/

**Автор**:
David C.Kay

**Теги:**
computer science
computer calculations
machine education
tensor calculus

**ISBN**: 0-07-033484-6

**Год**: 1988

Текст

• I ' I Coverage of all course fundamentals Effective problem-solving techniques 300 fully worked problems with complete solutions Supplements all standard texts \* ч \ ' ч V Ν4 Use With these COUPSes: S'Tensors if Electromagnetic Theory \£ Theoretical Physics Lif Aerodynamics j£ FluW* ,ЛШ#

SULU'M'S (Ч"ШХ1-1 OF THEORY AND PROBLEMS TENSOR CALCULUS UAVIU C> KAY, Ph.D. Pmfi'sfi···- of Mathematirs i'tiitrrviiy vf North Ctirnhnn tit AkAi'sm/A1 McGRAW-HILL .Sn.ii ftrtiictfro \H\^hitigtany If.t.',. Auf.'kianti hogata Caraiias Ijtmiioit Minimi M-e.xim ί'ίίν AMon Mttntrcal New Dehli io.4 J milt Нтр.иръге Sydney Ttp'kya Toronto

DAVID C. KAY is currently Professor and Chairman of Mathematics at the University of North Carolina at Asheville; formerly he taught in the graduate program at the University of Oklahoma for 17 years. He received his Ph.D. in geometry at Michigan State University in 1963. He is the author of more than 30 articles in the areas of distance geometry, convexity theory, and related functional analysis. Schaum's Outline of Theory and Problems of TENSOR CALCULUS Copyright © 1988 by The McGraw-Hill Companies, Inc. All rights reserved. Printed in the United States of America. Except as permitted under the Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher. 14 15 16 17 18 19 20 CUS CUS 09 08 07 06 05 ISBN D-D7-D33Mfi4-b Sponsoring Editor, David Beckwith Production Supervisor, Denise Puryear Editing Supervisor, Marthe Grice Library of Congress Cataloging-in-Publication Data Kay, David С Schaum's outline of theory and problems of tensor calculus. (Schaum's Outline series) 1. Calculus of tensors—Problems, exercises, etc. I. Title. II. Title: Theory and problems of tensor calculus. QA433.K39 1988 515'.63 87-32515 ISBN 0-07-033484-6 McGraw-Hill Л Division of The McGraw-Hill Companies

Preface This Outline is designed for use by both undergraduates and graduates who find they need to master the basic methods and concepts of tensors. The material is written from both an elementary and applied point of view, in order to provide a lucid introduction to the subject. The material is of fundamental importance to theoretical physics (e.g., field and electromagnetic theory) and to certain areas of engineering (e.g., aerodynamics and fluid mechanics). Whenever a change of coordinates emerges as a satisfactory way to solve a problem, the subject of tensors is an immediate requisite. Indeed, many techniques in partial differential equations are tensor transformations in disguise. While physicists readily recognize the importance and utility of tensors, many mathematicians do not. It is hoped that the solved problems of this book will allow all readers to find out what tensors have to offer them. Since there are two avenues to tensors and since there is general disagreement over which is the better approach for beginners, any author has a major decision to make. After many hours in the classroom it is the author's opinion that the tensor component approach (replete with subscripts and superscripts) is the correct one to use for beginners, even though it may require some painful initial adjustments. Although the more sophisticated, noncomponent approach is necessary for modern applications of the subject, it is believed that a student will appreciate and have an immensely deeper understanding of this sophisticated approach to tensors after a mastery of the component approach. In fact, noncomponent advocates frequently surrender to the introduction of components after all; some proofs and important tensor results just do not lend themselves to a completely component-free treatment. The Outline follows, then, the traditional component approach, except in the closing Chapter 13, which sketches the more modern treatment. Material that extends Chapter 13 to a readable introduction to the geometry of manifolds may be obtained, at cost, by writing to the author at: University of North Carolina at Asheville, One University Heights, Asheville, NC 28804-3299. The author has been strongly influenced over the years by the following major sources of material on tensors and relativity: J. Gerretsen, Lectures on Tensor Calculus and Differential Geometry, P. Noordhoff: Goningen, 1962. I. S. Sokolnikoff, Tensor Analysis and Its Applications, McGraw-Hill: New York, 1950. Synge and Schild, Tensor Calculus, Toronto Press: Toronto, 1949. W. Pauli, Jr., Theory of Relativity, Pergamon: New York, 1958. R. D. Sard, Relativistic Mechanics, W. A. Benjamin: New York, 1970. Bishop and Goldberg, Tensor Analysis on Manifolds, Macmillan: New York, 1968. Of course, the definitive work from the geometrical point of view is L. P. Eisenhart, Riemannian Geometry, Princeton University Press: Princeton, N.J., 1949. The author would like to acknowledge significant help in ferreting out typographical errors and other imperfections by the readers: Ronald D. Sand-

PREFACE strom, Professor of Mathematics at Fort Hays State University, and John K. Beem, Professor of Mathematics at the University of Missouri. Appreciation is also extended to the editor, David Beckwith, for many helpful suggestions. David С Kay

Contents Chapter 1 THE EINSTEIN SUMMATION CONVENTION 1 1.1 Introduction 1 1.2 Repeated Indices in Sums 1 1.3 Double Sums 2 1.4 Substitutions 2 1.5 Kronecker Delta and Algebraic Manipulations 3 Chapter 2 BASIC LINEAR ALGEBRA FOR TENSORS 8 2.1 Introduction 8 2.2 Tensor Notation for Matrices, Vectors, and Determinants 8 2.3 Inverting a Matrix 10 2.4 Matrix Expressions for Linear Systems and Quadratic Forms 10 2.5 Linear Transformations 11 2.6 General Coordinate Transformations 12 2.7 The Chain Rule for Partial Derivatives 13 Chapter 3 GENERAL TENSORS 23 3.1 Coordinate Transformations 23 3.2 First-Order Tensors 26 3.3 Invariants 28 3.4 Higher-Order Tensors 29 3.5 The Stress Tensor 29 3.6 Cartesian Tensors 31 Chapter 4 TENSOR OPERATIONS: TESTS FOR TENSOR CHARACTER 43 4.1 Fundamental Operations 43 4.2 Tests for Tensor Character 45 4.3 Tensor Equations 45 Chapter 5 THE METRIC TENSOR 51 5.1 Introduction 51 5.2 Arc Length in Euclidean Space 51 5.3 Generalized Metrics; The Metric Tensor 52 5.4 Conjugate Metric Tensor; Raising and Lowering Indices 55 5.5 Generalized Inner-Product Spaces 55 5.6 Concepts of Length and Angle 56

CONTENTS Chapter 6 THE DERIVATIVE OF A TENSOR 68 6.1 Inadequacy of Ordinary Differentiation 68 6.2 Christoffel Symbols of the First Kind 68 6.3 Christoffel Symbols of the Second Kind 70 6.4 Covariant Differentiation 71 6.5 Absolute Differentiation along a Curve 72 6.6 Rules for Tensor Differentiation 74 Chapter 7 RIEMANNIAN GEOMETRY OF CURVES 83 7.1 Introduction 83 7.2 Length and Angle under an Indefinite Metric 83 7.3 Null Curves 84 7.4 Regular Curves: Unit Tangent Vector 85 7.5 Regular Curves: Unit Principal Normal and Curvature 86 7.6 Geodesies as Shortest Arcs 88 Chapter 8 RIEMANNIAN CURVATURE 101 8.1 The Riemann Tensor 101 8.2 Properties of the Riemann Tensor 101 8.3 Riemannian Curvature 103 8.4 The Ricci Tensor 105 Chapter 9 SPACES OF CONSTANT CURVATURE; NORMAL COORDINATES .... 114 9.1 Zero Curvature and the Euclidean Metric 114 9.2 Flat Riemannian Spaces 116 9.3 Normal Coordinates 117 9.4 Schur's Theorem 119 9.5 The Einstein Tensor 119 Chapter 10 TENSORS IN EUCLIDEAN GEOMETRY 127 10.1 Introduction 127 10.2 Curve Theory; The Moving Frame 127 10.3 Curvature and Torsion 130 10.4 Regular Surfaces 130 10.5 Parametric Lines; Tangent Space 132 10.6 First Fundamental Form 133 10.7 Geodesies on a Surface 135 10.8 Second Fundamental Form 136 10.9 Structure Formulas for Surfaces 137 10.10 Isometries 138

