(Brand) Vector and Tensor Analysis (1947)

VECTOR and TENSOR

ANALYSISBy

LOUIS BRAND, Ch.E., E.E., Ph.D.PROFESSOR OF MATHEMATICS

UNIVERSITY OF CINCINNATI

New York JOHN WILEY & SONS, Inc.

London CHAPMAN & HALL, Limited

VECTOR AND TENSOR ANALYSIS.By Louis Brand. 439 pages. 5Y2 by8%. Cloth.

VECTORIAL MECHANICS. By LouisBrand. 544 pages. 5% by 8%.Cloth.

Published by John Wiley & Sons, Inc.

VECTOR and TENSOR

ANALYSISBy

LOUIS BRAND, Ch.E., E.E., Ph.D.PROFESSOR OF MATHEMATICS

UNIVERSITY OF CINCINNATI

New York JOHN WILEY & SONS, Inc.

London CHAPMAN & HALL, Limited

COPYRIGHT, 1947

BY

Louis BRAND

All Rights Reserved

This book or any part thereof must notbe reproduced in any form withoutthe writ+en permission of the putlisher.

THIRD PRINTING, NOVEMBER, 1948

PRINTED IN THE UNITED STATES OF AMERICA

To My Wife

PREFACE

The vector analysis of Gibbs and Heaviside and the moregeneral tensor analysis of Ricci are now recognized as standardtools in mechanics, hydrodynamics, and electrodynamics. Thesedisciplines have also proved their worth in pure mathematics,especially in differential geometry. Their use not only materiallysimplifies and condenses. the exposition, but also makes mathe-matical concepts more tangible and easy to grasp. Moreovertensor analysis provides a simple automatic method for construct-ing invariants. Since a tensor equation has precisely the sameform in all coordinate systems, the desirability of stating physicallaws or geometrical properties in tensor form is manifest.. Theperfect adaptability of the tensor calculus to the theory of rela-tivity was responsible for its original renown. It has since won afirm place in mathematical physics and engineering technologyThus the British analyst E. T. Whittaker rates the discovery ofthe tensor calculus as one of the three principal mathematicaladvances in the last quarter of the 19th century.

The first volume of this work not only comprises the standardvector analysis of Gibbs, including dyadics or tensors of valencetwo, but also supplies an introduction to the algebra of motors,which is apparently destined to play an important role in mechanicsas well as in line geometry. The entire theory is illustrated bymany significant applications; and surface geometry and hydro-dynamics * are treated at some length by vector methods inseparate chapters.

For the sake of concreteness, tensor analysis is first developedin 3-space, then extended to space of n dimensions. As in the caseof vectors and dyadics, I have distinguished the invariant tensorfrom its components. This leads to a straightforward treatmentof the affine connection and of covariant differentiation; and alsoto a simple introduction of the curvature tensor. Applications oftensor analysis to relativity, electrodynamics and rotating elec-

* For a systematic development of mechanics in vector notation see theauthor's Vectorial Mechanics, John Wiley & Sons, New York, 1930.

vii

viii PREFACE

tric machines are reserved for the second volume. The presentvolume concludes with a brief introduction to quaternions, thesource of vector analysis, and their use in dealing with finiterotations.

Nearly all of the important results are formulated as theorems,in which the essential conditions are explicitly stated. In thisconnection the student should observe the distinction betweennecessary and sufficient conditions. If the assumption of a certainproperty P leads deductively to a condition C, the condition isnecessary. But if the assumption of the condition C leads deduc-tively to the property P, the condition C is sufficient. Thus we-have symbolically

P -* C (necessary), C (sufficient) -> P.

When P C, the condition C is necessary and sufficient.The problems at the end of each chapter have been chosen not

only to develop the student's technical skill, but also to introducenew and important applications. Some of the problems are mathe-matical projects which the student may carry through step by stepand thus arrive at really important results.

As very full cross references are given in this book, an articleas well as a page number is given at the top of each page. Equa-tions are numbered serially (1), (2), . . . in each article. A ref-erence to an equation in another article is made by giving articleand number to the left and right of a point; thus (24.9) meansarticle 2.4, equation 9. Figures are given the number of the articlein which they appear followed by a serial letter; Fig. 6d, for ex-ample, is the fourth figure in article 6.

Bold-face type is used in the text to denote vectors or tensorsof higher valence with their complement of base vectors. Scalarcomponents of vectors and tensors are printed in italic type.

The rich and diverse field amenable to vector and tensor methodsis one of the most fascinating in applied mathematics. It is hopedthat the reasoning will not only appeal to the mind but also im-pinge on the reader's aesthetic sense. For mathematics, whichGauss esteemed as "the queen of the sciences" is also one of thegreat arts. For, in the eloquent words of Bertrand Russell:

"The true spirit of delight, the exaltation, the sense of beingmore than man, which is the touchstone of the highest excellence,is to be found in mathematics as surely as in poetry. What is

PREFACE is

best in mathematics deserves not merely to be learned as a task,but also to be assimilated as a part of daily thought, and broughtagain and again before the mind with ever-renewed encourage-ment. Real life is, to most men, a long second-best, a perpetualcompromise between the real and the possible; but the world ofpure reason knows no compromise, no practical limitations, nobarrier to the creative activity embodying in splendid edifices thepassionate aspiration after the perfect from which all great worksprings."

The material in this book may be adapted to several shortcourses. Thus Chapters I, III, IV, V, and VI may serve as acourse in vector analysis; and Chapters I (in part), IV, V, and IXas one in tensor analysis. I But the prime purpose of the authorwas to cover the theory and simpler applications of vector andtensor analysis in ordinary space, and to weave into this fabricsuch concepts as dyadics, matrices, motors, and quaternions.

The author wishes, finally, to express his thanks to his col-leagues, Professor J. W. Surbaugh and Mr. Louis Doty for theirhelp with the figures. Mr. Doty also suggested the notation usedin the problems dealing with air navigation and read the entirepage proof.

Louis BRANDUniversity of CincinnatiJanuary 15, 1947

CONTENTSPAGE

PREFACE . . . . . . . . . . . . . . . . . . . . . . . . . . . . Vii

CHAPTER I

VECTOR ALGEBRAARTICLE

1. Scalars and Vectors . . . . . . . . . . . . . . . . . . . . . 1

2. Addition of Vectors . . . . . . . . . . . . . . . . . . . . . 33. Subtraction of Vectors . . . . . . . . . . . . . . . . . . . 54. Multiplication of Vectors by Numbers . . . . . . . . . . . . . 65. Linear Dependence . . . . . . . . . . . . . . . . . . . . . 76. Collinear Points . . . . . . . . . . . . . . . . . . . . . . 87. Coplanar Points 128. Linear Relations Independent of the Origin. . . . . . . . . . 189. Centroid . . . . . . . . . . . . . . . . . . . . . . . . . 19

10. Barycentric Coordinates . . . . . . . . . . . . . . . . . . . 2311. Projection of a Vector . . . . . . . . . . . . . . . . . . . . 2412. Base Vectors . . . . . . . . . . . . . . . . . . . . . . . . 2413. Rectangular Components . . . . . . . . . . . . . . . . . . 2614. Products of Two Vectors . . . . . . . . . . . . . . . . . . . 2915. Scalar Product . . . . . . . . . . . . . . . . . . . . . . . 2916. Vector Product . . . . . . . . . . . . . . . . . . . . . . . 3417. Vector Areas . . . . . . . . . . . . . . . . . . . . . . . . 3718. Vector Triple Product . . . . . . . . . . . . . . . . . . . . 4019. Scalar Triple Product . . . . . . . . . . . . . . . . . . . . 4120. Products of Four Vectors . . . . . . . . . . . . . . . . . . 4321. Plane Trigonometry . . . . . . . . . . . . . . . . . . . . 4422. Spherical Trigonometry . . . . . . . . . . . . . . . . . . . 4423. Reciprocal Bases . . . . . . . . . . . . . . . . . . . . . . 4624. Components of a Vector . . . . . . . . . . . . . . . . . . . 4825. Vector Equations . . . . . . . . . . . . . . . . . . . . . . 5026. Homogeneous Coordinates . . . . . . . . . . . . . . . . . . 5127. Line Vectors and Moments . . . . . . . . . . . . . . . . . 5528. Summary: Vector Algebra . . . . . . . . . . . . . . . . . . 57

Problems . . . . . . . . . . . . . . . . . . . . . . . . . 59

CHAPTER II

MOTOR ALGEBRA

29. Dual Vectors . . . . . . . . . . . . . . . . . . . . . . . . 6330. Dual Numbers . . . . . . . . . . . . . . . . . . . . . . . 6431 Motors . . . . . . . . . . . . . . . . . . . . . . . . . . 65

xi

xii CONTENTSARTICLE PAGE32. Motor Sum . . . . . . . . . . . . . . . . . . . . . . . . 6733. Scalar Product . . . . . . . . . . . . . . . . . . . . . . . 6834. Motor Product . . . . . . . . . . . . . . . . . . . . . . . 7035. Dual Triple Product . . . . . . . . . . . . . . . . . . . . . 7236. Motor Identities . . . . . . . . . . . . . . . . . . . . . . 7337. Reciprocal Sets of Motors . . . . . . . . . . . . . . . . . . 7438. Statics . . . . . . . . . . . . . . . . . . . . . . . . . . 7539. Null System . . . . . . . . . . . . . . . . . . . . . . . . 7840. Summary: Motor Algebra . . . . . . . . . . . . . . . . . . 80

Problems . . . . . . . . . . . . . . . . . . . . . . . . . 82

CHAPTER III

VECTOR FUNCTIONS OF ONE VARIABLE

41. Derivative of a Vector . . . . . . . . . . . . . . . . . . . . 8442. Derivatives of Sums and Products . . . . . . . . . . . . . . 8643. Space Curves . . . . . . . . . . . . . . . . . . . . . . . 8844. Unit Tangent Vector . . . . . . . . . . . . . . . . . . . . 9045. Frenet's Formulas . . . . . . . . . . . . . . . . . . . . . 9246. Curvature and Torsion . . . . . . . . . . . . . . . . . . . 9547. Fundamental Theorem . . . . . . . . . . . . . . . . . . . 9748. Osculating Plane . . . . . . . . . . . . . . . . . . . . . . 9849. Center of Curvature . . . . . . . . . . . . . . . . . . . . 9950. Plane Curves . . . . . . . . . . . . . . . . . . . . . . . 10051. Helices . . . . . . . . . . . . . . . . . . . . . . . . . . 10552. Kinematics of a Particle . . . . . . . . . . . . . . . . . . . 10853. Relative Velocity . . . . . . . . . . . . . . . . . . . . . 11054. Kinematics of a Rigid Body . . . . . . . . . . . . . . . . . 11455. Composition of Velocities . . . . . . . . . . . . . . . . . . 12056. Rate of Change of a Vector . . . . . . . . . . . . . . . . . 12157. Theorem of Coriolis . . . . . . . . . . . . . . . . . . . . . 12358. Derivative of a Motor . . . . . . . . . . . . . . . . . . . . 12659. Summary: Vector Derivatives . . . . . . . . . . . . . . . . 128

Problems . . . . . . . . . . . . . . . . . . . . . . . . . 130

CHAPTER IV

LINEAR VECTOR FUNCTIONS

60. Vector Functions of a Vector . . . . . . . . . . . . . . . . . 13561. Dyadics . . . . . . . . . . . . . . . . . . . . . . . . . . 13662. Affine Point Transformation . . . . . . . . . . . . . . . . . 13863. Complete and Singular Dyadics . . . . . . . . . . . . . . . . 13964. Conjugate Dyadics . . . . . . . . . . . . . . . . . . . . . 14165. Product of Dyadics . . . . . . . . . . . . . . . . . . . . . 14266. Idemfactor and Reciprocal . . . . . . . . . . . . . . . . . . 14467. The Dyadic 4) x v . . . . . . . . . . . . . . . . . . . . . . 14668. First Scalar and Vector Invariant . . . . . . . . . . . . . . . 14769. Further Invariants . . . . . . . . . . . . . . . . . . . . . 148

CONTENTS xiiiARTICLE PAGE

70. Second and Adjoint Dyadic . . . . . . . . . . . . . . . . . 15171. Invariant Directions . . . . . . . . . . . . . . . . . . . . 15372. Symmetric Dyadics . . . . . . . . . . . . . . . . . . . . . 15673. The- Hamilton-Cayley Equation . . . . . . . . . . . . . . . 16074. Normal Form of the General Dyadic . . . . . . . . . . . . . 16275. Rotations and Reflections . . . . . . . . . . . . . . . . . . 16476. Basic Dyads . . . . . . . . . . . . . . . . . . . . . . . . 16677. Nonion Form . . . . . . . . . . . . . . . . . . . . . . . 16778. Matric Algebra . . . . . . . . . . . . . . . . . . . . . . . 16979. Differentiation of Dyadics . . . . . . . . . . . . . . . . . . 17180. Triadics . . . . . . . . . . . . . . . . . . . . . . . . . . 17281. Summary: Dyadic Algebra . . . . . . . . . . . . . . . . . . 172

Problems . . . . . . . . . . . . . . . . . . . . . . . . . 174

CHAPTER V

DIFFERENTIAL INVARIANTS

82. Gradient of a Scalar . . . . . . . . . . . . . . . . . . . . . 17883. Gradient of a Vector . . . . . . . . . . . . . . . . . . . . 18184. Divergence and Rotation . . . . . . . . . . . . . . . . . . 18385. Differentiation Formulas . . . . . . . . . . . . . . . . . . 18686. Gradient of a Tensor . . . . . . . . . . . . . . . . . . . . 18787. Functional Dependence . . . . . . . . . . . . . . . . . . . 19088. Curvilinear Coordinates . . . . . . . . . . . . . . . . . . . 19189. Orthogonal Coordinates . . . . . . . . . . . . . . . . . . . 19490. Total Differential . . . . . . . . . . . . . . . . . . . . 19791. Irrotational Vectors . . . . . . . . . . . . . . . . . . . . . 19892. Solenoidal Vectors . . . . . . . . . . . . . . . . . . . . . 20193. Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . 20394. First Fundamental Form . . . . . . . . . . . . . . . . . . 20495. Surface Gradients . . . . . . . . . . . . . . . . . . . . . . 20696. Surface Divergence and Rotation . . . . . . . . . . . . . . . 20797. Spatial and Surface Invariants . . . . . . . . . . . . . . . . 20998. Summary: Differential Invariants . . . . . . . . . . . . . . . 211

Problems . . . . . . . . . . . . . . . . . . . . . . . . . 213

CHAPTER VI

INTEGRAL TRANSFORMATIONS

99. Green's Theorem in the Plane . . . . . . . . . . . . . . . . 216100. Reduction of Surface to Line Integrals . . . . . . . . . . . . 218101. Alternative Form of Transformation . . . . . . . . . . . . . 221102. Line Integrals . . . . 222103. Line Integrals on a Surface . . . . . . . . . . . . . . . . . 224104. Field Lines of a Vector . . . . . . . . . . . . . . . . . . . 226105. Pfaff's Problem . . . . . . . 230106. Reduction of Volume to Surface Integrals . . . . . . . . . . . 233107. Solid Angle . . . . . . . . . . . . . . . . . . . . . . . . 236

xiv CONTENTSARTICLE PAGE

108. Green's Identities . . . . . . . . . . . . . . . . . . . . . . 237109. Harmonic Functions . . . . . . . . . . . . . . . . . . . . 239110. Electric Point Charges . . . . . . . . . . . . . . . . . . . 241111. Surface Charges . . . . . . . . . . . . . . . . . . . . . . 242112. Doublets and Double Layers . . . . . . . . . . . . . . . . . 243113. Space Charges . . . . . . . . . . . . . . . . . . . . . . . 245114. Heat Conduction . . . . . 247115. Summary: Integral Transformations . . . . . . . . . . . . . . 248

Problems . . . . . . . . . . . . . . . . . . . . . . . . . 250

CHAPTER VII

HYDRODYNAMICS

116. Stress Dyadic . . . . . . . . . . . . . . . . . . . . . . . 253117. Equilibrium of a Deformable Body . . . . . . . . . . . . . . 255118. Equilibrium of a Fluid . . . . . . . . . . . . . . . . . . . . 257119. Floating Body . . . . . . . . . . . . . . . . . . . . . . . 257120. Equation of Continuity . . . . . . . . . . . . . . . . . . . 258121. Eulerian Equation for a Fluid in Motion. . . . . . . . . . . . 260122. Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . 262123. Lagrangian Equation of Motion . . . . . . . . . . . . . . . . 263124. Flow and Circulation . . . . . . . . . . . . . . . . . . . . 266125. Irrotational Motion . . . . . . . . . . . . . . . . . . . . . 267126. Steady Motion . . . . . . . . . . . . . . . . . . . . . . . 268127. Plane Motion . . . . . . . . . . . . . . . . . . . . . . . 270128. Kutta-Joukowsky Formulas . . . . . . . . . . . . . . . . . 273129. Summary: Hydrodynamics . . . . . . . . . . . . . . . . . . 278

Problems . . . . . . . . . . . . . . . . . . . . . . . . . 280

CHAPTER VIII

GEOMETRY ON A SURFACE

130. Curvature of Surface Curves . . . . . . . . . . . . . . . . . 283131. The Dyadic On . . . . . . . . . . . . . . . . . . . . . . . 285132. Fundamental Forms . . . . . . . . . . . . . . . . . . . . 289133. Field of Curves . . . . . . . . . . . . . . . . . . . . . . . 293134. The Field Dyadic . . . . . . . . . . . . . . . . . . . . . . 293135. Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . 297136. Geodesic Field . . . . . . . . . . . . . . . . . . . . . . . 299137. Equations of Codazzi and Gauss . . . . . . . . . . . . . . . 300138. Lines of Curvature . . . . . . . . . . . . . . . . . . . . . 302139. Total Curvature . . . . . . . . . . . . . . . . . . . . . . 306140. Bonnet's Integral Formula . . . . . . . . . . . . . . . . . . 308141. Normal Systems . . . . . . . . . . . . . . . . . . . . . . 313142. Developable Surfaces . . . . . . . . . . . . . . . . . . . . 314143. Minimal Surfaces . . . . . . . . . . . . . . . . . . . . . . 316144. Summary: Surface Geometry . . . . . . . . . . . . . . . . . 319

Problems . . . . . . . . . . . . . . . . . . . . . . . . . 321

CONTENTS xv

CHAPTER IX

TENSOR ANALYSISARTICLE PAGE145. The Summation Convention . . . . . . . . . . . . . . . . . 328146. Determinants . . . . . . . . . . . . . . . . . . . . . . . 329147. Contragredient Transformations . . . . . . . . . . . . . . . 332148. Covariance and Contravariance . . . . . . . . . . . . . . . . 334149. Orthogonal Transformations . . . . . . . . . . . . . . . . . 336150. Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . 337151. The Metric . . . . . . . . . . . . . . . . . . . . . . . . 339152. Relations between Reciprocal Bases . . . . . . . . . . . . . . 339153. The Affine Group . . . . . . . . . . . . . . . . . . . . . . 340154. Dyadics . . . . . . . . . . . . . . . . . . . . . . . . . . 342155. Absolute Tensors . . . . . . . . . . . . . . . . . . . . . . 343156. Relative Tensors . . . . . . . . . . . . . . . . . . . . . . 345157. General Transformations . . . . . . . . . . . . . . . . . . 346158. Permutation Tensor . . . . . . . . . . . . . . . . . . . . 350159. Operations with Tensors . . . . . . . . . . . . . . . . . . . 350160. Symmetry and Antisymmetry . . . . . . . . . . . . . . . . 352161. Kronecker Deltas . . . . . . . . . . . . . . . . . . . . . . 353162. Vector Algebra in Index Notation . . . . . . . . . . . . . . . 354163. The Affine Connection . . . . . . . . . . . . . . . . . . . 356164. Kinematics of a Particle . . . . . . . . . . . . . . . . . . . 359165. Derivatives of e` and E . . . . . . . . . . . . . . . . . . . 360166. Relation between Affine Connection and Metric Tensor . . . . . 361167. Covariant Derivative . . . . . . . . . . . . . . . . . . . . 362168. Rules of Covariant Differentiation . . . . . . . . . . . . . . . 365169. Riemannian Geometry . . . . . . . . . . . . . . . . . . . 366170. Dual of a Tensor . . . . . . . . . . . . . . . . . . . . . . 370171. Divergence . . . . . . . . . . . . . . . . . . . . . . . . 371172. Stokes Tensor . . . . . . . . . . . . . . . . . . . . . . . 372173. Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . 374174. Relation between Divergence and Curl . . . . . . . . . . . . 376175. Parallel Displacement . . . . . . . . . . . . . . . . . . . . 376176. Curvature Tensor . . . . . . . . . . . . . . . . . . . . . 380177. Identities of Ricci and Bianchi . . . . . . . . . . . . . . . . 384178. Euclidean Geometry . . . . . . . . . . . . . . . . . . . . 385179. Surface Geometry in Tensor Notation . . . . . . . . . . . . . 388180. Summary: Tensor Analysis . . . . . . . . . . . . . . . . . . 392

Problems . . . . . . . . . . . . . . . . . . . . . . . . . 396

CHAPTER X

QUATERNIONS

181. Quaternion Algebra . . . . . . . . . . . . . . . . . . . . . 403182. Conjugate and Norm . . . . . . . . . . . . . . . . . . . . 406183. Division of Quaternions . . . . . . . . . . . . . . . . . . . 409

xvi CONTENTSARTICLE PAGE

184. Product of Vectors . . . . . . . . . . . . . . . . . . . . . 410185. Roots of a Quaternion . . . . . . . . . . . . . . . . . . . . 412186. Great Circle Arcs . . . . . . . . . . . . . . . . . . . . . . 414187. Rotations . . . . . . . . . . . . . . . . . . . . . . . . . 417188. Plane Vector Analysis . . . . . . . . . . . . . . . . . . . . 421189. Summary: Quaternion Algebra . . . . . . . . . . . . . . . . 426

Problems . . . . . . . . . . . . . . . . . . . . . . . . . 427,

INDEX . . . . . . . . . . . . . . . . . . .

.... . . . . . . . 431

CHAPTER I

VECTOR ALGEBRA

1. Scalars and Vectors. There are certain physical quantities,such as length, time, mass, temperature, electric charge, that maybe specified by a single real number. A mass, for example, maybe specified by the positive number equal to the ratio of the givenmass to the unit mass. Similarly an electric charge may be speci-fied by a number, positive or negative, according as the charge is"positive" or "negative." Quantities of this sort are called scalarquantities; and the numbers that represent them are often calledscalars.

On the other hand, some physical quantities require a directionas well as magnitude for their specification. Thus a rectilineardisplacement can only be completely specified by its length anddirection. A displacement may be represented graphically by asegment of a straight line having a definite length and direction.We shall call such a directed segment a vector. Any physicalquantity that involves both magnitude and direction, so that itmay be represented by a line segment of definite length and direc-tion, and which moreover conforms to the parallelogram law ofaddition (§ 2), is called a vector quantity. Velocity, acceleration,force, electric and magnetic field intensities are examples of vectorquantities. It is customary, however, in applied mathematics, tospeak of vector quantities as vectors.

DEFINITIONS: A vector is a segment of a straight line regarded ashaving a definite length and direction. Thus we may represent avector by an arrow. The vector directed from the point A to the

point B is denoted by the symbol AB. With this notation AB--and BA denote different vectors; they have the same length butopposite directions.

Besides the proper vectors just defined, we extend the term vectorto include the zero vector, an "arrow" of length zero but devoidof direction.

1

2 VECTOR ALGEBRA §1

If the initial point of a vector may be chosen at pleasure, thevector is said to be free. If, however, its initial point is restrictedto a certain set of points, the vector is said to be localized in thisset. When the set consists of a single point (initial point fixed),the vector is said to be a bound vector. If the vector is restrictedto the line of which it forms a part, it is called a line vector. Forexample, the forces acting upon rigid bodies must be regarded asline vectors; they may only be shifted along their line of action.

Two vectors are said to be equal when they have the same lengthand direction.

Vectors are said to be collinear when they are parallel to the sameline. In this sense two parallel vectors are collinear.

Vectors are said to be coplanar when they are parallel to the sameplane. In this sense any two vectors are coplanar.

A unit vector is a vector of unit length.In addition to the foregoing notation, in which we denote a

vector by giving its end points, we also shall employ single lettersin heavy (bold-face) type to denote vectors. Thus, in Fig. 2b thevectors forming the opposite sides of the parallelogram are equal(they have the same length and direction) and may therefore berepresented by the same symbol,

AB=DC=u, AD=BC=v.A vector symbol between vertical bars, as I AB I or I u 1, de-

notes the length of the vector. We shall also, on occasion, denotethe lengths of vectors, u, v, F by the corresponding letters u, v, Fin italic type. Bars about real numbers denote their positive mag-nitudes: thus I -3 1 = 3.

The preceding definition of a vector is adequate for the elemen-tary applications to Euclidean space of three dimensions. Forpurposes of generalization, however, it is far better to define avector as a new type of number-a hypernumber-which is givenby a set of real numbers written in a definite order. Thus, in ourordinary space it will be -seen that, when a suitable system ofreference has been adopted, a vector can be represented by a setof three real numbers, [a, b, c], called the components of the vector.With this definition the zero vector is denoted by [0, 0, 0]. Inorder to complete this definition of a vector, a rule must be givento enable us to compute the components when the system of refer-ence is changed.

§ 2 ADDITION OF VECTORS 3

2. Addition of Vectors. To obtain a rule for adding vectors, letus regard them, for the moment, as representing rectilinear dis-placements in space. If a particle is given two rectilinear displace-ments, one from A to B, and a second from B to C, the result isthe same as if the particle were given a-single displacement from Ato C. This equivalence may be represented by the notation,

(1) AB + BC = AC.

We shall regard this equation as the definition of vector addition.The sum of two vectors, u, v, therefore is defined by the followingtriangle construction (Fig. 2a) :

Draw v from the end of u; then the vector directed from the begin-ning of u to the end of v is the sum of u and v and is written u + v.

C

FIG. 2a Fia. 2b

Since any side of a triangle is less than the sum of the othertwo sides,

lu+vl _ lul +lvl,the equal sign holding only when u and v have the same direction.

Vector addition obeys both the commutative and associative laws:

(2) u+v = v+u,(3) (u + v) + w = u + (v + w).

In the parallelogram formed with u and v as sides (Fig. 2b),

u+v=AB+BC=AC, v+u=AD+DC=AC.This proves (2). In view of this construction, the rule for vectoraddition is called the parallelogram law.

To find u + v when u and v are line vectors whose lines intersectat A, shift the vectors along their lines so that both issue from A(Fig. 2b) and complete the parallelogram ABCD; then

---3 --3 ---9u+v=AB+AD= AC.


Since the diagonal of the parallelogram on u, v gives the line ofaction of u + v, the term "parallelogram law" is especially appro-priate for the addition of intersecting line vectors. We shall speakof the addition of line vectors as statical addition.

The associative law (3) is evident from Fig. 2c:

(u + v) + w = (AB -i- BC) -f- CD = AC -i- CD = AD,

u + (v -F w) = AB -f- (BC + CD) = AB + BD = AD.Since the grouping of the vectors is immaterial, the preceding swais simply written u + v + w.

Fia. 2c

B

FIG. 2d

From the commutative and associative laws we may deduce thefollowing general result: The sum of any number of vectors is inde-pendent of the order in which they are added, and of their groupingto form partial sums.

To construct the sum of any number of vectors, form a brokenline whose segments, in length and direction, are these vectorstaken in any order whatever; then the vector directed from thebeginning to the end of the broken line will be the required sum.The figure formed by the vectors and their sum is called a vectorpolygon. If A, B, C, , G, H are the successive vertices of avector polygon (Fig. 2d), then

(4)

When the vectors to be added are all parallel, the vector "polygon"becomes a portion of a straight line described twice.

If, in the construction of a vector sum, the end point of the lastvector coincides with the origin of the first, we say that the sumof the vectors is zero. Thus, if in (4) the point H coincides with A,we write

-3 -4 -- 3(5) AB+BC++GA=O.

§ 3 SUBTRACTION OF VECTORS 5

This equation may be regarded as a special case of (4) if we agree--->

that AA = 0.

The zero vector AA (or BB, etc.) is not a vector in the propersense since it has no definite direction; it is an extension of ouroriginal vector concept. From

--- f ----* --3 -* -3 -3AB+BB=AB, AA+AB=AB

we have, on writing AB = u,

u+0=u, 0+u=u.We shall refer to vectors which are not zero as proper vectors.3. Subtraction of Vectors. The sum of two vectors is zero when,

and only when, they have the same length and opposite directions:

AB + BA = 0. If AB = u, it is natural to write BA = -u inorder that the characteristic equation for negatives,

(1) u + (-u) = 0,will hold for vectors as well as for numbers. Hence by definition:

The negative of a vector is a vector of the same length but oppositedirection.

Note also that - (-u) = u.The difference u - v of two vectors is defined by the equation,

(2) (u - v) + v = U.Adding -v to both sides of (2), we have

(3) u-v=u+(-v);that is, subtracting a vector is the same as adding its negative. Theconstruction of u - v is shown in Fig. 3a.

A

Fia. 3a0

Fia. 3b

If 0 is chosen as a point of reference, any point P in space may

be located by giving its position vector OP. Any vector AB may


be expressed in terms of the position vectors of its end points(Fig. 3b),

AB = AO + OB = OB + (-OA),and, from (3),

(4)

-4 -- -4AB = OB - OA.

4. Multiplication of Vectors by Numbers. The vector u + u isnaturally denoted by 2u; similarly, we write -u + (-u) = -2u.Thus, both 2u and -2u denote vectors twice as long as u; theformer has the same direction as u, the latter the opposite direc-tion. This definition is generalized as follows:

The product au or ua of a vector u and a real number a is definedas a vector a times as long as u, and having the same direction as u,or the opposite, according as a is positive or negative. If a = 0,au=0.

In accordance with this definition,

a(-u) = (-a)u = -au, (-a)(-u) = au.These relations have the same form as the rules for multiplyingnumbers. Moreover, the multiplication of a vector by numbers iscommutative (by definition), associative, and distributive:

(1) au = ua,

(2) (a/3)u = a(/3u),

(3) (a + a)u = au + (3u.The product of the sum of two vectors by a given number is alsodistributive:

(4) a(u + v) = au + av.The proof of (4) follows immediately from the theorem that thecorresponding sides of similar triangles are proportional. Figure 4

applies to the case when a > 0.The quotient u/a of a vector by a number

u+V av a (not zero) is defined as the product of uvU au by 1/a.

Fia. 4 The developments thus far show that:As far as addition, subtraction, and multi-

plication by numbers are concerned, vectors may be treated formallyin accordance with the rules of ordinary algebra.

§5 LINEAR, DEPENDENCE 7

5. Linear Dependence. The n vectors u1, u2i , un are saidto be linearly dependent if there exist n real numbers X1, X2, ,

An, not all zero, such that

(1) X1U1 + X2U2 + ... + XnUn = 0.

If the vectors are not linearly dependent, they are said to belinearly independent. Consequently, if a relation (1) exists be-tween n linearly independent vectors, all the constants must bezero.

If m vectors ul, U2, , um are linearly dependent, any greaternumber n of vectors including these are also linearly dependent. Forif u1i u2, , um satisfy

X1U1 + A2U2 + ... + XrUm = 0,

we can give ?1, X2, , Xm the preceding values (at least one ofthese is not zero) and take X,n+1 = X m+2 = _ X n = 0. Then(1) is satisfied, and the n vectors ui are linearly dependent.

If Xu = 0 and X 0, u = 0; hence one vector is linearly depend-ent only when it is the zero vector. Hence the vectors of any setthat includes the zero vector are linearly dependent. Conse-quently, we need only consider sets of proper vectors in the theo-rems following.

If X1u1 + X2u2 = 0 and X1 0 0, we can write ul = au2 i henceul and u2 are collinear. Conversely, if ul and u2 are collinear,U1 = au2 (a 0 0). Therefore:

A necessary and sufficient condition that two proper vectors be lin-early dependent is that they be collinear.

If X1u1 + A2u2 + X3u3 = 0 and X1 0, we can write ul =aU2 + ,13u3; the parallelogram construction (Fig. 5a) now showsthat ul is parallel to the plane of u2 and u3. Conversely, ifU1, U2, u3 are coplanar, they are linearly dependent. For (a) iftwo of the vectors are collinear, they are linearly dependent, andthe same is true of all three; and (b) if no two vectors are col-linear, we can construct a parallelogram on ul as diagonal whosesides are parallel to u2 and u3 (Fig. 5a), so that

ul =AC=AB+BC=au2+$u3.Therefore:

A necessary and sufficient condition that three proper vectors belinearly dependent is that they be coplanar.


In space of three dimensions, four vectors ul, u2i u3, u4 are al-ways linearly dependent. For (a) if three of the vectors are co-planar, they are linearly dependent, and the same is true of allfour; and (b) if no three vectors are coplanar", we can construct a

FIG. 5a FIG. 5b

parallelepiped on ut as a diagonal whose edges are parallel toU2i n3, U4 (Fig. 5b), so that

-3 ---3 --a - 3ut = AD = AB + BC -}- CD = au2 + 9u3 -I- yu4.

Therefore: Any four vectors are linearly dependent.6. Collinear Points. If A, B, P are points of a straight line,

P is said to divide the segment AB in the ratio X when

(1)

--> - 3AP=APB.

As P passes from A to B (Fig. 6a), X increases through all positivevalues from 0 to infinity. If P describes the line to the left of A,

0FIG. 6a

X varies from 0 to -1; and, as P de-scribes the line to the right of B, Xvaries from -oo to -1. Thus X = 0,X = =L00 , A = -1 correspond, respec-tively, to the points A, B, and the in-finitely distant "point" of the line. Theratio A is positive or negative, accordingas P lies within or without the segmentAB.

To find the position vector of P, relative to an origin 0, write(1) in the form,

Then

(2)

OP - OA = X(OB - OP).

-- OA + A OBOP =

1 + A

§6 COLLINEAR POINTS

or, if we write X = (3/a,

(3)

- aOA +aOBOP = -

9

a+0In particular, if X = 1, P is the mid-point of AB.

In the following we shall denote the position vectors of thepoints A, B, C, , P by a, b, c, , p. Thus if C divides ABin the ratio /3/a,

(4)c=ash-(3b

a+0Thus the mid-point of AB has the position vector

2(a + b).

When the points C, D divide a segment AB internally and ex-ternally in the same numerical ratios ±X, we have

a+Ab a - Xbc =1+x, d= 1-xIf we solve these equations for a and b, we find that the pointsA, B also divide the segment CD in the same numerical ratiosf(1 - X)/(l + X). Pairs of points A, B and C, D having thisproperty are said to be harmonic conjugates; either pair is theharmonic conjugate of the other.

A useful test for collinearity is given by the following:

THEOREM. Three distinct points A, B, C lie on a straight linewhen, and only when, there exist three numbers a, /3, y, different fromzero, such that

(5) aa+/3b-+-yc=0, a+$+y=0.Proof. If A, B, C are collinear, C divides AB in some ratio /3/a;

hence on putting y = - (a + ,B) in (4) we obtain (5). Conversely,from (5) we can deduce (4) since a + -y 0; hence C lieson the line AB.

From (5) we conclude that C, A, B divide AB, BC, CA, re-spectively, in the ratios S/a, y/l3, a/y whose product is 1.

If an equation of the form (5) subsists between three distinct non-collinear points, we must conclude that a = (3 = y = 0. For atleast one coefficient y = 0; and from

as+3b=0, a+/3=0,we have a = b (A coincides with B) unless a = 0 = 0.


Another criterion for collinearity may be based on the staticaladdition of line vectors (§ 2).

THEOREM. The points A, B, C are collinear when the line vectors

AB, BC, CA are statically equal to zero:

(6) AB +BC +CA =O.

Proof. If we use = to denote statical equivalence, AB + BC

BD, a vector through B; and, since BD + CA = 0, B, C, Aare collinear.

Example 1. In the parallelogram ABCD, E and F are the middle pointsof the sides AB, BC. Show that the lines DE, DF divide the diagonal ACinto thirds and that AC cuts off a third of each line (Fig. 6b).

E " B

Fia. 6b Fia. 6c

The hypotheses of our problem are expressed by the equations:

d - a=c - b, 2e=a+b, 2f =b+c.Let DE cut AC at X. To find x, eliminate b from the first and second equa-

tions. Thus+ c2e

=2a =a;d-a+2e=a+c and d +

3 3

for the first member represents a point on DE, the second member a pointon AC, and, since the points are the same, the point is at the intersection Xof these lines. Comparison with (4) now shows that X divides DE in theratio 2/1, AC in the ratio 1/2.

Let DF cut AC at Y. To find y, eliminate b from the first and third equa-tions. Thus

d-a+2f=2c and d+2f-a+2c_3 3

Hence Y divides DF in the ratio 2/1, AC in the ratio 2/1.

Example 2. In a plane quadrilateral ABCD, the diagonals AC, BD inter-sect at P, the sides AB CD intersect at Q (Fig. 6c). If P divides AC andBD in the ratios 3/2 and 1/2, respectively, in what ratios does Q divide thesegments A B, CD?

§6 COLLINEAR POINTS 11

By hypothesis

hence

2a + 3c 2b + d

p 5 3

6a + 9c = 10b + 5d and - 5d = lOb - 6a _

for the first fraction represents a pointon CD, the second a point on AB,and both points are the same, that is,the point Q. Therefore Q divides CDin the ratio -5/9, AB in the ratio-10/6.

Example 3. Prove that the mid-points of the diagonals of a completequadrilateral are collinear.

In the complete quadrilateralABCDLM (Fig. 6d) let P, Q, R be themid-points of the diagonals AC, BD,LM. The sum of the line vectors,

4

--, ---> --, --- --i --> --->AB + AD = 2 AQ, CB + CD = 2 CQ, QA + QC = 2 QP;

hence we have the statical equivalence,

AB +AD +CB +CD =4PQ,

for the quadrilateral ABCD with diagonals AC, BD. Similarly, for the ad-rilateral BLDM with diagonals BD, LM,

--> --9 - 4 --+BL + BM + DL + D31 = 4QR;

and, for the quadrilateral LA.11C with diagonals LM, AC,

--4 --4 ----+ --+ -,LA + LC + MA + MC = 4 RP.

On adding these three equations, we find that the entire left member is stati-cally equal to zero; for

--+ --- --> ---* -9 ---=0, DM+MA = 0,

-+ - 4 -+CB +BM +MC-0, CD +DL +LC -O,

are statical equations, since the vectors in each are collinear. Hence

PQ +QR+RP =0,and I', Q, R are collinear.


7. Coplanar Points. THEOREM. If no three of the points A, B,C, D are collinear, they will lie in a plane when, and only when,there exist four numbers a, /3, y, S, different from zero, such that

(1) as+$b+yc+Sd = 0, a+0+y+S = 0.Proof. If A, B, C, D are coplanar, either AB is parallel to CD,

or AB cuts CD in a point P (not A, B, C, or D). In the respectivecases, we have

b - a=K(d - c);a+Ab c+A'd1+A = 1+A, -P,

where A, A' are neither 0 nor -1. In both cases, a, b, c, d are con-nected by a linear relation of the form (1). Conversely, let usassume that equations (1) hold good. If a + /3 = 0 (and hencey + S = 0), we have

a(a - b) + y(c - d) = 0

and the lines AB and CD are parallel. If a + /3 5,16 0 (and hencey+6 0),

as+/3b yc+Sd(2) a+/i y+6 -Pwhere P is a point common to the lines AB and CD. In bothcases, A, B, C, D are coplanar.

Note that (2) states that the point P in which AB, CD inter-sect divides AB and CD in the ratios #/a, 6/y. Similarly, ifa+y; 0,

as + yc /3b + 6d(3)

a -f-.y #+ 6 - qi

thus Q, the point in which AC and BD intersect, divides AC andBD in the ratios y/a, S//i.

What conclusion can be drawn if a + 3 F6 0?If an equation of the form (1) subsists between four distinct non-

coplanar points, we must conclude that a = a = y = S = 0. Forat least one coefficient 6 = 0; then, from

as+/3b+yc = 0, a+0+y = 0,and the fact that A, B, C are not collinear, we deduce (§ 6) thata=0=y=0.

§ 7 COPLANAR POINTS 13

Example 1. The Trapezoid. If ABCD is a trapezoid with AB parallel toDC (Fig. 7a), then

AB=XDC or b-a=A(c-d).Hence we have

b+Xd=a+Xc or b-Xc=a-Xd.These equations may be written

b+ad_a+Xc_ b - Ac_a - Ad+a 1+a -P -q;

for the former expressions represent the point P where the diagonals BD, ACmeet, and the latter the point Q where the sides BC, AD meet. EvidentlyP divides both BD and AC in the same ratio A; and Q divides both BC andAD in the same ratio -A. In what ratio does the line PQ divide AB?

In particular, if A = 1, the trapezoid becomes a parallelogram. The diag-onals then bisect each other at P, while Q recedes to infinity.

Q

C

Fia. 7a

Example 2. Theorem of Menelaus. If a line s cuts the sides BC, CA, AB ofthe triangle ABC in the points P, Q, R, respectively, the product of the ratios inwhich P, Q, R divide these sides equals - 1. Conversely, if P, Q, R divide thesides of the triangle in ratios whose product is - 1, the points are collinear.

Proof. (Fig. 7b.) We lose no generality if we assume that P, Q divideBC, CA in the ratios -y//3, -a/y:

(i) (/3-y)P=$b-yc,(ii) (y-a)q=yc-aa.In order to locate R, which lies on the lines PQ, AB, we seek a linear relationbetween p, q, a, b. Add (i) and (ii) to eliminate c and divide by S - a; then

(R - y)P + (y - a)q Sb - as

Thus R divides AB in the ratio -S/a. The product of the division ratios-y//3, -a/y, -/3/a is -1.


Conversely, let us assume that P, Q, R divide BC, CA, AB in the ratios-y/S, -a/y, -a/a whose product is -1. Then we have equations (i), (ii)and also

(iii) (a - fl)r = as - 13b.

From these we deduce the linear relation,

(iv) (0-y)p+(y-a)q+(a-$)r=0,in which the sum of the coefficients is zero. The points P, Q, R are thereforecollinear. *

Note. From (i), (ii), (iii) it is easily proved that the three pairs of linesBQ, CR; CR, AP; AP, BQ meet in the points A', B', C' given by

(-a+0+y)a' _ -aa+ftb+yc,(a-S+y)b' =aa-$b+yc,(a+9-y)c'=as+9b-yc.

If we add 2aa, 2gb, 2yc, respectively, to these equations, we find that thepoint S given by

(v) s =as +Ab+yca+#+y

is common to the lines AA', BB', CC'. Thus to every line s given by (iv)

C

Hence

(i)_ Sb + ye as + SS

a 0+y a+sRb + asyc as _

(ii) b' = y+a +b 'c'

=as+Sb -yc+Ss.a+R y+S

*This conclusion is obvious; for the line PQ must meet AB in the point Rfor which the product of the division ratios is -1.

we have a corresponding point S given by (v)

-the pole of s relative to the triangle ABC.

Example 3. Theorem of Ceva. If S is apoint in the plane of the triangle ABC, and thelines SA, SB, SC cut the sides opposite in thepoints A', B', C', then the product of the ratiosin which A', B', C' divide the sides BC, CA,AB equals 1. Conversely, if A', B', C' dividethe sides BC, CA, AB in ratios whose product

B is 1 the lines AA' BB' CC meet in a oint,C, , , p .

Proof. (Fig. 7c.) Since A, B, C, S areFte.7c coplanar,

aa+pb+yc+as =0, a+0+y+S =0.

§7 COPLANAR POINTS 15

These equations state that A', B', C' divide BC, CA, AB in the ratios -y/fl,a/y, fl/a, whose product is 1. Incidentally, A', B', C' divide SA, SB, SC inthe ratios a/a, fl/S, y/3 whose sum is -1.

Conversely, let us assume that A', B', C' divide BC, CA, AB in the ratiosy/$, a/-j, fl/a whose product is 1. Then we have equations (i), (ii), (iii).From these we find that the vectors as + (0 + y)a', fib + (y + a)b', yc +(a + fl)c' are all equal to as + lb + yc; the point,

as+pb+yc(iv) s

a + fl + yis therefore common to the lines AA', BY, CC'.

Note. From (i), (ii), (iii) it is easily proved that the three pairs of linesBC, B'C'; CA, C'A'; AB, A'B' meet in the points P, Q, R given by

(Q-y)p=Rb-yc,(v) (y - a)q = yc - aa,

(a-fl)r=as-/3b.From these equations we deduce the linear relation,

(vi) (fi-y)p+(y-a)q+(a-a)r=0,in which the sum of the coefficients is zero. The points P, Q, R therefore lieon a line s. Thus to every point S given by (iv) we have a correspondingline s whose points P, Q, R are given by (v)-the polar of S relative to thetriangle ABC.

Example 4. Let ABC and A'B'C' be two triangles, in the same or differentplanes, so that the vertices A, B, C correspond to A', B', C', and the sidesAB, BC, CA correspond to A'B', B'C', C'A'. We then have (Fig. 7d)

P

FIG. 7d

DESARGUES' THEOREM. If the lines joining the corresponding vertices of twotriangles are concurrent, the three pairs of corresponding sides intersect in col-linear points, and conversely.

Let the lines AA', BY, CC' intersect at S; then

as + a'a' = fib + fl'b' = yc + y'c' = s,a+a =t3+fl' ='Y+y'=1.


From these equations we find in the usual manner the points P Q, R in whichBC, B'C'; CA, C'A'; AB, A'B' intersect:

(i) =fib

- yc =fl'b'

- y'c/

(ii)

Jr fl_y

yC - as=

'6',y'C' - a'a' 1

q y_a y,_«,

(iii)_«a-3b «a'-fl'b'

Hence

at - of

(iv) (l; - 'y)p + (-y - a)q + (a - 6)r = 0,

(v) 'y')p + ('y' - a')q + (a' - 6')r = 0;either equation shows that P, Q, R are collinear.

To prove the converse, we may start with the expressions (i), (ii), (iii) forp, q, r. These ensure that P, Q, R are collinear; but, in order that (iv) and(v) determine the same division ratios for P, Q, R, we must have

,-.y' 'y'-a'= h,

y y - a a -or

(vii) a'-ha =fl'-hfi=y'-hy=k,h and k representing the values of the equal members of (vi) and (vii). Inthe usual way we now find from (i), (ii), (iii) that

aa'-haa=6'b'-hfib=y'c'-hyc=ks,where S is a point common to AA', BB', CC'.

FIG. 7e

Example 5. The Complete Quadrangle. A complete quadrangle consists offour coplanar points, its vertices, no three collinear, and the six lines, its sides,which join them. The three pairs of sides which do not meet at a vertex

3 7 COPLANAR POINTS 17

arc said to be opposite; and the three points in which they meet are calledrliagon<zl points. In the complete quadrangle ABCD (Fig. 7e) the pairs ofopposite sides (BC, AD), (CA, BD), (AB, CD), meet at the diagonal pointsL, :11, N, respectively.

The properties of this configuration of points and lines must all be con-sequences of the fact that A, B, C, D are coplanar points, that is,

(i) CA +(3b+yc+Sd =0, a+$+y+a =0.Since no three vertices are collinear, none of the scalars a, ,, y, S are zero.From equations (i) we locate at once the diagonal points at the intersectionsof the opposite sides:

Go +yc-aa+ad0+y a+S

_yc+aa /3b+5dm y+a 13+S

,

_aa+13b yc+Sda+13 'Y+s

We here assume that no two of the scalars a, fl, y, S have a zero sum; thediagonal points L, M, N are then all in the finite plane. But if for example,a + 0 = y + S = 0, AB and CD are parallel, and N is the point at infinityin their common direction.

To find the points X1, X2 where LM cuts AB and CD, we seek linear rela-tions connecting 1, m, a, b and 1, m, c, d, respectively.

Thus, from (ii) and (iii),

(vi) X2

('Y+a)m- (S +y)1 as -/3ba-/3 a-0 '

(0+y)1-(S+S)m yc - Sdy-S y-S

Equations (iv) and (v) show that N and X1 divide AB in the ratios '6/aand -/3/a; hence N, X1 are harmonic conjugates of A, B.

Equations (iv) and (vi) show that N and X2 divide CD in the ratios S/yand -S/y; hence N, X2 are harmonic conjugates of C, D.

Equation (v) shows that X1 divides LM in the ratio -(y + a)/(3 + y);and (vi) shows that X2 divides LM in the ratio -(S + 5)/(3 + y) _(y + a)/(3 + y); hence X1, X2 are harmonic conjugates of L, M.

We may now state the following harmonic properties of the complete quad-rangle in the

THEOREM. Two vertices of a complete quadrangle are separated harmonicallyby the diagonal point on their side and by a point on the line joining the othertwo diagonal points.

Two diagonal points are separated harmonically by points on the sides passingthrough the third diagonal point.


From (ii), (iii), (iv) we next find the points X1, Y1, Z1 in which AB, BC, CAare cut by LM, MN, NL, respectively; thus

as - $b 3b - yc yc - as$1 = I y1 = , Z1 =a-/3 0-y y - a

Since

(a-/3)xl+(f -y)yl+(y-a)Z1 =0, a-8+0-y+y-a =0,the points Xl, Y1, Zl are collinear. This is also a consequence of Desargues'Theorem applied to the triangles ABC, LMN, which are in perspectivefrom D.

Each vertex of the quadrangle is the center of perspective of the trianglesformed by the other three vertices and by the three diagonal points. Thus,with careful regard to exact correspondence, the triangle LMN is in perspec-

tive with DCB, CDA, BAD, ABC from A, B, C, D, respectively. Hence thecoliesponding sides of these triangles intersect in four lines a, b, c, d (only d,the line X1Y1Z1, is shown in Fig. 7e). These lines are the polars of A, B, C, Dwith respect to the triangle LMN (ex. 3).

Corresponding to the complete quadrangle given by four points A, B, C, D,we now have a complete quadrilateral given by four lines a, b, c, d. In theseconfigurations the roles of points and lines are interchanged (Fig. 7f):

The quadrangle has four vertices A, B, C, D and six sides consisting of threeopposite pairs (BC, AD), (CA, BD), (AB, CD) which meet in the three diag-onal points L, M, N.

The quadrilateral has four sides a, b, c, d and six vertices consisting of threeopposite pairs (bc, ad), (ca, bd), (ab, cd) which lie on the three diagonal lines1, m, n.

The quadrangle and quadrilateral have the same diagonal triangle LMN orlmn; the sides 1, m, n are opposite the vertices L, M, N.

8. Linear Relations Independent of the Origin. We have seenthat the position vectors of collinear or coplanar points satisfy a

§9 CENTROID 19

linear equation in which the sum of the scalar coefficients is zero.The significance of such relations is given by the

THEOREM. A linear relation of the form,

(1) XiP1 + X2P2 + ... + XnPn = 0,

connecting the position vectors of the points P1, P2, , Pn will beindependent of the position of the origin 0 when, and only when, thesum of the scalar coefficients is zero:

(2) X1 + X2 -{- ... + an = 0.

Proof. Change from 0 to a new origin 0'. Writing OPi = pi,

O'Pi = pi, 00' = d, we have pi = d + pi; hence (1) becomes(X1 + t2 +. +Xn)d+X1p1 +X2P2 XnPn' = 0.

This equation will have the same form as (1) when and only when(2) is satisfied.

For two, three, and four points relations independent of theorigin have a simple geometric meaning; namely, the points arecoincident, collinear, or coplanar, respectively. The question nowarises: What geometric property relates five or more points whoseposition vectors satisfy a linear relation independent of the origin?

9. Centroid. We shall encounter problems in which each pointof a given set is associated with a certain number. The points,for example, may represent particles of matter and the numbers,their masses or electric charges. In the latter case, the numbersmay be positive or negative. We shall now define a point P*called the centroid of a set of n points P1, P2, , Pn associatedwith the numbers ml, m2i , mn, respectively. Denote any oneof these "weighted" points by the symbol m.iPi; then, if the sumof the numbers mi is not zero, the centroid of the entire set is defined

as the point for which the sum of all the vectors m1P*Pi is zero. Thedefining equation for the centroid is thus

(1) 2;miP*Pi = 0, provided Mmi Fl- 0.

As we wish P* to be uniquely defined, the case Emi = 0 mustbe excluded; for then (1) is a relation independent of the positionof P* (j 8).

20 VECTOR ALGEBRA i9

When 2 mi 0 0 there is always a unique point P* which satis-fies (1). For if we choose an origin 0 at pleasure, (1) can bewritten

---4 -4F.mi(OPi - OP*) = 0,

or

(2) (2mi)OP* = 1m1OPi.

This relation, independent of the origin, fixes the position of P*relative to 0. The point P* thus determined is the only pointthat satisfies (1) ; for if Q is a second point for which

EmiQPi = 0,

we have, on subtraction from (1),2;mi(P*Pi

- QPi) = Timi(P*Pi + PQ) = (2;mi)P*Q = 0;

hence P*Q = 0, and Q coincides with P*.If the position vectors of P*, Pi are written p* and pi, (2) be-

comes

(3) (2;mi)P* = Emipi.

The centroid of the points miP1 is not altered when the numbersmi are replaced by any set of numbers cmi proportional to them;for, in (2), the constant c may be canceled from numerator anddenominator. In particular, if the numbers mi are all equal, wemay replace them all by unity; the centroid of the n points thenis called their mean center and is given by the equation:

-3 1 -- 1(4) OP* _ - ZOP1, or p* _ - Zpi.

n n

In finding the centroid P* of any set of weighted points miP1,we may replace any subset of points for which the sum of theweights is a number m' 0 0 by their centroid P with the weightm'. For, if V and 1", respectively, denote summations extendedover the points of the subset and over all the remaining points,we may write (3) in the form,

(2;'mi + Z"mi)P* = 7,'mipi + E"miPi,or

(m' + I"mi)P* = m'p' + E"mipi.

§9 CENTROID 21

This equation shows that P* is also the centroid of the point m'P'and the points miPi not included in the subset.

Finally, let us consider the nature of a set of n points miPi(?Pi F4- 0) for which Ximi = 0. If we attempt to find a point P*which satisfies

(5) ZmiP*Pi = 0 when Xmi = 0,

we find that there are two possibilities.(a) For any arbitrary choice of the points miPi, subject only to

the condition Emi = 0, there will in general be no point P* thatwill satisfy (5). Thus in the cases n = 2, 3, 4, equations (5) imply,respectively, that P1 and P2 coincide; P1, P2, P3 are collinear;P1, P2, P3, P4 are coplanar. Hence, if these conditions are notfulfilled, P* does not exist.

(h) If, however, a point P* can be found which satisfies (5), anypoint whatever will serve (§ 8). In fact, we see, from (2), thatthe existence of P* implies that

(6) ,6dmiOPi = 0 1mi = 0,

for any choice of 0. In particular, we may take any one of thegiven points Pi as origin. Thus, if we take 0 at P1, (6) becomes

miP1 Pi = 0,i=2

V

2; mi = -m1 0.i=2

Hence, from the defining equation (1), we see that P1 is the cen-troid of the points m2P2, m3P3, . . . , mnPn. Precisely the sameconclusion may be drawn for P2, P3, , Pn. We now can answerthe question raised at the end of § 8. Any set of weighted pointsmiPi (mi F6 0) whose position vectors satisfy a linear relation (6)independent of the origin has the intrinsic property that any point ofthe set is the centroid of all the remaining weighted points.

A set of n points having this property may be readily con-structed. Take any set of n - 1 weighted points miPi, such thatm1 + m2 + + mn_1 0, and adjoin to the set their centroidP* = Pn with the weight Mn = -(m1 + m2 + + mn-1).Then, since

-mnPn = m1P1 + m2P2 + + mn-1Pn-1,we have

n nX mipi = 0, 2;mi = 0.

1 t

22 VECTOR ALGEBRA

Thus the relation,

(7) as+Sb+yc=0, a+$+-y=0,

§9

between three collinear points shows that each point of the setaA, ,BB, yC is the centroid of the other two. Any point C on theline of A and B satisfies a relation of the type (7) and hence is thecentroid of these points when suitably weighted.

The relation,

(8) as+,9b+yc+8d = 0, a+a+7+S = O,between four coplanar points shows that each point of the setaA, ,BB, yC, 8D is the centroid of the other three. Any point Din the plane of A, B, C satisfies a relation of the type (8) and henceis the centroid of these points when suitably weighted.

Example 1. Centroid of Two Points. The centroid of aA and pB is given by

* as+RbP a+13

Hence P* divides AB in the inverse ratio f/a of their weights. In particular,if a = S, P* divides AB in half. The mean center of two points lies midwaybetween them.

Example 2. Mean Center of Three Points A, B, C. Let L, M, N be the mid-points of BC, CA, AB (Fig. 9a). Then the mean center P* of A, B, C is thecentroid of A and 2L, B and 2M, C and 2N. Hence (ex. 1) the segmentsAL, BM, CN are all divided by P* in the ratio of 2/1.

If A, B, C are not collinear, they are the vertices of a triangle. Its mediansAL, BM, CN intersect at P* and are there divided in the ratio of 2/1.

C

N a M2

Fia. 9a FIG. 9b

Example 3. Mean Center of Four Points A, B, C, D. Let L1, L2; M1, M2;N1, N2 be the mid-points of A B, CD; BC, DA; AC, BD, respectively (Fig. 9b).To find P* we may replace A, B, C, D by 2L1 and 2L2, or by 2M1 and 2M2,or by 2N1 and 2N2. Therefore P* is the mid-point of the segments L1L2,M1M2, N1N2. The bisectors of the three pairs of opposite sides of the quad-

§ 10 BARYCENTRIC COORDINATES 23

rangle ABCD (which may be plane or skew) intersect at P* and are theredivided in half.

If A, B, C, D are not coplanar, they determine a tetrahedron. Let A', B',C', D' be the mean centers of the triads BCD, CDA, DAB, ABC. Then P*is the centroid of A and 3A', B and 3B', C and 3C', D and 3D', and hencedivides each of the segments AA', BB', CC', DD' in the ratio of 3/1. Thepreceding result also shows that the bisectors of the three pairs of oppositesides of the tetrahedron meet at P*.

Example 4. The sum of n vectors AiB1 (i = 1, 2, , n) is given by

Z(bi-a;) =nb*-na* = n(b* - a*)where A* and B* are the mean centers of the initial points Ai and the terminalpoints Bi, respectively; hence

In particular,2;A;Bi = n A*B*.

-4 --A1B1 + A2B2 = 2 A*B*,

where A*, B* are the mid-points of A1A2 and B1B2, respectively.

10. Barycentric Coordinates. If P is any point in the plane of-3-3 --3the reference triangle ABC, the vectors AP, BP, CP are linearlydependent (§ 5) ; hence

«(p - a) + /3(p - b) + y(p - c) = 0,

(1)p=as+9b+-ic

«+6 + yThe denominator is not zero; for, if

a+a+y=0, then aa+(3b+yc=0,and A, B, C would be collinear, contrary to hypothesis. Thus Pis the centroid of the weighted points aA, $B, yC. The threenumbers a, /3, y are called the barycentric coordinates of P; as theyall may be multiplied by the same number without altering p,their ratios determine P. The vertices of the reference trianglehave the coordinates A(1, 0, 0), B(0, 1, 0), C(0, 0, 1); the unitpoint (1, 1, 1) is the mean center of A, B, C.

If P is a point in space and ABCD a reference tetrahedron, the

vectors AP, BP, CP, DP are linearly dependent (§ 5); hence

(2)

«(p-a)+$(pp-b)+y(p-c)+a(p-d) =0,as+Ab+yc+Sd

p a+O+y+S .

24 VECTOR ALGEBRA

The denominator is not zero; for, if

§ 12

a+8+y+S = 0, then as+,lib+yc+Sd =0and the points A, B, C, D would be coplanar, contrary to hypoth-esis. Thus P is the centroid of the weighted points aA, $B, yC, W.The four numbers a, 0, y, S are called barycentric coordinates of P;as they all may be multiplied by the same number without alter-ing p, their ratios determine P. The vertices of the reference tetra-

Fla. 11

hedron have the barycentric co-ordinates A(1, 0, 0, 0), B(0, 1, 0,0), C(0, 0, 1, 0), D(0, 0, 0, 1);the unit point (1, 1, 1, 1) is themean center of A, B, C, D.

11. Projection of a Vector. Tofind the projection of a vector

PQ upon a line x, a director plane7r must be specified. Pass planes

through P and Q parallel to 7r and let them cut x in the points

P1i Ql (Fig. 11). Then the vector P1Q1 is the 7r-projection of PQupon x.. If no director plane 7r is specified, we tacitly assume that IT isperpendicular to x; the projection is then orthogonal.

Let the vectors PQ, QR and their sum PR have the projections

P1Q1, Q1R1 and P1R1 on the line x; then, since

P1Q1 + Q1R1 = P1R1,

the projection of the sum of two vectors on a line is equal to the sumof their projection on this line.

12. Base Vectors. Let e1, e2, e3 be three linearly independentvectors; they are then non-coplanar. If the vectors are drawnfrom a common origin 0 (Fig. 12a), we may pass a closed planecurve through their end points El, E2, E3. If this curve is viewedfrom the side of its plane opposite to that on which 0 lies, theorder E1E2E3 defines a sense of circulation. If this sense is counter-clockwise, the set e1, e2, e3 is said to be right-handed or dextral; forit is then possible to extend the thumb, index, and middle fingersof the right hand so that they have the directions of e1, e2, e3,

§ 12 BASE VECTORS 2'r

respectively. If the sense defined by E1E2E3 is clockwise the setis said to be left-handed or sinistral.

Any vector u = PQ may be expressed as the sum of three vec-tors parallel to el, e2, e3, respectively. For, if we construct aparallelepiped on PQ as diagonal by passing planes through P and

R

Fia. 12a FIG 12b

Q parallel to e2 and e3, e3 and el, el and e2 (Fig. 12b), its edgeswill be parallel to el, e2, e3, and

PQ = PQ1 + Q1R + RQ = PQ1 + PQ2 + PQ3_ y

Since PQi is a scalar multiple of ei, say u`ei,

(1) u = ulel + u2e2 + u3e3,

where ul, u2, u3 are numbers called the components of u with re-spect to the basis el, e2, e3. Their indices are not exponents butmere identification tags; they are written as superscripts for reasonsgiven in Chapter IX. The vector u is often written [ul, u2, u3],with brackets to enclose its components.

If u is the position vector OP, the components ui are called theCartesian coordinates of the point P with respect to the basise1, e2, e3-

A straight line upon which two directions are distinguished iscalled an axis. One direction is called positive, the other negative.In a figure the positive direction is marked with an arrowhead.With a given basis, lines drawn through the origin 0 parallel to

26 VECTOR ALGEBRA § 13

el, e2, e3 and with their directions positive are called the coordi-nate axes. The numbers u', u2, u3 often are called the componentsof u on these axes.

The components of a zero vector are all zero; for, since el, e2, e3are linearly independent, u = 0 implies u' = u2 = u3 = 0.

If A is any scalar,

(2) Au = Aulel + Au2e2 + Au3e3.

If v = vie1 + v2e2 +.v3e3,

(3) u + v = (u1 + v')el + (u2 + v2)e2 + (u3 + v3)e3

With the bracket notation, (2) and (3) become

(4) A[ul, u2, u3] = [Au', Au2, Au3],

(5) [ul, u2, u3] + [v', v2, v3] = [u' + v1, u2 + v2, u3 + v3].

To multiply a vector by a number, multiply its components by thatnumber; to add vectors, add their corresponding components. In par-ticular,

(6) -[u', u2, u3] = [-u', -u2, -u3],

(7) [ul, u2, u3] - [v', v2, v3] = [u' - v', u2 - v2, u3 - v3].

If u = v, then u - v = 0; hence

(8) U=V implies u' = v', u2 = v2, u3 = v3.

When referred to the same basis, equal vectors have their correspondingcomponents equal.

13. Rectangular Components. Let i, j, k denote a dextral sys-tem of mutually perpendicular unit vectors. From an origin 0draw the coordinate axes x, y, z with positive directions given byi, j, k (Fig. 13). Any vector u now is determined by giving itscomponents u1, u2i u3 on these rectangular axes:

(1) u = uli + u2j + u3k.

Draw OP = u from the origin, and let OP1, OP2i OP3 be its

orthogonal projections on the axes. Then ul is the length of OPI,

taken positive or negative according as OPI has the direction ofi or -i; hence

u1 = I u I cos (i, u),

§ 13 RECTANGULAR COMPONENTS 27

where j u j denotes the length of u and (i, u) the angle betweeni and u. The rectangular components of u thus are obtained bymultiplying its length by the corresponding direction cosines:

(2) ul = I u I cos (i, u), u2 = I U I cos (j, u),

u3 = I u I cos (k, u).

R

Fic. 13

The Pythagorean Theorem gives the length of u in terms of itscomponents; from

0P2 = (OR)2 + (RP)2 = (OPl)2 + (OP2)2 + (OP3)2,

(3) I U 12 = (ul)2 + (u2)2 + (u3)2.

If we substitute from (2) in (3), we obtain the relation,

cost (i, u) + cost (j, u) + cost (k, u) = 1,

satisfied by the direction cosines of any vector.The rectangular coordinates of any point P are defined as the

components of its position vector OP on the x-, y-, and z- axes.Thus, if

(4) OP=xi+yj+zk,P has the rectangular coordinates (x, y, z). Since P1P2 = OP2

28 VECTOR ALGEBRA

- OPI, this gives

(5) PIP2 = (x2 - xl)i + (Y2 - yl)j + (z2 - zl)k

§13

if (xi, y,, zi) are the coordinates of Pi. The components of PIP2are found by subtracting the coordinates of PI from the correspondingcoordinates of P2.

With an orthogonal basis i, j, k, the components may be writtenwith subscripts (as previously) or superscripts. Separately, theyare called the x-, y-, and z- components and often are writtenul, U2, U3- Just as with a general basis, vectors are specified bygiving their components: u = -[ul, u2, u3]. Thus the point(x, y, z) has the position vector [x, y, z]; and (5) may be written

PIP2 = [x2 - xl, Y2 - yl, z2 - Z11-

The unit vectors i, j form a basis for all vectors u in their plane:

(6) u = uli + u2j.

If e is a unit vector in the plane and 0 is the angle (i, e), reckonedpositive in the sense from i to j, we define cos 0 and sin 0 as thecomponents of e :

(7) e=icos0+jsin0.Any vector in tine plane may be written

(8) u= Iule= Jul {icos(i,u)+jsin(i,u)};its rectangular components are

(9) ul = I U I cos (i, u), u2 = I u I sin (i, u), u3 = 0.

Example. Addition Theorems for the Sine and Cosine. Let a and b be twounit vectors such that the angles (i, a) = a, (a, b) = 0; then the angle (i, b)= a + 14, and, from (7),

a =icosa+jsina,b = i cos (a + fl) + j sin (a + 0).

If we refer b to the new basis i = a, j (a unit vector 90° ahead of a), we have

b=icos0+isin 0= (i cos a + j sin a } cos l3 + I i cos (a a + 2) + j sin (a + 2)1 sin 6.

15 SCALAR PRODUCT 29

Comparing the components of b in the two preceding expressions, we find

cos (a + 13) = cos a cos 8 + cos (a +2)

sin [3,

sin (a + /3) = sin a cos l3 + sin (a + 2) sinWith a =7r/2, these equations give

cos (/3 + 2) sin $, sin (p + 2) = cos 0;

hencecos (a + 0) = cos a cos 0 - sin a sin 0,

sin (a + )3) = sin a cos g + cos a sin g.

These addition theorems hold for all values of a and fl, positive or negative.

14. Products of Two Vectors. Hitherto we have consideredonly the products of vectors by numbers. Next we shall definetwo operations between vectors, which are known as "products,"because they have some properties in common with the productsof numbers. These products of vectors, however, will also proveto have properties in striking disagreement with those of numbers.

Since one of these products is a scalar and the other a vector,they are called the scalar product and vector product, respectively.The definitions of these new products may seem rather arbitraryto one unfamiliar with the history of vector algebra. We presentthis algebra in the form and notation due to the American mathe-matical physicist, J. Willard Gibbs (1839-1903).t It is an offshootof the algebra of quaternions, adapted to the uses of geometry andphysics. In Chapter X quaternion algebra is developed briefly,and the origin of the foregoing products is revealed.

15. Scalar Product. The scalar product of two vectors u and v,written u v, is defined as the product of their lengths and the cosineof their included angle:

(1) Jul IvI cos(u,v).

The scalar product is therefore a number which for proper (non-zero) vectors is positive, zero, or negative, according as the angle(u, v) is acute, right, or obtuse. Hence, for proper vectors,

(2) means u1v.t Professor of mathematical physics at Yale University. His pamphlet on

the Elements of Vector Analysis was privately printed in 1881. A more com-plete treatise on Vector Analysis (New Haven, Yale University Press, 1901)based on Gibbs's lectures, was written by Professor E. B. Wilson.

30 VECTOR ALGEBRA

When u and v are parallel,

Jul lvl, or -Jul lvI,

115

according as the vectors have the same or opposite directions; thus

From (1) we see that

(3) (au) ($v) = ag u . V.

The last result is obvious when a and i3 are positive numbers; theother cases then follow from the equations preceding.

Besides the components of a vector on the coordinate axes(§§ 12, 13) we shall also use the orthogonal projection and compo-nent of a vector on an arbitrary directed line 1. If the positivedirection of 1 is given by the unit vector e, the projection of uupon 1 (proj 1 u) is a scalar multiple of e. This scalar is called thecomponent of u upon 1 (comp1 u) ; its defining equation is therefore

(4) e comp j u = proj 1 u.

As in § 13, we compute comp j u as

(5) comps u = l u i cos (e, u);

since I e 1, this may also be written

(6) comps u = e u.

From the projection theorem of § 11,

(7) proj 1 (u + v) = proj 1 u + proj 1 v;

hence, from (3),

(8) comps (u + v) = comp] u + comps v.

The operations expressed by proj1 and comp] are distributive withrespect to addition.

The definition (1) of u v now may be written

(9) U V= I u l compu V=IV l comp u.

Scalar or "dot" multiplication is commutative and distributive:

(10)

(11) w(u+v) =wu+wv.

§ 15 SCALAR, PRODUCT

The last equation is nothing more than (8) multiplied bywhen 1 is taken along w.

If c 0 in the equation,

or

31

we can conclude either that a - b = 0 or that a - b and c areperpendicular. We cannot "cancel" c to obtain a = b unlessa - b and c are not perpendicular.

It is obvious that, in general, (u v)w u(v w).Since i, j, k are mutually perpendicular unit vectors,

(12) i

Hence, if we expand the product,

u ' v = lull + u2j + u3k) (v1i + v2j + v3k)we obtain

(13) u v = u1v1 + u2v2 + u3v3

The scalar product of two vectors is equal to the sum of the productsof their corresponding rectangular components.

Example 1. If u = [2, -1, 3], v = [0, 2, 4], u v = 0 - 2 + 12 = 10.Moreover,

comp,, u =u v

=10

= 2 236, compu v =uuv = 10 = 2.673;rv7

/20 luI 14

Uv 10

cos (u, v) = =I u v /280 = 0.5979, angle (u, v) = 530 18'.

Example 2. Identities involving scalar products may be given a geometrici.iterpretation. Thus (Fig. 15a)

gives cdcosw=a2-b2,and, if we write c = a + b, d = a - b, w = an-L,Ie (c, d). If a = b, c d = 0; then PQRS is arhombus and the angle PRT may be inscribedin a semicircle about Q. We thus have two geo-metric theorems:

1. The diagonals of a rhombus cut at rightangles.

2. An angle inscribed in a semicircle is a right P a Q

angle. Fro. 15aMoreover, from

(a-b) (a - b) =we have the cosine law: d2 = a2+0 - 2ab cos 0.


Example 3. We also may interpret identities involving scalar products by

regarding the vectors a, b, as position vectors OA, OB, from an

arbitrary origin O. Then a - b = BA and a + b = 2 011 where M is themid-point of AB.

Consider, for example, the identity,

(b-a)2+(c-b)2+(d-c)2+(a-d)2= (c-a)2+(d-b)2+(a+c-b-d)2,

which can be verified on expansion; u2 means u u. If we regard a, b, c, das the position vectors of the vertices of a space quadrilateral ABCD, p =

(a + c), q =2

(b + d) locate the mid-points P, Q of the diagonals AC, BD;hence

(AB)2 + (BC)2 + (CD)2 + (DA)2 = (AC)2 + (BD)2 + 4(QP)2.

The sum of the squares of the sides of any space quadrilateral equals the sum of

C

Fia. 15b

the squares of its diagonals plus fourtimes the square of the segment joiningtheir middle points.

Example 4. The identity,

(a-b) (h-c)0,

Bshows that the altitudes of a triangleABC meet in a point H, the ortho-

center of the triangle (Fig. 15b); for, if two terms of this equation are zero, thethird is likewise.

Similarly the identity,

\(a-b) k- a2

+ b b + c2

C

c+a\

shows that the perpendicular bisectors of the triangle ABC meet in a point K,the circumcenter of the triangle.

For the orthocenter H and circumcenter K the individual terms of the fore-going equations vanish; for example,

(a - 0, (a - 0.

On adding these, we obtain

=0;

or on writing g = 3 (a + b + c) for the position vector of the mean center Gof the triangle ABC.

(a-b) (h+2k-3g) =0.

§ 15 SCALAR PRODUCT 33

Since this equation also holds when a - b is replaced by b - c and c - a,we conclude that

h + 2k - 3g = 0.Therefore the mean. center of a triangle lies on the line joining the orthocenter tothe circumcenter and divides it in the ratio of 2/1. This line is called the Eulerline of the triangle.

Example 5. In order to interpret the identity,

(a + b - c - d)2- (a - b - c+ d)2 = 4(a - c) - (b-d),with reference to the plane quadrilateral ABCD, let P, Q, R, S denote themid-points of AB, BC, CD, DA; then (Fig. 15e) we have

(PR)2 - (QS)2 = CA DB.

Hence, if the diagonals of a quadrilateral cut at right angles, the lines joiningthe mid-points of opposite sides are equal. Since the lines PR, QS intersect

D

P R'Fta. 15c Fia.15d

in the mean center N of the points A, B, C, D and are bisected there (§ 9,ex. 3), a circle with N as center will pass through P, Q, R, S. If perpendicularsfrom P, Q, R, S are dropped upon the sides opposite, their feet, P', Q', R', Salso will lie on this circle. Thus we have proved the

THEOREM. When the diagonals of a quadrilateral are perpendicular, the mid-points of its sides and the feet of the perpendiculars dropped from them on theopposite sides all lie on a circle described about the mean center of the vertices.

In the figure formed by a triangle ABC and its three altitudes meeting atthe orthocenter H (Fig. 15d), three quadrilaterals, ABCH, BCAH, CABH,all have perpendicular diagonals. Their three eight-point circles are all thesame. This circle, whose center is at the mean center N of A, B, C, H, is thefamous nine-point circle of the triangle.

THEOREM. For any triangle ABC whose orthocenter is H, a circle whose centeris the mean center of A, B, C, H, passes through the nine points: the mid-pointsof the sides, the feet of its altitudes, and the mid-points of the segments joining Hto the vertices.


The center N of the nine-point circle has the position vector n given by

4n =a+b-}-c+h=3g+h=2h+2k,in view of ex. 4. Therefore the center N of the nine-point circle is collinearwith the mean center G of the triangle, its orthocenter H, and circumcenter K.Moreover N bisects the segment HK.

16. Vector Product. The vector product of two vectors u and v,written u x v, is defined as the vector,

(1) uxv= Jul Ivlsin(u,v)e,where e is a unit vector perpendicular to both u and v and formingwith them a dextral set u, v, e. If u and v are not parallel, a right-handed screw revolved from u towards v will advance in its nuttowards u x v.

When u and v are parallel, e is not defined; but in this casesin (u, v) = 0 and u x v = 0. Moreover, if u and v are not zero,u x v = 0 only when sin (u, v) = 0. Hence, for proper vectors,

(2) uxv = 0 means u I I v.

In particular u x u = 0.From (1) we see that

(-u)xV = ux(-v) _ -uxv, (-u). (-v) = uxv,(3) (au) x ((3v) = a# u x v.

The last result is obvious when a and /3 are positive numbers; the

FIG. 16

other cases then follow from the equationspreceding.

If u and v are interchanged in (1), thescalar factor is not altered, but e is reversed;hence

(4) vxu = -uxv.Vector multiplication is not commutative.

Draw u and v from the point A, and let pbe a plane perpendicular to u at A (Fig. 16).

Then u x v may be formed by a sequence of three operations:

(P) Project v on p, and obtain v';

(M) Multiply v' by I u I, and obtain I u I v';

(R) Revolve I u I v' about u through +90°

316 VECTOR PRODUCT

The resulting vector agrees with u x v in magnitude, for I V,I v I sin (u, v), and also in direction (upward in the figure). We isdicate this method of forming u X v by the notation,

(5) u X v = RMPV.

This means that v is projected, and the projection multiplied, anfinally revolved as previously described. Now each of these opeators is distributive: operating on the sum of two vectors is the sarr.as operating on the vectors separately and adding the result:hence

RMP(v + w) = RM(Pv + Pw)

= R(MPv + MPw) = RMPV + RMPvcThus, from (5),

(6) ux (v + w) = uxv + uxw, (V + W) Xu = vxu + wxu

Vector or "cross" multiplication is distributive. By repeated applications of (6) we may expand the vector product of two vectosums just as in ordinary algebra, provided that the order of th,factors is not altered. For example,

(a+b)x(c+d) =axc+axd+bxc+bxd.If c 7d 0 in the equation,

axc=bxc, or (a-b)xc=0,we can conclude either that a - b = 0 or that a - b and c areparallel. We cannot "cancel" c to obtain a = b unless a - b andc are not parallel.

We shall see in § 18 that in general, (u x v) x w P6 u x (v x w).Since the unit vectors i, j, k form a dextral orthogonal set, we

have the cyclic relations.

(7) ixj=k, jxk=i, kxi=j; ixi=jxj=kxk=0.Hence, if we expand the product,

u x v = (u1i + u2j + u3k) x (v1i + v2j + v3k),

we obtain

(8) u x V = (u2v3 - u3v2)i + (u3v1 - u1v3)j + (ulv2 - u2v1)k.


The components of u x v are the determinants formed by columns

°l and 3, 3 and 1 (not I and 3), 1 and l of the arrayul U2 u3

(VI V2 V3

hence we may write

(9)

i j k

u1 U2 U3

V1 V2 V3

For example, if u = [2, -3, 5], v = [-1, 4, 2], we compute thecomponents of u x v from the array,

C

2 -3 5-1 4 2 '

-3 542

= - 26,5 2

2 -1 _ -9,2 -3

-1 4=5.

Thus u x v = [ -26, -9, 5 ]. As a check, we verify. that u x v isperpendicular to both u and v: -52 + 27 + 25 = 0, 26 - 36 +10 = 0.

Example 1. To find the shortest distance d from a point A to the line BC.Method. Let e be the unit vector along BC. Then, if u is any vector from

A to the line (as AB or AC),

d = I uI sin(u,e) =Iuxe1.

Computation. If the points are A(3, 1, -1), B(2, 3, 0), C(-1, 2, 4),

u=AB=[-1,2,1], BC=[-3,-1,4], a=[-3, -1,4]x/26

[9, 1, 7]u x e = i s _ 2.245

x/ , d = IuxeI - -26

-4Check. If we take u = AC = [-4, 1, 5], u x e again has the preceding value.

Example 2. To find the shortest distance d from a point A to the planeBCD.

Method. Find a vector normal to the plane, such as BC x BD, and let nbe the unit vector in its direction. Then, if u is any vector from A to the

plane (as AB or AC), d = n uComputation. If the points are A(1, -2, 1), B(2, 4, 1), C(-1, 0, 1),

D(-1, 4, 2),

-- 4 -, ----> ---4BC = [-3, -4, 0], BD = [-3, 0, 1], BC-BD = [-4,3, -12];

-i [-4,3, -121 _ 14u=AB=[1,6,0]; n=13

-4Check. If we take u = AC = [-2, 2, 0], n u =.

17 VECTOR AREAS 37

Example 3. To find the shortest distance d between two non-parallel lines

AB, CD; and to locate the shortest vector PQ from AB to CD.

Method. Find the vector AB x CD which is perpendicular to both lines,and let n be a unit vector in its direction. Then, if u is any vector from AB

to CD (as AC orBD), d = In-

To find P and Q, write AP = a AB, CQ = y CD, and find the scalars a, y

from the condition that PQ = PA + AC + CQ is parallel to AB x CD. Thelength PQ = d.

Computation. If the lines AB, CD are given by the points A(1, -2, -1),B(4, 0, --3); C(1, 2, -1), D(2, -4, -5);

-4 --4 --> --->AB = [3, 2, -2], CD = [1, -6, -4], AB x CD = 10[-2, 1, -2];

n = 3[-2, 1, -21, AC = [0, 4, 0], d = n AC =To find P and Q, we have

-- --4 -+ - 4PQ = -aAB +AC +yCD

= -a[3, 2, -2] + [0, 4, 0] + y[l, -6, -4]= [-3a+y, -2a+4-6y,2a-4y];

and, since PQ is parallel to [-2, 1, -2],

-3a + y -2a + 4 - 6y 2a - 4y-2 1 -2

These equations give a = y = $; hence

OP=OA+AP=[1,-2,-1]+$[3,2,-2]=$[21,-10,-17],

OQ = OC + CQ = [1, 2, -1] + $[1, -6, -4] _ $[13, -6, -25].

Check. The distance PQ = .

17. Vector Areas. Consider a plane area A whose boundary istraced in a definite sense-shown by arrows in Fig. 17a. We shall

FIG. 17a

V

U

Fia. 17b

associate such an area with a vector of magnitude A, normal to itsplane, and pointing in the direction a right-handed screw would moveif turned in the given sense. Thus the parallelogram whose sides

38 VECTOR ALGEBRA $ 17

are the vectors u, v (Fig. 17b) and whose sense agrees with therotation that carries u into v is associated with the vector u x v;for its area is

A = Iul Ivlsin(u,v) = Iuxvl.All plane areas associated with the same vector will be regarded

as equal. The sum of two plane areas associated with the vectors,

Fm. 17c

u, v, is defined as the planearea associated with u + v.

A directed plane is a planeassociated with a definite nor-mal vector, say the unit normaln. A circuit in a directed planeis positive when it is counter-clockwise relative to n; a clock-wise circuit is negative.

Consider now a plane area Aassociated with the vector u.The projection of A on a di-rected plane p is A' = A cos 8(Fig. 17c). The circuital sense

of A is projected on A'. We shall give A' the sign correspondingto its circuit on p, and call this signed area the component of Aon p. Since I u = A and (n, u) = 8 or ir - 0,

n- u = u I cos (n, u) = A cos 0 or -A cos 8according as (n, u) is acute or obtuse. In both cases n u givesthe component of A on p:

(1) compp A = n u.THEOREM. When the vector areas of the faces of a closed poly-

hedron are all drawn in the direction of theoutward normals, their sum is zero.

Proof. Any polyhedron may be subdi-vided by planes into a finite number of A

tetrahedrons. Let ABCD (Fig. 17d) be onesuch tetrahedron. If we imagine ABC to

C

FIG. 17d

lie in the plane of the paper while D is above the paper, the out-ward vector areas of the triangular faces,

DAB, DBC, DCA, ABC,

§ 17 VECTOR AREAS 39

are given by one half of the respective vectors,

DA X DB, DB X DC, DC X DA, -AB-AC.If we choose D as origin, the sum of these vectors is given by

axb+bxc+cxa - (b - a)X(c - a) = 0.Thus the theorem is true for a tetrahedron.

Now write such an equation for all the tetrahedrons that makeup the polyhedron, and add the results. The vector areas overall inner faces cancel, for each appears twice but with opposeddirections. The net result on the left is double the sum of theoutward vector areas of the polyhedron's faces. Since this sum iszero, the theorem follows.

Consider now a polygon P1P2 . . Ph of area A lying in a di-rected plane of unit normal n (Fig. 17e), and suppose that thecircuit P1P2 . Ph is positive. In the D

figure n points up from the plane ofthe paper, so that a positive circuit iscounterclockwise. Choose an origin 0at pleasure above the plane (towards Pthe reader), and let r1, r2, , rh ''

denote the position vectors of the ver-tices. These vectors are the edges of P,

a pyramid having 0 as vertex and the Fla. 17egiven polygon as base. The triangularfaces of this pyramid have as their outward vector areas 2r1 x r2i

ire X r3, , irh x r1, while the outward vector area of the base is-An. From the preceding theorem the sum of these vector areasis zero; hence

(2) An = i (r1 x r2 + r2 x r3 + ... + rh x r1).

As the vertex 0 approaches the plane of the base, the terms onthe right of (2) vary continuously, but their sum is always An.Hence when 0 is in the plane of the polygon, (2) remains valid.If 0 is any origin of rectangular coordinates in the plane, let thevertices of the polygon be (x1, yl), (x2, y2), , (xh, yh) taken incounterclockwise order. Now n = k, and

r1 X r2 = (x1i + yll) X (x2i + y2i) = k,x1 x2

Y1 Y2

40 VECTOR ALGEBRA

so that (2) gives

(3) 2A = I xl x2

I

+I x2 x3 -+ -. ... + Xh xl

Y1 Y2 f y2 Y3 I I Yh Y1 J

§ 18

Example 1. The sum of the outward vector areas of the triangular prism(Fig. 17f) is

uxw +vXw - (u +v) xw = 0.The vector areas of the triangular bases cancel, forthey are equal in magnitude but opposite in direction.

Example 2. To find the area A of a triangleu whose vertices are (a, 0, 0), (0, b, 0), (0, 0, c), put

FIG. 17f r1 = ai, r2 = bj, r3 = ck in (2); then

An = i (abk + bci + caj), A = 11/a2b2 + b2c2 + OA

Example 3. To find the area of the polygon whose vertices in counter-clockwise order are (1, 2), (5, 4), (-3, 7), (-5, 5), (-1, -3), we form thearray,

(x) 1 5 -3 -5 -1 1

(y) 2 4 7 5 -3 2,

repeating the first column. Then, from (3),

2A = -6+47+20+20+1 =82, A =41.

18. Vector Triple Product. The vector (u x v) x w is perpendic-ular to u x v and therefore coplanar with u and v; hence (§ 5),

(u x v) x w = au + /3v.

But, since (u x v) x w is also perpendicular to w,

a U 0.

All numbers a, that satisfy this equation must be of the forma = -A v w, = X u w, where X is arbitrary. Thus we have

(uxv)xw =In order to determine X, we use a special basis in which i is col-

linear with u, j coplanar with u, v; then

u = u1i, V = v1i + v2j, w = wli + w2j + w3k.

Substituting these values gives, after a simple calculation, A = 1We therefore have the important expansion formulas,

(1) (uxv)xw = u-WVwx(uxv) = w-vu -wuv.

§ 19 SCALAR TRIPLE PRODUCT 41

In the left-hand members of (1), one of the vectors in parenthesisis adjacent to the vector outside, the other remote from it. Theright-hand members may be remembered as

(Outer dot Remote) Adjacent - (Outer dot Adjacent) Remote.

