Physics 621 & 622, 2008-2009

Advanced Electrodynamics

Instructor: Dr. Marco CavagliaNotes Compiled by: Phil Blom

Text Used:Classical Electrodynamics by J.D. Jackson

Last updated: May 26, 2009

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Contents

1 Introduction - Foundations of Electromagnetism 12

1.1 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

I Electrostatics 12

2 The Basic Assumptions of Electrostatics 13

3 The Dirac Delta Function 14

3.1 Properties of the Dirac Delta Function . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4 The Greens Function 16

4.1 More Solutions Using the Greens Function Solution . . . . . . . . . . . . . . . . . . . 18

4.2 Special Case - Point Like Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4.3 More Difficult Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

5 Interpretation of the Potential 20

6 The Electric Field on a Surface - Deriving Boundary Conditions 20

7 Definition of Potential Energy in a Charge Distribution 21

8 Electrostatics in the Presence of Continuous Media - Conductors 22

8.1 Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

9 Dipole Layer 23

10 Poisson Equation With Boundary Conditions - Greens Identity 25

10.1 Dirichlet Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

10.2 Neumann Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

10.3 Example of a Greens Formula Problem - Planar Conductor . . . . . . . . . . . . . . 26

11 Mean Value Theorem 27

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12 Methods for 2 - Dimensional Problems 28

13 Method of Images 29

13.1 Point Charge Near A Conducting Sphere . . . . . . . . . . . . . . . . . . . . . . . . . 29

14 Finding the Greens Function For a Sphere From the Image Solution 31

14.1 Conductive Sphere With Fixed Potential . . . . . . . . . . . . . . . . . . . . . . . . . 31

14.2 Sphere With Two Charges On A Line - Sphere In External Field . . . . . . . . . . . 32

14.3 Spherical Conductor with Different Potentials on Hemispheres . . . . . . . . . . . . . 32

15 Orthogonal Functions and Expansions 33

15.1 Fourier Series Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

15.2 Polynomial Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

15.3 Infinite Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

15.4 Applications and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

15.5 Two Dimensional Problems Solved With Expansions . . . . . . . . . . . . . . . . . . 36

16 Poisson’s Equation In Other Coordinate Systems - Spherical Coordinates 38

16.1 Boundary Value Problems With Azimuthal Symmetry . . . . . . . . . . . . . . . . . 39

16.2 Conic Holes and Sharp Points - Note: Will not be on tests or homework . . . . . . . 41

16.3 Azimuthal Variations - The Spherical Harmonics . . . . . . . . . . . . . . . . . . . . 42

16.4 The Addition Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

16.5 Greens Function Expansion For a Sphere . . . . . . . . . . . . . . . . . . . . . . . . 44

16.6 Examples of Problems Solved With Spherical Harmonics . . . . . . . . . . . . . . . . 46

17 Poisson’s Equation In Other Coordinate Systems - Cylindrical Coordinates 49

17.1 Simplifications For Different Geometries . . . . . . . . . . . . . . . . . . . . . . . . . 49

17.2 An Example - A Hollow Conductive Cylinder . . . . . . . . . . . . . . . . . . . . . . 49

17.3 Modified Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

18 Some Mathematics - Getting Greens Functions From A Generic Operator 51

19 Multipole Expansions In Electrostatics 52

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19.1 Multipole Treatment of a Dipole Source . . . . . . . . . . . . . . . . . . . . . . . . . 53

20 Transition To Non-negligible Media 56

20.1 An Alternate Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

20.2 More On Electrostatics In Real Media . . . . . . . . . . . . . . . . . . . . . . . . . . 58

21 Energy of a Dielectric in an External Field 61

22 A Theoretical Model For Polarization 62

23 Summary of Equations For Problems Involving Dielectrics 63

24 Solving Problems With Dielectrics 64

24.1 Method of Images - Planar Boundary With a Point Charge . . . . . . . . . . . . . . 64

24.2 Dielectric Sphere In External Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

II MAGNETOSTATICS 68

25 Introduction to Magnetostatics 68

25.1 Gauge Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

26 Deriving the Magnetic Field 69

26.1 Force on a Charged Particle From a Magnetic Field . . . . . . . . . . . . . . . . . . 70

26.2 Magnetic Field From a Circular Loop of Wire . . . . . . . . . . . . . . . . . . . . . . 71

26.3 First Order Approximation of the Magnetic Field . . . . . . . . . . . . . . . . . . . . 74

27 Current Density For a Group of Moving Charges 75

28 Force and Torque From an External Magnetic Field 76

29 Transition To Macroscopic Magnetostatics 77

29.1 Relations between H, B, and M - Boundary Conditions . . . . . . . . . . . . . . . . . 78

29.2 Some Important Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

30 Energy In An External Magnetic Field 80

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30.1 Special Case - Change in Energy of a Region When a Material Is Inserted . . . . . . 81

31 Behavior of Various Materials In a Magnetic Field 82

31.1 Paramagnets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

31.2 Ferromagnets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

32 Summary of Problem Solving Methods 85

32.1 Free Space Vector Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

32.2 Magnetization Known With No Current . . . . . . . . . . . . . . . . . . . . . . . . . 85

32.3 No Current, but no given magnetization . . . . . . . . . . . . . . . . . . . . . . . . . 85

32.4 Example: Uniformly Magnetized Sphere . . . . . . . . . . . . . . . . . . . . . . . . . 86

33 Additional Topics In Magnetostatics 87

33.1 Magnetic Shielding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

33.2 Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

33.3 Quasistatics Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

33.4 Faraday’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

III TRANSITION TO NON-STATICS 91

34 Modifications To Potentials For Non-Statics 91

35 Simplifications By Gauge Transformations 92

35.1 Gauge Transforms in Non-Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

35.2 Separating the Transverse and Longitudinal Current Density . . . . . . . . . . . . . 93

35.3 Summary of Gauge Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

36 Magnetic Monopoles 94

37 Special Symmetries - Parity and Time Inversion 96

37.1 Example - Expansion for Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . 97

38 Energy in an Electromagnetic Field - The Poynting Vector 97

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39 Momentum in the Electromagnetic Field 98

39.1 Angular Momentum in the Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

39.2 Radiation Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

40 Time Average of Vector Products 100

IV SPECIAL RELATIVITY AND TENSORS IN ELECTROSTATICS 102

41 Transition to Semester II 102

41.1 Review of First Semester: Maxwell’s Equations and Gauge Transforms . . . . . . . . 102

41.2 Introduction of Gaussian Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

42 Propagation in a Vacuum - Introduction to Special Relativity 104

43 The Lorentz Transform 105

43.1 Deriving the Lorentz Transform Group . . . . . . . . . . . . . . . . . . . . . . . . . . 105

43.2 Properties of the Lorentz Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

43.3 The Lorentz Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

43.4 New Notation - The 4 Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

43.5 The Boost Parameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

44 Space-Time Diagrams 110

44.1 The Line Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

44.2 Some More Interesting Cases - Moving Frames . . . . . . . . . . . . . . . . . . . . . 111

44.3 Length Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

45 Introduction to Tensors 114

45.1 Matrix Representation of Lorentz Transformations - Infinitesimal Generators . . . . 114

45.2 Tensor Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

45.3 Building Tensors from Other Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

45.4 Raising and Lowering Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

45.5 Some Special Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

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46 Back to Electromagnetism 122

46.1 Maxwell’s Equations in Tensor Form . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

46.2 Lorentz Transform of the Electromagnetic Field . . . . . . . . . . . . . . . . . . . . . 126

46.3 Lorentz Force in Invariant Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

47 The Lagrangian In Electrodynamics 129

47.1 The Action and Lagrangian For A Relativistic Free Particle . . . . . . . . . . . . . . 130

47.2 The Lagrangian For A Charged Particle With Minimal Coupling . . . . . . . . . . . 132

48 Hamiltonian Dynamics in Electromagnetism 133

48.1 The Minimal Coupling Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

48.2 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

48.3 Lagrangian Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

48.4 Results of Hamiltonian Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

49 Solving Problems Using Tensor Notation 136

49.1 Relativistic Doppler Shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

49.2 Particle in an Uniform Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . 137

50 The Lagrangian of the Electromagnetic Field 139

50.1 Gauge Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

51 Massive Photons and the Proca Lagrangian Density 141

51.1 The Proca LagrangianA Phenomenological Description of Superconductivity . . . . . . . . . . . . . . . . . 141

52 Symmetries and Invariant Equations of Motion 144

52.1 More on Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

52.2 Generalize The Result for Fields - The Stress Energy Tensor . . . . . . . . . . . . . 146

52.3 Deriving the Components of the Stress Energy Tensor for Electromagnetic Fields . . 147

53 Finding a General Solution to the Maxwell Equations 149

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V SPECIAL CASES 153

54 Plane Waves 153

54.1 Plane Wave Solutions Of Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . 153

54.2 Polarization of Plane Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

54.3 Plane Waves In Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

54.4 Energy Contained in Plane Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

54.5 Plane Waves Incident on a Boundary Between Media . . . . . . . . . . . . . . . . . . 159

54.6 Plane Wave Phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

54.7 Wave Packets - Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

54.8 Wave Packet Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

54.9 Modeling the Index of Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

54.10Relating the Displacement Field to the Electric Field . . . . . . . . . . . . . . . . . . 170

55 Geometric Optics 175

55.1 The Approximations and Assumptions of Geometric Optics . . . . . . . . . . . . . . 175

55.2 Consequences of Geometric OpticsFermat’s Theorem and Generalization of Snell’s Law . . . . . . . . . . . . . . . . . . 176

55.3 Optical Fibers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

56 Wave Guides 181

56.1 General Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

56.2 Solving problems in a waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

56.3 Energy Flow in a Wave Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

56.4 Resonant Cavities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

57 Solutions with Sources 190

57.1 Deriving the Electric and Magnetic Field for Relativistic Particles . . . . . . . . . . 195

57.2 Interesting Cases - Linear and Circular Motion . . . . . . . . . . . . . . . . . . . . . 198

57.3 Power Loss Per Unit Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

57.4 Application - Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

57.5 Back Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

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58 Localized Sources 206

58.1 The Near (Static) Zone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

58.2 The Intermediate Zone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

58.3 The Far Zone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

59 Scattering 212

59.1 Oscillating Scatterers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214

59.2 Scattering In A Medium - Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

VI Summary of Semester I 218

60 Electrostatics 219

60.1 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

60.2 Methods for Working Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

61 Magnetostatics 220

61.1 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

61.2 Methods for Working Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

VII Summary of Semester II 223

62 Special Relativity 223

63 Tensors and Lorentz Invariant Forms 223

64 Hamiltonian Dynamics in Electromagnetism 224

65 Special Cases 225

65.1 Plane Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225

65.2 Geometric Optics and Wave Guides . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

65.3 Solutions with Sources (Radiation) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228

VIII Greens Functions and Reference Equations 231

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66 Greens Functions for Nabla Squared 231

66.1 Free Space Greens Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231

66.2 Unique Geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231

67 Greens Function for the Helmholtz Operator 232

68 Expansions 233

IX Homework Assignments (Problems only) 234

69 Semester 1 234

69.1 Homework 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234

69.2 Homework 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

69.3 Homework 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236

69.4 Homework 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

69.5 Homework 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238

69.6 Homework 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

69.7 Homework 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240

69.8 Homework 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

69.9 Homework 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242

69.10Homework 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

69.11Homework 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244

70 Semester 2 245

70.1 Homework 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245

70.2 Homework 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246

70.3 Homework 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248

70.4 Homework 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249

70.5 Homework 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250

70.6 Homework 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251

70.7 Homework 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252

70.8 Homework 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

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70.9 Homework 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254

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Monday - 8/25

1 Introduction - Foundations of Electromagnetism

1.1 Maxwell’s Equations

Solve them by the end of the semester...

∇ · E =ρ

ε0

∇ · B = 0

∇× E +∂B

∂t= 0

∇× B = µ0j +1c2

∂E

∂t

Additional Equations include the conservation of charges and the Lorentz Force.

∂ρ

∂t+∇ · j = 0

F = q(E + v × B

)Variable definitions:E(x, y, z, t) = electric fieldB(x, y, z, t) = magnetic fieldρ(x, y, z, t) = charge densityj(x, y, z, t) = current densityε0, µ0 = constants

Electromagnetism is a Gauge Theory meaning that there is a hidden symmetry that can be foundwith some analysis of the equations. In the language of group theory, the equations forms anAbelian group. From the equations above, it is easily seen that Electromagnetism is a linear first-order theory as well. There is no dependence on terms such as E2, 1/E, or other nonlinear functions.This means that the equations for many problems are solvable without strong computational needsor numerical approximations.

12

Part I

Electrostatics

2 The Basic Assumptions of Electrostatics

In general, it is understood that the terms ρ and j are the sources for fields defined as E and B. Ifwe neglect one of these sources, we can begin by studying one of the simplest of cases: Electrostatics.If there is no current, then we can assume that the fields are static in time, thus:

j =∂E

∂t=∂B

∂t= 0

These assumptions yield a new set of equations, the Electric and Magnetic fields are decoupled, andif we assume the magnetic field is zero, the resulting equations for the electric field are:

∇ · E =ρ

ε0

∇× E = 0

F = qE

Next, we assume that the electric field has both direction and magnitude dependent on its geometriclocation and therefore it can be written as E = [Ex(x, y, z), Ey(x, y, z), Ez(x, y, z)]. Then, if werotate the coordinate system so that the electric field at some point (x, y, z) points only in the newx direction. Now the electric field can be written as: E′ = [E′x(x, y, z), 0, 0]. Then, taking the curlof this electric field (which by the equations above must be zero), one has:

∇× E = det

∣∣∣∣∣∣i j k∂∂x

∂∂x

∂∂x

Ex(x, y, z) 0 0

∣∣∣∣∣∣ =∂Ex∂z− ∂Ex

∂y= 0

⇒ ∂Ex∂z

=∂Ex∂y

= 0

And so the electric field can be written as a function depending only on the x coordinate. And sowe can write for some scalar function φ that

E = [Ex(x), 0, 0]⇒ E = −∇φ

and this leads to the new problem of solving for the potential φ:

∇× (−∇ · φ) = 0

∇ · (−∇ · φ) =ρ

ε0→ ∇2φ = − ρ

ε0

13

This new equation is the Poisson Equation and states the general problem involved in solvingelectrostatics:

∇2φ(x) = −ρ(x)ε0

3 The Dirac Delta Function

The dirac function is defined as an improper function such that:

∫ ∞−∞

δ(x− a)f(x)dx = f(a)

and if we let f(x) = 1 and a = 0, we have the normalization of the delta function:

∫ ∞−∞

δ(x)dx = 1

Mathematician’s Definition: δ can be defined by a distribution of functions. Consider the functionsg1(x), g2(x), . . . , gn(x) and let n −→ ∞. In most cases, this limit will not exist, however, if weconsider instead:

limn→∞

∫ b

af(x)gn

This value is bounded as the limit is taken. Because of this, we can define the delta function as thelimit of any distribution of functions gn(x) where:

δ(x) = limn→∞

gn(x)dx; gn(x) = nG(xn)

Where G(x) is a function that is bounded, even, and normalized to unity. Two examples of thisare:

G(x) =1π

11 + x2

→ gn(x) =n

π

11 + (nx)2 → lim

n→∞

∫ ∞−∞

f(x)n

π

11 + (nx)2dx = f(0)

G(x) =1π

sin(x)x→ gn(x) =

1π

sin(nx)x

δ(x) = limn→∞

1π

sin(nx)x

dx = limn→∞

12πix

(einx − e−inx

)=

12π

∫ ∞−∞

eikxdk

14

3.1 Properties of the Dirac Delta Function

Some properties of the dirac delta function include:

δ(x) = δ(−x)

δ(ax) =1|a|δ(a)

δ(x− x0)f(x) = δ(x− x0)f(x0)

xδ(x) = 0

δ(f(x)) =∑i

δ(x− xi)y′(xi)∫ ∞

−∞δ′(x)f(x)dx = −

∫ ∞−∞

δ(x)f ′(x)dx = −f ′(0)

Homework: A Problem To Try:

d

dxln(x) =

1x

; x 6= 0

What about at zero?

Wednesday - 8/27Solution to problem:

Integrate whole equation around zero and try to ”smooth” out the singularity point:

∫ ε

−ε

d

dxln(x)dx =

∫ ε

−ε

1xdx

The right hand side evaluates to zero on the region from −ε to +ε and the left hand side is:

∫ ε

−ε

d

dxln(x)dx = ln(x)|ε−ε = ln(ε)− ln(−ε)

But we can write the −ε term instead as: −ε = −1|ε| = |ε|eiπ and then the above is simply:

ln(ε)− ln(|ε|)− ln(eiπ) = 0− iπ

this leads to the result that iπ = 0 which is obviously not possible, so the right hand side of theabove must have a missing contribution at x = 0. That is:

d

dxln(x) =

1x− iπδ(x)

15

Next, we need to generalize the Dirac Delta function into three dimensions. this is easy enough inCartesian coordinates.

∫ ∫ ∫f(x, y, z)δ(x)δ(y)δ(z)dxdydz = f(0, 0, 0)→

∫δ3(x− x′)f(x)d3x = f(x′)

In other coordinate systems, we need only to account for additional terms in the volume element.Of particular interest are Cylindrical and Spherical Coordinates:

∫d3cylx =

∫ ∞0

rdr

∫ 2π

0dφ

∫ ∞−∞

dz → δ3cyl(r) =

δ(r − r′)r

δ(φ− φ′)δ(z − z′)

∫d3sphx =

∫ ∞0

r2dr

∫ π

0d cos θ

∫ 2π

0φ→ δ3

sph(r) =δ(r − r′)

r2δ(cos θ − cos θ′)δ(φ− φ′)

4 The Greens Function

Recall the Poisson Equation that we derived previously and consider writing φ in an integral rep-resentation:

∇2φ = −ρ(r)ε0→ φ =

∫ρ(r)ε0

f(x′, x)d3x′

Then, plugging this into the Poisson Equation, we can move the derivatives inside the integral sincethey are different variables x and x′ which are being used.

∇2xφ =

∫ρ(x′)ε0∇2xf(x′, x)d3x′ → −ρ(x)

ε0=∫ρ(x′)ε0∇2xf(x′, x)d3x′

This leads to the condition that:

−ρ(x) =∫ρ(x′)∇2

xf(x′, x)d3x′

Under closer examination, this result is the same as that of the Dirac Delta function. Therefore, ifwe choose f(x, x′) to be a Greens Function of the operator ∇2 then it satisfies:

∇2xf(x, x′) = −δ(x− x′)

thus the potential function φ(x) is can be found by the above relation that:

φ =∫ρ(r)ε0

f(x′, x)d3x′

16

By setting x′ = 0 and noting the rotational invariance of the forces, we can write the above as:

∇2 =1r2

∂

∂r

(r2 ∂

∂r

)+ Lθ

∇2f(r) =1r2

∂

∂r

(r2 ∂

∂r

)f(r) = −δ(r); r 6= 0

Checking the region away from r = 0 we can see that:

1r2

d

dr

(r2 d

dr

)f(r) = 0→ r2 d

drf(r) = k → df

dr=

k

r2

f(x, x′) =k

|x− x′|; x 6= x′

Finally, using the same trick as with the homework problem above, we can integrate around theinhomogeneous area (in this case, around a sphere at 0 or radius ε). That is:

∫b0ε

(∇2f = −δ(x)

)d3x

R.H.S. = −1

L.H.S. =∫b0ε

∇2fd3x =∫∂b0ε

∇f · ndA

The integral is now over only the surface of the sphere and we can recall that f = −kr so then

∇ · f = kr2 which leaves the integral as:

∫∂b0ε

k

R2R2dΩ = 4πk ⇒ k = − 1

4π

Therefore the Greens function for ∇2x is:

f(x, x′) =1

4π1

|x− x′|

And finally, the potential is:

φ(x) =1

4πε0

∫ρ(x′)|x− x′|

d3x′

17

Friday - 8/29

4.1 More Solutions Using the Greens Function Solution

Recall the Poisson Equation and the Greens Function solution we derived previously,

∇2φ(x) =ρ(x)ε0−→ φ(x) =

14πε0

∫ρ(x′)|x− x′|

d3x′

This solution is for the inhomogeneous problem. In order to get the most general solution, we needto solve the homogeneous problem:

∇2φ(x) = 0

In order to get the vector field E we need to refer to the definition:

E = −∇φ = −∇x[

14πε0

∫ρ(x′)|x− x′|

d3x′]

= − 14πε0

∫ρ(x′)∇x

1|x− x′|

d3x′

This derivative is simply shown to be:

∇x1

|x− x′|= − x− x′

|x− x′|3

Which gives:

E =1

4πε0

∫ρ(x′)

x− x′

|x− x′|3d3x′

4.2 Special Case - Point Like Charges

For point like charges, the charge density distribution can be written as:

ρ(x) =∑i

qiδ(x− xi)

for each of i charges. Let’s assume there is only one charge in the universe and it is located at apoint x0. Then,

ρ(x) = qδ(x− x0)

18

In this case, we have:

E =1

4πε0

∫qδ(x′ − x0)

x− x′

|x− x′|3d3x′ =

q

4πε0x− x0

|x− x0|3

In the case that x0 is the origin of the coordinate system, then this can be written as:

E =q

4πε0rer|r|3

=q

4πε01r2er

Alternately, we can plug the definition of ρ(x) straight into the Poisson Equation, this gives:

∇2φ(x) = − q

ε0δ(x− x0)→ ∇2

(ε0qφ(x)

)= δ(x− x0)

So, the general solution here is the Greens Function for above.

4.3 More Difficult Cases

Next, we’ll consider how to write ρ(x) for non-point like charge distributions. Consider a rod ofcharge from x = −a to x = a. If the rod has total charge Q then we need to take only values atz = 0 and y = 0 and x from −a to a. So that means:

ρ = Aδ(z)δ(y)Θ(a− |x|)

The coefficient A can be found by integrating the charge density. If we integrate over all space,then we should get the total charge.

Q =∫Aδ(z)δ(y)Θ(a− |x|)d3x = A(1)(1)(2a) = LA→ A = Q/L

From this information we can see that the units of the delta function are inverse length and thestep function is unitless. So, the charge distribution does indeed have units of charge over volume.

ρ =Q

Lδ(z)δ(y)Θ(a− |x|)

From this result, we can derive a simplified method for finding the E field,

∫ [∇ · E =

ρ

ε0

]=⇒

∫∂VE · ndA =

Q

ε0

19

5 Interpretation of the Potential

Finally, we can discuss exactly what φ is. Consider a field E with a charged particle moving in it.The work done in moving the particle from point a to point b is:

W = −∫ b

aF · dl = −q

∫ b

aE · dl = −q

∫ b

a−∇φ · dl

= q

∫ b

a∇φ · dl = [φ(b)− φ(a)]q

Therefore, as expected, the work in moving a charged particle in a complete loop is zero becausewhen a = b, W = 0.

Monday - 9/1 - No Class, Labor Day.

Wednesday - 9/3

6 The Electric Field on a Surface - Deriving Boundary Conditions

Consider Gauss’ Law for a surface of charged material. Firstly, we can define the charge density onthe surface as ρ(x):

ρ(x) = δ(x · n)σ(x× n, (x× n)× n)

The delta function picks out only points which lie on the surface and the function σ characterizesthe charge distribution on the surface. For reference, in the case that the surface of interest is aplane, we can define the plane to be orientated in the XP plane and the above becomes simply

ρ(x) = δ(z)σ(x, y)

We can plug this into Gauss’ Law and get:

∫SE · ndS =

1ε

∫Vδ(x · n)σ(x× n, (x× n)× n)dV

The right hand side becomes an integral over the surface because of the delta function, so this canbe written as:

∫SE · n− 1

ε0σ(x× n, (x× n)× n)dS = 0

20

The integral must be zero, and therefore the integrand itself must be zero. However, the E field onthe surface is actually dependent on both the field above and below the surface. So finally, we havethat for a charged surface the normal components of the field are discontinuous:

(E2 − E1

)· n =

1ε0σ

So what about the tangential components? Consider a complete path passing through a chargedsurface. In such a case, we can state that

∫E · dl = 0 because we are completing a full loop in

the field. Because of this, we can take the four sides of the rectangular path into considerationseparately. This integrates to: L1E2 +L2E

′+L3E1 +L4E′′ = 0 where E1 and E2 are the projected

fields along the path (that is the tangential component). If we shrink the height of the path, thenthe lengths of L2 and L4 will go to zero. Further, because we are integrating along the loop in aconstant direction we can state that L1 = −L3 and so the above will become L1(E1 −E2) = 0 andbecause the length is not zero, we have the solution that the tangential components are continuousacross such a boundary. So, to summarize:

(E2 − E1

)· n =

1ε0σ;

(E2 − E1

)× n = 0

This result summarizes the behavior of the E field on either side of a surface which contains acharge distribution σ.

We can plug this definition for the charge distribution into the integral form of φ:

φ(x) =1

4πε0

∫V

ρ(x′)|x− x′|

d3x′ =1

4πε0

∫S

σ(x′)|x− x′|

d2x′

7 Definition of Potential Energy in a Charge Distribution

We can define the Potential Energy of a charge distribution as

W =12

∫ρ(x)φ(x)d3x =

18πε0

∫ρ(x)d3x

∫ρ(x′)|x− x′|

d3x′

where we’ve included the factor of 12 to account for the fact that we’re considering the potential of

a distribution from itself. In the case of a distribution in some external potential, the factor wouldnot be necessary. Consider a system of point charges. In the system of n point like charges, wecan write the charge distribution as ρ(x) =

∑ni=1 δ(x− xi) where xi is the location of the ith point

charge. Plugging this into the above equation for the potential yields:

W =1

8πε0

∑ij

qiqj|xi − xj |

21

This seems alright at first, however, consider the case where i = j. In this case, the denominatorblows up. This is due to the fact that classical electrodynamics is not equipped to properly dealwith point sources. In order to get a physical results, we have to take the potential and subtractout the self energy. The physical potential of a charge distribution is then:

Wphys =1

8πε0

∑i 6=j

qiqj|xi − xj |

We can use a separate analysis to find the energy density w for a distribution of charge. We knowfrom definition that E = −∇φ, and that ∇2φ = −ρ/ε0. From this we can write:

W =12

∫−∇2φε0φd

3x =ε02

∫∇φ · ∇φd3x =

ε02

∫|E|2d3x

w =ε02|E|2

8 Electrostatics in the Presence of Continuous Media - Conductors

Inside of a conductor, the E field is 0 (this will be proved next). Because of this, we can considerthe field on either side of the boundary (shell) of the conductor and see that:

(E2 − E1

)· n =

σ

ε0=(E2 − 0

)· n =

σ

ε0⇒ w =

ε02

(σ

ε0

)2

Thus, the potential energy density for a conductor with surface charge σ can be written as:

w =σ2

2ε0

PROOF : The E field inside of a conductor is zero.Consider a complete path near a conductor which passes out of and in to the conductor at pointsa and b respectively. For this loop, it must hold that:

∫E · dl = 0→

∫ b

aEinside · dl +

∫ a

bEsurface · dl = 0

Because the charge on the conductor is evenly distributed (there are no isolated areas of excesscharge) it is possible to choose the locations of a and b so that φ(a) = φ(b) and the second integralis zero. In this case, regardless of the path chosen for the inside portion of the conductor, the Efield inside must be zero. Thus the field inside of a conductor is zero.

22

8.1 Capacitance

Consider a series of point charges as before: ρ =∑

i qiδ(x − xi). Plugging this into the definitionfor φ one gets the relation:

φ(x) =∑i

qiαi, αi =1

4πε01

|x− xi|

In the case that there are multiple conductors, the sum of the above can be written:

Vj =∑i

QiPij −→ Qi =∑j

CijVj , C = P−1

Friday - 9/5

9 Dipole Layer

Consider a thin separation between conducting surfaces. Assume that the charge density on thetop and bottom surface is σ(x) and −σ(x) respectively and the surfaces are separated by a distanced(x). If we define the origin to be some other location (see notes for illustrations) then we candefine the following vectors:

x = location of observer.x′ = location of upper layer edge.x′′ = x′ − dn location of lower layer edge.x− x′ = separation of observer and upper edge of Dipole Layer.d(x)n = x− x′ = separation between the two layers (with normal n)θ = angle between x− x′ and dn.

The potential from this arrangement can be written as:

φ(x) =1

4πε0

∫S

σ(x′)|x− x′|

d2x′

φ(x) =1

4πε0

∫S′

σ(x′)|x− x′|

da′ − 14πε0

∫S′′

σ(x′′)|x− x′′|

da′′

Using the above relation for the vector x′′ we can write this as:

φ(x) =1

4πε0

∫S′σ(x′)

(1

|x− x′|− 1|x− x′ + dn|

)da′

If we assume that dn is much small than x− x′, then we can expand this as:

23

1|v − a|

=1√

(v1 + a1)2 + (v2 + a2)2 + (v3 + a3)2

=1√

v21 + v2

2 + v23 + a2

1 + a22 + a2

3 + 2(v1a1 + v2a2 + v3a3)

=1√

v2 + a2 + 2v · a=

1

|v|√

1 + 2 v·a|v| +(|a||v|

)2

And approximating for small |a| we can write:

1|v − a|

≈ 1|v|

(1 +

a · v|v|2

+O(|a|2)

1|v − a|

≈ 1|v|

(1 + a · ∇ 1

|v|+O(|a|2

)This turns the above expression for the potential into:

φ(x) ≈ 14πε0

∫S′σ(x′)

(1

|x− x′|− 1|x− x′|

− dn · ∇x1

|x− x′|

)da′

= − 14πε0

∫S′σ(x′)

(dn · ∇x

1|x− x′|

)da′

Another trick we can use is to change the variable we’re differentiating by: ∂zf(z−z0) = −∂z0f(z−z0). This leaves us with,

φ =1

4πε0

∫S′σ(x′)

(dn · ∇x′

1|x− x′|

)da′

We can then evaluate the derivative and also define a new function D(x) to simplify our equation:

σ(x′)d = D(x′)

∇x′1

|x− x′|=

x− x′

|x− x′|3

φ =1

4πε0

∫S′D(x′)

n · (x− x′)|x− x′|3

da′

We know from our information above that the angle between n and x − x′ is θ. So then we canwrite:

n · (x− x′) = |x− x′| cos θ

24

Finally, we can use the geometry of the observer’s location we can write the solid angle which theobserver would see as:

dΩ =cos θ|x− x′|2

da′

So the potential from the Dipole layer at some observer’s point x can be written as:

φ(x) = − 14πε0

∫D(x′)dΩ

10 Poisson Equation With Boundary Conditions - Greens Identity

Starting with the divergence theorem:

∫V∇ · Ad3x =

∫∂Vn · Ad2x

one can define A1 = φ∇ψ, A2 = ψ∇φ. We can then apply both to the above and subtract the twoequations:

∫V

(∇ · φ∇ψ −∇ · ψ∇φ) d3x =∫∂V

(n · φ∇ψ − n · ψ∇φ) d2x∫V

(φ∇2ψ +∇φ · ∇ψ −∇φ · ∇ψ − ψ∇2φ

)d3x =

∫∂V

(φ (n · ∇ψ)− ψ (n · ∇φ)) d2x

∫V

(φ∇2ψ − ψ∇2φ

)d3x =

∫∂V

(φ∂ψ

∂n− ψ∂φ

∂n

)d2x

Now, we’ll assume that φ in this equation is the potential, and ψ is a Greens function of the operator∇2.

∇2φ = − ρε0

∇2ψ = −4πδ(3)(x− x′)

So then the above becomes:

∫V−4πφδ(3)(x− x′)− ψ ρ

ε0d3x =

∫∂V

(φ∂ψ

∂n− ψ∂φ

∂n

)d2x

And collecting terms, we obtain Green’s Formula:

25

φ(x) =1

4πε0

∫Vρ(x′)G(x− x′)d3x′ +

14π

∫∂V

(G(x− x′)∂φ

∂n− φ(x′)

∂G

∂n

)d2x′

10.1 Dirichlet Boundary Conditions

If φ on the surface of the boundary is known, then we can use the Dirichlet Greens function whichis defined as GD(x− x′) = 0 for x on ∂V . In this case,

φ = φV ol −1

4π

∫∂Vφ(∂V ′)

∂GD∂n

d2x′

The Dirichlet Greens Function can be derived by:

GD =1

|x− x′|+ λ(x− x′); ∇2λ = 0

10.2 Neumann Boundary Conditions

If the E field on the surface is known, then ∇φ is known and we can use Neumann conditions. Theresultant for this case is:

φ = φV ol +1

4π

∫∂VGN (x− x′)∂φ(x′)

∂nd2x′ +

1S

∫φ(x′)d2x′

φ = φV ol +1

4π

∫∂VGN (x− x′)∂φ(x′)

∂nd2x′+ < φ∂V >

10.3 Example of a Greens Formula Problem - Planar Conductor

Monday - 9/8 Consider a distribution of charge ρ near a conducting plane where the potential isequal to V0 (which could be zero if the conductor is grounded). The Greens function associatedwith this condition is a Dirichlet Greens Function:

GD(x, x′) =1

|x− x′|− 1|x− x′′|

x′ = (x′, y′, z′); x′′ = (x′, y′,−z′)

This Greens function is zero on the surface. Therefore in this case we can write the potential as:

φ(x) =1

4πε0

∫Vρ(x′)

(1

|x− x′|− 1|x− x′′|

)− 1

4π

∫∂VV0∂GD∂n

d2x′

26

Let’s check a simple case of a point charge over a conducting plane at z = d. The charge distributioncan then be written as: ρ(x) = qδ(x− d); d = (0, 0, d). The potential comes out to be:

φ(x) =1

4πε0

∫Vqδ(x− d)

(1

|x− x′|− 1|x− x′′|

)d3x− 1

4π

∫∂VV0∂GD∂n

d2x′

φ(x) =q

4πε0

(1√

x2 + y2 + (z − d)2− 1√

x2 + y2 + (z + d)2

)

−V0

4π

∫∂V

∂

∂z′

(1√

x′2 + y′2 + (z′ − d)2− 1√

x′2 + y′2 + (z′ + d)2

)dx′dy′

In the case that the conducting plane is grounded, then V0 = 0 and the above simply becomes:

φ(x) =q

4πε0

(1√

x2 + y2 + (z − d)2− 1√

x2 + y2 + (z + d)2

)

Which is a method known as ”image charge” since there is a second ”charge” located as a distanced below the plane and the resulting potential is the combination of the two.

11 Mean Value Theorem

Proof: Consider the potential around a point p. If there are no charges nearby, then we can writethe potential as only the second (boundary) terms.

φ(x) =1

4π

∫∂V

(G(x− x′)∂φ

∂n− φ(x′)

∂G

∂n

)d2x′

Using the free space Greens Function G = 1|x−x′| = 1

r then we can get more analysis from thisscenario. Consider a spherical shell of radius R around the point. In this case r → R and thepotential can be written:

φ(x) =1

4π

∫∂V

(1R

∂φ

∂n− φ(x′)

∂

∂n

1R

)R2dΩ′

The first term makes no contribution since:

∂φ

∂n′= ∇′φ · n′ = −E′ · n′ →

∫∇ · EdΩ′ = 0

And the second term can be simplified to make the potential:

27

φ(x) =1

4π

∫∂V

(−φ(x′)

∂

∂R

1R

)R2dΩ′

φ(x) =1

4π

∫∂V

(φ(x′)

1R2

)R2dΩ′ =

14π

∫∂Vφ(x′)dΩ′

So, the potential at any point p can be found by taking the mean potential along every point on aspherical shell around that point. Mathematically, that is:

φ(x) =1

4π

∫∂Vφ(x′)dΩ′

12 Methods for 2 - Dimensional Problems

Consider a two dimensional area with no charges, Poisson’s equation becomes:

∇2φ = 0

Making a change of coordinate we can write:

z = x+ iy; z = x− iy

∇2 =∂2

∂x2+

∂2

∂y2= 4

∂

∂z

∂

∂z

Poisson’s Equation then becomes:

4∂

∂z

∂

∂zφ = 0

The solution of this is any function that is only dependent on z or z. So, we can write the potentialas any function φ(x+ iy). For some function φ(z) we can perform the analysis:

12∂

∂zφ =

12

[∂

∂x+ i

∂

∂y

]φ = 0→ φ = U(x, y) + iV (x, y)

∂U

∂x+ i

∂V

∂x+ i

∂U

∂y− ∂V

∂y= 0

∂U

∂x=∂V

∂y;

∂V

∂x= −∂U

∂x

Then, by taking ∂∂x of the first and ∂

∂y of the second equation and then sum or subtract the results,we get the new conditions on U and V :

28

∇2U = 0; ∇2V = 0

From this information we can write the functions U and V from any function f(z). Consider thefunction φ(z) = z2. In this case,

z2 = (x+ iy)2 = x2 − y2 + 2ixy → U(x, y) = x2 − y2; V (x, y) = 2xy

Plotting out various constant values of U and V, we see that x2 − y2 = U generate equipotentiallines of a wedge shaped conductor while y = V

2x plots out the E field lines of a charged conductorof the same geometry.

This same analysis can be done with any f(z). The difficulty is in that it is unclear what geometrywill result from a generic function of z.

Wednesday - 9/10

13 Method of Images

Consider a bounded region we’ll label as ”region 1” with a charge distribution ρ(x) in it. Thepotential in the region can be found without considering the boundary directly by creating a virtualcharge density ρ′(x) in the second region. This is easiest to understand through a few examples.One example of this was seen last class in the virtual charge below the conducting plane. Consideranother:

13.1 Point Charge Near A Conducting Sphere

For today, consider a conducting sphere of radius a with a single point charge at some location x′

such that |x′| > a. If we take the physical region outside the sphere to be the real region and thearea inside the sphere as being the virtual area where we’ll use image charges, we can assume thatthere will be some virtual charge q′ along the same direction as x′ but located inside the sphere.Calling this point y we can the write:

x′ = x′n′; x′ > a

y = yn′; y < a

The potential can then be written as:

φ(x) =1

4πε0q

|x− x′|+

14πε0

q′

|x− y|

If the sphere is grounded (that is, the potential across it is zero) then we can write a condition onthe above that when x = a the potential must be zero.

29

φ(a) =1

4πε0

(q

|a− x′|+

q′

|a− y|

)= 0

q

a

1|n− x′

a n′|

+q′

y

1|n′ − a

y n|= 0

The solution with this condition is:

x′

a=a

y;

q

a=−q′

y

⇒ q′ = −q ax′

; y =a2

x′

So, the potential for this system can be written:

φ(x) =q

4πε0

(1

|xn− x′n′|− a

x′1

|xn− a2

x′ n′|

)

We can get an expression for the charge distribution on the surface by integrating ρ on the surfaceof the sphere. We can take the angle between the real charge n′ and the observer n as γ.

σ = −ε0∇φ · n|x=a =∂

∂x

q

4πε0

(1

|xn− x′n′|− a

x′1

|xn− a2

x′ n′|

)

σ = − q

4πa2

a

x′1− a2

x′2(1 + a2

x′2 −2ax′ cos γ

)3/2

This has maximums at γ = 0 and π and minima at γ = π

2 and 3π2 . This is as expected, the most

polarized angles are those that line up with the charge point and that on the opposite side. Thepresence of the charge polarizes the sphere.

If we integrate σ over the entire solid angle we can get the total charge on the sphere. This hasbeen left to an exercise for the student.

Q = −2π∫ 1

−1

q

4πa2

a

x′

(1− a2

x′2

)d cos θ(

1 + a2

x′2 −2ax′ cos θ

)3/2

= −q ax′

= q′

30

14 Finding the Greens Function For a Sphere From the ImageSolution

It should be possible to express the potential we found by the Greens Identity and from that, findthe Greens Function used.

φ(x) =1

4πε0

∫Vρ(x′)G(x− x′)d3x′ +

14π

∫∂V

(G(x− x′)∂φ

∂n− φ(x′)

∂G

∂n

)d2x′

φ(∂V ) = 0; GD(x, x′)|x=∂V = 0

This leaves,

φ(x) =1

4πε0

∫Vqδ(x− x′)GD(x− x′)d3x′ =

q

4πε0GD(x− x′)

So, comparing this to the solution above we have the Dirichlet Greens Function for a sphere ofradius a.

GD(x, x′) =1

|x− x′|− a

|x′|1

|x− a2

|x′|2 x′|

Friday - 9/12Recall the Dirichlet Greens Function for a sphere of radius of a and an observer outside the spherederived above. Next we’ll look at a few examples of its use.

14.1 Conductive Sphere With Fixed Potential

Consider a spherical conductor with a fixed potential V (a, θ, φ) on the surface but no charge distri-bution. The Greens Identity is the source of our potential equation, however only one term remainssince there are no charges and the Dirichlet Greens function goes to zero on the surface of thesphere. This leaves us with an expression for the potential of:

φ(x) = − 14π

∫∂VV (a, θ′, φ′)

∂GD∂n′

d2x′

So we just need to find the value of the term ∂GD∂n′ at x′ = a. This turns out to be:

∂GD∂n

=1a

a2 − x2

(x2 + a2 − 2ax cos γ(θ, φ))3/2

Where γ is the angle between the observer and the location x′ on the sphere surface. The potentialcan then be written as:

31

φ(x) = − 14π

∫SV (a,Ω′)

a(a2 − x2

)(x2 + a2 − 2ax cos γ(Ω′))3/2

dΩ′

14.2 Sphere With Two Charges On A Line - Sphere In External Field

Consider a conductive sphere at the origin with point charges of ±Q located at points R1, R2

respectively where

R1 = (0, 0, R); R2 = (0, 0,−R)

The charge density for this distribution is then:

ρ(x) = Qδ(x)δ(y) (δ(z −R)− δ(z +R))

And with the solution from the method of images, we have the potential is:

φ(x) =Q

4πε0

[ 1√|x|2 +R2 + 2R|x| cos γ

− 1√|x|2 +R2 − 2R|x| cos γ

+1√

|x|2R2

a2 + a2 − 2R|x| cos γ− 1√

|x|2R2

a2 + a2 + 2R|x| cos γ

]

Next, consider the case that we let R → ∞ but increase the charge Q such that the quantity QR2

remains constant. The above can be written as a Taylor expansion in such a case and the leadingorder term found is of order Q

R2 . This is actually a constant electric field surrounding the conductor.From this we have found that the potential can be written in terms of such a field as:

φ(x) = −E0 cos(γ)(|x| − a2

|x|3+ . . .

)+ V0

From this we can further compute the charge distribution on the sphere shell and the total inducedcharge.

σ(Ω) = −ε0∇φ · n|sph = 3ε0E0 cos(γ(Ω))

Q =∫σ(Ω′)dΩ′ = 0

14.3 Spherical Conductor with Different Potentials on Hemispheres

Consider a spherical conductor where an insulating ring at the equator allows the upper and lowerhemispheres to be charged with different potentials. Again because there are no charges to integratein the volume, there is only the boundary terms. The potential is then written as:

32

φ(x) = − 14π

∫∂VV (a, θ′, φ′)

∂GD∂n′

d2x′

This result is similar to the first example, but to solve it requires separating the integrated surfaceinto two pieces and doing each separately.

∫∂Vd2x′ →

∫ 2π

0dϕ′

∫ 0

−1d cosϑ′ +

∫ 2π

0dϕ′

∫ 1

0d cosϑ′

φ(x) = −V0

4π

∫ 2π

0

∫ 0

−1

∂GD∂n′

dϕd cosϑ+V0

4π

∫ 2π

0

∫ 0

−1

∂GD∂n′

dϕd cosϑ

The integrand is a function of γ which we have not defined. It is straightforward to visualize. Theangle between two vectors in spherical coordinates can be expressed as a difference in ϑ and ϕ.After some geometric analysis, it’s straightforward to show that:

cos γ = cos(ϑ) cos(ϑ′) + sin(ϑ) sin(ϑ′) cos(ϕ− ϕ′

)We can get an analytic solution if we assume that the observer is along the vertical axis, in thiscase ϑ′ = 0 or π and the above can be integrated to:

φ(x) = V0

(1− |x|2 − a2

|x|√|x|2 + a2

)

Far from the sphere, the solution can be approximated by the Legendre Polynomials series.Monday - 9/15

15 Orthogonal Functions and Expansions

Consider some coordinate x on a region (a, b). We can define a set of functions Um(x) (m =1 . . . N) such that

∫ b

aUm(x)U∗m′(x)dx = δmm′

In some cases, the term dx can contain an additional weighting function w(x) to assist in normalizingthe above relation. If the dimension N is infinite. Then any function can be expanded in terms ofthe functions.

f(x) =∑m

amUm(x); am =∫ b

aU∗m(x)f(x)dx = 〈Um, f〉

33

This leads to another useful relation:

∑m

Um(x)U∗m(y) = δ(x− y)

15.1 Fourier Series Expansions

On an interval from −a2 to a

2 we can expand any function in a series of sines and cosines:

f(x) =A0

2+∞∑m=1

(Am sin

(2πmxa

)+Bm cos

(2πmxa

))

Am =2a

∫ a2

−a2

f(x) cos(

2πmxa

)dx; Bm =

2a

∫ a2

−a2

f(x) sin(

2πmxa

)dx

15.2 Polynomial Series

On an interval from −1 to 1 the Legendre Polynomials can be used. These are polynomials suchthat Um(x) = Pm(x). These can be looked up in any math reference book.