CONTENTS Chapter 11 TENSORS IN CLASSICAL MECHANICS 154 11.1 Introduction 154 11.2 Particle Kinematics in Rectangular Coordinates 154 11.3 Particle Kinematics in Curvilinear Coordinates 155 11.4 Newton's Second Law in Curvilinear Coordinates 156 11.5 Divergence, Laplacian, Curl 157 Chapter 12 TENSORS IN SPECIAL RELATIVITY 164 12.1 Introduction 164 12.2 Event Space 164 12.3 The Lorentz Group and the Metric of SR 166 12.4 Simple Lorentz Matrices 167 12.5 Physical Implications of the Simple Lorentz Transformation 169 12.6 Relativistic Kinematics 169 12.7 Relativistic Mass, Force, and Energy 171 12.8 Maxwell's Equations in SR 172 Chapter 13 TENSOR FIELDS ON MANIFOLDS 189 13.1 Introduction 189 13.2 Abstract Vector Spaces and the Group Concept 189 13.3 Important Concepts for Vector Spaces 190 13.4 The Algebraic Dual of a Vector Space 191 13.5 Tensors on Vector Spaces 193 13.6 Theory of Manifolds 194 13.7 Tangent Space; Vector Fields on Manifolds 197 13.8 Tensor Fields on Manifolds 199 ANSWERS TO SUPPLEMENTARY PROBLEMS 213 INDEX 223

Chapter 1 The Einstein Summation Convention 1.1 INTRODUCTION A study of tensor calculus requires a certain amount of background material that may seem unimportant in itself, but without which one could not proceed very far. Included in that prerequisite material is the topic of the present chapter, the summation convention. As the reader proceeds to later chapters he or she will see that it is this convention which makes the results of tensor analysis survey able. 1.2 REPEATED INDICES IN SUMS A certain notation introduced by Einstein in his development of the Theory of Relativity streamlines many common algebraic expressions. Instead of using the traditional sigma for sums, the strategy is to allow the repeated subscript to become itself the designation for the summation. Thus, η axxx + a2x2 + агхъ + ■ ■ ■ + anxn = 2j aixi /=1 becomes just а{х(, where l^ii/i is adopted as the universal range for summation. EXAMPLE 1.1 The expression a(jxk does not indicate summation, but both aHxk and ai]xj do so over the respective ranges l^iSn and 15; 5л. If«=4, then avSk = βιΑ + a2iXk + a33xk + a44xk aijXj = β,.,χ, + ai2x2 + 0.3*3 + fl,4*4 Free and Dummy Indices In Example 1.1, the expression a^Xj involves two sorts of indices. The index of summation, j, which ranges over the integers 1, 2, 3,. . . , n, cannot be preempted. But at the same time, it is clear that the use of the particular character/ is inessential; e.g., the expressions airxr and alvxv represent exactly the same sum as aijxi does. For this reason, j is called a dummy index. The index i, which may take on any particular value 1,2, 3, . . . , η independently, is called a free index. Note that, although we call the index i "free" in the expression a^*., that "freedom" is limited in the sense that generally, unless i = k, aljXj Φ akixs EXAMPLE 1.2 If η = 3, write down explicitly the equations represented by the expression yi = airxr. Holding i fixed and summing over r = 1,2,3 yields У ι = anx, + ai2x2 + ai3x3 Next, setting the free index ί =1,2,3 leads to three separate equations: y, = a11x1 + a12x2 + a13x3 y3 = a31x, + a32x2 + a33x, Einstein Summation Convention Any expression involving a twice-repeated index (occurring twice as a subscript, twice as a superscript, or once as a subscript and once as a superscript) shall automatically stand for its sum 1

2 THE EINSTEIN SUMMATION CONVENTION [CHAP. 1 over the values 1,2,3, . . . , η of the repeated index. Unless explicitly stated otherwise, the single exception to this rule is the character n, which represents the range of all summations. Remark 1: Any free index in an expression shall have the same range as summation indices, unless stated otherwise. Remark 2: No index may occur more than twice in any given expression. EXAMPLE 1.3 (a) According to Remark 2, an expression like aiixi is without meaning, (b) The meaningless expression a'jxtxl might be presumed to represent a^x^f, which is meaningful, (c) An expression of the form al(xl + yt) is considered well-defined, for it is obtained by composition of the meaningful expressions aizi and xi + y. = zr In other words, the index i is regarded as occuring once in the term {xi + y.). 1.3 DOUBLE SUMS An expression can involve more than one summation index. For example, агхьу, indicates a summation taking place on both i and j simultaneously. If an expression has two summation (dummy) indices, there will be a total of n2 terms in the sum; if there are three indices, there will be η terms; and so on. The expansion of aijxiyj can be arrived at logically by first summing over i, then over j: аахгУ) = «i^iJ/ + ЧргУ) + <*3]х3У] + ■■■ + ап!.хпУ] [summed over i] = (я11#1)'1 + al2xly2 + ■ ■ ■ + allJx1yIJ) [summed over j] + (a2lx2yl + a22x2y2 + ■■■ + a2nx2yn) + О3ЛУ1 + a32x3y2 + ■■■ + a3nx3yn) + 0„Λ>Ί + ап2хпУ2 + ■■■ + аппхпуп) The result is the same if one sums over j first, and then over /. EXAMPLE 1.4 If η = 2, the expression yt = c'iarsxs stands for the two equations: yl — c1a11x1 + c1a21x1 + c1a12x2 + c1a22x2 _ ! ι 2 ι ! ι 2 1.4 SUBSTITUTIONS Suppose it is required to substitute yt = ai]xj in the equation Q = b^y^Xj. Disregard of Remark 2 above would lead to an absurd expression like Q = b^a^xjx-. The correct procedure is first to identify any dummy indices in the expression to be substituted that coincide with indices occurring in the main expression. Changing these dummy indices to characters not found in the main expression, one may then carry out the substitution in the usual fashion. step 1 <2 = bjjy^j, yt = a(jXj [dummy index j is duplicated] step 2 yi = airxr [change dummy index from j to r] step 3 <2 = b^a^x^Xj = airbijxrxJ [substitute and rearrange] EXAMPLE 1.5 If y. = a^Xj, express the quadratic form Q = g^y^j in terms of the x-variables. First write: y. = airxt, yf = ajsxs. Then, by substitution, Q = 8iMirxr)(aisxs) = g,jairajsxrxs or Q = hrsxrxs, where hrs = gnairajs.