In general (u x v) x w -/- u x (v x w); for the former is coplanarwith u and v, the latter with v and w. Cross multiplication of vec-tors is not associative.

From (1) we see that the sum of a vector triple product and itstwo cyclical permutations is zero:

(2) (axb)xc -{- (bxc) x a + (cxa)xb = O.

If 1 is a directed line carrying the unit vector e, we may expressany vector u as the sum of its orthogonal projections on 1 and ona plane p perpendicular to 1:

(15.6),(3) projl u = e comps u = (e u) e

(4) projp u = u - (e u) e = e x (u x e).

19. Scalar Triple Product. The scalar product of u x v and wis written u x v - w or [uvw]. No ambiguity can arise from the pa-rentheses being omitted, since u x (v w)is meaningless.

THEOREM. The product u x v - w isnumerically equal to the volume V of aparallelepiped having u, v, w as concur-rent edges. Its sign is positive or negativeaccording as u, v, w form a right-handedor left-handed set.

uXv

Fic. 19

Proof. The volume V (Fig. 19) may be computed by multiply-ing the area of a face parallel to u and v,

A = J u l H v l sin (u, v) = l u x e 1,

by the corresponding altitude h = I w I cos B:

V= IuxvIIwIcos6.

Now the angle between u x v and w is B or 7r - B according asu, v, w form a dextral (as in the figure) or a sinistral set (§ 12).The definition of a scalar product now shows that

V dextral.(1) u x v w = when the set u, v, w is

V} sinistral.


The dextral or sinistral character of a set u, v, w is not altered

by a cyclical change in their order. Hence, from (1),

(2)

But a dextral set becomes sinistral and vice versa, when the cy-clical order is changed :

(3)

Thus, if the set u, v, w is dextral, the products in (2) all equal V,while the products,

U"W V =all equal - V.

On account of the geometric meaning of the scalar triple product,we shall call it the "box product." $

If u, v, w are proper vectors, V = 0 only when the vectors arecoplanar. Therefore: three proper vectors are coplanar (parallel tothe same plane) when and only when their box product is zero. Inparticular, a box product containing two parallel vectors is zero;for example u x v u = 0.

The value of a box product is not altered by an interchange of thedot and cross. For

(4)

as dot multiplication is commutative. The notation [uvw] oftenis used for the box product, as the omission of dot and cross causesno ambiguity.

From (15.3) and (16.3), we have

(5) (au) x (av) . (tiw) = a(3y u x v w.

Finally, the distributive law for scalar and vector productsshows that a box product of vector sums may be expanded justas in ordinary algebra, provided that the order of the vector fac-tors is not altered. Thus, if we expand the product u x v w whenthe vectors are referred to the basis i, j, k, we obtain 27 terms ofwhich all but six vanish as they contain box products with twoor three equal vectors. The remaining six terms are those con-taining

[ijk] = [jki] = [kij] = 1, [ikj] = [kji] = [jik] = -1,$ The name proposed by J. H. Taylor, Vector Analysis, New York, 1939,

p. 46.

3 20 PRODUCTS OF FOUR VECTORS

and constitute the expansion of the determinant,

261 U2 U3

(6) [uvw] = v1 v2 v3 .

W1 W2 W3

This result also follows at once from (16.8).

43

Example. To find the point P where the line AB pierces the plane CDE.

Method. AP = it AB; the scalar it is then determined by the equation,

(i) CP CD X CE = (CA + it AB) . CD x CE = 0

which expresses that P is coplanar with C, D, E.Computation. With the points,

A(1, 2, 0), B(2, 3, 1); C(2, 0, 3), D(0, 4, 2), E(-1, 2, -2);

CP =CA+XAB =[-1,2, -3]+1`[1,1,1] =[7,-1,it+2,A-3];

CD xCE = [-2,4, -1]x[-3,2, -5] = [-18, -7,8];

hence, from (i),

-18(X - 1) -7(X+2)+8(x-3) =17x-20=0, 7;OP = OA + X AB = [1, 2,0] - -4[1, 1, 1] = .[-3, 14, -20].

-> --4 ---,Check. DP CD X CE = 0; for

DP = --[1,18,18],

20. Products of Four Vectors. Since

(a x b) (c x d) = a b x (c x d) = a (b d c - b c d),

(1)

(a x b) x x d) may be regarded as a triple productof a x b, c, d or of a, b, c x d. We thus are led to the two expan-sions,

(2) (a x b) x (c x d) = [acd]b - [bcd]a = [abd]c - [abc]d,

which give, in turn, the following equation connecting any fourvectors:

(3) a[bcd] - b[cda] + c[dab] - d[abc] = 0.

44 VECTOR ALGEBRA

When [abc] 0,

(4) [abc]d = [dbc]a + [adc]b + [abd]c

gives d explicitly in terms of the basis a, b, c.

§ 22

21. Plane Trigonometry. If the vectors a, b, c form a closedtriangle when placed end to end (Fig. 21),

(1) a+b+c=0.We shall denote the lengths of the sides by a, b, c and the interior

angles opposite them by A, B, C; then theangles (b, c), (c, a), (a, b) are equal, respec-tively,to7r- A,7r-B,,r-C.

From the identity,a

FIG. 21 (a+b) (a+b) =we obtain the cosine law for plane triangles:

(2) c2 = a2 + b2 - tab cos C.

Cyclical interchanges of the letters in (2) give two other forms ofthis law.

On multiplying (1) first by a x, then by b x, we find

(3) bxc = cxa = axb.Division by abc gives the sine law for plane triangles:

(4)

sin A sin B sin C

a b c

Since each product in (3) isdouble the vector area of thetriangle, its

(5) Area =2

be sin A

= I ca sin B = 2 ab sin C.

22. Spherical Trigonometry.Consider a spherical triangleABC on a sphere of unit radius,and let a, b, c, denote the position

FIG. 22

vectors of the vertices referred to its center (Fig. 22). The nota-tion is so chosen that a, b, c, form a dextral set: then [abc] > 0.

22 SPHERICAL TRIGONOMETRY 45

Let a, /3, y denote the sides (arcs of great circles) opposite thevertices, A, By C. The interior dihedral angles at these verticesalso are denoted by A, B, C. We shall consider only sphericaltriangles in which the sides and angles are each less than 180°.Now

(1)

(2) bxc = sin a a', cxa = sin a b', axb =

where a', b', c' are unit vectors. The vectors b' and c' are perpen-dicular to the planes of c, a and a, by respectively, and include anangle a' = it - A, the exterior dihedral angle at A. Moreover,from (2),

b'xc' _(c x a) x (axb) [abc]

a,sin /3sin y sin f3sin y

a positive multiple of a. Hence if a', 0', y' denote the exteriordihedral angles at A, By C, we have

(3) b' c' = cos all Ca= cos a' b' = cos y'

(4) b'xc' = sin a' a, c'xa' = sin#'b, a' b' = sin y'c.

Thus there is complete reciprocity between the vector setsa, by c, and a', b', c', and also between the spherical triangles theydetermine. While ABC has a, S, y for sides and a', (3', y' for ex-terior angles, A'B'C' has a', 0', y' for sides and a, /3, y for exteriorangles. Since the vertices of one triangle are poles of the corre-sponding sides of the other, the triangles are said to be polar.

From (2) and (4), we have

[abc] = sin a a a' = sin/3b - b' =

sin a' sin /3'b - by = sin y' c. c';

hence, on division,

(5)[abc] _ sin a _ sin / _ sin y

[a'b'c'] sin a' sin S' sin y'

This is the sine law for spherical triangles.Again, from (2), we have

sin /3 sin y by c' = (c x a) (axb) = (c a) (a b) - b c;

46 VECTOR ALGEBRA

hence, from (1) and (3),

(6) cos a = cos 0 cos y - sin 0 sin y cos a'.

Similarly, from (4), we deduce

(7) cos a' = cos 0' cos y' - sin f3' sin y' cos a.

§ 23

Equations (6) and (7) and the four others derived from them bycyclical permutation constitute the cosine laws for spherical tri-angles.

By replacing a', 0', y' in (5), (6), and (7) by 7r - A, 7r - B,it - C we may express the sine and cosine laws in terms of thesides and interior angles. Thus, we find, for the sine law,

sin a sin j sin y

sin A sin B sin C '

and, for the cosine laws,

cosa = cos,3cosy + sin $ sin -y cos A,

cos A = - cos B cos C + sin B sin C cos a.

In this version of the cosine laws, the structural similarity ex-hibited by (6) and (7) is lost.

23. Reciprocal Bases. Two bases, el, e2, e3 and e', e2, e3, aresaid to be reciprocal when they satisfy the nine equations:

el e'=1, el e2=0, el e3=0,(1) e2 e' = 0, e2 e2 = 1, e2 e3 = 0,

e3 e' = 0, e3 e2 = 0, e3 e3 = 1.

The superscripts applied to the base vectors are not exponents,but mere identification tags. By use of the Kronecker delta 6j, de-fined as

(2) ' 0

when i = jwhen i 5 j,

equations (1) condense to

(3) ei e' = Si (i, j = 1, 2, 3).

Consider the three equations in the first column of (1). Thesecond and third state that e' is perpendicular to both e2 and e3,

§ 23 RECIPROCAL BASES 47

that is, parallel to e2 x e3. Hence el = X e2 x e3; and, from thefirst equation, 1 = X el e 2 x e3. We thus obtain

(4) el = e2 x e3 e2 = e3 x el e3 = el x e2[ele2e3] [ele2e3] [ele2e3]

e2 and e3 being derived from el by cyclical permutation. Fromthe symmetry of equations (1) in the two sets e1, ei, we have also

e2xe3 e3xel elxe2(5) el = [ele2e3] , e2 = [ele2e3] , e3 = [ele2e3] .

Thus, either basis is expressed in terms of the other by preciselythe same formulas.

From (4) and (5), we have

1(e2 x e3) (e2 x e3)

e el =[ele2e3][ele2el]

_[eIe2e3][ele2e3]

on making use of (20.1) and equations (1); hence

(6) [ele2e3][ele2e3] = 1,

an equation which gives further justification for the name recip-rocal applied to the sets. Since the box-products in (6) must havethe same sign, a basis and itsreciprocal are both right-handed or both left-handed.

As to the orientation of re-ciprocal sets, we see from (1)that e' is perpendicular to theplane of e2 and e3 in the di-rection which makes an acuteangle with el. Similar state-ments apply to e2 and e3. Ifthe vectors et, et are all drawnfrom the same point 0 and cutby a sphere s about 0 in thepoints Ei, Et, respectively,

Fla. 23

the three planes OEiE; cut a spherical triangle E1E2E3 from s; andthe three planes OEE' cut out a second spherical triangle E'E2E3(Fig. 23). Either triangle is the polar of the other; for E', E2, E3are poles of the great-circle arcs, E2E3, E3E1, E1E2i and similarlyfor El, E2, E3.


On dividing the identity,

ei x (e2 x e3) + e2 x (e3 x el) + e3 x (el x e2) = 0 (18.2),

by [ele2e3], we obtain the relation,

(7) elxel +e2xe2 +e3xe3 = 0.

This has an interesting geometric interpretation. The great circleEIEI is perpendicular to both great circles E2E3 and E2E3; it thuscontains the altitudes of both triangles through the vertex labeled 1.Similarly, the great circles E2E2, E3E3 contain the altitudes ofboth triangles through the vertices labeled 2 and 3. The planesof the great circles EIEI, E2E2, E3E3 are perpendicular to el x el,e2 x e2, e3 x e3, respectively; and these vectors, in view of (7), liein a plane p. The poles of this plane (the ends of the diameter ofs perpendicular to p) are points common to all of the great circlesEiEi. Therefore, the three altitudes EiEi of the polar trianglesEIE2E3, E1E2E 3 meet in a pole P of the plane p.

When a basis and its reciprocal are identical, the basis is calledself-reciprocal. The equations (3) then become

(8) ei - e; = bi;. j'

These equations characterize an orthogonal triple of unit vectors.Hence a basis is self-reciprocal when and only when it consists of amutually orthogonal triple of unit vectors. The triple will be dex-tral if [ele2e3] = 1 (§ 19), sinistral if [ele2e3] = -1. Thus i, j, kand i, j, -k are typical dextral and sinistral orthogonal bases.

24. Components of a Vector. We now supplement the notationof § 12 by writing the components of a vector u ui or ui accordingas the basis is ei or et. The components ui are called contravariant,the components ui covariant, for reasons given in § 148. Any vec-tor now may be written in two forms:

(1) (2) u = ulel + u2e2 + u3e3i u = ulel + u2e2 + u3e3.

From (1) and (2), we obtain equations of the type u - e' = ul,u - el = ul ; all six are included in

(3) (4) ui = u ei, ui = u - ei (i = 1, 2, 3).

When a basis is given, a vector u is completely specified bygiving its components written in order. With the notation of

t With a self-reciprocal basis there is no need for superscripts; we thereforewrite S as Sii.

§ 24 COMPONENTS OF A VECTOR 49

§ 12, we write a vector u as a number triple, [ul, u2, u31 or[ul, U2, u31 according as it is referred to the basis ei or et. Wedenote the (non-zero) box products of the base vectors by

(5) E = [ele2e3l, 1/E = [ele2e31.

Addition and multiplication by scalars follow the formulas of§ 12; thus, with covariant components,

(6)

(7)

X [u1, u2, u31 = [Xu1, Xu2, Xu3]

[u1, u2, u31 + [v1, v2, u31 _ [u1 + v1, u2 + v2, u3 + vJ

A simple expression for the scalar product u v is obtained whenu and v are referred to different, but reciprocal, bases. Thus, ifwe compute

u . V =J (u'e1 + u2e2 + u3e3) . (vie1 + v2e2 + v3e3)

l(ulel + u2e2 + u3e3) - (vlel + v2e2 + v3e3)

we obtain, by virtue of equations (23.1),

(8) u V = u1V1 + u2V2 + u3e3 = u1v1 + u2e2 + u3V3.

We next compute the components of u x v, relative to bases e2and ei. Using covariant components gives

u x v = (uie' + u2e2 + u3e3) x (v1e1 + v2e2 + v3e3)

U2 U3

V2 V3

U2 U3

V2 V3

e2 x e3 +

e1

or, more compactly,

u3 u1

V3 V1

U3 u1

V3 V1

(9) u x v = E-1

e3xel +

e2

E

u1 U2

V1 V2

e3

E

e1xe2

Using contravariant components, we find, in similar fashion,

(10) u x v = E


From (9) and (10), we see that the components of q = u x v are

(11) q' = E-1Uj uk I

Vi vk= Eqt

u'

v'

11k 1

ak

where the indices ijk form a cyclical permutation of 123.From (9) and (10), we next obtain two expressions for the box

product u x v w:

(12) [uvw] = E-1U1 u2 U3

V1 V2 V3

W1 w2 w3

=Eu1 u2 u3

V1v2 v3

W1 w2 w3

The first formula of (12) may be written

(13) [uvw][e1e2e3] = v el v . e2 v - e3 f

whereas the second gives the analogous equation obtained by re-placing ei by ei. Since e1i e2, e3 may be any linearly independentset of vectors a, b, c, we also have

(14) [uvw][abc] =

ua ucva vb vcwa wb wc

When the basis ei is self-reciprocal (an orthogonal triple of unitvectors) e i = ei and E = +1. The two sets of components of uthen coalesce into a single set which we arbitrarily write U. Ifthe self-reciprocal basis is dextral, we can write el = i, e2 = j,e3 = k, E = 1; then

U V = U1V1 + U2V2 + u03 ,

U X S7 = [uvw] =

in agreement with our previous results.25. Vector Equations. A vector is uniquely determined when

its scalar and vector products with two known non-perpendicularvectors are given. Thus let

(1) ua=a, uxb=c, (ab-/- 0);

§ 26 HOMOGENEOUS COORDINATES 51

the scalar a and the vectors a, b, c are regarded as known, andb c = 0. We have

(2)

ax(uxb) =a -c, or

abBy direct substitution, (2) is seen to be a solution of equations (1).

A vector is also determined when its scalar products with threeknown non-coplanar vectors are given. For example, if

(3) u - e1 = u1, u - e2 = u2, u - e3 = u3,

and the sets et, e i are reciprocal, we have, from (24.2),

(4) u = ulel + 112e2 + u3e3.

By direct substitution, (4) is seen to be a solution of equations (3).26. Homogeneous Coordinates. Coordinates of a point, line or

plane are called homogeneous if the entity they determine is notaltered when the coordinates are multiplied by the same scalar.A coordinate that depends upon the choice of origin 0 bears thesubscript 0; such coordinates are written after those independentof the origin. In equations r denotes any position vector from 0to the entity in question, whereas r1i r2 denote position vectorsto points R1, R2 given in advance. In this article we use Latinletters for vectors, Greek for scalars.

Point Coordinates. If r = OA, we write r = ao/a. The "equa-tion" of the point A is then

(1) ra = ao,

and its homogeneous coordinates are (a, ao). Note that (Xa, Xao)determine the same point; hence the coordinates (a, ao) dependon three independent scalars: "there are o03 points in space."

Plane Coordinates. The equation of a plane through R1 andperpendicular to the vector a is (r - r1) a = 0; or, on writingao=r1-a,(2) r a = ao.

The homogeneous coordinates of the plane are (a, ao). Since(Xa, Xao) determine the same plane, the coordinates (a, ao) de-pend on three independent scalars: "there are oo 3 planes in space."


Line Coordinates. The equation of a line through Rl and par-allel to the vector a is (r - r1) x a = 0; or, on writing ao = rl x a,

(3) r x a = ao.

The homogeneous (or Plucker) coordinates of the line are (a, ao)and are connected by the relation,

(4) a ao = 0.In view of. (4) and the fact that (Xa, Xao) determine the same line,vile coordinates (a, ao) depend on four independent scalars: "thereare ooa lines in space."

We thus have the homogeneous coordinates:

Point (a, ao), Plane (a, ao), Line (a, ao).

The first coordinate cannot vanish. When the second coordinate(subscript 0) is zero, the origin 0 is on the point, plane, or line,respectively.

The distances of the point, plane, or line from the origin are,respectively,

(5)ao! laol aolIai' Jai ' Ia!

The first result is obvious. If p is the vector from 0 perpendicularto the plane r a = ao,

p =

p ao a p is the vector from0 perpendicular to the line r x a = ao,

axa0p x a = ao, p a= 0; hence p =

from (25.2), and I p I = ( ao I/I a IIf the origin is shifted from 0 to P, we must replace r in (1),

(2), (3) by PO + r to obtain the second coordinate referred to P.We thus obtain the shift formulas:

(6)

(7)

(8)

in the respective cases.

ap = ao + P0 a,

ap =-3

ap=ao+POxa,

§ 26 HOMOGENEOUS COORDINATES 53

Two Elements Determine a Third. We have the following cases:Two distinct points determine a line:

(9) (a, ao), (a, bo) (abo - $ao, ao x bo).

Two non-parallel planes determine a line:

(10) (a, ao), (b, $o) --> (a x b, 00a - aob).

A point and line determine a plane:

(11) (a, ao), (b, bo) -> (abo - aoxb, ao bo),

unless abo - ao x b = 0; then also ao bo = 0, and the point lieson the line.

A line and plane determine a point:

(12) (a, ao), (b, $o) -* (a b, 0oa - aoxb),

unless a b = 0. If 0oa - ao x b = 0 (and hence a b = 0), theline lies in the plane.

Proofs in outline.For (9) : The line AB has the equation,

Cras

/x(

a

as

J= 0, or r x (abo - gao) = ao x b0.

For (10) : The line is parallel to a x b; and

rx(axb) = $oa -aob.

For (11) : Refer the line (b, bo) to the point A(a, ao); then,from (8),

bA=bo - OAxb=bo - aoxba

The plane through A normal to bA has the equation

ao(r - - (abo - ao x b) = 0, or r (abo - ao x b) = ao bo.a

For (12) : The equations of line and plane, r x a = ao and r b13o, have the solution (§ 25) :

Qoa - ao x br=-a b


Two Lines. If Rl and R2 are points on the lines (a, ao), (b, bo),the lines are coplanar when and only when the vectors r1 - r2,a, b are coplanar. Since

(13) (r1

a necessary and sufficient condition that the lines be coplanar is

(14)

If the lines are not coplanar, let e be a unit vector in the direc-tion a x b of their common normal; then, from (13),

(15)a bo + b ao

IaxbI

This is the component of R2R1 in the direction of e; its numericalvalue gives the shortest distance between the lines.

Equations in Point and Plane Coordinates. If the point (a, so)lies on the plane (t, To) we have

(16)

on eliminating r from ra = so and r t = To. This may be re-garded as the equation of the point (a, so) in plane coordinates; oras the equation of the plane (t, To) in point coordinates.

If the line (p, po) contains the point (a, so) we have

(17) soxP -apo = 0,

on eliminating r from ra = so, r x p = po. This is the equation ofthe line (p, po) in point coordinates.

If the line (p, po) lies on the plane (t, TO) we have p t = 0 andhence

(18) Top - Po x t = 0,

on eliminating r from r x p = po and r t = To. This is the equa-tion of the line (p, po) in plane coordinates.

Example. If three points (a, ao), (S, bo), (y, co) lie on the line (p, po), wehave, from (17),

ao bo co-xP = XP = _XP = po;16 Y

hence

(i) y)+µ(0 y) =0,

§ 27 LINE VECTORS AND MOMENTS 55

since both vectors in parenthesis are parallel to p. This is a linear relationbetween ao/a, bo/)3, co/y in which the sum of the coefficients is zero (§ 6).

If the three planes (a, ao), (b, ,o), (c, yo) pass through the line (p, po), wehave, from (18),

a b cPOX-=POX-=Pox-=P,

ao lio yo

provided ao, 00, yo are not zero; hence

\ao yoi + µ \ljo -Yo -Op

since both vectors in parenthesis are parallel to po. This is a linear relationconnecting a/ao, b/3o, C/-Yo in which the sum of the coefficients is zero.

A shift in origin does not alter the coefficients in (i). In (ii), however, thecoefficients are changed but their sum still remains zero; if we write v =-A - IA and shift the origin to P,

X o+ AA -+ v-= 0 becomesPo -Yo

X' +lz'-+v -=0,pop 'YP

where X' = Xap/ao, u' = -flP/0o, d = vyp/yo Let the student prove thatx'+A '+v'=0.

27. Line Vectors and Moments. A vector which is restricted tolie in a definite line is called a line vector. If f is the vector, and

(1) r x f = fo, (f fo = 0)

the equation of its line of action, the line vector is completelyspecified by the Pliicker coordinates f, fo.

The vector fo is called the moment of f about the point 0. If

f=AB,(2) fo = OA x AB

is twice the vector area of the triangle OAB (§ 17); fo remains con-stant as f is shifted along its line of action.

If the origin is shifted from 0 to P, the moment of f about P

is PA x AB = (PO + OA) x AB; hence

(3) fp = fo + PO x f.

If s is any axis through P with the unit vector e, the componentof fp on s, namely, e fp (15.6), is called the moment of f about theaxis s. We speak of moment about an axis because e fp is inde-


pendent of the position of P on this axis; for, if the axis is givenby the unit line vector (e, eo),

(4)

e fp as the component of twice the vectorarea PAB on a directed plane of unit normal e (§ 17). In Fig. 27,

A

Fro. 27

A1B1 is the projection of AB on the plane;the moment of f about s is numericallyequal to twice the area PA1B1, that is, tothe product of the length A1B1 and the per-pendicular distance h of A1B1 from P; itssign is plus or minus according as a turn inthe sense PA1B1 would advance a right-handed screw in the direction of s or -s.

We repeat these important definitions.

The moment of the line vector f = ABabout a point P is

fp=PAxAB=rxf,(5)

where r is a vector from P to any point on f's line of action. Themoment of f about an axis is the component of r x f on the axis,where r is a vector from any point on the axis to any point on theline of action. The moment about a point is a vector; about anaxis, a scalar.

Example. The line of action of the force f = [1, -1, 2] passes through thepoint A (2, 4, -1). Find its moment about an axis through the pointP(3, -1, 2) and having the direction of the vector [2, - 1, 2].0

Since PA = [-1, 5, -3] and e = 3[2, -1, 21 is a unit vector along the axis,

fp = PA x f =i i k

-1 5 -3 = [7, -1, -4]1 -1 2

is the moment of f about P; and

is the moment of f about the axis.

§ 28 SUMMARY: VECTOR ALGEBRA 57

28. Summary: Vector Algebra. Equal vectors have the samelength and direction. Free vectors are added by the triangle con-struction. The negative of a vector is the vector reversed. Tosubtract a vector, add its negative. The product Au is a vectorA I times as long as u; its direction is the same as u if X > 0, the

reverse if X < 0.Vectors may be added, subtracted, and multiplied by real num-

bers in conformity with the laws of ordinary algebra.The n vectors ui are linearly dependent if there exist n real num-

bers Ai, not all zero, such that EAiui = 0. When n = 2, 3, lineardependence implies that the vectors are collinear or coplanar, re-spectively; and conversely. In space of three dimensions any fourvectors are linearly dependent.

A point P divides a segment AB in the ratio X = /3/a when----3 -3AP = APB; then

-3 --> -3(a + $)OP = aOA + SOB.

----3A linear relation2;A1OP1 = 0 connecting n position vectors will

hold for any origin when and only when :Ai = 0. When n =2, 3, 4, the points Pi are coincident, collinear, or coplanar, re-spectively.

A set of weighted points miPi has a unique centroid P if 2;mi$ 0; its defining equation is

--3-* XmiOPilmiP*Pi = 0; and OP* =

Emi

If all mi = 1, P* is called the mean center.The scalar product u v is defined as

Jul Ivlcos(u,v).Laws:

V. Up U. (v+w) =

u v Iu v

u v = l u v l

is a u ve a dextral set.

58 VECTOR ALGEBRA

Laws :

§ 28

UXV = `VxU, Ux(V+w) = uxv+uxw.If u, v 0, u x v = 0 implies that u l l v and conversely.

Expansion rule:

ux(vxw) =

Cross multiplication is not associative.The box product, u x v w or [uvw], is numerically equal to the

volume of a "box" having u, v, w as concurrent edges; its sign is+ or - according as u, v, w form a dextral or sinistral set. Thevalue of u x v w is not affected by a change in cyclical order ofthe vectors, or by an interchange of dot and cross. If u, v, w 5-4- 0,[uvw] = 0 implies that u, v, w are coplanar, and conversely.

A set of three vectors ej forms a basis if E = [ele2e3] 0. Toevery basis there corresponds a unique basis ei such that ej e'= S1; such bases are called reciprocal. If the indices i, j, k forma cyclical permutation of 12 3,

ei =e,xek

[e1e2e3]

ei x ekei = [ele2e3] ; [ele2e3][e'e2e3] = 1.

Both bases are dextral if E > 0; sinistral, if E < 0.Given a basis e1, a vector u may be written as Euiei or ruiei;

the numbers ui are contravariant components of u, ui covariantcomponents.

u+v = E(ui'+vi')ei = E(ui+vt)ei;Xu = 2:xuiei = 2;xuie`;

u V = uIv1 + u2v2 + u3v3 = u1v1 + u2v2 + u3v3;

el e2 e3 I Iel

e2 e3

U x v = E-1U2 E ul

2 3

vlV3 v v

U1 u2 U3Ut

u2 u3

[uvw] = E-1 I V1

W1

V2

W2

V3

W3

E 1vi

w2

w3

I

When the basis ej is self-reciprocal (ei = ei), its vectors form amutually orthogonal set of unit vectors; E = 1, if the basis is dex-tral, E = -1, if sinistral. A dextral self-reciprocal basis is written

PROBLEMS 59

i, j, k. For such a basis ui = ui; these rectangular components aregiven by

ui = I u I cos (ei, u), and I U I2 = (ul)2 + (u2)2 + (u3)2.

The two formulas previously given for u + v, Au, u - v, u x v,u x v w in each case become identical.

The equations of a point, line, and plane, in terms of theirhomogeneous coordinates (a, ao), (b, bo), (c, -yo), are

ra=ao, rxb=bo, r - c=yo,

respectively. When the origin is shifted to P,

ap = ao+POa, by = bo+POxb, yP =

The moment of the line vector (f, fo) about the point P isfp = r x f, where r is any vector from P to its line of action. Themoment of (f, fo) about an axis through P is the component of fpon this axis; if the axis is given by the unit line vector (e, eo), thisaxial moment

PROBLEMS

1. If ABC is any triangle and L, M, N are the mid-points of its sides, showthat, for any choice of 0, a

-- 3 - 3 - 3 -4 --3 - 3OA + OB + OC = OL + OM + ON.

--- ---, --> - 92. If OA' = 3 OA, OB' = 2 OB, in what ratio does the point P in which

AB and A'B' intersect divide these segments?3. Show that the mid-points of the four sides of any quadrilateral (plane

or skew) form the vertices of a parallelogram.4. P and Q divide the sides CA, CB of the triangle ABC in the ratios

--4 - 3x/(1 - x), y/(1 - y). If PQ = X AB, show that x = y = X.

5. E, F are the mid-points of the sides A B, BC of the parallelogram A BCD.Show that the lines DE, DF divide the diagonal AC into thirds and that ACcuts off a third of each line.

6. OAA', OBB', OCC', ODD' are four rays of a pencil of lines through 0cut by two straight lines ABCD and A'B'C'D'. If C and D divide AB inthe ratios r and s, C' and D' divide A'B' in the ratios r' and s', prove thatr/s = r'/s'.

7. The points A, B, C and A', B', C' lie, respectively, on two intersectinglines. Show that BC', CB'; CA', AC'; AB', BA' intersect in the collinearpoints P, Q, R (Pascal's Theorem).

60 VECTOR ALGEBRA

8. Lines drawn through a point P and the vertices A, B, C, D of a tetra-hedron cut the planes of the opposite faces at A', B', C', D'. Show that thesum of the ratios in which these points divide the segments PA, PB, PC, PDis -1. [Equation (10.2), with e = - (a + 0 + y + S) may be written

as+8b+yc+ad+ep =0, a+i9+y+s+e =0.From this we conclude, as in § 7, ex. 3, that a' _ (aa + ep)/(a +,E), etc.]

9. The line DE is drawn parallel to the base AB of the triangle ABC andis included between its sides. If the lines AE, BD meet at P, show that theline CP bisects AB.

10. The points P, Q, R divide the sides BC, CA, AB of the triangle ABCin the ratios a/(1 - a), l3/(1 - 0), y/(1 - y). If P, Q, R are collinear,

xp+yq+zr=0, x+y+z=0;putting p = (1 - a)b + ac, etc. in this equation, we obtain a linear relationin a, b, a whose coefficients have a zero sum. Show that this implies that theseparate coefficients vanish; hence deduce the Theorem of Menelaus (§ 7, ex. 2):

afy/(1 - a) (1 - p)(1 - y) = -1.11. Using an argument patterned after that in Problem 10, prove Carnot's

Theorem:If a plane cuts the sides AB, BC, CD, DA of a skew quadrilateral ABCD

in the points P, Q, R, S, respectively, the product of the ratios in whichP, Q, R, S divide these sides equals 1.

12. Id a plane quadrilateral ABCD, the point P in which the diagonalsAC, BD intersect divides these segments in the ratios 4/3 and 2/3, respec-tively. In what ratio does the point Q, in which the sides AB, CD meet,divide these segments? 0

13. If el = O 1, e2 = OE2i e3 = 03E

form a basis, the reciprocal setel, e2, e3 are vectors perpendicular to the planes OE2E3, OE3E1i OE1E2 andhaving lengths equal to the reciprocals of the distances of El, E2, E3 fromthese planes, respectively. Prove this theorem.

14. Prove that a necessary and sufficient condition that four pointsA, B, C, D be coplanar is that

[dbc] + [add] + [abd] - [abc) = 0.

15. Show that the shortest distance from the point A to the line BC isIaxb +bxc +cxal/Ib - cI.

16. If the vectors el, e2, e3; el, e2, e3 form reciprocal sets, show thatthe vectors e2 x e3, e3 x el, el x e2; e2 x e3, e3 x e1, el x e2 do likewise.

17. Prove the formulas:(a) [a x b, bxc, c x a] = [abc]2,

(b) (bxc) (a x d) +(c x a) (b x d) +(a x b) (c x d) =0,

(c) (bxc) x (a x d) +(c x a) x (b x d) +(a x b) x (c x d) = -2[abc]d,

(d) (d-a) (b-c)+(d-b) (c-a)+(d-c) (a-b)=0,

(e) (a-d) x (b-c)+(b-d) x (c-a)+(c-d) x (a-b) =2(a x b+b x c+c x a).

PROBLEMS 61

18. Find the shortest distance between the straight lines AB and CD when

(a) A(-2, 4, 3), B(2, -8, 0); C(1, -3, 5), D(4, 1, -7),

(b) A(2, 3, 1), B(0, -1, 2); C(1, 2, 5), D(-3, 1, 0).

19. Show that the lines AB, CD are coplanar, and find the point P in whichthey meet:

A(-2, -3, 4), B(2, 3, 0); C(-2, 3, 2), D(2, 0, 1).

--4[CD is parallel to CP = AP - AC = XAB - AC; find X and then P from---> -4OP = OA + XAB.]

20. If the points P, Q, R divide the sides BC, CA, AB of the triangle ABCin the same ratio, show that A, B, C and P, Q, R have the same mean center.

21. If the vectors OA, OB, OC lie in a plane and are equal in length, provethat

OA + OB + OC = OH,

where H is the orthocenter of the triangle ABC. [§ 15, ex. 4.]22. The power of a point P with respect to a sphere of center C and radius r

is defined as (CF)2 - r2. Prove that the power of P with respect to a sphere

having AB as diameter is PA I'B.23. Prove that the sum of the n2 powers of n given points, P1, P21 , P,,,

with respect to the n spheres having for diameters the n segments joining thegiven points to a variable point P in space, is constant. [Am. Math. Monthly,vol. 51, p. 96.1

24. The lines (a, ao), (b, bo)'are coplanar; then a bo + b ao = 0 (26.14).Show that they determine the plane (a x b, ao b) if a x b 0 0; and the point

aoxbo) if 00.Solve Problem 19 using these results.25. Show that the three planes (a, ao), (b, $o), (c, yo) meet in the point,

([abc], aobxc +Rocxa +yoaxb),

provided [abc] 0 0.26. Show that the three points (a, ao), (0, bo), (y, co) determine the plane,

(a bo x co + 0 co x ao + y ao x bo, [aobocol),

provided aboxco +3coxao +yaoxbo 0 0.27. The points P, Q, R divide the sides BC, CA, AB of the triangle ABC

in the ratio of 1/2. The pairs of lines (AP, BQ), (BQ, CR), (CR, AP) inter-sect at X, Y, Z, respectively. Show that the area of the triangle XYZ is1 /7 of the area of ABC. [Twice the vector area of XYZ is y x z + z x a +xxyl

When P, Q, R divide the sides in the ratio t/1, the area of XYZ is(1 - t)2/(1 + t + t2) times the area of ABC.

62 VECTOR ALGEBRA

28. Using (26.16), prove that(a) If four points (a, ao), (0, bo), (y, co), (S, do) lie on a plane, the vectors

ao/a, , do/S are connected by a linear relation in which the sum of thecoefficients is zero;

(b) If four planes (a, ao), (b, 3o), (c, yo), (d, So) pass through a point, thevectors a/ao, , d/So are connected by a linear relation in which the sumof the coefficients is zero, provided ao, Qo, 'yo, So 54 0.

29. If a sphere S with a fixed center cuts two concentric spheres S1, S2iprove that the distance between the planes of intersection SS, and SS2 isindependent of the radius of S. Extend this theorem to cover the case whenS fails to cut one or both of the concentric spheres. [Consider the radicalplanes SS1, SS2.1

30. Prove that the radical planes of the three spheres (r - c1)2 = P1 (1 =1, 2, 3) meet in the line

J`C1 X C2 + C2 X C3 + C3 X c1, 2c1(e2 - e3 - P2 + A3) + cycl.;,

their radical axis. Consider the case when e1 X C2 + C2 X C3 + C3 X cl = 0.31. A plane system of forces Fi acting through the points ri is equivalent

to a single force F = 2;Fi. When all the forces Fi are revolved about thesepoints through an angle B, show that their resultant F also revolves throughB about the point f (the astatic center) given by

F X M - (11ri Fi)Ft = 2 whereF

M = 'ri x Fi.

When the forces are parallel (Fi = Xie), show that the astatic center is thecentroid of the weighted points Xiri.

32. P* is the centroid of a set of n weighted points miPi for which Tmi = in.Prove the Theorems of Lagrange:

(a) 1mi(OPi)2 = m(OP*)2 + 2:mi(P*Pi)2.

(b) Vmim1(PPi)2 = in mi(P*Pj)2,

where ij ranges over 2n(n - 1) combinations.

[OPi = OP* + P*Pi; PiP7 = P*Pi - P*Pi.]

33. Deduce the following results from Prob. 32.(a) If ABCD are the vertices of a square in circuital order, (OA)2 + (OC)2

_ (OB)2 + (OD)2 for any 0. Generalize.(b) If r is the radius of a sphere circumscribed about a regular tetrahedron

of side a, r2 = 3x2/8.(c) The mean square of the mutual distances of all the points within a

sphere of radius r is 6r2/5.34. If P* and Q* are the mean centers of p points Pi and q points Q re-

spectively, prove that

Mean (PA)2 = Mean (P*Pi)2 + Mean (Q*Q1)2 + (P*Q*)2.

CHAPTER II

MOTOR ALGEBRA

29. Dual Vectors. The vector f, bound to the line whose equa-tion, referred to the origin 0, is

(1) rxf = fo,is completely determined by the two vectors f, fo, its Plucker co-ordinates. These obviously satisfy the relation,

(2) f fo = 0.

The vector f does not depend upon 0; but fo, the moment of fabout 0, becomes

(3) fp=fo+POxf,when the origin is shifted from 0 to P.

We now amalgamate f and fo into a dual vector.

(4) F=f+efowhere a is an algebraic unit having the property e2 = 0. If

f = 1, F is called a unit line vector. The unit line vectors,

A=a+eao,stand in one-to-one correspondence with the c0 4 lines of space.A line vector F of length X always may be written F = XA, whereA is a unit line vector. Unit line vectors depend upon four in-dependent scalars, general line vectors upon five.

Finally, for applications in mechanics, we consider dual vectorsF without the restriction (2). The vectors f, fo then involve sixindependent scalars. The dual vector (or the entity it represents)then is called a motor, provided its resultant vector f is independentof the choice of 0, while its moment vector fo changes in accord-ance with (3) when the origin is shifted to P.

Line vectors are thus special motors for which f fo = 0; forunit line vectors also f f = 1.

63

64 MOTOR ALGEBRA § 30

30. Dual Numbers. In analogy with the complex numbersx + ix', W. K. Clifford introduced dual numbers x + ex', in whichx, x' are real and e is a unit with the property e2 = 0.

In x + ex', x is called the real part and x' the dual part. Wewrite

x-{-ex'=y+ey' when x=y,x'=y';x+ex'=0 when x = 0, x' = 0.

Addition and multiplication of dual numbers are defined by theequations :

(1) (x+ex')+(y+ey') =x+y+e(x'+y'),(2) (x + ex') (y + ey') = xy + e(xy' + x'y)Observe that (2) may be obtained by distributing the product onthe left and putting e2 = 0. From these definitions we see thataddition and multiplication are commutative and associative andthat multiplication is distributive with respect to addition. Infact, the formal operations are precisely those of ordinary algebrafollowed by setting e2 = e3 = . . . = 0.

The negative of x + ex' is defined as -x - ex'.If A = a + ea', B = b + eb', the difference X = A - B and

quotient Y = A/B satisfy, by definition, the equations B + X= A, BY = A. We find that(3) A - B = a - b + e(a' - b');and that BY = A has the unique solution,

A a a'b - ab'(4) B=b+ e

b2

when and only when b F6 0. Division by a pure dual number eb'is not defined. The quotient (4) may be remembered by means ofthe device (a + ea') (b - eb')/ (b + eb') (b - eb') used in complexalgebra.

A dual number a + ea' in which a 5=1 0 is said to be proper; theproduct and quotient of proper dual numbers are also proper.

If the dual product (2) is zero, there are three alternatives:

x=x'=0; y=y'=0; x=y=0.The last shows that a dual product can vanish when neither factoris zero; for any two pure dual numbers have zero as their product.

§ 31 MOTORS 65

If the function f(x) has the derivative f'(x), we define its valuefor the dual argument X = x + Ex' by writing down its formalTaylor expansion and setting E2 = E3 = = 0; thus

(5) Ax + Ex') = AX) + Ex'f' (x)

In particular,

(6) sin (x + Ex') = sin x + Ex' cos x,

(7) cos (x + ex') = cos x - ex' sin x.

When x = 0, we have sin Ex' = Ex', cos ex' = 1; consequently (6)and (7) have the form of the usual addition theorems of the sineand cosine. Note also that sine X + cost X = 1.

31. Motors. We now can characterize a motor as a dual multipleof a unit line vector. Thus, on multiplying the unit line vectorA = a + Eao (a a = 1, a ao = 0) by the dual number X + EX',we obtain the dual vector,

(1) M = m + Emo = (X + EX') (a + Ea.o).

On equating the real and dual parts of both members, we obtain

(2) m = Xa, mo = Xao + A'a.

To show that M is a motor we need only verify that mo transformsin accordance with (29.3):

(3) MP =mo+POxm.In view of (2), this result follows from

mp = Nap + A'a = X (ao + PO -a) + X'a

_ (Xao + X'a) + PO x (Xa).

From (3), m mp = m mo; hence the scalars m m,and m moare invariants in the sense that they are not altered by a shift oforigin. From (2),

(4)

m 0, we call the invariant,

(5)

X' mmomm

66 MOTOR ALGEBRA §31

the pitch of the motor. Choosing A > 0, we have the uniquesolution,

A=ImI, A'=,zim1,Aa=m, Aao=mo-µm.

When m 0 0, M has the dual length,

(6) A + EX, = I m { (1 + eµ)

and its axis is along the line vector,

(7) ImIA=m+e(mo-µm).

The equation of the axis is therefore

(8)

(mxmo)Xmrxm =mo -gym= ;

this shows that the axis passes through the point Q given by--f mxmo

(9) r=OQ= ;

and, from (3),

(10) mQ=mo - =µm.

At all points on its axis M has the same moment µm, a fact alsoapparent from (7). Only for points P on the axis is the moment mpparallel to m.

Motors for which m 0 (A 0) are called proper. Propermotors are screws if m mo 0 0 (A, A' 0 0) ; line vectors if m mo =0 (A 0 0, A' = 0). Only proper motors have a definite axis.

If m = 0, mo 0 0 (A = 0, A' 0), M = emo is pure dual; thenmo is not altered by a change of origin. In this case M is calleda couple of moment mo. A couple may be regarded as a screw- ofinfinite pitch with an axis of given direction but arbitrary positionin space.

Finally if m = 0, mo = 0, the motor M = 0, then A = A' = 0from (2). Hence M = 0 only when its dual length A + eA' = 0.

Our classification of motors is therefore as follows:

Screw 0;1} Proper

Line vector

(mxmo)Xm

Couple (m=0,Zero (m=0, m0 =0):A=0,A' =0.

§ 32 MOTOR SUM 67

If two proper motors M, N are connected by a linear relationwith proper coefficients,

(a + ia')M + ($ + ia')N = 0, as 34 0,

either may be expressed as a dual multiple of the other; hence bothM and N are multiples of the same unit line vector and are there-fore coaxial. Conversely, if M and N are coaxial, they satisfy alinear relation with proper coefficients.

32. Motor Sum. The sum of the motors M = m + em0,N = n + eno is defined as the motor,

(1) M + N = m + n + e(mo + no).

That M + N is a motor follows from (29.3) :

mp+nr = mo+no+POx(m+n)

THEOREM 1. The sum of two line vectors is a line vector only whentheir axes are coplanar and their vectors have anon-zero sum.

Proof. If M and N are line vectors, M + N is a line vectorwhen and only when

m + n 0, and (m + n) (mo + no) = 0.

Since m mo = 0, n no = 0, the latter condition reduces to

(2)

which is precisely the condition (26.14) that the axes be coplanar.

THEOREM 2. If two line vectors intersect, their sum is a line vectorthrough the point of intersection.

Proof. If r1 is the position vector of the point Pl in which theline vectors M and N intersect, we may take mo = rl x m, no =r1 x n. Since the axis of M + N has the equation,

r- (m + n) = mo + no = r1 x (m + n),

it passes through the point P1.

THEOREM 3. If the line vectors M and N are parallel and n =Xm (X F6 -1), the axis of their sum divides any segment from M toN in the ratio X/1.


Proof. If Pl, P2 are points on the axes of M and N, we maytake mo = rl x m, no = r2 x n. The line vector,

M + N = (1+A)m+E(r,+Ar2)xm;and the equation of its axis,

r x (1 + A)m = (r1 + Xr2) x m,

is satisfied by the point r = (r1 + Ar2)/(1 + X) which dividesPiP2 in the ratio A/1. The division is internal when m and n havethe same direction (X > 0), external when they have opposite di-rections (A < 0).

THEOREM 4. The sum of two couples, if net zero, is anothercouple.

Proof. If M and N are pure dual, M + N, if not zero, is alsopure dual.

THEOREM 5. If M and N are line vectors such that n = -m andP1, P2 are points on their respective axes, M + N is a couple ofmoment (r1 - r2) x m.

Proof. Since M = m + er1 x m, N = -m + ere x (-m),

M + N = e(rl - r2) x m.

33. Scalar Product. The scalar product of the motors M =m + emo, N = n + eno is defined as the result of distributing theproduct,. (m + emo) + (n + eno), namely,

(1)

This dual number is independent of the choice of origin; for oncomputing mp, np from (29.3), we find that

(2)

The definition (1) shows that

(3)

The definition (1) for M M. N differs from that given by R. von Mises("Motorrechnung, ein neues Hilfsmittel der Mechanik," Z. angew. Math.Mech., vol. 4, 1924, p. 163) who defines M N as the invariant real scalar (2).The definition (2) is suggested by applications in mechanics; its consequencesare (a) the familiar rules of vector algebra do not all carry over into motoralgebra, and (b) the elegant generalization of the scalar product of unit vectors,given in (5), is lost.

§ 33 SCALAR PRODUCT 69

In order to compute the scalar product of two unit line vectorsA = a + ea0i B = b + - ebo, we shall suppose first that they arenot parallel and write

axb = esin <p,

where e is a unit vector. There is, then, a definite unit line vectorE in the direction of e which cuts both A and B at right angles.If we choose the origin at the point of intersection of A and E,and let denote the perpendicular distance between A and B,taken positive if a X b points from A to B, we may write

A = a, exb, E = e.Then we have

cos cp - ep sin cp.

If we now define the dual angle t between A and B as

(4) 'P + Ecp

a reference to (30.7) shows that

(5)

in complete analogy with a b = cos <p.If A and B are parallel, let e be any unit vector perpendicular

to both A and B. If the origin is chosen on A, the foregoing com-putation still applies; since sin <p = 0, we find A B = cos p = f 1.If A and B are collinear, we choose the origin on their commonline; then A = a, B = b, and A B = a b = -4-1 as before.Formula (5) covers these cases; in both cos 4) = cos gp = f 1 ac-cording as a and b have the same or opposite directions.

We note that

=cos(p-ecp'sin(p=0,when and only when

acos,p=0, rp'sin<p=0, or cp=2, =0;

that is, when the line vectors cut at right angles.

t The concept of dual angle is due to Study, Geometrie der Dynamen, Leipzig,1903, p. 205.


THEOREM. In order that the axes of two proper motors cut at rightangles, it is necessary and sufficient that their scalar product vanish.

Proof. We express the motors in the form,

M = (a + ea )A, N = (0 + E#')B,

where A, B are unit line vectors along their axes and a0 - 0.Then

M - N = [a$+

and, since the first factor is neither zero nor pure dual, M N = 0implies A B = 0, and conversely.

We denote the dual part of M N by

(6)

As previously noted, this is von Mises' definition of the scalarproduct. In common with M - N, the product M c N is commu-tative and distributive with respect to addition. The precedingcalculation shows that

AoB = -hence A o B = 0 implies p' = 0 or sin p = 0, that is, the linevectors intersect or are parallel, and conversely. The same cri-terion applies to line vectors aA, OB of arbitrary length. Conse-quently a necessary and sufficient condition that two line vectors1V ,JI be coplanar is that M e N = 0. This is the condition (26.14).

Writing the motor M = (a + ea')A gives

(a+ea')2=a2+2eaa'as the square of its dual length; and M o M = 2aa'. EvidentlyM o M = 0 implies that M is a line vector (a' = 0) or a couple(a = 0).

34. Motor Product. The motor product of the motors M =m + ano, N = n + eno is defined as the dual vector obtained bydistributing the product (m + emo) x (n + eno), namely,

(1) MxN = mxn + e(mxno +moxn).

When the origin is shifted to P, the dual part of M x N becomes

mxnp + mp x n = m x (no + POxn) + (mo + POxm) xn

= mxno + moxn + POx (mxn),

§ 34 MOTOR PRODUCT 71

in view of the identity,

PO x (m x n) + m x (n x PO) + n x (PO x m) = 0.

This transformation conforms with (29.3) and shows that M x Nis a motor.

The definition (1) shows that

(2) MxN = -NxM, Lx(M+N) =LxM+LxN.In order to compute the motor product of two unit line vectors

A = a + eao, B = b + ebo, we shall suppose, first, that they arenot parallel. With the same notation and choice of origin as in§ 33, we have

AxB =a xb + &p ax (exb)= e(sin (P + ecp' cos (p)

or, in terms of the dual angle 4 = c + eqp ,

(3) AxB = Esin4

in complete analogy with a x b = e sin cp.When A and B are parallel, let e be any unit vector perpendicular

to both A and B. If the origin is chosen on A, the foregoing com-putation still applies, and

AxB = e

e + Eeo, an arbitrary line vector in the direction of e.This conforms with the fact that A x B, being pure dual a couple),has its axis determined in directionlbut not as tolposition in space.When A and B are collinear, we choose an origin on their commonline;thenA=a,B bandAxB=axb=0andEisentirelyarbitrary. Formula (3) covers these cases in which sin 4 = f Ev'and 0, respectively.

We note that

A x B = sin (D E = (sin p + erp cos cp) E = 0,

(or sin 4) = 0) when and only when

sin(p=0, gyp'=0,

that is, when the line vectors are collinear.

THEOREM. In order that two proper motors be coaxial, it is neces-sary and sufficient that their motor product vanish.


Proof. With the notation of § 33,

M x N = [a/3 + e(af' + a'f )]A x B, (a/3 X 0).

Since the first factor is neither zero nor pure dual,,M x N = 0 im-plies sin 4) = 0, and conversely.

Finally let us consider the motor product,

(4) M x N = [a$ + E(a3' + a'/3) ] [sin gp + cos go] E,

for any proper motors M, N. If their axes coincide, M x N = 0.If their axes are parallel, sin <p = 0, the coefficient of E is puredual, and the axis of the couple M x N is any line parallel to thecommon normal to the axes of M and N. If their axes are non-parallel, sin cp 0, and M x N is a proper motor whose axis (alongE) is the common normal to the axes of M and N.

35. Dual Triple Product. The dual triple product of threemotors L = I + do, M = m + em0i N = n + eno is defined asthe dual number obtained by distributing the product L x M N;thus

(1)

This definition shows that

(2)

LxM N = LxM N = 0,

in exact analogy with vector products.With von Mises' definition of a scalar product, the preceding

triple product becomes

(5) LxMoN = (lxm) - no + (lxmo +loxm) - n;

this is the dual part of L x M N. Equations (2), (3), (4) alsoapply to L x M o N.

We now propose to find under what conditions L x M N = Cwhen L, M, N are proper motors. If the axes of L, M, N are allparallel, l x m= m x n= n x 1= 0 and L xM N= 0. If theiraxes are not all parallel, choose the notation so that 1 x m 0.Then L x M is a proper motor whose axis cuts the axes of L and Mat right angles. The condition L x M N = 0 then holds whenand only when the axis of N cuts the axis of L x M at right angles.

§ 36 MOTOR IDENTITIES 73

In this case the axes of L, M, N have a common normal (the axisof L x M). We therefore may state the

THEOREM. The dual triple product of three proper motors willvanish only when (a) their axes are all parallel, or (b) their axes havea common normal.

If the proper motors L, M, N satisfy a linear relation with propercoefficients,

(6) (a+ecx')L+($+E$')M+(y+ey')N=0, aIiy 0,

L x M N = 0, and we can apply the preceding theorem. Moreprecise information, however, is given by the equations,

MxN NxL LxM

a+ea'=a+e0' = +e1''

which follow from (6). These show that the three motors M x N,N x L, L x M are all proper or all couples. When proper thesemotors are all coaxial; when couples, their moments are allparallel. Hence from (6) we conclude that either (a) the axesof L, M, N, no two of which are parallel, have a common normal;or (b) their axes are all parallel and coplanar. Thus the linearrelation (6) implies that the axes of L, M, N have a common normal.Conversely, if the axes of L, M, N have a common normal, we caninfer a linear relation (6), provided the case in which two axes areparallel to each other but not to the third is excluded. We omitthe proof.

36. Motor Identities. All the identities of vector algebra arestill valid when the vectors are replaced by motors. For the motorproducts are defined as the results of applying the distributive lawto dual vectors and subsequently equating e2, e3, . . . to zero.Now the identities apply to the ordinary vectors forming the dualvectors, so that the results before equating e2, e3, to zero aretrue for any value of the scalar e. In particular, the identitieshold when e2, e3, are replaced by zero. Thus we have themotor identities:

(1) Lx(MxN)

(2) Lx (M N) + Mx (NHL) +Nx (L M) = 0,

(3) (L M) (NxP) =LNMP -LPMN.


The identity (2) has an interesting geometric interpretationwhen all nine motors,

L,M,N;MxN,NxL,LXM;Lx(MxN),Mx(NxL),Nx(LxM),are proper. In this case, the axes of the three last motors have acommon normal (§ 35). This common normal and the axes of thepreceding nine motors form a configuration of ten lines, each oneof which is normal to three others. This is essentially the theoremof Peterson and Morley. $ When L, M, N are line vectors alongthe sides of a plane triangle, this theorem implies the concurrenceof its altitudes. The theorem of Peterson and Morley thereforemay be regarded as a generalization of this result.

37. Reciprocal Sets of Motors. Consider two sets of propermotors M1i M2, M3; M1, M2, M3 for which ml m2 x m3 0,m1 m2 x m3 /- 0. The sets are said to be reciprocal when the nineequations,

(1) Mi Mi = S (i, j = 1, 2, 3),

are satisfied. The indices are merely labels, and S is the Kro-necker delta.

Let the set Mi be given; we then may compute the reciprocalset M' uniquely. Since M2 M1 = 0, M3 M1 = 0, the axis ofM1 is the common normal to the axes of M2 and M3; henceM1 = (X + eX')M2 x M3i and, since MI-M' = 1, A + eX' =1/M1 M2 X M3; the condition ml m2 x m3 0 0 ensures thatM1 M2 X M3 is not pure dual. Thus we find

MX Mk M'xMk(2) M2 = M1 M2 x M3 , Mi =

M1M2 x M3

where ijk represent any cyclical permutation of 123. That themotors (2) satisfy equations (1) is shown by direct substitution.Equations (2) have the same form as the corresponding equationsin vector algebra. Moreover, in

(M2xM3) . (M2xM3)1,M M1 =(M1 M2 x M3) (M1 M2 x M3)

the left member is 1, and, from (30.3) the numerator on the rightis 1; hence

(3) (M1 M2 x M3) (Ml M2 x M3) = 1.$ E. A. Weiss, Einfuhrung in die Linien-geometrie and Kinematik, Leipzig

1935, p. 85.

§ 38 STATICS 75

The axes of two reciprocal sets taken in the orderM1M3M2M'M3M2 form a skew hexagon with six right angles.Since

(4) M1 x M1 + M2 x M2 + M3 x M3 = 0,

the common normals of its opposite sides themselves admit a com-mon normal. The six sides, the three common normals of theopposite sides, and their common normal, thus form a Peterson-Morley configuration (§ 36).

If we put Mi = Mi in (1), we find that the only self-reciprocalmotor sets are formed by three mutually orthogonal unit line vec-tors through a point.

Any motor M may be expressed linearly with dual coefficientsin terms of M1, M2, M3, provided m1 m2 x m3 X 0. To showthat the representation is possible, we note that

(5) M = (a + ea')M1 + ($ + ea')M2 + (y + ey')M3

is equivalent to the two vector equations,

(6) m = amt + /3m2 + 'Ym3,

(7) mo = amlo + /3m20 + 'Ym30 + a'm1 + /3'm2 + 7'm3.

Since m1 m2 x m3 ; 0, the vectors ml, m2, m3 have a reciprocalset m1, m2, m3. From (6),

a = m- m1, /3 = m-m2, y = m-m3;

with a, 0, y known, we may determine a', 0', y' in the same wayfrom (7). The actual computation, however, can be effected mostsimply from (5) by using the reciprocal set m1, m2, m3. Thus wefind

(8) M = M M1M1 +M M2M2 +M M3M338. Statics. The statics of rigid bodies may be developed in-

dependently of dynamics from four fundamental principles: tPrinciple I (Vector Addition of Forces). Two forces acting on

the same particle may be replaced by a single force, acting on theparticle, equal to their vector sum.

Principle II (Transmissibility of a Force). A force acting on arigid body may be shifted along its line of action so as to act onany particle of that line.

f Sec Brand, L., Vectorial Mechanics, New York, 1930, Chapter II.


Principle III (Static Equilibrium). If the forces acting on arigid body, initially at rest, can be reduced to zero by means ofprinciples I and II, the body will remain at rest.

Principle IV (Action and Reaction). This need not concern ushere.

Principle II states in effect that a force acting on a rigid body isa line vector. Such a force is determined by the vector f givingits magnitude and direction and by its moment fo about 0 (§ 27).Thus the dual vector,

(1) F=f+eforepresents a force f acting along the line whose equation is r x f = fo.

If two forces P, Q act on a particle with position vector r,

P=p+Erxp, Q=q+E1xq;principle I asserts that both forces may be replaced by a singleforce, their resultant, whose vector and moment are p + q,r x (p + q) ; that is, the intersecting line vectors P and Q areequivalent to the line vector P + Q, their motor sum (§ 32).

Two parallel forces P, Q also have a resultant P + Q providedp, + q 0. To see this, we need only introduce a pair of opposedforces F, -F (equivalent to zero) acting along any line cutting thelines of P and Q. We may now apply principle I to find the re-sultants of P + F and Q - F. These coplanar forces intersect,since

(p+f)x(q-f) =fx(p+q) 96 0,and therefore have the resultant,

(P + F) + (Q - F) = P + Q.

If Q = XP (X 96 -1), the line of P + Q divides all segments fromP to Q in the ratio X/1 (§ 32, theorem 3).

If p + q = 0, po + qo 0, the parallel forces P, Q are equalin magnitude and opposite in direction and are said to form acouple of moment po + qo.

In any case, when a system of forces Fa acting on a rigid bodyis changed into an equivalent system G; by the application ofprinciples I and II, the force-sum and moment-sum remain un-altered :

21,; = 2;g1, lfto = 2;gio.

§ 38 STATICS 77

For each application of principle I leaves these sums unaltered;and shifting a force in accordance with principle II does not alterits vector or moment.

Now it can be shown that any system of forces Fi acting on arigid body can be reduced by means of principles I and II to twoforces, * say P= p + epo, Q = q + eqo. Hence, if EF1 = m +emo, we have

p+q=m, po+qo=mo.If m 0, m mo 0, the system Fi is equivalent to a screw

M = m + emo (§ 31) which may be expressed in many ways astwo non-coplanar forces. A screw cannot be reduced to a singleforce. The screw M may be regarded as a force m acting throughthe origin and a couple of moment mo.

If m 5,4- 0, m mo = 0, the forces P and Q are coplanar; form mo = 0 implies p qo + q po- = 0 (26.14). The forces P, Qnow may be reduced to the single force M = m + ano, theirresultant.

If m = 0, mo 0, the system F1 is equivalent to the coupleP, Q of moment mo.

Finally, if m = 0, mo = 0, the system F, reduces to zero, and,in accordance with principle III, the rigid body is in equilibrium.

In brief, the system of forces Ft acting on a rigid body is equivalentto the motor M = EFz j this may be a screw, a single force, a couple,or zero; only in the last case is the rigid body in equilibrium.

The moment of a force F = f + do about an axis along the unitline vector A = a + eao was defined (27.4) as a fp, P being apoint on the axis of A. The moment of a system of forces, equiva-lent to the motor M = m + emo, about this axis is therefore

Now

a

and, since OP * a = ao,

(2)

We call this the moment of the motor M about the axis A. Thismoment is independent of the choice of P on the axis.

* Brand, L., Vectorial Mechanics, p. 143.


39. Null System. A line is called a null line with respect to amotor M if the moment of M about this line is zero. In order thata line, given by the unit line vector A = a + eao, be a null linewith respect to M, it is necessary and sufficient that

(1)

If L = XA, L o M = XA o M; hence the axis of any line vector Lwill be a null line when and only when L o M = 0. The totalityof such lines constitutes the null system of the motor M.

A motor M can be replaced in many ways by two line vectorsF, G. When M is a screw (m mo F6 0), we shall prove that theline of one may be chosen at pleasure, provided it is not a nullline or parallel to the axis of M; after this choice F and G areuniquely determined.

Write F = aA, G = oB as multiples of unit line vectors, andchoose A as any line not in the null system. Now M = F + G isequivalent to

m = as + ob, mo = aao + obo

Since a ao = b bo = 0,

(2) a =

thus F = aA is known, and G = M - F. Since A is not a nullline, a mo + m ao 7 0; and, since A is not parallel to m,o 7 0, and G is actually a line vector.

All null lines through a point P are perpendicular to mp andtherefore lie in a plane p, the null plane of P. Moreover all thenull lines on any plane p pass through a point P, the null pointof p, where mp is perpendicular to p.

Analytic Proof. Consider a null line through the points (a, ao),(o, bo); the coordinates are (abo - aoo, ao x bo) from (26.9). Sinceit is a null line,

(3) m ao X bo + mo (abo - aoo) = 0.

This may be written

(4) (amo - ao x m) bo = ao moo,

§ 39 NULL SYSTEM 79

hence, if the point (a, ao) is fixed and ($, bo) varies over all nulllines through (a, ao), (,l, bo) always will lie in the plane whose co-ordinates are (amo - ao x m, ao mo).

Next, let the null line be common to the planes (a, ao), (b, Rio);its coordinates are (a x b, a00 - aob) from (26.10). Since it is anull line,

(5) m (a(3o - aob) + mo a x b = 0.

This may be written

(6) (mao - mo x a) b = m a $o;

hence, if the plane (a, ao) is fixed and (b,,30) turns about all nulllines in (a, ao), (b, Qo) always will pass through the point whosecoordinates are (m a, mao - mo x a).

Equations (3) and (5) are not altered by an interchange of thetwo points or two planes; hence the same is true of (4) and (6).Therefore:

If the null plane of point A passes through B, the null plane of Bpasses through A.

If the null point of plane a lies on b, the null point of b lies on a.From the preceding results:

(7) Point (a, ao) - null plane (amo - ao x m, ao mo) ;

(8) Plane (b, Ro) - null point (m b, mao - mo x b).

Indeed, if we solve

b=amo - aoxm,

for (a, ao), we obtain the null point in (8). Note also that ao b= a0o; let the student interpret this relation in the light of (26.16).

When M is a line vector, the null lines are lines that intersectits axis. Then (7) gives the plane through the point and M (26.11);and (8) 'the point on the plane and M (26.12).

If M is a screw (m mo F!5 0), not only are points associatedwith planes, but also lines with lines in general. For if A is anyunit line vector not in the null system of M or parallel to its axis,we can determine uniquely a unit line vector B, its conjugate, suchthat

(9) M = aA + (3B,


where a = m mo/A o M from (2). Any line vector C that cuts twoconjugate lines is a null line; for

MoC = aAoC+$B°C = 0.

In view of this property we may characterize conjugate lines asfollows: The conjugate of a line A is the line B: (i) which is commonto the null planes of all points on A, or (ii), which contains the nullpoints of all planes through A. With this point of view, null linesare self-conjugate. Along a line parallel to the axis of M, themoment of M is constant; since the null planes of its points areall parallel, we may say that its conjugate is at infinity.

40. Summary: Motor Algebra. The number X = x + ex' iscalled dual if x and x' are real and a is a unit with the propertyE2 = 0. If x 0, the dual number is proper. Operations withdual numbers are carried out as in ordinary algebra and then2, E3e , . . set equal to zero. Division, however, is only defined for

proper dual numbers. The function f(x + ex') is defined by meansof. its formal Taylor expansion; only the first two terms, f (x) +Ex' f'(x), appear since E2 = e3 = . . 0.

A line vector with the Pliicker coordinates a, ao (a ao = 0) iswritten as a dual vector, A = a + eao; for a unit line vectorI a I = 1. When the origin is shifted to P, ao becomes

ap = ao + PO- a.

A motor M = m + emo is a dual multiple of a unit line vector,

M = (X + EA')A = (X + EA') (a + eao)

now m mo = 0 only when A' = 0. A shift of origin to P changesthe moment vector as before:

mp=mo+P0xm,but m mp = m mo is invariant.

Motors for which X F 0 are said to be proper. Proper motorsare screwf when X' 0 0, line vectors when A' = 0. Proper motorshave a definite axis given by the unit line vector A. When X = 0,A' 0 0 we have a pure dual motor emo; it is called a couple ofmoment mo. The moment of a couple is a free vector, determinedin direction but not in position.

When X = A' = 0 (X + EX' = 0), the motor reduces to zero.

§ 40 SUMMARY: MOTOR ALGEBRA S1

The sum of the motors M = m + em0, N = n + en0 is definedas

M + N = m + n + e(mo + no).

If M and N are line vectors, M + N is a line vector only whenm no+n mo=0.

The scalar and motor products, M N and M x N are obtainedby distributing the products as in ordinary vector analysis andthen setting e2 = 0:

M N = m - n+MxN = mxn+ e(mxno+moxn).

If A and B are line vectors making an angle p and at a normaldistance apart (reckoned positive if a x b points from A to B),the dual angle between A and B is defined as F + &p'. Forunit line vectors,

A B= cos A x B= sin -tE

where E is a unit line vector in the direction a x b and cuttingA, B at right angles. These results are perfect analogues of

If M and N are non-parallel proper motors, the common normalto their axes is the axis of M x N. When M and N are parallel,M x N is a couple whose moment is perpendicular to their axes.

If M and N are proper motors, M N = 0 implies that theiraxes cut at right angles; M x N = 0 that their axes coincide; andconversely.

For three motors, the dual number L x M N is obtained by dis-tributing the product (1 + el0) x (m + emo) (n + eno) :

If L, M, N are proper, L x M N = 0 implies that their axes areall parallel or have a common normal, and conversely.

Two sets of three motors Mti, M' whose triple products areproper are said to be reciprocal when M; M' = S (i, j = 1, 2, 3).All formulas for reciprocal vectors have exact analogues for recip-rocal motors.

The forces acting on a rigid body are line vectors. A system F;of such forces is equivalent to their motor sum M = SFj, which

82 MOTOR ALGEBRA

may be a screw, a single-force, a couple, or zero; only when M = 0is the body in equilibrium.

All the rules of calculation in "real" vector algebra have exactanalogues in motor algebra.

PROBLEMS

1. If A is a unit line vector, show that the screw M = (A + ea')A may beexpressed as the sum of a line vector and a couple whose moment is parallelto the line vector, namely,

M = aA + eµm,

where A= X'/X is the pitch of the screw.2. Prove that the axes of the motors M, N and M + N either are parallel

or have a common normal.3. Express the screw M = 2i + e(i + j + k)(a) As the dual multiple of a unit line vector.(b) As the sum of a line vector and a couple whose moment is parallel to

the line vector.(c) Find the equation of the axis of M. -- ---,4. In the tetrahedron OABC, el = OA, e2 = OB, e3 = OC. If the forces

Pet, Qe2, Re3; P'(e3 - e2), Q'(e1 - e3), R'(e2 - el)

acting along its six edges are equivalent to the motor m + emo, show thatthey are uniquely determined by the six equations:

P1 = mo e1 ... P + Q' - R' = m , e1,[eie2e31

5. Show that the n forces P1Q1, P2Q2, , P. Q are equivalent to themotor,

n(q* -p*) +expi gi,

where p*, q* give the mean centers of the points pi and q;, respectively. [Cf.(9.4)].

6. In order that the axes of the motors, M = m + emo, N = a + eao, be_coplanar, it is necessary and sufficient that

where µ and v denote the pitch of M and N.7. If three line vectors, a + eao, b + ebo, c + eco, are parallel to a plane,

as +,6b + -yc = 0 (§ 5). Prove that any line vector d + edo meeting themis parallel to a fixed plane normal to aao + Rbo + yco.

8. The forces Xi, Yj, -Zk act through the points (0, 0, 0), (a, b, c), (0, b, 0),respectively. Show that they reduce to a single force if

X/a + Y/b + Z/c = 0.

PROBLEMS 83

9. If ABCD are the vertices of a skew quadrilateral and PQRS are the mid-points of its sides taken in order, show that the motor equivalent to the forces

AB, BC, CD, DA is also equivalent to four times the couple formed by PQ

and RS.10. If the moment of the motor M about each of the six edges of a tetra-

hedron is zero, show that M = 0. [Cf. (38.2)].11. If the origin 0 is taken on the axis of a screw M of pitch µ show that

the equation of the null plane of the point Ti is

If the axis of M is chosen as z-axis, this equation becomes xyl - yxl =A(z - zi).

CHAPTER III

VECTOR FUNCTIONS OF ONE VARIABLE

41. Derivative of a Vector. Let u(t) denote a vector functionof a scalar variable t over the interval a < t < b; that is, when tis given, u(t) is uniquely determined.* The function u(t) is saidto be continuous for the value t if u(t + At) -+ u(t) as At -* 0.

To obtain a clear idea of the way in which u varies with t, we

may regard u(t) = OP as a position vector issuing from the origin.Then as. t varies, the point P traces a certain curve r in space.

For the values t and t + At, let u(t) = OP, u(t + At) = OQ; thechange in u for the increment At is

--4 -* --4Du=u(t+At) -u(t) =OQ-OP=PQ.

The average change per unit of t is Au/At. When At > 0, Au/At

is a vector (PA' in Fig. 41) in

(1)

' R R * the direction of Au and 1/At times

Fia. 41

as long; but if At < 0, Au/At has

the direction of -Au (then PA'in Fig. 41 must be reversed). If

PA' approaches a limiting vector

PA as At -+ 0, we call PA thederivative of u(t) with respect to tand denote it by du/dt or u'(t).The equation defining the de-rivative is therefore

du = lim u(t + At) - u(t)= lim

Du

dt w -o At 'U - o At

* The word function implies a single-valued function.84

§ 41 DERIVATIVE OF A VECTOR 85

If the limit du/dt exists for the value t, u(t) is said to be differen-tiable at t. Then we may write Au/.t = du/dt + h where h --> 0as At -* 0; hence

0u =At (du

-dt + h) --). 0, as At --> 0,

and u(t + At) ---> u(t). Thus if u(t) is differentiable at t, it is alsocontinuous there.

If u(t) is differentiable at P, Q describes the arc QP of the curver as At -* 0, and the limiting direction of the chord PQ, and henceof the vector PA', is along the tangent at P. The limiting vectorPA = du/dt is therefore tangent to I' at P. Since Du/At has thesame direction as Au when At > 0, the opposite when At < 0, du/dtis a vector tangent t o r in the direction of increasing t. We re-state this important result as follows:0

If the vector u(t) = OP varies with t, so that P describes the curver when 0 is held fast, the derivative du/dt, for any value of t, is avector tangent to r at P in the direction of increasing t.

Example 1. If u = OP is a variable vector of constant direction, P willmove on a straight line when 0 is held fast; hence du/dt, being tangent to thisline, will be parallel to u.

Example 2. If u = OP is a variable vector of constant length, P will describea curve r on the surface of a sphere when 0 is held fast; hence du/dt, beingtangent to r at P, will be perpendicular to the radius OP of the sphere. Inbrief : If j u I is constant, du/dt is perpendicular to u.

If u is a constant vector, that is, constant in both length anddirection,

Du = 0, Ot = 0, and dt = 0.

The derivative of a constant vector is zero.When u is a function of a scalar variable s, and s in turn a

function of t, a change of At in t will produce a change As in sand therefore a change Au in u. On passing to the limit At --* 0in the identity,

Du Du As du du ds-, we get - = ---At As At dt ds dt

the familiar "chain" rule.

86 VECTOR FUNCTIONS OF ONE VARIABLE § 42

The higher derivatives of u(t) are defined as in the calculus:

du' du"u"(t) =dt

, U"'(t) =dt

, etc.

42. Derivatives of Sums and Products. Let u(t) and v(t) betwo differentiable vector functions of a scalar t. When t changesby an amount At, let Au, Av, and O(u + v) denote the vectorialchanges in u, v, and u + v. Then

u + v + O(u+v) = u+ Du+v+ Ov,O(u + v) = Du + AV,

A(u + v) Du AV

At At At

and passing to the limit At - 0 gives

d du dv

(1) (u +v)

+dt dt dt

Consequently, the derivative of the sum of two vectors is equal to thesum of their, derivatives. This result may be generalized to the sumof any number of vectors.

Consider next the product f(t)u(t) of a differentiable scalar anda vector function. When t changes by an amount At, let Af, Au,and A(fu) denote the increments of f, u, and fu, respectively.Then, since multiplication of vectors by scalars is distributive (§ 4),

fu + o(fu) = (f + Af)(u + Au) = fu + f Du + Af U + Of Au,

O(fu)=fAu+Afu+OfAu,

A(otu)

= fAU-u+Of-;

passing to the limit At - 0, and noting that Af -* 0, we have

(2) dt (f u) = f dt + dt u.

This is formally the same as the rule for differentiating a productof scalar functions.

§ 42 DERIVATIVES OF SUMS AND PRODUCTS 87

Important special cases of (2) arise when either f or u is constant:

d du d df-(cu) =cdt, -(fc) =atc.

If the components of a vector u(t) are ui(t) when referred to aconstant basis ei,

(3) u(t) = 2;ui(t) e,,du dug

dt dtei.

The components of the derivative of a vector are the derivatives of itscomponents.

If we pass now to the products u v and u x v, where u and vare vector functions of t, the same type of argument used in prov-ing (2) shows that

d dv du(4) -(u.v)=u-at+at-V,

(5)

dx(u v

dv dux - - x v- U

+dt.

dt dt

The proofs depend essentially upon the distributive laws for thedots and cross product. In (5) the order of the factors must bepreserved.

If 1 is an axis with unit vector e, the orthogonal component of uon 1 is e u (15.6) ; hence

d du ducomp1 u = e

- = compsdt dt dt

THEOREM 1. A necessary and sufficient condition that a propervector u be of constant length is that

(6)

Proof. Since

duu

dt=0.

Iu 12 = u u, we have, from (4),

d du

-1u12=2u'dt

If I u I is constant, the condition follows: conversely, the conditionimplies that I u I is constant.


THEOREM 2. A necessary and sufficient condition that a propervector always remain parallel to a fixed line is that

du(7) u X

dt= 0.

Proof. Let u = u(t)e where e is a unit vector; then

uXdu

= ue xCdu

ede\

= u2e Xde- - + u - J --

dt dt dt dt

If e is constant, de/dt = 0, and the condition follows. Conversely,since u 0 0, the condition implies that

de deex-=0. Alsodt dt

from theorem 1. These equations are contradictory unless de/dt= 0; that is, e is constant.

43. Space Curves. Consider a space curve whose parametricequations are

(1) x = x(t), y = y(t), z = z(t),

where x(t), y(t), z(t) are analytic functions of the real variable tdefined in a certain interval T of t values. To avoid having thecurve degenerate into a point, we explicitly exclude the case inwhich all three functions are constants. We also restrict the in-terval T so that there is just one value of t corresponding to eachpoint P of the curve. Then equations (1) set up a one-to-onecorrespondence between the points of the curve and the values oft in the interval T. The requirement that the functions x(t),y(t), z(t) be analytic in T ensures the continuity of the functionsand their derivatives of all orders and also guarantees a Taylorexpansion for each function about any point of T. A curve whichadmits a representation (1) with functions thus restricted is calledan analytic space curve. Moreover, if all three derivatives dx/dt,dy/dt, dz/dt do not vanish simultaneously for any value of t in theinterval T, the curve is said to be regular, and the parameter tis said to be a regular parameter.

bet us make a change of parameter,

t = P(u),

where jp(u) is an analytic function of u in a certain interval U; we

§ 43 SPACE CURVES 89

assume also that t just covers T as u ranges over U. Then theequations,

x = x[,P(u)l, y = y[O(u)l, z = z[co(u)]

constitute a new parametric representation of the curve. Writingdt/du = <p'(n), we have

dx dx dy dy dz dz

du = atlp

(u)' du = at (u)' du =a regular parameter, u is a regular parameter when and

only when (u) 0 in U. When this condition is fulfilled, theimplicit-function theorem ensures the existence of the inverse func-tion u = 4'(t) which is also single valued and analytic in T. Thusthe points of the curve not only correspond one to one to the tvalues in T, but also to the u values in U:

P -* t -> fi(t) = U, u --> (P(u) = t --+ P.

Let equations (1) define a regular analytic space curve. Theposition vector of the point P(t) is

(2) r = i x(t) + j y(t) + k z(t) = r(t);and, if we write t, z for dr/dt, dx/dt,

(3) t = it(t) + j y(t) + k i(t) F6 0.If Po(t = to) is a fixed point on the curve, the length of the arc sfrom Po to P is defined as

(4) s =f V dt,to

an analytic function of t which is positive or negative, accordingas t > to or t < to; moreover

(5) ds/dt = VT-t > 0,

The inverse function t = <p(s) is single valued and analytic, gnddt/ds > 0. Putting t = <p(s) in (2), we obtain r = r(s), in whichthe arc s is a regular parameter.

The integrand in (4) is unity only when t = 1; hence we havethe

THEOREM. For the curve r = r(t) the parameter t is the length ofarc measured from a fixed point when and only when dr/dt is a unitvector.


44. Unit Tangent Vector. Let r = r(t) be a regular analyticspace curve on which s = arc POP is reckoned positive in thedirection of increasing t. Then (Fig. 44a)

dr Ar PQ-- = lim - = limds A -o As Q -. P arc PQ

is a unit vector (§ 43), and hence

Emchord PQ

Q -. P arc PQ

The tangent line to the curve at P is defined as the limiting posi-tion of the secant PQ as Q approaches P. Hence from its definition

FIG. 44a FIG. 44b

we conclude that dr/ds is a unit vector T tangent to the curve atP and pointing in the direction of increasing arcs;

dr(1) = T.

ds

An important special case arises when r is a unit vector revolvingin a plane. If we imagine this vector R always drawn from thesame initial point 0, its end point will describe a circle of unitradius (Pig. 44b) and s = 0, where 0 denotes the angle in radiansbetween a fixed line OA and R. If P is a unit vector perpendicularto R in the direction of increasing angles, (1) now becomesdR/dO = P. If k is a unit vector normal to the plane and pointingin the direction a right-handed screw advances when revolved inthe positive sense of 0, P = k x R, and

dR(2)

dB= k x R = P.

§ 44 UNIT TANGENT VECTOR 91

In Fig. 44b, k points upward from the paper. Moreover, sincedP/d6 = k x (dR/d6), we have

(3)

dP-=kxP= -R.d6

We state these results in the

THEOREM. The derivative of a unit vector revolving in a plane,with respect to the angle that it makes with a fixed direction, is anotherunit vector perpendicular to the first in the direction of increasingangles.

If P(x, y) is a variable point on a plane curve, r = xi + yj,

dr dx dyT =

as = as' + ds ,

and i T = dx/ds, j T = dy/ds; hence

(4) ds = cos (i, T), ds = sin 01 T).

If (r, 0) are the polar coordinates of P, we write r = rR, wherethe polar distance r and the unit vector R are functions of 0. Now

dr dR d9 dr d6T = -- R+r --- = - R+r--P,ds d6 ds ds ds

and R T = dr/ds, P T = r d6/ds; hence

(5)

dr do

- = cos (R, T), r--

= sin (R, T).

Example 1. If 0 is measured counterclockwise from the x axis,

R=icos0+jsin0, P=icos(0+ 117r)

From (2), the corresponding components of dR/d9 and P must be equal; hence

dBcos 0 = cos (0 + 4r) sin B, d sin B = sin (0 + fir) = cos 0.

Example 2. If rl, r2 are the distances of a point P on an ellipse from the foci,rl + r2 = const. On differentiating this equation with respect to a, we have,from (5),

drl dr2+

ds= (RI + R2) T = 0.

ds


Since RI + R2 is perpendicular to T, the normal to the ellipse at P has thedirection of RI + R2. The normal therefore bisects the angle between the focalradii.

45. Frenet's Formulas. Let r = r(s) be a space curve, s =are POP, and T = dr/ds the unit tangent vector at P. Since T isof constant length, dT/ds, if not zero, must be perpendicular to T.A directed line through P in the direction of dT/ds is called theprincipal normal of the curve at P. Let N denote a unit vectorin the direction of the principal normal; then we may write

(1) dT= KN,

ds

where K is a non-negative scalar called the curvature of the curveat P.

Now B = T x N is a third unit vector perpendicular to both Tand N. Thus at each point of the curve we have a right-handedset of orthogonal unit vectors, T, N, B, such that

TxN = B, NxB = T, BxT = N.

As P traverses the curve, we speak of the moving trihedral TNB.A directed line through P in the direction of B is called the bi-normal to the curve at P.

Since B is a unit vector, dB/ds, if not zero, must be perpendicularto B. Differentiating B T x N, we have

dB dT dN dN

ds dsxN+Txds = Txdsin view of (1). Hence dB/ds is perpendicular to T as well as B

and therefore must be parallel to-B x T = N. We therefore may write

dB(2) = -TN,__

ds7-7-I/ where T is a scalar called the torsion

of the curve at P. The minus signTorsion positive is introduced in (2) so that, when 7-

Fla. 45 is positive, dB/ds has the directionof - N; then, as P moves along the

curve in the positive direction, B revolves about T in the same senseas a right-handed screw advancing in the direction of T (Fig. 45).

§ 45 FRENET'S FORMULAS 93

We may now compute dN/ds from N = by use of (1) and(2) ; thus

(3)

dN dB dT-xT+Bx-= -rNxT+KBxN.ds ds ds

Collecting (1), (2), and (3),

(4)

dT

ds

dN

ds

dB

ds

we have the set of equations:

known as Frenet's Formulas, which are fundamental in the theoryof space curves.

If we write (3) in the form,

dN

ds= (rT + KB) x N,

and introduce the Darboux vector,

(5)

we have

S = rT + KB,

SxT = KN, SxB = -TN.

Hence Frenet's Formulas may be put in the symmetric form:

dT(6)

ds= S x T,

dN- =SxNds

dBSxB.

ds

Since S N = 0, we may write the Darboux vector in the form:

(7)

dNS = Nx(SxN) =

Nxd8.

From (5), we see that the curvature K and torsion r are thecomponents of the Darboux vector on the binormal and tangent,respectively. From (4), it is clear that both K and r have thedimensions of the reciprocal of length; hence

p=1/K, a=1/r


have the dimensions of length and are called the radius of curva-ture and the radius of torsion, respectively.

Since N, by definition, has the same direction as dT/ds, thecurvature K is never negative. If K vanishes identically,

dT dr

ds =0,T=ds=a, r=as+b,

where a and b are vector constants. The curve is then a straightline. Conversely, for a straight line, T is constant and K = 0.The only curves of zero curvature are straight lines. For a straightline the preceding definition fails to determine N. We thereforeagree to give N any fixed direction normal to T, and as beforedefine B = T x N. Since B is also constant, dB/ds = 0 and T = 0.For a straight line the Darboux vector is zero.

The torsion may be positive or negative. As P traverses thecurve in the positive direction, the trihedral TNB will revolve aboutT as a right-handed or left-handed screw, according as r is posi-tive or negative. The sign of r is independent of the choice ofpositive direction along the curve; for, if we reverse the positivedirection, we must replace

dT dN ' dB dT dN dBs,

T, ds ,N,

ds ,By

ds by-8) -T,

ds ,N,

ds ,- B,

ds '

and equations (4) maintain their form with unaltered K and r.If r vanishes identically, dB/ds = 0, and B is a constant vector;

hence, fromdr

B T = Bds

0, B (r - ro) = 0,

that is, the curve lies in a plane normal to B. Conversely, for aplane curve, T and N always lie in a fixed plane, while B is a unitvector normal to that plane; hence dB/ds = 0 at all points whereN and B are defined (K* 0) and -r = 0. The only curves of zerotorsion are plane.

Example. Parallel Curves. Two curves r and rl are called parallel if aplane normal to one at any point is also normal to the other. The common

normal plane cuts r and rl in corresponding points P and PI; then PP, = rl - rlies in the plane of N and B and may be written XN + µB, where X, µ arescalars. Thus we have

(i) rl=r+XN+pB;

§ 46 CURVATURE AND TORSION 95

on differentiating (i) with respect to s and using Frenet's Formulas, we have

ds1 _ dX du(ii) Ti-

ds=T+dN +X(-KT +rB)+ds B -uTN

_ (1 - AK)T + (A' - µr)N + (A' + Xr)B.

Since T1 and T are normal to the same plane, we may choose the positive senseon t1 so that T1 = T; hence, from (ii), we conclude that

Moreover

^ = 1 - AK, ds = Jlr,ds

= -Ar.

dX du 1 d2ds()2+u2) =0,

and A2 + µ2 is constant. Consequently, the distance PP1 between correspondingpoints of parallel curves is always the same.

On differentiating T1 = T with respect to s, we have also

(iv) KjN1d3

1 = KN; hence ds =Kl ,

and N1 = N, B1 = B. Comparison with (iii) gives

K/KJ = 1 - AK or pl = p - A.

Finally, on differentiating B1 = B with respect to s, we have

r(v) -r1N1

dsi = -rN; hence, if r 0 0, - =7-1

From (iv) and (v), K/r = K1/7-1 at corresponding points.

46. Curvature and Torsion. By use of Frenet's Formulas, thecurvature and torsion are readily computed from the parametricequations of the curve. Thus, on differentiating r = r(t) threetimes and denoting t derivatives by dots, we have

_drdsds dt =

$T,

SZ dTf=ST+ -ds

= ST + S2KN,

dT

ds+(2saK+a2K)N+S3K

dN

CAS

= ST + SSKN + (2SSK + S2K)N + S3K(-KT + TB)

= (S - S3K2)T + (3MK + S2K)N + S3KTB.


Hencet x y= S3KB, t x It y= S6K2T,

and, since t = IsI 0,

(1), (2) K = I

It tx

t I 3T =

It x

r

I2

If the positive direction on the curve is that of increasing t,ds/dt > 0; and the preceding equations show that

T has the direction of t,

B has the direction of t x i',

and, since N = B x T,

N has the direction of (f x Y) x t.

