1 2 3 4 5y / δ

Figure 8.5Tangential velocity profile across the boundary layers located above and below a flatplate of negligible thickness located at y = 0.

the upper one, the tangential velocity profile across the both layers is written

vx(x, y) = U0 f ′(η), (8.64)

where

η =( U0

2 ν x

)1/2|y|. (8.65)

Here, f (η) is the solution off ′′′ + f f ′′ = 0 (8.66)

that satisfies the boundary conditions

f (0) = f ′(0) = 0, (8.67)

and

f ′(∞) = 1, (8.68)

f ′′(∞) = 0. (8.69)

Equation (8.66) is known as the Blasius equation.It is convenient to define the so-called displacement thickness of the upper bound-

ary layer,

δ(x) =∫ ∞

0

[1 −

vx(x, y)U0

]dy, (8.70)


which can be interpreted as the distance through which streamlines just outside thelayer are displaced laterally due to the retardation of the flow within the layer. (Ofcourse, the thickness of the lower boundary layer is the same as that of the upperlayer.) It follows that

δ(x) =(ν xU0

)1/2 √2∫ ∞

0[1 − f ′(η)] dη. (8.71)

In fact, the numerical solution of Equation (8.66), subject to the boundary condi-tions (8.67)–(8.69), yields

δ(x) = 1.72(ν xU0

)1/2. (8.72)

Hence, the thickness of the boundary layer increases as the square root of the distancefrom the leading edge of the plate. In particular, the thickness at the trailing edge ofthe plate is

δ(L)L=

1.72Re1/2 , (8.73)

where

Re =U0 Lν

(8.74)

is the appropriate Reynolds number for the interaction of the flow with the plate.Note that if Re 1 then the thickness of the boundary layer is much less than itslength, as was previously assumed.

The tangential velocity profile across the both boundary layers, which takes theform

vx(x, y) = U0 f ′[1.22

|y|δ(x)

], (8.75)

is plotted in Figure 8.5. In addition, the vorticity profile across the layers, which iswritten

ω(x, y) = −sgn(y) 1.22U0

δ(x)f ′′[1.22

|y|δ(x)

], (8.76)

is shown in Figure 8.6. The vorticity is negative in the upper boundary layer (i.e.,y > 0), positive in the lower boundary layer (i.e., y < 0), and discontinuous acrossthe plate (which is located at y = 0). Finally, the net viscous drag force per unit width(along the z-axis) acting on the plate in the x-direction is

D = 2∫ L

0σxy

∣∣∣y=0 dx, (8.77)

where the factor of 2 is needed to take into account the presence of boundary layersboth above and below the plate. It follows from Equation (8.59) (with m = 0) that

D = ρU 20

(ν

U0

)1/2 √2 f ′′(0)

∫ L

0x−1/2 dx = ρU 2

0

(ν LU0

)1/22√

2 f ′′(0). (8.78)


−0.6

−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

ω/

(U0/

δ)

−5 −4 −3 −2 −1 0 1 2 3 4 5y / δ

Figure 8.6Vorticity profile across the boundary layers located above and below a flat plate ofnegligible thickness located at y = 0.

In fact, the numerical solution of (8.66) yields

D = 1.33ρU 2

0 L

Re1/2 = 1.33 ρU0 (νU0 L)1/2. (8.79)

The previous discussion is premised on the assumption that the flow in the up-per (or lower) boundary layer is both steady and z-independent. It turns out thatthis assumption becomes invalid when the Reynolds number of the layer, U0 δ/ν, ex-ceeds a critical value which is about 600 (Batchelor 2000). In this case, small-scalez-dependent disturbances spontaneously grow to large amplitude, and the layer be-comes turbulent. Because δ ∝ x1/2, if the criterion for boundary layer turbulence isnot satisfied at the trailing edge of the plate, x = L, then it is not satisfied anywhereelse in the layer. Thus, the previous analysis, which neglects turbulence, remainsvalid provided U0 δ(L)/ν < 600. According to Equation (8.73), this implies that theanalysis is valid when 1 Re < 1.2 × 105, where Re = U0 L/ν is the Reynoldsnumber of the external flow.

Consider, finally, the situation illustrated in Figure 8.7 in which an initially ir-rotational fluid passes between two flat parallel plates. Let d be the perpendiculardistance between the plates. As we have seen, the finite viscosity of the fluid causesboundary layers to form on the inner surfaces of the upper and lower plates. Theflow within these layers possesses non-zero vorticity, and is significantly affected by


U0

boundary layer

potential flow

plate

l

d

Figure 8.7Flow between two flat parallel plates.

viscosity. On the other hand, the flow outside the layers is irrotational and essen-tially inviscid—this type of flow is usually termed potential flow (because it can bederived from a velocity potential satisfying Laplace’s equation). The thickness of thetwo boundary layers increases like x1/2, where x represents distance, parallel to theflow, measured from the leading edges of the plates. It follows that, as x increases, theregion of potential flow shrinks in size, and eventually disappears. (See Figure 8.7.)Assuming that, prior to merging, the two boundary layers do not significantly affectone another, their thickness, δ(x), is given by formula (8.72), where U0 is the speedof the incident fluid. The region of potential flow thus extends from x = 0 (whichcorresponds to the leading edge of the plates) to x = l, where

δ(l) =d2. (8.80)

It follows thatld= 11.8 Re, (8.81)

where

Re =U0 dν. (8.82)

Thus, when an irrotational high Reynolds number fluid passes between two parallelplates then the region of potential flow extends a comparatively long distance be-tween the plates, relative to their spacing (i.e., l/d 1). By analogy, if an irrotationalhigh Reynolds number fluid passes into a pipe then the fluid remains essentially irro-tational until it has travelled a considerable distance along the pipe, compared to itsdiameter. Obviously, these conclusions are modified if the flow becomes turbulent.


8.6 Wake Downstream of a Flat PlateAs we saw in the previous section, if a flat plate of negligible thickness, and finitelength, is placed in the path of a uniform high Reynolds number flow, directed par-allel to the plate, then thin boundary layers form above and below the plate. Outsidethe layers, the flow is irrotational, and essentially inviscid. Inside the layers, the flowis modified by viscosity, and has non-zero vorticity. Downstream of the plate, theboundary layers are convected by the flow, and merge to form a thin wake. (SeeFigure 8.4.) Within the wake, the flow is modified by viscosity, and possesses finitevorticity. Outside the wake, the downstream flow remains irrotational, and effectivelyinviscid.

Because there is no solid surface embedded in the wake, acting to retard the flow,we would expect the action of viscosity to cause the velocity within the wake, a longdistance downstream of the plate, to closely match that of the unperturbed flow. Inother words, we expect the fluid velocity within the wake to take the form

vx(x, y) = U0 − u(x, y), (8.83)

vy(x, y) = v(x, y), (8.84)

where|u| U0. (8.85)

Assuming that, within the wake,

∂

∂x∼ 1

x, (8.86)

∂

∂y∼

1δ, (8.87)

where δ x is the wake thickness, fluid continuity requires that

v ∼ δx

u. (8.88)

The flow external to the boundary layers, and the wake, is both uniform and essen-tially inviscid. Hence, according to Bernoulli’s theorem, the pressure in this regionis also uniform. [See Equation (8.22).] However, as we saw in Section 8.3, there isno y-variation of the pressure across the boundary layers. It follows that the pressureis uniform within the layers. Thus, it is reasonable to assume that the pressure isalso uniform within the wake, because the wake is formed via the convection of theboundary layers downstream of the plate. We conclude that

p(x, y) p0 (8.89)

everywhere in the fluid, where p0 is a constant.


The x-component of the fluid equation of motion is written

vx∂vx

∂x+ vy

∂vx

∂y= −1

ρ

∂p∂x+ ν

(∂ 2vx

∂x 2 +∂ 2vx

∂y 2

). (8.90)

Making use of Equations (8.83)–(8.89), the previous expression reduces to

U0∂u∂x ν ∂

2u∂y 2 . (8.91)

The boundary conditionu(x,±∞) = 0 (8.92)

ensures that the flow outside the wake remains unperturbed. Note that Equation (8.91)has the same mathematical form as a conventional diffusion equation, with x play-ing the role of time, and ν/U0 playing the role of the diffusion coefficient. Hence,by analogy with the standard solution of the diffusion equation, we would expectδ ∼ (ν x/U0)1/2 (Riley 1974).

As can easily be demonstrated, the self-similar solution to Equation (8.91), sub-ject to the boundary condition (8.92), is

u(x, y) =Q√π δ

exp(−y

2

δ 2

), (8.93)

where

δ(x) = 2(ν xU0

)1/2, (8.94)

and Q is a constant. It follows that∫ ∞

−∞u dy = Q, (8.95)

because, as is well-known,∫ ∞−∞ exp(−t 2) dt =

√π (Riley 1974). As expected, the

width of the wake scales as x1/2.The tangential velocity profile across the wake, which takes the form

vx(x, y)U0

= 1 − QU0 δ

1√π

exp(−y 2/δ 2), (8.96)

is plotted in Figure 8.8. In addition, the vorticity profile across the wake, which iswritten

ω(x, y)U0/δ

= − QU0 δ

2√π

y

δexp(−y 2/δ 2) (8.97)

is shown in Figure 8.9. It can be seen that the profiles pictured in Figures 8.8 and8.9 are essentially smoothed out versions of the boundary layer profiles shown inFigures 8.5 and 8.6, respectively.

Suppose that the plate and a portion of its trailing wake are enclosed by a cuboid


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

v x/U

0

−3 −2 −1 0 1 2 3y / δ

Figure 8.8Tangential velocity profile across the wake of a flat plate of negligible thicknesslocated at y = 0. The profile is calculated for Q/(U0 δ) = 0.5.

−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5

ω/

(U0/

δ)

−3 −2 −1 0 1 2 3y / δ

Figure 8.9Vorticity profile across the boundary layers above and below a flat plate of negligiblethickness located at y = 0. The profile is calculated for Q/(U0 δ) = 0.5.


U0 − u(y)

wakeplate

y = h

y = −h

U0

x = −l x = l

v(x)

v(x) y

x

Figure 8.10Control volume surrounding a flat plate and its trailing wake.

control volume of unit depth (in the z-direction) that extends from x = −l to x = +land from y = −h to y = h. (See Figure 8.10.) Here, l L and h δ(l), where Lis the length of the plate, and δ(x) the width of the wake. Hence, the control volumeextends well upstream and downstream of the plate. Moreover, the volume is muchwider than the wake.

Let us apply the integral form of the fluid equation of continuity to the controlvolume. For a steady state, this reduces to (see Section 1.9)∮

Sρ v · dS = 0, (8.98)

where S is the bounding surface of the control volume. The normal fluid velocityis −U0 at x = −l, U0 − u(y) at x = l, and v(x) at y = ±h, as indicated in the figure.Hence, Equation (8.98) yields

−∫ h

−hρU0 dy +

∫ h

−hρ [U0 − u(y)] dy + 2

∫ l

−lρ v(x) dx = 0, (8.99)

or ∫ h

−hu(y) dy = 2

∫ l

−lv(x) dx. (8.100)

However, given that u→ 0 for |y| δ, and because h δ, it is a good approximationto replace the limits of integration on the left-hand side of the previous expression by


±∞. Thus, from Equation (8.95),∫ h

−hu(y) dy = 2

∫ l

−lv(x) dx Q, (8.101)

where Q is independent of x. Note that the slight retardation of the flow inside thewake, due to the presence of the plate, which is parameterized by Q, necessitates asmall lateral outflow, v(x), in the region of the fluid external to the wake.

Let us now apply the integral form of the x-component of the fluid equation ofmotion to the control volume. For a steady state, this reduces to (see Section 1.11)∫

Sρ vx v · dS = Fx +

∫Sσx j dS j, (8.102)

where Fx is the net x-directed force exerted on the fluid within the control volumeby the plate. It follows, from Newton’s third law of motion, that Fx = −D, where Dis the viscous drag force per unit width (in the z-direction) acting on the plate in thex-direction. In an incompressible fluid (see Section 1.6),

σi j = −p δi j + ρ ν

(∂vi∂x j+∂v j

∂xi

). (8.103)

Hence, we obtain

−∫ h

−hρU 2

0 dy +∫ h

−hρ[U0 − u(y)

]2 dy

+2∫ l

−lρU0 v(x) dx = −D − 2 ρ ν

ddl

∫ h

−hu(y) dy, (8.104)

because the pressure within the fluid is essentially uniform, the tangential fluid ve-locity at y = ±h is U0, and v is assumed to be negligible at x = ±l. Making use ofEquation (8.101), as well as the fact that Q is independent of l, we get

D = ρU0 Q. (8.105)

Here, we have neglected any terms that are second order in the small quantity u. Acomparison with Equation (8.79) reveals that

Q = 1.33 (νU0 L)1/2, (8.106)

orQ

U0 δ= 0.664

(Lx

)1/2. (8.107)

Hence, from Equations (8.96) and (8.97), the velocity and vorticity profiles acrossthe layer are

vx(x, y)U0

= 1 − 0.375(L

x

)1/2exp(−y 2/δ 2), (8.108)


and

ω(x, y)U0/δ

= −0.749(L

x

)1/2 yδ

exp(−y 2/δ 2), (8.109)

respectively, where δ(x) = 2 (ν x/U0)1/2. Finally, because the previous analysis ispremised on the assumption that |1−vx/U0| = |u|/U0 1, it is clear that the previousthree expressions are only valid when x L (i.e., well downstream of the plate).

The previous analysis only holds when the flow within the wake is non-turbulent.Let us assume, by analogy with the discussion in the previous section, that this is thecase as long as the Reynolds number of the wake, U0 δ(x)/ν, remains less than somecritical value that is approximately 600. Because the Reynolds number of the wakecan be written 2 Re1/2 (x/L)1/2, where Re = U0 L/ν is the Reynolds number of theexternal flow, we deduce that the wake becomes turbulent when x/L >∼ 9 × 104/Re.Hence, the wake is always turbulent sufficiently far downstream of the plate. Ouranalysis, which effectively assumes that the wake is non-turbulent in some region,immediately downstream of the plate, whose extent (in x) is large compared with L,is thus only valid when 1 Re 9 × 104.

8.7 Von Karman Momentum Integral

Consider a boundary layer that forms on the surface of a rigid stationary obstacleof arbitrary shape (but infinite length and uniform cross-section) placed in a steady,uniform, transverse, high Reynolds number flow. Let x represent arc length alongthe surface, measured (in the direction of the external flow) from the stagnation pointthat forms at the front of the obstacle. (See Figure 8.11.) Moreover, let y representdistance across the boundary layer, measured normal to the surface. Suppose that theboundary layer is sufficiently thin that it is well approximated as a plane slab in theimmediate vicinity of a general point on the surface. In this case, writing the velocityfield within the layer in the form v = u(x, y) ex + v(x, y) ey, it is reasonable to modelthis flow using the slab boundary layer equations [see Equations (8.28) and (8.29)]

∂u∂x+∂v

∂y= 0, (8.110)

u∂u∂x+ v

∂u∂y− U

dUdx= ν

∂ 2u∂y 2 , (8.111)

subject to the standard boundary conditions

u(x,∞) = U(x), (8.112)

u(x, 0) = v(x, 0) = 0. (8.113)


Here, U(x) is the external tangential fluid velocity at the edge of the layer. Integrating(8.111) across the layer, making use of the boundary conditions (8.113), leads to

ν∂u∂y

∣∣∣∣∣y=0=

∫ ∞

0

(U

dUdx− u

∂u∂x− v ∂u

∂y

)dy

=

∫ ∞

0

[(U − u)

dUdx+ u

∂(U − u)∂x

+ v∂(U − u)∂y

]dy

=

∫ ∞

0

[(U − u)

dUdx+ u

∂(U − u)∂x

− (U − u)∂v

∂y

]dy

=

∫ ∞

0

[(U − u)

dUdx+ u

∂(U − u)∂x

+ (U − u)∂u∂x

]dy

=dUdx

∫ ∞

0(U − u) dy +

ddx

∫ ∞

0u (U − u) dy. (8.114)

Here, we have integrated the final term on the right-hand side by parts, making useof Equations (8.110), (8.112), and (8.113). Let us define the displacement thicknessof the layer [see Equation (8.70)]

δ1(x) =∫ ∞

0

(1 −

uU

)dy, (8.115)

as well as the so-called momentum thickness

δ2(x) =∫ ∞

0

uU

(1 −

uU

)dy. (8.116)

It follows from Equation (8.114) that

ν∂u∂y

∣∣∣∣∣y=0= U 2 dδ2

dx+ U

dUdx

(δ1 + 2 δ2). (8.117)

This important result is known as the von Karman momentum integral, and is fun-damental to many of the approximation methods commonly employed to calcu-late boundary layer thicknesses on the surfaces of general obstacles placed in highReynolds number flows. (See Section 8.10.)

8.8 Boundary Layer SeparationAs we saw in Section 8.5, when a high Reynolds number fluid passes around astreamlined obstacle, such as a slender plate that is aligned with the flow, a relativelythin boundary layer form on the obstacle’s surface. Here, by relatively thin, we meanthat the typical transverse (to the flow) thickness of the layer is δ ∼ L/Re1/2, whereL is the length of the obstacle (in the direction of the flow), and Re the Reynolds


stagnation point

potential flow streamlines

boundary layer

wake

separation point

obstacle

Figure 8.11Boundary layer separation.

number of the external flow. Suppose, however, that the obstacle is not streamlined:that is, the surface of the obstacle is not closely aligned with the streamlines of theunperturbed flow pattern. In this case, the typically observed behavior is illustratedin Figure 8.11, which shows the flow pattern of a high Reynolds number irrotationalfluid around a cylindrical obstacle (whose axis is normal to the direction of the un-perturbed flow). It can be seen that a stagnation point, at which the flow velocity islocally zero, forms in front of the obstacle. Moreover, a thin boundary layer coversthe front side of the obstacle. The thickness of this layer is smallest at the stagnationpoint, and increases towards the back side of the obstacle. However, at some pointon the back side, the boundary layer separates from the obstacle’s surface to form avortex-filled wake whose transverse dimensions are similar to those of the obstacleitself. This phenomenon is known as boundary layer separation.

Outside the boundary layer, and the wake, the flow pattern is irrotational andessentially inviscid. So, from Section 5.8, the tangential flow speed just outside theboundary layer (neglecting any circulation of the external flow around the cylinder)is

U(θ) = 2 U0 sin θ, (8.118)

where U0 is the unperturbed flow speed, and θ is a cylindrical coordinate definedsuch that the stagnation point corresponds to θ = 0. Note that the tangential flowaccelerates (i.e., increases with increasing arc-length, along the surface of the obsta-cle, in the direction of the flow) on the front side of the obstacle (i.e., 0 ≤ θ ≤ π/2),and decelerates on the back side. Boundary layer separation is always observed to


take place at a point on the surface of an obstacle where there is deceleration of theexternal tangential flow. In addition, from Section 5.8, the pressure just outside theboundary layer (and, hence, on the surface of the obstacle, because the pressure isuniform across the layer) is

P(θ) = p1 + ρU 20 cos(2 θ), (8.119)

where p1 is a constant. The tangential pressure gradient is such as to acceleratethe tangential flow on the front side of the obstacle—this is known as a favorablepressure gradient. On the other hand, the pressure gradient is such as to deceleratethe flow on the back side—this is known as an adverse pressure gradient. Boundarylayer separation is always observed to take place at a point on the surface of anobstacle where the pressure gradient is adverse.

Boundary layer separation is an important physical phenomenon because it givesrise to a greatly enhanced drag force acting on a non-streamlined obstacle placed ina high Reynolds number flow. This is the case because the pressure in the compara-tively wide wake that forms behind a non-streamlined obstacle, as a consequence ofseparation, is relatively low. To be more exact, in the case of a cylindrical obstacle,Equation (8.119) specifies the expected pressure variation over the obstacle’s surfacein the absence of separation. It can be seen that the variation on the front side of theobstacle mirrors that on the back side: that is, P(π − θ) = P(θ). (See Figure 8.12.) Inother words, the resultant pressure force on the front side of the obstacle is equal andopposite to that on the back side, so that the pressure distribution gives rise to zeronet drag acting on the obstacle. Figure 8.12 illustrates how the pressure distributionis modified as a consequence of boundary layer separation. In this case, the pressurebetween the separation points is significantly less than that on the front side of theobstacle. Consequently, the resultant pressure force on the front side is greater inmagnitude than the oppositely directed force on the back side, giving rise to a signif-icant drag acting on the obstacle. Let D be the drag force per unit width (parallel tothe axis of the cylinder) exerted on the obstacle. It is convenient to parameterize thisforce in terms of a dimensionless drag coefficient,

CD =D

ρU 20 a, (8.120)

where ρ is the fluid density, and a the typical transverse size of the obstacle (in thepresent example, the radius of the cylinder). The drag force that acts on a non-streamlined obstacle placed in a high Reynolds number flow, as a consequence ofboundary layer separation, is generally characterized by a drag coefficient of orderunity. The exact value of the coefficient depends strongly on the shape of the obstacle,but only relatively weakly on the Reynolds number of the flow. Consequently, thistype of drag is termed form drag, because it depends primarily on the external shape,or form, of the obstacle. Form drag scales roughly as the cross-sectional area (perunit width) of the vortex-filled wake that forms behind the obstacle.

Boundary layer separation is associated with strong adverse pressure gradients,or, equivalently, strong flow deceleration, on the back side of an obstacle placed in a


P (θ) − p1

θ

separation points

3π/2ππ/20

stagnation point

Figure 8.12Pressure variation over surface of a cylindrical obstacle in a high Reynolds numberflow both with (dashed curve) and without (solid curve) boundary layer separation.

high Reynolds number flow. Such gradients can be significantly reduced by stream-lining the obstacle: that is, by closely aligning its back surface with the unperturbedstreamlines of the external flow. Indeed, boundary layer separation can be delayed,or even completely prevented, on the surface of a sufficiently streamlined obstacle,thereby significantly decreasing, or even eliminating, the associated form drag (es-sentially, by reducing the cross-sectional area of the wake). However, even in thelimit that the form drag is reduced to a negligible level, there is still a residual dragacting on the obstacle due to boundary layer viscosity. This type of drag is calledfriction drag. As is clear from a comparison of Equations (8.79) and (8.120), thedrag coefficient associated with friction drag is O(Re−1/2), where Re is the Reynoldsnumber of the flow. Friction drag thus tends to zero as the Reynolds number tends toinfinity.

The phenomenon of boundary layer separation allows us to resolve d’Alembert’sparadox. Recall, from Section 5.8, that an idealized fluid that is modeled as inviscidand irrotational is incapable of exerting a drag force on a stationary obstacle, despitethe fact that very high Reynolds number, ostensibly irrotational, fluids are observedto exert significant drag forces on stationary obstacles. The resolution of the paradoxlies in the realization that, in such fluids, viscosity can only be neglected (and theflow is consequently only irrotational) in the absence of boundary layer separation.In this case, the region of the fluid in which viscosity plays a significant role is local-ized to a thin boundary layer on the surface of the obstacle, and the resultant friction


drag scales as Re−1/2, and, therefore, disappears in the inviscid limit (essentially, be-cause the boundary layer shrinks to zero thickness in this limit). On the other hand,if the boundary layer separates then viscosity is important both in a thin boundarylayer on the front of the obstacle, and in a wide, low-pressure, vortex-filled, wakethat forms behind the obstacle. Moreover, the wake does not disappear in the in-viscid limit. The presence of significant fluid vorticity within the wake invalidatesirrotational fluid dynamics. Consequently, the pressure on the back side of the obsta-cle is significantly smaller than that predicted by irrotational fluid dynamics. Hence,the resultant pressure force on the front side is larger than that on the back side, anda significant drag is exerted on the obstacle. The drag coefficient associated with thistype of drag is generally of order unity, and does not tend to zero as the Reynoldsnumber tends to infinity.

8.9 Criterion for Boundary Layer SeparationAs we have seen, the boundary layer equations (8.110)–(8.113) generally lead to theconclusion that the tangential velocity in a thin boundary layer, u, is large comparedwith the normal velocity, v. Mathematically speaking, this result holds everywhereexcept in the immediate vicinity of singular points. But, if v u then it follows thatthe fluid moves predominately parallel to the surface of the obstacle, and can onlymove away from this surface to a very limited extent. This restriction effectivelyprecludes separation of the flow from the surface. Hence, we conclude that separationcan only occur at a point at which the solution of the boundary layer equations issingular.

As we approach a separation point, we expect the flow to deviate from the bound-ary layer towards the interior of the fluid. In other words, we expect the normal ve-locity to become comparable with the tangential velocity. However, we have seenthat the ratio v/u is of order Re−1/2. [See Equation (8.18).] Hence, an increase ofv to such a degree that v ∼ u implies an increase by a factor Re1/2. For sufficientlylarge Reynolds numbers, we may suppose that v effectively increases by an infinitefactor. Indeed, if we employ the dimensionless form of the boundary layer equations,(8.23)–(8.27), the situation just described is formally equivalent to an infinite valueof the dimensionless normal velocity, Vy, at the separation point.

Let the separation point lie at x = x0, and let x < x0 correspond to the region ofthe boundary layer upstream of this point. According to the previous discussion,

v(x0, y) = ∞ (8.121)

at all y (except, of course, y = 0, where the boundary conditions at the surface ofthe obstacle require that v = 0). It follows that the deriviative ∂v/∂y is also infiniteat x = x0. Hence, the equation of continuity, ∂u/∂x + ∂v/∂y = 0, implies that(∂u/∂x)x=x0 = ∞, or ∂x/∂u = 0, if x is regarded as a function of u and y. Letu(x0, y) = u0(y). Close to the point of separation, x0 − x and u − u0 are small. Thus,


we can expand x0 − x in powers of u − u0 (at fixed y). Because (∂x/∂u)u=u0 = 0, thefirst term in this expansion vanishes identically, and we are left with

x0 − x = f (y) (u − u0)2 + O[(u − u0)3

], (8.122)

oru(x, y) u0(y) + α(y)

√x0 − x, (8.123)

where α = 1/√

f is some function of y. From the equation of continuity,

∂v

∂y= −∂u

∂x α(y)

2√

x0 − x. (8.124)

Upon integration, the previous expression yields

v(x, y) β(y)√

x0 − x, (8.125)

where

β(y) =12

∫ y

α(y′) dy′. (8.126)

The equation of tangential motion in the boundary layer, (8.111), is written

u∂u∂x+ v

∂u∂y= U

dUdx+ ν

∂ 2u∂y 2 . (8.127)

As is clear from Equation (8.123), the derivative ∂ 2u/∂y 2 does not become infiniteas x → x0. The same is true of the function U dU/dx, which is determined from theflow outside the boundary layer. However, both terms on the left-hand side of theprevious expression become infinite as x → x0. Hence, in the immediate vicinity ofthe separation point,

u∂u∂x+ v

∂u∂y 0. (8.128)

Because ∂u/∂x = −∂v/∂y, we can rewrite this equation in the form

−u∂v

∂y+ v

∂u∂y= −u 2 ∂

∂y

( vu

) 0. (8.129)

Because u does not, in general, vanish at x = x0, we conclude that

∂

∂y

(v

u

) 0. (8.130)

In other words, v/u is a function of x only. From Equations (8.123) and (8.125),

v

u=

β(y)u0(y)

√x0 − x

+ O(1). (8.131)


Hence, if this ratio is a function of x alone then β(y) = (1/2) A u0(y), where A is aconstant: that is,

v(x, y) A u0(y)2√

x0 − x. (8.132)

Finally, because Equation (8.126) yields α = 2 dβ/dy = A du0/dy, we obtain

u(x, y) u0(y) + Adu0

dy√

x0 − x. (8.133)

The previous two expressions specify u and v as functions of x and y near the pointof separation. Beyond the point of separation, that is for x > x0, the expressionsare physically meaningless, because the square roots become imaginary. This im-plies that the solutions of the boundary layer equations cannot sensibly be continuedbeyond the separation point.

The standard boundary conditions at the surface of the obstacle require that u =v = 0 at y = 0. It, therefore, follows from Equations (8.132) and (8.133) that

u0(0) = 0, (8.134)

du0

dy

∣∣∣∣∣y=0= 0. (8.135)

Thus, we obtain the important prediction that both the tangential velocity, u, and itsfirst derivative, ∂u/∂y, are zero at the separation point (i.e., x = x0 and y = 0). Thisresult was originally obtained by Prandtl, although the argument we have used toderive it is due to L. D. Landau (1908–1968) (Landau and Lifshitz 1987).

If the constant A in expressions (8.132) and (8.133) happens to be zero then thepoint x = x0 and y = 0, at which the derivative ∂u/∂y vanishes, has no particularproperties, and is not a point of separation. However, there is no reason, in general,why A should take the special value zero. Thus, in practice, a point on the surface ofan obstacle at which ∂u/∂y = 0 is always a point of separation.

Incidentally, if there were no separation at the point x = x0 (i.e., if A = 0) then wewould have ∂u/∂y < 0 for x > x0. In other words, u would become negative as wemove away from the surface, y being still small. That is, the fluid beyond the pointx = x0 would move tangentially, in the region of the boundary layer immediatelyadjacent to the surface, in the direction opposite to that of the external flow: that is,there would be “back-flow” in this region. In practice, the flow separates from thesurface at x = x0, and the back-flow migrates into the wake.

The dimensionless boundary layer equations, (8.23)–(8.27), are independent ofthe Reynolds number of the external flow (assuming that this number is much greaterthan unity). Thus, it follows that the point on the surface of the obstacle at which∂u/∂y = 0 is also independent of the Reynolds number. In other words, the locationof the separation point is independent of the Reynolds number (as long as this numberis large, and the flow in the boundary layer is non-turbulent).

At y = 0, the equation of tangential motion in the boundary layer, (8.111), iswritten

ν∂ 2u∂y 2

∣∣∣∣∣∣y=0= − 1

UdUdx=

1ρ

dPdx, (8.136)


where P(x) is the pressure just outside the layer, and use has been made of Equa-tion (8.6). Because u is positive, and increases away from the surface (upstream ofthe separation point), it follows that (∂ 2u/∂y 2)y=0 > 0 at the separation point itself,where (∂u/∂y)y=0 = 0. Hence, according to the previous equation,(

dUdx

)x=x0

< 0, (8.137)(dPdx

)x=x0

> 0. (8.138)

In other words, we predict that the external tangential flow is always decelerating atthe separation point, whereas the pressure gradient is always adverse (i.e., such as todecelerate the tangential flow), in agreement with experimental observations.

8.10 Approximate Solutions of Boundary Layer EquationsThe boundary layer equations, (8.110)–(8.113), take the form

∂u∂x+∂v

∂y= 0, (8.139)

u∂u∂x+ v

∂u∂y− U

dUdx= ν

∂ 2u∂y 2 , (8.140)


u(x,∞) = U(x), (8.141)

u(x, 0) = 0, (8.142)

v(x, 0) = 0. (8.143)

Furthermore, it follows from Equations (8.140), (8.142), and (8.143) that

ν∂ 2u∂y 2

∣∣∣∣∣∣y=0= −U

dUdx. (8.144)

The previous expression can be thought of as an alternative form of Equation (8.143).As we saw in Section 8.4, the boundary layer equations can be solved exactly whenU(x) takes the special form U0 x m. However, in the general case, we must resort toapproximation methods.

Following Pohlhausen (Schlichting 1987), let us assume that

u(x, y)U(x)

= f (η), (8.145)


where η = y/δ(x), and ∂/∂x 1/δ. In particular, suppose that

f (η) =

a + b η + c η 2 + d η 3 + e η 4 0 ≤ η ≤ 11 η > 1

, (8.146)

where a, b, c, d, and e are constants. This expression automatically satisfies theboundary condition (8.141). Moreover, the boundary conditions (8.142) and (8.144)imply that a = 0, and

f ′′(0) = −Λ(x), (8.147)

where ′ = d/dη, and

Λ =δ 2

ν

dUdx. (8.148)

Finally, let us assume that f , f ′, and f ′′ are continuous at η = 1: that is,

f (1) = 1, (8.149)

f ′(1) = 0, (8.150)

f ′′(1) = 0. (8.151)

These constraints corresponds to the reasonable requirements that the velocity, vor-ticity, and viscous stress tensor, respectively, be continuous across the layer. Giventhat a = 0, Equations (8.146), (8.147), and (8.149)–(8.151) yield

f (η) = F(η) + ΛG(η) (8.152)

for 0 ≤ η ≤ 1, where

F(η) = 1 − (1 − η) 3 (1 + η), (8.153)

G(η) =16η (1 − η) 3. (8.154)

Thus, the tangential velocity profile across the layer is a function of a single param-eter, Λ, which is termed the Pohlhausen parameter. The behavior of this profile isillustrated in Figure 8.13. Note that, under normal circumstances, the Pohlhausenparameter must lie in the range −12 ≤ Λ ≤ 12. For Λ > 12, the profile is such thatf (η) > 1 for some η < 1, which is not possible in a steady-state solution. On the otherhand, for Λ < −12, the profile is such that f ′(0) < 0, which implies flow reversalclose to the wall. As we have seen, flow reversal is indicative of separation. Indeed,the separation point, f ′(0) = 0, corresponds to Λ = −12. Expression (8.152) is onlyan approximation, because it satisfies some, but not all, of the boundary conditionssatisfied by the true velocity profile. For instance, differentiation of Equation (8.140)with respect to y reveals that (∂ 3u/∂y 3)y=0 ∝ f ′′′(0) = 0, which is not the case forexpression (8.152).

It follows from Equations (8.115), (8.116), and (8.152)–(8.154) that

δ1(x) = δ∫ 1

0(1 − f ) dη = δ

(3

10−Λ

120

), (8.155)

δ2(x) = δ∫ 1

0f (1 − f ) dη = δ

(37315− Λ

945− Λ 2

9072

). (8.156)


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

f

0 0.2 0.4 0.6 0.8 1η

Figure 8.13Pohlhausen velocity profiles for Λ = 12 (solid curve) and Λ = −12 (dashed curve).

Furthermore,

∂u∂y

∣∣∣∣∣y=0=

Uδ

f ′(0) =Uδ

(2 +

Λ

6

). (8.157)

The von Karman momentum integral, (8.117), can be rearranged to give

Uνδ2

dδ2

dx+δ 2

2

ν

dUdx

(2 +

δ1

δ2

)=δ2

U∂u∂y

∣∣∣∣∣y=0. (8.158)

Defining

λ(x) =δ 2

2

ν

dUdx, (8.159)

we obtain

Ud

dx

(λ

dU/dx

)= 2 [F2(λ) − λ 2 + F1(λ)] = F(λ), (8.160)


0

1

F

−0.15 −0.1 −0.05 0 0.05 0.1λ

Figure 8.14The function F(λ) (solid curve) and the linear function 0.47 − 6 λ (dashed line).

where

λ =

(37

315− Λ

945− Λ 2

9072

)2Λ, (8.161)

F1(λ) =δ1

δ2=

(3

10− Λ

120

) /(37

315− Λ

945− Λ 2

9072

), (8.162)

F2(λ) =δ2

U∂u∂y

∣∣∣∣∣y=0=

(2 +

Λ

6

) ( 37315−Λ

945−Λ 2

9072

), (8.163)

F(λ) = 2(

37315−Λ

945−Λ 2

9072

) [2 −

116315

Λ +

(2

945+

1120

)Λ 2 +

29072

Λ 3].

(8.164)

It is generally necessary to integrate Equation (8.158) from the stagnation point atthe front of the obstacle, through the point of maximum tangential velocity, to theseparation point on the back side of the obstacle. At the stagnation point we haveU = 0 and dU/dx 0, which implies that F(λ) = 0. Furthermore, at the point ofmaximum tangential velocity we have dU/dx = 0 and U 0, which implies thatΛ = λ = 0. Finally, as we have already seen, Λ = −12 at the separation point, whichimplies, from Equation (8.161), that λ = −0.1567.

As was first pointed out by Walz (Schlichting 1987), and is illustrated in Fig-ure 8.14, it is a fairly good approximation to replace F(λ) by the linear function


0.47 − 6 λ for λ in the physically relevant range. The approximation is particularlyaccurate on the front side of the obstacle (where λ > 0). Making use of this approxi-mation, Equations (8.159) and (8.160) reduce to the linear differential equation

ddx

U δ 22

ν

= 0.47 − 5dUdx

δ 22

ν, (8.165)


δ 22

ν=

0.47U 6

∫ x

0U 5(x′) dx′, (8.166)

assuming that the stagnation point corresponds to x = 0. It follows that

λ =0.47U 6

dUdx

∫ x

0U 5(x′) dx′. (8.167)

Recall that the separation point corresponds to x = xs, where λ(xs) ≡ λs = −0.1567.Suppose that U(x) = U0, which corresponds to uniform flow over a flat plate.

(See Section 8.5.) It follows from Equations (8.166) and (8.167) that

δ2(x)x=

0.69Re1/2 , (8.168)

where Re = U0 x/ν, and λ = 0. Moreover, according to Equations (8.148) and(8.162), Λ = 0 and δ1/δ2 = 2.55. Hence, the displacement width of the boundarylayer becomes

δ1(x)x=

1.75Re1/2 . (8.169)

This approximate result compares very favorably with the exact result, (8.73).Suppose that x = a θ and U(θ) = 2 U0 sin θ, which corresponds to uniform trans-

verse flow around a circular cylinder of radius a. (See Section 8.8.) Equation (8.167)yields

λ(θ) = 0.47cos θ

sin6 θ

∫ θ

0sin5 θ′ dθ′. (8.170)

Figure 8.15 shows λ(θ) determined from the previous formula. It can be seen thatλ = λs = −0.1567 when θ = θs 108. In other words, the separation point islocated 108 from the stagnation point at the front of the cylinder. This suggests thatthe low pressure wake behind the cylinder is almost as wide as the cylinder itself,and that the associated form drag is comparatively large.

Suppose, finally, that U = U0 x m. If m is negative then, as illustrated in Fig-ure 8.16, this corresponds to uniform flow over the back surface of a semi-infinitewedge whose angle of dip is

θ = − m1 + m

π

2. (8.171)

(See Section 8.4.) It follows from Equation (8.167) that

λ =0.47 m1 + 5 m

= − 0.47 θπ/2 − 4 θ

. (8.172)


0

0.1

0.2λ−

λs

0 20 40 60 80 100 120θ()

Figure 8.15The function λ(θ) for flow around a circular cylinder.

θ

x

Figure 8.16Flow over the back surface of a semi-infinite wedge.


We expect boundary layer separation on the back surface of the wedge when λ <λs = −0.1567. This corresponds to θ > θs, where

θs =π

2(−λs)

0.47 + 4 (−λs) 13. (8.173)

Hence, boundary layer separation can be prevented by making the wedge’s angle ofdip sufficiently shallow: that is, by streamlining the wedge, which has the effect ofreducing the deceleration of the flow on the wedge’s back surface. The critical valueof m (i.e., ms = −0.0125) at which separation occurs in our approximate solutionis very similar to the critical value of m (i.e., m∗ = −0.0905) at which the exactself-similar solutions described in Section 8.4 can no longer be found. This suggeststhat the absence of self-similar solutions for m < m∗ is related to boundary layerseparation.

8.11 Exercises8.1 Fluid flows between two non-parallel plane walls, towards the intersection

of the planes, in such a manner that if x is measured along a wall from theintersection of the planes then U(x) = −U0/x, where U0 is a positive constant.Verify that a solution of the boundary layer equation (8.35) can be found suchthat ψ is a function of y/x only. Demonstrate that this solution yields

u(x, y)U(x)

= F[(U0

ν

)1/2 yx

],

where u = ∂ψ/∂y, andF′′ − F 2 = −1,

subject to the boundary conditions F(0) = 0 and F(∞) = 1. Verify that

F(z) = 3 tanh2(α +

z√

2

)− 2

is a suitable solution of the previous differential equation, where tanh2 α =2/3.

8.2 A jet of water issues from a straight narrow slit in a wall, and mixes withthe surrounding water, which is at rest. On the assumption that the motion isnon-turbulent and two-dimensional, and that the approximations of boundarylayer theory apply, the stream function satisfies the boundary layer equation

ν∂ 3ψ

∂y 3 −∂ψ

∂x∂ 2ψ

∂y 2 +∂ψ

∂y

∂ 2ψ

∂x ∂y= 0.

Here, the symmetry axis of the jet is assumed to run along the x-direction,


whereas the y-direction is perpendicular to this axis. The velocity of the jetparallel to the symmetry axis is

u(x, y) = −∂ψ∂y,

where u(x,−y) = u(x, y), and u(x, y)→ 0 as y → ∞. We expect the momen-tum flux of the jet parallel to its symmetry axis,

M = ρ∫ ∞

−∞u 2 dy,

to be independent of x.

Consider a self-similar stream function of the form

ψ(x, y) = ψ0 x p F(y/x q).

Demonstrate that the boundary layer equation requires that p+q = 1, and thatM is only independent of x when 2 p − q = 0. Hence, deduce that p = 1/3and q = 2/3.

Suppose thatψ(x, y) = −6 ν x 1/3 F(y/x 2/3).

Demonstrate that F(z) satisfies

F′′′ + 2 F F′′ + 2 F′ 2 = 0,

subject to the constraints that F′(−z) = F′(z), and F′(z)→ 0 as z→ ∞. Showthat

F(z) = α tanh(α z)

is a suitable solution, and that

M = 48 ρ ν 2 α 3.

8.3 The growth of a boundary layer can be inhibited by sucking some of the fluidthrough a porous wall. Consider conventional boundary layer theory. As aconsequence of suction, the boundary condition on the normal velocity at thewall is modified to v(x, 0) = −vs, where vs is the (constant) suction velocity.Demonstrate that, in the presence of suction, the von Karman velocity integralbecomes

ν∂u∂y

∣∣∣∣∣y=0= U 2 dδ2

dx+ U

dUdx

(δ1 + 2 δ2) + U vs.

Suppose that

u(x, y) = U(x)

sin(α y) 0 ≤ y ≤ π/(2α)1 y > π/(2α)

,


where α = α(x). Demonstrate that the displacement and momentum widthsof the boundary layer are

δ1 = (π/2 − 1)α−1,

δ2 = (1 − π/4)α−1,

respectively. Hence, deduce that

ν (π/2 − 1)2

δ1= U (1 − π/4)

dδ1

dx+

dUdx

δ1 + (π/2 − 1) vs.

Consider a boundary layer on a flat plate, for which U(x) = U0. Show that,in the absence of suction,

δ1 = (π/2 − 1)(

84 − π

)1/2 (ν xU0

)1/2,

but that in the presence of suction

δ1 =(π/2 − 1) ν

vs.

Hence, deduce that, for a plate of length L, suction is capable of significantlyreducing the thickness of the boundary layer when

vs

U0 1

Re1/2 ,

where Re = U0 L/ν.

9Incompressible Aerodynamics

9.1 IntroductionThis chapter investigates the forces exerted on a stationary obstacle situated in a uni-form, high Reynolds number, subsonic (and, therefore, effectively incompressible—see Section 1.17) wind, on the assumption that the obstacle is sufficiently streamlinedthat there is no appreciable separation of the boundary layer from its back surface.Such an obstacle is termed an airfoil (or aerofoil). Obviously, airfoil theory is fun-damental to the theory of flight. Further information on this subject can be found inMilne-Thomson 1958.

The flow around an airfoil is essentially irrotational and inviscid everywhere apartfrom a thin boundary layer localized to its surface, and a thin wake emitted by itstrailing edge. (See Sections 8.5 and 8.6.) It follows that, for the flow external to theboundary layer and wake, we can write

v = −∇φ, (9.1)

which automatically ensures that the flow is irrotational. Assuming that the flow isalso incompressible, so that ∇ · v = 0, the velocity potential, φ, satisfies Laplace’sequation: that is,

∇ 2φ = 0. (9.2)

The appropriate boundary condition at the surface of the airfoil is that the normalvelocity be zero. In other words, n · ∇φ = 0, where n is a unit vector normal to thesurface. In general, the tangential velocity at the airfoil surface, obtained by solving∇2φ = 0 in the external region, subject to the boundary condition n · ∇φ = 0 onthe surface, is non-zero. Of course, this is inconsistent with the no slip condition,which demands that the tangential velocity be zero at the surface. (See Section 8.2.)However, as described in the previous chapter, this inconsistency is resolved by theboundary layer, across which the tangential velocity is effectively discontinuous, be-ing non-zero on the outer edge of the layer (where it interfaces with the irrotationalflow), and zero on the inner edge (where it interfaces with the airfoil). The disconti-nuity in the tangential velocity across the layer implies the presence of bound vorticescovering the surface of the airfoil (see Section 9.7), and also gives rise to a frictiondrag acting on the airfoil in the direction of the external flow. However, the mag-nitude of this drag scales as Re−1/2, where Re is the Reynolds number of the wind.(See Section 8.5.) Hence, such drag becomes negligibly small in the high Reynolds

257


number limit. In the following, we shall assume that any form drag, due to theresidual separation of the boundary layer at the back of the airfoil, is also negligiblysmall. Moreover, for the sake of simplicity, we shall initially restrict our discussionto two-dimensional situations in which a high Reynolds number wind flows trans-versely around a stationary airfoil of infinite length (in the z-direction) and uniformcross-section (parallel to the x-y plane).

9.2 Theorem of Kutta and ZhukovskiiConsider a two-dimensional airfoil that is at rest in a uniform wind of speed V whosedirection subtends a (clockwise) angle α with the negative x-axis. It follows thatthe wind velocity is V = −V cosα ex + V sinα ey, and the corresponding complexvelocity is dF/dz = V e iα. (See Section 6.4.) The air velocity a great distance fromthe airfoil must tend toward this uniform velocity. Thus, for sufficiently large |z|, wecan write (see Section 6.4)

dFdz= V e iα +

Az+

Bz 2 + · · · . (9.3)

According to Equation (6.172), the circulation, Γ, of air about the airfoil is deter-mined by performing the integral

−Re∮

C

dFdz

dz = Γ (9.4)

around a loop C that lies just above the airfoil surface. However, as discussed inSection 6.10, the value of this integral is unchanged if it is performed around anyloop that can be continuously deformed onto C, while not passing through the airfoilsurface, or crossing a singularity of the complex velocity, dF/dz (i.e., a line source ora z-directed vortex filament). Because (in the high Reynolds number limit in whichthe boundary layer and the wake are infinitely thin) there are no line sources or z-directed vortex filaments external to the airfoil, we can evaluate the integral arounda large circle of radius R, centered on the origin. It follows that z = R e i θ anddz = i R e i θ dθ = i z dθ. Hence,

Γ = −Re(i∮ [

V R e i (θ+α) + A + O(R−1)]

dθ)= 2π Im(A), (9.5)

which implies thatdFdz= V e iα + i

Γ

2π z+

Bz 2 + · · · (9.6)

at large |z|.As discussed in Section 6.11, the net force (per unit length) acting on the airfoil,

Incompressible Aerodynamics 259

L = X ex + Y ey, is determined by performing the Blasius integral,

12

i ρ∮

C

(dFdz

) 2

dz = X − i Y, (9.7)

around a loop C that lies just above the airfoil surface. However, as before, the valueof the integral is unchanged if instead we perform it around a large circle of radiusR, centered on the origin. Far from the airfoil,(

dFdz

) 2

= V 2 e 2 iα + iV Γ e iα

π z+

8π2 V B e iα − Γ 2

4π2 z 2 + O(z−3). (9.8)

So, we obtain

X − i Y =12

i 2 ρ

∮ [V 2 R e i (θ+2α) + i

V Γ e iα

π+ O(R−1)

]dθ = −i e iα ρV Γ, (9.9)

orX + i Y = i e−iα ρV Γ = e i (π/2−α) ρV Γ. (9.10)

In other words, the resultant force (per unit length) acting on the airfoil is of magni-tude ρV Γ, and has the direction obtained by rotating the wind vector through a right-angle in the sense opposite to that of the circulation. This type of force is known aslift, and is responsible for flight. The result (9.10) is known as the theorem of Kuttaand Zhukovskii, after the German scientist M. W. Kutta (1867–1944), and the Rus-sian scientist N. E. Zhukovskii (1847–1921), who discovered it independently. Notethat (at fixed circulation) the lift is independent of the shape of the airfoil. Further-more, according to the Kutta–Zhukovskii theorem, there is zero drag acting on theairfoil (i.e., zero force acting in the direction of the wind). In reality, there is alwaysa small friction drag due to air viscosity, as well as a (hopefully) small form dragdue to residual separation of the boundary layer from the back of the airfoil. There isactually a third type of drag, known as induced drag, that is discussed in Section 9.8.

As discussed in Section 6.11, the net moment per unit length (about the origin),M, acting on the airfoil is determined by performing the integral

Re

−12ρ

∮C

(dFdz

) 2

z dz

= M (9.11)

around a loop C that lies just above the airfoil surface. As before, we can deform Cinto a circle of radius R, centered on the origin, without changing the value of theintegral. Hence, we obtain

M = Re(−1

2i ρ∮ [

V 2 R 2 e 2 i (θ+α) + iV Γ R e i (θ+α)

π

+8π2 V B e iα − Γ 2

4π2 + O(R−1)]

dθ), (9.12)

orM = Re

[2π ρV B e i (α−π/2)

]. (9.13)


9.3 Cylindrical AirfoilsFor the moment, let us work in the complex ζ-plane, where ζ = ξ + i η. Considera cylindrical airfoil with a circular cross-section of radius a, centered on the origin,that is situated in a uniform, high Reynolds number wind of speed V whose directionsubtends a (clockwise) angle α with the negative ξ-axis. Let Γ be the circulation ofair around the airfoil. A slight generalization of the analysis of Section 6.4 revealsthat the appropriate complex velocity potential is

F(ζ) = V(ζ e iα +

a 2

ζe−iα

)+ i

Γ

2πln(ζ

a

), (9.14)

whereas the associated stream function takes the form

ψ(r, θ) = V(r − a 2

r

)sin(θ + α) +

Γ

2πln( ra

), (9.15)

where ζ = r e i θ. It follows that

dFdζ= V e iα + i

Γ

2π ζ− V a 2 e−iα

ζ 2 . (9.16)

Comparison with Equation (9.6) (with z = ζ) reveals that

B = −V a 2 e−iα. (9.17)

Hence, Equations (9.10) and (9.13) yield

V = V(− cosα eξ + sinα eη

), (9.18)

L = ρV Γ(sinα eξ + cosα eη

), (9.19)

M = 0, (9.20)

where V is the wind vector, L the lift vector, and M the moment of the lift vectorabout the origin. We conclude that, for a cylindrical airfoil of circular cross-section,the lift vector is normal to the wind vector, and the line of action of the lift passesthrough the centroid of the cross-section (because the lift generates zero momentabout the origin). (See Figure 9.1.)

Of course, a cylindrical airfoil of circular cross-section is completely unrealistic,because its back side (i.e., the side opposite to that from which the wind is incident)is not sufficiently streamlined to prevent boundary layer separation. (See Chapter 8.)However, as described in Section 6.7, we can use the conformal transformation

z = ζ +l 2

ζ(9.21)


α ξ

η

a

V

LΓ

Figure 9.1A cylindrical airfoil of circular cross-section.

to transform a cylinder of circular cross-section in the ζ-plane to a cylinder of ellipti-cal cross-section in the z-plane. (Note that both cross-sections have centroids locatedat the origin.) Moreover, a cylindrical airfoil of elliptical cross-section that is suffi-ciently elongated, and whose major axis subtends a sufficiently small angle with theincident wind direction, constitutes a realistic airfoil, because its back side is, for themost part, closely aligned with the external flow.

An elliptical airfoil of width c and thickness δ < c, as shown in Figure 9.2, isobtained when the parameters a and l are given the following values:

a =14

(c + δ), (9.22)

l =14

(c 2 − δ 2)1/2. (9.23)

In this case, the surface of the airfoil satisfies the parametric equations

x =c2

cos θ, (9.24)

y =δ

2sin θ. (9.25)

In particular, the airfoil’s leading and trailing edges correspond to θ = 0 and θ = π,respectively.

According to Equations (9.16) and (9.21), the complex velocity in the z-plane is


α d

Γ

M

y

xδ

c

L

V

FC

Figure 9.2A cylindrical airfoil of elliptical cross-section.

given bydFdz=

dFdζ

dζdz=

[V e iα + i

Γ

2π ζ− V a 2 e−iα

ζ 2

] (ζ 2

ζ 2 − l 2

). (9.26)

Thus, on the airfoil surface, where ζ = a e i θ, we obtain

dFdz= i[V sin(α + θ) +

Γ

π (c + δ)

](c + δ)

(δ cos θ + i c sin θ). (9.27)

A long way from the airfoil, ζ z − l 2/z, so that Equation (9.26) reduces to

dFdz V e iα + i

Γ

2π z+

V (l 2 e iα − a 2 e−iα)z 2 . (9.28)

A comparison with Equation (9.6) reveals that the circulation of air around the airfoiltakes the same value, Γ, in both the complex ζ- and z-planes. In other words, theconformal transformation (9.21) does not modify the circulation. The transformationalso does not modify the external wind speed or direction [because, from (9.16) and(9.28), dF/dζ = dF/dz = V e iα at very large |ζ | and |z|]. On the other hand, it isclear that the constant B, which takes the value zero in the complex ζ-plane, takesthe value

B = V (l 2 e iα − a 2 e−iα) (9.29)


in the complex z-plane. Hence, Equations (9.10) and (9.13) reveal that

V = V e‖, (9.30)

L = ρV Γ e⊥, (9.31)

M =π

8ρV 2 (c 2 − δ 2) sin(2α), (9.32)

where V is the wind vector, L the lift vector, and M the moment of the lift vectorabout the origin. Here,

e‖ = − cosα ex + sinα ey (9.33)

is a unit vector parallel to the incident wind direction, and

e⊥ = e‖ × ez = sinα ex + cosα ey (9.34)

is a unit vector perpendicular to the wind direction. We conclude that, for a cylindri-cal airfoil of elliptic cross-section, the lift vector is normal to the wind vector, but theline of action of the lift intersects the major axis of the airfoil a distance

d =M

L cosα=

14

(c − δ) sinα

Γ(9.35)

in front of the cross-section’s centroid, C, where Γ = Γ/[πV (c+δ)]. (See Figure 9.2.)Incidentally, the point, F, at which the line of action of the lift intersects the airfoil’smajor axis is conventionally termed the focus of the airfoil.

9.4 Zhukovskii’s HypothesisAccording to the previous analysis, the lift acting on a cylindrical airfoil of ellipticcross-section, situated in a uniform, high Reynolds number wind, depends on thecirculation, Γ, of air about the airfoil. But, how can we determine the value of thiscirculation?

Figure 9.3 shows the boundary layer and wake of a streamlined airfoil. Theboundary layer, which is localized on the surface of the airfoil, has a vortex intensityper unit length in the z-direction equal to U, where U is the tangential air speedimmediately above the layer. [See Equation (9.64).] Moreover, the wake is emittedby the airfoil’s trailing edge, and subsequently convected by the external air flow.(See Section 8.6.) Note that the flow is irrotational everywhere apart from inside theboundary layer and the wake. According to the analysis of Section 5.8, the rate ofchange of the circulation, Γ, around some curve C that encloses the airfoil is equalto minus the flux of z-directed vorticity across this curve: that is,

dΓdt= −

∮ωz v · dS. (9.36)


AirfoilTrailing Edge

Boundary Layer

Wake

CdS

V

U

Figure 9.3The boundary layer and wake of a streamlined airfoil. Only shaded regions possesnon-zero vorticity.

Here, v is the wind velocity,ω the wind vorticity, and dS an outward surface element(of unit depth in the z-direction) lying on C. We expect the vorticity flux to beindependent of the size and shape of C, otherwise the circulation of the flow, Γ,about the airfoil would not have a unique value. In the limit that C becomes verylarge, v→ V, where V is the incident wind velocity. Thus,

dΓdt= −V Ωz, (9.37)

where V is the wind speed, and Ωz the vortex intensity per unit length in the wake(at the point where it crosses the curve C). Here, we are assuming that the vorticitywithin the wake is convected by the flow, giving rise to a net flux of vorticity across C.Because the wake is essentially an extension of the boundary layer, it is reasonable toassume that its vortex intensity per unit length is proportional to that in the boundarylayer at the airfoil’s trailing edge, where the wake and boundary layer intersect. Inother words, Ωz = k U0, where U0 is the tangential velocity immediately above thetrailing edge of the airfoil, and k is a constant. It follows that

dΓdt= −k V U0. (9.38)

According to Equation (9.27), the tangential velocity just above the surface of the


airfoil is

U(θ) = Re[

c sin θ − i δ cos θ(c 2 sin2 θ + δ 2 cos2 θ)1/2

dFdz

]|ζ |=a

=

[V sin(α + θ) +

Γ

π (c + δ)

](c + δ)

(c 2 sin2 θ + δ 2 cos2 θ)1/2. (9.39)

Hence, given that the airfoil’s trailing edge corresponds to θ = π, we obtain

U0 =

[Γ

π (c + δ)− V sinα

] (c + δδ

). (9.40)

Thus, Equation (9.38) yields

dΓdt= −Γ + sinα, (9.41)

where Γ = Γ/[πV (c+ δ)], t = t/t0, and t0 = π δ/(k V). Assuming that the circulationof the flow about the airfoil is initially zero (i.e., Γ = 0 at t = 0), the previousequation can be solved to give

Γ = sinα[1 − exp(−t)

]. (9.42)

Clearly, as t → ∞ the normalized circulation Γ asymptotes to the constant value

Γ∞ = sinα. (9.43)

The corresponding constant value of the unnormalized circulation is

Γ∞ = πV (c + δ) sinα. (9.44)

According to Equation (9.40), when the circulation,Γ, takes the value Γ∞ the tangen-tial velocity at the airfoil’s training edge, U0, is zero. In other words, the steady-statecirculation set up around the airfoil is such as to render its trailing edge a stagna-tion point of the flow. This conclusion is known as Zhukovskii’s hypothesis, after itsdiscoverer N. E. Zhukovskii (1847–1921).

Incidentally, it should be clear, from the previous discussion, that the air circula-tion about the airfoil is only able to change its value because of the presence of theboundary layer, and the associated wake that trails from the airfoil’s trailing edge.This follows because the flow is irrotational everywhere except within the boundarylayer and the wake. Moreover, as we have seen, a change in circulation is necessarilyassociated with a net vorticity flux away from the airfoil, and such a flux cannot becarried by an irrotational wind. Thus, in the absence of the boundary layer and thewake, the air circulation about the airfoil would be constrained to remain zero (as-suming that it was initially zero), in accordance with the Kelvin circulation theorem.(See Section 5.8.) This implies, from Equation (9.31), that zero lift would act onthe airfoil, irrespective of its shape, and irrespective of the incident wind speed or


−2

−1

0

1

2

y/c

−2 −1 0 1 2x/c

Figure 9.4Streamlines around a slender cylindrical airfoil of elliptic cross-section situated ina uniform, high Reynolds number wind. The parameters for this calculation areδ/c = 0.1, α = π/12, and Γ = 0.

direction. In other words, flight would be impossible. Fortunately, as long as the tan-gential air velocity at the trailing edge of the airfoil is non-zero, the wake that trailsbehind the airfoil carries a net flux of z-directed vorticity, which causes the airfoilcirculation to evolve in time. This process continues until the circulation becomessuch that the tangential velocity at the airfoil’s trailing edge is zero: that is, suchthat the trailing edge is a stagnation point. Thereafter, the circulation remains con-stant (assuming that the wind speed and direction remain constant). Figures 9.4 and9.5 show the streamlines of the flow around a slender cylindrical airfoil of ellipticcross-section situated in a uniform, high Reynolds number wind whose direction ofincidence is slightly inclined to the airfoil’s major axis. In the first figure, the aircirculation about the airfoil is zero. In the second figure, the circulation is such as tomake the trailing edge of the airfoil a stagnation point.

According to Equations (9.31) and (9.44), when the air circulation about the air-foil has attained its steady-state value, Γ∞, the lift acting on the airfoil becomes

L = ρV Γ∞ = π ρV 2 (c + δ) sinα. (9.45)

The lift is positive (i.e., upward) when α > 0 (i.e., when the wind is incident on theairfoil’s bottom surface), negative (i.e., downward) when α < 0 (i.e., when the windis incident on the airfoil’s top surface), and zero when α = 0 (i.e., when the wind isincident parallel to airfoil’s major axis). Incidentally, the angle α is conventionally


−2

−1

0

1

2

y/c

−2 −1 0 1 2x/c

Figure 9.5Streamlines around a slender cylindrical airfoil of elliptic cross-section situated ina uniform, high Reynolds number wind. The parameters for this calculation areδ/c = 0.1, α = π/12, and Γ = Γ∞.

termed the angle of attack. Finally, from Equations (9.35) and (9.43), the focus ofthe airfoil is located a distance

d =14

(c − δ)sinα

Γ∞=

14

(c − δ) (9.46)

in front of the centroid of its cross-section. In the limit that the airfoil becomes verythin (i.e., δ c), this distance asymptotes to c/4. Thus, we conclude that the focusof a thin airfoil, which is defined as the point of action of the lift, is located onequarter of the way along the airfoil from its leading edge.

The previous analysis is premised on the assumption that there is no appreciableseparation of the boundary layer from the back of the airfoil, which implies the ne-glect of form drag. We can check that this assumption is reasonable by calculatingthe approximate locations of the boundary layer separation points using the analysisof Section 8.10. Let s represent arc-length along the surface of the airfoil, measuredfrom the front stagnation point. Assuming that, in accordance with Zhukovskii’shypothesis, the circulation is such that Γ = Γ∞, this stagnation point is located atθ = θ0, where θ0 = −2α. [See Equation (9.49).] It follows that

ds = (dx 2 + dy 2)1/2 =12

h(θ) dθ, (9.47)


0

180

θ ±(

)

0 10 20α()

Figure 9.6The angular locations of the boundary layer separation points, θ±, calculated as afunction of the angle of attack, α, for a cylindrical airfoil of elliptic cross-sectionsituated in a uniform, high Reynolds number wind. The solid, dashed, short-dash–dotted, and long-dash–dotted curves correspond to airfoils of ellipticity δ/c = 1.0,0.5, 0.25, and 0.125, respectively. The trailing edge of the airfoil is located at θ =180.

whereh(θ) = (c 2 sin2 θ + δ 2 cos2 θ)1/2. (9.48)

Moreover, from Equations (9.39) and (9.44), the tangential air speed just above thesurface of the airfoil can be written

U(θ) = V (c + δ)f (θ)h(θ)

, (9.49)

withf (θ) = sin(α + θ) + sinα. (9.50)

In addition, it can be shown that

ddθ

ln(

fh

)= g(θ), (9.51)

where

g(θ) =cos(α + θ)

f (θ)−

(c 2 − δ 2) cos θ sin θh 2(θ)

. (9.52)


vz(y = 0−)dz

ω

vz(y = 0+)

y

z

Figure 9.7Side view of a vortex sheet.

According to the analysis of Section 8.10, the separation points are located at θ = θ±,where λ(θ±) = −0.1567, and

λ(θ) = 0.47h 4(θ) g(θ)

f 5(θ)

∫ θ

θ0

f 5(θ′)h 4(θ′)

dθ′. (9.53)

Here, π > θ+ > θ0 and θ0 > θ− > −π. Moreover, the x- and y-coordinates of the sep-aration points are x± = (c/2) cos θ± and y± = (δ/2) sin θ±, respectively. Figure 9.6shows the angular locations of the separation points, calculated as a function of theangle of attack, for cylindrical airfoils of various different ellipticity, δ/c. (Note thatθ− has been re-expressed as an angle in the range 0 to 2π.) It can be seen that for abluff airfoil (e.g., δ/c = 1) the angular distance between the separation points is large,indicating the presence of a wide wake, and a high associated form drag (because themagnitude of form drag is roughly proportional to the width of the wake). On theother hand, for a slender airfoil (e.g., δ/c = 0.125) the angular distance between theseparation points is much smaller, indicating the presence of a narrow wake, and alow associated form drag. However, in the latter case, as the angle of attack is grad-ually increased from zero, there is an initial gradual increase in the angular distancebetween the separation points, followed by an abrupt, and very large, increase. Wewould expect there to be a similar gradual increase in the form drag, followed by anabrupt, and very large, increase. The value of the angle of attack at which this abruptincrease occurs is termed the critical angle of attack. We conclude that the previousanalysis, which neglects form drag, is valid only for slender airfoils whose angles ofattack do not exceed the critical value (which is generally only a few degrees).

9.5 Vortex SheetsA vortex sheet is defined as a planar array of parallel vortex filaments. Consider auniform vortex sheet, lying in the x-z plane, in which the vortex filaments run parallelto the x-axis. (See Figure 9.7.) The vorticity within the sheet can be written

ω = Ωx δ(y) ex, (9.54)


where δ(y) is a Dirac delta function. Here, Ω = Ωx ex is the sheet’s vortex intensityper unit length. Let vz(y = 0+) and vz(y = 0−) be the z-component of the fluid velocityimmediately above and below the sheet, respectively. Consider a small rectangularloop in the y-z plane that straddles the sheet, as shown in the figure. Integration ofω = ∇ × v around the loop (making use of the curl theorem) yields

∆vz ≡ vz(y = 0+) − vz(y = 0−) = Ωx. (9.55)

In other words, a vortex sheet induces a discontinuity in the tangential flow acrossthe sheet. The previous expression can easily be generalized to give

Ω = n × ∆v, (9.56)

where Ω is the sheet’s vortex intensity per unit length, n is a unit vector normal tothe sheet, and ∆v is the jump in tangential velocity across the sheet (traveling in thedirection of n). Furthermore, it is reasonable to assume that the previous relationholds locally for non-planar and non-uniform vortex sheets.

9.6 Induced FlowA vortex filament is necessarily associated with fluid flow circulating about the fila-ment. Let us determine the relationship between the filament vorticity and the flowfield that it induces. This problem is mathematically identical to determining themagnetic field generated by a current filament. In the latter case, the Maxwell equa-tion

µ0 j = ∇ × B (9.57)

can be inverted to give the well-known Biot-Savart law (Fitzpatrick 2008)

B(r) =1

4π

∫j(r′) × (r − r′)|r − r′| 3

d 3r′. (9.58)

Here, j is the current density, and B the magnetic field-strength. By analogy, giventhat vorticity is related to fluid velocity via the familiar relation

ω = ∇ × v, (9.59)

we can writev(r) =

14π

∫ω(r′) × (r − r′)|r − r′| 3

d 3r′. (9.60)

This expression allows us to determine the flow field induced by a given vorticitydistribution. In particular, for a vortex filament of intensity Γ the previous expressionreduces to

v(r) =1

4π

∫Γ(r′) × (r − r′)|r − r′| 3

dl′, (9.61)


where dl is an element of length along the filament. Likewise, for a vortex sheet ofintensity per unit length Ω, we obtain

v(r) =1

4π

∫Ω(r′) × (r − r′)|r − r′| 3

dS ′, (9.62)

where dS is an element of area of the sheet.

9.7 Three-Dimensional AirfoilsLet us now take into account the fact that realistic three-dimensional airfoils are offinite size. Consider Figure 9.8, which shows a top view of a stationary airfoil offinite size, situated in a (predominately) horizontal wind of velocity V = V e‖. In thefollowing, we shall sometimes refer to such an airfoil as a wing (although it actuallyrepresents a pair of wings on a standard fixed wing aircraft). Let us adopt the coor-dinate system shown in the figure, which is such that the x-z plane is horizontal, thewind is incident predominately from the x-direction, and the y-axis points verticallyupward. The wing is assumed to lie in the x-z plane. Let b be the wingspan, andlet c(z) and δ(z) be the width and thickness, respectively, of the wing cross-section(parallel to the x-y plane). (See Figure 9.9.) Suppose that the wing is symmetricabout the median plane, z = 0, so that c(−z) = c(z) and δ(−z) = δ(z). It follows thatc(z > b/2) = δ(z > b/2) = 0: that is, the wing extends from z = −b/2 to z = b/2.

Suppose that air circulation is set up around the wing parallel to the x-y planein such a manner as to produce an upward lift. It follows that the average pressureon the lower surface of the wing must exceed that on its upper surface. ConsiderFigure 9.9, which shows a back view of the airfoil shown in Figure 9.8. As we gofrom the median plane (z = 0) to a wing tip, Y, whether along the upper or the lowersurface of the wing, we must arrive at the same pressure at Y. It follows that thereis a drop in pressure as we move outward, away from the median plane, along thewing’s bottom surface, and a further drop in pressure as we move inward, towardthe median plane, along the upper surface. Because air is pushed in the direction ofdecreasing pressure, it follows that the air that impinges on the wing’s leading edge,and then passes over its upper surface, deviates sideways toward the median plane.Likewise, the air that passes over the wing’s lower surface deviates sideways awayfrom the median plane. (See Figure 9.8.)