CHAP. 1] THE EINSTEIN SUMMATION CONVENTION 3 1.5 KRONECKER DELTA AND ALGEBRAIC MANIPULATIONS A much used symbol in tensor calculus has the effect of annihilating the "off-diagonal" terms in a double summation. *<,■=*>-«Hi Wj (^) Kronecker Delta Clearly, δ;/ = δ,,, for all ι, j. EXAMPLE 1.6 Ifn = 3, δ/jXiXj = lx1X1 + Qx,X2 + 0XjX3 + Qx2xl + \х2хг + Qx2x3 + Qx3xl + 0x3JC2 + \хъХъ = (x1)2 + (x2f + (x,)2 = xixi In general, 8lpclxj = xixi and 8r/atrxl = atjxt. EXAMPLE 1.7 Suppose that V = g'rarsys and yt = birxr. If further, airbrj = d:j, find V in terms of the xr. First write ys - bs,x,. Then, by substitution, T = SrarsbsiXi = g'AS, = g'rXr Algebra and the Summation Convention Certain routine manipulations in tensor calculus can be easily justified by properties of ordinary sums. However, some care is warranted. For instance, the identity (1.2) below not only involves the distributive law for real numbers, a(x + y) = ax + ay, but also requires a rearrangement of terms utilizing the associative and commutative laws. At least a mental verification of such operations must be made, if false results are to be avoided. EXAMPLE 1.8 The following nonidentities should be carefully noted: Я/А+У/^я^ + Я-Л- (ail + aji)xiyj^2aijx,yj Listed below are several valid identities; they, and others like them, will be used repeatedly from now on. aiMs + У,) = aijXj + aiiyi (1.2) aijxiy]^aiiyixi (1.3) ai.xix] = ajixixj (1.4) (ai, + ап)х,х} = 2aijXiXj (1.5) («v-«,,>/*, =0 (1.6)

4 THE EINSTEIN SUMMATION CONVENTION [CHAP. 1 Solved Problems REPEATED INDICES 1.1 Use the summation convention to write the following, and assign the value of η in each case: (a) allbll + a2lbl2 + a3lbl3 + a4lbl4 (b) anbn + al2b12 + a13b13 + aubu + alsbls + al6bl6 (c) c'n+ 4 + 4 + 4 + 4 + 4 + 4 + 4 (1 =£*=£ 8) (β) β,A, (n = 4);(b)aublt (n = 6); (с) с), (и = 8). 1.2 Use the summation convention to write each of the following systems, state which indices are free and which are dummy indices, and fix the value of n: (a) cllx1 + c12x2 + c13x3 = 2 (b) ajxl + a2x2 + α^χ3 + α*χ4 = bj 21 1 22 2 Ί."\ "\ \/ J / 311 42 2 314 (a) Set d, =2, ίί2 = -3, and d3 = 5. Then one can write the system as ctjx; = d; (n = 3). The free index is i and the dummy index is /. (b) Here, the range of the free index does not match that of the dummy index (и = 4), and this fact must be indicated: a% = b, (J = 1,2) The free index is / and the dummy index is i. 1.3 Write out explicitly the summations Φι + У ι) cixi + скУк where η = 4 for both, and compare the results. с,(х, + У,) = ^(*! + У г) + с2(х2 + у2) + с3(х3 + Уз) + с4(*4 + У а) = СЛ + С1У1 + С2*2 + С2У2 + С3*3 + СзУз + С4*4 + С аУ А СЛ + С*У* = С1*1 + С2*2 + С3*3 + С4*4 + С1У1 + С2>Ί +/ъУъ + С аУ А The two summations are identical except for the order in which the terms occur, constituting a special case of (1.2). DOUBLE SUMS 1.4 If η = 3, expand Q = ά'χμ,^. Q = a^XjXy + α2'χ2χ} + a3/x3^ = α χγχγ + α XjX2 + a ~xlx3 + α x2xl + a x2x2 + a x2x3 + a" x3Xj + a" x3x2 + α33χ3χ3 1.5 Use the summation convention to write the following, and state the value of η necessary in each case: (θ) at Al + Я2А2 + йзАз + u12^21 + a22^22 + u32^23 + u13^31 + ^23^32 + u33^33 ? Ψ) ' g\t + 'g\2 + Ϋ2ι +'gL +'gj, +fc?2 +^221 + '*22

CHAP. 1] THE EINSTEIN SUMMATION CONVENTION 5 (fl) β,Α,-+ ei2fr2i + fli3fr3i = «Λ« (" = 3)- (b) Set c, =1 for each i (n = 2). Then the expression may be written g'llCi + g'l2Ci + #21С/ + S'll^i = (g'n + g'l2 + #21 + g^)^ = (8'jkcic,c)ci=S'ikcicjck 1.6 If и = 2, write out explicitly the triple summation crstxrysz. Any expansion technique that yields all 23 = 8 terms will do. In this case we shall interpret the triplet rst as a three-digit integer, and list the terms in increasing order of that integer: crstxrysz' = с111д:1)'121 + c112x1y1z2 + c121x1y2z1 + cl22x1y2z2 + c2^x2yxzl + c212x2y1z2 + c-,2lx2y2z'i + c222x2y2z2 1.7 Show that ailxixj = 0 if ац = i - /. Because, for all ι and j, αί; = —ajt and ж,.*. = хрс,, the "off-diagonal" terms а^х{х} (i <j; no sum) and tijfXjXi (/>«"; no sum) cancel in pairs, while the "diagonal" terms α,,(χ,)2 are zero to begin with. Thus the sum is zero. The result also follows at once from (1.5). 1.8 If the ai; are constants, calculate the partial derivative 3 ( \ JxTk (W,} Reverting to Σ-notation, we have: Σ at1xixj = Σ a^xpcj + Σ α,,*,*, + Σ aijxix) + Σ aipcixi i,j i*k I-k i^k i=k )*k f^k j = k j=k = C + (Σ akjx.)xk + [Σ aikXl)xk + akk(xk)2 'j^k ' i^k ' where С is independent of xk. Differentiating with respect to xk, τ— (Σβ,Λ-^)=0 + Σ akiXj + Σ aikx, + 2akkxk "лк vi,j ' j¥-k l^k = Σα„χ, + Σο,Λ- f ί or, going back to the Einstein summation convention, — (ait.xiXj) = ahixt + aikx, = (aik + aki)Xi SUBSTITUTIONS, KRONECKER DELTA 1.9 Express tf'yjj in terms of x-variables, if у ι — cijxj and b''cik = 8[. tfy.y. = b'\cirxr)(cjsxs) = (bi]cir)xrcjsxs = 3irx,.cjsxs = *,с,.Л = cijxixl 1.10 Rework Problem 1.8 by use of the product rule for differentiation and the fact that dxp ~ϊχΖ~8™

6 THE EINSTEIN SUMMATION CONVENTION [CHAP. 1 = a>j(Xj3ik + x.8jk) = akjXj + aikx, 1.11 If fly = Oji are constants, calculate (PijXiXj) dxkdxt Using Problem 1.8, we have Э2 dxkdxt (W/) = ^ [^ (W,)] = ~tk IN + *,>,] <9* (2α,Λ.) = 2e/V — (*,.) = 2β,7δΑι. = 2aw 1.12 Consider a system of linear equations of the form у = a''xj and suppose that (btj) is a matrix of numbers such that for all i and j, bird' = δ ■ [that is, the matrix (ft;.) is the inverse of the matrix (a1')]. Solve the system for xt in terms of the y'. Multiply both sides of the /th equation by bki and sum over i: bkiy' = bkia'ixJ = 8Jkx] = xk отх, = Ъуу\ 1.13 Show that, generally, aijk(xl + y})zk Φ а1]кхьгк + aijkyfzk. Simply observe that on the left side there are no free indices, but on the right, ;" is free for the first term and i is free for the second. 1.14 Show that cf/(*,. + y^Zj = cVjxiz] + cijyiz]. Let us prove (1.2); the desired identity will then follow upon setting atj = cJt. aaxi + аиУ; = Σ aijXj + Σ ацУ} = Σ (a^Xj + а1]У]) i i i = Еву(^-+)'7)=в,у(-Ку + Уу) Supplementary Problems 1.15 Write out the expression aibl (n =6) in full. 1.16 Write out the expression R'jki (n = 4) in full. Which are free and which are dummy indices? How many summations are there? 1.17 Evaluate 3'jxl (n arbitrary). 1.18 For η arbitrary, evaluate (α) δ„, (b) δ δϋ, (с) 8i}8[clk.