15.3 Infinite Intervals

If we define a new constant k = 2πma which we keep constant as we increase a to infinity, then wecan change the above Fourier Series into an Integral form.

∞∑m=−∞

cme2πimxa → 1√

2

∫ ∞−∞

c(k)eikxdk

This leads to the Fourier Transform:

f(x) =1√2

∫ ∞−∞

c(k)eikxdk

c(k) =1√2

∫ ∞−∞

f(x)eikxdk

15.4 Applications and Examples

15.4.1 Finite Box With Potentials on Different Walls

Consider a box of dimensions (a, b, c). If there are no charges, but the top panel of the box (z = c)has a voltage given by V (x, y) then we can use the above expansions to find the potential. Sincethere are no charges, the Poisson equation is simply:

34

∇2φ(x) = 0

and we’ll use a Fourier Integral to solve for the potential.

φ(x) =∫A(k)eik·xd3k

x = (x, y, z); k = (k1, k2, k3)→ k · x = k1x+ k2y + k3z

Inserting this into the equation, we have the condition that:

k21 + k2

2 + k23 = 0→ k3 = ±i

√k2

1 + k22

Because of the symmetry of the geometry, we can break the above into separate integrals. Thepotential is then:

φ(x) =∫A(k1, k2)eik1x+ik2ye∓

√k2

1+k22d2k =

∫A1(k1)eik1x

∫A2(k2)eik2ye∓z

√k2

1+k22dk1dk2

Because the solution must be zero at the boundaries of the box except for at z = c, we can writethe exponentials instead as sine functions. The conditions at the other edges of the box ( x = a,y = b) give the values of k1 and k2. These functions turn the integral back into a summation, sofinally, the potential is:

φ =∑m,n

Cm,n sin(mπx

a

)sin(nπy

b

)e∓z

q(mπa )2

+(nπb )2

Imposing the boundary condition at z = 0 we can write the other exponential as a hyperbolic sineand finally we have:

φ =∑m,n

Cm,n sin(mπx

a

)sin(nπy

b

)sinh

(z

√(mπa

)2+(nπb

)2)

Cm,n =4ab

∫ a

0

∫ b

0sin(mπx

a

)sin(nπy

b

) V (x, y)

sinh(z√(

mπa

)2 +(nπb

)2)dxdy

35

Wednesday - 9/17

15.5 Two Dimensional Problems Solved With Expansions

Consider a volume of space which allows the potential to be solved in two dimensions. In such aregion, one can write the Poisson Equation as:

∇2φ(x) = 0→[∂2

∂x2+

∂2

∂y2

]φ(x, y) = 0

Assume the potential can be written as a Fourier Integral Transform and then applying the aboveequation:

φ(x) =∫A(k)eik·x[

∂2

∂x2+

∂2

∂y2

]φ(x, y) =

∫A(k)(−|k|2)eik·x

Requiring this is zero yields the conditions on k1 and k2. Since |k|2 = k21 + k2

2 = 0 must hold, weknow that k2 = ±ik1. Therefore the most general solution is

φ(x, y) =∫A′(k1)eik1(x±iy)dk1 → φ(z) =

∫A(k)eikzdk

15.5.1 The Infinite Square Well

Consider an infinite square well of length a with the walls grounded and a potential V on the base.For such a geometry, we can solve the problem in two dimensions. Define the vertical axis of thewell to be the y axis and the horizontal to be x. This leads to a number of boundary conditions onthe above solution.

φ(x, y) =∫A(k)eik(x±iy)dk

limy→∞

φ(x, y) = 0; φ(x, 0) = V

φ(0, y) = 0; φ(a, y) = 0

The condition at y → ∞ kills the positive exponential. The combination of the two conditions forx = 0, x = a transform the oscillating exponential into sin

(nπa x)

and the integral into a summationover the index n. Finally, setting y to zero and then taking the inner product of the result withsin(mπa x)

will yield the value for the coefficient An. If n is even, then the coefficient is zero.Otherwise:

36

An(odd) =2anπ

Combining the result:

φ(x, y) =4Vπ

∞∑n=1,odd

1n

sin(nπax)e−

nπay

This result can be transformed a number of ways:

1n

sin(nπax)e−

nπay = Im

(Xn

n

); X = ei

πa

(x+iy

⇒ φ(z) =4VπIm

∞∑n=1,odd

Xn

n

However, the summation is actually a natural log:

∞∑n=1

Xn

n= ln (1 +X)

Or, to take only the odd values of n,

∞∑n=1,odd

Xn

n= ln

(1 +X

1−X

)

Plugging in the definition above for X and after some algebra we have the general form for theresult:

φ(z) =4VπIm

(ln

(1 + ei

πa

(x+iy)

1− eiπa

(x+iy)

))

Expanding this:

φ(z) =2Vπ

arctan

(sin(πxa

)sinh

(πxa

))

37

16 Poisson’s Equation In Other Coordinate Systems - SphericalCoordinates

Consider Poisson’s Equation in spherical coordinates:

∇2φ(r, ϑ, ϕ) = 0

1r

∂2

∂r2(rφ) +

1r2 sinϑ

∂

∂ϑ

(sinϑ

∂

∂ϑ

)φ+

1r2 sin2 ϑ

∂2

∂ϕ2φ = 0

We can separate this into 3 differential functions as:

φ(r, ϑ, ϕ) =∑∫

l,mu(r, l)P (ϑ, l,m)Q(ϕ,m)

Where the integral and sum denote that it is unclear if the values of l and m will be integer indices orcontinuous variables of the functions. This behavior is determined by the geometry of the problems.For our purposes currently, assume that both are integers and so this is:

φ(r, ϑ, ϕ) =∑l,m

u(r, l)P (ϑ, l,m)Q(ϕ,m)

Plugging this into the equation above we find that:

ul(r) =1r

(clr

l+1 + dlr−l)

Qm(ϕ) = Ameimϕ = Bme

−imϕ

In the case that m = 0 the equation for ϑ is simply the Legendre equation and so the function P ismerely the lth Legendre polynomial:

Pl,0(ϑ) = Pl(cosϑ)

38

Friday - 9/19

16.1 Boundary Value Problems With Azimuthal Symmetry

For a case where the geometry of the problem is symmetric about some axis, the geometry resultsin a potential that is only dependent on the radial distance and the secondary angle. This can bewritten as:

φ(r, ϑ) =∞∑l=0

1r

(Alr−l +Blr

l+1)Pl(cosϑ)

In the case that r = 0 is in the volume of interest, the first term must be ignored as it diverges atthat point. In this case the potential is written as:

φ(r, ϑ) =A0

r+∞∑l=0

1rBlr

l+1Pl(cosϑ)

where the term A0 is nonzero only for a point charge located at the origin.Alternately, if the volume of interest extends to r →∞, then the second term diverges and becausethe zeroth order term of that summation is a constant, it can be neglected entirely,

φ(r, ϑ) =∞∑l=0

Alr−(l+1)Pl(cosϑ)

As an example consider the hemisphere problem we worked previously. This problem can be solvedas a summation by noting that the potential on the sphere is a function only of ϑ. If we areinterested in the potential outside the sphere, we take the appropriate solution of the above twoand we can then write the potential as:

φ(r, ϑ) =∞∑l=0

Alr−(l+1)Pl(cosϑ); Al =

2l + 12

al+1

∫V (ϑ)Pl(cosϑ)d cosϑ

And the coefficients are determined by:

∫Pl(cosϑ)Pl′(cosϑ)d cosϑ = δll′

22l + 1

⇒∫ [

φ(r, ϑ) =∞∑l=0

Alr−(l+1)Pl(cosϑ)

]Pl′(cosϑ)d cosϑ

Another trick we can use with these expansions is to take a solution for an azimuthal symmetricproblem along the axis of symmetry and by expanding that solution we can find the solution in allspace. Given some problem which results in a potential which is symmetric about an axis where

39

the potential is known, we can write that potential as φ(z). However, consider the expansion whichwe are working with:

φ(r, 0) =∞∑l=0

1r

(Alr−l +Blr

l+1)Pl(cos(0))

however, it can be shown that for all values of l the term Pl(1) is simply 1. Therefore the potentialis merely,

φ(r, 0) =∞∑l=0

Alr−l − 1 +Blr

l =∞∑l=0

Alz−l − 1 +Blz

l

However, consider expanding the original unknown φ(z) in terms of powers of either r or 1r . The

choice between the two is dependent on whether we are looking a close cavity about the origin oran extended region that includes r →∞. In either case:

φ(z)→ φ(r) =∞∑l=0

C1l

rl+1+ C2lr

l → φ(r, ϑ) =∞∑l=0

(C1l

rl+1+ C2lr

l

)Pl(cosϑ)

In this manner, the solution for any point can be found by simply multiplying each term by theappropriate Legendre Polynomial of cosϑ. In most cases only one of the two coefficients is relevantfor a single solution but depending which region is being investigated will determine which.

16.1.1 Example - Ring of Charge

Consider a ring of charge (total charge = Q) some distance b above the x, y plane with radius a.The potential for this problem is difficult to find, however, we can easily see that along the axis ofsymmetry (x = y = 0) the solution is just:

φ(z) =1

4πε0Q

D

Where D is the separation of that location with the z axis. In this case we can define the dis-tance in terms of z, α the angle from the plane up to the ring, and c where c2 = a2 + b2:D =

√r2 + c2 − 2rc cosα. So then, the potential is;

φ(z) =1

4πε0Q√

z2 + c2 − 2zc cosα

Depending what value of r we are interested in we can either expand this about rc or c

r . Considerr c,

40

φ(r) =Q

4πε01r

1√1 + c2

r2 − 2 cr cosα⇒ φ(r, ϑ) =

Q

4πε0

∞∑l=0

clrl+1

Pl(cosα)Pl(cosϑ)

Note that the value of cosα can be written as b√a2+b2

where b is the location of the ring above thex, y plane and a is the radius of the ring.

16.1.2 Greens Function Solution For Problems With Azimuthal Symmetry

One last trick is to write the free space Greens Function for a point charge as an expansion. Consider:

G(x, x′); x′ = (0, 0, z′)→ G =1√

x2 + y2 + (z − z′)2

And so along the axis this simplifies to:

G(z, z′) =1

z − z′

which can be expanded about powers of either zz′ or z′

z depending which is larger. Out of this weget that:

1|r − r′|

=∞∑l=0

rl<

rl+1>

Pl(cos γ)

This is of great help if we are given some charge distribution in terms of the above expansion, sincethe potential can then be written as:

φ(r, ϑ) =1

4πε0

∞∑l=0

∫ρ(r′, ϑ′)

rl<

rl+1>

Pl(cos γ)r′2dr′dϕd cosϑ

φ(r, ϑ) =1

2ε0

∞∑l=0

∫ρ(r′, ϑ′)

r′l<r′>l+1

Pl(cos γ)r′2dr′d cosϑ′

Monday - 9/22

16.2 Conic Holes and Sharp Points - Note: Will not be on tests or homework

In the case that a problem includes azimuthal symmetry but the angle ϑ does not sweep out itsentire range of (0, π) but instead some angle (0, α), the radial solution is still valid, however theangular solution is changed. In this case, the general solution can be written as:

41

φ(r, ϑ) =∞∑k=1

AkrkPνk(cosϑ)

Where the values of νk are selected by the condition that:

Pνk(cosα) = 0

16.3 Azimuthal Variations - The Spherical Harmonics

If the geometry of a problem allows the variable m to remain an integer but isn’t azimuthal sym-metric, then m will become an index for the functions and the general solution becomes:

φ(r, ϑ, ϕ) =∞∑l=0

l∑m=−l

(Al,mr

l +Bl,mr−(l+1)

)Yl,m(ϑ, ϕ)

Yl,m(ϑ, ϕ) =

√2l + 1

4π(l −m)!(l +m)!

Pml (cosϑ); Yl,−m(ϑ, ϕ) = (−1)mY ∗l,m(ϑ, ϕ)

Pml (x) = (−1)m(1 + x2)m2dm

dxmPl(x); P−ml (x) = (−1)m

(l −m)!(l +m)!

Pml (x)

The orthonormality condition for the spherical harmonics is:

∫Y ∗l,m(ϑ, ϕ)Yl′,m′(ϑ, ϕ)dΩ = δl,l′δm,m′

Therefore any angular function g(ϑ, ϕ) can be expanded in spherical harmonics:

g(ϑ, ϕ) =∞∑l=0

l∑m=−l

Cl,mYl,m(ϑ, ϕ); Cl,m =∫Y ∗l,m(ϑ, ϕ)g(ϑ, ϕ)dΩ

16.4 The Addition Theorem

For some angle γ between angle x and x′ the relation between γ and the angles ϑ, ϑ′, ϕ, and ϕ′ canbe written as:

Pl(cos γ) =4π

2l + 1

l∑m=−1

Y ∗l,m(ϑ′, ϕ′)Yl,m(ϑ, ϕ)

42

We can use this result to write the free space Greens function G(x, x′) as:

GD(x, x′) =1

|x− x′|=∞∑l=0

rl<

rl+1>

Pl(cos γ)

GD(x, x′) =∞∑l=0

l∑m=−1

4π2l + 1

rl<

rl+1>

Y ∗l,m(ϑ′, ϕ′)Yl,m(ϑ, ϕ)

Which yields the form for the potential:

φ(r, ϑ) =1ε0

∞∑l=0

l∑m=−1

12l + 1

∫ρ(r′, ϑ′, ϕ′)

rl<

rl+1>

Y ∗l,m(ϑ′, ϕ′)Yl,m(ϑ, ϕ)r′2dr′dϕ′d cosϑ′

One important simplification to the above is that if we return to the case of azimuthal symmetry,then the spherical harmonics simplify and only the m = 0 term is relevant. It can be shown thatthis leads to the simplification that:

GD(x, x′) =1

|x− x′|=∞∑l=0

rl<

rl+1>

Pl (cosϑ)Pl(cosϑ′

)

43

Wednesday - 9/24

16.5 Greens Function Expansion For a Sphere

Using the above form for the free space Greens function in 3 dimensions one can write the GreensFunction for a sphere of radius a by extending this definition to the form:

GD(x, x′) =1

|x− x′|− a

|x′|1

|x− a2

|x′|2 x′|

The first term can immediately be written in the above form, however we need to write the secondin a form suitable to 1

|v−v′| . This can be done by bringing the coefficients inside and obtaining:

a

|x′|1

|x− a2

|x′|2 x′|

=1

| |x′|a x−

a|x′| x

′|

Therefore,

GD(x, x′) =1

|x− x′|− 1|y − y′|

y =|x′|ax; y′ =

a

|x′|x′

Consider the magnitudes of each of these vectors:

|x| = r; |x′| = r′

|y| = rr′

a; |y′| = r′a

r′= a

So, with the above expression, we can write the Greens Function for a sphere of radius a as anexpansion of the form:

GD(x, x′) = 4π∞∑l=0

l∑m=−l

12l + 1

Y ∗lm(ϑ′, ϕ′)Ylm(ϑ, ϕ)

(rl

r′l+1− al(

rr′

a

)l+1

)

This simplifies down to:

GD(x, x′) = 4π∞∑l=0

l∑m=−l

12l + 1

[rl<

rl+1>

(1− 1

a

(a2

r>r<

)l+1)]

Y ∗lm(ϑ′, ϕ′)Ylm(ϑ, ϕ)

44

16.5.1 Greens Function Expansion for Concentric Spherical Shells

Consider a pair of spherical conductive shells of radii a and b such that a < b. If we want to calculatethe potential for a < r < b then we need to solve the equation:

∇2GD(r, ϑ, ϕ, r′, ϑ′, ϕ′) = −4πδ(3)(x− x′) = −4πδ(r − r′)

r2δ(Ω− Ω′)

GD|r,r′=a = 0; GD|r,r′=b = 0

Where we’ve defined δ(Ω − Ω′) = δ(cosϑ − cosϑ′)δ(ϕ − ϕ′). We can immediately see that we cansatisfy the angular conditions if we choose:

GD(r, ϑ, ϕ, r′, ϑ′, ϕ′) =∞∑l=0

∞∑l′=0

l∑m=−l

l′∑m′=−l′

Alml′m′(r, r′)Y ∗l′m′(ϑ′, ϕ′)Ylm(ϑ, ϕ)

From this, we can find the conditions to determine Alml′m′(r, r′):

∇2(r,ϑ,ϕ)GD =

∞∑l=0

∞∑l′=0

l∑m=−l

l′∑m′=−l′

(∇2

(r,ϑ,ϕ)A)Y ∗Y + 2

(∇(r,ϑ,ϕ)A · ∇(r,ϑ,ϕ)Y

)Y ∗ +AY ∗∇2

(r,ϑ,ϕ)Y

It is clear that the term ∇A ·∇Y is zero since A is only a function of radial values while the sphericalharmonics are not. The dot product will then be

(r · Ω

)which is zero. The third term is solved by

noting that it is merely an eigenvalue equation.

∇2GD =∞∑l=0

∞∑l′=0

l∑m=−l

l′∑m′=−l′

(1r

∂2

∂r2

(rAlml′m′(r, r′)

)− l(l + 1)

r2Alml′m′(r, r′)

)Y ∗Y

It is clear from this equation that Alml′m′(r, r′) is actually only dependent on a single index l.Therefore we can write this function as Alml′m′(r, r′) = cl(r, r′)δl,l′δm,m′ . Thus we have an initialform for the Greens Function:

GD(r, ϑ, ϕ, r′, ϑ′, ϕ′) =∞∑l=0

l∑m=−l

cl(r, r′)Y ∗lm(ϑ′, ϕ′)Ylm(ϑ, ϕ)

1r

∂2

∂r2

(rcl(r, r′)

)− l(l + 1)

r2cl(r, r′) = −4π

δ(r − r′)r2

This has solutions of:

cl(r, r′) = Al(r′)rl +Bl(r′)r−(l+1)

45

cl(a, r′) = 0→ cl(r, r′) = Al(r′)[rl − a2l+1

rl+1

]cl(r, r′) = cl(r′, r); cl(r, b) = 0→ cl(r, r′) = c

[rl − a2l+1

rl+1

] [r′l − b2l+1

r′l+1

]The constant c can be found by integrating around r = r′. That is:

∫ r′+ε

r′−εc

[rl< −

a2l+1

rl+1<

][1rl+1>

−rl>b2l+1

]=∫ r′+ε

r′−ε−4π

δ(r − r′)r2

This gives:

c =4π

2l + 11

1−(ab

)2l+1

And thus the Greens Function for this geometry:

GD(r, ϑ, ϕ, r′, ϑ′, ϕ′) = 4π∞∑l=0

l∑m=−l

Y ∗lm(ϑ′, ϕ′)Ylm(ϑ, ϕ)

(2l + 1)(

1−(ab

)2l+1) [rl< − a2l+1

rl+1<

][1rl+1>

−rl>b2l+1

]

Friday - 9/26 - NO CLASS - DEBATE DAYMonday - 9/29

Note on a common mistake in the homework: For a derivative of a Greens Function

defined as ∂GD∂n

the direction of the normal is dependent on the volume and volume

boundary. However, for the surface charge density σ = −ε ∂φ∂n

the normal is always

pointing out of the conductive surface.

16.6 Examples of Problems Solved With Spherical Harmonics

Consider a charged ring of radius a inside of a grounded conductive sphere of radius b. Because thesphere is grounded, the potential on the sphere is 0 and therefore the potential in the entire volumeoutside the sphere is also 0. Inside the sphere, we can solve for the potential as:

φ(x) =1

4πε0

∫Vρ(x′)G(x− x′)d3x′ +

14π

∫∂V

(G(x− x′)∂φ

∂n− φ(x′)

∂G

∂n

)d2x′

Choosing a Dirichlet Greens function for a sphere (which we know), we can drive one boundaryterm to zero. The second boundary term is zero since the potential on the sphere is zero. Thisleaves us with only the volume integral:

46

φ(x) =1

4πε0

∫Vρ(x′)GD(x, x′)d3x′

The charge distribution can then be written as:

ρ(x) =Q

2πa2δ(r − a)δ(cosϑ− 1)

Plugging this all in we can immediately see the solution is:

φ(r, ϑ) =Q

4πε0

∞∑l=0

Pl(0)Pl(cosϑ)

[rl<

(1rl+1>

−rl>b2l+1

)]

The values of r< and r> are dependent on where inside the sphere we are. For r > a the values arer> = r and r< = a. For r < a the values are reversed.

16.6.1 Dipole Solved With Two Methods

An alternate method to find this solution would be to take the exact solution for the charge dis-tribution given, then add on a homogeneous solution and rig the coefficients so that the potentialgoes to zero at the desired location. As an example of this method, consider a geometry where twoopposite point charges are separated by a distance 2a.

φ(x) =q

4πε0

[1

|x− x′|− 1|x− x′′|

]; x′ = (0, 0, a), x′′ = (0, 0,−a)

Expanding for r > a:

φ(r, ϑ) =q

4πε0

∞∑l=0

[1r

(ar

)lPl(cos (ϑ))− 1

r

(ar

)lPl(cos (ϑ+ π))

]

And since we can write Pl(cos (ϑ+ π)) = Pl(− cosϑ) = (−1)lPl(cosϑ), The above simply becomes:

φ(r, ϑ) =q

4πε0

∞∑l=0

1r

(ar

)l (1− (−1)l

)Pl(cos (ϑ))

If we define a variable p = 2qa and take the limit as a goes to zero, but require that qa remainsconstant and finite, then we can drop terms of order higher than l = 1 and noting that the term oforder zero is a constant (which is irrelevant for the potential), we can write the familiar expressionfor a dipole:

47

φdipole(r, ϑ) =p

4πεcosϑr2

Now, consider putting the dipole in a spherical grounded conductor of radius b (like the ring ofcharge) and assume that b a. We could do as we did before using the Greens function for thesphere and the charge density as a sum of two delta functions.

φ(x) =1

4πε0

∫Vρ(x′)GD(x, x′)d3x′ =

q

4πε0

∫V

[δ(x− x′)− δ(x− x′′)

]GD(x, x′)d3x′

Alternately, we could use the solution from above and require the homogeneous solution to drivethe over all potential to zero on the surface of the sphere. That is:

φ = φdipole + φhom. =q

4πε0

∞∑l=0

1r

(ar

)l (1− (−1)l

)Pl(cos (ϑ)) +

q

4πε0b

a

[1

|x− b2

a2 x′|− 1|x− b2

a2 x′′|

]

φ =q

4πε0

∞∑l=0

1r

(ar

)l (1− (−1)l

)Pl(cos (ϑ)) +

1b

(rab2

)l (1− (−1)l

)Pl(cos (ϑ))

Again, letting a go to zero but keeping aq constant leaves only the l = 1 term:

φ(r, ϑ) =p

4πεcosϑ

(1r2− r

b3

)Alternately, we could have simply taken the result and added a homogeneous solution;

φ = φdipole + φhom. =q

4πε0

∞∑l=0

1r

(ar

)l (1− (−1)l

)Pl(cos (ϑ)) + Clr

lPl(cosϑ)

Cl = − q

4πε01bl+1

(ab

)l (1− (−1)l

)And in the limit of a→ 0 we once again find that:

φ(r, ϑ) =p

4πεcosϑ

(1r2− r

b3

)

48

17 Poisson’s Equation In Other Coordinate Systems - CylindricalCoordinates

Consider the cylindrical coordinates form for the Homogeneous Poisson Equation:

∇2φ(ρ, ϕ, z) =[∂2

∂ρ2+

1ρ

∂

∂ρ+

1ρ2

∂2

∂ϕ2+

∂2

∂z2

]φ(r, ϕ, z) = 0

This can be separated into phi(r, ϕ, z) = R(r)Q(ϕ)Z(z) where these functions satisfy:

d2

dz2Z(z)− k2Z(z) = 0→ Z = e±kz

d2

dϕ2Q(ϕ) + ν2Q(ϕ) = 0→ Q = e±νϕ

d2

dρ2R(ρ) +

1ρ

d

dρR(ρ) +

(k2 − ν2

ρ2

)R(ρ) = 0

The solutions of the last equation are Bessel Functions of the first time. Thus:

R(ρ) = C1Jν(kρ) + C2Yν(kρ)

So, for a problem where ϕ sweeps out a total circle of 2π, the general solution can be written as:

φ(ρ, ϕ, z) =∞∑n=0

∫dk (An(k)Jn(kρ) +Bn(k)Yn(kρ))

(Cn(k)einϕ +Dn(k)e−inϕ

) (E(k)ekz + F (k)e−kz

)

Wednesday - 10/1

17.1 Simplifications For Different Geometries

In the case that the region of interest contains ρ = 0, the Bessel Yn blows up and the term Bn mustbe zero. If the region contains z →∞ then ekz blows up and thus E(k) = 0. If z → −∞ then e−kz

blows up and F (k) = 0.

17.2 An Example - A Hollow Conductive Cylinder

Consider the example of a conductive cylinder of radius a and length l. Find the potential inside thecylinder under given boundary conditions. If the potential on the base (z = 0) and sides ρ = a) arezero and the top piece is described by some function V (ρ, ϕ) then we can write the above subjectto these boundary conditions. The condition at z = 0 means that E(k) = −F (k) and then the z

49

dependence becomes ekz − e−kz = sinh(kz). Also, because ρ = 0 is within our region, Bn(k) = 0and our potential is given by:

φ(ρ, ϕ, z) =∞∑n=0

∫dkAn(k)Jn(kρ) sinh(kz)

(Cn(k)einϕ +Dn(k)e−inϕ

)Imposing the boundary condition that φ(ρ = a) = 0 gives the values of k. Requiring that Jn(ka) = 0means that k = xn,m

a where xn,m gives the mth zero of Jn(x). This kills the integral and gives thevalues of k allowed for our potential. The angular terms are more easily dealt with in trig form, sofinally our potential is:

φ(ρ, ϕ, z) =∞∑n=0

∞∑m=1

Jn

(xn,ma

ρ)

sinh(xn,m

az)

(Cn,m cos (nϕ) +Dn,m sin (nϕ))

The two coefficients are found by the remaining boundary condition:

V (ρ, ϕ) =∞∑n=0

∞∑m=1

Jn

(xn,ma

ρ)

sinh(xn,m

al)

(Cn,m cos (nϕ) +Dn,m sin (nϕ))

∫ρdρdϕ

[V (ρ, ϕ) =

∞∑n=0

∞∑m=1

Jn

(xn,ma

ρ)

sinh(xn,m

al)

(Cn,m cos (nϕ) +Dn,m sin (nϕ))

]

×Jn′(x′n′,m′

aρ

)sin (n′ϕ)cos (n′ϕ)

The choice of sin or cos is determined by which of the two coefficients is being solved for.

17.3 Modified Bessel Functions

In the case that the geometry is periodic in z, the values of the variable k will become imaginaryand the solutions are then the Modified Bessel Functions.

In(x) = i−nJn(ix)

Kn(x) =π

2in+1H(1)

n (ix) =π

2in+1 (Jn(ix) + iYn(ix))

50

18 Some Mathematics - Getting Greens Functions From A GenericOperator

Consider the operator O and its eigenfunction and eigenvalue solutions, Oψ = λψ. Given such anoperator, if the set of its eigenfunctions ψn are known, then we can expand any function as eithera sum or integral transform (depending on the nature of the spectrum of the operator):

f(x) =∫dkc(k)ψk(x) =

∑n

cnψn(x)

Note also the properties of orthonormality and completeness (respectively):

∫ψn(x)ψn′(x)dx = δn,n′ ;

∑n

ψn(x)ψn(x′) = δ(x− x′)

Now, consider building a Greens function in this nature:

G(x, x′) =∑n

cn(x′)ψn(x)

Using the above relation we can find the coefficient and we get:

cn(x′) = −4πλnψ∗n(x′)

Thus we end with our two possible solutions for a three dimensional Greens function where theexact form is dependent on whether the spectra is discrete or continuous:

G(x, x′) = −4π∑n

ψ∗n(x′)ψn(x)λn

; G(x, x′) = −4π∫dkψ∗k(x

′)ψk(x)λ(k)

As an example, consider the Greens Function in Cartesian Coordinates for the Laplacian in a boxof dimensions a, b, c with Dirichlet boundary conditions. With Dirichlet conditions, we have:

ψl,m,n(x, y, z) =23/2

√abc

sin(lπx

a

)sin(mπy

b

)sin(nπz

c

)Applying the above relation gives the Dirichlet Greens Function for a rectangular box:

GD(x, x′) =8abc

∑l,m,n

sin(lπxa

)sin(lπx′

a

)sin(mπy

b

)sin(mπy′

b

)sin(lπzc

)sin(nπz′

c

)l2

a2 + m2

b2+ n2

c2

51

Friday - 10/3

19 Multipole Expansions In Electrostatics

Consider some charge distribution contained within a volume of space of scale L. At some point xsuch that L

|x| 1 we can write the potential as a series of the form:

φ(r, ϑ, ϕ) =1

4πε0

∞∑l=0

l∑m=−l

12l + 1

qlmYlm(ϑ, ϕ)rl+1

; ql,m =∫dr′r′2dΩ′ρ(r′,Ω′)r′lY ∗l,m(Ω′)

This is known as a multipole expansion, since each of the terms qlm is determined by a geometricaldistribution of point like charges. Of particular interest are the first three terms. They are themonopole, dipole, and quadrupole terms. Let’s consider them:

l = 0,m = 0

q0,0 =∫dr′r′2dΩ′ρ(r′,Ω′)r′2Y ∗0,0(Ω′) =

1√4π

∫d3x′ρ(x′) =

Q√4π

l = 1,m = −1, 0, 1

q1,0 =∫d3x′ρ(x′)r′Y ∗0,1(Ω′) =

√3

4π

∫d3x′ρ(x′)r′ cosϑ′ =

√3

4π

∫d3x′ρ(x′)z′

q1,1 =∫d3x′ρ(x′)r′Y ∗1,1(Ω′) =

√3

8π

∫d3x′ρ(x′)r′

(− sinϑ′e−iϕ

)=

−√

38π

∫d3x′ρ(x′)r′ sinϑ′

(cosϕ′ − i sinϕ′

)= −

√3

8π

∫d3x′ρ(x′)

(x′ − iy′

)q1,−1 = −q∗1,1 =

√3

8π

∫d3x′ρ(x′)

(x′ + iy′

)If we define a new quantity p =

∫d3x′ρ(x′)x′ Then we can write the above as:

q1,0 =

√3

4πpz; q1,1 =

√3

8π(px − ipy) ; q1,−1 =

√3

8π(px + ipy)

The quadrupole term can be written in terms of the quadrupole moment tensor :

Qij =∫ (

3x′ix′j − r′2δij

)ρ(x′)d3x′

52

With all of this, we can write the potential as:

φ(x) =1

4πε0

Qr

+p · xr3

+12

∑ij

Qijxixjr5

+ . . .

Q =

∫d3x′ρ(x′)

p =∫d3x′ρ(x′)x′

Qij =∫ (

3x′ix′j − r′2δij

)ρ(x′)d3x′

Lastly, with the definition E = −∇φ we can write the electric field components in terms of themultipole expansion:

Er =∞∑l=0

l∑m=−l

l + 1(2l + 1) ε0

ql,mYl,m(ϑ, ϕ)rl+2

Eϑ = −∞∑l=0

l∑m=−l

1(2l + 1) ε0

ql,m1rl+2

∂

∂ϑYl,m(ϑ, ϕ)

Eϕ = −∞∑l=0

l∑m=−l

1(2l + 1) ε0

ql,mim

sinϑYl,m(ϑ, ϕ)rl+2

Monday - 10/6

19.1 Multipole Treatment of a Dipole Source

Consider the electric field from a dipole. We can write this as:

E(2) = −∇φ = − 14πε0

∇( p · xr3

)= − 1

4πε0∇(px

∂

∂x

x

r3+ py

∂

∂y

y

r3+ pz

∂

∂z

z

r3

)= − 1

4πε0∇(pxr3− 3pxx2

r5+pxr3− 3pyy2

r5+pzr3− 3pzz2

r5

)If we define a vector x = rn then the above can be written as:

E(2) =1

4πε03n (p · n)− p

r3=

14πε0

3n (p · n)− p|x− x0|3

This solution is valid for dipoles unless x = x0 (that is r = 0). In that case we have a solution ofthe form:

53

E(2) =1

4πε03n (p · n)− p

r3+ kδ(r)

The coefficient k can be found by integrating around the singularity at r = 0. That is:

∫V

[E(2) =

14πε0

3n (p · n)− pr3

+ kδ(r)]d3x

Because of symmetry, the term on the right is zero for a spherical shell about the origin, thus:

∫VE(2)d3x = k

We can write the electric field as an expansion and thus the left hand side becomes:

∫VE(2)d3x =

∫V∇φ(2)d3x = −

∫∂Vφ(2)nd2x

= − R2

4πε0

∫∂Vd3x′

∫dΩρ(x′)

n

|x− x′|= − R2

4πε0

∫∂Vd3x′

∫dΩρ(x′)

∞∑l=0

nrl<

rl+1>

Pl(cos γ)

We can rotate the system so that only the l = 1 term is nonzero. The r> term is then R2 whichcancels with the other R2 term. This leaves:

∫VE(2)d3x = − 1

4πε0

∫∂Vd3x′ρ(x′)nr′

4π3

= − 13ε0

p

And finally we have the electric field from a dipole:

E(2) =1

4πε03n (p · n)− p

r3− 1

3ε0p δ(r)

19.1.1 Energy of a Dipole in an External Field

Consider some charge distribution in an external field. The potential energy of such a system is:

W =∫ρ(x)Φext(x)d3x

The external potential can be expanded as long as it is slowly varying in the region containing thecharge distribution. That is:

54

Φext(x) ≈ Φext(x0) + x · ∇Φext(x0) +12

∑i,j

xixj∂2Φext

∂xi∂xj(x0) + . . .

Plugging this in, we find:

W = QΦext(x0)− p · E(x0)− 16

∑i,j

Qi,j∂E

∂xi(x0) + . . .

If we assume that the external field is from a second dipole as some location x2, then we can writethe energy of the first dipole at x1.

W (d1−d2) =1

4πε0p1 · p2 − 3 (n · p1) (n · p2)

|x1 − x2|3

55

Wednesday - 10/8

20 Transition To Non-negligible Media

Up until now we’ve been working with charge distributions or conductors with voltage distributionsin vacuums. The changes to ”real” matter aren’t incredibly drastic, but need to be understood.Recall that we have been examining the behavior of:

∇× E = 0; ∇ · E =ρ

ε0

We will now be interested in a macroscopic treatment of these equations. Consider the functionF (x) and its spacial average which we will denote as 〈F (x)〉. We can define this as:

〈F (x)〉 =∫d3x′f(x′)F (x− x′)

From this definition, it is clear that derivatives commute with the averaging. That is: ddx〈F (x)〉 =

〈 ddxF (x)〉. Applying this to the above equations of electrostatics:

∇× E = 0→ ∇× 〈E〉 = 0

∇ · E =ρ

ε0→ ∇ · 〈E〉 =

⟨ρ

ε0

⟩Because ε0 is a constant, we only need to derive what 〈ρ〉 is and we can work problems in a regionwhich is not a vacuum. There are two methods to define exactly what this average charge densityis. Firstly, let’s consider the multipole expansion.

φ(x) =1

4πε0

(Q

r+p · xr3

+ . . .

)Now, consider only a small (infinitesimal) part of a region, for this small volume we can write that:

dφ(x) =1

4πε0

(dq

r+dp · xr3

+ . . .

)

dφ(x) =1

4πε0

(ρ(x)rdV +

dpdV · xr3

+ . . .

)

If we define the infinitesimal dipole moment per volume to be dpdV = P , and then integrate this

equation over the entire region, we get a result of:

56

φ(x) =1

4πε0

∫d3x′

(ρ(x)|x− x′|

dV +P · (x− x′)|x− x′|3

+ . . .

)However, note that the second term can be written as a derivative of the separation, that is:

P · (x− x′)|x− x′|3

= P (x′) · ∇x′(

1|x− x′|

)Integrating this by parts gives −∇x′P (x′) · 1

|x−x′| which we can plug into the above form for thepotential and obtain:

φ(x) =1

4πε0

∫d3x′

(ρ(x)−∇x′P (x′)|x− x′|

dV + . . .

)And comparing this with the vacuum equation we find that we can treat the macroscopic chargedensity as:

ρ = ρexcess −∇ · P

Where we’ve defined the vector P as the polarization per unit volume of the medium. With this inmind we can define what is known as the ”electric displacement field” D = Eε0 + P . And with thisinformation, we can write the macroscopic equations of electrostatics,

∇× E = 0; ∇ · D = ρ

Where in this case the electric field in the first equation is the macroscopic field and the displacementfield is defined as above in terms of the macroscopic electric field and polarization ”field”.

20.1 An Alternate Derivation

An alternate derivation of these equations can be found in section 6.6 of Jackson. This involvesaccounting for the actual electric charges of particles in the material. In this case we write thecharge density in the medium as:

ρ(x) =∑i

q(free)i δ(3)(x− xi) +

∑n,j

q(bound)jn δ(3)(x− (xjn + xn))

Where we’ve defined xi to be the location of each free electron, xn to be the location of each nucleus,and xjn to be the location of the jth electron bound to the nth nucleus. Taking the spacial averageof the above yields:

〈ρ(x)〉 =∑i

q(free)i

⟨δ(3)(x− xi)

⟩+∑n,j

q(bound)j,n

⟨δ(3)(x− (xjn + xn))

⟩

57

The second term can be expanded in terms of the weight function in the definition of the spacial av-erage and expanding for |xjn| |xn|, |x| generates an additional correction term which is dependenton the dipole moment of the medium. This gives a final result of:

〈ρ〉 =

⟨∑i

q(free)i δ(3)(x− xi) +

∑n,j

q(bound)jn δ(3)(x− (xjn + xn))

⟩+∇ ·

⟨∑n

pnδ(x− xn)

⟩

The first term of this result in the excess free and bound charges in the medium, while the secondterm is the total dipole moment per unit volume. This result agrees with the above result. Thequadurpole term is worked out in Jackson as well.

Friday - 10/10

20.2 More On Electrostatics In Real Media

The boundary conditions on the equations for macroscopic media are found by simply plugging thevalues into the vacuum boundary relations:

(D2 − D1

)· n = σ(

E2 − E1

)× n = 0

Considering the definition above, in the case of a weak electric field, the displacement field isa function of the electric field. Further, in this same limit, the polarization of the material isdependent on the electric field, that is:

D = D(E)

; P = P(E)

As proof of the above, consider some region of a medium subject to an external electric field Eext.Inside the material, a given amount of polarization will occur which will cause a local electric fieldin a given region. Denoting this internal field as Eint, we can write the total field as some point xas a sum of the two. However, the internal field can be considered as a combination of the localdipole induced minus the field from the entire dipole moment within the media. That is:

Etot = Eext +(Enear − Ep

)Taking average of the internal field, the near dipoles average out to zero since the integral wouldcover the entire volume. The same goes for the dipole term, except recall that we derived the electricfield from a dipole as:

E(2) =1

4πε03n (p · n)− p

r3− 1

3ε0p δ(r)

58

The delta function is the only part which contributes. Therefore the total electric field can bewritten:

〈Etot〉 = Eext +p

3ε0

We can expand this out for weak electric fields in terms of a multipole expansion,

〈Etot〉 = c1〈pmol〉+ c2〈Q(i,j)mol 〉

The second term can be dropped for weak fields, and this leaves a relation of the form

〈pmol〉 = ε0γmolEext

Where the variable γmol describes how the molecules in the media respond to the electric field (howmuch they are polarized). Multiplying this by the number of particles in the media, we get that:

N〈Pmol〉 = Nε0γmol〈E〉

However, note that the left hand side of this equation is simply the dipole moment of the media.This implies that:

P = ε0γmolN

(Eext +

P

3ε0

)

P = ε0χeE; χe =Nγmol

1− Nγmol3

Then if we define the electric susceptibility of the material χe and the dielectric constant or relativeelectric permittivity ε

ε0= 1+χe. In actuality, any variation in the equations resulting from qualities

of a material are absorbed into these variables. That means that the variable χe and therefore εcan be functions of position, time, temperature, electric field, magnetic field, density, or any othervariable that can arise in a physical experiment. These definitions allow us to write the equationfor the electric displacement as D = εE. Then,

∇ · D = ρ→ ∇ ·(εE)

= ρ

In the case that the electric susceptibility of the material is independent of position (in any manner),then this simplifies to the familiar form:

∇ · E =ρ

ε

However, if ε = ε(x) then the divergence gives:

59

∇ · E =ρ

ε− E · ∇ε

ε=ρ

ε− E · ∇ ln ε

∇2φ = −ρε−∇φ · ∇ ln ε

For the case that there are no excess charges, but χe is dependent on position, we have that:

∇ · P = ε0χe(x)∇ · E + ε0∇χe(x) · E

In the case that the variation in the susceptibility is a step function of the form:

χe(x) = Θ(x)χe1 −Θ(−x)χe2

then we can write the derivative of χe in terms of the delta function ∇χe = δ(x) (χe2 − χe1) |s andthus,

∇χe|s =ε2 − ε1ε0

n

Plugging this into the above equation gives:

∇ · P = εχe(x) (0) + ε0

(ε2 − ε1ε0

n

)· E = (ε2 − ε1) n · E = −σpol

σpol = (p2 − p1) · n = (ε2 − ε1)∇φ · n

This value describes the induced polarization on the boundary of two materials with differentdielectric properties when there are no charges present in the system.

60

Monday - 10/13 - No Lecture, Midterm Day.Wednesday - 10/15

21 Energy of a Dielectric in an External Field

Recall that the potential energy of a charge configuration in an electric field can be written as:

W =12

∫d3xρ(x)φ(x)

Where the charge distribution of interest is the excess charge in the dielectric (since the total chargewould include internal energy in the dielectric and isn’t of interest to us). It can be shown that forsome function defined as:

F (x) =∫ x

d3x′G(x′)→ 〈F (x)〉 =∫d3x′′f(x′′)F (x′ − x′′) =

∫d3x′′f(x′′)

∫ x−x′

d3x′G(x′)

x′′ + x′ = y → 〈F (x)〉 =∫d3y

∫ d3x′′x−x′

f(x′′)G(y′ − x′′)

F (x) =∫d3x′G(x′)→ 〈F (x)〉 =

∫ x

d3y〈G(y〉

From this we can write the macroscopic energy as:

〈W 〉 =12

∫d3x〈ρ(x)φ(x)〉

And if we assume that the medium is exhibits linear response, then the potential is proportional tothe charge density and we can further approximate that:

〈ρ(x)2〉 ∝ 〈ρ(x)〉2

And since the excess charge in this case obeys the relation: ∇ · D, the macroscopic energy is:

〈W 〉 =12

∫d3xφ∇ · D

And integrating by parts we can easily see that:

Wmacro =12

∫d3xE · D

61

Consider moving a dielectric into a region with an initial electric field. If the initial field anddisplacement field are E1 and D1 then we can denote the resulting fields as E2 and D2. Then wecan write that the change in energy of this movement is:

∆Wm =12

∫d3x

(E2 · D2 − E1 · D1

)This can be expanded as:

∆Wm =12

∫d3x

[(D2 − D1

) (E1 + E2

)− D2 · E1 + D1 · E2

]There are no excess charges in the dielectric in this case, so the first term is zero and since the fieldsare weak we note that the displacement and electric fields are proportional to one another. Thenwe have the result that:

∆Wm =12

∫d3x

[D1 · E1 − D2 · E2

]=

12

∫d3x [ε1 − ε2] E1 · E2

In the case that the region of interest is a vacuum, then ε1 → ε0. And noting that the term issimply the polarization of the material, then the change in energy is:

∆Wm =12

∫d3xP · E1

Where the vector P is the resulting polarization of the dielectric material in the field and E1 is theinitial electric field (before the dielectric is introduced).