The air that leaves the trailing edge of the wing at some point Q must have im-pinged on the leading edge at the different points P and P′, depending on whetherit travelled over the wing’s upper or lower surfaces, respectively. Moreover, air thattravels to Q via the wing’s upper surface acquires a small sideways velocity directedtowards the median plane, whereas that which travels to Q via the lower surface ac-quires a small sideways velocity directed away from the median plane. On the otherhand, the air speed at Q must be the same, irrespective of whether the air arrivesfrom the wing’s upper or lower surface, because the pressure (which, according to


V

airfoil

z = −b/2 z = 0 z = b/2

c(z)

x

z

Figure 9.8Top view of a three-dimensional airfoil of finite size.

δ(z)

z

y

b

Y Y ′−+

Figure 9.9Back view of a three-dimensional airfoil of finite size, indication the pressure varia-tion over its surface.


z P ′P

Q Q

x

Figure 9.10Top view of the airflow over the top (left) and bottom (right) surfaces of a three-dimensional airfoil of finite size.

Bernoulli’s theorem, depends on the air speed) must be continuous at Q. Thus, weconclude that there is a discontinuity in the direction of the air emitted by the trail-ing edge of a wing. This implies that the interface, Σ, between the two streams ofair that travel over the upper and lower surfaces of the wing is a vortex sheet. (SeeSection 9.5.) Of course, this vortex sheet constitutes the wake that trails behind theairfoil. Moreover, we would generally expect the wake to be convected by the inci-dent wind. It follows that the vorticity per unit length in the wake can be written

ΩΣ = −I(z) e‖, (9.63)

where I(z) = −∆vz, and ∆vz is tangential velocity discontinuity across the wake. [SeeEquation (9.56).]

As we saw previously, the boundary layer that covers the airfoil is such that thetangential velocity U just outside the layer is sharply reduced to zero at the airfoil sur-face. Actually, the nature of the substance enclosed by the surface is irrelevant to ourargument, and nothing is changed in our analysis if we suppose that this region con-tains air at rest. Thus, we can replace the airfoil by air at rest, and the boundary layerby a vortex sheet, S , with a vortex intensity per unit lengthΩS that is determined bythe velocity discontinuity U between the air just outside the boundary layer and thatat rest in the region where the airfoil was previously located. In fact, Equation (9.56)yields

ΩS = n × U, (9.64)

where n is an outward unit normal to the airfoil surface.We conclude that a stationary airfoil situated in a uniform wind of constant ve-

locity is equivalent to a vortex sheet S , located at the airfoil surface, and a wakeΣ that trails behind the airfoil, the airfoil itself being replaced by air at rest. Thevorticity within S is largely parallel to the z-axis [because n and U are both essen-tially parallel to the x-y plane—see Equation (9.64)], whereas that in Σ is parallelto the incident wind direction. (See Figure 9.11.) The vortex filaments within S aregenerally termed bound filaments (because they cannot move off the airfoil surface).Conversely, the vortex filaments within Σ are generally termed free filaments. The


V

Σ

x

yS

z

Figure 9.11Vortex structure around a wing.

air velocity both inside and outside S can be written

v = V + vΣ + vS , (9.65)

where V is the external wind velocity, vΣ the velocity field induced by the free vortexfilaments that constitute Σ, and vS the velocity field induced by the bound filamentsthat constitute S .

Consider some point P that lies on S . Let P+ and P− be two neighboring pointsthat are equidistant from P, where P+ lies just outside S , and P− lies just inside S ,and the line P−P+ is normal to S . We can write

v(P+) = V + vΣ(P+) + vS (P+), (9.66)

v(P−) = V + vΣ(P−) + vS (P−). (9.67)

However, v(P+) = U(P), where U(P) is the tangential air velocity just above pointP on the airfoil surface, and v(P−) = 0, as the air within S is stationary. Moreover,vΣ(P+) = vΣ(P−) = vΣ(P), because we expect vΣ to be continuous across S . On theother hand, we expect the tangential component of vS to be discontinuous across S .Let us define

vS (P) =12

[vS (P+) + vS (P−)] . (9.68)

This quantity can be identified as the velocity induced at point P by the bound vor-tices on S , excluding the contribution from the local bound vortex at P (because thisvortex induces equal and opposite velocities at P+ and P−). Finally, taking half thesum of Equations (9.66) and (9.67), we obtain

12

U(P) = V + vΣ(P) + vS (P). (9.69)


9.8 Aerodynamic ForcesThe net aerodynamic force acting on an three-dimensional airfoil of finite size canbe written

A = −∫

Sp n dS , (9.70)

where the integral is taken over the surface of the airfoil, S . Here, n is an outwardunit normal vector on S , dS is an element of S , and p is the air pressure. FromBernoulli’s theorem (in an irrotational fluid), we can write p = p0− (1/2) ρ v 2, wherep0 is a constant pressure. Because a constant pressure exerts no net force on a closedsurface, we get

A =12ρ

∫S

U 2 n dS , (9.71)

where U is the tangential air velocity just above the surface of the airfoil. Now,

U × (n × U) = U 2 n − (n · U) U = U 2 n, (9.72)

because n · U = 0 on the surface. Hence,

A =12ρ

∫S

U × (n × U) dS . (9.73)

Making use of Equations (9.64) and (9.69), the previous expression can be written

A = ρ∫

S(V + vΣ + vS ) ×ΩS dS = L + D + F, (9.74)

where

L = ρV ×∫

SΩS dS , (9.75)

D = ρ∫

SvΣ ×ΩS dS , (9.76)

F = ρ∫

SvS ×ΩS dS . (9.77)

Here, V, vΣ , and vS are the incident wind velocity, the velocity induced by the freevortices in the wake, and the velocity induced by the bound vortices covering thesurface of the airfoil, respectively. The forces L and D are called the lift and theinduced drag, respectively. (Note, that L now represents a net force, rather than aforce per unit length.) We shall presently demonstrate that the force F is negligible.

Let us assume thatΩS Ωz ez : (9.78)

that is, that the bound vortices covering the surface of the airfoil run parallel to thez-axis. This assumption is exactly correct for an airfoil of infinite wingspan and


constant cross-section. Moreover, it is a good approximation for an airfoil of finitewingspan, provided the airfoil’s length greatly exceeds its width (i.e., b c). Now,the incident wind velocity is written V = V e‖. Moreover, dS = dl dz, where dl is anelement of length that runs parallel to the x-y plane whilst lying on the airfoil surface.Making use of the curl theorem, we can easily show that∮

CΩz dl = Γ(z), (9.79)

where the closed curve C is the intersection of the airfoil surface with the plane z = z,and Γ(z) is the air circulation about the airfoil in this plane. Thus, it follows fromEquation (9.75) that

L = ρV∫ b/2

−b/2Γ(z) dz e⊥. (9.80)

This expression is the generalization of Equation (9.31) for a three-dimensional air-foil of finite size. As before, the lift is at right-angles to the incident wind direction.

Let us make the further assumption—known as the lifting line approximation(because the lifting action of the wing is effectively concentrated onto a line)—that

vΣ = −w(z) e⊥ (9.81)

throughout S , where −w(z) e⊥ is the induced velocity due to the free vortices in Σ,evaluated at the trailing edge of the airfoil. Here, the velocity w(z) is called thedownwash velocity. It follows from Equation (9.76) that

D = ρ∫ b/2

−b/2w(z)Γ(z) dz e‖. (9.82)

Note that the induced drag is parallel to the incident wind direction. The origin ofinduced drag is as follows. It takes energy to constantly resupply free vortices tothe wake, as they are swept downstream by the wind (note that a vortex filamentpossesses energy by virtue of the kinetic energy of its induced flow pattern), and thisenergy is supplied by the work done in opposing the induced drag. The drag actingon a well-designed airfoil (i.e., an airfoil with an aerodynamic shape that minimizesform drag) situated in a high Reynolds number wind (which implies that the frictiondrag is negligible) is generally dominated by induced drag.

According to Equations (9.62) and (9.77), the force F is written

F =ρ

4π

∫S

∫S ′

[ΩS (r′) × (r − r′)] ×ΩS (r)|r − r′| 3

dS dS ′. (9.83)

We can interchange primed and unprimed variables without changing the value ofthe integral. Hence,

F =ρ

4π

∫S

∫S ′

[ΩS (r) × (r′ − r)] ×ΩS (r′)|r′ − r| 3

dS ′ dS . (9.84)


z

l O

z

P

Γx

r − r′φ

Figure 9.12Semi-infinite vortex filament.

Taking the half the sum of the previous two equations, we obtain

F =ρ

8π

∫S

∫S ′

[ΩS (r′) × (r − r′)] ×ΩS (r) + [(r − r′) ×ΩS (r)] ×ΩS (r′)|r − r′| 3

dS dS ′.

(9.85)However, (a × b) × c + (b × c) × a + (c × a) × b = 0. Thus, the previous expressionyields

F =ρ

8π

∫S

∫S ′

[ΩS (r′) ×ΩS (r)] × (r − r′)|r − r′| 3

dS dS ′. (9.86)

But, the assumption (9.78) implies that ΩS (r′) ×ΩS (r) 0. Hence, F is negligible,as was previously stated.

Consider a closed surface covering the small section of the airfoil lying betweenthe parallel planes z = z and z = z + dz. The flux of vorticity into the surface dueto bound vortices at z is Γ(z). The flux of vorticity out of the surface due to boundvortices at z+dz is Γ(z+dz). Finally, the flux of vorticity out of the surface due to thefree vortices in the part of the wake lying between z and z + dz is I(z) dz. However,the net flux of vorticity out of a closed surface is zero, because vorticity is divergencefree. Hence,

Γ(z) = Γ(z + dz) + I(z) dz, (9.87)

which implies that

I(z) = −dΓdz. (9.88)

Finally, consider a semi-infinite straight vortex filament of vortex intensity Γ =−Γ ex that terminates at the origin, O, as shown in Figure 9.12. Let us calculate theflow velocity induced by this filament at the point P = (0, 0, z). From the diagraml = z tanφ, dl = z sec2 φ dφ, |r − r′| = z sec φ, and Γ × |r − r′| = Γ z ey. Hence, fromEquation (9.61), the induced velocity at P is v = vy ey, where

vy =Γ

4π z

∫ π/2

0cosφ dφ =

Γ

4π z. (9.89)


This result allows us to calculate the downwash velocity, w(z) = −vy(z), induced atthe trailing edge of the airfoil by the semi-infinite free vortices in the wake. Thevortex intensity in the small section of the wake lying between z and z+ dz is I(z) dz,so we obtain

w(z) = − 14π

∫ b/2

−b/2

I(z′) dz′

z − z′=

14π

∫dΓ(z′)z − z′

, (9.90)

where use has been made of Equation (9.88).

9.9 Ellipsoidal AirfoilsConsider an ellipsoidal airfoil whose outer surface is specified by the parametricequations

x =c0

2sin φ cos θ, (9.91)

y =δ0

2sin φ sin θ, (9.92)

z = −b2

cosφ, (9.93)

where 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π. Here, b is the wingspan, c0 the maximum wingwidth, and δ0 the maximum wing thickness. Note that the wing’s cross-section iselliptical both in the x-y and the x-z planes. It is assumed that b > c0 δ0: that is,the wingspan is greater than the wing width, which in turn is much greater than thewing thickness. At fixed φ (i.e., fixed z), the width and thickness of the airfoil arec(φ) = c0 sin φ and δ(φ) = δ0 sin φ, respectively.

Assuming that the two-dimensional result (9.44) holds at fixed z, we deduce thatthe air circulation about the wing satisfies

Γ(z) = πV c(z) sinα = Γ0 sinφ, (9.94)

whereΓ0 πV c0 α. (9.95)

Here, the angle of attack, α, is assumed to be small. From Equations (9.90) and(9.94), the downwash velocity in the region |z| < b/2 is given by

w(φ) =Γ0

2π b

∫ π

0

cosφ′ dφ′

cosφ′ − cosφ=Γ0

2 b

(1 +

cosφπ

∫ π

0

dφ′

cosφ′ − cosφ

). (9.96)

The integrand appearing in the integral∫ π

0

dφ′

cosφ′ − cosφ(9.97)


is singular when φ′ = φ. However, we can still obtain a finite value for the integralby taking its Cauchy principal part: that is,

limε→0

(∫ φ−ε

0

dφ′

cosφ′ − cosφ+

∫ π

φ+ε

dφ′

cosφ′ − cosφ

). (9.98)

Physically, this is equivalent to omitting the contribution of the local free vortex ata given point on the airfoil’s trailing edge to the downwash velocity induced at thatpoint, which is reasonable because a vortex induces zero velocity at its center. Hence,we obtain ∫ π

0

dφ′

cosφ′ − cosφ= limε→0

(

1sin φ

ln[sin (1/2) (φ + φ′)sin (1/2) (φ − φ′)

])φ−ε0

+

(1

sinφln[sin (1/2) (φ′ + φ)sin (1/2) (φ′ − φ)

])φ+επ

= limε→0

(1

sin φln[sin(φ − ε/2)sin(φ + ε/2)

])= 0, (9.99)

which implies that

w(φ) =Γ0

2 b. (9.100)

In the region |z| > b/2, we can write η = 2 z/b, so that

w(η) =Γ0

2 b

(1 −

η

π

∫ π

0

dφ′

cosφ′ + η

)=Γ0

2 b

1 − |η|√η 2 − 1

. (9.101)

Hence, we conclude that the downwash velocity profile induced by an ellipsoidalairfoil takes the form

w(z) =Γ0

2 b

1 |z| < b/21 − |z|/(z 2 − b 2/4)1/2 |z| > b/2

. (9.102)

This profile is shown in Figure 9.13. It can be seen that the downwash velocity isuniform and positive in the region between the wingtips (i.e., −b/a < z < b/2),but negative and decaying in the region outside the wingtips. Hence, we concludethat as air passes over an airfoil subject to an upward lift it acquires a net downwardvelocity component, which, of course, is a consequence of the reaction to the lift.On the other hand, the air immediately behind and to the sides of the airfoil acquiresa net upward velocity component. In other words, the lift acting on the airfoil isassociated with a downwash of air directly behind, and an upwash behind and toeither side of, the airfoil. The existence of upwash slightly behind and to the side ofa flying object allows us to explain the V-formation adopted by wild geese—a birdflying in the upwash of another bird needs to generate less lift in order to stay in theair, and, consequently, experiences less induced drag.


−2

−1.6

−1.2

−0.8

−0.4

0

0.4

0.8

1.2

w/(

Γ0/

2b)

−2 −1 0 1 2z/b)

Figure 9.13Downwash velocity profile induced at the trailing edge by an ellipsoidal airfoil.

It follows from Equation (9.93) and (9.94) that∫ b/2

−b/2Γ(z) dz =

Γ0 b2

∫ π

0sin2 φ dφ =

π

4Γ0 b. (9.103)

Hence, Equation (9.80), (9.82), and (9.100) yield the following expression for the liftand induced drag acting on an ellipsoidal airfoil,

L =π

4ρV bΓ0, (9.104)

D =π

8ρΓ 2

0 . (9.105)

The surface area of the airfoil in the x-z plane is

S =π

4b c0. (9.106)

Moreover, the airfoil’s aspect-ratio is conventionally defined as the length to widthratio for a rectangle of length b that has the same area as the airfoil: that is,

A =b 2

S=

4π

bc0. (9.107)


It thus follows from Equation (9.95) that

L = L0 α, (9.108)

D =2A

L0 α2, (9.109)

whereL0 = π ρV 2 S . (9.110)

9.10 Simple Flight ProblemsFigure 9.14 shows a side-view schematic of a fixed wing aircraft flying in a straight-line at constant speed through stationary air. Here, x is a horizontal coordinate, andy a vertical coordinate. The center of mass of the aircraft is assumed to be movingwith some fixed velocity V that subtends an angle δ with the horizontal. Thus, thewind velocity in the aircraft’s rest frame is −V. Let the aircraft’s wings, which areassumed to be parallel to its fuselage, be inclined at an angle α+δ to the horizontal. Itfollows that α is the angle of attack. The aircraft is subject to four forces: the thrust,T, developed by its engine, which is assumed to act parallel to its fuselage; the liftL, which acts at right-angles to V; the induced drag, D, which acts in the oppositedirection to V; and the weight, W, which acts vertically downward.

Vertical force balance yields

T sin(α + δ) + L cos δ = W + D sin δ, (9.111)

whereas horizontal force balance gives

T cos(α + δ) = D cos δ + L sin δ. (9.112)

Let us assume that the angles α and δ are both small. According to Equations (9.108)and (9.109), L L0 α and D (2 L0/A)α 2. Thus, L ∼ O(α) and D ∼ O(α 2).Moreover, it is clear from Equations (9.111) and (9.112) that T ∼ O(α 2) and W ∼O(α). Thus, to lowest order in α, Equation (9.111) yields

α WL0, (9.113)

whereas Equation (9.112) gives

δ TW− 2

AWL0. (9.114)

Expression (9.113) relates the angle of attack to the ratio of the aircraft’s weight toits (theoretical) maximum lift (at a given airspeed). Expression (9.114) relates theaircraft’s angle of controlled (i.e., at constant airspeed) ascent to the thrust developed


α

δ

V

T

W

airfoil

x

y

L

D

Figure 9.14Side view of a fixed wing aircraft in flight.

by its engine. An unpowered aircraft, such as a glider, has zero thrust. For suchan aircraft, Equation (9.114) reveals that the angle of controlled decent—which isusually termed the glide angle—takes the value

g =2A

WL0. (9.115)

At fixed airspeed, V , and wing surface area, S , (which implies that L0 is fixed) thisangle can be minimized by making the wing aspect-ratio, A, as large as possible.This result explains accounts for the fact that gliders (and albatrosses) have long thinwings, rather than short stubby ones. For a powered aircraft, the critical thrust toweight ratio required to maintain level flight (i.e., δ = 0) is

TW= g. (9.116)

Hence, this ratio is minimized by minimizing the glide angle, which explains whylong-haul aircraft, which generally need to minimize fuel consumption, tend to havelong thin wings. Finally, as we saw in Section 9.4, if the angle of attack exceedssome (generally small) critical value αc then boundary layer separation occurs onthe back sides of the wings, giving rise to a greatly increased level of drag acting onthe aircraft. In aerodynamics, this phenomenon is called a stall. As is clear fromEquations (9.110) and (9.113), the requirement α < αc is equivalent to

V > Vs =

(W

π ρ S αc

)1/2. (9.117)


In other words, a stall can be avoided by keeping the airspeed above the criticalvalue Vs, which is known as the stall speed. Note that the stall speed decreases withdecreasing altitude, as the air becomes denser.

9.11 Exercises9.1 Consider the integral

In(φ) =∫ π

0

cos(n φ′) dφ′

cosφ′ − cosφ,

where n is a non-negative integer. This integral is defined by its Cauchyprincipal value

In(φ) = limε→0

[∫ φ−ε

0

cos(n φ′) dφ′

cosφ′ − cosφ+

∫ π

φ+ε

cos(n φ′) dφ′

cosφ′ − cosφ

].

As was demonstrated in Section 9.9,

I0 = 0.

Show thatI1 = π,

andIn+1 + In−1 = 2 cos φ In,

and, hence, that

In = πsin(n φ)

sin φ.

9.2 Suppose that an airfoil of negligible thickness, and wingspan b, has a widthwhose z variation is expressed parametrically as

c(φ) =∑

ν=1,3,5,···cν sin(ν φ),

for 0 ≤ φ ≤ π, where

z = −b2

cosφ.

Show that the air circulation about the airfoil takes the form

Γ(φ) =∑

ν=1,3,5,···Γν sin(ν φ),

where Γν = πV α cν. Here, α is the angle of attack (which is assumed to be


small). Demonstrate that the downwash velocity at the trailing edge of theairfoil is

w(φ) =∑

ν=1,3,5,···

ν Γν2 b

sin(ν φ)sinφ

.

Hence, show that the lift and induced drag acting on the airfoil take the values

L =π

4ρV bΓ1,

D =π

8ρ

∑ν=1,3,5,···

ν Γ 2ν ,

respectively. Demonstrate that the drag to lift ratio can be written

DL=

2Aα

1 + ∑ν=3,5,7,···

ν c 2ν

c 21

,where A is the aspect ratio. Hence, deduce that the airfoil shape (in the x-y)plane that minimizes this ratio (at fixed aspect ratio) is an ellipse (i.e., suchthat cν = 0 for ν > 1).

9.3 Consider a plane that flies with a constant angle of attack, and whose thrust isadjusted such that it cancels the induced drag. The plane is effectively subjectto two forces. First, its weight, W = −W ey, and second its lift L = −k v vy ex+

k v vx ey. Here, x and y are horizontal and vertical coordinates, respectively, vis the plane’s instantaneous velocity, and k is a positive constant. Note that thelift is directed at right angles to the plane’s instantaneous direction of motion,and has a magnitude proportional to the square of its airspeed. Demonstratethat the plane’s equations of motion can be written

dvx

dt= −

v vy

h,

dvydt=v vx

h− g,

where h = k g/W is a positive constant with the dimensions of length. Showthat

12v 2 + g y = E,

where E is a constant. Suppose that vx =√g h (1+u) and vy =

√g hw, where

|u|, |w| 1. Demonstrate that, to first order in perturbed quantities,

dudt −

√g

hw,

dwdt 2

√g

hu.


Hence, deduce that if the plane is flying horizontally at some speed v0, andis subject to a small perturbation, then its altitude oscillates sinusoidally atthe angular frequency ω =

√2 g/v0. This type of oscillation is known as a

phugoid oscillation.

10Incompressible Viscous Flow

10.1 IntroductionThis chapter investigates incompressible flow in which viscosity plays a significantrole throughout the bulk of the fluid. Such flow generally takes place at relativelylow Reynolds number. From Section 1.14, the equations governing incompressibleviscous fluid motion can be written

∇ · v = 0, (10.1)

ρDvDt= −∇P + µ∇ 2v, (10.2)

where the quantityP = p + ρΨ, (10.3)

which is a combination of the actual fluid pressure, p, and the gravitational potentialenergy per unit volume, ρΨ , is known as the effective pressure. Here, ρ is the fluidmass density, µ the fluid viscosity, and Ψ the gravitational potential. More informa-tion on this topic can be found in Batchelor 2000.

10.2 Flow Between Parallel PlatesConsider steady, two-dimensional, viscous flow between two parallel plates that aresituated a perpendicular distance d apart. Let x be a longitudinal coordinate measur-ing distance along the plates, and let y be a transverse coordinate such that the platesare located at y = 0 and y = d. (See Figure 10.1.)

Suppose that there is a uniform effective pressure gradient in the x-direction, sothat

dPdx= −G, (10.4)

where G is a constant. Here, the quantity G could represent a gradient in actualfluid pressure, a gradient in gravitational potential energy (due to an inclination ofthe plates to the horizontal), or some combination of the two—it actually makes no

287


vx(y)

y = d

y = 0

−∇P

x

y

Figure 10.1Viscous flow between parallel plates.

difference to the final result. Suppose that the fluid velocity profile between the platestakes the form

v = vx(y) ex. (10.5)

From Section 1.18, this profile automatically satisfies the incompressibility con-straint ∇ · v = 0, and is also such that Dv/Dt = 0. Hence, Equation (10.2) reducesto

∇ 2v =∇Pµ, (10.6)

or. taking the x-component,d 2vx

dy 2 = −Gµ. (10.7)

If the two plates are stationary then the solution that satisfies the no slip constraint(see Section 8.2), vx(0) = vx(d) = 0, at each plate is

vx(y) =G2 µ

y (d − y). (10.8)

Thus, steady, two-dimensional, viscous flow between two stationary parallel platesis associated with a parabolic velocity profile that is symmetric about the midplane,y = d/2. The net volume flux (per unit width in the z-direction) of fluid between theplates is

Q =∫ d

0vx dy =

G d 3

12 µ. (10.9)

Note that this flux is directly proportional to the effective pressure gradient, inverselyproportional to the fluid viscosity, and increases as the cube of the distance betweenthe plates.

Suppose that the upper plate is stationary, but that the lower plate is moving inthe x-direction at the constant speed U. In this case, the no slip boundary conditionat the lower plate becomes vx(0) = U, and the modified solution to Equation (10.7)is

vx(y) =G2 µ

y (d − y) + U(

d − yd

). (10.10)

Incompressible Viscous Flow 289

α

x

h

y

Figure 10.2Viscous flow down an inclined plane.

Hence, the modified velocity profile is a combination of parabolic and linear profiles.This type of flow is known as Couette flow, in honor of Maurice Couette (1858–1943). The net volume flux (per unit width) of fluid between the plates becomes

Q =G d 3

12 µ+

12

U d. (10.11)

10.3 Flow Down an Inclined PlaneConsider steady, two-dimensional, viscous flow down a plane that is inclined at anangle α to the horizontal. Let x measure distance along the plane, and let y be a trans-verse coordinate such that the surface of the plane corresponds to y = 0. Suppose thatthe fluid forms a uniform layer of depth h covering this surface. (See Figure 10.2.)

The generalized pressure gradient within the fluid is written

dPdx= −G = −ρ g sin α, (10.12)

where g is the acceleration due to gravity. In this case, there is no gradient in theactual pressure in the x-direction, and the flow down the plane is driven entirely bygravity. As before, we can write

v = vx(y) ex, (10.13)


and Equation (10.2) again reduces to

d 2vx

dy 2 = −Gµ, (10.14)

whereG = ρ g sinα. (10.15)

Application of the no slip condition at the surface of the plane, y = 0, yields thestandard boundary condition vx(0) = 0. However, the appropriate physical constraintat the fluid/air interface, y = h, is that the normal viscous stress there be zero (i.e.,σxy

∣∣∣y=h = 0), because there is nothing above the interface that can exchange mo-

mentum with the fluid (assuming that the finite inertia and viscosity of air are bothnegligible.) Hence, from Section 1.18, we get the boundary condition

dvx

dy

∣∣∣∣∣y=h= 0. (10.16)

The solution to Equation (10.14) that satisfies the boundary conditions is

vx(y) =G2 µ

y (2 h − y). (10.17)

Thus, the profile is again parabolic. In fact, it is the same as the lower half of theprofile obtained when fluid flows between two (stationary) parallel plates situated aperpendicular distance 2 h apart.

The net volume flux (per unit width in the z-direction) of fluid down the plane is

Q =∫ h

0vx dy =

G h 3

3 µ=g sinα h 3

3 ν, (10.18)

where use has been made of Equation (10.15). Here, ν = µ/ρ is the kinematicviscosity of the fluid. Thus, given the rate Q that fluid is poured down the plane, thedepth of the layer covering the plane becomes

h =(

3 νQg sinα

)1/3. (10.19)

Suppose that the rate at which fluid is poured down the plane is suddenly in-creased slightly from Q to Q + δQ. We would expect an associated change in depthof the layer covering the plane from h to h + δh, where

δh =(

dhdQ

)δQ. (10.20)

Let the interface between the layers of different depth propagate in the x-direction atthe constant velocity V . In a frame of reference that co-moves with this interface, thevolume fluxes (per unit width) immediately to the right and to the left of the interface


are Q − V h and Q + δQ − V (h + δh), respectively. However, in a steady state, thesefluxes must equal one another. Hence,

δQ = V δh, (10.21)

or

V =dQdh=

G h 2

µ=

(9 Q 2 g sinα

ν

)1/3. (10.22)

As can easily be verified, this velocity is twice the maximum fluid velocity in thelayer.

10.4 Poiseuille FlowSteady viscous fluid flow driven by an effective pressure gradient established betweenthe two ends of a long straight pipe of uniform circular cross-section is generallyknown as Poiseuille flow, because it was first studied experimentally by Jean Poiseuille(1797–1869) in 1838. Suppose that the pipe is of radius a. Let us adopt cylindricalcoordinates whose symmetry axis coincides with that of the pipe. Thus, z measuresdistance along the pipe, r = 0 corresponds to the center of the pipe, and r = acorresponds to the pipe wall. Suppose that

∇P = −G ez (10.23)

is the uniform effective pressure gradient along the pipe, and

v = vz(r) ez (10.24)

the time independent velocity profile driven by this gradient. It follows from Sec-tion 1.19 that ∇ · v = 0 and Dv/Dt = 0. Hence, Equation (10.2) reduces to

∇ 2v = −Gµ

ez. (10.25)

Taking the z-component of this equation, we obtain

1r

ddr

(r

dvzdr

)= −

Gµ, (10.26)

where use has been made of Equation (1.155). The most general solution of theprevious equation is

vz(r) = − G4 µ

r 2 + A ln r + B, (10.27)

where A and B are arbitrary constants. The physical constraints are that the flowvelocity is non-singular at the center of the pipe (which implies that A = 0), and is


zero at the edge of the pipe [i.e., vz(a) = 0], in accordance with the no slip condition.Thus, we obtain

vz(r) =G4 µ

(a 2 − r 2). (10.28)

The volume flux of fluid down the pipe is

Q =∫ a

02π r vz dr =

πG a 4

8 µ. (10.29)

According to the previous analysis, the quantity Q/a 4 should be directly proportionalto the effective pressure gradient along the pipe. The accuracy with which experi-mental observations show that this is indeed the case (at relatively low Reynoldsnumber) is strong evidence in favor of the assumptions that there is no slip at thepipe walls, and that the flow is non-turbulent. In fact, the result (10.29), which isknown as Poiseuille’s law, is valid experimentally provided the Reynolds number ofthe flow, Re = U a/ν, remains less than about 6.5 × 103. Here, U = Q/(π a 2) is themean flow speed. On the other hand, if the Reynolds number exceeds the criticalvalue 6.5 × 103 then the flow in the pipe becomes turbulent, and Poiseuille’s lawbreaks down.

10.5 Taylor-Couette FlowConsider two thin cylindrical shells with the same vertical axis. Let the inner andouter shells be of radius r1 and r2, respectively. Suppose that the annular regionr1 ≤ r ≤ r2 is filled with fluid of density ρ and viscosity µ. Let the inner and outercylinders rotate at the constant angular velocities Ω1 and Ω2, respectively. We wishto determine the steady flow pattern set up within the fluid. Incidentally, this type offlow is generally known as Taylor-Couette flow, after Maurice Couette and GeoffreyTaylor (1886–1975).

It is convenient to adopt cylindrical coordinates, r, θ, z, whose symmetry axiscoincides with the common axis of the two shells. Thus, the inner and outer shellscorrespond to r = r1 and r = r2, respectively. Suppose that the flow velocity withinthe fluid is written

v = vθ(r) eθ = rΩ(r) eθ, (10.30)

where Ω(r) = vθ(r)/r is the angular velocity profile. Application of the no slip con-dition at the two shells leads to the boundary conditions

Ω(r1) = Ω1, (10.31)

Ω(r2) = Ω2. (10.32)

It again follows from Section 1.19 that ∇ · v = 0 and Dv/Dt = 0. Hence, Equa-tion (10.2) reduces to

∇ 2v =∇Pµ. (10.33)


Assuming that ∇P = 0 within the fluid, because any flow is driven by the angularrotation of the two shells, rather than by pressure gradients or gravity, and againmaking use of the results quoted in Section 1.19, the previous expression yields

1r

ddr

(r∂vθdr

)− vθ

r2 = 0, (10.34)

or1r 2

ddr

(r 3 dΩ

dr

)= 0. (10.35)

The solution of Equation (10.35) that satisfies the boundary conditions is

Ω(r) =1r 2

Ω1 − Ω2

r −21 − r −2

2

+ Ω2 r −21 − Ω1 r −2

2

r −21 − r −2

2

. (10.36)

It can be seen that this angular velocity profile is a combination of the solid bodyrotation profile Ω = constant, and the irrotational rotation profile Ω ∝ r −2.

From Section 1.19, the only non-zero component of the viscous stress tensorwithin the fluid is

σrθ = µ rddr

(vθ

r

)= µ r

dΩdr. (10.37)

Thus, the viscous torque (acting in the θ-direction) per unit height (in the z-direction)exerted on the inner cylinder is

τ1 = 2π r 21 σrθ(r1) = −4π µ

Ω1 − Ω2

r −21 − r −2

2

. (10.38)

Likewise, the torque per unit height exerted on the outer cylinder is

τ2 = −2π r 22 σrθ(r2) = 4π µ

Ω1 − Ω2

r −21 − r −2

2

. (10.39)

As expected, these two torques are equal and opposite, and act to make the twocylinders rotate at the same angular velocity (in which case, the fluid between themrotates as a solid body).

10.6 Flow in Slowly-Varying ChannelsAccording to Section 10.1, the equations governing steady, incompressible, viscousfluid flow are

∇ · v = 0, (10.40)

ρ (v · ∇) v = −∇P + µ∇ 2v. (10.41)


As we saw in Sections 10.2 and 10.4, for the case of flow along a straight channel ofuniform cross-section, ∇ · v and (v · ∇) v are both identically zero, and the governingequations consequently reduce to the simple relation

∇ 2v =∇Pµ. (10.42)

Suppose, however, that the cross-section of the channel varies along its length. Aswe shall demonstrate, provided this variation is sufficiently slow, the flow is stillapproximately described by the previous relation.

Consider steady, two-dimensional, viscous flow, that is predominately in the x-direction, between two plates that are predominately parallel to the y-z plane. Let thespacing between the plates, d(x), vary on some length scale l d. Suppose that

−∇P = G(x) ex, (10.43)

where G(x) also varies on the same lengthscale. Assuming that ∂/∂x ∼ O(1/l) and∂/∂y ∼ O(1/d), it follows from Equation (10.40) that

vy

vx∼ O

(dl

). (10.44)

Hence,

[(v · ∇) v]x ∼ O(v 2

x

l

), (10.45)

(∇ 2v)x =∂ 2vx

∂y 2

1 + O (dl

)2 . (10.46)

The x-component of Equation (10.41) reduces to

∂ 2vx

∂y 2

1 + O (ρ vx d 2

µ l

)+ O

(dl

)2 = −Gµ. (10.47)

Thus, if

dl 1, (10.48)

ρ vx d 2

µ l 1 (10.49)

—in other words, if the channel is sufficiently narrow, and its cross-section variessufficiently slowly along its length—then Equation (10.47) can be approximated as

∂ 2vx

∂y 2 −Gµ. (10.50)

This, of course, is the same as the equation governing steady, two-dimensional, vis-cous flow between exactly parallel plates. (See Section 10.2.) Assuming that the


plates are located at y = 0 and y = d(x), and making use of the analysis of Sec-tion 10.2, the appropriate solution to the previous equation is

vx(x, y) =G(x)2 µ

y [d(x) − y]. (10.51)

The volume flux (per unit width) of fluid between the plates is thus

Q =∫ d

0vx dy =

G(x) d 3(x)12 µ

. (10.52)

However, for steady incompressible flow, this flux must be independent of x, whichimplies that

G(x) = −dPdx= 12 µQ d −3(x). (10.53)

Suppose that a constant difference in effective pressure, ∆P, is established betweenthe fixed points x = x1 and x = x2, where x2 − x1 = l. Integration of the previousequation between these two points yields

−∆P = 12 µQ l 〈d −3〉, (10.54)

where 〈· · · 〉 =∫ x2

x1(· · · ) dx/l. Hence, the volume flux (per unit width) of fluid between

the plates that is driven by the effective pressure difference becomes

Q = −∆Pl

112 µ 〈d −3〉

. (10.55)

Moreover, the effective pressure gradient at a given point is

−G(x) =dPdx=∆Pl

1d 3(x) 〈d −3〉

, (10.56)

which allows us to determine the velocity profile at that point from Equation (10.51).Thus, given the average effective pressure gradient, ∆P/l, and the variable separation,d(x), we can fully specify the flow between the plates.

Using analogous arguments to those employed previously, but adapting the anal-ysis of Section 10.4, rather than that of Section 10.1, we can easily show that steadyviscous flow down a straight pipe of circular cross-section, whose radius a variesslowly with distance, z, along the pipe, is characterized by

vz(r, z) =G(z)4 µ

[a 2(z) − r 2

], (10.57)

G(z) = −∆Pl

1a 4(z) 〈a−4〉

, (10.58)

Q = −∆Pl

π

8 µ 〈a−4〉. (10.59)

Here, Q is the volume flux of fluid down the pipe, ∆P = P(z2) − P(z1), l = z2 − z1,


U

d1 d d2

l

x

y α

Figure 10.3The lubrication layer between two planes in relative motion.

and 〈· · · 〉 =∫ z2

z1(· · · ) dz/l. The approximations used to derive the previous results are

valid provided

al 1, (10.60)

ρ vz a 2

µ l 1. (10.61)

10.7 Lubrication TheoryIt is well known that two solid bodies can slide over one another particularly easilywhen there is a thin layer of fluid sandwiched between them. Moreover, under certaincircumstances, a large positive pressure develops within the layer. This phenomenonis exploited in hydraulic bearings, whose aim is to substitute fluid-solid friction forthe much larger friction that acts between solid bodies that are in direct contact withone another. Once set up, the fluid layer in hydraulic bearings offers great resistanceto being squeezed out, and is often capable of supporting a useful load.

Consider the simple two-dimensional case of a solid body with a plane surface(that is almost parallel to the x-z plane) gliding steadily over another such body, thesurface of the gliding body being of finite length l in the direction of the motion (thex-direction), and of infinite width (in the z-direction). (See Figure 10.3.) Experienceshows that the plane surfaces need to be slightly inclined to one another. Supposethat α 1 is the angle of inclination. Let us transform to a frame of reference inwhich the upper body is stationary. In this frame, the lower body moves in the x-direction at some fixed speed U. Suppose that the upper body extends from x = 0to x = l, and that the surface of the lower body corresponds to y = 0. Let d(x) bethe thickness (in the y-direction) of the fluid layer trapped between the bodies, whered(0) = d1 and d(l) = d2. It follows that

d(x) = d1 − α x, (10.62)


whereα =

d1 − d2

l. (10.63)

As discussed in the previous section, provided that

dl 1, (10.64)

ρU d 2

µ l 1, (10.65)

the cross-section of the channel between the two bodies is sufficiently slowly varyingin the x-direction that the channel can be treated as effectively uniform at each pointalong its length. Thus, it follows from Equation (10.10) that the velocity profilewithin the channel takes the form

vx(x, y) =G(x)2 µ

y[d(x) − y

]+ U

[d(x) − y

d

], (10.66)

whereG(x) = −dp

dx(10.67)

is the pressure gradient. Here, we are neglecting gravitational forces with respect toboth pressure and viscous forces. The volume flux per unit width (in the z-direction)of fluid along the channel is thus

Q =∫ d

0vx dy =

G(x) d 3(x)12 µ

+12

U d(x). (10.68)

Of course, in a steady state, this flux must be independent of x. Hence,

dpdx= −G(x) = 6 µ

[U

d 2(x)− 2 Q

d 3(x)

], (10.69)

where d(x) = d1 − α x. Integration of the previous equation yields

p(x) − p0 =6 µα

U (1d− 1

d1

)− Q

1d 2 −

1d 2

1

, (10.70)

where p0 = p(0). Assuming that the sliding block is completely immersed in fluid ofuniform ambient pressure p0, we would expect the pressures at the two ends of thelubricating layer to both equal p0, which implies that p(l) = p0. It follows from theprevious equation that

Q = U(

d1 d2

d1 + d2

), (10.71)

andp(x) − p0 =

6 µUα

[d1 − d(x)] [d(x) − d2]d 2(x) (d1 + d2)

. (10.72)


Note that if d1 > d2 then the pressure increment p(x) − p0 is positive throughout thelayer, and vice versa. In other words, a lubricating layer sandwiched between twosolid bodies in relative motion only generates a positive pressure, that is capable ofsupporting a normal load, when the motion is such as to drag (by means of viscousstresses) fluid from the wider to the narrower end of the layer. The pressure incrementhas a single maximum in the layer, and its value at this point is of order µ l U/d 2,assuming that (d1 − d2)/d1 is of order unity. This suggests that very large pressurescan be set up inside a thin lubricating layer.

The net normal force (per unit width in the z-direction) acting on the lower planeis

fy = −∫ l

0[p(x) − p1] dx = −6 µU

α 2

[ln(d1

d2

)− 2

(d1 − d2

d1 + d2

)]. (10.73)

Moreover, the net tangential force (per unit width) acting on the lower plane is

fx =

∫ l

0µ

(∂vx

∂y

)y=0

dx = −2 µUα

[2 ln

(d1

d2

)− 3

(d1 − d2

d1 + d2

)]. (10.74)

Of course, equal and opposite forces,

f ′x = − fx =µU l 2

d 20

32 k2

[ln(

1 + k1 − k

)− 2 k

], (10.75)

f ′y = − fy =µU l

d0

1k

[2 ln

(1 + k1 − k

)− 3 k

], (10.76)

act on the upper plane. Here, d0 = (d1 + d2)/2 is the mean channel width, andk = (d1 − d2)/(d1 + d2). It can be seen that if 0 < d2 < d1 then 0 < k < 1. Theeffective coefficient of friction, C f , between the two sliding bodies is conventionallydefined as the ratio of the tangential to the normal force that they exert on one another.Hence,

C f =fx

fy=

f ′xf ′y=

43

d0

lH(k), (10.77)

where

H(k) = k[ln(

1 + k1 − k

)− 3 k

2

]/ [ln(1 + k1 − k

)− 2 k

]. (10.78)

The function H(k) is a monotonically decreasing function of k in the range 0 < k < 1.In fact, H(k → 0) → 3/(4 k), whereas H(k → 1) → 1. Thus, if k ∼ O(1) [i.e., if(d1 − d2)/d1 ∼ O(1)] then C f ∼ O(d0/l) 1. In other words, the effective coefficientof friction between two solid bodies in relative motion that are separated by a thinfluid layer is independent of the fluid viscosity, and much less than unity. This resultis significant because the coefficient of friction between two solid bodies in relativemotion that are in direct contact with one another is typical of order unity. Hence,the presence of a thin lubricating layer does indeed lead to a large reduction in thefrictional drag acting between the bodies.


10.8 Stokes FlowSteady flow in which the viscous force density in the fluid greatly exceeds the ad-vective inertia per unit volume is generally known as Stokes flow, in honor of GeorgeStokes (1819–1903). Because, by definition, the Reynolds number of a fluid is thetypical ratio of the advective inertia per unit volume to the viscous force density (seeSection 1.16), Stokes flow implies Reynolds numbers that are much less than unity.In the time independent, low Reynolds number limit, Equations (10.1) and (10.2)reduce to

∇ · v = 0, (10.79)

0 = −∇P + µ∇ 2v. (10.80)

It follows from these equations that

∇P = µ∇ 2v = −µ∇ × (∇ × v) = −µ∇ ×ω, (10.81)

where ω = ∇ × v, and use has been made of Equation (A.177). Taking the curl ofthis expression, we obtain

∇ 2ω = 0, (10.82)

which is the governing equation for Stokes flow. Here, use has been made of Equa-tions (A.173), (A.176), and (A.177).

10.9 Axisymmetric Stokes FlowLet r, θ, ϕ be standard spherical coordinates. Consider axisymmetric Stokes flowsuch that

v(r) = vr(r, θ) er + vθ(r, θ) eθ. (10.83)

According to Equations (A.175) and (A.176), we can automatically satisfy the in-compressibility constraint (10.79) by writing (see Section 7.4)

v = ∇ϕ × ∇ψ, (10.84)

where ψ(r, θ) is the Stokes stream function (i.e., v · ∇ψ = 0). It follows that

vr(r, θ) = −1

r 2 sin θ∂ψ

∂θ, (10.85)

vθ(r, θ) =1

r sin θ∂ψ

∂r. (10.86)


Moreover, according to Section C.4, ωr = ωθ = 0, and

ωϕ(r, θ) =1r∂ (r vθ)∂r

− 1r∂vr∂θ=L(ψ)

r sin θ, (10.87)

where (see Section 7.4)

L = ∂ 2

∂r 2 +sin θr 2

∂

∂θ

1sin θ

∂

∂θ. (10.88)

Hence, given that |∇ϕ| = 1/(r sin θ), we can write

ω = ∇ × v = L(ψ)∇ϕ. (10.89)

It follows from Equations (A.176) and (A.178) that

∇ ×ω = ∇ϕ × ∇[−L(ψ)]. (10.90)

Thus, by analogy with Equations (10.84) and (10.89), and making use of Equa-tions (A.173) and (A.177), we obtain

∇ × (∇ ×ω) = −∇ 2ω = −L 2(ψ)∇ϕ. (10.91)

Equation (10.82) implies thatL 2(ψ) = 0, (10.92)

which is the governing equation for axisymmetric Stokes flow. In addition, Equa-tions (10.81) and (10.90) yield

∇P = µ∇ϕ × ∇[L(ψ)]. (10.93)

10.10 Axisymmetric Stokes Flow Around a Solid SphereConsider a solid sphere of radius a that is moving under gravity at the constant ver-tical velocity V ez through a stationary fluid of density ρ and viscosity µ. Here,gravitational acceleration is assumed to take the form g = −g ez. Provided that thetypical Reynolds number,

Re =2 ρV aµ

, (10.94)

is much less than unity, the flow around the sphere is an example of axisymmetricStokes flow. Let us transform to a frame of reference in which the sphere is stationary,and centered at the origin. Adopting the standard spherical coordinates r, θ, ϕ, thesurface of the sphere corresponds to r = a, and the surrounding fluid occupies theregion r > a. By symmetry, the flow field outside the sphere is axisymmetric (i.e.,


∂/∂ϕ = 0), and has no toroidal component (i.e., vϕ = 0). The physical boundaryconditions at the surface of the sphere are

vr(a, θ) = 0, (10.95)

vθ(a, θ) = 0 : (10.96)

that is, the normal and tangential fluid velocities are both zero at the surface. A longway from the sphere, we expect the fluid velocity to asymptote to v = −V ez. In otherwords,

vr(r → ∞, θ)→ −V cos θ, (10.97)

vθ(r → ∞, θ)→ V sin θ. (10.98)

Let us writev = ∇ϕ × ∇ψ, (10.99)

where ψ(r, θ) is the Stokes stream function. As we saw in the previous section, ax-isymmetric Stokes flow is characterized by

L 2(ψ) = 0. (10.100)

Here, the differential operator L is specified in Equation (10.88). The boundaryconditions (10.95)–(10.98) reduce to

∂ψ

∂r

∣∣∣∣∣r=a= 0, (10.101)

∂ψ

∂θ

∣∣∣∣∣r=a= 0, (10.102)

ψ(r → ∞, θ)→12

V r 2 sin2 θ. (10.103)

Equation (10.103) suggests that ψ(r, θ) can be written in the separable form

ψ(r, θ) = sin2 θ f (r). (10.104)

In this case,

vr(r, θ) = −2 cos θ f (r)

r 2 , (10.105)

vθ(r, θ) =sin θ

rd fdr, (10.106)

and Equations (10.100)–(10.103) reduce to(d 2

dr 2 −2r 2

)2f = 0, (10.107)

f (a) =d fdr

∣∣∣∣∣r=a= 0, (10.108)

f (r → ∞)→ 12

V r 2. (10.109)


−3

−2

−1

0

1

2

3

z/a

−3 −2 −1 0 1 2 3x/a

Figure 10.4Contours of the stream function ψ in the x-z plane for Stokes flow around a solidsphere.

Let us try a test solution to Equation (10.107) of the form f (r) = α r n. We findthat

α [n (n − 1) − 2] [(n − 2) (n − 3) − 2] = 0, (10.110)

which implies that n = −1, 1, 2, 4. Hence, the most general solution to Equa-tion (10.107) is

f (r) =Ar+ B r + C r 2 + D r 4, (10.111)

where A, B, C, D are arbitrary constants. However, the boundary condition (10.109)yields C = (1/2) V and D = 0, whereas the boundary condition (10.108) givesA = (1/4) V a 3 and B = −(3/4) V a. Thus, we conclude that

f (r) =V (r − a)2 (2 r + a)

4 r, (10.112)

and the stream function becomes

ψ(r, θ) = sin2 θV (r − a)2 (2 r + a)

4 r. (10.113)

(See Figure 10.4.) From Equation (10.87), the fluid vorticity is

ωϕ(r, θ) =L(ψ)

r sin θ=

sin θr

(d 2

dr 2 −2r 2

)f =

3 V a sin θ2 r 2 . (10.114)


−3

−2

−1

0

1

2

3

z/a

−3 −2 −1 0 1 2 3x/a

Figure 10.5Contours of the vorticity, ωφ, in the x-z plane for Stokes flow around a solid sphere.Solid/dashed lines correspond to opposite signs of ωϕ.

(See Figure 10.5.) Moreover, from Equation (10.81),

∇P = −µ∇ × ω = µ∇φ × ∇(ωφ r sin θ). (10.115)

Hence,

∂P∂r= −3 µV a cos θ

r 3 , (10.116)

∂P∂θ= −3 µV a sin θ

2 r 2 , (10.117)

which implies that the effective pressure distribution within the fluid is

P(r, θ) = p0 +3 µV a cos θ

2 r 2 , (10.118)

where p0 is an arbitrary constant. (See Figure 10.6.) However, P = p + ρΨ , whereΨ = g z = g r cos θ. Thus, the actual pressure distribution is

p(r, θ) = p0 − ρ g r cos θ +3 µV a

2 r 2 cos θ. (10.119)

From Section 1.20, the radial and tangential components of the force per unit


−3

−2

−1

0

1

2

3

z/a

−3 −2 −1 0 1 2 3x/a

Figure 10.6Contours of the effective pressure, P − p0, in the x-z plane for Stokes flow around asolid sphere. Solid/dashed lines correspond to opposite signs of P − p0.

area exerted on the sphere by the fluid are

fr(θ) = σrr(a, θ) =(−p + 2 µ

∂vr

∂r

)r=a, (10.120)

fθ(θ) = σrθ(a, θ) = µ(

1r∂vr∂θ+∂vθ∂r− vθ

r

)r=a. (10.121)

Now, vr(a, θ) = vθ(a, θ) = 0. Moreover, because ∇ · v = 0, it follows from Equa-tion (1.168) that (∂vr/∂r)r=a = 0. Finally, Equation (10.87) yields (∂vθ/∂r)r=a =

ωφ(a, θ). Hence,

fr(θ) = −p(a, θ) = −p0 + ρ g a cos θ − 3 µV2 a

cos θ, (10.122)

fθ(θ) = µωϕ(a, θ) =3 µV2 a

sin θ. (10.123)

Thus, the force density at the surface of the sphere is

f(θ) = −3 µV2 a

ez + (−p0 + ρ g a cos θ) er. (10.124)


It follows that the net vertical force exerted on the sphere by the fluid is

Fz =

∮S

f · ez dS

= −3 µV2 a

4π a 2 + 2π a 2∫ π

0(−p0 + ρ g a cos θ) cos θ sin θ dθ, (10.125)

which reduces to

Fz = −6π a µV +4π3

a 3 ρ g. (10.126)

By symmetry, the horizontal components of the net force both average to zero. Wecan recognize the second term on the right-hand side of the previous equation asthe buoyancy force due to the weight of the fluid displaced by the sphere. (SeeChapter 2.) Moreover, the first term can be interpreted as the viscous drag acting onthe sphere. Note that this drag acts in the opposite direction to the relative motion ofthe sphere with respect to the fluid, and its magnitude is directly proportional to therelative velocity.

Vertical force balance requires that

Fz = M g, (10.127)

where M is the sphere’s mass. In other words, in a steady state, the weight of thesphere balances the vertical force exerted by the surrounding fluid. If the sphere iscomposed of material of mean density ρ then M = (4π/3) a 3 ρ. Hence, in the framein which the fluid a large distance from the sphere is stationary, the steady verticalvelocity with which the sphere moves through the fluid is

V =29

a 2 g

ν

(1 − ρ

ρ

), (10.128)

where ν = µ/ρ is the fluid’s kinematic viscosity. Obviously, if the sphere is moredense than the fluid (i.e., if ρ/ρ > 1) then it moves downward (i.e., V < 0), and viceversa. Finally, the typical Reynolds number of the fluid flow in the vicinity of thesphere is

Re =2 ρV aµ

=49

a 3 g

ν 2

∣∣∣∣∣1 − ρρ∣∣∣∣∣ . (10.129)

For the case of a grain of sand falling through water at 20 C, we have ρ/ρ 2and ν = 1.0 × 10−6 m2/s (Batchelor 2000). Hence, Re = (a/6 × 10−5) 3, where ais measured in meters. Thus, expression (10.128), which is strictly speaking onlyvalid when Re 1, but which turns out to be approximately valid for all Reynoldsnumbers less than unity, only holds for sand grains whose radii are less than about60 microns. Such grains fall through water at approximately 8 × 10−3 m/s. For thecase of a droplet of water falling through air at 20 C and atmospheric pressure, wehave ρ/ρ = 780 and ν = 1.5 × 10−5 m2/s (Batchelor 2000). Hence, Re = (a/4 ×10−5) 3, where a is measured in meters. Thus, expression (10.128) only holds for


water droplets whose radii are less than about 40 microns. Such droplets fall throughair at approximately 0.2 m/s.

At large values of r/a, Equations (10.105), (10.106), and (10.112) yield

vr(r, θ) = −V cos θ +32

V cos θar+ O

(ar

)2, (10.130)

vθ(r, θ) = V sin θ − 34

V sin θar+ O

(ar

)2. (10.131)

It follows that

[ρ (v · ∇) v]r = ρ

vr ∂ vr∂r+vθr∂vr∂θ−v 2θ

r

∼ ρV 2 ar 2 , (10.132)

and

(µ∇ 2v)r ∼ µ∂ 2vr

∂r 2 ∼µV a

r 3 . (10.133)

Hence,[ρ (v · ∇) v]r

(µ∇2v)r∼ ρV r

µ∼ Re

( ra

), (10.134)

where Re is the Reynolds number of the flow in the immediate vicinity of the sphere.[See Equation (10.94).] Our analysis is based on the assumption that advective in-ertia is negligible with respect to viscosity. However, as is clear from the previousexpression for the ratio of inertia to viscosity within the fluid, even if this ratio ismuch less than unity close to the sphere—in other words, if Re 1—it inevitablybecomes much greater than unity far from the sphere: that is, for r a/Re. In otherwords, inertia always dominates viscosity, and our Stokes flow solution thereforebreaks down, at sufficiently large r/a.

10.11 Axisymmetric Stokes Flow In and Around a Fluid SphereSuppose that the solid sphere discussed in the previous section is replaced by a spher-ical fluid drop of radius a. Let the drop move through the surrounding fluid at theconstant velocity V ez. Obviously, the fluid from which the drop is composed mustbe immiscible with the surrounding fluid. Let us transform to a frame of reference inwhich the drop is stationary, and centered at the origin. Assuming that the Reynoldsnumbers immediately outside and inside the drop are both much less than unity, andmaking use of the previous analysis, the most general expressions for the streamfunction outside and inside the drop are

ψ(r, θ) = sin2 θ(A

r+ B r +C r 2 + D r 4

), (10.135)


and

ψ(r, θ) = sin2 θ

Ar+ B r +C r 2 + D r 4

, (10.136)

respectively. Here, A, B, C, et cetera, are arbitrary constants. Likewise, the previousanalysis also allows us to deduce that

vr(r, θ) = −2 cos θ( Ar 3 +

Br+C + D r 2

), (10.137)

vθ(r, θ) = sin θ(− A

r 3 +Br+ 2 C + 4 D r 2

), (10.138)

σrθ(r, θ) = µ sin θ(6 Ar 4 + 6 D r

), (10.139)

σrr(r, θ) = −p0 + ρ g r cos θ + µ cos θ(

12 Ar 4 +

6 Br 2 + 12 D r

)(10.140)

in the region r > a, with analogous expressions in the region r < a. Here, µ, ρ, p0are the viscosity, density, and ambient pressure of the fluid surrounding the drop. Letµ, ρ, and p0 be the corresponding quantities for the fluid that makes up the drop.

In the region outside the drop, the fluid velocity must asymptote to v = −V ez atlarge r/a. This implies that C = (1/2) V and D = 0. Furthermore, vr(a, θ) = 0—that is, the normal velocity at the drop boundary must be zero—otherwise, the dropwould change shape. This constraint yields A/a 3 + B/a + (1/2) V = 0.

Inside the drop, the fluid velocity must remain finite as r → 0. This impliesthat A = B = 0. Furthermore, we again require that vr(a, θ) = 0, which yieldsC + D a 2 = 0.

Two additional physical constraints that must be satisfied at the interface be-tween the two fluids are, firstly, continuity of tangential velocity—that is, vθ(a−, θ) =vθ(a+, θ)—and, secondly, continuity of tangential stress—that is, σrθ(a−, θ) = σrθ(a+, θ).These constraints yield −A/a 3 + B/a − V = 2 C + 4 D a 2 and 6 µ A/a 4 = 6 µD a,respectively.

At this stage, we have enough information to determine the values of A, B, C,and D. In fact, the stream functions outside and inside the drop can be shown to takethe form

ψ(r, θ) =14

V a2 sin2 θ

[(µ

µ + µ

)ar−(

2 µ + 3 µµ + µ

)ra+ 2

( ra

)2], (10.141)

and

ψ(r, θ) =14

V a2 sin2 θ

(µ

µ + µ

) ( ra

)2 [1 −

( ra

)2], (10.142)

respectively. (See Figures 10.7 and 10.8.)The discontinuity in the radial stress across the drop boundary is

σrr(a+, θ) − σrr(a−, θ) = p0 − p0 + (ρ − ρ) g a cos θ

− 3 µVa

[µ + (3/2) µµ + µ

]cos θ. (10.143)


−2

−1

0

1

2

z/a

−2 −1 0 1 2x/a

Figure 10.7Contours of the Stokes stream function in the x-z plane for Stokes flow in and arounda fluid sphere. Here, µ/µ = 10.

−2

−1

0

1

2

z/a

−2 −1 0 1 2x/a

Figure 10.8Contours of the Stokes stream function in the x-z plane for Stokes flow in and arounda fluid sphere. Here, µ/µ = 1/10.


The final physical constraint that must be satisfied at r = a is

σrr(a+, θ) − σrr(a−, θ) =2 γa, (10.144)

where γ is the surface tension of the interface between the two fluids. (See Sec-tion 3.3.) Hence, we obtain

p0 − p0 =2 γa, (10.145)

and

V =a 2 g

3 ν

(1 − ρ

ρ

) [µ + µ

µ + (3/2) µ

], (10.146)

where ν = µ/ρ is the kinematic viscosity of the surrounding fluid. The fact that wehave been able to completely satisfy all of the physical constraints at the interfacebetween the two fluids, as long as the drop moves at the constant vertical velocity V ,proves that our previous assumptions that the interface is spherical, and that the dropmoves vertically through the surrounding fluid at a constant speed without changingshape, were correct. In the limit, µ µ, in which the drop is much more viscousthan the surrounding fluid, we recover Equation (10.128): that is, the drop acts like asolid sphere. On the other hand, in the limit µ µ, and ρ ρ, which is appropriateto an air bubble rising through a liquid, we obtain

V =a 2 g

3 ν. (10.147)

10.12 Exercises10.1 Consider viscous fluid flow down a plane that is inclined at an angle α to the

horizontal. Let x measure distance along the plane (i.e., along the path ofsteepest decent), and let y be a transverse coordinate such that the surface ofthe plane corresponds to y = 0, and the free surface of the fluid to y = h.Show that within the fluid (i.e., 0 ≤ y ≤ h)

vx(y) =g sinα

2 νy (2 h − y),

p(y) = p0 + g ρ cosα (h − y),

where ν is the kinematic viscosity, ρ the density, and p0 is atmospheric pres-sure.

10.2 If a viscous fluid flows along a cylindrical pipe of circular cross-section thatis inclined at an angle α to the horizontal show that the flow rate is

Q =π a 4

8 µ(G + ρ g sinα) ,


where a is the pipe radius, µ the fluid viscosity, ρ the fluid density, and G thepressure gradient.

10.3 Viscous fluid flows steadily, parallel to the axis, in the annular region betweentwo coaxial cylinders of radii a and n a, where n > 1. Show that the volumeflux of fluid flow is

Q =πG a 4

8 µ

[n4 − 1 − (n2 − 1)2

ln n

],

where G is the effective pressure gradient, and µ the viscosity. Find the meanflow speed.

10.4 Consider viscous flow along a cylindrical pipe of elliptic cross-section. Sup-pose that the pipe runs parallel to the z-axis, and that its boundary satisfies

x 2

a 2 +y 2

b 2 = 1.

Letv = vz(x, y) ez.

Demonstrate that (∂ 2

∂x 2 +∂ 2

∂y 2

)vz = −

Gµ,

where G is the effective pressure gradient, and µ the fluid viscosity. Showthat

vz(x, y) =G2 µ

a 2 b 2 − b 2 x 2 − a 2 y 2

a 2 + b 2

is a solution of this equation that satisfies the no slip condition at the bound-ary. Demonstrate that the flow rate is

Q =πG4 µ

a 3 b 3

a 2 + b 2 .

Finally, show that a pipe with an elliptic cross-section has lower flow ratethan an otherwise similar pipe of circular cross-section that has the samecross-sectional area.

10.5 Consider a velocity field of the form

v(r) = r 2Ω(r) sin2 θ ∇ϕ,

where r, θ, ϕ are spherical coordinates. Demonstrate that this field satisfiesthe equations of steady, incompressible, viscous fluid flow (neglecting advec-tive inertia) with uniform pressure (neglecting gravity) provided that

ddr

(r 4 dΩ

dr

)= 0.


Suppose that a solid sphere of radius a, centered at the origin, is rotating aboutthe z-axis, at the uniform angular velocity Ω0, in a viscous fluid, of viscosityµ, that is stationary at infinity. Demonstrate that

Ω(r) = Ω0a 3

r 3 ,

for r ≥ a. Show that the torque that the sphere exerts on the fluid is

τ = 8π µΩ0 a 3.

10.6 Consider a solid sphere of radius a moving through a viscous fluid of vis-cosity µ at the fixed velocity V = V ez. Let r, θ, ϕ be spherical coordinateswhose origin coincides with the instantaneous location of the sphere’s cen-ter. Show that, if inertia and gravity are negligible, the fluid velocity, and theradial components of the stress tensor, a long way from the sphere, are

vr 32

V cos θar,

vθ −34

V sin θar,

σrr −p0 −92µV cos θ

ar 2 ,

σrθ 0,

respectively. Hence, deduce that the net force exerted on the fluid lying insidea large spherical surface of radius r, by the fluid external to the surface, is

F = −6π a µV,

independent of the surface radius.

11Waves in Incompressible Fluids

11.1 IntroductionThis chapter investigates low amplitude waves propagating through incompressiblefluids. More information on this topic can be found in Lighthill 1978, and Milne-Thomson 2011.

11.2 Gravity WavesConsider a stationary body of water, of uniform depth d, located on the surface of theEarth. This body is assumed to be sufficiently small compared to the Earth that itsunperturbed surface is approximately planar. Let the Cartesian coordinate z measurevertical height, with z = 0 corresponding to the aforementioned surface. Supposethat a small amplitude wave propagates horizontally through the water, and let v(r, t)be the associated velocity field.

Because water is essentially incompressible, its equations of motion are

∇ · v = 0, (11.1)

ρ∂v∂t+ ρ (v · ∇) v = −∇p − ρ g ez + µ∇ 2v, (11.2)

where ρ is the (uniform) mass density, µ the (uniform) viscosity, and g the (uniform)acceleration due to gravity. (See Section 1.14.) Let us write

p(r, t) = p0 − ρ g z + p1(r, t), (11.3)

where p0 is atmospheric pressure, and p1 the pressure perturbation due to the wave.Of course, in the absence of the wave, the water pressure a depth h below the surfaceis p0 + ρ g h. (See Chapter 2.) Substitution into Equation (11.2) yields

ρ∂v∂t −∇p1 + µ∇ 2v, (11.4)

where we have neglected terms that are second order in small quantities (i.e., termsof order v 2).

313


Let us also neglect viscosity, which is a good approximation provided that thewavelength is not ridiculously small. [For instance, for gravity waves in water, vis-cosity is negligible as long as λ (ν2/g)1/3 ∼ 5 × 10−5 m.] It follows that

ρ∂v∂t −∇p1. (11.5)

Taking the curl of this equation, we obtain

ρ∂ω

∂t 0, (11.6)

where ω = ∇ × v is the vorticity. We conclude that the velocity field associatedwith the wave is irrotational. Consequently, the previous equation is automaticallysatisfied by writing

v = −∇φ, (11.7)

where φ(r, t) is the velocity potential. (See Section 4.15.) However, from Equa-tion (11.1), the velocity field is also divergence free. It follows that the velocitypotential satisfies Laplace’s equation,

∇ 2φ = 0. (11.8)

Finally, Equations (11.5) and (11.7) yield

p1 = ρ∂φ

∂t. (11.9)

We now need to derive the physical constraints that must be satisfied at the wa-ter’s upper and lower boundaries. It is assumed that the water is bounded from belowby a solid surface located at z = −d. Because the water must always remain in con-tact with this surface, the appropriate physical constraint at the lower boundary isvz|z=−d = 0 (i.e., the normal velocity is zero at the lower boundary), or

∂φ

∂z

∣∣∣∣∣z=−d= 0. (11.10)

The water’s upper boundary is a little more complicated, because it is a free surface.Let ζ represent the vertical displacement of this surface due to the wave. It followsthat

∂ζ

∂t= vz|z=0 = −

∂φ

∂z

∣∣∣∣∣z=0. (11.11)

The appropriate physical constraint at the upper boundary is that the water pressurethere must equal atmospheric pressure, because there cannot be a pressure discon-tinuity across a free surface (in the absence of surface tension—see Section 11.11).Accordingly, from Equation (11.3), we obtain

p0 = p0 − ρ g ζ + p1|z=0 , (11.12)

Waves in Incompressible Fluids 315

orρ g ζ = p1|z=0 , (11.13)

which implies that

ρ g∂ζ

∂t= −ρ g ∂φ

∂z

∣∣∣∣∣z=0=∂p1

∂t

∣∣∣∣∣z=0, (11.14)

where use has been made of Equation (11.11). The previous expression can be com-bined with Equation (11.9) to give the boundary condition

∂φ

∂z

∣∣∣∣∣z=0= −g−1 ∂

2φ

∂t 2

∣∣∣∣∣∣z=0. (11.15)

Let us search for a wave-like solution of Equation (11.8) of the form

φ(r, t) = F(z) cos(ω t − k x). (11.16)

This solution actually corresponds to a propagating plane wave of wave vector k =k ex, angular frequency ω, and amplitude F(z) (Fitzpatrick 2013). Substitution intoEquation (11.8) yields

d 2Fdz 2 − k 2 F = 0, (11.17)

whose independent solutions are exp(+k z) and exp(−k z). Hence, a general solutionto Equation (11.8) takes the form

φ(x, z, t) = A e k z cos(ω t − k x) + B e−k z cos(ω t − k x), (11.18)

where A and B are arbitrary constants. The boundary condition (11.10) is satisfiedprovided that B = A exp(−2 k d), giving

φ(x, z, t) = A[e k z + e−k (z+2 d)

]cos(ω t − k x), (11.19)

The boundary condition (11.15) then yields

A k(1 − e−2 k d

)cos(ω t − k x) = A

ω 2

g

(1 + e−2 k d

)cos(ω t − k x). (11.20)

which reduces to the dispersion relation

ω 2 = g k tanh(k d). (11.21)

The type of wave described in this section is known as a gravity wave.

11.3 Gravity Waves in Deep WaterConsider the so-called deep water limit,

k d 1, (11.22)


in which the depth, d, of the water greatly exceeds the wavelength, λ = 2π/k, of thewave. In this limit, the gravity wave dispersion relation (11.21) reduces to

ω = (g k)1/2, (11.23)

because tanh(x) → 1 as x → ∞. It follows that the phase velocity of gravity wavesin deep water is

vp =ω

k=

(g

k

)1/2. (11.24)

Note that this velocity is proportional to the square root of the wavelength. Hence,deep-water gravity waves with long wavelengths propagate faster than those withshort wavelengths. The phase velocity, vp = ω/k, is defined as the propagationvelocity of a plane wave with the definite wave number, k [and a frequency givenby the dispersion relation (11.23)] (Fitzpatrick 2013). Such a wave has an infinitespatial extent. A more realistic wave of finite spatial extent, with an approximatewave number k, can be formed as a linear superposition of plane waves having arange of different wave numbers centered on k. Such a construct is known as a wavepulse (Fitzpatrick 2013). As is well known, wave pulses propagate at the groupvelocity (Fitzpatrick 2013),

vg =dωdk. (11.25)

For the case of gravity waves in deep water, the dispersion relation (11.23) yields

vg =12

(g

k

)1/2=

12vp. (11.26)

In other words, the group velocity of such waves is half their phase velocity.Let ξ(r, t) be the displacement of a particle of water, found at position r and time

t, due to the passage of a deep water gravity wave. It follows that

∂ξ

∂t= v, (11.27)

where v(r, t) is the perturbed velocity. For a plane wave of wave number k = k ex, inthe limit k d 1, Equation (11.19) yields

φ(x, z, t) = A e k z cos(ω t − k x). (11.28)

Hence, [cf., Equations (11.45)–(11.48)]

ξx(x, z, t) = −a e k z cos(ω t − k x), (11.29)

ξz(x, z, t) = a e k z sin(ω t − k x), (11.30)

vx(x, z, t) = aω e k z sin(ω t − k x), (11.31)

vz(x, z, t) = aω e k z cos(ω t − k x), (11.32)

andp1 = ρ g ξz, (11.33)


x

z

surface

z = 0

Figure 11.1Motion of water particles associated with a deep water gravity wave propagating inthe x-direction.

where use has been made of Equations (11.7), (11.9), and (11.27). Here, a is theamplitude of the vertical oscillation at the water’s surface. According to Equa-tions (11.29)–(11.32), the passage of the wave causes a water particle located a depthh below the surface to execute a circular orbit of radius a e− k h about its equilibriumposition. The radius of the orbit decreases exponentially with increasing depth. Fur-thermore, whenever the particle’s vertical displacement attains a maximum value theparticle is moving horizontally in the same direction as the wave, and vice versa.(See Figure 11.1.)