CHAP. 1] THE EINSTEIN SUMMATION CONVENTION 7 1.19 Use the summation convention to indicate a13b13 + a23b23 + a33b3i, and state the value of n. 1.20 Use the summation convention to indicate αιι(ΛΓι )Z + а\2х\хг + а\ъх\хг + а21х2хг + a22{x2)2 + a23x2x3 + а31хъхг + a32x3x2 + a33(x3)2 1.21 Use the summation convention and free subscripts to indicate the following linear system, stating the value of n: У \ C\\X\ C\2X2 y2 C21X1 ' C22-^2 1.22 Find the following partial derivative if the a:j are constants: _d_ dX — (anx1 + al2x2 + a13x3) (A: =1,2,3) к 1.23 Use the Kronecker delta to calculate the partial derivative if the atj are constants: д ι ^ 1.24 Calculate ~ [а„х,(х,У where the atj are constants such that ац = ajt. 1.25 Calculate д д (Vi¥*) where the a... are constants. 1.26 Solve Problem 1.11 without the symmetry condition on air 1.27 Evaluate: (а) Ъ)у, if yt = T?, (b) a^y,. if y, = b;jXj, (c) aijkyiyjyk if y, = bijXj. 1.28 If ε, = 1 for all i, prove that (a) (flj + e2 + - ■ ■ + α„)2 s EiElalaj (b) e,.(l + χ J = aiei + afx( (c) aiJ(xi +xj)^2aijEjxl if ац = aj;

Chapter 2 Basic Linear Algebra for Tensors 2.1 INTRODUCTION Familiarity with the topics in this chapter will result in a much greater appreciation for the geometric aspects of tensor calculus. The main purpose is to reformulate the expressions of linear algebra and matrix theory using the summation convention. 2.2 TENSOR NOTATION FOR MATRICES, VECTORS, AND DETERMINANTS In the ordinary matrix notation (a,v), the first subscript, i, tells what row the number ar lies in, and the second, /, designates the column. A fuller notation is [a,]mn, which exhibits the number of rows, m, and the number of columns, n. This notation may be extended as follows. Upper-Index Matrix Notation KL»: < 2 «1 «2 2 «2 a3 . 2 a3 . .a .a, [a" 'Ί = 'a11 a11 ml a 12 a 22 a ν»2 a 13 a . 23 a . am\ „1" " . a „2b . a inn . a Note that, for mixed indices (one upper, one lower), it is the upper index that designates the row, and the lower index, the column. In the case of pure superscripts, the scheme is identical to the familiar one for subscripts. EXAMPLE 2.1 \P% c1 c1 c1 Lj (,2 L3 2 2 2 С С С с, с2 с3 KJ23 = άλ d2 d3 dx d2 d3 = [d'jL· [χΓΑκ = [χ\ χ1ι χ1 *'] [у"4] у" у21 у31 Ь41 у12] у22 f2 42 У . Vectors A real «-dimensional vector is any column matrix v= [х^]п1 with real components xt =*n; one usually writes simply v = (xt). The collection of all real и-dimensional vectors is the n-dimensional real vector space denoted R". Vector sums are determined by coordinatewise addition, as are matrix sums: if A = [fliy]m„ and ДИМ™, then A + B = [ai; + b ,,]„,„ Scalar multiplication of a vector or matrix is defined by Basic Formulas The essential formulas involving matrices, vectors, and determinants are now given in terms of the summation convention.

CHAP. 2] BASIC LINEAR ALGEBRA FOR TENSORS 9 Matrix multiplication. If A = [ai}]mn and В = [Ьи]пк, then AB = [airbr}]mk (2.1а) Analogously, for mixed or upper indices, ЛВ = [а)]тп[Ь)]пк = K^,]mfc ЛЯ - Ит(ГЬ'']вк = [a%ri]mk (2.1b) wherein i and j are not summed on. Identity matrix. In terms of the Kronecker deltas, the identity matrix of order η is which has the property IA = AI = A for any square matrix A of order и. Inverse of a square matrix. A square matrix A = [α^]„„ is invertible if there exists a (unique) matrix В = [bij\nn, called the inverse of A, such that Л5 = В A = I. In terms of components, the criterion reads: airbrj = biA, = δν (2.2a) or, for mixed or upper indices, а% = Ъ\аТгЬ\ a'rbrj = b'rari = 8ij (2.2b) Transpose of a matrix. Transposition of an arbitrary matrix is defined by Ат =[а^]'тп = [а'и]пт, where a'u = ap for all i, j. If AT = A (that is, ац = α;/ for all i, j), then Л is called symmetric; if Л = —Л (that is, аг. = — fly-,- for all г, /), then Л is called antisymmetric or skew-symmetric. Orthogonal matrix. A matrix Л is orthogonal if ЛГ = Л_l (or if ATA = AAT — I). Permutation symbol. The symbol eljk_ _w (with η subscripts) has the value zero if any pair of subscripts are identical, and equals (-\)p otherwise, where ρ is the number of subscript transpositions (interchanges of consecutive subscripts) required to bring (ijk. . . w) to the natural order (123. . .n). Determinant of a square matrix. If Л = [fliy]„„ is any square matrix, define the scalar det Л = e, ι ι , au a,, a3, ■ · · ani (2.3) Ί'2'З ■ ■ ■ 'n "1 £!2 J,3 nln v ' Other notations are |Л|, \ai}\, and det(a;y). The chief properties of determinants are |ЛВ| = |Л||5| |ЛГ| = |Л| (2.4) Laplace expansion of a determinant. For each i and j, let Ml} be the determinant of the square matrix of order η — 1 obtained from Л = [аи]пп by deleting the ith row and y'th column; Mtj is called the minor of α · in |Л|. Define the cofactor of atj to be the scalar Аи = (-1)кМ„ where k = i+j (2.5) Then the Laplace expansions of \A\ are given by \Α\ = αυΑν = ανΑν = --· = αη]Αηί [row expansions] \A\ = anAn = af2Ai2 = · ■ · = ainAjn [column expansions] Scalar product of vectors. If u = (χέ) and ν = (_y,.), then uv = u · ν = urv = xjyi (2.7) If u = v, the notation uu = u" = v" will often be used. Vectors u and ν are orthogonal if uv = 0. Norm (length) of a vector. If u= (x{), then ||u||=Vu5 = V3cpc: (2.8)

10 BASIC LINEAR ALGEBRA FOR TENSORS [CHAP. 2 Angle between two vectors. The angle θ between two nonzero vectors, u = (x() and ν = (y,), is defined by uv xji cos θ : (OS 0^77") НИ νχ-χ,νηη It follows that 0 = it/2 if u and ν are nonzero orthogonal vectors. Vector product in R3. If u = (jt,) and v= (y;), and if the standard basis vectors are designated i-(Sn) Js(Si2) k-(5i3) then (2.9) uX v = i •4 >1 j x2 Уг к л3 >3 — х2 х3 Уг Уз ι — χλ Ул х? Уз J + χλ У, Хо У, (2.10а) Expressing the second-order determinants by means of (2.3), one can rewrite (2.10a) in terms of components only: uXv = (ei]kXjyk) (2.10b) 2.3 INVERTING A MATRIX There are a number of algorithms for computing the inverse of Α = [αί}]ηη, where |Л| т^О (а necessary and sufficient condition for A to be invertible). When η is large, the method of elementary row operations is efficient. For small и, it is practical to apply the explicit formula 1 A" = ш И,] Thus, for η = 2, in which |Л| = ana22 - al2a2l; and, for η = 3, * 'уу \Л ух 1 \А\ 1 *22 *12 Ли Л21 А31 ■"12 22 -^32 .-^13 ^23 ^33. in which ji -i-i d-y-jd-j-i 71 11 ^T. ^ -ι I Wi^Mii 1 4 1") / (2.11a) (2.11b) (2.11c) 2.4 MATRIX EXPRESSIONS FOR LINEAR SYSTEMS AND QUADRATIC FORMS Because of the product rule for matrices and the rule for matrix equality, one can write a system of equations such as Ъх - 4y = 2 -5x + 8y = 7 in the matrix form 3 -4 -5 8 км