If the parametric equations of a plane curve are x = x(t),y = y(t), we have

r = xi + yj, t = + yJ, r = xi + yl,and, from (1), its curvature is

(3) K = I xy - yx I(x2 +y2)

If the curve has the Cartesian equation y = y(x), we can regardx as parameter: x = t, y = y(t). Then (3) becomes

(4) K = (1 + y,2)I'

where the primes indicate differentiation with respect to x.If the curve has the polar equation r = r(9), we may write

r = rR, where R is a unit radial vector. Then regarding 0 as theparameter t, we have, from (44.2) and (44.3),

t = rR + rP, f = (r - r)R + 2rP;hence, from (1),

Ire+2r2-rrI(5)

K_

(r2+t2)§Example. Find the vectors T, N, B and the curvature and torsion of the

twisted cubicx=2t, y=t2, z=t3/3

at the point where t = 1.

1 47 FUNDAMENTAL THEOREM

Since r(t) _ [2t, t2, 0/31, we have

f = [2, 2t, t2], 1' = [0, 2, 2t], 7 = [0, 0, 2];

hence, when t = 1,

f = [2, 2, 1], i = [0, 2, 2], f = [0, 0, 2],

t x 1' = [2, -4,41, fxi' f = 8.Since T, B are unit vectors in the directions of t, t x t, and N = B x T,

[2,2,11, N=1t2,-1,-2], B=1[1,-2,21.T = 13 3 3

Moreover, from (1) and (2), K = $, r = $

97

47. Fundamental Theorem. Two curves for which the curvatureand torsion are the same functions of the arc are congruent.

Proof. Let the curves have the equations r = r1(s), r = r2(s).Bring the origins of are and the trihedrals T1N1B1, T2N2B2 at thesepoints into coincidence. At the points P1i P2 of the curves, corre-sponding to the same value of s, consider the function,

(1) f(s) =By Frenet's Formulas,

d=

K1N1 T2 + K2N2 T1

ds(-K1T1 + 71B1) N2 + (-K2T2 -- T2B2) . N1

T1N1 - B2 - T2N2 B1;

and, since K1 = K2, T1 = T2 for the same value of s,

--= 0, f (s) = C, a const.

ds

But at the point s = 0, the trihedrals coincide, and

C=f(0)=1+1+1=3.Hence for any value of s, f(s) = 3, an equation that implies thateach scalar product in (1) equals one; consequently,

T1 = T2, N1 = N2, B1 = B2.

From the first of these equations,

dr1 dr2

d' rl = r2 + a;

ds s

and a, the vector constant of integration, is zero, since r1 = r2when s = 0. Therefore r1(s) = r2(s), and the curves coincide.


48. Osculating Plane. The osculating plane of a curve at apoint PI is defined as the limiting plane through three of its pointsP1, P2, P3 as P2 and P3 approach P1.

Let the points P1, P2, P3 of the curve r = r(s) correspond to theare values 81, s2i 83 (sI < s2 < s3). Then, if n is the unit normalto the plane P1P2P3, the function,

f(s) = [r(s) - rl] nvanishes when s = s1i 82, s3. Hence from Rolle's Theorem,

f(s) = r'(s) nvanishes twice, and

f"(s) = r"(s) nvanishes once in the interval s1 < s < 83. Consequently, as P2and P3 approach P1, n approaches a limiting vector n1i such that

r'(sl) n1 = T1 n1 = 0,

r"(sl) n1 = K1N1 nl = 0.

If ICI 76 0, n1 is perpendicular to both T1 and N1i that is, parallelto B1. The osculating plane at a point of a curve is the plane of thetangent and principal normal at this point. For a plane curveT and 'N lie in the plane of the curve-the osculating plane at allof i points.

F r the curve r = r(t), B is parallel to t x 'r (§ 46); hence theequation of the osculating plane at any point r is

(1)

her e R is the position vector to any point of the osculating plane.The osculating planes of a curve form a one-parameter family.

The osculating planes at the points P1 and P2 intersect in a straightline; and, as P2 approaches P1, the limiting position of this lineis called the characteristic of the osculating plane at P1. To findthe characteristics of the osculating planes to the curve r = r(s),we must adjoin to their equation (R - r) B = 0, the equation ob-tained by differentiating it with respect to s, namely,

or

if r 0 0. The two equations,

(R-r)B=0, (R-r)N=0,

§ 49 CENTER OF CURVATURE 99

represent planes through the point r of the curve perpendicularto B and 11, respectively; together, they represent a line throughthe point r parallel to N x B = T. The characteristics of the oscu-lating planes of a skew curve are tangents to the curve.

Example. For the twisted cubic

x=2t, y=t z=t3/3of the example in § 46, t x 'r = 2[1, -2, 2] at the point (2, 1,

3)for which

t = 1. Hence, from (1), the equation of the osculating plane to the curve atthis point is

(x-2)-2(y-1) }2(z-3)=0.49. Center of Curvature. The center of curvature of a curve

at the point P1 is defined as the center of the limiting circle throughthree of its points P1, P2, P3, as P2 and P3 approach P1.

Since the limiting plane of P1, P2, P3 is the osculating planethrough P1, the limiting circle lies in this plane. If c and a denotethe position vector of the center and the radius of the circle throughthe points P1, P2, P3 of the curve r = r(s), the function,

f(s) = [r(s) - c] [r(s) - c] - a2vanishes for s = 81, 82i 83 (s1 < 82 < s3). Hence

f(s)vanishes twice, and

f"(s)

vanishes once in the interval s1 < s < s3.As P2 and P3 approach P1, c and a approach limits c1, a1, such

that(r1 - c1) (rl - c1) - ai = 0,

(r1 - c1) Tl = 0,

(r1 - c1) Nl + pi = 0.

Since r1 - cl lies in the osculating plane, its components onT1, N1, B1 are 0, -pl, 0, respectively, and r1 - cl = -p1R1. Thusthe center and radius of the limiting circle are

cl = r1 + p1 N1 and a1 = p1.

The center of curvature at any point P of the curve has the posi-tion vector,

(1) c=r+pN.


This is a point on the principal normal at a distance p from P inits positive direction.

Example. Curves of Constant Curvature. Let r = r(s) be a twisted curve tof constant curvature K. The locus t1 of its centers of curvature has the equa-tion,

(2) rl=r+pN.Differentiating with respect to s, we have

ds1 1

T1ds

= T + P(rB - KT) =K

B.

Choose the positive direction on rl so that Ti = B; thendal

r(3)(8 K

Differentiating T1 = B with respect to s now givesds1_K1N1 ds -7,N, or K1N1 = - KN;

hence N1 = -N and K1 = K. Thus r1 is also a curve of constant curvature.Since (2) may be written

(4) r = r1 + p1N1,

r is also the locus of the centers of curvature of r1. Consequently from (3)we also have d8/ds1 = r,/K1; hence rr1 = KK1. The two curves thus have thefollowing relations:

(5) T1 = B, Ni = -N, B1 = T; K1 = K, 771 = KK1.

Twisted curves of constant curvature may be associated in pairs, each curvebeing the locus of the centers of curvature of the other.

50. Plane Curves. For plane curves the torsion is zero andFrenet's Formulas reduce to

dT dN(1)

ds= KN

ds= -KT.

Since dT/ds is always directed towards the concave side of a planecurve, the same is true of N. The osculating plane at any point Pis the plane of the curve; and, if K 0 0, the center of curvature P1is given by

(2) rl = r + pN.

At points of inflection dT/ds = 0, K = 0, and N ceases to be de-fined. As we pass through a point of inflection, N abruptly re-verses its direction, and hence B = T x N does the same.

150 PLANE CURVES 101

To remedy this discontinuous behavior of N and B at points ofinflection, the following convention often is adopted in the differ-ential geometry of plane curves. Take B once and for all as afixed unit vector normal to the plane of the curve, and defineN = B x T. As before T, N, B form a right-handed set of orthog-onal unit vectors. The curvature K is defined by (1) and is posi-

Fia. 50a FIG. 50b

tive or negative, according as N has the direction of dT/ds or theopposite. Equations (1) still hold good; for

dN dT

- =BxdB =BXKN=-KT.

Let 4, be the angle from a fixed line in the plane to the tangentat P, taken positive in the sense determined by B (Fig. 50a).Then, from (44.2),

dT dT do d4, d#= BxT- = N-ds = dt' ds ds ds

and hence, from (1),d4, ds

(3) K=a, p=d .

The locus of the centers of curvature of a plane curve is calledits evolute. Let P and P1 be corresponding points of a curve r andits evolute r1 (Fig. 50b); and s = AP and sl = A1P1 denote cor-responding arcs. On differentiating the equation (2) of r1 withrespect to s, we have

ds1 dN dp dp

T1=T+pds+dsN=-N.

102 VECTOR FUNCTIONS OF ONE VARIABLE §50

Choose the positive direction on F1 so that T1 = N; then

ds1 dp

ds ds'and s1 = p + const.

Hence the tangent to F1 is normal to r; and, since As1 = Op, anarc of F1 is equal to the difference in the radii of curvature of rto its end points. These properties show that a curve may betraced by a taut string unwound from its evolute; the string isalways tangent to F1 and its free portion equal to p. From thispoint of view, r is called the involute of Fl.

From T1 = N, we have '1 = 4, + 7r/2; hence, from (3),

ds1 dp d2s4,2p1=d4,1 = d d .

Example 1. The only plane curves of constant non-zero curvature are circles.For from

dN=

dr dN

ds -KT, ords

= -pds

,

we have, on integration,

r - c = -pN, or Ir-cl2 =p2

This is the equation of a circle of radius p and center c.

FIG. 50c

Example 2. An involute r1 of a plane curve r is generated by the point P1of a taut string unwound from r (Fig. 50c). If t is the curve r - r(e), theinvolute is given by

r1 =P r - sT.

§ 50 PLANE CURVES

Hence, on differentiation with respect to 8,

dsl

Tlds =T - T - sKN = -8rcN.

If we choose the positive direction on r, so that Ti = -N,

(i)dsl d4, dsld8

= sx = s - , ordp

= s.

103

The intrinsic equation of a plane curve is the relation connecting s and ',say s = s(4,). For the are sl along its involute we have, from (i),

sl = f provided sl = 0, when ¢ = VO.0

The intrinsic equation of a circle of radius r is s = ry& when >G is measuredfrom the tangent at the origin of axes. For its involute we have sl = rp2/2,provided sl = 0 when' = 0.

The intrinsic equation of a catenary of parameter c is s = c tan when ais measured from the vertex (V, - 0). t The are sl, measured along the involutefrom the vertex of the catenary, is

sl = c f 1k tan k dy, = clog see ¢.0

Example 3. Envelopes. Consider the plane vector function of two variables,r = f(u, v). The one-parameter family of curves u = c (constant) is given by

rl = f(u, v), u = const.

If this family has a curve envelope given by v = p(u), namely,

(ii) R = f[u,,p(u)],

the vectors,

dR u_ of of drl _ of

au + av `P(u) and

dv av

are parallel at the points of contact; that is,

of of- x -- = 0.au av

This condition must be fulfilled if the envelope exists. If (iii) leads to a rela-tion v = p(u) (which does not make of/au oraf/av zero), this relation gives theenvelope (ii).

t See Brand, Vectorial Mechanics, equation (97.3). In Fig. 97b (p. 202), thestring PA will unwrap into the position PL; as P moves along the catenary,the locus of L is the involute of the catenary. The line LQ is tangent to theinvolute (Tj = -N); and, since the x-axis intercepts a constant segmentLQ = c on this tangent, the involute of the catenary is a tractrix, a curvecharacterized by this property.


Consider, for example, the one-parameter family of normals to the planecurve r = r(s), namely,

rl = r(s) + v N(s) (8 = const.)

If they have an envelope,

or,xari = (T - v,cT)xN = (1 -vK)B = 0.as av

Hence the envelope is given by v = 1/K = p, or

R = r(s) +'N(s),

namely the locus of the centers of curvature of the curve.

Example 4. Plane Caustic by Reflection. Light issuing from a point 0 isreflected from a mirror r (Fig. 50d).

TR _ The curve enveloped by the reflected

Fia. 50d

rays is called a caustic. If R and e arethe unit vectors along the incident andreflected rays, the reflected rays are theone-parameter family of lines,

ri = r(s) + ve(s).

To find their envelope, we form theequation :

de)arl arl _ (T+ 0.(iv)

as x 49V v dsJ xe =

Denote the angles (i, R) = 0, (R, T) - y; then (T, e) = y by the. law ofreflection, and

Now4,=(i,T) -0+y, p=(i,e)=0+2y-24, -0.de de dy& d0\ sin yl.a =kxe(2'--)=kxe(2K- r /

from (44.5). Substitution in (iv) now gives

whence

(v)1 1 2

v r p sin y

This equation determines v and the caustic curve.Letting r --, co, we obtain a beam of parallel rays; then v - 1p sin y.When r is a circular are of radius p with center on Ox, the caustic has a cusp

on Ox at a distance vi from r given by 1/vl + 1/r - 2/p.

ksin y-kv(2K- sin y =0;r

§ 51 HELICES 105

Example 5. The tractrix is a plane curve for which the segment of anytangent between the point of contact and a fixed line is constant. In Fig. 50e

the fixed line is the x-axis, the constant length c, and, if r = OP, rl = OP1,

r1=r+cT.

Hence if we differentiate with respect to s,

ds,= T + cKN;i

F iu. 50e

on multiplying by j we have

sink,

or

dsd = c tan V,.

Ifs =Owhen# -0, this gives

s = c log sec

for the intrinsic equation of the tractrix (cf. ex. 2).

51. Helices. A helix is a twisted curve whose tangent makes aconstant angle with a fixed direction. If e is a unit vector in thefixed direction, and a is the constant angle, the defining equationof a helix is

(1) (0<a<Zr)

106 VECTOR FUNCTIONS OF ONE VARIABLE

Differentiating (1) twice with respect to s gives

dNe a 0,=0;

§ 51

hence the Darboux vector 6 = N x (dR/ds) is parallel to e. Con-versely, if S for a curve is always parallel to a fixed vector e,

dTe-N=0, a--=0, and e - T=cosa,ds

where cos a is a constant of integration; the curve is therefore ahelix. The only twisted curves whose Darboux vector has a fixeddirection are helices. Thus helices are characterized by the prop-erty,

(2)

The fixed direction of

dSSX-=0ds

(§ 42, theorem 2).

S = TT+KB

is along the axis of the helix. Now

dS dr dK

(3)ds ds T +

Bds

for T dT/ds + K dB/ds = 0 (45.4); hence

-dS

C

d- d-\ d (Ksx- = KT - T -- a = -'r2 N--Jds ds A ds T

Thus 6 x (dS/ds) = 0 implies that the ratio K/r is constant, andconversely. Helices are the only twisted curves for which the ratioof curvature to torsion is constant.

If we put T = dr/ds in (1) and integrate, we obtain

e - r=scosa+cfor the component of r in the direction of e. Hence, if r1 denotesthe projection of r on a plane perpendicular to e,

(4) r = r1 + (s cos a + c)e.

Every helix therefore lies on a cylinder with generators parallelto e. The plane curve r1 traced by r1 is a normal section of thecylinder.

§ 51 HELICES 107

If s = POP is the arc along the helix, and s1 = POP1 the arcalong the normal section r1 through P0, we have, on differentiating(4) twice with respect to s,

(5)

ds1T=T1-+cosaeds

dsl 2

(6) KN = K1N1C ds

From (5),

ds1 2 =(T - e cos a) - (T - e cos a)

ds

= 1 -2cos2a+C082a

and, if we choose the positive direction on r1 so that s1with s,

(7)

ds1- = sin a.ds

sing a;

increases

Remembering that K and K1 are essentially positive, we now have,from (6), N = N1i and

(8) K = K1 sin2 a.

On differentiating e = T cos a+B sin a we find K/r =tan a; hence

(9) r = K1 sin a cos a.

Finally, on integrating (7), we obtain

(10) s1 = s sin a,

the constant of integration being zero, since both s and s1 aremeasured from the-same point Po.

The only twisted curve of constant curvature and torsion is thecircular helix. For, since K/r is constant, such curves are helices;and, from (8), K1 is constant; that is, the normal section is acircle.

The circular helix is the only twisted curve for which the Darbouxvector is constant. For from (3) we see that 6 is constant whenand only when K and r are constant.


Example. The curve,

x = a cos t, y = a sin t, z = bt,

is a circular helix; for the curve lies on the cylinder x2 + y2 = a2, and, since

t=[-asint,acost,b],

cos (k, T) = k t/I t I = b/1/a2.+ b2 (const.).Moreover

[-a cos t, -a sin t, 0],

T = [a sin t, -a cost, 0],

and, from (46.1) and (46.2),

_ ItXrI _ a bs

II I3 a2 +b2'r ItxiI2 a2+ b2

We now find the Darboux vector S = k/1/a2 + b2; it is constant in magni-tude and direction.

52. Kinematics of a Particle. To define the position P of aparticle moving along a curve r, choose a point PO of r from whichto measure arcs, and take a definite direction along r as positive.Then, if the arc s = POP is given as a function of the time, s = f (t),the motion of the particle is determined; for its position is givenat every instant. The speed v of the particle at the instant t thenis defined as

(1) v=ds

dt

Thus the speed will be positive or negative, according as s is in-creasing or decreasing at the instant in question.

The speed measures the instantaneous rate at which the par-ticle is moving along its path, but gives no information about itsinstantaneous direction of motion. We therefore introduce avector quantity, the velocity, which gives the rate at which theparticle is changing its position in both magnitude and direction.Let 0 be a definite point of a reference frame a, say the origin of

a system of rectangular axes fixed in a rigid body. Then if r = OPdenotes the position vector of P, the velocity v of P, relative to ,

is defined as the time derivative of its position vector:

(2) v =

§ 52 KINEMATICS OF A PARTICLE 109

As P moves along r, r may be regarded as a function of thearc s; hence

dr dr ds ds

= dt =T -

dt ds t

where T is the unit tangent vector to r at P in the direction ofincreasing arcs. Thus the velocity and speed are connected bythe relation,

(3) V = VT.

The velocity of P is represented by a vector tangent to the path at Pin the direction of instantaneous motion and of length numericallyequal to the speed.

Finally we define the acceleration a of the particle as the timederivative of its velocity :

(4)

From (3), we find

dv d2ra - dt dt2

dv dTa=-T+V ;tor, since T may be regarded as a function of s,

(5)

dT dT ds v= (KN)V=-N

dt ds dt p

dv v2a =

dtT +

pN.

The components of the acceleration in the positive directionthe tangent, principal normal, and binormal, are therefore

(6)dv v2ae=- t, an- ,

Pas = 0.

of

The acceleration of a particle P is a vector lying in the plane tothe tangent and principal normal to the path at P. The tangentialcomponent is the time derivative of the speed, the normal componentthe square of the speed divided by the radius of curvature at P.


The acceleration will be purely tangential when the motion isrectilinear (p = x) ; it will be purely normal when speed is con-stant (dv/dt = 0).

The velocity and acceleration vectors are regarded as localizedat the moving particle.

With rectangular axes, r = xi + yj + zk, and

7dr dx dy dz

k( ) vdt

,dt 1 + dt

l +dt

8)dv d2x d2y , d2z

k( adt

.dt2

1 + dt2 J +dt2

The rectangular components of v and a are the first and secondtime derivatives of x, y, z.

Example 1. Circular Motion. In the case of motion in a circle of radius r.

de dv _ d20s=rO, v=rdt, dt=rdt2,

the angle 0 being expressed in radians. On writing w = dB/dt for the angularspeed, we have

v = rw, at = r dw, an = rw2.

These results may also be deduced directly from the equation r = rR of thecircle if we make use of (44.2) and (44.3).

Example 2. Uniformly Accelerated Motion. If a particle has a constantacceleration a, and r = TO, v - vo when t - 0, we have, on integrating dv/dt =a twice;

v = at + vo, r = jat2 + vot + TO.

The path is the result of superposing the displacement "'ate, due to the accelera-tion, upon TO + vot due to rectilinear motion at constant velocity vo. It iseasily shown to be a parabola having its axis parallel to a. At the vertex ofthe parabola, v is perpendicular to a; the condition v a = 0 gives t =-a vo/a a for the time of passing the vertex.

53. Relative Velocity. In § 52, we have seen how to find thevelocity v of a particle P relative to any given reference frame .

If ' is a second reference frame, in motion with respect to ,

how is the velocity v' of P relative to a' related to v?At any instant t let P coincide with the point Q of a'. At a

later.instant t1 = t + At, let P and Q have the positions P1, Q1;

§ 53 RELATIVE VELOCITY 111

here Q is a fixed point of a', and its motion is due to the motionof ' relative to a. Then

and

PP1 is the displacement of P relative to ,

QQ1 is the displacement of Q relative to a,

Q1P1 is the displacement of P relative to ',

PP1 = PQ1 + Q1P1 = QQ1 + Q1P1

If we divide this equation by At and pass to the limit At -* 0, weobtain

(1) VP = VQ + VP, -,-

where VQ, the velocity of the point Q of R' relative to a, is calledthe transfer velocity of P.

If we regard the frame as "fixed" and velocities referred to itas "absolute," while velocities referred to a' are "relative," wemay state (1) as follows:

The absolute velocity of a particle is equal to the sum of its transferand relative velocities.

In many applications all points of the frame ' have the samevelocity relative to a; then vQ is the velocity of translation of ',and we may write

(2) Vp = Vtr + VP-

Example 1. Wind Triangle. An airplane p has the velocity v relative tothe ground (the earth e), v' relative to the air; and the air (the wind w) hasthe velocity V relative to the ground; thenv = V + v', from (2). In Fig. 53a,

v = ep, V = ew, v' = up;

thus vectors from e and w represent velocitiesrelative to the earth and wind, respectively.The magnitudes of v and v' give the groundspeed (ep) and air speed (wp) of the plane. The

w

directions of v and v', given as angles a and B'measured from the north around through theeast (clockwise), determine the track and

Fla. 53a

heading of the plane. The plane is pointed along its heading but travelsover the ground along its track. The angle (v', v) from heading to trackis the drift angle.


Example 2. Interception. A plane p is flying over the track PX with theground speed ep (Fig. 53b). As plane p passes the point P, a plane q departsfrom Q to intercept plane p. If the air speed of plane q is given, over whattrack shall q fly in order to intercept p?

Solution. In order that plane q may intercept plane p, the velocity of q

relative to p must have the direction QP.

Draw the vectors ep and ew, giving the velocities of plane p and the windw relative to earth e. With to as center describe a circle having the known airspeed of plane q as radius. If a ray drawn through point p in the direction of

,,q

Fia. 53b

-4QP cuts the circle at point q, plane q will intercept plane p on flying with the

ground speed eq over the track QY parallel to eq; for pq, the velocity of plane

q relative to plane p, has the direction QP. Interception occurs at I after aflying time of QI/eq (or PI/ep) hours.

Interception is impossible if the air-speed circle of plane q fails to cut theray. If the circle cuts the ray in just one point, as in Fig. 53b, plane q canintercept p on only one track. But if the circle cuts the ray in two points,say ql and q2, plane q can intercept p along two different tracks, parallel to

eql and eq2, respectively.

Example 3. Plane Returning to a Carrier. An airplane p leaves a carriera at 0 and patrols along the track OY while the carrier follows the courseOX with constant speed of v miles per hour (Fig. 53c). If the fuel in the tankallows the plane T hours of flying time at a given air speed, at what point Bmust the plane turn in order to rejoin the carrier at A, T hours after its depar-ture? Find also the time ti, the heading, and the ground speed vl on the legout; and the time t2, the heading, and the ground speed v2 on the leg back.

Solution. Let the vectors es and ew give the velocities of carrier and wind.With to as center describe a circle having the known air speed of p as radius-the air-speed circle. If this circle cuts the ray through e in the direction OY

at pl, epl is the velocity of plane p on the leg out (ground speed vl = epl).

§ 53 RELATIVE VELOCITY 113

The course of plane p relative to the carrier s is out and back along the same

straight line. Now sp1 is the velocity of p relative to s on the leg out; hence

the velocity of p relative to s on the leg back is a vector sp2 whose direction is

opposed to sp1. Thus the point p2 is at the intersection of the air-speed circle

with the line pis prolonged, and ep2 is the velocity of p on the leg back (groundspeed V2 = ep2).

p2

Fro. 53c

After T hours the carrier will travel the course OA = T es. The return track

of p is along a line BA parallel to ep2. Thus p travels t1 = OB/ep1 hours onthe track OB, t2 = BA/ep2 on the track BA, rejoining the carrier aftert1 + t2 = T hours.

On the two legs the plane has the speeds ul = spli u2 = sp2i relative to thecarrier. At any time carrier and plane lie on a line parallel to p18p2. When theplane is at B, the farthest point out, the carrier is at C(BC 11 pisp2). The dis-tance r = CB is called the radius of action of the plane. Relative to the carrierthe plane travels the course CB out, BC back; hence

r r-+_=T,u1 U2

t1 = r =T U2

U1 u1 + u2

T UIU2r=U1 + u2

t2=r=T u1.

u2 u1 + u2

These times agree with those previously

CB OBt1=-=-,spl epl

given; for

BC BAt2 = - = _-8P2 ep2

Example 4. Alternative Airport. Let a plane p depart from the airport 0along the track OY (Fig. 53c). If a landing at an airport Y is rendered danger-

114 VECTOR FUNCTIONS OF ONE VARIABLE §54

ous by local bad weather, the plane may be directed to land at an alternativeairport A. If the fuel supply allows T hours flying time to the plane, how farmay the plane fly on the track OY in order still to reach the port A by changingits course?

This problem is reduced to the one preceding if the airport A is regarded

as a carrier traveling from 0 to A with the uniform velocity OA/T. Lay off

es = OA/T and proceed precisely as in the carrier problem. The line CB,drawn parallel to pisp2, fixes the farthest point B at which the plane can changeits course and still reach the airport A. The time t1 when the turning point isreached is given in ex. 3.

54. Kinematics of a Rigid Body. We shall now investigate thevelocity distribution of the particles of a rigid body moving in anymanner.

Consider first a rigid body having a fixed line or axis; its motionis then a rotation about this axis. The position of the body at

I

Fia. 54a Fia. 54b

any instant may be specified by the angle 0 between an axial planeo fixed in our frame of reference and an axial plane p fixed inthe body (Fig. 54a). By choosing a positive direction on the axis(unit vector e), we fix the positive sense of 0 by the right-handedscrew convention. Then the angular speed w of the body at anyinstant is defined as

de(1) dt

Thus w is positive or negative, according as 0 is increasing or de-creasing at the instant in question.

§ 54 KINEMATICS OF A RIGID BODY 115

The velocity distribution in the revolving body may be simplyexpressed if we define the angular velocity as the vector,

dow = -- e.

dt

Note that w always is related to the instantaneous sense of rota-tion by the rule of the right-hand screw. --

Choose an origin 0 on the axis and let r = OP be the positionvector of any particle of the body. Then (Fig. 54b)

r=OQ+QP=ze+pR,where R is a unit vector perpendicular to the axis and revolvingwith the body. Since ze and p are constant during the motion ofP, the velocity of P is

dr dR do do (do \V

dt = p do dt = p dt e x R = --e J x (ze + pR)

that is,

(2) v = wxr.

The velocity of any particle of a body revolving about a fixed axis isequal to the vector product of the angular velocity and the positionvector of the particle referred to any origin on the axis.

Let us next consider a rigid body having one fixed point 0. Leti be a unit vector fixed in the body, j a unit vector in the directionof di/dt (perpendicular to i) and k = i x j. Then we may write

di dk di dj dj=aj, -=-xj+ix-=ix-

dt dt dt dt dt

Hence dk/dt (perpendicular to k) is also perpendicular to i andtherefore parallel to k x i = j. Thus we have

dk dj d

dt = j3j,dt dt

(k x i) = /3j x i + k x aj,

or, on collecting results

di dj dk

- = aj,dt

= (ak - #i) x j,dt

= Rj.


If we now write

54

w=ak - ii,these equations assume the same form:

di dj dk

dt = wxl' dt = wxjfdt

=

NowNow the position vector r of any particle P in the body may bereferred to the orthogonal triple i, j, k fixed in the body; thus

r=OP=xi+yj+zk,where x, y, z remain constant during the motion. Hence the ve-locity of P is given by

dr di dj dk

dt=xd6+ydt+zdt = wx(xi+yj+zk),

(3) V = co x OP.

Thus, at any instant, the velocity distribution in a rigid body withone point 0 fixed is the same as if it were revolving about an axisthrough 0 with angular velocity co. The line through 0 in thedirection of co is called the instantaneous axis of rotation, and co iscalled, as before, the angular velocity. Now, however, co may changein direction as well as in magnitude. With a fixed axis of rotation,

do dw =

dte =

dt(oe),

so that co may be regarded as the time derivative of the vectorangle Be. But with a variable axis of rotation co no longer can beexpressed as a time derivative.

Finally let us consider the general motion of a free rigid body.If, at any instant, all points of the body have the same velocity v,the motion is said to be an instantaneous translation. When the

velocity distribution is given by v = co -A P, the motion is saidto be an instantaneous rotation about an axis through A in thedirection of co. We now shall show that, in the most generalmotion of a rigid body, the velocities may be regarded as com-pounded of an instantaneous translation and rotation.


Let A be any point of the rigid body, and denote its velocityrelative to by VA. Consider a second reference frame ' havinga translation of velocity VA relative to . Then the motion of thebody relative to a' is an instantaneous rotation about an axisthrough A, since A has zero velocity relative to '. The velocityof any particle P of the body is therefore

vy = wxAP,relative to a', and, consequently,

(4) VP = V.4 + wxAP,relative to R. Moreover, for any other point Q of the body,

(4)' VQ = vA + w x AQ;

and, on subtracting this from (4), we get

VP = VQ + 0) x QP.

The content of these equations is stated in the following

THEOREM 1. If A is any point of a free rigid body, the velocitiesof its points are the same as if they were compounded of an instan-taneous translation VA and an instantaneous rotation w about anaxis through A; and w is the same for any choice of A.

Thus the instantaneous velocity distribution of a rigid body isdetermined by two vectors, its angular velocity w and the velocityVA of any point A of the body. These may be combined into thevelocity motor,

(5)

for, from (4),

(6)

V = Co + EVA;

VP= VA +PA xw,

in accordance with (31.3).

If w VA 0 0, the motor Visa screw. Since w VP = CO VA,

the velocities of all particles of the body have the same projectionon Co. The axis of V is called the instantaneous axis of velocity;its equation, from (31.8), is

= (wxv4)xw(7) rxw (origin A),ww


a line parallel to co and passing through the point Q given by

AQ = Co x VA/0) - uo; and, from (4)',

_ w.VA(8) VQ = VA + w x AQ = -co = projw vA.

w - w

All points of the instantaneous axis have the velocity VQ; for, if Ris another of its points,

VR = VA -I- w x (AQ -{- QR) = vQ since w x QR = 0.

Moreover vQ is characterized by being parallel to co; and, since allparticle velocities have the same projection on Co, their least nu-merical value at any instant is j vQ I. Referred to a point Q onthe instantaneous axis, the velocity motor becomes

(9) V = to -f- evQ.

This form, in which co and vQ are parallel, symbolizes

THEOREM 2. At any instant, the particle velocities of a rigid bodymay be represented by a screw motion-a rotation about an instan-taneous axis combined with a velocity of translation along this axis.

If Co 0, VA = 0, we have vQ = 0. The motion is then a purerotation of angular velocity w about the instantaneous axis. V be-comes a line vector along this axis.

If w = 0, Co VA 0, we see from (4) that all points of thebody have the same velocity. The instantaneous motion is then apure translation, and V = EVA is a pure dual motor.

Finally let us consider the case of plane motion of a rigid body.If co 0, the plane of the motion is perpendicular to w and vQ = 0;the motion is then an instantaneous rotation about an axis. Thepoint Q where the instantaneous axis cuts the (reference) plane ofmotion is called the instantaneous center; and, as previously,

-- 9(10) AQ =

wxVA

w - w

If the velocities of the points A, B of the body are known andVA, VB are not parallel, Q is at the intersection of the lines AQ, BQdrawn perpendicular to VA and VB, respectively.


Example. Rolling Curve. Let the curve r roll without slipping over afixed curve r1 (Fig. 54c). The points A and Al were originally in contact;and, if I is the instantaneous point ofcontact, let s = arc Al, sl = arc All.The conditions for pure rolling are

(1) S = Si, T = Tj at I. Vt=V

If I is regarded as a moving particle, its ve-locity relative to fixed and moving framesattached to r2 and r is the same; for

Al

dsl dsVI =

dtTi =

dtT = Vj.

Fia. 54c

Hence, if I coincides with the fixed point Q of r, we have, from (53.1),

vl=VQ+Vj, VQ=0.

The instantaneous center of the rolling curve is at its point of contact with thefixed curve.

The speed of rolling,ds ds1

v_ _

dt dt

The angular speed is co = de/dt, where 0 is the angle between the tangents atA and A1. We now differentiate the equation Ti = T with respect to t. SinceT is a function of s and B, we have

dT1 dsl OT ds 8T de

dsl dt = as dt + ae dt '

or, since Nl = N,

(iii)

KgN1v = KNv + Nw,

1 1 w

P1 P v

Here N is ir/2 in advance of T and p and P1 are positive when N points to therespective centers of curvature (in Fig. 54c, P1 < 0, p > 0 and w < 0).

r

(0>0;p1>0,p<0 6)>O;p1<0,p<0

Fla. 54d


If r and ri are circles of radius r and rl (Fig. 54d),

w _ 1 1) = I + Ifor external contact,

V ri ` r) r ri

for internal contact.V ri \ r/ r ri

When ri -+ co, ri becomes a straight line and v = wr.

55. Composition of Velocities. Let a rigid body have the ve-locity motor,

V' = w' + EVA,

relative to a frame a'; and let the frame a' have the motor,

V = CO + EVB,

relative to a "fixed" frame , where B is the point of a' whichcoincides for the instant with the point A of the body. What isthe motion of the body relative to a?

The velocity of any point P of the body relative to a' is

(1) Vp = VA + W' X AP.

At the instant in question let P coincide with the point Q of a'.Then the velocity of Q relative to is

(2) vQ=vB+WXBQ.

Now the velocities of P and A relative to are (53.1)

Vp = VQ + Vp, VA = VB + VA.

On adding (1) and (2) and observing that AP = BQ at the instantconsidered, we get

VP = VA + (w+W')XAP.

Thus the motion of the body relative to a is compounded of thevelocity of translation VA and the angular velocity w + w' aboutan axis through A ; that is, the motion is given by the motor,

V+V'=w+w'+evp.THEOREM. If the motion of a rigid body relative to a' is given

by the motor V' = w' + evA, and the motion of a' relative to byV = w + eVB, the motion of the body relative to a- is given by the

§ 56 RATE OF CHANGE OF A VECTOR 121

motor sum V + V' provided A and B coincide at the instant inquestion.

Two translations, ev' and evB thus compound into a translatione(vA + VB).

If V' = co' + evA and V = w + evB represent pure rotations,these motors are line vectors; hence the motion of the body rela-tive to a will be a pure rotation when and only when w + w' 0and the axes of rotation are coplanar (§ 32, theorem 1). If theaxes of rotation intersect at A, the motors referred to A are

V'=w', V=w; and V+V'=w+w'.Angular velocities about intersecting axes may be compounded byvector addition.

If V and V' represent pure rotations about parallel axes andw + w' = 0, V + V' = e(vB + vi); the motion of the body rela-tive to is then a pure translation of velocity vB + V.

56. Rate of Change of a Vector. Referred to a fixed origin 0

in the frame a, the vector u = PQ = OQ - OP; hencedu

(1)dt

= VQ - VP1

where vp, vQ are velocities relative to .

Similarly, if a' is a second frame in motion with respect to ,

the rate of change of u relative to a' is

d'u(2)

dt= ve - vp,

where v p', vQ are velocities relative to '.Let the motion of a' relative to be given by the motor

co + EVA, A being a fixed point of a'; and, at the instant in ques-tion, let P and Q coincide with the points R and S of a'. Then,from (53.1) and (54.4),

Vp = oP+OR =VP+oA + WXAR,

VQ = VQ+Vs =VQ+vA+wxAS;

and, on subtraction,

du d'u -> -dt = dt + w X (AS - AR).

122 VECTOR FUNCTIONS OF ONE VARIABLE § 56- -- --> --->But, since AS - AR = RS = PQ = u,

du du(3) d =

dt+ 0).U;

in particulardu

(4)dt

= w x u if u is fixed in a'.

When u = w, the angular velocity of the frame a', we have,from (3),

dw d'w(5)

dt dt

The vector dw/dt is denoted by a and called the angular accelera-tion vector of a.

Example. Kinematic Interpretation of Frenet's Formulas. At any pointP of a space curve, the trihedral TNB may be used as a frame of reference.If P moves along the curve with unit speed, ds/dt = 1 and s = t if t = 0at the origin of arcs. Then the arc s may be interpreted as the time, anddr/ds = T is the velocity of P.

If w is the angular velocity of TNB referred to a "fixed" frame a, we have,from (4),

dT dN dB

ds = , x T,ds = p1 x

N, ds = " x B.

A comparison with Frenet's Formulas (44.6) shows that 0) = S, the Darbouxvector. The Darboux vector of a space curve is the angular-velocity vector of itsmoving trihedral TNB.

Since the vertex P of the trihedral has the velocity T, its motion is representedcompletely by the velocity motor,

V =S+eT=rT+KB +Er.

From (54.7) and (54.8), we see that the axis of this motor passes through thepoint Q given by

---> 8 x T K S T rPQ =

S 8= ,,2+,2N and vQ = S-S S = K2 + '28.

In general, the trihedral TNB has, at every instant, a screw motion: a combina-tion of the angular velocity S about an axis through Q and the velocity vQalong this axis. The velocity of translation vanishes when and only when the

curve is plane (r = 0); then S = KB, PQ = pN. For plane curves TNB has, atevery instant, a pure rotation of angular velocity KB about the center ofcurvature.

§ 57 THEOREM OF CORIOLIS 123

57. Theorem of Coriolis. Let v and a denote the velocity andacceleration of a particle P, relative to a frame , while v' anda' denote these vectors relative to a frame ' in motion with re-spect to . Then if 0 and A are origins fixed in and a', we have

dr dvr= OP, v= -- a= -;

dt,

dt

d'r' d'v'r' = AP v'=

dt ,a

dt

If w is the angular velocity of ' with respect to , we have, ondifferentiating,

-- 3r = OA + r',

twice with respect to the time, and, on making use of (56.3),

dr'V =VA+

dt

= VA +wxr'+v',dr' dv' / dwa=aA+axr'+wx+ (a Jdt dt dt

a A+ a x r' + w x ((o x r' + V) + w x v'+ a'

a4+axr'+wx(wxr')+2wxv'+a'.The velocity and acceleration of the point Q of the frame j' withwhich P momentarily coincides are called the transfer velocity andtransfer acceleration of P. To find them, put v' = 0, a' = 0 in theforegoing equations; thus

VQ = VA + wxr',

aQ = aA + ax r' +wx

Our equations now read

(1) V=vQ+V',(2) a=aQ+2wxv'+a'.Equation (1) restates the theorem on the composition of velocitiesalready proved in § 55. Equation (2) shows that an analogoustheorem for the composition of accelerations is not in general true;


we have, in fact, the additional term 2w x v', known as the Coriolisacceleration. If we regard the frame as "fixed" and rates ofchange referred to it as "absolute," while the corresponding ratesreferred to ' are "relative," we may state (2) as the

THEOREM OF CORIOLIS. The absolute acceleration of a particle isequal to the sum of its transfer acceleration, Coriolis acceleration, andrelative acceleration.

The Coriolis acceleration, 2w x v', vanishes in three cases only:(a) co = 0; the motion of ' relative to is a translation.(b) v' = 0; the particle is at rest relative to '.(c) v' is parallel to co.A particle of mass m, acted on by forces of vector sum F, has

the equation of motion F = ma. When themotion is referred to a rotating framethe equation of motion becomes

N P(Q) (3) ma' = F - maQ - 2mw x v'.

UIf we regard -maQ and -2mw x v' as ficti-V tious forces, ma' equals a sum of forces just

O(A) as in the case of a fixed frame. The termFio. 57a -2mw x v' is called the Coriolis force. When

' has the constant angular velocity w abouta fixed axis through A, a = 0, aA = 0, and aQ = w x (w x r').Taking 0 at A and the z-axis along w (Fig. 57a), we have

-maQ = mw2 (r' - k r' k) = mw2 (OP - ON) = mw2 NP;

this vector perpendicular to the axis and directed outward is calledthe centrifugal force on the particle.

Example. Particle Falling from Rest. Refer the particle, originally at 0,to the revolving frame Oxyz attached to the earth: Oz points to the zenith(along a plumb line), Ox to the south, Oy to the east (Fig. 57b). The earth

revolves from west to east about the axis SN and its angular velocity at northlatitude a is

_ 2r24 X 602

(k sin X - i cos a) radians/see.

When the particle is at rest relative to the earth, the force acting upon itis its local weight mg; hence in (3) F - maQ = mg. Therefore the equationof motion becomes

(i) a'=g-2o,xv',

§ 57 THEOREM OF CORIOLIS 125

where a' and v' are the acceleration and velocity relative to the earth. Weshall integrate (i) by successive approximations under the initial conditions

= 0, v' = 0 when t = 0 and with g regarded as constant.1. If we neglect the term in w,

a' = g, v' = gt,

The displacement r' is along the plumb line.

=zgt2.

Fla. 57b

2. With v' - gt, (i) gives

a' = g - 2tw x g,

v'=gt-12wxg,

' = zr gt2 - 3 taw x g.

Since g = -gk, the second term gives an eastward deflection,

- 3 taw x g = 3 t3wg cos Xj.

3. With the last value of v',

a' = g - 2tw x g + 2t2w x (w x g),

v' = gt - t2w x g + 3 taw x (w x g),

= 2gt2 - 3t3w x g + Gt4w x (w xg).


The last term is a deflection in the meridian or xz-plane. Since w x g = -wgcos X j,

w x (w X g) =wng cos x (i sin X + k cos X),

r = 1 tow" sin X cos X i + 1 taw cos X ' 1 t2 - 114w2 cos'- X k.s 9 3 9 J- (29 6 9 )

The first and second terms give deflections to the south and east; the thirdshows that the particle in t seconds falls through a distance

-z = 2gt2 - I t4w2g cost X.

58. Derivative of a Motor. Let the motor M= m + EmA bereferred to the frame a' in which A is a fixed point; and let 'itself have the motion V= o + EV A with respect to the frame .

If m and mA are functions of the real variable t, the derivative ofM relative to a' is

(1)d'M dm d'mA

dt dt +e

dt

In order to compute dM/dt relative to a we must rememberthat mA depends not only upon t but also upon the point A, whichis in motion relative to a. If A moves to B in the interval At,

Om.4

MA (t + At) + AB X m(t + At) - m.4 (t)

At

mA (t + At) - mA (t) OB - OAAt

+ --At

X m(t + At),

and hence, if rA ° OA,

AMA dmA drA

l-.o At dt - + dtxm.

c

Relative to the frame a, we therefore define

(2)dM dm dmA drA

dt dt dt + dtm )

To verify that dM/dt is a motor, we must show that

dmP drp dmA dr,4 -) dm+ x m = -

+ x m + PA Xdt dt dt dt dt

mB(t + At) - m4(t)At 'At

(31.3)

§ 58 DERIVATIVE OF A MOTOR 127

This equation, in fact, follows from

mp = mA + (rA - rp) x mon differentiation.

If t denotes time,drA

dt = VA;

and

dm d'm dmA d'mA

dt =wxm+

dt ' dt = dt '

from (56.3). Substituting these results in (2) gives

(3)

dM d'M= (0-M+ E((axmA + VA xm) +dt dt

or, in view of (34.1),dM d'M

(4) V X M +dt dt

This is the motor analogue of (56.3).Making use of the definition (2), we may verify that the follow-

ing rules of differentiation are valid :

5)d

M + NdM dN

+(

( ) 'dt dt dt

(6)

d dM dAM(XM) = X +

,dt dt dt

d dN dM7 (M -N) = M ' + N

( )dt dt dt '

8d dN dM

NM -NM x +)( x ) = .

dt(

dt dt

Example 1. If the motion of a rigid body is given by the velocity motorV = w + EVA (A a fixed point of the body), its acceleration motor is

dV = a+e(aA +VAXW).

The axis of this motor is called the instantaneous axis of acceleration; its equa-tion may be written from (31.8).


Example 2. Let F = f + EfA be a force acting on a rigid body whose velocitymotor is V = w + EvA. Then

(i)dF

dt + E ( dt + VA x f) .

If F acts at the point P of the body, fp = 0 and

(ii)

Hence, referred to A,

(iii)

dF= df + Evp x f, referred to P.

d do+E(vPxf+APxdt/

let the reader show that (i) and (iii) are consistent.

Example 3. The moving trihedral TNB at the point P of a space curver = r(s) may be regarded as a rigid body having the velocity motor,

V=S+Evp=8+ET,

as the point P traverses the curve with unit speed. The three line vectorsT = T, N = N, B = B through P are fixed in the trihedral; hence, from (4),

ds =VxT, dN =VxN,B

=VxB.

These are Frenet's Formulas in motor form. Written out in full they become

T = KN, dN = -7-T + KB + EB, d- _ -rN - EN;

for exampledB dB- =(S+ET)xB=- - EN=-TN-EN.

59. Summary: Vector Derivatives. The derivative du/dt of avector function u(t) is defined as the limit of Au/At as At approaches

zero. If u = OP is drawn from a fixed origin and P describes thecurve C as t varies, du/dt is a vector tangent to C at P in the direc-tion of increasing t. If I u I is constant, C is a circle and du/dt isperpendicular to u.

The derivative of a constant vector is zero. The derivatives ofthe sum u + v and the products fu, u v, u x v are found by thefamiliar rules of calculus; but for u x v the order of the factors mustbe preserved.

§ 59 SUMMARY: VECTOR DERIVATIVES 129

If s is the are along a curve r = r(s), dr/ds = T, a unit vectortangent to the curve the direction of increasing s. The unit prin-cipal normal ft to the curve has the direction of dT/ds; and theunit binormal B = T X N; then [TNB] = 1. The vectors of themoving trihedral TNB change conformably to Frenet's Formulas:

dTKft = SXT,

ds

dN

ds

dB

ds

-KT + TB = 6 X ft,

-rft = SXB;

K is the curvature, r the torsion of the curve; and the Darboux vectoro = TT + KB is the angular velocity of the moving trihedral as itsvertex traverses the curve with unit speed (ds/dt = 1). For planecurves r = 0.

If t denotes time, a particle P traveling along the curve r = r(t)has the velocity and acceleration,

dr dv d2rv

'a -dt dt dt2

If v = ds/dt is the speed, and p = 1/K the radius of curvature of thepath,

dv v2V = VT, a = -- T+ - ft.

dt p

If A is any point of a free rigid body, the velocities of its pointsare given by the velocity motor,

V = w + EvA ;

here co, the angular velocity vector, is the same for any choice of A.Since V is a motor, for any point P of the body,

Vp = VA + PA Xw = vA +(0 XAP.

If, at any instant, co 0, the body has a screw motion about theaxis of V, the instantaneous axis of velocity; this reduces to a purerotation if V is a line vector (co VA = 0). If co = 0 the motion isan instantaneous translation of velocity VA.


If the frame a' has the angular velocity relative to , the ratesof change of a vector u relative to these frames are connected by

du d'u

dt dt +w .U.

A particle P has the velocity and acceleration v', a' relative toframe a-', v, a relative to the frame a; then, if the motion of a'relative to a is given by the motor V = co + EvQ, where Q is thepoint of ' coinciding at the instant with P,

v = VQ +v', a = aQ +2wxv' + a'.

The term 2w X v' is the acceleration of Coriolis. When a' is in trans-lation relative to , the velocity equation may be written v =vl, + V'.

The derivative of the motor M = m(t) + emA(t) is defined as

dM dm dmA drA

at = at +at-_

+X m

at

If t denotes time, and d'M/dt refers to a frame a' having themotion V = co + evA, relative to a frame , then

dM d'M

dt dtV X M +

PROBLEMS

1. If r and x are the distances of a point on a parabola from the focus anddirectrix, r - x = 0. Show that (R - i) - T = 0, and interpret the equation.

2. Prove that the tangent to a hyperbola bisects the angle between the

focal radii to the point of tangency. [Cf. § 44, ex. 2.13. An equiangular spiral cuts all vectors from its pole 0 at the same angle a

(R T = cos a). If (r, 0) are the polar coordinates of a variable point P onthe spiral, show that

(a) ds/dr = sec a, s - so = (r - ro) sec a;

(b)1 dr

= cot a, log r = (9 - Bo) cot a;rd0 ro

(c) p = ds/dB = r/sin a;

and that the center of curvature is the point where the perpendicular to OPat 0 cuts the normal at P.

PROBLEMS 13 1

4. If r = r(s) is a plane curve. r, show that(a) r1 = r + eN (c cont.) is a parallel curve r1 (§ 45, ex.l at it ii: l dis-

tance c from r;(b) Si = s - c, provided Si = s = 0, when ,y = 0;(C) P1 = P - C.5. A curve r = r(s) has the property that the locus r1 = r + rT (e const.)

is a straight line. Prove that the curve is plane, in fact, a tractrix. [Cf. § 50,ex. 5.)

6. An involute of a curve r, r = r(s), is a curve r1 which cuts the tangentsof r at right angles. Prove that r has the one-parameter family of involutes,

r1 = r + (c - s)T (c = const.);

and, if we take T1 = N, ds1/ds = (c - s)K.7. Show that the involute of the circle,

x = a cos t, y = a sin t,

obtained by unwrapping a string from the point t = 0 is

x1 = a(cost +tsint), y1 = a(sint - tonst);and that s1 = Zat2 gives the arc along the involute.

8. The cylinders x2 + y2 = a2, y2 + z2 = a2 intersect in two ellipses, oneof which is

x = a cos t, y = a sin t, z = a cos t.

Show that its radius of curvature is

P = a(l + sine t)I/''2.9. The equation of a cycloid is

x = a(t - sin t), y = all - cos t).Prove that

[sin1

(a) T =t

2, cos

2J, ds/dt = 2a sin

2

(b) >y = (i, T) = 2 - 2 , P = -4a sin 2 (44.3);

(c) The equation of its evolute is

x1 = a(t + sin t), y1 = -a(1 - cost).

10. Find the vectors T, N, B and the curvature and torsion of the twisted

cubic,x = 3t, y = 312, z = 20,

at the points where t = 0 and t = 1. Write the equations for the normal andosculating planes to the curve at the point t = 1.

11. Show that curvature and torsion of the curve,

x = a(3t - t3), y = 3at2, z = a(3t + t3),are

K = r = 1/3a(1 + t2)2.


12. Find the envelope of the family of straight lines in the xy-plane,

r = p(O)R + XP,

where R and P = k x R are the unit vectors of § 44, 0 = angle (i, R), and pis the perpendicular distance from the origin to the line. Show that thecurve,

r, = pR + p'Pis the envelope, and that

Ti = T, dsl/de = p + p".

13. Find the envelope of the family of lines for which the segment includedbetween the x-axis and y-axis is of constant length c. (In Problem 11 putp = c sins cos 0.1

Show that the envelope has the parametric equations,

x = c sin3 0, y = C C083 0;

and that the entire length of the curve is 6c.14. Show that the curvature and torsion of the curve,

x = et, y = e-t, z = /2 t,are

Sc = -T + a-t)2.

15. Verify that the curve,

x = a sin2 t, y = a sin t cos t, z = a cos t,

lies on a sphere. Show that the curve has a double point at (a, 0, 0) for theparameter values t = ±r/2, and that the tangents to the curve at this pointare perpendicular.

16. If a curve r = r(s) lies on a sphere (r - c) (r - c) = a2, show that

Hence, prove that

I dpr - C = -pN - --B.T ds

pr+ d (ldP)0

ds r ds

is a necessary and sufficient condition that a twisted curve (r d 0) lie on asphere.

17. The points on two curves r and r, are in one-to-one correspondence.

If T = T, at corresponding points, prove that

N = N1, B = B1, dsl/ds = K/Kt = r/T,.

When both curves cut the rays from 0 at the same angle, show that r, = cr;and that s, = as, if both arcs are measured from the same ray.

18. A curve r is called a Bertrand curve if its principal normals are principalnormals of another curve r1; then r1 is also a Bertrand curve, and

rl = r + XN, Ni = eN, where e = 1 or -1.

PROBLEMS

Show, in turn, that

11

= (1 - cK)T + cTB;(a) X = c (const.), Ti d

(b) If = angle (B, Bl), taken positive relative to N,

B1 = B cos (P + T sin p, ET1 = T cos rp - B sin rp;

(c) = const., and

(d)

(e)

ds1 ds1-ET1 ds = K sin p - r COS gyp, Kl

dsCOS cp + T sin gyp;

ds1 1 - CK -CrE - = _ ;ds cos p sin

C2TT1 = Sin2 (P, K1 = -E(Ti cot (p + 1/c);

133

since K1 = 0, the last equation determines E.'(f) As to the Darboux vectors, Si ds1/ds = S.

19. The characteristics of a one-parameter family of planes whose homo-geneous coordinates (§ 26) are (a(t), ao(t)), are the limiting lines of inter-section of the planes corresponding to the parameter values t and t + h ash -- 0. The line of intersection of the two planes has the homogeneous coor-dinates (26.10),

{a(t) x a(t + h), ao(t + h)a(t) - ao(t)a(t + h) [;

and, if we divide both by h and pass to the limit h -+ 0, show that we obtain

(a x a', apa - aoa')

as the coordinates of the characteristics.20. From the result of Problem 19 show that the three families of planes

associated with the twisted (r 0 0) curve r = r(s), namely,

(T, r T), normal planes,

(N, r N), rectifying planes,

(B, r B), osculating planes,

have as characteristics the respective families of lines:

[B, (r + pN) x B], parallel to B through the centers of curvature;

(8, r x 8), parallel to S through points of the curve;

(T, r x T), tangent to the curve (§ 48).

21. If a particle P of mass m is subject to a central force F = mf (r)R, where

r = OP and R is the unit vector along OP, its equation of motion isz

= f (r)R.m dt2 = F ordv

Prove that r x v = h, a vector constant; hence, show that the motion is plane

and that r = OP sweeps out area at a constant rate (Law of Areas).


22. When mf(r) = --ymM/r2 in Problem 21 the particle P is attractedtowards a mass M at 0 according to the law of inverse squares:

dv k(a)

dt= - - R (k

Show, in turn, that

(b) rxv = h

(c) h = r2 R x dR ,

(d) dt xh=kdR,

(e) v x h = k(R + e),

where a is a constant vector. From (b) and (e),

r x v h = h2, r vxh = kr(1 +acose)

where 0 = angle (e, r); thus obtain the equation of the orbit,

hz/k(f)

1 + e Cos 0

a conic section of eccentricity a referred to a focus as pole. When e < 1,prove that the orbital ellipse is described in the periodic time,

7, - area of ellipse 2ah/2

_ k a'and, hence,

(g) T2/a3 = 4,r2/-yM (Kepler's Third Law).

(See Brand's Vectorial Mechanics, § 177.)23. If the plane Ax + By + Cz = 1, fixed in space, is referred to rectan-

gular axes rotating with the angular velocity w = [wl, w2, w3], show that

rdA dB dC]=l dt ' dt ' dt

[A, B, C] x [WI, (02Y w3].

24. A particle P has the cylindrical coordinates p, p, z (cf. § 89, ex. 1). IfP moves in the pz-plane so that dp/dt and dz/dt are constant while the planeitself revolves about the z-axis with the constant angular speed dip/dt = w,find the acceleration of P,

(a) By direct calculation.(b) By use of the Theorem of Coriolis.25. A real, everywhere convex, closed plane curve with a unique tangent at

each point is called an oval. It can be shown that an oval has just two tangentsparallel to every direction in the plane. The distance between these tangentsis the width of the curve in the direction of the perpendicular. Prove thatthe perimeter of an oval of constant width b is 7rb (Barbier's Theorem).

CHAPTER IV

LINEAR VECTOR FUNCTIONS

60. Vector Functions of a Vector. A vector v is said to be afunction of a vector r if v is determined when r is given; and wewrite v = f(r). Since r is determined by its components, f(r) is afunction of two or three scalar variables according as r varies in aplane or in space. A vector function may be given by a formula,as f(r) _ (a X r) x r; or it may be defined geometrically. Thus if

r = OP varies over the points P of a given surface, and v = OQis the vector perpendicular on the tangent plane to the surface atP, v = f(r).

A vector function f(r) is said to be continuous for r = ro if

(1) lim f(r) = f(ro).r - ro

This means that, when the components of r approach those of roin any manner, the components of f (r) approach those of f (ro).

A vector function is said to be linear when

(2) f(r + s) = f(r) + f(s),

(3) f (Xr) = Af (r),

for arbitrary r, s, X. For example, linear vector functions are de-fined by the formulas kr, a X r, a b r, in which k, a, b are constant.It can be shown that, when a continuous vector function satisfiesthe relation (2), it also satisfies (3) and is therefore linear.

Since we assume that (2) holds when r = s = 0, f(0) = 2f(0);hence, for any linear vector function,

(4) f(0) = 0.

A linear vector function is completely determined when f(a,), f(a2),f(a3) are given for any three non-coplanar vectors al, a2, a3. For, ifwe express r in terms of al, a2, a3 as a basis,

r = xlal + x2a2 + x3a3,135

136 LINEAR VECTOR FUNCTIONS

we have, from (2) and (3),

(5) f(r) = x'f(a,) + x2f(a2) + x3f(a3)

§ 61

Let a', a2, a3 denote the set reciprocal to a,, a2, a3 (§ 23); then,if we write

f(ai) = b1, xt = r at,

f(r) may be written in either of the forms:

(6) f(r) = r (albs + a2b2 + a3b3),

(7) f(r) = (b,al + b2a2 + b3a3) r.

These formulas represent the most general linear vector function.61. Dyadics. In linear vector functions of the form,

f(r) = alb, r+a2b2.r+...+anbn.r,we now regard f (r) as the scalar product of r and the operator,

(1) = alb, + a2b2 + ... + anbn.

Assuming the distributive law for such products, we now write

(2) f (r) = ' r.

Following Willard Gibbs, we call the operator 4) a dyadic and eachof its terms aib1 a dyad. The vectors ai are called antecedents, thevectors bi consequents.

While a b is a scalar and a x b a vector, the dyad ab representsa new mathematical entity. Gibbs regarded ab as a new speciesof product, the "indeterminate product." We shall find indeed thatthis product conforms to the distributive and associative laws, butis not commutative (in general ab 0 ba).

Since r follows. in (2), we call r a postfactor. If we use r as aprefactor, we get, in general, a different linear vector function:

g(r) = r 4).

We proceed to develop an algebra for dyadics, laying down forthis purpose definitions for equality, addition, and multiplicationof dyadics.

Definition of Equality. We write 4) = NY when

(3) 4> r = 'Y r for every vector r.

§ 61 DYADICS

If s is an arbitrary vector, we have from (3)

137

s - (4) - r) = s - or rSince s 1 and s ' are two vectors that yield the same scalarproduct with every vector r, these vectors must be equal; thus

(4) s iD = s - T for every vector s.

Conversely, from (4) we may deduce (3). Therefore 4) = T wheneither (3) or (4) is fulfilled.

The Zero Dyadic. We write = 0 when

(5) 4) r = 0 for every r.

The preceding argument shows that 4) = 0 also when

(6) s S. 4) = 0 for every s.

Definition of Addition. The sum 4) +' of two dyadics is de-fined by the property,

(7) (4) +'Y) r= for every r.

If is given by (1), and

' = c1d1 + C2d2 + ... + Cmd,n,

4) +' = a1b1 + ... + anbn + c1d1 + ... + c..d,,,..

In the sense of this definition, 1 (or ') is the sum of its dyadterms, thus justifying our notation. The order of these dyads isimmaterial; but the order of the vectors in each dyad must notbe altered, for, in general, ab r ba r and hence ab 54 ba.

The Distributive Laws,

(8), (9) a(b + c) = ab + ac, (a + b)c = ac + bc,

are valid. The proof follows at once from the definition of equality;thus from

a(b + c) r = (ab + ac) r,

we deduce (8). We may now perform expansions as in ordinaryalgebra, if the order of the vectors is not altered; for example:

(a+b)(c+d) =ac+ad+bc+bd.


If X is any scalar,

(10) (Xa)b = a(Xb),

and we shall write simply Xab for either member.

§ 62

From (60.6) or (60.7) we conclude that any dyadic D can bereduced to the sum of three dyads. To effect this reduction on 4)as given by (1), express each antecedent ai in terms of the basisei, e2, e3,

ai = ale, + aie2 + a;e3,

expand by the distributive law, and collect the terms which havethe same antecedent: thus

4) _ (a1 ei + a2e2 + a3e3)bi = elf, + e2f2 + e3f3,

where f;We also may reduce 4) to the sum of three dyads by expressing

each consequent bi in terms of the basis el, e2, e3, and then ex-panding and collecting terms which have the same consequent;then 4) assumes the form,

4) = giei + 92e2 + 93e3-

Thus it is always possible to express any dyadic so that its ante-cedents or its consequents are any three non-coplanar vectors chosenat pleasure.

If a,, a2, a3 are non-coplanar, a dyadic 4) is completely deter-mined by giving their transforms (§ 60). If

4) - a, =b,, 4> - a2 = b2, (D - a3= b3,

and the set a', a2, a3 is reciprocal to al, a2, a3, we have explicitly

(11) 4) = bla' + b2a2 + b3a3.

For a physical example of a dyadic the reader may turn to§ 116, where the stress dyadic is introduced. The name tensor, nowused in a much more general sense, originally was ajTplied to thisdyadic.

62. Affine Point Transformation. If we draw the position vec-tors,

r=OP, r'=4 r=OQ,from a common origin 0, the dyadic 4) defines a certain transfor-mation of the points of space: to each point P corresponds a defi-

§_63 COMPLETE AND SINGULAR DYADICS 139

nite point Q. If, when P ranges over all space, Q does likewise,this transformation is called affine; the dyadic 4) then is calledcomplete.

Important properties of an affine point transformation follow atonce from the equations,

4) (a+b) =

which characterize a linear vector function. Since - 0 = 0, thetransformation leaves the origin invariant. Lines and planes aretransformed into lines and planes. Thus, for variable x, the

Line r = a + xb - Line r' = a' + xb';

and, for variable x, y, the

Plane r = a + xb + yc - Plane r' = a' + xb' + yc'.

The transformed equations always represent lines and planes when4) is complete; for we shall show in § 70 that b 5-4 0 implies b' F6 0,and b x c 5x-I 0 implies b' x c' 54 0.

63. Complete and Singular Dyadics. Given an arbitrary basis,e1, e2, e3, we can express any dyadic in the form,

(1) 4) = glee + 92e2 + g3e3

If we express r in terms of the reciprocal basis,

r = xle' + x2e2 + x3e3,

then, by virtue of the equations, et e' = St,

4) r = x1g1 + x282 + x383

When g1, g2, g3 are non-coplanar, r' = 4) r assumes all possiblevector values as r ranges over the whole of space. If we put

---3 --ar = OP, r' = OQ, the dyadic defines an affine transformationr' _ ( r of space into itself; 3-dimensional P-space goes into3-dimensional Q-space. A dyadic having this property is said tobe complete. A complete dyadic cannot be reduced to a sum ofless than three dyads; if, for example, we could reduce 4) to thesum of two dyads, ab + cd, all vectors r would transform into.vectors r' = a b r + c d r parallel to the plane of a and c.

140 LINEAR VECTOR FUNCTIONS § 63

If, however, g1, g2, g3, are coplanar, but not collinear, we can ex-press each gi in terms of two non-parallel vectors f1, f2, and reducec to the sum of two dyads:

(2) 4, = f1h1 + f2h2.

This dyadic transforms all vectors r into vectors r' = 1 - r in theplane of f1 and f2; 3-dimensional P-space goes into 2-dimensionalQ-space. A dyadic having this property is said to be planar. Aplanar dyadic cannot be reduced to a single dyad ab; for then allvectors r would transform into vectors a b . r parallel to a.

If 91, g2, g3 are collinear, we can replace each gi by a multipleof a single vector f and reduce 4' to a single dyad:

(3) 4 = fh.

This dyadic transforms all vectors r into vectors r' = fi - r parallelto f; 3-dimensional P-space goes into 1-dimensional Q-space. Sucha dyadic is called linear.

Finally, if g1, g2, g3 are all zero, 4' = 0.Planar, linear, and zero dyadics collectively are called singular.

The point transformation,

OQ = (D -OP,

corresponding to a singular dyadic 4' reduces 3-dimensional P-spaceto a 2-, 1-, or 0-dimensional Q-space.

This discussion shows that, when - is reduced to the form (1)in which the consequents are non-coplanar, then 4, is complete,planar, linear, or zero, according as the antecedents are non-coplanar, coplanar but not collinear, collinear, or zero.

In particular, we have the

THEOREM. A necessary and sufficient condition that a dyadical + bm + cn be complete is that the antecedents a, b, c and conse-quents 1, m, n be two sets of non-coplanar vectors.

As a corollary,

(4) 4) - r = 0 implies r = 0 when - is complete.

For, if - = al + bm + cn,

al-r+bm-r+cn-r=0, 1 - r = m - r = n - r = 0,

and hence r = 0.

§ 64 CONJUGATE DYADICS

64. Conjugate Dyadics. The dyadics,

4) = alb, + a2b2 + ... + anon,

,D, = b1a1 + b2a2 + + bn.an,

141

are said to be conjugates of each other. In general 4) and (D, aredifferent dyadics; but evidently

(1) 4) r=rdefine the same linear vector function. If two dyadics are equal,their conjugates are equal; for 4, =' implies that

4)

r, and hence 4),A dyadic is called :

by definition.

Symmetric if 4), = 4),

Antisymmetric if 4 = - 4).

The importance of these special types of dyadics is due to the

THEOREM. Every dyadic can be expressed in just one way as thesum of a symmetric and an antisymmetric dyadic.

Proof. For any dyadic 4), Ave have the identity,

4)+4), 4)(1) =

2+

2= + 52;

since

4rc = = *I St, = = -Q,2 2

41 and Sl are, respectively, symmetric and antisymmetric. More-over 4) can be so expressed in only one way. For, if

41+Q_N,,+9,

gave two such decompositions, we have, on taking conjugates,

q/ - SZ =

and hence _ V, Sl = S2'.


65. Product of Dyadics. If the transformations correspondingto two linear vector functions,

v=4) - r,

are applied in succession, their resultant,

is a third linear vector function; for w1 + w2 corresponds tor1 + r2, and Xw to Xr. This function is written

w =

and 'I' ( is called the product of the dyadics ' and 4), taken inthis order. The defining equation for the product 4) is there-fore

(1) (* for every r.

From (1) we find that the distributive and associative laws holdfor the products of dyadics:

(2) 4) ('I' + S2) -{- 4 S2,

(3) ((F +') +(4)

Proofs. For every vector r,

(I' + l)] r = 4 [( + 0) r]

r+l r]r)+ (St r)

r+ r;

[( +'F) St] r = ((F +'I') (St r)

r)+ r)

r = [(`I' ' ) ' r]

r)

§ 65 PRODUCT OF DYADICS 143

Equations t2), (3), (4) follow from these results by the definitionof equality. In the proofs of (2) and (3), definition (1) justifiesthe first and last steps; in the proof of (3), (1) is applied in everystep.

In order to compute a product explicitly, we first find theproduct of two dyads. By definition,

for every r, and hence

(5) (ab) (cd) = (b c) ad.

The product of ab and cd, in this order, is the scalar b c timesthe dyad ad. Similarly,

(6) (cd) (ab) = (d a) cb,

which in general differs from (5).Making use of the distributive law, we now may form the prod-

uct of any two dyadics,n m

4) = E aibi, ' = E c;d;,i=1 i=1

by expanding into nm dyads.n m

(7) = E (bi c,)aidri=1 j=1

The conjugate of 4) ' ism n

(8) (I' ) = E 71 (c; bi)d;aij=1 i=1

The conjugate of the product of two dyadics is the product of theirconjugates taken in reverse order.

Making use of (8), we now find that

r ( 4 ) - *) = (4) 'P ) , r = ( * ,-( 1 ) , )-r

hence

(9) r ((D 'Y) = (r (D) 4, for every r,

an associative law analogous to (1) but with r as prefactor.If el, e2, e3 form a basis and e', e2, e3 the reciprocal basis, we

can express any two dyadics in the form,

4) = fie, + f2e2 + f3e3, ' = elg1 + e2g2 + e3g3


In the product 4 - ', six dyads vanish, and we find

(10) 4) `I' = figs + f2g2 + f393-

§ 66

If 4) and' are complete, fl, f2, f3 and g1, g2, g3 are non-coplanarsets; then (- 4, is also complete. But if (F or 4, is singular, oneof these sets must be coplanar, and 4) 'Y is likewise singular.Therefore the product of two dyadics is complete when and only whenboth dyadic factors are complete.

If 4) is complete, it may be "canceled" from equations such as4.4,=O,4, (F=0, to giveT =O. Thus, ifare non-coplanar in (10), and hence g1 = g2 = g3 = 0. We alsomay "cancel" - in (F = 4) St; for this is equivalent to

(4, - 9) = 0, and hence 4, - SZ = 0.66. Idemfactor and Reciprocal. The unit dyadic or idemfactor

I is defined by the equation,

(1) I r = r for every r.

The idemfactor is unique (§ 60). For any basis e1, e2, e3, we haveI e; = e I e' = e'; hence, from (60.7),

(2) I = e1e' + e2e2 + e3e3 = e'e1 + e2e2 + e3e3;

in particular the self-reciprocal basis i, j, k gives

(3) I = ii + jj + kk.

Evidently I is symmetric and complete.

THEOREM. In order that a, b, c and u, v, w form reciprocal sets,it is necessary and sufficient that

(4) au+bv+cw=I.When the sets are reciprocal, we have just proved (4). Con-

versely, if (4) holds, a, b, c are non-coplanar, since I is complete;let a', b', c' denote the reciprocal set. Using these vectors as pre-factors on (4) gives u = a', v = b', w = c'.

Multiplying a dyadic by I leaves it unaltered; for, from

r (I.4)) _we conclude that

(5) 4 I = I (F = 4.

§ 66 IDEMFACTOR AND RECIPROCAL 145

If 4 I = I, 4) and ' are both complete, since I is complete(§ 65). From

T or (4,

4) = I. Two dyadics 4), 4, are said to be recip-rocals of each other when

(6) 4) 41 _I (D =I,and we write 4, = 4)-1, 4) = T-1

A complete dyadic 4) has a unique reciprocal (D-1. If

(7) 4) = elf, + e2f2 + e3f3, (D-1 = flee + f2e2 + f3e3,

for their products in either order give I:

4) . 4>-I = e,e' + e2e2 + e3e3, 4)-1 4) = f If, + f'f2 + f3f3.

Since dyadic multiplication is associative

(4) F) (T-1 4,-I) _ 4) ('F -1) 4,-1 = 4) (D-1 = I;

(8) (4) . *)-I = 'F-I .,I>-,.

Making use of (8), we have

(4) . * . Q) -1 = 2-1 , (4) q') -1 =

and, in

ggeneral,

the reciprocal of the product of n dyadics is theproduct of their reciprocals taken in reverse order.

For positive integral n we define4 , n = (P .

4)-n = (4)-i)n = (4)n)-I;

4)°=I.

With these definitions,

(12) 4)m . 4,n = Pm+n, (4)m) n. _ 4)mn,

for all integral exponents. But owing to the non-commutativityof the factors in a dyadic product, (4>. ')n is not in general equal,0 cpn . *n.


By means of reciprocal dyadics, we readily may solve certainvector and dyadic equations. Thus, if (F is complete, the equations,

(F r = v , r 4 ) =v, 4 )-4 1=2, (F 2,

have the respective unique solutions:

r=4)-1 v,67. The Dyadic (1) x v. If (F = Maibi, we define the dyadics:

(1) - x v = Ea1bi x v, v x 4) = Tv x aibi.

From these definitions we have the relations:

(2) (4) x v) r = 4) (vxr), r (v x (F) = (rxv) (F;

(3) (r-(P)-v,

When 4, = I, these give

(4)

(Ixv) = r = vxr.

Since these equations hold for any r,

(6) Ixv = vxI,

(7) (I x v), _ -Ixv.

Thus I x v is antisymmetric and planar; it transforms all vectors finto vectors perpendicular to v.

Since

aibi (I x v) = aibi x v, (I x v) aibi = v x aibi,

the definitions (1) give

(8) (P (Ixv) _ xv, (Ixv) - 4) = vx(F.

Consequently, the operations v x and x v on vectors or dyadicsmay be replaced by dyadic products (I x v) and (I x v).

For any vectors u, v, r we have

[I x (u x v)] r = (u x v) x r = (vu - uv) - r,

and hence

(9) 1 x (u x v) = vu - uv.

§ 68 FIRST SCALAR AND VECTOR IN\'AItIANT 147

From this identity we arrive at the general form of any anti-symmetric dyadic.

THEOREM. If the dyadic f = 2;aibi is antisymmetric, and thevector co = tai x bi,

(10) l = -21xw.Proof. The conjugate of f is -St = 2;biai; hence

-2SZ = 2; (biai - aibi) = EI x (ai x bi) = I x w.

Corollary. Every antisymmetric dyadic is planar.68. First Scalar and Vector Invariant. We may express a dyadic

(P = 2;aibi in various forms by substituting vector sums for ai, bi,and expanding and collecting terms by applying the distributivelaw. For all these forms there are certain functions of the vectors,in terms of which is expressed, which remain the same. Thesefunctions, which may be scalar, vector, or dyadic, are called in-variants of the dyadic.

Each step of the process in changing = 2;aibi from one formto another may be paralleled by the same step in transforming thescalar and the vector,

(1) Cpl = tai bi,

(2) 4) = 1dai x bi,

obtained by placing a dot or cross between the vectors of each dyadof (D. Each application of the distributive law in the transforma-tion of 4) is also valid in the corresponding transformations of 01and 4; and, just as 1 is not altered by these changes, the same istrue of the scalar cpj and vector 4. These quantities are thereforeinvariants of 4 with respect to the transformations in question.They are called, respectively, the scalar (or first scalar invariant)and vector of the dyadic.

For example, if P = ij + jk + kk,

c1 1, 4)=ixj+jxk+kxk=k+i.For the idemfactor I = ii + fl + kk, the scalar is 3 and the vec-tor 0; and we obtain these same values if I is expressed in termsof an arbitrary basis: I = ele' + e2e2 + e3e3. Again, for thedyadic,

S2 = IxV = iixv+jjxv+kkxv,


we have co, = 0, and

§ 69

w=ix(ixv)+Jx(Jxv)+kx(kxv) -2v;thus S2 = -ZI x w, as in (67.10).

The scalar or vector of the sum of two dyadics is the sum oftheir scalars or vectors: thus, if

(3) 4) =`I'+0; 'Pi=4i+Wi, 4=4+w.It is principally to this property that these invariants owe theirimportance.

In (64.1), we have expressed any dyadic 4) as the sum of a sym-metric and antisymmetric dyadic:

(4)

The antisymmetric part S2 =z

(4) - 4) has the vector,

From the theorem of § 67 we now have

(5)

Hence, from (4),

SZ = -2Ix4.

24=4) +F,,-Ix4,(6) +Ix

From (6), we have the

THEOREM. A necessary and sufficient condition that a dyadic besymmetric is that its vector invariant vanish.

69. Further Invariants. We may obtain further invariants ofthe dyadic c = 1aibi by processes that are distributive with re-spect to addition. The most important of these are the dyadic,

(1) 4)2 = 2;aixajbixbj,i.l

called by Gibbs the second of I ; its scalar invariant,

(2) (P2 = 2 (ai x a3) (bi x bj);i,)

and the scalar,

(3) (P3 = * (aixaj .ak)(bixbj .bk).i.i.k

§ 69 FURTHER INVARIANTS 149

In (1) the summation is taken over all permutations i, j. Wheni = j, the dyad vanishes; when j - i, the permutations i, j andj, i give the same dyad, so that each dyad occurs twice in thefinal sum; this doubling is avoided by the factor 2.

In (3) the summation is taken over all permutations i, j, k.When two subscripts are the same, the term vanishes; wheni, j, k all differ, the 3! = 6 permutations of these subscripts givethe same term, so that each term occurs six times in the final sum;this multiplication of terms is avoided by the factor

Let CF be reduced to the three-term form:

is

(4) CF = al+bm+cn.The invariants considered thus far are now

(5)

(6) 4 = axl+bxm+cxn,(7) 42 = bxcmxn+cxanxl+axblxm,(8) IP2 = (b x c) (m x n) + (c x a) (n x 1) + (a x b) (l x m),

(9) (p3 = (a x b c) (1 x m n).

The numbers 01, P2, 03 often are called the first, second, andthird scalars of 4).

If CF is singular, the antecedents a, b, c or the consequents1, m, n in (4) will be coplanar, and p3 = 0. Conversely, if = 0,a, b, c or 1, m, n are coplanar sets and 4) is singular. Therefore adyadic is singular when and only when its third scalar is zero.

If V3 = 0, 4) must be planar, linear, or zero. * When CF is linearit can be reduced to the form al and cF2 = 0. Conversely, if4'2 = 0, 4) will be linear or zero; for, if we choose a non-coplanarset a, b, c as antecedents in (4), b x c, c x a, a x b are also non-coplanar, and `1'2 = 0 implies that

mxn = nxl = lxm = 0;then 1, m, n are parallel or zero.

Therefore we may state the

THEOREM. Necessary and sufficient conditions that a dyadic CF be

Complete 'P3 3-' 0,

Planar are that <P3 = 0, CF2 0,

Linear CF2 = 0, (1) 0.

150 LINEAR, VECTOR FUNCTIONS § 69

We next compute the invariants of the dyadic _ 'D2. From(7),

*2 = (c x a) x (a x b) (n x 1) x (1 x m) + cyclical terms

= [abc] [lmn] (al + bm + cn) ;

hence, from (9),

(10) *2 = t03-1-

We now may compute the three scalars of 4,:

(11) V1 = P2, 412 = '3'P1, 4'3 =2;

the first follows from (8), the second from (10), and the third from

(b x c) (c x a) x (a x b) = [abc]2, (m x n) (n x 1) x (1 x m) = [lmn]2.

Finally, the vector invariant of ' is

4 = (bxc) x (mxn) + (cxa) x (nx1) + (axb) x (lxm).

If we express 4r in terms of a, b, c, the term in a is

4ab lxm = al - (bxm+cxn) = alfrom (6) ; hence

(12) 4r = (al + bm + cn)

It can be shown that all scalar invariants of 4) may be expressedin terms of the six scalars,

(13) 'PI,'02,'3,'

This property is expressed by saying that these six scalars form acomplete system of invariants. When 4) is symmetric, 4 _ 4, = 0,and the last three scalars vanish.

The third scalar of the product of two dyadics is equal to the productof their third scalars. For any two dyadics 4), ' can be put in theform,

4) = f1e1 + f2e2 + f3e3, = e'g1 + e2g2 + e3g3,

where e1, e2, e3 and e1, e2, e3 are reciprocal sets (§ 65); hence

'P43 = [f1f2f3][e1e2e3][e'e2e3][919293]

= [f1f2f3][g1g2g3],

which is the third scalar of

4' ' = f1g1 + f2g2 + f3g3

§ 70 SECOND AND ADJOINT DYADIC 1.51

Example. For the idemfactor I = ii + jj + kk, the second 12 = I, the vectoris zero, and the three scalars are 3, 3, and 1. From

(i) I = eiel + e2e2 + e3e3,

we have, for 12,

(ii) I = e2xe3e2xe3 +e3xe1e3xel +elxe2e1xe2,

and the vector,elxel + e2xe 2 + e3 x e3 = 0.

The third scalar gives[ele2e3][ele2e3] = 1.

From (ii) we have, for example,

I I. e3 = el x e2[e'e2e3], e3 = el x e2/[ele2e3]

Thus (i) and (ii) give a handy compendium of the properties of reciprocalvector sets.

70. Second and Adjoint Dyadic. The second of the dyadic4) = Eaibi, namely,

(1) (D2 =ialai

x a; bi x b;,i.;

has the property,

(2) (4 u) x (4) v) = -4)2 (u x v).

Proof. We have

v)i ;

= Mai x a; (bi u) (b; v)i;

= Ma; x ai (b, . u) (bi v),i;

on interchanging i and j. The left member also equals half thesum of the two last expressions :

-u)ij

or, with regard to (20.1),

"M(aixa;)(bixb;) (uxv) = 4)2 (uxv).i;


The conjugate of 4'2 is called the adjoint of 4) and written 4'a.The adjoint satisfies the important relation,

(3) 4)=gC3I.

Proof. Write (D = al + bm + cn; then

(4) 4)a = mxnbxc+nx1cxa+lxmaxb.Choose for the antecedents a, b, c of 4) a non-coplanar set. (§ 61).

Then, if a', b', c' denote the reciprocal set, we have, by directmultiplication,

D 4)a = (1 mxn)(abxc +b cxa + caxb)_ (1 m x n) (a b x c) (aa' + bb' + cc')

=(P3I.

If we choose the consequents 1, m, n of 4) as a non-coplanar set(then a, b, c may or may not be coplanar), let 1', m', n' denote thereciprocal set. Then

(a b x c) x n) (1'l + m'm + n'n)

= (P31-

If 4) is complete, P3 0; then (3) shows that

(5) 4'-1 = 4'a/1P3

From (3) we also may show that, if u, v, w are any three vectors,

(6) [4) u, 4) v, 4) w] = ,P3[uvw].

Proof. Since 4'a is the conjugate of 4>2i we have, from (2),

(U V) 4)a,

(U V)

4)a 4) by P3I, we obtain (6).We now can deduce important properties of the affine trans.

formation (§ 62) :

(7) r' r ((P3 $ 0).

The vector area u x v (§ 17) is transformed into

(8) u' x v' = 4'2 (u x v).

§ 71 INVARIANT DIRECTIONS 153

From (4), the third scalars of ad and P2 equal [abc]2[lmn]2 =,p3 0. Since 4) and c2 are complete, r 7-4 0 and u X v 0 implyr' 0, u' x v' 0 (thus filling a gap in § 62). An affine trans-formation invariably changes lines into lines, planes into planes.

Moreover (7) transforms any parallelepiped [uvw] into another[u'v'w'] whose volume is p3[uvw] according to (6). As any volumecan be regarded as the limit of a sum of parallelepiped elements,the affine transformation alters all volumes in the constant ratio of93/1.

71. Invariant Directions. We next seek those vectors r whichare transformed by 4) into scalar multiples of r, say

(1) - r = Ar.

If we write this equation,

It - r = 0 where '1 = 4) - Al,

it is clear that the multiplier A must make the dyadic T singular;its third scalar 43 is then zero (§ 69). Conversely, if A is a rootof the equation 03 = 0, A is a multiplier of 4); for, when 4, is planaror linear, 4, - r = 0 for all vectors r normal to the plane of the con-sequents of T.

Let us write

4) = al + bm + cn, I = aa' + bb' + cc',

where a, b, c are non-coplanar and a', b', c' the reciprocal set;then

4, = a(l - Xa') + b(m - Xb') + c(n - Ac'),

4'3 = [abc] [(1 - Xa')(m - Xb')(n - Xc')].

The second box product in 43 gives, on expansion,

[lmn] - A { [a'mn] + [b'nl] + [c'lm] }

+ A2 { [b'c'l] + [c'a'm] + [a'b'n] l - A3[a'b'c'].

Substituting for the primed vectors gives

.I bxc axa = [abc] '... ' b c = [abc] '

.. .


and then multiplying the entire expansion by [abc] yields

L'3 = [abc][1mn]

§71

- A{(bxc) (mxn) + (cxa) (nx1) + (axb) (lxm)}

-A3,

or, on making use of (69.5), (69.8), (69.9),

(2) 43 = <P3 - A<P2 + A2,P1 - A3.

Denote the right member of (2) by f (A) ; then the cubic equation,

(3) 'P3 = f(A) = 0,

is called the characteristic equation of 4,. Since 4) - AI is singularwhen f (X) = 0, the coefficients of f (X) depend only upon the natureof the dyadic 4) and not upon the particular form in which it isexpressed. We thus have an independent proof of the invarianceof <Pl, 'P2, 'P3

The three roots A1, A2i A3 of (3) are called the multipliers orcharacteristic numbers of -4). From the relations between the rootsand coefficients of an algebraic equation, we have

(4) IP1 = Al + A2 + A3, <P2 = X2X3 + A3A1 + A1A2, P3 = A1A2A3.

The cubic (3) may have three real roots (not necessarily dis-tinct) or one real root and two conjugate complex roots. To findthe invariant direction corresponding to a root A1i we consider inturn the three cases in which 4) - X1I is singular.

1. If 4 - X11 is planar, let 4' - A1I = ch + dk; then the post-factor r1 = h x k reduces the right member to zero and gives theinvariant direction for A1. If 4) - X1I has more than two dyads,the cross product of any two non-parallel consequents will be nor-mal to their plane and give a vector parallel to r1.

If Al is real, r1 is a real vector. But if Al is complex, say Al =a + i(3, then r1 = a + ib is also complex. Then, on equating thereal and imaginary parts of

(5) 4) (a + Zb) _ (a + ia) (a + Zb),

we have

(6) (D a = as - $b, 4) b = #a + ab.

§ 71 INVARIANT DIRECTIONS

In this case we know a second multiplier,

X2= a-i$, with I2=a-2b,as invariant direction. For, from (6),

4) (a - ib) = (a - i(3) (a - ib).

155

From (5), we conclude that a and b are not parallel; for, if b = ka,4) a = (a + i(3)a, which is impossible unless # = 0.

2. If 4) - X1I is linear, it may be reduced to a single dyad ch.Then, if r1 is any vector in the plane perpendicular to h, 4, r1

= X1r1. In this case there is a whole plane of invariant directionscorresponding to X1.

3. If (D - X1I is zero, 4 r1 = X1r1 for any vector r1. All direc-tions are invariant with the multiplier X1.

If we write r' _ 4) r, we have, from (70.2),

(7) U" V' = (D2 (Uxv).

Now 4) transforms all vectors in the plane of u, v into vectors ofthe plane u', V. These planes will be the same if u' x v' = x u x v,that is, if u x v is an invariant direction of 42. Hence the invariantplanes of 4) are normal to the invariant directions of (P2.

Example 1. = ii + j(i + 2j) + k(j + 2k). Then

fit = i(4i - 2j + k) + j(2j - k) + 2kk,

pp1= 1+2+2=5, I'2=4+2+2=8, S3=4.The characteristic equation,

f(T)=4-8a+5X2-X3=(1-X)(2-A)2=0,has the roots, X1 = 1, X2 = X3 = 2.

For the root, X1 = 1,

- XiI = ji + jj + kj + kk = j(i + j) + k(j + k),

and the corresponding invariant direction is

r1= (i+j)x(j+k) =i-j+k.For the double root, X2 = X3 = 2,

'P -1\2I= -ii +ji+kj = (-i+j)i+kj,and r2 = i x j = k.

Example 2. 4) = ij + jk + Id. Then

'Pi=0, 92=0, IP3=1.


The characteristic equation, 1 - X3 = 0, has the roots,

+ -1 -X1 = 1, X2 = w

2= , X3 = 2 = w2.