Finally, if we define h(x, z, t) = ξz(x, z, t)− z as the equilibrium depth of the waterparticle found at a given point and time then Equations (11.3) and (11.33) yield

p(x, z, t) = p0 + ρ g h(x, z, t). (11.34)

In other words, the pressure at this point and time is the same as the unperturbedpressure calculated at the equilibrium depth of the water particle.

11.4 Gravity Waves in Shallow WaterConsider the so-called shallow water limit,

k d 1, (11.35)

in which the depth, d, of the water is much less than the wavelength, λ = 2π/k, ofthe wave. In this limit, the gravity wave dispersion relation (11.21) reduces to

ω = (g d)1/2 k, (11.36)


because tanh(x) → x as x → 0. It follows that the phase velocities and groupvelocities of gravity waves in shallow water all take the fixed value

vp = vg = (g d)1/2, (11.37)

irrespective of wave number. We conclude that—unlike deep water waves—shallowwater gravity waves are non-dispersive in nature (Fitzpatrick 2013). In other words,in shallow water, waves pulses and plane waves all propagate at the same speed. Itcan be seen that the velocity (11.37) increases with increasing water depth.

For a plane wave of wave number k = k ex, in the limit k d 1, Equation (11.19)yields

φ(x, z, t) = A [1 + k 2 (z + d)2/2] cos(ω t − k x). (11.38)

Hence, Equations (11.7) and (11.27) give [cf., Equations (11.45)–(11.48)]

ξx(x, z, t) = −a (k d)−1 cos(ω t − k x), (11.39)

ξz(x, z, t) = a (1 + z/d) sin(ω t − k x), (11.40)

vx(x, z, t) = aω (k d)−1 sin(ω t − k x) (11.41)

vz(x, z, t) = aω (1 + z/d) cos(ω t − k x). (11.42)

Here, a is again the amplitude of the vertical oscillation at the water’s surface. Ac-cording to the previous expressions, the passage of a shallow water gravity wavecauses a water particle located a depth h below the surface to execute an elliptical or-bit, of horizontal radius a/(k d), and vertical radius a (1− h/d), about its equilibriumposition. The orbit is greatly elongated in the horizontal direction. Furthermore, itsvertical radius decreases linearly with increasing depth such that it becomes zero atthe bottom (i.e., at h = d). As before, whenever the particle’s vertical displacementattains a maximum value the particle is moving horizontally in the same direction asthe wave, and vice versa.

11.5 Energy of Gravity WavesIt is easily demonstrated, from the analysis contained in the previous sections, that agravity wave of arbitrary wavenumber k, propagating horizontally through water ofdepth d, has a phase velocity

vp = (g d)1/2[tanh(k d)

k d

]1/2. (11.43)

Moreover, the ratio of the group to the phase velocity is

vg

vp=

12

[1 +

2 k dsinh(2 k d)

]. (11.44)


It follows that neither the phase velocity nor the group velocity of a gravity wavecan ever exceed the critical value (g d)1/2. It is also easily demonstrated that thedisplacement and velocity fields associated with a plane gravity wave of wavenumberk ex, angular frequency ω, and surface amplitude a, are

ξx(x, z, t) = −acosh[k (z + d)]

sinh(k d)cos(ω t − k x), (11.45)

ξz(x, z, t) = asinh[k (z + d)]

sinh(k d)sin(ω t − k x), (11.46)

vx(x, z, t) = aωcosh[k (z + d)]

sinh(k d)sin(ω t − k x), (11.47)

vz(x, z, t) = aωsinh[k (z + d)]

sinh(k d)cos(ω t − k x). (11.48)

The mean kinetic energy per unit surface area associated with a gravity wave isdefined

K = 〈∫ ζ

−d

12ρ v 2 dz〉, (11.49)

whereζ(x, t) = a sin(ω t − k x) (11.50)

is the vertical displacement at the surface, and

〈· · · 〉 =∫ 2π

0(· · · )

d(k x)2π

(11.51)

is an average over a wavelength. Given that 〈cos2(ω t−k x)〉 = 〈sin2(ω t−k x)〉 = 1/2,it follows from Equations (11.47) and (11.48) that, to second order in a,

K =14ρ a 2ω 2

∫ 0

−d

cosh[2 k (z + d)]sinh2(k d)

dz =14ρ g a 2 ω 2

g k tanh(k d). (11.52)

Making use of the general dispersion relation (11.21), we obtain

K =14ρ g a 2. (11.53)

The mean potential energy perturbation per unit surface area associated with agravity wave is defined

U = 〈∫ ζ

−dρ g z dz〉 + 1

2ρ g d 2, (11.54)

which yields

U = 〈12ρ g (ζ 2 − d 2)〉 + 1

2ρ g d 2 =

12ρ g 〈ζ 2〉, (11.55)

orU =

14ρ g a 2. (11.56)


In other words, the mean potential energy per unit surface area of a gravity wave isequal to its mean kinetic energy per unit surface area.

Finally, the mean total energy per unit surface area associated with a gravity waveis

E = K + U =12ρ g a 2. (11.57)

This energy depends on the wave amplitude at the surface, but is independent of thewavelength, or the water depth.

11.6 Wave Drag on ShipsUnder certain circumstances (see the following section), a ship traveling over a bodyof water leaves behind it a train of gravity waves whose wavefronts are transverse tothe ship’s direction of motion. Because these waves possess energy that is carriedaway from the ship, and eventually dissipated, this energy must have been producedat the ship’s expense. The ship consequently experiences a drag force, D. Supposethat the ship is moving at the constant velocity V . We would expect the transversewaves making up the train to have a matching phase velocity, so that they maintaina constant phase relation with respect to the ship. To be more exact, we wouldgenerally expect the ship’s bow to always correspond to a wave maximum (becauseof the pile up of water in front of the bow produced by the ship’s forward motion).The condition vp = V , combined with expression (11.43), yields

tanh(k d)k d

=V 2

g d. (11.58)

Suppose, for the sake of argument, that the wave train is of uniform transverse widthw. Consider a fixed line drawn downstream of the ship perpendicular to its path. Therate at which the length of the train is increasing ahead of this line is V . Therefore, therate at which the energy of the train is increasing ahead of the line is (1/2) ρ g a 2wV ,where a is the typical amplitude of the transverse waves in the train. As is wellknown (Fitzpatrick 2013), wave energy travels at the group velocity, rather than thephase velocity. Thus, the energy flux per unit width of a propagating gravity waveis simply E vg. Wave energy consequently crosses our fixed line in the direction ofthe ship’s motion at the rate (1/2) ρ g a 2w vg. Finally, the ship does work against thedrag force, which goes to increase the energy of the train in the region ahead of ourline, at the rate D V . Energy conservation thus yields

12ρ g a 2 wV =

12ρ g a 2 w vg + D V. (11.59)

However, because V = vp, we obtain

D =12ρ g a 2 w

(1 −

vg

vp

)=

14ρ g a 2 w

[1 − 2 k d

sinh(2 k d)

], (11.60)


where use has been made of Equation (11.44). Here, k d is determined implicitly interms of the ship speed via expression (11.58). Note that this expression cannot besatisfied when the speed exceeds the critical value (g d)1/2, because gravity wavescannot propagate at speeds in excess of this value. In this situation, no transversewave train can keep up with the ship, and the drag associated with such waves con-sequently disappears. In fact, we can see, from the previous formulae, that whenV → (g d)1/2 then k d → 0, and so D → 0. However, the transverse wave amplitude,a, generally increases significantly as the ship speed approaches the critical value.Hence, the drag due to transverse waves actually peaks strongly at speeds just belowthe critical speed, before effectively falling to zero as this speed is exceeded. Conse-quently, it usually requires a great deal of propulsion power to force a ship to travelat speeds faster than (g d)1/2.

In the deep water limit k d 1, Equation (11.60) reduces to

D =14ρ g a 2 w. (11.61)

At fixed wave amplitude, this expression is independent of the wavelength of thewave train, and, hence, independent of the ship’s speed. This result is actually rathermisleading. In fact, (at fixed wave amplitude) the drag acting on a ship travelingthrough deep water varies significantly with the ship’s speed. We can account for thisvariation by incorporating the finite length of the ship into our analysis. A real shipmoving through water generates a bow wave from its bow, and a stern wave fromits stern. Moreover, the bow wave tends to have a positive vertical displacement,because water naturally piles up in front of the bow due to the forward motion of theship, whereas the stern wave tends tends to have a negative vertical displacement,because water rushes into the void left by the stern. Very roughly speaking, supposethat the vertical displacement of the water surface caused by the ship is of the form

ζ(x) ∝ cos(π

xl

). (11.62)

Here, l is the length of the ship. Moreover, the bow lies (instantaneously) at x = 0[hence, ζ(0) > 0], and the stern at x = l [hence, ζ(l) < 0]. For the sake of simplicity,the upward water displacement due to the bow is assumed to equal the downwarddisplacement due to the stern. At fixed bow wave displacement, the amplitude oftransverse gravity waves of wave number k = g/V 2 (chosen such that the phasevelocity of the waves matches the ship’s speed, V) produced by the ship is

a ∝ 1l

∫ l

0cos

(π

xl

)cos(k x) dx =

sin(π − k l)π − k l

k lπ + k l

: (11.63)

that is, the amplitude is proportional to the Fourier coefficient of the ship’s verticaldisplacement pattern evaluated for a wave number that matches that of the wave train(Fitzpatrick 2013). Hence, (at fixed bow wave displacement) the drag produced bythe transverse waves is

D ∝ a 2 ∝[sin(π − Fr−2)π − Fr−2

11 + π Fr 2

] 2

, (11.64)


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

D(a

.u.)

0 0.2 0.4 0.6 0.8 1 1.2Fr

Figure 11.2Variation of wave drag with Froude number for a ship traveling through deep water.

where the dimensionless parameter

Fr =V

(g l)1/2 (11.65)

is known as the Froude number. (See Section 1.16.)Figure 11.2 illustrates the variation of the wave drag with Froude number pre-

dicted by Equation (11.64). As we can see, if the Froude number is much less thanunity, which implies that the wavelength of the wave train is much smaller than thelength of the ship, then the drag is comparatively small. This is the case becausethe ship is extremely inefficient at driving short wavelength gravity waves. It canalso be seen that the drag increases as the Froude number increases, reaching a rela-tively sharp maximum when Fr = Frc = 1/

√π = 0.56, and then falls rapidly. Now,

Fr = Frc corresponds to the case in which the length of the ship is equal to half thewavelength of the wave train. In this situation, the bow and stern waves interfereconstructively, leading to a particularly large amplitude wave train, and, hence, to aparticularly large wave drag. The smaller peaks visible in the figure correspond toother situations in which the bow and stern waves interfere constructively. (For in-stance, when the length of the ship corresponds to one and a half wavelengths of thewave train.) A heavy ship with large a displacement, and limited propulsion power,generally cannot overcome the peak in the wave drag that occurs when Fr = Frc.Such a ship is, therefore, limited to Froude numbers in the range 0 < Fr < Frc, which


implies a maximum speed of

Vc = 0.56 (g l)1/2 = 1.75 [l(m)]1/2 m/s = 3.4 [l(m)]1/2 kts. (11.66)

This characteristic speed is sometimes called the hull speed. It can be seen that thehull speed increases with the length of the ship. In other words, long ships tend tohave higher hull speeds than short ones.

11.7 Ship WakesLet us now make a detailed investigation of the wake pattern generated behind a shipas it travels over a body of water, taking into account obliquely propagating gravitywaves, in addition to transverse waves. For the sake of simplicity, the finite length ofthe ship is neglected in the following analysis. In other words, the ship is treated asa point source of gravity waves. Consider Figure 11.3. This shows a plane gravitywave generated on the surface of the water by a moving ship. The water surfacecorresponds to the x-y plane. The ship is traveling along the x-axis, in the negativex-direction, at the constant speed V . Suppose that the ship’s bow is initially at pointA′, and has moved to point A after a time interval t. The only type of gravity wavethat is continuously excited by the passage of the ship is one that maintains a constantphase relation with respect to its bow. In fact, as we have already mentioned, the bowshould always correspond to a wave maximum. An oblique wavefront associatedwith such a wave is shown in the figure. Here, the wavefront C′D′, which initiallypasses through the bow at point A′, has moved to CD after a time interval t, suchthat it again passes through the bow at point A. Of course, the wavefront propagatesat the phase velocity, vp. It follows that, in the right-angled triangle AA′E, the sidesAA′ and A′E are of lengths V t and vp t, respectively, so that

sin β =vp

V. (11.67)

This, therefore, is the condition that must be satisfied in order for an obliquely prop-agating gravity wave to maintain a constant phase relation with respect to the ship.

In shallow water, all gravity waves propagate at the same phase velocity: that is,

vp = (g d)1/2, (11.68)

where d is the water depth. Hence, Equation (11.67) yields

β = sin−1[(g d)1/2

V

]. (11.69)

This equation can only be satisfied when

V > (g d)1/2. (11.70)


E

D D′

C C ′

β

A A′

y

x

V t

vp t

Figure 11.3An oblique plane wave generated on the surface of the water by a moving ship.

In other words, the ship must be traveling faster than the critical speed (g d)1/2.Moreover, if this is the case then there is only one value of β that satisfies Equa-tion (11.69). This implies the scenario illustrated in Figure 11.4. Here, the shipis instantaneously at A, and the wave maxima that it previously generated—whichall propagate obliquely, subtending a fixed angle β with the x-axis—have interferedconstructively to produce a single strong wave maximum DAE. In fact, the wavemaxima generated when the ship was at A′ have travelled to B′ and C′, the wavemaxima generated when the ship was at A′′ have travelled to B′′ and C′′, et cetera.We conclude that a ship traveling over shallow water produces a V-shaped wakewhose semi-angle, β, is determined by the ship’s speed. Indeed, as is apparent fromEquation (11.69), the faster the ship travels over the water, the smaller the angle βbecomes. Shallow water wakes are especially dangerous to other vessels, and par-ticularly destructive of the coastline, because all of the wave energy produced by theship is concentrated into a single large wave maximum. Note, finally, that the wakecontains no transverse waves, because, as we have already mentioned, such wavescannot keep up with a ship traveling faster than the critical speed (ρ g)1/2.

Let us now discuss the wake generated by a ship traveling over deep water. In thiscase, the phase velocity of gravity waves is vp = (g/k)1/2. Thus, Equation (11.67)yields

sin β =vp

V=

(g

k V 2

)1/2. (11.71)

It follows that in deep water any obliquely propagating gravity wave whose wavenumber exceeds the critical value

k0 =g

V 2 (11.72)


Dy

x

C ′

B′

B′′

C ′′

A′ A′′A β

β

E

Figure 11.4A shallow water wake.

can keep up with the ship, as long as its direction of propagation is such that Equa-tion (11.71) is satisfied. In other words, the ship continuously excites gravity waveswith a wide range of different wave numbers and propagation directions. The wakeis essentially the interference pattern generated by these waves. As is well known(Fitzpatrick 2013), an interference maximum generated by the superposition of planewaves with a range of different wave numbers propagates at the group velocity, vg.Furthermore, as we have already seen, the group velocity of deep water gravity wavesis half their phase velocity: that is, vg = vp/2.

Consider Figure 11.5. The curve APD corresponds to a particular interferencemaximum in the wake. Here, A is the ship’s instantaneous position. Consider a pointP on this curve. Let x and y be the coordinates of this point, relative to the ship.The interference maximum at P is part of the plane wavefront BC emitted sometime t earlier, when the ship was at point A′. Let β be the angle subtended betweenthis wavefront and the x-axis. Because interference maxima propagate at the groupvelocity, the distance A′P is equal to vg t. Of course, the distance AA′ is equal to V t.Simple trigonometry reveals that

x = V t − vg t sin β, (11.73)

y = vg t cos β. (11.74)

Moreover,dydx= tan β, (11.75)

because BC is the tangent to the curve APD—that is, the curve y(x)—at point P. It


Dy

xA A′

β

vg t

V t

B

C

P

Figure 11.5Formation of an interference maximum in a deep water wake.

follows from Equation (11.71), and the fact that vg = vp/2, that

x = X(1 − 1

2sin2 β

), (11.76)

y =12

X sin β cos β, (11.77)

where X(β) = V t. The previous three equations can be combined to produce

dydx=

dy/dβdx/dβ

=(1/2) dX/dβ sin β cos β + (1/2) X (cos2 β − sin2 β)

dX/dβ [1 − (1/2) sin2 β] − X sin β cos β

=sin βcos β

, (11.78)

which reduces todXdβ=

Xtan β

. (11.79)

This expression can be solved to give

X = X0 sin β, (11.80)

where X0 is a constant. Hence, the locus of our interference maximum is determinedparametrically by

x = X0 sin β(1 − 1

2sin2 β

), (11.81)

y =12

X0 sin2 β cos β. (11.82)


−0.3

−0.2

−0.1

0

0.1

0.2

0.3

y/X

0

0 0.1 0.2 0.3 0.4 0.5 0.6x/X0

A

B

C

D

Figure 11.6Locus of an interference maximum in a deep water wake.

Here, the angle β ranges from −π/2 to +π/2. The curve specified by the previousequations is plotted in Figure 11.6. As usual, A is the instantaneous position of theship. It can be seen that the interference maximum essentially consists of the trans-verse maximum BCD, and the two radial maxima AB and AD. As is easily demon-strated, point C, which corresponds to β = 0, lies at x = X0/2, y = 0. Moreover,the two cusps, B and D, which correspond to β = ± tan−1(1/

√8) = ±19.47, lie at

x = (8/27)1/2 X0, y = ±(1/27)1/2 X0.

The complete interference pattern that constitutes the wake is constructed out ofmany different wave maximum curves of the form shown in Figure 11.6, correspond-ing to many different values of the parameter X0. However, these X0 values must bechosen such that the wavelength of the pattern along the x-axis corresponds to thewavelength λ0 = 2π/k0 = 2πV 2/g of transverse (i.e., β = 0) gravity waves whosephase velocity matches the speed of the ship. This implies that X0 = 2 j λ0, where jis a positive integer. A complete deep water wake pattern is shown in Figure 11.7.This pattern, which is made up of interlocking transverse and radial wave maxima,fills a wedge-shaped region—known as a Kelvin wedge—whose semi-angle takes thevalue tan−1(1/

√8) = 19.47. This angle is independent of the ship’s speed. Finally,

our initial assumption that the gravity waves that form the wake are all deep waterwaves is valid provided k0 d 1, which implies that

V (g d)1/2. (11.83)


−6

−5

−4

−3

−2

−1

0

1

2

3

4

5

6

y/λ

0

−1 0 1 2 3 4 5 6 7 8 9 10 11x/λ0

Figure 11.7A deep water wake.

In other words, the ship must travel at a speed that is much less than the critical speed(g d)1/2. This explains why the wake contains transverse wave maxima.

11.8 Gravity Waves in a Flowing FluidConsider a gravity wave traveling through a fluid that is flowing horizontally at theuniform velocity V = V ex. Let us write

v(r, t) = V + v1(r, t), (11.84)

p(r, t) = p0 − ρ g z + p1(r, t), (11.85)

where v1 and p1 are the small velocity and pressure perturbations, respectively, dueto the wave. To first order in small quantities, the fluid equations of motion, (11.1)and (11.2), reduce to

∇ · v1 = 0, (11.86)(∂

∂t+ V · ∇

)v1 = −

∇p1

ρ, (11.87)


respectively. We can also define the displacement, ξ(r, t), of a fluid particle due tothe passage of the wave, as seen in a frame co-moving with the fluid, as(

∂

∂t+ V · ∇

)ξ = v1. (11.88)

The curl of Equation (11.87) implies that ∇ × v1 = 0. Hence, we can writev1 = −∇φ, and Equation (11.87) yields(

∂

∂t+ V · ∇

)φ =

p1

ρ. (11.89)

Finally, Equation (11.86) gives∇ 2φ = 0. (11.90)

The most general traveling wave solution to Equation (11.90), with wave vectork = k ex, and angular frequency ω, is

φ(x, z, t) = [A cosh(k z) + B sinh(k z)] cos(ω t − k x). (11.91)


p1(x, z, t) = ρ k (V − c) [A cosh(k z) + B sinh(k z)] sin(ω t − k x), (11.92)

and from Equation (11.88) that

ξz(x, z, t, ) = (V − c)−1 [A sinh(k z) + B cosh(k z)] sin(ω t − k x). (11.93)

Here, c = ω/k is the phase velocity of the wave.

11.9 Gravity Waves at an InterfaceConsider a layer of fluid of density ρ′, depth d′, and uniform horizontal velocityV ′, situated on top of a layer of another fluid of density ρ, depth d, and uniformhorizontal velocity V . Suppose that the fluids are bounded from above and belowby rigid horizontal planes. Let these planes lie at z = −d and z = d′, and let theunperturbed interface between the two fluids lie at z = 0. (See Figure 11.8.)

Consider a gravity wave of angular frequencyω, and wavenumber k, propagatingthrough both fluids in the x-direction. Let

ζ(x, t) = ζ0 sin(ω t − k x) (11.94)

be the small vertical displacement of the interface due to the wave. In the lowerfluid, the perturbed velocity potential must be of the form (11.91), with the constants


z = d′

z

z = 0

ρ′

ρ

x

V ′

V

z = −d

Figure 11.8Gravity waves at an interface between two immiscible fluids.

A and B chosen such that vz|z=−d = 0 and ξz(x, 0, t) = ζ(x, t). It follows that A =(V − c) ζ0/ tanh(k d) and B = (V − c) ζ0, so that

φ(x, z, t) = (V − c) ζ0cosh[k (z + d)]

sinh(k d)cos(ω t − k x). (11.95)

In the upper fluid, the perturbed velocity potential must again be of the form (11.91),with the constants A and B chosen such that vz|z=d′ = 0 and ξz(x, 0, t) = ζ(x, t). Itfollows that A = −(V ′ − c) ζ0/ tanh(k d′) and B = (V ′ − c) ζ0, so that

φ(x, z, t) = −(V ′ − c) ζ0cosh[k (z − d′)]

sinh(k d′)cos(ω t − k x). (11.96)

Here, c = ω/k is the phase velocity of the wave. From Equations (11.85) and (11.92),the fluid pressure just below the interface is

p(x, 0−, t) = p0 − ρ g ζ + p1(x, 0−, t)

= p0 − ρ g ζ0 sin(ω t − k x) − ρ k (V − c) A sin(ω t − k x)

= p0 −[ρ g − ρ k (V − c)2

tanh(k d)

]ζ0 sin(ω t − k x). (11.97)

Likewise, the fluid pressure just above the interface is

p(x, 0+, t) = p0 − ρ′ g ζ + p1(x, 0+, t)

= p0 − ρ′ g ζ0 sin(ω t − k x) − ρ′ k (V ′ − c) A sin(ω t − k x)

= p0 −[ρ′ g +

ρ′ k (V ′ − c)2

tanh(k d′)

]ζ0 sin(ω t − k x). (11.98)

In the absence of surface tension at the interface, these two pressure must equal oneanother: that is, [

p]z=0+z=0−= 0. (11.99)


Hence, we obtain the dispersion relation

(ρ − ρ′) g = k ρ (V − c)2

tanh(k d)+

k ρ′ (V ′ − c)2

tanh(k d′), (11.100)

which takes the form of a quadratic equation for the phase velocity, c, of the wave.We can see that:

i. If ρ′ = 0 and V = 0 then the dispersion relation reduces to (11.43) (with vp = c).

ii. If the two fluids are of infinite depth then the dispersion relation simplifies to

(ρ − ρ′) g = k ρ (V − c)2 + k ρ′ (V ′ − c)2. (11.101)

iii. In general, there are two values of c that satisfy the quadratic equation (11.100).These are either both real, or form a complex conjugate pair.

iv. The condition for stability is that c is real. The alternative is that c is complex,which implies that ω is also complex, and, hence, that the perturbation grows ordecays exponentially in time. Because the complex roots of a quadratic equationoccur in complex conjugate pairs, one of the roots always corresponds to anexponentially growing mode. In other words, an instability.

v. If both fluids are at rest (i.e., V = V ′ = 0), and of infinite depth, then the disper-sion reduces to

c 2 =g (ρ − ρ′)k (ρ + ρ′)

. (11.102)

It follows that the configuration is only stable when ρ > ρ′: that is, when theheavier fluid is underneath.

As a particular example, suppose that the lower fluid is water, and the upper fluidis the atmosphere. Let s = ρ′/ρ = 1.225 × 10−3 be the specific density of air at s.t.p.(relative to water). Putting V = V ′ = 0, d′ → ∞, and making use of the fact that s issmall, the dispersion relation (11.100) yields

c (g d)1/2[tanh(k d)

k d

]1/2 1 − 1

2s [1 + tanh(k d)]

. (11.103)

Comparing this with Equation (11.43), we can see that the presence of the atmo-sphere tends to slightly diminish the phase velocities of gravity waves propagatingover the surface of a body of water.

11.10 Steady Flow over a Corrugated BottomConsider a stream of water of mean depth d, and uniform horizontal velocity V =V ex, that flows over a corrugated bottom whose elevation is z = −d + a sin(k x),


where a is much smaller than d. Let the elevation of the free surface of the water bez = b sin(k x). We wish to determine the relationship between a and b.

We expect the velocity potential, perturbed pressure, and vertical displacement ofthe water to be of the form (11.91), (11.92), and (11.93), respectively, withω = c = 0,because we are looking for a stationary (i.e., non-propagating) perturbation driven bythe static corrugations in the bottom. The boundary condition at the bottom is

ξz(x,−d) = a sin(k x), (11.104)

which yieldsV −1 [−A sinh(k d) + B cosh(k d)] = a. (11.105)

At the free surface, we have

ξz(x, 0) = b sin(k x), (11.106)

which givesb = V −1 B. (11.107)

In addition, pressure balance across the free surface yields

ρ g b sin(k x) = p1(x, 0) = ρ k V A sin(k x), (11.108)

which leads tog b = k V A. (11.109)

Hence, from Equations (11.105), (11.107), and (11.109),

b =a

cosh(k d) − (g/k V 2) sinh(k d), (11.110)

orb =

acosh(k d) (1 − c 2/V 2)

, (11.111)

where c = [(g/k) tanh(k d)]1/2 is the phase velocity of a gravity wave of wave numberk. [See Equation (11.21).] It follows that the peaks and troughs of the free surfacecoincide with those of the bottom when |V | > |c|, and the troughs coincide with thepeaks, and vice versa, when |V | < |c|. If |V | = |c| then the ratio b/a becomes infinite,implying that the oscillations driven by the corrugations are not of small amplitude,and, therefore, cannot be described by linear theory.

11.11 Surface TensionAs described in Chapter 3, there is a positive excess energy per unit area, γ, as-sociated with an interface between two immiscible fluids. The quantity γ can also


be interpreted as a surface tension. Let us now incorporate surface tension into ouranalysis. Suppose that the interface lies at

z = ζ(x, t), (11.112)

where |ζ | is small. Thus, the unperturbed interface corresponds to the plane z = 0.The unit normal to the interface is

n =∇(z − ζ)|∇(z − ζ)|

. (11.113)

It follows that

nx −∂ζ

∂x, (11.114)

nz 1. (11.115)

The Young-Laplace Equation (see Section 3.2) yields

∆p = γ∇ · n, (11.116)

where ∆p is the jump in pressure seen crossing the interface in the opposite directionto n. However, from Equations (11.114) and (11.115), we have

∇ · n −∂2ζ

∂x 2 . (11.117)

Hence, Equation (11.116) gives

[p]z=0+z=0−= γ

∂ 2ζ

∂x 2 . (11.118)

This expression is the generalization of Equation (11.99) that takes surface tensioninto account.

Suppose that the interface in question is that between a body of water, of densityρ and depth d, and the atmosphere. Let the unperturbed water lie between z = −dand z = 0, and let the unperturbed atmosphere occupy the region z > 0. In the limitin which the density of the atmosphere is neglected, the pressure in the atmospheretakes the fixed value p0, whereas the pressure just below the surface of the water isp0−ρ g ζ+ p1|z=0. Here, p1 is the pressure perturbation due to the wave. The relation(11.118) yields

ρ g ζ − p1|z=0 = γ∂ 2ζ

∂x 2 , (11.119)

where γ is the surface tension at an air/water interface. However, ∂ζ/∂t = −(∂φ/∂z)z=0,where φ is the perturbed velocity potential of the water. Moreover, from Equa-tion (11.9), p1 = ρ (∂φ/∂t). Hence, the previous expression gives

g∂φ

∂z

∣∣∣∣∣z=0+∂ 2φ

∂t 2

∣∣∣∣∣∣z=0=γ

ρ

∂ 3φ

∂z ∂ 2x

∣∣∣∣∣∣z=0. (11.120)


This relation, which is a generalization of Equation (11.15), is the condition satis-fied at a free surface in the presence of non-negligible surface tension. Applyingthis boundary condition to the general solution, (11.19) (which already satisfies theboundary condition at the bottom), we obtain the dispersion relation

ω2 =

(g k +

γ k 3

ρ

)tanh(k d), (11.121)

which is a generalization of Equation (11.21) that takes surface tension into account.

11.12 Capillary WavesIn the deep water limit k d 1, the dispersion relation (11.121) simplifies to

ω 2 = g k +γ k 3

ρ. (11.122)

It is helpful to introduce the capillary length,

l =(γ

ρ g

)1/2. (11.123)

(See Section 3.4.) The capillary length of an air/water interface at s.t.p. is 2.7×10−3 m(Batchelor 2000). The associated capillary wavelength is λc = 2π l = 1.7 × 10−2 m.Roughly speaking, surface tension is negligible for waves whose wavelengths aremuch larger than the capillary wavelength, and vice versa. It is also helpful to intro-duce the critical phase velocity

vc = (2 g l)1/2. (11.124)

This critical velocity takes the value 0.23 m/s for an air/water interface at s.t.p.(Batchelor 2000). It follows from Equation (11.122) that the phase velocity, vp =

ω/k, of a surface water wave can be written

vp

vc=

[12

(k l +

1k l

)]1/2. (11.125)

Moreover, the ratio of the phase velocity to the group velocity, vg = dω/dk, becomes

vg

vp=

12

[1 + 3 (k l)2

1 + (k l)2

]. (11.126)

In the long wavelength limit λ λc (i.e., k l 1), we obtain

vp

v0 1

(2 k l)1/2 , (11.127)


andvg

vp 1

2. (11.128)

We can identify this type of wave as the deep water gravity wave discussed in Sec-tion 11.3.

In the short wavelength limit λ λc (i.e., k l 1), we get

vp

vc(k l2

)1/2, (11.129)

andvg

vp 3

2. (11.130)

This corresponds to a completely new type of wave known as a capillary wave. Suchwaves have wavelengths that are much less than the capillary wavelength. Moreover,Equation (11.129) can be rewritten

vp (

k γρ

)1/2, (11.131)

which demonstrates that gravity plays no role in the propagation of a capillary wave.In fact, its place is taken by surface tension. Finally, it is easily seen that the phasevelocity (11.125) attains the minimum value vp = vc when λ = λc (i.e., when k l = 1).Moreover, from Equation (11.126), vg = vp at this wavelength. It follows that thephase velocity of a surface wave propagating over a body of water can never be lessthan the critical value, vc.

11.13 Capillary Waves at an InterfaceConsider a layer of fluid of density ρ′, depth d′, and uniform horizontal velocityV ′, situated on top of a layer of another fluid of density ρ, depth d, and uniformhorizontal velocity V . Suppose that the fluids are bounded from above and belowby rigid horizontal planes. Let these planes be at z = −d and z = d′, and let theunperturbed interface between the two fluids be at z = 0. Suppose that the elevationof the perturbed interface is z = ζ, where ζ = ζ0 sin(ω t − k x). Finally, let γ be thesurface tension of the interface. Equations (11.97), (11.98), and (11.118) yield thedispersion relation

(ρ − ρ′) g + γ k 2 =ρ k (V − c)2

tanh(k d)+ρ′ k (V ′ − c)2

tanh(k d′), (11.132)

which is a generalization of the dispersion relation (11.100) that takes surface tensioninto account. Here, c = ω/k is the phase velocity of a wave propagating along theinterface.


For the case in which both fluids are at rest, and of infinite depth, the previousdispersion relation simplifies to give

(ρ − ρ′) g + γ k 2 = (ρ + ρ′) k c 2. (11.133)

Suppose that s = ρ′/ρ is the specific gravity of the upper fluid with respect to thelower. In the case in which s < 1 (i.e., the upper fluid is lighter than the lower one),it is helpful to define

l =[

γ

ρ g (1 − s)

]1/2(11.134)

c0 =

[2 g l

(1 − s1 + s

)]1/2. (11.135)

It follows thatc 2

c 20

=12

(1k l+ k l

). (11.136)

Thus, we conclude that the phase velocity of a wave propagating along the interfacebetween the two fluids achieves its minimum value, c = c0, when k l = 1. Further-more, waves of all wavelength are able to propagate along the interface (i.e., c 2 > 0for all k). In the opposite case, in which s > 1 (i.e., the upper fluid is heavier than thelower one), we can redefine the capillary length as

l =[

γ

ρ g (s − 1)

]1/2. (11.137)

The dispersion relation (11.133) then becomes

c 2 = g l(

s − 1s + 1

) (k l − 1

k l

). (11.138)

It is apparent that c 2 < 0 for k l < 1, indicating instability of the interface for waveswhose wavelengths exceed the critical value λc = 2π l. On the other hand, waveswhose wavelengths are less than the critical value are stabilized by surface tension.This result is exemplified by the experiment in which water is retained by atmo-spheric pressure in an inverted glass whose mouth is closed by a gauze of fine mesh(the purpose of which is to put an upper limit on the wavelengths of waves that canexist at the interface.)

11.14 Wind Driven Waves in Deep WaterConsider the scenario described in the previous section. Suppose that the lower fluidis a body of deep water at rest, and the upper fluid is the atmosphere. Let the air above


the surface of the water move horizontally at the constant velocity V ′. Suppose thatρ is the density of water, s = ρ′/ρ the specific gravity of air with respect to water,and γ the surface tension at an air/water interface. With V = 0, k d → ∞, k d′ → ∞,the dispersion relation (11.132) reduces to

ρ (1 − s) g + γ k 2 = ρ k c 2 + s ρ k (V ′ − c)2. (11.139)

This expression can be rearranged to give

c 2 −(2 V ′ s1 + s

)+

V ′ 2 s1 + s

= c 21 , (11.140)

which is a quadratic equation for the phase velocity, c, of the wave. Here,

c 21 =

g

k

(1 − s1 + s

)+

γ kρ (1 + s)

, (11.141)

where c1 is the phase velocity that the wave would have in the absence of the wind.In fact, we can write

c 21 =

12

(λ

λc+λc

λ

)c 2

0 (11.142)

where λc = 2π l is the capillary wavelength, and l and c0 are defined in Equa-tions (11.134) and (11.135), respectively.

For a given wavelength, λ, the wave velocity, c, attains its maximum value, cm,when dc/dV ′ = 0. According to the dispersion relation (11.140), this occurs when

V ′ = cm = (1 + s)1/2 c1. (11.143)

If the wind has any other velocity, greater or less than cm, then the wave velocity isless than cm.

According to Equation (11.140), the wave velocity, c, becomes complex, indicat-ing an instability, when

V ′ 2 >(1 + s)2

sc 2

1 =(1 + s)2

2 s

(λ

λc+λc

λ

)c 2

0 . (11.144)

We conclude that if the wind speed exceeds the critical value

V ′c =(1 + s)

s1/2 c0 = 6.6 m/s = 12.8 kts (11.145)

then waves whose wavelengths fall within a certain range, centered around λc, areunstable and grow to large amplitude.

The two roots of Equation (11.140) are

c =V ′ s1 + s

±[c 2

1 −s V ′ 2

(1 + s)2

]1/2. (11.146)


Moreover, ifV ′ < (1 + s−1)1/2 c1 (11.147)

then these roots have opposite signs. Hence, the waves can either travel with thewind, or against it, but travel faster when they are moving with the wind. If V ′

exceeds the value given previously then the waves cannot travel against the wind.Because c1 has the minimum value c0, it follows that waves traveling against thewind are completely ruled out when

V ′ > (1 + s−1)1/2 c0 = 6.6 m/s = 12.8 kts. (11.148)

11.15 Exercises11.1 Find the velocity potential of a standing gravity wave in deep water for which

the associated elevation of the free surface is

z = a cos(ω t) cos(k x).

Determine the paths of water particles perturbed by the wave.

11.2 Deep water fills a rectangular tank of length l and breadth b. Show that theresonant frequencies of the water in the tank are

(g π)1/2 (n 2 l−2 + m 2 b−2)1/4,

where n and m are integers. Neglect surface tension.

11.3 Demonstrate that a sinusoidal gravity wave on deep water with surface eleva-tion

ζ = a cos(ω t − k x)

possesses a mean momentum per unit surface area

12ρω a 2.

11.4 A seismic wave passes along the bed of an ocean of uniform depth d suchthat the vertical perturbation of the bed is a cos[k (x − V t)]. Show that theamplitude of the consequent gravity waves at the surface is

a[(

1 −c 2

V 2

)cosh(k d)

]−1

,

where c is the phase velocity of waves of wavenumber k.


11.5 A layer of liquid of density ρ and depth d has a free upper surface, and liesover liquid of infinite depth and density σ > ρ. Neglecting surface tension,show that two possible types of wave of wavenumber k, with phase velocities

c 2 = k g,

c 2 =k g (σ − ρ)

σ coth(k d) + ρ,

can propagate along the layer.

11.6 Show that, taking surface tension into account, a sinusoidal wave of wavenum-ber k and surface amplitude a has a mean kinetic energy per unit surface area

K =14

(ρ g + γ k 2) a 2,

and a mean potential energy per unit surface area

U =14

(ρ g + γ k 2) a 2.

11.7 Show that in water of uniform depth d the phase velocity of surface wavescan only attain a stationary (i.e., maximum or minimum) value as a functionof wavenumber, k, when

k =[sinh(2 k d) − 2 k dsinh(2 k d) + 2 k d

]1/2kc,

where kc = (ρ g/γ)1/2. Hence, deduce that the phase velocity has just onestationary value (a minimum) for any depth greater than 31/2 k −1

c 4.8 mm,but no stationary values for depths less than that.

11.8 Unlike gravity waves in deep water, whose group velocities are half theirphase velocities, the group velocities of capillary waves are 3/2 times theirphase velocities. Adapt the analysis of Section 11.7 to investigate the gen-eration of capillary waves by a very small object traveling across the surfaceof the water at the constant speed V . Suppose that the unperturbed surfacecorresponds to the x-y plane. Let the object travel in the minus x-direction,such that it is instantaneously found at the origin. Find the present positionof waves that were emitted, traveling at an angle θ to the object’s direction ofmotion, when it was located at (X, 0). Show that along a given interferencemaximum the quantities X and θ vary in such a manner that X cos3 θ takesa constant value, X1 (say). Deduce that the interference maximum is givenparametrically by the equations

x = X1 sec θ(tan2 θ − 1

2

),

y =32

X1 sec θ tan θ.


Sketch this curve, noting that it goes through the points (−0.5 X1, 0) and (0,±1.3 X1), and asymptotes to y = ±1.5 X 1/3

1 x 2/3.

12Terrestrial Ocean Tides

12.1 IntroductionThis chapter outlines the classical theory of terrestrial ocean tides, the study of whichwas pioneered by Pierre-Simon Laplace (1749-1827). Our treatment of ocean tidestakes all major harmonics of the tide generating potential into account, as well asthe finite elasticity of the Earth, and the self-gravity of the oceans. As will becomeapparent, the theory of terrestrial ocean tides is extremely complicated. Hence, forthe sake of simplicity and brevity, we shall only consider two relatively idealizedscenarios. First, tides in a frictionless ocean of constant depth that covers the wholesurface of the Earth, and, second, tides in a frictionless ocean of constant depththat only covers one hemisphere of the Earth (and lies between two meridians oflongitude). It turns out that these two scenarios (in particular, the latter) exhibit mostof the observed features of terrestrial ocean tides. More information on terrestrialtides can be found in Lamb 1993 and Cartwright 1999.

12.2 Tide Generating PotentialConsider an (almost) spherical planet of mass m, and radius a, and a spherical moonof mass m′, that execute (almost) circular orbits about their mutual center of mass.Let r′ (where r′ a) represent the vector position of the center of the moon, relativeto the center of the planet. It follows that (Fitzpatrick 2012)

r′′ =(

m′

m + m′

)r′ (12.1)

is the vector position of the center of mass, relative to the center of the planet. More-over, the planet and moon both orbit the center of mass with angular velocity ω1,where (Fitzpatrick 2012)

ω 21 =

G (m + m′)r′ 3

. (12.2)

Here, G is the universal gravitational constant (Yoder 1995).Neglecting the planet’s axial rotation (which is introduced into our analysis in

341


Section 12.6), the planet’s orbital motion about the center of mass causes its con-stituent points to execute circular orbits of radius r′′ with angular velocity ω1. Thecenters of these orbits have the same spatial relation to the center of mass that theconstituent points have to the center of the planet. Moreover, the orbits lie in par-allel planes. Thus, the common centripetal acceleration of the constituent points is(Fitzpatrick 2012)

g centripetal = ω21 r′′ = −∇Φcentripetal. (12.3)

It follows, from the previous three equations, that

Φcentripetal(r) = −G (m + m′)r′ 3

r′′ · r = −G m′

r′ 3r′ · r = −G m′

r′ 2r cos γ

= −G m′

r′rr′

P1(cosγ), (12.4)

where r is a position vector relative to the center of the planet, and

cosγ =r · r′

r r′. (12.5)

Here,

Pn(x) =1

2 n n!d n

dx n [(x 2 − 1)n], (12.6)

for n ≥ 0, denotes a Legendre polynomial (Abramowitz and Stegun 1965). In partic-ular,

P0(x) = 1, (12.7)

P1(x) = x, (12.8)

P2(x) =12

(3 x 2 − 1). (12.9)

The gravitational force per unit mass exerted by the moon on one of the planet’sconstituent points, located at position vector r, is written

g moon(r) = −∇Φmoon, (12.10)

where (Fitzpatrick 2012)

Φmoon(r) = − G m′

|r − r′|= −G m′

r′∑

n=0,∞

( rr′

)nPn(cos γ), (12.11)

assuming that r′ > r. The tidal force per unit mass acting on the point (i.e., the forceper unit mass that is not compensated by the centripetal acceleration) is

g tide(r) = g moon(r) − g centripetal = −∇Φtide, (12.12)

whereΦtide(r) = Φmoon(r) −Φcentripetal(r) (12.13)

Terrestrial Ocean Tides 343

is termed the tide generating potential. Thus, according to Equations (12.4) and(12.11),

Φtide(r) = −G m′

r′

[1 +

( rr′

)2P2(cosγ) + O

( rr′

)3]. (12.14)

The truncation of the expansion is ultimately justified by the fact that a/r′ 1.

12.3 Decomposition of Tide Generating PotentialLet r, θ, ϕ be right-handed spherical coordinates in a non-rotating reference framewhose origin lies at the center of the planet, and whose symmetry axis coincides withthe planetary rotation axis. Thus, the vector position of a general point is

r = r sin θ cosϕ er + r sin θ sin ϕ eθ + r cos θ eϕ, (12.15)

where er = ∇r/|∇r|, et cetera. Let the coordinates of the moon’s center be r′, θ′, ϕ′.It follows that

r′ = r′ sin θ′ cosϕ′ er + r′ sin θ′ sin ϕ′ eθ + r′ cos θ′ eϕ. (12.16)

Hence, from Equation (12.5),

cos γ = cos θ cos θ′ + sin θ sin θ′ cos(ϕ − ϕ′). (12.17)

Now, according to the spherical harmonic addition theorem (Arfken 1985),

Pn(cosγ) = Pn(cos θ) Pn(cos θ′) (12.18)

+ 2∑

m=1,n

(n − m)!(n + m)!

P mn (cos θ) P m

n (cos θ′) cos[m (ϕ − ϕ′)], (12.19)

which implies that

P2(cosγ) = P 02 (cos θ) P 0

2 (cos θ′) +13

P 12 (cos θ) P 1

2 (cos θ′) cos(ϕ − ϕ′)

+1

12P 2

2 (cos θ) P 22 (cos θ′) cos

[2 (ϕ − ϕ′)

]. (12.20)

Here,

P mn (x) = (−1)m (1 − x 2)m/2 d m[Pn(x)]

dx m , (12.21)

for n ≥ 0 and m ≤ n, denotes an associated Legendre function (Abramowitz andStegun 1965). In particular,

P 02 (x) =

12

(3 x 2 − 1), (12.22)

P 12 (x) = −3 x (1 − x 2)1/2, (12.23)

P 22 (x) = 3 (1 − x 2). (12.24)


Note that P 0n (x) = Pn(x).

Let

R (−m)n (r, θ, ϕ) =

( ra

)nP m

n (cos θ) cos(mϕ), (12.25)

R (0)n (r, θ) =

( ra

)nP 0

n (cos θ), (12.26)

R (+m)n (r, θ, ϕ) =

( ra

)nP m

n (cos θ) sin(mϕ), (12.27)

where n ≥ m > 0. According to Equations (12.14), (12.20), and (12.25)–(12.27),

Φtide(r, θ, ϕ) = −G m′ a 2

r′ 3[P 0

2 (cos θ′)R (0)2 (r, θ)

+13

P 12 (cos θ′) cosϕ′ R (−1)

2 (r, θ, ϕ)

+13

P 12 (cos θ′) sin ϕ′ R (+1)

2 (r, θ, ϕ)

+112

P 22 (cos θ′) cos(2 ϕ′)R (−2)

2 (r, θ, ϕ)

+112

P 22 (cos θ′) sin(2 ϕ′)R (+2)

2 (r, θ, ϕ)]. (12.28)

Here, we have neglected the unimportant constant term in Equation (12.14).

12.4 Expansion of Tide Generating PotentialSuppose that, in the r, θ, ϕ frame, the moon’s orbit is a Keplerian ellipse of majorradius R, and eccentricity 0 < e 1, lying in a fixed plane that is inclined at anangle δ > 0 to the plane θ = π/2. Thus, the closest and furthest distance between thecenters of the moon and the planet are Rp = (1− e) R and Ra = (1+ e) R, respectively.It follows that (Fitzpatrick 2013)

cos θ′ = cosλ sin δ, (12.29)

tanϕ′ =tan λcos δ

, (12.30)

where

λ = ω1 t + 2 e sin[(ω1 − ω2) t] + O(e 2), (12.31)

r′ = R1 − e cos[(ω1 − ω2) t] + O(e 2)

. (12.32)


Here, it is assumed that the closest point on the moon’s orbit to the center of the planetcorresponds to λ = ω2 t, where ω2 is the uniform precession rate of this point. [It isnecessary to include such precession in our analysis because the Moon’s perigee pre-cesses steadily in such a manner that it completes an orbit about the Earth once every8.85 years. This effect is caused by the perturbing influence of the Sun (Fitzpatrick2013).] Suppose that the inclination of the moon’s orbit to the planet’s equatorialplane, θ = π/2, is relatively small, so that sin δ 1. It follows that

cos θ′ = sin δ cos(ω1 t) + O(e sin δ), (12.33)

ϕ′ = ω1 t + 2 e sin[(ω1 − ω2) t] + O(e 2) + O(sin2 δ). (12.34)

Thus, Equations (12.22)–(12.24), (12.28), and (12.32)–(12.34) can be combined togive the following expression for the tide generating potential due to a moon in alow-eccentricity, low-inclination orbit:

Φtide(r, θ, ϕ) =G m′ a2

R 3

[12

(1 + 3 e cos [(ω1 − ω2) t] − 3

2sin2 δ cos[2ω1 t]

)R (0)

2 (r, θ)

+12

sin δR (−1)2 (r, θ, ϕ)

+12

sin δR (−1)2 (r, θ, ϕ − 2ω1 t)

− 7 e8R (−2)

2 (r, θ, ϕ − [(3/2)ω1 − (1/2)ω2] t)

−14R (−2)

2 (r, θ, ϕ − ω1 t)

+e8R (−2)

2 (r, θ, ϕ − [(1/2)ω1 + (1/2)ω2] t)

+O(e 2) + O(e sin δ) + O(sin2 δ)]. (12.35)

Here, we have retained a term proportional to sin2 δ in the previous expression, de-spite the fact that we are formally neglecting O(sin2 δ) terms, because the term inquestion gives rise to important fortnightly tides on the Earth.

12.5 Surface Harmonics and Solid HarmonicsA surface harmonic of degree n (where n is a non-negative integer), denotedSn(θ, ϕ),is defined as a well-behaved solution to

r 2 ∇ 2Sn + n (n + 1)Sn = 0 (12.36)


on the surface of a sphere (i.e., r = constant). Here, r, θ, ϕ are standard sphericalcoordinates, and ∇ 2 is the Laplacian operator. It follows that (Love 1927)

Sn(θ, ϕ) =∑

m=0,n

[c m

n P mn (cos θ) cos(mϕ) + d m

n P mn (cos θ) sin(mϕ)

], (12.37)

where the c mn and d m

n are arbitrary coefficients, and the P mn (x) associated Legendre

functions.A solid harmonic of degree n (where n is a non-negative integer), denotedRn(r, θ, ϕ),

is defined as a well-behaved solution to

∇ 2Rn = 0 (12.38)

in the interior of a sphere (i.e., the region r < constant). It follows that (Love 1927)

Rn(r, θ, ϕ) ∝ r n Sn(θ, ϕ). (12.39)

In particular, the functions R (±m)n (r, θ, ϕ) and R (0)

n (r, θ), introduced in Section 12.3,are solid harmonics of degree n. Note that the Cartesian coordinates xi (where i runsfrom 1 to 3) are solid harmonics of degree 1. Moreover, ∂Rn/∂xi is a solid harmonicof degree n − 1. Finally, Φtide(r), specified in Equation (12.35), is a solid harmonicof degree 2.

The following results regarding solid harmonics are useful (Love 1927):

xi∂Rn

∂xi= r

∂Rn

∂r= nRn, (12.40)

∇ 2(xi Rn) = ∇ · (Rn ∇xi + xi ∇Rn) = 2∇xi · ∇Rn = 2∂Rn

∂xi, (12.41)

∇ 2(r m Rn) = ∇ 2(r m+n Sn) =1r 2

ddr

[r 2 d

dr(r m+n Sn)

]− n (n + 1) r m+n−2 Sn

= m (m + 2 n + 1) r m−2Rn. (12.42)

Here, use has been made of the Einstein summation convention (Riley 1974).

12.6 Planetary RotationSuppose that the planet is rotating rigidly about the axis θ = 0 at the angular velocityΩ (where Ω ω1). The planet’s rotational angular velocity vector is thus

Ω = Ω cos θ er − Ω sin θ eθ. (12.43)

Letφ = ϕ − Ω t. (12.44)

It follows that r, θ, φ are spherical coordinates in a frame of reference that co-rotateswith the planet.


12.7 Total Gravitational PotentialThe planet is modeled as a solid body of uniform mass density ρ whose surface liesat

r = a + ζb(θ, φ), (12.45)

whereζb(θ, φ) =

∑n=1,∞

ζb n(θ, φ), (12.46)

and ζb n is a surface harmonic of degree n. Suppose that the planetary surface iscovered by an ocean of uniform mass density ρ whose surface lies at

r = a + d + ζt(θ, φ), (12.47)

whereζt(θ, φ) =

∑n=1,∞

ζt n(θ, φ), (12.48)

and ζt n is a surface harmonic of degree n. Here, d is the constant unperturbed depthof the ocean, whereas ζb and ζt are the radial displacements of the ocean’s bottomand top surfaces, respectively, generated by planetary rotation and tidal effects. It isassumed that |ζb|, |ζt| d a.

The net gravitational acceleration in the vicinity of the planet takes the form

g(r) = −∇Φ, (12.49)

where Φ(r) is the total gravitational potential. In the limit d/a 1, we can write

Φ(r, θ, φ) = Φ0(r) +∑

n=1,∞Φn(r, θ, φ) +Φtide(r, θ, φ), (12.50)

where

∇ 2Φ0 =

4πG ρ + 4πG ρd δ(r − a) r ≤ a0 r > a

, (12.51)

and∇ 2Φn>0 = 4πG (ρ ζb n + ρ ζn) δ(r − a). (12.52)

Here,ζ(θ, φ) = ζt(θ, φ) − ζb(θ, φ), (12.53)

is the net change in ocean depth due to planetary rotation and tidal effects, and ζn =

ζt n − ζb n. Moreover, δ(x) is a Dirac delta function (Riley 1974). The boundaryconditions are that the Φn be well behaved as r → 0, and

Φn(r)→ 0 (12.54)


as r → ∞. It follows that

Φ0(r ≤ a) = −g

2 a(3 a 2 − r 2) − 3 g d

ρ

ρ, (12.55)

Φ0(r > a) = −g a 2

r

(1 + 3

daρ

ρ

), (12.56)

and

Φn>0(r ≤ a, θ, φ) = −g(

32 n + 1

) (ζb n +

ρ

ρζn

) ( ra

)n, (12.57)

Φn>0(r > a, θ, φ) = −g(

32 n + 1

) (ζb n +

ρ

ρζn

) ( ra

)−(n+1), (12.58)

where

g =G ma 2 =

4π3

G ρ a (12.59)

is the mean gravitational acceleration at the planet’s surface. Note that, inside theplanet (i.e., r ≤ a), Φn(r) is a solid harmonic of degree n. We can identify thethree terms appearing on the right-hand side of Equation (12.50) as the equilibriumgravitational potential generated by the planet and the ocean, the potential generatedby non-spherically-symmetric radial displacements of the planet and ocean surfaces,and the tide generating potential, respectively.

12.8 Planetary ResponseThe interior of the planet is modeled as a uniform, incompressible, elastic, solidpossessing the stress-strain relation (Love 1927)

σi j = −p δi j + µ

(∂ξi

∂x j+∂ξ j

∂xi

), (12.60)

and subject to the incompressibility constraint

∇ · ξ = 0. (12.61)

Here, σi j(r) is the stress tensor, δi j the identity tensor, ξ(r) the elastic displacement,p(r) the pressure, and µ the (uniform) rigidity of the material making up the planet(Riley 1974).

The planet’s equation of elastic motion in the co-rotating frame is (Fitzpatrick2012)

∂ 2ξi

∂t 2 + 2 εi jk Ω j∂ξk

∂t=

1ρ

∂σi j

∂x j− ∂Φ∂xi− ∂χ∂xi, (12.62)


where εi jk is the totally antisymmetric tensor (Riley 1974), and (Fitzpatrick 2012)

χ(r, θ) = −12Ω 2 r 2 sin2 θ = χ0(r) + χ2(r, θ), (12.63)

χ0(r) = −Ω 2 a 2

3

( ra

)2, (12.64)

χ2(r, θ) =Ω 2 a 2

3

( ra

)2P 0

2 (cos θ). (12.65)

Note that χ2(r) is a solid harmonic of degree 2. The second term on the left-hand sideof Equation (12.62) is the Coriolis acceleration (due to planetary rotation), whereasthe final term on the right-hand side is the centrifugal acceleration (likewise, due toplanetary rotation). The first two terms on the right-hand side are the forces per unitmass due to internal stresses and gravity, respectively.

Equations (12.60), (12.61), and (12.62) can be combined to give

∂ 2ξ

∂t 2 + 2Ω × ∂ξ∂t=µ

ρ∇ 2ξ − ∇

(pρ+Φ + χ

). (12.66)

Let us assume that

Ω √µ/ρ

a: (12.67)

that is, the typical timescale on which the tide generating potential varies (in the co-rotating frame) is much longer than the transit time of an elastic shear wave throughthe interior of the planet. In this case, we can neglect the left-hand side of Equa-tion (12.66), and write

µ

ρ∇ 2ξ − ∇

(pρ+Φ + χ

) 0, (12.68)

which is equivalent to saying that the interior of the planet always remains in anequilibrium state.

Let

p(r, θ, φ) = p0(r) +∑

n=1,∞

[pn(r, θ, φ) − ρΦn(r, θ, φ)

]− ρΦext(r, θ, φ), (12.69)

whereΦext(r, θ, φ) = Φtide(r, θ, φ) + χ2(r, θ), (12.70)

is the sum of the tide generating potential and the fictitious centrifugal potential dueto planetary rotation. Note that Φext is a solid harmonic of degree 2. It follows fromEquations (12.50), (12.63), and (12.68) that

ddr

(p0

ρ+ Φ0 + χ0

)= 0, (12.71)

andµ∇ 2ξ −

∑n=1,∞

∇pn = 0. (12.72)


Taking the divergence of the previous equation, and making use of Equations (12.61),we find that ∇ 2 pn = 0, which implies that pn(r) is a solid harmonic (of degree n).

Equation (12.71) can be integrated to give

p0(r) ρ

2 a

(g −

23Ω 2 a

)(a 2 − r 2) + ρ g d + patm, (12.73)

where use has been made of Equations (12.55) and (12.64), as well as

p0(a) = ρ g d + patm. (12.74)

Here, patm is the atmospheric pressure at sea level. The previous boundary conditionensures that the mean pressure at the surface of the planet is able to support themean weight of the ocean, as well as the weight of the atmosphere. Incidentally, it isassumed that g Ω 2 a, and we have neglected terms of order ρ Ω 2 a d with respectto terms of order ρ g d.

The elastic displacement in the interior of the planet satisfies Equations (12.61)and (12.72). It is helpful to define the radial component of the displacement

ξr =xi ξi

r, (12.75)

as well as the stress acting (outward) across a constant r surface,

Xi = −x j σi j

r= p

xi

r− µ

r

[r∂ξi

∂r− ξi +

∂(r ξr)∂xi

], (12.76)

where use has been made of Equation (12.60). The radial displacement at r = a isequivalent to the displacement of the planet’s surface. Hence, according to Equa-tion (12.45),

ξr(a, θ, φ) = ζb(θ, φ). (12.77)

The stress at any point on the surface r = a must be entirely radial (because a fluidocean cannot withstand a tangential stress), and such as to balance the weight of thecolumn of rock and ocean directly above the point in question. In other words,

Xi(a, θ, φ) = g[ρ ζb + ρ (d + ζ)

] ( xi

r

)r=a. (12.78)

It follows from Equation (12.57), (12.69), (12.70), (12.74), (12.76), and (12.78) that ∑n=1,∞

pnxi

r− µ

r

[r∂ξi

∂r− ξi +

∂(r ξr)∂xi

]r=a

= ρ g∑

n=1,∞

[−ζ2 δn 2 +

(2 n − 22 n + 1

)ζb n +

(2 n − 22 n + 1

)ρ

ρζn

] ( xi

r

)r=a, (12.79)

whereζ2(θ, φ) = −g−1 Φext|r=a = −g−1 (Φtide + χ2)r=a (12.80)


is a surface harmonic of degree 2. Here, ζ2(θ, φ), which has the dimensions of length,parameterizes the perturbation due to tidal and rotational effects at the planet’s sur-face. Moreover, δi j is a Kronecker delta symbol (Riley 1974). Hence, we need tosolve Equations (12.61) and (12.72), subject to the boundary conditions (12.77) and(12.79).

Let us try a solution to Equations (12.61) and (12.72) of the form

ξi =∑

n=1,∞ξi n, (12.81)

where (Love 1927)

ξi n = An r 2 ∂pn

∂xi+ Bn pn xi +

∂φn

∂xi. (12.82)

Here, An and Bn are spatial constants, and φn(r) is an arbitrary solid harmonic ofdegree n. It follows that

r ξr n ≡ xi ξi n = (n An + Bn) r 2 pn + n φn, (12.83)

where use has been made of Equation (12.40). Moreover,

∂ (r ξr n)∂xi

= (n An + Bn) r 2 ∂pn

∂xi+ 2 (n An + Bn) pn xi + n

∂φn

∂xi, (12.84)

and

r∂ξi n

∂r− ξi n = n An r 2 ∂pn

∂xi+ n Bn pn xi + (n − 2)

∂φn

∂xi. (12.85)

Thus, the boundary conditions (12.77) and (12.79) become[(2 An + Bn) r 2 pn + n φn

]r=a= a ζb n, (12.86)

and

(1 − [2 n An + (2 + n) Bn] µ) (pn xi)r=a−µ[(2 n An + Bn) r 2 ∂pn

∂xi+ 2 (n − 1)

∂φn

∂xi

]r=a

= ρ g

[−ζ2 δn 2 +

(2 n − 22 n + 1

)ζb n +

(2 n − 22 n + 1

)ρ

ρζn

](xi)r=a, (12.87)

respectively. Suppose that φn(r) is such that

(2 n An + Bn) a 2 pn(r) + 2 (n − 1) φn(r) = 0. (12.88)

In this case, the boundary conditions (12.86) and (12.87) reduce to[−4 An + (n − 2) Bn

2 (n − 1)

]a pn|r=a = ζb n, (12.89)


and

(1 − [2 n An + (2 + n) Bn] µ) pn|r=a =

ρ g

[−ζ2 δn 2 +

(2 n − 22 n + 1

)ζb n +

(2 n − 22 n + 1

)ρ

ρζn

], (12.90)

respectively.The expression for ξi n given in Equation (12.82) satisfies Equations (12.61) and

(12.72) provided that

2 n An µ + (3 + n) Bn µ = 0, (12.91)

2 (2 n + 1) An µ + 2 Bn µ = 1, (12.92)

respectively, where use has been made of Equations (12.40)–(12.42). It follows that

An µ =(3 + n)

2 (2 n + 3) (n + 1), (12.93)

Bn µ = −2 n

2 (2 n + 3) (n + 1). (12.94)

Hence, the boundary conditions (12.89) and (12.90) yield

ζb n = h2 ζ2 δn 2 + h′n αn ζn, (12.95)

where

αn =

(3

2 n + 1

)ρ

ρ, (12.96)

hn =

(2 n + 12 n − 2

)/ [1 +

(2 n + 1) (2 n 2 + 4 n + 3)(6 + n 2)

µ

ρ g a

], (12.97)

h′n = −(

2 n + 13

)/ [1 +

(2 n + 1) (2 n 2 + 4 n + 3)(6 + n 2)

µ

ρ g a

]. (12.98)

Here, the αn parameterize the self-gravity of the ocean. Let

δΦ = Φext +∑

n=1,∞Φn. (12.99)

This quantity can be recognized as the total perturbing potential at the planet’s sur-face due to the combination of tidal, rotational, and ocean self-gravity, effects. Itfollows from Equation (12.57), (12.80), and (12.95) that

−(δΦ

g

)r=a=∑

n=1,∞

[(1 + k2) ζ2 δn 2 + (1 + k′n)αn ζn

], (12.100)


where

kn =

(3

2 n − 2

)/ [1 +

(2 n + 1) (2 n 2 + 4 n + 3)(6 + n 2)

µ

ρ g a

], (12.101)

k′n = −[1 +

(2 n + 1) (2 n 2 + 4 n + 3)(6 + n 2)

µ

ρ g a

]−1

. (12.102)

Here, the dimensionless quantities hn, h′n, kn, and k′n are known as Love numbers(Love 1911).

12.9 Laplace Tidal EquationsIn the co-rotating frame, the linearized (in v) equations of motion of the ocean thatcovers the surface of the planet are (Fitzpatrick 2012)

∇ · v = 0, (12.103)

∂v∂t+ 2Ω × v = −∇

(pρ+Φ + χ

), (12.104)

where v(r) is the fluid velocity, and p(r) the fluid pressure. Here, Φ(r) is the totalgravitational potential, χ(r) the centrifugal potential (due to planetary rotation), and2Ω × v the Coriolis acceleration (likewise, due to planetary rotation). Furthermore,we have neglected the finite compressibility of the fluid that makes up the ocean. Letr, θ, φ be spherical coordinates in the co-rotating frame, and let z = r−a. It is helpfulto define u = vθ, v = vφ, and w = vr. Assuming that |z| a, the previous equations ofmotion reduce to

1a sin θ

∂

∂θ(sin θ u) +

1a sin θ

∂v

∂φ+∂w

∂z 0, (12.105)

and

∂u∂t− 2Ω cos θ v = −1

a∂

∂θ

(pρ+ Φ + χ

), (12.106)

∂v

∂t+ 2Ω cos θ u + 2Ω sin θ w = − 1

a sin θ∂

∂φ

(pρ+ Φ + χ

), (12.107)

∂w

∂t− 2Ω sin θ v = − ∂

∂z

(pρ+Φ + χ

). (12.108)

In accordance with Equations (12.45), (12.47), and (12.53), the lower and uppersurfaces of the ocean lie at

z = ζb(θ, φ, t), (12.109)


andz = d + ζb(θ, φ, t) + ζ(θ, φ, t), (12.110)

respectively, where|ζb|, |ζ | d a. (12.111)

It follows that

w|z=0 =∂ζb

∂t, (12.112)

w|z=d =∂(ζb + ζ)∂t

. (12.113)

Let us assume that u = u(θ, φ, t) and v = v(θ, φ, t): that is, the horizontal componentsof the fluid velocity are independent of z. Integration of Equation (12.105) from z = 0to z = d, making use of the previous two boundary conditions, yields

∂ζ

∂t= − d

a sin θ

[∂

∂θ(sin θ u) +

∂v

∂φ

]. (12.114)

Equations (12.112)–(12.114) imply that

w ∼(

da

)u,(

da

)v u, v. (12.115)

In accordance with the analysis of Section 12.8, we can write

Φ + χ =

(g −

23Ω 2 a

)(z − d) + δΦ(θ, φ, t), (12.116)

where δΦ is the total perturbing potential due to tidal, rotation, and ocean self-gravity, effects. We have neglected an unimportant constant term. We have alsoneglected the z-dependence of the perturbing potential, because the variation length-scale of this potential is a, rather than d, so that

∆(δΦ) ∼ d∂δΦ

∂z∼

daδΦ δΦ, (12.117)

where ∆(δΦ) = δΦ|z=d − δΦ|z=0. Let us assume that the ocean is in approximatevertical force balance: that is,

∂

∂z

(pρ+ Φ + χ

) 0. (12.118)

It follows thatp − patm

ρ+Φ + χ = P(θ, φ, t), (12.119)

where patm is the (uniform and constant) pressure of the atmosphere. However, pres-sure balance at the ocean’s upper surface requires that

p|z=d+ζb+ζ= patm. (12.120)


Hence, we deduce that

P(θ, φ, t) = (Φ + χ)z=d+ζb+ζ g (ζb + ζ) + δΦ, (12.121)

to first order in small quantities [i.e., ζb/a, ζ/a, δΦ/(g a), and Ω 2 a/g]. Thus, Equa-tions (12.106), (12.107), and (12.115) yield

∂u∂t− 2Ω cos θ v = −1

a∂

∂θ

[g (ζb + ζ) + δΦ

], (12.122)

∂v

∂t+ 2Ω cos θ u = − 1

a sin θ∂

∂φ

[g (ζb + ζ) + δΦ

]. (12.123)

Let us justify our previous assumption that the ocean is in approximate verticalforce balance. Equations (12.108), (12.115), and (12.119) imply that

p − patm

ρ+Φ + χ = P(θ, φ, t) + O(dΩ v), (12.124)

where we have assumed that ∂/∂t ∼ O(Ω) and ∂/∂z ∼ d −1. However, Equa-tions (12.121)–(12.123) yield

P ∼ g ζ ∼ aΩ v dΩ v. (12.125)

Hence, Equation (12.124) becomes

p − patm

ρ+Φ + χ = P(θ, φ, t), (12.126)

which is equivalent to Equation (12.119).Let us justify our previous assumption that u only depends weakly on z. It follows

from Equations (12.117) and (12.122) that

u ∼ δΦaΩ

, (12.127)

∆u ∼ ∆(δΦ)aΩ

∼ daδΦ

aΩ∼ d

au u, (12.128)

where ∆u = u|z=d − u|z=0. In a similar manner, it can be shown that ∆v v.According to Equations (12.95) and (12.100),

ζb =∑

n=1,∞

[h2 ζ2 δn 2 + h′n αn ζn

], (12.129)

−g−1 δΦ =∑

n=1,∞

[(1 + k2) ζ2 δn 2 + (1 + k′n)αn ζn

]. (12.130)


Hence, Equations (12.114), (12.122), and (12.123) give

∂ζ

∂t= − d

a sin θ

[∂

∂θ(sin θ u) +

∂v

∂φ

], (12.131)

∂u∂t− 2Ω cos θ v = −g

a∂

∂θ

ζ − (1 + k2 − h2) ζ2 −∑

n=1,∞(1 + k′n − h′n)αn ζn

,(12.132)

∂v

∂t+ 2Ω cos θ u = − g

a sin θ∂

∂φ

ζ − (1 + k2 − h2) ζ2 −∑

n=1,∞(1 + k′n − h′n)αn ζn

.(12.133)

We can writeζ(θ, φ) =

∑n=1,∞

ζn(θ, φ), (12.134)

where [cf., Equation (12.37)]

ζn(θ, φ) =∑

m=0,n

c mn P m

n (cos θ) e i mφ, (12.135)

and the c mn are arbitrary complex constants. Now, (Abramowitz and Stegun 1965)∫ π

0P m

n (cos θ) P mn′ (cos θ) sin θ dθ =

(2

2 n + 1

)(n + m)!(n − m)!