CHAP. 2] BASIC LINEAR ALGEBRA FOR TENSORS 11 In general, any w x η system of equations aijxi = bi (l^i^m) can be written in the matrix form Лх=Ь (2.12a) (2.12 b) where А = [а^]тп, x = (xt), and b = (&,). One advantage in doing this is that, if m = η and A is invertible, the solution of the system can proceed entirely by matrices: x = А~1Ъ. Another useful fact for work with tensors is that a quadratic form Q (a homogeneous second-degree polynomial) in the η variables xx, x2, . . . , xn also has a strictly matrix representation: Q = α^χ,.χ} = xTAx (2.13) where the row matrix x' is the transpose of the column matrix χ = (xt) and where A = [а,.]лл. EXAMPLE 2.2 yx1 x2 x3\ 12 22 32 fl13" fl23 fl33_ ~*l~ X2 _*з_ \X j X2 X^\ = k-K*,)] = aiS^ The matrix A that produces a given quadratic form is not unique. In fact, the matrix В = \(A + AT) may always be substituted for A in (2.13); i.e., the matrix of a quadratic form may always be assumed symmetric. EXAMPLE 2.3 Write the quadratic equation 2>x2 + y2- 2z2 - 5xy - 6yz = 10 using a symmetric matrix. The quadratic form (2.13) is given in terms of the nonsymmetric matrix "3 -5 (Г A= 0 1-6 _0 0 -2. The symmetric equivalent is obtained by replacing each off-diagonal element by one-half the sum of that element and its mirror image in the main diagonal. Hence, the desired representation is [x у z\ 3 -5/2 0 -5/2 1 -3 0" -3 -2. x~ У -Z. = 10 2.5 LINEAR TRANSFORMATIONS Of utmost importance for the study of tensor calculus is a basic knowledge of transformation theory and changes in coordinate systems. A set of linear equations like У: 5x — 2y 3x + 2y (I) defines a linear transformation (or linear mapping) from each point (x, y) to its corresponding image (x, y). In matrix form, a,linear transformation may be written χ = Лх; if, as in (/), the mapping is one-one, then |Л|^0. There is always an alias-alibi aspect of such transformations: When (x, y) is regarded as defining new coordinates (a new name) for (x, y), one is dealing with the alias aspect; when (x, y) is regarded as a new position (place) for (x, y), the alibi aspect emerges. In tensor calculus, one is generally more interested in the alias aspect: the two coordinate systems related by χ = Лх are referred to as the unbarred and the barred systems.

12 BASIC LINEAR ALGEBRA FOR TENSORS [CHAP. 2 EXAMPLE 2.4 In order to find the image of the point (0, -1) under (/), merely set χ = 0 and у result is 1; the * = 5(0)-2(-l) = 2 y = 3(0) + 2(-l) = -2 Hence, (0, -1) = (2, -2). Similarly, we find that (2,1) = (8, 8). If we regard (x, y) merely as a different coordinate system, we would say that two fixed points, Ρ and Q, have the respective coordinates (0, -1) and (2,1) in the unbarred system, and (2, —2) and (8, 8) in the barred system. Distance in a Barred Coordinate System What is the expression for the (invariant) distance between two points in terms of differing aliases? Let χ = Ax (| A\ ¥^ 0) define an invertiblc linear transformation between unbarred and barred coordinates. It is shown in Problem 2.20 that the desired distance formula is d(x, y) = A/(x - У f G(x - у) = Уg,y Δ*,. Δ*,. (2.14) where [g,7]„„ = G = (AAT)"г and х-у = (Дх,). If Л is orthogonal (a rotation of the axes), then Sij = <V and (2.14) reduces to the ordinary form d(x,y) = \\x-y\\=VM^xi [cf. (2.8)]. EXAMPLE 2.5 Calculate the distance between points Ρ and Q of Example 2.4 in terms of their barred coordinates. Verify that the same distance is found in the unbarred coordinate system. First calculate the matrix G = (AAT)~1 = (A~~l)TA~~l (see Problem 2.13): and G = 16 2 2 -3 5 5 -2 3 2 τ Λ| = 10-(-6) = 16 φ 2 2] 1 -3 5.1 256 "2 -3" .2 5. *-'-h 2 2" .-3 5. 2 2 2" .-3 5. 1 56 13 .-И -11" 29. Hence gu — 13/256, g12 ■ (2.14) gives: -11/256, and g22= 29/256. Now, with i-y= [2- 8 -2-8]7=[-6 -10]', d2 = gij Ax, Axj 13 —11 29 = 25б(-6)2+2-Ж(-6)(-10)+25б(-10)2 13(36)-22(60)+29(100) _Q ~ 256 ~8 In the unbarred system, the distance between P(0, -1) and Q(2, 1) is given, in agreement, by the Pythagorean theorem: d2 = (0-2)2 + (-l-l)2=8 2.6 GENERAL COORDINATE TRANSFORMATIONS A general mapping or transformation Τ of R" may be indicated in functional (vector) or in component form: ΓΟΟ or T,(xux2,. ..,*„) In the alibi description, any point χ in the domain of Τ (possibly the whole of R") has as its image the point T(x) in the range of T. Considered as a coordinate transformation (the alias description), Γ sets up, for each point Ρ in its domain, a correspondence between (χέ) and (*■), the coordinates of Ρ in two different systems. As explained below, Γ may be interpreted as a coordinate transformation only if a certain condition is fulfilled.

CHAP. 2] BASIC LINEAR ALGEBRA FOR TENSORS 13 Bijections, Curvilinear Coordinates A map Τ is called a bijection or a one-one mapping if it maps each pair of distinct points χ Φ у in its domain into distinct points T(x) Φ T(y) in its range. Whenever Τ is bijective, we call the image χ = T(x) a set of admissible coordinates for x, and the aggregate of all such coordinates (alibi: the range of T), a coordinate system. Certain coordinate systems are named after the characteristics of the mapping T. For example, if Τ is linear, the (x^-system is called affine; and if Γ is a rigid motion, (x;) is called rectangular or cartesian. [It is presumed in making this statement that the original coordinate system (хг) is the familiar cartesian coordinate system of analytic geometry, or its natural extension to vectors in R".] Nonaffine coordinate systems are generally called curvilinear coordinates; these include polar coordinates in two dimensions, and cylindrical and spherical coordinates in three dimensions. 2.7 THE CHAIN RULE FOR PARTIAL DERIVATIVES In working with curvilinear coordinates, one needs the Jacobian matrix (Chapter 3) and, therefore, the chain rule of multivariate calculus. The summation convention makes possible a compact statement of this rule: If w = f{xx, x2, x3,. . . , xn) and xt = xt(u1, u2,. . . , um) (i = 1,2,... , n), where all functions involved have continuous partial derivatives, then dw df dxt du- дХ, ди: (l^j^m) (2.15) Solved Problems TENSOR NOTATION 2.1 Display explicitly the matrices (а) [Ц]42, (b) [b)]24, (c) [Sij]33. (a) [b{U = (b) [b)] δ11 δ21 δ31 δ12 δ22 δ32 δ13" δ23 δ3ϊ_ = "1 0 0 0 1 0 0" 0 1 (с) [δ4]33 = From (α) and (b) it is evident that merely interchanging the indices i and j in a matrix A = [a,7],„„ does not necessarily yield the transpose, AT. 2.2 Given a —a ~a 2b b -b Ac 2c -2c. verify that ΑΒΦΒΑ. AB 2a + fl-3fl 4a + 2α-6α 4b-b-3b 8b-2b-6b 8c-2c-6c 16c-4c-12c B = 4 -6' -2 3 6 -9. -6a — 3a + 9a 126 + 3b + 9b 24c -ι- 6c + 18c_ = "0 0 0" 0 0 0 .0 0 0_ = o