For X1 = 1,4, -all=i(j-i)+j(kj)+k(i-k)

is planar- its consequents are all normal to

r1=(j-i)x(k-j)=i+j+k,which is the corresponding invariant direction.

For X2 = w, the consequents of

4 -X21=i(j-wi)+j(k-wj)+k(i-wk)are all normal to

r2 = (j-wi)x(k-wj) =i+wj+w2k.For X3 = w2, the consequents of

4' -X3I=i(j-w2i)+j(k-w2j)+k(i-w2k)are all normal to

§ 72

r3 = (j - w2i) x (k - w2j) = i + w2j + wk.

72. Symmetric Dyadics. The characteristic numbers of a sym-metric dyadic are all real.

Proof. If X1 = a + i(3 is a complex multiplier and rl = a + 2bthe corresponding invariant direction of the symmetric dyadic 4),we have (§ 71)

,D (a+ib) = (a+ii)(a+ib);c a = as -(3b, 4 ) b = 0a+ab.

Since cD is symmetric, b (D a = a 4) b; hence

or =0.But a and b are real vectors, and a a + b b is positive; hence(3 = 0, and X1 is real.

The invariant directions corresponding to two distinct character-istic numbers of a symmetric dyadic are perpendicular.

Proof. From the equations,

4 rl = X1rl, `k r2 = X2r2 (X1 0 X2),and

r2,we deduce

(X1 - X2)rl r2 = 0, rl r2 = 0.

§ 72 SYMMETRIC DYADICS

Consider now the following cases.

1,57

1. If f (X) = 0 has three distinct roots, the corresponding in-variant directions are mutually perpendicular and may be denotedby i, j, k. Then

4, - i = X1i, 4 'j = X2j, 4) - k = X3k,

and, from (61.11),

(1) 4) = X1ii + X2JJ + X3kk.

2. If two roots of f (X) = 0 are equal, let X1 5-' X2 = X3. Then,if i, j are the perpendicular invariant directions corresponding toXl and X2, we may write

4, - i = X1i, (D 'j = X2j, 4, - k = v,

v being the (unknown) transform of k; then, from (61.11),

4) = X1ii + X2jj + vk.

Now the symmetry of 4) requires that vk = kv, and hence v mustbe a multiple of k, say yk. Thus

4' = X1ii + X2jj +ykk, <PI = X1 +X2+y;

but, since Cpl = X1 + 2X2 by hypothesis, -y = 1\2, and

(2) 'D = X1ii + X2jj + X2kk.

Corresponding to the double root X2 = X3, we have a whole plane ofinvariant directions.

3. If all three roots of f(X) = 0 are equal, X1 = X2 = X3, let idenote an invariant direction corresponding to X1, and write

4 ) i = Xli, 4> j = u, 4 ) k = v;

then, from (61.11),= X1ii + uj + vk.

Now the symmetry of 4) requires that uj + vk = ju + kv, andthis in turn shows that u and v have the form,

u = aj + yk, v = yj + $k.Thus

4) = X1ii + (ai + yk)j + (yj + $k)k,

(Pi = X1 + a + 0, o3 = Xl (a$ - y2);


but, since <pl = 3X1, X03 = X1 by hypothesis,

a + /3 = 2X1, a(3 - y2 = ai

Elimination of X1 gives the relation,

(a + 0)2 - 4(a/3 - y2) = (a - (3)2 + 4y2 = 0;

§ 72

and, since neither (a - 0)2 nor 4y2 can be negative, their sum canvanish only if a-i3=0,y=0. Hence a = 0 = X1, and 4> re-duces to

(3) 4, = X1ii + A1jj + X1kk = XII.

In the case of a triple root all directions are invariant.Evidently all cases are included in (1); by making two or three

of the roots equal, we obtain (2) and (3).

THEOREM. Every symmetric dyadic may be reduced to the form,

(4) 4, = aii + $jj + 7kk,

in which a, 0, y are the real multipliers and i, j, k corresponding in-variant directions.

The reciprocal of 4) is

(15)1 1 1=-ii+-jj+-kk;a a y

for .4 4)-1 = I. Moreover,

(6) 4'2 = f7ii + yajj + a$kk

and, by direct multiplication,

(7) 4)" = a"n + a"jj + y"kk.

If 4) is symmetric and 4)" = 0, then 4) = 0. Moreover, ifr 4 r = 0 for every r, - = 0; for, if we choose r = i, j, k inturn, a=/3=y=0.

Example 1. For the symmetric dyadic,

(D = i(j +k) +j(k +i) +k(i +j),

w1 = 0, v2 = -3, 93 = 2, and the characteristic equation,

2+3X-X3=(2-a)(1+x)2=0,has the roots, X1 = 2, X2 = X3 = -1.

§ 72 SYMMETRIC DYADICS

For X1 = 2,

-XiI=i(-2i+j+k)+j(-2j+k+i)+k(-2k +i+j),The consequents are all normal to

(-2i+j+k)x(-2j+k+i) =3(i+j+k);hence r1 = i + j + k.

For a2=X3= -1,

159

-X2I= (i + j + k) (i + j + k)

is linear. Hence any vector in the plane perpendicular to i + j + k is aninvariant direction.

If we choose the unit vectors,

i' = (i + j + k)/V'3, j' = (i - j)/V'2,as invariant directions for the roots 2 and -1, respectively, then the unitvector,

k' =i'xj' = (i+j -2k)/V,gives a second invariant direction for -1. Therefore

4) =2i'i'-j'j'-kk',as we may readily verify.

Example 2. Inertia Dyadic. Let 0 be any point of a rigid body and s anaxis through 0 in the direction of the unit vector e. Then, if dm is an elementof mass at P, at a distance p from the axis s, the moment of inertia of the body

about s is defined as the integral f p2 dm over the body.

p2 =r2- =e (r2Ihence, if we introduce the dyadic,

(8) K = I fr2dm - frr dm,

known as the inertia dyadic of the body for the point 0,

(9) e- fp2dm.

For example,

If r = OP,

i. f(r2-x2)dm= f(y2 +z2)dm

is the moment of inertia about the x-axis. Thus K effects a synthesis of themoments of inertia of a body about all axes through 0.

This dyadic also has the property that, for any pair of perpendicular unitvectors el, e2,

- el . K . e2 = f (el r) (r . e2) dm


is the product of inertia for the corresponding axes; thus

j = f j) dm = fxydm.

§ 73

The inertia dyadic K is evidently symmetric. Hence we always can findthree mutually perpendicular axes x, y, z through 0 such that

(10) K=Ali+Bjj+Ckk.These axes are called the principal axes of inertia at 0, and A, B, C are themoments of inertia of the body about these principal axes. The principalaxes are characterized by the property that the product of inertia for any pairis zero.

The ellipsoid,

(11) r K . r = 1, or Axe + Bye + Cz2 = 1,

is called the ellipsoid of inertia at 0; its principal axes are the principal axes ofinertia at O. It has the property that the moment of inertia about any axis s

through 0 and cutting the ellipsoid at P is 1/(OP)2. For, if OP = r = re,

73. The Hamilton-Cayley Equation. The identity (70.3) ap-plied to the dyadic ' = - AI gives

(1) `y 'Ya = f(X)I.

The form of ' given in § 71 shows that we may write

(2) 'a = A + BA + CA2,

where A, B, C are dyadics independent of A; hence, from (1),

(3) (4 - XI) (A + BA + CA2) = (-P3 - V2A + V1A2 - A3)I.

Since (3) is an identity in A, the dyadic coefficients of like powersof A in the two members must be equal, hence

4) A = (P3I,

(4)-V217

-C= -I.If we multiply these equations in order by I, 4), 42, (3 and add,the first members camel, and we get

(5) te3I - (P24) + (P1 cy - (1,3 = 0.

§ 73 THE HAMILTON-CAYLEY EQUATION 161

Every dyadic 4) satisfies this cubic equation, the Hamilton-C,ayleyEquation; it evidently is formed by replacing X by 4) in the char-acteristic equation,

(6) f(X) = V3 - cP2X + P1 X2 - X3 = 0,

and inserting I in the constant term. If the X1, X2, X3 are the char-acteristic numbers of 4),

AX) = (X1 - X) N - X) (X3 - X)

hence (5) also may be written

(7) (4) - X1I) . (1 - X2I) (4 - X31) = 0,

in which the dyadic factors are commutative.From equations (4), we find

C=I, B=4) -q'1I, A=4)2-914) +92I;hence, from (2),

(8) q,. = 4)2 (,P1 - X)4) + (V2 - (PI X + X2)I.

When X = X1, a characteristic number,

91 - X1 = X2 + X3, 'P2 - 'P1X1 + X1 = X2X3,

and (8) becomes

(9) (4) - a1I)a = (4) - X21) (4) - X3I).

Although every dyadic 4) satisfies the cubic (5), 4) will satisfyan equation of lower degree when `"a = (4) - XI)a vanishes for acharacteristic number X j. The equation of lowest degree satisfiedby 4) is called its minimum equation. *

If 4, = 4) - XI = 0 for a characteristic number X1 (then alsoTa = 0), the minimum equation is linear, namely,

(10) 4) -X1I=0.

When 4) = X11, 4, = (A1 - X)I, and 03 = f(A) = (X1 - X)3; theHamilton-Cayley Equation is therefore

(11) (4) - X11)3 = 0.

* For its formation and properties see Maeduflee, C. C., An Introduction toAbstract Algebra, New York, 1940, pp. 224-6. This treatment for n X nmatrices also applies to dyadics, regarded as 3 X 3 matrices.


If 4, does not vanish for any characteristic number, but *a, = 0when X = X1, we see from (9) that the minimum equation is thequadratic,

(12) (4, - X21) (,b - X31) = 0.

When Fd = 0, 4'2 = 0, and, consequently, 4, is linear (theorem,§ 69) ; hence we may write 4) = AiI + uv. If we take u = i,v = ai + /3j ± 'Yk, I = ii + jj + kk, we have

-A)I+aIi+flij+yik,and the determinant of 'F's matrix is

¢3 = f(A) = (A1 - X)2(X1 + a - A).

Thus the characteristic numbers of b are X1, A2 = A1, A3 = Al + a,and its Hamilton-Cayley Equation is

(13) ((D - A1I)2 . (4, - A3I) = 0.

From 'I = A1I + uv, we see that all vectors perpendicular to vhave the multiplier A1i whereas vectors parallel to u have themultiplier Xi + u v = Al + a. When u v = a = 0, (13) re-duces to (11).

We note that, in every case, the minimum equation and theHamilton-Cayley Equation have the same linear factors and differonly in their degree of multiplicity.

74. Normal Form of the General Dyadic. Every complete dy-adic transforms at least one set of mutually orthogonal directions(its principal directions) into another set of the same kind.

To find the principal directions of the complete dyadic -1,, con-sider the dyadic (D. The latter is complete (§ 65) and sym-metric; for, from (65.8), '(c. 4)). =

We therefore may write (§ 72)

(1) (DC = A1u + A2jj + A3kk, (Ai 0).

Consequently,

i.cc.4).j = j.4)a.(D .k = k.4)a.4) .i = 0,or

(-t i) =0.

§ 74 NORMAL FORM OF THE GENERAL DYADIC 163

The vectors 4 i, 4) j, 4 k are thus mutually orthogonal, andhence i, j, k give a set of principal directions of 4). If we write

(2) 4 i = ai', 4, j = #j', 4, . k = yk',

where i', j', k' is a second dextral set of unit vectors, we have,from (63.11),

(3) 4) = ai'i + (3j'j + yk'k.

Moreover, we always can arrange so that a, fl, y have the samesign. If, for example, a and 0 have one sign, -y the opposite, wecan replace i', j' by -i', -j', and the set -i', -j', k' still willbe dextral.

From (3) and,Pc = aii' + 3jj' + ykk',

we have, by direct multiplication,

(4)

(5)

4rD' 4) = a2ii + a2jj + y2kk,

= a2i'i' + 132j'j' + y2k'k'.

These symmetric dyadics, which in general are different, have thesame multipliers, evidently all positive.

We have therefore proved the

THEOREM. If (P is complete, any three invariant directions of,I)c 4) that are mutually orthogonal are principal directions of (D, andconversely. The principal directions of 4) transform into invariantdirections of 4' 4,. Any complete dyadic 4) can be reduced to thenormal form (3) in which the scalars a, (3, y are square roots of themultipliers of (D, 4) (or 4) (D,) having the same sign.

Example. Homogeneous Strain. In distinction to the ideal rigid body, theparticles of a deformable body are capable of displacements relative to oneanother. The totality of such relative displacements is said to constitute itsstate of strain.

Suppose that a particle at P moves to P' under the strain; then r' = OP'

is a continuous function of r = OP. The simplest type of strain occurs whenr' is a constant linear vector function of r,

(6) r' = 4' r (4 complete);

the strain is then said to be homogeneous. We have seen in § 62 and § 70 thata homogeneous strain transforms lines into lines and planes into planes; and


evidently parallelism is preserved. Moreover all volumes are altered in theconstant ratio of (P3/1.

Since 4' is complete, r = 4-1 r'. The particles originally on a sphereabout 0 are displaced so as to lie upon an ellipsoid; for r r = a2 transformsinto

When 4' is reduced to the form (3),

14'-1 +jj'+ kk',

i'i' + 12 j'j' + 2 k'k'a ,ti

and the foregoing strain ellipsoid (with a = 1) has the equation,x'2 ,2 z,2

cr2+ -2 +

r2= 1.

The principal directions i, j, k of 4) are called the principal axes of strain;they transform into ad', /3j' 7k', the principal semiaxes of the strain ellipsoid.

75. Rotations and Reflections. In order that a dyadic 4 trans-form all vectors so that their lengths are unchanged, it is necessary andsufficient that its inverse be equal to its conjugate:

(1) -1 = b'.

Proof. If, for any vector r,

(2) (4) r)then

r 4 ) c cb r = r I r , r -I) r= 0,and, since I 4) - I is symmetric, it must be zero:

4), 4' = I, or 4), _ -1.

Conversely, if (1) is fulfilled,

(1

A dyadic that preserves the lengths of vectors also preserves theangles between them: for, by virtue of (1),

(4) r)(4) - s) =r I I

and, since lengths are unaltered,

cos (c r, 4) s) = cos (r, s).

§ 75 ROTATIONS AND REFLECTIONS 165

Condition (1), although sufficient to ensure preservation of angles,is by no means necessary. Thus the dyadic XI preserves anglesbut multiplies all lengths by X.

Since fi preserves lengths and angles, any orthogonal set of unitvectors is transformed into another such set. Thus there are twopossible cases: the dextral set i, j, k is transformed into anotherdextral set i.', j', k', or into a sinistral set i', j', -k'. Hence wehave two types of length-preserving dyadics:

(3) fi = i'i + j'j ± k'k.

Their third scalar is V3 = =L1. Moreover,

fit = ±i'i ± j'j + k'k = -fi.This shows that the vector invariant of fit is f+; but, from(69.12), this vector is fi Equating these values, we obtain

(4) fi + _+;the vector invariant of fi gives an invariant direction of multiplier,'P3 = f 1. If we choose k in this direction, (3) becomes

(5) fi = Vi + j'j + p3kk, IP3 = f 1.

When V3 = 1, fi r transforms i, j, k into i', j', k. But a rota-tion 0 about k as axis, through an angle X such that i, j revolveinto i', j', transforms i, j, k in the same way; and, since a is linearvector function, fi = e (§ 60). Thus

(6) 0=i'i+j'j+kkis a rotation about the axis k through an angle X determined byits scalar and vector invariants:

(7) 01 = 1 + 2 cos X, 0 = -2 sin X k.When P3 = -1,

fi = i'i + j'j - kk = (i'i + j'j + kk) (ii + jj - kk).

The dyadic in the first factor is the rotation e. The dyadicii + jj - kk in the second factor transforms i, j, k into i, j, -k.But a reflection E in the plane of i, j transforms i, j, k in the sameway; and, since E is a linear vector function,

(8) E = ii + jj - kk = I - 2kk.Thus, when P3 = -1, fi = 0 E.

166 LINEAR VECTOR FUNCTIONS 76

THEOREM. A dyadic 4) that preserves lengths is a rotation e when(p3 = 1, and a rotation O followed by a reflection in the plane per-pendicular to its axis when c03 = -1. In the latter case n E willreduce to Z, a pure reflection, when O = I.

76. Basic Dyads. If we express both antecedents and conse-quents of a dyadic 4) in terms of a given basis e1, e2, e3, we obtainupon expansion 3 X 3 = 9 types of basic dyads eie; (i, j = 1, 2, 3).On collecting terms we may write 4:

(1) 4) _ (p11e1e, +(p12e1e2 + P13ele3 +

c'21e2e, + cp22e2e2 + P23e2e3 +

w e3e, +032e3e2 + (p33e3e3.

The nine coefficients 9 are called the contravariant components of4) relative to the basis ei (cf. § 23). If we drop the nine basicdyads e;e; in (1), we can represent 4) by the 3 X 3 matrix,

(P11 1p12

V13

= p21 (p22 (p23

031 032 (p33

a skeleton of numbers arranged in a definite order, which standsfor the full expression (1). This is analogous to the use of anumber triple (u', u2, u3), or 1 X 3 matrix, to represent the vectora = ule, + u2e2 + u3e3.

If we express the vectors of 4) in terms of the reciprocal basise we write

(2) 4)

The nine numbers pij are called the covariant components of (brelative to the basis ei. Just as before, we may represent 4) by amatrix of the components cpjj. This corresponds to the use of thenumber triple (ul, u2, u3) to represent the vector u = u,e1 +u2e2 + u3e3.

But with the same basis e1, we can represent 4) in two otherways. First, we may express the antecedents of 4) in terms of ei,the consequents in terms of e'; the basic dyads are then eie', and

(3) c _ Z2;oz j eie'.

3 77 NONION FORM 167

Or we may express the antecedents in terms of ei and consequentsin terms of e;, so that the basic dyads are eie;; then

(4) = XMv 5eie;.

The components gyp';, cpti' are called mixed. If we represent 1 bymatrices of mixed components, we must indicate the order of thesubscripts as shown in the preceding notation; for, in general,e. p, i However, if we use the full notations (3) or (4), thecomponents can be written gyp;; for the order of the indices then isshown by the base vectors.

77. Nonion Form. When the self-reciprocal orthogonal seti, j, k is used as a basis, all four representations of 4) given in § 76become the same. Since upper and lower indices no longer areneeded, we write the orthogonal components of D arbitrarily asVij. When no basis is indicated, the components of the matrix(<pij) shall be regarded as orthogonal, and 4) itself is said to be inits nonion form:

'P11 'P12 'P13

(1) _ 'P11ii + 'P121j + + 'P33kk = rP21 'P22 923

'P31 'P32 'P33

The conjugate of I corresponds to the transpose of this matrix:

(2)

'11 V21 'P31

'P12 'P22 'P32

'P13 'P23 'P33

Hence c is symmetric when 'Pij = 'Pji, antisymmetric when 'Pij =c° ('i1 = 0).The first scalar and vector invariant of P are readily computed:

(3) 'P1 = 911 + V22 + 'P33,

(4) = ('P23 - 'P32)1 + ('P31 - VP13)i + ('P12 - (p21)k

In order to compute 'P2 and <P3i write 4 in three-term form:

(5) = i(<P11i + 'P12j + (P13k) +

j (921i + 'P22i + <P23k) +

k((P311 + 'P32j + p033k)


Then, since [ijk] = 1, <P3 is the box product of three consequents,namely the determinant of matrix ():

(6) <P3=

'P11 'P12 'P13

'P21 'P22 'P23

'P31 'P32 'P33

Thus is singular when the determinant of its matrix vanishes.As to 4)2i the terms with antecedent i = j X k have as consequents,

('P211 + 'P223 + 'P23k) x (cP311 + <P32i + 'P33k) =(11i + (p1) + (D13k,

where 'ij denotes the cofactor t of 9i j in the determinant (6) ;hence

,11 ,12 ,13

(7) 24)21 ,22 X23

X31 ,32 ,33

(8) 'P2 =(D11 + X22 + X33,

and (% is the transpose of (7). Moreover, since (D-1 = 4)a/cp3(70.5),

(P11 'P21 4031

-1

12

22 32(9)

=

('P

P 0

'P13 '23 033

where (P''j = (I)ij/(p3 is the reduced cofactor of 'ij. Making use ofthe well-known relations in determinant theory,

'Pi1'Pj1 + <Pi2'Pj2 + 'P1i'P1j + <P2i(P2j + ati

we may verify that 4-' or 4)-1 4) give the idemfactor:

1 0 0

(10) I= 0 1 0

0 0 1

The characteristic equation of 4) is obtained by equating thethird scalar of 4' - XI to zero; hence, with 4) in nonion form, it

t The cofactor of vii in the determinant I pif I is defined as the coefficientof Pii in the expansion of the determinant; it equals the minor obtained bystriking out the ith row and jth column with the sign (-1)'+i affixed.

§ 78 MATRIC ALGEBRA

becomes

(11) f(X) =

(P11 - X 'P12 'P13

'P21 'P22 - X 'P23

(P31 <P32 IP33 - X

= 0.

169

Example. If the dyadic 4) in (1) is symmetric, and r = xi + yj + k, then

(i) r $ r = Pllx2 + 2co12xy + P22y2 + 2w13x + 2-P23Y + V33 = 0

represents a conic section-an ellipse, parabola, or hyperbola, according asV12 - IPiisv2z < 1, = 1, or >1. Let this conic cut the sides BC, CA, AB ofa triangle ABC in the points R1, Ri; R2, R2; R3, R3, respectively, and letthe corresponding division ratios be Plr p l; P2r P2i Par P3'- If we put r =(b + Pc)/(1 + p) in (i) we find that P1, P1 are the roots of the quadraticequation

hence

b 4) b +2Pb 4) - C +P2c (D c = 0;

c'4) 'c a'4 aand

b

follow in the same way. On multiplying these equations, we have

(ii) P1P1P2P2P3P3 = 1

Now let R2'R3i R3R1, R'R2 meet BC, CA, AB in the points S1, S2, S3 whichdivide the respective sides in the ratios al, 02, 03. Then by the Theorem ofMenelaus (§ 7, ex. 2)

P2P3°1 = - 1r P3P1Q2 = - 1r P1P20'3 = - 1

On multiplying these equations together and making use of (ii), we have0.10.20.3 = -1; hence S1, S2,83 are collinear. We thus have proved Pascal'sTheorem: $ The opposite sides of a hexagon (R1R'1R2RZR3R3) inscribed in a conicmeet in three collinear points (Si, S2, S3)-

78. Matric Algebra. The sum of two dyadics A + B in nonionform obviously is obtained by adding corresponding elements oftheir matrices. As to the product C = A B, let us consider theformation of a certain dyad of C, say c12ij. This evidently resultsfrom the product of terms in A with antecedent i and terms in Bwith consequent j: thus

C12 = a11b12 + a12b22 + a13b32 = 2;alb,2-r

The general result is therefore

Cij = 2;airbrj ;r

$ This proof is due to Wedderburn, Am. Math. Monthly, vol. 52, 1945, p.383.


the element in the ith row and jth column of A B is the sum of theproducts of the elements in the ith row of A by the correspondingelements in the jth column of B-the "row-column rule."

The foregoing rules for the sum and product of dyadics in non-ion form are precisely the classic definitions for the sum and prod-uct of square matrices. These definitions may be extended to rec-tangular matrices with m rows and n columns (m X n matrices).

1. The sum A + B of two m X n matrices is them X n matrixobtained by adding their corresponding elements.

2. The product of an m X p matrix A and a p X n matrix B isthe m X n matrix C = AB, whose element in the ith row and jthcolumn is

(1)

Pcjj = E airbrj.

r=1

Note that only similar matrices can be added; whereas in aproduct the second matrix must have the same number of rowsas the first has columns.

This extension enables us to interpret scalar products of dyadicsand vectors in terms of matric algebra. We regard a vector uwith rectangular components ui either as 1 X 3 matrix (row vector)or a 3 X 1 matrix (column vector). Since a dyadic A in nonionform is a 3 X 3 matrix, u must be a row vector in u A, a columnvector in A u. With this proviso, the rules of matric algebragive values of vector components in full agreement with vectoralgebra:

(2) v = u A, vi = u1a1i + u2a2i + u3a3i;

(3) w = A u, wi = ai1u1 + ai2u2 + ai3u3

The matric product of a row vector u into a column vector v isa 1 X 1 matrix consisting of a single element, the scalar product:

(4) u v = u1v1 + u2V2 + u3v3

But the matric product vu of a column vector into a row vectoris a 3 X 3 matrix, namely the dyad,

(5)

v1u1 v1u2 v1u3

Vu = v2u1 v2u2 v2u3

v3u1 v3u2 V3u3

§ 79 DIFFERENTIATION OF DYADICS 171

in nonion form. Matric multiplication, although associative and dis-tributive with respect to addition, is in general not commutative. Theproofs follow readily from (1).

79. Differentiation of Dyadics. If a dyadic (F is a function of ascalar variable t, we define

d(=lim

(F(t + At) - (F(t)

dt it -o At

For a single dyad ab, let a and b become a + Aa, b + Ab when tbecomes t + At; then, since

(a + Aa) (b + Ab) - ab Ab Aa Aa= a-+-b+-Ab,At At At At

we have, on passing to the limit At - 0,

(1)d db da

- (ab) =adt + dt

b.

This formula suggests the usual product rule with the order of thefactors preserved.

The derivative of the dyadic (D = 2;aibi is evidently the sum ofthe derivatives of its dyads. If 4, is given in the nonion form(77.1))

'P12

(2)d(dt

'P22

'P32

where the primes denote derivatives with respect to t. Note thatthe derivative of <03, the determinant of the matrix ('i;), is notthe determinant of the matrix

The derivatives of products such as - r, fi x r, s (- r, whichconform to the distributive law, are computed just as in the cal-culus when the order of the factors is preserved. For example,

d d(F dr

(3) dtr+(D

dt;

d ds d(F dr(4)

dt dt dt dt


80. Triadics. A triadic is defined as a sum of triads, Eaib,ci.A triad abc consists of three vectors written in a definite order.We may regard a triadic as an operator which converts vectors rinto dyadics; thus

4) r = Eaibici r, r 4) = 2;r aibici.

If 4) and' are two triadics, we write - = T, when

(1) 4) r = 4, r for every vector r.

Then, for any vector s, s (D r = s 4, r, and, from the defini-tion of equality for dyadics (61.3),

(2) s 4) = s - ' for every vector s.

Conversely, from (2) we may deduce (1). Thus from either (1) or(2) we may conclude that 4) = T.

Using a given basis ei and its reciprocal ei to form basic dyads,we have seen in § 76 that the 32 = 9 components of a dyadic areof 22 = 4 types. For a triadic 4), the 33 = 27 components are of23 = 8 types; for, for each vector in the basic, triads may bechosen from the set ei or e' giving 23 types; and, for a given type,each index on the base vectors may be chosen in three ways, giving33 components.

Similarly, we define a tetradic as the sum of tetrads Maibicidi.Two tetradics 4), I are equal when either (1) or (2) holds good.A tetradic has 34 = 81 components of 24 = 16 types.

We shall speak of scalars, vectors, dyadics, triadics, collec-tively as tensors of valence 0, 1, 2, 3, The equality of twotensors of valence n, say 4) = 'Y, depends upon the equality oftwo tensors of valence n - 1, as required by equations (1) or (2).

81. Summary: Dyadic Algebra. A linear vector function f(r) ischaracterized by the properties,

f(a + b) = f(a) + f(b), f(Xa) = Xf(a).

A dyadic 4) = X.aibi is the sum of dyads aibi; the vectors ai areantecedents, bi are consequents. The conjugate of 4) is 4 = 2;biai.Any linear vector function may be expressed as 4 r (or rand, if [ala2a3] 0 and f (ai) = bi (i = 1, 2, 3),

2f(r) _ 4 r, where 4) = blal + b2a + b3a3.

81 SUMMARY: DYADIC ALGEBRA 173

Basic definitions (r an arbitrary vector) :

NP : 4)

(D r0;(+'1')r=

r = (I' r);

I (idemfactor) : I r = r;

-i (reciprocal) : c (I)-1 = 4'-1 4 = I;4) xv: Taibixv if 4) = Eaibi;

V x (D: My x aibi.

Every dyadic may be reduced to the sum of three dyads,

4) =al+bm+cn,in which either antecedents or consequents may be an arbitrarynon-coplanar set. In this form, the principal invariants of 4) are:

Dyadic: 4, = bxcmxn+cxanxl+axblxm =Vector: = axl+bxm+cxn,

*= (b x c) x (m x n) + cycl

Scalar: (P1 =

'P2 = `I'1 = (b x c) (m x n) + cycl,

(P3 = [abc][lmn],

4PO

(D2;

4) is complete if 'P3 0 0; only complete dyadics have reciprocals.4) is planar if <P3 = 0, 4)2 0 0; a planar dyadic may be reduced totwo dyads. (D is linear if '1)2 = 0, 4) 0 0; a linear dyadic may bereduced to one dyad. Planar and linear dyadics are called singular.

Fundamental identities :

(4, p), = *' 4)" ((D ,F)-1 = j-1 4)-1;

Ixv = vxl, (Ixv)c = -Ixv, I x(uxv) = vu -uv;(Da = (Da `I' = (P3I (the adjoint 4)a = (h,);

1431 - 'P24) + tP1 (D2 - 4)3 = 0 (Hamilton-Cayley Equation).


4) is symmetric if 4), = 4), antisymmetric if 4), = - 4). Everydyadic 4) can be expressed uniquely as the sum of a symmetricand antisymmetric dyadic; the latter is - 2I x .

The vector r1 is an invariant direction of - with multiplier Al if4) r1 = A1r1. The multipliers of 4) satisfy the cubic,

503 - <02X +P1A2 - A3 = 0 (characteristic equation).

The invariant direction rl makes (4) - X11) r1 = 0.If 4) is symmetric its multipliers Ai are all real; and, if Al A2i

rl 1 r2. A symmetric 4) always may be reduced to the form,

`1) = A1u + A2JJ + X3kk;

the multipliers Ai need not be distinct.Any dyadic 4) can be reduced to the normal form,

(P = ai'i + $j'j + yk'k,

in which a, 0, y all have the same sign; i, j, k and i', j', k' are twodextral sets of orthogonal unit vectors.

When - is complete, the point transformation, r' = 4) r,changes lines into lines and planes into planes, preserves paral-lelism, and alters all volumes in the ratio 'P3/1. It preserveslengths when and only when 4-1 = 4,; it is then a rotation if03 = 1, a rotation followed by a reflection if p3 = -1.

PROBLEMS

1. Prove thatabxc +bcxa +caxb = [abc)I.

2. For any dyadic 4> show that

U.4> v - v 4> u =4. uxv.3. If 4> has the characteristic numbers X1, X2, X3 and the corresponding in-

variant directions r1i r2, r3, prove that 4)" (n an integer) has the characteristicnumbers X', 02, 03 corresponding to the salve invariant directions.

4. If 4> has the characteristic numbers X1, X2, 1\3 for the directions r1r r2, r3,prove that 4>2 has the characteristic numbers X2X3, X3X1, X1X2 corresponding tothe directions r2 x r3i r3 x r1, r1 x r2. [Cf. (70.2).]

5. If 4, = 4,, prove that *2 = 4>2,,, + = -4 and h = cir >k2 = v2, 4,3 = v36. Prove that the scalar invariants of 4' = 4>2 are

01 = 'c1 - 2'P2, 2 = 'P2 - 2'vl'P3, '3 = 'P3

[Use Prob. 3 and (71.4).]

PROBLEMS 17 5

7. Given the invariants c2, +, , , v3 of 4), find the corresponding invar-iants for

(a) k4), (b) P x u, (c) 4-1.

8. Compute (P2, +, and the six scalar invariants of (69.13), namely,

'Pl, V2, 4'3; + ' +, + ' 4) ' +, + ' p2 - +,

for the dyadic1 3 -2

= 2 0 4

-1 2 3

9. If D = I x e and e is a unit vector, prove that4)2 = - (I - ee), 4'3 = -4), D4 = I - ee, 4)5 = (P.

10. Show that the symmetric part of (P, namely, 4' = 1(4) +4,), has thescalar invariants

11 1h = c1, 42 = V2 - 4+' 413 = V3 4

11. Prove that ('t - ')2 = `1'2 *2-12. Prove that the first three scalar invariants of 4 and 4 4) are the

same.m n.

13. Ifs a;b,, 4' c;d1,1=1 J=1

we define the double-dot product 'F:4' as the scalarm n

E(a`.c1)(b`-d,).1-I j_1Prove that

(a) ++4':(b) (uv) :4' = u 4) v, 4 ) :1

(c) If 4) is given in the nonion form (77.1),

'P11 + V12 +'P13 +'P2I + P22 + V23 +'P21 +'P32 +P2

Hence 4, = 0 when and only when 4):4' = 0.14. Any central quadric surface with its center at the origin has an equa-

tion of the form,

(1) r- 'F r = 1,

where 4, is a symmetric dyadic; for, if 4) is reduced to the standard form(72.4), we have

axe + $y2 + yz2 = 1.

This represents an ellipsoid, an hyperboloid of one sheet, or an hyperboloidof two sheets, according as a, j3. y include no, one, or two negative constants.


If p 5*1 0, the equation (1) associates with every point (1, p) the plane(4' p, 1), its polar plane. Prove that:

(a) If the point (1, p) lies on the quadric surface, its polar plane (4' p, 1)is tangent to the quadric at (1, p). [Find the points where the line r = p + Xecuts the quadric when e 4' p = 0.]

(b) If the polar plane of (1, p) passes through (1, q), the polar plane of(1, q) passes through (1, p).

15. The diametral plane of any point (1, p) on the quadric surface (1) isthe locus of the mid-points of all chords parallel to p. Show that the diam-etral plane of (1, p) is (4). p, 0).

16. If three points (1, u), (1, v), (1, w) on the ellipsoid r 4) r = 1, satisfythe equations,

U.

the position vectors u, v, w are said to form a conjugate set. Show that:(a) The vectors u, v, w and 4' - u, 4) v, 4) w form reciprocal sets.(b) 4'-1 = uu + vv + ww.(c) For any conjugate set u, v, w, of the ellipsoid r (D r = 1, the sum

u u + v- v + w- w and the product u x v- w are constant.17. Verify by direct computation that the dyadic 4' in Problem 8 satisfies

its Hamilton-Cayley Equation.18. A rigid body with one point 0 fixed has the inertia dyadic K relative

to 0 (§ 72, ex. 2). If the angular velocity of the body at any instant is w,

show that its moment of momentum H (defined as f r x v dm) and kinetic

energy T (defined as i f v v dm) are given by

19. If the forces acting on the body of Problem 18 have the moment sumM about 0, the equation of motion is dH/dt = M. Show, from (56.3), that

dH dw

dt=K

dt+wxK w.

Let K = Aii + Bjj + Ckk (72.10) when referred to the principal axes ofinertia (fixed in the body). Then if w = [WI, w2i w3], M = [Ml, M2, M3] re-ferred to these axes, deduce Euler's Equations of Motion:

Awl - (B - C)w2w3 = M1,

Bm2 - (C - A )w3w1 = M2,

(A - B)w1w2 = M3-

20. In Problem 19 show that d7'/dt = M w (the energy equation).21. In Problem 18, suppose that the only forces acting on the body are its

weight W and the reaction R at the support 0; then if the center of mass is at 0,W passes through 0 and M = 0. Prove in turn that

(a) H is a constant vector.(b) T is a constant scalar.

PROBLEMS 17 7

(C) If w = OP, the locus of P in space is the invariable plane w H = 2T;:uul the locus of P in the body is the energy ellipsoid w K w = 27'.

(d) The energy ellipsoid is always tangent to the invariable plane at P.(c) The body moves so that its energy ellipsoid rolls without slipping on

the invariable plane (Poinsot's Theorem). [See Brand's Vectorial Mechanics,§ 219.]

22. The vectors of a dyadic 'F are fixed in a rigid body having the instan-taneous angular velocity w, relative to a "fixed" frame. Show that, relativeto this frame,

d'F/dt=ca x4) -4) xw.

23. A rigid body revolving about its fixed point 0 has the angular velocityw = [wl, W2, w3] referred to fixed rectangular axes through O. If I1, 12, I3 arethe moments of inertia about the fixed axes x, y, z and 123, I31, 112 are theproducts of inertia for yz, zx, xy, show that

dIl/dt = 2(I12w3 - 113-2), .. ,

dI23/dt = (13 - 12)wl + 113W3 - 112-2, ... ,

[If K is the inertia dyadic of the body relative to 0, Il = i K i, 123 =-j K K. k, etc. See § 72, ex. 2.]

24. The axis of a homogeneous solid of revolution has the direction of theunit vector e. If its principal moments of inertia at the mass center G areA, A, C, show that the inertia dyadic at G is

KG =AI+(C-A)ee.Hence find the moments and products of inertia with respect to fixed rectan-gular axes x, y, z through G if e = [1, in, n].

25. If G is the mass center of a rigid body of mass in and r* = OG, showthat the inertia dyadic at 0 is

Ko = m(r* r* I - r*r*) + KG.

Hence compare moments and products of inertia for parallel axes at 0 and G.26. If X is arbitrary parameter < a2 but b2 or c2, the dyadic,

x211 +b2

J1 {c2kk

(a > b > c),

defines a one-parameter family of confocal quadric surfaces r % r = 1.These are ellipsoids if X < c2, hyperboloids of one sheet if c2 < \ < b2, hyper-bolas of two sheets if b2 < X < a2.

Prove that(a) The central quadrics r T r = 1 and r O r = 1 are confocal when

and only when I-' - 0-1 = kI.(b) Two confocal central quadrics of different species intersect at right

angles. [O - 4, = ke *.]

CHAPTER V

DIFFERENTIAL INVARIANTS

82. Gradient of a Scalar. The points P of a certain region may

be specified by giving their position vectors r = OP; and we shallon occasion refer to P as the "point r." A scalar, vector, or dyadicwhich is uniquely defined at every point P of a certain region iscalled a point function in this region and will be denoted by f (r).For example, the temperature and velocity of a fluid at the pointsof a three-dimensional region are scalar and vector point functionsrespectively.

A scalar point function f (r) is said to be continuous at a pointP1 if to each positive number e, arbitrarily small, there correspondsa positive number S such that

If(r) - f(r1) <e when jr - r1 l <6;

then, as r approaches r1 in any manner, limf(r) = f(r1).From a point P1 draw a ray in the direction of the unit vector,

e =icosa+jcos/3+kcosy.Along this ray r = r1 + se, where s denotes the distance P1P, andf (r) is a function of s. We now define

(1) of = limf (r, + se) - f(r1)

ds s -.o s

as the directional derivative of f (r) at P, in the direction e. If thislimit exists on all rays issuing from P1i f (r) is said to be diferen-tiable at P1.

The rectangular coordinates of any point P on the ray r =r1 -}- se (s > 0) are

x = x1 + s cos a, y = y1 -}- s cos z = z1 -{- s cos y.178

82 GRADIENT OF A SCALAR

If f(r) is given as a function f(x, y, z),

df of dx of dy of dz(2) -- _ - - --F - - + -- --

ds ax ds ay ds az ds

= a fcos a +

ofcos i3 +

ofcos y;

ax ay az

or, since

(3)

179

e i = cos a, e j = cos a, e - k = cosy,

A=

e\1 ax+iaf +kaz/y

The vector in parenthesis is called the gradient of f (r) and is writ-ten grad f or

49 19

(4) Vfax+jay+kc3z

Vf is a vector point function; thus, when (3) is written as

(5)

df

ds

the direction enters only through the factor e. The directionalderivative of a scalar function at a point P is the component of itsgradient at P in the given direction. The gradient Vf at P effects asynthesis of all the directional derivatives of f at P. In effect, thevector of replaces the infinity of scalars df/ds.

When P varies along a curve tangent to e at P1, f(r) is a func-tion of the are s = P1P along the curve, and df/ds is still givenby (2) ; for, at P1, dr/ds = e (44.1), and hence

dx/ds = cos a, dy/ds = cos a, dz/ds = cos y.

Since df/ds, as defined by (1), does not depend upon any specificchoice of coordinates, (5) shows that the gradient Vf has the sameproperty. In fact, we determine Vf by giving the directional de-rivatives df /dsi in three non-coplanar directions e1. For, sinceei Of = df/dsi are the covariant components of Vf, we have,from (24.2),

(6) of = eldf

+ e2 df + e3 df .

ds1 ds:2 ds3

180 DIFFERENTIAL INVARIANTS § 82

We proceed to specify the length and direction of Vf independ-ently of the coordinate system. The points for which f has a con-stant value lie on a level surface of f. In any direction e tangentto the level surface at P, df/ds = e - Vf = 0; hence Vf is normalto the level surface at P. If n is a unit vector normal to the levelsurface and directed towards increasing values of f, n Vf > 0.Hence Vf has the direction of n, and its magnitude is the value ofdf,/ds in this direction. Writing this normal derivative df/dn, wehave

(7)

For example, if f = r, the distance OP from the origin, the levelsurfaces are spheres about 0 as center, and n = R, the unit radialvector; the normal derivative dr/dn = 1, and

(8) Vr=R.

When f is a function f (x, y, z) of rectangular coordinates, Vf isgiven by (4). In particular,

(9) Vx=i, Vy = j, Vz = k.

If f is a function f(u, v, w) of variables which themselves are func-tions of x, y, z, we have

d f of du of dv of dw

ds au ds av ds aw ds

or, in view of (5),

Vu Of+Vvaf +Vwa-f1\ au av aw

As this equation holds for every e,

.(10) Vf = Vu

of+ Vv a f + Vw Of

Ou av aw

When f = f (u, v) or f = f (u),

Vf = Vu of + Vv Of, Vf = Du

Of,

au av au

§ 83 C1t.ADIENT OF A VECTOR

respectively; for example,

181

1V(uv) = v Vu + it Vv, Vu' = nu"-1 Vu, V log u = - Vu.

It

When f is constant, Vf = 0; conversely, Vf = 0 implies of/ax =of/ay = of/az = 0, and f is constant.

If f (r) is differentiable in a region R, Vf is defined at all pointsof R. If moreover Vf is continuous in R, we say that f (r) is con-tinuously differentiable in R.

Example. Gradients in a Plane. The gradient of a point function f(x, y)in the xy-plane is

vf=afi+Of j.ax ay

At any point P, Vf is normal to the level curve f(x, y) = c through P.For a function f(r, 0) of plane polar coordinates,

=of

R

of-

orCf +00r'in the notation of § 44; for Vr = R and V0 = P/r (from (44.5) dB/ds = 1/r inthe direction perpendicular to R).

For a function f(rl, r2) of bipolar coordinates,

Vf =OrlR1+_R2,

where R1, R2 are unit radial vectors from 01, 02 directed to the point in ques-tion.

The families of curves u = c, v = c cut at right angles when Cu Cv = 0.For example, the ellipses rl + r2 = c and hyperbolas r, - r2 = c cut orthogo-nally since (R1 + R2) (R1 - R2) = 0-

83. Gradient of a Vector. Let f(r) denote a vector point func-tion in a certain region. It is said to be continuous at a point P1if to each e > 0 there corresponds a S > 0 such that

I f (r) - f(r1) I < e when I r - r1 I< S.

If f(r) is given by its rectangular components,

f(r) = f1i + f2j + f3k,

f(r) is continuous at P1 when the three scalar point functionsfi, f2, f3 are continuous there.


Let P1P be a ray drawn from P1 in the direction of the unitvector,

e = icosa+jcos$+kcosy.Along the ray r = r1 + so (s > 0) and f(r) is a function of s. Wenow define

df f(rl + se) - f(r1)-- = limds s-.o s

as the directional derivative of f (r) at P1 in the direction e. If thislimit exists on all rays issuing from P1, f(r) is said to be differen-tiable at Pl. Evidently f (r) is differentiable if its rectangular com-ponents f, are differentiable.

If f(r) is given as a function f(x, y, z) of rectangular coordinates,we have. just as in § 82,

df of dx of dy of dz(2)

ds axds+ayds+azdsof of of

= - cosa + cosa+ - cosy.ay

On replacing the cosines by e i, e j, e k,

df of of of

(3) ds = e1-+ja +kaz ,y

a formula entirely analogous to (82.3). The dyadic in parenthesisis called the gradient of f(r) and is written grad f or

of of of(4) Vf = i-+ j - +k --

ax ay az

Vf is a dyadic point function; thus, when (3) is written

df(5)

the direction enters only through the prefactor e. The gradientVf at P effects a synthesis of all the directional derivatives of fat P. In effect, the dyadic Vf replaces the infinity of vectors df/ds.

Since df/ds, as defined by (1), does not depend upon any specificchoice of coordinates, (5) shows that Vf has the same property.

§ 84 DIVERGENCE AND ROTATION

In fact Vf is determined by giving the directional derivativesdf/dsi in three non-coplanar directions ei:

(6) of _ e'df + e2 df + e3 df

ds1 ds2 ds3

For, if we put df/dsi = ei Vf, the right member becomes

(e'ei + e2e2 + e3e3) Vf = I I. Vf = Vf.

When f is function of rectangular coordinates f (x, y, z), Vf isgiven by (4). More generally, for f(u, v, w) we have

df of du of dv of dw

ds au ds av ds aw d s

C of of of+ VV

au av aw

for any e; and, from the definition of dyadic equality,

(7)of of ofVf = Vu-+Vv--+Vw-au av aw

84. Divergence and Rotation. The first scalar invariant of thegradient Vf of a vector f is called the divergence of f and is written

The vector invariant of Vf is called the rotation or curl of f andis written V x f, rot f, or curl f.

In terms of rectangular coordinates, we therefore have the de-fining equations:

of of of(1) Vf=gradf=i - +j - +k - ,

ax ay az

(2) V f = divfof of of= i - + j

3) xf = rotf

,ax ay az

of of of= ix-+ jx-+kx__.ax ay az

If f is resolved into rectangular components,

f=ifl+jf2+kf3,


we obtain from (1) its gradient of in nonion form with the matrix:

afl/ax af2/ax af3/ax

(4) of = afl/ay af2/ay af31ay

8fl/az af2/az af3/az

The first scalar and vector invariants of of now become

(5)v

f =axl+ a19h +

az,

Gayf3 af2 afl af3 af2 afl(6) vxf=i --- -f-j --- -}-k -- /az (az ax (ax ay)

The last expression is easily remembered when written in determi-nant form:

(7) Vxf =

j k

a a a

ax ay az

fl f2 f3

These expressions for divergence and rotation may be written downat once, if we regard V - f and V x f as products of the vectoroperator del (or nabla),

(8)

a a aV = i-+j-+k--,ax ay az

with the vector f = if, + jf2 + kf3.For the position vector r = xi + yj + zk, we have

Or Or Oryr = i-+ j - +k -- = ii+ jj +kk,ax ay az

the idemfactor; hence

(9) Vr = I, div r = 3, rot r = 0.

When f = the gradient of a scalar,

(9 (P a4'fl=ax, f2=a , f3=az-

y

§ 84 DIVERGENCE AND ROTATION

Using these components, (5) and (6) gives

10v

.

va2

(P a2V

++a2,p

( )'P

ax2 ay 2 az2

(11) V X VV = 0.

185

The differential operator V V or div grad is called the Laplacianand often is written

(12)__ a2 a2 a2

v2y2ax2 + a +az2

When f = V x g, the rotation of a vector g,

C193 092 _a91 a93 a92 C191

Aay az f2 az ax ' f3 ax ay

In this case (5) and (6) give

(13) 0,

(14) Vx(Vxg) _ V(V.g) - O29.

Proof of (14) : From (6),

r a292 - x291 - a291 a293

ay ax aye az2 + az ax Jt

+ .. .

= i j a tag, + a92 + a931 - v291 I + .. .lax ax ay az J

a=iax

(O. g) - V291 +...

= V(V . g) - V2g.

Note that, if we regard V as an actual vector and expandV x (V x g), but keep the V's to the left of g, we obtain V(V g) -(V V)g, the correct result.

The proofs of (11), (13), and (14) depend upon changing theorder of differentiations in mixed second derivatives; this is alwaysvalid when the derivatives in question are continuous.


The foregoing differential relations of the second order also maybe written

(10) div grad p = V2<p,

(11) rot grad cp = 0,

(13) div rot g = 0,

(14) rot rot g = grad div g - V2g.

Example. The velocity distribution in a rigid body is given by (54.4):

vP = vo -}- w x OP = vo + w x r.Hence, from (3),

rotvp = rot(wxr) = ix(wxi) + =3w- w =2w;the rotation of the velocity of the particles of a rigid body at any instant is equalto twice its instantaneous angular velocity.

85. Differentiation Formulas. Starting from the defining equa-tions (84.1), (84.2), (84.3), we now can deduce some useful iden-tities.

If X is a scalar point function:

(1) V(Xf) = (OX)f + XOf,

(2) div (hf) = (VX) f + X div f,

(3) rot (Xf) = (VX) x f + X rot f.

For the vector point function f x g:

(4) V (f x g) = (of) x g - (Vg) x f,

(5) div (f x g) = g rot f - f rot g,

(6) rot (f x g) = g Of - f Vg + f div g - g div f.

We give the proof of (6) :

Caf ag/rot(fxg) =1x axxg+ fxax

+...

g=g i -- (i -g+(i axg f -f.iax

+.ax \ ax! \

..

= g - Vf - (div f)g + (div g)f - f . Vg.

§ 86 GRADIENT OF A TENSOR

For the scalar point function f g:

(7) 0(f - g) _ (of) g+ (7g) . fFrom (68.6), D, = + I x hence

(8) (of), = Vf + I x rot f,

(9) (Vf) - a = a - Vf + a x rot f.

Making use of (9), we also can write (7) as

(10)

If '1 is a constant dyadic,

(11) v(r -1)) =I.4) =4);(12) div (r D) = cpl,

(13) rot (r 4) _+.Finally, if 4) is a constant symmetric dyadic,

(14) r.

From this result we can prove the

187

THEOREM. A symmetric dyadic I transforms any vector r = OPinto a vector r' = 4) r normal to the real central quadric surfacer 4) r = f1 * at the point P; and r' = 1/p where p is the distancefrom 0 to the tangent plane to the quadric at P.

Proof. From (72.4), we see that

ax2+(3y2+yz2 = f1represents a central quadric (an ellipsoid if a, y have the samesign). From (14), r' has the direction of n, a unit normal to thequadric at P; hence

r'n=r' = 1/I 1/p.

86. Gradient of a Tensor. If f(r) is a tensor point function ofvalence v (cf. § 80), the derivative of f(r) at Pl in the directionof the unit vector e is defined by

df f(rl + se) - f(rl)(1) -- = lim

ds s -o s

* The sign is chosen so that the quadric is real.


If this limit exists for all rays drawn from P1, f(r) is said to bedifferentiable at P1. When referred to a constant basis, f(r) is dif-ferentiable when all its scalar components are differentiable.

If f(r) is given as a function f(x, y, z) of rectangular coordinates,

df of dx of dy of dz

ds Ox ds+ay ds+azdsSince dx/ds = e - i, etc., we have, just as in § 83,

df(2)

ds= e vf,

of of of of(3) Vf=i-+j-+k-=i,.- t

ax ay az axr

Here Vf, the gradient of f (grad f) is tensor point function of va-lence v + 1; of effects a synthesis of all the directional derivatives(valence v) of the tensor f. Thus, if f is dyadic, the triadic ofreplaces the infinity of dyadics df/ds.

of is independent of the choice of coordinates. At any point, ofis completely determined when df/dsi is given for three non-coplanar directions ei:

df df df(4) of = e1 - + e2 - + e3 -

ds1 ds2 dsi '

for, if we put df/dsi = ej Vf, the right member becomes I of= Vf.

When f(r) is a vector, we obtain from the dyadic of the invari-ants div f and rot f by putting dots and crosses, respectively, be-tween the vectors of each term. Similarly, if f(r) is any tensor ofvalence v > 0, we obtain from of the further invariants,

(5)of=i - +j of- +k of of

ax

of

ay

of

az axr

of Of6

kV f i

i %

)(x

x

+ j x +x =

_ t ,

ax ay Oz axr

of valence v - 1 and v, respectively. In forming these invariantsfrom Vf, the dots and crosses are inserted between the first andsecond vectors to the left-we dot and cross in the first position.

t Here i, j, k; x, y, z are written il, i2, i3; xl, x2, x3, and we employ the sum-nmation convention: a repeated index (as r) denotes summation over the indexrange 1, 2, 3. See § 145.

86 GRADIENT OF A TENSOR 1S9

For example consider the dyadic

f(r) = fiiii + fi2ij + ... = fstisitwhere s and t are summation indices. Then if D, denotes a/8x

Vf = i,D,f = D,fst i,isit, a triadic;

V f = i, D,f = Dfst i, is it

= (Difit + D2f2t + D3f3t)it, a vector;

here i, is = 6,s, the Kronecker delta (23.2) ;

Vxf = i,>D,f = Drfsti,xisit= (D2f3t - D3f2t)iiit + (D3fit - Dif3t)i2it

+ (Dlf2t - D2f3t)i3it, a dyadic.

Again, if X(r) is a variable scalar and 4) a constant dyadic,

(7) V(X4)) = (VX)4), a triadic;

(8) V V. (a4)) = (VX) 4), a vector;

(9) Ox (A4)) = (VX) x 4), a dyadic.

For any tensor f(r) with continuous second derivatives we haveidentities which are generalizations of (10), (11), (13), (14) of § 84:

(10) V Vf = V2f,

(11) V X Vf = 0;

and, if f(r) is not a scalar,

(12)

V x x f) _ V(V f) - V2f.The proofs are straightforward applications of (3), (5) and (6):

V Vf = i, D,(isDsf) = i, is DrDsf = DrDrf = V2f;

V x Vf = it x D,(isDsf) = i, x is DrDsf = 0;

V (V x f) = it Dr(ls x Dsf) = i, x is DrDsf = 0;

V x (V x f) = i, x Dr(is x Dsf) = i, x (is x DrDsf)

= is i, DrDsf - i, is DrDsf

= i8D5(i, Df) - D,D,f= V(V f) - V2f.

Since i, x is = -is x i,, DrDsf = DsD,f, all non-zero terms inV x of and V (V x f) cancel in pairs.


87. Functional Dependence.

THEOREM 1. A necessary and sufficient condition that two con-tinuously differentiable functions u(x, y), v(x, y) satisfy identically afunctional relation f(u, v) = 0 is that their Jacobian vanish:

a(u, v) au/ax au/ay(1) = I

I= 0 or Vu x Vv = 0.

a(x, y) av/a. av/ay

Proof. The condition is necessary ; for, from f(u, v) = 0, wehave

of of of-Vu+--Vv = 0, Vu. Vv = 0.au av av

If it is constant, Vu = 0, and Vu x Vv = 0. If u is not constant,f (u, v) must contain v, of/ft is not identically zero, and againVuxVV=0.

Conversely, suppose that Vu x Vv = 0. This relation is satisfiedif either u or v is constant; thus, if u = c, we may take f (u, v) =it - c. If u and v are not constant, Vv is parallel to Vu; hence Vvis normal to the curves u = c. Along these curves, dv,/ds = 0, andv is constant. In other words, a level curve it = a is also a levelcurve v = b; when u is given, v is determined, and v is a functionof it.

THEOREM 2. A necessary and sufficient condition that two con-tinuously differentiable functions u(x, y, z), v(x, y, z) satisfy identi-cally a functional relation f (u, v) = 0 is that

i j k

(2) Vu x Vv = au/ax au/ay au./az = a.

av/ax av/ay av/az

The proof is essentially the same as in theorem 1. Instead oflevel curves of u and v, we now have level surfaces.

THEOREM 3. A necessary and sufficient condition that three con-tinuously differentiable functions u(x, y, z), v(x, y, z), w(x, y, z) sat-isfy identically a functional relation f (u, v, w) = 0, is that theirJacobian vanish:

a(au/ax au/ay au/az

u, v, w)(3) = av/ax av/ay av/az = Vu x Vv Vw = 0.

a(x, y, z)Ow/ax aw/ay aw/az

88 CURVILINEAR COORDINATES 191

Proof. The condition is necessary for, from f (u, v, w) = 0, wehave

of-Vu+ -ofVv+ -of Vw = O,

of0.

au av aw aw

If u and v alone satisfy a relation f (u, v) = 0, Vu x Vv = 0, andalso Vu x Vv Vw = 0. If this is not the case, f (u, v, w) must con-tain w, of/aw is not identically zero, and again Vu x Vv Vw = 0.

Conversely, suppose that Vu x Vv Vw = 0. If Vu x Vv = 0,f(u, v) = 0 from theorem 2. If Vu x Vv F- 0, consider the curve ofintersection of the level surfaces u = a, v = b. Vu and Ov arenormal to this curve; and, since Vu, Vv, Vw are coplanar, Vw isalso normal to the curve. Therefore dw/ds = 0 and w = c alongthe curve u = a, v = b; in other words, when u and v are given,w is determined: w is a function of u and v.

88. Curvilinear Coordinates. In a given region let

(1) u = u(x, y, z), v = v(x, y, z), w = w(x, y, z)

be three continuously differentiable functions whose JacobianVu x Vv Vw F4- 0 at all points. The functions, therefore, are notconnected by a relation f (u, v, w) = 0. Since the Jacobian is con-tinuous, its sign cannot change in the region; and, to be explicit,we shall suppose that

(2)

This involves no loss in generality; for, if the Jacobian were nega-tive, an interchange of v and w (for example) would make itpositive.

Under the foregoing hypotheses a well-known theorem $ statesthat in the neighborhood of any point (x0, yo, zo) the equations(1) have a unique inverse

(3) x = x(u, v, w), y = y(u, v, w), z = z(u, v, w);

and that these functions are also continuously differentiable. Atleast in a suitably restricted region of uvw space, each set of valuesit, v, w yields, through equations (3), a unique set x, y, z. In thisregion the correspondence (x, y, z) - (u, v, w) effected by (1) and

$ Cf. Sokolnikoff, Advanced Calculus, New York, 1939, p. 434.

192 DIFFERENTIAL INVARIANTS 488

(3) is one-to-one. Instead of specifying a point Po by the coordi-nates (xo, yo, zo), we may use instead the three numbers,

uo = u(xo, yo, zo), v0 = v(xo, yo, zo), wo = w(xo, yo, zo)

When (uo, vo, wo) are given, Po is located at the point of inter-section of the three coordinate surfaces,

u(x, y, z) = uo, v(x, y, z) = vo, w(x, y, z) = wo.

These, in general, will intersect in three curves, the coordinatecurves, along which only one of the quantities u, v, w can vary.For this reason u, v, w are called curvilinear coordinates, in dis-tinction to the rectangular coordinates x, y, z, for which the co-ordinate curves are straight lines.

If the position vector r is regarded as a function of x, y, z,Vr = I (84.9). But, if r is regarded as a function of u, v, w,

ar ar carVr = Vu-+ Vv-+ Vw-,au av aw,

from (83.7). We therefore have the fundamental relation,

(4) Vur,,

on writing ru = ar/au, etc. From the theorem of § 66, we nowconclude that:

The vector triples Vu, Vv, Vw and ru, r, rw form reciprocal sets.The vectors ru are tangent to the u curves, the coordinate curves

along which v and w are constant. Thus at any point P(u, v, w),r, r, rw are tangent to the three coordinate curves meeting there;and, from (23.6),

(5) Vv Vw] = 7

Here [rurrw] is the Jacobian a(x, y, z)/a(u, v, w) of the inversetransformation (3); and, from (2),

(6) J = 0.

Moreover, from the properties of reciprocal sets,

x x x

(7) Vu = rvJrw,

Vv = rJ ru,Vw =

ru

J r = J Vw J x Vv.

§ 88 CURVILINEAR COORDINATES 193

The volume of the parallelepiped whose edges are ru du, r dv,rw dw is called the element of volume dV; thus

(9) dV = [rurvrw] du dv dw = J du dv dw.

To find the gradient of a tensor f (u, v, w), we first compute

df of du of dv of dw(10)

+ +ds au ds av ds aw ds

= e (Vufu + Cvfv+ Vwfw).

But, from (86.2), df/ds = e Vf for every e; hence

(11) of = Vu fu + Vv fv + Vw fw.

Thus, in curvilinear coordinates the operator del becomes

a a a(12) V = Vu-+ Vv -- + Vw-

au ov aw

From (11), we obtain V f and V x f by dotting and crossing:

(13)

(14) Vxf = Vuxfu+ Cvxfv+ VwXfw.

For purposes of computation, it is usually more convenient toeliminate Vu, Vv, Vw from these formulas by use of equations (7).We thus obtain

1

J{rvxrwfu + rwxrufv + ruxrvfw} ,

and corresponding equations for V f and V x f. In view of theidentity,

(rv x rw)u + (rw x ru) v + (ru x rv) w = 0,

these also may be written

1(16) Of = J { (rv x r f) u + (rw x ru f) v + (ru x rv f)w },

1

(17) =J

(18) V x f = J 1 [(rv x rw) X flu + [(rw x ru) x f]v + [(ru x rv) x f]w }.


When the triple products in (18) are expanded, the brace becomes

(rw r71 f).u - (r rw f)u. + (r.u rw f) - (rw ru f)v

+ (rv ru f) w - (ru r f)w

= rw(rv f)u - rv(rw f), + ru(rw f) - rw(ru f) 1,

+ f)w - ru(rv f)w.

With this value for the brace, (18) may be written compactly indeterminant form:

1

ru

a

ra

rw

a(19) Vxf J au av aw

This equation reduces to (84.7) when ru, r,,, rw are replaced byrx = i, rv = j, rz = k.

89. Orthogonal Coordinates. The curvilinear coordinatesu, v, w are said to be orthogonal if the coordinate curves (alongwhich one coordinate only can vary) cut at right angles. Sinceru, r, r.v, are tangent to the coordinate curves, the coordinates areorthogonal when and only when

(1)

We choose the notation so that

(2) ru = Ua, r = Vb, rw = Wc,

where a, b, c are a dextral set of orthogonal unit vectors andU, V, W are all positive; then [abc] = 1, and

(3) J = [rurvrv,] = UVW.

The set of vectors reciprocal to ru, rv, rw are the gradients of thecoordinates:

a b c(4) Vu =

U, Vv =

V, Vw =

W.

From the general formulas (11), (17), (19) of the precedingarticle we obtain Vf, V f and V x f in orthogonal curvilinear co-ordinates:

§ 89 ORTHOGONAL COORDINATES 195

1 1 1(5) Vf =

Uafu +

Vbf +

Wcf

1 i18 a(6) V f =UVW

Lou(VW a f) -F a (WU b f)

+_

(7) Vxf =1

UVW

Ua Vb We

a a 8

au av aw

If we put f = Vg in (6), we obtain the Laplacian V2g = V. Vgin orthogonal coordinates,

1 I/V\) a1/

W U \1 a1/

UV \I(8) o2g

UVTV

laau \ U gu + av \ V gv/ + aw \ W gw/ 1

When the curvilinear coordinates are given by the equations,

x = x(u, v, w), y = y(zt, v, w), z = z(u, v, w),

we may compute the (positive) functions U, V, TV from equationsof the type,

(9) U = I ru _ xui + yuJ + zuk = 1/x2z z2u + yu + uThe element of volume (88.9) is now

(10) dV = UVW du dv dw.

Example 1. Cylindrical Coordinates. The point P(x, y, z) projects into thepoint Q(x, y, 0) in the xy-plane. If p,p are polar coordinates of Q in the xy-plane, u = p, v = gyp, uw = z are called the cylindrical coordinates of P (Fig.89a). They are related to rectangular coordinates by the equations:

(10) x = p cos (p, y = p sin gyp, z = Z.

From r = ix + j y + kz, we have

rp = [cos v, sin gyp, 0],

r, = [ -p sin rp, p cos gyp, 0],

r. = [0, 0, 11.

Since these vectors are mutually perpendicular and [rrr=] > 0, cylindrical


coordinates form an orthogonal system which is dextral in the order P, c, Z.Moreover,

(11) U = rn = 1, V = rw I = p, W

(12) J = UVW = P.

The level surfaces p = a, v = b, z = c are cylinders about the z-axis, planesthrough the z-axis, and planes perpendicular to the z-axis. The coordinate

FIG. 89a

curves for p are rays perpendicular to the z-axis; for p, horizontal circles cen-tered on the z-axis; for z, lines parallel to the z-axis. The element of volumedV =Pdpdpdz.

From (8), the Laplacian is

(13) V2g =1- 8

- (Pgp) +1

P aP Pa

g = log P, Pgv = 1; hence log p satisfies Laplace's Equation V2g = 0. Sucha function is called harmonic.

Example 2. Spherical Coordinates. The spherical coordinates of a pointP(x, y, z) are its distance r = OP from the origin, the angle 0 between OPand the z-axis, and the dihedral angle ' between the xz-plane and the planez OP (Fig. 89b). They are related to rectangular coordinates by the equations:

(14) x = r sin 0 cos p, y = r sin 0 sin gyp, z = r cos 0.

From r = ix + jy + its, we haverr = [sin 8 cos p, sin B sin gyp, cos 0],

rs = [r cosBcosgyp,rcos©singyp, -rsin0],

r, = rsin0cos p, 0].

90 TOTAL DIFFERENTIAL 197

since these vectors are mutually perpendicular and [rrror,] > 0, spherical co-ordinates form an orthogonal system which is dextral in the order r, 0,Moreover,

(15) U=jrr1, V=Irol=r, Iiir,rsin o;(16) J = UVIV = r2 sin 0.

The level surfaces r = a, 0 = b, = c are spheres about 0, cones about thez-axis with vertex at 0, and planes through the z-axis. The coordinate curves

Fia. 89b

for r are rays from the origin; for B, vertical circles centered at the origin;for gyp, horizontal circles centered on the z-axis. The element of volume dV= 7-2sin0drdodrp.

From (8), we now have

isin(17) V2g 0 (r2gr) + (sin 0 go) +1

9r2 sin 0 or aB sin 0 0`°

If g = 1 /r, r2gr = -1; hence 1 /r satisfies Laplace's Equation V2g = 0.

O. Total Differential. In passing from the point P to P, the---4 --->

position vector r = OP changes by the increment Ar = PP.Then, if f(r) is any differentiable tensor point function, the totaldifferential of f (r) is defined by the equation :

(1)

In particular, when f = r, Vf = I, and

(2) dr = Or.

The differential of the position vector is the same as the incrementThe defining equation (1) therefore may be written

(3) df=drVf.

198 DIFFERENTIAL INVARIANTS

If f is a function f (u, v, w) of curvilinear coordinates,

Vf = Vufu+VvfV+Vwfw,

and, from (3), since dr Vu = du, etc.,

(4) df = fu du + f dv + fw dw.

§91

91. Irrotational Vectors. A vector function f (r) is said to beirrotational in a region R when rot f = 0 in R.

If V(r) is a scalar function with continuous second derivatives,its gradient VV is irrotational; for, from (84.10),

(1) rot grad ,p = 0.

Conversely, if rot f = 0, and f is continuously differentiable in R,we shall show that f may be expressed as the gradient of a scalarV(r). Using rectangular coordinates, let f = f1i + f2j + f3k; then,if rot f = 0, the three determinants of the matrix,

(2)

(fl,

aax a/ay a/az\

f2 f3

all vanish (84.7). Under this condition we shall determine ascalar function ,p(r), so that Vv = f, that is,

(3)ax

= f1(x, y, z),ay

= f2(x, y, z),az

= f3 (x, y, z)

Let (xo, yo, zo) be an arbitrary point of R. On integratingapp/ax = f1 with respect to x and regarding y and z as constantparameters, we have

x

(4) P = J f1(x, y, z) dx + «(y, z),o

where «(x, y) is a function of x and y as yet undetermined. Hence

app x afl a« /' af2 a«_ -dx+-=I

-dx+-, oray ,, ay ay xo ax ay

af2(x, y, z) = f2(x, y, z) - f2(xo, y, z) +

ay-_J xafl a« f xaf3 -- dx + - _ - dx + , or

49Z 1o az az ax aza«

f3 (x, y, z) = f3 (x, y, z) - f3 (xo, y, z) + - -

§ 91 IRROTATIONAL VECTORS 199

Instead of three equations (3) for gp(x, y, z), we now have twoequations,

a« a«(5)

ay= f2(xo, y, z),

az= f3(xo, y, z)

for «(x, y), with the condition aft/az = af3/ay. The problem hasbeen reduced from three to two dimensions.

On integrating a«/ay = f2 with respect to y, we havev

(6) a =f f2(xo, y, z) dy + #(z),bo

where $(z) is a function of z to be determined. Hence

a«fY aaz

f2 d

dz

(3 fY a af3 -dy + - dy + , or

az vo y dz

(7)

f3(xo, y, z) = f3(xo, y, z) - f3(xo, yo, z) + dda

dz = f3 (xo, y0, z)

We now have one equation (7) to determine $(z); and

(8) J f3 (xo, yo, z) dz.o

On collecting the results (4), (6), and (8), we have finallyY z

(9) ' = ffi (x, y, z) dx +f f2 (xo, y, z) dy + f f3 (xo, yo, z) dz.Yo

irect substitution shows that Vip = f; and, if is a second func-Dtion for which V4, = f, V(¢ - cp) = 0, and 4 - (p is a constant.Thus (9) gives the solution of equations (3), determined uniquelyexcept for an additive constant.

There are evidently five other forms for ,p which may be obtainedfrom (9) by permuting 1, 2, 3 and making the corresponding per-mutation on x, y, z; for example,

Y

z

(10) c = ff2(x y , z) dy + ffs(x, yo, z) dz + ffi(x, yo, zo) dx.

Moreover, xo, yo, zo may be given any values that do not make theintegrands infinite.

200 DIFFERENTIAL INVARIANTS §91

In mathematical physics, it is customary to express an irrota-tional vector f as the negative of a gradient. Thus, if 4,we have

(11) f = -V ';¢ is then called the scalar potential of f.

Example 1. When

f = 2xzi + 2yz2j + (x2 + 2y2z - 1)k,

we find that rot f = 0; hence f = V. With xo = yo = zo = 0, we have,from (9),

=f X2xz dx +f 2yz2 dy - f Zdz = x2z + y2z2 - z.0 0 0

Example 2. Exact Equation. The differential equation,

(i)

is said to be exact when f dr = dp, the differential of a scalar. If (i) is exact,we have, from (90.3), dr f = dr vv; and, since dr is arbitrary, f =When f is continuously differentiable, f = V implies rot f = 0, and con-versely. Therefore we may state the

THEOREM. If f is a continuously differentiable vector, in order that f dr = 0be exact it is necessary and sufficient that rot f - 0.

Thus, in view of ex. 1, the equation,

2xz dx + 2yz2 dy + (x2 + 2y2z - 1) dz = 0,

is exact and may be put in the form dp = 0. Its general solution is = c,that is,

x2z + y2z2 - z = C.

If (i) is not exact, a scalar X which makes Xf dr = 0 exact is called an inte-grating factor of (i). The preceding theorem shows that X must satisfy theequation,

(ii) rotaf = 0 or VX x f + X rot f = 0.

On multiplying (ii) by f , we have

(iii) f rot f = 0.

Hence, when f dr = 0 admits an integrating factor, f rot f = 0. Con-versely, when f is continuously differentiable, the condition (iii) ensures theexistence of an integrating factor. We shall prove this in § 105. For thisreason f rot f = 0 is called the integrability condition for f dr = 0.

When f dr = 0 is integrable, ?f = V (p. Then the vector field f(r), beingparallel to Dip, is everywhere normal to the level surfaces rp = c. The con-dition (iii) therefore implies the existence of a one-parameter family of sur-faces everywhere normal to f. Thus the geometrical content of the conditionf rot f = 0 is that the vector field f (r) is surface-normal.

§ 92 SOLENOIDAL VECTORS 201

92. Solenoidal Vectors. A vector function f(r) is said to besolenoidal in a region R when div f = 0 in R.

If g(r) is a vector function whose components have continuoussecond derivatives, rot g is solenoidal, for, from (84.13),

(1) div rot g = 0.

Conversely, if div f = 0, we shall show that f may be expressedas the rotation of a vector g.

Using rectangular coordinates, let f = f1i + f2j + f3k, and (84.5)

aft aft af3(2) div f = -+-+- = 0.

ax ay az

We now shall determine a vector g = g1i + g2j + g3k, so that

i j k

(3) f = rot g = a/ax c1/ay a/az

91 92 93

We first find a particular solution of (3) for which g3 = 0; then(3) is equivalent to the scalar equations:

(4)

4992 a91 4992 1991fl = - , f2 = , f3 =_

az az ax ay

The first two equations of (4) are satisfied when

z

z

(5) 92 = - f f i (x, y, z) dz, 91 = f f2 (x; y, z) dz + 49(x, y) ;w Zo

in these integrations x and y are regarded as constant parameters,and a(x, y) is a function as yet undetermined. In order that thesefunctions satisfy the third equation of (4),

Zp

Zaf . + aJ2 J dz - as = f3,-(ax ay/ ay

or, in view of (2),

I.Zafs as

dz - - = f3 (x, y, z).az ay

When we perform the integration, this reduces to

as-f3 (x, y, z0) - = 0,

ay

202 DIFFERENTIAL IN VARIANTS

an equation which is satisfied by taking

v

a (X, y) = - f3 (x, y, zo) dy.Yo

Hence the vector g, whose components arez v

(6) gi = ff2(x, y, z) dz - ff3(x, y, zo) dy,a

z

92 = - f fi (x, y, z) dz,zo

§ 92

93 = 0,

is a particular solution of our problem.If G is any other solution, rot G = rot g = f, and hence

(7) rot (G - g) = 0.

But any irrotational vector may be expressed as the gradient of ascalar p (§ 91); hence the general solution of (7) is G - g = o,p,where ,p(r) is an arbitrary twice-differentiable scalar. Whendiv f = 0, the general solution of rot g = f is therefore g + Vip;its rectangular components are obtained by adding c /cx, 8,p/cy,8,p/8z to the components of g given in (6).

In mathematical physics, the solenoidal vector f = rot g is saidto'be derived from the vector potential g.

Example 1. When

f = x(z - y)i + y(x - z)j + z(y - x)k,

we find that div f = 0. Therefore f = rot g; if we take zo - yo = 0, the par-ticular solution (6) is

91 = f Zy(x - z) dz =xyz - Zyz2,

0

z

g2 f x(z - y) dz = - Zxz2 + xyz, ga = 0.0

Example 2. If u and v are continuously differentiable scalars, the vectorVu x Vv is solenoidal, for, from (85.3),

Vu x Vv = rot (uVv).

Conversely, we shall show in § 104 that a continuously differentiable sole-noidal vector always can be expressed in the form Vu x Vv.

§ 93 SURFACES 203

93. Surfaces. A surface is represented in parametric form bythe equations:

(1) x = x(u, v), y = y(u, v), z = z(u, v).

We assume that the three functions of the surface coordinates u, vare continuous and have continuous first partial derivatives, arequirement briefly expressed by saying that the functions are con-tinuously differentiable. In order that equations (1) represent aproper surface, we must exclude the two cases:

(i) the functions x.(u, v), y(u, v), z(u, v) are constants: equations(1) then represent a point;

(ii) these functions are expressible as functions of a single vari-able t = t(u, v); equations (1) then represent a curve.

In case (i) all the elements of the Jacobian matrix,

(2)

vanish. In case (ii), the rows of the matrix are dx/dt, dy/dt, dz/dtmultiplied by at/au and at/av, respectively, and all of its two-rowed determinants vanish identically. We exclude these casesby requiring that the matrix (2) be, in general, of rank two; then,at least one of its two-rowed determinants,

(3) Ayu zu

Bzu xu

C_ I xu

Yu= = ,yv zv Zv xv xv yv

is not identically zero.Even when the matrix in general is of rank two, the three de-

terminants A, B, C, all may vanish for certain points u, v. Suchpoints are called singular, in contrast to the regular points, whereat least one determinant is not zero.

If we introduce the position vector to the surface,

(4) r = xi + yj + zk = r(u, v),

(5)

and the condition for a regular point may be written

(6) ru x r, /- O.

204 DIFFERENTIAL INVARIANTS §94

If we introduce new parameters u, v by means of the equations,

(7) u = u(u, v), v = v(u, v),

we shall require that the Jacobian of this transformation,

a (u, v)(8) J =

a(u, i)FK 0.

Then equations (7) may be solved for u and v, yielding

(9) u = u(u, v), v = 17(u, v),

and the correspondence between u, v and u, v will be one to one.Since

X _ au av) x (ru au av\ X

(lp) rU ru - - rU - - - r -- J = J ru ry,au au av aU

the requirement J 0 0 makes ru x rv 0 a consequence of (6).Thus a point which is regular with respect to the parameters u, vis also regular with respect to u, D.

94. First Fundamental Form. A curve on the surface r(u, v)may be obtained by setting u and v equal to functions of a singlevariable t:

(1) u = u(t), v = v(t).

A tangent vector along the curve (1) is given by

dr

dt

and

(2)

The are s along the curve is defined as in (43.4) :

fro' dst t dt; anddt

=

Since du = ft dt, dv = v dt by definition, on multiplying (2) by dt2,we have

(3) ds2 = ru ru du2 + 2ru r du dv + r r, dv2.

This first fundamental quadratic form usually is written

(4) ds2 = E du2 + 2F du dv + Gdv2,

§ 94 FIRST FUNDAMENTAL FORM

where

(5) E = ru ru, F = ru r,,, G = r r,,.

Moreover, from (20.1),

(ru x rv) (ru x rv) _

hence

ru - ru ru rv

ru x rv 12 = EG - F2

is positive at every regular point.The curves v = const (u-curves) and u = const (v-curves) are

called the parametric curves on the surface. For these curvesdv = 0 and du = 0, respectively, and the corresponding elementsof arc are

(7) dsl = 1/E du, ds2 = N/G dv.

Since the vectors ru, rv are tangent to the u-curves and v-curves,respectively, the parametric curves will cut at right angles whenand only when

.(8)

The vector ru x rv is normal to the surface. The parallelogramformed by the vectors ru du, rv dv, tangent to the parametric curvesand of length dsl, ds2i has the vector area (§ 16),

(9) dS = ru x rv du dv.

We shall call this the vector element of area; the scalar element ofarea is

(10) dS = I ru x rv I du dv = E du dv.

The unit normal n to the surface will be chosen as

ruxrt ruxrvn= =E H

then dS = n dS. At every regular point the vectors ru, r, n forma dextral set; for

(12) [rurvn] = H = VEG > 0.

206 DIFFERENTIAL INVAItIANTS § 95

95. Surface Gradients. Let f(u, v) be a differentiable function,scalar, vector, or dyadic, which is defined at the points of thesurface r = r(u, v). We shall compute the derivative of f(u, v)with respect to the are s along a surface curve u = u(t), v = v(t).Along this curve,

(1)df du dv

ds fu ds + fv ds '

where du/ds = is/s, dv/ds = v/s, and, from (93.4),

-(2)

dsds = V L ic2 + 2Ficv + Gv2.dt

If we apply (1) to the position vector r(u, v), we obtain the unittangent vector e to the curve:

(3)

du dve=r,,,+rvds ds

Let a, b, c denote the set reciprocal to ru, r,,, n; then, since[rur ,n] = H,

rvxn nxru ruxrv,

b_

II , c =H

= n.(4) a

_H

Now from (3) a e = du/ds, b e = dv/ds; hence (1) may bewritten

df

ds

We now define afu + bf as the surface gradient of f and denote itby V8f t or Grad f:

(5) V8f = Grad f = afu + b f,,.

Grad f has the characteristic properties:

df(6) e- Grad f =

ds, n Grad f = 0;

(7) ru Grad f =of- , r Grad f =

of-au av

t In surface geometry we shall write of for the surface gradient.

96 SURFACE DIVERGENCE AND ROTATION 207

If f (r) is a tensor of valence v, Grad f is of valence v + 1. Atany point (u, v) of the surface, Grad f is in effect a synthesis ofall the values of df/ds for surface curves through this point. SinceGrad f depends only on the point.(u, v), df/ds is the same for allsurface curves having the unit tangent vector e at this point.

At any point (u, v) of the surface Grad f is completely deter-mined when df/dsi is given for two directions ei in the tangentplane at (u, v). If the set reciprocal to el, e2, n is e', e2, n,*

(8) Grad f = eldf + e2 df

dsi ds2

for, by virtue of equations (6), the right member of (8) may bewritten

(elel + e2e2 + nn) Grad f = I Grad f

where I is the idemfactor (§ 66). In particular if el and e2 = n x elare perpendicular unit vectors, el = el, e2 = e2. In view of (8),Grad f is independent of the coordinates x, y, z and of the surfaceparameters u, v.

96. Surface Divergence and Rotation. If f(r) is a tensor pointfunction defined over the surface r = r(u, v), its surface gradient,

(1) V8f=afu+bf,,,has the invariants,

(2)

(3) V,-f = axfu +bxf,,.We recall that the set a, b, n is reciprocal to ru, rv, n.

When f (r) is a vector, V8f is a planar dyadic; then the scalarand vector invariants of oaf are called the surface divergence andsurface rotation and are written

(4) V8 f = Div f, V8 x f = Rot f.

The second of V8f is the linear dyadic,

1(V3f)2 = axbfuxfv = Hnfuxfv;

the second scalar of V8f is therefore n f,, x f v/H.

* Since el x e2 = An, e3 = n, we have e3 = Xn/X = n.

208 DIFFERENTIAL INVAItIANTS

For the position vector r to the surface, we have

(6) V8r = ar.u + br v = I - nn,

(7) Div r = 2, Rot r = 0.

For the unit normal n,

(8) Rot n = 0.

196

To prove this, put f = n, a = r x n/H, b = n x ru/H in (3); then

H Rot n = (rz, x n) x nu + (n x ru) x n,,.

=

Now, from n n = 1, we have, on differentiation with respect tou and v,

(9) n nu = n nv = 0;

and, from ru n = r n = 0,

(10) ru nv = rv nu = -n ruv;hence H Rot n = 0.

From the defining equations (1), (2), (3), we may derive variousexpansion formulas. Thus, if X is a scalar, f a tensor,

(11) V3(af) = (V3X)f + XV3f,

(12) V. (Af) _ (V8A) f + XV8 f,

(13) V8 x (Xf) _ (V8X) x f + XV3 x f.

If g and f are vectors,

(14) Div (g x f) _ (Rot g) f - g Rot f;

and, in particular,

(15) Div (n x f) = -n Rot f.

Proof of (14) :

Div(gxf)

(Rot g) f - g Rot f.

§ 97 SPATIAL AND SURFACE INVARIANTS 209

97. Spatial and Surface Invariants. If f(r) is a tensor functionof valence v defined over a 3-dimensional region including thesurface r = r(u, v), its spatial gradient at a point (u, v) of thesurface may be computed from (86.4). If el, e2 are vectors in thetangent plane at (u, v) and e3 = n, then, at all points of the sur-face,

df df df df(1) Vf = e'-+e2-+n- = V8f +n - ;dsl ds2 do do

here the set e', e2, n is reciprocal to el, e2, n and df/dn denotesthe derivative of f in the direction of n. Moreover, if v > 0,

df df df df(2) V

ds.l ds2 do do

df df df df(3) Vxf = elx-+e2x-+nx-= V8xf +nx-.

dsl ds2 do do

From (1), we see that the tensors of valence v + 1,

(4) nx Vf = nx V j,

are the same over the surface; therefore both may be writtenn x Vf. Thus n x Vf may be computed solely from the values of fon the surface.

The same is true of the invariants obtained from n x Vf by dot-ting and crossing in the first position. Since

(5) nxVf = nxafu+nxbft,,from (96.1), this process yields

(n x a) fu + (nxb) f = n (a x fu + b x f,) = n V. x f,

n208f -nV8 f.Here n V8f means that n is dotted into the second vector fromthe left in each term of V8f. These invariants remain the samewhen computed from the corresponding spatial quantities. Thus weverify at once, from (3),

(6)

and, from (1) and (2),

(7) nVf-nV.f=n2V8f-nV8f.


When f is a vector, these invariants become

(6)' n rot f = n Rot f,(7)' (grad f) n - n div f = (Grad f) n - n Div f.

§ 97

We now express n x Vf and its invariants (6) and (7) in termsof the surface coordinates u, v. Since a, b, n and ru, r,,, n are re-ciprocal sets, we have, from (5),

1 1

(8) nx Vf - (r,fu - ruf II{(rvf)u - (ruf),,},

since ruv = r,,,, if these derivatives are continuous. Dotting andcrossing now yields

1

(9) n Vxf II

(10) II{(rvxf)u-(ruxf)v}.

Since ru, rv are tangent to the surface, (9) shows that: If avector f is everywhere normal to the surface, n rot f 0.

With f = V3g in (9) we obtain the important identity,

(11) =0;for then

rv f = gv, ru f = gu (95.7).

When f = pq, a dyad, we have, from (9),1

n V x (pq) = II {

(r,, Pq) u - (ru pq)

1 1= H

{ (r,, P)u - (ru P) v } q +H

p (rvqu - rugv),

that is,

(12) n V x (pq) = (n rot p)q + p n x Vq.

For future use we compute the invariant (10) when f = ng andg is an arbitrary tensor:

n2 Vf - n V f =H

1(rvxng)u - (ruxng)v}

1

= agu + bg,, +H

{(rvxn)u - (ruxn)v} g.

§ 98 SUMMARY: DIFFERENTIAL INVARIANTS 211

NowGrad g = agu + bg (95.5),

n 2 Grad n = (an. + n = 0 (96.9) ;

hence, on putting f = n in (10), we have

1-n Div n =

H{ (r x n)u - (ru x n) v } .

Thus, with f = ng,

(13) n? Vf - n V f = Grad g - (Div n) ng.

98. Summary : Differential Invariants. Let f (r) be a tensorpoint function of valence v; its derivative at the point P and inthe direction of the unit vector e is defined as

df f (r + se) - f (r)-- = limds s -- o s

(s > 0).

If df/dsi denote the directional derivatives corresponding to threenon-coplanar unit vectors ei, the gradient of f, namely,

grad f = Vf = e'ds

+ e2df

+ e3 ,1 2 63

has the property,df-- =e V.ds

Vf, a tensor of valence v + 1, thus gives a synthesis of all thedirectional derivatives of f at point r. If the vectors ei are i, j, k,

Vf = if., + jfv + kfz (fz = of/ax, etc.).

From Vf (valence v + 1) we derive the invariants V f (valencev - 1) and V x f (valence v) by dotting and crossing in the firstposition. When f is a vector (v = 1),

V f = div f, the divergence of f,

V x f = rot f, the rotation of f.

When r is a function r(u, v, w) of curvilinear coordinates

Vf = Vu fu + Vv fv + Vw f.W,

Vr = Vu ru + Vv r + Vw rw = I.


The sets Vu, Vv, Vw and ru, r',,, ru, are reciprocal. With J =we have

J Vf = (r x ru, f)u + cycl,

J V f = (r, x ru, f)u + cycl,ru r ru,

J V x f = ((r x ru,) x f)u + cycl = a/au a/av a/aw

ru, ru, are mutually orthogonal, the coordinates u, v, ware called orthogonal. Then if a, b, c denote a dextral set of or-thogonal unit vectors, ru = Ua, r = Vb, ru, = Wc, and J =UVW in the preceding formulas.

For any tensor f (r),

V Vf = V2f, V. Vf = 0,

V (V f) = 0, Vx (Vxf) = V(V _ f) - V2f.