δn n′ . (12.136)

Hence,∑n=1,∞

(1 + k′n − h′n)αn ζn =

∫ π

−π

∫ π

0GF (θ, φ; θ′, φ′) ζ(θ′, φ′) sin θ′ dθ′ dφ′, (12.137)

where

GF(θ, φ; θ′, φ′) =∑

n=1,∞

∑m=0,n

(2 n + 1)4π

(n − m)!(n + m)!

(1 + k′n − h′n)

αn P mn (cos θ) P m

n (cos θ′) e i m (φ−φ′). (12.138)

is termed the Farrell Green’s function (Farrell 1972). It follows that

∂ζ

∂t= − d

a sin θ

[∂

∂θ(sin θ u) +

∂v

∂φ

], (12.139)

∂u∂t− 2Ω cos θ v = −g

a∂

∂θ

[ζ − (1 + k2 − h2) ζ2 −

∮GF ζ dΩ′

], (12.140)

∂v


a sin θ∂

∂φ

[ζ − (1 + k2 − h2) ζ2 −

∮GF ζ dΩ′

], (12.141)


j ζ j m j σ j Classification0 −(ζ1 + ζ2) 0 0 Equilibrium1 −3 e ζ2 0 ω1 − ω2 Long Period2 +(3/2) sin2 δ ζ2 0 2ω1 Long Period3 − sin δ ζ2 1 Ω − 2ω1 Diurnal4 − sin δ ζ2 1 Ω Diurnal5 +(7 e/4) ζ2 2 2Ω − 3ω1 + ω2 Semi-diurnal6 +(1/2) ζ2 2 2Ω − 2ω1 Semi-diurnal

Table 12.1The principal harmonics of the forcing term in the Laplace tidal equations.

where dΩ′ ≡ sin θ′ dθ′ dφ′. Equations (12.139)–(12.141) are known collectively asthe Laplace tidal equations, because they were first derived (in simplified form) byLaplace (Lamb 1993).

The Laplace tidal equations are a closed set of equations for the perturbed oceandepth, ζ(θ, φ, t), and the polar and azimuthal components of the horizontal oceanvelocity, u(θ, φ, t), and v(θ, φ, t), respectively. Here, a is the planetary radius, g themean gravitational acceleration at the planetary surface, d the mean ocean depth, Ωthe planetary angular rotation velocity, and hn, h′n, kn, k′n are Love numbers definedin Equations (12.97), (12.98), (12.101), and (12.102), respectively. Amongst otherthings, the Love numbers determine the elastic response of the planet to the forcingterm, ζ2. Finally, the Farrell Green’s function parameterizes the self-gravity of theocean.

12.10 Harmonics of Forcing Term in Laplace Tidal Equations

Making use of Equations (12.25)–(12.27), (12.35), (12.44), (12.65), and (12.80), wecan write the forcing term in the Laplace tidal equations in the form

ζ2(θ, φ, t) = Re∑

j

ζ j(θ, φ, t), (12.142)

where

ζ j(θ, φ, t) = ζ j P mj

2 (cos θ) e i (mj φ+σ j t). (12.143)


j ζ j m j σ j Classification Symbol0 −(ζΩ + ζM + ζE) 0 0 Equilibrium S 0, M0

1 +(3/2) sin2 ε ζE 0 2ωE Long Period S sa

2 −3 eM ζM 0 ωM − ΠM Long Period Mm

3 +(3/2) sin2 ε ζM 0 2ωM Long Period M f

4 − sin ε ζM 1 Ω − 2ωM Diurnal O1

5 − sin ε ζE 1 Ω − 2ωE Diurnal P1

6 − sin ε (ζM + ζE) 1 Ω Diurnal K1

7 +(7 eM/4) ζM 2 2Ω − 3ωM + ΠM Semi-diurnal N2

8 +(1/2) ζM 2 2Ω − 2ωM Semi-diurnal M2

9 +(1/2) ζE 2 2Ω − 2ωE Semi-diurnal S 2

Table 12.2The principal harmonics of the forcing term in the Laplace tidal equations for theEarth.

The amplitudes, ζ j, azimuthal mode numbers, m j, and frequencies, σ j, of the princi-pal harmonics of the forcing term are specified in Table 12.1, where

ζ1 =Ω 2 a 2

3 g, (12.144)

ζ2 =12

m′

ma 4

R 3 . (12.145)

Here, m is the planetary mass, m′ the mass of the moon, and R the moon’s orbitalmajor radius. As indicated in the table, the harmonics of the forcing term can bedivided into four classes: an “equilibrium” term that is time independent; two “longperiod” terms that oscillate with periods much longer than the planet’s diurnal (i.e.,rotational) period; two “diurnal” terms that oscillate at periods close to the planet’sdiurnal period; and two “semi-diurnal” terms that oscillate at periods close to halfthe planet’s diurnal period. Here, we have neglected a semi-diurnal term that issignificantly smaller than the other two.

For the Earth-Moon system, m = 5.97 × 1024 kg, m′ = 7.35 × 1022 kg, a =6.37 × 106 m, R = 3.84 × 108 m, Ω = 7.29 × 10−5 rad. s−1 (Yoder 1995). It followsthat g = 9.82 m s−2, as well as ζ1 = ζΩ, ζ2 = ζM, where

ζΩ = 7.32 × 103 m, (12.146)

ζM = 1.79 × 10−1 m. (12.147)

Moreover, δ = ε, e = eM, ω1 = ωM , and ω2 = ΠM, where ε = 23.44, eM = 0.05488,360/ωM = 27.322 (solar) days, 360/ΠM = 8.847 years (Yoder 1995). Here, m,a, Ω, and ε are the Earth’s mass, mean radius, (sidereal) rotational angular velocity,and inclination of the equatorial plane to the ecliptic, respectively. Furthermore, m′,


j ζ j(m) m j f j = σ j/2Ω Period Symbol

1 +1.93 × 10−2 0 2.73 × 10−3 182.62 days S sa

2 −2.95 × 10−2 0 1.81 × 10−2 27.555 days Mm

3 +4.25 × 10−2 0 3.65 × 10−2 13.661 days M f

4 −7.12 × 10−2 1 4.63 × 10−1 25.819 hours O1

5 −3.24 × 10−2 1 4.97 × 10−1 24.066 hours P1

6 −10.36 × 10−2 1 5.00 × 10−1 23.934 hours K1

7 +1.72 × 10−2 2 9.45 × 10−1 12.658 hours N2

8 +8.95 × 10−2 2 9.63 × 10−1 12.421 hours M2

9 +4.07 × 10−2 2 9.97 × 10−1 12.000 hours S 2

Table 12.3The principal harmonics of the forcing term in the Laplace tidal equations for theEarth, excluding the equilibrium harmonic.

R, eM , ωM , and ΠM are the Moon’s mass, orbital major radius, orbital eccentricity,mean orbital angular velocity, and rate of perigee precession, respectively. Here, wehave neglected the small (i.e., about 5) inclination of the Moon’s orbital plane to theecliptic.

For the Earth-Sun system, m′ = 1.99 × 1030 kg and R = 1.50 × 1011 m (Yoder1995). It follows that ζ1 = ζΩ and ζ = ζE , where

ζE = 8.14 × 10−2 m. (12.148)

Moreover, δ = ε, e = eE , ω1 = ωE , and ω2 = ΠE , where eE = 0.0167, 360/ω1 =

365.24 days, and 360/ΠE = 20940 years (Yoder 1995). Here, m′ is the solar mass.Furthermore, R, eE , ωE , and ΠE are the Earth’s orbital major radius, orbital eccen-tricity, mean orbital angular velocity, and rate of perihelion precession, respectively.

Combining lunar and solar effects, we can write the forcing term in the Laplacetidal equations for the Earth in the form (12.142)–(12.143). The properties of theprincipal harmonics of the forcing term are specified in Tables 12.2 and 12.3. Thesymbols are due to G.H. Darwin (1845-1912).

12.11 Response to Equilibrium Harmonic

The j = 0 harmonic of the forcing term, which is associated with the Earth’s ax-ial rotation, is special, because the associated oscillation frequency is zero. In thiscase, Equations (12.139)–(12.141) yield u = v = 0. Hence, it follows from Equa-


tions (12.129), (12.142) and (12.143), as well as Table 12.2, that

ζb(θ) = −[h2 +

h′2 (1 + k2 − h2)α2

1 − (1 + k′2 − h′2)α2

](ζΩ + ζM + ζE) P2(cos θ), (12.149)

ζ(θ) = −[

1 + k2 − h2

1 − (1 + k′2 − h′2)α2

](ζΩ + ζM + ζE) P2(cos θ). (12.150)

For the Earth-Moon-Sun system, ζΩ + ζM + ζE = 7.32 × 103 m. Given the relativelylarge size of ζΩ + ζM + ζE , we expect the steady-state response to the equilibriumharmonic to be fluid-like (otherwise, the elastic stress within the Earth would exceedthe yield stress) (Fitzpatrick 2012). In other words, µ/(ρ g a) = 0 for the j = 0harmonic, which implies from Equations (12.97), (12.98), (12.101), and (12.102)that h2 = 5/2, h′2 = −5/3, k2 = 3/2, and k′2 = −1. Thus, it follows from the previoustwo equations that

ζb(θ) = −52

(ζΩ + ζM + ζE) P2(cos θ) = −18.3 P2(cos θ) km, (12.151)

ζ(θ) = 0. (12.152)

We deduce that the Earth’s rotation causes a planetary equatorial (i.e., θ = π/2) bulgeof about 9.2 km, and a polar (i.e., θ = 0) flattening of 18.3 km, but does not give riseto any spatial variation in ocean depth. The observed equatorial bulge and polar flat-tening of the Earth are 7.1 km and 14.3 km, respectively (Yoder 1995). Our estimatesfor these values are too large because, for the sake of simplicity, we are treating theEarth as a uniform body. In reality, the Earth possesses a mass distribution that isstrongly concentrated in its core.

12.12 Global Ocean TidesSuppose that the surface of the Earth is completely covered by an ocean of uniformmean depth d. The motion of this ocean under the action of the tide generatingpotential is governed by the Laplace tidal equations, (12.139)–(12.141), which canbe written

1sin θ

[∂

∂θ(sin θ ξ) +

∂η

∂φ

]+ ζ = 0, (12.153)

..ξ − 2Ω cos θ .

η = −g da 2

∂(ζ − F)∂θ

, (12.154)

..η + 2Ω cos θ

.ξ = −

g da 2 sin θ

∂(ζ − F )∂φ

, (12.155)

where

F (θ, φ, t) = (1 + k2 − h2) ζ2(θ, φ, t) +∮

GF(θ, φ; θ′, φ′) ζ(θ′, φ′, t) dΩ′, (12.156)


and

u =ad∂ξ

∂t, (12.157)

v =ad∂η

∂t. (12.158)

Here, . ≡ ∂/∂t.It is convenient to write (Love 1913)

ξ = −∂Φ

∂θ−

1sin θ

∂Ψ

∂φ, (12.159)

η = − 1sin θ

∂Φ

∂φ+∂Ψ

∂θ. (12.160)


DΦ = ζ, (12.161)

where

D ≡1

sin θ

(∂

∂θsin θ

∂

∂θ+

1sin θ

∂ 2

∂φ 2

). (12.162)

Forming (sin θ)−1 ∂(12.154)/∂φ− (sin θ)−1 ∂[sin θ(12.155)/∂θ], we obtain

−(∂

∂tD + 2Ω

∂

∂φ

) .Ψ + 2Ω

(cos θD − sin θ

∂

∂θ

) .Φ = 0. (12.163)

Similarly, forming −(sin θ)−1 (∂/∂θ) [sin θ (12.154)]− (sin θ)−1 ∂(12.155)/∂φ, we get(∂

∂tD + 2Ω

∂

∂φ

) .Φ + 2Ω

(cos θD − sin θ

∂

∂θ

) .Ψ =

g da 2 D (ζ − F ). (12.164)

Let µ = cos θ. Consider the response, ζ j(µ, φ, t), of the ocean to a particularharmonic of the forcing term that takes the form [see Equation (12.143)]

ζ2(µ, φ, t) = ζ j P mj

2 (µ) e i (mj φ+σ j t), (12.165)

where ζ j is a constant. We can write

Φ(µ, φ, t) = −∑

n=1,∞

zn

n (n + 1)P mj

n (µ) e i (mj φ+σ j t), (12.166)

Ψ (µ, φ, t) = i∑

n=1,∞Ψn P mj

n (µ) e i (mj φ+σ j t), (12.167)

where the P mn (x) are associated Legendre functions, and the zn and Ψn are constants.

Now, by definition (Abramowitz and Stegun 1965),

D P mn (µ) e i m φ = −n (n + 1) P m

n (µ) e i m φ. (12.168)


Hence, Equations (12.161) and (12.166) give

ζ j(µ, φ, t) =∑

n=1,∞zn P mj

n (µ) e i (mj φ+σ j t). (12.169)

Equations (12.136) and (12.138) imply that∮GF (θ, φ; θ′, φ′) ζ j(θ′, φ′, t) dΩ′ =

∑n=1,∞

(1 + k′n − h′n)αn zn P mjn (µ) e i (mj φ+σ j t).

(12.170)It follows from Equations (12.163) and (12.164) that∑

n=1,∞

Ψn

[− f j n (n + 1) + m j

]P mj

n (µ) + zn

[µ −

(1 − µ 2)n (n + 1)

ddµ

]P mj

n (µ)= 0,

(12.171)and∑

n=1,∞

zn

[n (n + 1)

β(1 − [1 + k′n − h′n]αn) +

f j m j

n (n + 1)− f 2

j

]P mj

n (µ)

+ f j Ψn

[n (n + 1) µ − (1 − µ2)

ddµ

]P mj

n (µ)=

6β

(1 + k2 − h2) ζ j P mj

2 (µ), (12.172)

where (Lamb 1993)

f j =σ j

2Ω, (12.173)

β =4 a 2Ω 2

g d. (12.174)

However, as is well known (Abramowitz and Stegun 1965),

µ P mn (µ) =

(n − m + 1

2 n + 1

)P m

n+1(µ) +( n + m2 n + 1

)P m

n−1(µ), (12.175)

(1 − µ 2)dP m

n

dµ= −

[n (n − m + 1)

2 n + 1

]P m

n+1(µ) +[(n + 1) (n + m)

2 n + 1

]P m

n−1(µ). (12.176)

Equations (12.171), (12.172), (12.175), and (12.176) can be combined to give[m j − n (n + 1) f j

]Ψn+

(n + 1

nn − m j

2 n − 1

)zn−1+

(n

n + 1n + m j + 1

2 n + 3

)zn+1 = 0, (12.177)

and[n (n + 1)

β(1 − [1 + k′n − h′n]αn) +

m j f j

n (n + 1)− f 2

j

]zn

+ f j

[(n − 1) (n + 1) (n − m j)

2 n − 1

]Ψn−1

+ f j

[n (n + 2) (n + m j + 1)

2 n + 3

]Ψn+1 =

6β

(1 + k2 − h2) ζ j δn 2.

(12.178)


Finally, Equations (12.177) and (12.178) yield the tridiagonal matrix equation (Love1913)

a(n) zn−2 + b(n) zn + c(n) zn+2 =6β

(1 + k2 − h2) ζ j δn 2 (12.179)

for n = 2, 4, 6, · · · , where

a(n) = −f j n (n + 1) (n − m j − 1) (n − m j)

(2 n − 3) (2 n − 1) [m j − n (n − 1) f j], (12.180)

b(n) =n (n + 1)

β

(1 − [1 + k′n − h′n]αn

)+

m j f j

n (n + 1)− f 2

j

−f j (n − 1)2 (n + 1) (n − m j) (n + m j)

n (2 n − 1) (2 n + 1) [m j − n (n − 1) f j]

−f j (n + 2)2 n (n − m j + 1) (n + m j + 1)

(n + 1) (2 n + 1) (2 n + 3) [m j − (n + 1) (n + 2) f j], (12.181)

c(n) = −f j n (n + 1) (n + m j + 1) (n + m j + 2)

(2 n + 3) (2 n + 5) [m j − (n + 1) (n + 2) f j]. (12.182)

Note that z0 = 0. The solution of a tridiagonal matrix equation is, of course, arelatively trivial numerical exercise (Press, et al. 2007).

We can most conveniently characterize the jth tidal harmonic in terms of thefunction ζ j(ϑ), which is defined such that

ζ j(µ, φ, t) = ζ j(ϑ) cos(m j φ + σ j t). (12.183)

Here, ϑ = π/2 − θ = sin−1(µ) represents planetary latitude. It follows that

ζ j(ϑ) =∑

n=2,∞zn P mj

n (sinϑ). (12.184)

The ζ j, m j, and f j values for the various harmonics of the forcing term in the Laplacetidal equations for the Earth are specified in Table 12.3. Note that, on the longitudinalmeridian φ = 0, the Moon and Sun culminate simultaneously at t = 0. Moreover, atthis time, the Sun and Moon are both passing though the summer solstice, and theMoon is passing though its perigee. The mean density of the Earth is such that ρ/ρ =5.5 (Yoder 1995). Moreover, the mean depth of the Earth’s oceans is d = 3.8 km(Yoder 1995), which implies that β = 23.1. Finally, the Earth responds elastically tothe relatively low-amplitude lunar and solar tidal gravitational fields in such a mannerthat µ/(ρ g a) 0.35 (Fitzpatrick 2012).

Figures 12.1, 12.2, and 12.3 show the long-period, diurnal, and semi-diurnalsolutions of the Laplace tidal equations together with the corresponding equilibriumtides (i.e., the tides calculated in the absence of ocean inertia),

ζ j(ϑ) =[

1 + k2 − h2

1 − (1 + k′2 − h′2)α2

]ζ j P mj

2 (sinϑ). (12.185)


−0.01

−0.005

0

0.005

0.01

ζj

0 0.2 0.4 0.6 0.8 1sin(ϑ)

−0.03

−0.02

−0.01

0

0.01

0.02

0.03

0.04

ζj

0 0.2 0.4 0.6 0.8 1sin(ϑ)

Figure 12.1Long-period solutions to the Laplace tidal equations (left panel) together with thecorresponding equilibrium tides (right panel). The solid, dashed, and dash-dottedlines correspond to j = 1, j = 2, and j = 3, respectively. Here, ζ j is measured inmeters.

It can be seen from Figure 12.1 that the long-period solutions are direct (i.e.,they have the same signs as the corresponding equilibrium tides). However, the am-plitudes of these solutions are about 4 times lower than those of the equilibrium tides.The j = 1 (S sa) and j = 3 (M f ) solutions cause a slight increase in the equatorialbulge of the Earth’s oceans when the Sun and Moon, respectively, pass through theequinoxes, and a slight reduction when they pass through the solstices. The j = 2(Mm) solution causes a slight increase in the equatorial bulge of the Earth’s oceanswhen the Moon passes through its perigee, and a slight reduction when it passesthrough its apogee.

It can be seen from Figure 12.2 that the diurnal solutions are inverted (i.e., theyhave the opposite signs to the corresponding equilibrium tides). The amplitude ofthe j = 4 solution is about 6 times lower than that of the j = 4 equilibrium tide.The amplitudes of the j = 5 and j = 6 solutions are negligibly small. (In fact, theamplitude of the j = 6 solution is identically zero.) The j = 4 (O1) solution causesthe Earth’s oceans to displace slightly toward the Moon when it passes through theequinoxes, and away from the Moon when it passes through the solstices.

Finally, it can be seen from Figure 12.3 that the semi-diurnal solutions are in-verted in equatorial regions, and direct in polar regions. Moreover, the amplitudes ofthese solutions are similar to those of the corresponding equilibrium tides. The dom-inant j = 8 (M2) solution generates a high tide at a particular point on the Earth’ssurface every 12.421 hours. The j = 9 (S 2) solution causes an enhancement of thehigh tide—a so-called spring tide—at the new and full moons (i.e., every half a syn-odic month, or 14.77 days). Finally, the j = 7 (N2) solution causes an enhancement


0

0.03

0.06

0.09

0.12

ζj

0 0.2 0.4 0.6 0.8 1sin(ϑ)

−0.015

−0.01

−0.005

0

ζj

0 0.2 0.4 0.6 0.8 1sin(ϑ)

Figure 12.2Diurnal solutions to the Laplace tidal equations (left panel) together with the corre-sponding equilibrium tides (right panel). The solid, dashed, and dash-dotted linescorrespond to j = 4, j = 5, and j = 6, respectively. Here, ζ j is measured in meters.

−0.35

−0.3

−0.25

−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

ζj

0 0.2 0.4 0.6 0.8 1sin(ϑ)

0

0.1

0.2

ζj

0 0.2 0.4 0.6 0.8 1sin(ϑ)

Figure 12.3Semi-diurnal solutions to the Laplace tidal equations (left panel) together with thecorresponding equilibrium tides (right panel). The solid, dashed, and dash-dottedlines correspond to j = 7, j = 8, and j = 9, respectively. Here, ζ j is measured inmeters.


of the spring tide—a so-called perigean spring tide—whenever the new or full mooncoincides with the passage of the moon through its perigee.

The previously described global solutions to the Laplace tidal equations werefirst obtained (in simplified form) by Laplace (Lamb 1993). Let us consider howthese solutions compare with the observed terrestrial tides. One obvious predictionof the global solutions is that the time variation in ocean depth at a given point onthe Earth’s surface should be expressible as a superposition of terms that oscillate atconstant amplitude (but not necessarily in phase with one another) with the periodsspecified in Table 12.3. This is indeed found to be the case (Cartwright 1999). An-other prediction is that the amplitudes of the long-period solutions should be smallcompared to the amplitudes of the semi-diurnal solutions. This is also found to bethe case. Yet another prediction is that amplitudes of the diurnal solutions shouldbe small compared to the amplitudes of the semi-diurnal solutions. This turns outnot to be the case (Cartwright 1999). In fact, this predication is an artifact of thefact that we are assuming that the ocean covering the Earth’s surface is of constantmean depth. If we relax this assumption, and allow the ocean depth to vary in arealistic manner, then the amplitudes of the semi-diurnal solutions become compa-rable to those of the diurnal solutions (Lamb 1993). The final prediction is that agiven solution is such that the tide at a given point on the Earth either oscillates inphase or in anti-phase with the corresponding equilibrium solution (i.e., the tide iseither direct or inverted). In particular, for the dominant M2 tide, this implies that,at a given point on the Earth’s surface, there should be either a tidal maximum or atidal minimum whenever the Moon culminates. It turns out that this is not the case(Cartwright 1999). Indeed, the observed phase difference between the M2 tide andthe corresponding equilibrium tide varies from place to place on the Earth’s surface,and can take any value. In other words, high tide at a given point on the Earth canoccur three hours before the Moon culminates, or two hours after, et cetera, and thetime lag between high tide and the culmination of the Moon varies from place toplace. We conjecture that this effect is caused by the continents impeding the flow oftidal waves around the Earth. Hence, we now need to consider tides in oceans thatdo not cover the whole surface of the Earth.

12.13 Non-Global Ocean Tides

Suppose that the surface of the Earth is covered by an ocean of uniform depth d thatextends from θ = 0 to θ = π, and from φ = φ− to φ = φ+, where |φ+ − φ−| < 2π. Forconsistency with the previous analysis, we must imagine that the remaining surfaceis covered by a frozen (i.e., ζ = 0) ocean of uniform depth d. The motion of theocean under the action of the tide generating potential is governed by the Laplace


tidal equations, (12.153)–(12.155), which are written

1sin θ

[∂

∂θ(sin θ ξ) +

∂η

∂φ

]+ ζ = 0, (12.186)

..ξ − 2Ω cos θ .

η = −g da 2

∂

∂θ(ζ − F ), (12.187)

..η + 2Ω cos θ

.ξ = − g d

a 2 sin θ∂

∂φ(ζ − F ), (12.188)

where

F (θ, φ, t) = (1 + k2 − h2) ζ2(θ, φ, t) +∮

GF(θ, φ; θ′, φ′) ζ(θ′, φ′, t) dΩ′. (12.189)

The finite extent of the ocean in the φ direction introduces the additional constraint

η(θ, φ±, t) = 0 (12.190)

into the problem.

12.14 Useful LemmaLet X(θ, φ, t), Y(θ, φ, t), P(θ, φ, t), and Q(θ, φ, t) be well-behaved functions of θ andφ. Suppose that

D P = − 1sin θ

[∂

∂θ(sin θ X) +

∂Y∂φ

], (12.191)

where1

sin θ∂P(θ, φ±, t)

∂φ= −Y(θ, φ±, t), (12.192)

Suppose, further, that

D Q =1

sin θ

[∂

∂θ(sin θ Y) −

∂X∂φ

], (12.193)

whereQ(θ, φ±, t) = 0. (12.194)

We wish to demonstrate that

X = −∂P∂θ− 1

sin θ∂Q∂φ, (12.195)

Y = − 1sin θ

∂P∂φ+∂Q∂θ. (12.196)


Let

X′ = X +∂P∂θ+

1sin θ

∂Q∂φ, (12.197)

Y′ = Y +1

sin θ∂P∂φ− ∂Q∂θ. (12.198)

Note, from Equations (12.192) and (12.194), that

Y′(θ, φ±, t) = 0. (12.199)

It follows from Equations (12.191), (12.193), (12.197), and (12.198) that

∂

∂θ(sin θ X′) +

∂Y′

∂φ= 0, (12.200)

∂

∂θ(sin θ Y′) − ∂X′

∂φ= 0. (12.201)

Equations (12.199) and (12.201) imply that

X′ =∂P′

∂θ, (12.202)

Y′ =1

sin θ∂P′

∂φ, (12.203)

and1

sin θ∂P′(θ, φ±, t)

∂φ= 0. (12.204)

Furthermore, Equation (12.200) yields

D P′ = 0. (12.205)

Multiplying by P′, integrating over the ocean, and making use of the boundary con-dition (12.204), we obtain

∫Ω

(∂P′

∂θ

)2+

(1

sin θ∂P′

∂φ

)2 dΩ = 0. (12.206)

Here, Ω is the surface of the Earth that is covered by ocean, and dΩ = sin θ dθ dφ.It follows that P′ is a constant. Thus, Equations (12.202) and (12.203) imply thatX′ = Y′ = 0, and, hence, that Equations (12.195) and (12.196) are valid.


12.15 Transformation of Laplace Tidal EquationsLet X = ξ, Y = η, P = Φ, and Q = Ψ in Equations (12.191)–(12.196). It follows that[cf., Equation (12.161)]

DΦ = ζ, (12.207)

DΨ =1

sin θ

[∂

∂θ(sin θ η) −

∂ξ

∂φ

], (12.208)

where

1sin θ

∂Φ(θ, φ±, t)∂φ

= 0, (12.209)

Ψ (θ, φ±, t) = 0. (12.210)

Here, use has been made of Equations (12.186) and (12.190). Furthermore [cf.,Equations (12.159) and (12.160)],

ξ = −∂Φ∂θ− 1

sin θ∂Ψ

∂φ, (12.211)

η = −1

sin θ∂Φ

∂φ+∂Ψ

∂θ. (12.212)

Let X = cos θ η, Y = − cos θ ξ, P = Φ′, and Q = Ψ ′ in Equations (12.191)–(12.196). It follows that

DΦ′ = − 1sin θ

[∂

∂θ(sin θ cos θ η) − ∂

∂φ(cos θ ξ)

], (12.213)

DΨ ′ = − 1sin θ

[∂

∂θ(sin θ cos θ ξ) +

∂

∂φ(cos θ η)

], (12.214)

where

1sin θ

∂Φ′(θ, φ±, t)∂φ

= cos θ ξ(θ, φ±, t), (12.215)

Ψ ′(θ, φ±, t) = 0. (12.216)

Furthermore,

cos θ η = −∂Φ′

∂θ− 1

sin θ∂Ψ ′

∂φ, (12.217)

− cos θ ξ = − 1sin θ

∂Φ′

∂φ+∂Ψ ′

∂θ. (12.218)


Substitution of Equations (12.211), (12.212), (12.217), and (12.218) into Equa-tions (12.187) and (12.188) yields

∂A∂θ+

1sin θ

∂B∂φ= 0, (12.219)

∂B∂θ− 1

sin θ∂A∂φ= 0, (12.220)

where

A =..Φ − 2Ω

.Φ′ − g d

a 2 (ζ − F), (12.221)

B =..Ψ − 2Ω

.Ψ ′. (12.222)

Equations (12.219) and (12.220) can be combined to give

D A = 0, (12.223)

D B = 0. (12.224)

Moreover, it follows from Equations (12.188), (12.190), (12.209), (12.210), (12.215),and (12.216) that

1sin θ

∂A(θ, φ±, t)∂φ

= 0, (12.225)

B(θ, φ±, t) = 0. (12.226)

We have already seen that the solution of Equation (12.205), subject to the bound-ary condition (12.204), is P′ = constant. It follows that the solution of Equa-tion (12.223), subject to the boundary condition (12.225), is A = constant. Anal-ogous arguments reveal that the solution of Equation (12.224), subject to the bound-ary condition (12.226), is B = 0. Hence, we deduce that the Laplace tidal equations,(12.186)–(12.188), are equivalent to the following set of equations:

DΦ = ζ, (12.227)

..Φ − 2Ω

.Φ′ − g d

a 2 (ζ − F ) = constant, (12.228)..Ψ − 2Ω

.Ψ ′ = 0, (12.229)

where Φ, Ψ , Φ′, Ψ ′, and F are defined in Equations (12.211), (12.212), (12.217),(12.218), and (12.189), respectively.


12.16 Another Useful LemmaSuppose that g(θ, φ) and f (θ, φ) are two well-behaved functions that satisfy either

1sin θ

∂ f (θ, φ±)∂φ

= 0, (12.230)

1sin θ

∂g(θ, φ±)∂φ

= 0, (12.231)

or

f (θ, φ±) = 0, (12.232)

g(θ, φ±) = 0. (12.233)

It is easily demonstrated that∫Ω

gD f dΩ = −∫Ω

(∂g

∂θ

∂ f∂θ+

1sin2 θ

∂g

∂φ

∂ f∂φ

)dΩ =

∫Ω

f D g dΩ. (12.234)

12.17 Basis EigenfunctionsSuppose that the Φr(θ, φ) are well-behaved solutions of the eigenvalue equation

DΦr + λr Φr = 0, (12.235)


1sin θ

∂Φr(θ, φ±)∂φ

= 0. (12.236)

It immediately follows from Equations (12.234) and (12.235) that∫Ω

(Φ ∗r DΦr −Φr DΦ ∗r ) dΩ = −(λr − λ ∗r )∫Ω

|Φr | 2 dΩ = 0. (12.237)

Hence, we deduce that the λr are real. It follows that we can choose the Φr to be realfunctions. Equations (12.234) and (12.235) also yield∫

Ω

Φr DΦr dΩ = −λr

∫Ω

Φ 2r dΩ = −

∫Ω

(∂Φr

∂θ

)2+

(1

sin θ∂Φr

∂φ

)2 dΩ, (12.238)

which implies that the λr are positive. Integration of Equation (12.235), subject tothe boundary condition (12.236), gives

λr

∫Ω

Φr dΩ = 0. (12.239)


Because λr is positive, this implies that∫Ω

Φr dΩ = 0. (12.240)

Finally, Equations (12.234) and (12.235) yield∫Ω

(Φs DΦr −Φr DΦs) dΩ = (λs − λr)∫Ω

ΦsΦr dΩ = 0. (12.241)

It follows that ∫Ω

ΦsΦr dΩ = 0 (12.242)

if λs λr. As is well known (Riley 1974), if Φr and Φr′ are linearly independentsolutions of (12.235) corresponding to the same eigenvalue, λr, then it is alwayspossible to choose linear combinations of them that satisfy∫

Ω

Φr Φr′ dΩ = 0. (12.243)

This argument can be extended to multiple linearly independent solutions corre-sponding to the same eigenvalue. Hence, we conclude that it is possible to choosethe Φr such that they satisfy the orthonormality condition

λr

∫Ω

Φr Φs dΩ = λs

∫Ω

Φr Φs dΩ = δr s. (12.244)

Let F(θ, φ) be a well-behaved function. Suppose that

1sin θ

∂F(θ, φ±)∂φ

= 0. (12.245)

We can automatically satisfy the previous boundary condition by writing

F =∑

r=1,∞ar Φr. (12.246)

(Note that F is undetermined to an arbitrary additive constant which is chosen so asto ensure that

∫Ω

F dΩ = 0.) Here, λ1 is the smallest eigenvalue of Equation (12.235),λ2 the next smallest eigenvalue, and so on. It follows from Equation (12.244) that

ar = λr

∫Ω

F Φr dΩ. (12.247)

Suppose that the Ψr(θ, φ) are well-behaved solutions of the eigenvalue equation

DΨr + µr Ψr = 0, (12.248)


Ψr(θ, φ±) = 0. (12.249)


Using analogous arguments to those employed previously, we can show that the µr

are real and positive, and that theΨr can be chosen so as to satisfy the orthonormalityconstraint

µr

∫Ω

Ψr Ψs dΩ = µs

∫Ω

Ψr Ψs dΩ = δr s. (12.250)

Let F(θ, φ) be a well-behaved function. Suppose that

F(θ, φ±) = 0. (12.251)

We can automatically satisfy the previous boundary condition by writing

F =∑

r=1,∞ar Ψr . (12.252)


ar = µr

∫Ω

F Ψr dΩ. (12.253)

12.18 Auxilliary EigenfunctionsLet

X = −cos θsin θ

∂Φr

∂φ, (12.254)

Y = cos θ∂Φr

∂θ, (12.255)

P = Φ′r, (12.256)

Q = Ψ ′′r (12.257)

in Equations (12.191)–(12.196). It follows that

DΦ′r =1

sin θ

[∂

∂θ

(cos θ

∂Φr

∂φ

)− ∂

∂φ

(cos θ

∂Φr

∂θ

)], (12.258)

DΨ ′′r =1

sin θ

[∂

∂θ

(sin θ cos θ

∂Φr

∂θ

)+

1sin θ

∂

∂φ

(cos θ

∂Φr

∂φ

)], (12.259)

and

1sin θ

∂Φ′r(θ, φ±)∂φ

= − cos θ∂Φr

∂θ, (12.260)

Ψ ′′r (θ, φ±) = 0, (12.261)


as well as

−cos θsin θ

∂Φr

∂φ= −

∂Φ′r∂θ− 1

sin θ∂Ψ ′′r∂φ

, (12.262)

cos θ∂Φr

∂θ= − 1

sin θ∂Φ′r∂φ+∂Ψ ′′r∂θ

. (12.263)

Let

X = cos θ∂Ψr

∂θ, (12.264)

Y =cos θsin θ

∂Ψr

∂φ, (12.265)

P = Φ′′r , (12.266)

Q = Ψ ′r (12.267)

in Equations (12.191)–(12.196). It follows that

DΨ ′′r = −1

sin θ

[∂

∂θ

(sin θ cos θ

∂Ψr

∂θ

)+

1sin θ

∂

∂φ

(cos θ

∂Ψr

∂φ

)], (12.268)

DΨ ′r =1

sin θ

[∂

∂θ

(cos θ

∂Ψr

∂φ

)− ∂

∂φ

(cos θ

∂Ψr

∂θ

)], (12.269)

and

1sin θ

∂Φ′′r (θ, φ±)∂φ

= −cos θsin θ

∂Ψr(θ, φ±)∂φ

, (12.270)

Ψ ′r (θ, φ±) = 0, (12.271)

as well as

cos θ∂Ψr

∂θ= −

∂Φ′′r∂θ− 1

sin θ∂Ψ ′r∂φ

, (12.272)

cos θsin θ

∂Ψr

∂φ= −

1sin θ

∂Φ′′r∂φ+∂Ψ ′r∂θ. (12.273)

12.19 Gyroscopic CoefficientsLet

βr, s = λs

∫Ω

Φ′r Φs dΩ = −∫Ω

Φ′r DΦs dΩ

=

∫Ω

(∂Φ′r∂θ

∂Φs

∂θ+

1sin2 θ

∂Φ′r∂φ

∂Φs

∂φ

)dΩ, (12.274)


where use has been made of Equations (12.235) and (12.236). It follows from Equa-tions (12.262) and (12.263) that

βr, s =

∫Ω

cos θsin θ

(∂Φr

∂φ

∂Φs

∂θ− ∂Φr

∂θ

∂Φs

∂φ

)dΩ

−∫Ω

1sin θ

(∂Ψ ′′r∂φ

∂Φs

∂θ−∂Ψ ′′r∂θ

∂Φs

∂φ

)dΩ. (12.275)

However, the second term on the right-hand side of the previous equation integratesto zero with the aid of Equation (12.261). Hence, we are left with

βr, s = −∫Ω

cos θsin θ

(∂Φr

∂θ

∂Φs

∂φ− ∂Φr

∂φ

∂Φs

∂θ

)dΩ. (12.276)

Let

β−r, s = λs

∫Ω

Φ′′r Φs dΩ = −∫Ω

Φ′′r DΦs dΩ

=

∫Ω

(∂Φ′′r∂θ

∂Φs

∂θ+

1sin2 θ

∂Φ′′r∂φ

∂Φs

∂φ

)dΩ, (12.277)


β−r, s = −∫Ω

cos θ(∂Ψr

∂θ

∂Φs

∂θ+

1sin2 θ

∂Ψr

∂φ

∂Φs

∂φ

)dΩ

−∫Ω

1sin θ

(∂Ψ ′r∂φ

∂Φs

∂θ−∂Ψ ′r∂θ

∂Φs

∂φ

)dΩ. (12.278)


β−r, s = −∫Ω

cos θ(∂Ψr

∂θ

∂Φs

∂θ+

1sin2 θ

∂Ψr

∂φ

∂Φs

∂φ

)dΩ. (12.279)

Let

βr,−s = µs

∫Ω

Ψ ′′r Ψs dΩ = −∫Ω

Ψ ′′r DΨs dΩ

=

∫Ω

(∂Ψ ′′r∂θ

∂Ψs

∂θ+

1sin2 θ

∂Ψ ′′r∂φ

∂Ψs

∂φ

)dΩ, (12.280)


βr,−s =

∫Ω

cos θ(∂Φr

∂θ

∂Ψs

∂θ+

1sin2 θ

∂Φr

∂φ

∂Ψs

∂φ

)dΩ

+

∫Ω

1sin θ

(∂Φ′r∂φ

∂Ψs

∂θ−∂Φ′r∂θ

∂Ψs

∂φ

)dΩ. (12.281)



βr,−s =

∫Ω

cos θ(∂Φr

∂θ

∂Ψs

∂θ+

1sin2 θ

∂Φr

∂φ

∂Ψs

∂φ

)dΩ. (12.282)

Finally, let

β−r,−s = µs

∫Ω

Ψ ′r Ψs dΩ = −∫Ω

Ψ ′r DΨs dΩ

=

∫Ω

(∂Ψ ′r∂θ

∂Ψs

∂θ+

1sin2 θ

∂Ψ ′r∂φ

∂Ψs

∂φ

)dΩ, (12.283)


β−r,−s =

∫Ω

cos θsin θ

(∂Ψr

∂θ

∂Ψs

∂φ− ∂Ψr

∂θ

∂Ψs

∂φ

)dΩ

+

∫Ω

1sin θ

(∂Φ′′r∂φ

∂Ψs

∂θ−∂Φ′′r∂θ

∂Ψs

∂φ

)dΩ. (12.284)


β−r,−s = −∫Ω

cos θsin θ

(∂Ψr

∂θ

∂Ψs

∂φ− ∂Ψr

∂φ

∂Ψs

∂θ

)dΩ. (12.285)

Incidentally, the βr,s, β−r,s, βr,−s, and β−r,−s are known collectively as gyroscopiccoefficients (Proudman 1916).

12.20 Proudman EquationsWe can automatically satisfy the boundary condition (12.209) by writing

Φ(θ, φ, t) =∑

r=1,∞pr(t)Φr(θ, φ). (12.286)

Likewise, we can automatically satisfy the boundary condition (12.210) by writing

Ψ (θ, φ, t) =∑

r=1,∞p−r(t)Ψr(θ, φ). (12.287)


It follows from Equations (12.211), (12.212), (12.227), and (12.235) that

ξ = −∑

r=1,∞

(pr∂Φr

∂θ+ p−r

1sin θ

∂Ψr

∂φ

), (12.288)

η =∑

r=1,∞

(p−r

∂Ψr

∂θ− pr

1sin θ

∂Φr

∂φ

), (12.289)

ζ = −∑

r=1,∞λr prΦr . (12.290)

Let

Ar = λr

∫Ω

Φ′Φr dΩ = −∫Ω

Φ′ DΦr dΩ (12.291)

=

∫Ω

(∂Φ′

∂θ

∂Φr

∂θ+

1sin2 θ

∂Φ′

∂φ

∂Φr

∂φ

)dΩ, (12.292)


Ar =

∫Ω

cos θ(−η ∂Φr

∂θ+

ξ

sin θ∂Φr

∂φ

)dΩ

−∫Ω

1sin θ

(∂Ψ ′

∂φ

∂Φr

∂θ− ∂Ψ

′

∂θ

∂Φr

∂φ

)dΩ. (12.293)


Ar =

∫Ω

cos θ(−η ∂Φr

∂θ+

ξ

sin θ∂Φr

∂φ

)dΩ. (12.294)

Finally, Equations (12.276), (12.282), (12.288), and (12.289) can be combined withthe previous equation to give

λr

∫Ω

Φ′Φr dΩ = −∑

s=1,∞

(βr, s ps + βr,−s p−s

). (12.295)

Let

A−r = µr

∫Ω

Ψ ′ Ψr dΩ = −∫Ω

Ψ ′ DΨr dΩ

=

∫Ω

(∂Ψ ′

∂θ

∂Ψr

∂θ+

1sin2 θ

∂Ψ ′

∂φ

∂Ψr

∂φ

)dΩ, (12.296)

where use has been made of Equations (12.216) and (12.248). It follows from Equa-


tions (12.217) and (12.218) that

A−r = −∫Ω

cos θ(ξ∂Ψr

∂θ+

η

sin θ∂Ψr

∂φ

)dΩ

+

∫Ω

1sin θ

(∂Φ′

∂φ

∂Ψr

∂θ− ∂Φ

′

∂θ

∂Ψr

∂φ

)dΩ. (12.297)


A−r = −∫Ω

cos θ(ξ∂Ψr

∂θ+

η

sin θ∂Ψr

∂φ

)dΩ. (12.298)

Finally, Equations (12.279), (12.285), (12.288), and (12.289) can be combined withthe previous equation to give

µr

∫Ω

Ψ ′ Ψr dΩ = −∑

s=1,∞

(β−r, s ps + β−r,−s p−s

). (12.299)

Operating on Equations (12.228) and (12.229) with λr∫Ω

(· · · )Φr dΩ and withµr∫Ω

(· · · )Ψr dΩ, respectively, yields the so-called Proudman equations (Proudman1916):

..pr + 2Ω∑

s=1,∞(βr, s

.ps + βr,−s.p−s) +

g da 2 λr pr = −

g da 2 Fr, (12.300)

..p−r + 2Ω∑

s=1,∞(β−r, s

.ps + β−r,−s.p−s) = 0, (12.301)

whereFr = λr

∫Ω

F Φr dΩ. (12.302)

Here, use has been made of Equations (12.240), (12.244), (12.250), (12.286), (12.287),(12.290), (12.295), and (12.299). Finally, it follows from Equations (12.138), (12.189),and (12.290) that

Fr = Pr −∑

s=1,∞γr, s ps, (12.303)

wherePr = λr

∫Ω

(1 + k2 − h2) ζ2Φr dΩ, (12.304)

andγr, s =

∑n=1,∞

∑m=0,n

(1 + k′n − h′n)αn Γmn, r Γ

m ∗n, s , (12.305)

withΓm

n, r = λr

∫Ω

P mn (cos θ) e i mφ Φr(θ, φ) dΩ. (12.306)


Here,P m

n (x) = b mn P m

n (x), (12.307)

where

b mn =

[2 n + 1

4π(n − m)!(n + m)!

]1/2. (12.308)

It follows from Equation (12.136) that∫ π

0P m

n (cos θ) P mn′ (cos θ) sin θ dθ =

δn n′

2π. (12.309)

Consider the response of the ocean to a particular harmonic of the tide generatingpotential for which

ζ2(θ, φ, t) = ζ j P mj

2 (cos θ) e i (mj φ+σ j t). (12.310)

Assuming a common exp( iσ j t) time dependence of the pr and the p−r, Equations(12.300) and (12.301) yield

(β−1 λr − f 2j ) pr +

∑s=1,∞

[(i f j βr, s − β−1 γr, s) ps + i f j βr,−s p−s

]

= −β−1 (1 + k2 − h2) ζ j Γ

mj

2, r

b mj

2

, (12.311)

and− f 2

j p−r + i∑

s=1,∞( f j β−r, s ps + f j β−r,−s p−s) = 0, (12.312)

where f j and β are defined in Equations (12.173) and (12.174), respectively.The task in hand (see Section 12.13) is to solve the Laplace tidal equations,

(12.186)–(12.189), subject to the constraint (12.190). Our basic approach is to ex-pand the fields appearing in the Laplace tidal equations—namely, ξ, η, and ζ—interms of a set of basis eigenfunctions—the Φr and the Ψr (see Section 12.17)—that are defined in such a manner as to automatically satisfy the boundary condi-tions. The expansion is specified in Equations (12.288)–(12.290). The Laplace tidalequations then reduce to the Proudman equations, (12.300)–(12.301), which are aset of coupled ordinary differential equations for the weights in the expansion—thepr(t) and the p−r(t). For a particular harmonic of the tide generating potential, theweights all oscillate at the same frequency, and the Proudman equations reduce fur-ther to give two coupled matrix equations for the amplitudes of the weights—seeEquations (12.311) and (12.312). In order to solve the matrix equations, we needto calculate the gyroscopic coefficients, βr,s, β−r,s, βr,−s, β−r,−s, as well as the Γm

n,r.These quantities are integrals involving the eigenfunctions and the associated Leg-endre functions, and are defined in Equations (12.276), (12.279), (12.282), (12.285),and (12.306), respectively.


12.21 Hemispherical Ocean TidesConsider a hemispherical ocean for which φ− = 0 and φ+ = π. In general, thereare many linearly independent solutions to the eigenvalue problem (12.235) and(12.236), subject to the orthonormality constraint (12.244), that correspond to a givenvalue of λr. It is convenient to differentiate these solutions by means of two indices,n and m. Thus, Φr → Φm

n and λr → λmn , where the index n ranges from 1 to ∞.

Moreover, for a given value of n, the index m ranges from 0 to n. Now,

∫ π

0cos(m φ) cos(m′ φ) dφ =

π m = m′ = 0π/2 m = m′ 00 m m′

. (12.313)

Hence, it follows that

Φmn (θ, φ) = a m

n P mn (cos θ) cos(m φ), (12.314)

andλm

n = n (n + 1), (12.315)

where

a mn =

(4/λm

n )1/2 m > 0(2/λm

n )1/2 m = 0. (12.316)

Likewise, the solutions to the eigenvalue problem (12.248) and (12.249), subject tothe orthonormality constraint (12.250), is such that Ψr → Ψ m

n and µr → µmn , where

Ψ mn (θ, φ) = a m

n P mn (cos θ) sin(m φ), (12.317)

andµm

n = n (n + 1). (12.318)

We also have (see Section 12.19)

βr, s → βm,m′n, n′ = −

∫Ω

cos θsin θ

∂Φmn

∂θ

∂Φm′n′

∂φ−∂Φm

n

∂φ

∂Φm′n′

∂θ

dΩ, (12.319)

β−r, s → β−m,m′−n, n′ = −

∫Ω

cos θ∂Ψ m

n

∂θ

∂Φm′n′

∂θ+

1sin2 θ

∂Ψ mn

∂φ

∂Φm′n′

∂φ

dΩ, (12.320)

βr,−s → βm,−m′n,−n′ =

∫Ω

cos θ∂Φm

n

∂θ

∂Ψ m′n′

∂θ+

1sin2 θ

∂Φmn

∂φ

∂Ψ m′n′

∂φ

dΩ, (12.321)

β−r,−s → β−m,−m′−n,−n′ = −

∫Ω

cos θsin θ

∂Ψ mn

∂θ

∂Ψ m′n′

∂φ−∂Ψ m

n

∂φ

∂Ψ m′n′

∂θ

dΩ, (12.322)


which yields

βm,m′n, n′ =

∫ π

0

∫ 1

−1

∂Φmn

∂µ

∂Φm′n′

∂φ−∂Φm

n

∂φ

∂Φm′n′

∂µ

µ dµ dφ, (12.323)

β−m,m′−n, n′ = −

∫ π

0

∫ 1

−1

(1 − µ 2)∂Ψ m

n

∂µ

∂Φm′n′

∂µ+

11 − µ 2

∂Ψ mn

∂φ

∂Φm′n′

∂φ

µ dµ dφ,

(12.324)

βm,−m′n,−n′ =

∫ π

0

∫ 1

−1

(1 − µ 2)∂Φm

n

∂µ

∂Ψ m′n′

∂µ+

11 − µ 2

∂Φmn

∂φ

∂Ψ m′n′

∂φ

µ dµ dφ, (12.325)

β−m,−m′−n,−n′ =

∫ π

0

∫ 1

−1

∂Ψ mn

∂µ

∂Ψ m′n′

∂φ−∂Ψ m

n

∂φ

∂Ψ m′n′

∂µ

µ dµ dφ, (12.326)

where µ = cos θ. However,∫ π

0sin(m φ) cos(m′ φ) dφ =

2 m/(m 2 − m′ 2) m + m′ odd0 m + m′ even

. (12.327)

Hence, if m + m′ is even then the β±m,±m′±n,±n′ are zero. Otherwise, we obtain

βm,m′n, n′

c mn c m′

n′=

2m 2 − m′ 2

∫ 1

−1

m 2 P mn

dP m′n′

dµ+ m′ 2

dP mn

dµP m′

n′

µ dµ, (12.328)

β−m,m′−n, n′

c mn c m′

n′= − 2 m

m 2 − m′ 2

∫ 1

−1

(1 − µ 2)dP m

n

dµdP m′

n′

dµ+

m′ 2

1 − µ 2 P mn P m′

n′

µ dµ, (12.329)

β−m,−m′−n,−n′

c mn c m′

n′=

2 m m′

m 2 − m′ 2

∫ 1

−1

P mn

dP m′n′

dµ+

dP mn

dµP m′

n′

µ dµ, (12.330)

as well asβm,−m′

n,−n′ = −β−m′ ,m−n′ , n , (12.331)

wherec m

n = a mn b m

n . (12.332)

Let

I m,m′n, n′ =

∫ 1

−1P m

n (µ) P m′n′ (µ) dµ. (12.333)

It follows, from symmetry, that I m,m′n, n′ = 0 when n + n′ − m − m′ is odd. When

n + n′ − m − m′ is even (Wong 1998),

I m,m′n, n′ =

∑p=0,pmax

∑p′=0,p′max

αmn,p α

m′n′ ,p′

Γ[(n + n′ − m − m′ − 2p − 2p′ + 1)/2]Γ[(m+ m′ + 2p + 2p′ + 2)/2]Γ[(n + n′ + 3)/2]

, (12.334)


where Γ(x) denotes a gamma function (Abramowitz and Stegun 1965),

αmn,p =

(−1)m+p (n + m)!2 m+2p (m + p)! p! (n − m − 2p)!

, (12.335)

and

pmax =

(n − m)/2 n − m even(n − m − 1)/2 n − m odd

, (12.336)

with an analogous definition for p′max.It can be shown that (Longuet-Higgins and Pond 1970)

βm,m′n, n′

c mn c m′

n′=

[n′ (n′ + 1) + m′ (m′ + 1)

m′ + 1− 2 m′

n (n + 1) − n′ (n′ + 1) + m′

m 2 + m′ 2

]I m,m′n, n′

+

(1

m′ + 1

)I m,m′+2n, n′ , (12.337)

βm,−m′n,−n′

c mn c m′

n′=

2 m(m′ 2 − m 2) (2 n′ + 1)

[(n′ − 1) (n′ + 1) (n′ + m′) I m,m′

n, n′−1

+n′ (n′ + 2) (n′ − m′ + 1) I m,m′

n,n′+1

], (12.338)

β−m,−m′−n,−n′

c mn c m′

n′=

2 m m′

m′ 2 − m 2 I m,m′n, n′ . (12.339)

According to Equation (12.306), we can also write Γmn,r → Γm,m′

n,n′ , where

Γm,m′n,n′ = (a m

n )−1(δm,m′

n,n′ + i κm,m′n,n′

), (12.340)

andδm,m′

n, n′ = δn n′ δm m′ . (12.341)

Here, κm,m′n, n′ = 0 if m + m′ is even, and

κm,m′n, n′ =

(2 m

m 2 − m′ 2

)λm′

n′ c mn c m′

n′ I m,m′n, n′ , (12.342)

otherwise. It follows from Equation (12.305) that γr, s → γm,m′n, n′ , where

γm,m′n, n′ = (1 + k′n − h′n)αn (a m

n )−2 δm,m′n,n′ − i ε m,m′

n, n′ + θm,m′n, n′ . (12.343)

Here, ε m,m′n, n′ = 0 if m + m′ is even, and

ε m,m′n, n′ = (1 + k′n − h′n)αn (a m

n )−2 κm,m′n,n′ − (1 + k′n′ − h′n′ )αn′ (a m′

n′ )−2 κm′,mn′ ,n (12.344)


otherwise. Now, if m and m′ are both even then

θm,m′n, n′ =

∑n′′=1,∞

m′′ odd∑m′′=0,n′′

(1 + k′n′′ − h′n′′ )αn′′ (a m′′n′′ )−2 κm′′ ,m

n′′ ,n κm′′ ,m′n′′ ,n′ ; (12.345)

if m and m′ are both odd then

θm,m′n, n′ =

∑n′′=1,∞

m′′ even∑m′′=0,n′′

(1 + k′n′′ − h′n′′ )αn′′ (a m′′n′′ )−2 κm′′ ,m

n′′ ,n κm′′ ,m′n′′ ,n′ ; (12.346)

and if m + m′ is odd then θm,m′n, n′ = 0.

If we let pr → i n p mn and p−r → i n+1 p−m

−n then Equations (12.311) and (12.312)become∑

n′=1,∞

∑m′=0,n′

([(β−1 [λm

n − (1 + k′n − h′n)αn (a mn )−2] − f 2

j

)δm,m′

n, n′ + f j βm,m′n, n′

+β−1 (ε m,m′n, n′ − θ

m,m′n, n′ )

]p m′

n′ − f j βm,−m′n,−n′ p−m′

−n′)

=β−1 (1 + k2 − h2) ζ j

c mj

2

(δ

mj ,m2, n + κ

mj ,m2, n

), (12.347)

and ∑n′=1,∞

∑m′=0,n′

([− f 2

j δm,m′n, n′ + f j β

−m,−m′−n,−n′

]p−m′−n′ + f j β

−m,m′−n, n′ p m′

n′)= 0, (12.348)

respectively. Here,

δm,m′n, n′ = i n+n′ δm,m′

n, n′ , (12.349)

βm,m′n, n′ = i n+n′+1 βm,m′

n, n′ , (12.350)

βm,−m′n,−n′ = i n+n′ βm,−m′

n,−n′ , (12.351)

β−m,m′−n, n′ = i n+n′ β−m,m′

−n, n′ , (12.352)

β−m,−m′−n,−n′ = i n+n′+1 β−m,−m′

−n,−n′ , (12.353)

ε m,m′n, n′ = i n+n′+1 ε m,m′

n, n′ , (12.354)

θm,m′n, n′ = i n+n′ θm,m′

n, n′ , (12.355)

κm,m′n, n′ = i n+n′+1 κm,m′

n, n′ . (12.356)

By symmetry, δm,m′n, n′ and θm,m′

n, n′ are only non-zero when m + m′ is even, and n + n′ iseven; βm,m′

n, n′ , β−m,−m′−n,−n′ , ε m,m′

n, n′ , and κm,m′n, n′ are only non-zero when m + m′ is odd, and

n+n′ is odd; and βm,−m′n,−n′ and β−m,m′

−n, n′ are only non-zero when m+m′ is odd, and n+n′

is even. It follows that all quantities appearing in Equations (12.347) and (12.348)


are real. Once we have solved these equations to obtain the p mn (which is a relatively

straightforward numerical task), we can reconstruct the tidal elevation as follows:

ζ(θ, φ, t) = ζc(θ, φ) cos(σ j t) + ζs(θ, φ) sin(σ j t), (12.357)

where

ζc(θ, φ) =n even∑n=1,∞

∑m=0,n

ζ mn Φ

mn (θ, φ) cos(n π/2), (12.358)

ζs(θ, φ) = −n odd∑

n=1,∞

∑m=0,n

ζ mn Φ

mn (θ, φ) sin(n π/2), (12.359)

andζ m

n = −λmn p m

n . (12.360)

Thus, the tidal amplitude at a given point on the ocean is

|ζ |(θ, φ) =[ζ 2

c (θ, φ) + ζ 2s (θ, φ)

]1/2. (12.361)


∆t(θ, φ)T

=∆ϕ(θ, φ)

2π, (12.362)

where

∆ϕ(θ, φ) = m jπ

2+ tan−1

[ζs(θ, φ)ζc(θ, φ)

]. (12.363)

Here, T = 2π/σ j is the oscillation period of the harmonic of the tide generatingpotential under consideration, ∆t is the time-lag between the peak tide at a givenpoint on the ocean and the maximal tide generating potential at φ = π/2, and ∆ϕ isthe corresponding phase-lag.

Figures 12.4 and 12.5 show the amplitude and phase-lag of the ( j = 3) M f long-period tide in a hemispherical ocean of mean depth d = 3.8 km (which corresponds toβ = 23.1), calculated assuming that ρ/ρ = 5.5 and µ/(ρ g a) = 0.35. The calculationincludes all azimuthal harmonics up to nmax = 20. Note that only a quarter of theocean is shown, because the amplitude is symmetric about the meridians ϑ = 0and φ = π/2, whereas the phase-lag is symmetric about the meridian ϑ = 0, andantisymmetric about the meridian φ = π/2. Here, ϑ ≡ π/2 − θ. Given that ζ3 =

4.25 × 10−2 m (see Table 12.3), the maximum amplitude of the tide is about 3.5 cm,and occurs at the poles. Moreover, it is clear from a comparison with Figure 12.1that the tide is direct (i.e., it is in phase with the equilibrium tide). In fact, the M f

tide in a hemispherical ocean is about four times larger in amplitude than that in aglobal ocean (i.e., an ocean that covers the whole surface of the Earth) of the samedepth. Otherwise, the two tides have fairly similar properties.

Figures 12.6 and 12.7 show the amplitude and phase-lag of the ( j = 6) K1 diur-nal tide in a hemispherical ocean of mean depth d = 3.8 km (which corresponds to


β = 23.1), calculated assuming that ρ/ρ = 5.5 and µ/(ρ g a) = 0.35. The calculationincludes all azimuthal harmonics up to nmax = 20. Given that ζ6 = −10.36 × 10−2 m(see Table 12.3), the maximum amplitude of the tide is about 15.8 cm, and occurs atmid-latitudes. This is very different to the case of a global ocean, where the ampli-tude of the K1 tide is identically zero everywhere. (See Figure 12.2.) Note that thetidal phase-lag only exhibits a fairly weak dependence on the azimuthal angle, φ. Infact, the K1 tide in a hemispherical ocean essentially oscillates in anti-phase with theequilibrium tide at the ocean’s central longitudinal meridian (φ = π/2), except closeto the poles, where it oscillates in phase. Again, this is very different to the case of aglobal ocean, where the tidal maximum lies on a meridian of longitude, and rotatessteadily around the Earth from east to west.

Figures 12.8 and 12.9 show the amplitude and phase-lag of the ( j = 8) M2 semi-diurnal tide in a hemispherical ocean of mean depth d = 3.8 km (which correspondsto β = 23.1), calculated assuming that ρ/ρ = 5.5 and µ/(ρ g a) = 0.35. The calcula-tion includes all azimuthal harmonics up to nmax = 20. Given that ζ8 = 8.95×10−2 m(see Table 12.3), the maximum amplitude of the tide is about 60.7 cm, and occursat the poles. This is very different to the case of a global ocean, where the ampli-tude of the M f tide is zero at the poles. (See Figure 12.3.) Another major differenceis that, in a hemispherical ocean, the tidal maxima circulate around points of zerotidal amplitude—such points are known as amphidromic points. Of course, in thecase of a global ocean, the tidal maxima lie on opposite meridians of longitude, androtate steadily around the Earth from east to west. The systems of tidal waves, cir-culating around amphidromic points, that is evident in Figure 12.9, are known asamphidromic systems, and are one of the the major features of tides in real oceans(Cartwright 1999). Generally speaking, the sense of circulation of amphidromic sys-tems in real oceans is counter-clockwise (seen from above) in the Earth’s northernhemisphere, and clockwise in the southern hemisphere.

In conclusion, our investigation of tides in a hemispherical ocean suggests thatthe impedance of the flow of tidal waves around the Earth, due to the presence ofthe continents, is likely to have a comparatively little effect on long-period tides, buta very significant effect on diurnal and semi-diurnal tides. In particular, for semi-diurnal tides, the impedance gives rise to the formation of amphidromic systems.

12.22 Exercises

12.1 Consider a tidal wave whose wavelength is very much less than the radius ofthe Earth. This corresponds to the limit n 1, where n is an azimuthal modenumber. Suppose, however, that the ocean depth, d, is allowed to vary with


Figure 12.4Relative amplitude, |ζ |/ζ3, of the ( j = 3) M f long-period tide in a hemisphericalocean.

Figure 12.5Phase-lag, ∆ϕ, of the ( j = 3) M f long-period tide in a hemispherical ocean.


Figure 12.6Relative amplitude, |ζ |/|ζ6|, of the ( j = 6) K1 diurnal tide in a hemispherical ocean.

Figure 12.7Phase-lag, ∆ϕ, of the ( j = 6) K1 diurnal tide in a hemispherical ocean.


Figure 12.8Relative amplitude, |ζ |/ζ8, of the ( j = 8) M2 semi-diurnal tide in a hemisphericalocean.

Figure 12.9Phase-lag, ∆ϕ, of the ( j = 8) M2 semi-diurnal tide in a hemispherical ocean.


position. Show that, in this limit, the Laplace tidal equations are written

∂ζ

∂t= − 1

a sin θ

[∂

∂θ(sin θ u d) +

∂(v d)∂φ

],

∂u∂t− 2Ω cos θ v = −

g

a∂

∂θ(ζ − ζ′),

∂v


a sin θ∂

∂φ(ζ − ζ′),

whereζ′ = (1 + k2 − h2) ζ2.

12.2 Consider short-wavelength tidal waves in a region of the ocean that is suffi-ciently localized that it is a good approximation to treat cos θ and sin θ as con-stants. We can define local Cartesian coordinates, x, y, z, such that dx = a dθ,dy = a sin θ dφ, and dz = r − a. It follows that the x-axis is directed south-ward, the y-axis is directed eastward, and the z-axis is directed verticallyupward. Show that, when expressed in terms of this local coordinate system,the Laplace tidal equations derived in the previous exercise reduce to

∂ζ

∂t= −∂(u d)

∂x− ∂(v d)

∂y,

∂u∂t− f v = −g ∂(ζ − ζ′)

∂x,

∂v

∂t+ f u = −g ∂(ζ − ζ′)

∂y,

wheref = 2Ω cos θ

is a constant. In fact, the previous equations are the linearized equations ofmotion of a body of shallow water confined to a tangent plane that touchesthe Earth at the angular coordinates θ, φ. This plane is known as the f -planebecause, as a consequence of the Earth’s diurnal rotation, it rotates about ez

at the angular velocity f .

12.3 Demonstrate that the set of equations derived in the previous exercise can bewritten in the coordinate-free (in the x-y plane) form

∂ζ

∂t= −∇ · (d v),

∂v∂t+ f ez × v = −g∇(ζ − ζ′),

where v = u ex + v ey. Let ω = ez · ∇ × v and Π = ∇ · v. Assuming that d and


f are constants, demonstrate that the previous equations are equivalent to

∂ 2ω

∂t 2 + f 2 ω − g d∇ 2 ω = − f g∇ 2ζ′,

∂ 2Π

∂t 2 + f 2Π − g d∇ 2Π = g∇ 2(∂ζ′

∂t

),

where ∇ 2 denotes a two-dimensional Laplacian (in the x-y plane).

12.4 Consider free (i.e., ζ′ = 0) plane-wave solutions to the f -plane equations,derived in Exercise 12.2, of the form

ζ(r, t) = ζ0 e i (σ t−k·r),

u(r, t) = u0 e i (σ t−k·r),

v(r, t) = v0 e i (σ t−k·r).

Here, r = (x, y), and ζ0, u0, v0 are constants. Assuming that d is constant,show that

u0 = g

(σ kx − i f kyσ 2 − f 2

)ζ0,

v0 = g

(σ ky + i f kx

σ 2 − f 2

)ζ0, ,

andσ 2 = f 2 + c 2 k 2,

where c =√

d g. This type of wave is known as a Poincare wave.

12.5 Suppose that the region y ≤ 0 corresponds to an ocean of constant depth d,whereas the region y > 0 corresponds to land. Consider free solutions to thef -plane equations in the region y < 0. We can trivially satisfy the constraintv(x, 0, t) = 0 by searching for solutions which are such that v = 0 for all y ≤ 0.Show that the most general such solution takes the form

ζ(x, y, t) = Z1(x + c t, y) + Z2(x − c t, y),

where c =√

d g, and

Z1(x + c t, y) = Z1(x + c t, 0) e+s y/l,

Z2(x − c t, y) = Z2(x − c t, 0) e−s y/l.

Here, Z1(x, 0) and Z2(x, 0) are arbitrary functions, l =√

d g/| f |, and s =sgn( f ). These solutions are known as Kelvin waves. Deduce that Kelvinwaves propagate along coastlines, at the speed c, in such a manner as to keepthe coastline to the right of the direction of propagation in the Earth’s northernhemisphere, and to the left in the southern hemisphere.


12.6 We can take into account the latitude dependence of the parameter f by writ-ing f = f0−β x, where f0 = 2Ω cos θ and β = 2Ω sin θ/a. Let us assume thatthe ocean is of constant depth, d. Furthermore, let us search for an almost in-compressible, free solution of the Laplace tidal equations which is such thatu ∂ψ/∂y and v −∂ψ/∂x. By eliminating ζ from the final two f -planeequations, show that

∇ 2(∂ψ

∂t

)+ β

∂ψ

∂y 0.

By searching for a wavelike solution of the previous equation of the formψ = ψ0 e i (σ t−k·r), deduce that

σ k 2 + β ky = 0.

This is the dispersion relation of a so-called Rossby wave. Demonstrate thatRossby waves always travel with a westward component of phase velocity.Finally, show that it is reasonable to neglect compression provided that σ | f0|.