14 BASIC LINEAR ALGEBRA FOR TENSORS [CHAP. 2 but В А 2a + 86 - 24c -2a + \b - 12c -2a - \b + 12c' -a-46 +12c «-26 +6c a +26 -6c 3a + 126 -36c -3<r + 66-18c -3a-66 +18c. ^O Thus, the commutative law (AB = ,ΖΜ) fails for matrices. Further, ЛЛ = О does not imply that A = О or л = о. 2.3 Prove by use of tensor notation and the product rule for matrices that (AB)T = BTAT, for any two conformable matrices A and B. Let A s [atf]mi|i Л ^ [bv]nk, AB - [c;7]mi, and, for all i and /, 4 = S &y = Ь„ сц = c„ Hence, Лг = [%■]„„, βΓ= [A-,L,, and (Afi)r = [c',]b. We must show that BTAT= [с'и]кт. By definition of matrix product, BTAT=:[b'ilalj]hm, and since Ka'rj = *rt"fr = fl„ the desired result follows. 2.4 Show that any matrix of the form A = BTB is symmetric. By Problem 2.3 and the involutory nature of the transpose operation, AT = (BTB)T = BT(BT)T = B'B = A 2.5 From the definition, (2.3), of a determinant of order 3, derive the Laplace expansion by cofactors of the first row. In the case п — Ъ, (2.3) becomes flll fl12 fl13 аП fl22 fl23 fl31 fl32 fl33 Kyi = e„-kaua2ja3k Since e,yi. = 0 if any two subscripts coincide, we write only terms for which (ijk) is a permutation of (123): \aij\ ~ *-123fl11^22fl33 ' ^132flUfl23^32 ' *-213fll 7fl2 I fl33 ^231^12^23^31 ^312^13^21^32 ^371^13^22^31 — fl11fl22fl33 — flufl23fl17 — fllzfl21fl3i + fl12fl23fl31 + fl13fl2jfl32 — fl13fl22fl31 = flu(fl22fl33 - fl23fl32) - fl12(fl21fl33 - fl23fl31) + fl]3(fl71fl37 - fl22fl31) But, for η =2, (2.3) gives 22 fl23 — , Λ — ι _ _ 11 12 22"зз ' C2.'t23u3-, fl22fl33 fl2^fl32 52 "33 " " " and the analogous expansions of — Л12 and + -A13. Hence, \aij\ = flllAl + ff12^12 + fl13^13 = lljAlj as in (2.6). 2.6 Evaluate: (<0 («) ь -2c 6 - -2c -2a Ь -2a 6 (b) 5 -2 15 -10 0 10 15 0 30 6 - (-2e)(-2c) = 62 - \ac

CHAP. 2] BASIC LINEAR ALGEBRA FOR TENSORS 15 (h) Because of the zeros in the second column, it is simplest to expand by that column: 5 -2 15 -10 0 10 15 0 30 = - = 2 (~2) -10 15 -10 15 10 30 10 30 = 2( + 0 5 15 15 30 -0 -1 1 1 2 5 151 -10 101 300(-2 - 1) = -900 2.7 Calculate the angle between the following two vectors in R5: χ = (1,0,-2,-1,0) and у = (0,0, 2, 2,0) We have: xy = (1)(0) + (0)(0) + (-2)(2) + (-1)(2) + (0)(0) = -6 x2 = i2 + o2 + (-2)2 + (-1)2 + 02 = 6 v2 = o2 + 02 + 22 + 22 + 02 = 8 and (2.9) gives cos θ ■ V6-V8 V3 2 or 0 = 577 2.8 Find three linearly independent vectors in R which are orthogonal to the vector (3, 4,1, -2). It is useful to choose vectors having as many zero components as possible. The components (0,1, 0, 2) clearly work, and (1,0, —3, 0) also. Finally, (0, 0, 2, 1) is orthogonal to the given vector, and seems not to ^e dependent on the first two chosen. To check independence, suppose scalars x, y, and ζ exist such that x(Q) + y(l) + z(0) = 0 *(l) + y(0) + z(0) = 0 x(0) + y(-3) + z(2) = 0 x(2) + y(0) +z(l) = 0 This system has the sole solution χ = у = ζ = 0, and the vectors are independent. 0 1 0 bJ + У 1 0 -3 L oj + ζ 0 0 2 L ι J = 0 0 0 LoJ or 2.9 Prove that the vector product in R is anticommutative: xXy = -yXx. By (2.10b), x x У = (eijkXjyk) and у x χ = (ецкурск) But eikj = -eilk, so that ецкУ'jXk егк1УкХ1 ечкУкХ1 е1/кХ)Ук INVERTING A MATRIX 2.10 Establish the generalized Laplace expansion theorem: arjAsj= \A\ Srs. Consider the matrix A* = 41 "*12 ar, ar \-a„i a „2 row s

16 BASIC LINEAR ALGEBRA FOR TENSORS [CHAP. 2 which is obtained from matrix A by replacing its sth row by its rth row (гФ s). By (2.6), applied to row r oiA*, det A* = a A*rJ (not summed on r) Now, because rows r and s are identical, we have for all /, Л*=(-1)"Л*=(-1)рЛч. (p = r-i) Therefore, det A* = (—l)parjAbj. But it is easy to see (Problem 2,31) that, with two rows the same, det A* = 0, We have thus proved that and this, together with (2.6) for the case r = J, yields the theorem. 2.11 Given a matrix A = [fl,y]„„, with |Л| ^0, use Problem 2.10 to show that AB = I where В [Л,] Since the (t, y')-element of В is Л ,/|у4| ab = κ,μ,,/μΐ)]™ = -^ [μ|δ,]„„ = j^j [δ,ν]„„ = / [It follows from basic facts of linear algebra that also В A — I; therefore, Al = B, which establishes (2.11a).} 2.12 Invert the matrix ~-2 0 1" A= 3 1 0 .2-2 3. Use (2.11c). To evaluate |v4|. add twice the third column to the first and then expand by the first row: 0 0 1 3 1 0 8-2 3 1- 3 1 8 -2 -8=-14 Then, computing cofactors as we go, 1 3 9 8 -2 -8 -4 -r 3 -2. = -3/14 1/7 1/14 9/14 4/7 -3/14 4/7 2/7 1/7 2.13 Let A and В be invertible matrices of the same order. Prove that (a) (AT)~l = (A~l)T (i.e., the operations of transposition and inversion commute); (ft) (AB)~l = B~Al. (a) Transpose the equations AA~l = A~lA = /, recalling Problem 2.3, to obtain (A~l)TAT = AT(A l)T = IT = I which show Lhat AT is invertible, with inverse (AT)~' =(^4"')7. (b) By the associative law for matrix multiplication, (AB)(B~A~') = A(BB^1)A~' = ΑΙΑ"1 = AA"1 = / and, similarly, (B~1A~l)(AB) = I Hence, (AB)"1 = B'A"1.

CHAP. 2] BASIC LINEAR ALGEBRA FOR TENSORS 17 LINEAR SYSTEMS; QUADRATIC FORMS 2.14 Write the following system of equations in matrix form, then solve by using the inverse matrix 3x-4y = -lH -5x + 8y = 34 The matrix form of the system is 3 -4 -5 The inverse of the 2x2 coefficient matrix is: 3 -4 -5 8 -18 34 1 Premultiplying (1) by this matrix gives χ J or χ = —2, у = 3. 24-(+20) 2 1 5/4 3/4. 8 4] Г 2 1 .5 з] 1.5/4 3/4. -18 34 (1) 2.15 If [bij] = [atJ] \ solve the η x η system for the ху in terms of the yr Multiply both sides of (7) by bki and sum on I: Ьк<у^Ък1а^х^Ь)хГхк Therefore, xf = b''yr (1) 2.16 Write the quadratic form in R4 Q — lx\ — 4хххг + Ъхххл - x\ + 10x2x4 + x\ — 6x3x4 + Ъх2л in the matrix form xTAx with A symmetric. Q = [x1 x2 x3 x4] 7 0 -4 3' 0-1 0 10 0 0 1-6 0 0 0 3. Ι^Λ-ι Λ·2 л-j *^4_| 7 0-2 3/2 0-105 -2 0 1-3 3/2 5-3 3 LINEAR TRANSFORMATIONS 2.17 Show that under a change of coordinates xt = a^x^, the quadric hypersurface ctjxtx. = 1 transforms to ctjx,jc;- = 1, where Cn = c„bribsf with (blj) = (avy1 This will be worked using matrices, from which the component form can be easily deduced. The hypersurface has the equation xTCx = 1 in unbarred coordinates, and x = Ax defines a barred coordinate system. Substituting x = Bx (B = A"1) into the equation of the quadric, we have (Bi)TC(Bx) = 1 or iTB TCBx = 1 Thus, in the barred coordinate system, the equation of the quadric is xTCx = 1. where С = BTCB.