22 A Theoretical Model For Polarization

Consider the definitions:

D = εE; ε = ε0 (1 + χe) ; P = ε0χeE; χe = χe (γmol)

If we can derive a simple model for the behavior of γmol then we can get an idea of how differentmaterials will react to electric fields and why they do so. Consider a molecule modeled as a simpleharmonic oscillator. If we assume that the system is in static equilibrium, then the electrostaticforces must balance the restoring oscillatory force, that is:

m¨x = −mω20x+ eE = 0→ x =

eE

mω2

62

If we multiply this by e and note that the polarization is defined by the charge times location, weget the result that:

ex = P =∑k

e2

mkω20k

E → γmol =∑k

e2

mkω20kε0

This models the behavior of γmol for each molecule in a dielectric.

23 Summary of Equations For Problems Involving Dielectrics

∇ · D = ρexcess

∇× E = 0→ E = −∇φ(D2 − D1

)· n = σexcess;

(E2 − E1

)× n = 0

P = ε0χeE; χe = χe(x, t, T, E, . . .)

−∇ · P = −(P2 − P1

)· n∣∣surface

= σinduced

And for linearly reacting media

D = ε0 (1 + χe) E = εE

Friday - 10/17

Test Review - Test was extra credit.

A note about finding quadurpole moments: if the distribution is symmetric in x and

y and spherically symmetric (as problem 3 was), then you only need to find Q33 and

then all off diagonals are zero and Q11 = Q22 = 12Q33.

63

Monday - 10/20

24 Solving Problems With Dielectrics

We can use many of the methods from before to solve problems, but some additional steps andanalysis is involved. Consider a few examples.

24.1 Method of Images - Planar Boundary With a Point Charge

Consider the case of two dielectric materials separated by a planar boundary at z = 0. The materialabove the xy plane has ε1 and the region below the plane has ε2. If there is a single point chargeat a distance d above the boundary (in region I) then we can assume that there will be an inducedpolarization on the boundary and we can solve for the potential in the two regions by inserting asecond charge in region II and replacing the real charge with a second image charge to simulatethe field from the induced polarization in region II. That is:

φI =1

4πε1

(q

R1+

q′

R2

)z > 0

φII =1

4πε2q′′

R1z < 0

In the above equations we’ve put a the charge q′′ in the same location as q and chosen to solve forits charge value. We could have just as easily chosen to make it a defined charge q0 and solvedfor its location, but this method is easier. The lengths R1 and R2 are the distances to the pointof interest from the real and first image charge respectively. Noting this, we can write these twodistances as:

R1 =√

(z − d)2 + ρ2; R2 =√

(z + d)2 + ρ2

As the charge on the boundary is zero, the boundary conditions we must satisfy are:

(D2 − D1

)· n = 0;

(E2 − E1

)× n = 0

E⊥ = −∂φ∂z

; E‖ = −∂φ∂ρ

D⊥ = −εi∂φ

∂z; D‖ = −εi

∂φ

∂ρ

And taking derivatives of the inverse lengths above give the results:

∂

∂z

1R1

= − ∂

∂z

1R2

=d

(ρ2 + d2)3/2

64

∂

∂ρ

1R1

=∂

∂ρ

1R2

=−ρ

(ρ2 + d2)3/2

Plugging this in gives the results:

q − q′ = q′′; q + q′ =ε1ε2q′′ ⇒ q′ = q

ε2 − ε1ε1 + ε2

q′′ = q2ε2

ε1 + ε2

24.2 Dielectric Sphere In External Field

Consider a spherical dielectric in an external electric field initially described as E = E0z. Thesolution of this problem is azimuthally symmetric, therefore we have the mathematical conditionsthat:

φin =∞∑l=0

AlrlPl (cosϑ)

φout = −E0r cosϑ+∞∑l=0

Blr−(l+1)Pl (cosϑ)

And the boundary conditions are found by:

E = −∂φ∂r− 1r

∂φ

∂ϑ− 1r sinϑ

∂φ

∂ϕ

E × n = E · ϑ→ −1a

∂φin∂ϑ

∣∣r=a

= −1a

∂φout∂ϑ

∣∣r=a

E · n = E · r → −ε0∂φout∂r

∣∣r=a

= −ε∂φin∂r

∣∣r=a

These conditions lead to the result that only the l = 1 terms in the sums are nonzero. So, thepotential of the sphere in an external field is

φin(r, ϑ) = − 3εε0

+ 2E0r cosϑ

φout(r, ϑ) = −E0r cosϑ+εε0− 1

εε0

+ 2E0a3

r2cosϑ

Wednesday - 10/22 From this potential we can write the electric field inside the sphere. Note thatr cosϑ = z,

E = −∇φ→ φin = − 3εε0

+ 2E0r cosϑ = − 3

εε0

+ 2E0z → E = Ez =

3εε0

+ 2E0

65

φout(r, ϑ) = −E0z +εε0− 1

εε0

+ 2E0a3

r3z

The potential outside is like that of the original field plus a dipole term ( φdipole ∝ p·xr3 ). So only pz

is non-zero and we can write that:

(εε0− 1

εε0

+ 2

)E0a3

r3z =

14πε0

pzz

r3→ pz =

(εε0− 1

εε0

+ 2

)4πε0E0a

3

And thus the polarization P is:

P =p

V= 3ε0

εε0− 1

εε0

+ 2E0z

This leads to a general result that for some region with an electric field contained inside it, theresulting polarization is determined by the internal field and change in dielectric constants:

Pin = (εin − εout) Ein

Next, we can calculate the polarization distribution,

σpol = −[Pout − Pin

]· n∣∣S

and as there is no polarization on the outside of the sphere (not a sphere embedded in a seconddielectric), we have that

σpol = Pin · n∣∣S

= 3ε0

(εε0− 1

εε0

+ 2

)E0z · r

∣∣r=a

= 3ε0

(εε0− 1

εε0

+ 2

)E0 cosϑ

The net charge can be found by integrating this solution from ϑ = 0 to π. However, integratingcosine on these limits gives zero. That is as expected since the net charge on the dielectric is zero.The resulting field outside the sphere can be found from the potential described as:

φout =1

4πε0

∫σpoldS

|x− x′|

Rotating so that the angles of γ and ϑ coincide, we can write this as:

φout =2π

4πε0

∫d (cosϑ)

3ε0

(εε0−1

εε0

+2

)E0 cosϑ

√a2 + r2 − 2ar cosϑ

=3E0

2

(εε0− 1

εε0

+ 2

)∫cosϑd (cosϑ)√

a2 + r2 − 2ar cosϑ

66

And from this we can easily find the electric field components

Er = −∂φ∂r

; Eϑ = −1r

∂φ

∂ϑ

However, it isn’t necessary to evaluate these. We can simply look at the behavior of Er at themiddle of the sphere.

Er(r = 0) = 3ε0

(εε0− 1

εε0

+ 2

)E0 =

P

3ε0

But recall, the delta function addition for the electric field at r = 0 for a dipole was exactly theabove. Therefore we can write the field outside the sphere as the external field plus the additionalfield of a dipole oriented at the center of the sphere pointing antiparallel to the field.

67

Part II

MAGNETOSTATICS

25 Introduction to Magnetostatics

Recall the Maxwell Equations. We are still assuming no time dependence, but instead of solving thesystem for B = 0, let’s consider the case of E = 0. This is called ”Magnetostatics”. The equationswe need to consider are:

∇ · E =ρ

ε0−→ ∇ · E =

ρ

ε0

∇ · B = 0 −→ ∇ · B = 0

∇× E = −∂B∂t−→ ∇× E = 0

∇× B = µ0j +1c2

∂E

∂t−→ ∇× B = µ0j

∇ · j = −∂ρ∂t→ ∇ · j = 0

If the electric field is zero, then this reduces to a system of three equations.

∇ · B = 0; ∇× B = µ0j; ∇ · j = 0

As it turns out, all three of the above can be solved by writing the magnetic field in terms of avector potential A. This can be seen as:

∇ · B = ∇ ·(∇× A

)= 0

∇×∇× A = ∇(∇ · A

)−∇2A = µ0j

If we assume that the divergence of A is zero, then the above leaves only the familiar (but nowmultidimensional) Laplacian Equation:

∇2A = −µ0j

Which has solutions of the form:

A(x) =µ0

4π

∫d3x′

j(x′)|x− x′|

And by taking the divergence of the solution and integrating by parts it can be shown that thissolutions satisfies both the conditions of ∇ · A = 0 and ∇ · j = 0.

68

Friday - 10/24However, the above is only the special solution for this equation the most general solution wouldbe something of the form:

Agen(x) =µ0

4π

∫d3x′

j(x′)|x− x′|

+∇φ

25.1 Gauge Transforms

It can be shown that this additional term is a gauge transform which doesn’t alter the physicalfield. The addition of ∇φ to the vector potential gives a contribution of ∇ ×∇φ to the magneticfield, but this is zero. Such an additional term is known as a ”Coulomb Gauge” since it requiresthat ∇2φ = 0. Therefore the most general vector potential in the Coulomb Gauge is simply thespecial solution above.

So can we always make a statement such as ∇ · A = 0? Consider a vector potential A such that∇ · A 6= 0. In this case we can define some new potential A′ as:

A′ = A+∇φ; ∇2φ = −∇ · A

If the function φ satisfies this requirement, then we can immediately see that this rigs the vectorpotential so that ∇ · A′ = 0. Therefore, yes, we can always make the potential satisfy ∇ · A = 0.

26 Deriving the Magnetic Field

From the above definition for the vector potential we can write that the resulting B field is;

B = ∇× A = ∇x ×µ0

4π

∫d3x′

j(x′)|x− x′|

Moving the operator into the integral gives:

B =µ0

4π

∫d3x′j(x′)∇x ×

1|x− x′|

Switching ∇x to −∇x′ and integrating by parts (noting that the boundary values are zero since thepotential should be zero at infinity) we find the magnetic field to be:

B =µ0

4π

∫d3x′

j(x′)× (x− x′)|x− x′|3

If we consider some infinitesimal portion of this field we can write that:

69

dB =µ0

4πj(x′)× (x− x′) d3x′

|x− x′|3

However, we can write that d3x′ = dV = da dl and define the current as Jda dl→ Idl. Then, lettingx′ = 0, the infinitesimal field can be written:

dB =µ0

4πIdl × x|x|3

Now we can derive the integral form of Maxwell’s equation for magnetism (i.e. Ampere’s Law):

∫ (∇× B = µ0j

)da · n∮

B · dl = µ0I

26.0.1 An Example - Infinitely Long Wire

Consider an infinitely long wire with constant uniform current I traveling through it. The infinites-imal magnetic field at some point a distance R from the wire from the a small length dl of the wirecan be written as:

dB =µ0

4πIdl × x|x|3

Defining ϑ to be the angle the vector x makes with the axis of the wire, we can write that dl× x =dl x sinϑ = dl R and further note that |x| =

√l2 +R2 and thus,

dB =µ0

4πIR

dl

(l2 +R2)3/2

Integrating l over all values gives the familiar result for the magnetic field a distance R from acurrent carrying wire:

|B| = µ0I

2πR

26.1 Force on a Charged Particle From a Magnetic Field

Given that the force on a particle in an electromagnetic field is F = q(E + v × B

). We can write

that since the electric field is zero and qv = I dl = j × B d3x,

70

F =∫d3x j(x)× B(x)

Using this we can write the force on one wire from a second as:

dF = j × Bd3x = I1dl1 ×µ0

2πI2dl2 × x|x|3

If we assume that there are closed loops of current, then the force is:

F =µ0

4πI1I2

∮dl1 ×

∮dl2 × x|x|3

Using vector relations, we can write this in an expanded form

F =µ0

4πI1I2

(∮ ∮dl1 · dl2∇

1|x|

+∮ ∮

dl2 · dl1∇1|x|

)The second term proportional to

∮dl∇ 1

x and since it’s a closed loop, this is zero. So only the firstterm is left. This gives the force as:

F = −µ0

4πI1I2

∮ ∮dl1 · dl2∇

1|x|

Which in the case of two infinitely long parallel wires separated by d, gives a force per unit lengthof:

∣∣∣∣dFdl1∣∣∣∣ =

µ0

2πI1I2

d

Monday - 10/27

26.2 Magnetic Field From a Circular Loop of Wire

Consider the magnetic field from a circular loop of wire in the xy plane with current I in the ϕdirection. The vector potential from the loop is then,

A =µ0

4π

∫d3x′

j(x′)|x− x′|

j = Kδ(r − a)δ(cosϑ) (sinϕex + cosϕey)

The constant K can be found by noting that:

71

∫jd3x =

∮Idl = 2πa2I → K =

I

a

Next, we can expand the separation of two points in spherical coordinates as:

|x− x′| =√r2 + r′2 − 2rr′ (cosϑ cosϑ′ + sinϑ sinϑ′ cos (ϕ− ϕ′))

Plugging this in, we can note the azimuthal symmetry of the problem and assume ϕ = 0. Thedeltas then pick out only the values of r′ = a and ϑ = π

2 . This leaves the vector potential as:

A(r, ϑ) =µ0Ia

4π

∫ 2π

0

cos (ϕ′) dϕ′√a2 + r2 − 2ar sinϑ cosϕ′

eϕ

From this we can immediately write the magnetic field as:

B = Brer +Bϑeϑ +Bϕeϕ

Br =1

r sinϑ∂

∂ϑ(sinϑAϕ)

Bϑ = −1r

∂

∂rAϕ

Bφ = 0

The exact solution of the above are elliptical functions. However, if we assume that the point ofinterest is far from the loop of wire then we can expand the solution for a

r 1. In this case we canwrite the above vector potential as:

A(r, ϑ) =µ0I

4πa

r

∫ 2π

0

cos (ϕ′) dϕ′√1 + a2

r2 +−2ar sinϑ cosϕ′eϕ

A ≈ µ0Ia2r sinϑ

4 (a2 + r2)32

(1 +

15a2r2 sin2 ϑ

8 (a2 + r2)2 + . . .

)eϕ

And keeping only the lowest order terms we find that the resulting B field far from the loop of wireis:

B ≈[µ0

2π(Iπa2

) cosϑr3

]er +

[µ0

4π(Iπa2

) sinϑr3

]eϑ

72

26.2.1 Interpretation and Analysis - Defining the Dipole Moment

As an aside, recall the electric field from a dipole far from the charges,

Edipole ≈[

14πε0

2p cosϑr3

]er −

[1

4πε0p sinϑr3

]eϑ

We can define the magnetic dipole moment as m = IA for a circular loop of current with current Iand area A.

Alternately, if we return to the above and expand in terms of powers of 1|x| :

A =µ0

4π

∫d3x′

j(x′)|x− x′|

= A =µ0

4π

∫d3x′j(x′)

1|x|

(1 +

x · x′

|x|2+ . . .

)We can then write the above in terms of the monopole and dipole terms,

A =µ0

4π1|x|

[∫d3x′j(x′) +

1|x|2

∫d3x′

(x · x′

)j(x′) + . . .

]It is easy to show that the first term is zero for magnetostatics since we can write that for a localizedcurrent:

∫d3x′∇x′ ·

(j(x′)x′i

)= 0

And then expand this as:

∫d3x′∇x′ · j(x′)xi + j(x′)∇x′ · x′i

The first term goes to zero since the currents are infinity are required to be zero. The second termpicks out the i component of the current and thus we can be sure that for each component of thecurrent,

∫d3x′ji(x′) = 0

This leaves only the dipole term, which we can expand as:

∫d3x′

(x · x′

)j(x′) =

∑j

xi

∫d3x′ji(x′)x′j

We can easily see from the divergence of the current being zero that:

73

∫d3x′∇x′ ·

(xixj j(x′)

)= 0→

∫d3x′

(x′jji(x

′)− x′ijj(x′))

= 0

We can add this to the above integral without altering it,

∫d3x′

(x · x′

)ji(x′) =

∑j

xj

∫d3x′ji(x′)x′j −

12(x′jji(x

′)− x′ijj(x′))

= −12

∑j,k

εijkxj

∫d3x′

(x′ × j

)k

= −12

[x×

∫ (x′ × j

)]i

Thus, we can write the dipole term as:

A =µ0

4π

(m× x|x|3

+ . . .

); m =

12

∫d3x′ x′ × j(x)

Wednesday - 10/29

26.3 First Order Approximation of the Magnetic Field

From the above result, we can derive a first order approximation for the magnetic field.

B = ∇× A = ∇×[µ0

4π

(m× x|x|3

+ . . .

)]=µ0

4π

[∇(

1|x|3

)× (m× x) +

1|x|3∇× (m× x)

]=µ0

4π

[−3

x

|x|5× (m× x) +

1|x|3

(m (∇ · x)− (m · ∇) x)]

=µ0

4π

[−3

1|x|5

[(−x · m) x+ (x · x) m] +3m|x|3− m

|x|3

]=µ0

4π

[3

(x · m) x|x|5

− 3m

|x|5+

3m|x|3− m

|x|3

]B =

µ0

4π

(3 (n · m) n− m

|x|3+ . . .

)As with the electric field’s dipole term, the contribution at |x| = 0 is undefined due to the singularity.This can be accounted for in the same manner (by integrating over the singularity) and it is foundthat the full solution for the magnetic field of a dipole contribution is:

B =µ0

4π

(3 (n · m) n− m

|x|3+

8π3mδ(x) + . . .

)

As an aside, recall the relation we derived for current density and current.

74

j d3x′ = Idl→ m =I

2

∮x× dl

|m| = I ×Area

Thus, for any planar current distribution, the dipole moment of the distribution will be normal tothe plane or current and have magnitude given by the current times the area enclosed in the loop.

27 Current Density For a Group of Moving Charges

Consider if instead of a current density j or current I, one were working with moving chargedparticles, in this case we can write the current density from the position, charge, and velocity ofeach particle:

j = ρv =∑i

qiviδ(x− xi)

Then, writing the dipole moment of such a distribution,

m =12

∫d3x′ x′ ×

[∑i

qiviδ(x′ − xi)

]=∑i

qi2

(xi × vi)

However, the cross product is simply the angular momentum of the particle without the mass term(M), thus we can write:

m =∑i

qi2Mi

Li

This leads to the intrinsic ”spin” for a charged particle. Classically, this quantity for a single particleis expected to be:

mintrinsic =q

2MS

However, it was found experimentally that (at least for electrons) this quantity was incorrect. Theexperimentally found dipole moment was mintrinsic = g q

2M S where for the electron g = 2.000116.The factor of 2 was explained by the Dirac equation using relativistic quantum mechanics. Theadditional small factor is explained in Quantum Field Theory from interaction with a polarizationin the background.

75

28 Force and Torque From an External Magnetic Field

The force on a current distribution was found earlier to be:

F =∫d3x j × Bext

And from this we can find that the torque on the distribution can be written:

N =∫d3x x×

(j × Bext

)Expanding the magnetic field as B = B(0) + (x · ∇) B|0 + . . . we can find the dominant terms ofthe above force and torque vectors.

F =∫d3x j ×

(B(0) + (x · ∇) B|0 + . . .

)The first term is zero since the integral of j is zero for electrostatics. The remaining term can bewritten tensor notation as:

Fi =∑j,k

εijk (m×∇)j Bk

And expanding out the cross product,

F = (m×∇) B = ∇(m · B

)− m

(∇ · B

)And since ∇ · B is zero for all cases, we can simplify write that to first order,

F (dipole) = ∇(m · B

)As expected, the force can be written in terms of the gradient of some other function. The potentialenergy for a dipole in an external field is:

F = −∇U → U = −m · Bext

Similar analysis gives the dipole contribution to the torque as:

N (dipole) =∫x′ ×

[j × B(0)

]d3x′ = m× B(0)

76

Friday - 10/31: HAPPY HALLOWEEN!!!

29 Transition To Macroscopic Magnetostatics

For the macroscopic equations of magnetostatics, we make the transitions:

〈∇ · B = 0〉 → ∇ · 〈B〉 = 0

〈∇ × B = µ0j〉 → ∇× 〈B〉 = µ0〈j〉 = j +∇× M

Where we’ve defined the average magnetization per unit volume M . Further, if we denote theaverage magnetic induction 〈B〉 as simply B and define the magnetic field in the media H = 1

µ0B−M

then the above equations become:

∇B = 0; ∇× H = j

As proof of the above claim for 〈j〉, consider some volume of material with current j in it. Insidethe volume, the current will cause small loops of current to form, each with a single dipole moment.We can denote this as:

M(x) =∑i

Ni〈mi〉

where Ni is the density of particles in a volume around xi. The vector potential for such a materialcan be written as:

A =µ0

4π

∫d3x′

j(x′)|x− x′|

+µ0

4π

∑i

mi × (x− xi)|x− xi|3

And some infinitesimal volume has:

dA =µ0

4πj(x′)|x− x′|

dV ′ +µ0

4πdm

dV ′(x′)× (x− x′)|x− x′|3

dV ′

However, we’ve defined the average magnetization over the volume to be the macroscopic magneti-zation. Therefore the above can be integrated on the volume to give:

A =µ0

4π

∫d3x′

(j(x′)|x− x′|

+M(x)× (x− x′)|x− x′|3

)

Noting that the second term can be written as M(x)×(x−x′)|x−x′|3 = M(x′ ×∇x′

(1

|x−x′|

), and then inte-

grating by parts gives the result:

77

A =µ0

4π

∫d3x′

j(x′) +∇x′ × M(x′)|x− x′|

Which produces the expected result that in the macroscopic case the current can be written asabove.

29.1 Relations between H, B, and M - Boundary Conditions

If we assume that the magnetization is a function of H then we can write that:

B = µ0H + M(H)→ B = B(H)

B = µH; µ = µ(H)

The boundary conditions for the magnetic field in macroscopic media are:

(B2 − B1

)· n = 0(

H2 − H1

)× n = −K

Where K is the idealized surface current density in between the two materials.

29.2 Some Important Cases

29.2.1 Constant Magnetic Permeability

In a material such as a paramagnet or diamagnet, the value of µ is constant. In such a case, theequations work out exactly as before, only with µ0 replaced by µ. The vector potential is then:

A =µ

4π

∫d3x′

j(x′)|x− x′|

29.2.2 No Current In a Region

In the case that there is no current density in a region, then the equations for magnetostatics resultin:

∇× H = 0→ H = −∇φM∇ · B = 0→ ∇ · µ∇φM = 0

78

29.2.3 No Current In a Region, Magnetization Is Given

If there is no current density, the we have from the above,

∇ ·(µ0

(H + M

))= 0→ ∇2φM = ∇ · M

And for such a problem on can immediately write the solution as:

φM =1

4π

∫d3x′

∇ · M|x− x′|

Expanding the above and taking the first term as x′ → 0 and x x′, one finds:

φM (x) ≈ − 14π∇(

1r

)·∫M(x′)d3x′ =

m · x4πr3

+ . . .

In the case of a Dirichlet boundary for the above magnetic scalar potential, one can write:

φM = − 14π

∫Vd3x′

(∇ · M

)GD(x, x′) +

14π

∮∂VσMGD(x, x′)

Where we’ve defined the effective magnetic surface charge density:

σM = n′ · M(x′)

79

Monday - 11/3

30 Energy In An External Magnetic Field

Recall that the energy for an electrostatic configuration in an external field can be written in avacuum and medium as:

w =ε02|E|2 → w =

12E · D

Similarly, in magnetostatics, one can write that:

w =1

2µ0|B|2 → w =

12H · B

This is proven by calculating the variation in force of moving a current in a medium. Consider someregion with an external field Bext into which a current density j is inserted. If the charge densityis then nudged a small amount, a variation in the energy can be calculated. Consider some smallportion of the current density, the force on it can be written as:

dF = Idl × Bext = j × Bextd3x

Then, if we move the distribution by a distance dl2 the new force is written:

dF · dl2 = I(dl × Bext

)· dl2 = I

(dl × dl2

)· B = IdAB · n

Thus the force is proportional to the area bounded by dl and dl2. The variation of this can then betaken,

δ (dW ) = δ(IdAB · n

)= Ida δ(B) · n

Integrating by parts to find the total variation of the work in the system:

δ(W ) =∫Idaδ(B) · n =

∫Ida

(∇× δ(A

)· n

By Stokes’ theorem, we can write this as:

δ(W ) =∮Cδ(A) · Idl =

∫δ(A) · jd3x

But since j = ∇× H we can use the relation ∇ ·(a× b

)= b · (∇× a)− a

(∇× b

)we can write the

above as simply:

80

δ(W ) =∫d3xH · δ

(∇× A

)→ δ(W ) =

∫d3xH · δ

(B)

And in the case of linearly reacting media, B = µH and µ = constant. In this case the abovesimply becomes:

δ(W ) =12

∫d3xH · B =

12

∫d3xA · j

Where the second definition comes from B = ∇× A and ∇× H = j.

30.1 Special Case - Change in Energy of a Region When a Material Is Inserted

For some region with an initial magnetic field, the change in energy when a material is inserted canbe written as:

∆W =12

∫d3x

(B · H − B0 · H0

)Expanding this out, we find that:

B · H − B0 · H0 =(H − H0

)·(B − B0

)− H · B0 + H0 · B

The first two terms are zero, this leaves only

∆W =12

∫d3x

(H0 · B − H · B0

)And again, for linearly reacting media, B = µH, so the above simply becomes:

∆W =12

∫d3x

(1µ0− 1µ

)B0 · B

Alternately, one can note that the magnetization in a material can be written as: M =(

1µ0− 1

µ

)B

and therefore the above is:

∆W =12

∫d3xM · B0

81

31 Behavior of Various Materials In a Magnetic Field

Note that in general the magnetic moment of a material can be written as:

m =q

2meL+ ms =

q

2me(r × p) + ms

And noting that in a magnetic field the momentum changes from p to p− qA, we can include thisin the above and find that the magnetic moment in a material is:

m = − e

2mer ×

(p+ eA

)+ ms

and writing the vector potential instead as A = µ0

(H × r

2

)+µ0 (r · ∇) H2 . The second term is zero

and this leaves only:

m = − e

2mer × p− e2µ0

4mer ×

(H × r

)+ ms

And taking the average in the medium, we find that:

< m >=⟨− e

2mer × p

⟩− e2µ0

6me

⟨r2⟩H

Next, consider the reaction of a diamagnetic material in an external field. Since in a diamagnetic,there is no intrinsic magnetic moment, the first term above is zero. This means that < m > isnegative and thus the change in energy is positive when it is inserted. The system will try to getback to the lower energy state, so the material will be pushed out of the field. Further, it can beseen that the magnetization is:

M = −e2µ0N

6me

⟨r2⟩H = χMH

where we’ve defined the magnetic susceptibility χM as:

B = µ0 (1 + χM ) H → M = χMH

For a paramagnetic material, the intrinsic moment is non zero and in most cases the first termdominates over the second for < m >. In this case the change in energy is negative and the param-agnet is pulled into the field. Next, we’ll look at the behavior of a paramagnetic and ferromagneticmaterial as well.

82

Wednesday - 11/5

31.1 Paramagnets

For each atom in a paramagnetic material, the outermost shell of electrons can exhibit a totalangular momentum j (not to be confused with current density). When a magnetic field is switchedon, the energy in the material changes by a given amount. We can write this as:

E = µ0H · 〈l + gss〉 = µ0H · 〈j + (gs − 1) s〉

If we absorb this spin term into the total angular momentum, we can denote this in terms of theLande factor gl,s,j :

E = µ0H · 〈j〉gl,s,j = µ0mjgl,s,jHz

Taking the average of this value gives the magnetic moment m and we can write the BoltzmannDistribution for the energy PE =

[exp

(EkT

)]/[∑

n exp(EnkT )].

〈m〉 =∑mj

µ0mjgl,s,jPEmj

This can be summed over all atoms to obtain the total magnetization in terms of the BrillouinFunction B(x):

M = 〈m〉N = Nµ0gl,s,jjB(x)

B(x) =1j

(j +

12

)coth

[x

(j +

12

)]− 1

2coth

(x2

); x = gl,s,jµ0

H

kT

In the case of high temperatures, x → ∞ and so B(x) ∝ HT . In this case, the magnetization goes

like:

M = χMH → χM ≈ Ng2l,s,jµ

20j

(j + 1)3kT

This behavior of M ≈ CT H at high temperatures follows the Curie Law since χM = C

T in this limit.In the limit of low temperatures, the value of x→∞ and so χM becomes a constant Ngl,s,jµ0j.

NOTE: There are several other effects that can cause paramagnetic phenomena. First,

atoms with paired electrons will exhibit a weak (10−3 times the magnitude of that of the

electron) magnetic momenta which can be measured. Second, in a metal the electrons

are free to move about, however if a magnetic field causes some spin up and some

spin down imbalance, then the excess unpaired electrons will cause a paramagnetic effect.

83

31.2 Ferromagnets

It has been observed in ferromagnetic materials that χM obeys the relation:

χM =C

T − TC

Where TC is come critical temperature below which this relation no longer holds. Note that inthe case of high temperatures, T TC and this reverts to the Curie law of χM ≈ C

T . The aboverelation can be derived with a number of assumptions.

1) Domains in the material have uniform magnetization. These ares behave similar to dipoles inthat they influence one another by external and exchange fields.

2) The interaction Hamiltonian can be written in terms of the exchange and external fields as

HI = −mi · Hext +∑i 6=j

Uij(|xi − xj |)

Assuming that the function in the summation can be written in terms of the scalar product of ji · jjwe can write this instead in terms of the exchange field:

HI = −mi ·(Hext + Hexch

)3) If we assume that the exchange field is proportional to the magnetization of the other domains,then we can define some relation Hexch = DM and all together we get the result:

M = χM(Hext + Hexch

)=C

T

(Hext +DM

)M =

C

T − TCHext

One last note about this result, in the case that T → TC , we can consider the value of |H||M | .

|H||M |

∝ T − TCC

and in this case the right hand side goes to zero. This means that the left hand side also must go tozero. Now, in addition to this, consider the case that |H| →. We still require that the above hold,therefore even in the limit of no external field, |M | 6= 0 for low temperatures.

84

Friday - 11/7

32 Summary of Problem Solving Methods

We can now summarize several methods of solving problems in magnetostatics.

32.1 Free Space Vector Potential

If there are no boundary conditions and j(x) is given, then the vector potential can be calculatedas:

A(x) =µ

4π

∫d3x′

j(x′)|x− x′|

32.2 Magnetization Known With No Current

If a region in space has no current but some defined magnetization M , then we can use eithermagnetic scalar potential or vector potential to solve the problem. Defining the magnetic scalarpotential from H = −∇φM , we can write the equation as:

φM =1

4π

∫Vd3x′∇ · M(x′)|x− x′|

+1

4π

∫∂Vda′

n · M(x′)|x− x′|

This can be written more compactly if we define: ρM = −∇ · M and σM = n · M since it is thensimply,

φM = − 14π

∫Vd3x′

ρM (x′)|x− x′|

+1

4π

∫∂Vda′

σM (x′)|x− x′|

Alternately, we can assume that B = ∇× A and from this we get the vector potential A as:

A(x) =µ

4π

∫Vd3x′∇× M(x′)|x− x′|

+µ

4π

∫∂Vda′

n× M(x′)|x− x′|

32.3 No Current, but no given magnetization

In this case, we can refer back to the definition for the magnetic scalar potential and write that:

∇ · (µφM ) = 0→ ∇2φM = 0

85

32.4 Example: Uniformly Magnetized Sphere

Consider a sphere of radius a which uniform magnetization barM = M0z. In this case we cansolve the problem with either of the methods in part 2 above. Firstly, consider if we use the scalarpotential.

∇ · M =∂

∂zM0 = 0; n · M = M0r · z = M0 cosϑ

The first term in the result is zero and the second term leaves only:

φM =1

4π

∫∂Vda′

M0 cosϑ|x− x′|

=M0

4π

∫a2dΩ′

cosϑ|x− x′|

Using the expansion for the free space Greens Function for azimuthal symmetry quickly evaluatesthis integral:

φM =M0a

2

4π(2π)

∑l

Pl (cosϑ)rl<

rl+1>

∫d cosϑ′PL

(cosϑ′

)cosϑ′

The integral produces a δl,1 which leaves the scalar potential as:

φM =M0a

2

3r<r2>

→

φ

(r>a)M = M0a3

3r2 cosϑ = M0a3

3zr3

φ(r<a)M = M0

3 r cosϑ = M03 z

Note that from the solution outside the sphere we can see that the result is that of a magneticdipole with dipole moment m = M0

4π3 a

3. Which is the magnetization over the volume.

An alternate analysis could be performed using the vector potential, however it requires computingthe following:

∇× M = i∂

∂yM0 − j

∂

∂xM0 = 0

n× M = M0r × z = −M0 sinϑ cosφ′i+M0 sinϑ cosϕ′

We can drop the first term in the r × z result, and the remaining term can be expressed in termsof ϕ meaning that we can write the potential outside as

Aϕ =µ0

4πM0a

2

∫dΩ′

sinϑ′ cosϕ′

|x− x′|Expressing this instead as

sinϑ′ cosϕ′ = −√

8π3Re[Y1,1(ϑ′, ϕ′

]we get the exact B field as from above.

86

Monday - 11/10

33 Additional Topics In Magnetostatics

33.1 Magnetic Shielding

Consider a spherical Shell of inner and outer radii a and b respectively. If there is an externalfield B = B0z = µ0H0z, then the sphere can be used to shield the area contained inside of it.Firstly, we need to set up the problem. There are no charges, so we simply need to solve the systemof equations formed by the boundary conditions on the solutions in the three regions. Since theproblem is azimuthally symmetric, we can write the resulting magnetic potential as:

φ(I)M = −H0r cosϑ+

∞∑l=0

B(I)l r−(l+1)Pl (cosϑ); r > b

φ(II)M =

∞∑l=0

(A

(II)l rl +B

(II)l r−(l+1)

)Pl (cosϑ); a < r < b

φ(III)M =

∞∑l=0

A(III)l rlPl (cosϑ); r < a

The boundary conditions require:

µ0∂φ

(III)M

∂r

∣∣∣r=b

= µ∂φ

(II)M

∂r

∣∣∣r=b

; µ∂φ

(II)M

∂r

∣∣∣r=a

= µ0∂φ

(I)M

∂r

∣∣∣r=a

∂φ(III)M

∂ϑ

∣∣∣r=b

=∂φ

(II)M

∂ϑ

∣∣∣r=b

;∂φ

(II)M

∂ϑ

∣∣∣r=a

=∂φ

(I)M

∂ϑ

∣∣∣r=a

The source function causes only the l = 1 terms to be nonzero. The result inside the cavity is theprimary interest to our problem, the coefficient A(III)

1 is found to be:

A(III)1 = −H0

(9µ′

(2µ′ + 1)(µ′ + 2)− 2a3

b3(µ′ − 1)2

)Where we’ve defined µ′ = µ

µ0. In the limit that µ′ 1, the above reduces to a potential of the form:

φ(III)M ≈ −H0

92µ′

(1− a3

b3

)r cosϑ

The magnetic field inside the cavity is thus negligible for µ′ 1. The cavity is shielded in thismanner.

87

33.2 Inductance

Consider the energy of a system containing two loops of current I1 and I2.

W =12

∫d3xj(x) · A(x) =

12

∫d3xH · B

Using the definition for the vector potential, we can write the energy of the two currents as:

W =µ0

8π

∫d3xj1(x) ·

∫d3x′

j2(x)|x− x′|

For some generic system of n current loops, we can write the above instead as:

W =µ0

8π

∑i,j

∫d3x

∫d3x′

ji(xi) · jj(xj)|xi − x′j |

We can break this into two parts, containing the mutual inductance i 6= j and the self inductancei = j.

W =µ0

4π

∑i

∫d3x

∫d3x′

ji(xi) · ji(xi)|xi − x′i|

+µ0

8π

∑i,j

∫d3x

∫d3x′

ji(xi) · jj(xj)|xi − x′j |

We can then define the Inductance coefficients as:

Li =µ0

4πI2i

∑i

∫d3x

∫d3x′

ji(xi) · ji(xi)|xi − x′i|

Mij =µ0

4πIiIj

∑i,j

∫d3x

∫d3x′

ji(xi) · jj(xj)|xi − x′j |

W =12

∑i

LiI2i +

∑i>j

MijIiIj

In the case of a single loop of wire, Mij = 0 and we are left with only:

W =12LI2 → L =

1I2

∫d3x|B|2

µ

33.3 Quasistatics Fields

We can now consider the case of a slowly varying electric field. If we assume that ∂D∂t ≈ 0 then we

can write the relevant equations:

88

∇× H = j; ∇ · B = 0; ∇× E = −∂B∂t

; j = σE

And noting that we can still write B = ∇× A, we can write that:

E = −∂A∂t−∇Φ

Assuming there is negligible free charge, the changing B is the only source of the electric field andthus the scalar potential is constant which we can set to zero. This leaves the above as: E = −∂A

∂t .From Ampere’s law, we require:

∇× B = µj = µσE

Replacing B and E in the above with the vector potential, we are left with a diffusion equation forthe potential:

∇×∇× A = ∇(∇ · A

)−∇2A

∇2A = µσ∂A

∂t

Wednesday - 11/12

33.4 Faraday’s Law

Consider some current distribution j(x) localized to a region of space. It is immediately clear thatE = 0 and B 6= 0 for this system. However, if the current distribution and observer are not in thesame inertial frame, then the charge density will appear to be moving with some velocity. For ouranalysis, assume that v c and is a constant. For this system, the equations for the fields are:

∇× E = −∂B∂t

∇ · E =ρ

ε0∇ · B = 0 ∇× B = µ0j +

1c2

∂E

∂t

We can write the total time derivative of the magnetic field as:

dB

dt=∂B

∂t+ (v · ∇) B

And expand this by vector relations and dropping the term ∇ · B gives:

(v · ∇) B = v(∇ · B

)+∇×

(B × v

)→ dB

dt=∂B

∂t+∇

(B × v

)89

Combining this with the other equation above, we find that:

∇E = −dBdt

+∇×(B × v

)→ ∇×

(E + v × B

)= −dB

dt

We can take this result and integrate the projection over the surface enclosing the distribution:

∫ [∇×

(E + v × B

)= −dB

dt

]· ndA

The righthand side gives the total time derivative of the magnetic flux through the region, whilethe second side can be rewritten via Stokes’ theorem to give:

∮dl ·(E + v × B

)=∮

1qF

We can define this to be the electromotive force ξ and we get the familiar relation:

ξ = −dΦdt

Note that in this analysis we’ve assigned a new force F = qE′ where the new field is an “effective”field we have defined from E′ = E+ v× B. This demonstrates how E and B are actually connectedin non-static systems.

90

Part III

TRANSITION TO NON-STATICS

34 Modifications To Potentials For Non-Statics

Recall that in general, Maxwell’s equations and other relevant equations for solving problems canbe summarized as:

∇ · E =ρ

ε0∇ · B = 0 ∇× E = −∂B

∂t∇× B = µ0j +

1c2

∂E

∂t

F = q(E + v × B

) ∂ρ

∂t+∇ · j = 0

And in the cases of Electrostatics and Magnetostatics we assume that E = −∇φ and B = ∇× Arespectively. Now that we’re dealing with non-statics, these assumptions will have to be altered. Aswe still require ∇ · B = 0, we can still define a vector potential A. Plugging this into the equationabove, we find that the new scalar potential can be defined as:

∇× E = − ∂

∂t∇× A = −∇× ∂A

∂t

∇ ·(E +

∂A

∂t

)= 0→ E +

∂A

∂t= −∇φ

So, summarizing, we have satisfied two of the equations by writing the fields in terms of the poten-tials:

B = ∇× A E = −∇φ− ∂A

∂t

Plugging these into the remaining Maxwell equations yields:

∇ ·(−∇φ− ∂A

∂t

)=

ρ

ε0→ ∇2φ+∇ · ∂A

∂t= − ρ

ε0

∇×∇× A = ∇(∇ · A

)−∇2A = µ0j +

1c2

(−∇φ− ∂A

∂t

)

∇2A−∇(∇ · A

)− 1c2∇∂φ∂t− 1c2

∂2A

∂t2= −µ0j → ∇2A− 1

c2

∂2A

∂t2−∇

(∇ · A+

1c2

∂φ

∂t

)= −µ0j

91

35 Simplifications By Gauge Transformations

35.1 Gauge Transforms in Non-Statics

Recall how we defined Gauge transforms to be cases where altering the potential functions didn’talter the physical fields. In this non-static case, we can still define a Gauge transform. If wetransform A′ = A+∇λ this results in B = B′. However, the resulting electric field is:

E′ = −∇φ− ∂A′

∂t= −∇φ− ∂A

∂t− ∂

∂t∇λ = −∇

(φ+

∂λ

∂t

)− ∂A

∂t

So, then, we can see that this transform holds still if define the additional transform on the scalarpotential:

φ′ = φ− ∂λ

∂tA′ = A+∇λ

Finally, consider the case that we transform the scalars so that one of the terms ∂

∂t

(∇ · A

)and

∇(∇ · A+ 1

c2∂φ∂t

)go to zero. In general, the second term would be equal to some function we’ll

define as χ(x, t). So, we want to rig the scalar transform above so that:

∇ · A+1c2

∂φ

∂t= χ

becomes:

∇ · A′ −∇2λ+1c2

∂φ′

∂t+

1c2

∂2λ

∂t2= χ→

−∇2λ+ 1

c2∂2λ∂t2

= χ

∇ · A′ + 1c2∂φ′

∂t = 0

This is known as the Lorenz Gauge since the resulting equations are wave equations of the form:

∇2A− 1c2

∂2A

∂t2= −µ0j ∇2φ+

1c2

∂2φ

∂t2= − ρ

ε0

A = −µ0j φ = − ρε0

Friday - 11/14Alternately, suppose we can write the evolution of the charge distribution as ρ(x, t), in this case wecan refer back to electrostatics and apply the Coulomb Gauge (∇ · A = 0). Plugging this into theabove results give:

∇2φ = − ρε0

92

∇2A− 1c2

∂2A

∂t2− 1c2

∂

∂t(∇φ) = −µ0j

We can easily show that from any potential A we can reduce the equations to the above by thetransform A′ = A+∇λ as long as we satisfy that:

∇2λ = −∇ · A→ λ =1

4π

∫d3x′∇x′ · A(x′)|x− x′|

We can further simplify this result since from the above we can write the solution for the scalarpotential as:

ρ(x, t) =1

4πε0

∫d3x′

ρ(x′, t)|x− x′|

35.2 Separating the Transverse and Longitudinal Current Density

So, plugging this into the second equation and noting that ∂ρ∂t = −∇ · j we can write the equation

for the vector potential as;

A = −µ0j +1

4πε0c2∇∫d3x′

∂∂tρ(x′, t)|x− x′|

And since c2 = 1µ0ε0

, the above reduces to:

A = −µ0j −µ0

4π∇∫d3x′∇ · j(x′, t)|x− x′|

If we make the following definitions,

j = jl + jt

jl = − 14π∇∫d3x′

∇x′ · j|x− x′|

jt = − 14π∇×∇×

∫d3x′

j

|x− x′|

Then the above simply reduces to:

A = −jt

The components of j given above satisfy:

93

∇ · jt = 0 ∇× jl = 0

Which means that they are the transverse and longitudinal components of the current.