13Equilibrium of Compressible Fluids

13.1 Introduction

In this chapter, we investigate the equilibria of compressible fluids, such as gases.As is the case for an incompressible fluid (see Chapter 2), a compressible fluid inmechanical equilibrium must satisfy the force balance equation

0 = ∇p + ρ∇Ψ, (13.1)

where p is the static fluid pressure, ρ the mass density, and Ψ the gravitational po-tential energy per unit mass. In an ideal gas, the relationship between p and ρ isdetermined by the energy conservation equation, (1.89), which can be written

ργ

γ − 1DDt

(pργ

)=κMR∇ 2(

pρ

). (13.2)

Here, γ is the ratio of specific heats, κ the thermal conductivity, M the molar mass,and R the molar ideal gas constant. Note that the viscous heat generation term hasbeen omitted from the previous equation (because it is zero in a stationary gas). Thelimits in which the left- and right-hand sides of Equation (13.2) are dominant aretermed the adiabatic and isothermal limits, respectively. In the isothermal limit, inwhich thermal transport is comparatively large, so that Equation (13.2) can only besatisfied when ∇ 2(p/ρ) → 0, the temperature (recall that T ∝ p/ρ in an ideal gas)distribution in the gas becomes uniform, and the pressure and density are conse-quently related according to the isothermal gas law,

pρ= constant. (13.3)

On the other hand, in the adiabatic limit, in which thermal transport is negligible, sothat Equation (13.2) can only be satisfied when D/Dt(p/ργ) → 0, the pressure anddensity are related according to the adiabatic gas law,

pργ= constant. (13.4)

393


13.2 Isothermal AtmosphereThe vertical thickness of the atmosphere is only a few tens of kilometers, and is,therefore, much less than the radius of the Earth, which is about 6000 km. Conse-quently, it is a good approximation to treat the atmosphere as a relatively thin layer,covering the surface of the Earth, in which the pressure and density are only func-tions of altitude above ground level, z, and the gravitational potential energy per unitmass takes the form Ψ = g z, where g is the acceleration due to gravity at z = 0. Itfollows from Equation (13.1) that

dpdz= −ρ g. (13.5)

Now, in an isothermal atmosphere, in which the temperature, T , is assumed not tovary with height, the ideal gas equation of state (1.84) yields [cf., Equation (13.3)]

pρ=

R TM. (13.6)

The previous two equations can be combined to give

dpdz= −gM

R Tp. (13.7)

Hence, we obtainp(z) = p0 exp(−z/H), (13.8)

where p0 105 N m−2 is atmospheric pressure at ground level, and

H =R TgM

(13.9)

is known as the isothermal scale height of the atmosphere. Using the values T =273 K (0 C), M = 29 × 10−3 kg, and g = 9.8 m s−2, which are typical of the Earth’satmosphere (at ground level), as well as R = 8.315 J mol−1 K−1, we find that H =7.99 km. Equations (13.6) and (13.8) yield

ρ(z) = ρ0 exp(−z/H), (13.10)

where ρ0 = p0/(gH) is the atmospheric mass density at z = 0. According to Equa-tions (13.8) and (13.10), in an isothermal atmosphere, the pressure and density bothdecrease exponentially with increasing altitude, falling to 37% of their values atground level when z = H, and to only 5% of these values when z = 3 H.

13.3 Adiabatic AtmosphereIn fact, the temperature of the Earth’s atmosphere is not uniform, but instead de-creases steadily with increasing altitude. This effect is largely due to the action of

Equilibrium of Compressible Fluids 395

convection currents. When a packet of air ascends, under the influence of such cur-rents, the diminished pressure at higher altitudes causes it to expand. Because thisexpansion generally takes place far more rapidly than heat can diffuse into the packet,the work done against the pressure of the surrounding gas, as the packet expands,leads to a reduction in its internal energy, and, hence, in its temperature. Assumingthat the atmosphere is in a continually mixed state, while remaining in approximatevertical force balance (such a state is known as a convective equilibrium), and thatthe effect of heat conduction is negligible (because the mixing takes place too rapidlyfor thermal diffusion to affect the temperature), we would expect the adiabatic gaslaw, (13.4), to offer a much more accurate description of the relationship betweenatmospheric pressure and density than the isothermal gas law, (13.3).

Let p = p/p0, ρ = ρ/ρ0, and T = T/T0, where p0, ρ0, and T0 are the pressure,mass density, and temperature of the atmosphere, respectively, at ground level. Theadiabatic gas law, (13.4), can be combined with the ideal gas equation of state, (13.6),to give

p = ρ γ = T γ/(γ−1). (13.11)

The isothermal scale height of the atmosphere is conveniently redefined as [cf., Equa-tion (13.9)]

H =R T0

gM=

p0

g ρ0. (13.12)

Equations (13.5), (13.11), and (13.12) yield

dpdz= −p 1/γ, (13.13)

where z = z/H, or, from Equation (13.11),

dTdz= −γ − 1

γ. (13.14)

The previous equation can be integrated to give

T (z) = T0

(1 − γ − 1

γ

zH

). (13.15)

It follows that the temperature in an adiabatic atmosphere decreases linearly withincreasing altitude at the rate of [(γ − 1)/γ] (T0/H) degrees per meter. This rate isknown as the adiabatic lapse rate of the atmosphere. Using the values γ = 1.4,T0 = 273 K, and H = 7.99 km, which are typical of the Earth’s atmosphere, weestimate the lapse rate to be 9.8 K km−1. In reality, the lapse rate only takes thisvalue in dry air. In moist air, the lapse rate is considerably reduced because of thelatent heat released when water vapor condenses.

Equations (13.11) and (13.15) yield

p(z) = p0

(1 − γ − 1

γ

zH

)γ/(γ−1)

, (13.16)

ρ(z) = ρ0

(1 − γ − 1

γ

zH

)1/(γ−1)

. (13.17)


Because γ/(γ − 1) 3.5 and 1/(γ − 1) 2.5, it follows that pressure decreasesmore rapidly than density in an adiabatic atmosphere. Moreover, the previous threeequations imply that an adiabatic atmosphere has a sharp upper boundary at z =[γ/(γ − 1)] H 28 km. At this altitude, the temperature, pressure, and density allfall to zero. Of course, above this altitude, the temperature, pressure, and densityremain zero (because they cannot take negative or imaginary values). In contrast,an isothermal atmosphere has a diffuse upper boundary in which the pressure anddensity never fall to zero, even at extreme altitudes. It should be noted that, in reality,the Earth’s atmosphere does not have a sharp upper boundary, because the adiabaticgas law does not hold at very high altitudes.

13.4 Atmospheric Stability

Suppose that the atmosphere is static (i.e., non-convecting). Moreover, let p(z) andρ(z) be the pressure and density, respectively, as functions of altitude. Consider apacket of air that is in equilibrium with the surrounding air at some initial altitudez1, but subsequently moves to a higher altitude z2. Thus, the packet’s initial pressureand density are p1 = p(z1) and ρ1 = ρ(z1), respectively. Now, at the higher altitude,the packet must adjust its volume in such a manner that its pressure matches that ofthe surrounding air, otherwise there would be a force imbalance across the packetboundary. It follows that the packet pressure at altitude z2 is p2 = p(z2). Assumingthat the packet moves upward on a much faster time scale than that required forheat to diffuse across it (but still a sufficiently slow time scale that it remains inapproximate pressure balance with the surrounding air), we would expect its internalpressure and density to be related according to the adiabatic gas law, (13.4). Thus,the packet’s density at altitude z2 is ρ2 = (p2/p1)1/γ ρ1. Now, if ρ2 > ρ(z2) thenthe packet is denser than the surrounding air. It follows that the packet’s weightexceeds the buoyancy due to the atmosphere, causing the packet to sink back to itsoriginal altitude. On the other hand, if ρ2 < ρ(z2) then the packet is less densethan the surrounding air. It follows that the buoyancy force exceeds the packet’sweight, causing it to rise to an even higher altitude. In other words, the atmosphereis unstable to vertical convection when [p(z2)/p(z1)]1/γ ρ(z1) < ρ(z2) for any z2 > z1:that is, when

pργ

∣∣∣∣∣z2

<pργ

∣∣∣∣∣z1

, (13.18)

for any z2 > z1. It follows that the atmosphere is only stable to vertical convectionwhen p/ργ is a monotonically increasing function of altitude. As is easily demon-strated, this stability criterion can also be written

d ln pd ln ρ

< γ, (13.19)


or, making use of the ideal gas equation of state,

d ln Td ln ρ

< γ − 1. (13.20)

Convection is triggered in regions of the atmosphere where the previous stabilitycriterion is violated. However, such convection acts to relax these regions back toa marginally-stable state in which p/ργ is uniform: in other words, an adiabaticequilibrium.

13.5 Eddington Solar ModelLet us investigate the internal structure of the Sun, which is basically a self-gravitatingsphere of incandescent ionized gas (consisting mostly of hydrogen). Adopting aspherical coordinate system (see Section C.4), r, θ, φ, whose origin coincides withthe Sun’s geometric center, and making the simplifying (and highly accurate) as-sumption that the mass distribution within the Sun is spherically symmetric, we findthat

dmdr= 4π r 2 ρ, (13.21)

where m(r) is the total mass contained within a sphere of radius r, centered on theorigin, and ρ(r) the mass density at radius r. Now, as is well known, the gravitationalacceleration at some radius r in a spherically symmetric mass distribution is the sameas would be obtained were all the mass located within this radius concentrated at thecenter, and the remainder of the mass neglected (Fitzpatrick 2012). In other words,

dΨdr=

G mr 2 , (13.22)

whereΨ (r) is the gravitational potential energy per unit mass, and −dΨ/dr the radialgravitational acceleration. The force balance criterion (13.1) yields

dpdr+ ρ

dΨdr= 0, (13.23)

where p(r) is the pressure. The previous three equations can be combined to give

1r 2

ddr

(r 2

ρ

dpdr

)= −4πG ρ. (13.24)

In order to make any further progress, we need to determine the relationshipbetween the Sun’s internal pressure and density. Unfortunately, this relationship isultimately controlled by energy transport, which is a very complicated process in astar. In fact, a star’s energy is ultimately derived from nuclear reactions occurringdeep within its core, the details of which are extremely complicated. This energy is


then transported from the core to the outer boundary via a combination of convec-tion and radiation. (Conduction plays a much less important role in this process.)Unfortunately, an exact calculation of radiative transport requires an understandingof the opacity of stellar material, which is an exceptionally difficult subject. Finally,once the energy reaches the boundary of the star, it is radiated away. The followingingenious model, due to Eddington (Eddington 1926), is appropriate to a star whoseinternal energy transport is dominated by radiation. This turns out to be a fairly goodapproximation for the Sun. The main advantage of Eddington’s model is that it doesnot require us to know anything about stellar nuclear reactions or opacity.

Now, the temperature inside the Sun is sufficiently high that radiation pressurecannot be completely neglected with respect to conventional gas pressure. In otherwords, we must write the solar equation of state in the form

p = pg + pr, (13.25)

wherepg =

ρ k Tµmp

(13.26)

is the gas pressure (modeling the plasma within the Sun as an ideal gas of free elec-trons and ions), and (Chandrasekhar 1967)

pr =13αT 4 (13.27)

the radiation pressure (assuming that the radiation within the Sun is everywhere inlocal thermodynamic equilibrium with the plasma). Here, T (r) is the Sun’s internaltemperature, k the Boltzmann constant, mp the mass of a proton, and µ the relativemolecular mass (i.e., the ratio of the mean mass of the free particles making up thesolar plasma to that of a proton). Note that the electron mass has been neglectedwith respect to that of a proton. Furthermore, α = 4σ/c, where σ is the Stefan-Boltzmann constant, and c the velocity of light in a vacuum. Incidentally, in writingEquation (13.26), we have expressed M/R in the equivalent form µmp/k.

Let

pg = (1 − β) p, (13.28)

pr = β p, (13.29)

where the parameter β is assumed to be uniform. In other words, the ratio of the ra-diation pressure to the gas pressure is assumed to be the same everywhere inside theSun. This fairly drastic assumption turns out—perhaps, somewhat fortuitously (Mes-tel 1999)—to lead to approximately the correct internal pressure-density relation forthe Sun. In fact, Equations (13.26)–(13.29) can be combined to give

p = K ρ 4/3, (13.30)

where

K =

( kmp

)4 3α

β

µ 4 (1 − β)4

1/3 . (13.31)


0

1

2

θ, y

0 1 2 3 4 5 6 7ξ

Figure 13.1The functions θ(ξ) (solid) and y(ξ) (dashed).

It can be seen, by comparison with Equation (13.4), that the previous pressure-density relation takes the form of an adiabatic gas law with an effective ratio ofspecific heats γ = 4/3. Note, however, that the actual ratio of specific heats for afully ionized hydrogen plasma, in the absence of radiation, is γ = 5/3. Hence, the4/3 exponent, appearing in Equation (13.30), is entirely due to the non-negligibleradiation pressure within the Sun.

Let Tc = T (0), ρc = ρ(0), and pc = p(0), be the Sun’s central temperature,density, and pressure, respectively. It follows from Equation (13.30) that

pc = K ρ4/3c , (13.32)

and from Equations (13.26) and (13.28) that

Tc =pc

ρc

mp

k(1 − β) µ. (13.33)

Suppose that

TTc= θ, (13.34)

where θ(r) is a dimensionless function. According to Equations (13.26), (13.28), and


6

7

log 1

0(T

[K

])

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1m/M

Figure 13.2Solar temperature versus mass fraction obtained from the Eddington Solar Model(solid) and the Standard Solar Model (dashed).

(13.30),

ρ

ρc= θ 3, (13.35)

ppc= θ 4. (13.36)

Moreover, it is clear, from the previous expressions, that θ = 1 at the center of theSun, r = 0, and θ = 0 at the edge, r = R (say), where the temperature, density, andpressure are all assumed to fall to zero. Suppose, finally, that

r = a ξ, (13.37)

where ξ is a dimensionless radial coordinate, and

a =(

K

πG ρ 2/3c

)1/2. (13.38)

Thus, the center of the Sun corresponds to ξ = 0, and the edge to ξ = ξ1 (say), whereθ(ξ1) = 0, and

R = ξ1 a. (13.39)

Equations (13.35)–(13.38)can be used to transform the equilibrium relation (13.24)


1

2

3

4

5

log 1

0(ρ

[kg

m−3

])

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1m/M

Figure 13.3Solar mass density versus mass fraction obtained from the Eddington Solar Model(solid) and the Standard Solar Model (dashed).

into the non-dimensional form1ξ 2

ddξ

(ξ 2 dθ

dξ

)= −θ 3. (13.40)

Moreover, Equation (13.21) can be integrated, with the aid of Equations (13.35),(13.37), and (13.40), and the physical boundary condition m(0) = 0, to give

m = 4π ρc a 3 y, (13.41)

wherey(ξ) = −ξ 2 dθ

dξ. (13.42)

Equation (13.40) is known as the Lane-Emden equation (of degree 3), and can, unfor-tunately, only be solved numerically (Chandrasekhar 1967). The appropriate solutiontakes the form θ = 1 − ξ 2/6 + O(ξ 4) when ξ 1, and must be integrated to ξ = ξ1,where θ(ξ1) = 0. Figure 13.1 shows θ(ξ), and the related function y(ξ), obtained vianumerical methods. Note that ξ1 = 6.897, and y1 = y(ξ1) = 2.018.

According to Equation (13.41), the solar mass, M∗ = m(R), can be written

M∗ = 4π ρc a 3 y1, (13.43)

which reduces, with the aid of Equations (13.31) and (13.38), to

β

(1 − β)4 = ζ2, (13.44)


where

ζ = µ 2 M∗M0, (13.45)

and

M0 =

1(πG)3

(k

mp

)4 3α

1/2 4π y1 = 3.586 × 1031 kg. (13.46)

Moreover, it is easily demonstrated that

K =πG M 2/3

∗(4π y1)2/3 . (13.47)

According to Equations (13.44) and (13.45), the ratio, β/(1 − β), of the radiationpressure to the gas pressure in a radiative star is a strongly increasing function of thestellar mass, M∗, and mean molecular weight, µ. In the case of the Sun, for whichζ 1, Equation (13.44) can be inverted to give the approximate solution

β = ζ 2 −14ζ 4 + O(ζ 6). (13.48)

Using the observed solar mass M∗ = 1.989 × 1030 kg, and the value µ = 0.68 (whichrepresents the best fit to the Standard Solar Model mentioned in the following), wefind that β = 6.58 × 10−4. In other words, the radiation pressure inside the Sun isonly a very small fraction of the gas pressure. This immediately implies that radiativeenergy transport is much less efficient than convective energy transport. Indeed, inregions of the Sun in which convection occurs, we would expect the convective trans-port to overwhelm the radiative transport, and so to drive the local pressure-densityrelation toward an adiabatic law with an exponent 5/3. Fortunately, convection onlytakes place in the Sun’s outer regions, which contain a minuscule fraction of its mass.

Equations (13.32), (13.33), (13.39), (13.43), and (13.47) yield

Tc =ξ1

4 y1µ (1 − β)

G M∗ mp

R k= 1.34 × 107 K, (13.49)

ρc =ξ 3

1

4π y1

M∗R 3 = 7.63 × 104 kg m−3, (13.50)

pc =ξ 4

1

16π y 21

G M 2∗

R 4 = 1.24 × 1016 N m−2, (13.51)

where the solar radius R has been given the observed value 6.960 × 108 m. The ac-tual values of the Sun’s central temperature, density, and pressure, as determined bythe so-called Standard Solar Model (SSM),1 which incorporates detailed treatmentsof nuclear reactions and opacity, are Tc = 1.58 × 107 K, ρc = 15.6 × 104 kg m−3,and pc = 2.38 × 1016 N m−2, respectively. It can be seen that the values of Tc, ρc,

1http://en.wikipedia.org/wiki/Standard\_Solar\_Model


10

11

12

13

14

15

16

17

log 1

0(p

[Nm

−2])

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1m/M

Figure 13.4Solar pressure versus mass fraction obtained from the Eddington Solar Model (solid)and the Standard Solar Model (dashed).

and pc obtained from the Eddington model lie within a factor of two of those ob-tained from the much more accurate SSM. Figures 13.2, 13.3, and 13.4 show thetemperature, density, and pressure profiles, respectively, obtained from the SSM 2

and the Eddington model. The profiles are plotted as functions of the mass fraction,m(r)/M∗ = y(ξ)/y1, where ξ = r/a. It can be seen that there is fairly good agreementbetween the profiles calculated by the two models. Finally, Figure 13.5 comparesthe ratio, β/(1 − β), of the radiation pressure to the gas pressure obtained from theSSM and the Eddington model. Recall, that it is a fundamental assumption of theEddington model that this pressure ratio is uniform throughout the Sun. In fact, itcan be seen that the pressure ratio calculated by the SSM is not spatially uniform.On the other hand, the spatial variation of the ratio is fairly weak, except close to theedge of the Sun, where convection sets in, and the Eddington model, thus, becomesinvalid. We conclude that, despite its simplicity, the Eddington solar model does aremarkably good job of accounting for the Sun’s internal structure.

2The SSM data is obtained from http://www.ap.stmarys.ca/˜guenther/evolution/ssm1998.html


−4

−3.5

−3

−2.5

log 1

0[β/(

1−

β)]

0 0.2 0.4 0.6 0.8 1m/M

Figure 13.5Ratio of radiation pressure to gas pressure calculated from the Eddington SolarModel (solid) and the Standard Solar Model (dashed).

13.6 Exercises13.1 Prove that the fraction of the whole mass of an isothermal atmosphere that

lies between the ground and a horizontal plane of height z is

1 − e−z/H .

Evaluate this fraction for z = H, 2 H, 3 H, respectively.

13.2 If the absolute temperature in the atmosphere diminishes upwards accordingto the law

TT0= 1 − z

c,

where c is a constant, show that the pressure varies as

pp0=

(1 − z

c

)c/H.

13.3 If the absolute temperature in the atmosphere diminishes upward accordingto the law

TT0=

11 + β z

,


where β is a constant, show that the pressure varies as

pp0= exp

(− z

H− 1

2β z 2

H

).

13.4 Show that if the absolute temperature, T , in the atmosphere is any given func-tion of the altitude, z, then the vertical distribution of pressure in the atmo-sphere is given by

lnpp0= −T0

H

∫ z

0

dzT.

13.5 Show that if the Earth were surrounded by an atmosphere of uniform temper-ature then the pressure a distance r from the Earth’s center would be

pp0= exp

[a 2

H

(1r−

1a

)],

where a is the Earth’s radius.

13.6 Show that if the whole of space were occupied by air at the uniform tem-perature T then the densities at the surfaces of the various planets would beproportional to the corresponding values of

exp(gM aR T

),

where a is the radius of the planet, and g its surface gravitational acceleration.

13.7 Prove that in an atmosphere arranged in horizontal strata the work (per unitmass) required to interchange two thin strata of equal mass without distur-bance of the remaining strata is

1γ − 1

(ργ−12 − ργ−1

1

) ( p1

ργ1

− p2

ργ2

),

where the suffixes refer to the initial states of the two strata. Hence, show thatfor stability the ratio p/ργ must increase upwards.

13.8 A spherically symmetric star is such that m(r) is the mass contained within ra-dius r. Show that the star’s total gravitational potential energy can be writtenin the following three alternative forms:

U = −∫ M∗

0

G mr

dm =12

∫ M∗

0Ψ dm = −3

∫ R

0p dV

Here, M∗ is the total mass, R the radius, Ψ (r) the gravitational potential perunit mass (defined such that Ψ → 0 as r → ∞), p(r) the pressure, anddV = 4π r 2 dr.


13.9 Suppose that the pressure and density inside a spherically symmetric star arerelated according to the polytropic gas law,

p = K ρ (1+n)/n,

where n is termed the polytropic index. Let ρ = ρc θn, where ρc is the central

mass density. Demonstrate that θ satisfies the Lane-Emden equation

1ξ 2

ddξ

(ξ 2 dθ

dξ

)= −θ n,

where r = a ξ, and

ξ =

[(n + 1) K

4πGρ (n−1)/n

c

]1/2.

Show that the physical solution to the Lane-Emden equation, which is suchthat θ(0) = 1 and θ(ξ1) = 0, for some ξ1 > 0, is

θ = 1 −ξ 2

6

for n = 0,

θ =sin ξξ

for n = 1, and

θ =1

(1 + ξ 2/3)1/2

for n = 5. Determine the ratio of the central density to the mean densityin all three cases. Finally, demonstrate that, in the general case, the totalgravitational potential energy can be written

U = − 35 − n

G M 2∗

R,

where M∗ is the total mass, and R = a ξ1 the radius.

13.10 A spherically symmetric star of radius R has a mass density of the form

ρ(r) = ρc (1 − r/R).

Show that the central mass density is four times the mean density. Demon-strate that the central pressure is

pc =5

4πG M 2

∗R

,

where M∗ is the mass of the star. Finally, show that the total gravitationalpotential energy of the star can be written

U = −2635

G M 2∗

R.

14One-Dimensional Compressible Inviscid Flow

14.1 IntroductionThis chapter investigates one-dimensional, compressible, inviscid flow. Flow is saidto be one-dimensional when the fluid properties only depend on a single Cartesiancoordinate. Flow is said to be compressible when there is a significant variation inthe mass density along a given streamline. Generally speaking, compressible flow ismuch more common in gases than in liquids. As described in Section 1.17, subsonicgas dynamics (i.e., dynamics in which the typical flow speed is much less than thepropagation speed of sound waves in the gas) is essentially incompressible, and isgoverned by the same equations that govern the incompressible flow of liquids. Onthe other hand, sonic gas dynamics (i.e., dynamics in which the typical flow speed iscomparable with the sound speed) exhibits significant differences to subsonic dynam-ics. Hence, this chapter will concentrate on sonic gas dynamics. More informationon this subject can be found in Liepmann & Roshko 1957, Hughes & Brighton 1999,and Milne-Thomson 2011.

14.2 Thermodynamic ConsiderationsConsider a quantity of gas contained in some form of enclosure. If the system isleft alone for a sufficiently long time then it will attain an equilibrium state in whichall macroscopically measurable quantities become independent of time. Examplesof such quantities are the mass, m, the pressure, p, the volume, V , and the absolutetemperature, T . It is helpful to distinguish between so-called extensive and intensiveequilibrium quantities. A quantity is said to be extensive if it is proportional to themass of the gas present in the enclosure. On the other hand, a quantity is said tobe intensive if it is independent of the mass. Thus, the mass, m, and volume, V ,are extensive quantities, whereas the pressure, p, and the absolute temperature, T ,are intensive. For every extensive quantity (except the mass), we can introduce acorresponding intensive quantity by dividing by the mass. For instance, the intensivequantity that corresponds to the volume, V , is the specific volume, v = V/m, which issimply the volume per unit mass (i.e., the volume of a unit mass of gas). Of course,v = 1/ρ, where ρ is the mass density.

407


A gas is said to be ideal if it obeys the law (Reif 1965),

p v = R T, (14.1)

where the constant R is termed the specific gas constant. This quantity is related tothe molar gas constant, R, introduced in Section 1.15, according to R = R/M, whereM is the molar mass—that is, the mass of NA = 6.022 × 1023 gas molecules. Here,NA is Avogadro’s number (Reif 1965).

For all gases, whether in mean motion, or not, there exists an internal energyfunction, E, that is independent of the mean motion, and dependent only on thevariables of state, p, ρ, and T . Moreover, this function is such that when a smallquantity of heat, q, is added to the system (Reif 1965),

q = dE + p dV. (14.2)

Thus, the quantity dE is the excess of the energy supplied over the mechanical workdone by the pressure. Of course, internal energy is an extensive quantity. The cor-responding intensive quantity is the specific internal energy, E = E/m. In an idealgas, the specific internal energy,E, is a function of the absolute temperature, T , alone(Reif 1965). It follows that

dE = k(T ) dT, (14.3)

and Equation (14.2) becomes

q = m (k dT + p dv). (14.4)

The heat capacity at constant volume of the gas is defined

CV =q

dT

∣∣∣∣∣v. (14.5)

Thus, Equation (14.4) implies that

CV = m k. (14.6)

It is clear that CV is an extensive quantity. The corresponding intensive quantity isthe specific heat capacity at constant volume,

CV =CV

m= k. (14.7)

Finally, the molar specific heat capacity at constant volume, which is the heat capac-ity of Avogadro’s number of gas molecules, takes the form

cV = M CV . (14.8)

It follows, from the preceding equations, that

dE = CV dT =cV

MdT. (14.9)

One-Dimensional Compressible Inviscid Flow 409

In an ideal gas, cV is a constant, independent of the temperature (Reif 1965), in whichcase the previous equation can be integrated to give [cf., Equation (1.83)]

E = CV T =cV

MT. (14.10)

Equation (14.1) yieldsv dp + p dv = R dT. (14.11)

Combining the previous equation with Equations (14.4) and (14.7), we obtain

q = m[(CV + R) T − v dp

]. (14.12)

Now, the heat capacity at constant pressure of the gas is defined

Cp =q

dT

∣∣∣∣∣p

(14.13)

Thus, Equation (14.12) implies that

Cp = m (CV + R). (14.14)

It is clear that Cp is an extensive quantity. The corresponding intensive quantity isthe specific heat capacity at constant pressure,

Cp =C p

m= CV + R. (14.15)

Finally, the molar specific heat capacity at constant pressure, which is the heat ca-pacity of Avogadro’s number of gas molecules, takes the form [cf., Equation (1.86)]

cp = M Cp = cV + R. (14.16)

The molar specific heat capacity at constant pressure of an ideal gas is a constant,independent of the temperature.

The enthalpy of a gas is defined (Reif 1965)

H = E + p V. (14.17)

This quantity is obviously extensive. The corresponding intensive quantity is thespecific enthalpy,H = H/m. It follows that, for an ideal gas,

H = E + p v = E + R T = Cp T =cp

MT, (14.18)

where use has been made of Equations (14.1), (14.10), (14.15), and (14.16).In the preceding analysis, we denoted a small quantity of heat by q, rather than

dQ, because there is, in general, no function Q of which q is an exact differential.However, we can write (Reif 1965)

q = T dS , (14.19)


where dS is the differential of an extensive function, S , called the entropy. To justifythe previous expression, note from Equations (14.1), (14.4) and (14.7) that

dS = m(CV

dTT+

pT

dv)= m

(CV

dTT+ R

dvv

). (14.20)

However, Equation (14.1) yields

dpp+

dvv=

dTT. (14.21)

Hence, we obtain

dS = m(Cp

dvv+ CV

dpp

), (14.22)

where use has been made of Equation (14.15). In an ideal gas, CV and Cp are con-stants, independent of T , so the previous expression can be integrated to give

S = mCV ln(p v γ) = mCV ln(

pργ

), (14.23)

where

γ =Cp

CV(14.24)

is the constant ratio of specific heats. Obviously, the specific entropy is defined

S = Sm= CV ln

(pργ

). (14.25)

Furthermore, it follows from Equations (14.2), (14.17), and (14.19) that

dE = −p dV + T dS , (14.26)

dH = V dp + T dS , (14.27)

or, equivalently,

dE = −p dv + T dS, (14.28)

dH = v dp + T dS. (14.29)


14.3 Isentropic FlowIn the limit of vanishing viscosity and heat conduction, the equations of compressibleideal gas flow, introduced in Section 1.15, can be written

DρDt= −ρ∇ · v, (14.30)

DvDt= −∇p

ρ− ∇Ψ, (14.31)

DDt

(pργ

)= 0, (14.32)

where v is the flow velocity, and Ψ the potential energy per unit mass. It is clear fromEquations (14.25) and (14.32) that the specific entropy is constant along a streamline,but not necessarily the same constant on different streamlines. Such flow is said tobe isentropic. From Equation (14.29), isentropic flow is characterized by

dH = dpρ

(14.33)

along a streamline. More generally, isentropic flow is characterized by p/ργ, ρ/T γ−1,and T γ/p γ−1, constant along streamlines.

14.4 Sound WavesSuppose that v = u(x, t) ex, ρ = ρ(x, t), p = p(x, t), and Ψ = 0 in Equations (14.30)and (14.31). We obtain

∂ρ

∂t+ u

∂ρ

∂x+ ρ

∂u∂x= 0, (14.34)

∂u∂t+ u

∂u∂x+

1ρ

∂p∂x= 0. (14.35)

Equation (14.32) implies that the flow is isentropic. In other words, p/ργ = constant.Thus, it follows that

∂p∂x= c 2 ∂ρ

∂x, (14.36)

where

c 2 ≡dpdρ=γ pρ. (14.37)


Hence, Equations (14.34) and (14.35) become

∂ρ

∂t+ u

∂ρ

∂x+ ρ

∂u∂x= 0, (14.38)

∂u∂t+ u

∂u∂x+

c 2

ρ

∂ρ

∂x= 0. (14.39)

Unfortunately, these equations are difficult to solve exactly, because of the presenceof nonlinear terms such as u ∂u/∂x. (See Example 14.9.)

Let us write u(x, t) = u1(x, t), ρ(x, t) = ρ0 + ρ1(x, t), and p(x, t) = p0 + p1(x, t),where quantities with the subscript 1 are assumed to be much smaller that corre-sponding quantities with the subscript 0. Thus, we are now considering the evolutionof small-amplitude pressure and density perturbations in a stationary gas of uniformdensity and pressure ρ0 and p0, respectively. To first order in small quantities, theprevious two equations yield

1ρ0

∂ρ1

∂t+∂u1

∂x= 0, (14.40)

∂u1

∂t+

c 20

ρ0

∂ρ1

∂x= 0, (14.41)

where

c0 =

√γ p0

ρ0. (14.42)

The general solution to Equations (14.40) and (14.41) is well known (Fitzpatrick2013):

ρ1(x, t)ρ0

= F(x ∓ c0 t), (14.43)

u1(x, t) = ±c0 F(x ∓ c0 t), (14.44)

where F(x) is an arbitrary function. According to this solution, a small-amplitudedensity perturbation of arbitrary shape propagates through the gas, either in the pos-itive or negative x-directions (corresponding to the upper and lower signs, respec-tively), without changing shape, at the constant speed c0. This type of perturbationis known as a sound wave, and c0 is consequently referred to as the sound speed.

It is clear, from the previous analysis, that the general expression for the localsound speed in an isentropic ideal gas of pressure p, density ρ, and temperature T , is

c =√γ pρ=√γRT , (14.45)

where use has been made of Equation (14.1). It follows that the speed of sound issolely determined by the temperature of an ideal gas.

Let us now consider the effect of finite wave amplitude on the propagation of asound wave through an isentropic ideal gas. The previous analysis suggest that, in a


frame of reference moving with the local flow velocity, u, a sound wave propagatesat the speed c =

√γ p/ρ. Thus, the net wave propagation speed is

c′ = u + c. (14.46)

Using the isentropic law, p/ργ = constant, to eliminate p from c =√γ p/ρ, we

deduce that

c = c0

(ρ

ρ0

) (γ−1)/2

. (14.47)

Now, Equations (14.43) and (14.44) suggest that, in the presence of a sound wavepropagating in the +x-direction, the general differential relationship between the lo-cal flow velocity and the density is

du = cdρρ. (14.48)

Making use of Equation (14.47), we can integrate the previous expression to give

u =∫ ρ

ρ0

cdρρ=

2 c0

γ − 1

( ρρ0

) (γ−1)/2

− 1

= 2γ − 1

(c − c0), (14.49)

or

c = c0 +

(γ − 1

2

)u. (14.50)

It follows that

c′ = c + u = c0 +

(γ + 1

2

)u, (14.51)

or

c′ = c0

1 +(γ + 1γ − 1

) ( ρρ0

) (γ−1)/2

− 1

. (14.52)

Thus, we conclude that the net wave propagation speed is higher than c0 in regionsof compression (i.e., ρ > ρ0), and lower in regions of rarefaction (i.e., ρ < ρ0). Thisimplies that a finite-amplitude sound wave distorts as it propagates in such a man-ner that the regions of compression tend to catch up with the regions of rarefaction,leading to a monotonic increase with time of the velocity and temperature gradientsin the former regions, and a monotonic decrease in the latter regions. Eventually, thevelocity and temperature gradients in the regions of compression become so largethat it is no longer valid to neglect the effects of viscosity and heat conduction, lead-ing to a local breakdown of the isentropic gas law. Viscosity and heat conductionprevent any further steepening of the velocity and temperature gradients in the re-gions of compression, and the wave subsequently propagates without additional dis-tortion. However, it has now effectively been transformed into a shock wave. (SeeSection 14.8.)

It is clear, from the previous discussion, that, at finite amplitude, there is an im-portant difference in the behavior of compression and rarefaction waves propagating


through a gas. A compression wave tends to steepen (i.e., the temperature and ve-locity gradients tend to increase), and eventually becomes a shock wave, in whichcase it is no longer isentropic. A rarefaction wave, on the other hand, always re-mains isentropic, because it tends to flatten, thereby, further reducing the velocityand temperature gradients. Hence, there are no rarefaction, or expansion, shocks.

14.5 Bernoulli’s Theorem

According to Bernoulli’s theorem, which was introduced in Section 4.3, the quantityp/ρ + T is constant along a streamline in steady inviscid flow, where T is the totalenergy per unit mass. For the case of a compressible fluid, T = E + (1/2) v 2 + Ψ .Hence, we deduce that

E +pρ+

12v 2 + Ψ (14.53)

is constant along a streamline. In particular, for an ideal gas, we find that

H + 12v 2 + Ψ = Cp T +

12v 2 + Ψ (14.54)

is constant along a streamline, where use has been made of Equation (14.18).Consider a solid object moving through an ideal gas. Generally speaking, there is

at least one stagnation point in front of the object, where the gas comes to rest relativeto it. At this point, the gas is adiabatically compressed, and there is an associated risein temperature. In the frame of reference in which the object is at rest, and thegas a long way from it moves with constant speed and temperature, application ofBernoulli’s theorem yields

Cp T +12v 2 = Cp T0, (14.55)

where T0 is the stagnation point temperature, and T the temperature at some generalpoint where the flow speed is v. Thus, the total temperature rise due to adiabaticcompression is

∆T = T0 − T =v 2

2Cp, (14.56)

where v and T are the asymptotic flow speed and temperature, respectively. It canbe seen that the stagnation temperature rise only depends on the velocity differencebetween the object and the gas, and is independent of the gas’s density, temperature,or pressure. Note, however, that a lower molar mass implies a higher specific heat,and, hence, a smaller stagnation temperature rise.


14.6 Mach Number

In an ideal gas, the local Mach number of the flow is defined (see Section 1.17)

Ma =v

c, (14.57)

where c =√γR T is the local sound speed. [See Equation (14.45).] Setting v 2 =

Ma 2 c 2 in Equation (14.55), we obtain

TT0=

[1 +

12

(γ − 1) Ma2]−1

, (14.58)

where use has been made of Equations (14.15) and (14.24), which imply that

CV =

(1

γ − 1

)R, (14.59)

CP =

(γ

γ − 1

)R. (14.60)

Incidentally, the relation (14.58) is valid for any streamline, because the stagna-tion temperature, T0, can be defined, by means of Equation (14.55), even when thestreamline in question does not pass through a stagnation point. We can combineEquation (14.58) with the isentropic relation, p /ργ = constant along a streamline(see Section 14.3), as well as the ideal gas law, (14.1), to give

pp0=

(TT0

)γ/(γ−1)

=

[1 +

12

(γ − 1) Ma2]−γ/(γ−1)

, (14.61)

ρ

ρ0=

(TT0

)1/(γ−1)

=

[1 +

12

(γ − 1) Ma 2]−1/(γ−1)

. (14.62)

Here, p0 and ρ0 are the pressure and density, respectively, at the stagnation point. Inprinciple, the stagnation values, T0, p0, and ρ0, can be different on different stream-lines. However, if a solid object moves through a homogeneous ideal gas that isasymptotically at rest then the stagnation parameters become true constants, inde-pendent of the streamline. Such flow is said to be homentropic.

A point where the speed of a steadily flowing ideal gas equals the local speed ofsound, v = c, is termed a sonic point. The sonic temperature, T1, pressure, p1, anddensity, ρ1, are simply related to the stagnation values. In fact, setting Ma = 1 in


Equations (14.58), (14.61), and (14.62), we obtain

T1

T0=

(2

γ + 1

), (14.63)

p1

p0=

(2

γ + 1

)γ/(γ−1)

, (14.64)

ρ1

ρ0=

(2

γ + 1

)1/(γ−1)

. (14.65)

Finally, if we combine Equations (14.58), and (14.61)–(14.65), then we get

TT1=

[1 +

(γ − 1γ + 1

) (Ma 2 − 1

)]−1

, (14.66)

pp1=

[1 +

(γ − 1γ + 1

) (Ma 2 − 1

)]−γ/(γ−1)

, (14.67)

ρ

ρ1=

[1 +

(γ − 1γ + 1

) (Ma 2 − 1

)]−1/(γ−1)

. (14.68)

14.7 Sonic Flow through a NozzleConsider an ideal gas flowing steadily through a straight nozzle with a slowly-varyingcross-sectional area, A = A(x), where x measures distance along the nozzle. The tem-perature, T , density, ρ, pressure, p, and normal velocity, u = vx, are all assumed tobe slowly-varying functions of x that are constant across any cross section. Thus,in this quasi-one-dimensional problem, we are effectively disregarding the compara-tively small components of the flow velocity orthogonal to the x-axis. Furthermore,because all streamlines have the same parameter values in any cross section, the flowis homentropic.

Of course, the net mass flow rate, Q, must be constant along the nozzle, so

Q = ρ(x) A(x) u(x). (14.69)

In particular, ρ A u = ρ1 A1 u1, where the subscript 1 refers to the sonic point (assum-ing that such a point exists). Using u/c = Ma and u1/c1 = 1, we find that

AA1=ρ1

ρ

u1

u=

1Ma

c1

cρ1

ρ=

1Ma

(T1

T

)1/2 (ρ1

ρ

), (14.70)

where use has been made of Equation (14.45). It follows from Equations (14.66) and(14.68) that

AA1=

1Ma

[1 +

(γ − 1γ + 1

) (Ma 2 − 1

)] (γ+1)/[2 (γ−1)]

. (14.71)


0

1

2

3

0 1 2 3Ma

Figure 14.1Simple model of a de Laval nozzle for γ = 7/5. The solid, dashed, long-dashed,dash-dotted, and dotted curves show A/A1, p/p1, ρ/ρ1, T/T1, and u/u1, respectively.

Moreover, the previous two equations can be combined with Equation (14.68) to give

uu1= Ma

[1 +

(γ − 1γ + 1

) (Ma 2 − 1

)]−1/2

. (14.72)

Figure 14.1 shows A/A1, p/p1, ρ/ρ1, T/T1, and u/u1, plotted as functions of thelocal Mach number, Ma, for an ideal gas with γ = 7/5. Here, use has been madeof Equations (14.71), (14.67), (14.68), (14.66), and (14.72), respectively. Inspectingthe curves, we can see, somewhat surprisingly, that the cross-sectional area function,A/A1, attains a minimum value when Ma = 1. In fact, the figure indicates that,for subsonic flow (Ma < 1), a decreasing cross-sectional area of the nozzle in thedirection of the gas flow leads to an increasing flow speed, and decreasing pressure,density, and temperature. However, for supersonic flow (Ma > 1), this behavior isreversed, and an increasing cross-sectional area of the nozzle leads to an increasingflow speed, and decreasing pressure, density, and temperature. We conclude thatlocal Mach number of gas flowing through a converging nozzle (i.e., a nozzle whosecross-sectional area decreases monotonically in the direction of the gas flow) cannever exceed unity. Moreover, the maximum Mach number (i.e., unity) is achievedon exit from the nozzle, where the cross-sectional area is smallest. On the other hand,the local Mach number of gas flowing through a converging-diverging nozzle (i.e., anozzle whose cross-sectional area initially decreases in the direction of the gas flow,attains a minimum value, and then increases) can exceed unity. For this to happen, the


flow conditions must be arranged such that the sonic point corresponds precisely tothe narrowest point of the nozzle, which is generally known as the throat. In this case,as the gas flows through the converging part of the nozzle, the local cross-sectionalarea, A, travels down the left-hand, subsonic branch of the A/A1 curve shown inFigure 14.1, while the flow speed, u, simultaneously increases. After passing throughthe throat at the sonic speed, the gas flows through the diverging part of the nozzle,and the cross-sectional area travels up the right-hand, supersonic branch of the A/A1curve, while the flow speed continues to increase. The type of converging-divergingnozzle just described is known as a de Laval nozzle, after its inventor, Gustaf de Laval(1845–1913).

Consider a de Laval nozzle whose gas supply is derived from a large reservoir.Assuming that the gas in the reservoir is essentially at rest, it follows that the temper-ature, pressure, and density of the gas in the reservoir correspond to the stagnationtemperature, pressure, and density—T0, p0, and ρ0, respectively. Equation (14.63)–(14.65) then specify the temperature, pressure, and density of the gas at the throat ofthe nozzle—T1, p1, and ρ1, respectively—in terms of the temperature, pressure, anddensity of the gas in the reservoir.

Suppose that a de Laval nozzle exhausts gas into the atmosphere, whose pressureis patm. Now, for the case of incompressible flow, the pressure of the gas exhaustedfrom a nozzle, pe (say), must match the ambient pressure, patm. The reason for this isthat any mismatch between the exhaust and ambient pressures is instantly communi-cated to the whole fluid by means of sound waves that travel infinitely fast (becausean incompressible fluid corresponds to the limit γ → ∞). Of course, the sound speedis finite in a compressible gas. However, in subsonic compressible flow, upstreamcommunication is still possible, because the local sound speed exceeds the local flowspeed. However, in supersonic compressible flow, upstream communication is im-possible, because sound waves cannot catch up with the flow. Consequently, in thecase of a nozzle with a subsonic exhaust speed, we would generally expect the ex-haust pressure to match the ambient pressure. However, in the case of a nozzle witha supersonic exhaust speed, the exhaust pressure can be significantly different to theambient pressure.


Ma e =

√(2

γ − 1

) ( p0

pe

)(γ−1)/γ

− 1

, (14.73)

where Ma e is the Mach number of the gas exhausted by the nozzle, pe the exhaustpressure, and p0 the reservoir pressure. Suppose that M e < 1—that is, the exhaustspeed of the gas is subsonic. In this case, we would expect pe = patm, so that

Ma e =

√(2

γ − 1

) ( p0

patm

)(γ−1)/γ

− 1

. (14.74)

It follows that the critical value of p0/patm at which the exhaust speed becomes sonic


(i.e. Ma e = 1) is (p0

patm

)crit=

(γ + 1

2

)γ/(γ−1)

. (14.75)

Thus, in order for a de Laval nozzle to achieve supersonic exhaust speeds, p0/patmmust exceed this critical value. For the case of a gas with γ = 7/5, we find that(p0/patm)crit = 1.89. Note that if p0/patm does not exceed the critical value then,as the gas flows through the converging part of the nozzle, its local cross-sectionalarea, A, travels down the left-hand, subsonic branch of the A/A1 curve shown inFigure 14.1. However, when the gas passes through the throat of the nozzle, thearea turns around, and then backtracks up the left-hand branch while the gas passesthrough the diverging part of the nozzle. In this case, the Mach number never reachesunity. In fact, in the converging part of the nozzle, the flow speed increases, whilethe pressure, density, and temperature decrease. The flow speed attains its maximum,subsonic value at the throat of the nozzle, while the pressure, density, and pressuresimultaneously attain minimum values. Finally, in the diverging part of the nozzle,the flow speed decreases, while the pressure, density, and temperature increase.

Equations (14.64) and (14.67) yield[1 +

(γ − 1γ + 1

)(Ma 2

e − 1)]=

(2

γ + 1

) (p0

pe

)(γ−1)/γ

. (14.76)

Moreover, Equations (14.45) and (14.63) imply that

u1 =

√RT0

(2

γ + 1

). (14.77)

Hence, Equations (14.72), (14.73), (14.76), and (14.77) give the following expressionfor the exhaust speed from a de Laval nozzle,

ue =

√R T0

(2 γγ − 1

) 1 − ( pe

p0

)(γ−1)/γ. (14.78)

Because pe cannot be negative, we deduce that there is an upper limit to the exhaustspeed: namely,

ue max =

(2 γγ − 1

)1/2c0 =

√2Cp T0, (14.79)

where use has been made of Equation (14.60). Here, c0 is the reservoir sound speed.If the exhaust pressure of a de Laval nozzle is higher than the ambient pressure,

pe > patm, then the gas is said to be under-expanded. In this case, a pattern ofstanding shock waves, called shock diamonds, forms in the exhaust plume externalto the nozzle. On the other hand, if the exhaust pressure is lower than the ambientpressure, pe < patm, then the gas is said to be over-expanded. In this case, a staticshock front forms inside the diverging part of the nozzle. (See Section 14.8.) As thegas passes through the front, its speed drops abruptly from a supersonic to a subsonicvalue, whereas the pressure, density, and temperature all increase abruptly. As thesubsonic gas flows through the remainder of the nozzle, its velocity decreases further.


14.8 Normal ShocksAs previously described, there is an effective discontinuity in the flow speed, pres-sure, density, and temperature, of the gas flowing through the diverging part of anover-expanded Laval nozzle. This type of discontinuity is known as a normal shock.Let us investigate the properties of such shocks.

Our fundamental equations are the mass conservation equation [see Equation(14.30)],

DρDt= −ρ∇ · v, (14.80)

the momentum conservation equation [see Equation (14.31)],

DvDt= −∇p

ρ, (14.81)

and the energy conservation equation [see Equation (1.75)],

DEDt= − p

ρ∇ · v. (14.82)

In writing the previous equations, we have neglected viscosity, heat conduction, andpotential energy. Note that we have used the more fundamental energy conservationequation, (1.75), rather than Equation (14.32), because the latter equation incorpo-rates the ideal gas law, and this law is not valid inside the shock, because the gasthere is not in thermodynamic equilibrium.

Consider a compressible gas flowing steadily down a duct of constant cross-sectional area A. Let x measure distance along the duct. The gas’s temperature,T (x), density, ρ(x), pressure p(x), specific internal energy E(x), and normal velocityu(x) = vx, are all assumed to be constant across any cross-section, and independentof time. Let the shock be situated at x = 0. Suppose that in the region upstream of theshock, x < 0, the temperature, density, pressure, specific internal energy, and flowspeed of the gas take the constant values T1, ρ1, p1, E1, and u1, respectively. Like-wise, suppose that in the region downstream of the shock, x > 0, the temperature,density, pressure, specific internal energy, and flow speed of the gas take the constantvalues T2, ρ2, p2, E2, and u2, respectively. Of course, in the immediate vicinity ofthe shock, T (x), ρ(x), p(x), E(x), and u(x) are all rapidly-varying functions of x. Wewish to find the relationship between the upstream and downstream gas parameters.We can achieve this goal by invoking conservation of mass, momentum, and energyacross the shock—in other words, by making use of Equations (14.80)–(14.82).

The mass continuity equation (14.71) yields

udρdx= −ρ

dudx, (14.83)


ddx

(ρ u) = 0. (14.84)


The momentum conservation equation (14.72) yields

ududx= −1

ρ

dpdx, (14.85)


ρ ududx+

dpdx= 0, (14.86)

ord

dx

(ρ u 2 + p

)= 0, (14.87)

where use has been made of Equation (14.84). Finally, the energy conservation equa-tion (14.73) yields

udEdx= −

pρ

dudx, (14.88)


ρ udEdx+

ddx

(p u) = udpdx, (14.89)

ord

dx(ρ uE + p u) = −ρ u 2 du

dx= −

12ρ u

du 2

dx= −

ddx

(12ρ u 3

), (14.90)

where use has been made of Equations (14.84) and (14.86). Thus, we obtain

ddx

(ρ uE + p u +

12ρ u 3

)= 0, (14.91)

which reduces tod

dx

(E + p

ρ+

12

u 2)= 0 (14.92)

with the aid of Equation (14.84). The previous equation can also be written

ddx

(H +

12

u 2)= 0, (14.93)

whereH = E + p/ρ is the specific enthalpy of the gas. [See Equation (14.18).]Integrating Equations (14.84), (14.87), and (14.93) across the shock, we obtain

ρ1 u1 = ρ2 u2, (14.94)

ρ1 u 21 + p1 = ρ2 u 2

2 + p2, (14.95)

H1 +12

u 21 = H2 +

12

u 22 , (14.96)

where H1 = E1 + p1/ρ1, et cetera. We shall assume that the gas upstream and


downstream of the shock obeys the ideal gas law, and has the same ratio of specificheats, γ. It follows that

p1,2 = R ρ1,2 T1,2, (14.97)

H1,2 =

(γ

γ − 1

)R T1,2. (14.98)

where use has been made of Equations (14.1), (14.18), and (14.60). Let Ma1 and Ma2

be the Mach numbers upstream and downstream of the shock, respectively. Thus,

Ma1,2 =u1,2√γR T1,2

, (14.99)

where use has been made of Equations (14.45) and (14.57). Equations (14.96),(14.98), and (14.99) can be combined to give

T2

T1=

1 + (1/2) (γ − 1) Ma21

1 + (1/2) (γ − 1) Ma22

. (14.100)

Equations (14.94) and (14.99) yield

ρ2

ρ1=

u1

u2=

√T1

T2

Ma1

Ma2, (14.101)

orp2

p1=

√T2

T1

Ma1

Ma2, (14.102)

where use has been made of Equation (14.97). Finally, Equations (14.95), (14.97),and (14.99) give

p2

p1=

1 + γMa 21

1 + γMa 22

. (14.103)

Eliminating T2, ρ2, and p2 between Equations (14.100), (14.101), and (14.103),we obtain

Ma2 =

2 + (γ − 1) Ma21

1 − γ + 2 γMa 21

1/2 . (14.104)

Note thatdMa 2

2

dMa 21

= − 1 + γ

1 − γ + 2 γMa 21

2

< 0. (14.105)

Moreover, it is easily verified that Ma2 = 1 when Ma1 = 1. Hence, we deduce that

Ma2 1 as Ma1 1. (14.106)

In other words, if the downstream flow is subsonic then the upstream flow is super-sonic, and vice versa. Equations (14.1), (14.100), (14.101), (14.103), and (14.104)


can be combined to give

p2

p1=

1 − γ + 2 γM 21

γ + 1, (14.107)

ρ2

ρ1=

u1

u2=

(γ + 1) Ma 21

2 + (γ − 1) Ma21

, (14.108)

T2

T1=

p2

p1

ρ1

ρ2. (14.109)

The previous three equations completely describe the conditions downstream of theshock in terms of the upstream conditions. We can rearrange these equations to give

Ma 21 =

γ − 1 + (γ + 1) (p2/p1)2 γ

, (14.110)

ρ2

ρ1=

u1

u2=γ − 1 + (γ + 1) (p2/p1)γ + 1 + (γ − 1) (p2/p1)

, (14.111)

T2

T1=

(p2

p1

) [γ + 1 + (γ − 1) (p2/p1)γ − 1 + (γ + 1) (p2/p1)

]. (14.112)

The latter two equations are known as the Rankine-Hugoniot relations.According to Equation (14.25), the jump in specific entropy across the shock is

S2 − S1 = CV ln[

p2

p1

(ρ1

ρ2

)γ]. (14.113)

It follows from Equations (14.107) and (14.108) that

S2 − S1 = CV ln[G(Ma 2

1 )], (14.114)

where

G(x) =(

1 − γ + 2 γ xγ + 1

) [2 + (γ − 1) x

(γ + 1) x

]γ. (14.115)


dGdx= 2 γ (γ − 1) (1 − x) 2 [2 + (γ − 1) x] γ−1

[(γ + 1) x] γ+1 ≥ 0. (14.116)

Moreover, G(1) = 1. Hence, we deduce that

S2 − S1 0 as Ma1 1. (14.117)

Now, the second law of thermodynamics forbids a spontaneous decrease in the spe-cific entropy of the gas as it passes through the shock (Reif 1965). In other words, thesecond law of thermodynamics demands that S2 − S1 ≥ 0. It follows that Ma1 ≥ 0.However, it is easily seen from Equations (14.107)–(14.109) that p2/p1 = ρ2/ρ1 =


u2/u1 = T2/T1 = 1 when Ma1 = 1. In other words, there is no shock (i.e., no dis-continuity in the properties of the gas at x = 0) when the upstream Mach numberis exactly unity. Thus, we conclude that the only type of normal shock that is con-sistent with the second law of thermodynamics is one in which the upstream flowis supersonic, and the downstream flow subsonic—that is, Ma1 > 1 and Ma2 < 1.It is apparent from Equations (14.107)–(14.109) that if Ma1 > 1 then p2/p1 > 1,ρ2/ρ1 > 1, and T2/T1 > 1. In other words, the passage of the gas through the shockfront leads to both compression and heating.

The dimensionless parameter

∆pp1=

p2 − p1

p1(14.118)

is a convenient measure of shock strength. Thus, a shock is said to be weak if∆p/p1 1, and strong if ∆p/p1 1.

Consider a weak shock. According to Equations (14.111), (14.112), (14.114),and (14.116),

∆ρ

ρ1

1γ

∆pp1, (14.119)

∆TT1(γ − 1γ

)∆pp1, (14.120)

∆SR

(γ + 1)12 γ 2

(∆pp1

) 3

, (14.121)

where ∆ρ = ρ2 − ρ1, et cetera, and use has been made of Equation (14.59). It can beseen that the flow across the shock is isentropic (i.e., p/ργ and T γ/p γ−1 are constant)to first order in the shock strength. This is the case because the increase in specificentropy across the shock is only third order in the shock strength.

Consider a strong shock. It can be seen from Equations (14.111) and (14.112)that

ρ2

ρ1→ γ + 1γ − 1

, (14.122)

T2

T1→(γ − 1γ + 1

)∆pp1, (14.123)

as ∆p/p1 → ∞. Clearly, there is a finite limit to the degree of compression of gaspassing through a strong shock. Thus, in the case of a strong shock, the large increasein the pressure across the shock front is predominately caused by a large increase inthe temperature. In other words, the gas flowing through a strong shock is subjectmoderate compression, but intense heating.

Finally, let us transform to a frame of reference that is co-moving with the gas up-stream of the shock. In this reference frame, the shock appears to propagate througha stationary gas of pressure, density, and temperature, p1, ρ1, and T1, respectively,at the speed u1. In other words, in the new reference frame, our stationary shock is


transformed into a shock wave. (See Section 14.4.) The pressure, density, and tem-perature behind the wave are p2, ρ2, and T2, respectively. Also, the gas behind thewave follows the shock front at the speed u1 − u2. We can regard Ma1 as the Machnumber of the shock wave in the unperturbed gas. It follows from Equation (14.110)that

Ma 21 = 1 +

(γ + 12 γ

)∆pp1. (14.124)

Thus, for a weak shock wave (i.e., ∆p/p1 1),

Ma1 1. (14.125)

In other words, a weak shock wave propagates through the unperturbed gas at thelocal sound speed. Indeed, such a wave is essentially indistinguishable from a con-ventional sound wave. On the other hand, for a strong shock wave (i.e., ∆p/p1 1),

Ma1

√(γ + 12 γ

)∆pp1. (14.126)

We conclude that a strong shock wave propagates through the unperturbed gas at aspeed that greatly exceeds the local sound speed.

14.9 Piston-Generated Shock WaveConsider the situation illustrated in Figure 14.2 in which, at t = 0, a tight-fittingpiston is suddenly pushed into a stationary gas, contained in a uniform tube, at thesteady speed Vp, generating a shock front that propagates away from the piston, andinto the gas, at the speed Vs. Suppose that the piston is located at x = 0 at t = 0,and moves in the +x-direction. The gas can be divided into two regions. In region 1,which lies to the right of the shock wave (i.e., x > Vs t), the gas remains undisturbed.Hence, its velocity is u1 = 0. Let its pressure, density, and temperature be p1, ρ1,and T1, respectively. In region 2, which lies between the piston and the shock wave(i.e., Vp t < x < Vs t), the gas is co-moving with the piston, so its velocity is u2 = Vp.Let its pressure, density, and temperature, be p2, ρ2, and T2, respectively. Thus, thesystem is formally the same as that discussed in the final paragraph of the previoussection, provided that we make the identifications Vs = u1 and Vp = u1 − u2.

Let β = (p2 − p1)/p1 be our conventional measure of shock strength. It followsfrom Equations (14.111) and (14.112) that

ρ2

ρ1=

2 γ + (γ + 1) β2 γ + (γ − 1) β

, (14.127)

T2

T1= (1 + β)

[2 γ + (γ − 1) β2 γ + (γ + 1) β

], (14.128)


region 1region 2

x

Vp

x = Vp t

shoc

kpa

th

t

pist

onpa

th

Vs

x = Vs t

Figure 14.2A piston-generated shock wave.

where γ is the gas’s ratio of specific heats. We can define the Mach number of theshock wave as Mas = Vs/c1 = u1/c1, where c1 is the sound speed in the unperturbedgas. According to Equation (14.110),

Mas =

[1 +

(γ + 12 γ

)β

]1/2. (14.129)

Finally, we can define the Mach number of the piston as Map = Vp/c1. Thus,

Map =u1 − u2

c1=

u1

c1

(1 − u2

u1

)= Mas

(1 − ρ1

ρ2

), (14.130)

where use has been made of Equation (14.111). Equations (14.127) and (14.129) canbe combined with the previous equation to give

1γ 2 Ma 2

p

=1β 2 +

γ + 12 γ β

. (14.131)

Thus, Equations (14.127)–(14.129), and (14.131), fully determine the parameters ofthe shocked gas, as well as the speed of the shock front, in terms of the parametersof the unperturbed gas and the piston speed. In particular, for a weak shock (i.e.,


β 1),

β γMap, (14.132)

Mas 1 +(γ + 1

4

)Map, (14.133)

whereas for a strong shock (i.e., β 1),

β γ (γ + 1) Ma 2

p

2, (14.134)

Mas (γ + 1

2

)Map. (14.135)

Thus, a weak shock is associated with a subsonic piston (i.e., Map 1), whereas astrong shock is associated with a supersonic piston (i.e., Map 1).

14.10 Piston-Generated Expansion WaveConsider the situation illustrated in Figure 14.3 in which, at t = 0, a tight-fittingpiston is suddenly withdrawn from a stationary gas, contained in a uniform tube,at the steady speed Vp, generating an expansion wave that propagates away fromthe piston, and into the gas. Suppose that the piston is located at x = 0 at t = 0,and moves in the −x-direction. As described in Section 14.4, the expansion wave isisentropic.

The sudden withdrawal of the piston creates a step-function change in the flowspeed, pressure, density, and temperature, of the gas inside the tube, such that theimmediately adjacent gas moves with the piston. As described in Section 14.4, thisstep function flattens as the expansion wave begins to propagate. Locally, withinthe wave, the disturbance travels at the sound speed. Because the temperature variesacross the wave (which starts out as a step in this variable), the local speed of distur-bance propagation also varies across the wave. On the edge of the wave adjacent tothe region of undisturbed gas, which we shall denote region 1, the temperature andpropagation speed are greatest. On the opposite edge, the temperature and propaga-tion speed are least. Let the region lying between the piston and the left edge of theexpansion wave be denoted region 2. In region 1, the flow speed, pressure, density,and temperature, of the gas are u1 = 0, p1, ρ1, and T1, respectively. On the otherhand, in region 2, the flow speed, pressure, density, and temperature, of the gas areu2 = −Vp (i.e., the gas in region 2 is co-moving with the piston), p2, ρ2, and T2,respectively. As shown in Figure 14.3, the expansion wave expands, or “fans” out, asit propagates. In fact, the x-t plot is a fan of constant sonic speed lines that show thedevelopment of the wave. These lines are called characteristics, and follow the pathof local isentropic disturbances. The absolute speed of the disturbance is the sum of


region 2 region 1

x

t

Vp

pistonpath

expansion front

x = c1 t

c1

x = −Vp t

x = [c1 − (1/2)(γ + 1) Vp] t

Figure 14.3A piston-generated expansion wave.

the local sonic speed and the local gas flow speed. The terminating characteristic onthe right has slope dx/dt = c1, where c1 is the region-1 sound speed. The terminatingcharacteristic on the left has the slope dx/dt = c2−Vp, where c2 is the region-2 soundspeed. The latter slope may be either positive or negative, depending on whether c2

is greater than, or less than, Vp.Making use of isentropic relationships (see Section 14.3), we can explicitly eval-

uate the structure of the fan as follows. In terms of c1, the local sonic speed in thefan can be written [cf., Equation (14.47)]

c(x) = c1

(ρ

ρ 1

) (γ−1)/2

. (14.136)

Moreover, the relationship between the gas speed, u(x), and the density in the waveis [cf., Equation (14.48)]

du = cdρρ. (14.137)

The previous two expressions yield [cf., Equation (14.50)]

c(x) = c1 +12

(γ − 1) u(x). (14.138)

Hence, the absolute disturbance speed in the fan is [cf., Equation (14.51)]

c′(x) = c(x) + u(x) = c1 +12

(γ + 1) u(x). (14.139)


In particular, the terminating characteristic on the left (for which u = −Vp) has theslope

dxdt= c2 − Vp = c1 −

12

(γ + 1) Vp. (14.140)

It follows thatc2 = c1 −

12

(γ − 1) Vp. (14.141)

Standard isentropic relations (see Section 14.3) then yield the density and pressurechanges across the fan:

ρ2

ρ1=

[1 − 1

2(γ − 1)

Vp

c1

] 2/(γ−1)

, (14.142)

p2

p1=

[1 − 1

2(γ − 1)

Vp

c1

] 2 γ/(γ−1)

. (14.143)

Here, use has been made of the fact that c ∝ T 1/2. At the critical piston withdrawalspeed Vp = [2/(γ − 1)] c1, the pressure and density in region 2 are both reduced tozero, and the terminating characteristic on the left co-moves with the piston. Anyfurther increase in the withdrawal speed makes no difference to the flow.

The equation of motion of the right edge of the expansion wave is

x1 = c1 t. (14.144)

Likewise, the equation of motion of the left edge is

x2 =

[c1 −

12

(γ + 1) Vp

]t. (14.145)

Finally, the equation of motion of a general point in the expansion wave is

x =[c1 +

12

(γ + 1) u]

t. (14.146)

The previous three equations can be combined to give

u(x) =[−1 +

(x − x2

x1 − x2

)]Vp (14.147)

for x2 ≤ x ≤ x1. In other words, the flow speed varies linearly with x inside theexpansion wave. It follows from Equations (14.138) and (14.141) that

c(x) − c2

c1 − c2=

x − x2

x1 − x2(14.148)

for x2 ≤ x ≤ x1. Thus, the sound speed also varies linearly with x inside the expan-sion wave. However, according to Equation (14.136),

ρ (γ−1)/2 − ρ (γ−1)/22

ρ(γ−1)/21 − ρ (γ−1)/2

2

=x − x2

x1 − x2(14.149)


for x2 ≤ x ≤ x1. Furthermore, standard isentropic relations yield

p (γ−1)/2γ − p (γ−1)/2 γ2

p (γ−1)/2γ1 − p (γ−1)/2 γ

2

=x − x2

x1 − x2(14.150)

for x2 ≤ x ≤ x1. Thus, neither the density nor the pressure vary linearly with x insidethe expansion wave.

14.11 Exercises14.1 Prove the following useful theorems regarding partial derivatives:(

∂X∂Y

)Z= 1

/(∂Y∂X

)Z,(

∂X∂Y

)Z=

(∂X∂W

)Z

/(∂Y∂W

)Z,(

∂X∂Y

)Z= −

(∂Z∂Y

)X

/(∂Z∂X

)Y

14.2 (a) The specific internal energy of a (not necessarily ideal) gas is defined by

dE = T dS − p dv.

Demonstrate that

T =(∂E∂S

)v

, p = −(∂E∂v

)S,

and (∂T∂v

)S= −

(∂p∂S

)v

.

(b) The specific enthalpy of a gas is defined by

H = E + p v.

Demonstrate that

T =(∂H∂S

)p, v =

(∂H∂p

)S,

and (∂T∂p

)S=

(∂v

∂S

)p.


(c) The specific Helmholtz free energy of a gas is defined by

F = E − T S.

Demonstrate that

S = −(∂F∂T

)v

, p = −(∂F∂v

)T,

and (∂S∂v

)T=

(∂p∂T

)v

.

(d) The specific Gibbs free energy of a gas is defined by

G = H − T S.

Demonstrate that

S = −(∂G∂T

)p, v =

(∂G∂p

)T,

and (∂S∂p

)T= −

(∂v

∂T

)p.

14.3 Demonstrate that the specific heat at constant volume of a (not necessarilyideal) gas can be written

cv =(∂E∂T

)v

= T(∂S∂T

)v

.

Likewise, show that the specific heat at constant pressure takes the form

cp =

(∂H∂T

)v

= T(∂S∂T

)p.

14.4 The quantities

α =1v

(∂v

∂T

)p,

κT = −1v

(∂v

∂p

)T,

κS = −1v

(∂v

∂p

)S,


are known as the coefficient of thermal expansion, the isothermal compress-ibility, and the adiabatic compressibility, respectively. Demonstrate that for a(not necessarily ideal) gas,

cp − cv =T v α 2

κT,

κT − κS =T v α 2

cp. (14.151)

Hence, deduce thatcp

cv=κT

κS.

Show that for the special case of an ideal gas, α = 1/T , κT = 1/p, and κS =1/(γ p). Hence, obtain the following standard results for an ideal gas:

cp

cv= γ,

cp − cv = R,

cp =

(γ

γ − 1

)R,

14.5 Show that for an ideal gas

Ma =v

c0

1 − 12

(γ − 1)(v

c0

)2−1/2

,

TT0= 1 − 1

2(γ − 1)

(v

c0

)2,

pp0=

1 − 12

(γ − 1)(v

c0

)2γ/(γ−1)

,

ρ

ρ0=

1 − 12

(γ − 1)(v

c0

)2 1/(γ−1)

,

where Ma is the Mach number, v the flow speed, and T0, p0, ρ0, and c0, are thetemperature, pressure, density, and sound speed, respectively, at the stagnationpoint.

14.6 Consider the flow of an isentropic ideal gas down a straight nozzle with aslowly-varying cross-sectional area, A(x), where x measures distance alongthe nozzle. Let u(x), ρ(x), c(x), and Ma(x) be the flow speed, density, sonicspeed, and Mach number, respectively. Demonstrate that

c 2 dρρ+ u du = 0,


diaphragm

p4 p4 > p1

c4

region 2 region 3 region 4

contact surfaceshock expansion front

|u2||Vs|

region 1

p1

Figure 14.4A shock tube.

anddρρ+

dAA+

duu= 0.

Hence, show thatdAA=

duu

(Ma 2 − 1).

Deduce that the throat of the nozzle (where A attains its minimum value) eithercorresponds to a sonic point (i.e, Ma = 1), or a point of maximum or minimumflow speed. Finally, demonstrate that

dMaMa=

[1 + (1/2) (γ − 1) Ma2

1 −Ma 2

]dAA.