The operator V V. V is called the Laplacian.A vector f is called irrotational if rot f = 0, solenoidal if div f

= 0. An irrotational vector f can be expressed as the gradient ofa scalar (f = Dip) ; a solenoidal vector f can be expressed as therotation of a vector (f = rot g).

For the surface r = r(u, v), the fundamental quadratic form is

ds2 = E due + 2F du dv + G dv2,where

At a regular point,

E>0,and the unit surface normal is defined as n = ru x

The derivative of a tensor f (r) along any surface curve tangentto the unit vector e at the point (u, v) is

df

ds

the set a, b, n is reciprocal to ru, r, n.The surface gradient,

V8f = Grad f = a fu + b f,,,

PROBLEMS

has the properties,df

e. Grad f =ds

, n Grad f = 0;

df dfr.u Grad f = - , r, Grad f =

au av

213

From V3f we derive the invariants,

O3 f = Oaxf = axfu+bxfv,by dotting and crossing in the first position. When f is a vector,

V8 f = Div f, the surface divergence of f,

Og x f = Rot f, the surface rotation of f.

The tensor n x of and its dot and cross invariants,

n?ofare not altered when V is replaced by V8. They may be computedsolely from the values f(r) assumes on the surface. Thus, ifH =

1n x of =

H{ (ru.f) v } ;

and the invariants follow by dotting and crossing between r,, rand f.

PROBLEMS

1. If R = r/r is the unit radial vector, prove that

div R = 2/r, rot R = 0.

2. For any scalar function f (r) of r alone prove that

V2f(r) = frr + 2fr/r.

If v2f (r) = 0, show that f = A/r + B.S. If a is a constant vector, prove that

grad (a. r) = a, div (a x r) = 0, rot (a x r) = 2a.

4. If a is a constant vector, prove that

grad (a - f)

div (a x f) = -a rot f,rot(axf)

Specialize these results when f = r.


5. For any vector point function f prove that

6. If Sp and ¢ are scalar point functions, prove that

div (Vp x v) = 0

7. If e is a unit vector, prove that

div (e r)e = 1, rot (e r)e = 0;

div (e x r) x e = 2, rot (exr) xe = 0.

8. If a is a constant vector, prove that V(r x a) = I x a.9. Show that Laplace's Equation V2f = 0 in cylindrical and spherical co-

ordinates isa2f 1 of 1 a2f a2f

+ p (9p + p2 a22= 0,

aptd

a2(rf) 1 a / af\ 1 a2fr{ -- sin0-J+ =0.are sin 0 a0 a0 sine 0 5;i

10. Prove that

an

1 1

vC T = - (3RR - I)

where R is a unit radial vector.11. If f is a vector point function, prove that

V (Vf), = grad div f; V x (Of), _ (V rot f),.

12. If Vf is antisymmetric, prove that rot f is constant and that the dyadicVf itself is constant.

13. If X is a scalar point function, prove that

V(,I) = VXI, V V. (XI) = VX, V x (XI) = VX x I.

14. If f is a vector point function, prove that

V (I x f) = rot f, v x (I x f) = (Vf), - I div f.

15. For the dyad fg prove that

V V. (fg) = (div f)g + f Vg, V x (fg) _ (rot f)g - f x Vg.

In particular, if r is the position vector,

V V. (rr) = 4r, V x (rr) _ -I x r.

16. If is a constant dyadic, prove that

div 4 . r = 91; rot e r = -+ (§ 69).

17. If the scalar function f(p, gyp) in plane polar coordinates is harmonic,prove that f(p-', tip) is also harmonic.

18. If the scalar function f(r, 0, Sp) in spherical coordinates is harmonic,prove that r 'f (r ', 0, gyp) is harmonic.

PROBLEMS 215

19. If f, g, h are scalar point functions, prove that

V2(fg) = g02f + 2Vf Vg + fV2g;

V2(fgh) = gh V2f + hf V2g + fg v2h + 2fvg vh + 2gvh of + 2hvf vg.

20. If u, v, w are orthogona' coordinates and f(u), g(v), h(w) are scalar func-tions of a single variable, show that

V2(fgh) = fgh { f + egg + hh

21. If a is a constant vector and f = ar", prove that

of = nrs-2ra, e'-f = n(n + 1)ri-2a.

22. If f = r"r, prove that

of = r"I + nrn-err, V2f = n(n + 3)rn-2r.

23. If f = r"r, find a scalar function p such that f = vcp.24. Prove that rot f = 0, and find V so that f = vp:

(a) f = yzi + zxj + xyk; (b) f = 2xyi + (x2 + log z)j + y/z k;

(c) f = c r, cD a constant symmetric dyadic.

25. If a is a constant vector and f = r" a x r, prove that

div f = 0, rot f = (n + 2)rna - nrn-2(a r)r.

26. Find a vector g such that rot g = a x r (cf. § 92).27. Prove that a = - rot -- .

r r'

28. The position vectors from 0 to the fixed points P1, P2 are r1, r2.(a) If f = (r - r1) x (r - r2), show that

(b) Prove that

divf =0, rotf =2(r2 -r1).

V(r-rl) (r-r2) =2r-r1-r2.29. If f(u, v) is a vector function defined over the surface r = r(u, 1'), prove

thatH Div f =n (ruxf -r, xfu).

30. If the vector f(u, v) is always normal to the surface r = r(u, v), provethat Rot f is tangent to the surface or zero.

31. If u, v, w are curvilinear coordinates, prove the operational identities:

a a a[rurvru7'le = ruxrw-+ru,xru-+ruxrv-;au av aw

a a a(ruxr,,)xV =rv -- ru -; ru.V =_.au av au

CHAPTER VI

INTEGRAL TRANSFORMATIONS

99. Green's Theorem in the Plane. Let R be a region of thexj-plane bounded by a simple closed curve C which consists of afinite number of smooth arcs. Then, if P(x, y), Q(x, y) are con-tinuous functions with continuous first partial derivatives,

CIQ(1) J I1 x - y) dx dy = f(P dx F Q dy),

where the circuit integral on the right is taken in the positivesense; then a person making the circuit C will always have theregion R to his left.

Fra. 99a

Consider first a region R in which boundary C is cut by everyline parallel to the x- or y-axis in two points at most (Fig. 99a).Then, if a horizontal line cuts C in the points (x1, y), (x2, y),

a

f f aQ dx dy =f {Q(x2, Y) - Q(xi, y)) dy

a

=fQ(X2,

y) dy + fQ(xiy) dy

= fQ(x, y) dy.

216

§ 99 GREEN'S THEOREM IN THE PLANE 217

And, if a vertical line cuts C in the points (x, yl), (x, y2),b

I f aydy dx = f I P(x, Y2) - P(x, Y1)) dx

fb(yl) dx - J P(x, y2) dx

6

= - fP(x, y) dx.

On subtracting this equation from the preceding, we get (1).We now can extend this formula to more general regions that

may be divided into a finite number of subregions which have theproperty that a line parallel to the x- or y-axis cuts their boundary

Frc. 99b Fia. 99c

in at most two points. For in each subregion formula (1) is valid,and, when these equations are added for all the subregions, thesurface integrals add up to the integral over the entire region; butthe line integrals over the internal boundaries cancel, since eachis traversed twice, but in opposite directions, .leaving only the lineintegral over the external bounding traversed in a positive direc-tion (Fig. 99b).

The boundary of R may even consist of two or more closedcurves: thus in Fig. 99c the region R is interior to Cl but exterior toC2; we may now make a cut between Cl and C2 and traverse theentire boundary of R in the positive sense as shown by the arrows.The line integrals over the cut cancel, for it is traversed twiceand in opposite directions; and the resultant line integral over Cconsists of a counterclockwise circuit of Cl and a clockwise circuitof Cam.

218 INTEGRAL TRANSFORMATIONS § 100

Although Green's Theorem is commonly stated for scalar func-tions P, Q, it is evidently true when P and Q are tensors withcontinuous first partial derivatives; for, when P and Q are referredto a constant basis, the theorem holds for each scalar component.

100. Reduction of Surface to Line Integrals. A surface is saidto be bilateral if it is possible to distinguish one of its sides fromthe other. Not all surfaces are bilateral. The simplest unilateral

b a,surface is the Mobius strip; thismay be materialized by taking a

a b'

FIG. 100arectangular strip of paper, givingit one twist, and pasting the ends

ab and a'b' together (Fig. 100a). This surface has but one side;if we move a point P along its median line and make a completecircuit of the strip it will arrive at the point P directly under-neath. Since we can travel on a continuous path from one sideof the Mobius Strip to the opposite, these sides cannot be distin-guished. This is not the case with a spherical surface, which hasan inside and an outside, or any portion S of the surface botindedby a simple closed curve C; for we cannot pass from one side of Sto the other without crossing C.

Let S be a portion of a bilateral surface r = r(u, v) bounded by'a simple closed curve C which consists of a finite number of smootharcs. The surface itself is assumed to consist of a finite numberof parts over which the unit normal n is continuous. The positivesense on C is such that a person, erect in the direction of n, willhave S to the left on making the circuit C. If such an orientedsurface is continuously deformed into a portion of the xy-planebounded by a curve C', n becomes k (the unit vector in the direc-tion +z) and the positive circuit on C' forms with n = k a right-handed screw.

Let f (r) be a tensor point function (scalar, vector, dyadic, etc.)whose scalar components have continuous derivatives over S.Such a tensor is said to be continuously differentiable. When theforegoing conditions on S and C are fulfilled, we then have the

BASic THEOREM I. If f(r) is a continuously differentiable tensorpoint function over S, the surface integral of n x Of over S is equalto the integral of Tf taken about its boundary C in the positive sense:

(1) fn Vf dS = JTf ds.

§ 100 REDUCTION OF SURFACE TO LINE INTEGRALS 219

Proof. Since

1n x Vf =

II{ (ref),}, dS = H du dv,

fn x Vf dS =IS,

{ (r ,f).u - du dv,s'

where S' is the region of the uv-plane which contains all parametervalues u, v corresponding to points of S. We now may apply

U

FIG. 100b

Green's Theorem in the plane to the last integral, letting u and vplay the roles of x and y; thus

fs, { (rtf)u - du dv = f(rtf du + dv),

where the circuit integral is taken about the curve C', which formsthe boundary of S', in the positive sense, that is, in the directionof a rotation of the positive u-axis into the positive v-axis. If weregard u and v as the functions of the are s on the curve C, itbecomes

/ du dvTfds,I ruf-+rf-)ds =ic

\ ds ds

du dv dr= T,ru

ds +ry

ds A

the positive unit tangent along C. Formula (1) thus is established.

220 INTEGRAL TRANSFORMATIONS

If we introduce the vector elements of surface and of arc,

dS = n dS, dr = T ds,

§100

the basic theorem for transforming surface to line integrals be-comes

(2) fdS X Vf =fdr f,

in which dS and dr must be written as prefactors.If f is a vector, dyadic, etc., we may obtain the other integral

transformations from (1) by placing a dot or a cross between thefirst two vectors in each term (dyad, triad, etc.) of the integrands;for both of these operations are distributive with respect to addi-tion and therefore may be carried out under the integral sign.This process applied to the tensor n x Vf gives n V X f and n z Vf

- nV f; we thus obtain the important formulas:

(3) ffl.VxfdsfT.fds,

(4) f (n z Vf - n V f) dS = if T X f ds.

When f is a vector function, (3) becomes Stokes' Theorem:

,(3)' fn.rotfds=fT.fdsa result of first importance in differential geometry, hydrodynamicsand electrodynamics.

If S is a closed surface which is divided into two parts S1, S2by the curve C, we may choose n as the unit external normal;then, from (1),

fnxVfds = fTlf ds,c

fn,< Vf dS = fT2f ds,

where T2 = - T1, since the positive sense on C regarded asbounding S1 is reversed when regarded as bounding S2. On addingthese integrals, we find that the integral of n X Vf over the entiresurface is zero. We indicate this by the notation,

(5) fnx Of dS = 0,

§ 101 ALTERNATIVE FORM OF TRANSFORMATION 321

which denotes an integral over a closed surface. On taking dotand cross invariants in (5), we have also

(6) fn.VxfdS=0,

(7) f(n'4Vf_nV.f)dS=0.

Example 1. Put f = r in (4) and (7); then, since grad r = I, div r = 3,

J'ndS = 2irxdr, fnds = 0.

The last result, which states that the vector area of a closed surface is zero,generalizes the polyhedron theorem of § 17.

Example 2. When div f = 0, we have seen that f = rot g (§ 92); hence

fThus, if f = Vu x Vv, div f = 0, and we may take g as uVv or -vVu; hence

Jn. Vu x Vv dS = it dv = - Jv du = 2 f(u dv - v du).

In particular, when u = x, v = y, n = k, this formula expresses an area in the

xy-plane as a circuit integral 2 J(x dy - y dx) over its boundary.

101. Alternative Form of Transformation. If we replace f byng in (100.4), where g is any continuously differentiable tensor,we have, from (97.13),

(1) f { Grad g - (Div n) ng } dS = fT x n g is.

This integral transformation is quite as general as that given bybasic theorem I. For, if we replace g by of and then cross in thefirst position, the integrand on the left becomes

a x (nuf + n f.u) + b X n fv) _ -n x Grad f,

since Rot n = 0; whereas, on the right,

(Txn) Xnf = -Tf.

We thus retrieve the basic transformation (100.1).


We now shall express (1) in slightly different notation. On theright the vector T x n = m is the unit normal to C tangent to Sand pointing outward. If we write f instead of g and J = - Div n(J is not the Jacobian of § 88), (1) now becomes

(2) f (vsf + J nf) dS = fm f ds.

On dotting and crossing in the first position of the tensor inte-grands, we obtain the additional formulas:

(3) f(vs.f+Jn.f)ds =

ff f f =

Div n is called the mean curvature of the surface (§ 131).Surfaces for which J = 0 are called minimal surfaces; for suchsurfaces the integrands on the left reduce to the first term.

On the plane, n = k, a constant vector, and Div n = 0; thenJ = 0 in (2), (3), and (4). When f is a vector, these equationsbecome (if we write dA for dS) :

(5) fGrad f dA = i(m f ds,

(6) fDivfdA

(7) JRotfdA = fm -f ds,

where m is the external unit normal to the closed plane curveforming the boundary. We may deduce (6) from (7) by replacingfbykxf;for

Rot (k x f) = k Div f, m x (k x f) = km f.

102. Line Integrals. When f(r) is a vector, consider the lineintegral,

(1)rf dr

dt,J'f.dr= -o to dt

ta ken over a curve C: r = r(t) from the point P0 to P. The valueof this integral depends on the curve C and the end points Po and

§ 102 LINE INTEGRALS 223

P, but not on the parameter t. For, if we make the change ofparameter t = t(T),

t dr pT dr dt r drf - dt= f---drf - dr.Jcu dt

TOdt dr ,, ar

Let us consider under what conditions the line integral (1) is in-dependent of the path C-that is, when its value is the same forall paths from Po to P.

We shall call a closed curve reducible in a region R if it can beshrunk continuously to a point without passing outside of R.Thus in the region between two concentric spheres all closedcurves are reducible. However in the region composed of thepoints within a torus, all curves that encircle the axis of the torusare irreducible. A region in which all closed curves are reducibleis called simply connected. Thus the region between two concen-tric spheres is simply connected, whereas the interior of a torusis not.

If f is continuously differentiable, and rot f = 0 in a region R,the line integral of f around any reducible closed curve in R is zero;for we can span a surface S over R that lies entirely within R(shrinking the curve to a point generates such a surface), andthen, by Stokes' Theorem,

ff . dr = fn. rotfdS=0.

We may now prove the

THEOREM. If f is a continuously differentiable vector, and rot f

= 0 in a simply connected region, the line integral J'f dr between

any two points of the region is the same for all paths in the regionjoining these points.

Proof. If C1 and C2 are any two curves POAP, POBP joiningPO to P, the line integral of f around the circuit POAPBPO is zero;for all closed curves in a simply connected region are reducible.Hence

JoA P PfPBP,

Pof dr=0.


Under the conditions of this theorem, the line integral (1) froma fixed point Po to a variable point P of the region depends onlyupon P; in other words the integral defines a scalar point function,

(2) p(r) = Jf dr.o

Let us compute d<p/ds = e Vg for an arbitrary direction e. Wehave

,p(r + se) - p(r) =f+8e 8f0

since dr = e ds along the ray from r to r + se. By the law of themean for integrals,

ff.eds = sef(r+e),

and hence

d(P= lim p(r

+ se) - p(r)ds s -o s

Since e - Vp = e f for any e,

(3)

(0 < 9 < s),

= e f(r).

Thus an irrotational vector f may be expressed as the gradientof scalar gyp, given by (2). The determination of p given in § 91 isthe line integral of f taken over a step path from (xo, yo, zo) to(x, y, z). Thus in (91.9) the integrals, in reverse order, are takenover the straight segments from

P0(xo, Yo, zo) - (x0, Yo, z) - (x0, y, z) - P(x, y, z).

There are six such step paths; for the first segment can be chosenin three ways and the second in two.

In mathematical physics it is customary to express an irrota-tional vector as the negative of a gradient:

(4) f = - V4,; then ¢(r) =jTO

dr

is called the scalar potential of f.103. Line Integrals on a Surface. We next consider line and

circuit integrals over curves restricted to lie on a given surface,or within a certain region S of the surface. A region S of a surface

§ 103 LINE INTEGRALS ON A SURFACE 225

is called simply connected if all closed surface curves in S arereducible-that is, can be shrunk continuously to a point withoutpassing outside of S. In the plane, for example, the region withina circle is simply connected, but the region between two concentriccircles is not.

If the vector f is continuously differentiable, the circuit integral

if dr about any reducible curve of S is zero when rot f is every-

where tangential to the surface; forfWhen S is a portion of the level surface u(x, y, z) = c of a scalarfunction, Vu on the surface is parallel to the surface normal andthe condition n rot f = 0 may be written

(1) 0.

Just as in § 101, we may now prove the

THEOREM 1. If f is a continuously differentiable vector, andn rot f = 0 in a simply connected portion S of a surface, the line

integral ff dr between any two points of S is the same for all

surface curves joining these points.

Under the conditions of this theorem the line integral,r

(2) p(r) = f dr,ra

is the same over all surface curves from Po to P and thereforedefines a scalar point function V(r) in S. Along any definite curver = r(s) issuing from P0 (s = 0) c is a function of the are s,

s dr 8

p(s) =J'ff Tds,

where T is the unit tangent vector; hence at any point P d(p/ds= f T. Since the curve may be varied so that T assumes anydirection e at P, we have the relation,

d(3)

s


for all unit vectors e in the tangent plane at P. Now if e1, e2are two such vectors, and el, e2, n the set reciprocal to e1, e2, n,we have, from (95.8),

Grad cp = eld

+ e2d

= ele1 f + e2e2 f.ds1 ds2

Since f = (ele1 + e2e2 + nn) f,

(4) Grad 'P = ft,

the projection of f on the tangent plane at P. We have thus

proved the

THEOREM 2. If f is a continuously differentiable vector, andn rot f = 0 in a simply connected region of a surface, the tangentialprojection of f on the surface is

ft = Grad 9 where 9 = f dr.ro

104. Field Lines of a Vector. When f(r) is a continuously dif-ferentiable vector function, the curves tangent to f(r) at all theirpoints are called the field lines of f. If r = r(t) is a field line,dr/dt is tangent to the curve (§ 41) and consequently a multipleof f. The field lines have therefore the differential equation:

(1) fxdr = 0.

If f = f1i + f2j + f3k, (1) is equivalent to the system,

(2)dxdydzfl f2 f3

Any integral u(x, y, z) = a of this system represents a surface locusof field lines. For, from du = dr Vu = 0 and (1), we have

(3)

and, since Vu is normal to the surface u = a, the vector f at anypoint of the surface is in the tangent plane.

If v(x, y, z) = b is a second integral of (2), we have also

(4)

If v is independent of u, Vu x Vv 0 0 (§ 87, theorem 1). From (3)and (4) we conclude that f is parallel to Vu x Vv, and hence

(5) Vu x Vv = Xf.

104 FIELD LINES OF A VECTOR, 227

But the curve in which the surfaces u = a, v = b intersect is every-where tangent to Vu X Vv. Hence we have the

THEOREM 1. If u = a and v = b are independent integrals of thesystem f x dr = 0, the surfaces they represent intersect in the field linesof f.

From (3) and (4), we see that u and v are independent solutionsof the partial-differential equation:

aw aw caw

(6) f- Vw =flax + f2 a + f3 az = 0-ax

In view of (5), this equation is equivalent to

(7) Vu X Vv Vw = 0.

From § 87, theorem 3, (7) is satisfied when and only when u, v andw are connected by a functional relation. Hence we have

THEOREM 2. The general solution of the partial differential equa-tion f Vw = 0 is w = cp(u, v), where cp is an arbitrary function andu = a, v = b are independent integrals of f X dr = 0.

When the vector field f(r) is solenoidal, div f = 0. In § 92 wefound that we could express f as rot g. We now deduce anotherform for f which gives at once its field lines.

THEOREM 3. A solenoidal vector f which is continuously differ-entiable can be expressed in the form:

(8) f = Vu X Vv.

Proof. If we assume the relation (8), we have f Vu = 0,f Vv = 0; thus both u and v are solutions of f Vw = 0. If wechoose for u some integral of the system (2), f Vu = 0. Now,if dr is a differential on the level surface u = a., we have, from (8),

(9) f X dr = Vv du - Vu dv = - Vu dv.

In the function u(x, y, z), at least one variable is actually present.If z is present, au/az = uz is not identically zero; hence, on multi-plying (9) by k , we have k f X dr = -uz dv and

kxf dr(10) v= - ,

uz


the integral being taken over a curve on the surface u = a. Thisintegral is independent of the path; for, from (85.3) and (85.6),

kxf 1\ 1rot = O - J X (kXf) + -rot (kXf)

uz uz U,

2(Vuz)X(kxf) -kOf

uZ uZ

-(f V i )k + (k Vu,)f f22

ui uz

kXf fZ - cu -1 aVu -rot = _-- (f - Vu) = 0.

uZ uZ uZ az

If u(x, y, z) contains x or y, we may replace k and uz in (10) byi and ux or j and uv.

With the values of it and v thus obtained, we now have, fromtheorem 2 of § 103,

x -1Vu X Vv = Vu x Grad v = - V it X

(k f)c = - Vu x (k X f) = f,uZ uZ

where Grad v refers to the surface u = a, and (k X f)t is a tangen-tial projection upon it. Moreover f - Vv = 0.

The field lines of the vector Vu X Vv are the curves in which thesurfaces u = a, v = b intersect.

Example 1. The field lines of the vector,

f = xzi + yzj + xyk,

have the differential equations:

(i)dx dy dz

From these, we obtain the equations,

ydx-xdy=0, y dx + x dy - 2z dz = 0,

which have the integrals:

y/x = a, xy - z2 = b.

These one-parameter families of surfaces intersect in the two-parameter familyof field lines.

§ 104 FIELD LINES OF A VECTOR 229

When one integral, as y/x = a, is known, we may find a second by obtain-ing the field lines on the surface y/x = a. These must satisfy the equationobtained from (i) by putting y = ax, namely, ax dx = z dz. Its integralax2 - z2 = b gives the field lines on the surface y/x = a. On replacing a byy/x, we obtain a second family of surfaces xy - z2 = b which intersects thefirst family y/x = a in the field lines.

Example 2. The field lines of the vector,

f = xi + 2x 2j + (y + z)k,

have the differential equations:

(ii)dx dy dz

x 2x2 y + z

From dy = 2x dx, we obtain the family of surfaces,

y-x2=a.As no other integrable combination is evident, we put y = a + x2 in (ii); then

dx _ dz dz z ax a+x2+z' or dx-x=x-I-x.

This linear equation, with the integrating factor 11x, has the solution,

z = -a+x2+bx.A second family of integral surfaces is therefore

z +Y - 2x = b.x

Example 3. The vector,

f = x(y - z)i + y(z - x)j + z(x - y)k,

is solenoidal; for div f = 0. In order to express f as Vu x Vv we choose for uan integral of the system,

dx _ dy _ dz

x(y - z) y(z - x) z(x - y)

Since the sum of the denominators is zero,

dx+dy+dz=0, x+y+z=a;hence we may take

u = x + y + Z.

We now use (10) to compute v. On the surface u = a, z = a - x - y, and

-kxf dr = -x(y -z)dy+y(z - x)dx= y(a-2x-y)dx+x(a-x-2y)dy.


Since this is an exact differential, the method of § 91 gives

v = f(aY_2xY_Ydx=axY_?J_x?J

= xy(a - x - y) = xyz.

With u = x + y + z, v = xyz, we may readily verify that f = Vu x vv.

105. Pfaff 's Problem. Since rot f is solenoidal, we have fromtheorem 3 of § 104,

(1) rot f = Vu x Vv = rot (u Vv),

rot (f - uVv) = 0;

hence (§ 91) f - uVv is the gradient Vw of a scalar, and

(2) f = Vw + uvv.

When f is a continuously differentiable vector, we may find threescalars u, v, w so that (2) holds good. The determination of thesescalar functions is known as Pfaff's Problem. We proceed to givea simple and direct solution.

Assuming the truth of (2), we at once deduce (1); hence

(3) Vu rot f = 0, Vv rot f = 0,

so that both It and v satisfy the same partial differential equation :V. rot f = 0. Let v = a be some integral of the system dr x rot f= 0; then Vv rot f = 0. Now on the surface v = a we have,from (2),

this integral, taken over a curve of the surface v = a, is inde-pendent of the path since Vv rot f = 0. On substituting the func-tions v and w thus obtained in (2), this equation uniquely deter-mines u.

We now must show that, with It, v, w thus determined, Vw +uVv = f. Let Grad w and n denote the gradient and unit normalon a surface of the family v = const; then (103.4)

dw dwVw=Grad w+ndn ft+ndn

§ 105 PFAFF'S PROBLEM 231

and the projection of Vw + uVv tangential to the surface is ft.Moreover u was chosen to make the normal projections of Vw +uVv and f the same, that is,

dwn-+uVv=f-ft.do

From (1) and (2),

f- rot f =a (u, v, w)

C )(X' y, z)

hence f rot f = 0 implies that u., v, w are functionally dependent(§ 87, theorem 3). In this case,

(4) f dr = dw -i-- u(v, w) dv = 0

is an ordinary differential equation which admits solutions undergeneral conditions-as when u(v, w) and au/caw are continuous ina region R. * Hence the condition

(5) f-rot f=0shown in § 91, ex. 2, to be necessary for the integrability of f dr= 0, is also sufficient.

When (5) is fulfilled, let (p(w, v) = C be the general solution of(4). Then

d(p =amp

dw + - dv = 0aw av

must yield (4) upon division by app/8w. In other words, X =a,p/8w is an integrating factor of (4) :

Then Xf = VV, and f is everywhere normal to the surfaces p =const. The field lines of f are then the orthogonal trajectories ofthese surfaces.

A family of curves is said to form a field in a region R if justone curve of the family passes through every point of R. The unitvectors T, N) B along the curves are then vector point functionsin R; and the first Frenet Formula dT/ds = KN can be written

(6) T- VT = KN.

* See Agnew, Differential Equations, New York, 1942, p. 310, et seq.


Now, from (85.7) and (85.9),

T T =

T = -KN.

§ 105

For a surface-normal field

T rot T = 0, (T x rot T) X T = rot T,

and, from (7), we have

(8) rot T = KB, rot T I = K.

The Darboux vector of a surface-normal field of curves is there-fore 8 = rT + rot T.

Example. Find u, v, w so that

f = [2yz, zx, 3xy] = Vw + u Vv.

Solution. rot f = [2x, -y, -z]; since the system,

dx _ dy _ dz

2x -y -zhas the solution z/y = a, we take v = z/y. On the surface v = a, z = ay;hence

w = f f dr = f (2yz dx + zx dy + 3xy dz)

= f (2ay2 dx + 4axy dy)

2axy2 = 2xyz.Now u is given by

[2yz, zx, 3xy] = [2yz, 2zx, 2xy] + u[0, -z/y2, 1/y],

whence u = xy2. Thus our solution is

u = xy2, v = z/y, w = 2xyz.

Since f U. rot f = 0, u, v, and w must be functionally related; in fact w = 2uv.The total differential equation,

(i) 2yzdx+zxdy+3xydz =0,

is thus equivalent to

dw + udv = 3udv + 2vdu = 0,

and has the integral u9 = C, or x2yz3 = C.

106 REDUCTION OF VOLUME TO SURFACE INTEGRALS 233

106. Reduction of Volume to Surface Integrals. Let S be aclosed surface and V the volume it encloses. Then S has twosides, an inside and an outside, and at all regular points of S wehave a definite unit normal n directed towards the outside. Weshall suppose that S consists of a finite number of parts over whichn is continuous.

Consider, first, a surface S which is cut in at most two pointsby a line parallel to the z-axis; denote them by (x, y, z1) and(x, y, z2), where zl < z2. Then S has a lower portion Sl con-

z

*_y

FIG. 106

silting of all the points (x, y, z1), and an upper portion S2 con-sisting of the points (x, y, z2). We suppose also that the pointsfor which zl = z2 form a closed curve separating Sl from S2 (Fig.106). The equations of Sl and S2 may be taken as z = zl(x, y),z = z2(x, y).

Now let f (x, y, z) be a tensor function of valence v whose scalarcomponents have continuous partial derivatives throughout V.Then, if the volume integral of of/8z over V is written as a tripleintegral with the element of volume dV = dx dy dz, we may effecta first integration with respect to z,

(i) ff az dx dy dz

=fff( x, y, z2) dx dy -fff (x, y, z1) dx dy,

whe re the double integrals are taken over the common projectionA of Sl and S2 on the xy-plane.

234 INTEGRAL TRANSFORMATIONS §106

If we regard x, y as the parameters u, v on the surfaces S1 andS2, the vector element of surface is ry x ry dx dy (94.9). Now from

r = ix + jy + kz(x, y), rz = i + kzz, ry = j -}- kzy,

Over S2 (z = z2) this vector has the direction of the external nor-mal n; but over Sl (z = zl) it has the direction of the internalnormal -n. Hence, if we denote the vector element of area inthe direction of the external normal by dS = n dS,

n dS = ± (k - izz - jzy) dx dy, k . n dS = fdx dy,

where the plus sign applies to S2 and the minus to S1. The twointegrals over A now may be combined into a single integral over8, so that we may write

(1) j/,f dx dy dz =f I

This formula is also valid when S is bounded laterally by a partof a cylinder parallel to the z-axis and separating Sl from S2. For(i) holds as before; and, in (1), k n = 0 over the cylinder, so thatit contributes nothing to the integral over S.

We now may remove the condition that S is cut in only twopoints by a line parallel to the z-axis. For, if we divide V intoparts bounded by surfaces which do satisfy this condition andapply formula (1) to each point and add the results, the volumeintegrals will combine to the left member of (1); the surface inte-grals over the boundaries between the parts cancel (for each ap-pears twice but with opposed values of n), whereas the remainingsurface integrals combine to the right member df (1).

Finally we may extend (1) to regions bounded by two or moreclosed surfaces, that is, regions with cavities in them, by this sameprocess of subdivision. Additional surfaces must be introduced sothat the parts of V are all bounded by a single closed surface, andthe surface integrals over these will cancel in pairs as before.

When x, y, z form a dextral system of axes, the same is true ofy, z, x and z, x, y. Hence, if in (1) we make cyclic interchanges inx, y, z, we obtain the corresponding formulas:

(2) fff a dx dy dz = y, z) i n dS,

§ 106 REDUCTION OF VOLUME TO SURFACE INTEGRALS 235

(3) fff ay dx dy dz =f ff(x, y, z) j n dS.

If we insert the prefactors k, i, j in the integrands of (1), (2), and(3), respectively, add the resulting equations, and note that

iof +jof +kof = Vf,ax ay az

we obtain finally

(4) f of d V =in f dS,

using J( to denote integration over a closed surface. The inte-

grands are tensors of valence v + 1. We have thus proved

BASic THEOREM II. If f(r) is a continuously differentiable tensorpoint function over the region V bounded by a closed surface S, whoseunit external normal n is sectionally continuous, then the integral ofof over the volume V is equal to the integral of of over the surface S.

If f(r) is a vector or tensor of higher valence, we may obtainfrom (5) other integral transformations by placing a dot or a crossbetween the first two vectors in each term of the integrands; forboth of these operations are distributive with respect to additionand therefore may be carried out under the integral sign. We thusobtain the important formulas:

(5) fv.fdv=Jn.fds,

(6) fv<fdv = JnxfdS.

When f is a vector, (5) is known as the divergence theorem; the

integral fn f dS then is called the normal flux of f through the

surface.

Example. If rot f = 0 in a simply connected origin, f = 102), and

ffdV =f = fncodS.

Thus the volume integral of an irrotational vector can be expressed a surfaceintegral over the boundary.


For example, the center of mass of a homogeneous body of mass m is fixed bythe position vector,

r*=mfrdm=yfrdV;

for m = pV, where p denotes the constant density. Since rot r = 0, we have,from (102.2),

rer=V, where P=

U

Hence

Vr* = i 0 nr2 dS.

107. Solid Angle. The rays from a point 0 through the pointsof a closed curve generate a cone; and the surface of a unit sphereabout 0 intercepted by this cone is called the solid angle & of thecone.

The reciprocal of the distance r = OP is a harmonic function(§ 89, ex. 2) ; hence

1 1 Rdiv grad = 0, and grad = -

r r r2

is a solenoidal vector.Let us now apply the divergence theorem to the vector f = R/r2

in the region interior to a cone of solid angle 0 and limited exter-nally by a surface S, internally by a small sphere a about 0 ofradius a. Within this region f is continuously differentiable,

div f = 0 and fn f dS = 0. The external normal n = - R over

0', and over the conical surface n R = 0; hence

a R 1 Sa(1) J r2 dS =,a2 dS = a2 = S2,

where Sa is the area cut from or by the cone. Note that the ratioSa/a2 is independent of the radius and may be computed witha=1.

If S is a closed surface, we have

(2)

n R 14r, 0 inside of S2

dS =r l0, 0 outside of S.

When 0 is outside of S, f is continuously differentiable throughoutits interior, and the foregoing result is immediate. In this case

§ 108 GREEN'S IDENTITIES

the elements of solid angle,

(3) d1t =2

dS,r

237

corresponding to the same ray cancel in pairs.108. Green's Identities. We now apply the divergence theorem

(106.5) to the vector f = p0j,, where <p and ' are scalar functionshaving continuous derivatives of the first and second order, re-spectively, in a region R bounded by a closed surface S. Now

div (,p0') _ V p V4, + p div V ' (85.2),

n d¢/dn,

where the normal derivative d¢/dn is in the direction of the externalnormal to the bounding surface. Hence, on writing the operatordiv grad as V2, we have

(1) V p V. V4, dV +f g' V24,dV do dS.

In case ' is not defined outside of S, we replace d ,/dn by thenegative of the derivative along the internal normal -n. Formula(1) is known as Green's first identity.

If both cp and 4, have continuous derivatives of the first andsecond orders, we may interchange (p and 4, in (1). On subtract-ing this result from (1), we obtain Green's second identity:

d4,(2) f(v2G - 4'V2 d

v) dV = \ do - dS.41 dnl

We now take ,' = 1/r, where r is the distance OP. If 0 is inte-rior to S, we cannot apply (2) to the entire region enclosed since4, becomes infinite at 0. We therefore exclude 0 by surroundingit by a small sphere v of radius e and apply (2) to the region R'between S and v; then

1 dl ldc dl 1dipV24pdV = co----- dS- cP----- dS.

JS \ dn r r dn J, dn r r dn

On the sphere r = e,

d (1l d (1l 1 1 dp

do \ r/ dr \ r) r2 e2 ' do ar '2ddS = eQ,


dil denoting the solid angle subtended by dS. The integral over vis therefore

J( + 1 ate) dSdSt + Ef a

M = E adg,

Eo 2 E ar ar f or

where p is a value of <p at some point of o. Now, as e -* 0,

d 1 1dS -

do r r do

-v2(p dV-- -f-v2(P dVR1r

R rfor the integrand remains finite if we use the spherical element ofvolume dV = r2 sin 0 dr dB d4p. We thus obtain Grreen's third iden-tity:

r v2 i dip d it(3) J

r dV +j

\r do do r/dS.

When 0 is exterior to S we may put ¢ = 1/r directly in (2); inthis case the right member of (3) equals zero.

We may deduce three analogous identities from (101.6),

fDivfdA

the divergence theorem in the plane. With f = (p Grad ¢, we find

(4) fv.vdA+fv2dA = f (P do ds,

where we now interpret V and V2 as Grad and Div Grad anddi/dn as a derivative in the direction of the external normal m.By an interchange of cp and , we find as before

dip(5) f(cv24, - ,'tv2

dp) dA = /C do - # d.n) ds.

To obtain the third identity from (5), we take 4, = log p, wherep is the distance OP in the plane. Since log p is a solution ofLaplace's Equation in the plane,

Div Grad # = 0,

§ 109 HARMONIC FUNCTIONS 239

v24, = 0 in (5). As before, we must exclude the origin from theregion by surrounding it by a small circle y of radius E. Applying(5) to the region remaining, we have

- flog p v2cp dA =R

J\ d d

c dolog p - log p

dods + ,. (P do log p - log p

dods.

On the circle p = E,

d d 1 1 dip a P

dolog p

dplog p

p E ' do= - ap ' ds = E d6;

the integral over y is therefore

f\ acvlog E - - - E dB = E log c- - d8 - 27r gyp,ap ap

where is a value of p at some point of y. When e -> 0, E log t--* 0, and we obtain

d duo(0) 27r p(0) = log p V2,p dA + (`p do

log p - log p -) ds.R C /

109. Harmonic Functions. A solution of Laplace's Equation,

(1) div grad cp = 0,

is called a harmonic function. A function <p is said to be harmonicin a region R if it has continuous derivatives of the first andsecond orders and V2p = 0 in R.

If a vector f is both irrotational and solenoidal, its scalar poten-tial 4, is harmonic; for, from (102.4),

f = -Vi, divf = -V2' = 0.If in Green's second identity (108.2) cp and 4, are harmonic

throughout R, we have

(2) f\odn - dn/dS=0.

Thus, if ' = 1, we have, for any harmonic function cp,

(3) do dS = 0.


If cp is harmonic in the region bounded by a closed surface S,Green's third identity (108.3) gives the value v(P) at any interiorpoint P,

v l(4) is \r do - do r/dS,

where r = PQ, the distance from P to points Q on the surface.Thus a function gyp, harmonic in the region enclosed by S, is deter-mined completely at any interior point P by the values of <p anddcp/dn on the boundary.

When P is exterior to the surface S, we have

(5) 0 - .Js (r do do r) dS'

on putting i = 11r in (2).If S is a sphere of radius r about P as center and lying entirely

within the region R, we have, from (4),

4irv(P) =r do dS

+- - JCp dS;f r2

or, in view of (3),

(6) rc'(P) = 4r2f p dS.

Since the surface of the sphere is 47rr2, we have the

MEAN VALUE THEOREM. The value of a function, harmonic in aregion R, at any point P is equal to the mean of its values on anysphere about P as center and lying entirely within R.

This theorem shows that a function which is harmonic in aclosed bounded region R, but not constant, attains its extremevalues only on the boundary. For let P be a frontier point of theset for which cp attains its minimum value m. If P were an interiorpoint of R, there would be a sphere about P, lying within R, onwhich V > m at some points; hence the mean value of c over thesphere would be greater than (p(P) = m.

If in Green's first identity (108.1) we take cp = 4' and assumethat (p is harmonic in R, we have

fi =(7) o1 2 dV - dS.

is do

§ 110 ELECTRIC POINT CHARGES 241

Hence, if either cp = 0 or dcp/dn = 0 on S, the volume integral in(7) vanishes; and, since I nip 12 is continuous and never negative,we must conclude that V p = 0, and cp has a constant value through-out R. If 'p = 0 on S, 'p = 0 in R; and, if d'p/dn = 0 on S, p = Cin R.

Now let 'i and 'P2 be two harmonic functions in R; then theirdifference <p = V, - 'P2 is also harmonic. If 01 = 'P2 on S, cp = 0on S and cpl = 'p2 throughout R. If d'pl/dn = dSP2/dn on S, dcp/dn= 0 on S and <pl = IP2 + C throughout R.

If a harmonic function 'p has a constant value C on S, 'p = Cthroughout R; for the harmonic function cp - C is 0 on S.

110. Electric Point Charges. By Coulomb's Law, an electriccharge el at 0 exerts a force of

(1) F = el2 R fr2

upon a charge e2 at P; r = OP, and R is a unit vector in the direc-tion OP. Thus the charges repel when elel > 0 (charges of samesign) and attract when ele2 < 0. The force exerted by a charge eat 0 upon a unit charge at P, namely,

(2)

eE = a R,

r

is called the electric intensity at P due to the charge e. SinceR = Vr,

(3)

eE _ -V-,r

so that E has the scalar potential (102.4),

(4)

The vector E is both irrotational and solenoidal; for

1(5) rot E = 0, div E _ -eV2 - = 0.

r

t In a vacuum, if the charges are measured in statcoulombs and the distancein centimeters, the force is given in dynes.

242 INTEGRAL TRANSFORMATIONS § III

If S is a closed surface, we have, from (107.2),

0 inside of S,J n R 147re(6) n E dS = e dS =

r2 0 0 outside of S.

We assume that the intensity due to a system of point chargese1, e2, , e,, is the sum of their separate intensities: E = EEi.Since Ei = - Vei/ri, where ri is the distance from the charge eito P,

e1 e2E _ - -+-+r1 r2

and the potential of the system is

(7)

r,en/

el e2 er1 r2 rn

If this system of charges is within the closed surface S, we have,from (6),

(8) Jn.EdS = 47r1ei.

The normal flux of the electric intensity through a closed surface isequal to 4ir times the sum of the enclosed charges.

111. Surface Charges. If a surface S carries a distributedcharge a per unit of area, the potential at a point P due to thecharged element dS at Q (regarded as a point charge a dS) isa dS/r, where r is the distance QP. If we assume that the surfacedensity a is continuous or piecewise continuous over S, the totalpotential at P is

dS

The electric intensity at a point P outside of S, due to thecharge element a dS at Q, is -a dS Vp 1/r; the total electric in-tensity at P is therefore

(2) E fov- dS = -Vpp.r

The notation Op means that P must be varied in computing thegradient. Since P is outside of S, 1/r and its first partial deriva-

3 112 DOUBLETS AND DOUBLE LAYERS 243

tives are continuous for all positions of Q on S; hence in computingE = -Vpcp, we may differentiate (1) under the integral sign.

In differentiating functions of r = PQ, we may vary either Por Q, holding the other point fixed. Thus, if R is a unit vector in-4the direction PQ,

VQr = R, Vpr = - R,

when Q and P, respectively, are varied; and, in general,

(3) Vpf(r) = f'(r)Vpr = -f'(r) VQr = -VQf(r)Consequently, we also may write (2) as

(2)' E _ 1 dS.r

In integrals, such as this, the subscript on V may be omitted onthe understanding that the variation is at dS.

From (1) and (2), we see that cp and E are continuous at allpoints P not on the surface. At such points cp is harmonic; for

/'vp"p = I crop dS = 0.

s r

At a point P on the surface, the integrals for <p and E are im-proper since r passes through zero. However it can be shown tthat if v is piecewise continuous on S, p is defined on S and iseverywhere continuous. Moreover, as P approaches a surfacepoint Q from the positive side (toward which n points) or the nega-tive side, under general conditions E approaches limiting valuesE+ and E_, such that

(4) E+ - E_ = 47rv n,

where o- and n are the surface density and unit normal at Q. Thusthe normal component of E experiences a jump of 4irv as P passesthrough the surface.

112. Doublets and Double Layers. The potential at P due toa charge -e at Q and a charge +e at Q' is (110.7)

//1 1-e e

QQ'PG r

t See O. D Kellogg, Foundations of Potential Theory, Berlin, 1929, Chapter3, §5.


If Q' approaches Q and at the same time the charges increase, sothat the product,

e QQ' = m,

remains constant, the limiting result is called a doublet of moment m.The potential of this doublet is

1 1

1

(1) P=QimQje(QQ')PQQQ,PQ} =m - OQ - ;r

for the limit of the

frallction

in the second member is the directionalderivative of 1/r in the direction of m.

A continuous distribution of doublets over a surface with mo-ments everywhere in the direction of the normal n is called adouble layer. If /in dS is the moment of the doublet at the surfaceelement dS at Q, the potential of the double layer at an outsidepoint P is

(2)

where r = PQ. Since

1_

fs r

1dS=dSl

r r2

is the solid angle subtended by dS (107.3), we also may write

(3) 1P _ -f Adas

When µ is constant, this reduces to -µf2, where Sl is the total solidangle subtended by S at P.

When µ, the moment density, t is piecewise continuous, p is con-tinuous at all points P not on the surface. At such points cp isharmonic; for

s r

At a point Q where µ is continuous and the surface has contin-uous curvature, it can be shown * that cp has definite limits cp+, (p_

t In the case of a magnetic shell, u is called the density of magnetization.* See 0. D. Kellogg, op. cit., Chapter 6, § 6.

113 SPACE CHARGES 245

according as P approaches Q from the positive or negative side ofthe surface, and that

(4) p+ - Cpl = 4rµ.

At a point P outside of S, the electric intensitycontinuously differentiable. Since pn is a functionwe have, from (2),

(5) E cep = - µn CQ VP dS.fs

1

r

But, from (85.6),

(-1

\ = 1GP X n x G'Q -n VP6Q - ,

r

so that we also may write

(6) E = Vp x 1 µn x VQ -1 dS.s r

E = -Vpca isof Q (not P),

Consequently the intensity due to a double layer has, besides thescalar potential gyp, also a vector potential A (§ 92) :

(7)

(8)

E = rot pA,

A = fLn VQ dS.r

When ja is constant, A may be transformed into a circuit integralabout the boundary of S; for, from the basic theorem (100.1),

(9) A=AfnxVs r c r

Example. Let (P be a function harmonic in a region bounded by a closedsurface S; then, at any interior point P (109.4),

d (1w(P)

47rrdS

do r/ dti.

Comparison with (111.1) and (112.2) shows that p may be regarded as thepotential due to a surface change of density a = (d'/dn)/4a and a doublelayer of moment density IA = -c/4,r.

113. Space Charges. If a region V carries a distributed chargep per unit of volume, the potential at a point P due to the chargedelement dV at Q (regarded as a point charge p dV) is p dV/r, where


r is the distance QP. If we assume that volume density p is piece-wise continuous over V, the total potential at a point P outsideof V is

(1) (P= I aT

The electric intensity at P due to the charge element p dV atQ is -p dV Vp 1/r; the total electric intensity at P is therefore

(2) E fp ©p dV = - Vpp.XSince P lies outside of V, 1/r and its first partial derivatives arecontinuous for all positions of Q within V; hence, in computingVpcp, we may differentiate (1) under the integral sign.

When P lies within the charged region V, the integrals forand E are improper since r passes through zero. But it can beshown ¶ that, when p is piecewise continuous, (p and E exist at thepoints of V and are continuous throughout space. Moreover thepotential p is everywhere differentiable, and E = - Vptp.

The equation (110.8) also may be proved for space charges;namely,

(3) fn.EdS = 47rfp dV,

the integral on the right covering all space charges within S. Theclosed surface S may either completely enclose the charged regionV or cut through it. If S encloses no charges, the integral (3) iszero.

When p is continuously differentiable, it can be shown that Ehas the same property. Now if Sl is any closed surface enclosinga subregion Vl of V, the divergence theorem shows that

fn.EdS = fdiv E dV.Yi

But, from (3),

fdiv Ed V = 47r J pdV;Yi

¶ See O. D. Kellogg, op. cit., Chapter 6, § 3.

§ 114 HEAT CONDUCTION

and, since this holds for any subregion V1 of V,

(4)

or, if we put E

(5)

div E = 47rp,

-4Trp.

247

This partial-differential equation is called Poisson's Equation. Atpoints outside of the charged region V, p = 0, and the potential Psatisfies Laplace's Equation:

(6) v2 P = 0.

114. Heat Conduction. In mathematical physics the integraltheorems often are used in setting up differential equations. Asan illustration, consider the flow of heat at a point P of a body.According to Fourier's Law, the direction of flow is normal to theisothermal surface through P; and the flow F (calories per second)per unit of surface is proportional to the temperature gradient at P.Thus F may be represented by the vector,

(1) F = -k VT cal./sec./cm.2,

where T is the temperature and k the thermal conductivity of thebody at P. Since k is positive, the minus sign in (1) expresses thefact that heat flows in the direction of decreasing temperature.

Let p and c denote the density and specific heat of the body at P.Then the rate at which heat is being absorbed in a region R boundedby a closed surface S is

acpT dV = fcp aT

dV.at at

If n is the outward unit normal to the surface S, the rate at whichheat flows into R through S is

- in FdS =fdiv (kVT) dV.

Hence, if heat is being generated in R at the rate of h calories perunit of volume,

fcp!dva=f{div (kVT) + h} dV.


Since this holds for an arbitrary region R within the body,

(2)

OTcp - = div (kVT) + h.*

at

This is the differential equation of heat conduction.If the body is homogeneous, k is constant, and (1) becomes

aT(3) cp

at= k V2T + h;

and, if there is no internal generation of heat, we have

(4)aT

= K V2 T,at

§ 115

where K = k/cp is called by Kelvin the "diffusivity."When the heat flow becomes steady the temperature distribu-

tion is constant in time. Hence in the steady state (4) reduces toLaplace's Equation V2T = 0.

Example. Find the temperature distribution in a homogeneous hollowsphere whose inner and outer surfaces are held at constant temperatures.

When the flow is steady, T is a function of r alone; hence, from (89.17),

V 2T r dr (r2T r), dr(r2)__O, T = r + B.

A and B may be determined from the conditions T = Tl when r = rl, T = T2when r = r2.

115. Summary : Integral Transformations. If f (r) is a continu-ously differentiable tensor point function, the basic integral trans-formations are:

(A) fvfdv=fnfds (fdV Vf =ids f);

(B) fri x Vf dS = fTf ds (fdS x Vf = fdr f).

* If f(r) is a continuous scalar function in a region V and ff(r) dV = 0

for an arbitrary subregion of V, then f(r) = 0 throughout V. For, if f(r) 96 0at a point P, we can surround P with a sphere a so small that f does not change

sign in a, and hence f f (r) dV 0, contrary to hypothesis.0

§ 115 SUMMARY: INTEGRAL TRANSFORMATIONS 249

In addition,

(C)

J"T2ofds= Jl rl

may be regarded as a third basic type. From these, other integraltransformations may be deduced by dotting and crossing withintensor functions. If we dot and cross in the first position, we get,from (A) :

fv.fdv=fn.fds,

(Ax) fvxfdv =inxfdS.Similarly, from (B),

fn.vxfds=fT.f,

(Bx) f(n? Vf - nV -f) dS = fT f ds,

in which V may be replaced by V8. In the B transformations thesurface integrals vanish when taken over a closed surface.

When f is a vector function, (A - ) is known as the divergencetheorem and (B - ) as Stokes' Theorem.

If we replace f by of in (B x) we obtain

(B') f (osf + J nf) dS = fm f ds,

a transformation of the same scope as (B); J = - Div n is themean curvature of the surface, and m = T x n is the unit externalnormal to the bounding circuit. From (B') we derive

(B'.) f(vs.f-i-Jn.f)ds=fm.fds,

(BI X) f(Vsxf+Jllxf)dS =fmxfds.

On a plane, n is constant, and J = 0; if we write dA for dS, theB' transformations become

(P) fV8f dA = fm f ds,


fv3.fdA

(Px) fvsxfdA=fmxf(1s.

When f is a vector, (P - ) is the divergence theorem in the plane.If f (r) is a continuously differentiable vector and rot f = 0 in a

simply connected region of 3-space, the line integral of f - dr is in-dependent of the path, and

rf = V , p(r) = f dr.

ro

If f (r) is a continuously differentiable vector, and a rot f = 0in a simply connected portion of a surface, the line integral off - dr over surface curves is independent of the path, and

ft = Grad (p, (r) =ra

rf .

dr=frf, .

dr,

where ft = f - (n f)n is the tangential projection of f on thesurface.

PROBLEMS

1. Show that a closed curve lying in a plane with unit normal a encloses anarea A given by

nA = frxdr

2. Compute the integral / n rot f dS over that portion of the sphere

r - a lying above the xy-plane when f = p(r)c.

3. Prove thatJ'r x n dS = 0 over any closed surface S. If a body bounded

by S is subjected to a uniform pressure -pn per unit of area, show that theseforces have zero moment about any point in space.

4. If the closed curve C encloses a portion of a surface S, show that

In x r dS = I JTr2ds.

5. If f = ia(x, y) + jv(x, y), prove that

ff x dr = kf fdiv f dx dy

where C is closed curve in the xy-plane enclosing the region A.

PROBLEMS 251

6. If c is a constant vector, prove that

is

where V is the volume enclosed by the surface S.7. The closed surface S encloses a volume V. If the vector f is everywhere

normal to S, prove that rot f dV = 0.V

8. Show that the centroid of a volume V bounded by a closed surface Sis given by

Vr* = z_fr2n dS.

Apply this formula to find the centroid of a hemisphere of radius a, center0, lying above the xy-plane.

9. If p denotes the distance from the center of the ellipsoid r 4) r = 1to the tangent plane at the point r, show that the integrals of p and 1/p overits surface have the values

fPds = 3V, IdS/p = cv1V,

where V is the enclosed volume [p = r n, 1/p = r 4) n]. Express theseresults in terms of the semiaxes a, b, c of the ellipsoid.

10. Find a vector v = f(r)r such that div v = rm (in 96 -3). Prove that

frmdv11. Show that

(a) rrmR dV = -M

1 rm+in dS

(b)R

dV = flog r n dS.r

12. If p and q are vector functions, prove that

(a)

(m76 -1);

13. If r = PQ, the solid angle subtended at P by the surface S (over whichQ ranges) is

nx(cxr)dS=2Vc

rmr n dS.

iSlp = - n VQ - dS (107.1),

is r


where V0l /r is computed at Q. Prove that

Vp12p 1TXVQdSr ra

where r = PQ and the curve C is the boundary of S.

14. If C and C' are two unlinked closed curves, show that if r = PP' thedouble-circuit integral,

r TxT'ds (is' = 0 .

ar

If C and C' are two simple loops, linked as in a chain, show that the precedingdouble integral is ±4a, the sign depending on the sense of C'.

15. Show that tangential forces of constant magnitude o- acting along a closedplane curve C are equivalent to a couple of moment 2QA, where A is the vectorarea enclosed by C. What is the moment of the couple when C is a twistedcurve? [Put f = r in (100.4).]

16. Let f(r) be a solenoidal vector function. Prove that f = rot g (§ 92)where

t"OD

g = -r X fxi(,r) dX or r x j Xf(ar) dXo t

provided the integrals exist. (The integrand of the former may become infi-nite at the lower limit.)

In particular if f(r) is homogeneous of degree n 0 -2, f(,\r) = X"f(r) andg is given by g = (f x r) /(n + 2). Prove this formula directly by making useof r Vf = of and (85.6).

CHAPTER VII

HYDRODYNAMICS

116. Stress Dyadic. Let S be any closed surface inside a de-formable body. It divides the body into two portions, A within S,and B without S. The forces acting on A are of two kinds: (i)body or mass forces which act on the interior particles, and (ii)surface forces, which act on the bounding surface S.

Let k denote the average body force on the element of massAm; then, if Am shrinks to zero while always enclosing a point P,the limit of P/Om is defined as the body force R, per unit of mass,at P.

Let F,, denote the average surface force acting on A over thevector element of surface nOS of the boundary S; then, if ASshrinks to zero while always enclosing a point P, the limit ofFn/AS is defined as the surface force F, per unit of area, actingon A at P. Here n denotes a directed line normal to S at P anddirected from A to B. The notation Fn thus associates a surfaceforce with a surface element of unit normal n. The force F dS isexerted on the element n dS by the matter toward which n points.

The surface force acting on B on the same element n dS is, fromthe law of action and reaction,

(1) F_n = -Fn.Consider now a small tetrahedron (Fig. 116) with three faces

parallel to the coordinate planes; and let outwardly directed linesnormal to the faces be denoted by -x, -y, -z, n. If the areaof the inclined face is A, the faces normal to x, y, z have areasA cos (n, x), A cos (n, y), A cos (n, z), and the volume of thetetrahedron is 3Ah where h is the altitude of P above the base A.If body and surface forces are replaced by averages and p denotesthe average density, the equilibrium of the tetrahedron requiresthat

F,,A + F_IA cos (n, x) + F_NA cos (n, y)

+ F_ZA cos (n, z) + 3Ahpr2 = 0.253

254 HYDRODYNAMICS §I II

If we divide out A and let h ---. 0 so that the tetrahedron shrink,to the vertex P, we have in the limit

F,, + F_x cos (n, x) + F_ cos (n, y) + F_Z cos (n, z) = 0.

If n is a unit vector along the line n, cos (n, x) = n i, etc.; and,since F_x = -Fr, we have

(2) Fn = n (i Fx + i Fy + k Fe).

All surface forces in (2) now refer to the point P. Thus a plan(through P normal to n divides the body into two parts; and thE

-x

Fla. 116

part toward which n points acts upon the other part with the fort(F,, per unit of surface at P. The dyadic.,

(3) c= i Fx + j Fy + k F=,

is called the stress dyadic at P; it effects a synthesis of all surfac(forces at P by means of the relation,

(4)

If the deformable body is a fluid, we assume as an experimentafact that the stress across any surface element of a fluid in equilibriumis a pressure normal to the element; hence

Fn = -pn; F. = - 711, Fv = - p2J, Fz = - p3k.

With these values, (2) becomes

pa=nipli-f-nj P2j+nkp3k;

117 EQUILIBRIUM OF A DEFORMABLE BODY 255

and, on multiplying by i , j , k , in turn, we have

(5) p=P1=P2=P3At any point within a fluid the pressure is the same in all directions.At any point where the pressure is p, the stress dyadic is

(6) `, = -p(ii + ii + kk) = -pI.

For fluids in motion it is no longer true that only normal stressesexist. Viscous fluids in motion do exert tangential stresses; but inmany problems these tangential stresses are small and may beneglected. We shall develop the mechanics of fluids on the hypoth-esis of purely normal stress. We imply this assumption by speak-ing of perfect or non-viscous fluids; for a perfect fluid the stressdyadic is -pI.

117. Equilibrium of a Deformable Body. If a body in a strainedstate is in equilibrium, any portion of it bounded by a closed sur-face S is in equilibrium under its body forces and the surface forcesacting on S. Let the body forces be R per unit of mass, and let thesurface forces Fn per unit of area be given by the stress dyadic 4.Then, from (116.4), F,, = n 4, and the equations of equilibrium are

,(1) fRdv+fn.ds=o

(2) J'r x Rp dV + jr x (n (b) dS = 0,

where (2) is the moment equation about the origin.If we transform the surface integrals by means of (106.5),

in - -1) dS =fv 4dV,

frx(n.4)ds= xrdS= -

(1) and (2) become

f(R + V )dV=0,

f[PRxr+ V.(xr)]dV = 0.

256 HYDRODYNAMICS § 117

Since these integrals vanish for an arbitrary choice of S, their inte-grands must be identically zero. The equations of equilibrium are,therefore

(3) =0,(4) pRxr+V.(cxr) =0.

On eliminating pR from these equations, we obtain

(5) V.(4)xr) - (V -4)) -r = 0.

From (5) we conclude that 4), the vector invariant of c, is zero;for, if we write 4) = i F,z + j Fy + k F2,

a

ya - (aFx

+ a

a yy aFZ\

ax(Fxxr) +a (F,-r) +az (FZxr)

ax + az xr-0,Fx'< +Fyx-+F2xar= Fxxi+Fyxj+FLxk=0,

ax ay az

and hence 4) = 0. In view of the theorem of § 68, we see that themoment equation (2) requires that the stress dyadic (D be symmetric.

If we writeFx = i Xx -f - j Yx + k Zx, . . .

the stress dyadic may be written in the nonion form,Xx Yx Zx

(6) c = Xy Y, Zy ,

XZ YZ ZZ

where, by virtue of its symmetry,

(7) Xy = Yx, YZ = Zy, Zx = XZ.

The normal components of stress Xx, Yy, ZZ occur in the principaldiagonal, whereas the tangential components, or components ofshear, are equal in pairs. More specifically, in perpendicularplanes, the components of shear perpendicular to their line of inter-section are equal.

Since 4) is symmetric, we always can find three mutually or-thogonal vectors i, j, k, so that

(8) 4) = a ii + j3 jj + y kkThen i, j, k give the principal directions of stress, and a, arethe principal stresses at the point in question.

§ 119 FLOATING BODY 257

118. Equilibrium of a Fluid. For a fluid the stress dyadic is

4_ -pI and -Vp,from (86.8); the equation of equilibrium (117.3) is therefore

(1) Vp = pR.

Consider a liquid of constant density p in equilibrium undergravity. The body force per unit mass is then g, the accelerationof gravity, and

(2) Op = pg = pgk = pg Oz = V (pgz)

when the z-axis is directed downward along a plumb line Hence

(3) p = p.9z + po,

where p = po when z = 0. If the origin is at a free surface, po isthe atmospheric pressure, and pgz is called the hydrostatic pressure.

119. Floating Body. Let the sur-face S of a floating body V be di-vided by its plane section A at thewater line into two parts: S1 sub-merged, S2 in air (Fig. 119). If po isthe atmospheric pressure, pi = pgzthe hydrostatic pressure at depthz, po + pi acts on S, po on S2; orwe may say that po acts over S,while p1 acts on the closed surfaceS1 + A (pl = 0 on A) enclosing the

Fia. 119

volume V1 below the water line. Hence the equations of equi.librium of the body are:

(1) fgdm -ispondS -i+ p1ndS = 0,is A

(2) frxgdrn -frxnpodS - f rnpl dS = 0.S S,+A

We consider in turn the integrals in (1) :

fgdm = W, the weight of the body;

fpondS = poin dS = 0;

258 HYDRODYNAMICS

Vpi fpgn pl dS = dV = dV = W1,

§ 120

the weight of the displaced liquid. Therefore (1) states thatW = W1i this is the

PRINCIPLE OF ARCHIMEDES: A floating body in equilibrium dis-places its own weight of liquid.

In order to interpret (2), we note that the center of mass of adiscrete set of particles Pi of mass mis defined as a centroid (9.3);its position vector r* is given by

(3) mr* =

where m is the total mass. Similarly, the center of mass of acontinuous body is defined by

-(4) mr* =fr dm.

If we change signs throughout in (2), we have the followingintegrals to consider:fgxrdm = gxfrdm = gxmr* = Wxr*,

vvv

n x pir dSs+A

ojnxrdSs

= po Ifrot r dV = 0,s

= Trot (p1r) dV = j(Opl) x r dVvt v

= g x pr dV = g x m1r= W1 x r.vt

Thus (2) reduces to W x (r* - r) = 0; this states that the centerof mass of the body and of the displaced liquid are on the same vertical.

120. Equation of Continuity. Let p and v denote the densityand velocity of a fluid at the point P and at the instant t. Con-

sider the mass of fluid dV within a fixed but arbitrary closed

surface S. This mass is increasing at the rate,

a ap

atdV = J at dV'

§ 120 EQUATION OF CONTINUITY 259

the time differentiation being local. This rate must equal the rate

at which fluid is entering S, namely, - in pv dS, where n de-

notes the outward unit normal. Hence

V = - fn pv dS = - fdiv (pv) dV,datIap

when the divergence theorem (106.5) is applied. Since the inte-gral of ap/at + div (pv) over any closed region within the fluid iszero, we conclude that

(1)

ap

at +div (pv) = 0.

This equation of continuity may be put in another form by intro,ducing the substantial rate of change dp/dt instead of the local timerate Op/at. Along the actual path, or line of motion, of a fluidparticle,

x = x(t), y = y(t), z = z(t),

the density p(t, x, y, z) becomes a function of t alone, and

dp OP OP dx ap dy ap dz

dt at + Ox dt + ay dt + az dt

OP OP+

ap

+ kOP

at +° \l OZ

gives the time rate at which the density of a moving fluid particleis changing. We thus obtain the important relation,

(2)

dp OP+ v - op'

dt at

connecting the substantial and local changes of p.For any tensor function f(t, r) of time and position, associated

with a fluid particle moving with the velocity v, we have, in thesame way,

(3)df of

dt at + v Of.

260 HYDRODYNAMICS

From (85.2),

div (pv) = v - Vp + p div v;

hence the equation of continuity (1) also may be written

(4)

dpdt+pdivv=0.

§ 121

Since div v = - (dp/dt)/p we can interpret div v as the relativetime rate of decrease of density of a fluid particle having thevelocity v. Thus a positive value of div v implies a negativedp/dt and, consequently, an attenuation of the fluid at the pointconsidered; hence the term divergence.

For an incompressible fluid, dp/dt = 0, and

(5) div v = 0.

This equation has the integral equivalent,

(6) in - v dS = fdiV v dV = 0;

the "flux" of an incompressible fluid across the boundary of afixed closed surface within the fluid is zero. If an incompressiblefluid is also homogeneous, p is constant.

121. Eulerian Equation for a Fluid in Motion. In the Eulerianor statistical method of treatment we aim at finding the velocity,density, and pressure (v, p, p) of the fluid as functions of the timeat all points of space (r) occupied by the fluid. Consider the fluidwithin a fixed closed surface S at any instant t. By D'Alembert'sPrinciple, the body and surface forces, together with the reversedinertia forces (-ma), may be treated as a system in statical equi-librium. In a perfect fluid the surface force is a normal pressure:Fn = -pn (§ 116). Hence, if the body force per unit mass is de-noted by R, D'Alembert's Principle applied to the fluid enclosedby S gives the equation,

(1) f(R_a)Pdv_fnpds=o,

(2) frx(R - a)pdV -frXn pdS = 0,

5 121 EULERIAN EQUATION FOR A FLUID IN MOTION 261

where a denotes the acceleration of the fluid particles. On trans-forming the surface integrals in (1) and (2), we have

f {(R-a)p-Vp}dV=0,

f {r x (R - a)p + rot (pr) } dV = 0.

Since these integrals vanish for any choice of S, the integrands areidentically zero. From (85.3), rot (pr) = (Vp) x r; the equationsof motion are therefore

(3) (R - a)p - Vp = 0,

(4) rx(R-a)p-rxVp=0.When (3) holds good, (4) is identically satisfied and may beomitted. The Eulerian Equation of Motion is therefore

1(5) a=R--Vp.

P

Here a = dv/dt is the acceleration of a moving fluid particle andmust be distinguished from av/at, the rate of change of fluid ve-locity at a fixed point. These substantial and local rates of changeare connected by the relation (120.3) :

dv av

(6) a dt at+v - Vv.

When the density p is a function of p only, we introduce thefunction:

P dp dP 1(7) P = fe - ; then VP = - Vp = - Vp.

P dp P

Moreover, if the body force R has a single-valued scalar potentialQ (§ 102),

(8) R = - VQ;

such forces are said to be conservative. Under these conditions(5) becomes

dv(9)

dt= - v (Q + P).


In the Eulerian method r and t are independent variables, sothat ar/at = 0. Equation (120.3) applied to r,

dr or

dtI=v,

is simply an identity.The.lines of the fluid which at any instant are everywhere tan-

gent to v are called its stream-lines. They are not the actual pathsof the fluid particles except when the flow is steady, that is, whenv is constant in time (av/at = 0). The stream-lines have the dif-ferential equation v x dr = 0.

Example. Revolving Fluid. If the fluid is revolving with constant angularvelocity w = wk about a vertical axis, its velocity distribution is that of arigid body; hence, with the origin on the axis of revolution, we have, from-4(54.3), vp = w x OP, or simply v = w x r. Then a = w x v, and (9) becomes

(i) wx(wxr) = -V(Q+P).

Now

hence, from (i),

co x (w x r) = r - (ww - w2I)

V[Q + P -

- 2v{r (w2I - ww) r} (85.14)

- 1 w2 v(r2 - z2)

- 2w2 0(x2 + y2);

2W2(x2 + y2)1= 0,

Q + P - 21w2(x2 + y2) = cont.

For gravitational body forces, R = g, and Q = gz, if the z-axis is directedupward; and, if the density is constant, P = p/p. In this case, (ii) becomes

gz + p/p - iw2(x2 + y2) = cont.

At a free surface p is constant; a free surface is therefore a paraboloid ofrevolution.

122. Vorticity. Starting with the Eulerian Equation in the form.

(1)

av

at + v vv = - V (Q + P),

we transform v Vv by means of (85.9) and (85.7) :

v Vv = (Vv) v - v x rot v = 2 V(v v) - v x rot v;

§ 123 LAGRANGIAN EQUATION OF MOTION

hence (1) may be written

av(2) - - vxroty =

at

Now, from (84.11),

avrot -- vxroty = 0,at

arot V

-o(Q + P + 2v2).

263

at= rot (v x rot v)=

(rot v) Vv - v - V rot v - (rot v) (div v),

when we make use of (85.6) and (84.13) ; and, since

arot y drot yat dt '

drot y(3) dt + (rot v) (div v) = (rot v) Vv.

When v gives the velocities of a rigid body having the instan-taneous angular velocity w, rot v = 2w (§ 84, ex.). For a liquidwe may regard

(4) w=yrot Vas the molecular rotation or vorticity of the fluid particles.

In (3) we now replace rot v by 2w; and, from the equation ofcontinuity (120.4),

dp+

d

dt

/1\(5) p diV v =0, diV v = p

dt p

Then (3), after division by p, becomes

ldw d(1\=

w

p dt+wdt\p/ P

vv,

d (w _ co

This is known as Helmholtz's Equation.123. Lagrangian Equation of Motion. In the Lagrangian or his-

torical method of dealing with a moving fluid, the motion of theindividual fluid particles is followed from their initial positions ro


to their position r after a time t. Thus the history of each fluidparticle is traced. If the function f(r) is associated with a fluidparticle, r is a function of the independent variables ro and t.

In space differentiation we may form gradients relative to r orro; these are denoted by the symbols V and vo. Thus we maycompute df (§ 90) as either

df = dro vo f or df = dr Vf;and, since

(1) dr = dro vor,

and

dro vof = dro Dor offor arbitrary displacements dro, we have

(2) vol = vor vf.

In particular, when f = ro, we have

(3) I = vor vro,

so that the dyadics vor and vro are reciprocal. Consequently, ifwe multiply (2) by vro as prefactor, we have

(4) of = vro vofAn element of fluid volume dVo is altered by the transformation

(1) in the ratio J/1, where J denotes the third scalar invariant ofvro (§ 70) : thus

(5) dV = J dVo.

If we use rectangular coordinates,

axo/ax ayo/ax azo/ax

(6) vro = axo/ay ayo/ay azo/ay

axo/az ayo/az azo/az

and J is the determinant of this matrix, namely, the Jacobiana(xo, yo, zo)/a(x, y, z). The element of mass Po dVo becomesp dV = pJ dVo; and, since mass is conserved,

d(7) PJ = Po or

dt(pJ) = 0.

This is the equation of continuity in the Lagrangian method. Since

§ 123 LAGRANGIAN EQUATION OF MOTION 265

dp/dt = -p div v from (120.4),d (dJ

dt (PJ) = P dtJ div v = 0,

and (7) is equivalent todJ

(8) - = J div v.dt

The Dynamical Equation of Lagrange corresponding to theEulerian Equation (121.9) is obtained by multiplying the latterby Vor as prefactor; thus

dv(9) Vor

dt= -Vor V(Q + P) _ -oo(Q + P),

from (2). Since ro and t are the independent variables, the sym-bols Vo and d/dt commute; hence the left member of (9) may bewritten

d d

- {(Vor) v} - (Vov) v =

dt{ (cor) v} - vozv2,

and (9) becomes

(10) d ( (oor) v} = - Vo(Q + P - zv2).

With the aid of the dyadic Vor we may integrate Helmholtz'sEquation (122.6). If we make use of (4), this becomes

d/ w\ w w d w d) Vor,

dt p p p dt p dt

since, from (3), the time derivative of Vro Vor is zero. Multiply-ing this equation by Oro as postfactor now gives

d

dt \p / 'Oro + p \dt orb/ dt \p

fro/ - 0,10

w (00- Vro = -,P PO

the constant wo/Po being the value of (w/p) Vro when t = 0.Multiplication by Vor gives finally

P PO

This is Cauchy's integral of Helmholtz's Equation.

266 HYDRODYNAMICS 124

To verify this integral, we need only differentiate (11) with re-spect to t: thus

d(w = wo pr = 0 Ov= cooG'r

P P0 Po Po P

The lines of a fluid which are everywhere tangent to w arecalled its vortex-lines. Their differential equation is cox dr = 0.Consider a vortex line at the instant t = 0, and let its differentialequation be wo x dro = 0. After a time t, coo and dro become

w =P

coo Vor, dr = dro Vor.Po

Hence, if coo = X dro, then co = (p/po) X dr; that is, wo X dro = 0implies w X dr = 0. Thus we have proved the

THEOREM. If the body forces have a potential and p = f (p) orconstant, the vortex-lines move with the fluid. Vortex-lines alwaysconsist of the same fluid particles.

124. Flow and Circulation. The tangential line integral of thevelocity of a fluid along any path is called the flow along that path.If the path is closed, the flow is called the circulation.

If the path AB at time t was AoBo when t = 0, the flow overAB is

fB Bo

v dr f dro (Vor) v;Ao

and, since ro and t are independent variables,

6f v dr = { (Vor) vjOdroa

When the body forces have a potential Q and p = f(p), we have,from (123.10),

BpJ.Bdr= - vo(Q-- I+ ' -v)

Ao

Bo

fo

LBdr V( Q+P-2v2);

§ 125 IRROTATIONAL MOTION

or, since the last integrand is a perfect differential,

d fv.dr=(1) -- [-Q - P + v2]and,

in particular,d

(2)dt

fv.dr=0.The last result is Kelvin's

267

CIRCULATION THEOREM. If the body forces have a potential andp = f(p) or constant, the circulation over any closed curve movingwith the fluid does not alter with the time.

125. Irrotational Motion. When the body forces have a poten-tial and p = f(p) or constant, Cauchy's Equation (123.11) showsthat, if the vorticity of a fluid vanishes at any instant, it willremain zero thereafter. Then rot v = 0 in space and time, andthe motion is termed irrotational. In any simply connected por-tion of the fluid, we may express v as the gradient of a scalar(§ 102) ; thus

fr

(1) v = -Vs , P = - v dr,ro

and (p is called the velocity potential. Then the velocity is every-where normal to the equipotential surfaces (p = const and is di-rected toward decreasing potentials. Hence, along any line ofmotion, cp continually decreases; in a simply connected region thelines of motion cannot form closed curves.

By Stokes' Theorem "the circulation iv dr = 0 over any re-

ducible curve (§ 102) ; and, as this curve moves with the fluid, thecirculation around it remains zero (circulation theorem, § 124).From this fact we again may deduce that, if the motion is irrota-tional at any instant, it remains irrotational thereafter.

The equation of continuity (120.4) now becomes

(2) dt -p02(p = 0.

For an incompressible fluid, dp/dt = 0, and V2(P = 0; then <p is aharmonic function.

268 HYDRODYNAMICS

The equation of motion (122.2) reduces to

(3) v (Q + P + 2v2 - ail = 0,

§ 126

since av/at = - V (a(p/at) ; for a local time differentiation, a/at (rconstant) and a space differentiation V (t constant) commute witheach other. Hence

(4) Q + P + 12 - at = f (t)

an arbitrary function of the time; and, if the flow is steady,Q + P +

2v2 is an absolute constant.

Every harmonic function cp represents some irrotational flow ofan incompressible liquid whose velocity v = -VV. The problemconsists in finding a solution of V2V = 0 which conforms to thegiven conditions. For example, in a flow with central symmetry,<p must be a function of r alone; hence, from (89.17),

v2 V 1 dC

2 d`pl 2 d`p =a, ar2dr r drJl

0, r dr=b-,rNow v = aR/r2 where R is a unit radial vector; and the flux persecond through any sphere of radius r about the origin is constant:47rr2 (a/r2) = 47ra. When this flux is given, a is determined. Thevalue of b is immaterial, since it disappears in v = - vcp; as it iscustomary to have <p vanish at infinity, we choose b = 0.

126. Steady Motion. A flow is said to be steady when it is in-variable in time. Then all local time derivatives are zero. Thusap/ at = 0, and the equation of continuity (120.1) is simply

(1) div (pv) = 0.

Also av/at = 0 and the stream-lines are also lines of motion. TheEulerian Equation (122.2) now becomes

(2) vxroty = v(Q+P+1v2).If T is a unit tangent along a stream-line (T x v = 0) or a vortex-

line (TX rot v = 0), T v X rot v = 0; hence T V(Q + P + 2v2) _0, or

d(3)

ds(Q + P + Zv2) = 0 along a stream- or vortex-line.

3126 STEADY MOTION 269

We thus have proved

BERNOULLI'S THEOREM. Let the body forces have a potential andP = f(p) or constant; then, for a steady flow,

(3) Q + P +z

v2 = const

along any 'stream-line or vortex-line.The constant will vary, in general, from one line to another.

However, if the motion is irrotational as well as steady,V(Q+P+zv2) = 0,

(4) Q+P+zv2=C,where C is an absolute constant-the same throughout the fluid.

Now suppose that the fluid has a constant density p; this isnearly fulfilled in the case of a liquid. The equation of continuity

is div v = 0 or fn - v dS = 0. Along any tube whose surface

consists of stream-lines, let the normal cross section be denoted

by A. If the tube is sufficiently thin, and we apply -f n - v dS

= 0 to the portion between A 1 and A2, we have approximately

v1A1 - v2A2 = 0, or vA = const

as the equation of continuity along a tube of flow.If the liquid is subject only to gravitational body forces, their

potential is gz, where z is measured upward from a horizontalreference plane. Thus with

P=p/P, Q=gz,we have

(5) gz + p +2v2 = const

P

along a stream-line. If z, p, v are known at the section A0(zo, po, vo), we have, from (5),

(6) Pg(z-Z)+(p-po)+zP(v2 -vo)=0.For a thin tube we may take vA = v0A0, and (6) becomes

2(7) p - po = Pg(zo - z) -

2Pvo C2 - 1) .


Thus the pressure is least at the narrowest part of the tube.Example. Torricelli's Law. When liquid escapes from an orifice near the

bottom of a vessel which is kept filled to a constant level, the flow may beregarded as steady. Consider a stream tube extending from the orifice of thearea A to the upper surface where its area is Ao. At this surface PO is theatmospheric pressure, and vo = vA/Ao; at the orifice p = po, and the outflowspeed is v. With these values we have, from (6),

v2(1 - A2/A0) = 2g(zo - z) = 2gh,

where h is the distance from the free surface to the orifice. When Ao is largecompared with A, we have approximately v2 = 2gh, a result known as Torri-celli's Law.

127. Plane Motion. When the flow is the same in all planesparallel to a fixed plane it and the velocity has no componentnormal to Tr, the motion is said to be plane. Such motion is com-pletely determined by the motion in 7r, which may be taken asthe xy-plane. Velocity, density, and pressure are all functions ofx and y alone; and, for any tensor function f(x, y), we have, from(97.1),

Ofgrad f = Grad f + k - = Grad f;

49Z

and, similarly,

div f = Div f, rot f = Rot f.

The flux across any curve C in the xy-plane is defined as thevolume of liquid per second crossing a right cylindrical surface of

unit height based on C. For an observerwho travels the curve in the positive senseof s, the flux crossing C from (his) right toleft is

(1) fkxT.vds=fvxk.dr.

x here k, T, k x T are unit vectors along theFIG. 127 z-axis, the tangent and normal to the curve

(Fig. 127).In any simply connected portion of the fluid the flux across a

plane curve joining two points will be independent of the path,provided

k rot (v x k) = 0 (§ 103, theorem 1).

§ 127 PLANE MOTION 271

From (85.6),

rot (v x k) = k - Vv - k div v = -k div v.

For an incompressible fluid, div v = 0 (120.5) and F will be inde-pendent of the path. Hence

(2) ¢(r)ro

defines a scalar point function; and, from § 103, theorem 2,Grad 4, = v x k, since v x k lies in the xy-plane. Writing V forGrad, we have

(3) V,k=vxk, v=kxVik.

V,' is everywhere normal and v everywhere tangent to thecurves t = const; these curves are therefore stream-lines (§ 121).Consequently, the function is called the stream function.

If we set g = -4k, we have

(4) rot g=kxV,,=v, divg=0;thus the velocity has the vector potential -,pk (§ 92).

From (85.6), we have

(5) rot v = rot (k x k V2VI.

Therefore the plane motion of an incompressible fluid will be irrota-tional when and only when the stream function is harmonic:

(6) V2W = 0.

In this case v = - Vq: the velocity has a scalar potential (p; and,since

(7) V2(p= -dive=0,the velocity potential is also harmonic.

Fromv= -V<p=kxV#,

we have the relations,

(8) Vp = (VP) x k, VL. = k x Vsp;

or, in terms of rectangular coordinates,

i 3

app a,k 491P--=i--j-ax ay ay ax

272 HYDRODYNAMICS

Thus (8) is equivalent to the equations:

(9)

at 49 (P alp

5 127

ax ay ' ay ax

But these are precisely the Cauchy-Riemann Equations whichconnect the real and imaginary parts of the analytic functionw = <p + i>y of the complex variable z = x + iy. We thus haveproved the important

THEOREM. In any plane, irrotational motion of an incompres-sible fluid, the velocity potential p and the stream function ' are twoharmonic functions which combine into an analytic function V + i¢of a complex variable x + iy.

Since -,y + icp = i1G), we see that, if cp + iy is analytic,-4, + i<p is also; consequently, if cp and ' are the velocity potentialand stream function for an irrotational plane flow, -4, and V arethe corresponding functions for another flow of this type.

From (8), we have Ocp 0¢ = 0: the stream-lines (4i = const)cut the equipotential lines (,p = const) at right angles.

Since the complex potential w = cp + i>G is an analytic functionof z,

dw app a .9,P ag'w'=-=-+i-=--i-;dz ax ax ax ay

or, since the velocity has the components,

ag appvx -1 Vc , vY -j Vc

ax a,

y

(10) w' _ -vz+ivy, - w' = vx+ivy)

where w' denotes the conjugate of w'. Thus the velocity at anypoint is given by the complex vector -w'; its magnitude is I w' 1.

Example 1. Assume the complex potential w = azn (a real); then, if wewrite z = rete,

w = arner'ne = arn(COS n9 + i sin no);

,p = amcosn9, 4, = amsinn9.

The stream-lines are the curves whose polar equations are rn sin n9 = const.For the cases n = 1, 2, -1, we put z = x + iy, using rectangular coordi-

nates.

(a) n = 1: w = az = ax + iay.

§ 128 KUTTA-JOUKOWSKY FORMULAS 273

The stream-lines are the lines y = const, and the flow has the constant ve-locity -t3' = -a in the direction of -x.

(b) n = 2: w = az2 = a(x2 - y2) + i2axy.

The equipotential and stream-lines are the two families of equilateral hyper-bolas,

x2 - y2 = const, xy = const.

Since the stream-line xy = 0 may be taken as the positive halves of the x-axisand y-axis, these may be considered as fixed boundaries and the motion re-garded as a steady flow of liquid in the angle between two perpendicularwalls. The velocity at any point is -zb' _ -2a2; its magnitude varies directlyas the distance from the origin.

(c) n = -1: w = a/z = a(x - iy)/(x2 + y2).

The equipotential and stream-lines are two families of circles:

x/(x2 + y2) = const, y/(x2 + y2) = const,

tangent to y-axis and x-axis, respectively, at the origin. The velocity -w'= a/22 becomes infinite at the origin.

Example 2. With the complex potential,

w = V(z +a2/Z) = V(reie + a2r 'e-'°), (V real),

a2= V r + a2 cos B,

r >G=a2V r-a2 sin 0.r

The stream-line 4, = 0 includes the circle r = a and the x-axis sin 0 = 0.Since w' = V(1 - a2/z2), the complex velocity,

/ -2b'=-V(1-a22/-.-V

as z --+ oo.

Therefore we may regard the motion as a flow to the left about an infinitecylindrical obstacle of radius a. At a great distance from the obstacle, theflow has a sensibly uniform velocity - Vi.t

When body forces are neglected, the Bernoulli Equation (126.4) givesp/p + 1v2 = const. From the symmetry of the flow about the cylinder, it isclear that the total pressure exerted by the fluid on the cylinder is zero. Weshall consider this matter for cylinders of arbitrary section in the next article.

128. Kutta-Joukowsky Formulas. Consider an incompressiblefluid flowing past an infinite cylindrical obstacle of arbitrary crosssection. The flow, in a plane perpendicular to the generators ofthe cylinder, is assumed to have the sensibly uniform velocity vat a great distance from the obstacle. Then, if the motion is

t See Lamb, Hydrodynamics, Cambridge, 1916, p. 75, for the stream-lines.


steady and irrotational, and the body forces on the fluid are neg-lected, we have, from (126.4),

(1) p=K --v - v,

where K is an absolute constant.If n is a unit internal normal to the boundary C of the obstacle

(Fig. 128),

(2) F =jnpds, M =frxnpdsc c

give the resultant force and moment about the origin exerted by

Ay

0FIG. 128

x

fluid pressure on a unit length of cylinder. Substituting p from(1) in these integrals and noting that

in ds = 0, Jr x n ds =J rot r dA = 0,

from (101.5), (101.7), we have

(3) F=2 J

When the flow is given by the complex potential w = + i4,, weshall compute these integrals after converting them into circuitintegrals in the complex z-plane.

The vectors r = xi + yj and r = xi - yj correspond to z andits conjugate 2:

r z = x + iy = r(cos d + i sin 0) = re'6,

7 2 = x - iy = r(cos 0 - i sin 0) = re-t0.

§ 128 KUTTA-JOUKOWSKY FORMULAS

By definition,

r1 r2 = r1r2 cos (02 - 01), r1 x r2 = rlr2 sin (02 - 01)k;

hence

275

r r2 it x rk = rre,(B2--B,)= 202-

Thus

1 ' 1 2 1 2 -r1 r2 and r1 x r2 k correspond to the real and imaginary

parts of 21z2:

(4) r1 r2 - (R(21x2) = 2 (21z2 + x122),

(5) r1 x r2 . k -9(202) = 2 (21z2 -

Note also that k x r ti iz; for the multiplications by k x andi (= eia/2) both revolve a vector in the xy-plane through 7r/2.

When r1 and r2 are perpendicular or parallel we have, respec-tively,

(6) r1 r2 = 0 - 21z2 + z122 = 0,

(7) rl x r2 = 0 - 21z2 - z122 = 0.

Turning now to the integrals (3), we have the internal normaln = k x T for a counterclockwise circuit of C; hence

nds = kxTds = kxdr-idz.Moreover, if v'' v, then v v ,..' v v, and

c

The moment M, normal to the plane, is completely specified bythe scalar,

P-c2

and, since

k r x (k x T) ds = (k x r) (k x T) ds = r dr '-' 01(2 dz),

M P v v 0? (2 dz)P G1

vi 2 dz.2 c 2 c

t The real and imaginary parts of any complex number z are (R(z) _2(z + 2) and 4(z) = 2(z - 2). Note also that the conjugate of a product isthe product of the conjugates: z = 2122.