35.3 Summary of Gauge Transforms

So then summarizing, the vector and scalar potentials can be transformed as:

A→ A+∇λ; φ→ φ− ∂λ

∂t

and the physical fields are left unchanged. And from this, two specific cases can be defined:

35.3.1 Lorentz-Gauge

Let ∇ · A+ 1c2∂φ∂t = 0, then the resulting decoupled equations are:

A = −µ0j φ = − ρε0

35.3.2 Coulomb Gauge

Let ∇ · A = 0. If in addition to this we have the conditions ρ = 0 and j = 0 then φ = 0 results inthe equation:

A = 0

36 Magnetic Monopoles

If magnetic monopoles do exist in the universe, then Maxwell’s equations will be altered as shownbelow:

∇ · E = ρ→ ∇ · D = ρe

∇ · B = 0→ ∇ · B = ρm

∇× E = −∂B∂t→ ∇× E = −∂B

∂t− jm

∇× H =∂H

∂t+ j → ∇× H =

∂H

∂t+ je

With the conditions of:

94

∂ρe∂t

+∇ · je = 0∂ρm∂t

+∇ · jm = 0

However, these equations show a global symmetry in that we can change any pair of field definitions(E and H for example) as:

E = E′ cosα+ Z0H′ sinα

H = −Z0H′ cosα+ E′ sinα

Or alternately written:

(EZ0H

)=(

cosα sinα− sinα cosα

)(E′

Z0H′

)But this is simply an SO(2) symmetry which is isomorphic to a symmetry in U(1) meaning thisis merely adding a phase to the field. This means that we can actually rig the charge densitydefinitions so that only electric monopole sources exist, only magnetic monopoles exist, or somewherein between. Since we physically only see electric monopoles, we rig the system to only dependon such sources. However, it is interesting to note that the Dirac equation predicts a quantizedelectronic-magnetic charge condition of eg

4π~ = n2 where e is the electric charge and g is the magnetic

charge.

95

Monday - 11/17

37 Special Symmetries - Parity and Time Inversion

We can summarize a few other symmetries which the Maxwell equations obey. For instance, spacialtranslation (x → x′ + c), time translation (t → t′ + c), and rotations (x → xR) all conserve thefields. In all, the Maxwell Equations are invariant under all Poincare Transformations (all relativistictransforms). Of particular interest to us are parity transformations (x → −x), and time inversiontransformations (t → −t). We can see that the electric and magnetic fields are still vector fieldsunder such transforms, however, let’s consider how the fields behave under transforms. By the firstof Maxwell’s Equations, the electric field changes as:

∇ · E =ρ

ε0

Under the parity transformation, the following changes occur:

∇ → −∇ ⇒ ∇ · E → −∇ · E′

since we require that the electronic charge be conserved in such a transformation, the scalar functionρ must also stay the same. Therefore in order to conserve the field we must assume that E′ = −E.Thus the electric field is a “real vector”.We can find the behavior of the magnetic field from:

∇× E =∂B

∂t→ (−∇)×

(−E)

=∂B′

∂t

Thus the vector field B remains the same in a parity transformation. This behavior for B is due to itbeing defined by a cross product of two real vectors. Such an object is known as a “pseudo-vector”.However, to check that our results make sense, we’ll check what happens in the remaining MaxwellEquation:

∇× B = µ0j +1c2

∂E

∂t→ (−∇)×

(B)

= µ0j′ +

1c2

∂(−E)

∂t

This requires that the current density change sign in parity, which is what we expect since the vectorfor the velocity will change direction. Thus we have the results that in a parity transformation:

x→ −x⇒

ρ→ ρj → −jE → −EB → B

Considering the time reversal, we can check the result by applying the transform to:

96

∇× E = −∂B∂t

∇× B = µ0j +1c2

∂E

∂t

We can see immediately that this causes the changes of:

t→ −t⇒

ρ→ ρj → −jE → EB → −B

37.1 Example - Expansion for Polarization

We can find a form for the expansion of P = ε0χeE by applying the above relations. Since we needto add a function of the magnetic field which is invariant under the above transformations, fromthis we can immediately see that up to second order, the only allowed terms are:

1ε0P = χeE + χ1

∂E

∂t× B0 + χ2

(B0 · B0

)E + χ3

(E · B0

)B0 + . . .

38 Energy in an Electromagnetic Field - The Poynting Vector

We’ve looked at the energy in a field in the case of Electrostatics and Magnetostatics, now we’llderive an expression for the general case. We can start from the force equation:

F = q(E + v × B

)And note that the mechanical work done in moving a particle along some infinitesimal distance dlcan be written as:

dWmech = F · dl→ dWmech

dt= F · dl

dtdt

And noting that the second term is simply the force dotted with the velocity, we can drop thesecond term (as expected, the magnetic field does no work), and this leaves:

dWmech

dt= qE · v =

∫d3xρv · E

And given that Wmech = −WEM since energy is conserved, and the charge density times the velocityis just the current density, the above gives that the work done by the field is:

97

dWEM

dt= −

∫d3x j · E

Applying the relation j = 1µ0

(∇× B − 1

c2∂E∂t

)we can write the above in terms of only the fields:

dWEM

dt= − 1

c2

∫d3x

(∇× B − 1

c2

∂E

∂t

)· E = ε0

∂

∂t

∫d3x

12|E|2 − 1

µ0

∫d3x E ·

(∇× B

)And using vector relations and Maxwell’s Equations:

E ·(∇× B

)= −∇ ·

(E × B

)+ B ·

(∇× E

)= −∇ ·

(E × B

)− 1

2∂

∂t|B|2

So, combining this with the above gives the result:

dWEM

dt=

∂

∂t

∫d3x

(ε02|E|2 − 1

2µ0|B|2

)+

1µ0

∫d2x n ·

(E × B

)From this result, we can also gain some information about energy conservation. Defining the scalarenergy density u and flux vector s as:

u ≡ 12(E · D + B · H

)=ε02|E|2 − 1

2µ0|B|2

s ≡ E × H =1µ0E × B

The vector field s is defined to be the Poynting Vector and is used in a number of other derivations.The above gives the continuity relation that:

∂u

∂t+∇ · s = −j · E

Which when there is no current is simply an energy conservation equation. In the case of a currentin a region, the energy is altered some amount by the moving charges.

Wednesday - 11/19

39 Momentum in the Electromagnetic Field

Last time we defined the energy in an electromagnetic field in terms of the energy density u andthe Poynting vector s which described the flux of energy. Today we’ll work with the momentum inthe field. Starting from Newton’s law and the equation for the force on a charged particle.

98

dPmechdt

= F = q(E + v × B

)= −dPEM

dt

So, the momentum in the field can be written as:

dPEMdt

= −∫d3x

(ρE + j × B

)We can replace ρ with ε0∇ · E and j with 1

µ0∇× B − ε0 ∂E∂t to get the momentum in terms of only

the fields.

dPEMdt

= −ε0∫d3x

(E(∇ · E

)− c2B ×

(∇× B

)+ B × ∂E

∂t

)We can expand the last term as:

B × ∂E

∂t=

∂

∂t

(B × E

)− ∂B

∂t× E

And replacing the derivative of B with −∇× E and noting that the resulting term is the PoyntingVector, we can write the variation in the momentum as:

dPEMdt

=∂

∂t

[1c2

∫d3x s

]+ ε0

∫d3x

[−E

(∇ · E

)+ c2B ×

(∇× B

)+ E ×

(∇× E

)]Using vector relations, the larger term can be reduced to the divergence of some tensor, thus theabove is simply,

dPEMdt

=∂

∂t

[1c2

∫d3x s

]+∑i

∫d3x

∂

∂xiT i

In the case that there are no charges in the system and the field is the only source of momentum,the change goes to zero and the above results in a continuity equation of the form:

∂

∂t

[1c2

∫d3x s

]+∇ · T = 0

From this we can see that the momentum density is given by pEM = 1c2s. Thus the total momentum

in the field is:

pEM =1c2s PEM =

∫d3pEM

99

39.1 Angular Momentum in the Field

From this definition we can write the angular momentum density at some location in the field andthe total angular momentum in a volume of the field as:

lEM = r × pEM =1c2r × s→ LEM =

1c2

∫d3x r × s

39.2 Radiation Pressure

From the momentum we can define the radiation pressure as the force per unit area

PEM =∣∣∣∣dFdA

∣∣∣∣ =∣∣∣∣dPEMdAdt

∣∣∣∣ =1c2

dV

dAdt

∣∣E × H∣∣This can be reduced further since dV

dA = dl and dldt = c for the field. Thus we are left with only

PEM =1c|E × H|

In order to get the macroscopic pressure, we have to take the time average of this result. So,concluding this portion we’ll write out to time average such a quantity.

40 Time Average of Vector Products

Consider two vectors C(x, t) and D(x, t). We can write either of these in terms of its frequencycomponents via a Fourier Integral Transform:

C(x, t) =∫dωe−iωtC ′(x, ω)→ dC(x, t)

dω= C ′(x, ω)e−iωt

And the combination of two vectors can then be written in terms of their complex conjugates:

dC(x, t)dω

=12(C ′(x, ω)e−iωt + C ′∗(x, ω)eiωt

)dD(x, t)dω

=12(D′(x, ω)e−iωt + D′∗(x, ω)eiωt

)Thus any combination C ⊗ D can be consider a combination:

C ⊗ D ≡ 14(C ′(x, ω)e−iωt + C ′∗(x, ω)eiωt

)⊗(D′(x, ω)e−iωt + D′∗(x, ω)eiωt

)

100

=14(C ′ ⊗ D′e−2iωt + C ′∗ ⊗ D′∗e2iωt + C ′ ⊗ D′∗ + C ′∗ ⊗ D′

)Taking the time average of this, the first two terms drop out and we are left with only the last two.Combining these, we get the result that:

〈C ⊗ D〉t ≡12Re(C∗ ⊗ D

)Thus, the above relations can be time averaged to give:

〈s〉t =12Re(E∗ × H

)〈u〉t =

14Re(E · D∗ + B · H∗

)

101

Part IV

SPECIAL RELATIVITY AND TENSORS INELECTROSTATICS

Wednesday - 1/21

41 Transition to Semester II

41.1 Review of First Semester: Maxwell’s Equations and Gauge Transforms

Maxwell’s Equations in a vacuum are:

∇ · E =ρ

ε0; ∇ · B = 0

∇× E +∂B

∂t= 0; ∇× B = µ0j +

1c2

∂E

∂t∂ρ

∂t+∇ · j = 0; F = q

(E + v × B

)We can use the scalar and vector potentials φ and A where the electric and magnetic fields aredescribed as E = −∇φ − ∂A

∂t and B = ∇× A. From these definitions for the potentials Maxwell’sEquations can be expressed as:

∇2φ+∇ · ∂A∂t

= − ρε0

∇2A− 1c2

∂2A

∂t2−∇

(∇ · A+

1c2

∂φ

∂t

)= −µ0j

Further, under gauge transformations, the potentials can be modified in specific ways that leave thephysical fields unchanged. Those changes are described as:

A→ A+∇λ; φ→ φ− ∂λ

∂t

The two most useful gauge transforms are the Coulomb and Lorenz transformations:

41.1.1 Lorentz-Gauge

Let ∇ · A+ 1c2∂φ∂t = 0, then the resulting decoupled equations are:

A = −µ0j φ = − ρε0

102

41.1.2 Coulomb Gauge

Let ∇ · A = 0. If in addition to this we have the conditions ρ = 0 and j = 0 then φ = 0 results inthe equation:

A = 0

41.2 Introduction of Gaussian Units

For the rest of this semester we will be using Gaussian units. We can write Maxwell’s Equations interms of generic scalars relating the fields and sources as:

∇ · E = 4πα1ρ; ∇ · B = 0

∇× E = −α3∂B

∂t; ∇× B = 4πα2α4j +

α2α4

α1

∂E

∂t

previously we’ve used the SI Units relations of:

α1 =1

4πε0; α2 =

µ0

4πα3 = 1; α4 = 1

From this point forward we will be using Gaussian Units, where:

α1 = 1; α2 =1c2

α3 =1c

; α4 = c

This unit system is useful since it makes the only scalar constant the speed of light c. The resultingequations are:

∇ · E = 4πρ; ∇ · B = 0

∇× E = −1c

∂B

∂t; ∇× B =

4πcj +

1c

∂E

∂t

∂ρ

∂t+∇ · j = 0; F = q

(E +

1cv × B

)For more conversions, see page 782 in Jackson.

103

42 Propagation in a Vacuum - Introduction to Special Relativity

Consider electromagnetic waves propagating in a vacuum without sources. The solution is thatdescribed by:

A = 0 φ = 0

= − 1c2

∂2

∂t2+∇2

This is the solution of waves moving at speed c. However, what is this speed relative to? There areseveral possible results. Either (1) the speed c is that relative to some preferred reference frame,(2) the speed c is that seen in any frame, or (3) for some specific set of frames, the waves propagateat speed c. This is the problem of invariance and its solution leads to the main postulate of SpecialRelativity.Classical mechanics refers to the Galilean Transformation for coordinate systems moving at constantvelocity to one another as:

x→ x′; x′ = x− vt

In component form that is:

x′ = x− vxt

y′ = y − vyt

z′ = z − vzt

t = t′

From this assumption, we can write the transformation of the potentials in terms of some transfor-mation scalars ki

φ′ = k0φ+3∑i=1

kiAi; A′j = k′0φ+3∑i=1

k′iAi

Next, consider how the wave operator changes in this transformation. The vector operator ∇remains unchanged since each component transforms as:

∂

∂x′=∂x

∂x′∂

∂x+∂y

∂x′∂

∂y+∂z

∂x′∂

∂z+

∂t

∂x′∂

∂t=

∂

∂x→ ∇′ = ∇

However, the transformation of the time derivative generates extra terms as:

∂2

∂t′2=

∂2

∂t2+ 2v · ∇ ∂

∂t+ (v · ∇) (v · ∇)

104

This result implies that there is some preferred reference frame for propagation of light. However,experiments showed otherwise. Clearly something was incorrect. The equations of Electromag-netism seemed to be correct and no gauge transformation would produce more correct results. Theonly thing remaining was a new transformation between reference frames. This leads to the primarypostulate of special relativity:

The laws of nature and the results of all experiments performed in a given frame of reference areindependent of the translational motion of the system as a whole. More precisely, there exists a triplyinfinite set of equivalent Euclidean reference frames moving with constant velocities in rectilinearpaths relative to one another in which all physical phenomena occur in an identical manner.

However, changing the way that two frames are related requires rewriting much of classical physics.One particularly troubling result of a new transform is that it requires t 6= t′. This results indifferent reference frames observing different rates of time.

43 The Lorentz Transform

Friday - 1/23

43.1 Deriving the Lorentz Transform Group

Our goal is to find a change in coordinates for which Maxwell’s Equations remain invariant. In thiscase, we want the potentials φ and A to remain unchanged in the different coordinate frames. Todo this, we must find transformation in which the wave equation is unchanged for some potential.That is,

ψ = 0→ ′ψ = 0

To do so, we must define new operators and vectors which will incorporate time into the coordinatesystem. The position 4-vector and the 4-vector nabla operator are defined as:

x =

ctxyz

=(ctx

); ∇ =

(1c∂∂t ,

∂∂x ,

∂∂y ,

∂∂z

)=(

1c∂∂t , ∇

)

From these definitions, we can show that the wave operator can be rewritten as:

= ∇g∇T ; g =

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

=(

1 00 −1

)

105

where we’ve shortened notation of the 4 × 4 matrix g to a 2 × 2. This statement can quickly beshown by expanding the above:

=(

1c∂∂t , ∇

)(1 00 −1

)(1c∂∂t ,∇

)=(

1c∂∂t , ∇

)(1c∂∂t ,−∇

)=

1c2

∂2

∂t2−∇2

So, now we need to define a transformation x′ = Λx which keeps the operator ∇g∇T invariant.Consider the structure of this matrix Λ.

ct′

x′

y′

z′

=

Λ0

0 Λ01 Λ0

2 Λ03

Λ10 Λ1

1 Λ12 Λ1

3

Λ20 Λ2

1 Λ22 Λ2

3

Λ30 Λ3

1 Λ32 Λ3

3

ctxyz

It would be time consuming and difficult to determine each component of the matrix via the equa-tions generated by the above statement. An alternate way to continue is to show that ∇′ = ∇Λ−1.This can be shown by noting that ∇x = 4. And also, ∇′x′ = 4. Then, making multiple substitutions,

∇′x′ = ∇′ (Λx) = ∇Λ−1Λx

So, with these conditions, we can make the invariance condition:

∇′g∇′T = ∇Λ−1g(Λ−1

)T ∇T = ∇g∇T → Λ−1g(Λ−1

)T = g

Multiplying by Λ, taking the transpose and multiplying by Λ again yields the condition:

ΛgΛT = g

This defines the Lorentz Transformation Group which is a subgroup of the more general PoincareGroup. The Poincare Group consists of transformations of the form:

x′ = Λx+ A

where A is some 4 vector defining a translation in space or time. The total dimensionality of thePoincare transformation is 10. Those are 3 from rotation, 3 from velocity, and 4 from the translation.From this we can see that the Lorentz transform we’re deriving will have 6 dimensions from therotations and velocities.

43.2 Properties of the Lorentz Transform

With the definition above, several properties of the Lorentz transform can be defined.

106

• DeterminantThe determinant of the transform is always ±1. This can be shown by:

det(ΛgΛT

)= det (g)→ det (Λ) (−1)

(ΛT)

= −1

[det (Λ)]2 = 1→ det (Λ) = ±1

Transforms with determinant +1 are called “Proper Transforms”.

• Making The Transform Self AdjointGiven some form for Λ, it is always possible to rig the coefficients of the transform so thatΛ = ΛT . This is because we can easily show that both ΛgΛT = g and ΛT gΛ = g are validstatements. The result of this is that the components of the transform can always be writtenas Λji = Λij .

• Allowed Values of Time ElementThe first element (Λ0

0) can take only certain values. This is evident from the first equationresulting from multiplying out the matrices.

1 =(Λ0

0

)2 − 3∑i=1

(Λ0i

)2 → (Λ0

0

)2 ≥ 1

thus the allowed values of this element are Λ00 ≥ 1 or Λ0

0 ≤ −1. The positive solutionsare “orthochronous transforms” and this is the second characteristic of the transform we arelooking for.

One Last NoteFurther, one can use the velocities to set Λji with i = 0 or j = 0 equal to zero (obviously except thecomponent with i = j = 0). This results in a transform of the form:

Λ =

1 0 0 00 Λ1

1 Λ12 Λ1

3

0 Λ21 Λ2

2 Λ23

0 Λ31 Λ3

2 Λ33

=(

1 00 R

)

where the sub-matrix R describes some rotation.Monday - 1/26

43.3 The Lorentz Transformation

In order to actually find the form for the transformation, consider a problem in only 2 dimensions.In this case we have that:

x =(ctx

)

107

And the condition equation becomes:

ΛgΛT = g →(α1 α2

α3 α4

)(1 00 −1

)(α1 α3

α2 α4

)=(

1 00 −1

)Combining this with the requirement that the transformation be proper (det(Λ) = +1) gives:

α1α4 − α2α3 = 1

α21 − α2

2 = 1α3

α1− α2

α1= 0

α23 −

1α2

1

(1 + α2α3)2 = −1

the first equation can be solved to eliminate α4 from the system and since we assume α1 6= 0 thethird equation gives an immediate result of α2 = α3 which is expected. The last equation combinedwith the second equation gives no additional information. The resulting system leaves 3 quantitiesand only 2 equations. However, letting α2 = −βα1 simplifies the equations and gives a result onlyin terms of the new scalar quantity β. The transform is then defined as:

Λ =

1√1−β2

−β√1−β2

−β√1−β2

1√1−β2

If we use the notation that: γ = 1√

1−β2then we can write the transform more simply as:

ct′ = γ (ct− βx)

x′ = γ (x− βct)

The reverse transform is given by:

ct = γ(ct′ + βx′

)x = γ

(x′ + βct′

)These results can be generalized to multiple dimensions quickly:

t′ = γ(t− v · x

c2

)x′ = x+

γ − 1β2

(β · x

)β − γβct

β =v

c; γ =

1√1− |β|2

108

Consider the system at the location x = 0. In this case, the results give the value of β.

t′(x = 0) = γt; x′(x = 0) = −γβct→ x′

t′= −βc

the frame x′ is traveling away from the frame x with speed v, so from the above result it is clearthat β = v

c . Further analysis shows that the transformation is only valid for β ≤ 1. That meansthat velocities greater than c are not allowed by the transform. Consider a wave propagating. Thelocation of the wave can be described as c2t2 − x2. It can quickly be shown that in either frame,the relation is equal, that is:

c2t2 − x2 = c2t′2 − x′2 = 0

This means that the quantity above is invariant under a Lorentz transform. Taking the differentialof this results in the definition for the proper time τ .

dt′ = γ(dt− v

c2dx)

; dx′ = γ (dx− v dt)

c2dτ2 = c2dt2 − dx2

The next question is that of how velocities sum. In classical mechanics if a person is riding on someobject with velocity v and throws or shoots some other particle or object with velocity vobj , thensomeone in the rest frame sees the object moving at a velocity which is simply the sum of the two.However, given these relations, consider the result of this same problem

v′obj =dx′

dt′=

dx− vdtdt− v

c2dx

=dxdt − v

1− vc2dxdt

=vobj − v

1− v·vobjc2

v′obj =vobj − βc1− β

c vobj

43.4 New Notation - The 4 Vector

Consider if we define the 4-vector as:

x =

ctxyz

=

x0

x1

x2

x3

In this notation, the transform is easily summarized as:

x′0 = γ (x0 − βx1)

109

x′1 = γ (x1 − βx0)

43.5 The Boost Parameter

Because of the behavior of the variable β, it is possible to define it instead as a hyperbolic trigono-metric function. Consider the definitions:

β = tanh (χ)

γ = cosh (χ)

γβ = sinh (χ)

The new variable χ is the rapidity or boost parameter of the transformation. Consider the form ofthe transform in terms of this new variable:

x′0 = x0 cosh (χ)− x1 sinh (χ)

x′0 = −x1 sinh (χ) + x0 cosh (χ)

Or in matrix form:

(x′0x′1

)=(

cosh (χ) − sinh (χ)− sinh (χ) cosh (χ)

)(x0

x1

)note that if one changes χ→ iχ and x0 → ix0 then the above becomes simply the familiar rotationmatrix. This is because the Lorentz transformation is actually a hyperbolic rotation. Just asthe rotation matrices are a group of SO(3), the Lorentz Transform are a group of SO(1, 3) whichcontains the group SO(3).Wednesday - 1/28

44 Space-Time Diagrams

44.1 The Line Element

Last time we showed that the differential element dτ was invariant under a Lorentz transformationas:

c2dτ2 = c2dt2 − dx2

This can be alternately defined as the line element c2dτ2 = ds2 and distances in spacetime canbe determined from it. Let’s consider the basic geometry of spacetime. Consider the examples inFigure 1.

110

Figure 1: Spacetime Diagrams - Example of determining separations in space time diagrams. The length of thelines AC, DE, and EF are calculated in the notes.

The length of AC can be determined as:

ds2 = 52 − 32 = 25− 9→ |AC| = 4

And the lengths of DE and EF are similarly calculated as:

ds2 = 32 − 32 = 0→ |DE| = |EF | = 0

From this result we can see that the null separated lines (DE and EF ) define paths of zero length.These lines are “null lines” and define the edges of light cones. As another example, let’s considertwo observers both at rest shining flashlights at one another. This is shown in Figure 2. When Philshines his flashlight at Marco, the light travels along the line with slope 1 and reaches Marco aftersome time ∆t. Then Marco shines his flashlight back at Phil, who sees the light at some later time2∆t. This is simple and straightforward when no one is moving.

44.2 Some More Interesting Cases - Moving Frames

When one frame is moving relative to another, the transforms we’ve derived describe how theobservers view events. Note that we refer to “events” as something that is designated with aspecific 4-vector (that is, it has a specific location and specific time that it occurred). The hyperbolictransformation in spacetime diagram form is shown in Figure 3. Note that due to the time variablebeing essentially an imaginary space variable, the axes don’t transform as one would expect.

Consider for instance if Phil were to observe two events (A and B) which occurred in the samelocation, but at separate times. If another individual (Sky) were to observe the same events from aframe moving at speed v relative to Phil, then she would observe something different. The time andlocation of the first event would remain unchanged, however, Sky would see the second event behind

111

Figure 2: Stationary Observers - Light signals being passed between two stationary observers.

Figure 3: Stationary Observers - Light signals being passed between two stationary observers.

112

her and the two events would be separated by a different amount of time. The exact difference isgiven as:

∆t′ = γ∆t; ∆x′ = −γv∆t

The diagram of these events is shown in Figure 4.

Figure 4: Events viewed From Different Frames - Two events take place in the same location but differenttimes for one observer, but in different places and with a different time separation for another observer.

Consider also if two events take place at the same time in one frame. How does this appear inanother? The space time diagrams for this case are shown in Figure 5.

Figure 5: Simultaneous Events - Two events take place at the same time but in different locations for oneobserver, but at different times and with a different spacial separation for another observer.

One interesting thought experiment in special relativity is the Twins Paradox. If two identical twinswere to undergo different experiences, could this alter their ages? Suppose one of the twins remainson Earth while the other enters a spacecraft and travels at some velocity v (which is a large fraction

113

of c) out to Alpha Centauri and then returns. Because the second individual had traveled at highspeeds for an extended period of time, time dilation would cause the twins to be different ages. Theparadox is due to the fact that from the second individual’s frame of reference, the first twin is theone who should be younger. This can be resolved by noting that the second twin had to acceleratemultiple times in the trip and special relativity only deals with constant velocities.Friday - 1/30

44.3 Length Contraction

We’ve seen how time dilation occurs in special relativity, next consider how the lengths of objectsappear in different frames. If an object is of length L in the rest frame, in a frame traveling atspeed v relative to that frame, the object will seem to change by a factor of:

L′ =1γL

Because the value of γ is always greater than or equal to one, the length of the object is alwayscontracted in the observer’s frame. This is seen in the spacetime diagram in Figure 6.

Figure 6: Length Contraction - The length of an object in the frame at rest and a frame traveling at somevelocity relative to it.

45 Introduction to Tensors

45.1 Matrix Representation of Lorentz Transformations - Infinitesimal Gener-ators

Before we begin delving into tensor notation, consider what can be determined about the LorentzTransformations from matrix representation analysis. Note that for some pair of 4-vectors a and b,the matrix scalar product is defined as:

114

(a, b) = ab = aT b

and the scalar product can be written as:

a · b = (a, gb) = (ga, b) = agb = aT gb

where we’ve defined the covariant vector gx as:

gx =

x0

−x1

−x2

−x3

The Lorentz transform is one such that x′ = Ax for some 4×4 transformation matrix A. We requirethat the norm of the vector (x, gx) is left invariant, that is:

x′T gx′ = xT gx→ xTAT gAx = xT gx

and since this must hold for any 4-vector, the transformation matrix A must obey the relation;

AT gA = g

Constructing the transform A can be done by assuming that it has the form A = eL for some4× 4 matrix L. In this case, the determinant (which we already know must be equal to 1 for suchtransforms) becomes:

det(eL)

= eTr(L)

This requires that the matrix L be traceless. Next, plugging this definition into the condition on Agives the result that:

AT = eLT

; gAT g = egLT g; A−1 = e−L

gLT g = −L; (gL)T = − (gL)

Thus the condition on A results in a condition on L that it be a real, traceless, antisymmetric 4× 4matrix. The most general form of the solution is then:

L =

0 L01 L02 L03

L01 0 L12 L13

L02 −L12 0 L23

L03 −L13 −L23 0

115

This can be generalized in terms of several rotation matrices (S, and boost matrices K.

S1 =

0 0 0 00 0 0 00 0 0 −10 0 1 0

; S2 =

0 0 0 00 0 0 10 0 0 00 −1 0 0

; S3 =

0 0 0 00 0 −1 00 1 0 00 0 0 0

K1 =

0 1 0 01 0 0 00 0 0 00 0 0 0

; K2 =

0 0 1 00 0 0 01 0 0 00 0 0 0

; K3 =

0 0 0 10 0 0 00 0 0 01 0 0 0

Of particular interest are the squares of these different matrices:

S21 =

0 0 0 00 0 0 00 0 −1 00 0 0 −1

; S22 =

0 0 0 00 −1 0 00 0 0 00 0 0 −1

; S23 =

0 0 0 00 −1 0 00 0 −1 00 0 0 0

K21 =

1 0 0 00 1 0 00 0 0 00 0 0 0

; K22 =

1 0 0 00 0 0 00 0 1 00 0 0 0

; K23 =

1 0 0 00 0 0 00 0 0 00 0 0 1

From this it is seen that any combination

(ε · S

)3 or(ε′ · K

)3 can be written as −ε · S or ε′ · Krespectively. That is, any power of the matrices can be written as a multiple of the matrix or itssquare. The general result of the transform can be summarized now in terms of some rotation ωand some boost ζ.

L = −ω · S − ζ · K

A = e−ω·S−ζ·K

Note two limiting cases which we are already familiar with. First, consider the case of ω = 0 andζ = ζx. Then the transform is the familiar hyperbolic transform:

A =

cosh (ζ) − sinh (ζ) 0 0− sinh (ζ) cosh (ζ) 0 0

0 0 1 00 0 0 1

While in the case that ζ = 0 and ω = omegaz, the transform becomes:

116

A =

1 0 0 00 cos (ωz) − sin (ωz) 00 − sin (ωz) cos (ωz) 00 0 0 1

Generalizing the result above for a rotationless boost in some direction, ζ = β tanh−1 β the boostis written as:

Λboost(β) =

γ −γβ1 −γβ2 −γβ3

−γβ1 1 + (γ−1)β21

β2(γ−1)β1β2

β2(γ−1)β1β3

β2

−γβ2(γ−1)β1β2

β2 1 + (γ−1)β22

β2(γ−1)β2β3

β2

−γβ3(γ−1)β1β3

β2(γ−1)β2β3

β2 1 + (γ−1)β23

β2

Lastly, consider the commutator relations between the matrices derived above. It can easily beshown that:

[Si, Sj ] = εijkSk

[Si,Kj ] = εijkKk

[Ki,Kj ] = −εijkSk

The first relation is the known rotation commutation. The second shows that the K transforms asa vector under rotations, and the final shows that in general, boosts do not commute. These specifythat the Lorentz group is SL(2, C) or O(3, 1).Monday - 2/2

45.2 Tensor Notation

In order to simplify notation, we’ll transition from matrix and vector notations to tensors. This issimple at first, but one must be careful to follow a few simple rules of working with tensors. Theserules are called “Summation Conventions” and are summarized as:

1. The location of the indices must be respected: in two dimensional tensors, superscripts denotecolumns while subscripts denote rows.

2. Repeated indices always appear in superscript-subscript pairs and imply summation. Theseindices are referred to as ”dummy indices“ since they can be replaced by any other index.Such as:

gαβaαbβ = gγδa

γbδ

notations such as gααaαbβ or gαβgβγ are incorrect.

117

3. Indices not repeated must match indices on the opposite side of the equation. These are notsummed over and are called ”free indices“ since the value of such an index can be changed aslong as it is changed on both sides of the equation. For instance, the transform xα′ = Λα βxβ

is correct. However, the equation xγ′ = Λα βxβ makes no sense.

With those rules in place, we can define a few objects. First, the column vector:

x =

ctxyz

=

x0

x1

x2

x3

can be more compactly written as xα where α can take values of 0, 1, 2, or 3. The column vectortransforms as:

xα′

= Λα βxβ

such an object is a tensor of rank 1 called a contravariant vector. Consider also the row vectoroperator defined as:

∇ =(

1c∂∂t

∂∂x

∂∂y

∂∂z

)This can be compactly written as ∂

∂xα where again α can take the allowed values stated above. Thisobject transforms as:

∂

∂xα′=(Λ−1

)βα∂

∂xβ= Λα β

∂

∂xβ

where we’ve introduced the notation(Λ−1

)βα = Λα β for the inverse of a matrix. An object that

transforms in this way is called a covariant vector. Next, consider how the matrix g will be writtenin tensor notation. It is chosen that g → gαβ so that the relation:

g = gαβ; g−1 = gαβ → gαβgβγ = δγ α

With these new notations, let’s check how the transformation condition and the Lorentz transfor-mation itself will be written. First, consider the relations:

(Λ−1

)βα = Λα β; ΛΛ−1 = I → Λα βΛα γ = δβ γ

and the condition ΛgΛT = g then can be derived from:

Λα β = gαγgβδΛγ δ → Λε β

[gαγg

βδΛγ δ]

= δε α

118

Multiplying both sides by gψη and then change the value of η to α changes the above relation into:

gψαΛε βgαγgβδΛγ δ = gψαδε α

changing the order and simplifying indices by gψαgαγ = δψ γ which eliminates some of the indicesand changes Λγ δ → Λψ δ. Changing the order of the values gives the form of the condition equationin tensor notation:

ΛgΛT = g → Λψ δgδβΛε β = gψε

Thus any Lorentz Transform will obey the relation:

Λα γ = gαβgβδΛγ δ

Lastly, consider how to construct higher rank tensors and other objects. As an example, considerthe rank 5 tensor Λαβγδε. This object can be constructed as:

Λαβγδε = Λα µΛβ νΛγ ρΛδ σΛε τΛµνρστ

Consider the scalar function f(x). The transform of this function can be written as:

f(x)→ f ′(x′) = f(x′)

thus the scalar should not change under the transform. Alternately, consider a vector potentialA(X), this quantity should transform as:

A→ A′(x′) = ΛA(x′)

Lastly, it is quickly seen that the quantity xαxβ is invariant in the case that α = β since:

x′αx′β =(

Λα βxβ)

(Λα γxγ) = δβγxβxγ = xβxβ

119

Wednesday - 2/4

45.3 Building Tensors from Other Tensors

There are several methods to build a tensor out of other tensors. These are linear combinations oftensors of equal rank and form, direct products of tensors (multiplication), Contraction of tensors,and Differentiation. Consider each:

1. Linear Combinations of TensorsThe linear combination of tensors with matching upper and lower indices is a new tensor withthose same indices. For instance:

Tα β = aRα β + bSα β

and Tα β is a tensor since it can be shown that it transforms as:

T ′α β = Λα γΛβ δ (aRγ δ + bSγ δ) = Λα γΛβ δT γ δ

2. Direct ProductsThe products of the components of two tensors produces a tensor with the indices of all indicesof the original tensors. For instance:

Tα βγ = Aα βB

γ

and it’s simple to show that this new object transforms as a tensor:

T ′α βγ = Λα δΛβ εAδ εΛγ ηBη = Λα δΛβ εΛγ ηT δ ε η

3. ContractionSetting the upper and lower indices equal and summing over the values 0, 1, 2, 3 yields a tensorwith rank decreased by 2. Consider the example:

Tαγ = Tα βγβ

and again it’s easily shown that:

T ′αγ = Λα δΛβ εΛγ ηΛβ κT δ ε ηκ = Λα δΛγ ηδεκTδεηκ = Λα δΛγ ηT δ η

4. DifferentiationThe derivative of any tensor is a tensor with an additional lower index. Again, as an exampleconsider:

Tαβγ =

∂

∂xαT βγ

120

and one last time we can show that this is indeed a new tensor since it transforms as:

T ′αβγ = Λα δ

∂

∂xδΛβ ηΛγ εT ηε = Λα δΛβ ηΛγ ε

∂

∂xδT ηε = Λα δΛβ ηΛγ εTδ ηε

45.4 Raising and Lowering Indices

The Minkowski Metric (g) can be used to raise and lower indices of tensors. Recall the definitionof the metric,

gαβ = gαβ =

1, α = β = 0

−1, α = β = 1, 2, 30, else

By multiplying by g and contracting it is possible to raise or lower indices of tensors. A fewexamples:

xα → xβ by gαβxα

Vα → V β by gαβVα

Tαβγ → Tα δγ by gβδT

αβγ

45.5 Some Special Tensors

1. Minkowski MetricThe Minkowski metric which we’ve been using is a special tensor which is the same in anyframe of reference. This can easily be seen by:

g′α′β′ = Λα

′αΛβ

′βg

αβ = gα′β′

2. Levi-Civita TensorThe Levi-Civita tensor is defined as εαβγδ where:

εαβγδ =

1, if αβγδ even permutation of 0123−1, if αβγδ odd permutation of 0123

0, else

3. Zero Tensor

It is always possible to define a tensor with an arbitrary pattern of indices and all componentsbeing zero. This is a zero tensor.

121

46 Back to Electromagnetism

Now that we’ve covered some of the basics of tensor notation, let’s turn our attention back tothe topic of electromagnetism. Recall that we’ve defined a scalar potential and a vector potential(which has 3 components). That gives us 4 quantities to calculate, which we can define in terms ofa 4-vector. Consider the definition:

Aµ =(A0, A

)=(φ(x), Ax(x), Ay(x), Az(x)

)From this we can increase the rank to 2 by differentiating,

Sνµ = ∂νA

µ

Then, raising the index so the two are both up,

gνσSνµ = gνσ∂νA

µ = Sσµ

where we’ve defined,

∂µ ≡(

1c

∂

∂t,∇)

∂µ ≡(

1c

∂

∂t,−∇

)So then we can define the antisymmetric tensor F νµ as:

F νµ ≡ ∂µAν − ∂νAµ

As an example of what this tensor represents, consider the components with ν = 0,

F 01 = ∂0A1 − ∂1A0 =

1c

∂

∂tAx +

∂

∂xφ = −Ex

So, the components of F νµ with ν = 0 are the negatives of the components of the electric field.

F 0i = −Ei

122

Friday - 2/6 Last time we saw that some of the elements of the F νµ tensor were the electric field.Let’s summarize what we’ve found and check the remaining elements.

F ii = 0

F 0i = −Ei

F ij = ∂iAj − ∂jAi = − ∂

∂xiAj − ∂

∂xjAi = εijkBk

So, then the rank two tensor Fµν can be written in matrix form as:

Fµν =

0 −Ex −Ey −EzEx 0 −Bz ByEy Bz 0 −BxEz −By Bx 0

=

0 −E1 −E2 −E3

E1 0 −B3 B2

E2 B3 0 −B1

E3 −B2 B1 0

and it is straightforward to show that lowering the indices changes the signs of the each of thecomponents:

Fµν = gµγFγβgβν =

0 Ex Ey Ez−Ex 0 −Bz By−Ey Bz 0 −Bx−Ez −By Bx 0

=

0 E1 E2 E3

−E1 0 −B3 B2

−E2 B3 0 −B1

−E3 −B2 B1 0

It is also useful to define what is called the ”Dual Tensor“ as ∗F νµ = 1

2εαβγδFγδ. It is easily shown

that this switches the locations of the magnetic and electric field components and changing severalsigns (E → B, B → −E). This gives:

∗F νµ =

0 −B1 −B2 −B3

B1 0 E3 −E2

B2 −E3 0 E1

B3 E2 −E1 0

From all of this, we can immediately see that we can write the electromagnetic fields in some inertiaframe 0′ relative to another 0 as:

F ′νµ = Λν γΛµ δF γδ

from this it is clear that the components of the E and B fields mix in the moving frame, whichinfers that the two fields are actually a single object.

123

46.1 Maxwell’s Equations in Tensor Form

By taking the derivative of the potentials tensor, it is possible to build the Maxwell Equations.However, simply writing ∂ρF νµ makes an object with 4× 4× 4 free elements, while we are lookingfor something with only 4 dependent dimensions.

1. Contraction of IndicesIf we use gαβ to raise the index on the derivative and then contract over it, we end up withthe quantity ∂µFµν which is a set of 4 equations. Consider a few possible values of the indices:

∂iFi0 = ∂1F10 + ∂2F20 + ∂3F30 = −∇ · E

∂iFji = ∂1F1j + ∂2F2j + ∂3F3j =1c

∂

∂tEj +

(∇× B

)jThese are the Maxwell Equations proportional to the source functions ρ and j. As we didwith the potentials, it is possible to build a tensor out of the sources.

jν = (cρ, j)

and with this definition, the remaining Maxwell Equations can easily be written as:

∂µFµν =

4πcjν

2. Linear Combination to Reduce OrderConsider if we expand the above object:

∂ρFνµ = ∂ρ (∂νAµ − ∂µAν) = ∂ρ∂νAµ − ∂ρ∂µAν

from this expanded form we can see that adding on additional terms of the form ∂νFµρ and∂µFρν then all the elements would cancel and the resulting condition is:

∂ρFνµ + ∂νFµρ + ∂µFρν = 0

but this is satisfied identically if Fνµ = ∂νAµ−∂µAν , so what does it actually mean? Considera few possible elements:

ν = 1ρ = 2µ = 3

→ ∂2F13 + ∂1F32 + ∂3F21 = − ∂

∂yBy −

∂

∂xBx −

∂

∂zBz

⇒ ∇ · B = 0

Repeating for another set of indices yields each component of the condition ∇× E = 0. So,we have shown that our scalar tensor definition alone gives two of Maxwell’s equation in theform:

124

∂ρFνµ + ∂νFµρ + ∂µFρν = 0

Alternately, this can be written in terms of the dual tensor as:

∂∗νFνµ = 0

Recall that the continuity of charge relation can be derived from the Maxwell Equations. To getthe tensor form of this, one simply differentiates the appropriate equation:

∂ν

(∂µF

µν =4πcjν)→ ∂νj

ν = 0

At this time it is possible to summarize Maxwell’s Equations in tensor form:

∂µFµν =

4πcjν

∂ρFνµ + ∂νFµρ + ∂µFρν = 0

It was useful to write Maxwell’s Equations in terms of the potentials φ and A, so we’ll do thatagain.

∂µ (∂µAν − ∂νAµ) =4πcjν

but we can write that:

∂µ∂µ =

1c2

∂2

∂t2−∇2 =

∂µAµ =

1c

∂

∂tφ+∇ · A

recall that in the Lorenz Gauge, this last term is set to zero. Thus in the Lorenz Gauge, Maxwell’sEquations in term of the potentials are simply:

Aν =4πcjν

Monday - 2/9

125

46.2 Lorentz Transform of the Electromagnetic Field

Recall that we’ve derived the form of the electromagnetic fields in terms of the tensor F νµ. This isgiven as:

F ′νµ = Λν ν′Λµ µ′F ν′µ′

where,

Fµν =

0 −Ex −Ey −EzEx 0 −Bz ByEy Bz 0 −BxEz −By Bx 0

=

0 −E1 −E2 −E3

E1 0 −B3 B2

E2 B3 0 −B1

E3 −B2 B1 0

Λν µ(β) =

γ −γβ1 −γβ2 −γβ3

−γβ1 1 + (γ−1)β21

β2(γ−1)β1β2

β2(γ−1)β1β3

β2

−γβ2(γ−1)β1β2

β2 1 + (γ−1)β22

β2(γ−1)β2β3

β2

−γβ3(γ−1)β1β3

β2(γ−1)β2β3

β2 1 + (γ−1)β23

β2

Consider first the electric field. The transform is given as:

−E′i = F ′0i = Λ0µΛi νFµν = Λ0

0Λi kF 0k + Λ0jΛi 0F

j0 + Λ0jΛi kF jk

Plugging in values from the above matrix representations,

−E′i = (γ)(δi k +

γ − 1β2

βiβk

)(−Ek

)+ (−γβj)

(γβi) (Ej)

+ (γβj)(δi k +

γ − 1β2

βiβk

)(εjklBl

)= −γEi − γ γ − 1

β2βiβkE

k − γ2βjβiEj + γβjε

jilBl + γγ − 1β2

βiβkβjεjklBl

The last term is zero since βkβjεjkl is an antisymmetric tensor multiplied by a symmetric one.

Writing the summed terms as dot products and the remaining Levi-Civita term as a cross product,we find:

−E′i = −γEi − γ γ − 1β2

(β · E

)βi − γ2

(β · E

)βi + γ

(β × B

)iLastly, we can expand this in vector terms and find that the electric field components transform as:

E′ = γE +(γγ − 1β2

+ γ2

)(β · E

)β − γ

(β × B

)

126

Then, simplifying the γ term using the fact that |β|2 = γ2+1γ2 ,

γ

(−γ − 1|β|2

+ γ

)= γ

(γ − γ − 1

γ2 − 1γ2

)= γ2

(1− γ − 1

(γ − 1) (γ + 1)γ

)=

γ2

(1− γ

γ + 1

)=

γ2

γ + 1

Thus,

E′ = γ(E + β × B

)− γ2

γ + 1(β · E

)β

The magnetic field can be derived the same way and the result is:

B′ = γ(B − β × E

)− γ2

γ + 1(β · B

)β

One important result from these equations is to consider some system where in the frame O, themagnetic field is zero. How will the system appear in another frame O′ moving with respect to thefirst. Then, in each system the electromagnetic field will be:

O →B = 0E 6= 0

O′ →

B′ = −γβ × E

E′ = γE − γ2

γ+1

(β · E

)β

An additional feature of this system is that if we consider the term E′ × β:

E′ × β = γE × β − γ2

γ + 1(β · E

)β × β = −γβ × E

however, in the frame O′, the velocity is seen as being in the opposite direction. Thus we can writethat:

B′ = β′ × E

46.2.1 Example: Point Charge In Moving Frame

Consider a point charge at rest with respect to some frame O′. If the observer is in another frameO which O′ is moving with respect to at velocity v (that is, β′ = −β = − v

c ), then in the movingframe,

E′ =q

r′3

[xi+ yj + zz

]127

B′ = 0

And in the observer’s frame,

E = γ(E′ + β′ × B

)− γ2

γ + 1(β′ · E′

)β = γE′ − γ2

γ + 1(β′ · E′

)β

B = γ(B′ − β′ × E′

)− γ2

γ + 1(β′ · B

)β = −γβ′ × E

If the velocity is oriented only along the i direction, then the above simplifies to:

E =

E′xγE′yγE′z

B =

0γ vxc Ez−γ vxc Ey

Thus the electric field lines are more intense along the axes perpendicular to the direction of prop-agation. This is seen in Figure 11.9 of Jackson.Wednesday - 2/11 No ClassFriday - 2/13 No ClassMonday - 2/16

46.3 Lorentz Force in Invariant Form

As it is currently written, the only parts of the Lorentz Force which is invariant are the charge qand the speed of light c.