14.7 As indicated in Figure 14.4, a shock tube is a tube of uniform cross section thatis divided by a diaphragm into two chambers that contain different gases at dif-ferent pressures. Let x measure distance along the tube, and let the diaphragmbe located at x = 0. Suppose that the chamber to the left of the diaphragm(which lies at x < 0) is filled with stationary gas of pressure, density, temper-ature, and ratio of specific heats, p1, ρ1, T1, and γ1, respectively. Likewise,suppose that the chamber to the right of the diaphragm (which lies at x > 0) isfilled with stationary gas of pressure, density, temperature, and ratio of specificheats, p4, ρ4, T4, and γ4, respectively. It is assumed that p4 > p1. At t = 0,the diaphragm is ruptured. As indicated in the figure, a shock wave subse-quently travels to the left with speed |Vs|, and an expansion wave to the rightwith speed c4. The so-called contact surface marks the boundary betweenthe two different gases that were originally on either side of the diaphragm.Neglecting diffusion, the gases do not mix, but are permanently separated bythe contact surface. On either side of the contact surface, which moves to theleft with speed |u2|, the temperatures and densities can be different, but thepressures and flow velocities must be the same. We can divide the gas in the


tube into four regions. Regions 1 lies to the left of the shock wave. Region 2lies between the shock wave and the contact surface. Region 3 lies betweenthe contact surface and the expansion wave. Region 4 lies to the right of theexpansion wave. Thus, we expect the flow velocity, pressure, density, tem-perature, and ratio of heats to be u1 = 0, p1, ρ1, T1, and γ1, respectively, inRegion 1; u2, p2, ρ2, T2, and γ1, respectively in Region 2; u2, p2, ρ3, T3, andγ4, respectively in Region 3; and u4 = 0, p4, ρ4, T4, and γ4, respectively, inRegion 1. Note that u2 and Vs are negative. Demonstrate that

|u2| = c1

(p2

p1− 1) [

2/γ1

2 γ1 + (γ1 + 1) (p2/p1 − 1)

] 1/2

,

and

|u2| =2 c4

γ4 − 1

1 − ( p2

p4

)(γ4−1)/γ4 ,

where c1 and c4 are the sound speeds in Regions 1 and 4, respectively. Hence,obtain

p4

p1=

p2

p1

1 − (γ4 − 1) (c1/c4) (p2/p1 − 1)√2 γ1

√2 γ1 + (γ1 + 1) (p2/p1 − 1)

−2 γ4/(γ4−1)

.

This expression give the shock strength, p2/p1, implicitly as a function ofthe diaphragm pressure ratio, p4/p1. All other quantities of interest can beexpressed in terms of the shock strength. Show that

|Vs| = c1

[1 +

(γ1 + 12 γ1

) (p2

p1− 1)] 1/2

,

ρ2

ρ1=

2 γ1 + (γ1 + 1) (p2/p1 − 1)2 γ1 + (γ1 − 1) (p2/p1 − 1)

,

T2

T1=

p2

p1

[2 γ1 + (γ1 − 1) (p2/p1 − 1)2 γ1 + (γ1 + 1) (p2/p1 − 1)

],

ρ3

ρ4=

(p2/p1

p4/p1

) 1/γ4

,

andT3

T4=

(p2/p1

p4/p1

) (γ4−1)/γ4

.

14.8 Show that the maximum shock strength and shock speed attainable in a (uni-form) shock tube, in the limit p4/p1 → ∞, are√

p2

p1=

c4

c1

√2 γ1 (γ1 + 1)(γ4 − 1)

,


and

|Vs| =(γ1 + 1γ4 − 1

)c4,

respectively. Show, further, that the contact surface moves at the speed

|u2| =(

2γ4 − 2

)c4.

14.9 Show that Equations (14.38) and (14.39) can be written in the form

2γ − 1

(∂c∂t+ u

∂c∂x

)+ c

∂u∂x= 0,

∂u∂t+ u

∂u∂x+

2 aγ − 1

∂c∂x= 0,

where c is the sound speed. By adding and subtracting the previous equations,obtain [

∂

∂t+ (u + c)

∂

∂x

] (u +

2 cγ − 1

)= 0,[

∂

∂t+ (u − c)

∂

∂x

] (u − 2 c

γ − 1

)= 0.

These equations indicate that the quantities P = u + 2 c/(γ − 1) and Q =u − 2 c/(γ − 1) are constant on curves that have the slope dx/dt = u + c anddx/dt = u − c, respectively. These curves are called characteristics, and Pand Q are known as Riemann invariants. In situations in which all the Qcharacteristics originate from regions where the gas is at rest, we expect Q tobe constant throughout the gas. Deduce that [cf., Equation (14.50)]

c = c0 +

(γ − 1

2

)u,

where c0 is the stagnation sound speed. Show that[∂

∂t+ (u + c)

∂

∂x

](u + c) = 0.

Hence, conclude that the P characteristics are straight-lines.

14.10 Consider a one-dimensional sound wave propagating through an ideal gaswhose unperturbed sound speed is c0. At time t = 0, the velocity perturba-tion, u(x, t), due to the wave, has the form u0 sin(k x), where u0 and k are bothpositive. Demonstrate that shocks form after a time

t∗ 2

(γ + 1) k u0


has elapsed. [Hint: Shocks are associated with the crossing of different Pcharacteristics.] Show that in the time interval t = 0 to t = t∗ a local maximumof u(x, t) travels a distance [

1 +2 c0

(γ + 1) u0

]1k,

14.11 An ideal gas is initially at rest in a uniform tube, and occupies the region tothe right of a tight-fitting piston whose position is X(t) = (1/2) a t 2 for t > 0.Here, a > 0. Show that a shock first forms at t = t∗ and x = x∗, where

t∗ =(

2γ + 1

)c0

a,

x∗ =(

2γ + 1

)c 2

0

a.

Here, c0 is the stagnation sound speed, and γ the ratio of specific heats.

15Two-Dimensional Compressible Inviscid Flow

15.1 Introduction

This chapter investigates two-dimensional, supersonic, compressible, inviscid flow.Flow is said to be two-dimensional when the fluid properties only depend on twoCartesian coordinates. More information on this subject can be found in Liepmann& Roshko 1957, Hughes & Brighton 1999, Emanuel 2000, and Anderson 2003.

15.2 Oblique Shocks

Consider a normal shock in which the flow upstream and downstream of the shockfront is parallel to the x-axis, and front itself lies in the y-z plane. (See Section 14.8.)Suppose that u1 and u2 are the upstream and downstream flow speeds, respectively.Let us now view this shock in a frame of reference that moves with respect to ouroriginal frame at the constant velocity −v ey. As illustrated in Figure 15.1, viewingthe shock in the new reference frame has the effect of adding a component of velocityof magnitude v, directed parallel to the y-axis, to both the upstream and the down-stream flow. The resultant upstream velocity is now of magnitude w1 = (u 2

1 + v2)1/2,

and subtends an angle β = tan−1(u1/v), known as the wave angle, with the shockfront. It is evident that β may be adjusted to any value via a suitable choice of v.Now, because u2 is not the same as u1, the inclination of the downstream flow to theshock front is different to that of the upstream flow. In other words, the directionof the flow turns abruptly as it passes through the shock. Because u2 is always lessthan u1 (see Section 14.8), the deviation is always towards the shock front. In otherwords, the deflection angle, θ, defined in Figure 15.1, is positive.

The relationship between the conditions upstream and downstream of the shock iseasily obtained from the analysis of Section 14.8, because viewing a normal shock ina moving frame of reference does not change the relationship between the upstreamand downstream conditions in the original reference frame. The only modification isthat the upstream Mach number is now defined as Ma1 = w1/c1, rather than u1/c1.Here, c1 is the upstream sound speed. Thus, given that u1 = w1 sin β, we simplyneed to make the transformation Ma1 → Ma1 sin β in the previous analysis. Hence,

437


x

v w 1β

u2

y

u1

vw 2

β − θ

Figure 15.1An oblique shock.

Equations (14.107), (14.108), (14.109), and (14.113), yield

p2

p1=

1 − γ + 2 γMa 21 sin2 β

γ + 1, (15.1)

ρ2

ρ1=

u1

u2=

(γ + 1) Ma21 sin2 β

2 + (γ − 1) Ma 21 sin2 β

, (15.2)

T2

T1=

[1 − γ + 2 γMa 2

1 sin2 β] [

2 + (γ − 1) Ma21 sin2 β

](γ + 1) 2 Ma 2

1 sin2 β, (15.3)

S2 − S1

R= ln

1 − γ + 2 γMa 2

1 sin2 β

γ + 1

11−γ (γ + 1) Ma2

1 sin2 β

2 + (γ − 1) Ma21 sin2 β

− γ1−γ , (15.4)

respectively. Here, p1, ρ1, T1, and S1 are the upstream pressure, density, temper-ature, and specific entropy, respectively, p2, ρ2, T2, and S2 are the correspondingdownstream quantities, γ is the ratio of specific heats, R the specific gas constant,and use has been made of Equation (14.59). As before (see Section 14.8), the secondlaw of thermodynamics demands that S2−S1 ≥ 0, which implies that Ma1 sin β ≥ 1.This sets a minimum inclination of the upstream flow to the shock front for a givenupstream Mach number. The maximum inclination is, of course, β = π/2. Thus,

µ ≤ β ≤π

2, (15.5)

Two-Dimensional Compressible Inviscid Flow 439

where

µ = sin−1(

1Ma1

)(15.6)

is termed the Mach angle. The downstream Mach number can be calculated by not-ing that Ma2 = w2/c2, rather that u2/c2, where c2 is the downstream sound speed,and u2 = w2 sin(β − θ). In other words, we simply need to make the transforma-tion Ma2 → Ma2 sin(β − θ) in the previous analysis. Thus, it follows from Equa-tion (14.104) that

Ma 22 sin2(β − θ) =

2 + (γ − 1) Ma21 sin2 β

1 − γ + 2 γMa 21 sin2 β

. (15.7)

According to Figure 15.1,tan β =

u1

v, (15.8)

andtan(β − θ) = u2

v. (15.9)

Eliminating v, and then making use of Equation (15.2), we obtain

tan(β − θ)tan β

=u2

u1=ρ1

ρ2=

2 + (γ − 1) Ma21 sin2 β

(γ + 1) Ma 21 sin2 β

. (15.10)

Now,

tan(β − θ) ≡ tan β − tan θ1 + tan β tan θ

, (15.11)

so Equation (15.10) can be rearranged to give

tan θ =2 cot β

(Ma 2

1 sin2 β − 1)

2 +Ma 21[γ + cos(2 β)

] . (15.12)

Here, use has been made of the identity cos(2 β) ≡ cos2 β − sin2 β. The previousexpression implies that θ = 0 at β = µ and β = π/2, which are the limits of the rangeof allowed values for θ defined in Equation (15.5). Within this range, θ is positive,and, must, therefore, have a maximum value, θmax. This is illustrated in Figure 15.2,where the relationship between θ and β, for an ideal gas with γ = 1.4, is plotted forvarious values of Ma1.

If θ < θmax then, for each value of θ and Ma1, there are two possible solutions,corresponding to two different values of β. The larger value of β corresponds to thestronger shock [because, according to Equation (15.1), the shock strength, p2/p1−1,is a monotonically increasing function of β].

Also shown in Figure 15.2 is the locus of solutions for which Ma2 = 1. In thesolution with the stronger shock, the downstream flow always becomes subsonic. Onthe other hand, in the solution with the weaker shock, the downstream flow remainssupersonic, except for a small range of values of θ that are slightly smaller that θmax.


0

10

20

30

40

θ( )

0 10 20 30 40 50 60 70 80 90β( )

Figure 15.2Oblique shock solutions for γ = 1.4. The solid curves show solutions for Ma1 = 1.2,1.4, 1.6, 2, 3, 4, 5, 10, and ∞, in order from the innermost to the outermost curve.The dashed curve shows the locus of solutions for which θ = θmax. The dash-dottedcurve shows the locus of solutions for which Ma2 = 1. Solutions to the left and rightof the dash-dotted curve correspond to Ma2 > 1 and Ma2 < 1, respectively.

Equation (15.10) can be rearranged to give

1Ma 2

1 sin2 β=

(γ + 1

2

)tan(β − θ)

tan β−(γ − 1

2

), (15.13)

which can be further reduced to

Ma 21 sin2 β − 1 =

(γ + 1

2

)Ma 2

1sin β sin θcos(β − θ)

. (15.14)

For small deflection angles, θ, the previous expression can be approximated by

Ma 21 sin2 β − 1

[(γ + 1

2

)Ma 2

1 tan β]θ. (15.15)


β

θwall

wall

Ma1

β

shock front

Ma2

θ

Figure 15.3Supersonic flow in concave corner.

15.3 Supersonic Flow in Corner or over Wedge

In practice, what determines β and θ? In other words, how are oblique shocks gen-erated? In non-viscous flow, any streamline can be replaced by a solid boundary.Thus, the oblique shock flow, described in the previous section, provides the solutionto the problem of supersonic flow in a concave corner, as illustrated in Figure 15.3.For a given values of Ma1 and θ, the values of β and Ma2 are determined. [See Exer-cise 15.1 and Equation (15.7).] For the present, we shall only consider cases in whichMa2 > 1. This restricts the oblique shock solution to the weaker of the two possiblesolutions. We also require that θ be less than θmax. The stronger shock solution, aswell as cases in which θ > θmax, will be discussed later. (See Section 15.7.)

By symmetry, supersonic flow in a concave corner that turns though an angleθ is equivalent to one half of the flow pattern that results when a supersonic fluidis normally incident on a symmetric wedge of nose angle 2 θ. This is illustrated inFigure 15.4. It is clear, from the figure, that the presence of the wedge only affectsthe flow in the region that lies to the right of the two shock fronts that are attached tothe wedge apex, P. In particular, the presence of the apex at P only affects the flowin the region that lies to the right of the shock fronts.

The part of Figure 15.2 that we are presently considering (i.e., Ma2 > 1) is suchthat a decrease in the wedge angle, θ, corresponds to a decrease in the wave angle,β. When θ decreases to zero (causing the wedge to disappear), β decreases to thelimiting value µ. Moreover, the shock strength (which parameterizes the jump inquantities across the shock front) becomes zero. (See Section 15.4.) In fact, thereis no disturbance in the flow in the limit θ → 0. Thus, in Figure 15.4, there is nolonger anything unique about the point P; indeed, this point might correspond to anypoint in the flow. The angle µ is simply a characteristic angle associated with the


Ma1

θ

β

shock front

Ma1

Ma2

θβ

Ma2

wedge

P

shock front

Ma1

Figure 15.4Supersonic flow over symmetric wedge.

local Mach number of the flow, according to the relation (15.6). As we have alreadymentioned, µ is known as the Mach angle. Note that 0 ≤ µ ≤ π/2 for Ma1 ≥ 1.

The lines of inclination µ (with respect to the upstream directed streamline) thatmay be drawn at any point in the flow are known as Mach lines. At a general point,P, there are always two lines that intersect the local streamline at the angle µ. (Inthree-dimensional flow, the Mach lines define a conical surface with apex P, knownas a Mach cone.) Thus, a two-dimensional supersonic flow pattern is associated withtwo families of Mach lines. These are conventionally distinguished by the labels(+) and (−). Those lying in the (+) set run to the right of the (upstream directed)streamline, whereas those in the (−) set run to the left. (See Figure 15.5.) Mach linesare also sometimes called characteristics. It is clear, from our previous discussion,that the flow conditions at point P can only affect those at some other point Q if thelatter point lies between the (+) and (−) Mach lines passing through point P. (SeeFigure 15.5, as well as Exercise 15.2.) Hence, we deduce that, unlike subsonic flow,there is no upstream influence in supersonic flow.

Oblique shocks can also be distinguished by the labels (+) and (−), according towhich set of characteristics they asymptote to in the limit of zero shock strength.


streamline

(+)P

Q

(−)

µ

µ

Figure 15.5Right- and left-running Mach lines [labeled (+) and (−), respectively] emanatingfrom an arbitrary point, P, in a supersonic flow field.

15.4 Weak Oblique ShocksFor small deflection angles, θ, the oblique shock equations reduce to relatively simpleexpressions. In fact, we already saw in Equation (15.15) that if θ 1 then

Ma 21 sin2 β − 1

[(γ + 1

2

)Ma 2

1 tan β]θ. (15.16)

It follows that β is either close to µ or π/2, depending on whether Ma2 > 1 orMa2 < 1. (See Figure 15.2.) For the present, we are only considering the former case(Ma2 > 1), for which we may use the approximation [see Equation (15.6)]

tan β tan µ =1√

Ma 21 − 1

. (15.17)

Equation (15.16) then reduces to

Ma 21 sin2 β − 1

(γ + 1

2

) Ma 21√

Ma 21 − 1

θ. (15.18)

This is the basic relation that is needed to obtain all other approximate expressions,because these expressions only depend on the normal component of the upstreamflow, Ma1 sin β. Thus, Equation (15.1) yields

∆pp1

γMa 21√

Ma 21 − 1

θ, (15.19)


where ∆p = p2 − p1. It is apparent that the shock strength, ∆p/p1, is directly propor-tional to the deflection angle, θ.

The changes in the other flow quantities, except the specific entropy, across theshock front are also directly proportional to θ. (See Exercise 15.3.) The change in theentropy, on the other hand, is proportional to the third power of the shock strength(see Section 14.8), and, hence, to the third power of the deflection angle.

To explicitly find the deviation of the wave angle, β, from the Mach angle, µ, wewrite

β = µ + ε, (15.20)

where ε µ. Thus, we have

sin β = sin(µ + ε) sin µ + ε cos µ. (15.21)

However, by definition, sin µ = 1/Ma1, and cos µ = (Ma 21 − 1)1/2/Ma1 [see Equa-

tion (15.6)], so we obtain

Ma1 sin β 1 + ε√

Ma 21 − 1, (15.22)

or

Ma 21 sin 2 β 1 + 2 ε

√Ma 2

1 − 1. (15.23)

Finally, comparison with Equation (15.18) reveals that

ε (γ + 1

4

) Ma 21

Ma 21 − 1

θ. (15.24)

In other words, for a finite deflection angle, θ, the wave angle, β, differs from theMach angle, µ, by an amount, ε, that is of the same order of magnitude as θ.

The change in flow speed across the shock front is obtained from the ratio

w 22

w 21

=u 2

2 + v2

u 21 + v

2=

(u2/v)2 + 1(u1/v)2 + 1

=tan2(β − θ) + 1

tan2 β + 1=

cos2 β

cos2(β − θ), (15.25)

where use has been made of Equations (15.8) and (15.9). However,

cos2 β = 1 − sin2 β Ma 2

1 − 1

Ma 21

1 − 2 ε√Ma 2

1 − 1

, (15.26)

where use has been made of Equation (15.23). Similarly,

cos2(β − θ) Ma 2

1 − 1

Ma 21

1 − 2 (ε − θ)√Ma 2

1 − 1

. (15.27)


(a) (b)

(c)

Ma1

Ma1 Ma1

wall

shock

θ∆θ

θ

Mach lines

weak shocks

Figure 15.6Compression of supersonic flow that is turned through an angle θ. Case (a) shows asingle turn through an angle θ. Case (b) shows the turn subdivided into a number ofsmaller turns of magnitude ∆θ. Case (c) shows a smooth continuous turn through anangle θ.

Hence, to first order in θ,

w2

w1 1 − θ√

Ma 21 − 1

, (15.28)

∆w

w1 − θ√

Ma 21 − 1

, (15.29)

where ∆w = w2 − w1.

15.5 Supersonic Compression by TurningA shock front increases the pressure and density of the fluid that passes through it.In other words, the front compresses the flow. A simple method for compressing asupersonic flow is to turn it through an oblique shock, by deflecting the wall throughan angle θ, as shown in Figure 15.6(a).

The wall deflection can be subdivided into several segments, which make smaller


corners of angle ∆θ, as shown in Figure 15.6(b). Compression then occurs throughsuccessive oblique shocks. These shocks divide the flow field close to the wall intosegments of uniform flow. Further out, the shocks must intersect one another, be-cause they are convergent. However, for the present, we shall only consider the flowclose to the wall. In this region, each segment of the flow is independent of the fol-lowing segment. In other words, the flow pattern can be constructed step by step,proceeding downstream. This property of limited upstream influence exists as longas the deflection does not become so large that the flow is rendered subsonic.

To compare the compression in the two cases, (a) and (b), of Figure 15.6, we shallmake use of the approximate expressions for weak shocks obtained in the precedingsection. For each shock in case (b), we have

∆p ∼ ∆θ, (15.30)

∆S ∼ (∆θ)3. (15.31)

Thus, if the complete turn is subdivided into n equal segments, so that

θ = n∆θ, (15.32)

then the overall pressure and specific entropy changes are

pn − p1 ∼ n∆θ ∼ θ, (15.33)

Sn − S1 ∼ n (∆θ)3 ∼ θ (∆θ)2 ∼θ 3

n 2 . (15.34)

Hence, if the compression is achieved by means of n 1 weak shocks then thenet specific entropy increase can be reduced very significantly, compared to thatgenerated by a single shock that produces the same deflection. In fact, the reductionis by a factor of order 1/n 2.

In the limit that n → ∞, the smooth turn of Figure 15.6(c) is obtained. In thiscase, the net increase in specific entropy becomes vanishingly small. In other words,the compression becomes isentropic. We can also deduce the following results. First,the shocks become vanishingly weak, their limiting positions being straight Machlines emanating from the wall. Second, each segment of uniform flow becomes van-ishingly narrow, and eventually coincides with a Mach line. Thus, on each Machline, both the flow inclination and the Mach number are constant. Third, the limitedupstream influence is preserved. That is, the flow upstream of a given Mach lineis not affected by any downstream modifications in the wall. Finally, the approxi-mate expression for the change in flow speed across a weak shock, Equation (15.29),becomes the differential expression

dww= − dθ√

Ma 2 − 1. (15.35)

Let us now consider what happens to the flow further away from the wall, wherethe weak shocks and Mach lines of Figure 15.6(b) and Figure 15.6(c), respectively,


shock front

Ma2envelope ofMach lines

b

a

Ma2

wall

Ma1

Ma1

Figure 15.7Convergence of Mach lines in continuous compression by turning.

converge. Consider Figure 15.7. Because of the convergence of the Mach lines,the change from the initial Mach number, Ma1, to the final Mach number, Ma2, isachieved in a much shorter distance on streamline b than on streamline a. Thus,the velocity gradients on streamline b are higher than those on streamline a. Now,an intersection of Mach lines would imply infinitely high velocity gradients, becausethere would be two different values of the Mach number at a single point. In practice,this does not occur, because, in the region where the Mach lines converge, and beforethey cross, the velocity gradients become high enough that the conditions are nolonger isentropic. In fact, a shock wave develops, as illustrated in the figure.

15.6 Supersonic Expansion by Turning

Up to now, we have only considered turns that are concave. That is, turns in whichthe wall is deflected into the oncoming flow. Let us now consider what happens in aconvex turn, where the wall is deflected away from the oncoming flow. In particular,let us investigate supersonic flow over a convex corner, such as that illustrated inFigure 15.8(a).

A convex turn through a single oblique shock, like that illustrated in Figure 15.8(a),is not possible. In such a turn, the deviation of the flow, as it passes through shock, isinevitably away from the plane of the shock front. Such a deviation would require thedownstream normal flow speed to exceed the upstream normal flow speed (becausethe upstream and downstream tangential flow speeds are equal). (See Section 15.2.)


shock

wall

Ma2

Ma1

µ2µ1

Mach lines

Ma1

Ma2

µ1µ2

Ma1

Ma2

(a) (b)

(c)

Figure 15.8Expansion of supersonic flow via turning. Case (a) shows a single turn through anoblique shock, which is forbidden on thermodynamic grounds. Case (b) shows aPrandtl-Mayer expansion fan. Case (c) shows smooth continuous expansion.

However, although this would satisfy the equations of fluid motion, it would leadto a spontaneous decrease in specific entropy across the shock front, and is, thus,forbidden by the second law of thermodynamics. (See Section 14.8.)

What actually happens is as follows. The nonlinear mechanism that tends tosteepen a compression (see Sections 14.4 and 15.5) produces the opposite effect inan expansion. In particular, instead of being convergent, the Mach lines are divergent,as shown in Figure 15.8(b) and (c). Consequently, there is a tendency for velocitygradients to decrease with increasing distance from the wall. Thus, an expansion isisentropic throughout the fluid.

The expansion at a corner takes place via a fan of straight Mach lines that isgenerally known as a Prandtl-Mayer expansion fan. [See Figure 15.8(b).] The ar-guments that lead to this conclusion are as follows. First, the flow up to the corneris uniform, at Mach number Ma1 (say), and, thus, the leading Mach line must bestraight, and inclined to the flow at the Mach angle µ1 = sin−1(1/Ma1). Because ofthe limited upstream influence in supersonic flow, the same argument may be appliedto each succeeding portion of the flow. The terminating Mach line stands at the angleµ2 = sin−1(1/Ma2) to the wall. Here, Ma2 is the Mach number of the downstreamflow. Second, because there is no characteristic length to define a scale in this prob-lem, any variation of flow parameters can only be with respect to angular position,


measured relative to the corner. That is, flow conditions must be constant along raysemanating from the corner.

Figure 15.8(c) shows a typical expansion over a continuous convex turn. Becausethe flow is isentropic, it is also reversible (i.e., if the direction of the flow is reversedthen the fluid is isentropicaly compressed).

Equation (15.35), which specifies the relationship between θ and Ma in an isen-tropic compression or expansion by turning, can be written

−dθ =√

Ma 2 − 1dww, (15.36)

or

−θ + const. =∫ Ma √

Ma 2 − 1dww≡ ν(Ma). (15.37)

In order to evaluate the integral, and, thus, derive an explicit form for the functionν(Ma), we need to rewrite w in terms of Ma using the definition

w = c Ma. (15.38)

Here, c is the local sound speed. Given that c ∝√

T [see Equation (14.45)], it followsfrom Equation (14.58) that

c 20

c 2 = 1 +(γ − 1

2

)Ma 2, (15.39)

where c0 is the stagnation sound speed. Thus,

dww=

dMaMa+

dcc=

(1

1 + [(γ − 1)/2] Ma2

)dMaMa

. (15.40)

The function ν(Ma), which is known as the Prandtl-Mayer function, is then given by(see Exercise 15.10)

ν(Ma) ≡∫ Ma

1

√M 2 − 1

1 + [(γ − 1)/2] M 2

dMM

(15.41)

=

(γ + 1γ − 1

)1/2tan−1

[(γ − 1γ + 1

)(Ma 2 − 1)

]1/2 − tan−1([

Ma 2 − 1]1/2)

.

Here, the constant of integration has been arbitrarily chosen such that ν = 0 cor-responds to Ma = 1. A supersonic Mach number, Ma, is thus associated with adefinite value of the function ν. In fact, as Ma increases from 1 to ∞, ν increasesmonotonically from 0 to νmax, where

νmax =π

2

(γ + 1γ − 1

)1/2− 1

. (15.42)


−θ = ν(Ma) − ν(Ma1), (15.43)


θθ

Ma > 1

Ma < 1

Ma > 1

a

d

βd

θ

π/2µ

a

(a) (b)

Ma1

Ma = Ma1

Ma = 1

θ = θmax

bc

b

c

shoulder

afterbody

Figure 15.9Detached shock waves. (a) Detached shock on wedge with afterbody. [At letteredpoints on the shock, flow deflection is given by corresponding points in (b).] (b)Deflection angle versus wave angle for fixed Ma1.

as the relationship between the deflection angle, θ, and the Mach number, Ma, in anisentropic compression or expansion by turning. In a compressive turn, ν decreases,whereas, in an expansive turn, it increases: in each case, by an amount equal to thechange in θ. Here, we have set θ1 = 0, because only the change in θ matters. Notethat θ is positive in compressive turns, and negative in expansive turns.

15.7 Detached ShocksLet us return to the problem of supersonic flow, of Mach number Ma1, over a sym-metric wedge of nose angle 2 θ, that was previously discussed in Section 15.3. Whathappens when the wedge angle, θ, is greater than θmax?

In fact, there is no rigorous analytical treatment for cases where θ exceeds θmax.Experimentally, it is observed that the flow configurations are like those sketched inFigure 15.9(a). The flow is compressed through a curved shock front, known as abow shock, that is detached from, and stands some distance upstream of, the wedgeapex. The shape of the bow shock, as well as its detachment distance, depend on thegeometry of the wedge, as well as the upstream Mach number, Ma1.

On the central streamline, where the shock is normal, as well as on the nearbyones, where it is nearly normal, the flow is compressed to subsonic speeds. Fartherout, as the shock becomes weaker, its inclination becomes less steep, approachingthe upstream Mach angle asymptotically. Thus, conditions along the bow shock runthe whole range of oblique shock solutions for the given Mach number. This is


illustrated in Figure 15.9(b), which shows one curve from Figure 15.2. The completedescription of the bow shock is complicated by the fact that its location cannot befound explicitly, because of the subsonic influence upstream, which implies that theentire flow field must be solved as one. There is an absolute limit of θ, just greaterthan 45, beyond which the shock will always detach, no matter how large the valueof the upstream Mach number, Ma1. Hence, if the wedge is replaced by a blunt-nosedobstacle then the shock will always detach (because θ = 90 at the nose of such anobstacle).

For a given wedge angle, θ, the sequence of events with decreasing upstreamMach number, Ma1, is as follows. When Ma1 is sufficiently high, the shock frontis attached to the wedge apex, and its straight portion is independent of both theshoulder and the afterbody. [See Figure 15.9(a).] The straight portion of the shockfront lies between the wedge apex and the point where the first Mach line, emanatingfrom the shoulder, intersects it. As Ma1 decreases, the wave angle, β, increases. Witha further decrease in the Mach number, a value is reached at which the fluid behindthe shock becomes subsonic. The shoulder now has an effect on the whole shockfront, which may become curved, although still remaining attached to the wedgeapex. These conditions correspond to the region between the lines Ma2 = 1 andθ = θmax in Figure 15.2. At the Mach number corresponding to θ = θmax, the shockfront starts to detach from the apex. This Mach number is called the detachmentMach number. With a further decrease of Ma1, the detached shock moves upstreamof the wedge apex.

15.8 Shock-Expansion TheoryIt is possible to solve many problems in two-dimensional supersonic flow by patch-ing together appropriate combinations of the oblique shock wave, described in Sec-tion 15.2, and the Prandtl-Mayer expansion fan, described in Section 15.6. For ex-ample, let us consider the flow over a simple two-dimensional airfoil section.

Figure 15.10 shows a flat plate inclined at an angle of attack α0 to the oncomingsupersonic flow. The streamline ahead of the leading edge is non-inclined, becausethere is no upstream influence. Moreover, the flow streams over the upper and lowersurfaces are completely independent of one another. Thus, the flow on the uppersurface is turned through an expansion angle α0 by means of a Prandtl-Mayer expan-sion fan attached to the leading edge of the airfoil, whereas the flow on the lower sideis turned through a compression angle α0 by means of an oblique shock. The flowon the upper surface is recompressed to the upstream pressure p1 by means of anoblique shock wave attached to the trailing edge of the airfoil. Likewise, the flow onthe lower surface is re-expanded to the upstream pressure by means of an expansionfan. The uniform pressures, p2 and p′2, respectively, on the upper and lower surfacesof the airfoil can easily be calculated by means of oblique shock theory and Prandtl-Mayer expansion theory. Given the pressures, the lift and drag per unit transverse


α0

p2p1

p′2

shock

shock expansion fan

expansion fan

p1

slipstream

plateMa1

c

Figure 15.10A flat lifting plate. Ma1 is the upstream Mach number, p1, p2, et cetera, denotepressures, and α0 is the angle of attack.

length of the airfoil are simply

L = (p′2 − p2) c cosα0, (15.44)

D = (p′2 − p2) c sinα0, (15.45)

respectively, where c is the chord-length (i.e., width). (See Figure 15.10.) The in-crease in entropy of the flow along the upper surface of the airfoil is not the same asthat for the flow along the lower surface, because the upper and lower shock wavesoccur at different Mach numbers. Consequently, the streamline attached to the trail-ing edge of the airfoil is a slipstream—that is, it separates flows with the same pres-sures, but slightly different speeds, temperatures, and densities—inclined at a smallangle relative to the free stream. (The angle of inclination is determined by the re-quirement that the pressures on both sides of the slipstream be equal to one another.)

Note that the drag that develops on the airfoil is of a completely different natureto the previously discussed (see Chapter 9) drags that develop on subsonic airfoils,such as friction drag, form drag, and induced drag. This new type of drag is termedsupersonic wave drag, and exists even in an idealized, inviscid fluid. It is ultimatelydue to the trailing shock waves attached to the airfoil.

Comparatively far from the airfoil, the attached shock waves and expansion fansintersect one another. The expansion fans then attenuate the oblique shocks, mak-ing them weak and curved. At very large distances, the shock waves asymptote tofree-stream Mach lines. However, this phenomenon does not affect the previous cal-culation of the lift and drag on a flat-plate airfoil.


15.9 Thin-Airfoil TheoryThe shock-expansion theory of the previous section provides a simple and generalmethod for computing the lift and drag on a supersonic airfoil, and is applicableas long as the flow is not compressed to subsonic speeds, and the shock waves re-main attached to the airfoil. However, the results of this theory cannot generally beexpressed in concise analytic form. In fact, the theory is mostly used to obtain nu-merical solutions. However, if the airfoil is thin, and the angle of attach small, thenthe shocks and expansion fans attached to the airfoil become weak. In this situa-tion, shock-expansion theory can be considerably simplified by using approximateexpressions for weak shocks and expansion fans.

The basic approximate expression [cf. Equation (15.19)]

∆pp γMa 2√

Ma 2 − 1

∆θ, (15.46)

specifies the relative change in pressure across either a weak oblique shock (see Sec-tion 15.4) or a weak expansion fan (see Exercise 15.6) that deflects flow of Machnumber Ma through an angle ∆θ. Because, in the weak wave approximation, thepressure, p, never greatly differs from the upstream pressure, p1, and the Mach num-ber, Ma, never differs appreciably from the upstream Mach number, Ma1, we canwrite

p − p1

p1

γMa 21√

Ma 21 − 1

θ, (15.47)

which is correct to first order in θ. Here, θ is the deflection angle relative to theupstream flow.

It is convenient to define a dimensionless quantity known as the pressure coeffi-cient:

Cp =p − p1

q1, (15.48)

where q1 = (1/2) ρ1 w21 , and ρ1 and w1 are the upstream density and flow speed,

respectively. Given that the upstream sound speed is c1 =√γ p1/ρ1, and Ma1 =

w1/c1, we obtain

Cp =2

γMa 21

p − p1

p1, (15.49)

which yields

Cp =2 θ√

Ma 21 − 1

. (15.50)

This is the fundamental formula of thin-airfoil theory. It states that the pressure coef-ficient is proportional to the local deflection of the flow from the upstream direction.

Consider the flat-plate airfoil shown in Figure 15.10. The deflection angle is −α0


on the upper surface of the airfoil, and +α0 on the lower surface. (A positive deflec-tion angle corresponds to compression, and a negative deflection angle to expansion.)Thus, the pressure coefficients on the upper and lower surfaces are

CpU = −2α0√

Ma 21 − 1

(15.51)

and

CpL =2α0√

Ma 21 − 1

, (15.52)

respectively. It is convenient to define the dimensionless coefficient of lift,

CL =L

q1 c, (15.53)

and the dimensionless coefficient of drag,

CD =D

q1 c. (15.54)

It follows that

CL =(pL − pU) c cosα0

q1 c= (CpL −CpU ) cosα0, (15.55)

CD =(pL − pU) c sinα0

q1 c= (CpL − CpU) sinα0. (15.56)

Making use of Equations (15.51) and (15.52), as well as the conventional small angleapproximations cosα0 1 and sinα0 α0, we obtain

CL =4α0√

Ma 21 − 1

, (15.57)

CD =4α 2

0√Ma 2

1 − 1. (15.58)

The focus (see Section 9.3) of the airfoil is at the midchord. Moreover, the ratioD/L 2 = (Ma 2

1 − 1)1/2/4 is independent of α0.As a second example, consider the diamond-section airfoil pictured in Figure 15.11.

This airfoil has a nose angle 2 ε, and zero angle of attack. The pressure coefficienton the front face of the airfoil is

CpF =2 ε√

Ma 21 − 1

, (15.59)


2 ε2 εMa1

p3

p3p2

p2

shock expansion fan

c

t

Figure 15.11A diamond-section airfoil. Ma1 is the upstream Mach number. p2 and p3 denotepressures.

whereas that on the rear face is

CpR = −2 ε√

Ma 21 − 1

. (15.60)

It follows that the pressure difference is

p2 − p3 =4 ε√

Ma 21 − 1

q1, (15.61)

giving a drag

D = (p2 − p3) t = (p2 − p3) ε c =4 ε 2√

Ma 21 − 1

q1 c, (15.62)

where t and c are the thickness and chord-length of the airfoil, respectively. (SeeFigure 15.11.) Thus,

CD =D

q1 c=

4 ε 2√Ma 2

1 − 1=

4√Ma 2

1 − 1

( tc

)2. (15.63)

Figure 15.12 shows the cross-section of an arbitrary airfoil. The cross-section


xR

yU(x)

y

x

yC(x)

yL(x)

xL

Figure 15.12An arbitrary airfoil.

is assumed to be uniform in the z-direction, with the upstream flow parallel to thex-axis. The upper surface of the airfoil corresponds to the curve y = yU(x), the lowersurface to the curve y = yL(x), and the camber line (i.e., the centerline) to the curvey = yC(x). Furthermore, the leading and trailing edges of the airfoil lie at x = xL

and x = xR, respectively. Hence, yU(xL) = yC(xL) = yL(xL) and yU(xR) = yC(xR) =yL(xR). By definition,

yC(x) =yU(x) + yL(x)

2, (15.64)

for xL ≤ x ≤ xR. It is helpful to define the half-width of the airfoil,

h(x) =yU(x) − yL(x)

2, (15.65)

for xL ≤ x ≤ xR. Note that h(xL) = h(xR) = 0. We can also define the mean angle ofattack of the airfoil:

α0 =yU(xL) − yU(xR)

xR − xL=yL(xL) − yL(xR)

xR − xL=yC(xL) − yC(xR)

xR − xL. (15.66)

Thus, we can write

yC(x) = yC(xL) − α0 (x − xL) − αC(x), (15.67)

where αC(xL) = αC(xR) = 0. Here,

α(x) = α0 + αC(x) (15.68)


is the local angle of attack of the camber line. Thus, the camber function, αC(x),parameterizes the deviations of α(x) from α0 across the width of the airfoil. Equa-tions (15.64), (15.65), and (15.67) yield

yU(x) = yC(x) + h(x) = yC(xL) − α0 (x − xL) − αC(x) + h(x), (15.69)

yL(x) = yC(x) − h(x) = yC(xL) − α0 (x − xL) − αC(x) − h(x). (15.70)

Hence, the airfoil shape is completely specified by the thickness function, h(x), thecamber function, αC(x), and the mean angle of attack, α0.

The pressure coefficients on the upper and lower surfaces of the airfoil are [seeEquations (15.50)]

CpU =2√

Ma 21 − 1

dyU

dx, (15.71)

CpL =2√

Ma 21 − 1

(−dyL

dx

), (15.72)

respectively. It follows from Equations (15.68)–(15.70) that

dyU

dx= −α(x) +

dhdx, (15.73)

dyL

dx= −α(x) − dh

dx. (15.74)

Now, the lift and drag per unit transverse length acting on the airfoil are given by[cf., Equations (15.53)–(15.56)]

L = q1

∫ xR

xL

(CpL −CpU

)dx, (15.75)

D = q1

∫ xR

xL

[CpL

(−dyL

dx

)+CpU

(dyU

dy

)]dx. (15.76)

Thus, it follows from Equations (15.71)–(15.74) that

D =4 q1√

Ma 21 − 1

∫ xR

xL

α(x) dx, (15.77)

L =2 q1√

Ma 21 − 1

∫ xR

xL

(dyL

dx

)2+

(dyU

dx

)2 dx

=4 q1√

Ma 21 − 1

∫ xR

xL

(dhdx

)2+ α 2

dx. (15.78)


It is helpful to define the chord-average operator:

y ≡ 1c

∫ xR

xL

y(x) dx, (15.79)

where c = xR − xL is the chord-length. Taking the average of Equation (15.73),making use of Equation (15.66), as well as the fact that h(xL) = h(xR) = 0, we obtain

α = α0. (15.80)

However, the average of Equation (15.68) yields

α = α0 + αC , (15.81)

which implies that αC = 0. We can also write

α 2 = (α0 + αC)2 = α 20 + 2α0 αC + α

2C = α

20 + α

2C . (15.82)

Hence, the coefficients of lift and drag, L/(q1 c) and D/(q1 c), are written

CL =4α√

Ma 21 − 1

=4α0√

Ma 21 − 1

, (15.83)

CD =4√

Ma 21 − 1

(

dhdx

)2+ α 2

=

4√Ma 2

1 − 1

(

dhdx

)2+ α 2

0 + α2C

, (15.84)

respectively. Thus, in thin-airfoil theory, the lift only depends on the mean angleof attack, whereas the drag splits into three components. Namely, a drag due tothickness, a drag due to lift, and a drag due to camber.

15.10 Crocco’s TheoremBernoulli’s theorem for the steady, inviscid flow of an ideal gas, in the absence ofbody forces, implies that, on a given streamline,

H + 12v 2 = H0, (15.85)

whereH is the specific enthalpy,H0 the stagnation enthalpy, and v the flow velocity.(See Section 14.5.) The fluid equation of motion, (14.31), reduces to

(v · ∇) v = −∇pρ. (15.86)


However, according to Equation (A.171),

(v · ∇) v ≡ ∇(v 2/2) − v × ω, (15.87)

where ω = ∇ × v. Hence, we obtain

v ×ω = ∇(v 2/2) +∇pρ. (15.88)

Now, Equation (14.29) implies that

∇H = ∇pρ+ T ∇S, (15.89)

where T is the temperature, and S the specific entropy. Moreover, Equation (15.85)yields

∇H + ∇(v 2/2) = ∇H0. (15.90)

Thus, we arrive at Crocco’s theorem:

v ×ω = ∇H0 − T ∇S. (15.91)

In most aerodynamic flows, the fluid originates from a common reservoir, whichimplies that the stagnation enthalpy,H0, is the same on all streamlines. Such flow istermed homenergic. It follows from Equation (14.18) that the stagnation temperature,T0, is the same on all streamlines in homenergic flow. According to Crocco’s theo-rem, an irrotational (i.e., ω = 0 everywhere) homenergic (i.e., ∇H0 = 0 everywhere)flow pattern is also homentropic (i.e., ∇S = 0 everywhere). Conversely, a homener-gic, homentropic flow pattern is also irrotational (at least, in two dimensions, wherev and ω cannot be parallel to one another).

15.11 Homenergic Homentropic FlowConsider a steady, two-dimensional, homenergic, homentropic flow pattern, in theabsence of body forces. Suppose that all quantities are independent of the Cartesiancoordinate z, and that the flow velocity, q, is confined to the x-y plane. Let q =u ex + v ey. Equations (14.25), (14.30), and (14.31) reduce to

∂

∂x(ρ u) +

∂

∂y(ρ v) = 0, (15.92)

u∂u∂x+ v

∂u∂y= −1

ρ

∂p∂x, (15.93)

u∂v

∂x+ v

∂v

∂y= −1

ρ

∂p∂y, (15.94)

pp0=

(ρ

ρ0

)γ, (15.95)


where p0 and ρ0 are the uniform stagnation pressure and density, respectively. (Notethat p0 and ρ0 must be uniform because the stagnation specific entropy, S0 ∝ p0/ρ

γ0 ,

and the stagnation temperature, T0 ∝ p0/ρ0, are both uniform.) Equation (15.95)implies that

∇p = c 2 ∇ρ, (15.96)

where c =√γ p/ρ is the sound speed. Hence, Equations (15.92)–(15.94) yield

uρ

∂ρ

∂x+v

ρ

∂ρ

∂y= −∂u

∂x− ∂v∂y, (15.97)

u 2 ∂u∂x+ u v

∂u∂y= −c 2

ρu∂ρ

∂x, (15.98)

u v∂v

∂x+ v 2 ∂v

∂y= −

c 2

ρv∂ρ

∂y. (15.99)

Summing the previous two equations, and then making use of Equation (15.97), weobtain (

u 2 − c 2) ∂u∂x+(v 2 − c 2

) ∂v∂y+ u v

(∂u∂y+∂v

∂x

)= 0. (15.100)

Finally, given that c ∝√

T , Equation (14.58) implies that

c 2 = c 20 −

12

(γ − 1)(u 2 + v 2

), (15.101)

where c0 is the stagnation sound speed.

15.12 Small-Perturbation TheoryA great number of problems of interest in compressible fluid mechanics are con-cerned with the perturbation of a known flow pattern. The most common case is thatof uniform, steady flow. Let U denote the uniform flow velocity, which is directedparallel to the x-axis. The density, pressure, and temperature are also assumed to beuniform, and are denoted ρ∞, p∞, and T∞, respectively. The corresponding soundspeed is c∞, and the Mach number is Ma∞ = U/c∞. Finally. the velocity field of theunperturbed flow pattern is

u = U, (15.102)

v = 0. (15.103)

Suppose that a solid body, such as an airfoil, is placed in the aforementioned flowpattern. The cross-section of the body is assumed to be independent of the Cartesian


coordinate z. The body disturbs the flow pattern, and changes its velocity field, whichis now written

u = U + u′, (15.104)

v = v′, (15.105)

where u′ and v′ are known as induced velocity components. We are interested insituations in which u′/U 1 and v′/U 1.

Equation (15.101) can be combined with the previous two equations to give

c 2 = c 2∞ −

12

(γ − 1)(2 U u′ + u′ 2 + v′ 2

). (15.106)

It then follows from Equation (15.100) that

(1 −Ma 2∞)∂u′

∂x+∂v′

∂y= Ma 2

∞

(γ + 1)u′

U+

(γ + 1

2

) (u′

U

) 2

+

(γ − 1

2

) (v′

U

)2 ∂u′

∂x

+Ma 2∞

(γ − 1)u′

U+

(γ + 1

2

) (v′

U

) 2

+

(γ − 1

2

) (u′

U

)2 ∂v′∂y+Ma 2

∞v′

U

(1 +

u′

U

) (∂u′

∂y+∂v′

∂x

). (15.107)

The previous equation is exact. However, if u′/U and v′/V are small then it becomespossible to neglect many of the terms on the right-hand side. For instance, neglectingterms that are third-order in small quantities, we obtain

(1 −Ma 2∞)∂u′

∂x+∂v′

∂y Ma 2

∞ (γ + 1)u′

U∂u′

∂x+Ma 2

∞ (γ − 1)u′

U∂v′

∂y

+Ma 2∞v′

U

(∂u′

∂y+∂v′

∂x

). (15.108)

Furthermore, if we neglect terms that are second-order in small quantities then weget the linear equation

(1 −Ma 2∞)∂u′

∂x+∂v′

∂y 0. (15.109)

Note, however, than in so-called transonic flow, where Ma∞ 1, the coefficient of∂u′/∂x on the left-hand side of Equation (15.108) becomes very small. In this situ-ation, it is not possible to neglect the first term on the right-hand side. However, thecondition Ma∞ 1 does not affect the term ∂v′/∂y on the left-hand side of Equa-tion (15.108), and so the other terms on the right-hand side can still be neglected.Thus, transonic flow is governed by the non-linear equation

(1 −Ma 2∞)∂u′

∂x+∂v′

∂y Ma 2

∞ (γ + 1)u′

U∂u′

∂x. (15.110)


On the other hand, subsonic (i.e., Ma∞ < 1) and supersonic flow (i.e., Ma∞ > 1)are both governed by Equation (15.109). Another situation in which certain termson the right-hand side of Equation (15.108) must be retained is hypersonic flow (i.e.,Ma∞ 1). This follows because, although u′/U and v′/V are small, their productswith Ma 2

∞ can still be non-negligible. Roughly speaking, Equation (15.109) is validfor 0 ≤ Ma∞ ≤ 0.8 and 1.2 ≤ Ma∞ ≤ 5. In other words, transonic flow correspondsto 0.8 < Ma∞ < 1.2, and hypersonic flow to Ma∞ > 5 (Anderson 2003).

The pressure coefficient is defined

Cp =p − p∞

(1/2) ρ∞U 2 =2

γMa 2∞

p − p∞p∞

. (15.111)

(See Section 15.9.) Equation (15.101) implies that

c 2 +12

(γ − 1) q 2 = c 2∞ +

12

(γ − 1) U 2, (15.112)

where q 2 = (U + u′) 2 + v′ 2. Moreover, Equation (14.61) yields

pp∞=

[2 + (γ − 1) Ma 2

∞

2 + (γ − 1) Ma2

] γ/(γ−1)

, (15.113)

where Ma = q/c. The previous two equations can be combined to give

pp∞=

[1 +

12

(γ − 1) Ma2∞

(1 − q 2

U 2

)] γ/(γ−1)

. (15.114)

Hence, we obtain

Cp =2

γMa 2∞

[1 + 12

(γ − 1) Ma2∞

(1 − q 2

U 2

)] γ/(γ−1)

− 1

, (15.115)

which reduces to

Cp =2

γMa 2∞

[1 − 12

(γ − 1) Ma 2∞

(2 u′

U+

u′ 2 + v′ 2

U 2

)] γ/(γ−1)

− 1

. (15.116)

Using the binomial expansion on the expression in square brackets, and neglectingterms that are third-order, or higher, in small quantities, we obtain

Cp −2 u′

U+ (1 −Ma 2

∞)(u′

U

) 2

+

(v′

U

)2 . (15.117)

For two-dimensional flows, in the limit in which Equation (15.109) is valid, it isconsistent to retain only first-order terms in the previous equation, so that

Cp −2 u′

U. (15.118)


Letf (x, y) = 0 (15.119)

be the equation of the surface of the solid body that perturbs the flow. At the surface,the velocity vector of the flow must be perpendicular to the local normal: that is, theflow must be tangential to the surface. In other words,

q · ∇ f = 0, (15.120)

which reduces to(U + u′)

∂ f∂x+ v′

∂ f∂y= 0. (15.121)

Neglecting u′ with respect to U, we obtain

v′

U −

∂ f /∂x∂ f /∂y

=dydx, (15.122)

where dy/dx is the slope of the surface, and v′/U the approximate slope of a stream-line.

Now, the body has to be thin in order to satisfy our assumption that the inducedvelocities are relatively small. This implies that the coordinate y differs little fromzero (say) on the surface of the body. Hence, we can write

v′(x, y) = v′(x, 0) +(∂v′

∂y

)y=0y + · · · (15.123)

in the immediate vicinity of the surface. Within the framework of small-perturbationtheory, it is consistent to neglect all terms on the right-hand side of the previousequation after the first. Hence, the boundary condition (15.122) reduces to

v′(x, 0) = U(

dydx

)body

, (15.124)

respectively.Because a homenergic, homentropic flow pattern is necessarily irrotational (see

Section 15.10), we can writeq = U ex − ∇φ, (15.125)

where φ is the perturbed velocity potential. (See Section 4.15.) It follows that

u′ = −∂φ∂x, v′ = −∂φ

∂y. (15.126)

Hence, Equations (15.109), (15.118), and (15.124) become

(1 −Ma 2∞)∂ 2φ

∂x 2 +∂ 2φ

∂y 2 0, (15.127)

Cp 2U∂φ

∂x, (15.128)

∂φ(x, 0)∂y

−U(

dydx

)body

, (15.129)

respectively.


15.13 Subsonic Flow Past a Wave-Shaped WallThe following simple example serves to clarify many of the concepts introducedin the previous section. Consider flow past a straight wall with a small-amplitudesinusoidal modulation whose surface is specified by

y = ε sin(α x). (15.130)

Here, ε is the amplitude of the modulation, whereas l = 2π/α is the wavelength. Inthe limit ε → 0, the unperturbed flow is of uniform speed U, directed parallel tothe x-axis. The corresponding Mach number is Ma∞. The fluid occupies the regiony > ε sin(α x).

The perturbed flow is governed by Equation (15.127),

(1 −Ma 2∞)∂ 2φ

∂x 2 +∂ 2φ

∂y 2 0, (15.131)

and is subject to the boundary condition (15.129), which can be written

∂φ(x, 0)∂y

= −Udydx= −U ε α cos(α x). (15.132)

We also expect the perturbed flow to become vanishingly small far from the wall,which implies that

∂φ(x,∞)∂x

=∂φ(x,∞)∂y

= 0. (15.133)

Consider subsonic flow, for which 1 −Ma 2∞ ≡ m 2 > 0. Equation (15.131) is of

the elliptic type∂ 2φ

∂x 2 +1

m 2

∂ 2φ

∂y 2 = 0 (15.134)

(Arfken 1985). Let us search for a separable solution of the form

φ(x, y) = F(x) G(y). (15.135)

We obtainF′′

F= − 1

m 2

G′′

G, (15.136)

where ′ denotes derivative with respect to argument. Given that the left-hand side ofthe previous equation is a function of x only, whereas the right-hand side is a functionof y only, the two sides must equal the same constant. In other words,

F′′

F= −k 2, (15.137)

G′′

G= +k 2. (15.138)


Here, the sign of the constant has been chosen so as to allow the boundary condition(15.133) to be satisfied. The most general solution of Equation (15.137) is

F(x) = A1 sin(k x) + A2 cos(k x), (15.139)

where A1 and A2 are arbitrary constants. Likewise, the most general solution ofEquation (15.138) is

G(y) = B1 e−m k y + B2 e+m k y, (15.140)

where B1 and B2 are arbitrary constants.The boundary condition (15.133) implies that B2 = 0. The boundary condi-

tion (15.132) then yields

−m k B1 [A1 sin(k x) + A2 cos(k x)] = −U ε α cos(α x). (15.141)

This condition can only be satisfied at all x if A1 = 0, k = α, and m B1 A2 = U ε.Thus, we obtain

φ(x, y) =U εm

cos(α x) e−mα y, (15.142)

u′ = −∂φ∂x=

U ε αm

sin(α x) e−mα y, (15.143)

v′ = −∂φ∂y= U ε α cos(α x) e−mα y. (15.144)

According to Equation (15.128), the pressure coefficient at the wall can be written

Cp 2U∂φ

∂x

∣∣∣∣∣y=0= − 2 ε α√

1 −Ma 2∞

sin(α x). (15.145)

It immediately follows that there is zero net drag force acting on the wall, because thepressure variations are in phase with the wall’s sinusoidal modulations, and, hence,symmetrical about the crests and troughs.

Let us now consider under what circumstances the approximations made in thesmall-perturbation theory developed in the previous section are valid for the problemunder investigation. In fact, small-perturbation theory is premised on three assump-tions. First, that u′/U 1 and v′/V 1. It is evident from Equations (15.143) and(15.144) that this assumption is valid provided

ε α√1 −Ma 2

∞

1. (15.146)

Second, in Equation (15.110) it is assumed that

(1 −Ma 2∞)∂u′

∂x Ma 2

∞ (γ + 1)u′

U∂u′

∂x, (15.147)


which implies thatMa 2∞ (γ + 1) ε α

(1 −Ma 2∞) 3/2

1. (15.148)

Finally, in Equation (15.123), it is assumed that the second term on the right-handside is negligible compared to the first, when evaluated at the surface of the wall,which implies that

ε α

√1 −Ma 2

∞ 1. (15.149)

Hence, we deduce that the analysis described in this section is valid provided

ε α√

1 −Ma 2∞,

(1 −Ma 2∞) 3/2

Ma 2∞ (γ + 1)

,1√

1 −Ma 2∞

. (15.150)

15.14 Supersonic Flow Past a Wave-Shaped WallSuppose that the unperturbed flow in the problem considered in the previous sec-tion is supersonic, so that Ma 2

∞ − 1 ≡ λ 2 > 0. The perturbed flow is governed byEquation (15.127), which is now of the hyperbolic type

∂ 2φ

∂x 2 −1λ 2

∂ 2φ

∂y 2 0 (15.151)

(Arfken 1985). As is easily verified, the previous equation has the general solution

φ(x, y) = f (x − λ y) + g(x + λ y), (15.152)

where f and g are arbitrary functions (Fitzpatrick 2013). It is clear that f is constantalong lines x−λ y = const., whereas g is constant along lines x+λ y = const.. Theselines are inclined at the Mach angle, µ = cot−1[(Ma 2

∞ − 1) 1/2], to the undisturbedflow. They are, in fact, the Mach lines, or characteristics, of the unperturbed flow.(See Section 15.3.) The former characteristics are inclined downstream. In otherwords, they originate at the wall. The latter characteristics are inclined upstream. Inother words, they originate at infinity. Because there are no sources at infinity, thelatter characteristics carry no perturbation, which implies that g = 0.

The boundary condition at the wall, (15.129), yields

−λ f ′(x) = −U ε α cos(α x), (15.153)

where ′ denotes derivative with respect to argument. It follows that

f (x) =U ελ

sin(α x), (15.154)


and, hence, that

φ(x, y) = f (x − λ y) = U ελ

sin[α (x − λ y)], (15.155)

u′ = −∂φ∂x= −U ε α

λcos[α (x − λ y)], (15.156)

v′ = −∂φ∂y= U ε α cos[α (x − λ y)]. (15.157)

Note that, unlike the case of subsonic flow, the amplitude of the perturbed velocitydoes not decrease with increasing distance from the wall.

The pressure coefficient at the wall is

Cp 2U∂φ

∂x

∣∣∣∣∣y=0=

2 ε α√Ma 2∞ − 1

cos(α x). (15.158)

It can be seen, by comparison with Equation (15.145), that the maxima and minimaof the pressure are now phase-shifted by π/2 radians with respect to the correspond-ing maxima and minima in the subsonic case. Hence, the pressure distribution atthe wall is anti-symmetric with respect to the crests and troughs of the wall. Con-sequently, a net drag force is exerted on the wall. The mean coefficient of drag perwavelength is [see Equation (15.56)]

CD =1l

∫ l

0Cp

dydx

dx, (15.159)

which is obtained by replacing the sine of the slope of the wall by the tangent, dy/dx.This approximation is valid within the context of the small-perturbation theory. How-ever, according to Equations (15.130) and (15.158), Cp can be written

Cp =2√

Ma 2∞ − 1

dydx, (15.160)

which implies that

CD =2√

Ma 2∞ − 1

(dydx

)2. (15.161)

Here, the bar denotes a period average of the form(dydx

)2=

1l

∫ l

0

(dydx

)2dx. (15.162)

Equations (15.160) and (15.161) are valid for any small-amplitude, periodic modula-tion of the wall. For the particular sinusoidal modulation under consideration in thissection, we find that

CD =(ε α) 2√Ma 2∞ − 1

. (15.163)


The discussion of the range of validity of the approximations used in deriving theprevious results follows the same lines as in the subsonic case.

15.15 Linearized Subsonic FlowThe aim of this section is to modify the two-dimensional, incompressible, subsonicaerodynamic theory discussed in Chapter 9 so as to take the finite compressibility ofair into account.

Consider compressible, subsonic flow over a thin airfoil at a small angle of attack.Such a situation can be analyzed using the small-perturbation theory introduced inSection 15.12. As before, the unperturbed flow is of uniform speed U, directedparallel to the x-axis, and the associated Mach number is Ma∞. The perturbed flow,due to the presence of the airfoil, is governed by Equation (15.127), which can bewritten in the elliptic form (see Section 15.13)

∂ 2φ

∂x 2 +1

m 2

∂ 2φ

∂y 2 0, (15.164)

wherem =

√1 −Ma 2

∞. (15.165)

Equation (15.164) can be transformed into a familiar incompressible form byintroducing a transformed coordinate system (ξ, η), such that

ξ = x, (15.166)

η = m y. (15.167)

In this transformed space, we define a transformed perturbed velocity potential,φ(ξ, η), such that

φ(ξ, η) = m φ(x, y). (15.168)

Now,

∂ξ

∂x= 1,

∂ξ

∂y= 0,

∂η

∂x= 0,

∂η

∂y= m, (15.169)

so

∂

∂x=∂ξ

∂x∂

∂ξ+∂η

∂x∂

∂η=∂

∂ξ, (15.170)

∂

∂y=∂ξ

∂y

∂

∂ξ+∂η

∂y

∂

∂η= m

∂

∂η. (15.171)

It follows that

∂φ

∂x=

1m∂φ

∂ξ,

∂φ

∂y=∂φ

∂η,

∂ 2φ

∂x 2 =1m∂ 2φ

∂ξ 2 ,∂ 2φ

∂y 2 = m∂ 2φ

∂η 2 . (15.172)


Thus, Equation (15.164) transforms to give

∂ 2φ

∂x 2 +∂ 2φ

∂y 2 0. (15.173)

However, this is simply Laplace’s equation, which governs incompressible flow. (SeeSection 4.15.) Hence, φ(ξ, η) represents an incompressible flow in (ξ, η) space thatis related to a compressible flow, represented by φ(x, y), in (x, y) space.

Suppose that the surface of the airfoil is given by y = f (x) in (x, y) space, andη = q(ξ) in (ξ, η) space. According to Equation (15.129), the boundary condition onthe flow at the airfoil surface is

Ud fdx −∂φ(x, 0)

∂y(15.174)

in (x, y) space. The analogous boundary condition in (ξ, η) space is

Udqdξ −∂φ(ξ, 0)

∂η. (15.175)

However,∂φ

∂y=∂φ

∂η, (15.176)

which implies thatd fdx=

dqdξ. (15.177)

In other words, the shape of the airfoil in (x, y) space is the same as that in (ξ, η)space. Hence, the coordinate transformation (x, y)→ (ξ, η) transforms compressibleflow over an airfoil in (x, y) space to incompressible flow over the same airfoil in(ξ, η) space.

According to Equation (15.128), the pressure coefficient in (x, y) space is written

Cp 2U∂φ

∂x=

2U m

∂φ

∂ξ. (15.178)

The corresponding incompressible pressure coefficient in (ξ, η) space is

Cp0 2U∂φ

∂ξ. (15.179)

A comparison of the previous two equations reveals that

Cp =Cp0√

1 −Ma 2∞

. (15.180)

This result is known as the Prandtl-Glauert rule: it is a similarity rule that relatesincompressible flow over a given two-dimensional airfoil to subsonic compressible


flow over the same airfoil. Given that the coefficient of lift is directly related to thepressure coefficient (see Section 15.9), we can also deduce that

CL =CL0√

1 −Ma 2∞

, (15.181)

where CL is the coefficient of lift for compressible flow, and CL0 the correspondingcoefficient for incompressible flow. It is clear that compressibility has the effect ofincreasing the lift on a given airfoil.

Equation (15.181) indicates that the coefficient of lift goes to infinity as Ma∞ ap-proaches unity, which is an impossible result. The quandary is resolved by recallingthat linearized theory breaks down in the transonic regime (where Ma∞ 1). In fact,the Prandlt-Glauert rule is only accurate up to Mach numbers of approximately 0.8.

The effect of compressibility on subsonic flow-fields can be seen by noting that

u′ = −∂φ∂x= − 1

m∂φ

∂ξ=

u√1 −Ma 2

∞

, (15.182)

where u = −∂φ/∂ξ is the incompressible velocity perturbation parallel to the unper-turbed flow. It can be seen that as Ma∞ increases, the parallel velocity perturbation,u′, also increases relative to u. In other words, compressibility strengthens the dis-turbance of the flow introduced by the airfoil.

In conventional inviscid, incompressible aerodynamic theory, a two-dimensionalairfoil experiences zero aerodynamic drag (assuming that there is no separation ofthe boundary layer). (See Chapters 8 and 9.) This is due to the fact that, in theabsence of friction, the pressure distributions over the forward and rearward portionsof the airfoil exactly cancel in the direction of the unperturbed flow. Note that thecompressible pressure coefficient, Cp, only differs from the incompressible pressurecoefficient, Cp0 , by a constant scale-factor. Thus, if the distribution of Cp0 over thesurface of the airfoil results in zero drag then the distribution of Cp will also resultin zero drag. In other words, in subsonic flow, compressibility does not give rise toadditional aerodynamic drag.

15.16 Linearized Supersonic FlowThe aim of this section is to re-examine the problem of supersonic flow past athin, two-dimensional airfoil using the small-perturbation theory developed in Sec-tion 15.12.

As before, the unperturbed flow is of uniform speed U, directed parallel to thex-axis, and the associated Mach number is Ma∞. The perturbed flow, due to thepresence of the airfoil, is governed by Equation (15.127), which can be written in the


hyperbolic form (see Section 15.14)

∂ 2φ

∂x 2 −1λ 2

∂ 2φ

∂y 2 0, (15.183)

whereλ =

√Ma 2∞ − 1. (15.184)

As we saw in Section 15.14, the general solution of Equation (15.183) is

φ(x.y) = f (x − λ y) + g(x + λ x), (15.185)

where f and g are arbitrary functions. Because disturbances only propagate alongdownstream-running Mach lines (i.e., Mach lines that originate at the airfoil), weonly need the function f for the upper surface, and the function g for the lowersurface. Thus,

φ(x, y) = f (x − λ y) y > 0, (15.186)

φ(x, t) = g(x + λ y) y < 0. (15.187)

On the upper surface, the boundary condition (15.129) yields

U(

dydx

)U= −∂φ(x, 0+)

∂x= λ f ′(x), (15.188)

which implies that

f ′(x) =Uλ

(dydx

)U. (15.189)

Similarly,

g′(x) = −Uλ

(dydx

)L. (15.190)

According to Equation (15.128), the pressure coefficient on the surface of the airfoilis given by

Cp 2U

(∂φ

∂x

)y=0. (15.191)

Hence, we obtain

Cp =2U

f ′(x) (15.192)

on the upper surface, and

Cp =2Ug′(x) (15.193)

on the lower surface. Thus,

CpU =2√

Ma 2∞ − 1

(dyU

dx

), (15.194)

CpL =2√

Ma 2∞ − 1

(−dyL

dx

). (15.195)


This is the same result as that obtained in Section 15.9, where the pressure on thinairfoils was obtained by an approximation to the shock-expansion method. In fact,for the purposes of calculating the velocity and pressure perturbations on the surfaceof the airfoil, the linearized theory discussed in this section is equivalent to the weakwave approximations of Section 15.9.

15.17 Exercises

15.1 Show that Equation (15.12) can be written in the form

X 3 + b X 2 + c X + d = 0, (15.196)

where

X = sin2 β,

b = −Ma 2

1 + 2

Ma 21

+ γ sin2 θ

,c =

2 Ma21 + 1

Ma 41

+

(γ + 12

)2+γ − 1Ma 2

1

sin2 θ,

d = −cos2 θ

Ma 41

.

Let

Q =13

c − 19

b 2,

R =16

(b c − 3 d) −127

b 3,

D = Q 3 + R 2,

Demonstrate that the oblique shock solution only exists for D < 0 [i.e., whenEquation (15.196) possesses three real roots.] Show that the strong shock so-lution, βs, and the weak shock solution, βw, are given by

βs = tan−1(√

χs

1 − χs

),

βw = tan−1(√

χw1 − χw

),


where

χs = −13

b + 2√−Q cosφ,

χw = −13

b −√−Q

(cosφ −

√3 sin φ

),

φ =13

tan−1 √−D

R

+ ∆ .Here, ∆ = 0 if R ≥ 0, and ∆ = π if R < 0.

15.2 Assuming that information propagates with respect to a two-dimensional su-personic flow pattern at the local sound speed, show that, in order for the flowat some point P to affect the flow at some other point Q, the latter point mustlie between the (+) and (−) characteristics that pass through P.

15.3 Show that for a weak oblique shock with β µ,

∆ρ

ρ1

1γ

∆pp1,

∆TT1(γ − 1γ

)∆pp1,

∆SR

(γ + 1)12 γ 2

(∆pp1

)3,

where ∆ρ = ρ2 − ρ1, et cetera, and

∆pp1

γMa 21√

Ma 21 − 1

θ.Here, β is the wave angle, µ the Mach angle, θ 1 the deflection angle, γ theratio of specific heats, R the specific gas constant, and Ma1 the upstream Machnumber. Furthermore, p1, ρ1, T1, and S1 are the upstream pressure, density,temperature, and specific entropy, respectively, whereas p2, ρ2, T2, and S2 arethe corresponding downstream quantities. Show, also, that

∆MaMa1

−

1 + [(γ − 1)/2] Ma21√

Ma 21 − 1

θ.15.4 Show that for a weak oblique shock

β µ1 + ε,

β − θ µ2 − ε,


|θ2|wall

wall

η

Mach line

Ma1

Figure 15.13Flow of Mach number Ma1 over convex corner of deflection angle |θ2|.

where ε 1. Here, µ1 and µ2 are the Mach angles upstream and downstreamof the shock front, respectively. Moreover, β is the wave angle, and θ 1the deflection angle. Hence, deduce that the shock front subtends the sameangle, ε, with the Mach lines upstream and downstream of it. In other words,the shock position is the “average” of the Mach line positions on either sideof it. Consider supersonic flow incident on a wedge of small nose angle, withan afterbody, as illustrated in Figure 15.9(a). Assume that the shock front isattached to the apex of the wedge, and that the flow downstream of the shockis supersonic. Use the result just proved to show that the shape of the shockfront in the region of attenuation by expansion (i.e., in the region in whichthe shock front is intersected by Mach lines emanating from the shoulder) isparabolic. [Hint: Use the well-known optical result that a parabolic mirrorperfectly focuses a parallel beam of light rays.] (Leipmann & Roshko 1957.)