But v and dz are parallel along C, which forms part of a stream,line; hence v dz = v d2, from (7), and we have

F = -P

iJV2

dz, M dz2 2 c

for the corresponding complex force and moment. Finally we formF and replace v2z dz in M by its conjugate v2z dz; thus

dw 2jJ02 dz= P-z dz,(8)P

2 J C 2 c dz

(dwPa?

2

(9) M = - 2 6l v2 z dz = -2 \ dz) z dz,

C C

since v = -dw/dz (127.10). These integrals exist if dw/dz remainsfinite over C.

Now, outside of C, v is an analytic function which approachesv at infinity. The Laurent series for v therefore has the form:

(10) 2' = P. + al/z + a2/z2 + a3/z3 + .. .

If C. is a large circle of radius R about the origin and enclosing C-we have, by Cauchy's Integral Theorem,

ff(z) dz = ff(z) dz, if f (z) is analytic between C and C . $

Thus the counterclockwise circulation about C is

y = dr ti i J(P dz - v dz) dz dz.c cWfcv

Moreover the static moment of the circulation is

µ = fry dr-fzvdz = fz3dz.C

If we replace v by the series in (10), all integrals vanish exceptthose involving

dz Reie i dB 21r

JC. z - Reie - i f dB = 2ri.

t See, for example, Franklin, A Treatise on Advanced Calculus, New York,1940, §§ 267, 268.

§ 128 KUTTA-JOUKOWSKY FORMULAS

We thus find

(11) y = 27ri al, µ = 27ri a2.

277

We now can compute the integrals in (8) and (9) in terms ofv,,, y, and /2. From (10), we have

2v2 = v2 + 2a1v /z + (2a25 + a1) /z2 + .. .

When this series is substituted in (8) and (9), all integrals vanish

except fdz/z = 2rri; hence

(12) F = 2 i ipyv,,, F = -ipyv,,;

l)2-7ri = -pR?()u0)(13) M = - 2 6?(2a2v + a2

since a2 = -y2/4r is real, and ff?(2iri a2) = 0. In vector notationthese give

(14) F = -pykxv,,,

(15) M= -pL xF,y

for the resultant force and moment on a unit length of cylinder.From (15) we see that F acts through the point,

(16) µ/y = Jr v dr/ Jv - dr,c

which may be called the centroid of the circulation.The equations (14) and (15) are called the Kutta-Joukowsky

Formulas. From (14) we see that, for a counterclockwise circula-tion (y > 0), the force F is upward if the flow is horizontal and tothe left (v = - Vi); we then have a lift of pyVj per unit lengthof cylinder. The counterclock circulation diminishes the flow ve-locity below the cylinder and increases it above; by Bernoulli'sTheorem this results in an excess of pressure below with a conse-quent upward lift.

In the absence of circulation about the cylinder (y = 0), F = 0,and there can be no lift. This is the case in § 127, ex. 2, where

w = V(z + a2/z), v = -dw/dz = - V(1 - a2/z2),

and a1 = 0 in the Laurent series.


129. Summary: Hydrodynamics. For a perfect (non-viscous)fluid the stress dyadic is -pI; at any point within a fluid thepressure is the same in all directions and exerted normal to a sur-face element.

If f (t, r) is any tensor function associated with a fluid particlemoving with the velocity v, its substantial rate of change is

df of

dt at + V Vf,

where of/8t is the local rate of change.In the Eulerian method of dealing with fluid motion, the aim is

to compute v, p, and p as functions of the independent variablesr, t. The (kinematic) equation of continuity is

8p dp

at +div(pv) = dt+pdivv=0;

and, for an incompressible fluid (dp/dt = 0), becomes div v = 0.The Eulerian Equation of Motion is

(E) dt=-V(Q+P), P=f dp,

when the body forces have a potential Q and the density a func-tion of p only. From this we can deduce the Differential Equa-tion of Helmholtz for the vorticity co = 1 rot v:

dt \p/ = PVv.

In the Lagrangian method the aim is to follow the motion of thefluid particles from their initial positions ro (t = 0) to their posi-tions r after a time t. The independent variables are now ro andt; the position r of a particle at time t is a function of ro and t.We may take gradients relative to ro(V0) or r(V); and these con-form to the relations,

Vof = Vor Of, Vf = Vro Vof; Vor Vro = I.

If J is the third scalar of the dyadic Vro, the equation of continuitybecomes

pJ = po, or dJ/dt = J div v.

§ 129 SUMMARY: HYDRODYNAMICS 279

The Lagrangian Equation of Motion corresponding to the EulerianEquation (E) is

dv(L) Vor

dt= - Vo(Q + P).

From this we deduce Cauchy's integral of Helmholtz's Equation,

(0 0)0_=-.Vor;P Po

and this in turn shows that vortex-lines (co x dr = 0) move with

the fluid. We find, moreover, that the circulation fv dr over

any closed curve moving with the fluid does not alter with the time.When the motion is irrotational (rot v = 0), v = - VV, where

p is the velocity potential, and iv dr = 0 over any reducible

curve. When the body forces are conservative and p is a functionof p alone,

Q + P + Zv2 - a = f (t), an arbitrary function of t.

For a steady-motion equation (E) gives

vxroty=V(Q+P+Zv2);whence Bernoulli's Theorem: Q + P + Zv2 is constant along astream-line or vortex-line, and this constant is absolute if thesteady motion is also irrotational.

For an incompressible fluid in plane motion, the integral inde-pendent of the path,

¢(r) = x k dr (k I. plane of motion),

defines the stream function. The curves 4, = const are the stream-lines (v x dr = 0). If the motion is also irrotational, both streamfunction and velocity potential are harmonic (V2# = 0, V2V = 0),and V4, = k x VV. This is the vector equivalent of the Cauchy-Riemann Equations which guarantee that w = p + iii is an ana-lytic function of the complex variable z = x + iy in the plane ofmotion. The complex potential w therefore has a unique derivativedw/dz; and the negative of its conjugate gives the complex velocityvector: v = -w'.

280 HYDRODYNAMICS

PROBLEMS

1. A mass of liquid is revolving about a vertical axis with the angular speedf(r), where r is the perpendicular distance from the axis. With cylindricalcoordinates r, 0, z (we replace p, p in § 89, ex. 1 by r, 0 to avoid conflict withthe notation of Chapter VII), let a, b, c be unit vectors in the directions ofr,., rg, rZ (Fig. 89a). If the angular velocity w = f(r)c, prove that

v = rf(r)b, rot v = r dr [ref (r)]c. [Cf. (89.7).]

2. If the motion in Problem 1 is irrotational, show that

a aPC, v=rb,

and that the velocity potential v = - ao is not single valued (a, $ areconstants).

For a liquid of constant density p under the action of gravity alone, showthat the pressure is given by

gz+p+-const.

3. If a fluid is bounded by a fixed surface F(r) = 0, show that the fluid mustsatisfy the boundary condition v VF = 0.

More generally, if the bounding surface F(r, t) = 0 varies with the time,show that the fluid satisfies the boundary condition

F + v 7F = 0 [Cf. (120.3).]

4. A sphere of radius a is moving in a fluid with the constant velocity u.Show that at the surface of the sphere the velocity of the fluid satisfies thecondition

!v-u) (r - ut) = 0.5. A fluid flows through a thin tube of variable cross section A. Show that

for a tube PoP of length s

aJ8PA ds + pAv 0;

hence deduce the equation of continuity

at (pA) + as (pAv) = 0.

6. From the equation of fluid equilibrium (118.1) show that when p is afunction of p alone, rot R = 0 and that the potential of R is -P.

If p is not a function of p alone, show that equilibrium is only possible whenR R. rot R = 0. [Cf. (121.7) for P.]

PROBLEMS 281

7. A gas flows from a reservoir in which the pressure and density are po, pointo a space where the pressure is p. If the expansion takes place adiabatically,p/py = const. (y is the ratio of specific heats), show that

P = y py-1p

Neglecting body forces and the velocity of the gas in the reservoir, show thatthe velocity v of efflux when the motion becomes steady is given by

y=1v2 = 2y 7?0 1 - (p/po) 7 } .

y - 1 PO

The velocity of sound in a gas is c = 1'yp/p; hence show that

2v2 _

y- 1 (c0 - c2).

8. If a body of liquid rotates as a whole from r = 0 to r = a with the con-stant angular velocity coo, and rotates irrotationally with the angular velocityto = woa2/r2 when r = a, we have the so-called "combined vortex" of Rankine.If z = za at the free surface when r = a, show that the free surface is given by

Z = za.+-(r2-a2) r 5 a,

w2a2 ! a2`Z = Z. + - 1 1 - 2J , r a.2 \ r

Prove that the bottom of the vortex (r = 0) is a distance w2a2/g below thegeneral level (r = oc) of the liquid.

9. If the vorticity is constant throughout an incompressible fluid, provethat V2v = 0. [Cf. (84.14).]

10. When the motion of an incompressible fluid is steady, deduce fromHelmholtz's Equation (122.6) that w Vv = v Vw.

11. If an incompressible liquid in irrotational motion occupies a simplyconnected region, show that

fv2dv=f dodS'

where p is the velocity potential and the normal derivative dp/dn is in thedirection of the external normal to the bounding surface. [Cf. (108.1).]

12. Prove Kelvin's theorem: The irrotational motion (v) of an incompres-sible fluid occupying a simply connected region S with finite boundaries has

less kinetic energy (T =z

pfv v dV) than any other motion (vl) satisfying

the same boundary conditions.[Put vi = v + v'; then div v' = 0 and n v' = 0, where n is the unit nor-

mal vector to the boundary S. Show that Ti = T + T'.]

282 HYDRODYNAMICS

13. Under the conditions of Problem 11, show that if(a) v = const. over the boundary, or(b) dip/dn = 0 over the boundary,

then V is constant throughout the region and v = 0.Hence show that the irrotational motion of a liquid occupying a simply

connected region is uniquely determined when the value of either w or d'p/dnis specified at each point of the boundary.

14. If the body forces have a potential Q, the integrals,

T f a pv2 dV, U= f pQ dV,

represent the kinetic and potential energy of the fluid within the region of

integration. Show that for an incompressible fluid

d(T-[-U)_-

15. Assuming that the earth is a sphere of incompressible fluid of constantdensity p and without rotation, show that the pressure at a distance r from itscenter is

p = 2gpa(1 - r2/a2),

where a is the radius of the earth.

[If -y is the constant of gravitation, the attraction on a unit mass at thedistance r is

47rtipr3/r2 = gr/a

where g is the attraction when r = a; hence the body force -grR/a has thepotential 'gr2/a.]

Compute the pressure at the center if p is taken equal to the mean densityof the earth, (pg = 5.525 X 62.4 lb./ft.3).

16. In example 1 of § 127, consider the motion when n = r/a(0 < a < r).Since the lines 0 = 0, 0 = a are parts of the same stream-line = 0, we havethe steady irrotational motion of a liquid within two walls at an angle a. Findthe radial and transverse components of v at any point (r, 0).

17. Discuss the plane, irrotational motion when the complex potentialw = rp + i¢ (§ 127) is given by z = cosh w. Show that the equipotential linesand stream-lines are the families of confocal ellipses and hyperbolas:

xz

+y2

c2 cosh' v c2 sinh2 ,p -1,

x2 2

C2 COS2 ' c2 sin2 - 1'with foci at (fc, 0).

Show that the stream-lines 4, = nor, where n is any positive integer, corre-spond to the part of the x-axis from x = fc to x = f co. If we regard this asa wall; we have the case of a liquid streaming through a slit of breadth 2c in aninfinite plane.

CHAPTER VIII

GEOMETRY ON A SURFACE

130. Curvature of Surface Curves. Let C be any curve on thesurface r = r(u, v) with unit normal n (94.11). As the point Ptraverses C with unit speed, the Darboux vector,

(1) 6 = TT+KB,

gives the angular velocity of the moving trihedral TNB.Consider now the motion of the dextral trihedral Tnp (p = T x n)

associated with the curve (Fig. 130). Since the trihedrals Tnp and

T

T points upwardfrom paper

FIG. 130

TNB have T in common, the motion of Tnp relative to TNB is arotation about T. If p = angle (N, n), taken positive in the senseof T, the angular velocity of Tnp relative to TRB is (d<p/ds)T.Therefore the angular velocity of Tnp is

(2) w = Sds \T + ds /

-{- KB.

Now N X n = T sin p from the definition of p and n x (N x n)- p sin gyp; hence

N=ncosV - psingyp, B =TXN=pcosp+nsincp.Substituting this value of B in (2) gives

(3) (T -f dco/ds) T + K sin p n + K cos (p p.283

284 GEOMETRY ON A SURFACE § 130

The three scalar coefficients in (3) are written t, y, k and arenamed as follows:

(4) t = r -}- d<p/ds, the geodesic torsion,

(5) y = K sin (p, the geodesic curvature,

(6) k= K cos rp, the normal curvature.

With this notation,

(3)' w = tT+-yn+kp.If we reverse the positive sense on the curve, we must replace

T, s, <p by - T, -S7 -P (since T determines the sense of sp), but Kand r are unaltered (§ 45). Therefore a change of positive senseleaves t and k unaltered, but reverses the sign of the geodesic curva-ture y.

Since P is moving along the curve with unit speed, ds/dt = 1,and we take s = t. For any vector u fixed in Tnb we have du/ds= w X u (56.4) ; hence

do(7) -=wxn= -kT-tnxT.

ds

Now n is uniquely defined by (94.11) at all regular points of thesurface and do/ds = T Grad n is the same for all surface curvesthrough a point having a common unit tangent T. From (7),

do do(8) k = -- T, t = -- nxT,

ds ds

we may therefore state

THEOREM 1. At a given point of a surface, the normal curvatureand geodesic torsion are the same for all surface curves having acommon tangent there.

Since cos p = n N, we see from (6) that the curvature K =k/cos rp is completely determined by T and N at the point con-sidered, provided p 5-6 a/2. But, since T and N determine theosculating plane at the point (§ 48), we have

THEOREM 2. All curves of a surface passing through a point andhaving a common osculating plane, not tangent to the surface, havethe same curvature there, namely, the curvature of the plane curve cutfrom the surface by the osculating plane.

§ 131 THL DYADIC vn 285

In view of this theorem, we now confine our attention to thecurvature of plane sections of the surface.

Consider now the normal section Cn of the surface cut by a planethrough n and T at P. For this curve Nn = fn, cos cp = =E-1, andKn = ±k, according as N,, and n have the same or opposite direc-tions. If k 0, the center of curvature of C at P, namely,

Cn = r + N. 1K. = r + n/k,

is called the center of normal curvature for the direction T. Butany surface curve C tangent to T at P has c = r + N/K for itscenter of curvature. Hence c - cn = N/K - n/k, and

1 cost(C - Cn.) T = 0, (c - C,,) N = - - = 0,

K k

from (6), so that c - cn is perpendicular to the osculating planeof C.

THEOREM 3 (Meusnier). The center of curvature of any surfacecurve is the projection of the corresponding center of normal curvatureupon its osculating plane if the latter is not tangent to the surface.

Curves on a surface along which t, y, or k vanish are named asfollows:

t = 0: lines of curvature,

-y = 0: geodesic lines or geodesics,

k = 0: asymptotic lines.

When t = 0,. do/ds = -k T, from (7); hence

dr do(9) - X - = 0, along a line of curvature.

ds ds

When k = 0, do/ds = -t n x T, from (7); hence

dr do(10) - - = 0, along an asymptotic line.

ds ds

The parametric line v = const is a line of curvature when ru x nu= 0, and an asymptotic line when r,, n,, = 0.

131. The Dyadic Vn. Let e and e' = n X e be two perpendicularunit tangent vectors to the surface r = r(u, v) at the point Pl.

286 GEOMETRY ON A SURFACE

If k, t and k', t' are the normal curvature and geodesic torsionassociated with these directions, we have, from (130.7),

do _ -ke - tnxe = -ke - te',ds

do_ - k'e' - t' n X e' k'e' + t'e;

ds'

do do(1) Grad n = e

ds+ e' -- = e(-ke - te') + e'(-k'e' + t'e).

ds'

This dyadic is symmetric, since its vector invariant, Rot n = 0(96.8). We have, in fact,

(2) Rot n = - (t + t')n = 0, t' = - t;(3) Vn = -k ee - k'e'e' - t(ee' + e'e).

For brevity, here and elsewhere in this chapter, we use V insteadof Grad since no misunderstanding is possible. We state the re-sult (2) as

THEOREM 1 (Bonnet). The geodesic torsions associated with anytwo perpendicular tangents at a point of a surface are equal in magni-tude but opposite in sign.

From (1) we find that the first and second scalar invariants ofOn are - (k + k') and kk' + tt'. We write

(4) J = k + k' = - Div n,

(5) K = kk' - t2;

J is called the mean curvature and K the total curvature of the sur-face at P. Since On and its invariants are point functions overthe surface, the values of J and K are independent of the choiceof e; we therefore have

THEOREM 2. For any pair of perpendicular tangents at P, k + k'and kk' + tt' have the same value.

From (3), we have

(6) k= t= -eVne';

§ 131 THE DYADIC C-n 287

hence k and t are not altered when e is replaced by -e. If weassume that k remains finite at P and is not constant, there arecertain directions for which k attains its extreme values. To findthese we examine the variation of k as e revolves about P in thetangent plane. If the angle 0 between e and a fixed line in thetangent plane is taken positive in the sense determined by n, wehave (§ 44)

de de'=nXe = e', -=nXe'= -e;

do do

and, from (6),

dk(7) de

The extremes of k therefore occur when t = 0. If the directionel gives an extreme value k1, tl = 0. Then the perpendiculardirection e2 = n x e1, for which t2 = 0 from (2), gives another ex-treme value k2. But since k + k' is constant, if k = k1 is a maxi-mum, k' = k2 is a minimum, and vice versa.

If we take e = e1, e' = e2 in (3), we have

(8) Vn = -k1e1e1 - k2e2e2,

and the symmetric dyadic Vn appears in the standard form (72.4).Evidently el and e2 are the invariant directions of Vn with multi-pliers -k1, -k2. Moreover, if k is not constant, ±e1 and feeare the only invariant directions at P. We state these results in

THEOREM 3. If the normal curvature is finite and not constant ata point of the surface, it attains its maximum and minimum valuesfor just two normal sections, at right angles to each other, and char-acterized by the vanishing of the geodesic torsion.

The orthogonal directions e1, e2 are called the principal direc-tions at P, and the corresponding normal curvatures k1, k2, theprincipal curvatures. From (4) and (5),

(9) J = kl + k2, K = k1k2;

consequently k1 and k2 are roots of the quadratic,

(10) k2-Jk+K=0,the characteristic equation (§ 71) for the symmetric planar dyadicOn.


Consider finally the case when k is constant at P; from (7),t = 0 for all directions, and (3) becomes

(11) On = -k(ee + e'e') = -k(I - nn).

Any direction in the tangent plane at P is an invariant directionwith multiplier - k, and P is called an umbilical point or simplyan umbilic.

If k is constant over the entire surface, we have

Vn = -k or, V (n + kr) = 0,

and n + kr is constant over the surface. If k = 0, n is a constant,and the surface is a plane. If k 0,

n--kr=c, Ir-c12 = 1/k2;

and the surface is a sphere (center c, radius 1/k). Therefore theonly surfaces whose points are all umbilical are the plane and sphere.All curves on the plane or sphere are lines of curvature (t = 0).

Example 1. Formulas (6) give k and t for any direction e. If the angle(ele) = 0 is reckoned positive in the sense of n (el x e = n sin 0),

e =elcos0+e2sin0, e' = -elsin0+e2cosO.

Taking Vn in the form (8), we have

(12) k = e (klelel + k2e2e2) e = ki eos2 0 + k2 sin2 0,

(13) t = e (k1elel + k2e2e2) e' _ (k2 - k1) sin 0 cos 0.

These equations are due, respectively, to Euler and Bonnet. Since 2t =(k2 - ki) sin 20, it is clear that t attains its extreme values ±(k2 - kl)/2 forthe directions 0 = fir/4.

Example 2. At a surface point the directions for which k = 0 are calledasymptotic. From (12), the asymptotic directions are given by tan20 =-ki/k2 and are real and distinct only when K = klk2 < 0. When kl = 0,k2 0 0, both asymptotic directions coalesce with the principal direction el.In an asymptotic direction, t2 = -K, from (5).

Along an asymptotic line, k = K cos p = 0; and, if K , 0, cos v = 0, and,p = =1= 7r/2. Along a curved asymptotic line, the osculating plane remainstangent to the surface; moreover -y = ±K, and t = T.

Referred to the principal direction el the asymptotic directions have theslopes f and are perpendicular when and only when these slopesare ±1; then -kl/k2 = 1, and J = kl + k2 = 0.

§ 132 FUNDAMENTAL FORMS 289

Example 3. If e is the unit tangent vector of a surface curve, the trihedralee'n has the angular velocity,

(14) w = to - ke' + yn

along the curve, and

de de'ds

= cu x e = kn + ye',ds

= w x e' = to - ye.

On differentiating equations (6), we now have

(130.3)',

dVn e,ds U /

ds d-s

Since dVn/ds is the same along all surface curves having the common tangente at a point, the same is true of the expressions,

(15) ds - 27t andd-s

+ y(k - k'),

respectively discovered by Laguerre and Darboux.

132. Fundamental Forms. On the surface r = r(u, v), the gra-dient of a tensor function f(u, v) is given by (95.5),

(1) Vf = afu + bf,

where a, b, n is the set reciprocal to ru, r, n:

(2) aru + br + nn = I.From (1), we have, in particular,

(3) Vu=a, Vv=b;(4) Vr = aru + br = I - nn.The dyadic Vr is symmetric and may also be written rua + rub;it transforms any vector f(u, v) into its projection ft on the tan-gent plane at (u, v) :

(5)

Vr acts as an idemfactor on vectors tangent to the sur-face at the point (u, v) and on dyadics whose vectors lie in thetangent plane.

If we form the product,

(Vr) (Vr) = (aru + (rua + r,;b),


we obtain

(6) yr = E as + Flab + ba) + G bb,

where

(7) E = ru ru, F = ru rv, G = rv r,are the coefficients of the first fundamental form,

(8) dr dr = E due + 2F du dv + G dv2,

§ 132

defined in § 94. In fact (8) follows from (6) when we formdr yr dr and note that a . dr = du, b dr = dv, from (3).

If we compute

(vn) - (vr) = (anu + bn,) (rua +

we obtain, in similar fashion,

(9) - Vn = L as + M(ab + ba) + N bb,

whereL = - nu ru = n ruu,

(10) M = -nu r _ -nv ru = n ruv,

The relations between scalar products in (10) are obtained bydifferentiating the equations,

=0,with respect to u and v :

nu ru + n ruu = 0, nv ru + n ruv = 0,

nu rv+n rvv=0, n,,

From (9) we obtain the second fundamental form,

(11) -dr do = L du2 + 2M du dv + N dv2,

by computing the product dr Vn dr.Remembering that Vn is symmetric, we next compute

(Vn) (Vn) = (anu + bnv) (nua + nab)to obtain

(12) (Vn)2 = e as + flab + ba) + g bb,

§ 132 FUNDAMENTAL FORMS

where

(13) e = nu nu, f = nu n,,, 9 = n nFrom (12), we obtain the third fundamental form,

(14) do do = e due + 2f du dv + g dv2,

by computing the product dr - (Vn)2 dr.

291

The quantities (7), (10), and (13) are known, respectively, asthe fundamental quantities of the first, second, and third orders.

If e1, e2 are unit vectors in the principal directions at P,

Vr = e1e1 + e2e2,

Vn = -k1 e1e1 - k2 e2e2,

(on)2 = ki e1e1 -1- k2 e2e2,

and, on multiplying these equations in turn by K = k1k2, J. _k1 + k2, 1, and adding, we get

(15) (on)2+JVn+KVr=0.This is the Hamilton-Cayley Equation for the planar dyadic Vn;for its first and second scalar invariants are -J, K, and Vr is theidemfactor in the tangent plane. Substituting from (6), (9), and(12) for the dyadics in (15), we obtain the following relations be-tween the fundamental quantities:

(16) e-JL+KE=0, f -JM+KF=0,g-JN+KG=0.

Example 1. From the reciprocity of a, b, n and ru, r,., n,

(17)r,, x

,Hbnxru,

ru x r,, n; hence

(18)

We now may compute the invariants -J and K of Vn:

Vn = anu +bn,,, (On)2 = axbnuxn,,;

ruxrv n

K = (a x b) . (nu x ne,) = nu x nv aruxr,- n


Since the cross products in these equations are all parallel to n, these equa-tions may be written also in the vector form:

(19) -J ru x r,. = nu x r,, + ru x n,.,

(20) K ru x rn = nu x nv.

If we multiply these equations by (ru x r,;) , apply (20.1), and introducethe fundamental quantities from (7) and (10), we have

(21) J(EG - F2) = GL - 2F31 + EN,

(22) K(EG - F2) = LN - 312.

These values of J and K also may be found from

(9) -Vn = a(La + Mb) + b(Ma + Nb),

by making use of (18).

Example 2. The surface z = z(x, y), with the position vector,

r=xi+yl+z(x,y)k,can be written in the parametric form

x=u, y=v, z=z(u,v).If we denote the partial derivatives z.,, zy, zxx, zxy = zyx, zyy by p, q, r, s, t,respectively, we have

ru=rx=i+pk, r,, =ry=l+qk;hence, from (7),

E = 1 + p2, F = pq, G = 1 + q2;

H=EG-F2=1+ p2+g2;Hn=ruxru= -pi - gl+k.

Furthermore,ruu = rk, ru = sk, r,.y. = tk;

hence, from (10),L=r/H, M=s/H, N=t/H.

We may now compute the mean curvature J from (21), the total curvatureK from (22):

(23)

(24)

T- (1+q2)r-2pgs+(1+p2)t(1 + p2 + q2)

rt - s2K = (1+p2+q2)2

The principal curvatures k1, k2 are the roots of the quadratic (131.10):

(25) k2 - Jk + K = 0.

§ 134 TI-HE' FIELD DYADIC 293

133. Field of Curves. A one-parameter family of curves on asurface r = r(u, v) is said to form a field over a portion S of thesurface if one and only one curve of the family passes throughevery point of S. After the positive direction on one of the curveshas been chosen, the positive direction on the others is taken sothat the unit tangent vector to the curves is a continuous vectorpoint function over S.

If the curves of the field have the equation p(u, v) = const,their differential equation is

ccudu

From this we may compute Al = dv/du at all points of S. Thenthe vector,

dr ar ar av

+ au - ru -I Alrv,du a u av

is tangent to the field curve through the point (u, v).Similarly if A2 = dv/du for a second one-parameter family of

curves, the vector ru + A2r is a tangent to the curve through(u, v). When the second family cuts the first everywhere at rightangles,

(ru + Air,.) (ru + A2r,.) = 0,

or, in terms of fundamental quantities,

(1) E + (A1 + A2)F + A1A2 G = 0.

Since Al = -,pu/p is known, (1) is the differential equation of theorthogonal trajectories of the field curves. Standard existence theo-rems for differential equations of the first order guarantee, undervery general conditions, a solution of (1).

Since ru.,r are tangent vectors to the curves v = const, u =

const, the parametric curves cut at right angles when

(2)

134. The Field Dyadic. Consider a field of curves C1 and theirorthogonal trajectories C2 on a portion S of the surface r = r(u, v).At every point (u, v) their unit tangents e1, e2 = n x e1, and thesurface normal n form a dextral trihedral of unit vectors e1e2n.As a point traverses CI with unit speed, e1e2n has the angularvelocity (130.3),

(1) wl = t1e1 - k1e2 + yin;


since p1 = el x n = -e2. Similarly, along C2,

(2) (02 = t2e2 + k2e1 + y2n,

since p2 = e2 x n = e1. Therefore

de1= w1 x e1 = kin + yle2,

ds1

de1= w2 x e1 = -t2n + y2e2

ds2

We now have, from (95.8),

de1 de1Grad e1 = e1

ds1 + e2 ds2

= e1(k1n + y1e2) + e2(t1n + 72e2)

= (k1e1 + t1e2)n + (y1e1 + y2e2)e2,

since t2 = -t1 by Bonnet's Theorem (§ 131). Now, from (130.7),

do-k1e1 - t1 n x e1;

ds1and, if we write

(3) R = y1e1 + 72e2,

the field dyadic Grad e1 for the curves Cl becomes

do(4) vet = -

ds1n + R e2-

If we replace e1 and e2 in (3) by n x e1 = e2, n x e2 = -e1,R is unchanged, since y2e2 + (-y1)(-e1) = R. Therefore thefield dyadic for the curves C2 is

(5)

We have moreover

dovet =- -n - Re,.ds2

do do do do(6) Vn=e1-+e2-=-e1+-e2.

ds1 ds2 ds1 ds2

With the notation,do do

(7) P =ds1

ds2

§ 1.34 THE FIELD DYADIC

the preceding equations become

(4)' Vet = Re2 - Pn,

(5)' De2 = - Re 1 - Qn,

(6)' Vn = Pel + Qe2.

From (4), we have

or, in view of (3),

295

y1 = n rot el.

The geodesic curvature is therefore a surface invariant.

THEOREM. A field of curves having the unit tangent vector e(u, v)has at every point the geodesic curvature,

1(8) y = n rot e =

H{ (r e),, - (ru

The last expression follows from (97.9).Consider now a, curve C which cuts the curves Cl at an angle

6 = (eli e), reckoned positive in the sense of n. Along C the tri-hedral ee'n has the angular velocity

(9) w = to - ke' + yn (131.14);

and, since the trihedral e1e2n has the angular velocity - (de/ds) nrelative to ee'n, the angular velocity of e1e2n along C is

dB dB\(10) w -asn=to - ke'-F y -ds n'hence

de1 / - \w -n Xe1,ds ds

dO \ dO

The left member of (11) shows that y - dB/ds is the same for allsurface having e as common tangent vector.


Since Vel e2 = R, from (4), we may write (11) in the form:

(12)de

e R =ds

If C is a member of a field of curves, the angle 0 = (el, e) is apoint function over this field, and

(13) R = =ye+y'e'-OB.If the field curves C, and C cut everywhere at the same angle,VO=0,and R=ye+y'e'.

Example. For a field of surface curves p(u, v) = c, the unit tangent vec-tor is

e =dr/ds where puu.+cvv =0.

Thereforecv (Pa- _ _ ` ; u _ - , v

'Pv - 'Pu X A

U y 1

and, if we substitute these values in

e e =Eci2+2Fuv+Gi,2 = 1,we have

= E Vn - 2F vupt, + G woto find X. Thus

e =

where the sign is chosen to give the positive sense desired. The geodesiccurvature now is given by (8) :

(14) H lau(FP,,

Gpu) _ ava (EPv - Fvu)1

This formula is due to Bonnet.Let us apply it to find the geodesic curvature of the parametric curves in

the sense of increasing u and v. For the curves v = const, vu = 0, pv = 1,a = 1/E, el = ru/N/-r,,; for the curves u = const, vu = 1, 'RU = 0, ae3 = r/1'G; hence in the respective cases,

\ 1-G 1 }(15) Y1 =H S av E} ya = l iau G av

These formulas also follow at once from (8) when the foregoing values of eland e3 are used.

The subscript 3 refers to the curves u = const when they cut the curvesv = const at an arbitrary angle; when this angle is it/2, we use the subscript 2.

§ 135 GEODESICS 297

135. Geodesics. The geodesic curvature of a curve C has beendefined (§ 130) as

(1) K sin gyp,

where p, the angle (N, n), is taken positive in the sense of T. SincedT/ds = KN,

(2)

dT-xn = KNXn = KSin VT = yT;ds

and, as dT/ds is unaltered by a change in positive direction, y mustchange sign with T (§ 130).

A surface curve for which y = 0 is called a geodesic. Thus, if astraight line can be drawn on a surface, it is necessarily a geodesicsince K = 0. If a geodesic is curved (K 0), y = 0 implies sin <p= 0 and <p = 0 or 7r. We thus have

THEOREM 1. Along a curved geodesic the principal normal isalways normal to the surface.

Along a curved geodesic,

(3)

dipk = K COS p = ±K, t = T + - = T;

ds

and, if the geodesic is straight (K = r = 0) and the same equationsapply (k = t = 0).

THEOREM 2. Along a geodesic the normal curvature is numeri-cally the same as the curvature and the geodesic torsion equals thetorsion.

On putting y = 0 in (2), we have

d2r d2r(4) - x n = 0, or or

ds2= 0,

ds2

since the tangential projection of d2r/ds2 is zero (132.5). If wereplace Vr by aru + br,,, we obtain the scalar differential equa-tions of a geodesic:

(5)d2

r d2rru - = 0, r - = 0.

d82 dS2

These differential equations of the second order show that the geo-desics on a surface form a two-parameter family. In general, ageodesic may be determined by two conditions, by specifying (a)

298 GEOMETRY ON A SURFACE a 1:35

that it shall pass through a given point in a given direction, or (b),that it shall pass through two given points.

That a geodesic on a surface in general can be found to fullillconditions of the type (a) or (b) is plausible in view of the follow.ing theorems from mechanics.

THEOREM 3. A particle constrained to move on a surface and fieffrom the action of any tangential forces will describe a geodesic writiconstant speed.

Proof. In view of (52.5), the equation of motion ma = F maybe written

m

(ddt

T + Kv2 N) = pn,where v is the speed and pn the normal force. Multiplying by Tgives dv/dt = 0, v = const. If K = 0, the particle describes astraight line. If K F4- 0, N = ±n, and sin cp = 0. Since eitherK or sin (p must vanish, y = 0 in both cases, and the particle de-scribes a geodesic. The pressure I p I = mKV2.

THEOREM 4. If a weightless flexible cord is stretched over a smoothsurface between two of its points, its tension is constant, and the lineof contact is a geodesic.

Proof. Since the cord is in equilibrium, the vector sum of allthe forces acting upon any portion of it vanishes. Let F denote

FT

Fia. 135

the magnitude of the tension at P, distant s (arc POP) from thefixed end of the cord, and pn the normal reaction of the surfaceper unit length. Then from the equilibrium of the length s of thecord (Fig. 135),

sFT - FOTO + o pn ds = 0.0

§ 136 GEODESIC FIELD

On differentiating this equation with respect to s, we have

dF- T+FKN+pn = 0;ds

299

hence dF/ds = 0, and F is constant. If K = 0, the line of contactis straight; if K n= ±n, and I p I = KF. In either case,-y = 0, and the line of contact is a geodesic.

As a cord can be stretched between any two points of a convexsurface, the line of contact is a geodesic fulfilling condition (b).On concave surfaces we must imagine the cord replaced by a thinstrip of spring steel laid flatwise.

136. Geodesic Field. Consider now a one-parameter family ofgeodesics that form a field of tangent vector el over a portion Sof the surface. Then, since

(1) y1 0,

we have, from Stokes' Theorem,

fT e1 ds = fn rote,dS=0,

for any sectionally smooth closed curve lying entirely within S.Writing 0 = angle (e1, T) gives

(2) fcos 0 ds = 0

as the integral equivalent of (1). This formula leads to

THEOREM 1. An arc of a geodesic that is one of the curves of ageodesic field is shorter than any other surface curve joining its endpoints and lying entirely within, the field.

Proof. Let AI'B be a geodesic arc of the field and AQB anyother curve of surface covered by the field (Fig. 136a). Applying(2) to the circuit APBQB formed by these arcs, we have

f ds +PB BQA

f PB(ls = f QBcos 0 ds < /QBI cos 0 1 ds < fQBds;

that is, are APB < are AQB.


We next consider the orthogonal trajectories of a geodesic field.Their basic property is given by

THEOREM 2 (Gauss). The orthogonal trajectories of a geodesicfield intercept equal arcs on the geodesics.

Q

_A'

-A

FIG. 136a FIG. 136b

B!

B-

Proof. Let AB, A'B' be geodesic arcs intercepted between thecurves AA', BY cutting them at right angles (Fig. 136b). Thenapplying (2) to the circuit ABB'A'A, and, noting that

cos0 = 1,0, -1,0

over AB, BY, B'A', A'A, we have are AB = arc A'B'.137. Equations of Codazzi and Gauss. These celebrated equa-

tions in the theory of surfaces simply state that

(1) n - Qx(Vn) = 0,

(2) n - V x (Ve1) = 0.

If we remember that V means V8, both identities follow from(97.11).

In order to express these identities in terms of the vectors,

(3) P = do/dsl, Q = dn/ds2, R = y1e1 + '2e2,

we make use of (97.12) :

(4) n - Vx(pq) _

On substituting

(5) Vn = Pe1 + Qe2

in (1), we have, from (4),

(n - rotP)e1 + 0;

and, if we put (§ 134)

(6) Vet = Re2 - Pn, Ve2 = -Re1 - Qn,

§ 137 EQUATIONS OF CODAZZI AND GAUSS 301

we have

(n rot P - Q n x R)e, + (n rot Q + P n x R)e2 = 0.

This equivalent to the two equations:

(7) n- rot P = -n - QxR,

(8) n rot Q = -n R x P.These are the Equations of Codazzi.

We next put Ve, = Re2 - Pn in (2) and obtain

(n - rot R)e2 - (n -

Replacing vet and Vn by the foregoing values, we have

This is also equivalent to two equations. One of these is the sameas (7) ; the other,

(9)

is the Equation of Gauss. In the right-hand member,

K,

the second scalar invariant of Vn or the total curvatureGauss's Equation thus becomes

(10) n rot R = -K.

(§ 131).

This is perhaps the most important result in the theory of surfaces.The equations of Codazzi and Gauss can be expressed in terms

of the quantities ki, tt;_,yi along the field curves.t From (130.7),

do do-

= -kie, - tin x e,, as = -k2e2 - t2n x e2,i 2

and, since t.2 _ n , , -ei,(11) -P = kie, + tie2i -Q = tie, + k2e2 (134.7).

In order to compute the left members of (7), (8), (10), we applythe identity,

(12)

t Now k1 and k2 denote the normal curvatures of the orthogonal field curves;only when t1 = t2 = 0 are k1 and k2 principal curvatures (§ 131).

302 GEOMETRY ON A SURFACE § 13S

which follows at once from (85.3), to both terms of -P, -Q andR. Remembering that n rot ei = yt, we thus obtain

dk1 dt1-n rot P = y1k1 + y2t1 - - + -

ds2 dsl

-n rot Q = 'Y1t1 + -12k2 -dt1

+dk2

ds2 dsl

dyl dyeyi+y2 - +ds ds2 1

and, since

n Q x R = ylk2 - y2tl, n R x P = -yltl + y2k1,the Equations of Codazzi and Gauss become

(13) y1(k1 - k2) + 2724 +dt1 dk1- - - = 0,ds1 ds2

dk2 dt1(14) 72(k2 - k1) + 2y1t1 +

dsl-

ds= 0,

2

2 dye d-11(15) yi+yi+---+h'=0.

ds1 ds2

Equation (14) states the same property for the e2-field that (13)states for the el-field. To show this, change the subscripts 1, 2in (13) into 2, -1 (n x e2 = -e1) ; then, since

k-1 = k1i t2 = -t1, y-1 = -71, ds_1 = -ds1,

we obtain (14).138. Lines of Curvature. The curves on the surface which are

everywhere tangent to the principal directions (§ 131) are calledlines of curvature. Along a line of curvature the geodesic torsionis zero (t = 0), and the torsion z = - dcp/ds.

From (130.7), we have do/ds = -kT along lines of curvature.Their differential equation is therefore

dr do dr do

(1) dsxds= 0, or

n'dsxds= 0;

for (dr/ds) x (dn/ds) is always parallel to n. In particular, theparametric curves are lines of curvature, if

ruxnu = 0, 0;

§ 138 LINES OF CURVATURE 303

then, if k1, k2 are the principal curvatures,

(2) nu = -klru, n = -k2r,.On the plane and sphere, all curves are lines of curvature (§ 131).

On other surfaces there are in general two orthogonal principaldirections at each point; and the lines of curvature form two fieldscutting each other at right angles. If el and e2 = n X el are thecorresponding unit field vectors, the Equations of Codazzi become

dk2(3)

dk1= 'Y1

//\ki - k2), = y2(kl - k2).

dsl

These lead to a simple proof of

THEOREM 1. If K = 0, J 0 0, the lines of curvature along whichthe normal curvature vanishes are straight lines.

Proof. Since K = klk2 = 0, J = k1 + k2 0 0, we may supposethat k1 = 0, k2 0 0. Hence yl = 0, from (2); and, since

kl = K1 COS cp = 0, yj = Kj sin c = 0,

we conclude that K1 = 0.Along the rulings of the surface (which are asymptotic lines as

well as lines of curvature) t1 = 0, kl = 0, and do/dsl = 0 (130.7);hence n has a fixed direction along a ruling. This also followsfrom (134.1); for co = 0, and the trihedral ele2n has a fixed orien-tation along a ruling. In general a surface has a different tangentplane at each point and is therefore the envelope of a two-param-eter family of planes. However the ruled surface under considera-tion has the same tangent plane along an entire ruling and istherefore the envelope of a one-parameter family of planes. Such asurface is called developable.

Consider now the lines of curvature cutting the rulings orthog-onally. Their tangent vector e2 = n x el is constant along a ruling.From (137.14) and (137.15),

'Y2dk2 dye 2. _d _ 0.=ds1

122)dsl

= -y2idsl k2

But y2/k2 = tan P2; hence X02 = (N2, n) is constant along a ruling,and N2 as well. This property lends plausibility to the fact thatthe developable surface may be bent into a plane (after certaincuts are made) without stretching or tearing.

304 GEOMETRY ON A SURFACE §138

We now may amplify theorem 1: If K = 0, J 54 0, the surfaceis developable. We shall see in § 142 that developable surfaces areof three types; cylinders (including planes), cones, and tangent sur-faces (generated by the tangents to a curve which is not a straightline).

THEOREM 2. The normals to a surface along one of its curves havean envelope when and only when the curve is a line of curvature; inthis case the envelope is the locus of the centers of normal curvaturealong the curve.

Proof. Let s denote the are measured along the curve C of thesurface S, and r(s) and n(s) the position vector and unit surfacenormal at a point of C. The points on the family of normals haveposition vectors r(s) + X n(s), where X is a variable scalar. If wetake X as a function of s, say X = X(s), the curve C,

f = r(s) + X(s)n(s)

will be the envelope of the normals if

df do dX do

- =T+X+

dsn orT+X --

is parallel to n. But since both T and do/ds are perpendicular ton, the envelope will exist only when

doT + X = (1 - Xk)T - Xt n x T = 0 (130.7),

that is, when t = 0 and X = 1/k. Therefore the normals have anenvelope only along lines of curvature; and then the envelope isthe curve f = r + n/k, the locus of the centers of normal curva-ture along C.

We turn now to some theorems relative to the lines of intersec-tion of surfaces.

THEOREM 3. If two surfaces S1, S2 cut under a constant angle 0,the curve of intersection has the same geodesic torsion whether regardedas a curve of Sl or of S2.

Proof. If N is the unit principal normal along the curve ofintersection,

P2 - Pl = (N, n2) - (N, nl) = (n1, N) + (N, n2) = (n1, n2) = 0,

§ 138 LINES OF CURVATURE

and

(lg' dot2-tl =T -}---- T - -_- =O.d.s ds ds

305

From this result, we have at once

THEOREM 4 (Joachimsthal). If two surfaces cut under a constantangle, their curve of intersection is a line of curvature of both or ofneither; and, conversely, if the line of intersection is a line of curva-ture of both, the surfaces cut under a constant angle.

We are now in position to prove the celebrated

THEOREM 5 (Dupin). Two surfaces belonging to different fami-lies of a triple orthogonal system cut one another in lines of curvatureof each.

Proof. Denote the geodesic torsion at a point P on the curveof intersection of the surfaces Si and Sj by tij or tji, according asthe curve is regarded as belonging to Si or Sj. Now at the pointP we have

tij = tji (theorem 3), tij = -tik (§ 131, theorem 1),

where i, j, k represents any permutation of the indices 1, 2, 3.Then

tij = - tik = - tki = tkj = tik = - tji = - tij,

so that tij = 0.

Example. A surface of revolution about the z-axis has the parametricequations,

z = u cos v, y = u sin v, z = z(u),

where u, v are the plane polar coordinates (p, gyp).equation,

r = u R(v) + k z(u),

where R is a unit radial vector (R k = 0). Now

ru=R+kz', r =uP;ruXr k - z'R

n = _ru x r (t + z,2)

These give the vector

z" z'nu = - ru, n , r,,.(1 + z,2)2 u(1

+z'2)=

306 GEOMETRY ON A SURFACE § l39

The last equations show that the parametric lines (the parallels and meridiansof our surface) are the lines of curvature, and that the principal curvaturvare

Consequently,

(4)

kl =

Z/1

k2 =z'

(1 + z) 1' u(1 + z,2)z

z' z"

(5)

K = k k2 =l u(l + z,2)2

uz" + z'(l + z'2)J=k1+k2=u(l + z'2)2

Let us apply these results to find K and J for the torus generated by revolv-ing the circle,

(u-a)2+z2=b2,

about the z-axis. On differentiation, we find

zt =a-u b2

1 +z'2z z2

b2z" _ - 3

2

hence, from (4) and (5),

_u - a a-2uK b2u

Jbu

If we introduce the latitude 0 on the generating circle, u = a + b cos 0, and

Cos0 1 cos0(6) K _

b(a+bcose)'-J

+ba+bcos0

139. Total Curvature. The Gauss Equation (137.15),

(1)

dyl1Y1 f- yz

+- + K = 0,ds1 ds2

shows that, if orthogonal geodesics can be chosen as fielt ^urves,71 = 72 = 0, and K = 0; hence

THEOREM 1. Orthogonal geodesic fields can exist only onof zero total curvature.

We now compute K from the Gauss Equation (137.10); using(97.9), we have

1(2) - K = n rot R = - { (r R)u - (ru

H

§ 139 TOTAL CURVATURE 307

Let the parametric curves cut at an angle 0 = (el, e3) where el =ru/-/E, e3 = r, Then, if e2 = n x el, e4 = n x e3i we have,from (134.13),

and

(95.7).av

Substitution in (2) now gives the elegant Formula of Liouville forthi' total curvature,

1 ja a(-I"-\/G' ) +

0201/E) --(3) K

- H 1av (yl au au av.

in which II = 1/EG - F2, yl and 73 are given by (134.15), and

(4)F

-\/EG EG

When the parametric curves are orthogonal, 0 = it/2 and F = 0,II = EG. Equations (134.15) now give

1 aE/av 1 aG/au(5) yl = - 21i NIT ' 73 - 211 \/Gand Liouville's Formula becomes

1

(6) h'211

aau (il_ aG a / 1 aE)lIau + av \II a v f

From (3), we obtain an immediate proof of

THEOREM 2 (Gauss). The total curvature K of a surface dependsonly upon ' e coefficients E, F, G of the first fundamental form andtheir fir f .,and second partial derivatives with respect to u and v.

This iu.idamental theorem was first proven by Gauss after longand t' iious calculations. Gauss's name of "Theorema egregium"(La' n egregius means literally "out of the herd") for this resultshows that he was fully aware of its importance.

Example. To find a surface of revolution of constant negative total curva-ture, we set K = -1/a2 in (138.4). The resulting differential equation,

2z'z" 2u(1 + z'2)2 a2 '

R = ylel + y2e2 = y3e3 + y4e4 - V0;

/ -/ aeyj

-Vi, y3VG--


has the first integral,

§ 140

1 u2 u2

1 + z'2 a2A or cost i = a2 + A,

where >G = tan -l z' is the inclineftion of the tangent to the u-axis (Fig. 139).If we impose the condition u = -a when 4, = 0, A = 0 and u = -aNow

dz=

dz du= tank a sin 0 = a(sec ¢ - cos 0,

d,k du dik

z = a log (sec V, + tan p) - a sin ¢ + B;

FIG. 139

and B = 0 if z = 0 when ¢ = 0. Therefore the meridian of our surface ofconstant negative K has the parametric equations:

u = -a cos ¢, z = a log (sec V, + tan ik) - a sin ik.

This curve, asymptotic to the z-axis as 4, --> a/2, is called a tractrix. Theequation u = -a cos yG shows that the segment of any tangent between thecurve and z-axis has the constant length a. This property is characteristic ofthe tractrix (ef. § 50, ex. 5).

140. Bonnet's Integral Formula. The differential formula forthe total curvature,

(1) K = -n rot R,

has an important integral equivalent. If we integrate K over aportion of a surface S bounded by a simple closed curve whichconsists of a finite number of smooth arcs, we have, by Stokes'

§ 140 BONNET'S INTEGRAL FORMULA 309

Theorem,

fKdS= -

f(. - ds,

from (134.12); hence

(2) fKds =fdo -fy ds.

If the bounding curve has a continuously turning tangent, fdo= 2ir and (2) becomes

(3) fKdS = 2ir fds.This very important formula was discovered by the French geom-

eter, Bonnet, in 1848. The integral fK dS over S is called the

integral curvature of S.Let us first apply (2) to the figure bounded by two geodesic arcs

APB and AQB meeting at A and B. Then denoting the interiorangles at the corners by A and B (Fig. 140a), we have

(4) fK dS = fdo = 2ir - (ir - A) - (ir - B) =A+ B.

Since A + B > 0 this equation is impossible when K = 0; hence,on a surface of negative or zero total curvature, two geodesic arcs

Fu}. 140a FIG. 140b

310 GEOMETRY ON A SURFACE 3 14t)

cannot meet in two points so as to enclose a simply connected area.We may state (4) as follows:

THEOREM 1. The integral curvature of a geodesic lune is equal tothe sum of its interior angles.

We next apply (2) to a geodesic triangle ABC (Fig. 140b) : thatis, the figure enclosed by three geodesic arcs. Again denoting theinterior angles at the corners by A, B, C, we have

fKdS = fdO=27r- (7r-A) - (ir-B) - (rr-C),

(5) fKds=A+B+C_.THEOREM 2 (Gauss). The integral curvature of a geodesic triangle

is equal to its "angular excess," that is, the excess of the sum of itsangles over 7r.

When K is constant, the integral curvaturefK dS is the prod-

uct of K by the area S. If K is identically zero, as on a plane, thesum of the angles of a geodesic triangle is 7r. If K 34 0, we have

THEOREM 3 (Gauss). On a surface of constant non-zero total curva-ture, the area of a geodesic triangle is equal to the quotient of itsangular excess by the total curvature.

Example 1. A sphere of radius r has the constant total curvature K = I /r2.For all normal sections at a point are great circles of curvature 1/r; all pointsare umbilical points and k1k2 = 1/r2. Formula (5) now states that: The areaof a spherical triangle is r2 times its angular excess in radians.

If e denotes the angular excess in right angles, we therefore have the formula,

4,rr2Area = r2 2 e =

8e,

that is, the area of a spherical triangle equals the area of a spherical octant timesthe angular excess in right angles.

Example 2. Let us decompose a closed bilateral surface S, consisting en-tirely of regular points (§ 93), into a number f of curvilinear "polygons" Siwhose sides are analytic arcs. These sides meet in vertices at which at leastthree arcs come together. Since S is regular and bilateral, the surface has acontinuous unit external normal n which defines a positive sense of circuit oneach polygon by the rule of the right-handed screw. When positive circuits

§ 140 BONNET'S INTEGRAL FORMULA 311

are made about two adjoining polygons, their common side is traversed in oppo-site directions. Now for each Si we have, by Bonnet's Theorem (2),

J(i KdS=27r- (7r-ai;)-Js

where ail are the interior angles at the vertices of Si. If these equations forall of the f polygons Si are added, we have

i j

fKdS 2irf - (a -

for the integrals f y ds over adjoining sides cancel in pairs since y changes

sign with change of direction. In each polygon the number of interior anglesequals the number of sides; hence 2;2:7r = 2eir, where e is the total number ofsides or edges. Moreover the sum of the interior angles about any vertex is27r; hence TdIai; = 2v7r, where v is the total number of vertices. We thusobtain

(6) f-e+v=1 KdS.is,

From the right member of (6) we see that the number on the left is independentof the manner in which S is subdivided into polygons. From the left member

we conclude that cb K dS is not altered when S is deformed into another

completely regular surface.If S can be continuously deformed into a sphere, we have

(7)

for a sphere, and

(8)

fK dS = 47rr2 = 4a

f - e + v = 2.

Evidently f - e + v is not altered by any continuous deformation of S, eventhough the resulting surface is not completely regular. Thus, if S is trans-formed into a polyhedron (a closed solid with plane polygons for faces), therelation (8) still holds good and constitutes the famous Polyhedron Formulathat Euler discovered in 1752: Faces + vertices - edges = 2.

Next suppose that S can be continuously deformed into a torus. From(138.6), for a torus,

2 2"

(9) f"KdS=ffKb(a+bcoso)dedv = f cos e do dv, = 0;

0 0

hence f - e + v = 0 for any surface continuously deformable into a torus.This relation holds, for example, for any ring surface with polyhedral faces.


Example 3. Parallel Displacement. A vector f, that always remains tangentto a surface S, is said to undergo a parallel displacement along a surface curveC when the component of df/ds tangential to the surface is zero; then

(10)

Since

df dfnx--=0 or ds=>`n.

the length of f must remain constant during a parallel displacement.Now the trihedral Tnp associated with the curve C has the angular velocity,

w = tT + yn + kp (130.3').

If the angle 0 = (T, f) is reckoned positive in the sense of n, the trihedralfng (g = f x n) has the angular velocity n dB/ds relative to Tnp. Hence, asf moves along C with unit speed, the trihedral fng revolves with the angularvelocity,

SZ = to + n - = tT + (y + dB/ds)n + kp,

hence df/ds = St x f, and

dfxn = (12 xf) xn = (n - fl)f =C'Y

+do> f.

Hence f will undergo a parallel displacement along the curve C when and onlywhen

de(11) y+ - =0.

If f is tangent to C (or, more generally, when 0 remains constant), f willundergo a parallel displacement along C only when C is a geodesic.

THEOREM. If a tangential surface vector is given a parallel displacement abouta smooth closed curve C, it will revolve through an angle,

(12) 27r - if y ds = f K dS.

Proof. From (11) we see that f revolves through the angle,

ds - fy dsifrelative to T, the unit tangent vector of C. Since T itself revolves through 2r

in passing around C, the total rotation of f is 27r - fy ds, orJK dS by

Bonnet's Integral Formula (3). JJJJJ

§ 141 NORMAL SYSTEMS 313

141. Normal Systems. At every point of the surface S, r =r(u, v), a straight line is defined by the unit vector m(u, v). Thepoints of this two-parameter family of lines are given by

(1) rl = r + Am.

Under what circumstances do these lines admit an orthogonal sur-face? That is, when do they form the system of normals to asurface?

When A is a definite continuous function A(u, v), r1 is the posi-tion vector of a surface S1. If V denotes a surface gradient rela-tive to S, we have, from (1),

Vr1 = Or + (VX)m + AOm,

+VA,

since m m = 1. In Vr1 = arlu + brl,,, the postfactors are tan-gent to S1; hence, in order that m be normal to S1, it is neces-sary and sufficient that (Or1) m = 0, that is,

(2) - VA = (Or) m = m - (m n)n (132.5).

But (2) implies that

(3) n rot m = 0;

conversely, from § 103, theorem 2, (3) implies that

(4) mg = (Or) m = -VA where X= - f m dr,ro

the integral being taken over any path from ro to r(u, v) on S. Wehave thus proved

THEOREM 1. In order that a two-parameter family of linesr(u, v) + Xm(u, v) form a normal system, it is necessary and suffi-cient that n rot m = 0, where n is the unit normal to the surfacer = r(u, v).

When m fulfils condition (3), we can find a one-parameter familyof surfaces normal to the lines r + Am. We need only determineA(u, v) from (4) with some definite choice of ro. Since ro may bechosen at pleasure, all possible values of A are then given by


X + C, where C is an arbitrary constant. We thus obtain a one-parameter family of surfaces,

r1 = r + (X + C)m,

normal to the lines.Consider now a geodesic field over S with the unit tangent vec-

tor e. Since -y = n rot e = 0 at every point of S, we have

THEOREM 2 (Bertrand). The tangents to the geodesics of a fieldform a normal system of lines.

We conclude with an important theorem in geometrical optics.

THEOREM 3 (Malus and Dupin). If a normal system of lines isreflected or refracted at any surface, it still remains a normal system.

n

FIG. 141

Proof. Let el denote the unit vectors along the incident raysand e2 the unit vectors along the refracted (or reflected) ray (Fig.141). If µi denotes an absolute refractive index, µ1 sin 01 =µ2 sin 02 (Snell's Law) and e1, e2, n are coplanar; hence,

(µ2e2 - µ1e1) x n = 0 and n-rot (µ2e2 - µ1e1) = 0,

by the theorem following (97.10). For the reflected ray (§ 75),

e2 = (I - 2nn) e1 and n - rot (e2 - e1) = 0.

In either case: n rot e1 = 0 implies n - rot e2 = 0-142. Developable Surfaces. A developable surface (§ 138) is the

envelope of one-parameter family of planes. The limiting line ofintersection of two neighboring tangent planes is a line, or ruling,on the surface. A developable is therefore a ruled surface. Theequation of a ruled surface may be written

(1) r = p(u) + ve(u),

§ 142 DEVELOPABLE SURFACES 315

where p = p(u) is a curve C crossing the rulings and e(u) is aunit vector along the rulings. The surface normal is parallel to

ru x rv = (Pu + veu) x e.

When the normal maintains the same direction along a ruling, thesurface is developable. For the same plane is tangent to the sur-face along an entire ruling; and, as each value of u corresponds toa ruling and hence to a tangent plane, the surface is enveloped bya one-parameter family of planes. Consequently the surface (1)is developable when and only when pu x e and eu x e are parallel;that is, when pu, e and eu are coplanar:

(2) pu e x eu = 0.

Case 1: e x eu = 0. Since e e = 1, we have e . eu = 0 andhence eu = 0. Therefore e is constant, and (1) represents a gen-eral cylinder.

Case 2: e x eu 34 0. We then may write (§ 5)

(3) Pu = a(u)e + 5(u)eu.

With a function A(u), as yet undetermined, let us write (1) as

(4) r= (P+Xe)+(v-X)e=q+(v-X)e,where

(5)

Then

q(u) = p(u) + X(u)e(u).

qu = pu + Xue + Xeu = (a + Xu)e + (S + x)eu;

and, if we choose X = -,6,

(6) qu = (a - {3u)e.

If a = l3u along C, qu = 0 and q is constant. Then (4) repre-sents a cone of vertex q if the vectors e(u) are not coplanar, a planeif they are coplanar.

In general, however, a 0 $u; then (6) shows that q(u) traces acurve having e as tangent vector. The surface (4) then is gener-ated by the tangents of the curve q = q(u); it is the tangent surfaceof this curve.

Developable surfaces are planes, cylinders, cones or tangent surfaces.


143. Minimal Surfaces. In § 131 the mean curvature of a sur-face was defined as J = - Div n = kl + k2.

With f = 1, the integral theorem (101.2) becomes

(1) fJnds=Jmde.Next put f = r in (101.4) ; then, since Or = I - nn, Rot r = 0,and we have

(2) fr x Jn dS =Jrxmds.

These equations admit of a simple mechanical interpretation:

THEOREM 1. The normal pressures Jn (per unit of area) over anysimply connected portion of surface S bounded by a closed curve Care statically equivalent to a system of unit forces (per unit of length)along the external normals to C and tangent to S.

Consider, now, a soap film with a constant surface tension q perunit of length and subjected to an unbalanced normal force pnper unit of area. Since any portion of the film bounded by aclosed curve C is in equilibrium, we must have

fpn dS + 2qJmds= f (p + 2Jq)n dS = 0;

the surface tension q is doubled, because it is exerted on both sidesof the film. We therefore conclude that

(3) p = -2Jq.Thus, for a soap bubble of radius r, the normal curvature is every-where

k = «x - n = rc cos Tr =- 11r and J = -2/r;thus the pressure inside of a soap bubble exceeds the pressure out-side by an amount p = 4q/r.

In particular, if the film is exposed to the same pressure on bothsides, p = 0; then J = 0 at all points of the film. A surface whosemean curvature J is everywhere zero is called a minimal surface.Thus a soap film spanned over a wire loop of any shape and withatmospheric pressure on both sides materializes a minimal surface.

Let So be a minimal surface bounded by a closed curve C. Itsexistence is guaranteed by its physical counterpart, the soap film

§ 143 MINIMAL SURFACES 317

spanned over C. Imagine now that So is embedded in a field ofminimal surfaces, that is, a one-parameter family of surfaces in acertain region R enclosing So, such that through every point of Rthere passes one and only one surface of the family. Such a fieldmay be generated, for example, by the parallel translation of So.At each point of R, the unit normal no to the minimal surfacesmay be so chosen that no is a continuous vector-point function;and, from (97.2),

div no = Div no + nodno

do= -J + 0 = 0.

Now let S be any other surface in R spanning the curve C andenclosing with So a volume V of unit external normal n. Fromthe divergence theorem,

fn = n no over

n no cos dS < fdS.

We have thus proved

THEOREM 2. A minimal surface spanning a closed curve and be-longing to a field of minimal surfaces has a smaller area than anyother surface spanning the same curve and lying entirely in the field.

Example 1. In order to find a surface of revolution which is also minimal,we set J = 0 in (138.5). The resulting differential equation,

may be writtenuz" + z'(1 + z'2) = 0

z'z" z'z" 1

z72 +z'2+u=0

and has the first integral uz'/1/1 + z'2 = C. Solving for z', we have z' _

C/1/u2 - C2; hence

z + k =Ccosh-r C , u=Ccoshz Ck.

This represents a catenary in the uz-plane; when revolved about the z-axisit generates a surface known as the catenoid. Catenoids are the only minimalsurfaces of revolution.


Example 2. The right helicoid is a surface generated by a line which alwayscuts a fixed axis at right angles while revolving about and sliding along theaxis at uniform rates. In other words, the right helicoid is a spiral ramp.

A right helicoid about the z-axis has the parametric equations,

(4) x = u cos v, y = u sin v, z = as,

where u, v are plane polar coordinates p, p in the xy-plane. The lines v =const are the horizontal rulings on the surface; the lines u = const are circularhelices on the cylinder x2 + y2 = u2.

Equations (4) give the vector equation,

(5) r = u R(v) + av k,

where R is a unit radial vector (R . k = 0). Now

ru = R, r = uP + ak;

ruxr uk - aPn - =_ I ru x rv I a2

a anu = (u2 + a2)2 rv, nv = (u2 + a2)2 ru.

Since ru r, = 0, the parametric lines are orthogonal; and, since ru . nu = 0,r n = 0, they are also asymptotic lines. From nu x n = K ru x r (132.20),we have

K= - a2(u2 + a2)2

and, since nu x r + ru x n = 0, J = 0 (132.19). The right helicoid is a mini-mal surface. Its negative total curvature is constant along any helix u - const.

Along the lines of curvature,

(rudu +rvdv) x (nudu ruxnudu2 +rvxnndv2 = 0,

or du2 - (u2 + a2) dv2 = 0. The two families of these lines therefore satisfythe differential equations:

du/ u2 + a2 = fdv.On integration, these give

usinh-1- = c f v or u = a sinh (c f v)a

as the finite equations of the lines of curvature.

Example 3. On a minimal surface, the asymptotic lines form an orthogonalsystem (§ 131, ex. 2). If we choose them as our field curves, k1 = k2 = 0,and the Codazzi Equations (137.13), (137.14) become

27211 + i = 0, 2y1t1 - - = 0.ds2

144 SUMMARY: SURFACE GEOMETRY 319

Since K = -ti,dK dt1 dK dt1= -2t1- _ -472K, = -2t1- = 4y1K;dsl ds1 ds2 ds2

dK dKVK = el - + e2 - = 4K(y1e2 - 72e1) = 4K n x R (134.3).

dsl ds2

144. Summary: Surface Geometry. On a surface r = r(u, v) ofunit normal n, the angular velocity of the trihedral Tnp (p =T x n) along a curve is w = tT + yn + kp. If (p = angle (N, n),positive in the sense of T,

t = r+ dv/ds, y= K sin cp, k = K cos V

are the geodesic torsion, geodesic curvature, and normal curvature, re-spectively. From

do(1)

ds=wxn= -kT - tnxT,

we conclude that k and t are the same for all surface curves havinga common tangent at a point.

Important surface curves and their differential equations:dr do

Lines of curvature (t = 0) : -x- = 0;ds ds

dr doAsymptotic lines (k = 0) : - - = 0;

ds dsd2r d2r

Geodesics (y = 0) : n xds2

= 0 or Vrds2

= 0.

The equations for geodesics follow from

dT-xn=KNxn=yT.ds

If e and e' = n x e are perpendicular directions in the tangentplane,

-Vn = k ee + k' e'e' + t ee' + t' e'e.

Since Rot n = 0, Vn is symmetric, and t + t' = 0. The first andsecond scalar invariants of - Vn,

J = - Div n = k + k', the mean curvature;

K = kk' + tt' = kk' - t2, the total curvature,


have the same value for any pair of perpendicular directions. Ifthe point is not an umbilic (k constant for all directions), k attainsits extreme values in the orthogonal principal directions e1, e2 forwhich t1 = 0, t2 = 0. In terms of the principal curvatures k1, k2,J = k1+'k2iK=k1k2.

If a, b, n are reciprocal to ru, r,,, n, the first and second funda-mental forms are

Or = E as + F(ab + ba) + G bb,

-Vn = Las+M(ab+ba) -}-Nbb;and the fundamental quantities are given by

L = -ru nu, M = -ru nv = -rv nu, N = -rv nv.

If e1i e2 = n x e1 are the unit tangent vectors along a field ofsurface curves and their orthogonal trajectories, we have

Vet = Re2 - Pn, Ve2 =-Re1 - Qn, Vn = Pet + Qe2,

where

P = do/ds1, Q = dn/ds2, R = y1e1 + y2e2.

Codazzi Equations: n V x (Vn) = 0; or

Gauss Equation: n V x (De1) = 0; or

n rot R = -K.For any field of curves of tangent vector e,

y = n- rot e.

The last two equations show that y and K may be expressed interms of E, F, G and their partial derivatives.

When embedded in a geodesic field, the arc of a geodesic isshorter than any other curve lying within the field and having thesame end points.

If a field of curves of tangent vector e cuts the field e1 at theangle 0 = (eli e), positive in sense of n,

R = y1e1 + y2e2 = ye + y'e' - V0.

PROBLEMS 321

Bonnet's Integral Formula,

fK d S= fdo - ds,

is the integral ("Stokian") equivalent of the Gauss Equation,n - Rot R = -K. If the simple closed curve C has a continu-

ously turning tangent fdo = 27r, but if the path has angles (in-

terior angles ai),

fdo = 2ir - - as).

If C is a geodesic triangle, fK dS = al + a2 + a3 - ir; and, on

a surface of constant K 3-1 0, the area of a geodesic triangle is(al + a2 + a3 - ir)/K. A sphere of radius a has the constantpositive curvature 1/a2; a tractrix of revolution for which thetangential distance to the asymptote is a has the constant negativecurvature -1/a2.

A minimal surface (J = 0) embedded in a field of minimal sur-faces and spanning a closed curve C has a smaller area than anyother surface lying within the field and spanning C. Soap filmsspanned over wire framework materialize minimal surfaces. Theright helicoid is a ruled minimal surface. The catenoid is the onlyminimal surface of revolution.

PROBLEMS

1. If u, v, are plane polar coordinates r, 0, prove that the surface obtained byrevolving the curve z = f(x) about the z-axis has the parametric equations:

x = u cos v, y = u sin v, z = z(u).

2. A straight line, which always cuts the z-axis at right angles, is revolvedabout and moved along this axis. The surface thus generated is called a conoid.If u and v are plane polar coordinates in the xy-plane, show that the conoidhas the parametric equations:

x = u cos v, y = u sin v, z = z(v).

In particular when dz/dv is constant the conoid is a right helicoid:

x = U cos v, y = u sin v, z = av.

3. The central quadric surface,

x2 y2 z2 _a2 ± b2 ± 1,


is an ellipsoid, a hyperboloid of one sheet, or a hyperboloid of two sheets accordingas the terms on the left have the signs (+, +, +), (+, +, -), (+, -, -).Show that the corresponding parametric equations are

x = a sin u cos v, y = b sin u sin v, z = c cos u;

x = a cosh u cos v, y = b cosh u sin v, z = c sinh u;

x = a cosh u cosh v, y = b cosh u sinh v, z = c sinh u.

4. The paraboloid,x2 y2 2z-=-,a2t

b2 c

is elliptic or hyperbolic, according as the terms on the left have the signs (+, +)or (+, -). Show that the corresponding parametric equations are

x = au cos v, y = bu sin v, z = Zcu2;

x = au cosh v, y = bu sinh v, z =. 4cu2.

5. The position vector,r = f(u) + g(v),

traces a surface of translation. Show that any curve u = a (const.) may beobtained by giving the curve rl = g(v) a translation f(a); and that any curvev = b may be obtained by giving the curve r2 = f(u) a translation g(b).

What are the surfaces,

r=f(u)+bv, r=au+bv+c?6. Show that a right circular cylinder of radius a has the constant mean

curvature J = 1/a.7. For the surface xyz = a3 show that

(X2y2 + y2z2 + 22x2)2 ' (x2y2 + y2z2 + 22x2)7

[Cf. § 132, ex. 2.]8. For the elliptic paraboloid,

x2/a2 + y2/b2 = 2z/c

prove that the total curvature K 5 c2/a2b2.9. For the surface of revolution about the z-axis,

r = iu cos v + ju sin v + kz(u),show that

E=1+z'2 F=O, G=u2;L = 2'2)1, M = 0, N = uz'/(1 + 2'2)1. [Cf. § 132.1

From (132.21) and (132.22) deduce the values of K and J and from (132.25)find the principal curvatures. Check with (138.4) and (138.5).

3a6 Jr = 2a3(x2 + y2 + z2)

PROBLEMS 323

10. For the conoid,

r = iu cos v + ju sin v + kz(v),show that

E=1, F=0, G=u2+z'2=H2;L = 0, M = -z'/H, N = uz"/H;

K = -z'2/H4, J = uz"/H3. [Cf. § 132.1

When z = av, show that the resulting right helicoid is minimal (§ 143);and that its principal curvatures are fa/(u2 + a2).

11. Show that the curvatures J and K have the dimensions of (length) -1and (length) -22. Verify this in Problems 6, 7, 8, 9, 10.

12. The position vector of a twisted curve r is given as a function ro(s) of theare. The surface generated by its tangents is

r=rc(v)+uT(v)

where v = s and u is the distance along a tangent measured from r. Show thatthis tangent surface has the curvatures K = 0, J = z/i u Jx. What are theprincipal curvatures?

13. If a parabola is revolved about its directrix, show that the principalcurvatures of the surface of revolution satisfy the relation 2k1 + k2 = 0.

[With the notation of § 138, ex., the parabola has the equation z2 _4a(u - a) when the directrix is the z-axis and u = 2a at the focus.]

14. The vectors a, b, n and ru, r,,, n form reciprocal sets of vectors over thesurface r = r(u, v) [Cf. § 95]. Hence show that any vector f(u, v), defined overthe surface, may be written

15. Prove that

f

H2a = Gr - Fr,,, H2b = -Fru + Er,,.

[Use Prob. 14 and (132.18).]16. Prove that

FM - GL

II, a

FL - EM' b =H2 ' H2 ;

FN - GMH2

I

FM - ENH2

[Use Prob. 15.]

17. Prove the derivative formulas of Weingarten:

HZnu = (FM - GL)ru + (FL - EM)r,,,

H2n _ (FN - GM)ru + (FM - EN)r,,.

[Use Prob. 14 and Prob. 16.]


18. Show that the asymptotic lines of the surface r = r(u, v) have the dif-ferential equation,

(ru du + r dv) (nu du + n dv) = 0

[Cf. (130.10)]; or, in terms of fundamental quantities,

Ldu2+2Mdudv+Ndv2 = 0.19. Prove that the asymptotic lines on a right helicoid (Prob. 10) are the

parametric curves and form an orthogonal net.20. For the ruled surface,

r = p(u) + ve(u), (142.1).

prove that M = pu e x eu/H, N = 0.21. Prove that a ruled surface is developable when, and only when, M = 0.

[Cf. (142.2).]22. Prove that the asymptotic lines of a developable surface, not a plane, are

the generating lines (counted twice). [Differential equation: du 2 = 0.]23. Prove that, along a curved asymptotic line,(a) The osculating plane is tangent to the surface.(b) The geodesic torsion equals the torsion: t = r.(c) The geodesic curvature is numerically equal to the curvature: y = ±K.(d) The square of the torsion is equal to the negative of the total curvature:

2 = -K.24. Prove that

(a) If an asymptotic line is a plane curve other than a straight line, it isalso a line of curvature.

(b) An asymptotic line of curvature is plane.25. Prove that the following conditions are necessary and sufficient in order

that the surface be(i) A plane: L = M = N = 0.(ii) A sphere: E/L = F/M = GIN.26. Show that the lines of curvature of the surface r = r(u, v) have the

differential equation,

n (r, du + r dv) x (nu du + n,, dv) = 0

[Cf. (138.1)]; or, in terms of fundamental quantities,

I dv2 -du dv du2 I

E F G

L M N= 0.

27. Show that the parametric curves are lines of curvature when, and onlywhen, F = 0, M = 0.

28. In the notation of § 132, ex. 2, prove that the lines of curvature on thesurface z = z(x, y) have the differential equation,

dy2 -dx dy dx2

1 + p2 pq 1 + q2 = 0.

PROBLEMS 325

29. Show that the hyperbolic paraboloid of Prob. 4 has the parametricequations:

x = 2a(u + v), y = 2b(u - v), z = cuv.

Prove that the differential equations of its asymptotic lines and lines ofcurvature are, respectively,

dvl2 a2 + b2 + c2y2du dv = 0, \du / a2 + b2 + c2u2

and find the uv-equations of the asymptotic lines and lines of curvature.30. If the parametric curves on a surface are its lines of curvature, show that

nu = -kiru, nv = -k2rv.

31. Two surfaces that have the same normal lines are called parallel. Sur-face points on the same normal are said to correspond; thus

r(u, v) = r(u, v) + An(u, v)

determines corresponding points on the parallel surfaces S and S. Prove thatthe distance X between the parallel surfaces is constant.

[If we take the lines of curvature of S as parametric curves,

ru = (1 - Aki)ru + Xun, rv = (1 - Xk2)rv + Xn

(Prob. 30); and, on cross multiplication,

ffn- = (1 - Xki)(1 - Ak2)Hn + Xu(1 - Xk2)n x rv + X ,(l - Xkl)ru x n.

Now ii = En where e = ±1; and as n, ru, r are mutually orthogonal,

H = E(1 - Xki)(1 - Xk2)H, Xu = Xv = 0.1

32. Prove that the lines of curvature on parallel surfaces correspond andthat the corresponding principal curvatures satisfy the equations,

_ Ekl Eke

T2 =kl1 - Xkl ' 1 - xk2

Hence the mean and total curvature of S are given by

_ E(J - 2XK) _J - KK

1rJ+X2K' 1-xJ+x2K33. Show that surfaces parallel to a surface of revolution are also surfaces

of revolution.34. Prove the "Theorema egregium" of Gauss by establishing Baltzer's

formula for the total curvature K:

0 U Ei, GuFuv - UGuu -Evv 2Eu Fu - 2Ev

E F ZEv E FH4K=I Fv--2GuF G -'Gu F G


Method. From (132.10) and (132.22) show that

H4K = [ruurx][rv,;r r,;] - [ruvrurv]2.

Using (24.14), the right member becomes

ruu rvv

ru rvv

rv rvv

ruu ru ruu rv

E F

F G

ruv ruv ruv ru ruv rv

ruv ru E F

ruv rv F G

In both determinants the upper left element has the cofactor EG - F2; hencewe may replace these elements by ruu rvv - ruv ruv and 0, respectively.Now, by differentiating

we may show that

rut,, ru i2F!u, ruv ru = 2iEv, rvv ru = F,, i2Gu,

rvv rv = zGv, ruv rv = 2Gu> ruu r,, - Fu - 2Ev;

ruu rvv - ruv ruv - Fuv 1G !Evv.uu - 2 vv

35. On the surface r = r(u, v) let us write

u1 = u, u2 = v; D1 = a , D2 = a ;au av

el = ru, e2 = rv; el = a, e2 = b; then e" ep = SQ,

where the Greek indices have the range 1, 2. Moreover let

gap = ea ep, 9ap = ea e$, g = det gap =911 912

921 922

hap = n DaDpr = hpa; h = det ha p.

Prove that

(a) 9u= E,g12=F,922=G;g=EG-F2.(b) 911 =

922 g12 - -912 922 = 922

9 9 9that is, g"p is the reduced cofactor of gap in det gap.

(c) h11=L,h12=M,h22=N;h=LN-M2;(d) Daep - Dpea;(e) Dagpy + Dog-y. - D.rgap = 2ey Daep.(f) Daep = ey(Dagp , + Dpgya - Dygap) + nhap (summed over y = 1, 2).(g) e" = g"pep (summed over (3 = 1, 2). [Cf. § 145.]36. The Christoffel symbols Papa, g, y = 1, 2) for a surface r = r(ul, u2)

having gap du" dup as first fundamental form (§ 132) are computed from theequations:

Pap = ey . Daep.

PROBLEMS 327

With reference to Prob. 35 prove that(a) ra0 = r'a(b) rP# = Diggya - Dyga#) (summed over p = 1, 2).(c) 29r1i = 922 D1911 - 2g12 D1912 + 912 D29uu,

29ri2 = 922 D2911 - 912 D1922,29r22 = -912 D2922 + 2g22 D2912 - 922 D1922

[Use the formula of part b; for example,

21'1ii = 911(D1911 + D1g11 - D1911) + 91Y(D1912 + D1921 - D2911),129t1 = 922 Dig,, - 912(2D1912 -

(d) From the formulas for r,# obtain the formulas for rap.(e) From Prob. 35 (f) prove the derivative formulas of Gauss:

XDaep = raryea + heron (summed over a = 1, 2).

37. From n efl = 0 show that eo Dan = -han. Hence prove the deriva-tive formulas of Weingarten;

Dan = -hafieO (summed over S = 1, 2).

Check these results with Prob. 17.38. With the notations of Prob. 35 show that equations (132.21) and (132.22)

may be writtenJ = g0h,,#, K = h/g (a, 0 summed over 1, 2).

39. Show that the tangential projection of the vector curvature dT/ds = KNiof a surface curve is

d2r

(--dua duDr

= ey ds2 + raA ds ds

summed over a, 0, y = 1, 2. Hence prove that the equations of a geodesicon a surface are

y dua B

d3 + radds ds

= 0 (Y = 1, 2).

a 8

1 T = ep ds , = ep ds2 + Daeg d , Vr = eyes. Cf. Prob. 36. ]

CHAPTER IX

TENSOR ANALYSIS

145. The Summation Convention. Expressions which consistof a sum of similar terms may be condensed greatly, without anyessential loss of clarity, by indicating summation by means of re-peated indices. This usage, originally due to Einstein, is statedprecisely in the following

SUMMATION CONVENTION. Any term, in which the same index(subscript or superscript) appears twice, shall stand for the sum ofall such terms obtained by giving this index its complete range ofvalues. This range of values, if not understood, must be specifiedin advance.

By way of illustration, we repeat some of the formulas of § 24,using the summation convention. The index range is 1, 2, 3.

The two forms of a vector (24.1) and (24.2) become

(1) U = uiei, u = u;e'.

In these equations i and j are summation (or dummy) indices. Butany other letter will do as well; thus

u = carer = ule1+ u2e2 + u3e3.

In order to compute u - v, we first recall the defining equationfor reciprocal sets,

(2) ei e' = Si

Now

(3) u . v = (uiei) . (v,e') = uiv; 'Vi' = uiei;

(23.3).

we here use different indices in expressing u and v in order to getthe 32 = 9 terms in the expanded product (six are zero). Simi-larly,

(4) u v = (uiei) . (vie;) = uiv' S = u1vi.328

§ 146 DETERMINANTS 329

Note that the summation indices in the preceding examples ap-pear once as subscript and once as superscript. The significanceof this arrangement (not required by the summation convention)soon will be apparent.

146. Determinants. A permutation of the first n natural num-bers is said to be even or odd, according as it can be formed from123 n by an even or odd number of interchanges of adjacentnumbers. The total number of permutations of n different num-bers is n!; one half of these are even and one half odd. For ex-ample, when n = 3, 123, 231, 312 are even permutations, 213, 132,321 are odd.

Consider now a permutation ijk r of the numbers 123 n.We then define the permutation symbols ijk...r and eijk""r asequal to 1 or -1, according as ijlc . r is an even or an odd per-mutation of 123 . . . n; but, if any index is repeated, the epsilon iszero. For example, when n = 3,

E123 = 231 = 312 = 1 E213 = 132 = 321 = - 1

112 = 122 = 222 = 0.

The epsilons just defined are useful in dealing with determi-nants. To be specific, we shall take n = 3; but all the formulasapply without change of form for any value of n.

From the definition of a determinant :

(1) a = det a _

1 2 3a1 a1 a1

as a22 a2

3

1 2 3a3 a3 a3

ei jka 12a3

k 1 2 3e' ai aj ak

The implied summations on i, j, k produce 33 = 27 terms; of these,21 involve repeated indices and therefore give a zero epsilon,whereas 6 involve permutations of 123 and give precisely the3! = 6 terms of the determinant. (Write them out!)

The theorems relative to an interchange of rows and columnsare given by i '

(2) eijkarasataer8t = ijkarL

a8atk

Proof. When r, s, t are 1, 2, 3, equations (2) reduce to (1).Since rst changes sign when two adjacent indices are interchanged,

330 TENSOR ANALYSIS § 146

(2) will beestablished when the right members are shown to havethis same property. Now

Eijkarasat = eijkasarat = - ejikasa r aki

and, if we interchange the summation indices i, j, the last expres-sion becomes - eijkasajat .

If we multiply the first equation of (2) by e" and sum on rst,we obtain

(3) 3! a = eijkerstarasae.

On the left er8tergt, summed over the 3! permutations of 123, equals3!; on the right, the summation extends over all six indices andproduces 36 = 729 terms. Many with zero epsilons vanish; and

a3 appears six times, corre-each non-vanishing term such as a1la22 3sponding to the 3! permutations of its factors. Equation (3) thusis not useful for computation; its importance rests on the informa-tion it gives about the determinant when its elements are trans-formed.

The cofactor of an element az in the determinant (1) is definedas its coefficient in the expansion of the determinant. The co-factor of a; is denoted by A'. To find A',* strike out the row andcolumn in which ai stands; then A equals the resulting minortaken with the sign (-1) i+j. If the elements of any row or col-umn are multiplied by their respective cofactors and added, weobtain the determinant-its Laplace expansion; but, if the ele-ments of any row (column) are multiplied by the correspondingcofactors of another row (column) and added, the sum is zero.These important properties both are included in

(4) aiA; = a'Ai = a S .

Here r is the summation index. If in (4) we put j = i, both r andi are summation indices, and we get

(5) a7AT = a; A, = a Si,

where S = Si + S2 + S3 = 3 (n = 3). But, if we put j = i in(4) and suspend summation on i, we obtain the Laplace expansions,

(6) a7,Ar = arAi = a (i fixed),

for S = 1 for a fixed i.

§ 146 DETERMINANTS 331

A cofactor divided by the value of the determinant is called areduced cofactor. The reduced cofactor of a2 is therefore

(7) a; = A,/a (a F6 0).

In terms of reduced cofactors, equations (4) become

(8) aTc = 8a.

When the elements of a determinant are written aid, its defini-tion becomes

(9) a = E'.'kalia2Ja3k = E1Jkailaj2ak3-

The reduced cofactor of ai3 then is written ai and equations (8)become

(10) aira'r = aria'' = SL.

If the elements ai; are functions of a variable x, we have, from(9),

da _ ilk (dali dal; da3k\

=E

xa2ja3k + a1i

xa3k + aiia2i

dx dx )

= dal-Ali+da2jA21+da3kA3kdx dx dx '

where A 'j denotes the cofactor of aij. Summing on two indices,we may write

da daii(11) - = Ai' (i, j = 1, 2, 3).

d.r, dx

The derivative of a determinant is the sum of the products formed bydifferentiating each clement and multiplying by the cofactor of theelement.

The properties (8) of reduced cofactors enable us to solve asystem of linear equations when the determinant of the coefficientsis not zero. Consider, for example, the equations:

(12) ax' = yt (i, j = 1, 2, 3; a 0).

To solve these for x', multiply (12) by the reduced cofactor akand sum on i : the left member becomes S; x' = xk, and we obtainxk = a;yi, or, on replacing k by j,

(13) x' = aiyi (i, j = 1, 2, 3).

332 TENSOR ANALYSIS 5 147

To find the product of the determinants a = det a, b = det b;,we have, from (1) and (2),

(14) ab = EiJkaia2a3b

s t= a1

ia2 ja3 kEr8tbsrbjbk

= Erst(albi)(aabj)(a3kbk)

r 8= ErstC1C2Ct3,

where

(15) c% = aabj = aib1 -F a2b2 -f- a2b33.

From (15), we see that the element in the ith row and jth columnof the product ab is given by the sum of the products of the corre-sponding terms in the ith row of a and the jth column of b-theso-called "row-column" rule.

Since the value of a determinant is not altered by an interchangeof rows and columns, we also can compute ab by "row-row,""column-row," and "column-column" rules. For example, theproduct

(16)

1 2 3a1 al al

a2 a2 a2

1 z 3a3 a3 a3

1 1 1

a1 a2 a3

2ai a2 a3

3 3 3a1 a2 a3

1 0 0

0 1 0

0 0 1

= 1.

by use of the row-row rule.

THEOREM. If a determinant a 0, the determinant formed by re-placing each element by its reduced cofactor is 1/a.

If we solve equations (13) for yi by using the reduced cofactorsin det a, we will obtain (12); in other words, the reduced cofactorof a in its determinant is a. The two determinants in (16) thusare reciprocally related: each is formed from the reduced cofactorsof the other. .'.

147. Contragredient Transformations. Let us now introduce anew basis e1i e2, e3 by means of the linear transformation,

61 = cie1 + ciez + Cie3,

(1) e2 = c2e1 + c2e2 + 4e3,

e3 = c3e1 + c3e2 + c3e3,

§ 147 CONTRAGREDIENT TRANSFORMATIONS 333

where the coefficients ci are real constants whose determinantc 0. In brief,

(1) ei = c'ej, c = det ca /- 0.

The condition c 0 ensures that e1, e2, e3 are linearly independ-ent. For, if there were three constants At such that

Aiei = 0, then Aicej = 0;

the linear independence of the vectors ej requires that c'Ai = 0,and, since c P 0, these equations only admit the solution Ai = 0.This argument applies without change to space of n dimensions,in which a basis consists of n linearly independent vectors.

In the present case (n = 3), we also may argue as follows.From (1), we have

161 . e2 x e3 = cic2c3 ei ej x ek = C'iC2iC3Eijk e1 ' e2 x e3;

hence, on writing E = el e2 x e3, we have

(2) E _ (det c')E = cE.