F = q

(E +

1cv × B

)If we instead write the force as a change in momentum dp

dt , then we can write this in terms of theinvariant proper time.

c2dτ2 = c2dt2 + dx2 = c2dt2

(1 +

1c2

(dx

dt

)2)

=c2

γ2dt2 → dτ =

1γdt

Using this result we can write dpdt = 1

γdpdτ . And so the Lorentz Force equation looks like:

dp

dτ= q

(γE +

γ

cv × B

)Further, note that the cross product term has the form γv = γ dxdt = dx

dτ . Pulling the factor c out ofthe equation,

128

dp

dτ=q

c

(γcE +

dx

dτ× B

)This can be simplified by defining a 4-velocity uµ = dxµ

dτ which gives it components:

uµ =(dx0

dτ,dx

dτ

)=(cdt

dτ,dx

dτ

)= (cγ, u)

considering only the ith component,

dpi

dτ=q

c

(u0Ei + εijkujBk

)=q

c

(u0F i0 + ujF

ij)

dpi

dτ=q

cF iνuν

This three dimensional vector can be defined as a 4-vector if we add a zero component. Considerthe result:

dpµ

dτ=q

cFµνuν

This defines the minimal coupling of a particle with the electromagnetic field. But what exactly isp0? Let’s evaluate it:

dp0

dτ=q

cF 0νuν = −q

cEiγvi =

qγ

cE · v

This is the variation of the energy normalized to c. So then, we can see that p0 = 1c |E|. This defines

the 4-momentum as:

pµ =(

1c|E|, p

)

47 The Lagrangian In Electrodynamics

47.0.1 Review of the Lagrangian in Classical Mechanics

Recall that in classical mechanics, the Lagrangian is some function L (qi(t), qi(t), t) which definesthe action of a particle S as:

S =∫ t2

t1

L (qi(t), qi(t), t) dt

129

The minimal variation of the action over the interval (t1, t2) is described by:

δS =∫ t2

t1

[∂L

∂qiδqi +

∂L

∂qiδqi

]dt =

∫ t2

t1

[∂L

∂qiδqi +

∂L

∂qi

d

dt(δqi)

]dt

δS =∫ t2

t1

δqidt

[∂L

∂qi+d

dt

∂L

∂qi

]= 0⇒ d

dt

∂L

∂qi=∂L

∂qi

47.1 The Action and Lagrangian For A Relativistic Free Particle

In order to have the Lagrangian be an invariant scalar, we’ll define it as

L = L

(xµ(τ),

dxµ

dτ, τ

)However, we could use some other parameter than τ . One possible choice is ds2 = gµνdx

νdxµ. Withthis we can write that the action is proportional to:

S ∝∫ds =

∫ √gµνdxµdxν

There are multiple other ways to expand this in terms of other differential variables. The two we’llconsider are:

S ∝∫dτ

√gµν

dxµ

dτ

dxν

dτ=∫dτ√gµν xµxν

S = −mc∫ s2

s1

ds

√gµν

dxµ

ds

dxν

ds

Wednesday - 2/18

130

Review of Group Theory In Physics: For any group, each element can be repre-

sented by some object G = eL where G is an element of the group and L is an elementof the Lie Algebra of the group. Consider the two examples from the homework:

• The Orthogonal Group O(n)The orthogonal group are all matrices which satisfy RRT = I. From the rep-resentation given above, it is easily seen that this requires the Lie Algebra tosatisfy:

eL“eL

”T= eLeL

T

= I → L = −LT

Therefore each matrix element of L can be given by lij = −lji. This means thatof the n2 elements of L, n of them are zero along the diagonal, and half of theremaining terms are determined by the above condition. Therefore, there aren2−n

2= n(n−1)

2free elements of the Lie Algebra

• The Special Unitary Group SU(n)The special unitary group are all matrices that satisfy R†R = 1 which have de-terminant equal to 1. The characteristics of L can be determined as before andthe total number of elements are then determined to be half of the total elementsminus 1 along the diagonal so that the determinant is 1. Therefore there are atotal of n2 − 1 free elements in L.

From the action derived last time, we can immediately see that the Lagrangian for a free relativisticparticle is given by:

L = −mc√gµν xµxν

Deriving the equations of motion from this:

d

ds

∂L

∂xµ=

∂L

∂xµ

The right hand side is zero since the given Lagrangian doesn’t depend on position. The remainingterm is:

∂L

∂xµ=

−mc2√gρσxρxσ

∂

∂xµ(gρσxρxσ)

∂

∂xµgρσx

ρxσ = gρσ

(∂xρ

∂xµxσ +

∂xσ

∂xνxρ)

= gρσ(δρµx

σ + δσν xρ)

= gµσxσ + gρσx

ρ = 2gµν xν

Plugging this result in and canceling the powers of 2 leaves,

∂L

∂xµ=−mcgµν xν√gρσxρxσ

Then, applying the derivative with s, the equation of motion is:

131

d

ds

gµνdxν

ds√gρσ

dxρ

dsdxσ

ds

= 0

this is a set of 4 equations (µ = 0, 1, 2, 3) which determine the exact location of the free particle inspacetime. Consider though, if we redefine the parameter s as some new value s′ = ks, then theabove equation doesn’t change. Further, note that c2dτ2 = gµνdx

µdxν . With this notation, if weredefine s = cτ , then the square root term is equal to 1. This leaves the expected result of:

d

dτgµν

dxν

dτ= 0→ d2xν

dτ2= 0

The 4-acceleration is zero for a free particle. That means that it maintains a constant velocity andtravels in a straight line (as expected).

47.2 The Lagrangian For A Charged Particle With Minimal Coupling

We can use the scalar Aρxρ as the additional term for a particle coupled to the electromagneticfield. With this assumption, the Lagrangian becomes:

L = −mc√gµν xµxν −

q

cAµxµ

We can find the equations of motion as before, but more terms are non-zero. Consider:

d

ds

∂L

∂xµ=

∂L

∂xµ→ d

ds

(− mcgµν x

ν√gµν xµxν

− q

cAµ

)= −q

c

∂Aρ

∂xµxρ

d

ds

(mcgµν x

ν√gµν xµxν

)+q

c

∂Aµ∂xρ

dxρ

ds=q

c

∂Aρ∂xµ

dxρ

ds

Making a change of parameter from s→ τ turns the square root into c and this gives:

md2xµdτ2

= −qc

dxρ

dτ

(∂Aµ∂xρ

− ∂Aρ

∂xµ

)= −q

c

∂xρ

∂τFρµ

d2xµ

dτ2=

q

mc

∂xρ∂τ

F ρµ

Friday - 2/20 For this interaction Lagrangian, it is possible to show that the Action integral forthe second term can be alternately written in the form:

132

S = Sfree + Sint

Sint =∫ s2

s1

dsAµdxµ

ds= −1

c

∫d4xjµAµ

this can be accomplished by writing the tensor jµ as:

jµ = q

∫dτ u′µ δ4

(x− x′(τ)

)

48 Hamiltonian Dynamics in Electromagnetism

Recall from classical mechanics that given some Lagrangian L = L (q, q, t), the Hamiltonian can bederived by:

H = piqi − L pi =∂L

∂xi

48.1 The Minimal Coupling Hamiltonian

So, using our Lagrangian for the relativistic particle with minimal coupling, we can define the tensorpµ = ∂L

∂xµ . Then, defining the Hamiltonian in tensor notation,

H = pµxµ − L

Plugging in our value for L,

pµ =∂L

∂xµ=−mcgµν xν√gνµxν xµ

− q

cAµ

H =−mcgµν xν xµ√

gνµxν xµ− q

cAµxµ −

[−mc

√gµν xµxν −

q

cAµxµ

]= 0

The Hamiltonian is zero since the system is constrained. That is, some of the elements aren’tindependent of the others.

48.2 Constraints

Consider the constraint:

C =(pµ +

q

cAµ

)(pµ +

q

cAµ)−m2c2 = 0

133

this constraint demonstrates how p0 and p aren’t independent of one another since it can be solvedto get one in terms of the other. We can rewrite this in the case that the particle is free:

pfreeµ pµ free −m2c2 = 0

where we can make the change(pµ = pµ free + q

cAµ)

to recover the interacting case. From thisdefinition, it is easily shown that the above statement is true since:

pfreeµ pµ free −m2c2 =m2c2√g2µν x

ν2gνµx

ν −m2c2 = 0

48.3 Lagrangian Multipliers

Consider if we add a Lagrangian Multiplier l to the Action:

S =∫dsL =

∫ds (pµxµ −H)→

∫ds(pµx

µ −H − lH + pl l)

Here we can define the new Hamiltonian H ′ = H−lH and the equations of motion for the coordinatel are:

l =∂H ′

∂pl= 0→ l = constant

pl = −∂H′

∂l= −H = 0→ H = 0

With this result, we can make the choice for Hamiltonian:

H =√(

pµ +q

cAµ

)(pµ +

q

cAµ)−m2c2

This generates the equations of motion

xµ =∂H

∂pµpµ = − ∂H

∂xµ

Which can be shown to give the resulting equation of motion from last class,

d2xµ

dτ2=

q

mc

∂xρ∂τ

F ρµ

134

48.4 Results of Hamiltonian Analysis

There are several results regarding the momentum tensor pµ which we can make somewhat quickly.Recall that we’ve shown that:

pfreeµ =−mc√gνµxν xµ

gνµxν

If we make the change of parameter from s→ τ , then the square root is equal to c and we are leftwith:

pfreeµ = −mgνµdxν

dτ→ pfreeµ −mdxµ

dτ

From this it is evident that:

• pν is a constant of the system. This can be seen since pµpµ −m2c2 = 0. So, all componentsof the tensor are conserved quantities.

• It can be shown that pµ transforms as:

pµ′ = Λµ νpν

and therefore we know that pµ is indeed a tensor.

• The relation pµpµ = m2c2 is valid in all systems. This requires some additional analysis.

Consider if we take γ = 1 + ε where ε is some very small quantity. In this case, the relationcan be written as:

pµ =(p0, p

)= (mcγ, mγv)

And with velocity much smaller than c, γ can be approximated as:

γ ≈ 1 +12v2

c2+ . . .

Plugging these in gives the results:

p0 = mc+12m

cv2 + . . . p = mv

(1 +

12v2

c2+ . . .

)≈ mv

The second result is straightforward in that it states the momentum vetoer p is the classicallyknown p = mv relation in the limit of small velocities. Additionally, the first term can quicklybe seen to represent the rest energy plus the kinetic energy short of a factor of c. Thus,

pµ =(

1cE, p

)=(mc2γ,mγv

)135

Lastly, consider the above relation pµpµ = m2c2. Plugging in the above results,

p02 − p2 = m2c2 → 1c2E2 = p2 +m2c2

E2 = c2p2 +m2c4

Which is the familiar result for energy of a relativistic particle. Thus we can see that inrelativistic mechanics, both momentum and energy are conserved for a system with no externalforces.Monday - 2/23

49 Solving Problems Using Tensor Notation

49.1 Relativistic Doppler Shift

Consider a particle emitting light of frequency ω′ in the frame O′. If an observer in a frameO observes the light, what kind of Doppler shift will occur according to the laws of specialrelativity? Consider the phase of the emitted light:

ϕ = ωt− k · x

This scalar quantity should be invariant under a Lorentz transform, therefore we should beable to write it as ϕ = kµx

µ. The tensor kµ is known as the wave vector and from the aboverelation we can see that it has components:

kµ =(ωc,−k

)kµ =

(ωc, k)

The transform of the wave vector is then given by:

k0′ = γ(k0 − β · k

)k′ = k +

γ − 1β2

(β · k

)β − γβk0

In terms of the vector k and the frequency, the transform is written:

ω′ = γ(ω − β · k

)k′ = k +

γ − 1β2

(β · k

)β − γ

cβω

Taking the dot product and using the angle between the observer and direction of the emittedlight as ϑ, the above relations can be reduced to:

136

β · k′ = β · k + (γ − 1) β · k − γ

cβ2ω = γ

(β · k − β2ω

c

)βk′ cosϑ′ = γβ

(k cosϑ− βω

c

)Using the fact that |k| = ω

c , this can be further written as:

ω′ cosϑ′ = γω (cosϑ− β)

Repeating this process with the cross product generates

ω′ sinϑ′ = ω sinϑ

which combined with the other result gives the relation between the emitted frequency andobserved frequency along with the emitted angle and observed angle.

ω′ = γω (1− β cosϑ)

tanϑ′ =sinϑ

cosϑ− β

Written in terms of the emitted and observed frequencies, this is

ωobsωemit

=

√1− v2

c2

11− v

c cosϑ

one interesting result of this is that for the relativistic Doppler Shift, even transverse lightis shifted by an amount depending only on how fast the frames are moving relative to oneanother.

49.2 Particle in an Uniform Magnetic Field

Consider the trajectory of a particle in a uniform magnetic field described by B = Be3. Thesystem is described by:

dpµ

dτ=e

cFµνuν

Where uµ = (cγ, γv), pµ = muµ, e is the charge of the particle, m is the mass of the par-ticle, and Fµν is the rank two tensor we’ve seen before. Since there is no electric field, thecomponents F 0i are zero and thus our equation above becomes:

dp0

dτ=e

cF 0νuν = 0→ p0 = constant

137

This result shows that the energy is constant in time. From this it is also clear that ddτmcγ = 0

and thus |v|2 and γ are constant as well. The remaining components are:

dpi

dτ=e

cF ijuj =

e

cεijkBkuj =

e

c

(u× B

)iReplacing dτ = 1

γdt and u = γv, the factors of γ cancel and leave:

dp

dt=e

cv × B =

d

dtmγv

dv

dt=

e

mγcv × B

This result can be written instead in terms of the frequency ωB as:

dv

dt= v × ωB

Then, if we separate the components of the velocity parallel and perpendicular to the magneticfield, we get that:

d

dt

(v‖ + v⊥

)=(v‖ + v⊥

)× ωB

The cross product with the component parallel to B is zero and this leaves only:

dv‖

dt+dv⊥dt

= v⊥ × ωB

Since the parallel component has no component on the right side, it must be zero, thus theparallel velocity is a constant of the motion. The remaining terms can be determined byassuming that the velocity has the form:

v = v1e1 + v2e2 + v‖e3

The above differential equation the reduces to the coupled equations:

dv1

dt= v2ωB

dv2

dt= −v1ωB

This has solutions of sines and cosines, but it is more informative to write it as:

v1 = A1eiωBt +B1e

−iωBt v2 = A2eiωBt +B2e

−iωBt

Plugging into the equations above yields the relations iA2 = −A1 and iB2 = B1.

v = A1eiωBt (e1 + ie2) +B1e

−iωt (e1 − ie2) + v‖e3

138

Lastly, requiring that the final result be real gives the condition A1 = B1, plugging in thisgives the velocity as

v = 2A1 [cos (ωBt) e1 − sin (ωBt) e2] + v‖e3

Integration gives the trajectory as:

x = x0 +2A1

ωB[sin (ωBt) e1 + cos (ωBt) e2] + v‖t e3

Wednesday - 2/25

50 The Lagrangian of the Electromagnetic Field

Recall that we’ve derived Maxwell’s Equations in tensor form as:

∂µFµν =

4πcjµ

Fµν = ∂µAν − ∂νAµ

We can write a Lagrangian for the fields and check that it is correct by showing that theresulting equations of motion are Maxwell’s equations for the fields. Consider the action ofthe field:

S =∫d4x£ =

∫dt

∫d3x£ =

∫dtL

Here we’ve defined the Lagrangian density £ and the Lagrangian L by L =∫d3x£. The

Lagrangian density can then be written as the field contribution and the matter contribution.Note that the matter term will contain the free particle and coupling components. Considerthe definitions:

£ = £field + £matter

£field = − 116π

FµνFµν £matter = −1cjµAµ + £free

where we can neglect the free particle term from our analysis. Using this definition, we cancheck that the resulting equations of motion are indeed the Maxwell Equations. Consider:

δS = 0→ δ

∫d4£ =

∫d4x

[∂£∂φ(i)

δφ(i) +∂£

∂(∂φ(i)

)δ (∂µφ(i))]

= 0

∫d4x

[∂£∂φ(i)

δφ(i) − ∂µ

(∂£

∂(∂φ(i)

)) δφ(i) + ∂µ

(∂£

∂(∂φ(i)

)δφ(i)

)]= 0

139

The third term is evaluated only on the boundaries of the region, which we require have zerovariation, so that term can be neglected. The remaining terms give the condition:

∂µ∂£

∂(∂µφ(i)

) =∂£∂φ(i)

Consider each side separately. The left hand side of the equation gives:

∂µ∂

∂ (∂µAν)

[− 1

16πF ρσFσρ −

1cjµAµ

]The second term is zero since it’s independent of ∂µAν . This leaves only the first term whichhas certain symmetries we can exploit to solve:

− 116π

∂

∂ (∂µAν)F ρσFσρ = −Fρσ

16π2

∂

∂ (∂µAν)F ρσ = −Fρσ

8π∂

∂ (∂µAν)(∂ρAσ − ∂σAρ) = −Fρσ

8π2 (gρµδσ ν)

∂µ∂£

∂ (∂µAµ)= − 1

4π∂µF

µν

And the right hand side is simple:

∂

∂Aν

(−1cjµA

µ

)= −1

cjµδ

µν = −1

cjν

therefore we’ve recovered Maxwell’s Equations,

14π∂µF

µν =

1cjν ⇒ ∂µF

µν =

4πcjν

50.1 Gauge Invariance

Let’s check if the action integral is gauge invariant. Consider the generic change of gaugedescribed by Aµ → A′µ + ∂µλ. Plugging this in, the field term of the Lagrangian densitydoesn’t change, but the coupling term becomes:

−1cjµAµ → −

1cjµ(A′µ − ∂µλ

)Plugging this into the integral, we get that these terms are:

S =∫. . .− 1

c

∫d4xjµA′µ +

1c

∫jµ∂µλ

And integrating by parts to separate the last term gives:

140

∫jµ∂µλ =

∫d4x [∂µ (jµλ)− λ∂µjµ]

These two resulting terms are zero under most conditions. The first term is zero as long asthe boundary of the region contains no currents and the second it zero by definition since itis the continuity condition. Thus for these definitions, the action integral is indeed Lorentzinvariant.

51 Massive Photons and the Proca Lagrangian Density

We have seen that in the Lorenz Gauge the resulting tensor equation for a source free regionis:

Aµ = 0

The solutions of this are waves that propagate at speed c and carry zero mass. Considerwhat would change in the Lagrangian if the field wasn’t massless. The resulting field equationwould look like:

( +m2

)φ = 0

This would require an additional term of order φ2 in the Lagrangian density, which leads tothe proposed Proca Lagrangian Density of:

£field = − 116π

FµνFµν +µ2

8πAµA

µ

the factor µ is proportional to the mass of the field and has units of inverse length. Theresulting modified Maxwell Equations are then:

∂µFµν + µ2Aν =

4πcjν

Friday - 2/27

51.1 The Proca LagrangianA Phenomenological Description of Superconductivity

Recall the Proca Lagrangian Density which we wrote out last class,

£Proca = − 116π

FµνFµν +µ2

8πAµA

µ − 1cjµA

µ

From this we get the equation of motion

141

∂µFνµ + µ2Aµ =

4πcjµ

where µ has units of inverse length and can be written as µ = mγc~ . If we substitute in the

definition F νµ = ∂νAµ − ∂µAν then the equation of motion becomes:

Aν − ∂ν∂µAµ + µ2Aν =4πcjν

By applying ∂ν to both sides of this equation we get the result,

∂νAν −∂µA

µ + µ2∂νAν =

4πc∂νj

ν

µ2∂νAν =

4πc∂νj

ν

The term ∂νjν = 0 by continuity of charge, and therefore we have the result that ∂νAν = 0.

This leaves the equation of motion:

Aν + µ2Aν =4πcjν

If we assume the system is static, then → −∇2 and the equation of motion becomes:

∇2Aν − µ2Aν = −4πcjν

If we further consider only the electrostatic case, the only non zero component of Aµ is A0 = φ.This gives the equation for the potential as:

∇2φ− µ2φ = −4πρ

Consider if the charge distribution is a single point charge. In the case that there is noadditional term, the solution of this equation took the form φ = q

r . However, the additionalterm requires using the Greens function of the Helmholtz Operator:

(∇2 − µ2

)φ = −4πqδ(3) (x− x0)⇒ φ =

q

re−µr

For reference, the Greens Function of the Helmholtz Operator is given as:

G(x, x′) =1

|x− x′|e−k|x−x

′|

This resulting potential is exponentially damped and acts only at short range. In particlephysics, it is known as the Yukawa Potential. At this time, one can make several assumptionsabout the system and use this result as a method to phenomenologically treat superconductors.Consider the postulates:

142

– The current on the surface of a superconductive material can be described by:

j = nQv

where n is the particle density, Q is the charge of the individual particles, and v is thevelocity of the particles.

– The coupling of the particles to the electromagnetic field is written as:

pµ → pµ +Q

cAµ

Separating the spacial and components we get the change in energy and momentum as:

E → E +Q

cφ p→ p− Q

cA

combining this with the above gives the current on the surface of the superconductor:

j =nQ

mQcp− Q2n

mQcA

We can make the assumption that in the superconductor the particles carry no individualmomentum, which leaves the result for the current on the surface as:

j = −Q2n

mQcA

This result is similar to Ohm’s Law (j = σE).

Taking the curl of both sides of this equation gives the current in terms of the magnetic fieldon the surface,

∇× j = −Q2n

mQcB

The primary results of this treatment can be found by returning to the original MaxwellEquations. Consider first:

Aµ =4πcjµ

And taking only the spatial part,

A =4πc

(− nQ

2

mQc

)A→

( +

4πnQ2

mQc2

)A = 0

Similarly, using the relation ∇× B = 4πc j = −4πnQ2

mQc2A and taking the curl of both sides:

143

∇×∇× B = −4πnQ2

mQc2∇× A

∇(∇ · B

)−∇2B = −4πnQ2

mQc2B

But by definition ∇ · B = 0, which leaves the equation for B on the surface of the supercon-ductor:

∇2B =4πnQ2

mQc2B

The solutions of this are exponentials with coefficients mQc2

4πnQ2 , this gives the penetration depthof the magnetic field into the structure of the superconductor.

52 Symmetries and Invariant Equations of Motion

Consider two generic conditions which lead to a change in the Lagrangian of a system butleave the equations of motion unchanged.

– ScalingIt is easily shown that a change of L → kL leaves the equations of motion unchangedsince: [

d

dt

∂(kL)∂q

=∂(kL)∂q

]=[d

dt

∂L

∂q=∂L

∂q

]– Addition of a function

Consider the change of Lagrangian given by L→ L+ ddtF (q(t), t). It can be shown that

the equations of motion remain unchanged in this case as well.

The results of this analysis is that certain symmetries cause invariances in the equations ofmotion. An example of this is the Lagrangian for a free particle.

L =1

2m|q|2

The exact direction of ¯q is independent of the Lagrangian, therefore all vectors which are radiiof the sphere in velocity space with magnitude matching the above are solutions. This leadsto an invariance in the rotation of the system about the origin.

Monday - 3/2

144

52.1 More on Symmetries

We are considering the consequences of having symmetries and invariances in the Lagrangian Den-sity, Equations of Motion, and Actin Integral. Consider a system described by:

L (q(t), q(t), t) ; S =∫dtL

If both q(t) and q′(t) are solutions of the equations of motion, then what can be determined aboutthe system? Rewriting the action as an action density dS = Ldt, we can describe the symmetriesas:

L(q′(t′), q′(t′), t′

)dt′ = dt

L (q(t), q(t), t) +

d

dtf (q(t), t)

Note that the change in variable has only effected the coordinates qi and not the Lagrangian functionL. To proceed, it is sufficient to require that the transforms be continuous. That is, we want changesthat can be written as:

q(t)→ q′(t′) = q′(t, ε) t→ t′ = t′(t, ε)

If we assume that ε is infinitesimal, then these transitions can be written as:

q′(t) = q(t) + δq = q(t) + εδq t′ = t+ δt = t+ εδt

Where we’ve denoted (. . .) as a finite quantity. We can then write the Action Density to find theunderlying symmetries δdS = d

dtf(q(t), t)dt. Beginning from the Action Density,

d (δS) = δ (Ldt) = δ (dt)L+ dtδL

Consider a few terms of this expansion separately,

dtδL = dt

(∂L

∂qδq +

∂L

∂qδq +

∂L

∂tδt

)

dt′

dt=

d

dt(t+ δt) = 1 +

d

dtδt→ δ (dt) = dt′ − dt = dt

(dt′

dt− 1)

= dtd

dtδt

Then, we can combine δtdLdt + L ddtδt = d

dt (Lδt) and find:

δ (dS) =df

dtdt =

∂L

∂q+∂L

∂qδq +

∂

∂t(Lδt)

dt

145

At this point it is useful to reorder δ ∂∂tq, this gives:

df

dt=∂L

∂q+d

dt

(∂L

∂qδq

)− d

dt

(∂L

∂q

)δq +

∂

∂t(Lδt)

The first and third terms cancel if the function q(t) is a solution of the equations of motion. Thisleaves that:

d

dt

[∂L

∂qδq + Lδt− f(q(t), t)

]= 0

This result is known as Noether’s Theorem and can be generalized to a system with n variationswhich leave the equations of motion invariant. For each variation εi and coordinate qj(t) there isan underlying result of

∂L

∂qjδqj

i + Lδti − f i(q(t), t) = constant

52.1.1 Example - Time Translations

Consider a transform of t→ t′ = t+ ε. The equations of motion q(t) transform into new functionsq′(t′) and it can be quickly shown that:

q(t)→ q′(t′) = q′(t+ ε) = q(t)

q′(t) = q(t− ε) = q(t)− εδq = q(t)− εq → q = −δq

this gives the constant of the motion as:

∂L

∂q(−q) + L (1)− f i(q(t), t) = constant

−∂L∂qq + L = −H = constant

Here we see that the conserved quantity is the negative of the Hamiltonian of the system. Note theimportant assumption we’ve made that in the case of ∂L

∂t = 0, then f(q(t), t) = 0.

52.2 Generalize The Result for Fields - The Stress Energy Tensor

Consider the case of the action for a field. The results above can be applied and one finds that forthe action defined as S =

∫d4x£(φ, ∂µφ, xµ), the Lagrangian density is invariant under transforms

of the form:

146

£→ £′ = k£ + ∂νfν

The transformations can be written as:

φ(x)→ φ′(x′) = φ(x) + δφ xµ → x′µ = xµ + δxµ

Repeating the above process, one finds:

∂µjµ = 0 jµ =

∂£∂ (∂µφi)

δφi + £δxµ − fµ

Using the above definitions for the transforms, it is possible to write the variations as:

δ (xµ) = ενδµ ν δ (φ) = −εµ∂µφ

Plugging these into the above result, we find some new rank 2 tensor Tµν :

Tµν =∂£

∂ (∂µφ)(∂νφ) + £gµν

This result in known as the Stress Energy Tensor and we’ll look more closely at it next class.Consider first the time derivative term ν = 0, This term must contain the energy of the field:

Tµ0 =∂£

∂ (∂µφ)(∂0φ

)+ £gµ0

Using the fact that ∂µTµν = −∇ · T i it is clear that T 00 is a constant of the system and can bewritten as:

T 00 =∂£

∂ (∂0φ)(∂0φ

)+ £g00 =

∂L

∂φiφi −£ = H = constant

Wednesday - 3/4

52.3 Deriving the Components of the Stress Energy Tensor for ElectromagneticFields

Last times we found that the Stress Energy Tensor can be written as:

Tµν =∂£

∂ (∂µφ)(∂νφ) + £gµν

147

Recall that the Lagrangian Density for E & M fields is:

£ = − 116π

FνµFνµ; Fµν = ∂µAν − ∂νAµ

From this, we can quickly find that:

∂

∂ (∂σAτ )£ = − 1

16π· 2Fαβ · gασgβτ = − 1

4πF στ

Tµν = − 14πFµγ∂νAγ +

116π

gνµFαβFαβ

It can be shown that the Stress Energy Tensor satisfied ∂µTµν = 0 given that the source free field

equations are also satisfied (∂µFµν = 0). However, we desire to find a symmetric Stress EnergyTensor. This can be done by adding a divergentless quantity (that is, any term which will leave theresulting field equations unchanged). Consider making the change of:

T ′µν = Tµν − 14π∂λ

(F λµAν

)Then, expanding the new term,

T ′µν = − 14πFµγ∂νAγ +

116π

gνµFαβFαβ − 1

4π

[Aν(∂λF

λµ)

+ F λν (∂λAν)]

The term ∂λFλµ is zero and the remaining term can be combined with the first term as:

T ′µν =1

4π

[Fµλ (∂νAλ − ∂λAν)

]+

116π

gµνFαβFαβ

And finally, removing the prime since this is our new Stress Energy Tensor, this gives:

Tµν =1

4πFµγFγ

ν +1

16πgµνFαβF

αβ

Let’s consider the components:

• T 00

Consider the element ν = µ = 0. Plugging in these values to the above gives:

T 00 =1

4πF 0γFγ

0 +1

16πg00FαβF

αβ =1

4π

((−Ei

) (−Ei

)+

14

(1)[−2(|E|2 − |B|2

)])T 00 =

18π(|E|2 + |B|2

)148

The resulting term is the energy density w which we derived in the previous semester withthe corrected units for our current analysis.

• T 0i

Consider the elements µ = 0, ν = 1, 2, 3. Again plugging in values,

T 0i = − 14πF 0γFγ

i + 0 =1

4π(−Ei

) (−εjikBk

)=

14πεijkEjBk

T 0i =1

4π(E × B

)iThese values are the components of the Poynting vector (which we can define here as T =(T 01, T 02, T 03

)) we saw last semester. It can be shown that the condition ∂µTµν = 0 actually

gives the energy conservation equation:

∂0T00 +

3∑i=1

∂iTi0 = 0→ 1

c

∂

∂tw +∇ · T = 0

Integrating both sides over all space gives:

1c

∂

∂tW −

∫d3x∇ · T = 0

Which we wrote last semester as 1c∂W∂t +∇ · s = 0.

53 Finding a General Solution to the Maxwell Equations

Now that we have the basics of tensors in electromagnetism finished, we can derive a generalsolution to the Maxwell Equations using this notation. Recall that in the Lorentz Gauge, the tensorequations reduce to:

∂µ = Fµν =4πcjν −→ Aν =

4πcjν

So, in order to get a general solution, we need only find the Greens Function of the operator.The general solution would then be written as:

Aν = AνHom +4πc

∫d4x′D

(x− x′

)jν(x′); AνHom = 0; D(x, x′) = δ(4)

(x− x′

)We can write the Greens Function as a Fourier Transform in 4 dimensions and note that the deltafunction can be written similarly as:

δ(x0 − x′0

)δ(3)

(x− x′

)=

12π

∫dk0e

ik0(x0−x′0) 1(2π)3

∫d3keik·(x−x

′)

149

Then, we can write the Greens function as:

D(x, x′) =1

(2π)4

∫d4kD(kµ)e−ikµ(xµ−x′µ)

If we operate on this with = ∂ν∂ν , then we can determine D(kµ),

D(x, x′) =1

(2π)4∂ν∂ν

∫d4kD(kµ)e−ikµ(xµ−x′µ)

The left hand side is merely the delta functions defined about, while the right hand side can beevaluated by moving the operator inside the integral:

∂ν∂νe−ikµ(xµ−x′µ) = (−kνkν) e−ikµ(xµ−x′µ)

⇒ D(k) = − 1kνkν

And so,

D(x, x′) = − 1(2π)4

∫d4k

1kνkν

e−ikµ(xµ−x′µ)

Friday - 3/6To solve the problematic singularity at kνkν = 0 in the Greens function, consider separating theterms and integrating over k0 first:

D(x, x′) = − 1(2π)4

∫d3ke−ik·(x−x

′)

∫dk0 1

k02 − |k|2e−ik

0(x0−x′0)

We can evaluate the integral by Residue theory by using the expansion at each pole. Considerintegrating along the path given by the real part of k0 except at the points k0 = ±|k| where wecover a small half circle into the positive imaginary part of the complex plain. The path can becomplete by looping over the upper or lower complex plane. The choice of upper or lower dependson the sign of x0−x′0. If x0−x′0 < 0, then we close the loop on the upper half plane. If x0−x′0 > 0,then we close on the lower half plane. Note that our choice of avoiding the poles by extending thepath into the upper half plane causes the upper plane loop to contain no singularities and thereforethe integral is zero for that loop. This leaves:

D(x, x′) = − i

(2π)3 Θ(x0 − x′0

) ∫d3ke−ik·(x−x

′)Res

[∫dk0 1

k02 − |k|2e−ik

0(x0−x′0), k0 = ±|k|]

The residue gives a contribution of:

150

Res

[∫dk0 1

k02 − |k|2e−ik

0(x0−x′0), k0 = ±|k|]

=1

2|k|

(e−i|k|(x

0−x′0) − ei|k|(x0−x′0))

=i

|k|sin(|k|(x0 − x′0

))And this leaves the Greens Function:

D(x, x′) = − 1(2π)3 Θ

(x0 − x′0

) ∫d3k

1|k|

sin(|k|(x0 − x′0

))e−ik·(x−x

′)

If we take the plane defined by k and x − x′ as a reference, then we can change to sphericalcoordinates:

d3k → d|k||k|2dΩ k ·(x− x′

)→ |k||x− x′| cosϑ

The result is independent of ϕ and therefore that integral can be evaluated. Further the integralover d (cosϑ) can be evaluated quickly:

∫ 1

−1d (cosϑ) ei|k||x−x

′| cosϑ =1

i|k||x− x′|

[ei|k||x−x

′| − e−i|k||x−x′|]

Writing the result as a sine function, the Greens Function is:

D(x, x′) =Θ(x0 − x′0

)2π2

1|x− x′|

∫ ∞0

d|k| sin(|k|(x0 − x′0

))sin(|k|(x− x′

))At this point, it is useful to write the two trigonometric functions as exponentials and note thatthey expand to 4 terms which can be rewritten as two terms by changing in range of integration,

D(x, x′) =Θ(x0 − x′0

)2π2

1|x− x′|

∫ ∞−∞

d|k|(−1

4

)[ei|k|[(x

0−x′0)+|x−x′|] − ei|k|[(x0−x′0)−|x−x′|]]

But these integrals can be written as delta functions of the arguments,

∫ ∞−∞

d|k|ei|k|[(x0−x′0)+|x−x′|] = 2πδ[(x0 − x′0

)+ |x− x′|

]∫ ∞−∞

d|k|ei|k|[(x0−x′0)−|x−x′|] = 2πδ[(x0 − x′0

)− |x− x′|

]The first of these two results is always zero since both quantities inside the delta must always bepositive. This leaves the Greens Function:

151

D(x, x′) =1

4πΘ(x0 − x′0

)|x− x′|

δ[(x0 − x′0

)− |x− x′|

]Recall the property of the delta function:

δ(x2 − a2

)=

12|a|

[δ (x+ a) + δ (x− a)]

The two terms here are actually equivalent to the above two (the first of which is zero). From this,we can write the final result of this analysis, the “Retarded Greens Function” for the wave operator:

D(x, x′) =1

2πΘ(x0 − x′0

)δ[(x0 − x′0

)2 − |x− x′|2]Consider some characteristics of this result. Firstly, the step function causes the observer at x0 toonly see an electromagnetic field after the source has emitted the wave. Recalling that x0 = ct, thismakes sense. The observer shouldn’t see the field until the source has caused it to exist. The deltafunction causes no waves to be seen until:

c(t− t′

)= |x− x′| → c =

|x− x′|t− t′

This requires that all electromagnetic waves travel at the constant speed c. Again, this is somethingwe expect and understand.

152

Part V

SPECIAL CASES

54 Plane Waves

Monday - 3/9

54.1 Plane Wave Solutions Of Maxwell’s Equations

Recall the general solution to Maxwell’s Equations which we’ve derived recently.

Aµ = Aµhom +4πc

∫d4xD(x− x′)jµ(x′)

Consider the case of a source free problem, the solution reduces to only the homogeneous solution.We can write this as a Fourier expansion:

Aµ = 0→ Aµ =∫d4kcµ(k)e−ikµx

mu

Applying the wave operator on this form gives some additional conditions:

∫d4kcµ(k)e−ikµx

mu =∫d4kcµ(k)kµkµe−ikµx

mu = 0

The trivial solution cµ(k) = 0 can be ignored. Therefore, we require kµkµ = k02 − |k|2 = 0. Thiscan be written as the condition:

ω2 = c2|k|2

We further require that our potentials satisfy the Lorentz condition ∂µAµ = 0. This requires:

∫d4kcµ(k) (−kµ) e−ikµa

µ= 0→ cµ(k)km = 0 c0ω

c= c · k

Given this form for the potentials, it is possible to determine the form of the electric and magneticfields. Consider the magnetic field:

B = ∇× A = ∇×∫d4k c(k)e−ikµx

µ=∫d4k

(∇× c(k)e−ikµx

µ)

=

=∫d4k

(∇e−ikµxµ × c(k) + e−ikµx

µ∇× c(k))

153

B = i

∫d4k e−i(ωt−k·x)

(k × c

)And the electric field:

E = −∇φ− 1c

∂

∂tA = −∇

∫d4k c0(k)e−i(ωt−k·x) − 1

c

∂

∂t

∫d4k c(k)e−i(ωt−k·x)

=∫d4k c0(k)

(ik)e−i(ωt−k·x) − 1

c

∫d4k c(k)

(−ik0c

)e−i(ωt−k·x)

E = −i∫d4k e−i(ωt−k·x)

(c0k − ω

cc)

It can be shown that these result satisfy the Maxwell Equations. Consider though, the alternatedefinition for the electric field given by:

E = −i∫d4k

c

ωk ×

(k × c

)e−i(ωt−k·x)

It is straight forward to show that this is indeed the electric field derived above by noting the vectorrelation:

k ×(k × c

)=(k · c

)k −

(k · k

)c = c0k0k − |k|2c = k0

(c0k − k0c

)From this, it is evident that E and B can be written in terms of c and therefore the two fields arenot independent of one another.

54.1.1 Special Case - Monochromatic Waves

Consider what the above solution becomes for the case of monochromatic waves. The vector cbecomes:

c(k, ω) = G(k, ω)δ (ω − ωc) δ(3)(k − kc

)Given this assumption, the integrals collapse in the definitions of the fields and one finds:

B = i(kc × G

)e−i(ωct−k·x)

E = −iωc

[kc ×

(kc × G

)]e−i(ωct−k·x)

Thus, the directionality of E and B are orthogonal to one another and to the vector kc. If we definea few of these terms, these fields become simpler to deal with:

154

ikc × G ≡ Bcε2 iω

ckc ×

(kc × G

)≡ Ecε1 kc ≡ |kc|ε3

Then, the electric and magnetic fields are simply:

E = Ece−i(ωct−k·x)ε1 B = Bce

−i(ωct−k·x)ε2

Taking the first of the above definitions and crossing it with kc gives:

ikc ×(kc × G

)= Bckc × ε2

Plugging in the other definitions,

−Ecω

cε1 = Bc|kc|ε3 × ε2 = Bc|kc| (−ε1)

ω

cEc = Bc|kc| → Ec = Bc

Plugging these back into the above, it is possible to derive the relation:

Bc =c

ωckc × Ec

Finally, the general solution can be written as:

E = Ece−i(ωct−kc·x) B = Bce

−i(ωct−kc·x)

Where Ec and Bc are related as defined above. It is important to note that these solutions are onlydependent on two parameters. Consider propagation described by:

Ec = Ec1ε1 + Ec2ε2 → Bc = Bc1ε1 +Bc2ε2 = −Ec2ε1 + Ec1ε2

The only independent parameters are Ec1 and Ec2. The physical solution is the real part of thisresult. Note that the coefficients are complex and therefore can be written as:

Ecn = Reiϑ

The additional phase ϑ can add a phase to the propagating plane waves which is dependent on thevalue of ϑ. For a generic one, the induced change is known as “elliptical polarization”.Wednesday - 3/11

155

54.2 Polarization of Plane Waves

Last time, we derived a relation between the two coefficients of the electric field in a plane wave.

E = Ec1

[ε1e−i(ωt−k·x) + ε2Re

−i(ωt−k·x−ϑ)]

The physical solution is written as the real part of the above function. That is,

Ephy = Ec1[ε1 cos

(k · x− ωt

)+ ε2R cos

(k · x− ωt+ ϑ

)]Consider the behavior of this function as x and t approach zero. The resulting equation gives:

Ephy(x = 0, t = 0) = Ec1 [ε1 + ε2R cosϑ]

At this point in spacetime, the wave is directed along a line that traces out an elliptical path in theε1, ε2 plane. The path that this line covers is described by the polarization of the electromagneticwave. There are several special cases of these polarizations:

• Circular PoliarizationConsider the case of R = 1 and ϑ = π

2 . The resulting equation can be described as:

Ec = Ec1 (ε1 + iε2)

Bc = −iEc1 (ε1 − iε2)

which allows the definition of ε± ≡ ε1 ± iε. This gives the final result of:

Ex = Ec1 cos(ωt− k · x

)Ey = Ec1 sin

(ωt− k · x

)These results trace out a circle in the plane.

• Linear PolarizationConsider the case of R = 1 and ϑ = 0. In this case only the single polarization occurs andthe resulting wave is said to be “linearly polarized” since its orientation doesn’t change in theε1, ε2 plane.