15.5 Show that, to second order in the deflection angle, θ, the relative change inpressure across a weak oblique shock is written

∆pp1

γMa 21√

Ma 21 − 1

θ + γMa 21

4 (Ma21 − 1) 2

[(γ + 1) Ma4

1 − 4 (Ma21 − 1)

]θ 2,

where ∆p = p2 − p1. Here, p1 is the upstream pressure, p2 the downstreampressure, Ma1 the upstream Mach number, and γ the ratio of specific heats.(Leipmann & Roshko 1957.)

15.6 An ideal gas of pressure, density, temperature, and Mach number p1, ρ1, T1,and Ma1, respectively, flows over a convex corner that turns through an angle


|θ2|, as shown in Figure 15.13. Consider a particular Mach line in the Prandtl-Mayer expansion fan that subtends an angle η with the continuation of theupstream wall, as shown in the figure. Let p, ρ, T , Ma, |θ|, µ, and ν be thepressure, density, temperature, Mach number, magnitude of the deflection an-gle, Mach angle, and Prandtl-Mayer function, respectively, on the Mach linein question. Furthermore, let

η = ν(Ma1) +π

2−(γ + 1γ − 1

)1/2z,

z1 = tan−1

(γ − 1γ + 1

)1/2(Ma 2

1 − 1)1/2

,where γ is the ratio of specific heats. Show that, inside the fan,

µ = cot−1[(Ma 2

1 − 1)1/2 tan ztan z1

],

|θ| = tan−1[(Ma 2

1 − 1)1/2]− tan−1

[(Ma 2

1 − 1)1/2 tan ztan z1

]+

(γ + 1γ − 1

)1/2(z − z1),

Ma 2 = 1 + (Ma 21 − 1)

tan 2 ztan 2 z1

,

ν = ν(Ma1) + tan−1[(Ma 2

1 − 1)1/2]− tan−1

[(Ma 2

1 − 1)1/2 tan ztan z1

]

+

(γ + 1γ − 1

)1/2(z − z1),

p = p1

(cos zcos z1

)2 γ/(γ−1)

,

ρ = ρ1

(cos zcos z1

)2/(γ−1)

,

T = T1

(cos zcos z1

)2,

where z1 ≤ z ≤ z2. Here, z2 is defined implicitly by |θ|(z2) = |θ2|. Demonstratethat the fan extends over the range of angles η2 ≤ η ≤ η1, where

η1 = cot−1[(Ma 2

1 − 1)1/2],

η2 = η1 −(γ + 1γ − 1

)1/2(z2 − z1).

Show that µ, p, ρ→ 0, Ma→ ∞,

ν→ νmax ≡π

2

(γ + 1γ − 1

)1/2− 1

,


reflected shock

Ma2

β′1

Ma1

Ma3

wall

β1

incident shock

Figure 15.14Reflection of oblique shock by wall. Here, Ma1, Ma2, et cetera, are Mach numbers.

and|θ| → |θmax| ≡ νmax − ν(Ma1),

in the limit that z → π/2. Hence, deduce that if |θ2| > |θmax| then the the fanonly extends over the region −|θmax| ≤ η ≤ η1, and the region −|θ2| < η <−|θmax| is occupied by a vacuum (i.e., a gas with zero pressure and density).

Assuming that |θ| 1, show that

|θ| 2γ − 1

(γ − 1γ + 1

)1/2 Ma 21 − 1

Ma 21

(z − z1),

η1 − η γ − 1

2

Ma 21

Ma 21 − 1

|θ|,and

∆pp1 −

γMa 21√

Ma 21 − 1

|θ|,∆ρ

ρ1 1γ

∆pp1,

∆TT1 γ − 1

γ

∆pp1,

∆MaMa1

−1 + [(γ − 1)/2] Ma2

1

γMa 21

∆pp1,

where ∆p = p − p1, et cetera. Of course, the previous four relations are the


same as those for a weak shock. (See Exercise 15.3.) Why is this not surpris-ing? (Hint: The jump in specific entropy across a weak shock is third order inthe deflection angle.) Deduce that to second order in the deflection angle,

∆pp1 −

γMa 21√

Ma 21 − 1

|θ| + γMa 21

4 (Ma21 − 1) 2

[(γ + 1) Ma4

1 − 4 (Ma21 − 1)

]|θ| 2

for a weak Prandtl-Mayer fan. (See Exercise 15.5.)

15.7 If an oblique shock is intercepted by a wall then it is reflected, as illustrated inFigure 15.14. Calculate β′1−β1, assuming that the shocks are sufficiently weakthat the approximate expressions of Section 15.4 can be used. Demonstratethat, in this limit,

β′1 − β1 2(γ + 1

4

) Ma 21

Ma 21 − 1

− 1 θ1,

where θ1 1 is the deflection angle of the incident shock, and γ the ratioof specific heats. Show that β′1 − β1 > 0 if 1 < Ma1 < [4/(3 − γ)]1/2 andβ′1 − β1 < 0 if Ma1 > [4/(3 − γ)]1/2. Demonstrate that Ma2 Ma1 (1 − δ) andMa3 Ma1 (1 − 2 δ), where

δ =

1 + [(γ − 1)/2] Ma21√

Ma 21 − 1

θ1.

15.8 Figure 15.15 shows a situation in which two oblique shocks of the same family[in this case, the (−) family], produced by successive concave corners in a wall,merge together to form a single stronger shock [of the (−) family]. Assumingthat the shocks are sufficiently weak that the approximate expressions of Sec-tion 15.4 can be used (which implies that θ1 1 and θ2 1), demonstratethat

δ ε1 + ε2,

where

ε1 =

(γ + 1

4

) Ma 21

Ma 21 − 1

θ1,

ε2 =

(γ + 1

4

) Ma 21

Ma 21 − 1

θ2,

and γ is the ratio of specific heats. Show that

β3 β1 + ε2,


slipstream

Ma2

wall θ1

θ2

δMa3

β1

shockshock

Ma4shock

β3

Ma1

Ma1

Figure 15.15Merging of two oblique shocks of the same family produced by successive concavecorners of deflection angles θ1 and θ2. Here, Ma1, Ma2, et cetera, are Mach numbers.

and also that Ma2 Ma1 (1 − δ1) and Ma3 Ma4 Ma1 (1 − δ1 − δ2), where

δ1 =

1 + [(γ − 1)/2] Ma21√

Ma 21 − 1

θ1,

δ2 =

1 + [(γ − 1)/2] Ma21√

Ma 21 − 1

θ2.

Demonstrate that the strength of the merged shock is approximately the sumof the strengths of the two component shocks, and, hence, that the pressureson either side of the slipstream shown in the figure are equal (at least, to firstorder in θ1 and θ2). Finally, show that

S4 − S3

R γ (γ + 1)

4Ma 6

1

(Ma 21 − 1)3/2

θ1 θ2 (θ1 + θ2),

where S is specific entropy, and R the specific gas constant. It is, thus, clearthat the specific entropy is not quite the same on either side of the slipstream.

15.9 If two shocks of opposite families intersect then they pass through one another,but are slightly bent in the process, as illustrated in Figure 15.16. Assumingthat the shocks are sufficiently weak that the approximate expressions of Sec-tion 15.4 can be used (which implies that θ1 1, θ2 1, et cetera), show


(+) (−)

δ

(−) (+)

θ1

θ2

Ma1

θ′1

θ′2 Ma′3

Ma2

Ma′2

Ma3

Figure 15.16Crossing of two oblique shocks of different families. Here, Ma1, Ma2, et cetera, areMach numbers, and θ1, θ2, et cetera, are deflection angles.

that

p2 − p1

p1

γMa 21√

Ma 21 − 1

θ1,

p3 − p1

p1

γMa 21√

Ma 21 − 1

(θ1 + θ2),

p′2 − p1

p1

γMa 21√

Ma 21 − 1

θ′1,p′3 − p1

p1

γMa 21√

Ma 21 − 1

(θ′1 + θ′2),

where p denotes pressure. Hence, deduce that

θ2 θ′1,

θ′2 θ1,

δ θ′1 − θ1.

Show, that the respective strengths of the two shocks are unaffected by theintersection (at least, to first order in the deflection angles).


15.10 Show that Equation (14.66) can be written in the form

sin2 µ +

(γ − 1γ + 1

)cos2 µ =

c 21

w 2 ,

where µ is the Mach angle, γ the ratio of specific heats, c1 the sound speed atthe sonic point, and w the flow speed. Deduce that√

Ma 2 − 1dww=

(b 2 − 1

b 2 + tan2 µ

)dµ,

whereb 2 =

γ − 1γ + 1

,

and, hence, that∫ √Ma 2 − 1

dww= µ − 1

btan−1

(1b

tan µ)+ const.

Finally, demonstrate that

ν(Ma) ≡∫ Ma

1

√Ma 2 − 1

dww=

(γ + 1γ − 1

)1/2tan−1

[(γ − 1γ + 1

)(Ma 2 − 1)

]1/2− tan−1

([Ma 2 − 1

]1/2).

(Leipmann & Roshko 1957.)

15.11 Show that for a thin, symmetrical airfoil, with zero angle of attack, whoseprofile is a lens defined by two circular arcs, the drag coefficient is

CD =16

3√

Ma 21 − 1

( tc

)2,

where Ma1 is the upstream Mach number, t the maximum thickness, and cthe chord-length. Demonstrate that for a given thickness ratio, t/c, the airfoilwith the minimum drag is a symmetric diamond profile. (Leipmann & Roshko1957.)

15.12 Prove that on a supersonic swept-back wing of infinite span the thin-airfoilpressure coefficient, (15.50), is multiplied by the sweepback factor,

1√

1 − n 2,

wheren =

tanΛ√Ma 2

1 − 1.


Here, Λ is the sweepback angle, and Ma1 the upstream Mach number. [Hint:Resolve Ma1 into components normal and parallel to the leading edge. Theflow may then be studied in the plane normal to the leading edge using standardthin-airfoil theory.] (Leipmann & Roshko 1957.)

15.13 Consider the problem of subsonic flow past a wave-shaped wall that was dis-cussed in Section 15.13. Show that if the flow is bounded by a second wallthat lies at y = b (where b > 0) then

φ(x, y) =U εm

cos(α x)cosh[mα (b − y)]

sinh(mα b).

Show, on the other hand, that if the flow is bounded by a free surface at y = b(where p = p∞) then

φ(x, y) =U εm

cos(α x)sinh[mα (b − y)]

cosh(mα b).

(Leipmann & Roshko 1957.)

AVectors and Vector Fields

A.1 Introduction

This appendix outlines those aspects of vector algebra, vector calculus, and vectorfield theory that are useful in the study of fluid dynamics. Most of the material ap-pearing in this appendix is reproduced from Fitzpatrick 2008. For more informationon vector algebra, vector calculus, and vector field theory see Schey 1992 and Milne-Thomson 2011.

A.2 Scalars and Vectors

Many physical entities (e.g., mass and energy) are entirely defined by a numericalmagnitude (expressed in appropriate units). Such entities, which have no directionalelement, are known as scalars. Moreover, because scalars can be represented by realnumbers, it follows that they obey the laws of ordinary algebra. However, there exitsa second class of physical entities (e.g., velocity, acceleration, and force) that are onlycompletely defined when both a numerical magnitude and a direction in space arespecified. Such entities are known as vectors. By definition, a vector obeys the samealgebra as a displacement in space, and may thus be represented geometrically by a

straight-line,→PQ (say), where the arrow indicates the direction of the displacement

(i.e., from point P to point Q). (See Figure A.1.) The magnitude of the vector isrepresented by the length of the straight-line.

It is conventional to denote vectors by bold-faced symbols (e.g., a, F) and scalarsby non-bold-faced symbols (e.g., r, S ). The magnitude of a general vector, a, isdenoted |a|, or just a, and is, by definition, always greater than or equal to zero. Itis convenient to define a vector with zero magnitude—this is denoted 0, and has nodirection. Finally, two vectors, a and b, are said to be equal when their magnitudesand directions are both identical.

483


A.3 Vector Algebra

Suppose that the displacements→PQ and

→QR, shown in Figure A.2, represent the

vectors a and b, respectively. It can be seen that the result of combining these two

displacements is to give the net displacement→PR. Hence, if

→PR represents the vector

c then we can writec = a + b. (A.1)

This defines vector addition. By completing the parallelogram PQRS , we can alsosee that

→PR=

→PQ +

→QR=

→PS +

→S R . (A.2)

However,→PS has the same length and direction as

→QR, and, thus, represents the

same vector, b. Likewise,→PQ and

→S R both represent the vector a. Thus, the previous

equation is equivalent toc = a + b = b + a. (A.3)

We conclude that the addition of vectors is commutative. It can also be shown thatthe associative law holds: that is,

a + (b + c) = (a + b) + c. (A.4)

The null vector, 0, is represented by a displacement of zero length and arbitrarydirection. Because the result of combining such a displacement with a finite lengthdisplacement is the same as the latter displacement by itself, it follows that

a + 0 = a, (A.5)

where a is a general vector. The negative of a is defined as that vector which has thesame magnitude, but acts in the opposite direction, and is denoted −a. The sum of aand −a is thus the null vector: that is,

a + (−a) = 0. (A.6)

P

Q

Figure A.1A vector.

Vectors and Vector Fields 485

c = a + b

P

S

R

Q

b

a

a

b

Figure A.2Vector addition.

c = a − b

b a

c

−b

Figure A.3Vector subtraction.

We can also define the difference of two vectors, a and b, as

c = a − b = a + (−b). (A.7)

This definition of vector subtraction is illustrated in Figure A.3.

If n > 0 is a scalar then the expression n a denotes a vector whose direction isthe same as a, and whose magnitude is n times that of a. (This definition becomesobvious when n is an integer.) If n is negative then, because n a = |n| (−a), it followsthat n a is a vector whose magnitude is |n| times that of a, and whose direction is


Py

O x

z

Figure A.4A right-handed Cartesian coordinate system.

opposite to a. These definitions imply that if n and m are two scalars then

n (m a) = n m a = m (n a), (A.8)

(n + m) a = n a + m a, (A.9)

n (a + b) = n a + n b. (A.10)

A.4 Cartesian Components of a VectorConsider a Cartesian coordinate system Oxyz, consisting of an origin, O, and threemutually perpendicular coordinate axes, Ox, Oy, and Oz. (See Figure A.4.) Sucha system is said to be right-handed if, when looking along the Oz direction, a 90

clockwise rotation about Oz is required to take Ox into Oy. Otherwise, it is said tobe left-handed. It is conventional to always use a right-handed coordinate system.

It is convenient to define unit vectors, ex, ey, and ez, parallel to Ox, Oy, and Oz,respectively. Incidentally, a unit vector is a vector whose magnitude is unity. Theposition vector, r, of some general point P whose Cartesian coordinates are (x, y, z)is then given by

r = x ez + y ey + z ez. (A.11)

In other words, we can get from O to P by moving a distance x parallel to Ox, thena distance y parallel to Oy, and then a distance z parallel to Oz. Similarly, if a is anarbitrary vector then

a = ax ex + ay ey + az ez, (A.12)

where ax, ay, and az are termed the Cartesian components of a. It is conventional towrite a ≡ (ax, ay, az). It follows that ex ≡ (1, 0, 0), ey ≡ (0, 1, 0), and ez ≡ (0, 0, 1).Of course, 0 ≡ (0, 0, 0).


According to the three-dimensional generalization of the Pythagorean theorem,the distance OP ≡ |r| = r is given by

r =√

x 2 + y 2 + z 2. (A.13)

By analogy, the magnitude of a general vector a takes the form

a =√

a 2x + a 2

y + a 2z . (A.14)

If a ≡ (ax, ay, az) and b ≡ (bx, by, bz) then it is easily demonstrated that

a + b ≡ (ax + bx, ay + by, az + bz). (A.15)

Furthermore, if n is a scalar then it is apparent that

n a ≡ (n ax, n ay, n az). (A.16)

A.5 Coordinate TransformationsA Cartesian coordinate system allows position and direction in space to be repre-sented in a very convenient manner. Unfortunately, such a coordinate system alsointroduces arbitrary elements into our analysis. After all, two independent observersmight well choose Cartesian coordinate systems with different origins, and differentorientations of the coordinate axes. In general, a given vector a will have differentsets of components in these two coordinate systems. However, the direction andmagnitude of a are the same in both cases. Hence, the two sets of components mustbe related to one another in a very particular fashion. Actually, because vectors are

represented by moveable line elements in space (i.e., in Figure A.2,→PQ and

→S R rep-

resent the same vector), it follows that the components of a general vector are notaffected by a simple shift in the origin of a Cartesian coordinate system. On the otherhand, the components are modified when the coordinate axes are rotated.

Suppose that we transform to a new coordinate system, Ox′y′z′, which has thesame origin as Oxyz, and is obtained by rotating the coordinate axes of Oxyz throughan angle θ about Oz. (See Figure A.5.) Let the coordinates of a general point P be(x, y, z) in Oxyz and (x′, y′, z′) in Ox′y′z′. According to simple trigonometry, thesetwo sets of coordinates are related to one another via the transformation

x′ = x cos θ + y sin θ, (A.17)

y′ = −x sin θ + y cos θ, (A.18)

z′ = z. (A.19)

Consider the vector displacement r ≡→OP. Note that this displacement is represented


P

y′y

x′

xθ

z

O

Figure A.5Rotation of the coordinate axes about Oz.

by the same symbol, r, in both coordinate systems, because the magnitude and di-rection of r are manifestly independent of the orientation of the coordinate axes. Thecoordinates of r do depend on the orientation of the axes: that is, r ≡ (x, y, z) inOxyz, and r ≡ (x′, y′, z′) in Ox′y′z′. However, they must depend in a very specificmanner [i.e., Equations (A.17)–(A.19)] which preserves the magnitude and directionof r.

The components of a general vector a transform in an analogous manner to Equa-tions (A.17)–(A.19): that is,

ax′ = ax cos θ + ay sin θ, (A.20)

ay′ = −ax sin θ + ay cos θ, (A.21)

az′ = az. (A.22)

Moreover, there are similar transformation rules for rotation about Ox and Oy. Equa-tions (A.20)–(A.22) effectively constitute the definition of a vector: in other words,the three quantities (ax, ay, az) are the components of a vector provided that theytransform under rotation of the coordinate axes about Oz in accordance with Equa-tions (A.20)–(A.22). (And also transform correctly under rotation about Ox and Oy.)Conversely, (ax, ay, az) cannot be the components of a vector if they do not trans-form in accordance with Equations (A.20)–(A.22). Of course, scalar quantities areinvariant under rotation of the coordinate axes. Thus, the individual components ofa vector (ax, say) are real numbers, but they are not scalars. Displacement vectors,and all vectors derived from displacements (e.g., velocity and acceleration), automat-ically satisfy Equations (A.20)–(A.22). There are, however, other physical quantitiesthat have both magnitude and direction, but are not obviously related to displace-ments. We need to check carefully to see whether these quantities are really vectors.(See Sections A.7 and A.9.)


A.6 Scalar ProductA scalar quantity is invariant under all possible rotational transformations. The in-dividual components of a vector are not scalars because they change under transfor-mation. Can we form a scalar out of some combination of the components of one, ormore, vectors? Suppose that we were to define the “percent” product,

a % b ≡ ax bz + ay bx + az by = scalar number, (A.23)

for general vectors a and b. Is a % b invariant under transformation, as must be thecase if it is a scalar number? Let us consider an example. Suppose that a ≡ (0, 1, 0)and b ≡ (1, 0, 0). It is easily seen that a % b = 1. Let us now rotate the coordinateaxes through 45 about Oz. In the new coordinate system, a ≡ (1/

√2, 1/

√2, 0) and

b ≡ (1/√

2, −1/√

2, 0), giving a % b = 1/2. Clearly, a % b is not invariant underrotational transformation, so the previous definition is a bad one.

Consider, now, the dot product or scalar product:

a · b ≡ ax bx + ay by + az bz = scalar number. (A.24)

Let us rotate the coordinate axes though θ degrees about Oz. According to Equa-tions (A.20)–(A.22), a · b takes the form

a · b = (ax cos θ + ay sin θ) (bx cos θ + by sin θ)

+ (−ax sin θ + ay cos θ) (−bx sin θ + by cos θ) + az bz

= ax bx + ay by + az bz (A.25)

in the new coordinate system. Thus, a · b is invariant under rotation about Oz. Itis easily demonstrated that it is also invariant under rotation about Ox and Oy. Weconclude that a · b is a true scalar, and that the definition (A.24) is a good one.Incidentally, a · b is the only simple combination of the components of two vectorsthat transforms like a scalar. It is readily shown that the dot product is commutativeand distributive: that is,

a · b = b · a,

a · (b + c) = a · b + a · c. (A.26)

The associative property is meaningless for the dot product, because we cannot have(a · b) · c, as a · b is scalar.

We have shown that the dot product a · b is coordinate independent. But what isthe geometric significance of this property? In the special case where a = b, we get

a · b = a 2x + a 2

y + a 2z = |a| 2 = a 2. (A.27)

So, the invariance of a · a is equivalent to the invariance of the magnitude of vector aunder transformation.


b − a

O

B

A

b

a

θ

Figure A.6A vector triangle.

Let us now investigate the general case. The length squared of AB in the vectortriangle shown in Figure A.6 is

(b − a) · (b − a) = |a| 2 + |b| 2 − 2 a · b. (A.28)

However, according to the “cosine rule” of trigonometry,

(AB)2 = (OA)2 + (OB)2 − 2 (OA) (OB) cos θ, (A.29)

where (AB) denotes the length of side AB. It follows that

a · b = |a| |b| cos θ. (A.30)

In this case, the invariance of a ·b under transformation is equivalent to the invarianceof the angle subtended between the two vectors. Note that if a · b = 0 then either|a| = 0, |b| = 0, or the vectors a and b are mutually perpendicular. The anglesubtended between two vectors can easily be obtained from the dot product:

cos θ =a · b|a| |b|

. (A.31)

The work W performed by a constant force F that moves an object through adisplacement r is the product of the magnitude of F times the displacement in thedirection of F. If the angle subtended between F and r is θ then

W = |F| (|r| cos θ) = F · r. (A.32)

The work dW performed by a non-constant force f that moves an object throughan infinitesimal displacement dr in a time interval dt is dW = f · dr. Thus, the rateat which the force does work on the object, which is usually referred to as the power,is P = dW/dt = f · dr/dt, or P = f · v, where v = dr/dt is the object’s instantaneousvelocity.


S

Figure A.7A vector area.

A.7 Vector AreaSuppose that we have planar surface of scalar area S . We can define a vector areaS whose magnitude is S , and whose direction is perpendicular to the plane, in thesense determined by a right-hand circulation rule (see Section A.8) applied to the rim,assuming that a direction of circulation around the rim is specified. (See Figure A.7.)This quantity clearly possesses both magnitude and direction. But is it a true vector?We know that if the normal to the surface makes an angle αx with the x-axis thenthe area seen looking along the x-direction is S cosαx. This is the x-component of S(because S x = ex · S = ex · n S = cosαx S , where n is the unit normal to the surface).Similarly, if the normal makes an angle αy with the y-axis then the area seen lookingalong the y-direction is S cosαy. This is the y-component of S. If we limit ourselvesto a surface whose normal is perpendicular to the z-direction then αx = π/2−αy = α.It follows that S = S (cosα, sinα, 0). If we rotate the basis about the z-axis by θdegrees, which is equivalent to rotating the normal to the surface about the z-axis by−θ degrees, so that α→ α − θ, then

S x′ = S cos (α − θ) = S cosα cos θ + S sinα sin θ = S x cos θ + S y sin θ, (A.33)

which is the correct transformation rule for the x-component of a vector. The othercomponents transform correctly as well. This proves both that a vector area is a truevector, and that the components of a vector area are the projected areas seen lookingdown the coordinate axes.

According to the vector addition theorem, the projected area of two plane sur-faces, joined together at a line, looking along the x-direction (say) is the x-componentof the resultant of the vector areas of the two surfaces. Likewise, for many joined-upplane areas, the net area seen looking down the x-axis, which is the same as the areaof the outer rim seen looking down the x-axis, is the x-component of the resultant ofall the vector areas: that is,

S =∑

i

Si. (A.34)


If we approach a limit, by letting the number of plane facets increase, and their areasreduce, then we obtain a continuous surface denoted by the resultant vector area

S =∑

i

δSi. (A.35)

It is clear that the area of the rim seen looking down the x-axis is just S x. Similarly,for the areas of the rim seen looking down the other coordinate axes. Note that itis the rim of the surface that determines the vector area, rather than the nature ofthe surface spanning the rim. So, two different surfaces sharing the same rim bothpossess the same vector area.

In conclusion, a loop (not all in one plane) has a vector area S which is theresultant of the component vector areas of any surface ending on the loop. Thecomponents of S are the areas of the loop seen looking down the coordinate axes. Asa corollary, a closed surface has S = 0, because it does not possess a rim.

A.8 Vector ProductWe have discovered how to construct a scalar from the components of two generalvectors, a and b. Can we also construct a vector that is not just a linear combinationof a and b? Consider the following definition:

a ∗ b ≡ (ax bx, ay by, az bz). (A.36)

Is a ∗ b a proper vector? Suppose that a = (0, 1, 0), b = (1, 0, 0). In this case,a ∗ b = 0. However, if we rotate the coordinate axes through 45 about Oz thena = (1/

√2, 1/

√2, 0), b = (1/

√2, −1/

√2, 0), and a ∗ b = (1/2, −1/2, 0). Thus,

a ∗ b does not transform like a vector, because its magnitude depends on the choiceof axes. So, previous definition is a bad one.

Consider, now, the cross product or vector product:

a × b ≡ (ay bz − az by, az bx − ax bz, ax by − ay bx) = c. (A.37)

Does this rather unlikely combination transform like a vector? Let us try rotating thecoordinate axes through an angle θ about Oz using Equations (A.20)–(A.22). In thenew coordinate system,

cx′ = (−ax sin θ + ay cos θ) bz − az (−bx sin θ + by cos θ)

= (ay bz − az by) cos θ + (az bx − ax bz) sin θ

= cx cos θ + cy sin θ. (A.38)

Thus, the x-component of a × b transforms correctly. It can easily be shown that theother components transform correctly as well, and that all components also transform


b

middle finger

index finger

thumb

θ

a × b

a

Figure A.8The right-hand rule for cross products. Here, θ is less that 180.

correctly under rotation about Ox and Oy. Thus, a × b is a proper vector. Inciden-tally, a × b is the only simple combination of the components of two vectors thattransforms like a vector (which is non-coplanar with a and b). The cross product isanti-commutative,

a × b = −b × a, (A.39)

distributive,a × (b + c) = a × b + a × c, (A.40)

but is not associative,a × (b × c) (a × b) × c. (A.41)

The cross product transforms like a vector, which means that it must have a well-defined direction and magnitude. We can show that a × b is perpendicular to both aand b. Consider a · a× b. If this is zero then the cross product must be perpendicularto a. Now,

a · a × b = ax (ay bz − az by) + ay (az bx − ax bz) + az (ax by − ay bx)

= 0. (A.42)

Therefore, a × b is perpendicular to a. Likewise, it can be demonstrated that a × b isperpendicular to b. The vectors a, b, and a × b form a right-handed set, like the unitvectors ex, ey, and ez. In fact, ex × ey = ez. This defines a unique direction for a × b,which is obtained from a right-hand rule. (See Figure A.8.)

Let us now evaluate the magnitude of a × b. We have

(a × b)2 = (ay bz − az by)2 + (az bx − ax bz)2 + (ax by − ay bx)2

= (a 2x + a 2

y + a 2z ) (b 2

x + b 2y + b 2

z ) − (ax bx + ay by + az bz)2

= |a| 2 |b| 2 − (a · b)2

= |a| 2 |b| 2 − |a| 2 |b| 2 cos2 θ = |a| 2 |b| 2 sin2 θ. (A.43)


Thus,

|a × b| = |a| |b| sin θ, (A.44)

where θ is the angle subtended between a and b. Clearly, a × a = 0 for any vector,because θ is always zero in this case. Also, if a× b = 0 then either |a| = 0, |b| = 0, orb is parallel (or antiparallel) to a.

Consider the parallelogram defined by the vectors a and b. (See Figure A.9.) Thescalar area of the parallelogram is a b sin θ. By convention, the vector area has themagnitude of the scalar area, and is normal to the plane of the parallelogram, in thesense obtained from a right-hand circulation rule by rotating a on to b (through anacute angle). In other words, if the fingers of the right-hand circulate in the directionof rotation then the thumb of the right-hand indicates the direction of the vector area.So, the vector area is coming out of the page in Figure A.9. It follows that

S = a × b, (A.45)

Suppose that a force F is applied at position r, as illustrated in Figure A.10.The torque about the origin O is the product of the magnitude of the force and thelength of the lever arm OQ. Thus, the magnitude of the torque is |F| |r| sin θ. Thedirection of the torque is conventionally defined as the direction of the axis through Oabout which the force tries to rotate objects, in the sense determined by a right-handcirculation rule. Hence, the torque is out of the page in Figure A.10. It follows thatthe vector torque is given by

τ = r × F. (A.46)

The angular momentum, l, of a particle of linear momentum p and position vectorr is simply defined as the moment of its momentum about the origin: that is,

l = r × p. (A.47)

S

b

θa

b

a

Figure A.9A vector parallelogram.


τ

O

θ

P

Q

F

r sin θ

r

Figure A.10A torque.

A.9 RotationLet us try to define a rotation vector θ whose magnitude is the angle of the rotation,θ, and whose direction is parallel to the axis of rotation, in the sense determined bya right-hand circulation rule. Unfortunately, this is not a good vector. The prob-lem is that the addition of rotations is not commutative, whereas vector additionis commuative. Figure A.11 shows the effect of applying two successive 90 rota-tions, one about Ox, and the other about the Oz, to a standard six-sided die. In theleft-hand case, the z-rotation is applied before the x-rotation, and vice versa in theright-hand case. It can be seen that the die ends up in two completely different states.In other words, the z-rotation plus the x-rotation does not equal the x-rotation plusthe z-rotation. This non-commuting algebra cannot be represented by vectors. So,although rotations have a well-defined magnitude and direction, they are not vectorquantities.

But, this is not quite the end of the story. Suppose that we take a general vectora, and rotate it about Oz by a small angle δθz. This is equivalent to rotating thecoordinate axes about Oz by −δθz. According to Equations (A.20)–(A.22), we have

a′ a + δθz ez × a, (A.48)

where use has been made of the small angle approximations sin θ θ and cos θ 1.The previous equation can easily be generalized to allow small rotations about Oxand Oy by δθx and δθy, respectively. We find that

a′ a + δθ × a, (A.49)

whereδθ = δθx ex + δθy ey + δθz ez. (A.50)


x

x-axisz-axis

x-axis z-axis

z

y

Figure A.11Effect of successive rotations about perpendicular axes on a six-sided die.

Clearly, we can define a rotation vector, δθ, but it only works for small angle rotations(i.e., sufficiently small that the small angle approximations of sine and cosine aregood). According to the previous equation, a small z-rotation plus a small x-rotationis (approximately) equal to the two rotations applied in the opposite order. The factthat infinitesimal rotation is a vector implies that angular velocity,

ω = limδt→0

δθ

δt, (A.51)

must be a vector as well. Also, if a′ is interpreted as a(t+ δt) in Equation (A.49) thenit follows that the equation of motion of a vector that precesses about the origin withsome angular velocity ω is

dadt= ω × a. (A.52)

A.10 Scalar Triple ProductConsider three vectors a, b, and c. The scalar triple product is defined a ·b× c. Now,b × c is the vector area of the parallelogram defined by b and c. So, a · b × c is thescalar area of this parallelogram multiplied by the component of a in the directionof its normal. It follows that a · b × c is the volume of the parallelepiped defined by


c

b

a

Figure A.12A vector parallelepiped.

vectors a, b, and c. (See Figure A.12.) This volume is independent of how the tripleproduct is formed from a, b, and c, except that

a · b × c = −a · c × b. (A.53)

So, the “volume” is positive if a, b, and c form a right-handed set (i.e., if a lies abovethe plane of b and c, in the sense determined from a right-hand circulation rule byrotating b onto c), and negative if they form a left-handed set. The triple product isunchanged if the dot and cross product operators are interchanged,

a · b × c = a × b · c. (A.54)

The triple product is also invariant under any cyclic permutation of a, b, and c,

a · b × c = b · c × a = c · a × b, (A.55)

but any anti-cyclic permutation causes it to change sign,

a · b × c = −b · a × c. (A.56)

The scalar triple product is zero if any two of a, b, and c are parallel, or if a, b, and care coplanar.

If a, b, and c are non-coplanar then any vector r can be written in terms of them:that is,

r = α a + βb + γ c. (A.57)

Forming the dot product of this equation with b × c, we then obtain

r · b × c = α a · b × c, (A.58)

soα =

r · b × ca · b × c

. (A.59)

Analogous expressions can be written for β and γ. The parameters α, β, and γ areuniquely determined provided a · b × c 0: that is, provided the three vectors arenon-coplanar.


A.11 Vector Triple ProductFor three vectors a, b, and c, the vector triple product is defined a × (b × c). Thebrackets are important because a×(b×c) (a×b)×c. In fact, it can be demonstratedthat

a × (b × c) ≡ (a · c) b − (a · b) c (A.60)

and(a × b) × c ≡ (a · c) b − (b · c) a. (A.61)

Let us try to prove the first of the previous theorems. The left-hand side andthe right-hand side are both proper vectors, so if we can prove this result in oneparticular coordinate system then it must be true in general. Let us take convenientaxes such that Ox lies along b, and c lies in the x-y plane. It follows that b ≡(bx, 0, 0), c ≡ (cx, cy, 0), and a ≡ (ax, ay, az). The vector b × c is directed alongOz: in fact, b × c ≡ (0, 0, bx cy). Hence, a × (b × c) lies in the x-y plane: in fact,a × (b × c) ≡ (ay bx cy, −ax bx cy, 0). This is the left-hand side of Equation (A.60) inour convenient coordinate system. To evaluate the right-hand side, we need a · c =ax cx + ay cy and a · b = ax bx. It follows that the right-hand side is

RHS = ( [ax cx + ay cy] bx, 0, 0) − (ax bx cx, ax bx cy, 0)

= (ay cy bx, −ax bx cy, 0) = LHS, (A.62)

which proves the theorem.

A.12 Vector CalculusSuppose that vector a varies with time, so that a = a(t). The time derivative of thevector is defined

dadt= limδt→0

[a(t + δt) − a(t)

δt

]. (A.63)

When written out in component form this becomes

dadt≡(dax

dt,

daydt,

daz

dt

). (A.64)

Suppose that a is, in fact, the product of a scalar φ(t) and another vector b(t).What now is the time derivative of a? We have

dax

dt=

ddt

(φ bx) =dφdt

bx + φdbx

dt, (A.65)

which implies thatdadt=

dφdt

b + φdbdt. (A.66)


.

O x

y

P

Q

l

P

f

Q l

Figure A.13A line integral.

Moreover, it is easily demonstrated that

ddt

(a · b) =dadt· b + a · db

dt, (A.67)

andddt

(a × b) =dadt× b + a × db

dt. (A.68)

Hence, it can be seen that the laws of vector differentiation are analogous to those inconventional calculus.

A.13 Line IntegralsConsider a two-dimensional function f (x, y) that is defined for all x and y. Whatis meant by the integral of f along a given curve joining the points P and Q in thex-y plane? Well, we first draw out f as a function of length l along the path. (SeeFigure A.13.) The integral is then simply given by∫ Q

Pf (x, y) dl = Area under the curve, (A.69)

where dl =√

dx 2 + dy 2.For example, consider the integral of f (x, y) = x y 2 between P and Q along the

two routes indicated in Figure A.14. Along route 1, we have x = y, so dl =√

2 dx.Thus, ∫ Q

Px y 2 dl =

∫ 1

0x 3√

2 dx =

√2

4. (A.70)


y

P = (0, 0)

Q = (1, 1)

1

2

2

x

Figure A.14An example line integral.

The integration along route 2 gives∫ Q

Px y 2 dl =

∫ 1

0x y 2 dx

∣∣∣∣∣∣y=0+

∫ 1

0x y 2 dy

∣∣∣∣∣∣x=1

= 0 +∫ 1

0y 2 dy =

13. (A.71)

Note that the integral depends on the route taken between the initial and final points.The most common type of line integral is that in which the contributions from dx

and dy are evaluated separately, rather that through the path-length element dl: thatis, ∫ Q

P

[f (x, y) dx + g(x, y) dy

]. (A.72)

For example, consider the integral∫ Q

P

(y dx + x 3 dy

)(A.73)

along the two routes indicated in Figure A.15. Along route 1, we have x = y + 1 anddx = dy, so ∫ Q

P

(y dx + x 3 dy

)=

∫ 1

0

[y dy + (y + 1)3 dy

]=

174. (A.74)

Along route 2,∫ Q

P

(y dx + x 3 dy

)=

∫ 1

0x 3 dy

∣∣∣∣∣∣x=1+

∫ 2

1y dx

∣∣∣∣∣∣y=1=

74. (A.75)

Again, the integral depends on the path of integration.


O P = (1, 0)

Q = (2, 1)

1

x

y2

2

Figure A.15An example line integral.

Suppose that we have a line integral that does not depend on the path of integra-tion. It follows that ∫ Q

P( f dx + g dy) = F(Q) − F(P) (A.76)

for some function F. Given F(P) for some point P in the x-y plane,

F(Q) = F(P) +∫ Q

P( f dx + g dy) (A.77)

defines F(Q) for all other points in the plane. We can then draw a contour map ofF(x, y). The line integral between points P and Q is simply the change in height inthe contour map between these two points:∫ Q

P( f dx + g dy) =

∫ Q

PdF(x, y) = F(Q) − F(P). (A.78)

Thus,

dF(x, y) = f (x, y) dx + g(x, y) dy. (A.79)

For instance, if F = x 3 y then dF = 3 x 2 y dx + x 3 dy and∫ Q

P

(3 x 2 y dx + x 3 dy

)=[x 3 y

]QP

(A.80)

is independent of the path of integration.It is clear that there are two distinct types of line integral—those that depend only

on their endpoints and not on the path of integration, and those that depend both ontheir endpoints and the integration path. Later on, we shall learn how to distinguishbetween these two types. (See Section A.18.)


A.14 Vector Line IntegralsA vector field is defined as a set of vectors associated with each point in space. Forinstance, the velocity v(r) in a moving liquid (e.g., a whirlpool) constitutes a vectorfield. By analogy, a scalar field is a set of scalars associated with each point in space.An example of a scalar field is the temperature distribution T (r) in a furnace.

Consider a general vector field A(r). Let dr ≡ (dx, dy, dz) be the vector elementof line length. Vector line integrals often arise as∫ Q

PA · dr =

∫ Q

P(Ax dx + Ay dy + Az dz). (A.81)

For instance, if A is a force-field then the line integral is the work done in going fromP to Q.

For example, consider the work done by a repulsive inverse-square central field,F = −r/|r 3|. The element of work done is dW = F · dr. Take P = (∞, 0, 0) andQ = (a, 0, 0). The first route considered is along the x-axis, so

W =∫ a

∞

(− 1

x 2

)dx =

[1x

]a∞=

1a. (A.82)

The second route is, firstly, around a large circle (r = constant) to the point (a,∞, 0),and then parallel to the y-axis. (See Figure A.16). In the first part, no work is done,because F is perpendicular to dr. In the second part,

W =∫ 0

∞

−y dy(a 2 + y 2)3/2 =

[1

(y 2 + a 2)1/2

]0∞=

1a. (A.83)

In this case, the integral is independent of the path. However, not all vector lineintegrals are path independent.

A.15 Surface IntegralsLet us take a surface S , that is not necessarily co-planar, and divide it up into (scalar)elements δS i. Then ∫ ∫

Sf (x, y, z) dS = lim

δS i→0

∑i

f (x, y, z) δS i (A.84)

is a surface integral. For instance, the volume of water in a lake of depth D(x, y) is

V =∫ ∫

D(x, y) dS . (A.85)


x

2

1

2

Q P

y

a ∞O

Figure A.16An example vector line integral.

To evaluate this integral, we must split the calculation into two ordinary integrals.The volume in the strip shown in Figure A.17 is[∫ x2

x1

D(x, y) dx]

dy. (A.86)

Note that the limits x1 and x2 depend on y. The total volume is the sum over allstrips: that is,

V =∫ y2

y1

dy[∫ x2(y)

x1(y)D(x, y) dx

]≡∫ ∫

SD(x, y) dx dy. (A.87)

Of course, the integral can be evaluated by taking the strips the other way around:that is,

V =∫ x2

x1

dx∫ y2(x)

y1(x)D(x, y) dy. (A.88)

Interchanging the order of integration is a very powerful and useful trick. But greatcare must be taken when evaluating the limits.

For example, consider ∫ ∫S

x y 2 dx dy, (A.89)

where S is shown in Figure A.18. Suppose that we evaluate the x integral first:

dy(∫ 1−y

0x y 2 dx

)= y 2 dy

[x 2

2

]1−y0=y 2

2(1 − y)2 dy. (A.90)

Let us now evaluate the y integral:∫ 1

0

(y 2

2− y3 +

y 4

2

)dy =

160. (A.91)


dy

x2x1

y1

y2

y

x

Figure A.17Decomposition of a surface integral.

We can also evaluate the integral by interchanging the order of integration:∫ 1

0x dx

∫ 1−x

0y 2 dy =

∫ 1

0

x3

(1 − x)3 dx =1

60. (A.92)

In some cases, a surface integral is just the product of two separate integrals. Forinstance, ∫ ∫

Sx 2 y dx dy (A.93)

where S is a unit square. This integral can be written∫ 1

0dx∫ 1

0x 2 y dy =

(∫ 1

0x 2 dx

) (∫ 1

0y dy

)=

13

12=

16, (A.94)

because the limits are both independent of the other variable.

A.16 Vector Surface IntegralsSurface integrals often occur during vector analysis. For instance, the rate of flow ofa liquid of velocity v through an infinitesimal surface of vector area dS is v · dS. Thenet rate of flow through a surface S made up of very many infinitesimal surfaces is∫ ∫

Sv · dS = lim

dS→0

[∑v cos θ dS

], (A.95)

where θ is the angle subtended between the normal to the surface and the flow veloc-ity.


1 − y = x

x(1, 0)

(0, 1)

y

(0, 0)

Figure A.18An example surface integral.

Analogously to line integrals, most surface integrals depend both on the surfaceand the rim. But some (very important) integrals depend only on the rim, and noton the nature of the surface which spans it. As an example of this, consider incom-pressible fluid flow between two surfaces S 1 and S 2 that end on the same rim. (SeeFigure A.23.) The volume between the surfaces is constant, so what goes in mustcome out, and ∫ ∫

S 1

v · dS =∫ ∫

S 2

v · dS. (A.96)

It follows that ∫ ∫v · dS (A.97)

depends only on the rim, and not on the form of surfaces S 1 and S 2.

A.17 Volume IntegralsA volume integral takes the form∫ ∫ ∫

Vf (x, y, z) dV, (A.98)

where V is some volume, and dV = dx dy dz is a small volume element. The volumeelement is sometimes written d 3r, or even dτ.

As an example of a volume integral, let us evaluate the center of gravity of asolid pyramid. Suppose that the pyramid has a square base of side a, a height a, andis composed of material of uniform density. Let the centroid of the base lie at theorigin, and let the apex lie at (0, 0, a). By symmetry, the center of mass lies on the


line joining the centroid to the apex. In fact, the height of the center of mass is givenby

z =∫ ∫ ∫

z dV/ ∫ ∫ ∫

dV. (A.99)

The bottom integral is just the volume of the pyramid, and can be written∫ ∫ ∫dV =

∫ a

0dz∫ (a−z)/2

−(a−z)/2dy∫ (a−z)/2

−(a−z)/2dx =

∫ a

0(a − z)2 dz

=

∫ a

0(a 2 − 2 a z + z 2) dz =

[a 2 z − a z 2 + z 3/3

]a0=

13

a 3. (A.100)

Here, we have evaluated the z-integral last because the limits of the x- and y- integralsare z-dependent. The top integral takes the form∫ ∫ ∫

z dV =∫ a

0z dz

∫ (a−z)/2

−(a−z)/2dy∫ (a−z)/2

−(a−z)/2dx =

∫ a

0z (a − z)2 dz

=

∫ a

0(z a 2 − 2 a z 2 + z 3) dz =

[a 2 z 2/2 − 2 a z 3/3 + z4/4

]a0

=1

12a 4. (A.101)

Thus,

z =1

12a4/

13

a3 =14

a. (A.102)

In other words, the center of mass of a pyramid lies one quarter of the way betweenthe centroid of the base and the apex.

A.18 GradientA one-dimensional function f (x) has a gradient d f /dx which is defined as the slopeof the tangent to the curve at x. We wish to extend this idea to cover scalar fields intwo and three dimensions.

Consider a two-dimensional scalar field h(x, y) that represents (say) height abovesea-level in a hilly region. Let dr ≡ (dx, dy) be an element of horizontal distance.Consider dh/dr, where dh is the change in height after moving an infinitesimal dis-tance dr. This quantity is somewhat like the one-dimensional gradient, except thatdh depends on the direction of dr, as well as its magnitude. In the immediate vicin-ity of some point P, the slope reduces to an inclined plane. (See Figure A.19.) Thelargest value of dh/dr is straight up the slope. It is easily shown that for any otherdirection

dhdr=

(dhdr

)max

cos θ, (A.103)


O

y contours of h(x, y)

θ

x

P

high

low

direction of steepest ascentdr

Figure A.19A two-dimensional gradient.

where θ is the angle shown in Figure A.19. Let us define a two-dimensional vector,grad h, called the gradient of h, whose magnitude is (dh/dr)max, and whose direc-tion is the direction of steepest ascent. The cos θ variation exhibited in the previousexpression ensures that the component of grad h in any direction is equal to dh/drfor that direction.

The component of dh/dr in the x-direction can be obtained by plotting out theprofile of h at constant y, and then finding the slope of the tangent to the curve atgiven x. This quantity is known as the partial derivative of h with respect to x atconstant y, and is denoted (∂h/∂x)y. Likewise, the gradient of the profile at constantx is written (∂h/∂y)x. Note that the subscripts denoting constant x and constant y areusually omitted, unless there is any ambiguity. It follows that in component form

grad h ≡(∂h∂x,∂h∂y

). (A.104)

Now, the equation of the tangent plane at P = (x0, y0) is

hT (x, y) = h(x0, y0) + α (x − x0) + β (y − y0). (A.105)

This has the same local gradients as h(x, y), so

α =∂h∂x, β =

∂h∂y. (A.106)

For small dx = x − x0 and dy = y − y0, the function h is coincident with the tangentplane. It follows that

dh =∂h∂x

dx +∂h∂y

dy. (A.107)


But, grad h ≡ (∂h/∂x, ∂h/∂y) and dr ≡ (dx, dy), so

dh = grad h · dr. (A.108)

Incidentally, the previous equation demonstrates that grad h is a proper vector, be-cause the left-hand side is a scalar, and, according to the properties of the dot product,the right-hand side is also a scalar provided that dr and grad h are both proper vectors(dr is an obvious vector, because it is directly derived from displacements).

Consider, now, a three-dimensional temperature distribution T (x, y, z) in (say)a reaction vessel. Let us define grad T , as before, as a vector whose magnitude is(dT/dr)max, and whose direction is the direction of the maximum gradient. Thisvector is written in component form

grad T ≡(∂T∂x,∂T∂y,∂T∂z

). (A.109)

Here, ∂T/∂x ≡ (∂T/∂x)y,z is the gradient of the one-dimensional temperature profileat constant y and z. The change in T in going from point P to a neighboring pointoffset by dr ≡ (dx, dy, dz) is

dT =∂T∂x

dx +∂T∂y

dy +∂T∂z

dz. (A.110)

In vector form, this becomes

dT = grad T · dr. (A.111)

Suppose that dT = 0 for some dr. It follows that

dT = grad T · dr = 0. (A.112)

So, dr is perpendicular to grad T . Because dT = 0 along so-called “isotherms”(i.e., contours of the temperature), we conclude that the isotherms (contours) areeverywhere perpendicular to grad T . (See Figure A.20.)

It is, of course, possible to integrate dT . For instance, the line integral of dTbetween points P and Q is written∫ Q

PdT =

∫ Q

Pgrad T · dr = T (Q) − T (P). (A.113)

This integral is clearly independent of the path taken between P and Q, so∫ Q

P grad T ·dr must be path independent.

Consider a vector field A(r). In general, the line integral∫ Q

P A · dr depends onthe path taken between the end points. However, for some special vector fields theintegral is path independent. Such fields are called conservative fields. It can beshown that if A is a conservative field then A = grad V for some scalar field V . Theproof of this is straightforward. Keeping P fixed, we have∫ Q

PA · dr = V(Q), (A.114)


dr

isotherms

T = constant gradT

Figure A.20Isotherms.

where V(Q) is a well-defined function, due to the path independent nature of the lineintegral. Consider moving the position of the end point by an infinitesimal amountdx in the x-direction. We have

V(Q + dx) = V(Q) +∫ Q+dx

QA · dr = V(Q) + Ax dx. (A.115)

Hence,

∂V∂x= Ax, (A.116)

with analogous relations for the other components of A. It follows that

A = grad V. (A.117)

The force field due to gravity is a good example of a conservative field. Now, ifA(r) is a force-field then

∫A · dr is the work done in traversing some path. If A is

conservative then ∮A · dr = 0, (A.118)

where∮

corresponds to the line integral around a closed loop. The fact that zero network is done in going around a closed loop is equivalent to the conservation of energy(which is why conservative fields are called “conservative”). A good example of anon-conservative field is the force field due to friction. Clearly, a frictional systemloses energy in going around a closed cycle, so

∮A · dr 0.


A.19 Grad OperatorIt is useful to define the vector operator

∇ ≡(∂

∂x,∂

∂y,∂

∂z

), (A.119)

which is usually called the grad or del operator. This operator acts on everythingto its right in a expression, until the end of the expression or a closing bracket isreached. For instance,

grad f = ∇ f ≡(∂ f∂x,∂ f∂y,∂ f∂z

). (A.120)

For two scalar fields φ and ψ,

grad (φψ) = φ gradψ + ψ gradφ (A.121)

can be written more succinctly as

∇(φψ) = φ∇ψ + ψ∇φ. (A.122)

Suppose that we rotate the coordinate axes through an angle θ about Oz. Byanalogy with Equations (A.17)–(A.19), the old coordinates (x, y, z) are related tothe new ones (x′, y′, z′) via

x = x′ cos θ − y′ sin θ, (A.123)

y = x ′ sin θ + y′ cos θ, (A.124)

z = z′. (A.125)

Now,∂

∂x′=

(∂x∂x′

)y′ ,z′

∂

∂x+

(∂y

∂x′

)y′,z′

∂

∂y+

(∂z∂x′

)y′ ,z′

∂

∂z, (A.126)

giving∂

∂x′= cos θ

∂

∂x+ sin θ

∂

∂y, (A.127)

and∇x′ = cos θ∇x + sin θ∇y. (A.128)

It can be seen, from Equations (A.20)–(A.22), that the differential operator ∇ trans-forms in an analogous manner to a vector. This is another proof that ∇ f is a goodvector.


y + dy

z

zx + dx

x

z + dzy

x

y

Figure A.21Flux of a vector field out of a small box.

A.20 DivergenceLet us start with a vector field A(r). Consider

∮S A · dS over some closed surface

S , where dS denotes an outward pointing surface element. This surface integral isusually called the flux of A out of S . If A represents the velocity of some fluid then∮

S A · dS is the rate of fluid flow out of S .If A is constant in space then it is easily demonstrated that the net flux out of S

is zero. In fact, ∮A · dS = A ·

∮dS = A · S = 0, (A.129)

because the vector area S of a closed surface is zero.Suppose, now, that A is not uniform in space. Consider a very small rectangular

volume over which A hardly varies. The contribution to∮

A · dS from the two facesnormal to the x-axis is

Ax(x + dx) dy dz − Ax(x) dy dz =∂Ax

∂xdx dy dz =

∂Ax

∂xdV, (A.130)

where dV = dx dy dz is the volume element. (See Figure A.21.) There are analo-gous contributions from the sides normal to the y- and z-axes, so the total of all thecontributions is ∮

A · dS =(∂Ax

∂x+∂Ay∂y+∂Az

∂z

)dV. (A.131)

The divergence of a vector field is defined

divA = ∇ · A =∂Ax


∂z. (A.132)


.

exterior contributions survive

S

interior contributions cancel

Figure A.22The divergence theorem.

Divergence is a good scalar (i.e., it is coordinate independent), because it is the dotproduct of the vector operator ∇ with A. The formal definition of ∇ · A is

∇ · A = limdV→0

∮A · dSdV

. (A.133)

This definition is independent of the shape of the infinitesimal volume element.One of the most important results in vector field theory is the so-called divergence

theorem. This states that for any volume V surrounded by a closed surface S ,∮S

A · dS =∫

V∇ · A dV, (A.134)

where dS is an outward pointing volume element. The proof is very straightforward.We divide up the volume into very many infinitesimal cubes, and sum

∫A · dS over

all of the surfaces. The contributions from the interior surfaces cancel out, leavingjust the contribution from the outer surface. (See Figure A.22.) We can use Equa-tion (A.131) for each cube individually. This tells us that the summation is equivalentto∫∇·A dV over the whole volume. Thus, the integral of A·dS over the outer surface

is equal to the integral of ∇ · A over the whole volume, which proves the divergencetheorem.

Now, for a vector field with ∇ · A = 0,∮S

A · dS = 0 (A.135)

for any closed surface S . So, for two surfaces, S 1 and S 2, on the same rim,∫S 1

A · dS =∫

S 2

A · dS, (A.136)


S

rim

S2

S1

Figure A.23Two surfaces spanning the same rim (right), and the equivalent closed surface (left).

as illustrated in Figure A.23. (Note that the direction of the surface elements onS 1 has been reversed relative to those on the closed surface. Hence, the sign of theassociated surface integral is also reversed.) Thus, if ∇ · A = 0 then the surfaceintegral depends on the rim, but not on the nature of the surface that spans it. On theother hand, if ∇ · A 0 then the integral depends on both the rim and the surface.

Consider an incompressible fluid whose velocity field is v. It is clear that∮

v ·dS = 0 for any closed surface, because what flows into the surface must flow outagain. Thus, according to the divergence theorem,

∫∇ · v dV = 0 for any volume.

The only way in which this is possible is if ∇·v is everywhere zero. Thus, the velocitycomponents of an incompressible fluid satisfy the following differential relation:

∂vx

∂x+∂vy

∂y+∂vz∂z= 0. (A.137)

It is sometimes helpful to represent a vector field A by lines of force or field-lines.The direction of a line of force at any point is the same as the local direction of A.The density of lines (i.e., the number of lines crossing a unit surface perpendicular toA) is equal to |A|. For instance, in Figure A.24, |A| is larger at point 1 than at point2. The number of lines crossing a surface element dS is A · dS. So, the net numberof lines leaving a closed surface is

∮S

A · dS =∫

V∇ · A dV. (A.138)

If ∇ ·A = 0 then there is no net flux of lines out of any surface. Such a field is calleda solenoidal vector field. The simplest example of a solenoidal vector field is one inwhich the lines of force all form closed loops.


A.21 Laplacian OperatorSo far we have encountered

∇φ =(∂φ

∂x,∂φ

∂y,∂φ

∂z

), (A.139)

which is a vector field formed from a scalar field, and

∇ · A = ∂Ax


∂z, (A.140)

which is a scalar field formed from a vector field. There are two ways in which wecan combine gradient and divergence. We can either form the vector field ∇(∇ · A)or the scalar field ∇ · (∇φ). The former is not particularly interesting, but the scalarfield ∇ · (∇φ) turns up in a great many physical problems, and is, therefore, worthyof discussion.

Let us introduce the heat flow vector h, which is the rate of flow of heat energyper unit area across a surface perpendicular to the direction of h. In many substances,heat flows directly down the temperature gradient, so that we can write

h = −κ ∇T, (A.141)

where κ is the thermal conductivity. The net rate of heat flow∮

S h · dS out of someclosed surface S must be equal to the rate of decrease of heat energy in the volumeV enclosed by S . Thus, we have∮

Sh · dS = − ∂

∂t

(∫c T dV

), (A.142)

where c is the specific heat. It follows from the divergence theorem that

∇ · h = −c∂T∂t. (A.143)

21

Figure A.24Divergent lines of force.


Taking the divergence of both sides of Equation (A.141), and making use ofEquation (A.143), we obtain

∇ · (κ∇T ) = c∂T∂t. (A.144)

If κ is constant then the previous equation can be written

∇ · (∇T ) =cκ

∂T∂t. (A.145)

The scalar field ∇ · (∇T ) takes the form

∇ · (∇T ) =∂

∂x

(∂T∂x

)+∂

∂y

(∂T∂y

)+∂

∂z

(∂T∂z

)=∂ 2T∂x 2 +

∂ 2T∂y 2 +

∂ 2T∂z 2 ≡ ∇

2T. (A.146)

Here, the scalar differential operator

∇ 2 ≡ ∂ 2

∂x 2 +∂ 2

∂y 2 +∂ 2

∂z 2 (A.147)

is called the Laplacian. The Laplacian is a good scalar operator (i.e., it is coordinateindependent) because it is formed from a combination of divergence (another goodscalar operator) and gradient (a good vector operator).

What is the physical significance of the Laplacian? In one dimension, ∇ 2T re-duces to ∂ 2T/∂x 2. Now, ∂ 2T/∂x 2 is positive if T (x) is concave (from above), andnegative if it is convex. So, if T is less than the average of T in its surroundings then∇ 2T is positive, and vice versa.

In two dimensions,

∇ 2T =∂ 2T∂x 2 +

∂ 2T∂y 2 . (A.148)

Consider a local minimum of the temperature. At the minimum, the slope of Tincreases in all directions, so ∇ 2T is positive. Likewise, ∇ 2T is negative at a localmaximum. Consider, now, a steep-sided valley in T . Suppose that the bottom of thevalley runs parallel to the x-axis. At the bottom of the valley ∂ 2T/∂y 2 is large andpositive, whereas ∂ 2T/∂x 2 is small and may even be negative. Thus, ∇ 2T is positive,and this is associated with T being less than the average local value.

Let us now return to the heat conduction problem:

∇ 2T =cκ

∂T∂t. (A.149)

It is clear that if ∇ 2T is positive in some small region then the value of T there is lessthan the local average value, so ∂T/∂t > 0: that is, the region heats up. Likewise,if ∇ 2T is negative then the value of T is greater than the local average value, andheat flows out of the region: that is, ∂T/∂t < 0. Thus, the previous heat conductionequation makes physical sense.


A.22 CurlConsider a vector field A(r), and a loop that lies in one plane. The integral of Aaround this loop is written

∮A · dr, where dr is a line element of the loop. If A is

a conservative field then A = ∇φ and∮

A · dr = 0 for all loops. In general, for anon-conservative field,

∮A · dr 0.

For a small loop, we expect∮

A · dr to be proportional to the area of the loop.Moreover, for a fixed-area loop, we expect

∮A·dr to depend on the orientation of the

loop. One particular orientation will give the maximum value:∮

A · dr = Imax. If theloop subtends an angle θ with this optimum orientation then we expect I = Imax cos θ.Let us introduce the vector field curl A whose magnitude is

|curl A| = limdS→0

∮A · drdS

(A.150)

for the orientation giving Imax. Here, dS is the area of the loop. The direction ofcurl A is perpendicular to the plane of the loop, when it is in the orientation givingImax, with the sense given by a right-hand circulation rule.

Let us now express curl A in terms of the components of A. First, we shallevaluate

∮A · dr around a small rectangle in the y-z plane, as shown in Figure A.25.

The contribution from sides 1 and 3 is

Az(y + dy) dz − Az(y) dz =∂Az

∂ydy dz. (A.151)

The contribution from sides 2 and 4 is

−Ay(z + dz) dy + Ay(z) dy = −∂Ay∂y

dy dz. (A.152)

So, the total of all contributions gives∮A · dr =

(∂Az

∂y−∂Ay∂z

)dS , (A.153)

where dS = dy dz is the area of the loop.Consider a non-rectangular (but still small) loop in the y-z plane. We can divide it

into rectangular elements, and form∮

A · dr over all the resultant loops. The interiorcontributions cancel, so we are just left with the contribution from the outer loop.Also, the area of the outer loop is the sum of all the areas of the inner loops. Weconclude that ∮

A · dr =(∂Az

∂y−∂Ay∂z

)dS x (A.154)

is valid for a small loop dS = (dS x, 0, 0) of any shape in the y-z plane. Likewise, wecan show that if the loop is in the x-z plane then dS = (0, dS y, 0) and∮

A · dr =(∂Ax

∂z−∂Az

∂x

)dS y. (A.155)


z

z + dz

1

4

3

zy 2 y + dy

y

Figure A.25A vector line integral around a small rectangular loop in the y-z plane.

Finally, if the loop is in the x-y plane then dS = (0, 0, dS z) and∮A · dr =

(∂Ay∂x−∂Ax

∂y

)dS z. (A.156)

Imagine an arbitrary loop of vector area dS = (dS x, dS y, dS z). We can constructthis out of three vector areas, 1, 2, and 3, directed in the x-, y-, and z-directions,respectively, as indicated in Figure A.26. If we form the line integral around all threeloops then the interior contributions cancel, and we are left with the line integralaround the original loop. Thus,∮

A · dr =∮

A · dr1 +

∮A · dr2 +

∮A · dr3, (A.157)

giving ∮A · dr = curl A · dS = |curl A| |dS| cos θ, (A.158)

where

curl A =(∂Az

∂y−∂Ay∂z,∂Ax

∂z− ∂Az

∂x,∂Ay∂x− ∂Ax

∂y

), (A.159)

and θ is the angle subtended between the directions of curl A and dS. Note that

curl A = ∇ × A. (A.160)

This demonstrates that ∇×A is a good vector field, because it is the cross product ofthe ∇ operator (a good vector operator) and the vector field A.

Consider a solid body rotating about the z-axis. The angular velocity is given byω = (0, 0, ω), so the rotation velocity at position r is

v = ω × r. (A.161)


dS

yx

3

2 1

z

Figure A.26Decomposition of a vector area into its Cartesian components.

[See Equation (A.52).] Let us evaluate∇×v on the axis of rotation. The x-componentis proportional to the integral

∮v · dr around a loop in the y-z plane. This is plainly

zero. Likewise, the y-component is also zero. The z-component is∮

v ·dr/dS aroundsome loop in the x-y plane. Consider a circular loop. We have

∮v ·dr = 2π rω r with

dS = π r2. Here, r is the perpendicular distance from the rotation axis. It followsthat (∇ × v)z = 2ω, which is independent of r. So, on the axis, ∇ × v = (0 , 0 , 2ω).Off the axis, at position r0, we can write

v = ω × (r − r0) +ω × r0. (A.162)

The first part has the same curl as the velocity field on the axis, and the second parthas zero curl, because it is constant. Thus, ∇ × v = (0, 0, 2ω) everywhere in thebody. This allows us to form a physical picture of ∇ × A. If we imagine A(r) asthe velocity field of some fluid then ∇ × A at any given point is equal to twice thelocal angular rotation velocity: that is, 2ω. Hence, a vector field with ∇ × A = 0everywhere is said to be irrotational.

Another important result of vector field theory is the curl theorem:∮C

A · dr =∫

S∇ × A · dS, (A.163)

for some (non-planar) surface S bounded by a rim C. This theorem can easily beproved by splitting the loop up into many small rectangular loops, and forming theintegral around all of the resultant loops. All of the contributions from the interiorloops cancel, leaving just the contribution from the outer rim. Making use of Equa-tion (A.158) for each of the small loops, we can see that the contribution from all of


the loops is also equal to the integral of ∇ × A · dS across the whole surface. Thisproves the theorem.

One immediate consequence of the curl theorem is that∇×A is “incompressible.”Consider any two surfaces, S 1 and S 2, that share the same rim. (See Figure A.23.) Itis clear from the curl theorem that

∫∇ × A · dS is the same for both surfaces. Thus,

it follows that∮∇ × A · dS = 0 for any closed surface. However, we have from the

divergence theorem that∮∇×A · dS =

∫∇ · (∇×A) dV = 0 for any volume. Hence,

∇ · (∇ × A) ≡ 0. (A.164)

So, ∇ × A is a solenoidal field.

We have seen that for a conservative field∮

A · dr = 0 for any loop. This isentirely equivalent to A = ∇φ. However, the magnitude of ∇ × A is lim dS→0

∮A ·

dr/dS for some particular loop. It is clear then that ∇ × A = 0 for a conservativefield. In other words,

∇ × (∇φ) ≡ 0. (A.165)

Thus, a conservative field is also an irrotational one.

A.23 Useful Vector Identities

Notation: a, b, c, d are general vectors; φ, ψ are general scalar fields; A, B are generalvector fields; (A ·∇) B ≡ (A ·∇Bx, A ·∇By, A ·∇Bz) and ∇ 2A = (∇ 2Ax, ∇ 2Ay, ∇ 2Az)


(but, only in Cartesian coordinates—see Appendix C).

a × (b × c) = (a · c) b − (a · b) c, (A.166)

(a × b) × c = (c · a) b − (c · b) a, (A.167)

(a × b) · (c × d) = (a · c) (b · d) − (a · d) (b · c), (A.168)

(a × b) × (c × d) = (a × b · d) c − (a × b · c) d, (A.169)

∇(φψ) = φ∇ψ + ψ∇φ, (A.170)

∇(A · B) = A × (∇ × B) + B × (∇ × A) + (A · ∇) B + (B · ∇) A, (A.171)

∇ · ∇φ = ∇ 2φ, (A.172)

∇ · ∇ × A = 0, (A.173)

∇ · (φA) = φ∇ · A + A · ∇φ, (A.174)

∇ · (A × B) = B · ∇ × A − A · ∇ × B, (A.175)

∇ × ∇φ = 0, (A.176)

∇ × (∇ × A) = ∇ (∇ · A) − ∇ 2A, (A.177)

∇ × (φA) = φ∇ × A + ∇φ × A, (A.178)

∇ × (A × B) = (∇ · B) A − (∇ · A) B + (B · ∇) A − (A · ∇) B. (A.179)

A.24 ExercisesA.1 The position vectors of the four points A, B, C, and D are a, b, 3 a+ 2 b, and

a − 3 b, respectively. Express→AC,

→DB,

→BC, and

→CD in terms of a and b.

A.2 Prove the trigonometric law of sines

sin aA=

sin bB=

sin cC

using vector methods. Here, a, b, and c are the three angles of a planetriangle, and A, B, and C the lengths of the corresponding opposite sides.

A.3 Demonstrate using vectors that the diagonals of a parallelogram bisect oneanother. In addition, show that if the diagonals of a quadrilateral bisect oneanother then it is a parallelogram.

A.4 From the inequalitya · b = |a| |b| cos θ ≤ |a| |b|

deduce the triangle inequality

|a + b| ≤ |a| + |b|.


A.5 Find the scalar product a · b and the vector product a × b when

(a) a = ex + 3 ey − ez, b = 3 ex + 2 ey + ez,

(b) a = ex − 2 ey + ez, b = 2 ex + ey + ez.

A.6 Which of the following statements regarding the three general vectors a, b,and c are true?

(a) c · (a × b) = (b × a) · c.

(b) a × (b × c) = (a × b) × c.

(c) a × (b × c) = (a · c) b − (a · b) c.

(d) d = λ a + µb implies that (a × b) · d = 0.

(e) a × c = b × c implies that c · a − c · b = c |a − b|.(f) (a × b) × (c × b) = [b · (c × a)] b.

A.7 Prove that the length of the shortest straight-line from point a to the straight-line joining points b and c is

|a × b + b × c + c × a||b − c|

.

A.8 Identify the following surfaces:

(a) |r| = a,

(b) r · n = b,

(c) r · n = c |r|,(d) |r − (r · n) n| = d.

Here, r is the position vector, a, b, c, and d are positive constants, and n is afixed unit vector.

A.9 Let a, b, and c be coplanar vectors related via

α a + βb + γ c = 0,

where α, β, and γ are not all zero. Show that the condition for the pointswith position vectors u a, v b, and w c to be colinear is

α

u+β

v+γ

w= 0.

A.10 If p, q, and r are any vectors, demonstrate that a = q + λ r, b = r + µp,and c = p + νq are coplanar provided that λ µ ν = −1, where λ, µ, and ν arescalars. Show that this condition is satisfied when a is perpendicular to p, bto q, and c to r.


A.11 The vectors a, b, and c are non-coplanar, and form a non-orthogonal vectorbase. The vectors A, B, and C, defined by

A =b × c

a · b × c,

plus cyclic permutations, are said to be reciprocal vectors. Show that

a = (B × C)/(A · B × C),

plus cyclic permutations.