Since the vectors ei form a basis E 96 0, hence c 0 impliesE 0 and the linear independence of the vectors e1.

For any vector u,

(3) u = 'uiei = ujej;

hence, on substitution from (1),

ujej,

or, since the vectors ej are linearly independent,

(4) u.l = Ci,u

If y is the reduced cofactor of c in the determinant c, we haveon solving equations (4),

(5) ut =or, written out in full,

ul = riu1 + 'Y2u2 + Y3u3,

(5) u2 = 'Ylui + Y2u2 + Y3 u3,

u3 = 'iu1 + Y32 3u2 + 'Y3u3.


The linear transformations (1) and (5) are said to be contra-gredient. Their matrices,

Cl C1 c1 71 72 73

1 2 3 2 2C = C2 C2 C2 , r = 71 72 73

1

2 3 3 3 3C3 C3 C3 71 72 73

are so related that any element of r is the reduced cofactor of thecorresponding element of C in its determinant c; and any elementof C is the reduced cofactor of the corresponding element of r inits determinant y.

In order to state this definition analytically, we remind thereader of the following definitions from matrix algebra. The prod-uct AB of two square matrices A and B is the matrix whose elementin the ith row and jth column is the sum of the correspondingproducts of the elements in the ith row of A and jth column of(row-column rule). The transpose of a matrix A is a matrix A'obtained from A by interchanging rows and columns. The unitmatrix I has the elements Si (ones in the principal diagonal, zeroselsewhere).

If we compute Cr' or rC' and make use of the equations,

(6) cz7r = S, 7rc1 = bj,

we find that

(7) Cr'=rC'=I.These equations characterize contragredient transformations.

Two matrices whose product is I are said to be reciprocal. ThusC and r' or C' and r are reciprocal matrices. From (7), we seethat contragredient matrices are so related that the transpose of eitheris the reciprocal of the other.

148. Covariance and Contravariance. When new base vectorsare introduced by means of the transformation,

(1) ei = c e5, det c 0.

the components of any vector u are subjected to the contra-gredient transformation,

(2) ui = 7;u'.

In view of (24.3), this may be written

u-ei=7jue';

§ 148 COVARIANCE AND CONTRAVARIANCE 335

and, as this holds for every vector u,

(3)

The basis ei, reciprocal to ei, thus is transformed in the same wayas ui. To find the transformation for the components ui, multiply(1) by u ; then, from (24.4),

(4) ui = cua transformation cogredient with (1). Thus quantities writtenwith subscripts transform in the same way (cogrediently); andquantities written with superscripts also transform in the sameway-but the latter transformations are contragredient to the former.Thus the position of the index indicates the character of thetransformation.

A vector u is said to have the contravariant components ui, thecovariant components ui. These terms suggest variation unlike andlike that of the base vectors ei; in other words, the transformation(1) is regarded as a standard. Thus [u', u2, u3] and [U1, u2i u3]are two ways of representing the same vector u; in the first it isreferred to the basis ei, in the second to the reciprocal basis e'

'[u1, u2, u3J often is called a contravariant vector, [u1, u2, u3] a co-variant vector. Actually, the vector u is neither contravariant orcovariant, but invariant.

By using the properties of the reduced cofactors,i r i

(5)eir'Yr = Yi = ai,

we may solve equations (1) to (4) for the original base vectors andcomponents (§ 146). Thus the equations,

(6) ei = c,ei, ui = ciuj, et = 'Yie', uz - yiu',have the solutions,

(7) ei = -?j-6j, ui = Yiui; e' = c ,%J, ut = c ui.

Note that the matrices c, y in (7) are the matrices cz, -y' of (6)transposed. To be quite clear on this point the reader shouldwrite out equations ei = (.'e; and et = c;ei in full.

From (1) and (3), we find that

ei ' e' = cier yie8 = ciYaSr = c 'Yr = 31j,

which shows that the new bases have the fundamental propertyof reciprocal sets.


An expression, such as uiv i, uie i, ei e i, summed over the sameupper and lower indices, maintains its form under the transforma-tion. Take, for example, the scalar product uivi: making use of(2) and (4), we have

uiv2 = Ciur78vs = bryurv8 = u'1T.

The index notation thus automatically indicates quantities thatare invariant to affine transformations of the base vectors.

149. Orthogonal Transformations. When the basis ei and thenew basis ei both consist of mutually orthogonal triples of unitvectors, the transformation,

(1) ei = ciei,

is called orthogonal. Since both bases are self-reciprocal (§ 23),the corresponding transformation (148.3) between the reciprocalbases,

ei = 7e', now becomes ei = 7e;.

Since this transformation must be the same as (1),

C11 ci ci 7i 72 732 q(2) C = c2 c2 c3 = 71 72 7s = r

1

C 3 C33 3 3 3

C3 71 72 73

The matrix of an orthogonal transformation is called an orthogonalmatrix. From (2), we have the

THEOREM. Every element of an orthogonal matrix is its own re-duced cofactor in the determinant of the matrix.

In view of the properties of reduced cofactors, the coefficientsof an orthogonal transformation satisfy the equations:

(3) crc; = Cic; = 8L.

The matric equation (147.7) characterizing contragredient trans-formations becomes

(4) CC' = Ifor orthogonal transformations. Thus a real matrix is orthogonalwhen its transpose equals its reciprocal.

If c denotes the determinant of C, we have, from (4),

(5) c2 = 1, c = f 1.

§ 150 QUADRATIC FORMS 337

If the bases ei and ei are both dextral or both sinistral, we canbring the trihedral e1e2e3 into coincidence with e1e2e3 by continu-ous motion-in fact, by a rotation about an axis through theorigin. At any stage of the motion, the base vectors are relatedby equations of the form (1); and, as the motion progresses, thecoefficients c change continuously from their initial to their finalvalues S , and c = det ci becomes det S = 1. But, since this de-terminant equals =E1 at all stages of the motion and must changecontinuously if at all, c = 1 when the bases have the same orien-tation.

If one basis is dextral, the other sinistral, e1e2e3 and ele2(-e3)have the same orientation. Hence the determinant of the trans-formation,

el = c1er, e2 = C3er, -e3 = -caer,namely, -c, must equal 1; thus c = -1 when the bases have dif-ferent orientations.

150. Quadratic Forms. A real quadratic polynomial,

(1) A(x, x) = aijxixj (i, j = 1, 2, , n),

for which aij = aji is called a real quadratic form in the variablesx1, x2, , x"`. The symmetry requirement aij = aji entails noloss in generality; for, if aij 0 aji, the form is not altered if wereplace aij and aji by

2(aij + aji).

The determinant of the coefficients, a = det aij, is called thediscriminant of the form. The form is said to be singular if det aij= 0, non-singular if det aij 0 0.

Associated with the quadratic form A (x, x) is the bilinear form,

(2) A(x, y) = aijx'y' = A(y, x),

known as its polar form.The expansion of aij(xi + Xyi) (x' + Xyj) shows that a quad-

ratic form and its polar are related by the identity,

(3) A (x + X y, x + Xy) = A (x, x) + 2XA (x, y) + X2A (y, y).

If we make the linear transformation,

(4) xi=c yr, c=detc;F-- 0,

the form (1) becomes

(5) B(y, y) = br4yry8 where bra = aijcTc'.

338 TENSOR ANALYSIS

From the rule for multiplying determinants,

§150

det bra = det (ai;cT) det c8 = det ai; det cr det c8;

the discriminant of B(y, y) is therefore

(6) b = det bra = c2a.

Linear transformations of non-zero determinant do not alter the singelar or non-singular character of a quadratic form.

If the form A (x, x) is singular, it can be expressed in terms offewer than n variables which are linear functions of x. For sincedet ai, = 0, the system of n linear equations ai;xi = 0 has a solu-tion (xa, xo, , xo) which does not consist entirely of zeros; thuswe may assume that xo 5-4- 0. Now A(xo, y) = ai;xoy' = 0, and,from (3),

A (x + Axo, x + Axo) = A (x, x),

irrespective of the value of A. If we now choose A = -xl/xoand write

y i = x, i + Axp = x i - x4x1 /x0,

we have A (y, y) = A (x, x). Since y' = 0, A (y, y) is expressed interms of the n - 1 variables y2, y3, , y".

Henceforth we shall consider only non-singular forms. A non-singular form is said to be definite when it vanishes only forxl = x2 = = x" = 0. For all other sets of values, the sign ofa definite quadratic form is always the same. To prove this, sup-pose that A(x, x) > 0, for the set x1, x2, , x" and A(y, y) < 0for the set y', y2, ,

y". Then, from (3),

(7) A (x + Ay, x + Ay) = 0

is a quadratic equation in A having two distinct real roots A1i A2; for

{A(x, y) 12 - A(x, x)A(y, y) > 0.

Consequently, the form vanishes for two different sets of values,xi + Alyi and xi + A2yi and therefore cannot be definite.

For a definite quadratic form, the quadratic equation (7) in Amust have a pair of either complex roots or equal roots; hence, fora definite form,

(8) JA (x, y) 12 - A(x, x)A(y, y) < 0,

the equal sign corresponding to the case xi + Ayi = 0, (i = 1., n), in which the sets xi and yi are proportional.

§ 152 RELATIONS BETWEEN RECIPROCAL BASES 339

A definite quadratic form is called positive definite or negativedefinite according as its sign (for non-zero sets) is positive or nega-tive. A non-singular form which is not definite is called indefinite:such a form may vanish for values x` other than zero.

When A (x, x) is positive definite, it is well known that we canfind a real linear transformation (4) which will reduce A (x, x) toa sum of squares:

(9) B(x, x) = Siyiyj = ylyl + y2y2 + ... + ynyn

The discriminant of a positive definite form is therefore positive;for, from (6), b = 1 = c2a, and a = 1/c2 > 0.

151. The Metric. Using the notation,

(1) 9i; = ei ej = gji,

for the nine scalar products of the base vectors, we have

(2) u . v = (uiei) . (viej) = giptivj, u u = gi;uiu'.

Thus u u is a real quadratic form,

giiU'u1 + g22u2u2 + 933u3u3 + 2g12u'u2 + 2923u2u2 + 2931u3u1r

in the variables u1, u2, u3; and, since u u > 0 when u 0, thisform is positive definite. Moreover, u v is given by the associatedbilinear (polar) form in ui, vj.

From (2), we have

(3) I U I = gijl(itljr

(4) Cos (u, V) =U V _ gijuw'

lUl VI VgijuiujVVgijvivi

Thus, when the six constants gij are known, we can find thelengths of vectors and the angles between them when a fixed unitof length is adopted. The quantities gjj, since they make measure-ments possible, are said to determine the metric of our 3-space; andgijuiuj is called the metric (or fundamental) quadratic form.

152. Relations between Reciprocal Bases. The discriminant c'the metric form is

(1) 9=911 912 913

921 922 923

931 932 933

:" Bother, Introduction to Higher Algebra, New York, 1907, p. 150.


The reduced cofactor of gij in this determinant is denoted by gij;and we have, from (146.10), the relations:

(2)girg.r = 9rigr' = Si.

Since any vector may be written

u = ei,

(3) ei = gije'We may solve these equations for ej by multiplying by gik andsumming on i; thus

ik ik j k j k9 ei = 9 gije = 3je = e ,or, if we interchange i and k,

(4) ei = 9kiek-

From (4), we have alsoei ej = gkiek ej = gkiak = gji

(5) gij = ez e' = 9'iMaking use of (3) and (4), we now have, for any vector u,

(6)

(7) ui = ei u = gijuj

The equations enable us to convert contravariant components of avector to covariant and vice versa.

Finally, from (24.14), we have

(8) g = det gij = det (e1 ej) = [eie2e312 = E2,

(9) det gij = det (ei ej) = [ele2e312 = E-2 = 19

153. The Affine Group. When an origin 0 is given, the con-stant basis el, e2, e3 defines a Cartesian coordinate system x1, x2,x3; for any point P is determined by the components of its positionvector :

OP = x'el + x2e2 + x3e3.

The components xi of OP are called the Cartesian coordinates of Prelative to the basis ei.

§ 153 THE AFFINE GROUP 341

When the base vectors ei are subjected to the transformation,

(1) et=ce;,

the invariance of OP, namely,

(2) i''es = xZei,

induces a contragredient transformation on the coordinates. For,if we substitute from (1) in (2), we find that x' = ci"xi; that is,

(3) xi = c;x'.

The matrix c? in (1) is transposed in (3); and, while (1) expressesthe new base vectors in terms of the old, (3) expresses the oldcoordinates in terms of the new. We now can solve (3) for thenew coordinates by multiplying by the reduced cofactors ti; andsumming; thus we find xiy, = xk, or

(4) x' = y2xi, -y = det yj = 1/c.

The transformation (3) is called affine, or, more specifically,centered affine, since the origin 0 has not been altered. Thecentered-affine transformations with non-zero determinant form agroup, that is,

(a) the set includes the identity transformation:

ri = S ; a ; ' = x`, det 6 = 1;

(b) each transformation of the set has an inverse;(c) the succession of two transformations of the set is equivalent

to a single transformation of the set.

Thus the transformations,

a x', x' = bj;xk, give xa = Ckx1,where

ck = a bk and det ck = (det a) (det bk) F4- 0.

Our present point of view is that a transformation of the basevectors induces a definite transformation of coordinates. But thereverse point of view is adopted when transformation groups moregeneral than the affine are under consideration : the coordinates aretransformed, and we then inquire as to corresponding transformationof the base vectors.

342 TENSOR ANALYSIS a 151

In the case of the affine group the transformation (3) of coordi-nates obviously entails the transformation (1) of base vector.,.For, if we put x' = x'cy in (2), we have

xi(czej - ei) = 0

for all xi, that is, (1) must hold.154. Dyadics. Under the affine transformation,

(1) ei = c ei, xi = y;xi (c 0),

we have also

(2) ei = y;ei, ti = e xi (148.6).

Contravariant and covariant components of a vector, ut and ui,transform like x i and xi, respectively. The sets of components u iand ui often are called contravariant and covariant vectors. Actu-ally they are different representations of the same invariant vector,

(3) U = uiei = uiei.

Nine (3 X 3) numbers Tii which transform like the nine prod-ucts of vector components uivi, namely,(4) Tii = yryBT'8

are called contravariant components of a dyadic. Similarly, ninenumbers Ti; which transform like uiei, namely,

(5) T,i = ('icjTra,

are called covariant components of a dyadic. Finally, nine num-bers Ti; or T;i which transform like u ivi or uivi, respectively,namely,(6) 7,'i =

yrc;TB, T:' =are called mixed components of a dyadic. Note that in Tti theupper index comes first, in Tii the lower index.

Just as ui and ui are two representations of a single invariantvector u, the components Tii, Tii, Ti, T;i are four representationsof one and the same dyadic,

(7) T = Tiieiei = Tiieiei = Ttieiei = Ti eiei.

Note that the indices on components and base vectors always areplaced in the same order from left to right. All forms of T given

§ 155 ABSOLUTE TENSORS 343

in (7) are invariant; for example, on using (1) and (4), we haveT'16x.e = Yr

iYs ijTrsc'eh ckek r= 3,h6kTr8e1,ek = Thkeheb,

from the properties of reduced cofactors (146.8).The coefficients gij of the metric quadratic form are tensor com-

ponents; for, from (151.1),

(8) 9ij = ei ' ej = (tier) ' (cjes) = c'- jug,.

in agreement with (5). The same is true of gij; for, from (152.5),

(9) 9i7 = e ' e' = (Yrer) - (Yses) = Y:Y39r8,

in agreement with (4). Again, if we define b, S transformsas a mixed tensor,

(10) ei ej = (Yrer) - (c;es) = Yrc;a8.

Indeed, gij, gij and b are all components of the same tensor,

(11) gijeiej = gijeiej = beiej = 8jeiej = eiei = eiei,

namely, the idemfactor (§ 66). In fact, if we use the formulas,

(12) ei = girer, ei = girer (§ 152),

and remember the properties of gij, g" as reduced cofactors, anytwo members of (11) can be shown to be equal; for example,

gijeie' = 9ij9irere' = b;ere-i = eje',i j i jr r ti igije e = gc,e g er = Sie er = e ei.

Since eiei and eiei represent the same tensor, the order of theindices in its mixed components, the Kronecker deltas, is im-material.

Any component of T may be expressed in terms of componentsof another type by means of the metric form. For example, wehave, from (7) and (12),

Tit = ei,T.ej = 9trer'T'ej = 9irT,J,= 9'r ei'T -er =

girgjs er - T - e8 = girgj8Trs

155. Absolute Tensors. We next define absolute tensors withrespect to the centered-affine group of transformations: xi = cxj

344 TENSOR ANALYSIS

Scalars p(x', x2, x3) which have one component in each coor(li-nate system given by

(1)gxl,

x2, x3) = P(x', x2, x3)

are said to be absolute tensors of valence zero.Vectors have three components, and these may be of two types

u,, ui. The laws of transformation,

(2) 2li = ciuj, ui = Yu

characterize absolute tensors of valence one. The vector itself,

u = uiei = uiei,

is invariant to the transformation: u = u.Dyadics have 32 = 9 components, and these may be of 22 = 4

types: Ti;, Tij, T'j, Tij. If these components transform, respec-tively, like the products ujv;, uivj, uiv;, uivj of components of twoabsolute vectors, they are said to form absolute tensors of valencetwo. The dyadic itself,

T = Tijeie' = Ti'eiej = TZjeie' = T'jetej,

is invariant to the transformation: T = T.In general, an absolute tensor of valence m is a set of 3' compo-

nents that transform like the product of m absolute-vector com-ponents; and, since each of these may be covariant or contra-variant, the components are of 2' types. One of these types ispurely contravariant (T1 .k), another purely covariant (Tij...k);the remaining types are mixed and have both upper and lowerindices.

Consider, for example, a tensor T of valence three (a triadic);its 33 = 27 components may be of 23 = 8 types. Its covariantcomponents Ti;k will transform like uiv;wk, namely,

(3) Tijk = c2cjckT,.t,

and the mixed components Tijk like uivjwk, namely,

jk r j k at(4) Ti = ea'Ya t Tr

The tensor itself,

T = T.;keiejek = Tijkeiejek = ,

§ 156 RELATIVE TENSORS

is invariant to the transformation; for example,

T = Tijkeiejek

(CiciCkTrst) (7ea) (' beb) (} cec)

SaSbS,'Trsteaebec

= Trstereset = T.

345

Note that the indices on the base vectors in T have the same orderas the indices on the component but occupy opposed positions.

We may solve the 27 equations (3) for the original componentsTrst by multiplying by yayby, and summing; we thus find

j k(5)- r s tyaybycTijk = SaSbScTrst = Tabc

In similar fashion we find, from (4),

(6) yaCbCkijk = brbbb -'r = Tab`.

156. Relative Tensors. When the coordinates are subjected toan affine transformation,

(1) xi = Cr, ;xti = yrxr (c = det c s 0),

the transformation equations, for all tensor components thus farconsidered contain only the coefficients c, y. More generally, wemay have equations of transformation, such as

(2) T.jk = CNCr ji t r ,

which involve the Nth power of the determinant c. In this casethe quantities in question are said to be components of a relativetensor of weight N. When N = 0, equations (2) reduce to (155.4)for an absolute tensor. As a consequence of (2), the relative tensorT = Trseere8et becomes

(3) T = cNT

when the transformation (1) is effected.The law of transformation of a tensor component is determined

by

(i) its valence: the number of its indices;(ii) its type: the position of its indices from left to right;(iii) its weight: the power of c that enters into the transforma-

tion equations.


If the weight of a tensor is not specified, it is assumed to be zero;the tensor is then absolute.

A relative scalar so of weight N (valence zero) is transformedaccording to

(4) gxl, x2, x3) = CNc,(x1) x2, x3).

The box product of the base vectors, E = el e2 X e3, is a relativescalar of weight 1; for, from (147.2),

(5) E = cE.

The box product of the reciprocal base vectors, el . e2. e3 = E-1,is a relative scalar of weight - 1; for E-1 = c-1E-1, from (5).

The determinant,

(6) g = det gij = E2 (152.8),

is a relative scalar of weight 2; for E2 = c2E2.

THEOREM. The permutation symbols Eijk and eijk, regarded as thesame set of numbers in all coordinate systems, are components of rela-tive tensors of weight - 1 and 1, respectively.

Proof. If the formulas (146.2) are applied to the determinants,

c = det c2j y = det -?j- = 1/c ,we have

CEijk = iCjCkerat' YEzlk = yr"Ys7 Brat.

Since Eijk = Eijk, Eijk = Eijk by definition, these equations assumethe form,

jarat, Eijk i j kerate = C8Ctk ryaytz _ Y

required by relative tensors of weight -1 and 1.157. General Transformations. Three equations,

(1) x' = fz(xls x2 x3) f

in which the functions f i are single-valued for all points of a regionR and which can be solved reciprocally to give the three equa-tions,

(2) xt = gi(xl 22, 3)

in which the functions gi also are single valued, determine a one-to-one correspondence between the sets of numbers xi and xi. Ifwe regard both sets of numbers as coordinates of the same point,

§ 157 GENERAL TRANSFORMATIONS 347

(1) defines a transformation of coordinates. The affine transforma-tion is the particular case of (1) in which the functions f i arelinear and homogeneous in x', x2, x3.

Consider now the totality of such transformations in which thefunctions f i(x', x2, x3) are analytic functions having a non-vanishingJacobian in R:

(3)

ax

OX

azi=det- 3,6 0.

ax'

The implicit-function theorem ensures the existence of the solu-tions (2) of equations (1) in a sufficiently restricted neighborhoodof any point. In order to have a transformation of coordinatesin R, we must assume that the solution (2) exists and is singlevalued throughout R.

The coordinate transformations (1) thus defined form a group;for

(a) they include the identity transformation xi = xi whoseJacobian is 1;

(b) each transformation (1) has an inverse (2) whose Jacobianax/ax I is the reciprocal of (3);

(c) the succession of two transformations of the set is equiva-lent to another transformation of the set.

As to (c), the two transformations,

xL = fi(xl x2 x3) xz = h'(x' x2 x3)

are equivalent to

xz = h'[f`(x),f2(x),f3(x)] = ji(xl,x2, x3),

for which the Jacobian,

axi /0xi axrdet - = det (- -/

= (detaxi- /

( det -)L 0.ax' \axr ax'/ \ axr \ ax'/ax

Moreover, since f i and hi are analytic functions, the functions jiare likewise analytic.

In the affine transformation,

(4)

we have

(5)

xi = i t i = crxi-ryrx , x ,

axi axi- z;

axryr

axr= r

348 TENSOR ANALYSIS

the Jacobians,

axi axi(6) det - = y, det - = c, and yc = 1.

axr axr

§ 157

A tensor of weight N, with respect to the affine group, trans-forms according to the pattern,

(7) TU'.k = CNyTy3CtkTrst.

In view of (5) and (6), this equation may be written

(8)ax

I N

axi ax' axtTi'k- at axr axs axk

Now, by definition, a set of 33 functions Tr. $t is said to form a tensorof valence 3 and weight N with respect to the general transforma-tions (1) provided the components transform according to thepattern (8). This equation becomes (7) when the transformationis affine and constitutes a natural generalization of (7). The cor-responding equation for a tensor of any valence or type is nowobvious. In particular, for contravariant and covariant compo-nents of an absolute vector,

axi _ axr

(9), (10) uy = - ur, ii = - ur.axr ax i

Since the new tensor components are linear and homogeneousin terms of the old components, we have the important result: Ifthe components of a tensor vanish in one coordinate system, theyvanish in all coordinate systems.

More generally, tensor equations maintain their form in all co-ordinate systems. If any geometrical or physical property is ex-pressed by means of an equation between tensors, this equationin an arbitrary coordinate system expresses the same property.

Although the coordinates themselves are not vector componentsin the general group (1), their differentials dxi are the componentsof a contravariant vector; for, from (1),

axi axi axi axi(11) dxi = - dxI + - dx2 + - dx3 = - dxr.

ax' ax2 axi axr

We regard the differentials dxr as the prototype of contravariantvectors; and the rule of total differentiation gives the correct pat-tern for their transformation.

§ 157 GENERAL TRANSFORMATIONS 349

The partial derivatives app/axi of an absolute scalar p(x', x2, x3)are the components of a covariant vector; for, since

0(.' x2, x3) = P(xl, x2,x3),

when we replace xi on the right by the values (2),a P ax2 a,p ax3 _ axr aV

(12) +1 i 2 i 3 i rax ax ax ax ax ax at ax ax

We regard the derivatives ap/axr as the prototype of covariantvectors; and the chain rule for partial differentiation gives thecorrect pattern for their transformation.

In the affine transformation, the position vector r = xiei, and

(13) ei = ar/axi.

We now adopt (13) as the definition of ei in any coordinate systemx'; then

ar ar axr axr(14) ei=-=--=-er.

axi axraxt axz

The base vectors ei thus transform after the pattern in (10) andtherefore merit a subscript.

The base vectors ei must transform after the pattern in (9),namely,

a. t(15) ey = - er;

axr

for, since er e8 = S8,

axr axj axr axj axjei

49V ax8 a.2'i ax' axi

and the sets ei, e' are also reciprocal.We now can show that

(16) gij = gij = S

transform as tensor components; we need only make the replace-ments indicated by (5) in the proofs of § 154. In fact gij, gi', S;are all components of the idemfactor ere' = erer; thus S; some-times is written gg. Moreover,

(17) ei = ei ere' = gire', ei = ei. erer = girer.

350 TENSOR ANALYSIS

158. Permutation Tensor. The box product E = el e2 x e3 isa relative scalar of weight 1:

ax

axE

for the proof of (147.2) applies when we replace c; by axe/a.V.Moreover, E-' = e' e2 x e3 is a relative scalar of weight -1.

The permutation tensor is defined by

(2) e1 - ej x ek eiejek = ei ej x ek eiejek.

These triadics are equal absolute tensors. For, if we putei = gireT) ej =

9ise3,

ek = gkter,

the left member becomes er es x e` ereset. Moreover, (157.14) and(157.15) show that

(3) C - = ei ej x ek eiejek.

From (2), we see that the covariant and contravariant compo-nents of the permutation tensor are

(4) e1 - ej x ek = EijkE, ei ej x ek = EijkE-1

Since E and E-' are relative scalars of weight 1 and -1, Eijk andEijk are components of relative tensors of weight - 1 and 1. Thisis the theorem of § 156; the former proof still holds when obviouschanges are made.

159. Operations with Tensors. The three basic operations ontensors are addition, multiplication, and contraction.

1. Addition of tensor components of the same valence, weight, andtype generates a tensor component of precisely the same char-acters.

Example. If Pij, Qij are both of weight N,

ax N axr ax"Fij + Qij _ -- (Pre + Qr.)-

ax a.ri azj

Hence Tit = Pij + Qij transforms in exactly the same manner as Pij and Qij.

2. Multiplication of tensor components of valence m1i m2, ofweight N1, N2, and of arbitrary type generates a tensor componentof valence m1 + m2i of weight N1 + N2, and of a type which isdefined by the position of the indices in the factors.

§ 159 OPERATIONS WITH TF''\SORS

Example. Let Pi' and Uk have the weights 2 and 1; then

P'uax ' ati a-' Pr8

k = I ai ly axr axe

ax

az

axt=axkut

Ox

as

3at'.

az' axtPT8 ut.

ax'' ax8 D.

351

Hence T'jk = P''uk transforms as a tensor of valence 3, of weight 3, and ofthe type indicated by its indices. Even though P''uk = ukPi', the producttensors,

Pu = Pi'ukeie?ek, uP = ukPz'ekeie

are not the same.

3. Contraction. In any mixed tensor component an upper andlower index may be set equal and summed over the index rangethis generates a tensor component of the same weight and of va-lence two less. Its type is determined by the remaining indices,not involved in the summation.

Example 1. In the absolute component T'!.k set j = k; then

u=_VV jT1.11'+'V.22+T133

is a contravariant vector, fori i t i t

Tt = a ax ax T3 = az ax.. -ax"" ax8 all

T ..t - Of ax8 T r t

all al'

ur.=-38T'st =ax,* axr

T"` = ax"

A tensor component may also be contracted with respect to twoindices on the same level; we need only raise or lower one indexand then contract as previously.

Example 2. To contract the absolute component V% on the indices i, jwe first lower the index j

4 !rT.jk = T..k9i*;

setting i = j now gives the covariant vector:: !Vk = T ..kgir = T ..kgij.

We may also obtain this vector by lowering the index i and setting i = j:

Tick = Tt"k = V" k9ri

Contraction on a pair of indices is equivalent to the scalar multi-plication of the corresponding base vectors in the complete tensor.

In the preceding examples, T = T!.ke,e ek,

u = Ti'.keiej ek = r'.ks;ei = T .,ei,

v = T:'.kei ejek = Tt'.kgi,ek


The proof that contraction produces tensors follows from the equa-tion :

T'.'.keie;ek = TY?.keie,ek.

When the new base vectors on the left are expressed in terms ofthe old, this becomes an identity; and, since scalar multiplicationis distributive with respect to addition, it remains an identity whenbase vectors in the same position are multiplied on both sides.Thus, in ex. 2,

V = T'.kgiiek = v.

As long as a contracted tensor has two or more indices, the fore-going process may be repeated, each contraction reducing the va-lence by two but leaving the weight unaltered. Thus the tensorT!.kh may be contracted twice to yield two different scalars 7".'.i;and T'!.;i, each consisting of nine terms. These are obtained fromthe invariant tensor TT'.kheiejekeh by the formation of ei - ek,e; eh and ei - eh, e; ek, respectively.

Contraction often is combined with multiplication. For ex-ample, if we multiply the vectors u, v.; and then contract, weobtain their scalar product u'vi. Again the product A B of twodyadics defined in § 65 is equivalent to tensor multiplication fol-lowed by contraction on the two inner indices. Thus, if A =Ati,eV , B = Bkhekeh,

A - B = Ai;Bkhe'e' . ekeh = Aj;B'he'eh.

The product of the idemfactor I = ere? and a tensor T of valencem is a tensor IT of valence m + 2. In general IT differs fromTI; but the contracted products reproduce T:

160. Symmetry and Antisymmetry. A tensor is said to be sym-metric in two indices of the same type (both covariant or bothcontravariant) if the value of any component is not changed bypermuting them. It is antisymmetric or alternating in two indicesof the same type if permuting them in any component merelychanges its sign. Thus, if

Tabc = Tbac, Tabc = -Tcba,

the tensor T is symmetric in a, b, alternating in a, c.

§ 161 KRONECKER DELTAS 353

Symmetry or antisymmetry are properties that subsist after ageneral transformation of coordinates. Thus, for the precedingexample, we have

axa axb axe axb axa a.T.eTijk =

aTabc = k Tbac = Tjik

ax ax axk ax ax £ ax

axe axb axa(-Tcba)

axk axj axi

A tensor cannot be symmetric or alternating in two indices ofdifferent types; for a property such as Ti j = Vi i does not subsistafter a transformation of coordinates.

A tensor is said to be symmetric in any set of upper or lowerindices if its components are not altered in value by any permuta-tion of the set. The subsistence of this property in one coordinatesystem ensures it in all.

A tensor is said to be alternating (or antisymmetric) in any setof upper or lower indices if its components are not altered in valueby any even permutation of the indices and are merely changed insign by an odd permutation of these indices.

A tensor Tij or Tijk (T ij, Tijk) which is alternating in all indicesis called a bivector or trivector, respectively. In such alternatingtensors, all components having two equal indices are zero. In atrivector Tijk, the non-zero components can have but two values,±T123; moreover, the contracted product,

EijkTijk = 3!T123

In this connection we remind the reader that a given permuta-tion of n indices from some standard order can be accomplishedby a succession of transpositions of two adjacent indices and thatthe total number of transpositions required to bring about a defi-nite permutation is always even or always odd. The permutationin question is said to be even or odd in the respective cases.

161. Kronecker Deltas. We now generalize the simple Kro-necker delta S by introducing two others, defined as follows:

(1)aiajbkc

- Eabr Eijk

ij ijr ijr(2) aab = EabrE = aabr

Since the epsilons have weights of 1 and -1, ba"bk, is an absolutetensor of valence six; hence b b, formed by contracting S, is anabsolute tensor of valence four.


Those definitions show that S b and bay can only assume thevalues 0, 1, -1; they are evidently alternating in both upper andlower indices. Their precise values in any case are as follows:

If both upper and lower indices of a generalized delta consist ofthe same distinct numbers chosen from 1, 2, 3, the delta is 1 or -1according as the upper indices form an even or odd permutationof the lower; in all other cases the delta is zero.

This rule is a direct consequence of the properties of the epsilons.We have, for example,

012 32 23 13 o.12 = 1, 023 - - 1, 011 = 021

;

123 231x123 = 0123 - 1,

6213 v321= 0123 8123 = 0

162. Vector Algebra in Index Notation. The three operationson tensors enable us to give a succinct account of vector algebra.Vectors are to be regarded as absolute unless stated to be relative;and we shall often speak of components as vectors.

If w = u + v, we have

vi =ui+vi, or wi=ui+vi;obviously vector addition is commutative and associative.

The tensor product of two vectors u, v is the dyad uv. On con-traction, uv yields the scalar product,

(1) u . v = uivi = uivi = gijuivj = giju1vj.

From (1), we have

U. (v+w) =The antisymmetric dyadic (bivector) P = uv - vu is called the

outer product of u and v; its covariant components are

(2) Pjk = UjVk - UkVj = 8jkuavb.

The dual of Pjk (cf. § 170) is the contravariant vector of weight 1:

(3) p 2=Z

e ijk Pik = E''ikujvk;

its components,

U2V3 - u02, U3V1 - u1V3, u1V2 - u2v1Y

are the same as the non-zero components Pik. Since E-1 is ascalar of weight -1, the components E-'p' are absolute. Thecorresponding vector,

§ 162 VECTOR ALGEBRA IN INDEX NOTATION 355

(4) E-lpiei = E-1EijkeiujVk = E-1

v1 V2 V3

Similarly, from the outer product,

(2)' P'k = bab26avb = u'vk - ukvJ,

= uXv (24.9).

we obtain as dual the covariant vector of weight - 1:

(3)' _ ipi = ZEijkPJk = eijklljvk.

The absolute components Epi again give u X v :

el e2 e3

(4)' Epiei = Eeijkeiujvk = E ul u2 u3 = uxv (24.10).

vl v2 v3

From (4) or (4)', we have

uXv = - V X u, U- (v + w) . u x v + U X W.

The components of the triple product u x (v X w) are

(u X (v X w))' = E-1EijkUj(V X w)k (4)

= E-hE'fkujEEkabvawb (4)'

= EijkEabkujvawb

= SabUjva'Uib (161.2)

= uj(viwj - vjwi)

= (u w)vi - (u - V)wi;hence

u X (v X w) = (u w)v - (u v)w.

The box product u x v w is given by either of the absolutescalars,

(5) (U X V)'wi = E-'Cijk,ujvkwi = E-1

(5)' (u X v)z-w = Eeijku'vkw' = E

ul u2 U3

Vi V2 V3

W1 W2 W3

U1 u2 u3

V1 V2

V3

w1 w2 w3

in agreement with (24.12).

356 TENSOR ANALYSIS § m3

163. The Afne Connection. Any vector can be expressed as alinear combination of the base vectors. When the 32 derivativesof the base vectors ae;/axi = Die; are thus expressed,

(1) Die; = ref + r e2 + r 'e3 = ri,er,

the 33 coefficients r, are called components of the affine connection.If we multiply (1) by ek , the right-hand member becomes rzJj r= r ; hence

(2) r = ek Die;.

The law of transformation for r is given by

ec -- DQ I I(3) r = ek Die; = eb ),c 'l iax / ax / \ax

wherea axa a axb

(4) Di = - _ - - _ - Da;axi axi axa axi

The differential operator Di transforms like a covariant vector (hencethe subscript). In (3), Da acts on both scalar and vector factorsfollowing; hence, from (1),

axk a2xb axa axbe ( eb+--raber

ax` axiax' axi axi

Since e° eb = bb, ec er = 6'r, this gives

a2xb axk axa axb axk(5) + - - - rabaxiaxj axb axi ax' ax`

for the desired law of transformation. Therefore r is a tensorcomponent when and only when

a2xb axb= 0, - = CJ (const),

49-Pi axi ax'

and hence xb = c;x', if the coordinates xi and xi have a commonorigin.

The components r of the affine connection (between the deriva-tives of the base vectors and the vectors themselves) are tensorcomponents only with respect to affine transformations.

axk \ /axb

§ 163 THE AFFINE CONNECTION 357

Although the Irk are not tensor components for general trans-formations, we shall see that the index notation still serves a use-ful purpose. j

Since e1 = D;r (157.13),

(6) P = ek DiD;r = ek D;Dir = Pk;

and, from (5), we see that the symmetry of P in the subscriptspersists after a transformation of coordinates.

If we transform coordinates from x to IF, we have, from (3),

(7) P4 - \axk k/ \ax1' \4924

or, on making the replacements,

kaxk (9x, axbe = - es, Di = - Da, e; eb,ax, axi ax'

_r _ axrl

axal

(axb

(8) PPQ - ax, ec/ \a2p Da/ \ai7 eb/

a relation of the same form as (3). Consequently, the successionof transformations r --p I` -> f produces a transformation r -- rof the same form as r -+ 1. We express this property by sayingthat the transformation (3), or its equivalent (5), is transitive.

If, in particular, x i = x', we may delete all the tildas (-) in (7) ;this equation then gives, on expansion,

a2xJ air a. a. axrk(9) P,,

axP axq ax' axP a:rq axkI'2J.

These equations constitute the transformation inverse to (5)and also may be derived by solving these equations for P'ab.

This solution may be effected by the multiplication of (5) byaxi ax' axr

axp ax4 axk

t The coordinates xi present a similar situation. For afflne transformationsthey are components of a vector; for general transformations this is not thecase, but the indices still serve to indicate that their differentials dxi are vectorcomponents.

358 T1E;NSOR ANALYSIS §163

On writing y = x, the equation of transformation (5) becomes:

i a2-' ax,axb 1

ayk(10) + - - rab

lay, ayj ayi ay' axc

If we now make the change of coordinates,

(11) xr = xo + yT - 2(rPq)oypy4,

where the gammas are computed for x = xo, the point x" = xocorresponds to yr = 0. Since ay'r/ayi =

ax, a2xrayi = Si -

ayi ay.i - - (r)o;

hence, at the point x' = xo (yr = 0), the brace in (10) becomes

-(r )0 +aj(rab)o = 0.

Consequently, all the gammas I'- = 0 vanish at the origin y'' = 0of the new coordinates. Such a system of coordinates is termedgeodesic. Since the gammas are not tensors, the equations riI; = 0(y = 0) do not imply r = 0 (x = x0).

Example 1. For cylindrical coordinates p, p, z, we have (§ 89, ex. 1)

r = [x, y, z] = [P cos ip, P sin ,p, z].

If we put x1 = p, x2 ='P,x3 =z,

el = Dlr = [cos p, sin p, 0],

e2 = D2r = p[- sin p, cos {p, 0],

e3=D3r=[0,0,1];

Die, = 0, D2ei = [- sin gyp, cos (a, 0] = 1 e2, D3ei = 0,P

D2e2 = -p[cos rp, sin p, 0] = -pei, D3e2 = 0,

D3e3 = 0.

Hence all gammas are zero except

rig = r2, = 1/P, r22 = -P-

Example 2. For spherical coordinates r, sp, B, we have (§ 89, ex. 2)

r = [x, y, z] = r[sin 0 cos p, sin 0 sin p, cos 0].

§ 164 KINEMATICS OF A PARTICLE

If we putxi =r,x2= V,x3=0,*

ei = Djr = [sin 0 cos gyp, sin 0 sin gyp, cos 01,

e2 = D2r = r sin 01 - sin gyp, cos p, 0],

e3 = D3r = r[cos 0 cos gyp, cos 6 sin gyp, - sin 0];

D1e1 = 0, D2e11

= 1 e2, D3e11

= I e3,r r

D2e2 = -r sin2 6 e1 - sin 0 cos 0 e3, D2e3 = cot 0 e2,

D3e3 = -rel.The non-zero gammas are therefore

r12=r21=1/r, r13=r31= 1/r, r22= -r sin2 B,

r22 = - sin 0 cos 0, r23 = 1132= Cot 0, r33 = -r.

359

164. Kinematics of a Particle. We now may find velocity andacceleration of a particle in general coordinates:

(1) V = vkek, a = dv/dt = akek.

In rectangular coordinates x1 = x, x2 = y, x3 = z we have vk =dxk/dt (52.7). Hence, in general coordinates 2i, these componentsbecome

8xk dxi dxk(2) vk = - - _ -

axi dt dt

The time derivatives of the coordinates, which we write ±k, thustransform as contravariant vector components.

For the acceleration we have, from (1),

dvk aek dxia =

dtek + vk

ax dt = vkek + vtvkr:kei;i

and, on interchanging the summation indices j, k in the last term,

(3) a = (vk + viv'r ,)ek

Therefore, in any coordinate system the velocity and accelerationcomponents are

(4) vk = xk ak = xk + x°'x'I').

Since the base vectors ei are not, in general, unit vectors, wemust distinguish between the components vk and ak and the nu-merical values of the terms in vkek, akek. Thus the terms v'e1,a'e1 have the numerical values vllell, a'lell.

* The order r, p, 0 is sinistral and [ele2e3] < 0; cf. p. 197.

360 TENSOR ANALYSIS § 1&

Example 1. Cylindrical Coordinates p, ,p, z. From § 163, ex. 1, el, e2, e3have the lengths 1, p, 1; hence the values of the velocity terms in (1) area, pp, z

From the non-zero gammas r22 = -p, rig = 1/p, we have

a1 = a - Psp2, a2 = sp + 2Pcp/P, a3 = z;

and the numerical values of the acceleration terms are

P - Pcp2, p' + 2p', z.

Example 2. Spherical Coordinates r, p, 0. From § 163, ex. 2, e1, e2, e3 havethe lengths 1, r sin 0, r; hence the velocity terms in (1) have the values,

r, rcc sin B, r9.

With the non-zero gammas r22 = -r sin 2 0, r33 = -r we have, from (4),

al = r - p2r22 + O2r33 = r - rrp2 sin2 B - r92;

with r12 = 1/r, r23 = cot 6,

a2 = + 2r,pr12 + 2,pOr23 = + 2r(p/r + 24 cot 0;

and, with r13 = 1/r, r22 = - sin 6 cos 0,

a3 = 9 + 2rOri3 + ,,2r22 = 9 + 2r9/r - ,p2 sin B cos 0.

Hence the values of the acceleration terms are

1' - r,p2 sin2 0 - r92, r,p sin 0 + 2i ,p sin 0 + 2r09 cos 0,

r9 + 2r9 - r02 sin B cos 0.

165. Derivatives of ei and E. From the relation e' er = S*,we have, on differentiation,

(Die') er = - e' Deer = - rsr,and hence

(1) Die' rrerFrom Die; = r jer and (1), we have

(2) Di(e,e') = rz,ere' - rreer = 0,

on interchanging the summation indices r, j in the second term.Since the product E = el e2 x e3 is distributive with respect to

addition, its partial derivatives may be found by the familiar rulefor differentiating a product,

DiE = (Diet) . e2 x e3 + el (Die2) x e3 + el e2 x (D:e3)

= r,r'1er e 2 x e3 + ri2e1 er x e3 + ri3e1 e2- er1 2 3_ (ri1 + ri2 + ri3)e1 . e2 x e3,

§ 166 AFFINE CONNECTION AND METRIC TENSOR 361

or, if we apply the summation convention,

(3) DiE = rirE.

The derivative of E-i = el e2 e3 is therefore(4) DiE-i = -E-2DiE rirE-i

From (3), we have

DjDiE = (Djrir + rarr .,)E;and, since DjDiE = DiDjE,

(5) Dj rir = Dir;r166. Relation between Affine Connection and Metric Tensor.

On differentiating gij = ei ej, we have

(1) Dk9ij = rkier - ej + rkjei - er = rkjJrj + rkjgir

Since Dkgij and F are both symmetric in the indices ij, there are3 X 6 = 18 quantities in each set. Equations (1), 18 in numberand linear in the 18 gammas, may be solved for the latter.

We first introduce the notation,

(2) F9rk = rij,k,

just as if r k were a tensor whose upper index was lowered. Thenwe have also

(3) rij,rgrk = r ,for the left member equals

ri,gar9rk = I't S = r

We note that Fij, k is also symmetric in the indices i, j. Moreover

(4) Diej = F jer = rij,se9,

if we put e,. = g,.ses.We may now write (1) in the form:

(5) rki,j + rkj,i = Dkgij.

If we permute i, j, k cyclically in (5), we obtain the equations,

rij,k + rik,j = Di9jk,

rjk,i + rji,k = Dj9ki;

362 TENSOR ANALYSIS

and, upon subtracting (5) from their sum, we obtain

(6) ri.i.k =Z

(Digjk + D;gki - Dkgi1)

We may now compute r from (3).

§ 167

In the older literature, the components of the affine connectionare denoted by

[ii,k] = rij,k, [ . V Z)} = r.,These Christoffel symbols of the first and second kind, respectively,therefore are defined by the equations:

J l 7

(7) [ij, k] =z

(Digik. + D7gki - Dkgii), [ } = gkr[ij, r].

167. Covariant Derivative. The gradient of an absolute tensorT is defined as

aT(1) vT=eh

axh

Since T = T,

aT axh aT ax8 aT aTaxr ;eh = _ r__ 8 r--

axh axra

ax8 axh= Sre

8= e

hence VT is a tensor of valence one greater than T.If the components of T are T ;.'.'.k (the order of the indices is

not specified), the components of VT are written

VTab

(2) h i;...k,

the index h on V corresponding to the differentiation a/axh. Thisis a covariant index, for the operator Dk = a/axh transforms likea covariant vector:

a axr a axr(3) Dh=----=-Dr.

axh axh axr axh

For this reason the components (2) are called covariant derivativesof Tab

If T is a relative tensor of weight N and valence m, E-NT is anabsolute tensor, whose gradient,

ehDh(E-NT),

is an absolute tensor of valence m + 1. If this is multiplied byEN, we again obtain a relative tensor of weight N; this tensor isdefined to be the gradient of T and written

(4) VT = ENehDh(E-NT).

§ 167 COVARIANT DERIVATIVE 363

When N = 0, (4) reduces to (1). The components of VT, denotedby prefixing Vh to the components of T as in (2), are called co-variant derivatives.

From (165.3), we haveDhE_N = -NE-'V-1DhE = -NrhrE-N,

and henceDh(E-'YT) = E-v(DhT - NrhrT),

(5) VT = eh(Dh - Nrhr)T.

Thus the operator,h rV = e (Dh - Nrhr),

applied to any invariant tensor T (with its complement of basevectors) generates another tensor VT of the same weight and va-lence one greater.

We next compute the components of VT explicitly, where

(6) T = Tq '...ceaeb ... e,eiej ... ek.

If T is a relative tensor of weight N, VT contains the term,kN rr Tab' '

ehe e,e iej e .- hr ij... k aab

It remains to compute the part of VT due to the operator ehDh.Now Dh acts on the "product" of T ;.'.'.k and a series of base vec-tors. Since this product is distributive with respect to addition,DhT may be computed by the usual rule for a product; hence

DhT = a b c iaj . . . ek(DhTi;...k)eaeb . . . eye

+ eceiej ... ek + .. .ab i j kec(Dhe )e ... e +

In the second line putDhea = rhaer, ...

, Dhe, =

and in the successive terms interchange the summation indicesr and a, r and b, , r and c. In the third line put

i i r k- k rDhe - -rhre , ... , Dhe - - rhre ,and in the successive terms interchange the summation indicesr and i, r and j, , r and k. When this is done, each term ofVT contains the same complex of base vectors,

eheaeb ... eceiej ... ek,


and the component VhTt6.'.k of VT equals the sum of their scalarcoefficients :

17 v Tab e = D Tab . Ck- q- k

aTij...krhr + + Tij.krchr

r rTrj...k rhi - - Tij...r rhk

- Nrr rTab chr

This is the general formula for the covariant derivative of anytensor component, absolute or relative. The last term is absentwhen T is absolute (N = 0). For every upper index

v Ta*: contains a term T°::.r:: r* yh i k t k hr

and, for every lower index *,

OhTa' *:: ° contains a term - T" ' a rr.k

We consider now some important special cases.If p is a relative scalar of weight N,

(8) vhcp = Dh-p - Nrhr(c

Since E and E-1 are scalars of weight 1 and -1, we have, from(165.3) and (165.4),

(9) vhE = DhE - rhrE = 0,(10) vhE-1 = DhE-' + rhrE-1 = 0.

Again, since g = det gij = E2 is a relative scalar of weight 2,

(11) Vhg = DhE2 - 2r;,rE2 = 0.

If v = vie1 = vie' is an absolute vector,

(12)

(13)

VhV = Dhvti + vrrhr,rvhvi = Dhvi - vrrhi.

These expressions are the mixed and covariant components of oneand the same dyadic Vv.

For an absolute dyadic T, the components of VT may have theforms :

(14) VhT2 = DhT'' + Tr1rhr + Tirrhr,

(15) VhT`j = DhVj + T'jrhr - Tirrhj,

(16) VhTij = DhTij - Trjrhi - T,rrhj.

§ 168 RULES OF COVARIANT DIFFERENTIATION 365

For the metric tensor,

(17) G = gijeiej = gijeiej = eiei = I,we have, from (165.2),

(18) VG = e'DhG = ehDh(eiei) = 0.

The components of VG therefore vanish:

(19) Chgij = 0, Vhgii = 0, Vhbi = 0.

Note that (166.1) is equivalent to Vkgij = 0.

Example. We can write

i20) VT = eh(Dh - Nrhr)T = ehvhT,

if we regard Vh as an operator that acts only on scalars. Covariant differentia-tion then is defined by this operational equation. With this convention, wehave also

VVT = ei(Di - Nrir)ej(Dj - Nrn)T = eiejVjVjT.

The second member may be written

eiej(Di - Nrit) (Dj - Nrjr)T - eirise'(Dj - Nrr )T= eiej ( (Di - Nrit) (Dj - Nrir) - rij(D. Nr )1T

when indices j and s are interchanged; hence

viv jT = (Di - N rir) (Dj - Nrrr,)T - rijV,T,

ViViT = (D, - Nrjr) (Di - Nrsr)T - r-! v.T,

and, on subtraction,

(21) (CiV1 - Vjvi)T = (D1Dj - DjDi)T,

in view of (165.5). On the left the operators Vr act only upon the scalar com-ponents of T. We note that (21) applies to relative as well as absolute tensors.

168. Rules of Covariant Differentiation.1. The covariant derivative of the sum or product of two tensors may

be computed by the rules for ordinary differentiation.If we introduce a system of geodesic coordinates yi 1163), the

corresponding gammas 1 , will all vanish at the point yi = 0;hence, from (167.7),

(1)OTab...r=DTab...r

h ij...k h ij...k,

at the origin of geodesic coordinates, which moreover can bechosen at pleasure. For example, we have the tensor equation,

Vh(Tt'uk) = (VhTz3)uk + T''Vhuk,


in geodesic coordinates and therefore in any coordinates (§ 157).Consequently, the sum and product rules of ordinary differentia-tion also apply in covariant differentiation.

2. The covariant derivatives of the epsilons and Kronecker deltasare zero.

Since these tensors have constant components, their covariant de-rivatives vanish at the origin of a system of geodesic coordinates;hence they vanish in all coordinate systems.

3. The operation of contraction is commutative with covariant dif-ferentiation. For example, if we contract T=jk on the indices i, jto form

T?ik = SiT -jk,we have

OhVik = BiZVO -jk.

4. The components of the metric tensor (gij, gij, S) may be treatedas constants in covariant differentiation.

This follows at once from (167.19). For example,

Civj = Vi(g7rvr) = gjiViv'.Thus we may find Vivj by lowering the index j in Divj; in otherwords, Divj and Vivj are components of one and the same tensor:

Vv = V vieiej = V1vjeiei.

Evidently the operation of raising or lowering an index is commu-tative with covariant differentiation.

5. The relative scalars E, E-1 and g may be treated as constantsin covariant differentiation.

Since E, E-1 and g = E2 are relative scalars of weight 1, -1, 2,respectively, we have, from (167.4),

VE = 0, V -'=0, Vg = 0;

for in each case Dh(E-NT) = DO = 0-169. Riemannian Geometry. Any set of objects which can be

placed in one-to-one correspondence with the totality of orderedsets of real numbers (.r.1, x2, .

, x") satisfying certain inequalities,

I xi - ai I < ki (a1 and ki > 0 are constants),

is said to form a region of space of n dimensions.$ We speak of(x1) x2, . , x") as a point; but the actual objects may be very

$ Veblen, Invariants of Quadratic Differential Forms, Cambridge, 1933, p.13. In some applications the numbers xi may be complex.

§ 169 RIEMANNIAN GEOMETRY 367

diverse. Thus an event in the space-time of relativity may bepictured as a point in four-space; and the configurations of a dy-namical system, determined by n generalized coordinates, often areregarded as points in n-space.

If we associate the space (x', x2, , x") with an arbitrarynon-singular quadratic form,

(1) 9ii dxi dx', (9ij = 9ii, g = det gij 0),

we have a Riemannian space with a definite system of measure-ment prescribed by this form (§ 151) ; and the geometry of thismetric space is called Riemannian geometry. We assume that thecoefficients gij are continuous twice-differentiable functions of x.The base vectors ei are not specified; but their lengths and theangles between them may be found from the relations:

(2) 9ij = ei e,.

The reciprocal base vectors now are given by

(3) ei =9irer,

where gii is the reduced cofactor of gij in det gij; for(4) ei e; = girgir = d (152.2).

Moreover

(5)ei e' = girS* = gij

If we now transform coordinates from xi to xi, we assume thatthe new base vectors are given by

axr axi(6) ei = - er, ei = - es;

axb ax8

then ei and e' still form reciprocal sets, for

axr a'V8r = o.

a:C i ax8

In view of (2) and (5), equations (6) show that gjj and gi'transform like absolute dyadics. These tensors often are calledthe fundamental covariant and contravariant tensors of Riemann-ian geometry. By use of the equations,

(7) e = g,rer, ei = girer,

they permit us to raise and lower indices (§ 154) and thus repre-sent any tensor of valence m by 2' types of components.

368 TENSOR ANALYSIS § lti'9

At a given point, gij gives the orientation of the base vectors eirelative to each other. In order to determine the relative orienta-tion of sets of base vectors at different points, we must know thecomponents r of the affine connection, defined by

(8) Diej = rer.Then, just as in § 165, we have also

(9) Die' Tier (165.1).

We now assume that the affine connection is symmetric (r = r ).The gammas then are determined by the metric tensor (§ 166) :

(10) rk = 1 kr(Digjr + Djgri - Drgij) IV 29

The epsilons in n-space, defined in § 146, have n subscripts orsuperscripts. The equations,

ax .. xi axj axk(11) - (9.t. . . - Eb...

ax axa axb ax` '

ax axa axb ax°

(12) axIEij ...k =

ax` 49V...axk Eab...ej,

t We shall call the base vectors constant if Die, = 0 (i, j = 1, 2,. , n);then raf = 0 and the functions gij = ei e; are also constant. Conversely,when gii are constants, (10) shows that r!J = 0, and hence Die, = 0.

In Riemannian space it is not, in general, possible to introduce coordinatesxi for which the base vectors ei are constant. Only flat space (§ 175) is com-patible with such Cartesian coordinates; then each point has the position vectorr = xiei and ei = ar/axi. Moreover, for any coordinates 2i in flat space, wehave (cf. § 163, ex. 1, 2)

ax' or ax'ar=

at i.(i) ei =

021 e'=

ax' &V

The geometry on a surface with the metric tensor gi; is Riemannian (n = 2).Unless the surface is flat (a plane, for example), constant base vectors cannotbe introduced. If we regard the surface as immersed in Euclidean 3-spaceeach surface point has the position vector r = xi + yj + zk, where

x = x(u, v), y = y(u, v), z = z(u, v)

are the Cartesian equations of the surface. If we write xl = u, x2 = v, thebase vectors on the surface may be taken as ei = ar/axi; for equation (i)shows that these vectors transform in the manner prescribed in (6). Herer is a position vector in Euclidean 3-space; but, in general, the surface pointshave no position vector in the Riemannian 2-space they define.

Any Riemannian space of n dimensions may be regarded as immersed ina Euclidean space (§ 178) of n(n + 1)/2 dimensions. This theorem has notas yet been rigorously proved; see Veblen, op. cit., p. 69.

§ 169 RIEMANNIAN GEOMETRY 369

generalized from (146.2), show that eij"'k and eij...k are relativetensors of weight 1 and -1; for these are the powers of the JacobianI ax/ax I when it is transferred to the right-hand side.

There are n types of Kronecker deltas in n-space: S,, S b, S bc,up to Sa'bc'...f with 2n indices. As in § 161, they are defined interms of the epsilons. In the case n = 4, for example,

i 1 ibcd 1 ijcdSa = - eabcde Sab - - eabcde ,

3! 2!

ijk 1 ijkd ijkh ijkhabc eabcde , aabcd = eabcde

The rule given in § 161 for the value (0, 1, or -1) of a generalizeddelta still applies. Moreover the preceding definitions show thatall the deltas are absolute tensors, alternating in both upper andlower indices. See Prob. 42.

The n-rowed determinant,

(13) g = 1t etii...ke,s...tgir9js ... gkt [cf. (146.3)],n!

generalized from (152.1), is the contracted product of two epsilonsof weight 1 and n absolute dyadics. Hence the discriminant g ofthe fundamental quadratic form is a relative scalar of weight 2.

The cofactor of gij in g is ggij; hence, from (146.11) and (166.5),we have

Dhg = gg"Dhgij = ggi (rhi, j + rhj,or, in view of (166.3),

(14) Dhg = grhi + gr,. = 2grhr-From (14), we have also

(15) DhV_ -\'g-r,.,,,

a result of the same form as (165.3) with E replaced by -/g.In defining the gradient of a tensor, we replace E by -\/g in

(167.4); thus, in Riemannian n-space,N N

(16) VT = g2ehDh(g 2T).The components of VT again are given by (167.7). Hence thisformula for the covariant derivative is still valid in Riemanniangeometry.


Since the metric tensor G = gjkejek = ekek still has the prop-erty DhG = 0, VG and its components vanish as before:

(17) Ohgij = 0, Ohg J = 0, Vhb = 0.

Moreover, from (16), vq = gehDh1 = 0, and

(18) Ohg = 0.

170. Dual of a Tensor. An m-vector is a tensor of valence mwhich is alternating in all indices (cf. § 160). For convenience inwording, we also regard scalars (m = 0) and vectors (m = 1)as m-vectors. In n-space we can associate with any m-vector(0 < m < n) an (n - m)-vector, its dual, defined as follows:

If and Qij...k are m-vectors of weight N, their duals arethe (n - m)-vectors,

(1)1 ab...c ij...kPj...k,

m.

(2) gab...c - li Qij...kEij...kab...c,

m.

of weights N + 1 and N - 1, respectively. Note that the epsilonshave n indices in n-space; and, in the contracted products, thecontravariant tensor is written first, and the summation indices areadjacent.

A scalar Sp has two duals, Eab .htp, Eab . hV, according as we use(1) or (2); they have the same numerical components but differentweights. The dual of an n-vector Tijk...h is the scalar:

1 ijk...hE Tijk...h = T123...n.

n!

THEOREM. If T is an m-vector (0 m < n),

(3) dual dual T = T.

Proof. Write T = P, dual T = p. Using (2) and (1), we have

(dual pab crs.

(n -- m) !1

= Eab...c ra...t(n - m) !m!

1 i;...k_ - brs tPi;...k [Cf. Prob. 42.]

m!

= Pra...t)

1

§ 171 DIVERGENCE 371

since Pip -.k is alternating in all indices. Hence dual p = P; and,similarly, dual q = Q.

When T is an n-vector, say Tijk...h, (1) gives the dual T123 . . nnow (2) gives

dual dual T = Eijk...hT123...n =

Thus the theorem holds in this case also, provided both dualizingequations (1), (2) are used.

171. Divergence. The gradient of v = viei is

Vv = vhviehei.

The first scalar of this dyadic is

(1) div v = %ib =V11)

If vi is an absolute vector, the divergence is the absolute scalarViv'; this definition applies in n-space and for any coordinatesystem.

When vi is a relative vector of weight 1,

VhV' = Dhvi + 17rrhr - vtrhr (167.7).

On contracting with h = i, the second and third terms cancel; forvrrsr = virir, since r and i are summation indices. Hence

(2) div v = Divi (wt. v = 1).

If vi is absolute, vi has the weight 1 imparted by the scalar/; hence

viva = vi(9-i91vi) = 9-lvd(91v=) = 9-1Di(91vi),

in view of (2). Thus

(3) div v =9

Di( / vi) (wt. v = 0).

The Laplacian V2p of the scalar p is defined as div Vp. If rp isabsolute, v = V p is an absolute vector; then

Vr = Drop, Vi = 9irDrcp,and, from (3),

(4) v2(p Di(V -J 9irDrcp)

The divergence of any tensor T is defined as the gradient VTcontracted on the first and last indices. Thus if T has the com-

372 TENSOR ANALYSIS 1 172

porients T 'j- kh of valence m and weight N,

(5)(divT)ij...k = VhTij...kh

is a tensor of valence m - 1 and weight N.

THEOREM. When T i' ., kh is an m-vector of weight 1, div T is the(m - 1)-vector,

(6) (div T) ij...k = DhTij...kh,

where V in the defining equation is replaced by D. Moreover,

(7) div div T = 0 (m > 1).

Proof. From (167.7),VhTij...kh = DhTij...kh + 1,hrTrj...kh + ... + rk Tij...rh

+ rhrT ij" . kr - rhrT ij... kh.

The two final terms cancel, as we see on" interchanging the sum-mation indices h, r in the last term. The remaining terms con-taining 1'hr vanish separately on summing over h and r; for rh,. issymmetric and T antisymmetric in these indices. Thus (6) isproved.

From (6) and the alternating character of T,

(div div T) 'j... = DkDhT'j... kh = 0.

172. Stokes Tensor. The gradient of the covariant vector vk isthe dyadic Vhvk. From this we form the antisymmetric dyadic,

(1) Pij = Sf V hvk,

known as the Stokes tensor. t When Vk is albsolute,

Ohvk = hVk -rrhkvr,

.

and, since az4r = 0 owing to the symmetry of rr , we have

(2) Pij = SfDhvk = Divj - Djvi (wt. v = 0).

In 2-space we can form from Pij the relative scalar of weight 1,

(3) 2eijPij = P12 = Dlv2 - D2v1 (wt. V = 0),

t Veblen, op. cit., p. 64.

§ 172 STOKES TENSOR

and from this the absolute scalar,

373

(Dlv2 - D2v1) (wt. V = 0).(4)1

-Vg

This is the absolute invariant on the surface with the fundamentalform gig dxi dx', written n rot v or n V X v $ in § 97.

In 3-space we can form from Pii the relative vector of weight 1,

(5) wi = 1EijkP = Ei'kD7'v-2 7k k,

having the components,

D2v3 - D3v2i D3v1 - D1v3, D1v2 - D2v2.

The absolute vector,

(6) rot v = 1 wiei = 1 Ei'keiDjvk,1/9

may be written as a symbolic determinant (cf. § 146):

el e2 e31

(7) rot v = D1 D2 D3V9

V1 V2 V3

Comparing this with (88.19) now shows that rot v is in fact therotation of v previously defined.*

In § 91 we proved that a vector v is the gradient of a scalar in3-space when and only when rot v = 0. In n-space we have thecorresponding

THEOREM. Let v be a continuously differentiable vector in a regionR. Then, in order that v be a gradient vector,

(8) V = OAP, vi = Dip,

it is necessary and sufficient that the Stokes tensor vanish in R:

(9) Pi,=Diva-D;vi=0.Proof. The condition is necessary; for, if vi = Dip, Pi; vanishes

identically, owing to the continuity of the second derivatives of cp.

In (97.9), H=-v19, u=x1, v=x2, ru=el, rv=e2, r,.f

= f2 in our present notation.* In (88.19), J u = x1, ru = el, ru f = fl, etc., in our present

notation.


The condition is sufficient. For the method of § 91, extended tocase of n coordinates, leads to the function,

z1

(10) (P =f vj (xl, x2) ... , xn) dxlal

x2

+I.

v2 (a', x2, ... xn) dx2a2

f 3

" f" v3 (ale a21 x3, xn) dx3 ...

nzn

v (ai a2 .. an-1 xn) dxn

where all a2, , an are the coordinates of a fixed but arbitrarypoint. In the ith integral xi is the variable of integration whilexi+1' ...

, xn are regarded as constant parameters. We now canshow from (9) and (10) that Dig = vi. Let us compute, for ex-ample, D3V. Only the first three integrals in (10) contain x3; theirderivatives with respect to x3 are, respectively,

v3(xl, x2) ... , xn) _ v3(al) x2, ..., x"),

v3(a1 , x2, ... , xn) - v3(al, a2, ... 'X n), v3(al, a2, ... , xn),

when we make use of D3vi = Dlv3, D3v2 = D2v3 in the first andsecond; hence D34P = v3(xl, x2, , xn).

173. Curl. We define the curl of a covariant tensor Tb...d ofvalence m (m < n) as the (m + 1)-vector:

1(1) (curl T)hij...k =m.-

bhij .k VaTbc. d

When m = 0, T = P (a scalar), 0! = 1, and

curl cp = ahV aco = V MPis the gradient of gyp.

When m = 1, T = v (a vector), curl v is the Stokes tensor(172.1).

In general, we have the

THEOREM. When Tab.. d is an absolute tensor of valence m < n,

b(2) (curl T)hij ... k =

Shaijc ......kD..Tbc...d,

M.

§ 173 CURL 375

where V in the defining equation is replaced by D. Moreover,

(3) curl curl T = 0 (m < n - 1).

Proof. From (167.7), we have

VaTbc...d = DaTbc...d - rabTrc...d 1'adTbc...r.

Hence in the right member of (1) there are m terms of the type,

1 Sabc...dl,r 7,ab rc...d,

m!

these all vanish separately when we sum over the subscripts of I'a*.Thus (2) is proved.

From (2) we have

(curl curl T) = 1 Sahi kD (-1-1 Sab d D Tg a

(m +1)'

.

(,'_.

1 Sfet...a DgDaTb...dm!

which vanishes when we sum over g and a.When Tbc...d is an m-vector, the summation in (1) or (2) may

be carried out in m + 1 stages by setting a = h, i, j, , k inturn and summing over the other m indices. Thus, from (1), wehave

(4) (curl T)hij...k = VhTij...k + (-1)mViTj...kh + ... ,

taking the m + 1 cyclical permutations of the subscripts hij .. kin order and placing (- 1)' before the second, fourth, terms.In the cases m = 1, 2, 3, we thus obtain

(curl T)ij = VjTj - ojTi,

(curl T)ijk = ViTjk + VJTki + VkTij,

(curl T)hijk = VhTijk - ViTjkh + VVTkhi - VkThij

When Tbc...d is an absolute m-vector, we obtain, from (2), an equa-tion of the form (4) with V replaced by D.

When S is an absolute tensor of valence m - 1, T = curl S isan absolute m-vector and curl T = 0. Then (4) becomes

(8) 0 = (-1)'DiTjk...h + .


Thus if vi is absolute, T i j = S Davb is an absolute bivector, and

(9) DiTjk + DjTki + DkTij = 0.

174. Relation between Divergence and Curl. For alternatingtensors, we have the

THEOREM. If T be d is an n2-vector (n2 < n),

(1) dual curl T = div dual T,

provided dual T is taken contravariant when T is a

Proof. From § 170,

(dual curl T)pq"'r

(m + 1)!m!

1

scalar.

ii ... k(CUrlT) ij...k

Epq...r ij...kaab ..kdOaTb...d

M!

pq... r ab... dVaTb...d

Va (1. E pq "r Tb ...d

= Va(dual T)P . ra

= (div dual T)pq*'*.

On taking duals of both members of (1), we have

(2) curl T = dual div dual T.

Moreover, if we replace T in (1) by dual T, we have also

(3) div T = dual curl dual T.

175. Parallel Displacement. A vector p is said to undergo aparallel displacement along a curve xi = pi(t) when dp/dt = 0along this curve. If p = pkek, we have

dp dpk aek dxi dpk dxi- ek + pk = - ek + pk rikej,dt dt axi dt dt dt

or, on interchanging summation indices j, k in the last term,

dp

(+pr)ek.dt dt

§ 175 PARALLEL DISPLACEMENT 377

Hence, if dp/dt = 0, the components pk satisfy the differentialequations:

dpk dx'

(1) dt +pj

dt r". = 0 (k = 1, 2, ... , ii).

A solution pk(t) of this system, satisfying the arbitrary initial con-ditions pk(0) = ak, defines a vector at each point t of the curve.The vector ak at the point PO (t = 0) is said to undergo a paralleldisplacement along the curve into the vector pk(t) at the point P.In (1) dxi/dt depends upon the functions pi(t) defining the curve;hence, in general, the solutions pk(t) will change when the curveis altered. In other words, the vector pk obtained by a paralleldisplacement of ak from PO to P depends, in general, upon the pathconnecting these points.

The length of a vector p and the angle between two vectors p, qremain constant during a parallel displacement; for, if dp/dt anddql dt vanish along a curve, we have also

d ddt(P.P) =0, dt(P - q) = 0.

We shall say that a vector remains constant during a parallel dis-placement.

If s is the are along the curve,

ds2 = gi; dxi dx1 = g;xix' dt2.

If we choose the are as parameter (t = s), we have gjjziz' = 1, anequation which states that the tangent vector dxi/ds to the curveis of unit length. If a curve has the property that its unit tangentvectors dxi/ds are parallel with respect to the curve, it is said tobe a path curve for the parallel displacement. With t = s, pk =dxk/ds, (1) gives the differential equations of the path curves,

d2xk dxi dx'

(2) ds2 t' ds ds -0.

The path curves are the "straightest" curves in our Riemannianspace-the analogues of straight lines in Euclidean geometry.

When equations (1) can be satisfied by functions pk(x', , x")of the coordinates alone, the parallel displacement is independentof the path, and the space is said to be flat. Then

dpk apk dxi

dt axe dt


and the ordinary differential equations (1) are replaced by thepartial-differential equations:

ap''(3) axi + p'1' = V pk = 0.

Since Vipk are the components of Vp = e'Dip (§ 167), we see thata flat space contains vectors p(x', , X'), such that Vp = 0, or

(4) Dip = 0 (i = 1, 2, ... , n).

Since pk = p ek, we see that (4) is equivalent to

(5) D,Pk = p Dzek.

For any fixed value of k, the n functions p D,ek are componentsof a gradient vector, and for this it is necessary and sufficient that

Di(p D,ek) - D,(p Diek) = 0 (§ 172, theorem) ;

or, since Dip = 0,

p (DiD, - D,Di)ek = 0.

These equations must hold independently of the initial conditionsimposed upon p and are therefore equivalent to

(6) (DiD; - D,Di)ek = 0.

Since Di transforms like a covariant vector (167.3), the operator

(7) Di, = DiD; - D,Di = a bDaDb,

transforms like a covariant dyadic; for

ax, axb a2xb axa axb

DiD' -(at.,

Da) \ax1 Dbl axiaxI Db + azs aD0Db,

Di, _

Moreover,

Di;ek

ax, axbDiD, - D,Di = -. Dab-

at, ax'

axa axb \1 (axc axa axb axc

axkDabec,

Caxi at-, Dab/ \axk ec/ at - 4921

since D,,Dkxc = 0; hence

eh axa axb axc axhed DabecDiJek = - -axk

at t at, axd

§ 175 PARALLEL DISPLACEMENT 379

This equation shows that e' Di;ek is an absolute mixed tensorof valence four, say

(8) Rijk-h = eh Di;ek.

The components of this curvature tensor R are thus the coefficientsin the equation,

(9) Dijek = Rijkheh;

the condition (6), necessary for flat space, now assumes the tensorform,(10) Rtijk h = 0.

If it holds in one coordinate system, it holds in all.When the space is flat, we can choose n linearly independent

vectors ai at a point P and, by giving them parallel displacementsto neighboring points obtain a set of constant base vectors ei = aiin a region about P. For these base vectors, we have

gi; = ai a; = const, I'k = 0 (166.6),

and the corresponding coordinate system x is called Cartesian.To determine the Cartesian coordinates y = x corresponding to

the base vectors ai, we haveayr

ek =axk a,.,

aek c12yr

ar,OX, ax' axk

and, on dot-multiplying (11), member for member, by

ay8- er = a8 ,axr

(12)

ays a2y3r

ax, rA axi axk

t

From (11), we have the necessary conditions for the integrabilityof equations (12):

D;ek = Dke;, Di,jek = 0;

t This equation also follows from (163.9) with Tj = 0.

380 TENSOR, ANALYSIS § 176

that is,r = r h

rjk rkj, Rijk = 0.

These conditions are also sufficient for the complete integrabilityof equations (12).$ When these conditions are fulfilled, (12) ad-mits solutions y(x', x2, , x") which with ay/axi take on arbi-trary values at a given point. If we place the origin of the Car-tesian coordinate system at the point x0, we have y = 0 whenx = x0; and, if we choose n linearly independent sets of initialvalues,

axi - pi,a

= p2, ... ,ax"

= pn (i = 1, 2, ... , n),

when x = x0, we obtain n corresponding solutions y'(x) whoseJacobian I ay/ax I = det p 0 when x = x0. In the region aboutx = x0 for which I ay/ax 19 0, the n functions yj(x) thus obtaineddefine a Cartesian coordinate system. In brief, we have the im-portant

THEOREM. A necessary and sufficient condition that a Riemannspace, with symmetric afne connection, be flat is that its curvaturetensor vanish identically.

We may readily compute the components Rijkh from (8):

*.k'; eh (DiDjek - D;Diek)

= eh {Di(r;ker) - Dj(riker)I

eh {(Dirjk)er - (Djrik)er} + eh {rjkrires - rikrjres},

and, on putting eh e, eh e8 = 887 we have

(13) Ri;kh = D2-r' - Djr k + r; r r -

176. Curvature Tensor. From the defining equation for thecurvature tensor,

Dijek = jirer,

we obtain the covariant components,

(1) eh Dijek = Rijk r9rh = Rijkh

$ Cf. Veblen, op. cit., p. 70-1.

§ 176 CURVATURE TENSOR :381

We now can express the covariant curvature tensor Rijkh in termsof the gammas :

Rijkh = eh (DIDjek - D;Diek)

= eh' {Di(rjk,re') - Dj(rik,rer)} (166.4)

Sh{Dirjk,r - Djrik,rI - 0h{ rjk,rria - rik,rrja};(2) Rijkh = Dirjk,h - Djrik,h - rjk,rrik + rik,rrjhSince

rjk,rrih = grerikrih = rjkrih,s,

we also may write (2) in the form:

(3) Rijkh = Dirjk,h - Djrik,h - rikrih,r + rr,krjh,r.

Rijkh has the following types of symmetry:

Rijkh + Rijkh = 0;

Rijkh + Rijhk = 0;

Rijkh + Rjkkh + Rkiih = 0;

Rijkh - Rkhij = 0.

Proofs. (I) follows from Dij = -Dji. Since the scalars ek eh= gkh have continuous second derivatives (§ 169),

gives (II). Thus Rijkh is alternating in its first two and last twoindices.

Since the affine connection is symmetric (§ 169), we have

(4) Die; = r er = r;ier = Djei.

Hence, on adding the identities,

Dk(D1ej - Djei) = 0,

Di(Djek - Dkej) = 0,

Dj(Dkei - Diek) = 0,we obtain

Dijek + Djkei + Dkjej = 0,

which, on multiplication by eh- , gives (III).


Now (IV) is a consequence of (I), (II), and (III). From (III),we have

Rijkh + Rikih + Rkiih = 0,

Rjkhi + Rkhji + Rhjki = 0,

-Rkhij - Rhhkj - Rikhj = 0,

- Rhijk - Rijkk - Rjhik = 0.

When we add these equations and make use of (I) and (II), onlythe underlined terms survive, namely, 2Rijkh - 2Rkhij, and weobtain (IV).

The symmetry relations (I) to (IV) reduce the number of inde-pendent components of Rijkh to i22n2(n2 - 1).

Proof. Rijkh = 0 when i = j or k = h (I, II); hence, in general,the number of non-zero components is (nC2)2 = n2(n - 1)2/4.But, when ij ; kh, these are paired, because Rijkh = Rkhij (III);hence, if we add the number nC2 of unpaired components Rjjij tothe preceding total, we obtain double the number of componentswith distinct values. The number of components with distinctvalues thus is reduced to

1n2(n - 1)2 n(n - 1)

4 + 21=*n(n-1)(n2-n+2).

These are still further reduced by the ,,C4 relations (III); fori, j, k, h must all be different in order to get a new relation. If,for example, i = j,

Riikh + Rikih + Rkiih = Rikih + Rkiih = 0

is already included in (I). The number of linearly independentcomponents is therefore

n(n - 1)(n2 - n + 2) -n(n - 1)(n - 2)(n - 3)

24= n2(n2 - 1).

For n = 2, 3, 4 this gives 1, 6, 20 linearly independent compo-nents Rijkh, respectively.

Example 1. When n = 2, the contracted product,

eijEkhR,,kh = 4R1212,

§ 176 CURVATURE TENSOR 383

is a relative scalar of weight 2 (§ 169); hence R1212/g is an absolute scalar.Now, from (2),

(5) R1212 = D1r21,2 - D2r11,2 - r21,.ri2 + rll,rr22

We shall compute this expression when the base vectors are orthogonal:g12 = 0. From (166.6),

rll,1 = !Digll, r11,2 = -iD2g11,

r12,1 = ZD2911, r12,2 = 2D1g22,

r22,, = -2D1922, r22,2 = 3Dzg22.

Moreover, since g = 911922, 911 = 1/911, g2 = 1/g22; hence

r21,rr12 = r21,,ri2 + r21,2r12

= r21,,r,2,1 911 + r21,2r,2,2 g22

- (D2g11)2 + (D1922 )2

4911 4922

r11,rr22 = rll,lr22 + r11,2r22

= r11.1r22,1 911 + rll,2r22,2 g22

(Dig,,) (D1922) (D2911) (D2g22)

491, 4922

Substituting these results in (5) gives

81212 - DiD1922 + 2 D2D2911

lDlgll + D1922 D19zz + Dw11 + D2922 D2g11 jSt l\ 911 922 911 922 J

2

The absolute scalar, \ J

_81212 1

(.\/g-

1 \(6) K

9 219(Dl D19zz +

D2D2 911 J ,

is precisely the total curvature of the surface whose fundamental differentialform is

ds2 = gll dx1 dx1 + 922 dx2 dx2;

for, if we put x1 = u, x2 = v, 911 = E, g22 = G, H =, (6) agrees with(139.6).

Example 2. We may contract the tensor,

(7) R41kh

= ghrRijkr,

in essentially two different ways.


With h = k, we have

(8) Rt1xk = gkrRiikr = 0,

since gkr and Rijkr are, respectively, symmetric and antisymmetric in k, r.From (175.13), we see that (8) is equivalent to the identity:

(9) Dirjr = Djr;,.

With h = i, we obtain the Ricci tensor,

(10) Rik = R0 ' = 9ihRijkhi

this is a symmetric dyadic; for, from (IV),

(11) Rkj = 9''`Rikih = 9'Rihik = ghiRhiki = Rik.

The first scalar of this dyadic,

(12) R = gikRik = gikgihRijkh,

is an absolute invariant. In the case n = 2, g12 = 0 considered in ex. 1, wehave(13) R = 2g11g22R2112 = -2R1212/9 = 2K.

177. Identities of Ricci and Bianchi.Ricci Identity. In analogy with

Dij = D,Dj - DjDj, we also write Vij = ViVj - V, Vi.

With this notation, (167.21) becomes

(1) VijT = DijT

for any tensor, absolute or relative, with its base vectors. On theleft Vii acts only on scalars (cf. § 167, ex.); on the right Dij actsonly on base vectors, for Dij<p = 0 when cp is a scalar. Equation(1) yields the Ricci identity when we evaluate DijT by making useof the formulas :(2) Dijek = Rij'heh (175.9),

(3)Dijek = - Rijh keh.

Equation (3) is proved as follows:

Dijek = Dijgkrer = gkrR,;r.aea =gkryh8Rijrhee

kr ha k h_ -g g Rijhrea = R;jh e .

For the vector vkek, we have

Dijvkek = vkDijek = vkRijk aea = vaRija kek,

(4) Vijvk = v8RtijB'k.

§ 178 EUCLIDEAN GEOMETRY

Similarly, for vkek,

Vijvkek = vkDijek = -vkRijs ke3 = -v3Rijk'ek,

(5) Vijvk = -v8Rijk8.

385

The general Ricci identity now is readily established. Thus, ifthe components of T are Thk:::;,a, for every upper index * in T,

DijTh::...m contains a term, Th..' ..mRij,;

and, for every lower index * in T,

vijTh::;:: ,, contains a term, ,Bianchi Identity. At the origin of geodesic coordinates (§ 163),

we have, from (174.13),

DiRjkhm = DiRjkhm = DiDjrkh - DiDkrjh.

By permuting ijk cyclically in this equation, we obtain two others.On adding the three equations, we find that the right memberscancel; we thus obtain the Bianchi identity:

(6) DiRjkhm + OjRkihm + OkRijh-m - 0.

Since this tensor equation holds at any point, it is also true forgeneral coordinates.

178. Euclidean Geometry. When the space is Rat, we can de-termine a Cartesian coordinate system xi (§ 175). The corre-sponding metric tensor gij has then constant components.. If inaddition the metric form gijxixj is positive definite, the space andits geometry are termed Euclidean. We can then always make areal linear transformation to coordinates yi for which gij = b, theKronecker delta, and the metric form becomes a sum of squares(§ 150) :

b yiyj = y'y' + y2y2 +.... + yny+l.

The corresponding base vectors ai then form an orthogonal set(ai - aj = b), and the coordinates yi are said to form an orthog-onal system.

Let xi denote a Cartesian coordinate system with the base vec-tors ai. If we transform to another Cartesian system xi with thebase vectors Ai, we have

8xt

aj = - ai

386 TENSOR ANALYSIS

and, since both a, and ai are constant,

aa; a2xiai = 0,

a2k aXk a2

exi

82k a2'= 0.

On integrating this equation twice, we have

(1) xi=c;2'+Ci,

§ 178

where c and Ci are sets of constants. The transformation betweenany two Cartesian coordinate systems is therefore linear with con-stant coefficients. As in the general transformations of § 157, werequire that the Jacobian,

(2)ax'

det - = det c 54 0.a2'

The transformations (1) with non-vanishing determinant form agroup-the affine group.

When the Cartesian coordinate systems y and y are both or-thogonal, the law of transformation,

ayaayb aya aya

S;.gi; = ayi - gab, becomes - =Ji09)

If we multiply this equation by ayi/ayk and sum with respect to i,we obtain

ayk ay'(3)

ay;= ayk

Since orthogonal coordinate systems are also Cartesian, the trans-formation between two orthogonal systems has the form:

(4) yk = c;4j' + Ck.

The inverse transformation is

(5) yj = Yk(yk - Ck),

where Yk is the reduced cofactor of c; in det c;. Equation (3) thusbecomes

(6)

This is precisely the condition that the matrix c; be orthogonal(§ 149, theorem); thus a transformation between orthogonal coordi-

§ 178 EUCLIDEAN GEOMETRY 387

nate systems is orthogonal (has an orthogonal matrix). Condition(6), which characterizes an orthogonal matrix, also implies that

(7) c'iCj = cic = Sj,

in view of the relations (146.8) between reduced cofactors. Con-ditions (7), in turn, imply (6); either (6) or (7) is a necessary andsufficient condition that the matrix c; be orthogonal.

Two orthogonal transformations,

yz=ay'+Ay, yj =bkyk+Bj,have an orthogonal resultant,

yis i i i, j;

we have, for example,r r r a r t s a t s a icic; = (aebi)(atbj) = Stbib; = bib; = S;.

Moreover (5), the inverse of (4), and the identity transformationyi = S;yj are orthogonal. Consequently, the orthogonal transforma-tions form a group.

In view of (3), the equations,

ayk ayjVi = ayj vk, U' =ayk tlk

show that covariant and contravariant vectors transform alikeunder orthogonal transformations. Within the orthogonal group,the distinction between covariance and contravariance vanishes, andtensor components may be written indifferently with upper orlower indices. For example, we may write Sij or Sij for the Kro-necker delta.

The orthogonal group of transformations admits as a subgroupthose transformations for which

a(8) det c = yy = 1.

a

If we regard (4) as a transformation between the points y and yin the same Cartesian coordinate system, the transformation iscalled a displacement or rigid motion. In fact in 3-space the trans-formation (5) may be written

(9) s=r-a,


where the dyadic CF in non ion form is given by the matrix (ck).Since this matrix is orthogonal and its determinant is 1, the trans-formation (9) is a translation followed by a rotation (§ 75, theo-rem), in brief, a displacement. The subgroup characterized bydet c = 1 is therefore called the displacement group. In view of(8), we see that, within the displacement group, the distinction be-tween absolute and relative tensors also vanishes.

A displacement which leaves the origin invariant is called arotation. Thus the transformation y' = c;y' is a rotation if thematrix (c) is orthogonal and has the determinant 1. Rotationsform a subgroup of the displacement group.

179. Surface Geometry in Tensor Notation. The equations,

x' = xi(ul) u2) (i = 1, 2, 3),

define a surface embedded in Euclidean 3-space. The space co-ordinates xi are rectangular Cartesian and are designated by italicindices (range 1, 2, 3); the surface coordinates u" are curvilinearand are designated by Greek indices (range 1, 2). If we writeal = i, a2 = j, a3 = k, the position vector to the surface isr = xiai. The metric tensor in space is then

(1) Si; = ai - a;.

If we limit the coordinate transformations x - x to the displace-ment group (§ 178) the distinction between covariance and contra-variance as well as the distinction between absolute and relativetensors does not exist in 3-space.

First fundamental form. The base vectors on the surface are

ar ar axi(2) a"=-=--=x.ai,

au" axi au"

where x" = axi/au". Since

e"-e#= =x'xp'si; =xaxs,

the metric tensor for the surface is

(3) gap = e., ep = x;.xP.

This tensor defines the first fundamental form on the surface:ds2 = g"g du"du#.

§ 179 SURFACE GEOMETRY IN TENSOR NOTATION 389

Note that x, is a covariant surface vector; for if we make thetransformation u -+ u, we have

axe axi auo

aic" aufl au"

If a vector v has the "surface components" v" (a = 1, 2),

v = vae« =

and vi = v"xa (i = 1, 2, 3) are the "space components" of v.

Unit surface normal. The space vector,

N = el x e2 = Eijkaixix2,

has the components Ni = Eijk xix2; moreover

N2 = EijkEist xix2 x1x2 =sit

xix2 XIX2

= xix2 xix2 - xix2 xix2 = 911922 - g12

Thus N2 = det ga# = g t; hence the unit normal n to the surfacehas the components Ni/\:

(4) ni = ni = eijk xix2.