E = Ec1 cos(k · x− ωt

)(ε1 + ε2)

156

54.3 Plane Waves In Media

Recall Maxwell’s Equations in a medium (without sources):

∇ · B = 0 ∇ · D = 0

∇× E = −1c

∂B

∂t∇× H =

1c

∂D

∂t

If we plug in the relations D = εE and H = µB, then the above can be written as:

∇ · B = 0 ∇ · E = 0

∇× E = −1c

∂B

∂t∇× B =

µε

c

∂E

∂t

If we define the speed of propagation in the medium as v = c√µε then these become:

∇ · B = 0 ∇ · E = 0

∇× E = − 1v√µε

∂B

∂t∇× B =

√µε

v

∂E

∂t

Lastly, if we define the new fields ˜E =√εE ˜B = 1√

µB. Then we can write:

∇ · ˜B = 0 ∇ · ˜E = 0

∇× ˜E = −1v

∂ ˜B∂t

∇× ˜B =1v

∂ ˜B∂t

This result shows that all results from the vacuum case can be extended to the case of propagationin media. It is only necessary to change:

c→ v =c√µε

E → ˜E =√εE

B → ˜B =1√µB

Note that this change also requires adjustments to the other condition equations. The new condi-tions require:

[B =

c

ωk × E

]→ B =

√εµn× E with ω =

|k|c√εµ

157

54.4 Energy Contained in Plane Waves

It is possible using the Poynting vector to derive how much energy is contained in an electromagneticplane wave. Consider the time-averaged Poynting vector:

S = Re( c

8πE × B∗

)Plugging in the above result for B,

S = Re( c

8πE ×

(n× E

))=

S =c

8π|E|2n

This result is the flux of energy per unit area per unit time in the wave. Consider deriving theenergy density:

energy density =energy

volume=

energy

area× c× time=flux

c→ u =

18π|E|2

Note that in a medium, the Poynting vector and resulting energy density is:

S =c

8π

√ε

µ|E|2n u =

ε

8π|E|2

158

Friday - 3/13

54.5 Plane Waves Incident on a Boundary Between Media

Consider a plane wave incident on a boundary between materials. Let the boundary exist onthe plane z = 0 and an incident wave with vector k directed towards the origin. The incidentelectromagnetic wave can be described by:

E = E0e−i(ωt−k·x) B =

√εµnk × E0e

−i(ωt−k·x)

Using these definitions, all of our analysis can be done with only the electric field. Consider thefield in the upper and lower half plane:

E(1) = E + E′ E(2) = E′′

Where we’ve denoted E′ as the reflected wave and E′′ as the transmitted/refracted wave. Thesegive:

E = E0e−i(ωt−k·x) Incident

E′ = E′0e−i(ω′t−k′·x) Reflected

E′′ = E′′0e−i(ω′′t−k′′·x) Refracted

Consider the known boundary conditions on regions with changes in media. The normal displace-ment field must be continuous across the region. This gives:

(ε1E

(1))⊥

=(ε2E

(2))⊥

ε1

[E0⊥e

−i(ωt−k·x) + E′0⊥e−i(ω′t−k′·x)

]= ε2E

′′0⊥e−i(ω′′t−k′′·x)

Consider this result at the origin ( x = 0),

ε1

[E0⊥e

−iωt + E′0⊥e−iω′t

]= ε2E

′′0⊥e−iω′′t

Multiplying this by its complex conjugate gives a condition of:

ε21

[E0⊥e

−iωt + E′0⊥e−iω′t

] [E∗0⊥e

iωt + E∗′0⊥eiω′t]

= ε22E′′0⊥e−iω′′tE∗′′0⊥e

iω′′t

ε21

[|E0⊥|2 + |E′0⊥|2 + 2Re

(E0⊥E

′0⊥e

i(ω−ω′)t)]

= ε22|E′′0⊥|2

159

This condition requires ω′ = ω. Further, consider the initial result and note that this requires thatω′′ = ω as well. Repeating this process for the system at time t = 0 gives a similar result for thewave vectors. These give;

ω = ω′ = ω′′;(k · x

)=(k′ · x

)=(k′′ · x

)This demonstrates how the frequency of light passing through media doesn’t change. Plugging theseinto the above definitions give:

E = E0e−i(ωt−k·x) Incident

E′ = E′0e−i(ωt−k·x) Reflected

E′′ = E′′0e−i(ωt−k·x) Refracted

Consider some incident electric field with amplitude |E0| incident on such a boundary. How can onedetermine the amount of energy transmitted through and reflected back? Consider the remainingboundary conditions given some additional geometrical information. Define the angle the incidentlight makes with the normal of the boundary to be ϑi, the angle of the refracted light to be ϑr andthe angle of the reflected light to be ϑR. These give the above dot products as:

k · x = kx cos(

3π2

+ ϑi

)= kx sinϑi

k′ · x = k′x cos(π

2− ϑR

)= k′x sinϑR

k′′ · x = k′′x cos(

3π2

+ ϑr

)= k′′x sinϑr

Recall that in each medium, ki = ωc

√εiµi. Therefore the first two of the above give:

kx sinϑi = k′x sinϑR → ϑi = ϑR

That is, the incident and reflected angles are equal. Additionally, for the incident and refractedwaves

kx sinϑi = k′′ sinϑr

Using the notation for the refractive index n = cv =√εµ, the result is Snell’s Law:

n1 sinϑi = n2 sinϑr

160

In the general case, what do the four boundary conditions give in terms of E0 on either side of theboundary region? Consider plugging in the definitions above to the 4 boundary conditions:

∆D⊥ = 0→[ε1(E0 + E′0

)− ε2E′′0

]· n = 0

∆E‖ = 0→[E0 + E′0 − E′′0

]× n = 0

∆B⊥ = 0→[(k × E0

)+(k′ × E′0

)−(k′′ × E′′0

)]· n = 0

∆H‖ = 0→[

1µ1

((k × E0

)+(k′ × E′0

))− 1µ2

(k′′ × E′′0

)]× n = 0

Any wave can be written as a linear position of the two cases of E0 parallel or perpendicular to theplane of incidence. Therefore, to determine the energy of the refracted and reflected waves, thesetwo special cases must be done separately and each will give different results of the above. Considerfirst the case of E0 perpendicular to the plane. This turns the above into:

0 = 0

|E0|+ |E′0| − |E′′0 | = 0

kE0 sinϑi + k′E′0 sinϑR − k′′E′′0 sinϑr = 0

1µ1

(kE0 cosϑi − k′E′0 cosϑR

)− 1µ2k′′E′′0 cosϑr = 0

Using the previous result that ϑi = ϑR and the definition for k in each medium, one finds:

√ε1µ1

(E0 − E′0

)cosϑi =

√ε2µ2E′′0 cosϑr

E0 + E′0 = E′′0

Which can be solved to give the ratio between the incident and reflected and refracted waves,

E′′0E0

=2n1 cosϑi

n1 cosϑi + n2µ1

µ2cosϑr

= Transmitted F ield

E′0E0

=n1 cosϑi − n2

µ1

µ2cosϑr

n1 cosϑi + n2µ1

µ2cosϑr

= Reflected F ield

For the components parallel to the plane of incidence, the boundary conditions require that

(E0 − E′0

)cosϑi = E′′0 cosϑr√

ε1µ1

(E0 + E′0

)=√ε2µ2E′′0

161

Which generates the conditions:

E′′0E0

=2n1n2 cosϑi

µ1

µ2n2

2 cosϑi + n1

√n2

2 − n21 sin2 ϑi

= Transmitted F ield

E′0E0

=µ1

µ2n2

2 cosϑi − n1

√n2

2 − n21 sin2 ϑi

µ1

µ2n2

2 cosϑi + n1

√n2

2 − n21 sin2 ϑi

= Reflected F ield

Monday - 3/23

54.6 Plane Wave Phenomena

Given all of this analysis of plane waves, it is useful to examine several special cases and interestingphenomena. Consider some special cases of the above relations.

54.6.1 The Brewster Angle

Consider the above solutions for the reflected and refracted intensity of the plane waves if there isno reflected wave parallel to the incident plane. This can be found by requiring:

E′0E0

= 0→ µ1

µ2n2

2 cosϑi − n1

√n2

2 − n12 sin2 ϑi = 0

If we consider that in most cases µ1 ≈ µ2, then we can square both sides and write the conditionas:

n22 cos2 ϑi = n1

2(n2

2 − n12 sin2 ϑi

)→ tanϑi =

n2

n1

Any incident waves on the boundary with incoming angle ϑi will contain no reflected componentparallel to the plane of incidence. This result is useful for photographers and optical sciences toeliminate glare from a surface or to generate uniformly polarized electromagnetic waves from asurface.

54.6.2 Total Internal Reflection

Consider the known result from Snell’s Law:

n1 sinϑi = n2 sinϑr → ϑr = arcsin(n1

n2sinϑi

)

162

For the case that the quantity inside the arcsin function is greater than unity, the refracted angleis undefined. Let’s consider what is actually going on. If the above condition is squared, we finds:

sin2 ϑr =n1

2

n22

sin2 ϑi = 1− cos2 ϑr cosϑr =

√1− n1

2

n22

sin2 ϑi

If the transition angle is defined as ϑi0 = arcsin(n2n1

), then this condition can be more simply stated:

cosϑr =

√1− n1

2

n22

sin2 ϑi =

√1− sin2 ϑi

sin2 ϑi0

Consider the resulting transmitted wave,

E′′ = E′′0e−iωt+ik′′·x

If we consider the penetration direction to be ε3, then the resulting wave is described as:

E′′ = E′′0e−iωt+i(k′′y y+k′′z z) = E′′0e

−iωt+i(k sinϑry+k cosϑrz)

Then plugging in the above result for cosϑr,

E′′ = E′′0e−iωt+ik sinϑry e

ikz

r1− sin2 ϑi

sin2 ϑi0

For any incident angle such that sin2 ϑi ≥ sin2 ϑi0 , then the above becomes:

E′′ = E′′0e−iωt+ik sinϑry e−

|z|δ δ−1 ≡ k

√sin2 ϑi

sin2 ϑi0− 1

This explains what actually occurs during total internal reflection. The refracting wave decaysexponentially on a scale determined by δ. Such a wave is referred to as an “evanescent” wave.Consider the flux of energy through the boundary, this is determined by taking the component ofthe Poynting Vector along the surface,

S · n =c

8πRe[E′′ × H ′′∗

]· n

and plugging in:

H =1µ2

; B =√ε2µ2nk′′ × E′′ → H ′′ =

√ε2µ2nk′′ × E′′

163

S · n =c

8π

√ε2µ2|E′′|2Re [n · nk′′ ] =

c

8π

√ε2µ2|E′′|2 cosϑr

Recall though, that the value of cosϑr is pure imaginary for angles larger than the critical anglefor total internal reflection. Therefore there is no energy flux into the second medium in this case.Two interesting methods of measuring these evanescent waves have been used. In a method similarto quantum tunneling, if the boundary layer containing n2 is of thickness smaller than δ, then thedecaying wave will penetrate and some signal can be measured on the other side of the boundary.Alternately, the Goos-Hanchen Effect has been observed where reflections from depths up to δ areobserved by interference patterns in the reflected signal.

54.7 Wave Packets - Introduction

Recall that we have determined the most general solution for Maxwell’s Equations as:

Aµ =∫d4kDµ

(k)eikµx

µ

In this case, consider the spherical coordinates solution:

Aµ =1c

∫dωk2dkdΩDµ (Ω, k) eiωt+ik·x

In addition to this definition, we have the relations from Maxwell’s Equations (kµkµ = 0) and thechosen gauge (Dµkµ = 0). These give the additional definitions:

k02= |k|2 =

ω2

c2→ ω = ω

(k)

Because we can write ω as a function of the wave vector, we can define some new coefficientDµ = Dµδ (ω − ω (k)). This eliminates the integral over dω and the general solution becomes:

Aµ =1√2π

∫dk ˜Dµ (ω(k), k) eiω(k)t+ikr

In the next section we’ll return to denoting ˜Dµ ≡ Dµ so as to simply notation. However, note thatthis is some new coefficient.Wednesday - 3/25

54.8 Wave Packet Velocity

Integrating the above determines the reverse transform:

164

∫ [Aµ =

1√2π

∫dkDµ (k, ω) eiωt+ikr

]eik′rdr

Dµ(k, ω) =1√2π

∫ ∞0

drAµ(r, t)eikr

Note that in the case of Dµ = Dµ0 δ(ω−ω0) then the result is the planewave solution we’ve previous

examined. Consider instead some case where ω(k) is approximately constant. In this case, the firstderivative is the only relevant value and the frequency can be approximated as:

ω(k) ≈ ω(k0) +dω

dk

∣∣∣k0

(k − k0) + . . .

Plugging this into the above equation gives:

Aµ(r, t) =1√2π

∫dkDµ(ω, k)e−i(ω0+ dω

dk|k0

(k−k0))t+ikr

Pulling out the constant phase factor:

Aµ(r, t) =1√2πe−iω0t+i

dωdk|k0k0t

∫dkDµ(ω, k)e−i[

dωdk|k0t−r]k

However, note that this is actually the initial transform evaluated at some modified position:

Aµ(r, t) =1√2πe−iω0t+i

dωdk|k0k0t

∫dkDµ(ω, k)e−i[

dωdk|k0t−r]k = e

−iω0t+idωdk

∣∣∣k0

k0t

Aµ(r − dω

dk|k0t, 0

)

Consider only the amplitude of this result:

|Aµ(r, t)| =∣∣∣∣Aµ(r − dω

dk

∣∣∣k0

t, 0)∣∣∣∣

This means that if a packet of waves are initially centered at some point r = 0, then after sometime ∆t, the packet will have moved to some new location centered at r = dω

dk

∣∣∣k0

∆t. Note that this

group result gives a group velocity for the wave packet, but doesn’t describe any distortion of thepacket. That is, if the initial packet is a given waveform, then the transmitted packet is the samewaveform.

Note that in a medium, the frequency of waves varies as:

ω(k) =ck√εµ

=ck

n(k)

165

From this, it is possible to determine the derivative as:

dω

dk=

c

n(ω) + ω dndω

Therefore the group velocity is given as:

vg =dω

dk

∣∣∣k0

=c

n(ω0) + ω dndω |ω0

In most cases the value of n(ω) ≥ 1 and for “normal dispersion” dndω > 0, this results in the above

quantity being less than c for all cases. This is expected since for the wave packet to travel at avelocity larger than c would violate the results of special relativity. However, solid state physicistshave found systems with dn

dω < 0 which results in group velocities larger than the speed of light.Such materials are said to have “anomalous dispersion”. These group velocities aren’t physicallyreal since the systems require dω

dk to vary quickly and the above approximation is inappropriate.

54.9 Modeling the Index of Refraction

In order to build a model of the refractive index as a function of frequency, consider the definitionn =

√εµ. The value of µ is approximately 1 for most materials, therefore we can determine the

behavior by modeling the polarization in a material and then finding ε(ω). Consider a model of amedium build out of atoms surrounded by bound electrons. Each electron is harmonically bound toa nucleus and the incoming wave causes displacement of the electrons which induces a dipole andtherefore a polarization. From this model, it is possible to get some idea of the behavior of ε(ω)and therefore the refractive index of the material.Setting the origin of the system at the nucleus of an atom, we can model the motion of a singleelectron (assuming there is no electrostatic background and the motion is non-relativistic):

L = T − V =12m ˙x2 − 1

2kx

Then the Hamiltonian including coupling the incoming waves can be written as:

12m

p2 +12kx p→ p′ = p+

e

cA

The equations of motion can then be determined as:

xi =∂H

∂pi→ ˙x =

1m

(p+

e

cA)

pi =∂H

∂xi→ ˙p = −kx

These can be combined to give the motion of the electron as:

166

m

(¨x+

k

mx

)=e

c˙A = −eE

where we’ve used the definition E = −∇φ − 1c∂A∂t . In addition to the above results, it is useful to

introduce a damping term, the final result for the motion is:

m

(¨x+

k

mx+ γ ˙x

)= −eE

The first step to solving this is to note that the incoming electric field can be described by E =E0e

−iωt, therefore the position must have similar form. Plugging in the definition for the electricfield and a position of x = x0e

−iωt, the above becomes:

m(−ω2x0 − iωγx0 + ω0

2x0

)= −eE0

Solving for x0,

x =e

m

E0

ω2 − ω02 + iγω

e−iωt =e

m

E

ω2 − ω02 + iγω

And using the definition for the dipole moment of p = −ex, the resulting dipole moment of theelectron is:

p = −e2

m

E

ω2 − ω02 + iγω

The total polarization is the written in terms of the density of atoms per unit volume N , and thenumber of electrons per molecule with binding frequency ωl which we’ll denote as fl.

P =Ne2

m

∑j

flωl2 − ω2 − iωγl

Finally, using the definition from last semester (in Gaussian units) for the polarization P = ε−14π ,

the refractive index can be modeled as:

n(ω) =√ε(ω)µ ε = 1 +

4πNe2

m

∑l

flωl2 − ω2 − iωγl

Note the result that for each molecule the sum∑

l fl gives the total number of electrons Z.Friday - 3/27Consider ordering the frequencies as ω1 < ω2 < . . . < ωl < . . .. If one assumes that γl ω thenquantity in the summation can be approximated as:

167

flωl2 − ω2 − iωγl

≈ flω2l − ω2

In the case that ω > ω1 then the above is positive. However, if the real and imaginary parts of theabove are considered separately, then the behavior is determined as:

ωl2 − ω2 + iωγl

(ωl2 − ω2)2 + ω2γl2

The real and imaginary parts exhibit the behavior:

The values of ωl are resonant absorption frequencies. This can be shown by examining the wavenum-ber k. Let k = β + iα2 give the real and imaginary parts of the wave number. Consider some wavepropagating along k = kε3. The wave can the be written as:

E ∝ e−iωt+ik·x = e−iωt+iβz−α2z

The last term in the exponential causes a damping. This means that any imaginary component ofthe wave number will result in a damping of the field. Using some other definitions, it is possibleto determine:

k2 = β2 − α2

4+ iαβ =

ω2

c2n2 =

ω2

c2εµ

α ∼=Im (ε)Re (ε)

β β ∼=ω

c

√Re (ε)

This result shows that there is maximum attenuation when the real part of epsilon goes to zero.Consider a few special cases of these results.

54.9.1 Free Electrons - Conductors

Consider some material where a number of electrons are free. That is, for some portion of the sumover l, the binding frequency ωl = 0 and γl is very small compared to the bound electrons. In thiscase, we can separate the summation to include only the bound electrons. All of the free electrons

168

have the same ωl = ω0 and γl = γ0, so the summation can be dropped for the term containing thesecontributions. This leaves:

ε = εb −4πNe2

m

f0

ω2 + iωγ0εb ≡ 1 +

4πNe2

m

∑l

flωl2 − ω2 − iωγl

Where f0 is the number of free electrons per molecule. Multiplying the free electron term by −iigives the relation:

ε = εb +4πiNe2

m

f0

ω (γ0 − iω)

Consider how this effects wave propagation.. Maxwell’s Equations give:

∇× H =4πcj +

1c

∂D

∂t

We are consider only the case of harmonic dependence and current free propagation, therefore thisbecomes:

∇× H0 = − iωcD0 = − iω

cεE0

Plugging in the result above for ε in a medium with free electrons,

∇× H0 = − iωcεbE0 −

iω

c

4πiNe2

m

f0

ω (γ0 − iω)E0

The first term describes the effect of the bound electrons in the material. Consider this second termmore closely:

+4πc

Ne2f0

m (γ0 − iω)E0 ≡

4πcσE0

This quantity σE0 acts as an effective current j′ in the conductor due to the incident electromagneticwave. This quantity σ is the conductivity of the material and this method we’ve used to model ε isknown as the “Drube Model”.

54.9.2 High Frequency Limit - Plasma Physics

Consider the limiting case of ω ω1 and ω γl. The permeability can then be approximated as:

ε ≈ 1− ωp2

ω2ωp

2 ≡ Ne2Z

m

169

This approximation is valid cases when ω is very large and also when all electrons are free (whichwould mean ωl = 0 for all values of l and all molecules). This second case describes the physics of aplasma. Because of this, the factor ωp is called the “plasma frequency”. So what is the attenuationin this case? Recall that for most cases µ ≈ 1 and therefore:

n =ck

ω≈√ε→ ε =

c2k2

ω2= 1− ωp

2

ω2

k =1c

√ω2 − ωp2 ε =

ω2 − ωp2

ω2

Consider this result in several cases. Recall that k = β + iα2 . So, for very high frequencies ω > ωp,k is real and thus,

α = 0 β =1c

√ω2 − ωp2

Alternately, for lower frequencies ω < ωp, k is pure imaginary and then,

α =2c

√ωp2 − ω2 ∼ 2ωp

cβ = 0

Monday - 3/30

54.10 Relating the Displacement Field to the Electric Field

Consider the relation between the displacement field D (x, t) and the electric field E (x, t). Ingeneral, the relation

D = εE

In the case of a vacuum, ε = 1 and the two fields are identical. In uniform materials with constantε, the above relation is simply a scalar relation between the fields. Consider the case we’ve beenexamining where the permeability is a function of the frequency. In this case, we can write thedisplacement as a Fourier Transform,

D (x, t) =1√2π

∫ ∞−∞

dω ˜D (x, ω) e−iωt

Using the above definition, we can make the substitution ˜D (x, ω) = ε (ω) ˜E (x, ω) and the abovebecomes:

D (x, t) =1√2π

∫ ∞−∞

dωε (ω) ˜E (x, ω) e−iωt

170

And since the function ˜E can be determined by the inverse transform of the electric field,

D (x, t) =1

2π

∫ ∞−∞

dωε (ω) eiωt∫ ∞−∞

dt′E(x, t′

)eiωt

′=

12π

∫ ∞−∞

dt′E(x, t′

) ∫ ∞−∞

dωε (ω) e−iω(t−t′)

Expanding ε(ω) as 1 + (ε(ω)− 1), we can write the first term as a delta function which simplifiesthe solution to:

D (x, t) = E (x, t) +1

2π

∫ ∞−∞

dt′E (x, t)∫ ∞−∞

dω [ε (ω)− 1] e−iω(t−t′)

Lastly, if we define some new variable τ ≡ t− t′ then the above becomes:

D (x, t) = E (x, t) +∫ ∞−∞

dτE (x, t− τ)G (τ) G (τ) ≡ 12π

∫ ∞−∞

dω [ε (ω)− 1] e−iωτ

There are several restrictions on the kernel function G (τ) which can be determined by this relation.Requiring that causality hold, the displacement field at some time t cannot be dependent on anythingoccuring at time larger than t. Additionally, it is required that the integral above be finite. Thesecondition give:

G (τ) = 0 for τ < 0 by causality

G (τ)→ 0 for τ →∞ by convergence

Consider the solution for the permeability we generated for a single resonance,

ε (ω)− 1 = 4πNe2Z

m

1ω0

2 − ω2 − iωγ0=

ωp2

ω02 − ω2 − iωγ0

Taking the inverse transform for the above definition for G (τ), one finds that for such a resonance,

G (τ) =1

2π

∫ ∞−∞

dω [ε (ω)− 1] e−iωτ =ωp

2

2π

∫ ∞−∞

dωeiωτ

ω02 − ω2 − iωγ0

This integral has poles located at:

ω = − iγ0

2±√ω0

2 − γ02

4

the integral can then be determined by Residue analysis since these poles lie in the complex plane.Consider the exponential for complex frequencies,

171

eiωτ = eiRe(ω)τ+Im(ω)τ

For τ < 0 the asymptotics of this require closing the path in the upper plane, which has no polesand therefore gives the expected result of:

G (τ) = 0 τ < 0

For τ > 0, the sum of the two residues give:

G (τ) =ωp

2e−τγ0

2√ω0

2 − γ02

4

sin

(τ

√ω0

2 − γ02

4

)

Combining these results and consider multiple resonances, the final form of the kernel is:

G (τ) =∑l

ωp2e−

τγl2√

ωl2 − γl2

4

sin

(τ

√ωl2 −

γl2

4

)Θ (τ)

Note that this solution decays at a rate described by 2γ , physically this turns out to be on the order

of 10−7 to 10−9 seconds.

54.10.1 Kramer-Kronig Relations

Using a bit more complex analysis it is possible to determine some additional relations for thepermeability. Consider the permeability and its complex conjugate,

ε (ω) = 1 +∫ ∞

0dτG (τ) eiωτ ε∗ (ω) = 1 +

∫ ∞0

dτG∗ (τ) e−iωτ

however, because the kernel function must be real, this gives that:

Re [ε (ω)] = Re [ε (−ω)] Im [ε (ω)] = −Im [ε (−ω)]

this shows that the real part of the permeability is an even function while the imaginary part is andodd function. Because of this, ε must be analytic for Im (ε) ≥ 0. Because of this, we can use theCauchy Integral formula to write:

ε (z) = 1 +1

2πi

∮C

ε (ω′)− 1ω′ − z

dω′

In we choose the path to be along the real axis and a semicircle in the upper half complex plane,and then separate the two sections, one finds:

172

ε (z) = 1 +1

2πi

∫ ∞−∞

dω′ε (ω′)− 1ω′ − z

+1

2πi

∫adω′

ε (ω′)− 1ω′ − z

Then, if we consider expanding the numerator by integrating by parts,

ε(ω)− 1 =∫ ∞

0dτG (τ) eiωτ =

∫ ∞0

dτ

[G (τ) eiωτ

iω

]′− G′ (τ) eiωτ

iω

Where the (. . .)′ = ∂∂τ (. . .). This process can be repeated for the second term,

ε(ω)− 1 =∫ ∞

0dτ

[G (τ) eiωτ

iω

]′+

1iω

(G′ (τ) eiωτ

iω

)′+G′′ (τ) eiωτ

(iω)2

Evaluating the integral gives only contributions at τ = 0 and τ = infty. Since it was requiredabove that the kernel converge, the contributions at τ =∞ are zero, this leaves:

ε− 1 = − 1iωG (0)− G′ (0)

ω2+ . . .

Since these terms are constant they can be moved out of the integral, this leaves:

G (τ) =1

2π

∫ ∞−∞

dωeiωτ[− 1iωG(0)− 1

ω2G′(0) + . . .

]The first term must be zero for the kernel to be convergent, this means that to leading order ε ∝ 1

ω2 .This shows that the contribution from the semicircle integral converges to zero. This leaves onlythe integral along the real line. This can be evaluated except at the pole at ω′ = z. We can avoidthis by integrating around it. Consider moving the pole off the real line by:

ε(z) = 1 +1

2πi

∫ ∞−∞

dω′ε(ω′)− 1ω′ − z − iδ

Then, the integral can be written as:

ε(z) = 1 +1

2πiP

∫ ∞−∞

dω′ε(ω′)− 1ω′ − z

+1

2πiπi (ε(ω)− 1)

Where the principle part of the integral represents:

P

∫ ∞−∞

(. . .) = limη→0

(∫ z−η

−∞. . .+

∫ ∞z+η

. . .

)

173

Replacing z with ω, we find the relations between the imaginary and real parts of the permeabilityare:

Re (ε(ω)) = 1 +1πP

∫ ∞−∞

dω′Im (ε(ω′))ω′ − ω

Im (ε(ω)) = − 1πP

∫ ∞−∞

dω′Re (ε(ω′))− 1

ω′ − ω

Multiplying numerator and denominator by ω′ + ω gives the simplified results:

Re (ε(ω)) = 1 +2πP

∫ ∞−∞

dω′Im (ε(ω′))ω′2 − ω2

Im (ε(ω)) = −2ωπP

∫ ∞−∞

dω′Re (ε(ω′))− 1ω′2 − ω2

174

Wednesday - 4/1

55 Geometric Optics

55.1 The Approximations and Assumptions of Geometric Optics

Recall that in the Lorentz Gauge, for a region without sources, Maxwell’s Equations are:

Aµ (x, t) = 0 = ∇2 − 1c2

∂2

∂t2

In a medium with variations in the refractive index, the speed of propagation changes. This givesv(x) = c

n(x) . Plugging this into the above for propagation in such a medium,

(∇2 − n2(x)

c2

∂2

∂t2

)Aµ(x, t) = 0

If we assume that the solution has the form Aµ(x, t) = Aµ0 (x)e−iωt then this becomes a problemonly in 3 spacial dimensions,

(∇2 +

ω2

c2n2(x)

)Aµ0 (x) =

(∇2 + k0

2n2(x))Aµ0 (x) = 0

Where we’ve defined k0 = ωc . Note that in the case of constant index of refraction this is a Helmholtz

Equation and the plane wave solutions we’ve seen previously are the exact solution. In the case ofnon-constant index, consider plugging in the form:

Aµ0 (x) ≡ A(x)eik0S(x)

Note that we’ve reduced the notation from Aµ0 to A. Then the condition above becomes

∇2A+ 2ik0∇A · ∇S + ik0A∇2S − k02A (∇S)2 + k0

2n2A = 0

Dividing by k02,

1k0

2∇2A+

2ik0∇A · ∇S +

i

k0A∇2S − A (∇S)2 + n2A = 0

And if we make the change of notation k0 = ωc = 2π

λ0= λ−0. If we assume that the variations in the

index of refraction are small, then it is possible to perform an expansion for λ−0 ∼ 0. In this case,the above can be broken into orders of λ−0 as:

(∇S) = n2 (x)

175

A∇2S + 2∇A · ∇S = 0

∇2A = 0

The term of order λ−20 can be dropped entirely. The first of the above describes the scalar function

S(x, y, z) and the vector field ∇S which we can define to be along the unit vector us. In this casethe above gives that

|∇S| = n(x)→ ∇S = n(x)us

Plugging this into the second equation gives:

A∇2S + 2n(x)∇A · us = 0

A∇2S + 2n(x)∂A

∂s= 0

Here we’ve defined the derivative operator ∂∂s to be the derivative along one of the curves ∇S(x) =

[∇S](s). Dividing this result by 2n gives the second equation of geometric optics,

∂A

∂s= − A

2n(x)∇2S

55.2 Consequences of Geometric OpticsFermat’s Theorem and Generalization of Snell’s Law

Consider some region which has A = 0. In this region there is no EM field and from the aboveequations it is clear that ∂A

∂s is also zero and therefore the regions remains “dark”. If two suchregions exist, then in a third region between them, ∇S defines a light ray and S(x) defines curvesof constant phase which are though of as wave fronts. The paths ∇S are perpendicular to suchcurves and therefore define the geometric path of propagation. Note that in the simple case ofn(x) = constant, the ray propagate in straight lines and the wave fronts are planes (as expected).

Consider some ray path S(s) and additionally some second curve γ(s). The length of each curvecan be determined as:

S(s) =∫ p2

p1

n(x)ds γ(s′) =∫ p2

p1

n(x′)ds′

Where we’ve denoted s and s′ as the two paths. However, from the above definitions, this becomes:

∫ p2

p1

n(x)ds =∫ p2

p1

∂S

∂sds = ∆S(s)|p2

p1

176

The travel time along each can then be defined as:

∫ p2

p1

n(x)vdt = c [t(p1)− t(p2)]→ ∆t =1c

∆S

From this information we can compare the travel time between the two paths. One finds that:

∫ p2

p1

∂γ

∂s′ds′ = γ(p1)− γ(p2)

Because of this we can see that ∂γ∂s′ = ∇S · uγ ≤ |∇S|. That is,

∂γ

∂s′≤ n(x)

Comparing this with the above result for the travel time,

1c

∆S ≤ 1c

∆γ → ∆tS ≤ ∆tγ

That is, the curve S(s) defines the shortest travel time for a light ray between two points p1 and p2

in the geometry defined by n(x). This result in known as Fermat’s Theorem.Lastly, consider taking the derivative of the first of the geometric optics equations along a ray,

d

ds[∇S = n (x) us]

The order of the operations on the left hand side can be reversed and us can be written as drds . This

gives the result:

∇dSds

=d

ds

(n(x)

dr

ds

)

∇n (x) =d

ds

(n(x)

dr

ds

)

This result describes how the unit vector us changes given the changes in n(x). This is the gener-alized result of Snell’s Law.

177

Friday - 4/3

55.3 Optical Fibers

From the generalized result for Snell’s law it is possible to determine the ray paths for light propa-gating in any medium with variations in refractive index. Consider some optical fiber:

n0 = ncoatingn1 = nfiber

In order for total internal reflection to occur the angle of incidence must be greater than somecritical angle ϑi0 = arcsin

(n0n1

). Since the angle of propagation is related to the incident angle by

ϑ = π2 − ϑi the condition for total internal reflection is

ϑ ≤ ϑmax = arccos(n0

n1

)Given this information, consider how many of the modes inside the fiber actually propagate overlong axial distances. To do so, consider the definition:

∆ ≡ n12 − n0

2

2n12

=(n1 + n0) (n1 − n0)

2n12

Since n1 and n0 are roughly of the same order, it is useful to make the approximation,

∆ ≈ n1 − n0

n1= 1− n0

n1→ n0

n1≈ 1− 0∆

For a typical fiber, ∆ ≈ 10−2 and then using the above definition ϑmax ≈ 5 − 10 degrees. Todetermine the propagating modes, consider the transverse electric field:

ET ∝ eikT ·xT kT · xT = kxx+ kyy

Since we want only the fiend that propagates we require that the field at the boundary of the fiberand coating vanish. This requires:

eikT a = ei2πj

Where is j is some integer. This then requires:

j =kTa

2π→ n =

#length

=j

a=kT2π

178

the number density is then,

d2n =d2kT

(2π)2

and integrating over the cross-sectional area of the fiber,

d2N = Area× 2d2kT

(2π)2

Where we’ve inserted a factor of 2 to account for both polarizations. At this point, let’s note thatkT = k sinϑ ≈ kϑ since the angles are so small. Then integrating this result over allowed values ofthe wave vector gives the number of modes as:

N = 2πa2 1(2π)2

∫d2kT =

2πa2

(2π)2 2π∫ ϑmax

0kT (ϑ)dkT (ϑ) =

12

(ak)2 ϑmax2

Using the relation derived above,

N =12

(ak√

2∆)2

=12V 2

Where we’ve defined V as the “fiber parameter”. Plugging in the approximations from above, thisresult shows that there are approximately 300 modes which propagate for a maximum angle of5 − 10 degrees. Let’s compare this result to the total number of propagating modes. The abovederivation then becomes:

N = a2

∫ π2

0kT (ϑ)dkT (ϑ) = a2k2

∫ π2

0cosϑ sinϑdϑ =

12k2a2

Noting that the factor of ϑmax2 is approximately .1 radians. This shows that only 1100 of the possible

modes propagate.

Consider a fiber with a radially varying index of refraction. Let the index decrease such that the rayspropagating are constantly refracted towards the middle of the fiber. We can use the generalizationof Snell’s Law to write:

d

dxn(x) =

d

ds

(n(x)

d

dsx

)=

d

ds(n(x) sin θ)

d

dzn(x) =

d

ds

(n(x)

d

dsz

)=

d

ds(n(x) cos θ)

The assumption above gives the immediate result:

179

d

dzn(x) = 0→ n(x) cos θ = constant

We can define this constant to be the maximum index which the ray reaches. This then gives:

n(x) cos θ = nmax →dz

ds=nmaxn(x)

Using this result, we can write the derivative dds = dz

dsddz = nmax

n(x)ddz . Plugging this into the radial

equation gives:

dn

dx=

d

ds

(n(x)

dx

ds

)=nmaxn(x)

d

dz

(n(x)

nmaxn(x)

dx

dz

)=nmax

2

n(x)d2x

dz2

n(x)d

dxn(x) =

12d

dxn2(x) = nmax

2d2x

dz2

Multiplying by dxdz and then integrating:

∫ [12dx

dz

d

dxn2(x) =

12nmax

2 d

dz

(dx

dz

)2]dz −→ nmax

2

(dx

dz

)2

= n2(x) + constant

In the case of x = xmax, the change in x is zero, this gives that the constant above is nmax2. So,integrating again,

dx

dz=

√n2(x)− nmax2

nmax2→ dz =

nmax√n2(x)− nmax2

dx

z(x) =∫ x

0

nmax√n2(x)− nmax2

dx

From what we saw last time and the result above for dsdx it is possible to write the travel time along

an optical path as:

T =2c

∫ xmax

0n(x)

ds

dxdx =

2c

∫ xmax

0

n2(x)√n2(x)− nmax2

dx

Where the factor of 2 is due the integral only being over half the path of the ray.

Monday - 4/6

180

56 Wave Guides

Previously we’ve seen that geometric optics is a useful approximation in an optical fiber. Considerthe case of propagation in a waveguide with wavelengths large enough that geometric optics is notapplicable. The full solutions of Maxwell’s Equations must then be determined. The resultingproblem can be solved by:

Aµ = 0 →E = −1

c∂∂tA−∇A

0

B = ∇× A

Applying these relation, we find that the fields satisfy:

E = 0 B = 0

If we assume that the solutions are built out of planewaves, then the wave operator can be written:

=1v2

∂2

∂t2−∇2 = −ω

2

v2−∇2

Finally, using the relation v = cεµ , the first of our system of equations is:

[∇2 +

ω2

c2εµ

]EB

= 0 with

∇ · B = 0∇ · E = 0

Consider some geometry with a waveguide extending to infinity. Denoting the axial direction as εz,the solution can be expanded in a Fourier transform in z,

E, B ∝ e±ikz−iωt

Then, breaking the differential operator into axial and transverse components, ∇2 = ∇T 2 + ∂2

∂z2

allows us to reduce the problem to 2 dimensions. This above then becomes:

[∇T 2 +

ω2

c2µε− k2

]EB

= 0 (0)

Here we can write the fields in terms of axial and transverse components, E =(ET , Ez

)and the

same for the magnetic field. Consider a few representations for the fields:

E = E(x, y)e±ikz−iωt = ET + Ez = Et + Ez εz =(εz × E

)× εz + Ez εz

B = B(x, y)e±ikz−iωt = BT + Bz = Bt +Bz εz =(εz × B

)× εz +Bz εz

Using these relations, Maxwell’s Equations give 6 additional requirements on the above:

181

∇ · E = 0→ ∇T · ET = −∂Ez∂z

(1)

∇ · B = 0→ ∇T · BT = −∂Bz∂z

(2)

[∇× E − 1

c

∂

∂tB = 0

]·×

εz →

(∇T × ET

)· εz = iω

c Bz (3)

∇TEz − ∂∂z ET = iω

c εz × BT (4)

[∇× B − µε

c

∂

∂tE = 0

]·×

εz →

(∇T × BT

)· εz = − iω

c εµEz (5)

∇TBz − ∂∂z BT = iω

c µεεz × ET (6)

56.0.1 Zero-Transverse Solution

Consider the solution of the above set of equations in the special case of no transverse field. Pluggingin Ez = Bz = 0 in the above gives the result:

∇T · ET = 0 ∇T · BT = 0

∇T × ET = 0 − ∂

∂zET =

iω

c

(εz × BT

)∇T × BT = 0 − ∂

∂zBT = − iω

cµε(εz × ET

)This reduces to a problem in 2 dimensional electrostatics. The fourth and sixth of the above give therelation between ET and BT so all that is necessary is to determine ET from remaining conditions.Solving for the relation between the electric and magnetic fields is simple if we recall the axialdependence,

ikBT = − iωcµε(εz × ET

)Solving for BT gives a factor ±ωµε

kc . However, note that v = c√εµ = ±ω

k . Combining this with theconditions on electric field:

∇T · ET = 0 ∇T × ET = 0 BT =√εµεz × ET

These solutions are known as “TEM” Modes of the waveguide since they are Transverse Electricand Magnetic waves. In the more general case we can break the solution into transverse electricand transverse magnetic modes.

182

56.1 General Solution

Consider the general solution for the wave guide. It is necessary to decouple the six equations above.Consider the fourth equation, differentiating with respect to the axial variable allows combinationwith equation (6).

∂

∂z

[∇TEz −

∂

∂zET =

iω

cεz × BT

]−→ ∇T

∂Ez∂z− ∂2

∂z2ET =

iω

cεz ×

∂BT∂z

∇T∂Ez∂z− ∂2

∂z2ET =

iω

cεz ×

(∇TBz +

iω

cµε[εz × ET

])Evaluating all of the axial derivatives:

ik∇TEz + k2ET =iω

cεz ×

(∇TBz +

iω

cµε[εz × ET

])Solving for the transverse electric field gives:

ET =i

ω2

c2µε− k2

(±k∇TEz −

ω

cεz ×∇TBz

)

Repeating this for the magnetic field one finds the similar relation:

BT =i

ω2

c2µε− k2

(±k∇TBz −

ω

cµε [εz ×∇TEz]

)

To determine the whole solution one finds a solution to the wave equation (0) from above. Thosesolutions must also satisfy equations (1) and (2) above. Finally, equations (3) - (6) are containedin the two above results.Alternately, one can build a solution of linear combinations of TE and TM modes. This doneby assuming that Bz and Ez are zero respectively. Then the remaining components are easilydetermined from the above.

56.1.1 Boundary Conditions

Consider the boundary conditions on the surface of the waveguide. In general, we require,

n× E = 0 n · B = 0

The first of these is satisfied by:

183

det

∣∣∣∣∣∣εx εy εznx ny 0Ex Ey Ez

∣∣∣∣∣∣ = 0

nyEz εx − nxEz εy + (nxEy − nyEx) εz = 0

n× ET |∂V = 0 Ez|∂V = 0

The second boundary condition requires:

∂Bz∂n|∂V = 0

For the choice of TM waves, the solutions already satisfy the conditions on Ez and Bz. Thereforeonly the condition on n× ET must be applied. Alternately, for TE waves, only the condition on Bzneeds to be applied. In either case, one can solve equation (0) for either Ez or Bz and determinean expansion for the solutions quickly.

184

Wednesday - 4/8

56.2 Solving problems in a waveguide

Consider how to solve a problem involving propagation through a waveguide. The general solutionwill consist of a linear combination of TE modes and TM modes. Let’s consider some alternatedefinitions for the fields in either case:

TM MODES:

Hz = 0 Ez = ψEe±ikz

ET = ± ikγ2∇TψE

Boundary Condition:

ψE |∂V = 0

TE MODES:

Hz = ψHe±ikz Ez = 0

HT = ± ikγ2∇TψH

Boundary Condition:

∂

∂nψH |∂V = 0

Along with some connection relations between the two cases:

(∇T 2 + γ2

)ψE,H = 0 HT =

4πcZεz × Ez

γ2 =ω2

c2εµ− k2

Where we’ve introduced the quantity Z which is the impedance. In either case the impedance hasthe form:

ZTE =4πωµc2k

ZTM =4πkεµ

In order to solve problems one first determines the appropriate eigenfunctions for the Helmholtzoperator above. Then, selecting either TE or TM modes, the appropriate boundary conditionsare applied to generate the exact form of ψH or ψE . Using this result, either ET or HT can bedetermined. From that, the remaining two components of the fields can be found as well.Consider also, the definition above for γ. Solving this for the coefficient k gives:

k =

√ω2

c2µε− γ2

If the quantity inside the square root is negative, then k is pure imaginary and there is no propagationalong the axial direction. The corresponding frequency for which k transitions from real to imaginaryis the so-called “cut-off frequency” and can be written:

185

ωλ =γλc√εµ

Here the value of γλ can be determined from the solutions of the Helmholtz operator. Let’s consideran example:

56.2.1 Example - Rectangular Wave Guide

Consider a rectangular wave guide bounded by planar surfaces at x = 0, x = a, y = 0, and y = b.Let’s determine the TE modes of such a wave guide. In this case we immediately can write:

Ez = 0(∂2

∂x2+

∂2

∂y2+ γ2

)ψH = 0

Given the geometry, the best choice for the general solution is:

ψH = ψx(x)ψy(y) = (ψ1x cos (Ax) + ψ2x sin (Ax))(ψ1y cos (By) + ψ2y sin (By)

)The boundary condition for TE modes requires that the first derivative goes to zero on the bound-aries. That is,

∂

∂xψx|0,a = 0

∂

∂yψy|0,y = 0

This means that only the cosine terms above will remain. This gives the function ψH as:

ψH(x, y) = ψH0 cos(mπax)

cos(nπby)

γ2 = π2

(m2

a2+n2

b2

)

Where the indices n and m can be any combination of positive integers. The lowest eigenvaluecorresponds to either n = 0,m = 1 or n = 1,m = 0 depending on the exact geometry of the waveguide (That is, if a < b or a > b). From the function above, one can determine the field by:

HT = ± ikγ2∇TψH HT =

4πcZεz × ET

56.3 Energy Flow in a Wave Guide

Consider the energy flowing through a waveguide. We can take the Poynting vector along the axialdirection to determine the flux of energy in the guide. Assuming we are working with TM modes,the Poynting vector in the guide can be written as:

186

S =c

8πE × H∗ =

c

8π

∣∣∣∣∣∣εx εy εzEx Ey EzHx∗ Hy

∗ 0

∣∣∣∣∣∣S =

c

8π[εxEzHy

∗ − εyEzHx∗ + εz (ExHy

∗ − EyHx∗)]

So then the flux through the waveguide is:

Φ = S · εz =c

8π

∣∣ET × HT∗∣∣

The power can be determined by integrating this over the cross-sectional area of the waveguide.This gives:

P =c

8π

∫AT

∣∣ET × HT∗∣∣ d2x

Since we are working in the case of TM modes, the above quantities have the forms:

ET = ± iαγ2∇TψE HT =

4πcZεz × ET

A little algebra leads to:

P =4πk2

γ2cZ

∫AT

dA |∇ψE |2

Consider only the integral factor of this result:

∫AT

dA (∇TψE) · (∇TψE∗) =∫AT

dA[∇T (ψE∗∇TψE)− ψE∗∇2ψE

]=∮∂AT

dl ψE∗∇TψE −

∫AT

ψE∗∇2ψE

Due to the boundary conditions, this first term is zero. The second term can be evaluated usingthe Helmholtz operator above to give ∇T 2ψE = −γ2ψE . This leaves:

PTM =εµk

8πγ2

∫AT

dA|ψE |2 =c

8π

√ε

µ

(ω

ωλ

)2√

1− ωλ2

ω2

∫AT

dA|ψE |2

Repeating the above for the TE modes, one finds the simple change of:

187

PTE =c

8π

õ

ε

(ω

ωλ

)2√

1− ωλ2

ω2

∫AT

dA|ψE |2

In order to determine the energy per unit length, the above result can be divided by the groupvelocity of the propagating waves. That is:

P =energy

time→ P

vg=

energytimelengthtime

=energy

length= u

The group velocity is given by:

vg =dω

dkω =

c√εµ

√k2 − γ2

vg =c√εµ

√1− ωλ2

ω2

This combined with the above give:

uTM =1

8π

(ω

ωλ

)2

ε

∫AT

|ψ|2dA uEM =1

8π

(ω

ωλ

)2

µ

∫AT

|ψ|2dA

188

Monday - 4/13

56.4 Resonant Cavities

Consider a waveguide which does not extend to infinity in both directions. If the ends of the waveguide are located at z = 0 and z = d, then our treatment from previous lectures is still viable, butwith the added boundary conditions at the ends:

TM MODES:

Hz = 0 Ez = ψEe±ikz

ET = ± ikγ2∇TψE

Boundary Conditions:

ψE |∂VT = 0∂

∂nEz|∂Vz = ET |∂Vz = 0

Resulting field for cavity of length d:

Ez = ψE(x, y) cos(pπdz)

p = 0, 1, 2, 3 . . .