18 BASIC LINEAR ALGEBRA FOR TENSORS [CHAP. 2 DISTANCE IN A BARRED COORDINATE SYSTEM 2.18 Calculate the coefficients gIJ in the distance formula (2.14) for the barred coordinate system in R2 defined by xi = atjxjy where an = a22 = 1, an = 0, and a2l = 2. We have merely to calculate G = (AAT) \ where A = (au): AA1 1 01Γ1 2 2 lJLo 1 1-1+0 1-2 + 0 2-1+0 2-2+1-1 By (2.11b), Thus, gu=5, g12=g2 1 2 2 5 (лл'Г = — ~~^> #22 " 1.2 5. = 1. _ ι 5-4 5 .-2 -2" 1. = 5 .-2 -2" 1. 2.19 Test the distance formula obtained in Problem 2.18 by finding the distance between the aliases of (xt) = (1, -3) and (yt) = (0, -2), which points are a distance V2 apart. The coordinates for the given points in the barred system are found to be 1 0 2 1 .;]-[.!] 1 0 2 1. Π or (*г) = (1, -1) and (y,.) = (0, -2). Using the giy calculated in Problem 2.18, d(x, y) = \/5(l ~ 0)2 - 2 ■ 2(1 -0)(-l + 2) + 1(-1 + if = V5 2.20 Prove formula (2J4). In unbarred coordinates, the distance formula has the matrix form d(x, y) = ||x - y|| = \/(х~у)Г(х~У) Now, χ = Ax or χ = Bx, where В — A"1; so we have by substitution, d(x, y) = V(5x - Byf(Bx - By) = У(В(х - y))TB(x - y) = V(x - y)TBTB(x - y) = V(x - У)г^(х ~ У) where G = ΰΓθ = (Α~λ)τΑ~λ =(A1)~1A~1 =(AAT)~\ the last two equalities following from Problem 2.13. RECTANGULAR COORDINATES 2.21 Suppose that (x ) = (x, у, ζ) and (χ') = (χ, у, ζ) (the use of superscripts here anticipates future notation) denote two rectangular coordinate systems at О and that the direction angles of the i'-axis relative to the х-, у-, and z-axes are (an β:,γ{), i = 1,2,3. Show that the correspondence between the coordinate systems is given by x=Ax, where x = (x, y, z), χ = (x, y, z), and where the matrix cos аг cos a2 cos a. cos β1 cos yx cos β2 cos γ2 cos j83 cos γ3 is orthogonal. Let the unit vectors along the x-, y-, and z-axes be i = OP, j = OQ, and к = OR, respectively (see Fig. 2-1). If χ is the position vector of any point W(x, y, z), then χ i + у j + ζ к

CHAP. 2] BASIC LINEAR ALGEBRA FOR TENSORS 19 W(x,y,z) VIEW OF AXES FOR BARRED COORDINATES Fig. 2-1 We know that the (x, y, z)-coordinates of Ρ are (cos a1? cos ft, cos уг). Similar statements hold for the coordinates of Q and R, respectively. Hence: i — (cos «j )i + (cos ft )j + (cos γ± )k j = (cos a2)i + (cos /32)j + (cos y2)k к = (cos a3)i + (cos ft)j + (cos y3)k Substituting these into the expression for χ and collecting coefficients of i, j, and k: x~(x cos «j + у cos a2 + ζ cos a3)i + (x cos ft + у cos β2 + ζ cos ft)j + (x cos yj + у cos γ2 + ζ cos γ3 )k Hence, the дг-coordinate of W is the coefficient of i, or χ — χ cos «j + у cos a2 + ζ cos a3 Similarly, у = χ cos ft + у cos ft + ζ cos ft ζ — χ cos 7j + у cos γ2 + f cos γ3 In terms of the matrix A defined above, we can write these three equations in the matrix form x = ATi (1) Now, the (i, /)-element of the matrix AAT is cos at cos a; + cos ft cos ft + cos yt cos γ for i, /= 1,2, 3. Note that the diagonal elements, (cos a,)2 + (cos ft)2 + (cosy,)2 (i =1,2,3) are the three quantities OP· OP, OQ ■ OQ, OR · OR; i.e., they are unity. If ΊΦ], then the corresponding element oi AAT\ъ either OP· OQ, OP· OR, or OQ· OR, and is therefore zero (since these vectors are mutually orthogonal). Hence, AAT = / (and also ATA = /), and, from (1), Ax = AATx = x

20 BASIC LINEAR ALGEBRA FOR TENSORS [CHAP. 2 CURVILINEAR COORDINATES 2.22 A curvilinear coordinate system (x, y) is defined in terms of rectangular coordinates (x, v) by χ = x' - xy y = xy (1) Show that in the barred coordinate system the equation of the line у = χ - 1 is у = χ - χ. [In the alibi interpretation, (i) deforms the straight line into a parabola.] It helps initially to parameterize the equation of the line as χ = t, у = t — 1. Substitution of χ = t, у = t— 1 in the change-of-coordinates formula gives the parametric equations of the line in the barred coordinate system: x = t -t(t- 1) = t у = ί(ί- 1) = t1 - t (2) Now t may be eliminated from (2) to give у = χ2 — χ. CHAIN RULE 2.23 Suppose that under a change of coordinates, xt = xi(xl, x2,. . . , xn) (1 S i S ή), the real- valued vector functions (Γ,) and (T;) are related by the formula dxr T, = Tr^ (i) Find the transformation rule for the partial derivatives of (Γ,)—that is, express the dfjdXj in terms of the dTrldxs—given that all second-order partial derivatives are zero. Begin by taking the partial derivative with respect to fj of both sides of (1), using the product rule: dt, д Эх Τ ~^ дх, Эх, I ' Эх дТг дХг д \дХг дх, дх, г дХ. У дх. By assumption, the second term on the right is zero; and, by the chain rule, dTL_dTr dxs dXj dXs dXj Consequently, the desired transformation rule is <9Г. дТг дх, дхг дх, dxs dXj дХ, Supplementary Problems 2.24 Display the matrices (a) [uiJ],5, (b) [u%5, (c) [u"]53, (d) [δ)]36. 2.25 Carry out the following matrix multiplications: И -1 2 1 -1 2 0 (b) [ 1 1 2 1

CHAP. 2] BASIC LINEAR ALGEBRA FOR TENSORS 21 2.26 Prove by the product rule and by use of the summation convention the associative law for matrices: (AB)C= A(BC) where A = (a ), В = (btj), and С = (c:j) are arbitrary matrices, but compatible for multiplication. 2.27 Prove: (a) if A and В are symmetric matrices and if А В = ΒΑ = С, then С is symmetric; (b) HA and В are skew-symmetric and if А В = ~BA = C, then С is skew-symmetric. 2.28 Prove that the product of two orthogonal matrices is orthogonal. 2.29 Evaluate the determinants И 3 -2 1 5 (b) 2 3 1 1 0 -1 -1 1 2 (c) -1 1 1 0 0 1 -1 1 1 1 -1 1 0 1 1 1 0 0 0 0 1 1 0 0 0 2.30 In the Laplace expansion of the fourth-order determinant |aj, the six-term summation e2ijkal2a2ia3Ja4k appears, (a) Write out this sum explicitly, then (b) represent it as a third-order determinant. 2.31 Prove that if a matrix has two rows the same, its determinant is zero. (Hint: First show that interchanging any two subscripts reverses the sign of the permutation symbol.) 2.32 Calculate the inverse of о [I \ Ф) 0 1 2 1 -1 1 2 0 -I 2.33 (a) Verify the following formulas for the permutation symbols eif and eijk (for distinct values of the indices only): t-ик (j-i)(k-i)(k-j) \j — i\ \k - i\ \k - j\ (b) Prove the general formula: = (i, - i\) (i3 - h) ■ ■ ■ (i„ - h) (i3 - i2) ■ ■ ■ (i„ - i2) ■ ■ ■ (i„ - i„_,) _ Ί'2 ' ■ ■ ln \ i — / / — / · · ■ 7 — 77 — 7 ■·. 7* — 7* · · * \ ί — 7* '2 ΊΙΙ'^ ΊΙ 1'и ΊΙ гз *2l l'« li\ r« ίη-\\ 2.34 Calculate the angle between the R6-vectors χ = (3, -1, 0,1, 2, -3) and у = (-2, 1, 0, 2.35 Find two linearly independent vectors in R1 which are orthogonal to the vector (3, - 2.36 Solve for χ and у by use of matrices: 3x - Ay = -23 5x +- 3y = 10 2.37 Write out the quadratic form in R3 represented by Q =xTAx, where "1 4 3" A= 4 2 0 Ъ 0 -lj Π l>~1" 1,0,0). •2,1). 2.38 Represent with a symmetric matrix A the quadratic form in R4 Q = -3*i - x\ + x\ - xvx2 - χγχ3 + 6*!*4