A.12 In the notation of the previous exercise, demonstrate that the plane passingthrough points a/α, b/β, and c/γ is normal to the direction of the vector

h = αA + βB + γC.

In addition, show that the perpendicular distance of the plane from the originis |h|−1.

A.13 Evaluate∮

A · dr for

A =x ex + y ey√

x 2 + y 2

around the square whose sides are x = 0, x = a, y = 0, y = a.

A.14 Consider the following vector field:

A(r) = (8 x 3 + 3 x 2 y 2, 2 x 3 y + 6 y, 6).

Is this field conservative? Is it solenoidal? Is it irrotational? Justify youranswers. Calculate

∮C A · dr, where the curve C is a unit circle in the x-y

plane, centered on the origin, and the direction of integration is clockwiselooking down the z-axis.

A.15 Consider the following vector field:

A(r) = (3 x y 2 z 2 − y 2, −y 3 z 2 + x 2 y, 3 x 2 − x 2 z).

Is this field conservative? Is it solenoidal? Is it irrotational? Justify youranswers. Calculate the flux of A out of a unit sphere centered on the origin.

A.16 Find the gradients of the following scalar functions of the position vectorr = (x, y, z):

(a) k · r,

(b) |r| n,(c) |r − k|−n,

(d) cos(k · r).


Here, k is a fixed vector.

A.17 Find the divergences and curls of the following vector fields:

(a) k × r,

(b) |r| n r,

(c) |r − k|n (r − k),

(d) a cos(k · r).

Here, k and a are fixed vectors.

A.18 Calculate ∇ 2φ when φ = f (|r|). Find f if ∇ 2φ = 0.

BCartesian Tensors

B.1 IntroductionAs we saw in Appendix A, many physical entities can be represented mathemati-cally as either scalars or vectors, depending on their transformation properties underrotation of the coordinate axes. However, it turns out that scalars and vectors are par-ticular types of a more general class of mathematical constructs known as tensors.In fact, a scalar is a tensor of order zero, and a vector is a tensor of order one. Influid mechanics, certain important physical entities (i.e., stress and rate of strain) arerepresented mathematically by tensors of order greater than one. It is therefore nec-essary to supplement our investigation of fluid mechanics with a brief discussion ofthe mathematics of tensors. For the sake of simplicity, we shall limit this discussionto Cartesian coordinate systems. Tensors expressed in such coordinate systems areknown as Cartesian tensors. For more information on Cartesian tensors see Jeffries1961, Riley 1974, and Temple 2004.

B.2 Tensors and Tensor NotationLet the Cartesian coordinates x, y, z be written as the xi, where i runs from 1 to 3. Inother words, x = x1, y = x2, and z = x3. Incidentally, in the following, any lowercaseroman subscript (e.g., i, j, k) is assumed to run from 1 to 3. We can also write theCartesian components of a general vector v as the vi. In other words, vx = v1, vy = v2,and vz = v3. By contrast, a scalar is represented as a variable without a subscript:for instance, a, φ. Thus, a scalar—which is a tensor of order zero—is representedas a variable with zero subscripts, and a vector—which is a tensor of order one—isrepresented as a variable with one subscript. It stands to reason, therefore, that atensor of order two is represented as a variable with two subscripts: for instance,ai j, σi j. Moreover, an nth-order tensor is represented as a variable with n subscripts:for instance, ai jk is a third-order tensor, and bi jkl a fourth-order tensor. Note that ageneral nth-order tensor has 3n independent components.

The components of a second-order tensor are conveniently visualized as a two-dimensional matrix, just as the components of a vector are sometimes visualizedas a one-dimensional matrix. However, it is important to recognize that an nth-order

525


tensor is not simply another name for an n-dimensional matrix. A matrix is merely anordered set of numbers. A tensor, on the other hand, is an ordered set of componentsthat have specific transformation properties under rotation of the coordinate axes.(See Section B.3.)

Consider two vectors a and b that are represented as ai and bi, respectively, intensor notation. According to Section A.6, the scalar product of these two vectorstakes the form

a · b = a1 b1 + a2 b2 + a3 b3. (B.1)

The previous expression can be written more compactly as

a · b = ai bi. (B.2)

Here, we have made use of the Einstein summation convention, according to which,in an expression containing lower case roman subscripts, any subscript that appearstwice (and only twice) in any term of the expression is assumed to be summedfrom 1 to 3 (unless stated otherwise). Thus, ai bi = a1 b1 + a2 b2 + a3 b3, andai j b j = ai1 b1 + ai2 b2 + ai3 b3. Note that when an index is summed it becomes adummy index and can be written as any (unique) symbol: that is, ai j b j and aip bp

are equivalent. Moreover, only non-summed, or free, indices count toward the orderof a tensor expression. Thus, aii is a zeroth-order tensor (because there are no freeindices), and ai j b j is a first-order tensor (because there is only one free index). Theprocess of reducing the order of a tensor expression by summing indices is known ascontraction. For example, aii is a zeroth-order contraction of the second-order tensorai j. Incidentally, when two tensors are multiplied together without contraction theresulting tensor is called an outer product: for instance, the second-order tensor ai b j

is the outer product of the two first-order tensors ai and bi. Likewise, when two ten-sors are multiplied together in a manner that involves contraction then the resultingtensor is called an inner product: for instance, the first-order tensor ai j b j is an innerproduct of the second-order tensor ai j and the first-order tensor bi. It can be seenfrom Equation (B.2) that the scalar product of two vectors is equivalent to the innerproduct of the corresponding first-order tensors.

According to Section A.8, the vector product of two vectors a and b takes theform

(a × b)1 = a2 b3 − a3 b2, (B.3)

(a × b)2 = a3 b1 − a1 b3, (B.4)

(a × b)3 = a1 b2 − a2 b1 (B.5)

in tensor notation. The previous expression can be written more compactly as

(a × b)i = εi jk a j bk. (B.6)

Here,

εi jk =

+1 if i, j, k is an even permutation of 1, 2, 3−1 if i, j, k is an odd permutation of 1, 2, 30 otherwise

(B.7)

Cartesian Tensors 527

is known as the third-order permutation tensor (or, sometimes, the third-order Levi-Civita tensor). Note, in particular, that εi jk is zero if one of its indices is repeated: forinstance, ε113 = ε212 = 0. Furthermore, it follows from Equation (B.7) that

εi jk = ε jki = εki j = −εk ji = −ε jik = −εik j. (B.8)

It is helpful to define the second-order identity tensor (also known as the Kroe-necker delta tensor),

δi j =

1 if i = j0 otherwise

. (B.9)

It is easily seen that

δi j = δ ji, (B.10)

δii = 3, (B.11)

δik δk j = δi j, (B.12)

δi j a j = ai, (B.13)

δi j ai b j = ai bi, (B.14)

δi j aki b j = aki bi, (B.15)

et cetera.The following is a particularly important tensor identity:

εi jk εilm = δ jl δkm − δ jm δkl. (B.16)

In order to establish the validity of the previous expression, let us consider the variouscases that arise. As is easily seen, the right-hand side of Equation (B.16) takes thevalues

+1 if j = l and k = m j, (B.17)

−1 if j = m and k = l j, (B.18)

0 otherwise. (B.19)

Moreover, in each product on the left-hand side of Equation (B.16), i has the samevalue in both ε factors. Thus, for a non-zero contribution, none of j, k, l, and mcan have the same value as i (because each ε factor is zero if any of its indices arerepeated). Because a given subscript can only take one of three values (1, 2, or 3),the only possibilities that generate non-zero contributions are j = l and k = m, orj = m and k = l, excluding j = k = l = m (as each ε factor would then have repeatedindices, and so be zero). Thus, the left-hand side of Equation (B.16) reproducesEquation (B.19), as well as the conditions on the indices in Equations (B.17) and(B.18). The left-hand side also reproduces the values in Equations (B.17) and (B.18)because if j = l and k = m then εi jk = εilm and the product εi jk εilm (no summation)is equal to +1, whereas if j = m and k = l then εi jk = εiml = −εilm and the product


εi jk εilm (no summation) is equal to −1. Here, use has been made of Equation (B.8).Hence, the validity of the identity (B.16) has been established.

In order to illustrate the use of Equation (B.16), consider the vector triple productidentity (see Section A.11)

a × (b × c) = (a · c) b − (a · b) c. (B.20)

In tensor notation, the left-hand side of this identity is written

[a × (b × c)]i = εi jk a j (εklm bl cm), (B.21)

where use has been made of Equation (B.6). Employing Equations (B.8) and (B.16),this becomes

[a × (b × c)]i = εki j εklm a j bl cm =(δil δ jm − δim δ jl

)a j bl cm, (B.22)

which, with the aid of Equations (B.2) and (B.13), reduces to

[a × (b × c)]i = a j c j bi − a j b j ci = [(a · c) b − (a · b) c]i . (B.23)

Thus, we have established the validity of the vector identity (B.20). Moreover, ourproof is much more rigorous than that given earlier in Section A.11.

B.3 Tensor TransformationAs we saw in Appendix A, scalars and vectors are defined according to their trans-formation properties under rotation of the coordinate axes. In fact, a scalar is in-variant under rotation of the coordinate axes. On the other hand, according to Equa-tions (A.49) and (B.6), the components of a general vector a transform under aninfinitesimal rotation of the coordinate axes according to

a′i = ai + εi jk δθ j ak. (B.24)

Here, the ai are the components of the vector in the original coordinate system, the a′iare the components in the rotated coordinate system, and the latter system is obtainedfrom the former via a combination of an infinitesimal rotation through an angle δθ1about coordinate axis 1, an infinitesimal rotation through an angle δθ2 about axis 2,and an infinitesimal rotation through an angle δθ3 about axis 3. These three rotationscan take place in any order. Incidentally, a finite rotation can be built up out of a greatmany infinitesimal rotations, so if a vector transforms properly under an infinitesimalrotation of the coordinate axes then it will also transform properly under a finiterotation.

Equation (B.24) can also be written

a′i = Ri j a j, (B.25)


whereRi j = δi j − δθk εki j (B.26)

is a rotation matrix (which is not a tensor, because it is specific to the two coordinatesystems it transforms between). To first order in the δθi, Equation (B.25) can beinverted to give

ai = R ji a′j. (B.27)

This follows because, to first order in the δθi,

Rik R jk = (δik − δθl εlik) (δ jk − δθm εm jk) = δik δ jk − δθl δ jk εlik − δθm δik εm jk

= δi j − δθl εli j − δθl εl ji = δi j, (B.28)

where the dummy index m has been relabeled l, and use has been made of Equa-tions (B.8), (B.10), and (B.12). Likewise, it is easily demonstrated that

Rki Rk j = δi j. (B.29)

It can also be shown that, to first order in the δθi,

εi jk Rli Rm j Rnk = εlmn. (B.30)

This follows because

εi jk Rli Rm j Rnk = εi jk (δli − δθa εali) (δm j − δθb εbm j) (δnk − δθc εcnk)

= εi jk

[δli δm j δnk − δθa

(εali δm j δnk + εam j δli δnk + εank δli δm j

)]= εlmn − δθa (εimn εial + εinl εiam + εilm εian)

= εlmn − δθa (δma δnl − δml δna + δna δlm − δnm δla + δla δmn − δln δma)

= εlmn. (B.31)

Here, there has been much relabeling of dummy indices, and use has been made ofEquations (B.10) and (B.16). It can similarly be shown that

εi jk Ril R jm Rkn = εlmn. (B.32)

As a direct generalization of Equation (B.25), a second-order tensor transformsunder rotation as

a′i j = Rik R jl akl, (B.33)

whereas a third-order tensor transforms as

a′i jk = Ril R jm Rkn almn. (B.34)

The generalization to higher-order tensors is straightforward. For the case of a scalar,which is a zeroth-order tensor, the transformation rule is particularly simple: that is,

a′ = a. (B.35)


By analogy with Equation (B.27), the inverse transform is exemplified by

ai jk = Rli Rm j Rnk a′lmn. (B.36)

Incidentally, because all tensors of the same order transform in the same manner, itimmediately follows that two tensors of the same order whose components are equalin one particular Cartesian coordinate system will have their components equal in allcoordinate systems that can be obtained from the original system via rotation of thecoordinate axes. In other words, if

ai j = bi j (B.37)

in one particular Cartesian coordinate system then

a′i j = b′i j (B.38)

in all Cartesian coordinate systems (with the same origin and system of units as theoriginal system). Conversely, it does not make sense to equate tensors of differentorder, because such an equation would only be valid in one particular coordinatesystem, and so could not have any physical significance (because the laws of physicsare coordinate independent).

It can easily be shown that the outer product of two tensors transforms as a tensorof the appropriate order. Thus, if

ci jk = ai b jk, (B.39)

and

a′i = Ri j a j, (B.40)

b′i j = Rik R jl bkl, (B.41)

then

c′i jk = a′i b′jk = Ril al R jm Rkn bmn = Ril R jm Rkn al bmn

= Ril R jm Rkn clmn, (B.42)

which is the correct transformation rule for a third-order tensor.The tensor transformation rule can be combined with the identity (B.29) to show

that the scalar product of two vectors transforms as a scalar. Thus,

a′i b′i = Ri j a jRik bk = Ri j Rik a j bk = δ jk a j bk = a j b j = ai bi, (B.43)

where use has been made of Equation (B.14). Again, the previous proof is more rig-orous than that given in Section A.6. The proof also indicates that the inner productof two tensors transforms as a tensor of the appropriate order.

The result that both the inner and outer products of two tensors transform as


tensors of the appropriate order is known as the product rule. Closely related to thisrule is the so-called quotient rule, according to which if (say)

ci j = aik b jk, (B.44)

where b jk is an arbitrary tensor, and ci j transforms as a tensor under all rotationsof the coordinate axes, then aik—which can be thought of as the quotient of ci j andb jk—also transforms as a tensor. The proof is as follows:

a′ik b′jk = c′i j = Ril R jm clm = Ril R jm alk bmk = Ril R jm alk Rpm Rqk b′pq

= Ril Rqk alk b jq = Ril Rkm alm b′jk, (B.45)

where use has been made of the fact that ci j and bi j transform as tensors, as well asEquation (B.28). Rearranging, we obtain

(a′ik − Ril Rkm alm) b′jk = 0. (B.46)

However, the b′i j are arbitrary, so the previous equation can only be satisfied, in gen-eral, if

a′ik = Ril Rkm alm, (B.47)

which is the correct transformation rule for a tensor. Incidentally, the quotient ruleapplies to any type of valid tensor product.

The components of the second-order identity tensor, δi j, have the special propertythat they are invariant under rotation of the coordinate axes. This follows because

δ′i j = Rik R jl δkl = Rik R jk = δi j, (B.48)

where use has been made of Equation (B.28). The components of the third-orderpermutation tensor, εi jk, also have this special property. This follows because

ε′i jk = Ril R jm Rln εlmn = εi jk, (B.49)

where use has been made of Equation (B.31). The fact that εi jk transforms as a properthird-order tensor immediately implies, from the product rule, that the vector productof two vectors transforms as a proper vector: that is, εi jk a j bk is a first-order tensorprovided that ai and bi are both first-order tensors. This proof is much more rigorousthat that given earlier in Section A.8.

B.4 Tensor FieldsWe saw in Appendix A that a scalar field is a set of scalars associated with everypoint in space: for instance, φ(x), where x = (x1, x2, x3) is a position vector. Wealso saw that a vector field is a set of vectors associated with every point in space:


for instance, ai(x). It stands to reason, then, that a tensor field is a set of tensorsassociated with every point in space: for instance, ai j(x). It immediately follows thata scalar field is a zeroth-order tensor field, and a vector field is a first-order tensorfield.

Most tensor fields encountered in physics are smoothly varying and differen-tiable. Consider the first-order tensor field ai(x). The various partial derivatives ofthe components of this field with respect to the Cartesian coordinates xi are written

∂ai

∂x j. (B.50)

Moreover, this set of derivatives transform as the components of a second-order ten-sor. In order to demonstrate this, we need the transformation rule for the xi, which isthe same as that for a first-order tensor: that is,

x′i = Ri j x j. (B.51)

Thus,∂x′i∂x j= Ri j. (B.52)

It is also easily shown that∂xi

∂x′j= R ji. (B.53)

Now,∂a′i∂x′j=∂a′i∂xk

∂xk

∂x′j=∂(Ril al)∂xk

R jk = Ril R jk∂al

∂xk, (B.54)

which is the correct transformation rule for a second-order tensor. Here, use has beenmade of the chain rule, as well as Equation (B.53). [Note, from Equation (B.26),that the Ri j are not functions of position.] It follows, from the previous argument,that differentiating a tensor field increases its order by one: for instance, ∂ai j/∂xk

is a third-order tensor. The only exception to this rule occurs when differentiationand contraction are combined. Thus, ∂ai j/∂x j is a first-order tensor, because it onlycontains a single free index.

The gradient (see Section A.18) of a scalar field is an example of a first-ordertensor field (i.e., a vector field):

(∇φ)i =∂φ

∂xi. (B.55)

The divergence (see Section A.20) of a vector field is a contracted second-order ten-sor field that transforms as a scalar:

∇ · a = ∂ai

∂xi. (B.56)

Finally, the curl (see Section A.22) of a vector field is a contracted fifth-order tensorthat transforms as a vector

(∇ × a)i = εi jk∂ak

∂x j. (B.57)


The previous definitions can be used to prove a number of useful results. Forinstance,

(∇ × ∇φ)i = εi jk∂

∂x j

(∂φ

∂xk

)= εi jk

∂ 2φ

∂x j ∂xk= 0, (B.58)

which follows from symmetry because εik j = −εi jk whereas ∂ 2φ/∂xk ∂x j = ∂2φ/∂x j ∂xk.

Likewise,

∇ · (∇ × a) =∂

∂xi

(εi jk

∂ak

∂x j

)= εi jk

∂ak

∂xi ∂x j= 0, (B.59)

which again follows from symmetry. As a final example,

∇ · (a × b) =∂

∂xi

(εi jk a j bk

)= εi jk

∂a j

∂xibk + εi jk a j

∂bk

∂xi

= bi εi jk∂ak

∂x j− ai εi jk

∂bk

∂x j= b · (∇ × a) − a · (∇ × b). (B.60)

According to the divergence theorem (see Section A.20),∮S

ai dS i =

∫V

∂ai

∂xidV, (B.61)

where S is a closed surface surrounding the volume V . The previous theorem iseasily generalized to give, for example,∮

Sai j dS i =

∫V

∂ai j

∂xidV, (B.62)

or ∮S

ai j dS j =

∫V

∂ai j

∂x jdV, (B.63)

or even ∮S

a dS i =

∫V

∂a∂xi

dV. (B.64)

B.5 Isotropic TensorsA tensor which has the special property that its components take the same valuein all Cartesian coordinate systems is called an isotropic tensor. We have alreadyencountered two such tensors: namely, the second-order identity tensor, δi j, and thethird-order permutation tensor, εi jk. Of course, all scalars are isotropic. Moreover,as is easily demonstrated, there are no isotropic vectors (other than the null vector).It turns out that the most general isotropic Cartesian tensors of second-, third-, andfourth-order are λ δi j, µ εi jk, and α δi j δkl + β δik δ jl + γ δil δ jk, respectively, where λ,µ, α, β, and γ are scalars. Let us prove these important results (Hodge 1961).


The most general second-order isotropic tensor, ai j, is such that

a′i j = Rip R jq apq = ai j (B.65)

for arbitrary rotations of the coordinate axes. It follows from Equation (B.24) that,to first order in the δθi,

δθm

(εmis as j + εm js ais

)= 0. (B.66)

However, the δθi are arbitrary, so we can write

εmis as j + εm js ais = 0. (B.67)

Let us multiply by εmik. With the aid of Equation (B.16), we obtain

(δii δks − δis δki) as j + (δi j δks − δis δk j) ais = 0, (B.68)

which reduces to2 ai j + a ji = ass δi j. (B.69)

Interchanging the labels i and j, and then taking the difference between the twoequations thus obtained, we deduce that

ai j = a ji. (B.70)

Hence,ai j =

ass

3δi j, (B.71)

which implies thatai j = λ δi j. (B.72)

For the case of an isotropic third-order tensor, Equation (B.67) generalizes to

εmis as jk + εm js aisk + εmks ai js = 0. (B.73)

Multiplying by εmit, εm jt, and εmkt, and then setting t = i, t = j, and t = k, respectively,we obtain

2 ai jk + a jik + ak ji = assk δi j + as js δik, (B.74)

2 ai jk + a jik + aik j = assk δi j + aiss δ jk, (B.75)

2 ai jk + ak ji + aik j = as js δik + aiss δ jk, (B.76)

respectively. However, multiplying the previous equations by δ jk, δik, and δi j, andthen setting i = i, j = i, and k = i, respectively, we obtain

2 aiss + asis + assi = assi + asis, (B.77)

2 asis + aiss + assi = assi + aiss, (B.78)

2 assi + aiss + asis = asis + aiss, (B.79)


respectively, which implies that

aiss = asis = assi = 0. (B.80)

Hence, we deduce that

2 ai jk + a jik + ak ji = 0, (B.81)

2 ai jk + a jik + aik j = 0, (B.82)

2 ai jk + ak ji + aik j = 0. (B.83)

The solution to the previous equation must satisfy

aik j = a jik = ak ji = −ai jk. (B.84)

This implies, from Equation (B.8), that

ai jk = µ εi jk. (B.85)

For the case of an isotropic fourth-order tensor, Equation (B.73) generalizes to

εmis as jkl + εm js aiskl + εmks ai jsl + εmls ai jks = 0. (B.86)

Multiplying the previous by εmit, εm jt, εmkt, εmlt, and then setting t = i, t = j, t = k,and t = l, respectively, we obtain

2 ai jkl + a jikl + ak jil + al jki = asskl δi j + as jsl δik + as jks δil, (B.87)

2 ai jkl + a jikl + aik jl + ail jk = asskl δi j + aisks δ jl + aissl δ jk, (B.88)

2 ai jkl + ak jil + aik jl + ai jlk = ai jss δkl + as jsl δik + aissl δ jk, (B.89)

2 ai jkl + al jki + ailk j + ai jlk = ai jss δkl + aisks δ jl + as jks δil, (B.90)

respectively. Now, if ai jkl is an isotropic fourth-order tensor then asskl is clearly anisotropic second-order tensor, which means that is a multiple of δkl. This, and similararguments, allows us to deduce that

asskl = λ δkl, (B.91)

as jsl = µ δ jl, (B.92)

as jks = ν δ jk. (B.93)

Let us assume, for the moment, that

ai jss = assi j, (B.94)

aisks = asisk, (B.95)

aissl = asils. (B.96)


Thus, we get

2 ai jkl + a jikl + ak jil + al jki = λ δi j δkl + µ δik δ jl + ν δil δ jk, (B.97)

2 ai jkl + a jikl + aik jl + ailk j = λ δi j δkl + µ δik δ jl + ν δil δ jk, (B.98)

2 ai jkl + ak jil + aik jl + ai jlk = λ δi j δkl + µ δik δ jl + ν δil δ jk, (B.99)

2 ai jkl + al jki + ailk j + ai jlk = λ δi j δkl + µ δik δ jl + ν δil δ jk. (B.100)

Relations of the formai jkl = a jilk = akli j = alk ji (B.101)

can be obtained by subtracting the sum of one pair of Equations (B.97)–(B.100)from the sum of the other pair. These relations justify Equations (B.94)–(B.96).Equations (B.97) and (B.101) can be combined to give

2 ai jkl + (ai jlk + aik jl + ailk j) = λ δi j δkl + µ δik δ jl + ν δil δ jk, (B.102)

2 aikl j + (aik jl + ailk j + ai jlk) = λ δik δ jl + µ δil δ jk + ν δi j δkl, (B.103)

2 ail jk + (ailk j + ai jlk + aik jl) = λ δil δ jk + µ δi j δkl + ν δik δ jl. (B.104)

The latter two equations are obtained from the first via cyclic permutation of j, k,and l, with i remaining unchanged. Summing Equations (B.102)–(B.104), we get

2 (ai jkl+aikl j+ail jk)+3 (ai jlk+aik jl+ailk j) = (λ+µ+ν) (δi j δkl+δik δ jl+δil δ jk). (B.105)

It follows from symmetry that

ai jkl+aikl j+ail jk = ai jlk+aik jl+ailk j =15

(λ+µ+ν) (δi j δkl+δik δ jl+δil δ jk). (B.106)

This can be seen by swapping the indices k and l in the previous expression. Finally,substitution into Equation (B.102) yields

ai jkl = α δi j δkl + β δik δ jl + γ δil δ jk, (B.107)

where

α = (4 λ − µ − ν)/10, (B.108)

β = (4 µ − ν − λ)/10, (B.109)

γ = (4 ν − λ − µ)/10. (B.110)

B.6 ExercisesB.1 Show that a general second-order tensor ai j can be decomposed into three

tensorsai j = ui j + vi j + si j,


where ui j is symmetric (i.e., u ji = ui j) and traceless (i.e., uii = 0), vi j isisotropic, and si j only has three independent components.

B.2 Use tensor methods to establish the following vector identities:

(a) a · b × c = a × b · c = −b · a × c.

(b) (a × b) × c = (a · c) b − (c · b) a.

(c) (a × b) · (c × d) = (a · c) (b · d) − (a · d) (b · c).

(d) (a × b) × (c × d) = (a × b · d) c − (a × b · c) d.

(e) ∇ · (φ a) = φ∇ · a + a · ∇φ.

(f) ∇ × (φ a) = φ∇ × a + ∇φ × a.

(g) ∇ × (a × b) = (b · ∇) a − (a · ∇) b + (∇ · b) a − (∇ · a) b.

(h) ∇(a · a) = 2 a × (∇ × a) + 2 (a · ∇) a.

(i) ∇ × (∇ × a) = ∇(∇ · a) − ∇ 2a.

Here, [(b · ∇)a]i = b j ∂ai/∂x j, and (∇ 2a)i = ∇ 2ai.

B.3 A quadric surface has an equation of the form

a x 21 + b x 2

2 + c x 23 + 2 f x1 x2 + 2 g x1 x3 + 2 h x2 x3 = 1.

Show that the coefficients in the previous expression transform under rota-tion of the coordinate axes like the components of a symmetric second-ordertensor. Hence, demonstrate that the equation for the surface can be writtenin the form

xi Ti j x j = 1,

where the Ti j are the components of the aforementioned tensor.

B.4 The determinant of a second-order tensor Ai j is defined

det(A) = εi jk Ai1 A j2 Ak3.

(a) Show thatdet(A) = εi jk A1i A2 j A3k

is an alternative, and entirely equivalent, definition.

(b) Demonstrate that det(A) is invariant under rotation of the coordinateaxes.

(c) Suppose that Ci j = Aik Bk j. Show that

det(C) = det(A) det(B).


B.5 IfAi j x j = λ xi

then λ and x j are said to be eigenvalues and eigenvectors of the second-ordertensor Ai j, respectively. The eigenvalues of Ai j are calculated by solving therelated homogeneous matrix equation

(Ai j − λ δi j) x j = 0.

Now, it is a standard result in linear algebra that an equation of the previousform only has a non-trivial solution when (Riley 1974)

det(Ai j − λ δi j) = 0.

Demonstrate that the eigenvalues of Ai j satisfy the cubic polynomial

λ 3 − tr(A) λ 2 + Π(A) λ − det(A) = 0,

where tr(A) = Aii and Π(A) = (Aii A j j − Ai j A ji)/2. Hence, deduce that Ai j

possesses three eigenvalues—λ1, λ2, and λ3 (say). Moreover, show that

tr(A) = λ1 + λ2 + λ3,

det(A) = λ1 λ2 λ3.

B.6 Suppose that Ai j is a (real) symmetric second-order tensor: that is, A ji = Ai j.

(a) Demonstrate that the eigenvalues of Ai j are all real, and that the eigen-vectors can be chosen to be real.

(b) Show that eigenvectors of Ai j corresponding to different eigenvaluesare orthogonal to one another. Hence, deduce that the three eigenvec-tors of Ai j are, or can be chosen to be, mutually orthogonal.

(c) Demonstrate that Ai j takes the diagonal form Ai j = λi δi j (no sum) ina Cartesian coordinate system in which the coordinate axes are eachparallel to one of the eigenvectors.

B.7 In an isotropic elastic medium under stress, the displacement ui satisfies

∂σi j

∂x j= ρ

∂ 2ui

∂t 2 ,

σi j = ci jkl12

(∂uk

∂xl+∂ul

∂xk

),

where σi j is the stress tensor (note thatσ ji = σi j), ρ the mass density (whichis a uniform constant), and

ci jkl = K δi j δkl + µ [δik δ jl + δil δ jk − (2/3) δi j δkl].

the isotropic stiffness tensor. Here, K and µ are the bulk modulus and shearmodulus of the medium, respectively. Show that the divergence and thecurl of u both satisfy wave equations. Furthermore, demonstrate that thecharacteristic wave velocities of the divergence and curl waves are [(K +4µ/3)/ρ]1/2 and (µ/ρ)1/2, respectively.

CNon-Cartesian Coordinates

C.1 IntroductionNon-Cartesian coordinates are often employed in fluid mechanics to exploit the sym-metry of particular fluid systems. For example, it is convenient to employ cylindricalcoordinates to describe systems possessing axial symmetry. In this appendix, weinvestigate a particularly useful class of non-Cartesian coordinates known as orthog-onal curvilinear coordinates. The two most commonly occurring examples of thisclass in fluid mechanics are cylindrical and spherical coordinates. (Incidentally, theEinstein summation convention is not used in this appendix.) For more informationon orthogonal curvilinear coordinates see Milne-Thomson 2011.

C.2 Orthogonal Curvilinear CoordinatesLet x1, x2, x3 be a set of standard right-handed Cartesian coordinates. Furthermore,let u1(x1, x2, x3), u2(x1, x2, x3), u3(x1, x2, x3) be three independent functions of thesecoordinates which are such that each unique triplet of x1, x2, x3 values is associatedwith a unique triplet of u1, u2, u3 values. It follows that u1, u2, u3 can be used asan alternative set of coordinates to distinguish different points in space. Because thesurfaces of constant u1, u2, and u3 are not generally parallel planes, but rather curvedsurfaces, this type of coordinate system is termed curvilinear.

Let h1 = |∇u1|−1, h2 = |∇u2|−1, and h3 = |∇u3|−1. It follows that e1 = h1 ∇u1,e2 = h2 ∇u2, and e3 = h3 ∇u3 are a set of unit basis vectors that are normal to surfacesof constant u1, u2, and u3, respectively, at all points in space. Note, however, that thedirection of these basis vectors is generally a function of position. Suppose that theei, where i runs from 1 to 3, are mutually orthogonal at all points in space: that is,

ei · e j = δi j. (C.1)

In this case, u1, u2, u3 are said to constitute an orthogonal coordinate system. Sup-pose, further, that

e1 · e2 × e3 = 1 (C.2)

at all points in space, so that u1, u2, u3 also constitute a right-handed coordinate

539


system. It follows thatei · e j × ek = εi jk. (C.3)

Finally, a general vector A, associated with a particular point in space, can be written

A =∑i=1,3

Ai ei, (C.4)

where the ei are the local basis vectors of the u1, u2, u3 system, and Ai = ei · A istermed the ith component of A in this system.

Consider two neighboring points in space whose coordinates in the u1, u2, u3

system are u1, u2, u3 and u1 + du1, u2 + du2, u3 + du3. It is easily shown that thevector directed from the first to the second of these points takes the form

dx =du1

|∇u1|e1 +

du2

|∇u2|e2 +

du3

|∇u3|e3 =

∑i=1,3

hi dui ei. (C.5)

Hence, from (C.1), an element of length (squared) in the u1, u2, u3 coordinate systemis written

dx · dx =∑i=1,3

h 2i du 2

i . (C.6)

Here, the hi, which are generally functions of position, are known as the scalefactors of the system. Elements of area that are normal to e1, e2, and e3, at agiven point in space, take the form dS 1 = h2 h3 du2 du3, dS 2 = h1 h3 du1 du3, anddS 3 = h1 h2 du1 du2, respectively. Finally, an element of volume, at a given point inspace, is written dV = h du1 du2 du3, where

h = h1 h2 h3. (C.7)

It can be seen that [see Equation (A.176)]

∇ × ∇ui = 0, (C.8)

and

∇ ·h 2

i

h∇ui

= 0. (C.9)

The latter result follows from Equations (A.175) and (A.176) because (h 21 /h)∇u1 =

∇u2×∇u3, et cetera. Finally, it is easily demonstrated from Equations (C.1) and (C.3)that

∇ui · ∇u j = h−2i δi j, (C.10)

∇ui · ∇u j × ∇uk = h−1 εi jk. (C.11)

Consider a scalar field φ(u1, u2, u3). It follows from the chain rule, and the rela-tion ei = hi ∇ui, that

∇φ =∑i=1,3

∂φ

∂ui∇ui =

∑i=1,3

1hi

∂φ

∂uiei. (C.12)

Non-Cartesian Coordinates 541

Hence, the components of ∇φ in the u1, u2, u3 coordinate system are

(∇φ)i =1hi

∂φ

∂ui. (C.13)

Consider a vector field A(u1, u2, u3). We can write

∇ · A =∑i=1,3

∇ · (Ai ei) =∑i=1,3

∇ · (hi Ai ∇ui) =∑i=1,3

∇· h

hiAi

h 2i

h∇ui

=∑i=1,3

h 2i

h∇ui · ∇

(hhi

Ai

)=∑i=1,3

1h∂

∂ui

(hhi

Ai

), (C.14)

where use has been made of Equations (A.174), (C.9), and (C.10). Thus, the diver-gence of A in the u1, u2, u3 coordinate system takes the form

∇ · A =∑i=1,3

1h∂

∂ui

(hhi

Ai

). (C.15)

We can write

∇ × A =∑k=1,3

∇ × (Ak ek) =∑k=1,3

∇ × (hk Ak ∇uk) =∑k=1,3

∇(hk uk) × ∇uk

=∑

j,k=1,3

∂(hk Ak)∂u j

∇u j × ∇uk, (C.16)

where use has been made of Equations (A.178), (C.8), and (C.12). It follows fromEquation (C.11) that

(∇ × A)i = ei · ∇ × A =∑

j,k=1,3

hi∂(hk Ak)∂u j

∇ui · ∇u j × ∇uk =∑

j,k=1,3

εi jkhi

h∂(hk Ak)∂u j

.

(C.17)Hence, the components of ∇ × A in the u1, u2, u3 coordinate system are

(∇ × A)i =∑

j,k=1,3

εi jkhi

h∂(hk Ak)∂u j

. (C.18)

Now, ∇ 2φ = ∇·∇φ [see Equation (A.172)], so Equations (C.12) and (C.15) yieldthe following expression for ∇ 2φ in the u1, u2, u3 coordinate system:

∇2φ =∑i=1,3

1h∂

∂ui

hh 2

i

∂φ

∂ui

. (C.19)

The vector identities (A.171) and (A.179) can be combined to give the followingexpression for (A · ∇) B that is valid in a general coordinate system:

(A · ∇) B =12

[∇(A · B) − ∇ × (A × B) − (∇ · A) B + (∇ · B) A

− A × (∇ × B) − B × (∇ × A)] . (C.20)


Making use of Equations (C.13), (C.15), and (C.18), as well as the easily demon-strated results

A · B =∑i=1,3

Ai Bi, (C.21)

A × B =∑

j,k=1,3

εi jk A j Bk, (C.22)

and the tensor identity (B.16), Equation (C.20) reduces (after a great deal of tediousalgebra) to the following expression for the components of (A · ∇) B in the u1, u2, u3

coordinate system:

[(A · ∇) B]i =∑j=1,3

(A j

h j

∂Bi

∂u j−

A j B j

hi h j

∂h j

∂ui+

Ai B j

hi h j

∂hi

∂u j

). (C.23)

Note, incidentally, that the commonly quoted result [(A · ∇) B]i = A · ∇Bi is onlyvalid in Cartesian coordinate systems (for which h1 = h2 = h3 = 1).

Let us define the gradient ∇A of a vector field A as the tensor whose componentsin a Cartesian coordinate system take the form

(∇A)i j =∂Ai

∂x j. (C.24)

In an orthogonal curvilinear coordinate system, the previous expression generalizesto

(∇A)i j = [(e j · ∇) A]i. (C.25)

It thus follows from Equation (C.23), and the relation (ei) j = ei · e j = δi j, that

(∇A)i j =1h j

∂Ai

∂u j−

A j

hi h j

∂h j

∂ui+ δi j

∑k=1,3

Ak

hi hk

∂hi

∂uk. (C.26)

The vector identity (A.177) yields the following expression for ∇ 2A that is validin a general coordinate system:

∇ 2A = ∇(∇ · A) − ∇ × (∇ × A). (C.27)

Making use of Equations (C.15), (C.18), and (C.19), as well as (C.21) and (C.22),and the tensor identity (B.16), the previous equation reduces (after a great deal oftedious algebra) to the following expression for the components of ∇ 2A in the u1, u2,


u3 coordinate system:

(∇ 2A)i = ∇ 2Ai +∑j=1,3

2

hi h j

(1hi

∂hi

∂u j

∂

∂ui−

1h j

∂h j

∂ui

∂

∂u j

)A j

+h

hi h 2j

A j

h 2i

∂h j

∂ui

∂

∂u j

h 2i

h

− Ai

h 2j

∂hi

∂u j

∂

∂u j

h 2j

h

+

A j

hi

hh 3

j

1h j

∂h j

∂ui

∂

∂u j

h 2j

h

+ hh 2

j

∂

∂ui

h 2j

h

∂

∂u j

h 2j

h

− ∂ 2

∂ui ∂u j

h 2j

h

− Ai

hi h 2j

2hi

(∂hi

∂u j

)2− ∂

2hi

∂u 2j

. (C.28)

Note, again, that the commonly quoted result (∇ 2A)i = ∇ 2Ai is only valid in Carte-sian coordinate systems (for which h1 = h2 = h3 = h = 1).

C.3 Cylindrical Coordinates

In the cylindrical coordinate system, u1 = r, u2 = θ, and u3 = z, where r =√

x 2 + y 2,θ = tan−1(y/x), and x, y, z are standard Cartesian coordinates. Thus, r is the perpen-dicular distance from the z-axis, and θ the angle subtended between the projection ofthe radius vector (i.e., the vector connecting the origin to a general point in space)onto the x-y plane and the x-axis. (See Figure C.1.)

A general vector A is written

A = Ar er + Aθ eθ + Az ez, (C.29)

where er = ∇r/|∇r|, eθ = ∇θ/|∇θ|, and ez = ∇z/|∇z|. Of course, the unit basis vectorser, eθ, and ez are mutually orthogonal, so Ar = A · er, et cetera.

As is easily demonstrated, an element of length (squared) in the cylindrical coor-dinate system takes the form

dx · dx = dr 2 + r 2 dθ 2 + dz 2. (C.30)

Hence, comparison with Equation (C.6) reveals that the scale factors for this systemare

hr = 1, (C.31)

hθ = r, (C.32)

hz = 1. (C.33)


Oθ

y

x

eθ

r

er

z

Figure C.1Cylindrical coordinates.

Thus, surface elements normal to er, eθ, and ez are written

dS r = r dθ dz, (C.34)

dS θ = dr dz, (C.35)

dS z = r dr dθ, (C.36)

respectively, whereas a volume element takes the form

dV = r dr dθ dz. (C.37)

According to Equations (C.13), (C.15), and (C.18), gradient, divergence, and curlin the cylindrical coordinate system are written

∇ψ = ∂ψ∂r

er +1r∂ψ

∂θeθ +

∂ψ

∂zez, (C.38)

∇ · A = 1r∂

∂r(r Ar) +

1r∂Aθ∂θ+∂Az

∂z, (C.39)

∇ × A =(1r∂Az

∂θ− ∂Aθ∂z

)er +

(∂Ar

∂z− ∂Az

∂r

)eθ +

(1r∂

∂r(r Aθ) −

1r∂Ar

∂θ

)ez,

(C.40)

respectively. Here, ψ(r) is a general scalar field, and A(r) a general vector field.According to Equation (C.19), when expressed in cylindrical coordinates, the


Laplacian of a scalar field becomes

∇ 2ψ =1r∂

∂r

(r∂ψ

∂r

)+

1r 2

∂ 2ψ

∂θ 2 +∂ 2ψ

∂z 2 . (C.41)

Moreover, from Equation (C.23), the components of (A · ∇) A in the cylindricalcoordinate system are

[(A · ∇) A]r = A · ∇Ar −A 2θ

r, (C.42)

[(A · ∇) A]θ = A · ∇Aθ +Ar Aθ

r, (C.43)

[(A · ∇) A]z = A · ∇Az. (C.44)

Let us define the symmetric gradient tensor

∇A =12

[∇A + (∇A)T

]. (C.45)

Here, the superscript T denotes a transpose. Thus, if the i j element of some second-order tensor S is S i j then the corresponding element of ST is S ji. According toEquation (C.26), the components of ∇A in the cylindrical coordinate system are

(∇A)rr =∂Ar

∂r, (C.46)

(∇A)θθ =1r∂Aθ∂θ+

Ar

r, (C.47)

(∇A)zz =∂Az

∂z, (C.48)

(∇A)rθ = (∇A)θr =12

(1r∂Ar

∂θ+∂Aθ∂r− Aθ

r

), (C.49)

(∇A)rz = (∇A)zr =12

(∂Ar

∂z+∂Az

∂r

), (C.50)

(∇A)θz = (∇A)zθ =12

(∂Aθ∂z+

1r∂Az

∂θ

). (C.51)

Finally, from Equation (C.28), the components of ∇ 2A in the cylindrical coordi-nate system are

(∇ 2A)r = ∇ 2Ar −Ar

r 2 −2r 2

∂Aθ∂θ, (C.52)

(∇ 2A)θ = ∇ 2Aθ +2r 2

∂Ar

∂θ− Aθ

r 2 , (C.53)

(∇ 2A)z = ∇ 2Az. (C.54)


r

Oxφ

θ

z

y

Figure C.2Spherical coordinates.

C.4 Spherical CoordinatesIn the spherical coordinate system, u1 = r, u2 = θ, and u3 = φ, where r =

√x 2 + y 2 + z 2,

θ = cos−1(z/r), φ = tan−1(y/x), and x, y, z are standard Cartesian coordinates. Thus,r is the length of the radius vector, θ the angle subtended between the radius vectorand the z-axis, and φ the angle subtended between the projection of the radius vectoronto the x-y plane and the x-axis. (See Figure C.2.)

A general vector A is written

A = Ar er + Aθ eθ + Aφ eφ, (C.55)

where er = ∇r/|∇r|, eθ = ∇θ/|∇θ|, and eφ = ∇φ/|∇φ| . Of course, the unit vectors er,eθ, and eφ are mutually orthogonal, so Ar = A · er, et cetera.

As is easily demonstrated, an element of length (squared) in the spherical coor-dinate system takes the form

dx · dx = dr 2 + r 2 dθ 2 + r 2 sin2 θ dφ2. (C.56)

Hence, comparison with Equation (C.6) reveals that the scale factors for this systemare

hr = 1, (C.57)

hθ = r, (C.58)

hφ = r sin θ. (C.59)


Thus, surface elements normal to er, eθ, and eφ are written

dS r = r 2 sin θ dθ dφ, (C.60)

dS θ = r sin θ dr dφ, (C.61)

dS φ = r dr dθ, (C.62)

respectively, whereas a volume element takes the form

dV = r 2 sin θ dr dθ dφ. (C.63)

According to Equations (C.13), (C.15), and (C.18), gradient, divergence, and curlin the spherical coordinate system are written

∇ψ = ∂ψ∂r

er +1r∂ψ

∂θeθ +

1r sin θ

∂ψ

∂φeφ, (C.64)

∇ · A = 1r 2

∂

∂r(r 2 Ar) +

1r sin θ

∂

∂θ(sin θ Aθ) +

1r sin θ

∂Aφ∂φ

, (C.65)

∇ × A =[

1r sin θ

∂

∂θ(sin θ Aφ) −

1r sin θ

∂Aθ∂φ

]er

+

[1

r sin θ∂Ar

∂φ− 1

r∂

∂r(r Aφ)

]eθ

+

[1r∂

∂r(r Aθ) −

1r∂Ar

∂θ

]eφ, (C.66)

respectively. Here, ψ(r) is a general scalar field, and A(r) a general vector field.According to Equation (C.19), when expressed in spherical coordinates, the Lapla-

cian of a scalar field becomes

∇ 2ψ =1r 2

∂

∂r

(r 2 ∂ψ

∂r

)+

1r 2 sin θ

∂

∂θ

(sin θ

∂ψ

∂θ

)+

1r 2 sin2 θ

∂ 2ψ

∂φ 2 . (C.67)

Moreover, from Equation (C.23), the components of (A · ∇)A in the sphericalcoordinate system are

[(A · ∇) A]r = A · ∇Ar −A 2θ + A 2

φ

r, (C.68)

[(A · ∇) A]θ = A · ∇Aθ +Ar Aθ − cot θ A 2

φ

r, (C.69)

[(A · ∇) A]φ = A · ∇Aφ +Ar Aφ + cot θ Aθ Aφ

r. (C.70)

Now, according to Equation (C.26), the components of ∇A in the spherical coor-


dinate system are

(∇A)rr =∂Ar

∂r, (C.71)

(∇A)θθ =1r∂Aθ∂θ+

Ar

r, (C.72)

(∇A)φφ =1

r sin θ∂Aφ∂φ+

Ar

r+

cot θ Aθr

, (C.73)

(∇A)rθ = (∇A)θr =12

(1r∂Ar

∂θ+∂Aθ∂r− Aθ

r

), (C.74)

(∇A)rφ = (∇A)φr =12

(1

r sin θ∂Ar

∂φ+∂Aφ∂r−

Aφr

), (C.75)

(∇A)θφ = (∇A)φθ =12

(1

r sin θ∂Aθ∂φ+

1r∂Aφ∂θ−

cot θ Aφr

). (C.76)

Finally, from Equation (C.28), the components of∇ 2A in the spherical coordinatesystem are

(∇ 2A)r = ∇ 2Ar −2Ar

r 2 −2r 2

∂Aθ∂θ− 2 cot θ Aθ

r 2 − 2r 2 sin θ

∂Aφ∂φ, (C.77)

(∇ 2A)θ = ∇ 2Aθ +2r 2

∂Ar

∂θ− Aθ

r 2 sin2 θ− 2

r 2 sin θ∂Aφ∂φ, (C.78)

(∇ 2A)φ = ∇ 2Aφ −Aφ

r 2 sin2 θ+

2r 2 sin2 θ

∂Ar

∂φ+

2 cot θr 2 sin θ

∂Aθ∂φ. (C.79)

C.5 ExercisesC.1 Find the Cartesian components of the basis vectors er, eθ, and ez of the cylin-

drical coordinate system. Verify that the vectors are mutually orthogonal.Do the same for the basis vectors er, eθ, and eφ of the spherical coordinatesystem.

C.2 Use cylindrical coordinates to prove that the volume of a right cylinder of ra-dius a and length l is π a 2 l. Demonstrate that the moment of inertia of a uni-form cylinder of mass M and radius a about its symmetry axis is (1/2) M a 2.

C.3 Use spherical coordinates to prove that the volume of a sphere of radius a is(4/3) π a 3. Demonstrate that the moment of inertia of a uniform sphere ofmass M and radius a about an axis passing through its center is (2/5) M a 2.

C.4 For what value(s) of n is ∇ · (r n er) = 0, where r is a spherical coordinate?


C.5 For what value(s) of n is ∇ × (r n er) = 0, where r is a spherical coordinate?

C.6 (a) Find a vector field F = Fr(r) er satisfying ∇ · F = r m for m ≥ 0. Here,r is a spherical coordinate.

(b) Use the divergence theorem to show that∫V

r m dV =1

m + 3

∫S

r m+1 er · dS,

where V is a volume enclosed by a surface S .

(c) Use the previous result (for m = 0) to demonstrate that the volume of aright cone is one third the volume of the right cylinder having the samebase and height.

C.7 The electric field generated by a z-directed electric dipole of moment p,located at the origin, is

E(r) =1

4π ε0

[3 (er · p) er − p

r3

],

where p = p ez, and r is a spherical coordinate. Find the components of E(r)in the spherical coordinate system. Calculate ∇ · E and ∇ × E.

C.8 Show that the parabolic cylindrical coordinates u, v, z, defined by the equa-tions x = (u 2 − v 2)/2, y = u v, z = z, where x, y, z are Cartesian coordi-nates, are orthogonal. Find the scale factors hu, hv, hz. What shapes are theu = constant and v = constant surfaces? Write an expression for ∇ 2 f inparabolic cylindrical coordinates.

C.9 Show that the elliptic cylindrical coordinates ξ, η, z, defined by the equa-tions x = cosh ξ cos η, y = sinh ξ sin η, z = z, where x, y, z are Cartesiancoordinates, and 0 ≤ ξ ≤ ∞, −π < η ≤ π, are orthogonal. Find the scale fac-tors hξ, hη, hz. What shapes are the ξ = constant and η = constant surfaces?Write an expression for ∇ f in elliptical cylindrical coordinates.

DEllipsoidal Potential Theory

D.1 IntroductionLet us adopt the right-handed Cartesian coordinate system x1, x2, x3. Consider ahomogeneous ellipsoidal body whose outer boundary satisfies

x 21

a 21

+x 2

2

a 22

+x 2

3

a 23

= 1, (D.1)

where a1, a2, and a3 are the principal radii along the x1-, x2-, and x3-axes, respec-tively. Let us calculate the gravitational potential (i.e., the potential energy of a unittest mass) at some point P ≡ (x1, x2, x3) lying within this body. More informationon ellipsoidal potential theory can be found in Chandrasekhar 1969.

D.2 AnalysisConsider the contribution to the potential at P from the mass contained within adouble cone, whose apex is P, and which is terminated in both directions at thebody’s outer boundary. (See Figure D.1.) If the cone subtends a solid angle dΩthen a volume element is written dV = r 2 dr dΩ, where r measures displacementfrom P along the axis of the cone. Thus, from standard classical gravitational theory(Fitzpatrick 2012), the contribution to the potential takes the form

dΨ = −∫ r′

0

G ρr

dV −∫ 0

r′′

G ρ(−r)

dV, (D.2)

where r′ = |PQ|, r′′ = −|PR|, and ρ is the constant mass density of the ellipsoid.Hence, we obtain

dΨ = −G ρ(∫ r′

0r dr +

∫ r′′

0r dr

)dΩ = −

12

G ρ (r′ 2 + r′′ 2) dΩ. (D.3)

The net potential at P is obtained by integrating over all solid angle, and dividing theresult by two to adjust for double counting. This yields

Ψ = −14

G ρ∮

(r′ 2 + r′′ 2) dΩ. (D.4)

551


PO x

r

Q

R

dΩ

Figure D.1Calculation of ellipsoidal gravitational potential.

From Figure D.1, the position vector of point Q, relative to the origin, O, is

x′ = x + r′ n, (D.5)

where x = (x1, x2, x3) is the position vector of point P, and n a unit vector pointingfrom P to Q. Likewise, the position vector of point R is

x′′ = x + r′′ n. (D.6)

However, Q and R both lie on the body’s outer boundary. It follows, from Equa-tion (D.1), that r′ and r′′ are the two roots of∑

i=1,3

(xi + r ni

ai

)2= 1, (D.7)

which reduces to the quadratic

A r 2 + B r + C = 0, (D.8)

where

A =∑i=1,3

n 2i

a 2i

, (D.9)

B = 2∑i=1,3

xi ni

a 2i

, (D.10)

C =∑i=1,3

x 2i

a 2i

− 1. (D.11)

Ellipsoidal Potential Theory 553

According to standard polynomial equation theory (Riley 1974), r′+r′′ = −B/A, andr′ r′′ = C/A. Thus,

r′ 2 + r′′ 2 = (r′ + r′′)2 − 2 r′ r′′ =B 2

A2 − 2CA, (D.12)

and Equation (D.4) becomes

Ψ = −12

G ρ∮ 2

(∑i=1,3 xi ni/a 2

i

)2(∑i=1,3 n 2

i /a2i

)2 +1 −

∑i=1,3 x 2

i /a2i∑

i=1,3 n 2i /a

2i

dΩ. (D.13)

The previous expression can also be written

Ψ = −12

G ρ∮ 2

∑i, j=1,3 xi x j ni n j/(a 2

i a 2j )(∑

i=1,3 n 2i /a

2i

)2 +1 −

∑i=1,3 x 2

i /a2i∑

i=1,3 n 2i /a

2i

dΩ. (D.14)

However, the cross terms (i.e., i j) integrate to zero by symmetry, and we are leftwith

Ψ = −12

G ρ∮ 2

∑i=1,3 x 2

i n 2i /a

4i(∑

i=1,3 n 2i /a

2i

)2 + 1 −∑

i=1,3 x 2i /a

2i∑

i=1,3 n 2i /a

2i

dΩ. (D.15)

LetJ =

∮dΩ∑

i=1,3 n 2i /a

2i

. (D.16)

It follows that1ai

∂J∂ai=

∮ 2 n 2i /a

4i(∑

i=1,3 n 2i /a

2i

)2 dΩ. (D.17)

Thus, Equation (D.15) can be written

Ψ = −12

G ρ

J −∑i=1,3

Ai x 2i

, (D.18)

whereAi =

Ja 2

i

− 1ai

∂J∂ai. (D.19)

At this stage, it is convenient to adopt the spherical angular coordinates, θ and φ(see Section C.4), in terms of which

n = (sin θ cosφ, sin θ sin φ, cos θ), (D.20)

and dΩ = sin θ dθ dφ. We find, from Equation (D.16), that

J = 8∫ π/2

0sin θ dθ

∫ π/2

0

sin2 θ cos2 φ

a 21

+sin2 θ sin2 φ

a 22

+cos2 θ

a 23

−1

dφ. (D.21)


Let t = tanφ. It follows that

J = 8∫ π/2

0sin θ dθ

∫ ∞

0

dta + b t2 = 4π

∫ π/2

0

sin θ dθ(a b)1/2 , (D.22)

where

a =sin2 θ

a 21

+cos2 θ

a 23

, (D.23)

b =sin2 θ

a 22

+cos2 θ

a 23

. (D.24)

Hence, we obtain

J = 4π a1 a2 a 23

∫ π/2

0

sin θ sec2 θ dθ(a 2

1 + a 23 tan2 θ)1/2 (a 2

2 + a 23 tan2 θ)1/2

. (D.25)

Let u = a 23 tan2 θ. It follows that

J = 2π a1 a2 a3

∫ ∞

0

du∆, (D.26)

where∆ = (a 2

1 + u)1/2 (a 22 + u)1/2 (a 2

3 + u)1/2. (D.27)

Now, from Equations (D.19), (D.26), and (D.27),

Ai =2π a1 a2 a3

a 2i

∫ ∞

0

du∆− 1

ai

∂

∂ai

(2π a1 a2 a3

∫ ∞

0

du∆

)= −

2π a1 a2 a3

ai

∫ ∞

0

∂

∂ai

(1∆

)du = 2π a1 a2 a3

∫ ∞

0

du(a 2

i + u)∆. (D.28)

Thus, Equations (D.18), (D.26), and (D.28) yield

Ψ = −34

G M

α0 −∑i=1,3

αi x 2i

, (D.29)

where

α0 =

∫ ∞

0

du∆, (D.30)

αi =

∫ ∞

0

du(a 2

i + u)∆. (D.31)

Here, M = V ρ and V = (4/3) π a1 a2 a3 are the body’s mass and volume, respec-tively.

Ellipsoidal Potential Theory 555

The total gravitational potential energy of the body is written (Fitzpatrick 2012)

U =12

∫Ψ ρ dV, (D.32)

where the integral is taken over all interior points. It follows from Equation (D.29)that

U = −38

G M 2

α0 −15

∑i=1,3

αi a 2i

. (D.33)

In writing the previous expression, use has been made of the easily demonstratedresult

∫x 2

i dV = (1/5) a 2i V . Now,

2d

du

( u∆

)= −

1∆+∑i=1,3

a 2i

(a 2i + u)∆

, (D.34)

so ∑i=1,3

αi a 2i =

∫ ∞

0

∑i=1,3

a 2i du

(a 2i + u)∆

=

∫ ∞

0

[2

ddu

( u∆

)+

1∆

]du = α0. (D.35)

Hence, we obtain

U = − 310

G M 2 α0. (D.36)

D.3 ExercisesD.1 Demonstrate that the volume of an ellipsoid whose bounding surface satis-

fiesx 2

1

a 21

+x 2

2

a 22

+x 2

3

a 23

= 1,

is V = (4/3) π a1 a2 a3.

D.2 Demonstrate that the moments of inertia about the three Cartesian axes of ahomogeneous ellipsoidal body of mass M, whose bounding surface satisfies(x1/a1)2 + (x2/a2)2 + (x3/a3)2 = 1, are

I1 =M5

(a 22 + a 2

3 ),

I2 =M5

(a 21 + a 2

3 ),

I3 =M5

(a 21 + a 2

2 ).


D.3 According to MacCullagh’s formula (Fitzpatrick 2012), the gravitational po-tential a relatively long way from a body of mass M whose center of masscoincides with the origin, and whose principal moments of inertial are I1, I2,and I3 (assuming that the principal axes coincide with the Cartesian axes),takes the form

Ψ (x1, x2, x3) −G M

r−

G (I1 + I2 + I3)2 r 3 +

3 G (I1 x 21 + I2 x 2

2 + I3 x 23 )

2 r 5 ,

where r = (x 21 + x 2

2 + x 23 )1/2. Demonstrate that if the body in question is a ho-

mogeneous ellipsoid whose bounding surface satisfies (x1/a1)2 + (x2/a2)2 +

(x3/a3)2 = 1 then

Ψ (x1, x2, x3) −G Mr− G M

5 r 5 a 21

[x 2

1 −12

(x 22 + x 2

3 )]

(D.37)

−G M5 r 5 a 2

2

[x 2

2 −12

(x 21 + x 2

3 )]−

G M5 r 5 a 2

3

[x 2

3 −12

(x 21 + x 2

2 )].

D.4 Show that the gravitational potential external to a homogeneous ellipsoidalbody of mass M, whose outer boundary satisfies (x1/a1)2+(x2/a2)2+(x3/a3)2 =

1, takes the form

Ψ (x1, x2, x3) = −34

G M

α0 −∑i=1,3

αi x 2i

, (D.38)

where

α0 =

∫ ∞

λ

du∆,

αi =

∫ ∞

λ

du(a 2

i + u)∆,

and ∆ = (a 21 + u)1/2 (a 2

2 + u)1/2 (a 23 + u)1/2. Here, λ is the positive root of

∑i=1,3

x 2i

a 2i + λ

= 1.

Demonstrate that, at large λ, Equation (D.38) reduces to Equation (D.37).

ECalculus of Variations

E.1 IndroductionThis appendix gives a brief overview of the calculus of variations. More informationon this topic can be found in Riley 1974.

E.2 Euler-Lagrange EquationIt is a well-known fact, first enunciated by Archimedes, that the shortest distancebetween two points in a plane is a straight-line. However, suppose that we wish todemonstrate this result from first principles. Let us consider the length, l, of variouscurves, y(x), which run between two fixed points, A and B, in a plane, as illustratedin Figure E.1. Now, l takes the form

l =∫ B

A(dx 2 + dy 2)1/2 =

∫ b

a[1 + y′ 2(x)]1/2 dx, (E.1)

where y′ ≡ dy/dx. Note that l is a function of the function y(x). In mathematics, afunction of a function is termed a functional.

In order to find the shortest path between points A and B, we need to minimizethe functional l with respect to small variations in the function y(x), subject to theconstraint that the end points, A and B, remain fixed. In other words, we need tosolve

δl = 0. (E.2)

The meaning of the previous equation is that if y(x) → y(x) + δy(x), where δy(x)is small, then the first-order variation in l, denoted δl, vanishes. In other words,l → l + O(δy 2). The particular function y(x) for which δl = 0 obviously yields anextremum of l (i.e., either a maximum or a minimum). Hopefully, in the case underconsideration, it yields a minimum of l.

Consider a general functional of the form

I =∫ b

aF(y, y′, x) dx, (E.3)

557


b

y

x

A

B

a

Figure E.1Different paths between points A and B.

where the end points of the integration are fixed. Suppose that y(x) → y(x) + δy(x).The first-order variation in I is written

δI =∫ b

a

(∂F∂yδy +

∂F∂y′

δy′)

dx, (E.4)

where δy′ = d(δy)/dx. Setting δI to zero, we obtain∫ b

a

(∂F∂yδy +

∂F∂y′

δy′)

dx = 0. (E.5)

This equation must be satisfied for all possible small perturbations δy(x).Integrating the second term in the integrand of the previous equation by parts, we

get ∫ b

a

[∂F∂y− d

dx

(∂F∂y′

)]δy dx +

[∂F∂y′

δy

]ba= 0. (E.6)

However, if the end points are fixed then δy = 0 at x = a and x = b. Hence, the lastterm on the left-hand side of the previous equation is zero. Thus, we obtain∫ b

a

[∂F∂y− d

dx

(∂F∂y′

)]δy dx = 0. (E.7)

The previous equation must be satisfied for all small perturbations δy(x). The onlyway in which this is possible is for the expression enclosed in square brackets in theintegral to be zero. Hence, the functional I attains an extremum value whenever

ddx

(∂F∂y′

)−∂F∂y= 0. (E.8)

Calculus of Variations 559

This condition is known as the Euler-Lagrange equation.Let us consider some special cases. Suppose that F does not explicitly depend on

y. It follows that ∂F/∂y = 0. Hence, the Euler-Lagrange equation (E.8) simplifies to

∂F∂y′= const. (E.9)

Next, suppose that F does not depend explicitly on x. Multiplying Equation (E.8) byy′, we obtain

y′ddx

(∂F∂y′

)− y′ ∂F

∂y= 0. (E.10)

However,d

dx

(y′∂F∂y′

)= y′

ddx

(∂F∂y′

)+ y′′

∂F∂y′. (E.11)

Thus, we getddx

(y′∂F∂y′

)= y′

∂F∂y+ y′′

∂F∂y′. (E.12)

Now, if F is not an explicit function of x then the right-hand side of the previousequation is the total derivative of F, namely dF/dx. Hence, we obtain

ddx

(y′∂F∂y′

)=

dFdx, (E.13)

which yields

y′∂F∂y′− F = const. (E.14)

Returning to the case under consideration, we have F =√

1 + y′ 2, according toEquation (E.1) and (E.3). Hence, F is not an explicit function of y, so Equation (E.9)yields

∂F∂y′=

y′√1 + y′ 2

= c, (E.15)

where c is a constant. So,y′ =

c√

1 − c 2= const. (E.16)

Of course, y′ = constant is the equation of a straight-line. Thus, the shortest distancebetween two fixed points in a plane is indeed a straight-line.

E.3 Conditional VariationSuppose that we wish to find the function y(x) which maximizes or minimizes thefunctional

I =∫ b

aF(y, y′, x) dx, (E.17)


subject to the constraint that the value of

J =∫ b

aG(y, y′, x) dx (E.18)

remains constant. We can achieve our goal by finding an extremum of the new func-tional K = I + λ J, where λ(x) is an undetermined function. We know that δJ = 0,because the value of J is fixed, so if δK = 0 then δI = 0 as well. In other words,finding an extremum of K is equivalent to finding an extremum of I. Application ofthe Euler-Lagrange equation yields

ddx

(∂F∂y′

)− ∂F∂y+

[ddx

(∂[λG]∂y′

)− ∂[λG]

∂y

]= 0. (E.19)

In principle, the previous equation, together with the constraint (E.18), yields thefunctions λ(x) and y(x). Incidentally, λ is generally termed a Lagrange multiplier. IfF and G have no explicit x-dependence then λ is usually a constant.

As an example, consider the following famous problem. Suppose that a uniformchain of fixed length l is suspended by its ends from two equal-height fixed pointsthat are a distance a apart, where a < l. What is the equilibrium configuration of thechain?

Suppose that the chain has the uniform density per unit length ρ. Let the x- andy-axes be horizontal and vertical, respectively, and let the two ends of the chain lieat (±a/2, 0). The equilibrium configuration of the chain is specified by the functiony(x), for −a/2 ≤ x ≤ +a/2, where y(x) is the vertical distance of the chain below itsend points at horizontal position x. Of course, y(−a/2) = y(+a/2) = 0.

According to standard Newtonian dynamics, the stable equilibrium state of aconservative dynamical system is one that minimizes the system’s potential energy(Fitzpatrick 2012). Now, the potential energy of the chain is written

U = −ρ g∫y ds = −ρ g

∫ a/2

−a/2y (1 + y′ 2)1/2 dx, (E.20)

where ds =√

dx 2 + dy 2 is an element of length along the chain, and g is the acceler-ation due to gravity. Hence, we need to minimize U with respect to small variationsin y(x). However, the variations in y(x) must be such as to conserve the fixed lengthof the chain. Hence, our minimization procedure is subject to the constraint that

l =∫

ds =∫ a/2

−a/2(1 + y′ 2)1/2 dx (E.21)

remains constant.It follows, from the previous discussion, that we need to minimize the functional

K = U + λ l =∫ a/2

−a/2(−ρ g y + λ) (1 + y′ 2)1/2 dx, (E.22)


where λ is an, as yet, undetermined constant. Because the integrand in the functionaldoes not depend explicitly on x, we have from Equation (E.14) that

y′ 2 (−ρ g y + λ) (1 + y′ 2)−1/2 − (−ρ g y + λ) (1 + y′ 2)1/2 = k, (E.23)

where k is a constant. This expression reduces to

y′ 2 =(λ′ +

y

h

)2− 1, (E.24)

where λ′ = λ/k, and h = −k/(ρ g).Let

λ′ +y

h= − cosh z. (E.25)

Making this substitution, Equation (E.24) yields

dzdx= −h−1. (E.26)

Hence,

z = −xh+ c, (E.27)

where c is a constant. It follows from Equation (E.25) that

y(x) = −h [λ′ + cosh(−x/h + c)]. (E.28)

The previous solution contains three undetermined constants, h, λ′, and c. We caneliminate two of these constants by application of the boundary conditions y(±a/2) =0. This yields

λ′ + cosh(∓a/2 h + c) = 0. (E.29)

Hence, c = 0, and λ′ = − cosh(a/2 h). It follows that

y(x) = h [cosh(a/2 h) − cosh(x/h)]. (E.30)

The final unknown constant, h, is determined via the application of the constraint(E.21). Thus,

l =∫ a/2

−a/2(1 + y′ 2)1/2 dx =

∫ a/2

−a/2cosh(x/h) dx = 2 h sinh(a/2 h). (E.31)

Hence, the equilibrium configuration of the chain is given by the curve (E.30), whichis known as a catenary (from the Latin for chain), where the parameter h satisfies

l2 h= sinh

( a2 h

). (E.32)


E.4 Multi-Function VariationSuppose that we wish to maximize or minimize the functional

I =∫ b

aF(y1, y2, · · · , yF , y′1, y

′2, · · · , y

′F , x) dx. (E.33)

Here, the integrand F is now a functional of the F independent functions yi(x), fori = 1,F . A fairly straightforward extension of the analysis in Section E.2 yields Fseparate Euler-Lagrange equations,

ddx

(∂F∂y′i

)− ∂F∂yi= 0, (E.34)

for i = 1,F , which determine the F functions yi(x). If F does not explicitly dependon the function yk then the kth Euler-Lagrange equation simplifies to

∂F∂y′k= const. (E.35)

Likewise, if F does not explicitly depend on x then all F Euler-Lagrange equationssimplify to

y′i∂F∂y′i− F = const, (E.36)

for i = 1,F .

E.5 ExercisesE.1 Find the extremal curves y = y(x) of the following constrained optimization

problems, using the method of Lagrange multipliers:

(a)∫ 1

0

(y′ 2 + x2

)dx, such that

∫ 10 y

2 dx = 2.

(b)∫ π

0 y′ 2 dx, such that y(0) = y(π) = 0, and

∫ π0 y

2 dx = 2.

(c)∫ 1

0 y dx, such that y(0) = y(1) = 1, and∫ √

1 + y′ 2 dx = 2π/3.

E.2 Suppose P and Q are two points lying in the x-y plane, which is orientatedvertically such that P is above Q. Imagine there is a thin, flexible wireconnecting the two points and lying entirely in the x-y plane. A frictionlessbead travels down the wire, impelled by gravity alone. Show that the shape


of the wire that results in the bead reaching the point Q in the least amountof time is a cycloid, which takes the parametric form

x(θ) = k (θ − sin θ) ,

y(θ) = k (1 − cos θ) ,

where k is a constant.

E.3 Find the curve y(x), in the interval 0 ≤ x ≤ p, which is of length π, andmaximizes ∫ p

0y dx.

565

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