TE MODES:

Hz = ψHe±ikz Ez = 0

HT = ± ikγ2∇TψH

Boundary Condition:

∂

∂nψH |∂VT = 0 Hz|∂Vz = 0

Resulting field for cavity of length d:

Hz = ψH(x, y) sin(pπdz)

p = 0, 1, 2, 3 . . .

Where we’ve introduced the notation of ∂VT as the boundary along the wave guide and ∂Vz for theboundary on the end caps if the waveguide is turned into a cavity. The connection relations remainthe same:

(∇T 2 + γ2

)ψE,H = 0 HT =

4πcZεz × Ez

γ2 =ω2

cεµ− k2 =

ω2

cεµ− p2π2

d2

189

57 Solutions with Sources

Up until now, we’ve considered general solutions without sources. Recall the general solution wederived:

Aµ(x) = AµHom(x) +4πc

∫d4x′D

(x− x′

)jµ(x′)

AµHom(x) = 0 Aµ(x) =4πcjµ(x)

xD(x− x′

)= δ(4)

(x− x′

)We derived the retarded Greens function to preserve causality,

D(x− x′

)=

14π

Θ(x0 − x0′) δ [(x0 − x0′)+ |x− x′|

]|x− x′|

Consider the simple case of a single point-like charge moving along some trajectory in spacetime.We can assume there are no additional contributions of the field (that is, AµHom = 0) and only theGreen function term will contribute to the field. The solution is then dependent on jµ(x) for thecharge moving along the trajectory. We can write this in terms of the familiar charge and currentdensities,

jµ(x) = (cρ, j)

ρ = q δ(3)(x− x′ (τ)

)j = q v(τ)δ

(x− x′(τ)

)Also, it is convenient to use the properties of the delta function and write:

DR(x, x′) =1

2πΘ(x0 − x0′) δ [(x− x′)2]

Lastly, jµ should be invariant, so the above relations must be rewritten as:

ρ = q

∫dx0′δ

(x0 − x0′) δ(3)

(x− x′

)= q

∫dx0′δ(4)

(x− x′

)And using the relation c2dτ2 = dx02 − dx2 = dx02 1

γ2 , we can write the differential dx0 = γcdτ =u0dτ . Then the charge density is:

ρ = q

∫dτ u0 δ(4)

(x− x′

)And repeating for the current density,

190

j = q

∫dx0v′δ(4)

(x− x′

)= qc

∫dτγv′δ(4)

(x− x′

)j = qc

∫dτu′(τ)δ

(x− x′(τ)

)And combining these results,

jµ = qc

∫dτ uµ(τ) δ(4)

(x− x′(τ)

)So, plugging this into the general solution,

Aµ(x) =4πc

∫d4x′DR

(x, x′

)jµ(x′)

=4πc

∫d4x′

[1

2πΘ(x0 − x0′) δ [(x− x′)2]] [qc∫ dτ uµ

[x′′(τ)

]δ(4)

(x′ − x′′(τ)

)]

Aµ(x) = 2q∫d4x′

∫dτΘ

(x0 − x0′) δ [(x− x′)2]uµ [x′′(τ)

]δ(4)

(x′ − x′′(τ)

)The integral in d4x and the 4-d delta function collapse one integral, this leaves:

Aµ(x) = 2q∫dτ uµ

[x′(τ)

]Θ(x0 − x0′) δ [(x− x′)2]

If we change notation so that the observer is located at x0, x and the source at location r0 (τ) , r (τ),then the above can be written more simply as:

Aµ(x) = 2q∫dτ uµ (τ) Θ

(x0 − r0 (τ)

)δ[(x− r (τ))2

]Note the behavior of this integral. It only takes values when (x− r(τ))2 = 0. That means there issome proper time τ = τ0 which satisfies the equation:

x0 − r0(τ0) = |x− r (τ0) |

191

Wednesday - 4/15 This result can be solved for x0 to show that the time at which the observersees the potential in terms of the time and location it the source caused it can be described by:

x0 = r0(τ0)− |x− r(τ0)| → tx = ts −1c|x− rs|

Next, by modifying the delta function remaining in the integral, it is possible to derive an alternateexpression for Aµ.

δ[(x− r(τ))2

]= δ [(xα − rα) (xα − rα)]

And by the delta relation:

δ [f(x)] =∑i

δ (x− xi)| ∂∂xf(x)|xi

This can be written instead as:

δ [(xα − rα) (xα − rα)] =δ (τ − τ0)

| ddτ (xα − rα) (xα − rα) |τ0

=δ (τ − τ0)

|2 (xα − rα) ddτ (xα − rα) |τ0

=δ (τ − τ0)

|2 (xα − rα)uα|τ0

The delta function collapses the integral over τ and the factor of 2 cancels. This leaves:

Aµ = quµ (τ0)

|[xα − rα (τ0)]uα (τ0)|

Expanding the term in the denominator gives

[xα − rα (τ0)]uα (τ0) = [x0 − r0(τ0)]u0 − [x− r(τ0)] · u(τ0)

If we define the vector from x to r to be R = Rn, then the above can be written as:

(x0 − r0(τ0))u0 − (x− r(τ0)) · u(τ0) = Rγc−Rn · γv = Rγc(1− n · β

)And so finally,

Aµ(x) = quµ

cγR(1− n · β

)∣∣∣Ret. time

Where the result is evaluated at the retarded time τ = τ0. Using the definitions for uµ, the actualcomponents of the potential can be determined as:

192

A0(x) = φ(x) =q

R(1− n · β

)∣∣∣τ=τ0

A(x) =qβ

R(1− n · β

)∣∣∣τ=τ0

As a special case, Jackson works out the potentials and fields for a point charge moving in a straightline. This can be found in chapter 14 of Jackson.

Consider the field from a moving source. We have previously determined that the electric andmagnetic fields can be written as:

Fαβ = ∂αAβ − ∂βAµ

The dependence which requires evaluation of the partial derivatives ∂α and ∂β have been hidden inthe above procedures, so the evaluation of these terms can problematic. Consider:

Aα = quα

uλ (xλ − rλ)→ ∂βAα = q

[1

u (x− r)∂βuα + uα∂β

(1

u (x− r)

)]Where we’ve shortened the notation of uλ (xλ − rλ)→ u (x− r). If the differential ∂β can be turnedinto a derivative with respect to τ then the above will be much easier to deal with. Consider makingthe change:

∂βAα = q

[∂uα

∂τ

u (x− r)∂βτ + uα

∂

∂τ

(1

u (x− r)

)∂βτ

]

Where we can define:

∂xβ

∂τ= uβ

∂xβ∂τ

= uβ ⇒∂τ

∂xβ= (uβ)−1 =

(x− r)β

u (x− r)= ∂βτ

and then plugging this result in,

∂βAα = q

[∂uα

∂τ (x− r)β

[u (x− r)]2+ uα

∂

∂τ

(1

u (x− r)

)(x− r)β

u (x− r)

]

= q1

u (x− r)

[∂uα

∂τ (x− r)β

u (x− r)+ uα (x− r)β ∂

∂τ

(1

u (x− r)

)]

Using the relation that ∂∂τ (x− r)β = −uβ, we can write that:

(x− r)β ∂uα

∂τ=

∂

∂τ

[(x− r)β uα

]− uα ∂

∂τ(x− r)β =

∂

∂τ

[(x− r)β uα

]+ uαuβ

193

Then,

∂βAα = q1

u (x− r)

[1

u (x− r)

(∂

∂τ

[(x− r)β uα

]+ uαuβ

)+ uα (x− r)β ∂

∂τ

(1

u (x− r)

)]The first and third terms combine to give the total derivative and thus finally we have that:

∂βAα = q1

u (x− r)

[∂

∂τ

((x− r)β uα

u (x− r)

)+

uαuβ

u (x− r)

]

changing the indices, we immediately write that:

∂αAβ = q1

u (x− r)

[∂

∂τ

((x− r)α uβ

u (x− r)

)+

uβuα

u (x− r)

]And combining them, the symmetric term drops and leaves the field tensor as:

Fαβ = q1

u (x− r)

[∂

∂τ

((x− r)β uα − (x− r)α uβ

u (x− r)

)]τ0

194

Friday - 4/17

57.1 Deriving the Electric and Magnetic Field for Relativistic Particles

Using the above result we can write the components of the electric and magnetic fields in a moreusable form. Recall that the components of the electric field are represented as Ei = −F 0i. Usingthe above,

Ei = − q

u (x− r)d

dτ

[(x0 − r0

)ui − (x− r)i u0

u (x− r)

]τ0

Using previous notation,

x0 − r0 = R (x− r)i = Rni

u (x− r) = γcR(1− β · n

)Then, the electric field can be written as:

Ei = − q

γcR(1− β · n

) ddτ

[Rγvi −RniγcγcR

(1− β · n

)]τ0

= − q

γcR(1− β · n

) ddτ

[βi − ni

1− β · n

]τ0

E =q

γcR(1− β · n

) ddτ

[n− β

1− β · n

]τ0

Distributing the differential,

E =q

γcR(1− β · n

) [ dndτ −

dβdτ

1− β · n− n− β(

1− β · n)2 d

dτ

(1− β · n

)]τ0

Evaluating the derivatives,

dβ

dτ=dt

dτ

dβ

dt= γ

dβ

dt

x− r = Rn→ −drdτ

= ndR

dτ+R

dn

dτ

dn

dτ= − 1

R

(dr

dτ+ n

dR

dτ

)= − 1

R

(u+ n

d

dτ

(x0 − r0

))= − 1

R

(γv + n

(−u0

))dn

dτ= −γc

R

(n− β

)195

d

dτ

(β · n

)= β · dn

dτ+ n

dβ

dτ= β ·

(−γcR

(n− β

))+ n · γ dβ

dt

d

dτ

(β · n

)=γc

R

(|β|2 − β · n

)+ γn · dβ

dt

Plugging these in gives the result:

E =

q n− βγ2R2

(1− β · n

)3 +q

c

n×[(n− β

)× ˙β

]γR(1− β · n

)3τ0

The magnetic field can be determined in the same manner and after some manipulation one findsthe expected result

B =[n× E

]τ0

Let’s consider what the physics of these results are. In the case of β = 0, the electric field reducesto:

E = qn

R2

this is just the Coulomb field we’ve seen many times. Next, if we consider the case of ˙β = 0 butβ 6= 0, then we are left with:

E = q

[n− β0

γ2R2(1− β0 · n

)3]τ0

This is just a static Coulomb field undergone a boost of β0 evaluated at the retarded time. Fromthese results we can note the contributions of each term in the electric field. The first term (thenon-zero term in these two cases) describes the contribution of a static field undergone some boost.The second term must then describe the remaining contributions (the radiation). Let’s considerthis radiation term more closely. We’ll start by writing the pointing vector,

S =c

4πE × B =

c

4π

∣∣E∣∣2 nHowever, using the above result, we can easily see that |E|2 will have the form:

|E|2 =κ

R2+

λ

R3+

σ

R4

If this result is integrated on all space, then the terms only the first term will contribute. That is,

196

limR→∞

∫|E|2R2dΩ =

∫κdΩ

This shows that only a particle moving with ˙β 6= 0 will radiate EM waves. We can determine thevalue of this flux by solving for this scalar quantity κ.

Erad =q

c

n×[(n− β

)× ˙β

]γR(1− β · n

)3τ0

And using the relation ˙β = γ dβdτ = γc a, then,

Erad =q

c

[n×

[(n− v

c

) γc × a

]γR(1− 1

c v · n)3

]τ0

We will solve this more exactly in the next lecture, but first consider the non-relativistic case. Inthe limit of |v| c, the fields reduce to:

Erad ≈q

c2

n× (n× a)R

Brad ≈ n× Erad

Then, in this case we can write:

S ≈ c

4πnφ|Erad|2 =

q2

4πc3nφ|n× (n× a)|2

R2=

q2

4πc3nφ|(n · a) n− a|2

R2

Where nφ is some new unit vector in a direction which an observer is measuring the flux. Usingthis result, we can determine the power density as:

P = limR→∞

∫ (S · nφ

)R2dΩ⇒ dP

dΩ= lim

R→∞

(S · nφ

)R2

So, simplifying the vector term, |(n · a) n− a|2 = a2 sin2 θ, the power can be written as:

dP

dΩ=

q2

4πc3a2 sin2 θ

P =q2a2

4πc32π∫ π

0sin2 θdθ =

23q2a2

c3

This result gives the total power radiated by a charged particle moving with acceleration a at non-relativistic speeds. This result is known as “Larmor’s Formula”. As stated above, we’ll derive the

197

relativistic relation next time.Monday - 4/20Last time we determined the non-relativistic power loss of a charged particle. This was given by:

dP

dΩ=q2a2

4πc3sin2 θ =

q2

4πm2c3

∣∣∣∣dpdt∣∣∣∣2 sin2 θ

This can be generalized for relativistic motion and the result gives:

P =2q2

3cγ6

[(dβ

dt

)2

−(β × dβ

dt

)2]τ0

57.2 Interesting Cases - Linear and Circular Motion

From this result, one can determine power loss for two general cases of motion. Any motion can bebroken into a linear and angular component. We’ll consider first linear motion:

57.2.1 Linear Motion

For linear motion in the non-relativistic limit,

Energy loss

unit distance=

∆W∆x

=dW

vdt

And using the non-relativistic relation for energy and momentum, W = p2

2m → dW = p dpm , the

above can be written as:

Energy loss

unit distance=dW m

pdt=dp

dt

This then gives the non-relativistic energy loss as:

P ≈ 2q2

3c3m2

(dW

dx

)2

|v| c

In the relativistic limit, the above changes by a factor γ6, however the relation between dpdt and dW

dxremains the same, therefore the relativistic relation is simply:

P =2q2

3c3γ6a2

198

57.2.2 Circular Motion

In the case of circular motion, the change in momentum is a change in direction ( since the magnitudeof p remains constant). Thus,

∆p∆t

=p dθ

dτ= pγ

dθ

dt= pγω

this then gives the loss as:

P =2q2

3c3m2γ2ω2p2

In the non-relativistic limit the above simplifies as γ → 1, p→ mv = mωR. Thus the non-relativisticlimit goes as:

P ≈ 2q2

3c3ω4R2 =

2q2

3cβ4

R2|v| c

The exact relation can be written instead as:

P =2q2c

3R2β4γ4

The change in the number of γ factors is from the form of(dβdt

)2−(β × dβ

dt

)2which expands to

gives something of the form∣∣∣dβdt ∣∣∣2 (1− β2

)=∣∣∣dβdt ∣∣∣2 γ−2.

57.3 Power Loss Per Unit Frequency

We’ve found a result for the power loss radiated per solid angle, let’s consider finding a relationfor the loss per frequency by Fourier transforming the results so far. Recall that the radiated fieldlooks like

Erad =q

c

n×[(n− β

)× dβ

dt

]R(1− n · β

)3τ0

Brad =[Erad × n

]τ0

So then the flux given by[S · nφ

]gives that the radiated energy per unit solid angle per unit time

is:

dW

dΩdt=

c

4πq2

c2

∣∣∣∣∣∣n×

[(n− β

)× dβ

dt

](1− n · β

)3∣∣∣∣∣∣2

τ0

199

Evaluating this instead at the emission time requires writing dWdΩdt as dW

dΩdtemdtemdt . Where t =

tem + 1c |R| (tem). This can be evaluated as:

dt

dtem= 1 +

1c

d

dtemR (tem)

R = Rn = x− r (tem)→ dR

dtem= −vem = −cβem

Therefore,

dtemdt

= 1 +1c

dR

dtem= 1− βem · n

And then finally,

dW

dΩdtem=

q2

4πc

∣∣∣n× [(n− β)× dβ

dt

]∣∣∣2(1− n · β

)5tem

From this we can immediately write:

dP

dΩ=

q2

4πc

∣∣∣n× [(n− β)× dβdt

]∣∣∣2(1− n · β

)5 ≡ |A(t)|2 =c

4π|Erad|2 ·R2

A(t) ≡√

c

4πRErad

So then the energy per solid angle over all time can be found by integrating this result:

dP

dΩ all time=∫ ∞−∞|A(t)|2dt

Then, writing A as its fourier transform,

dP

dΩ all time=∫ ∞−∞

dt1√2π

∫ ∞−∞

dω1√2π

∫ ∞−∞

dω′e−i(ω+ω′)t ˜A(ω) ˜A∗(ω)

Noting the Dirac delta form of the time integral, two of the integrals can be evaluated immediately,this leaves:

dP

dΩ all time=∫ ∞−∞

∣∣∣ ˜A (ω)∣∣∣2 dω =

∫ ∞0

[∣∣∣ ˜A (ω)∣∣∣2 +

∣∣∣ ˜A (−ω)∣∣∣2] dω

200

Because A must be real, the second term can be transformed as ˜A (−ω) = ˜A∗ (ω). This leaves:

d2I

dωdΩ= 2

∣∣∣ ˜A (ω)∣∣∣2 = 2

∣∣∣∣ 12π

∫ ∞−∞

dteiωt√

c

4πRErad

∣∣∣∣2Changing this integration variable to the emission time,

d2I

dωdΩ=

q2

4π2c

∣∣∣∣∣∫ ∞−∞

dt′eiω(t′− 1cn·r(t′)) n×

[n− β

]× dβ

dt′(1− β · n

)2∣∣∣∣∣2

Wednesday - 4/22To simplify this result, we can note that the vector term can be written as:

n×[n− β

]× ˙β(

1− β · n)2 =

∂

∂t

[n×

(n× β

)1− β · n

]

this means that the above is:

d2I

dωdΩ=

q2

4π2c

∣∣∣∣∣∫ ∞−∞

dt′eiω(t′− 1cn·r(t′)) ∂

∂t′

[n×

(n× β

)1− β · n

]∣∣∣∣∣2

This can be evaluated by integrating by parts. The boundary terms are negligible (we assume att′ = ±∞ there was no motion) and we can note that ∂

∂t′

(t′ − 1

c n · r (t′))

= 1− β · n. This leaves:

d2I

dωdΩ=q2ω2

4π2c

∣∣∣∣∫ ∞−∞

dt′eiω[t′− 1cn·r(t′)] [n× (n× β)]∣∣∣∣2

Consider the behavior of this result. It can be shown that this function has a peak value at afrequency that’s approximated by ωpeak ∼ q2γ

c . At relativistic speeds, the γ factor becomes largeand the peak output frequency becomes large as well. Further, it can be shown that the functionbehaves as:

d2I

dωdΩ∼ e−

ωωc ωc =

32γ3( cR

)Further, because electromagnetism is a linear theory, for a system of N particles, the result wouldbe written as:

d2I

dωdΩ=

N∑l=1

q2l ω

2

4π2c

∣∣∣∣∫ ∞−∞

dt′eiω[t′− 1cn·rl(t′)] [n× (n× βl)]∣∣∣∣2

201

And in the limit of continuous media, the summation becomes an integral over all space and theterms qlβl become a current,

d2I

dωdΩ=

ω2

4π2c3

∣∣∣∣∫ ∞−∞

dt′∫Vd3x eiω[t′− 1

cn·x] [n× (n× j)]

∣∣∣∣2

57.4 Application - Scattering

We can apply these results to scattering phenomena and determine energy flux of the scatteredparticles. Consider as an example non-relativistic Thomson Scattering. We have shown that theenergy lost per solid angle can be written as:

dP

dΩ=

c

4π

∣∣RE∣∣2 =q2

4πc

∣∣∣n× (n× ˙β)∣∣∣2

And in the non-relativistic limit, ˙β = ∂β∂t . Consider some single polarization of light. The outgoing

field can then be written as:

E = Eε

In this case, the energy loss is:

dP

dΩ=

q2

4πc

∣∣∣ε · [n× (n× ˙β)]∣∣∣2 =

q2

4πc3|ε · ˙v|2

Then, the time averaged result would be given by:

⟨dP

dΩ

⟩=

q2

8πc3|ε∗ · ˙v|2

Given some incoming electric field Ein, the force on the particle can easily be written as qEin. So,for a field of the form:

Ein = εinEinei(kin·x−ωt) ⇒ ˙v =

q

mEinεine

i(kin·x−ωt)

Plugging this result into the above gives that the time averaged power loss is:

⟨dP

dΩ

⟩=(q2

mc2

)2c

8π|Ein|2 |ε∗ · εin|2

It is useful to instead of considering the power loss per unit solid angle, to consider the differentialcross section. This quantity can be written as:

202

dσ

dΩ≡

energy radiatedtime × solid angleenergy incidentarea × time

=

⟨dPdΩ

⟩S · n

So, plugging in our values from above,

dσ

dΩ=(q2

mc2

)2

|ε∗ · εin|2

If we consider some incident field with wave vector kin along the z direction, then the polarizationlies in the xy plane. That gives:

εin = ax+ by

The outgoing field will then have some wave vector making an angle ϑ with the z axis and (we’llsuppose) some angle ϕ with the x axis. This means that εout = ε in the above will have componentsin a plane perpendicular to kout which can be written as:

ε = cε1 + dε2

ε1 = −z sinϑ+ cosϑ (x cosϕ+ y sinϕ) ε2 = −x sinϕ+ y cosϕ

In order to get the differential cross section, each component of the outgoing wave is consideredand then summed over, while (if there are multiples) the incoming polarizations are averaged. Thisresults in:

|εin · ε∗out|2 =

12(1 + cos2 ϑ

)Which gives the differential and total cross sections as:

dσ

dΩ=

12

(q2

mc2

)2 (1 + cos2 ϑ

)σ =

∫dΩ

dσ

dΩ=

8π3

(q2

mc2

)2

This is the familiar Thomson cross section which is the expected result. It can be shown that thissame result holds for the relativistic case.

203

Friday - 4/24

57.5 Back Reaction

Consider a point-like charged particle moving in a electromagnetic field. As the particle is acceler-ated, it radiates energy off into the field. In the case of a static (infinitely stable) external field, thedynamics of the system are not altered by this radiation. However, in the case that the radiatedenergy is of the same order as the energy in the field, the dynamics will be altered by what is calledthe “back reaction”. Consider the non-relativistic case of energy loss through radiation:

P =2q2

3c3

∣∣∣∣dvdt∣∣∣∣2 → ∆W = P∆t

The change in kinetic energy can be written as:

∆Wkin∼= m

(∆v2

)= m

(∆v∆t

)2

∆t2

Thus, if the variation of energy is on the scale of:

∆t ∼=2q2

3mc3

Then back reaction will alter the dynamics of the system. From this result, one can modify theLorenz Equation to include these effects. Consider the Lorenz equation:

muµ = mduµ

dτ= qFµνextuν

We can add a term (some function fµ) to this to account for back reaction. Consider a fewcharacteristics of such a term:

• In the non-relativistic case, we expect to recover dWdt from above. That is:

dW

dt∼= mc

du0

dt= − 2q2

3mc3

∣∣∣∣dvdt∣∣∣∣2

f0 ∼= −2q2

3mc3

∣∣∣∣dvdt∣∣∣∣2 non− relativistic

• By contracting the above equation, we find that:

[muµ = qFµνextuν ]umu = 0 since uµuµ = 0

fµuµ = 0

204

• Because we don’t want the physics to change or depend on new quantities. The only constantsallowed in the form of this additional term will be q, m, and c.

A form which meets all three above requirements is:

fµ ∝ uµ − (uνuν)uµ

Which then gives the modified Lorenz Equation as:

muµ = qFµνextuν +2q2

3[uµ − (uνuν)uµ]

Consider the energy term of this equation:

mu0 = qF 0νextuν +

2q2

3[u0 − (uνuν)u0

]= qF 0ν

extuν +2q2

3(u)2 u0 +mτ0u

0

Where we’ve defined τ0 ≡ 2q2

3mc3and used the relation uµuµ + u2 = 0. The momentum terms can be

written as:

m ( ˙v − τ0 ¨v) = Fext

However, by consider two special cases it becomes clear that this result generates many non-physicaland problematic solutions.

• Zero Field SolutionConsider the above for Fext = 0. The solutions lead to:

¨v =1τ0

˙v → ˙v = etτ0

This gives an infinite acceleration for the solution. Clearly this is non-physical.

• Integral Solution FormConsider the integral form:

m ˙v =∫ ∞

0ds e−sFext [t+ τ0s, x (t+ τ0s)]

By expanding in orders of τ0 it can be shown that this is a solution of the above since:

m ˙v = Fext (t, x(t)) + τ0dFextdt

+ . . .

However, this integral for generates problematic solutions for some cases. If a field is turnedon at a given time, the above becomes:

205

m ˙v =∫ ∞

0ds e−sF0Θ (t+ τ0s)

Making a substitution of the integration variable leads to:

m ˙v = F0

∫ ∞0

1τdye− y−t

τ0 Θ (y) = F0etτ0

This result gives non-zero acceleration before the field was turned on!

While this section will not be included in tests, homework, etc., the results are important to under-stand. The theory of classical electromagnetism begins to breakdown at certain limits. This caseof a point-like particle moving with back reaction is one such case.

58 Localized Sources

We spent much of last semester solving electro- and magneto- statics problems. Let’s consider thegeneral case of solving for the E and B fields from a localized source. We can begin by writing thefamiliar solution:

Aµ = Aµhom +4πc

∫d4x′DR

(x− x′

)jµ(x′)

and making the adjustments that Aµhom is zero and now jµ is some distribution of charges. Fromthese assumptions, we can write the general solution for the vector potential as:

A =4πc

∫d3x′

∫c dt′j

(x′, t′

) 14π

1c δ(t′ − t+ |x−x′|

c

)|x− x′|

by assuming that j and therefore also A have monochromatic time dependence, we can write thisas:

A =1c

∫d3x′

j (x′)|x− x′|

eik|x−x′| φ =

1c

∫d3x′

ρ (x′)|x− x′|

eik|x−x′|

however, because we can compute the fields as:

H = ∇× A E =i

k∇× H

We only need to find the vector potential of the above.Monday - 4/27We’ve seen that we can determine the electric and magnetic fields from only knowing the vectorpotential, A which we defined from:

206

A (x) =1c

∫d3x′

j (x′)|x− x′|2

eik|x−x′|

There is no analytical solution of this for general j, however by expanding for special values ofthe scale of the problem, a few approximations can be determined. Two scales determine theapproximation we take. First, the size of the source d will determine the range of values x′ will takein the integration. Second, the wavelength of the emitted wave λ will determine the behavior of theexponential term. Comparing these values to the observer’s distance from the source |x− x′| = r,one can determine three cases which have approximate solutions.

58.1 The Near (Static) Zone

In the case of d r λ, an approximation of kr 1 is appropriate. In this case we can expandthe vector potential first as:

A ∼ 1c

∫d3x′

j (x′)|x− x′|

(1 + kr +

12k2r2 + . . .

)Keeping only the first order term, we can then expand the remaining vector term in sphericalharmonics:

A ∼ 1c

∫d3x′

j (x′)|x− x′|

=1c

∫d3x′j

(x′)∑l,m

4π2l + 1

rl<

rl+1>

Y ∗l,m (Ω)Yl,m(Ω′)

58.2 The Intermediate Zone

In the intermediate zone, we have d r ∼ λ. There are no appropriate expansion methods in thiscase and most analysis leads to complications. One possible method is to use a WKB type methodto connect the asymptotics of the Near Zone and Far Zone and call those solutions the IntermediateZone solution.

58.3 The Far Zone

The Far Zone is defined as d λ r. In this case the exponential term is not small, and thereforewe expand the 1

|x−x′| term first. This is done easily by first writing the approximate form:

∣∣x− x′∣∣ ≈ r − n · x′Plugging this in, we find:

207

A ∼ 1c

∫d3x′

j (x)r(1− n·x′

r

)eikre−ikn·x′ =eikr

cr

∫d3x′

j (x)1− n·x′

r

e−ikn·x′

58.3.1 Zeroth Order Term

We can now expand in orders of 1r and get A = A0 + A1 + . . . and determine the form of the first

few terms. We can write this first term as:

A0 =eikr

cr

∫d3x′j (x) e−ikn·x

′

Next, assuming n · x′ is small, we can expand:

A0 ≈eikr

cr

∑l

(ik)l

l!

∫d3x′j

(x′) (n · x′

)lTaking only the l = 0 term,

A0,0 =eikr

cr

∫d3x′j

(x′)

We saw this integral in magnetostatics and found that it was zero (no magnetic monopoles). Nowthat we are not assuming static systems, we can use the same change of form to write:

ji = ∇ (xij)− xi∇ · j

Then, the ith component of the above approximation is written as:

A(i)0,0 =

eikr

cr

∫d3x′ji

(x′)

=eikr

cr

∫d3x′

[∇(x′ij)− x′i∇ · j

]The boundary term is zero again. However, since we are no longer dealing with a static case, wehave:

∇ · j +∂ρ

∂t= 0

And since we are assuming harmonic dependence, ρ ∝ e−iωt. This gives:

∇ · j − iωρ = 0

So, the ith component of the vector potential approximation is:

208

A(i)0,0 = − iω

c

eikr

r

∫d3x′x′iρ

(x′)

but this integral is just the electric dipole. Thus:

A0,0 = −ik eikr

rp

The resulting fields are defined as:

H = ∇× A =k2eikr

r

(1− 1

ikr

)n× p

E =i

k∇× H = k2 (n× p)× ne

ikr

r+ [3n (n · p)− p]

(1r3− ik

r2

)eikr

Considering the power emitted, only the terms of order 1r are relevant. This leaves:

Hrad =k2

reikr (n× p) Erad =

k2

reikr (n× p)× n = Hrad × n

Then, the time averaged power emitted per solid angle and the total power emitted are:

⟨dP

dΩ

⟩time

=c

8πk4 |(n× p)× n|2 =

c

8πk4 |p|2 sin2 θ

〈P 〉time =ck4

3|p|2

Note two important characteristics of this result. First, at observation angle θ = 0, there is noradiation emitted. Second, the radiated energy is polarized to n and p. Let’s consider an exampleof this result in a dipole antenna.

58.3.2 Example - Dipole Antenna

If we have an antenna of length d with cross-sectional area A carrying current I(z)eP iωt =I0

(1− 2|z|

d

)eiωt, then the current in the antenna has the form:

j =I(z)eiωt

A=I0

A

(1− 2

|z|d

)e−iωtz

Plugging this result into the continuity equation,

209

∇ · j = −∂ρ∂t→ ρ =

1iω

∂

∂zjz =

2iωd

I0

A

z

|z|

Plugging this into the integral form above:

p =∫d3x′x′ρ

(x′)

=1A

∫dxdy

∫dz

2iωdI0z

|z|

px = 0, py = 0, pz =iI0d

2ω

Plugging in:

⟨dP

dΩ

⟩time

=c

8πk4 I

20d

2

4ω4sin2 θ

〈P 〉time =ck4

3I2

0d2

4ω4=I2

0 (kd)2

12c

210

Wednesday - 4/29

58.3.3 First Order Term

Taking the second order terms of our vector potential

A ≈ eikr

cr

∫d3x′

j (x)1− n·x′

r

e−ikn·x′

A1 =1c

eikr

r

[1r− ik

] ∫d3x′j

(x′) (n · x′

)The vector term in the integral can be expressed alternately as:

(x′ × j

)× n =

(x′ · n

)j −

(x′ · j

)n

This allows us to write this first order term as:

A1 =1c

eikr

r

[1r− ik

] ∫d3x′

12(x′ × j

(x′))× n+

12[(x′ · j

(x′))n+

(x′ · n

)j(x′)]

Recalling the definition m ≡ 12

∫d3x′ [x′ × j (x′)], and simplifying the remaining terms with the

relation previously used in the zeroth order analysis, we can write this as:

A1 (x)ik

r

(1− 1

ikr

)n× m− ick

2

∫d3x′x′

(n · x′

)ρ(x′)

From this result we can compute the fields E1 and H1. Taking only the first term in the integral,this gives:

E(1)1 = −k

2

c(n× m)

eikr

r

(1− 1

ikr

)

H(1)1 =

1c

k2 (n× m)× ne

ikr

r+ eikr [3n (n · m)− m]

(1r3− ik

r

)note that if only the first order terms are considered (that is, the radiation terms), this result givesthe same form as the zeroth order fields only with p→ m and an additional factor of 1

c . The secondterm gives Addison contributions which can be written as:

H(2)1 =

−ik3

6eikr

rn× Q Q = Qα, α = 1, 2, 3

Qα = Qαβηβ Qαβ ≡∫

3(xαxβ − r2δα,β

)ρ (x) d3x

211

Using this, we find that the power contribution can be written as:

⟨dσ

dΩ

⟩=

k6c

288π

∣∣(n× Q)× n∣∣2From the above results we can summarize the behavior of the first order term in the expansion byits dependence on the magnetic dipole and the electric quadrupole.

58.3.4 Second Order and Higher Contributions

Without any further analytical work, we can immediately claim that the second order term will bedependent on the magnetic quadrupole and the electric octopole. The higher order terms continuethis behavior. However, to get the radiation field, only the zeroth and first order results are ofinterest in most cases.

59 Scattering

Consider some incident wave defined as:

Ein = ε0E0eiωt+ik·n0 Hin = n0 × Ein

The scattered field can then be written as:

Esc =k2

reikr−iωt

[(n× p)× n− 1

cn× m

]Hsc = n× Esc

Where p and m are the electric and magnetic dipole moments of the scattering body respectively.From these definitions we can define the differential cross-section of the scatterer.

dσ

dΩ≡ Energy Scattered/T ime/SolidAngle

Energy Incident/T ime/Area=

∣∣Ssc∣∣2 r2∣∣Sin∣∣2Plugging in fields from above gives the incident and scattered Poynting vectors.

dσ

dΩ=

c8πk

4∣∣ε∗ · [(n× p)× n− 1

c n× m]∣∣2

c8π

∣∣ε∗0 · Ein∣∣2And noting that ε · n = 0 this simplifies to:

dσ

dΩ=k4

E20

∣∣∣∣ε∗ · p+ (n× ε∗) 1cm

∣∣∣∣2

212

Note that if we recall that k = ωc this is Rayleigh’s Law since for increasing ω the scattering cross

section gets larger.

59.0.5 Example - Dielectric Sphere

A simple example using this result can be made using a dielectric sphere of radius a. Recall thatfor such a sphere in a uniform electric field:

p =ε− 1ε+ 2

a3Ein m = 0

Plugging this into the above immediately gives the scattering cross section as:

dσ

dΩ= k4a6

∣∣∣∣ε− 1ε+ 2

∣∣∣∣2 |ε∗ · ε0|2If we again average on the incident polarizations and sum over the scattered polarizations, thisgives:

dσ

dΩ= k4a6

∣∣∣∣ε− 1ε+ 2

∣∣∣∣2 12(1 + cos2 θ

)59.0.6 Generalization for Multiple Scatterers

Consider a system made up of several scattering bodies. We can transform our result to accountfor this by writing:

dσ

dΩ=k4

E20

∑i

∣∣∣∣[ε∗ · pi + (n× ε∗) 1cmi

]eiqi·xi

∣∣∣∣2Where we’ve introduced the notation of:

qi = kn0 − kn x = scatterer position

Also, we can separate terms for each polarization (instead of summing over outgoing polarizations)and get the polarization of the outgoing field:

Π =

dσdΩ

∣∣∣⊥− dσ

dΩ

∣∣∣‖

dσdΩ

∣∣∣⊥

+ dσdΩ

∣∣∣‖

For the dielectric sphere example above, this result is:

213

Π (θ) =sin2 θ

1 + cos2 θ

Friday - 5/1

59.1 Oscillating Scatterers

Consider some particle bound in a simple harmonic oscillator potential with an electromagneticfield incident on it. If we can determine the dipole moment p for this oscillating particle, we candetermine the scattering from it as well. The motion can be described as:

F = −mω20x+ eEin

¨x+ Γ ˙x+ ω20x =

e

mEin

If we assume harmonic behavior, then x(t) ∝ eiωt and the equation of motion gives:

x(t) =e

mE′in

eiωt

ω20 − ω2 − iΓω

Where we’ve separated the incident field as Ein = E′ineiωt. We can then get the dipole moment as:

p =∫d3x′ρ

(x′)x′ =

∫d3x′eδ

(x− x′

)x′ = ex(t)

So, plugging in our result from above:

p =e2

mE′in

eiωt

ω20 − ω2 − iΓω

Then, plugging this into the equation for the differential cross section (and noting that |E′in| = E0),

dσ

dΩ=k4e4

m2

∣∣∣∣ 1ω2

0 − ω2 − iΓω

∣∣∣∣2 |ε∗ε0|So, this can be alternately written as:

dσ

dΩ=(e2

mc2

)2ω4(

ω20 − ω2

)2 + Γ2ω2

12(1 + cos2 θ

)

214

And integrating to get the whole cross section,

σ =8π3

(e2

mc2

)2ω4(

ω20 − ω2

)2 + Γ2ω2

59.2 Scattering In A Medium - Diffraction

Consider electromagnetic waves incident on a medium. In this case, it would be possible to usethe summation of all the scattering bodies, but the number of terms would be very large. Instead,we can make what is known as the “Born Approximation”. However, we’ll first need to set up theproblem. Recall Maxwell’s Equations in a medium:

∇ · B = 0 ∇ · D = 0

∇× E = −1c

∂B

∂t∇× H =

1c

∂D

∂t

So, we can then combine these to build an equation for D as:

∇(∇ · D = 0

)→ ∇×

(∇× D

)+∇2D = 0

1c

∂

∂t

(∇× H =

1c

∂D

∂t

)→ ∇× 1

c

∂H

∂t=

1c2

∂2D

∂t2

Combining these results gives:

[∇2 − 1

c2

∂2

∂t2

]D = −∇×

(∇× D

)−∇× 1

c

∂H

∂t

The last term can be expanded as:

D = −∇×(∇× D

)− 1c∇× ∂

∂t

(H − B

)− 1c∇× ∂B

∂t

And using Maxwell’s Equations,

D = −∇×(∇× D

)− 1c∇× ∂

∂t

(H − B

)+∇×

(∇× E

)Collecting terms, this gives:

D = −∇×(∇×

[E − D

])− 1c∇× ∂

∂t

(H − B

)

215

The righthand side is like a diffraction source since the case of E = D and H = B it vanishes andwe are left with E = 0. Next, if we assume that the source is monochromatic, we can write thisas:

[∇2 + k2

0

]D = jdiff jdiff = ∇×

(∇×

[E − D

])+ ik0∇×

(H − B

)And then, we can simply write this as we have previously done for fields with sources,

D = Dincident −1

4π

∫d3x′jdiff

(x′) eik|x−x′||x− x′|

Where we’ve taken only the homogeneous solutions which correspond to an incident plane wave andthe Greens Function term is the outgoing scattered wave. In the limit of |x− x′| /gg1, this can beapproximated as:

D = Din −1

4πeikr

rAsc Asc ≡

∫d3x′

[jdiff

(x′)e−ikn·x

′+ . . .

]This can be evaluated by integrating by parts. The terms are similar in structure, so we’ll work outone in detail.

∫d3x′eikn·x

′∇×(H − B

)=∫d3x′

[∇×

[eikn×x

′ (H − B

)]+ ike−ik·x

′∇×(B − H

)]The boundary terms can be dropped since we assume E = D and B = H at large distances fromthe scatterer. This gives the result of:

Asc =k2

4π

∫d3x′e−ikn·x

′ [n×

(D − E

)]× n− n×

(B − H

)The cross section is then:

dσ

dΩ=

∣∣Asc · ε∗∣∣2∣∣Din

∣∣2This is the point where the Born Approximation is used. Consider if the material in the scatterercan be described by some small δε and δµ. In this case, we can write the fields in the medium as:

D = Ein + δεEin B = Hin + δµHin

Then, the scattering cross section simplifies to:

⟨dσ

dΩ

⟩Born

=(k2

4π

)2 ∣∣∣∣∫ d3x′eiq·x′[(δε) ε∗ · ε0 + (δµ) (n× ε∗) · (n× ε0)]

∣∣∣∣2216

If we recall that the magnetic dipole contribution is scaled by 1c then we can neglect it and find

that the approximate differential cross section is:

⟨dσ

dΩ

⟩Born

∼=(k2

4π

)2 ∣∣∣∣∫ d3x′eiq·x′[(δε) ε∗ · ε0]

∣∣∣∣2For a spherical scattering body of radius a, this can be approximated as:

⟨dσ

dΩ

⟩Born

∼=k4

6q|δε|2 |ε∗ · ε0|2 sin (qa)− qacos (qa)2

And in the limit of q → 0 (that is, k ∼ k′), this reduces to:

⟨dσ

dΩ

⟩Born,q→0

∼= k4a6

∣∣∣∣δε3∣∣∣∣2 |ε∗ · ε0|2

217

Part VI

Summary of Semester I

Friday - 11/21Maxwell Equations in vacuum:

∇ · E =ρ

ε0; ∇ · B = 0

∇× E +∂B

∂t= 0; ∇× B = µ0j +

1c2

∂E

∂t

∂ρ

∂t+∇ · j = 0; F = q

(E + v × B

)Transition to Media:

D = D(E, B); H = H(E, B)

If the media is linear then these can be written in terms of matrices. In the case that the mediumis linear and isotropic (at least piecewise) then the above reduce to:

D = εE; H =1µB

And, in the medium, the Maxwell equations are:

∇ · D = ρ ∇ · B = 0

∇× E + ∂B∂t = 0 ∇× H = j + ∂D

∂t

Polarization and Magnetization:

D = ε0χeP ; P = ε0χeE → ε = ε0(1 + χe)

H =1µB − M ; B = µ0χMM → µ = µ0(1 + χM )

Boundary Conditions:

(D2 − D1

)· n = σ

(B2 − B1

)· n = 0(

E2 − E1

)× n = 0

(H2 − H1

)× n = −K

218

Energy Density w and Energy Flux (Poynting Vector) s:

w =12(D · E + H · B

)s = E × H

60 Electrostatics

60.1 Equations

Assume ∂E∂t = ∂B

∂t = 0 and resulting equation(s) to solve are:

E = −∇φ→ ∇2φ = − ρε0

If there is a medium which is isotropic and ε is at least piecewise constant, then the above becomes:

D = εE =→ ∇2φ = −ρε

Solution of the above: Green’s Formula

φ(x) =1

4πε0

∫Vρ(x′)G(x− x′)d3x′ +

14π

∫∂V

(G(x− x′)∂φ

∂n− φ(x′)

∂G

∂n

)d2x′

Use reference pages for Greens Function forms.

Energy:Self Energy:

w =12ρ(x)φ(x)

Energy of distribution in external potential:

w = ρ(x)φext(x)

In both cases, the total energy is simply the integral over all space:

W =∫Vd3xw(x)

219

60.2 Methods for Working Problems

NO CHARGES:1) 2-D Solutions from Complex Plane

F (z) = U(x, y) + iV (x, y)

Solutions U(x, y) = c and V (x, y) = c′ are solutions of Poisson’s Equation. But there is no way totell what geometry will result from a given function F (z).