Second fundamental form. On the surface with metric tensorgang, the Christoffel symbols are given by (cf. § 166)

(5) is = ZgX7(D«goti + DRgy« - D7g«#)

Covariant derivatives are then computed from the formulas of§ 167. In particular we have, from (167.13),

(6) Vaxkp = DaDftxk - VPxQ.

Since the covariant derivative of the metric tensor gij is zero(167.19),

(7) Vagoti =V. xksxy = 0.

If the product rule (§ 168, 1) is used, this equation and its cyclicalpermutations give

(7a) xyV 4 + x0Vaxy = 0,

(7b) xgVVxy + xyVflxx = 0,

(7c) xkovyxa + xxVyx1 = 0.t This also follows from the expansion of (el x e2) (e1 x e2) given in (20.1).

390 TENSOR, ANALYSIS § 179

Subtract the third equation from the sum of the first two; then,in view of (6), we obtain

xyDax3 = 0.

This equation states that the space vector Vax is perpendicularto both el and e2 (whose space components are 4, x2); that isVax, is a multiple of the unit normal nk, say

(8) oax3 = ha#nk.

The symmetric covariant tensor,k

(9) ha0 = nkVaXQ = hha,

defines the second fundamental form on the surface: ha#duadus.

Derivative formulas. From (6) and (8) we have

(10) Daxkg = F xk\ + haonk;

these equations are the derivative formulas of Gauss. If we adjointhe (constant) base vector ak to each term they become

(10)' Dae# = raaea + hafln. f

On multiplying (10) by xry and summing on k, we have also

(11) 'xk Da X = raggay = rap,7.

Since nk is a space vector with no components along el or e2,Vank = Dank. On differentiating nknk = 1, we obtain

(12) nkVank = nkDank = 0;

hence Dank (1 nk) is a tangential surface vector. Similarly, fromnkx' = 0, we obtain

nkvaxp i xpvank = 0;or, in view of (9) and the symmetry of hao,

(13) hp = -xxVank = -xaV nk.Hence

hapxa = -9aavpnk,

90XhaflxX o Vpnk = - Yank,

(14) Dank = Dank - -ham xa.

t Note that the term hapn in this equation is in apparent disagreement with(163.1); this is due to the fact that our 2-dimensional geometry is not intrinsicbut that of a 2-space embedded in a 3-space.

§ 179 SURFACE GEOMETRY IN TENSOR NOTATION 391

These equations are the derivative formulas of Weingarten. If weadjoin the (constant) base vector ak to each term, they become

(14)' Dan =

Equations of Codazzi and Gauss. From (8) we have

Osxy = hsynk,

VaV#xky = (Vahs.y)nk + hsyVank,

= (Vahsy)nk - hsyhaIxk

in view of (14). Now form V$Vaxk, and subtract it from the lastequation; writing Vas for the operator VaVs - we thusobtain

(15) VasXY = (Vahsy - V$hay)nk + (hayha"' - hsyha")xx.

If we replace Vasx1' by the value,XVasxak = -Raay xkx,

given by Ricci's Identity (177.5) and adjoin ak to each term, (15)becomes

(16) -Rasyx ex = (Vahsy - oshay)n + (hayhft" - hsyhal')ex.

This vector equation is equivalent to the scalar equations :

(17) 0 = Vahsy - vshay,

(18) ROT" = hsyha)` -

Equations (17) are the Equations of Codazzi. When a = /3 theright member vanishes identically; and an interchange of a and 0repeats the same equation. Hence there are but two independentCodazzi equations; these may be written with a = 1, S = 2, y =1, 2;

(19) V1h21 - V2h11 = 0, V1h22 - V2h12 = 0.

On multiplying (18) by gas and summing on X, we obtain thecovariant curvature tensor:

(20) Rasys = hsyhas - hayhss.

From the symmetry has = hsa we may verify at once the foursymmetry relations of Rasys :

Rasys = - Rsays, Rasys = - Rassy,

Rasys + Rsyas + Ryass = 0, Rasys = R,ysas.

392 TENSOR ANALYSIS §180

As Greek indices range over 1, 2, these relations show that thereis but one independent equation (20). This Equation of Gaussmay be taken as

(21) - 81212 = h11h22 -22

= h,

where h = det han.

Total and mean curvature. The contracted product,

E"S Erya R,,,#,3 = 4 R1212 and g = 811922 - 912

are both relative scalars of weight 2; hence,

(22) K = -R1212/9 = h/9is an absolute scalar, namely the total curvature of the surface(§ 176, Ex. 1).

The mean curvature of the surface is defined as the absolutescalar J = g"0hg.

Since g"R is the cofactor of gag in det ga#,

(23)

hence911 = 922/9, 922 = 911/9,

912 = 921 = -912/9;

(24) J =9a,6haQ

= (922h11 - 2g12h12 + 911h22)/9

180. Summary: Tensor Analysis. Under general transforma-tions,

.tiJ

={i(x1, x2, . .,

xn),

ax

ax001

the component of a relative tensor of weight N transforms accord-ing to the pattern:

Ti'.k =ax

ax

N axi ax' axcb

axa axb axk

When N = 0, the tensor is absolute. For brevity, components oftensors often are called tensors.

The number of indices on a tensor component is called itsvalence. In n-space a tensor of valence m has nm components.

A tensor of valence zero is a scalar. A scalar p(x) has one com-ponent in each coordinate system given by

ax N,7p (.t) =

I ax I

'P(x)

§ 180 SUMMARY: TENSOR ANALYSIS 393

A tensor of valence one is a vector. The differentials of the co-ordinates and the gradients of an absolute scalar are the prototypesof absolute contravariant and covariant vectors :

axz a. axrdx' _ - dzr, -- _ - -

axr axi axraxi

Measurement is introduced into Riemannian geometry by thenon-singular quadratic form:

ds2 = gi; dxi dx.' (gi; = gji, g = det gi; F6 0).

The character of the geometry depends upon the choice of then(n + 1)/2 functions gij(x) of the coordinates. The relationsei e; = gij determine the lengths of the base vectors ei and theangles between them.

Any vector in n-space at the point x is linearly dependent uponthe n base vectors ei at this point. The reciprocal base vectorse' are defined by e' . e; = S . If g = det gij and g'' is the reducedcofactor of gij in g,

(1} ei = girer, ei = girer;

= gi'.(2) ei e; = gi;, ei e' = S, e' eIn passing from coordinates x to x, the transformation of the

base vectors is prescribed by

(3)

axr axiei=- er, ei=-er.axi axr

These equations show that gjj, M, g2' are components of an abso-lute dyadic, the metric tensor G = gi,eiel = e;e'.

Use of equations (1) permits indices to be raised or lowered ontensor components:

9irT.r., T.i. = girT*r.

If T is a tensor of weight N (say T = T?'.ke,ejek),

ax IN

T = - T.at l

Addition of tensor components of the same valence, weight, andtype produces a component of this same character.

394 TENSOR ANALYSIS §180

Multiplication of tensor components of valence ml, m2i of weightN1, N2, and of arbitrary type produces a component of valencem1 + m2 and of weight N1 + N2.

Contraction of a tensor of valence m > 1 results on forming thedot product of any two of its base vectors. If the vectors in ques-tion are ei e' = s, the components of the contracted tensor areobtained from the original components by putting i = j and per-forming the implied summation.

The components of the affine connection rk are functions of thecoordinates defined by

(4) Die; = I'er (rk = ek Diei).

Then also

(5) Die' = - rrer;(6) DiG = 0 (G = eye');

(7) Dig = 2grirBy definition,

ri,,k = gkrr, then rk. = gkrri;,r.

is symmetric in ij,When r k13

ri7,k = 2(Digik + Digki - Dkgi.l)The gradient of a tensor of weight N and valence m is defined

as the tensor,N N

(9) VT = g2ehDh(g 2T)

of weight N and valence m + 1. When T is absolute, VT =ehDhT.

The components of VT, denoted by prefixing Vh to the compo-nents of T, are called covariant derivatives. For any tensorwe have

(10) VhT : = DhT; - NrhrT:+ (T -rhr + ...) - (TK:rr +

the first parenthesis contains one term for every upper index, thesecond contains one term for every lower index.

The metric tensor G is absolute and VG = 0, from (6); hence

Vhgii = 0, Vhba = 0, Vhgii = 0.

§ 180 SUMMARY: TENSOR ANALYSIS 395

Since g is a relative scalar of weight 2, Vg = 0 from (9), andOhg=0.

The covariant derivatives of the epsilons and Kronecker deltasare zero.

The divergence of a tensor T is defined as the gradient VT con-tracted on the first and last indices: thus

(divT)ij...k = VhTij...kh;

when T is an m-vector of weight 1, we may replace Oh by Dh.The curl of a covariant tensor of valence m < n is de-

fined as the (m + 1)-vector:1 abc...d(curl T)hii...k = ' Shij...kVaTbc ...d.

m.

When T is absolute, we may replace Va by Da.A Riemannian n-space xi with the metric tensor gij and base

vectors ei has the associated curvature tensor,

RiJk h -eh (DiDj - DjDi)ek.

Its covariant components,

Rijkh = eh (DZDj - D;Di)ek,

have four types of symmetry:

Rijkh + Rjikh = 0, Rijkh + Riihi = 0,

Rijkh + Rikkh + Rki h = 0, Rijkh - Rkhij = 0.

These relations reduce the number of linearly independent com-ponents Rijkh to n2(n2 - 1)/12. When n = 2, there is but oneindependent component, say R1212; and the absolute scalar-R1212/g = K, the total curvature of a surface whose funda-mental form is gij dxi dxj.

A Cartesian coordinate system yi is one in which the componentsgij of the metric tensor are constants; then all t = 0, and thebase vectors ai remain invariable in space (aaj/aye = 0).

The Riemannian space xi with metric tensor gjj is said to beflat if it is possible to transform to a Cartesian coordinate system.When the affine connection is symmetric, a necessary and suffi-cient condition for a flat space is that the curvature tensor vanish.

If the space is flat and the metric form gijxixj is positive definite,the space and its geometry are termed Euclidean. We then can

396 TENSOR ANALYSIS

make a real linear transformation to orthogonal coordinates yi forwhich gij = S; the metric form then becomes a sum of squares.A transformation y i = c)y' + C' between orthogonal coordinatesystems is characterized by the relations:

ayi ay'or c; = y= (orthogonal matrix);

ay' ay,then

crx; = cTcT = S, and det c; _ ± 1.

Within this orthogonal group of transformations, the distinction be-tween covariance and contravariance vanishes. Orthogonal trans-formations for which det c) = 1 form the displacement subgroup inwhich the distinction between absolute and relative tensors van-ishes.

When yi and y' are regarded as points in the same coordinatesystem, the transformation yi = c)y' is a rotation when its matrixis orthogonal and its determinant +1.

PROBLEMS

Summation Convention. Index range is 1, 2, 3 unless otherwise stated.1. Prove the following:

(a) Etjk ETjk = 2! Si;

(b) Eijk Srst = 3! Erst;

(c) Si = 3, S = 3 2, Stjk = 3! [§ 161.1

2. Show thatu1 u2 143

V1 v2 v3

W 1 1U2,W3

= Eijk UiVjWk.

3. Show that the two-rowed determinant formed by columns i, j of thematrix,

xi x2 ... xn

Yy2 ... yn) is ST8 xrys.

4. For the dyadic vi, in (77.1) show that the scalar invariants are

'P1 = 'Pit, 'P2 = fit, 'P3 =3!1 Eijk Erat

'Pzr'Pjs'Pkt;

and that the vector invariant has the components z fi k'Pik.5. Show that the general solution of the equations:

aix' = 0, bix' = 0 is x' = X ajbk.

PROBLEMS 397

6. Show that the cofactor Arl of the element a= in a = det a is

At = 1 Eiik Eratr 2

At al = l E Eiik aPaa ' = l E E pat r[ i - 2 rat : f k = 'l rat a = b a.]

7. If yl, y2, y3 are functions of x1, x2, x3, det (ay'/axi), is called their Javabian and written

a(yl, y2, y3)more briefly,

,or,

a(xl, x2, x3)

If z1, z2, z3 are functions of yl, y2, y3, prove that

Oz

ax

If, in particular, the functions

z`show that

8. Prove that

a(y', y',xr'

x,'a(

9. Prove that the

ay

ax

ax

ay

ayx

= 1.

yk) = Stik a(yl, Y2, y3)xt) rat a(xl'

x2,x3) .

t

Cofactor of ay. inaxi

ay I ax' I ayis -.Ox ay' ax r

ay' axi ay=

i = t

ax' a yk a yk -ak

10. If the elements of a = det aii are functions of x1, x2, x3, prove that

as _ aa,t Aataxr axr

11. Prove thata

axr

ay

ax

02y' axi ay

axr axi ayi ax

[Apply Probs. 9 and 10.]12. Prove that the bordered determinant:

vl V2 V3 0

all a12 an ul= uiviAii.

a21 a22 a23 U2

a31 a32 a33 u3

az

ayayax

(yl, y2, y3) = x`,

This determinant is formed by bordering det aii with the vectors vi and ui.If aii is symmetric it also equals viuiA `i.

398 TENSOR ANALYSIS

13. When the index range is from 1 to n show that the bordered determi-nant (written compactly),

V, 0_ 1)n+lusy7iAi7 (i, = 1, 2,

aii ui

14. If Aij is the cofactor of aij in a = det aii, and A = det Aii, show thataA = a3 (and hence A = a2 when a 96 0). State the corresponding theoremwhen the index range is from 1 to n.

15. Show that the n-rowed determinant,

a = det apq = ij"'k aii a2i ... ank

... ii...k= j8E ari aaj ... aek,

n!and that the cofactor of ari is

Ari = 1 Erd... Eij...k a8. . . . atk(n - 1)! '

16. If det aii is symmetric (aii = aii), show that det Aii is also (A ii = Aii).17. If the n-rowed determinant a = det aii is antisymmetric (aii = -aii),

show that a = 0 when n is odd.18. Show that the linear equations,

aij xi = 0, (i, j = 1, 2, , n),

for which det aij = 0 and not all the cofactors Aij vanish has a non-zero solu-tion of the form xi = Aki for some value of k.

19. If aij and gii are symmetric dyadics and gijx`xi is a positive definitequadratic form (§ 150), prove that the roots of the cubic equation,

(1) det (aii - X gii) = 0, (i, j = 1, 2, ... , n),

are all real.[The system of n linear equations,

(aij - X gii)zj = 0,

must have a solution zi = xi + iyi other than (0, 0, , 0) when X = a + 1$is a root of (1). Hence

[aii - (a + i$)gii11l(x1 + iy') = 0;

and, on equating real and imaginary parts to zero, we have

(2) aii xi - a gii x' +,6 gii y' = 0,

(3) aij yi - a gii yj - 0 gii xi = 0.

On multiplying (2) by yi and (3) by xi, and subtracting, we find that

0(gii xx' + gii yy') = 0, and hence 0 = 0;

for if gii xx' = gij y'y' = 0, xi = yi = 0 and consequently zi = 0, contrary tohypothesis.]

PROBLEMS 399

Tensor Character.20. Prove the theorem: If aii, bii, cj are absolute dyadics,

a = det aii, b = det bii, c = det cj

are scalars of weight 2, -2, 0; the cofactors Aii, Bii, Ct in these determinantsare dyadics of weight 2, -2, 0; and the reduced cofactors A"/a, Bii/b, Cs/care all absolute dyadics. [Cf. Prob. 15.1

21. Show that, if u, v, w are absolute vectors,

ui u2 u3

Vi V2 V3

W1 W2 W3

and

u1 U2 u3

vl V2 V3

W1 W2 w3

are relative scalars of weight 1 and -1.22. If aii and b" are absolute antisymmetric dyadics in 3-space, show that(a) a23, a31, a12 are components of a contravariant vector ui of weight 1;.(b) b23, b31, b12 are components of a covariant vector vi of weight -1:

[ux. =Z1 Eijk aik; Vi = 2 Eiik bik].

23. If T1. ..,,, Sz1"'h are absolute n-vectors (§ 170) in n-space, prove thatT12...n and S12 ' are relative scalars of weight 1 and -1.

[n! T12... , = eti ...h T"...n.]

24. If vi, ui, aii are absolute tensors, prove that the bordered determinantin Prob. 13 is a relative scalar of weight 2.

25. Prove the "quotient law": If the set of functions viTi:: a are tensorcomponents of the type indicated by the indices for all absolute vectors vi,then Tz:: c is a tensor of the same weight.

26. If uk is a covariant vector, prove that the total differential equationuk dxk = 0 has the same form in all coordinate systems.

This equation is said to be integrable when there exists a function X suchthat Auk = aw/axk; for the equation is then equivalent to dp = 0 and _const is an integral. Show that

eilk uiDjuk = 0

is a necessary condition for the integrability of ilk dxk = 0 and that the formof this condition is the same in all coordinate systems.

27. Let u, v be quantities (scalars, base vectors, tensors) whose "product"uv is distributive with respect to addition but not necessarily commutative.Show that the differential operator Dii = DiDi - DiDi has the property,

Dii(uv) = (Disu)v + u(Di v).

28. Show by direct calculation that the operator Dii transforms like a co-variant dyadic:

ax, axbDii = . - Dab.

a2i a2i

400 TENSOR ANALYSIS

29. Deduce (177.3) from (177.2) by applying the operator Dii to eh ek

= Bt- [Cf. Prob. 27.130. If T is a tensor (complete with base vectors) of weight N and valence in,

prove that eiei DiiT is a tensor of weight N and valence m + 2.[Since E-NT is an absolute tensor, R5'-NT = E-NT; hence, from Prob. 28,

_i N_.

,.-axa -axb N a b NeieDi (R-,T) = e e e

axi ax'Thus (Prob. 26),

eie1 Dii(E-NT) = E-N eiei DiiT

is an absolute tensor, which multiplied by EN yields the relative tensoreiei DiiT of weight N.]

31. In Riemannian n-space with the metric tensor gii show that the gradientof an absolute scalar p(xl, . , xn) is given by

Vv = eiDi,p = eigii Div = 1 (Div)e7Gi1,9

or, in view of Prob. 13, by the bordered n-rowed determinant

(1) vw =(-1)n+i ei 0

9 19ii Disc

Show that the scalar product of (1) by VV = elDkv gives(2) (-1)n+1 Disc 0

9 19ii Div I

32. In Riemannian n-space with the metric tensor gii show that the diver-gence of an absolute vector v is given by

Gd t')iv v Di(9 D' (-!'L- V9-

or by the bordered n-rowed determinant,

Di 0(-1)n+l(1) dlv v = 1

1/9 9ii 9 Vi

where the determinant is to be expanded according to the elements of thefirst row and the operators Di applied to differentiate their cofactors.

In particular, if v = Vv, vi = Dip, we obtain the Laplacian,

Di 0(_1)n+11

V9- 9ii9

Di'v

33. The equations of a surface in 3-space are xi = xi(ul, u2). Show thatt

za =auu

(i = 1, 2, 3; « = 1, 2)a

PROBLEMS 401

is a contravariant vector in 3-space (1 = 1, 2, 3; a fixed) and a covariant vec-tor in the 2-space (a = 1, 2; i fixed) formed by the surface.

Prove that1 ao i kpi = 2 e ziik xa x9

is an absolute covariant space vector normal to the surface. Write out itsthree components in full.

34. In special relativity it is customary to use the independent variables,

xl=x, x2=y, x3=Z, x4=ctwhere c is the speed of light. The interval ds between two events in spacetime is defined by

(1) ds2 = CO dt2 - dx2 - dye - dz2 = gii dxi dxi.

Hence the metric tensor is given by

(2) 9i1 = g22 = 933 = -1, 941 = 1, gii = 0 (i j).

If v is the speed of a particle relative to a frame j5, we have (ds/dt)2 = c2 - v2from (1), and hence

ds c v' \\(3) dt=y where y=(

/1--)

In a rest frame ao attached to the particle a clock registers the proper time r.Putting v = 0, t = r in (3) we have

(4)ds dt_

dr = c anddr = y.

Corresponding to the position vector (x, y, z) in space we now have theevent vector xi = (x, y, z, ct) in space time. The velocity and acceleration four-vectors are now defined as

(5)dxi duiU = - ,dr dr

Denoting t derivatives with dots, prove that

Eli _ _y (1, y, Z, c), ui = 'Y(-x, -y, -Z, c); utui = c2;

w' 'Y(7x + -Yx, 'Yy + 'Yy, 'YZ + ?'z, -ic); wtui = 0.

Thus the velocity four-vector has the constant magnitude c and is alwaysperpendicular to acceleration four-vector.

Show also that in the rest frame ao (t = r),

uo = (0, 0, 0, c), uo = (x, y, z, 0).

W, = -

Covariant Differentiation.35. Prove the product rule (§ 168, 1) for covariant differentiation without

resort to geodesic coordinates.

402 TENSOR ANALYSIS

36. If Ti iis an absolute bivector, show that

1div T = vi 7'ij = - Dj(V'g Tij).

37. If u and v are absolute vectors, prove that

Dh(gij UV) = ui. Vhv2 + vi Vhuh.

If I u I is the length of u, prove that

DI I-38. Prove that

Ui Vh U

Vijvk + vjkvi + Vkivj = 0. [Cf. (177.5).39. Prove that

Rijkh + RJkih T RkjJh --0-

By contracting this identity with h = j, show that the Ricci tensor Sik (§ 176,ex. 2) is symmetric.

40. In the Bianchi identity (177.6) raise the index h and contract withj = h, k = m to obtain

vi Rjk'k + Vi Rki + vk Ri1'k = 0.

On introducing the mixed Ricci tensor,

Rij = R = Rii = Rr4i

(its first scalar), show that this equation implies that the divergence of thetensor R*j - 1SjR is zero.

41. Check the results of § 164, ex. 1, by differentiating r = pR(,p) + zktwice with respect to the time. [Apply (44.2) and (44.3).]

42. The generalized Kronecker delta 6a'.".' d has k subscripts and k super-scripts (1 5 k 5 n), each ranging from 1 to n. Its value is defined as follows:If both upper and lower indices consist of the same set of distinct numbers,chosen from 1, 2, , n, the delta is 1 or -1 according to the upper indicesfrom an even or an odd permutation of the lower; in all other cases the deltais zero.

Prove that if Som.': d has 2k indices,

(n - k)! Sab...d - E

where the epsilons necessarily have n indices. Hence show that the delta isan absolute tensor.

CHAPTER X

QUATERNIONS

181. Quaternion Algebra. The problem of extending 3-dimen-sional vector algebra to include multiplication and division wasfirst solved by Sir William Rolvan Hamilton in 1843. He foundthat it was necessary to invent an algebra for quadruples of num-bers, or quaternions, before a serviceable algebra for number triples,or vectors, was possible. Without attempting to motivate Ham-ilton's invention, we proceed to a brief account of quaternionalgebra.

A real quaternion is a quadruple of real numbers written in a def-inite order. We shall designate quaternions by single letters,p, q, r; thus q = (d, a, b, c), q' = d', a', b', c'). The fundamentaldefinitions are the following.

Equality: q = q' when and only when d = d', a = a', b = b',c=c'.

Addition:

(1) q+q' = (d+d',a+a',b+b') c+c').Multiplication by a Scalar X:

(2) Xq = (ad, Xa, Xb, Xc).

Negative: -q = (-1)q.Subtraction: q - q' = q + (- q'). Hence

q - q' = (d-d',a-a',b-b',c-c').The zero quaternion (0, 0, 0, 0) is denoted simply by 0.From these definitions it is obvious that, as far as addition, sub-

traction, and multiplication by scalars are concerned, quaternionsobey the rules of ordinary algebra:

(3) p + q = q+p, (p+q) +r = p+ (q+r);(4) Xq = qX, (Xu)q = X(µq);

(5) (X + µ)q = Xq +µq, X(p + q) = Xp + Xq403

404 QUATERNIONS § isi

In order to define the product qq' of two quaternions in a con-venient manner, we shall denote the four quaternion units as fol-lows:

1 = (1, 0, 0, 0), i = (0, 1, 0, 0), j = (0, 0, 1, 0), k = (0, 0, 0, 1).

Then, in view of the preceding definitions, we can write any qua-ternion in the form:

(6) q = (d, a, b, c) = dl + ai + bj + ck.

Definition of Multiplication: The quaternion product,

qq' = (dl + ai + bj + ck) (d'l +a'i+b'j+c'k),

is obtained by distributing the terms on the right as in ordinaryalgebra, except that the order of the units must be preserved, andthen replacing each product of units by the quantity given in thefollowing multiplication table:

(7) Firstfactor

1 i j k

1 i j k

i -1 k -jj -k -1 i

k j -i -1Note that i2 = j2 = k2 = -1, and the cyclic symmetry of theequations :

ij = k, jk = i, ki = j; ji = - k, kj = - i, ik = -j.With this definition we find that

(8) qq'=dd'-aa'-bb'-cc'+d(a'i+b'j+c'k)+d'(ai+bj+ck)

Ii j k

+ a b c

a' b' c'

If we form q'q by interchanging primed and unprimed letters, thefirst two lines above remain unchanged, but the interchange ofrows in the determinant is equivalent to changing its sign; hence

§ 181 QUATERNION ALGEBRA 4U3

q'q = qq' only when the determinant is zero. That q'q and qq' differin general was to be expected; for, from the table, ij = k, ji = - k.

The table shows that multiplying a unit by 1 leaves it unchanged ;hence (dl)q = q(dl) = dq, (dl)(d'l) = dd'l; and, from (1), dl +d'1 = (d + d')1. Quaternions of the form dl therefore behaveexactly like real scalars and may be identified with them. Hence-forth we shall write dl or (d, 0, 0, 0) simply as d; in particular,1 or (1, 0, 0, 0) is regarded as the real unit 1.

We also may identify i, j, k with a dextral set of orthogonal unitvectors. For, if we make the orthogonal transformation,

i = c11i + e12j + c13k,

c21i + c22j + c23k,

I6 = c3ii + c32i + c33k,

we have the relations cirri, = S,j (149.3) and

i2 = - Clr Clr = - 1,

li j k

3k = C2. C3r + C21 C22

C31 C32 C33

C23 = c11i + c12j + C13k = i,

since the elements of det cii (= 1) are their own cofactors (§ 149).Thus every quaternion q = d + ai + bj + ck is the sum of a

scalar d and vector v = ai + bj + ck. With Hamilton we use thesymbols Sq and Vq to denote the scalar and vector parts of q; thus

(9) Sq = d, Vq = ai + bj + ck, q = Sq + V q.

The operations S and V are evidently distributive with respect toaddition.

We are now in position to prove the fundamental

THEOREM. Quaternion multiplication is associative and distribu-tive with respect to addition; but the commutative law pq = qp holdsonly when one factor is a scalar, or the vector parts of both factors areproportional. In symbols:

(pq)r = p(qr);

p(q + r) = pq + pr, (p + q)r = pr + qr;(12) pq = qp only when Vp = 0, or Vq = 0, or Vp = XVq.

406 QUATERNIONS § 182

Proof of (10). It will suffice to verify (10) for all possible com-binations of the units i, j, k. Since the multiplication table is un-changed under a cyclical permution of i, j, k, we need examine (10)only for those products whose left factor is i; thus we find

Proof of (11). If we form p(q + r) and pq + pr by formal ex-pansion and without use of the multiplication table, the two ex-pressions will agree term for term; hence they will agree also afterthe table is used.

Proof of (12). We have seen that q'q = qq' only when the de.terminant in (8) vanishes. This occurs only in the three cases:

a b ca=b=c=0; a'=b'=c'=0; -_-_-;a' b' c'

that is, either q or q' must be scalar, or Vq and Vq' must be pro-portional.

From (8), we have

(13) S(qq') = S(q'q),

for both scalars equal dd' - aa' - bb' - cc'. From (13), we mayprove that, the value of the scalar part of a quaternion product is notchanged by a cyclical permutation of its factors. We have, for ex-ample,

S(p . qr) = S(qr . p) = S(q . rp) = S(rp - q),and hence

(14) S(pqr) = S(qrp) = S(ppq).

182. Conjugate and Norm. The conjugate of a quaternion q,written Kq, is defined as

(1) Kq = Sq - Vq.

§ 182 CONJUGATE AND NORM 407

The conjugate of a sum of quaternions is evidently the sum oftheir conjugates:

(2) K(q + q') = Kq + Kq'.Since the vector parts of q and Kq differ only in sign, q(Kq) _

(Kq) q. This product is known as the norm of q and is written Nq.If

q=d+ai+bj+ck, Kq=d-ai-bj-ck,we have, from (181.8),

(3) Nq = q(Kq) = (Kq)q = d2 + a2 + b2 + c2.Therefore Nq is a scalar; and Nq = 0 implies that a = b = c =d = 0, that is, q = 0. If Nq = 1, q is called a unit quaternion.

If v = ai + bj + ck, we have, from (181.8),

i j k

(4) vv' (aa' + bb' + cc') + a b c

a' b' c'

Since changing the sign of the determinant is equivalent to inter-changing the second and third rows,

(5) K(vv') = v'v.

On taking the conjugate of every term in

qq' _ (d+v)(d'+v') = dd' + dv' + d'v + vv',we see that

(6) K(qq') = dd' - dv' - d'v + vv = (d' - v') (d - v) = (Kq') (Kq)

Therefore the conjugate of the product of two quaternions is equal tothe product of their conjugates taken in reverse order. Since Kv =- v, (5) is a special case of (6).

We now use this property to compute the norm of a product.From (3), we have

N(pq) = pq - K(pq) = pq . Kq . Kp = p . Nq . Kp = pKp - Nq,

since Nq is a scalar and therefore commutes with Kp; hence

(7) N(pq) = Np Nq.

The norm of the product of two quaternions is equal to the product oftheir norms.


By mathematical induction we immediately may extend (6) and(7) to products of n quaternion factors:

(8) K(gig2 ... TO = Kgn . Kg.-1 ... Kq1,

(9) N(g1g2 ... qn) = Nq1 . Nq2 ... Nqn.'

From (6), we conclude that the product of two quaternions is zeroonly when one factor is zero. Thus, if pq = 0, Np - Nq = 0; and,since the norms are scalars, Np = 0 or Nq = 0, whence p = 0 orq = 0.

We now can appreciate Hamilton's exceptionally happy choiceof a multiplication table for the quaternion units. t For quater-nion algebra has a unique place among the algebras of hyper-numbers,

x = x1e1 + x2e2 + ... + xnen,

linear in the units ei, with coefficients xi in the field of real num-bers, and for which the associative law of multiplication holdsgood. For it can be shown $ that the most general linear associa-tive algebra over the field of reals, in which a product is zero only whenone factor is zero, is the algebra of real quaternions.

Quaternions include the real numbers (x, 0, 0, 0) with a singleunit 1, and the complex numbers (x, y, 0, 0) with two units 1, i.Both real and complex numbers form a field, that is, a set of num-bers in which the sum, difference, product, and quotient (thedivisor not being zero) of two numbers of the set must be definitenumbers belonging to the set. Moreover, quaternions includevectors (0, x, y, z) in space of three dimensions. But (4) showsthat the product of two vectors is not in general a vector, but a quater-

* This theorem applied to qq' gives Euler's famous identity:

(d2 + a2 + b2 + c2)(d,2 + a,2 + b12 + c,2)

(dd'-as'-bb'-cc')2+(ad'+da'+bc'-cb')2+(bd' + db' + ca' - ac')2 + (cd' + dc' + ab' - ba')2.

t As to the genesis of quaternions, Hamilton himself has written:"They started into life, or light, full grown, on the 16th of October, 1843,

as I was walking with Lady Hamilton to Dublin, and came up to BroughamBridge, which my boys have since called the Quaternion Bridge. That is tosay, I then and there felt the galvanic circuit of thought close; and the sparkswhich fell from it were the fundamental equations in i, j, k; exactly such as Ihave used them ever since."

t Dickson, Linear Algebras, London, 1914, p. 10.

§ 183 DIVISION OF QUATERNIONS 409

nion. Unlike addition-the sum of two vectors is always a vector-the multiplication of vectors leads outside of their domain.Vector multiplication is not closed, and, consequently, a pure vectoralgebra having all the desirable properties of quaternion algebrais not possible.

183. Division of Quaternions. If q is not zero, Nq is a non-zeroscalar; and we may write the defining equation for the norm (182.3)as

qKq

= 1, or qq = 1.

We therefore define Kq/Nq as the reciprocal of q and write

Kq(1)

q-1= - ; then qq-1 = q-'q = 1.

Nq

The last equations show that Nq - Nq-1 = 1, or

(2) Nq-1 = -Nq

In order to divide p by q ( 0), we must solve the equation

(3) (4) rq = p or qr = p

for r. This is easily done by multiplying by q-1 on the right in(3), on the left in (4). We thus obtain the two solutions,

(5)(6) r1 = pq-1, r2 = q-1p,

which are in general different. For this reason the symmetricalnotation p/q will not be used. The notations (5), (6) are unam-biguous: r1 satisfies (3) and may be called the left-hand quotientof p by q; and r2, the right-hand quotient, satisfies (4). Thesesolutions are unique; if, for example,

rq = r1q, (r - r1)q = 0,and, since q 0 0, r - r1 = 0, or r = r1.

On taking norms in (5) and (6), we have

(7) Nrl = Nr2 =Np

N q

The norm of either quotient of two quaternions is equal to the quotientof their norms.


From (182.8) and (182.9),

(8) (gig2 ... qn.)-i = N(gig2 ... q) = qn i. qn 11 ... qi 1.

nn

The reciprocal of the product of n quaternions is equal to the productof their reciprocals taken in reverse order.

The definition (1) shows that the reciprocal of a unit quaternionis its conjugate; the reciprocal of a unit vector is its negative.

Example. Solve the equations (3) and (4) when

p=1+3i-j+k, q=2-i-2k.From (1), we have q-' = (2 + i + 2k)/9; hence

ri=pq 1=1(1+3i-j+k)(2+i+2k) = 1(-3 + 5i - 7j + 5k);r2 =q1p = 1(2+i+2k)(1 +3i - j+k) = y(-3+9i+3j+3k).Note that Sri = Sr2, Nri = Nr2 in conformity with (181.13) and (182.7).

, .84. Product of Vectors. The product of two vectors vv' is thequaternion (182.4) whose scalar and vector parts are

(1) S(vv') = - (aa' + bb' + cc'),i j k

(2) V (vv') = a b c

a' b' c'

In order to find their geometric meaning we adopt a special basisi, j, k. Choose i as the unit vector along

v' v, j as the unit vector perpendicular to iv in the plane of v, v' and so directed that

° the angle (j, v') is not greater than 900k k i thFi 184 Th it t). e un vec or( g. en s

Fia. 184 which completes the right-handed orthogo-nal basis i, j, k. If the angle (v, v') = 0, we have

v vli, v' = Iv'I(icos0+jsin0),vv' = vIv'I(-cos0+ksin0);

(3) S(vv') = - I v I I v' I cos 0,

(4) V(vv') = IvI v' sin 8 k.

Noting that k is the unit vector perpendicular to the plane of vand v' and directed so that v, v', k form a right-handed set, we see

§ 184 PRODUCT OF VECTORS 411

that (3) and (4) are geometric expressions for the scalar and vec-tor parts of vv', entirely independent of the basis i, j, k. Theyform the cornerstones of the vector algebra of J. Willard Gibbs,who defined the scalar and vector product of two vectors u, v as

(5)(6) u- v=-S(uv), u X v = V(uv),

using the dot and cross to distinguish between these two types of"multiplication." Henceforth we shall use Suv, Vuv, Kuv to de-note S(uv), V(uv), K(uv); similarly Suvw = S(uvw), etc.

A change in the order of the vectors in (3) has no effect, but in(4) reverses the direction of k; hence

(7) (8) Svu = Suv, Vvu = - Vuv.

From u(v + w) = uv + uw and the distributive character of Sand V,

(9) (10) Su(v + w) = Suv + Suw, Vu(v + w) = Vuv + Vuw.

Since Kuv = (- v) (- u) = vu,

uv = Suv + Vuv, vu = Suv - V UV;

(11)(12) Suv = 2(uv + vu), Vuv = 1(uv - vu).

Turning now to products of three vectors, we have

(13) Suvw = Svwu = Swuv (181.14).

Since Kuvw = (-w)(-v)(--u) = -wvu (§ 182),

(14) Suvw = SKuvw = -Swvu

(15) Vuvw = - VKuvw = Vwvu.

Also from uvw = u(Svw + Vvw),

(16) Suvw = SuVvw,

(17) Vuvw = uSvw + VKVvw.

We now compute Vuvw in another way:

2Vuvw = uvw - Kuvw

= uvw + wvu

= uvw + vuw - vuw - vwu + vwu + wvu

= (uv + vu)w - v(uw + wu) + (vw + wv)u;

412 QUATERNIONS

hence, from (11),

(18) V uvw = uSvw + wSuv - vSuw.

Comparison with (17) now gives the important formula:

(19) VuVvw = wSuv - vSuw.

§ 185

Finally, we express these results in the dot and cross notationof Gibbs:

(5)(6) Jul Ivl cos(u,v), uXv = Jul lvlsin(u,v)k;(7)(8) v, vxu= -u-V;(9)

(10) u.(v+w) =uXv+uxw;(13)

(14)

(19) uX(vXw) =

From (5) and (6) we have also

(20) uv =

as the connecting link between quaternion and vector algebra.185. Roots of a Quaternion.t Every quaternion

q = d + ai + bj + ck

with real coefficients may be written as a real multiple of a unitquaternion :

(1)

Here

q = h(cos 0 + e sin B), 0 < 0 < 27r.

h = d2+a2+b2+c2,

cos B = d/h, sin 9 = f \/a2 + b2 + c2/h

and, when a2 + b2 + c2 0, e is the unit vector:

(2) eai+bj±ck= f

1/a2+b2+c2t In this and following articles we again denote vectors with bold-face

letters; but ab denotes the quaternion (not dyadic) product.

§ I35 ROOTS OF A 413

When q is a real number, sin 0 = 0, and e may be chosen at pleas-ure. Since e2 = - 1, we have, by De'Xloivre's Theorem,

(3) q' = h"(cos no + e sin no).

We now may find the nth roots of a real quaternion

(4) Q = H p -I- a sin gyp), 0 < (P < ,r;

the angle always may be taken in the interval from 0 to it bychoosing the appropriate e in (2). In solving q' = Q, we considertwo cases:

1. sin 0; we choose the e in q the same as in Q. Then

hn = H, cos no = cos cp, sin no = sin gyp,

and n nth roots of Q are given by (1), provided

(5) h = H' I', the positive root,

(6) 0 = (gyp + 2irm)/n (m = 0, 1, , n - 1).

These n values of 0 comprise all values in the interval 0 < 0 < 2irwhich satisfy the preceding equations.

2. sin p = 0: the e in q is then an arbitrary unit vector.

IfQ>0: o =0, 0 =2mir/n (m=0,1, ,n-1).When n = 2, the values 0 = 0, Tr give just two roots =LA/Q-, bothreal. When n > 2, some values of 0 (0 0 or ir) give non-real rootsq with which any e may be associated.

IfQ<0: p =ir, 0=(2m+1)ir/n (n=0,1, ,n-1).In every case some values of 0 (0 ir) give non-real roots q withwhich any e may be associated.

We summarize these results in the

THEOREM. A quaternion with real coefficients, but not a real num-ber, has exactly n nth roots. If Q is a positive real number, it hasjust two square roots ± /; in all other cases a real number hasinfinitely many quaternion roots with real coefficients.

In all cases the roots may be computed from (5) and (6). Forexample, if Q = 1 + i + j + k, we write

Q = 2(cos 60° + e sin 60°), e = (i + j + k)/1'3.

414 QUATERNIONS

The cube roots of Q are then

§186

q = (cos 0 + e sin 0), 0 = 20',140',260'.

186. Great Circle Arcs. Every unit quaternion

q=d+ai+bj+ck (Nq=1)

can be expressed in the form

(1) q = cos 0 + e sin 0.

Here e is given by (185.2); and 0 satisfies

(2) cos 0 = d, sine = a2 + b2 + c2.

If we choose the plus sign in these formulas, 0 < 0 < 7r. In par-ticular, if q = 1, -1, e, the angle 0 = 0, zr, 2 7r, respectively.

THEOREM. The unit quaternion cos 0 + e sin 0 may be expressedas the quotient ba-1 of any two vectors which satisfy the conditions:

(i)lal=lbl,(ii) Angle (a, b) = 0,(iii) Plane a, b is perpendicular to e,(iv) a, b, e form a dextral set.

Proof. In view of (i), we may write

Ial bIcos0+Ial bIsin0ecos9+esin9=

IaI2hence, if we choose the vectors a and b so that conditions (ii),(iii) and (iv) are fulfilled,

-Sab + Vab -Sba - Vba ba Kacosh+esin0= _ = --=b-

Na Na Na Na

or, in view of (183.1),

(3) cos 0 + e sin 0 = ba-1.

From Fig. 186 we see that, when a, b, e form a right-handedset, the angle (a, b), when less than 7r, is counterclockwise, viewedfrom the tip of e. We shall describe the sense of (a, b) as positiverelative to e. When the angle (a, b) = 0 or 7r, q is 1 or -1, re-spectively, and e is entirely arbitrary.

§.,%G'ItEAT CIRCLE ARCS 415

To every unit quaternion, q = cos 0 + e sin 0 corresponds to a

great circle are AB of a sphere centered at 0, provided OA = aand OB = b satisfy the preceding conditions (i) through (iv).Thus q corresponds to an are of a great circle whose plane is nor-mal to e and whose central angle 0has the positive sense relative to e.All such arcs of this great circle areequally valid representations. If theare AB represents q, AB is free tomove about in its great circle, providedits length and sense remain unaltered.

The cases q = 1 (B = 0) andq = -1 (0 = -7r), in which e is

q

FIG. 186

arbitrary, are exceptional. Any point of the sphere representsq = 1; and any great semicircle represents q = - 1. A unit vec-tor q = e (0 = 2 r) corresponds to a quadrantal are in the planethrough 0 normal to e.

If q = cos 0 + e sin 0 corresponds to the are AB,

(4) qI = Kq = cos 6 - e sin B

corresponds to an are having the same plane and angle, but whosesense is positive relative to -e. Hence q-' corresponds to theare BA. Moreover,

(5) - q = - cos 0 - e sin 0 = cos (ir - 0) - e sin (7r - 0) ;

hence, if q corresponds to the are AB, and AOA' is a diameter,-q corresponds to are A'B, the supplementary are reversed inSense.

Using the sign '' to denote correspondence, we sum up our find-ings as follows:

1 ti point, -1 - semicircle, e quart.ercircle;

q - are AB, q-I - are BA, -q '' are A'B.

The utility of this representation is due to a simple analyticalmethod of adding great circle arcs "vectorially." To add two arcs,shift them along their great circles until the terminal point of the first(AB) coincides with the initial point of the second (BC); then the

great circle arc AC is defined as their vector sum. If a = OA,

416 QUATERNIONS §186

b = OB, c = OC, the preceding theorem shows that the arcs AB,BC, AC are represented by the quaternions ba-1, cb-1, ca-1, re-spectively. Write

p = ba-1, q = cb-1; then ca-1 = cb-lba-1 = qp;

and the equation,

are AB + are BC = arc AC,

may be written

(6) are p + arc q = are qp.

For three arcs,

(7) arc p + are q + are r = arc qp + are r = arc rqp;

and, in general, the vector sum of any number of great circle arcs isgiven by the arc corresponding to the product of their representativequaternions taken in reverse order.

In interpreting such arc-quaternion equations, remember thatare q = 0 means that q = 1; and, if are q is any great semicircle,q = -1. For example, if arc p, are q, are r form the sides of aspherical triangle taken in circuital order,

arc p + arc q + arc r = 0, arc rpq = 0, rpq = 1.

In general, the arcs representing the quaternions q1, q2, , qn,taken in this circuital order, will form a closed spherical polygonwhen and only when

(8) qnqn-1 ' g2g1 = 1.

Example. Spherical Trigonometry. Consider again the spherical triangleABC of § 22. With the notation of this article

are BC ' cb-1 = cos a + a' sin a,

are CA ' ac-1 = cos $ + b' sin Q,

are AB - ba 1 = cos y + c' sin y.

The "vector" equation

are CA + arc A B = arc CB

corresponds to the quaternion equation (bat) (ac 1) = be 1, or

(i) (cosy + c' sin y) (cos i3 + b' sin 3) = cos a - a' sin a.

§ 187 ROTATIONS

If we expand the left member, put

c'b' = - b' c' - b' x c' cos a' - a sin a';

and equate scalar parts in both members, we have

(ii) cos 0 Cos y - sin 3 sin y cos a' = cos a.

417

This is the cosine law (22.6) of spherical trigonometry. On equating the vectorparts in both members of (i) we have,

(iii) a' sin a + b' sing cos y + c' cos l; sin y = a sin a' sin 0 sin y

and hence, on multiplication by a ,

sin a' sin /i sin y = a a' sin a = a bxc,

or

sin a' a- bxcsina sin a sin#sin y

Since the right member is unchanged by a cyclical permutation, we have

(iv)sin a' sin /3' sin y'sina sing sin y

the sine law (22.5) of spherical trigonometry.

187. Rotations. With the aid of quaternion algebra, finite rota-tions in space may be dealt with in a simple and elegant manner.This application depends upon the fundamental

THEOREM. If q and r are any non-scalar quaternions, then

(1) r' = qrq-1

is a quaternion whose norm and scalar are the same as for r. Thevector Vr' is obtained by revolving Vr conically about Vq throughtwice the angle of q. Thus if

q = Nq(cos 0 + i sin 6),

Vr' is obtained by revolving Vr conically about i through an angle 26.

Proof. The norm and scalar of r' are

(2) N(qrq-1) = Nq Nr Nq-1 = Nr (182.9),

(3) S(qrq-1) = S(q 'qr) = Sr. (181.14).

Moreover on writing r = Sr + Vr,qrq-1 = Sr + q(Vr)q-1;


Now, from (3), q(Vr)q-1 has the same scalar as Vr and is there-fore a vector; hence

(4)

Let us now write

Flo. 187a

then from (4),

Vr' = Nr sin e' where e' = qeq-1.

If we choose j in the plane of e and i(Fig. 187a) and k to complete the dextralset i, j, k, then

e = icosA+ Jsin A

e' = (qiq-1) cos A + (qjq 1) sin A.

Since Vq is parallel to i, qi = iq andqiq--1 = iqq-1 = i.

Moreover,qjq-1 = (cos 0 + i sin 0)j(cos 0 - i sin 0)

= (j cos 0 + k sin 0) (cos 0 - i sin 0)

= j (cost 0 - sine 0) + k(2 sin 0 cos 0) ;

hence j goes into

j' = j cos 20 + k sin 20,

a vector obtained by revolving j about i through an angle 20 inthe positive sense. Consequently

e = icosA+ jsinX -*e' = icosA + j'sinA

and also Vr -. Vr' by a conical revolution about i of the sameamount. This completes the proof.

We note that vectors transform into vectors. In particular, ifq = a, a unit vector,. 0 = 90°, and

e' = aea-1 = -aea

V (qrq-1) = q(Vr)q-1

r = Nr(cos c + e sin (p);

is obtained by revolving e conically through 180° about a. Thetransformation -a( )a thus gives vectors a half-turn about a.

§ 187 ROTATIONS 419

The transformation a( )a may be regarded as a half-turn fol-lowed by a reversal; a vector thus transformed is simply reflectedin a plane normal to a (Fig. 187b). Thus aea is the reflection of ein the plane normal to a.

A rotation through an angle a about aimplies that the angle turned has the positivesense relative to a. If

p = Cos. a+asin1a,q= cos

2S -{- b sin

z3

are unit quaternions, the operators p( )p-1and q( )q-1 effect rotations of a about a and

a

Fia. 187b

,3 about b; for brevity we call these the rotations p and q. Thesuccession of rotations p, q corresponds to the operator,

qp()p-'q-1 = qp( ) (qp)-1;

since qp is also a unit quaternion, say

qp = cos zy + c sin 21

'Y,

the resultant is equivalent to the single rotation qp, that is, a ro-tation through the angle y about c. Similarly, the resultant ofthe rotations q and p corresponds to the operator,

pq()q-'p-1 = pq( )(pq)-1

Since pq = qp only when Vp and Vq are parallel (the values p, q= -4-1 are excluded), the composition of rotations is non-commu-tative except when they have the same axis.

The rotation p followed by the rotation q is equivalent to the singlerotation qp. More generally, the succession of rotations q1, q2,qn is equivalent to the single rotation gnq,,-1 . g2g1.

Since (-q)-1 = -q-1, the rotations q( )q-1 and (-q)( )(-q)are the same. If q = cos B + a sin B,

-q = cos(Tr-9)+(-e)sin(7r-8);thus the rotation -q is a rotation through 27r - 20 about -e;this produces the same result as the rotation q, namely 20about e.

Since q-lq( )q-lq = 1( )1, the rotation q-1( )q is the reverse ofq( )q-1; this is also evident from q-1 = cos 0 - e sin 0.


Example 1. The rotation of 90° about j followed by a rotation of 90° abouti is represented by the quaternion product

(cos 45° + i sin 45°) (cos 45° + j sin 45°) = !(I + i + j + k);

that is, by1 i+j+kV' i + j + k+ - = cos 60° + - sin 60°.

2 V3 2 -\/3

The resultant rotation is therefore a rotation of 120° about an axis equallyinclined to the (positive) axes of x, y and z.

Example 2. The resultant of two reflections in planes normal to the unitvectors a, b corresponds to the operator

ba( )ab = ba( ) (ba)-1.

But if a b = cos 0, a x b = e sin 0, we have

ba = -a b - axb = -(cosB+esin0) (184.20).

Now the rotation q( )q-' = (-q)( )(-q)-1; hence successive reflections intwo plane mirrors is equivalent to a rotation about their line of intersectionof double the angle between them.

Example 3. From (181.13) we know that S(qp) = S(pq); moreover from (4)

V (qp) = V (gpgq-1) = qV (pq)q-1

Hence V(qp) is obtained by revolving V(pq) about Vq through double theangle of q. Thus, if u is a vector, V(up) is obtained by revolving V(pu) 180°about u; that is, the vector u bisects the angle between V(pu) and V(up).

Now let a, b, c be three radial vectors from the center of a sphere to itssurface. Then a, b, c bisect the angles between Vabc, Vbca; Vbca, Vcab;Vcab, Vabc respectively. In other words, if we form a spherical trianglewhose vertices are Vabc, Vbca, Vcab, the middle points of the sides oppositelie on the vectors b, c, a respectively.

Example 4. If the sides of a spherical polygon are represented by thequaternions qi, q2, , qn taken in this circuital order, gnqn-1 g2gt = 1(186.8). Hence the succession of rotations,

gnqn-1 ... q2ql( )gig2 ... q.-1qn = 1( )1,

about axes through a point 0 will restore a body to its original position. Westate this result for the case of a triangle as follows:

THEOREM (Hamilton and 1)onkin). If ABC is any spherical triangle, threesuccessive rotations represented by the directed arcs 2 BC, 2 CA, 2 AB (about theirpolar axes) will restore a body to its original position.

This same theorem applies to the polar triangle A'B'C'. Since the sideB'C' = a' = 7r - A (§ 22) and has OA as polar axis, successive rotations of2ir - 2A, 27r - 2B, 2ir - 2C about OA, OB, OC will restore a body to itsoriginal position. Since the rotations 2ir - 2A and -2A about OA give thesame displacement, we may state the

§ 188 PLANE VECTOR ANALYSIS 421

THEOREM (Hamilton). If ABC is any spherical triangle on a sphere centeredat 0, three successive rotations about OA, OB, OC through the angles 2A, 2B, 2Cin the sense of CBA will restore a body to its original position.

188. Plane Vector Analysis. The three-term quaternion c + ai+ bj has given rise to two types of vector analysis in the plane.The one interprets c + ai as a vector w in the complex plane; thenthe product wlw2 is always a complex vector. The other inter-prets ai + bj as a "real" vector w and decomposes the quaternionproduct w1w2 into its scalar and vector parts, which are usedseparately as "products."

To indicate the interpretation used, we denote complex and realvectors by italic and bold-face letters, respectively. Thus theplane vector whose components are u, v may be written as

w = u + iv, or w=ui+vj.In the first case,

(1) w1w2 = (ulna - vlv2) + (ulv2 + u2v0i;

in the second,

(2) w1w2 = - (ulu2 + vlv2) + (u1v2 - u2v1)k.

In Gibbs's notation,

ulu2 + V1V2 = w1 w2i u1v2 - u2v1 = k w1 x W2.

Let w = u - vi denote the conjugate of w; then, from (1) and (2),

w1W2 = W1 W2.+ik.wlxw2,

W1W2 = W1 W2 - i k W1 x w2,and hence

(3) W1 W2 = 2 (wlw2 + W1w2),

(4) 21i(w1w2 - wlw2)

The conditions (128.6) and (128.7) for perpendicular and parallelvectors may be read from these equations.

We next consider corresponding differential invariants. Theoperator,

a a a aV =i-+j--+i-ax ay ax ay

422 QUATERNIONS §188

(read - as corresponds to). If we introduce the conjugate variables,

z=x+iy, 2=x-iy,a a az az a az az a-+i- _ -+i- -+ -+i- -

ax ay ax ay az (ax ay a2

a a(1 + i2) + (1 - i2)

az a2,or

(5)

a a 0 a 8 a2

+ Z

_2 ' - Z

_ax a-y az ax ay az

follows in the same way.Corresponding to the gradient V of a real function p(x, y), we

have 2 a(p/az in the complex plane, in which the variables x, y inare replaced by the values,

x = 2 (2 + z), y = 2i(2 - z).

For example, x2 + y2 = z2, and hence 0(x2 + y2) = 2z in thecomplex plane.

The unit vector,

(6) e = icos0+jsin8NeiO = cos0+isin9.In view of (3), the operator for differentiation in the direction e,namely

(7)

Thereforedw aw aw

(8) - = e Vw -e t8 - + e-i0 -,

ds az az

and hence dz/ds ti eie. If w is a complex function, dw/dz in thedirection 0 corresponds to the ratio of dw/ds to dz/ds; hence

dw aw _2. aw

(9) dz az + eaz

in the direction 0.

When dw/dz is independent of 0, w is said to be an analytic func,tion of z; for this, it is necessary and sufficient that

aw a a=0, or -+i- (u+iv)=0,

az ax ay

§ 188 PLANE VECTOR ANALYSIS 423

in view of (5). The last condition is equivalent to the familiarCauchy-Riemann Equations :

(10)au av au av

=0, -+-=0.ax ay ay ax

We next find the correspondents for div w = V w and k rot w= k V x w by making use of (3), (4) and (5) :

3z' awdivwti- + -,a2 az

(12)/aw awk rot w tii(---\az az

Moreover, for the Laplacian V2 = V V, we have

a2 a2 a2V2=2(13)

azaz+2azaz - 4azazThus a real function cp(z, 2) is a harmonic when a2,p/3z ai = 0; forexample,

log I z I = 2 log zz =2

(log z + log z)

is harmonic.If the plane vector w - w(z, 2), the condition,

Ow aw(14) rotw=ON---=0.

az az

When rot w = 0,

w = VX, where a = w dr;:o

hence, when w(z, 2) is irrotational,

(15) w = z , where p f= (w dz + w dz)

is the real function 2X(z, 2). The field lines cut the curvesconst at right angles.

If the vector w is plane, the condition,

Ow aw(16) div w = 0 - - + - = 0.

az a2


From (85.6), rot (k x w) = k div w; hence, when w is solenoidal,k X w is irrotational. Since k X w is w revolved through +7r/2,k x w - iw. Thus when w(z, 2) is solenoidal, (15) applies when wis replaced by iw (and iv by -iw). From this result, we concludethat

(17) w = i where = ifZ(w dz - w d2)az zo

is a real function. The field lines are the curves = const.From these results we have

THEOREM 1. If <p(z, 2) is a real function with continuous partialderivatives, app/a2 and i app/a2 give, respectively, an irrotational fieldwith lines orthogonal to cp = const, and a solenoidal field with lines

= const.

The preceding results give a simple method for decomposing aplane vector function into an irrotational and a solenoidal part.For, if

w = W1 + w2i rot W1 = 0, div W2 = 0,

there exist real functions gyp,' such that

app a, aw=-+i-=-(o +ii');a2 a2 az

' + i¢ = Jw(z, 2) d2 (z const)

is determined to an arbitrary additive f (z) and w1 = app/az, w2 =i ay/az.

If the vector field w(z, 2) is both irrotational and solenoidal,rot w = 0 implies that

asp az=

pw=-; then w, divw=2 =0.az az dz dz

Hence p is a real harmonic function, and w is an analytic functionof z. Conversely, if w is an analytic function of z, (9w/a2 = 0;hence aw/az = 0, and (11) and (12) show that w is solenoidal andirrotational.

THEOREM 2. In order that the complex vector w(z, 2) be irrotationaland solenoidal, it is necessary and sufficient that its conjugate be ananalytic function of z.

§ 188 PLANE VECTOR. ANALYSIS

Example 1. The vector

w = 22 = (x2 - y2) - 2xy i,

425

is both irrotational and solenoidal; for its conjugate z2 is an analytic functionof Z.

Example 2. To decompose the vector,

w = z2 = (x2 - y2) + 2xy i,

into its irrotational and solenoidal parts, we may take

v + ik = fz2 d2 = z22.

Since

wl

= 12z2(2 + z), ¢ = Zi z2(2 - z);

av_ = z2 + 2z2 =

2(3x2 + y2) + xy it

a2

akw2 = i

a2= -z2 + 2z2 = - i (x2 + 3y2) + xy i.

Stokes' Theorem in the plane,

fk . rot w dA = fw . dr,

corresponds to

aw awi- - - dA = a (w d2 + w dz),

492 az

by virtue of (12) and (3). If we replace w by -iw (and iv by i4 b),we obtain, after canceling i,

f(afvaw

i - + - dA = 2 J(-w dz + w dz).az az

On adding these equations and then replacing w by w, we have

aw(18) fw dz = 2if - dA.

az

When w is analytic in the region within the circuit, aw/a2 = 0,

and (18) reduces to Cauchy's Integral Theorem:Jw dz = 0.

,P - i= 22Z,

426 QUATERNIONS

Example 3. When w = z in (18),

2iA = 12 dz = J(x - iy) (dx + i dy) = if(x dy - y dx).

§ 189

This gives the well-known circuit integral for a plane area.With w = z2, we obtain the static moments of a plane area about the axes,

expressed as circuit integrals.

189. Summary : Quaternion Algebra is a linear, four-unit(1, i, j, k) associative algebra over the field of reals. The unit 1has the properties of the real one; and

i2=j2=k2= -1, ij=k, ji= -k,the last equations admitting cyclical permutations. Quaternionsq = d + ai + bj + ck include real and complex numbers (d,d + ai). Since i, j, k may be interpreted as dextral set of orthog-onal unit vectors, quaternions also include vectors v = ai + bj+ ck. Thus q = d + v, a scalar plus a vector.

Quaternion multiplication is associative and distributive, butnot in general commutative; in fact pq = qp holds only when por q is a scalar or when the vector parts of p and q are propor-tional.

The product vv' of two vectors is the quaternion :

vv' = - (aa' + bb' + cc') +

= -v - v' + v x v' in Gibbs' notation.

The quaternion q = d + v has the conjugate Kq = d - v. Theconjugate of the product qq' is K(qq') = (Kq') (Kq).

The norm of a non-zero quaternion q is the positive real number,

Nq = q(Kq) = d2 + a2 + b2 + c2;

and N(qq') = (Nq) (Nq'). The equation q = 0 implies Nq = 0,and conversely; hence, if qq' = 0, either q = 0 or q' = 0.

A unit quaternion q (Nq = 1) may be put in the form

q = cos0+esin9 (IeI = 1),

and associated with the great circle are of angle 0 and pole e onthe unit sphere. On a fixed great circle, all arcs of the same length

PROBLEMS 427

and sense correspond to the same q and are denoted by arc q. Thescalars 1 and - 1 correspond, respectively, to any point and to anygreat semicircle of the sphere.

Great circle arcs may be added vectorially; and

are p + are q = arc qp.

If three arcs form a spherical triangle, say-

arcp+arcq+arcr = 0, then rpq= 1.

If q = Nq(cos 0 + e sin 0), the operator q( )q-1 effects a conicalrevolution of 20 about e on the vector of the operand; thus ifr' = qrq-1, then

Nr' = Nr, Sr' = Sr, T Y = Vr revolved about e through 20.

When q = e, a unit vector, q-1 = -e; the operator -e( )e givesvectors a half-turn about the axis e.

The operator e( )e reflects vectors in the plane normal to e.

PROBLEMS

1. Solve the quaternion equations, rq = p, qs = p, for r and s when

q=2-i-2k, p= 1+3i-j+k.Verify that Vr = Ns.

2. Every quaternion q satisfies the quadratic,

q2 - 2qSq + Nq = 0,

known as its principal equation. The conjugate Kq satisfies the same equation.3. Show that 2 + 5i and 2 + 3j + 4k have the same principal equation;

hence factor its left member in two different ways.4. Show that if we identify the quaternion units with the 2 X 2 matrices

1 - (0 1 ), ( 00)'

j\-1 0), - \-c0 0/'L

where L (iota) is the complex unit, L2 = -1, these matrices satisfy the multi-plication table (181.7).

5. If u, v, w are vectors prove the following identities:

(a) u22 _ -Nu; (u - v)(u + v) = u'- + 2Vuv - v2;

(b) 2 Suvw = uvw - wvu;

(c) 2 Vuvw = uvw + wvu;

(d) S(u + v) (v + w) (w + u) = 2 Suvw.

428 QUATERNIONS

6. Show that the multiplication table of the units i, j, k is completely givenby

i2=j2=k2=ijk= -1.7. Interpret the equations, ij = k, jk = i, ki = j and kji = 1, in terms of

"arc vectors."

8. Solve the equation aq + qb = c for the quaternion q if a, b, c are knownquaternions and Na 5x, Nb. [q = (ac - cb)/(Nb - Na).)

9. Solve the equation qa + bq = q2. [Reduce to aq-1 + q -1b = 1.]10. If a, b, c are vectors for which Vabc = 0, prove that a, b, c are mutually

orthogonal.

[Vabc = aSbc + VaVbc and VaVbc 1 a; hence Sbc = 0, VaVbc = 0.1

11. If a, b, c, d are vectors for which Vabcd = 0, prove that(a) bed is a vector parallel to a;(b) Vbcda = Vcdab = Vdabc = 0;(c) the vectors a, b, c, d are coplanar;(d) cda is a vector parallel to b; and thus cyclically.12. A body is revolved through 90 degrees about two axes e1, e2 which

intersect at an angle 0 (cos 0 = e1 e2). Show that the equivalent singlerotation is through an angle 2 cos' z (1 - cos 0) and about an axis parallelto e1 + e2 - e1 x e2.

13. If a and b are unit vectors along intersecting axes, show that the half-turns -a( )a, -b( )b in this order are equivalent to a rotation of twice theangle from a to b about their common perpendicular.

14. The quaternion q = a1a2 a,,_1 an is the product of it unit vectors.Show that if Sq = 0, q is the vector

q = ±anan-1 ... a2a1 (+ when it is odd, - when n is even).

Hence show that the successive reflections ai( )ai in the order i = 1, 2, 3,, n reduce to the single reflection q( )q when n is odd, and to a half-turn

-q( )q when n is even.15. If the successive reflections ai( )ai in the order i = 1, 2, 3,. , it re-

duce to a single reflection or half-turn, show that this is true for any cyclicalpermutation of this order.

16. Show that the succession of reflections in three coaxial planes reducesto a single reflection.

17. Prove the famous theorem of Euler (1776): Any displacement of a rigidbody which leaves the point 0 fixed is equivalent to a rotation about an axis through0. [Let the displacement move the trihedral ijk in the body to the new posi-tion ijj1k1. This may be accomplished by two reflections: the first takesi -+ i1, j -s j'; the second takes j' - ji, leaving i1 undisturbed.]

18. If q = cos 0 + e sin 0 (not scalar), show that the rotation q( )q-1 fol-lowed by the reflection a( )a reduces to a single reflection b( )b when andonly when a e = 0; moreover,

b = aq = a cos 0 + a x e sin 0.

19. If we define qn for arbitrary real n by (185.3), show that any quaternionmay be expressed as the power of a vector. When q = cos 0 + e sin 0, prove

PROBLEMS 429

that q = e2e/r; and that the r( ti tion of angle about the axis e is representedby the operator e ' n ( )e -" Oi*.

20. Show that successive h ili-turns about three mutually orthogonal andintersecting axes will restore a body to its original position.

21. ABC is a spherical triangle and P, Q are the middle points of the sidesAB, BC, respectively. Prove .hat two successive rotations represented by thearcs AB, BC are equivalent tc the rotation represented by twice the great cir-cle are PQ.

22. Prove the theorem: Th,ee successive rotations represented by the arcsAB, BC, CA of a spherical triangle ABC are equivalent to a rotation about OAthrough an angle equal to the spherical excess (A + B + C - 7r) of ABC.

23. Prove Rodriques' Cons'ruction for the composition of two rotationsthrough the angles a, f3 about he axes OA, OB:

Draw great circle arcs AC m.nd BC such that

angle (AB, AC) = -!a, angle (BA, BC) = 2

determining the spherical triangle ABC; then rotation a about OA followed by0 about OB is equivalent to the rotation

-y = 2(CA, CB) about OC.

117se Hamilton's Theorem given in §187, Ex. 4.]Give the construction when rotation 0 about OB is followed by rotation a

about OA. Draw a figure to illustrate both constructions.24. Prove that successive rotations through angles of gyp, ,r/2, o about the

axes of x, y and z respectively are equivalent to a rotation of 7r/2 about they-axis.

25. If r = xi + yj + zk and r' = x'i + y'j + z'k, are position vectors andq is an arbitrary quaternion, show that the transformation r' = qr(Kq) repre-sents the most general rotation and expansion of 3-dimensional space. Whatis the ratio of expansion?

If q = d + ai + bj + ck, express x', y', z' in terms of x, y, Z.

INDEX

The numbers refer to pages. A starred number locates the definition of theterm in question. The letter f after a number means "and following pages."

Terms under a noun (key word) are to be read in before this word, unlesspreceded by a preposition such as "of" or "for." Terms under an adjective(key word) are to be read in after this word; the repeated adjective is indicatedby dash.

Absolute tensors, 343fAcceleration, 109*

angular, 122*four-vector, 401general components of, 359normal component of, 109of Coriolis, 124*of gravity, 124rectangular components of, 110relative, 123tangential component of, 109transfer, 123*uniform, 110

Addition, of dyadics, 137*of forces, 75of great circle arcs, 415fof motors, 67*of quaternions, 403*of tensors, 350*of vectors, 3*statical, 4*

Addition theorems (sine, cosine), 28Adjoint dyadic, 152*Af ine connection, 356*f, 361, 368-group, 340f, 386-transformation, 138Alternative airport, 113Analytic function, of (real) coordi-

nates, 347of complex variable, 272, 422*

Angular acceleration, 122*-excess, 310-speed, 114*-velocity, 115*

Antecedents of a dyadic, 136*Antisymmetric dyadic, 141 *-tensor, 352*Archimedes, Principle of, 258Arc vector, 415fArea, of a polygon, 39

of a spherical triangle, 310of a triangle, 44vector, 37*f

Areas, Law of, 133Astatic center, 62Asymptotic lines, 285*, 288, 303Axes, rectangular, 26

principal, of a dyadic, 162of inertia, 160of strain, 164of stress, 256

Axis, 25*instantaneous, of velocity, 117*

of acceleration, 127*of a motor, 66*radical, 62

Barbier's Theorem, 134Base vectors, 24, 147, 349, 367, 368

dextral and sinistral, 24, 25, 41Basis, 25Bernoulli's Theorem, 279Bertrand, J., 132, 314Bianchi identity, 385Bilateral surface, 218Binormal, 92*Bivector, 353*B6cher, M., 339

431

432 INDEX

Bonnet, 0., 286, 288, 296, 309Bonnet's Integral Formula, 309Box product, 42*Bracket notation for vectors, 25

Carnot's Theorem, 60Cartesian coordinate system, 379 *f,

385fCatenary, 103, 317Catenoid, 317Cauchy-Riemann Equations, 272,

423Cauchy's integral, 265-Integral Theorem, 276, 425Caustic, 104Center, astatic, 62

instantaneous, 118*mean, 20*of curvature, 99*of mass, 258*, 236of normal curvature, 285*

Central forces, 133, 134Centrifugal force, 124Centroid, 9*Ceva, Theorem of, 14Characteristic equation, 154*, 168f-numbers, 154*, 156Characteristics, 98*, 99, 133Christoffel symbols, 326, 362*, 389Circle, nine-point, 33

of curvature, 99rolling, 120

Circular motion, 110Circulation, 266*fClifford, W. K., 64Codazzi, Equations of, 300f, 391Cofactor, 168*, 330*f

reduced, 331 *Cogredient transformations, 335Collinear points, 8f-vectors, 2*Complete quadrangle, 16-quadrilateral, 18Components, contravariant, 335, 342,

344covariant, 335, 342, 344mixed, 342, 344of acceleration, 164, 359of affine connection, 356

Components of a dyadic, 166, 342of stress, 256of a tensor, 344of a vector, 26, 48of velocity, 164, 359

Composition of velocities, 120fConditions, necessary and sufficient,

viii*Cone, 304, 315Conjugate, of a complex number 274,

422of a dyadic, 141of a quaternion, 406*

Cc njugate lines, 79, 80Conoid, 321 *, 323Consequents of a dyadic, 136Continuity, Equation of, 258f, 264Contraction, 351 *fCovariant derivative, 362*f, 365f, 401

of dyadics, 364of epsilons, 366of Kronecker deltas, 366of metric tensor, 366of scalars, 364, 366of tensor product, 365of tensor sum, 365of vectors, 364

Covariant vector, 335*Cross product, 34Curl, of a tensor, 374*, 376

of a vector, 183*Curvature (K), 92*, 96, 97

center of, 99*, 285circle of, 99geodesic (y), 284*integral, 309*flines of, 285*, 302fmean (J), 222, 286*, 287, 292, 392normal (k), 284*, 287of plane curves, 101of surface curves, 283fprincipal, 287*radius of (p), 93*total (K), 286*, 287, 292, 306f, 392

Curvature tensor, 379, 380 *fContragredient transformations, 334*Contravariant components, of a

dyadic, 342of a tensor, 344

I \ DEX

Contravariant component;;, c a ve:--tor, 48, 335

Contravariant vector, 334iCoordinates, barycentric-, '23*

Cartesian, 25*, 340*, 335curvilinear, 191cylindrical, 195geodesic, 358*homogeneous, 51fline, 52orthogonal, 89fplane, 51Pliicker, 52, 55, 63point, 51spherical, 196

Coplanar line vectors,-points, 12f-vectors, 2Coulomb's Law, 241Coriolis, acceleration of, 124"

Theorem of, 124Cosine law, for plane triangles, 44

for spherical triangles, 46 417Covariant components, of a ,ayadie,

342of a tensor, 344of a vector, 48, 335

Curves, Bertrand, 132*congruent, 97field of, 231*, 293*of constant curvature, '100parallel, 94*, 131parametric, 205*plane, 100freducible, 223*space, 88f

Cycloid, 131Cylinder, 304, 315

D'Alembert's Principle, 260Darboux, G., 289Darboux vector, 93*, 105, 1:!2, 232Del (V), 184*Deltas, Kronecker, 46*, 253:',366,

402Derivative, covariant, 36:2*f 365f

directional, 178, 182, :187normal, 180of a determinant, 331

433

Derivative, of a dyadic, 171of a motor, 126of a vector, 84f

Derivative formulas, of Gauss, 327,390

of Weingarten, 323, 327, 391Desargues' Theorem, 15Determinants, 329*f

bordered, 397, 400Developable surface, 303, 304, 314*fDickson, L. E., 408Differential, total, 90*Direction cosines, 27Discriminant of a quadratic form, 337Displacement group, 388*Distributive Laws, 30, 35, 137Divergence, of a vector (div f, V f),

183*f, 260*, 371of a tensor, 371 *f, 376surface (Div f, V f), 207*

Divergence Theorem, 235*Division of quaternions, 409*Dot product, 29Double-dot product, 175Double layer, 244Doublet, 244Dual angle, 69*-number, 64 *-vector, 63*Dual of a tensor, 354, 370*, 376Dupin, Theorem of, 305Dyad, 136*, 166Dyadic, 136*, 342

adjoint, 152*antisymmetric, 141complete, 139*conjugate, 141*field, 293finertia, 159*invariants of, 147, 148f, 150*linear, 140*planar, 140*second of a, 151*singular, 140*stress, 253fsymmetric, 140*, 156unit, 144*zero, 137*, 140

Dyadic equality, 136*

434 INDEX

E = [eie2ea], 49*E, F, G, 205*, 290*e,f,g,291*Einstein, A., 328Electric intensity, 241*Ellipsoid, of inertia, 160*Energy, of a fluid, 282

(kinetic) of a top, 176Energy ellipsoid, 177Envelopes, 103, 132Epsilons 329*, 346,

366,369Equation, exact, 200

integrable, 200, 231intrinsic, 103*

Equiangular spiral, 130Equilibrium, of a cord, 298

of a deformable body, 255fof a fluid, 257of a rigid body, 76

Euclidean geometry, 385f-space, 368, 385*Euler, L., 176, 260, 288, 311, 408, 428Eulerian Equations, for fluid motion,

260ffor a top, 176

Evolute, of a plane curve, 101 *

Falling body, 124Field dyadic, 293fField, of curves, 231*, 293*, 299

of geodesics, 299of numbers, 408*of surfaces, 317

Field lines, 226fFlat space, 377*f, 285Floating body, 257Flow, 266*Force, body, 253, 255

centrifugal, 124*conservative, 261Coriolis, 124*on rigid body, 75fsurface, 253, 255transmissibility of, 75

Form, definite, 338*first fundamental, 204*, 290*, 388indefinite, 339*metric, 339*

Form, polar, 337*quadratic, 337fsecond fundamental, 290*, 389singular, 337 *third fundamental, 291 *

Franklin, P., 276Frenet's Formulas, 93*, 100, 122Functional dependence, 190fFundamental quantities, 291-quadratic form, 339

g = det gij, 339*, 369*y (geodesic curvature), 284*, 295,

297Gammas (rte), 326, 356*, 368Gauss, C. F., 300, 301, 307, 310

derivative formulas of, 327, 390Equation of, 301, 302, 306, 320, 392Theorem of, 307

Geodesic, 285*, 297fGeodesic coordinates, 358*-curvature (y), 284*, 295, 297-field, 299-line, 285*-lune, 310-torsion (t), 284*, 286, 305-triangle, 310Gibbs, J. W., 29, 136, 421, 426Gradient, of a scalar, 179*

of a tensor, 187*of a vector, 181 *surface, 206 *

Green's identities, 237, 238-Theorem, 216fGroup (definition), 341*

displacement, 387fof affine transformations, 340fof general transformations, 347of orthogonal transformations, 387of rigid motions, 387f

H = [rrn] = E(, 205* 291Half-turn, 427*, 429Hamilton, W. R., 403, 405, 408, 420,

421Hamilton-Cayley Equation, 160f,

176, 291Harmonic function, 196, 197, 214,

239*, 244f, 247, 423f, 424

INDEX

Heat conduction, 247Helix, 105*f

circular, 107Helmholtz's Equation, 263Homogeneous coordinates, 51-strain, 163

i, j, k, 26*i, j, k, 404 *Idemfactor (I), 144*, 151, 34i3Inertia, dyadic, 159*

moment of, 159*product of, 160*

Integral, circuit, 216fline, 222f, 224fsurface, 218f, 221f, 233fvolume, 233f

Integrability condition, 20), 231Interception (of a plane), 11':Intrinsic equation, 103*Invariable plane, 177Invariant, differential, 181, :.83, 207,

209first scalar, 147*, 149 1'13

of a dyadic, 147*f, 149second scalar, 149*, 173third scalar, 149*, 17.3vector, 147*, 149, 173

Invariant directions, 153f, 156.'-planes, 155Inverse square law, 134Involute of a catenary, 103

of a circle, 103, 131of a plane curve, 102*of a space curve, 131 *

Irrotational vector, 198*f

J = 192*J (mean curvature), 286*Jacobian, 190*f, 264, 347Joachimsthal, Theorem of, ;305

K (total curvature), 286*k (normal curvature), 284*K (curvature), 92*, 96Kelvin's Theorem, on circu' ation,

267on minimum energy, 281

Kepler's Laws, 134

435

Kinematics, of it, particle, 108f, 359of a rigid body, 114E

Kinetic energy, of a fluid, 281, 282of a top, 176

Kronecker delta (Sji), 46*, 343, 366generalized, 353*, 366, 369, 402

Kutta-Joukowsky Formulas, 273f

L, M, N, 290*Lagrange's Theorems, 62Lagrangian Equations for fluid mo-

tion, 265Laguerre, E., 289Lamb, H., 273Laplace expansion of a determinant,

330Laplace's Equation, 196, 197, 214,

238, 247 *, 248Laplacian (V2), 185*, 400, 423Laurent series, 276Lift, 277Linear dependence of vectors, 7*-relation between vectors, 9, 12, 19,

21, 22-relation between motors, 73-vector function, 135*Line integral, 216f, 218f, 222f, 224fLiouville, Formula of, 307

Macduffee, C. C., 161Magnetic shell, 244Males and Dupin, Theorem of, 314Matric algebra, 169fMatrix, 167, 169f, 203, 334, 336Mean center, 20*-curvature (J), 222, 286*, 287, 292,

392-value theorem (harmonic func-

tions), 240Menelaus, Theorem of, 13Metric, 339*Metric quadratic form, 339*-tensor, 349, 366Meusnier's Theorem, 285Minimal surface, 316*fMinimum equation (of a dyadic),

161*Mises, von, R., 68, 70Mobius strip, 218*

436 INDEX

Moment, of inertia, 159*of momentum, 176*of a motor, about an axis, 77*f

about a point, 63*fof a vector, about an axis, 55*f

about a point, 55 *fMotion, of a fluid, 260f, 263f

irrotational, 267line of, 267, 268plane, 270fsteady, 268f

of a particle, 108fcircular, 110relative, 110f, 117, 123funder gravity, 124uniformly accelerated, 110

of a rigid body, 114f, 127Motor, 63, 65*

acceleration, 127*axis of, 66 *derivative of, 126*force, 77*pitch of, 66*proper, 66*velocity, 117*, 120, 128

Motor identities, 73f-product, 70*-sum, 67Moving trihedral, 92*, 122, 128, 283,

312m-Vector, 370*

Nabla (V), 184*, 363, 421Nonion form of a dyadic, 167*fNorm, of a quaternion, 407*Normal, principal, 92*Normal derivative, 180*-form of a dyadic, 162f-plane, 133-system of lines, 232, 313*f-vector to a surface, 205*, 208, 222,

286, 316, 389Null line, 78*-plane, 78*-system, 78*f

Operators:ta()a,418,419Di = a/axi, 356, 362

Operators:Dij = DID, - D,Di, 378, 384V (grad), 184, 215, 363V2, 185Vh, 362, 365Dij = vivj Vjvi, 384V. (Grad), 206q( )q-', 419

Orbit, of a planet, 134Orthogonal coordinates, 194*f-group, 387, 396-surfaces (triple system), 305-trajectories, 231, 300-transformations, 336*, 386, 387-triple of unit vectors, 26*, 48, 194,

336Outer product, of vectors, 354*

P, Q, R, 294*Parallel displacement, 312, 376f-curves, 94 *Parallelogram law, 3Parametric equations:

of a circular helix, 108of a conoid, 321, 323of an ellipsoid, 322of a general curve, 88of a general surface, 203of a hyperboloid, 322of a paraboloid, 322of a right helicoid, 318of a surface, of revolution, 305, 322

of translation, 322Particle, falling, 124Pascal's Theorem, 169Path curves, 377*Permutation symbols, 329, 346-tensor, 350 *Peterson and Morley, Theorem of, 74Pfaff's Problem, 230fPitch, of a motor, 66*Plane, diametral, 176

equation of, 51normal, 133osculating, 98*, 133polar, 176radical, 62rectifying, 133

Planet, motion of, 134

INDEX 437

PI(icker coordinates, , '.Poinsot's Theorem, 1;Point of division, 8*iPoisson's Equation, 2.t"'Polar form, 337-triangles, 45, 420Polygon, plane, 39

spherical, 416, 420Polyhedron, 38Polyhedron Formula )f l:, ler, 311 *Position vector, 5*, 8Postfactor, 136*Potential, complex, 27'4'

scalar, 200*, 241fvector, 202*, 245velocity, 267*

Prefactor, 136*

*

Pressure, fluid, 255, '& 5i'Principal axes, of ine 1t.a, .60*

of strain, 164*of stress, 256*

Principal curvatures, 28(1Principal directions, of a 1 yadic, 162*

on a surface, 287*Principal normal, 92'-stresses, 256 *Principles, fundamental c f statics,

75fProduct, box, 42

cross, 34dot, 29double-dot, 175*indeterminate, 13(.motor, 70*of dual numbers, 64''of dyadics, 142*of four vectors, 43of matrices, 170*of quaternions, 404"of tensor componcnl., 1150*of vectors, 410*

by numbers, 6*outer, 354*scalar, of motors, 6,s"

of vectors, 29*fscalar triple, 41 *vector, 29, 34*fvector triple, 40*

Pythagorean Theoroii-., 2 7

Quadrangle, complete, 16fQuadratic forms, 337*f, 367Quadric surfaces, 175, 321, 322Quadrilateral, complete, 18Quaternion, conjugate of (Kq),

406 *norm of (Nq), 407*real, 403reciprocal of, 409*roots of, 412fscalar part of (Sq), 405unit, 407*, 412, 414vector part of (Vq), 405zero, 403

Quaternion algebra, 403f-division, 409*-multiplication, 404*f-units (i, j, k), 404*, 408Quotient law, 399

R, 294 *R, 90*Radical axis, 62-plane, 62Radius, of curvature (p), 94*

of torsion (v), 94*Rank of a matrix, 203Rate of change, local, 259*, 261

of a vector, 121substantial, 259*, 261

Reciprocal bases, 46*f, 335, 367-dyadics, 145*-quaternions, 409*-sets of motors, 74*f-sets of vectors, 46*f, 144, 192, 207,

209Reduced cofactor, 331 *Reflections, 164f, 419Regular point of a surface, 203*Relative acceleration, 123*-motion, 117-scalar, 346*-tensor, 345*f-velocity, 110*f, 123*Relativity, special, 401Revolving fluid, 262-unit vector, 90, 91Ricci identity, 384-tensor, 384

438 INDEX

Riemannian geometry, 366f-space, 366*f, 377Rodrigues' construction, 429Rolling curve, 119Roots of a quaternion, 412Rotation (rot f, V x f), 183*f, 194,

195, 263 *surface (Rot f, V. x f), 207*

Rotations in space, 164f, 396, 417f,420, 421

Ruled surface, 303, 314Ruling, 303, 314

Scalar, 1Scalar product:

of two motors, 68*fof two vectors, 29*fof three vectors, 41f

Shear, components of, 256Shift formulas, 52Singular dyadic, 140*-point of a surface, 203*-quadratic form, 337Solenoidal vector, 201 *f, 227Solid angle, 236*Space charges, 245fSpatial invariants, 209Speed, 108*Spherical trigonometry, 44f, 416Static Equilibrium, Principle of, 76Statics, 75fStep path, 224Stokes tensor, 372*-Theorem, 220*, 309Strain, homogeneous, 163Stream-lines, 262*, 268, 269Stress dyadic, 253f

for a fluid, 255Sum, of dual numbers, 64*

of dyadics, 137*of matrices, 170*of motors, 67*of quaternions, 403*of tensors, 350*of vectors, 3*f

Summation convention, 188, 328*fSurface, developable, 303, 304,

314*fminimal, 316*f

Surface, of constant total curvature,307f, 310

of revolution, 305, 307, 322of translation, 322parallel, 325parametric equations of, 203*quadric, 175, 321, 322ruled, 303, 314*tangent, 315*, 323

Surface charges, 242f-divergence (Div f), 207*-gradient (Grad f), 206*f-integral, 218f, 233f-invariants, 209-normal, 205 *-rotation (Rot f),Symmetric dyadics, 141*, 156f-tensors, 352 *Symmetry, of affine connection, 357,

361of curvature tensor, 381of inertia dyadic, 160of stress dyadic, 256

T, N, B, 92*t (geodesic torsion), 284z (torsion), 92*, 96Tangent vector, unit (T), 90*, 92Taylor, J. 11., 42Tensor, 172, 344, 348*

absolute, 343falgebra, 350fantisymmetric, 352, 353contraction of, 351contravariant, 344covariant, 344covariant derivative of, 362*, 365*fcurl of, 374 *curvature, 379, 380*fdifferential invariants of a, 209fV f,Vxf,188*V. f, V. x f, 207*divergence of, 372*dual of, 370*gradient of, 362*Of, grad f, 188*V8f, Grad f, 206*metric, 339, 349, 361, 366mixed, 344

INDEX 139

Tensor, relative, 345fIticci, 384*Stokes, 372*ftransformation equations of a, in

afline group, 344, 345in general, 348*

Tensor equations, 348-operations, 350Tetradic, 172Theorems egregium (Gauss), 307Torricclli's Law, 270Torsion (r), 92*

geodesic (t), 284*, 286radius of (v), 93*

Total curvature (K), 286*, 287, 292,306f, 392

-differential, 197*Tractrix, 105*, 308Transformation of integrals, surface

to line, 216, 218f, 221fvolume to surface, 233f

Transpose, of a matrix, 167Triadic, 172*Trigonometry, plane, 21

spherical, 21f, 416Trivector, 353 *

U, V, IV, 194*Umbilical point, 288*Unit dyadic (I), 144*-quaternion, 407*, 412, 414Unit vector, 2*

along binormal (B), 92*along principal normal (N), 92*along surface normal (n), 205*along tangent (T), 90*, 92derivative of, 91*radial (R), 90*revolving, 90transverse (P), 90*

Uniformly accelerated motion, 110

Valence of a tensor, 172*, 344, 345Veblen, 0., 366, 368, 380Vector, 1 *

are, 415base, 24f

Vector, bound, 2*contravariant, 48*, 335covariant, 48*, 335Darboux, 93*, 106, 122, 133, 232dual, 63 *free, 2*irrotational, 198*fline, 2*, 55, 63, 66moment, 63*position, 5*, 8proper, 1rate of change of, 121resultant, 63*solenoidal, 201 *f, 227unit, 2*zero, 1 *, 4

Vector addition, 3*-algebra, 3f, 29f, 354-equations, 50-quantity, 1 *Velocity, 108*

absolute, 111angular, 115*, 116*fcomplex, 272*general components of, 359of sound, 281rectangular components of, 110relative, 110ftransfer, 111 *

Velocity four-vector, 401-motor, 117*, 120Vortex, combined, 281Vortex lines, 2fi6*, 269Vorticity, 263*

Wedderburn, J. 11. M., 169Weingarten, Derivative formulas of,

323, 327, 391Weight of a tensor, 345*Wind triangle, 111

Zero dual number, 64*-dyadic, 137*-motor, 66*-quaternion, 403*-tensor, 348-vector, 1*, 2*, 4

(Brand) Vector and Tensor Analysis (1947)

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Transcript of (Brand) Vector and Tensor Analysis (1947)