2) 2-D and 3-D Solutions via Series ExpansionExpansions for the potential are listed on Reference Pages

SYSTEM WITH CHARGES:For exact solutions, use Greens Formula Solution or Method of ImagesFor approximate solutions, use Multipole expansion methods.

ELECTROSTATICS IN MEDIA:Boundary conditions gives relations for potential in different mediaPolarization induce between materials with different ε values:

σinduced = −(P2 − P1

)· n|∂V = −∇ · P

61 Magnetostatics

61.1 Equations

Assume ∂E∂t = ∂B

∂t = 0 and resulting equation(s) for magnetostatics are:

∇ · B = 0; ∇× B = µ0j

or in a medium:

∇ · B = 0; ∇× H = j

Where for linear materials B = µ0

(M + H

)and for paramagnetics and diamagnetics specifically

B = µH. The resulting equations to solve depend on the problem given.

61.2 Methods for Working Problems

The first equation above gives the result that B can be written as the vector potential A whereB = ∇× A and choosing the gauge in which ∇× A = 0, we are left with the equation ∇2A = −µ0j.

220

The solution for this problem is:

A(x) =µ0

4π

∫Vd3x′

j(x′)|x− x′|

+ AHom.

Where the additional solution is that of the homogeneous problem. In the case of localized currentdensities, this term can be neglected. The method above works well for problems with a given j.However, if a problem gives j = 0 in some region, then the equations above reduce to:

∇× H = 0→ H = −∇φM ; ∇2φM = 0

Lastly, in the case that a region contains no current density, but some magnetization M then wecan refer to the definition above to write the equations:

B = µ0

(M + H

); H = −∇φM → ∇2φM = −∇ · M

Taking into account boundary terms as well, this becomes:

φM (x) = − 14π

∫Vd3x′∇′ · M(x′)|x− x′|

+1

4π

∫∂Vd2x′

n · M(x′)|x− x′|

If we instead want to merely approximate the field at some point, we can use the multipole expansionwe derived previously, The leading order term is m, the magnetic dipole contribution. We havedefined it to be:

m =12

∫d3x′ x′j(x′)

|m| = I ×A

Where the magnitude is proportional to the current times the area of the distribution. We can thenwrite the approximation for the vector potential and the magnetic scalar potential as:

φM =m · x|x|3

+ . . .

A =µ0

4πm× x|x|3

Lastly, there are several useful relations which are necessary for working some problems:Ampere’s Law:

∮∂SB · dl = µ0I;

∮∂SH · dl = I

221

Biot-Savart Law:

dB =µ0

4πIdl × r|r|3

And Faraday’s Law:

ξ = −dΦdt

222

Part VII

Summary of Semester II

62 Special Relativity

The transform between inertia frames can be generalized as:

t′ = γ

(t− 1

cβ · x

)x′ = x+

γ − 1β2

(β · x

)β − γβct

In the case that β is directed only along a single axis (let’s say x), then this simplifies to:

ct′ = γ (ct− βx)

x′ = γ (x− βct)

y′ = y z′ = z

The invariant line element is defined as:

c2dτ2 = c2dt2 − |dx|2 = c2dt2′ −∣∣dx′∣∣2

An object moving with velocity u in a frame described by β is seen as having velocity u′:

u′ =u− βc

1− betac u

63 Tensors and Lorentz Invariant Forms

In tensor notation, we defined:

∂µ ≡(

1c

∂

∂t,∇)

∂µ ≡(

1c

∂

∂t,−∇

)

Aµ ≡(φ, A

)jµ ≡ (cρ, j)

Fµν ≡ ∂µAν − ∂νAµ F 0i = −Ei F ij = εijkBk

223

Which lead to the tensor equations:

∂µFµν =

4πcjµ Aµ =

4πcjµ

Using the Lorentz transform from the special relativity section, we derived the transformations ofthe electric and magnetic fields:

E′ = γ(E + β × B

)− γ2

γ + 1(β · E

)β

B′ = γ(B − β × E

)− γ2

γ + 1(β · B

)β

64 Hamiltonian Dynamics in Electromagnetism

We defined the tensors:

uµ ≡(cdt

dτ,dx

dτ

)= (cγ, u)

pµ ≡(

1cE, p

)= (mcγ,mγv) = muµ

Using these tensor forms, the equation of motion for a charged particle in an electromagnetic fieldwas written:

d

dτpµ =

q

cFµνuν →

dEkindt = eE · v

dpdt = eE + e

c v × H

With,

E2 = c2 |p|2 +m2c4

p = γmv Ekin = γmc2 −→ v =pc2

Ekin

The relativistic doppler shift was determined by introducing kµ ≡(ω,−k

),

ωobsωemit

=1γ

11− βcosθ

The Stress-Energy tensor was derived using Noether’s Theorem:

224

d

dt

[∂L

∂qδq + Lδt− f(q(t), t)

]= 0

in the case of fields, this becomes:

∂µjµ = 0 jµ =

∂£∂ (∂µφi)

δφi + £δxµ − fµ

Which generates the Stress Energy tensor:

Tµν =∂£

∂ (∂µφ)(∂νφ) + £gµν =

14πFµγFγ

ν +1

16πgµνFαβF

αβ

65 Special Cases

65.1 Plane Waves

Under the assumption that the generated plane waves are monochromatic, Maxwell’s Equationsallow solutions of:

E = Ece−i(ωct−kc·x) B = Bce

−i(ωct−kc·x) Bc =c

ωckc × Ec

The polarization of the electric component can be written as:

E = Ec

[ε1e−i(ωct−kc·x) + ε2Re

−i(ωct−kc·x−θ)]

Two general cases are known:

• Circular Polarization - When R = 1 and θ = π2 the electric and magnetic components oscillate

around each other.

• Linear Polarization - when R = 1 and θ = 0 the components are separated.

For propagation through media, the following changes are made:

c→ v =c√εµ

E →√εE B → 1

õB

B0 =c

ωk × E0 →

√εµn× E0 with ω =

kc√µε

The energy in a plane wave can be determined by the Poynting vector,

225

S = Re( c

8π

∣∣E × B∗∣∣) =c

8π

∣∣E∣∣2 n (n along k)

And in media,

S =c

8π

√ε

µ

∣∣E∣∣2 nFor boundaries between materials with different refractive indices n = c

v =√εµ, denoting the

incident quantities i, reflected quantities R, and refracted quantities r.

θi = θR n1 sin θi = n2 sin θr

θB = arctan(n2

n1

)→ Er‖ = 0 θi0 = arcsin

(n2

n1

)→ Er = 0

And in general for propagation perpendicular and parallel to the plane of incidence,

(E′′0E0

)⊥

=2n1 cosϑi

n1 cosϑi + n2µ1

µ2cosϑr

= Transmitted F ield

(E′0E0

)⊥

=n1 cosϑi − n2

µ1

µ2cosϑr

n1 cosϑi + n2µ1

µ2cosϑr

= Reflected F ield

(E′′0E0

)‖

=2n1n2 cosϑi

µ1

µ2n2

2 cosϑi + n1

√n2

2 − n21 sin2 ϑi

= Transmitted F ield

(E′0E0

)‖

=µ1

µ2n2

2 cosϑi − n1

√n2

2 − n21 sin2 ϑi

µ1

µ2n2

2 cosϑi + n1

√n2

2 − n21 sin2 ϑi

= Reflected F ield

For wave packets, as long as ω (k) ≈ ω (k0)+ dωdk |k0 (k − k0)+ . . . is a valid approximation, the group

velocity and modulation envelope are given by:

vg =dω

dk|k0 |Aµ(x, t)| =

∣∣∣∣Aµ(x− dω

dk|k0t, 0

)∣∣∣∣Alternately, since ω(k) = ck

n(k) ,

vg =dω

dk|k0 =

c

n(ω) + ω dndω

It is possible to model the refractive index as a summation of harmonic oscillators (fl of which haveω = ωl), this gives:

226

n(ω) =√ε(ω)µ ε(ω) = 1 +

4πNe2

m

∑l

flω2l − ω2 − iωγl

For conductors this can be separated for bound electrons and free electrons:

εcond = εb +4πiNe2

m

f0

ω (γ0 − iω)

and for plasmas, ω >> ω1 and ω >> γl, so we can approximate the above as:

ωplasma ≈ 1−ω2p

ω2ωp ≡

Ne2Z

m

Lastly, the Kramer Kronig relations give the real and imaginary parts of ε in terms of the other,

Re (ε(ω)) = 1 +2πP

∫ ∞−∞

dω′Im (ε(ω′))ω′2 − ω2

Im (ε(ω)) = −2ωπP

∫ ∞−∞

dω′Re (ε(ω′))− 1ω′2 − ω2

65.2 Geometric Optics and Wave Guides

For geometric optics, the propagation is described by:

(∇2 − n(x)

c2

∂2

∂t2

)Aµ (x, t) = 0

And we make the assumption that:

Aµ (x, t) =(Aµ(x)eik0S(x)

)e−iωt

This results in a generalization of Snell’s Law:

∇n(x) =d

ds

(n(x)

dr

ds

)We also derived Fermat’s Principle that for any generic curve γ, the relation:

∆tS ≤ ∆tγ

is true. This means that the ray paths defined here are defined by curves with the smallest propertime between points. The paths are defined by:

227

z(x) =∫ x

0

nmax√n2x′ − nmax2

dx′

Travel time is then:

T =2c

∫ xmax

0

n2(x)√n2x′ − nmax2

dx′

For waveguides, we found that:

TM MODES:

Hz = 0 Ez = ψEe±ikz

ET = ± ikγ2∇TψE

Boundary Conditions:

ψE |∂VT = 0∂

∂nEz|∂Vz = ET |∂Vz = 0

Resulting field for cavity of length d:

Ez = ψE(x, y) cos(pπdz)

p = 0, 1, 2, 3 . . .

TE MODES:

Hz = ψHe±ikz Ez = 0

HT = ± ikγ2∇TψH

Boundary Condition:

∂

∂nψH |∂VT = 0 Hz|∂Vz = 0

Resulting field for cavity of length d:

Hz = ψH(x, y) sin(pπdz)

p = 0, 1, 2, 3 . . .

Where we’ve introduced the notation of ∂VT as the boundary along the wave guide and ∂Vz for theboundary on the end caps if the waveguide is turned into a cavity. The connection relations remainthe same:

(∇T 2 + γ2

)ψE,H = 0 HT =

4πcZεz × Ez

γ2 =ω2

cεµ− k2 =

ω2

cεµ− p2π2

d2

65.3 Solutions with Sources (Radiation)

Solving Maxwell’s Equations with sources, we found:

jµ = qc

∫dτuµ (τ) δ(4)

(x− x′ (τ)

)Aµ (x) =

q

c

[uµ

γR(1− n · β

)]τ=τ0

228

Fαβ =q

u (x− r)

[∂

∂τ

((x− r)β uα − (x− r)α uβ

u (x− r)

)]τ=τ0

Where the results are evaluated at the retarded time τ0. Taking only the 1R terms which contribute

to radiation,

Erad =q

c

n×[(n− β

)× ˙β

]γR(1− n · β

)3τ0

Brad = n× Erad

And in the non-relativistic case, we can make the simplification,

Erad ≈q

c2

n× (n× a)R

Brad = n× Erad

Using these results, we get the power radiated away per solid angle by a moving particle. In thenon-relativistic limit, Larmor’s Formula is found:

dP

dΩ≈ q2

4πc3|a|2 sin2 θ P =

2q2

3c3|a|2

In general, the power radiated is:

P =2q2

3cγ6

[(dβ

dt

)2

−(β × dβ

dt

)2]τ0

For specific motions:

Linear → P =2q2

3c3γ6a2

Circular → P =2q2c

3R2β4γ4

Further, we found the radiated power per unit frequency and solid angle for moving point chargesor a given current,

d2I

dωdΩ=q2ω2

4π2c

∣∣∣∣∫ ∞−∞

dt′eiω[t′− 1cn·r(t′)] [n× (n× β)]∣∣∣∣2 =

ω2

4π2c3

∣∣∣∣∫ ∞−∞

dt′∫Vd3x eiω[t′− 1

cn·x] [n× (n× j)]

∣∣∣∣2We then determined that the scattering cross section of an object could be written as:

dσ

dΩ=

⟨dPdΩ

⟩S · n

=(q2

mc2

)2

|ε∗ · εin|2 =12

(q2

mc2

)2 (1 + cos2 θ

)229

σ =8π3

(q2

mc2

)2

For localized sources, we determined solutions to the vector potential in the near and far field:

Anear ≈1c

∫d3x′j

(x′)∑l,m

4π2l + 1

rl<

rl+1>

Y ∗l,m (θ, φ)Yl,m(θ′, φ′

)In the far field, we considered only the radiation terms:

Efar =k2

reikr

[(n× p)× n− 1

c(n× m)

]+O

(1r2

)Hfar = n× Efar

because of the factor of 1c , the main contribution to the radiated field is the dipole moment of the

source.

Lastly, we can use this result to get some general forms for scattering cross sections. Consider theincident and scattered fields:

Ein = ε0E0eiωt−ik0·n0 Hin = n0 × Ein

Esc =k2

reikr

[(n× p)× n− 1

c(n× m)

]Esc = n× Esc

The differential cross section is then given by:

dσ

dΩ=

∣∣Ssc∣∣2 r2∣∣Sin∣∣2 =k4

E0

∣∣∣∣ε∗ · p+1c

(n× ε∗) m∣∣∣∣2

230

Part VIII

Greens Functions and Reference Equations

66 Greens Functions for Nabla Squared

66.1 Free Space Greens Functions

G(x, x′) =1

|x− x′|For azimuthal symmetry:

1|x− x′|

=∞∑l=0

rl<

rl+1>

Pl(cosϑ)Pl(cosϑ′)

And in general:1

|x− x′|=∞∑l=0

l∑m=−l

4π2l + 1

rl<

rl+1>

Y ∗l,m(ϑ′, ϕ′)Yl,m(ϑ, ϕ)

1|x− x′|

=2π

∞∑m=−∞

∫ ∞0

dk eim(ϕ−ϕ′) cos[k(z − z′

)]Im (kρ<)Km (kρ>)

66.2 Unique Geometries

Conducting Plane (located at z = 0):

GD(x, x′) =1

|x− x′|− 1|x− x′′|

x′ = (x′, y′, z′); x′′ = (x′, y′,−z′)

Conducting Sphere (of radius a):

GD(x>, x<) =1

|x> − x<|− a

|x<|1

|x> − a2

|x<|2 x<|

Expansion form:

GD(x, x′) = 4π∞∑l=0

l∑m=−l

12l + 1

[rl<

rl+1>

(1− 1

a

(a2

r>r<

)l+1)]

Y ∗lm(ϑ′, ϕ′)Ylm(ϑ, ϕ)

∂GD∂n|x→a = −1

a

a2 − x′2

(a2 + x′2 − 2ax′ cos(γ))3/2= − 1

a2

1− x′2

a2(1 + x′2

a2 − 2x′a cos(γ))3/2

231

∂GD∂n′|x′→a = −1

a

x2 − a2

(x2 + a2 − 2ax cos(γ))3/2= − 1

ax

1− a2

x2(1 + a2

x2 − 2ax cos(γ))3/2

Conducting concentric spheres of radii a and b (a ≤ r ≤ b )

GD(r, ϑ, ϕ, r′, ϑ′, ϕ′) = 4π∞∑l=0

l∑m=−l

Y ∗lm(ϑ′, ϕ′)Ylm(ϑ, ϕ)

(2l + 1)(

1−(ab

)2l+1) [rl< − a2l+1

rl+1<

][1rl+1>

−rl>b2l+1

]

Conducting Plane With Hemisphere Bulge of Radius a

GD(x, x′) =1

|x− x′|− 1|x− x′′|

− a

|x′|1

|x− a2

|x′|2 x′|

+a

|x′′|1

|x− a2

|x′′|2 x′′|

x′ = (x′, y′, z′); x′′ = (x′, y′,−z′)

L-Shaped Wedge Along x > 0 and y > 0 from z = 0 to ∞.

GD(x, x′) =1

|x− x′0|− 1|x− x′1|

− 1|x− x′2|

+1

|x− x′3|

x′0 = (x′, y′, z′); x′1 = (x′,−y′, z′); x′2 = (−x′, y′, z′); x′3 = (−x′,−y′, z′)

Parallel Conducting Plates at z = 0 and z = a

GD(x, x′) =∞∑

n=−∞

1|x− x′n|

− 1|x− x′′n|

x′n = (x′, y′, 2an+ z′); x′′n = (x′, y′, 2an− z′)

67 Greens Function for the Helmholtz Operator

(∇2 − k2

)→ G

(x, x′

)=eik|x−x

′|

|x− x′|

232

68 Expansions

Given φ on some surface ∂V :

φ(x, y, z) =∫ ∞−∞

d3k A(k1, k2, k3)eik1xeik2yeik3z

φ(r, ϑ) =∞∑l=0

[Alr

l +Blr−(l+1)

]Pl (cosϑ)

φ(r, ϑ, ϕ) =∞∑l=0

l∑m=−l

[Al,mr

l +Bl,mr−(l+1)

]Yl,m (ϑ, ϕ)

φ(ρ, ϕ, z) =∞∑n=0

∫dk (An(k)Jn(kρ) +Bn(k)Yn(kρ))

(Cn(k)einϕ +Dn(k)e−inϕ

) (E(k)ekz + F (k)e−kz

)

233

Part IX

Homework Assignments (Problems only)

69 Semester 1

69.1 Homework 1

69.1.1 Problem 1

Using the definition of the Dirac Delta function, prove the following properties:

δ (x) = δ (−x) xδ (x) = 0

δ [y(x)] =∑i

δ (x− xi)(dy

dx

)−1

x=xi

where y(xi) = 0

69.1.2 Problem 2

Show that the one dimensional integral representation of the Dirac Delta function given by:

δ(x) =1

2π

∫ ∞−∞

dkeikx

can be written as the distribution limn→∞ nG(x) where G(x) = e−πx2.

69.1.3 Problem 3

Using Dirac delta functions in the appropriate coordinates, express the following charge distributions as threedimensional charge densities ρ (x).

1. Total charge Q distributed evenly over a spherical shell of radius R.

2. Charge of λ per unit length distributed over an infinitely long cylindrical shell of radius R.

3. Total charge Q distributed evenly over a disc of radius R and negligible thickness.

4. Square loop with negligible radius, side length L and total charge Q.

234

69.2 Homework 2

69.2.1 Problem 1

The Greens function in two dimensions is defined by the equation:

∇2⊥G(x, x′) = −2πδ(2) (x− x′)

Where ∇⊥ ≡ ∂2

∂x2 + ∂2

∂y2 .

1. Find the Greens function G (x, x′).

2. Using the result of part (a), write the solution of the Poisson Equation ∇2φ = − 1ε0ρ (x) in two

dimensions with no boundary conditions.

69.2.2 Problem 2

Show by derivation (not by substitution!) that the Greens Function for the one-dimensional operator

L = −(d2

dx2+ k2

), LG (x, x′) = −δ (x− x′)

with boundary conditions G(0, x′) = dGdx (1, x′) = 0 is:

G(x, x′) = − sin(kx) cos [k (1− x′)]k cos k

Θ (x′ − x)− sin(kx′) cos [k (1− x)]k cos k

Θ (x− x′)

69.2.3 Problem 3

Show that for a region V bounded by conductors (not necessarily grounded), the following relation holds:

∫V

[ρ1φ2 − ρ2φ1] dV =∑n

[Q2V1 −Q1V2]n

where Q and V represent the net charge and voltage on each conductor, the sum is then over all conductorsand the subscripts 1 and 2 refer to tow different cases of charge distribution and the resulting potentialwith the same geometry for the conductors. Use the above result to prove that the potential of a neutralconducting sphere of radius a due to a point charge at a distance d > a is V = q

d .

235

69.3 Homework 3

69.3.1 Problem 1

Show that the potential outside of a sphere with radius a with two conducting hemispheres held at constantpotential ±V0 and separated by a thing insulating ring can be written as:

φ(r, θ, φ) =3V0a

2

2r2

[cos θ − 7a2

12r2

(52

cos3 θ − 32

cos θ)

+ . . .

]for r >> a and θ is the angle between the positive pole and the observation point.

69.3.2 Problem 2

Using the method of images, discuss the problem of a point charge q inside a thin hollow grounded sphere ofradius a. Determine:

1. The potential inside and outside the sphere.

2. the induced surface charge density and the total induced charge.

3. the magnitude and directions of the force acting on the point charge.

69.3.3 Problem 3

The insulating floor of a laboratory is covered with thin flat circular metal tiles of radius a, held at finitepotential. Assume that the surface of the laboratory is much larger than any measuring device.

1. Write the appropriate Green Function for the lab.

2. If a tile is held at constant potential φ = V0, while all others are grounded, use an integral expressionto write the potential at a generic point in the lab.

3. Show that along the axis of the tile,

φ(z) = V0

(1− z√

a2 − z2

)4. Show that at large distances (ρ2 + z2 >> a2), the potential is approximately:

φ ≈ V0a2

2z

(ρ2 + z2)32

[1− 3a2

4 (ρ2 + z2)+

5(3ρ2a2 + a4

)8 (ρ2 + z2)2 + . . .

]

236

69.4 Homework 4

69.4.1 Problem 1

A grounded conductor has the shape of an infinite plane except for a hemispherical bulge of radius a. If acharge q is placed above the center of the bulge at a distance d(> a) above the plane.

1. Find the potential in the region above the conductor.

2. Find the total induced charge on the conductor.

3. find the Dirichlet Green’s function.

69.4.2 Problem 2

A charge q is placed near the inner corner of a grounded, semi-infinite L-shaped conductor a distance d fromboth arms.

1. Determine the potential in the region containing the point charge.

2. Determine the Dirichlet Greens Function.

3. Determine the potential in the outer region if the conductor is held at constant potential V0.

69.4.3 Problem 3

Consider a pair of infinitely long parallel-plate conductors separated by a distance a. If a point charge isplaced a distance a from the top plate (a− d from the bottom plate):

1. Using the method of images, compute the potential generated by this configuration.

2. Write the Dirichlet Green’s function.

3. Compute the induced surface charge distribution.

237

69.5 Homework 5

69.5.1 Problem 1

A sphere of radius a has a charge uniformly distributed over its surface with charge density Q(4πa2

)−1,except for a spherical cap at the north pole defined by the cone θ = α. This cone is kept at zero potential.Show that the potential outside the sphere is:

φ(r, θ) =Q

8πε0a

∞∑l=0

Pl+1 (cosα)− Pl−1 (cosα)2l + 1

(ar

)l+1

Pl (cos θ)

Where P−1 (cosα) is defined to be equal to −1. Discuss the limiting form of the potential as the cap becomesvery large or very small.Hint: The relation (2l + 1)Pl (x) = P ′l+1 (x)− P ′l−1 (x) is necessary to solve this problem.

69.5.2 Problem 2

A thin flat conducting disc of radius a is maintained at constant potential V0. If the surface charge densityis proportional to

(a2 − ρ2

)− 12 ,

1. Show that the potential for r > a is:

φ =2V aπr

∞∑l=0

(−1)l

2l + 1

(ar

)2l

P2l (cos θ)

2. Find the potential for r < a.

3. Find the capacitance of the disc.

69.5.3 Problem 3

A flat conducting ring of infinitesimal thickness, inner radius a, and outer radius b is uniformly charged withtotal charge Q.

1. Write the 3-dimensional charge distribution density in cylindrical coordinates.

2. Find the potential inside the ring.

238

69.6 Homework 6

69.6.1 Problem 1

Show that the Greens Function for a flat 2-dimensional circular wedge of angle β and radius a is:

G(ρ, ρ′, φ, φ′) =∞∑n=1

4nρnπβ

−

[ρnπβ

+ −(ρ+

a

)nπβ

]sin(nπ

βφ

)sin(nπ

βφ′)

69.6.2 Problem 2

A hollow thin cylinder of length L and radius a has the end faces kept at zero potential. The potentials onthe other surface is described by φ(a, φ, z) = V (φ, z). Using the cylindrical coordinates with the end faces atz = 0, L, determine the series solution for the potential inside. How does this series simplify for a constantpotential V (φ, z) = V0?

69.6.3 Problem 3

An infinite thin flat sheet of conducting material has a circular thing cut of radius a. The part of theconducting sheet inside the cut is kept at constant potential V0, while the sheet outside the cut is grounded(zero potential).

1. Show that the potential at any point above the sheet is:

φ(ρ, φ, z) =∫ ∞

0

dk e−kz

[12B0J0 (kρ) +

∞∑n=1

Jn(kρ) (An(k) sin (nφ) +Bn(k) cos (nφ))

]

where, An(k)Bn(k)

=kV0

π

∫ ∞0

dρ ρ

∫ 2π

0

dφ Jn (kρ)

sin(nφ)cos(nφ)

2. Using the limit ρ→ 0 of this result, find the potential along the axis of the inner part of the sheet.

3. Show that the potential above the cut is:

φ(φ, z) = V a

∫ ∞0

dk e−kzJ0 (ka) J1 (ka)

239

69.7 Homework 7

69.7.1 Problem 1

A nucleus with quadrupole moment Q is placed in external electric field E = (Ex, Ey, Ez). Show that thequadrupole contribution to the energy is:

W (4) = −e4Q [∇zEz]x=0

(For notations, see discussion in Jackson Section 4.2)

69.7.2 Problem 2

Write the multipole expansion for the three dimensional localized charge distribution:

ρ (x) =1

64πr2e−r sin θ

how many non-vanishing multipoles does the expansion have? Find the exact potential outside the chargedistribution.

69.7.3 Problem 3

Compute the multipole momenta and the potential multipole expansion for an axial molecule H − O −H,where the distance between H and O is a = 0.95718A, and H and O are modeled by point charges |e| and−2|e| respectively. (note: this is not the correct description of the water molecule. In reality, the molecule isnot axial, but makes an angle of 104.474o)

240

69.8 Homework 8

69.8.1 Problem 1

Two concentric spheres of radii a and b carry charge −Q and +Q respectively. Half of the cavity betweenthe spheres is filled with a dielectric material with dielectric constant ε. Find:

1. The electric field between the spheres.

2. The surface charge distribution on the outer surface of the inner sphere.

3. The polarization charge density induced on the surface of the dielectric at r = a.

69.8.2 Problem 2 - Jackson 4.8 parts (a) and (c)

A long, right cylindrical shell of dielectric has constant εε0

and inner and outer radii a and b. The mediumoutside the dielectric has ε

ε0= 1. If the cylinder is placed in an external uniform electric field perpendicular

to the axis of the cylinder.a) Determine φ and E in all three regions.c) Discuss the limiting cases of a→ 0 and b→∞.

69.8.3 Problem 3

Consider a flat face dielectric with a point charge q at a distance d from the face.

1. Find the polarization charge density and total charge on the face.

2. Find the force exerted on the dielectric by the point charge.

241

69.9 Homework 9

69.9.1 Problem 1

Show that the magnetic induction B on the axis of a circular solenoid of length L, radius a, carrying arighthanded current I is:

Bz =µ0nI

2

1√1 + a2

z2

+1√

1 +[

aL−z

]2

Where n >> 1 is the number of turns per unit length.

69.9.2 Problem 2

A spherical conducting shell of radius a and surface charge σ rotates with constant angular velocity ω =(0, 0, ω). Compute the magnetic induction of the field B inside and outside the spherical shell.

69.9.3 Problem 3- Jackson 5.15 (a) and (b)

Consider a pair of long straight wires carrying current in the z direction of I and −I. Describe the magneticfield H in terms of φM ; H = ∇φM .(a) If the wires are located at x = ±d2 , y = 0 show that for d >> ρ, the potential is that of a 2-dimensionaldipole:

φM = − Id2π

sinφρ

+O

(d2

ρ2

)(b) Closely placed wires are enclosed in a centered cylinder of radii a and b with µ = µrµ0. Determine φMfor ρ < a, a < ρ < b, and b < ρ. Show that the field outside is the 2-dimensional dipole as in part (a), onlywith a scaling factor of:

F =4µrb2

(µ2r + 1) b2 − (µ2

r − 1)2a2

242

69.10 Homework 10

69.10.1 Problem 1 - Jackson 5.3

Right-circular solenoid of length L, radius a, and N turns per unit length carries current I. Show that theaxial magnetic inductance can be written as:

Bz =µ0NI

2(cos θ1 + cos θ2)

In the limit of NL→∞. The angles are defined by vectors from a point z along the axis to the edges of thesolenoid.

69.10.2 Problem 2 - Jackson 5.8

A region contains j = j(r, θ)φ. The distribution is ”hollow”, meaning that j(r → 0) = 0.(a) Show that:

Aφ(r, θ) = −µ0

4π

∑l

mlrlP ′l (cos θ) r inside hollow

Aφ(r, θ) = −µ0

4π

∑l

nlr−(l+1)P ′l (cos θ) r outside hollow

69.10.3 Problem 3 - Jackson 5.20

(a) Starting from the relation:

F =∫d3xj (x)× Bext (x)

where Bext is some external field and given that M inside of a region bounded by a surface S is equivalentto a surface current and volume current described as:

σM = M × n jM = ∇× M

show that the total magnetic force in the absence of macroscopic conductive currents is:

F = −∫V

(∇ · M

)Bextd

3x+∫∂V

(M · n

)Bextd

2x

(b) Consider a sphere of radius R with uniform magnetization M located at r = 0 with magnetization alongsome generic direction θ0, φ0. If the external field is:

Bext =(B0 (1 + βy)B0 (1 + βx)

)Calculate the components of the force on the sphere.

243

69.11 Homework 11

69.11.1 Problem 1- Jackson 6.20 (a) and (b)

A dipole source flashes at t = 0,

ρ (x, t) = δ(x)δ(y)δ′(z)δ(t) j = −δ(x)δ(y)δ(z)δ′(t)

Where δ′(a) = ∂∂aδ(a).

a) Show that the instantaneous Coulomb potential is

φ (x, t) = − 14πε0

δ(t)z

r3

b) Show that the transverse current jt is given by:

jt (x, t) = −δ′(t)[

23ε3δ(x)− ε3

4πr3+

34πr3

n (ε3 · n)]

69.11.2 Problem 2 - Jackson 6.8

A dielectric sphere (dielectric constant ε, radius a, centered at origin) has a uniform electric field E = E0xapplied. If the sphere rotates about z with angular velocity ω, show that there is a magnetic field H = −∇φMwhere:

φM =35

(ε− ε0ε+ 2ε0

)ε0E0ω

(a

r>

)5

xz

where r> is the larger of (a, r).

69.11.3 Problem 3

A localized electric charge distribution produces an electrostatic field E. Into this field is placed a smalllocalized time-independent current density j.(a) Show that the total momentum of the EM field is:

p =1c2

∫d3xφj

where φ is the potential of the localized charge distribution, provided the product φH falls off rapidly enoughat large distances.(b) Assuming that the current is localized to a small region (compared to scale of field variation), expandthe electrostatic potential in Taylor series and show that:

p =1c2E(0)× m m =

12

∫d3xx× j (x)

244

70 Semester 2

70.1 Homework 1

70.1.1 Problem 1

1. Show that the Lie Algebra of the orthogonal group O (n) is the set of all real skew-symmetric realmatrices. Show that the dimension of the group is n(n−1)

2 .

2. Show that the Lie Algebra of the SU(n) is the set of all complex matrices A that satisfy AT = −A andtrA = 0. (Hint: use the relation det

(eiA)

= eitr(A)). Show that the dimension of the group is n2 − 1.

70.1.2 Problem 2 - Jackson 11.5

A frame K ′ moves with velocity v relative to another frame K. If a particle has velocity and acceleration u′

and a′ in K ′, show that in K,

a‖ =

(1− v2

c2

) 32

(1 + v2

c2

)3 a′‖ a⊥ =1− v2

c2(1 + v2

c2

)3 [a′⊥ +v

c2× (a′ × u′)

]

70.1.3 Problem 3 - Jackson 11.6

A rocket leaves earth in 2100. One of a set of twins born in 2080 stays on Earth, the other rides the rocket.If the rocket accelerates at g in rest frame along a straight-line path for 5 years, decelerates for 5 years andthen turns and repeats the trip, the twin in the rocket is 40 years older.

1. What year is it on earth?

2. How far from Earth did the rocket travel?

245

70.2 Homework 2

70.2.1 Problem 1

A runner carries a 20-m pole so fast in the direction of it length that it appears to be only 10 m long in arest frame. As the runner carries the pole through the front door of a barn 10 m long, just at the instantthe head of the pole reaches the closed rear door. Then the front door is closed, enclosing the pole in thebarn. The other door opens and the runner goes through. From the runner’s point of view, however, thepole is 20 m long and the barn is only 5 m deep. Thus, the pole can never be enclosed in the barn. Explainqualitatively and quantitatively this apparent paradox.

70.2.2 Problem 2

In an inertial frame, to events occur simultaneously at a distance of 3 m apart. In a frame moving withrespect to the inertial frame, one event occurs later than the other by a time difference of 10−8s. By whatspatial distance are the two events separated in the moving frame?

70.2.3 Problem 3 - Jackson 11.9

An infinitesimal Lorentz transformation and its inverse can be written as:

x′α =(gαβ + εαβ

)xβ xα =

(gαβ + ε′αβ

)x′β

where ε and ε′ are infinitesimal quantities.

1. Show that from the definition of inverse, ε′αβ = −εαβ .

2. show that by preservation of the norm, εαβ = −εβα.

3. Write the transform in terms of contravariant components on both sides of the equation to show thatεαβ is equivalent to the matrix L in equation 11.93.

70.2.4 Problem 4

Compute explicitly the following quantities in terms of the electric and magnetic fields. Can you guess whichof them are zero before you compute them?

FνµFνµ Fµµ FνµF

νρ ενµρσF

νµF ρσ

70.2.5 Problem 5

If Sα and T βγ are tensors under Lorentz transformations, show explicitly that each of the following are alsotensors under Lorentz transformations:

1. Wαβγ ≡ SαT βγ

2. Xαβγ ≡ SαTβγ

3. Yα ≡ SβT βα4. Zαβγ ≡ ∂αTβγ

246

70.2.6 Problem 6

Prove the following relations are true:

− 13!εµνστ ενµσα = δτα − 1

2!εµνστ ενµαβ = δσταβ − εµνστ εναβγ = δνσταβγ

Where δτα is a Kronecker tensor and δα1...αnβ1...βn

is the complete antisymmetric product of n Kronecker tensors.

247

70.3 Homework 3

70.3.1 Problem 1

An observer measures E and B in a rest frame as:

E = E0n B = 2E0n′

A second observer sees the fields as parallel. Determine the relative velocity between the observers.

70.3.2 Problem 2

A particle of mass M at rest in a laboratory decays into a pair of particles with mass m1 and m2. Show thatthe total energy and kinetic energy of the first particle in the lab frame are:

E1 =M2 +m2

1 −m22

2Mc2 T = Qc2

(1− m

M− Q

2M

)Q = M −m1 −m2

70.3.3 Problem 3

Consider some tensor T νµ that satisfies ∂muT νµ = 0. Show that:

d2

dt2

∫d3xT 00xixj = 2

∫d3xT ij

where i, j = 1, 2, 3 and T νµ is localized in space (boundary terms can be neglected).

248

70.4 Homework 4

70.4.1 Problem 1 - Jackson 11.23

Two particles collide in a laboratory frame. Use center of momentum coordinates to solve the problem.

mass = m1 p1 int = plab, Energy = Elabmass = m2 p2 int = 0

−→

mass = m3 p3 at angle θ3

mass = m4 p4 at angle θ4

1. Use invariant scalar products to show that the total energy in the center of momentum frame and the3 momentum are:

W 2 = m21 +m2

2 + 2m2Elab p′ =m2

Wplab

2. Show that the Lorentz transform to the center of mass coordinate are:

βcm =pcm

m2 + Elabγcm =

m2 + ElabW

3. Show that in the non-relativistic limit, the above results reduce to the familiar:

W ≈ m1 +m2 +(

m2

m1 +m2

)p2lab

2m1p′ ≈ m2

m1 +m2plab βcm ≈

plabm1 +m2

70.4.2 Problem 2 - Jackson 11.28 (a) and (b)

An electric dipole instantaneously at rest at the origin in frame K ′ has only electric field and potentialsφ′ = p·r

|r|3 ; A′ = 0. If this frame moves with velocity v = 1c β with respect to some other frame K,

1. Show that in K:

φ =p · RR3

A = βp · RR3

R = x− x0(t) with v =dx0

dt

2. Show explicitly that the potentials in K satisfy the Lorentz Condition (∂nuAν = 0).

70.4.3 Problem 3

A particle of mass m and charge e moves in a uniform, static, electric field E0.

1. Solve for the velocity and position of the particle as an explicit function of time, assuming that theinitial velocity v0 is perpendicular to the electric field.

2. Using the previous result, derive the trajectory of the particle. Discuss the shape of the trajectory forshort and long times (define what ”short” and ”long” times are).

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70.5 Homework 5

70.5.1 Problem 1 - Jackson 12.6

Static electric and magnetic fields make angle θ between them.(a) By suitable choice of axes, solve for the motion of a particle with mass m and charge e in rectangularcoordinates.(b) For parallel fields, show that the results above reduce to:

x = AR sinφ y = AR cosφ z =R

ρ

√1 +A2 cosh (ρφ) ct =

R

ρ

√1 +A2 sinh (ρφ)

R =mc2

eBA = arbitrary constant ρ =

E0

B0

70.5.2 Problem 2

(a) Show that if the Lagrangian density for a set of fields is invariant under infinitesimal coordinate trans-formations xα → xα + εα, then

∂νTνµ = 0 Tµν =

∂£∂ (∂µφi)

∂νφi − gµν£

(b) Show that if the Lagrangian density for a set of fields is invariant under infinitesimal coordinate trans-formations xα → xα + εαβxβ where εαβ = −αβα, then

∂νMνµρ = 0 Mµνρ = (xρgµν − xνgµρ) £ +

∂£∂ (∂µφi)

(xν∂ρ − xρ∂ν)φi

This quantity is the relativistic generalization of a classical quantity. which one? What kind of infinitesimaltransformations are the above?

70.5.3 Problem 3 - Jackson 12.16

Starting with the Proca-Lagrangian, show that Tαβ for the case of a massive photons is:

Θαβ =1

4π

[gαγFγλF

λβ +14GαβFλνF

λν + ν2

(AαAβ − 1

2gαβAλA

λ

)]Show that for fields in interaction with some source jρ, the differential conservation laws take the form:

∂αΘαβ =1cjλF

λβ

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70.6 Homework 6

70.6.1 Problem 1

Two plane semi-infinite slabs of the same uniform, isotropic, nonpermeable, lossless dielectric with index ofrefraction n are parallel and separated by an air gap (n = 1) of width d. A plane EM wave of frequency ω isincident on the gap from one of the slabs with an incident angle θi. For linear polarization both parallel andperpendicular to the plane of incidence, calculate the reflection and transmission coefficients for the wholegap.

70.6.2 Problem 2

the time dependence of electric pulses in a good conductor is governed by the frequency-dependent conduc-tivity,

σ =f0Ne

2

m (γ0 − iω)

Consider longitudinal electric fields in a conductor. Using Ohm’s law, Gauss’ Law, and the continuityequation,(a) Show that the time-Fourier transformed charge density satisfies the equation (in Gaussian units)

[σ(ω)− iω

4π

]ρ (x, ω) = 0

(b) Using σ(ω) = σ01−iωτ where σ0 = ω2

pτ

4π and τ is a damping time, show that the for the approximationω0τ >> 1, any initial pulse will oscillate with the plasma frequency and decay in amplitude with a delayconstant λ = (2τ)−1.

70.6.3 Problem 3

The ionosphere can be roughly approximated by a medium with dielectric constant ε = 1 − ω2p

ω2 , whereω2p = 4πNZe2

m , starting from a height of 300 km and extending to infinity. (All are given in Gaussian units).For waves with polarization both parallel and perpendicular to the plane of incidence.(a) Show that for ω > ωp, reflection toward the earth is total for angle of incidence greater than some criticalangle. Determine this critical angle.(b) A radio amateur operating at λ = 21m can only receive stations at a distance of d > 1000km. Assumingthat the waves are reflected by the lowest layer of the ionosphere, estimate the density of electrons in theionosphere.

251

70.7 Homework 7

70.7.1 Problem 1 - Jackson 7.19

An approximately monochromatic wave packet has the form u(x, 0) = f(x)eik0x with f(x) the modulationenvelope. For each given form of f(x), calculate |A(k)|2, sketch both |u(x, 0)|2 and |A(k)|2. Evaluate theRMS derivations ∆x, ∆k normalized to the source functions and show that in each case ∆x∆k ≥ 1

2 .

1. f(x) = Ne−α2 |x|

2. f(x) = Ne−α24 x

2

3. f(x) =N (1− α |x|) α|x| < 1

0 α|x| > 1

4. f(x) =

N −α < x < α0 else

70.7.2 Problem 2 - Jackson 7.20

A homogeneous, isotropic, nonpermeable dielectric is characterized by n(ω) (possibly complex)(a) Show that the general solution for plane waves in 1-dimensions can be written as:

u(x, t) =1√2π

∫ ∞−∞

dωe−iωt[A(ω)ei

ωc n(ω)x +B(ω)e−i

ωc n(ω)x

](b) Show that given u(0, t) and u′(0, t), the coefficients above can be determined as:

A(ω)B(ω)

=

12

1√2π

∫ ∞−∞

dt eiωt[u(0, t)± ic

ωn(ω)u′(0, t)

]

70.7.3 Problem 3 - Jackson 7.22

Use the Kramer-Kronig relations to determine the real part of ε given the imaginary part below. Sketch boththe real and imaginary parts for each case.

(a) Im (ε) = λ [Θ (ω − ω1)−Θ (ω − ω2)] ω2 > ω1 > 0

(b) Im (ε) = λγω

(ω20−ω2)2

+γ2ω2

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70.8 Homework 8

70.8.1 Problem 1

Consider a perfectly conducting waveguide with circular cross section of radius a.

1. Find the TE and TM propagating modes and their cutoff frequencies.

2. The waveguide is turned into a coaxial cable by the introduction of a concentric cylindrical conductorof radius b inside the waveguide. Compute the modes and frequencies when the difference between theradii is smaller compared to the average radius.

3. The waveguide (without the coaxial contribution) is turned into a cavity of length L by putting twoflat end faces in the guide. Determine the resonant frequencies of the cavity.

70.8.2 Problem 2

Consider and infinitesimally thin, perfectly conducting right-triangular waveguide with wall lengths a, a, and√2a.

1. By analytical extension, show that each triangular TM (TE) mode generates a TM (TE) mode of asquare waveguide of side length a.

2. Find the TM and TE modes of the triangular waveguide and their propagation frequencies.

70.8.3 Problem 3

Consider an optical fiber with graded index of refraction for rays confined to the x − z plane, n(x) =n(0)sech (αx).

1. For transverse coordinate x(z) of the ray, show that:

αx = sinh−1 [sinh (αx0) sin (αz)]

where n(x0) = n(0) cos [θ(0)], and the origin in z is when the ray has x = 0.

2. Find the optical path and the half period of the ray.

253

70.9 Homework 9

70.9.1 Problem 1

Calculate the time-averaged power radiated per unit solid angle and the total power radiated for a non-relativistic particle with charge e moving (a) along the z axis with displacement z(t) = z0sin (ω0t) and (b)in a circle of radius (R) in the xy plane with constant angular frequency ω0.

70.9.2 Problem 2 - Jackson 14.11

A particle of charge Ze and mass m moves in external E and B fields. (a) Show that the classical relativisticresult for the instantaneous radiated energy per unit time is:

P =23Z4e4

m2c3γ2[(E + β × B

)2 − (β · E)2]Where E and B are evaluated at the particle and the γ factor is the particles instantaneous Lorentz factor.(b) Show that the result of part (a) can be expressed as

P =2Z4r2

0

3m2cFµνpνp

λFλµ r0 ≡e2

mc2

70.9.3 Problem 3

A non-relativistic particle of charge ze, mass m, and initial speed v0 scatters on a fixed nucleus of atomicnumber Z at an impact parameter b. Assuming there is no deflection, show that the total energy radiated is:

W =πz4Z2e6

3m2c3v0b3

254