BASIC LINEAR ALGEBRA FOR TENSORS [CHAP. 2 Given the hyperplane crxr = 1, how do the coefficients c, transform under a change of coordinates xi = aijX/? Calculate the glJ for the distance formula (2.14) in a barred coordinate system defined by x = Ax, with Test the distance formula of Problem 2.40 on the pair of points whose unbarred coordinates are (2, -1) and (2, -4). (a) Show that for independent functions xi = xi(xl, x2, ■ ■ ■ , ■*„)> дх, Эх, (b) Take the partial derivative with respect to xk of (1) to establish the formula d2xi dxr _ dzxr dxf dxs dxkdxr dxj ~ dxsdx dxr dxk ' >

Chapter 3 General Tensors 3.1 COORDINATE TRANSFORMATIONS At this point the notation for coordinates will be changed to that usual in tensor calculus. Superscripts for Vector Components The coordinates of a point (vector) in R" will henceforth be denoted as (x1, x2, x3, . . . , x"). Thus, the familiar subscripts are now replaced by superscripts, and the upper position is no longer reserved for exponents alone. It will be clear by context whether a character represents a vector component or the power of a scalar. EXAMPLE 3.1 If a power of some vector component is to be indicated, obviously parentheses are necessary; thus, (x3)2 and (χ"-1)5 represent, respectively, the square of the third component, and the (n - l)st component raised to the fifth power, of the vector x. If и is introduced as a real number, then u2 and и3 are powers of и and not vector components. If (c)*7 appears without explanation, the parentheses indicate the use of the superscript к as an exponent and not as the index of a vector component. Rectangular Coordinates Coordinates in R" are called rectangular (also rectangular cartesian or cartesian) if they are patterned after the usual orthogonal coordinate systems of two- and three-dimensional analytic geometry. A general definition that is workable in this setting is, in effect, an assertion of the converse of the Pythagorean theorem. Definition 1: A coordinate system (x) is rectangular if the distance between two arbitrary points P(x\ x2, . . . , x") and Q(y\ y2, . . . , y") is given by pq = VO1 - ylf + (*2 - y'f + ··· + (*"- ynf Ξ \4- Δ*; Δ*7 where Ax = χ' - у'. Under orthogonal coordinate changes, which are isometric, the above formula for distance is invariant (cf. Section 2.5). Hence, all coordinate systems (xl) defined by χ = a\x\ where (a'f) is such that а[аг; = 8if, are rectangular. It can be shown that these are the only rectangular coordinate systems "whose origin coincides with that of the (x1)-system. Curvilinear Coordinates Suppose that in some region of R" two coordinate systems are defined, and that these two systems are connected by equations of the form ST : χl = x\x\ x2,. . . , хя) (lgiS/i) (3.1) where, for each i, the function, or scalar field, x'(x , χ , , . . , x") maps the given region in R" to the reals and has continuous second-partial derivatives at every point in the region (is class C2). The transformation ST, if bijective, is called a coordinate transformation, as in Section 2.6. If (x) are ordinary rectangular coordinates, the (x') are called curvilinear coordinates unless ST is linear, in which case (xl) are called affine coordinates, For convenience, the three most common curvilinear-coordinate systems are presented below. In each case, a "reverse" notation is employed: the two- or three-dimensional curvilinear system {x) is defined by the mapping ST that takes it into a rectangular system (x') of the same dimension. Polar coordinates (Fig. 3-1). Let (χ1, χ2) = (x, y) and (x1, x2) = (r, Θ), under the restriction r> 0. 23

24 GENERAL TENSORS [CHAP. 3 P{r, Θ) Fig. 3-1 3 : X = Г COS θ у = r sin θ Then, (-1 1 2 Χ = X COS Χ -2 1 · 2 χ = χ sin χ χ' = V(*-1)2 + (*2)2 У = tan"1(f2/f1) (3.2) (The inverse given here is, in the equation for x~, valid only in the first and fourth quadrants of the XjXj-plane; other solutions must be used over the other two quadrants. Likewise for the 0-coordinate in the cylindrical and spherical systems.) Cylindrical coordinates (Fig. 3-2). If (jc1, χ2, χ ) = (x, y, z) and (jc\ x2, x3) = (r, 0, z), where r > 0, r -1 1 2 X = X COS X -2 1 · 2 χ = χ sin χ -3 _ 3 Э"1 У = V(fx)2 + (*2)2 x2 = tan-1 (i2/*1) У = *э (3.3) Spherical coordinates (Fig. 3-3). If (jc15 jc2, jc3) = (x, y, z) and (χ , χ , χ ) = (ρ, φ, 0), where ρ > 0 and Ο^φ ё π, x1 = V(^1)2 + (^2)2 + (^3)2 χ2 = cos-1 (хъ1\[{х1)2 + (χ2)2 + (χ3)2) Ρ·4) jc^tan-1^2/*1) (Caution: In an older but still common notation for spherical coordinates, 0 denotes the polar angle and φ the equatorial angle.) -1 1-2 3 x = χ sin χ cos χ -2 1 . 2 . 3 χ = χ sin χ sin χ -3 1 2 X = X COS X ST'1 : \ (r, Θ) - polar coordinates for Q in jry-plane 3 : * = r cos 0 у = г sin 0 г = г Fig. 3-2

CHAP. 3] GENERAL TENSORS 25 3 . rx = ρ sin φ cos θ ! у = ρ sin ψ sin θ Сг = ρ cos φ (г, 9)—polar coordinates for Q in xy-plane Fig. 3-3 The Jacobian The n2 first-order partial derivatives dx'ldx1 arising from (3.1) are normally arranged in an η x η matrix, (3.5) dX1 дХ1 дх' дХ1 дХ дХ1 дХ дХ' дХ' дХ' б?* б?*" б?*1 б?*" б?Х" 1 " дХ дх" Matrix / is the Jacobian matrix, and its determinant $ =det / is the Jacobian, of the transformation ST. EXAMPLE 3.2 In R2 let a curvilinear coordinate system (*') be defined from rectangular coordinates (xl) by the equations or . Ιχ' = χ'χ2 J ■ \x-2 = (xy Since дх^дх1 = χ2, дхх1дх2 = х\ дх21дхх = 0, and дх21дх2 = 2x2, the Jacobian of ST is ,1 X X' 0 2χΊ = 2(хГ A well-known theorem from analysis states that ?T is locally bijective on an open set Ш in R" if and only if $ Φ 0 at each point of 4. When β Φ 0 in 4 and 2Г is class С2 in %, then (3.1) is termed an admissible change of coordinates for Ш. EXAMPLE 3.3 The curvilinear coordinates of Example 3.2 are admissible for the regions x2 > 0 and x2 < 0 (both open sets in the plane). See Problem 3.1. In an admissible change of coordinates, the inverse transformation 5"1 (the local existence of which is guaranteed by the theorem mentioned above) is also class C2, on Ш, the image of % under ST. Moreover, if ST"1 has the form 5"-1 : x' = x'(x\x, , ..,xn) (li(gn) (3.6)

26 GENERAL TENSORS [CHAP. 3 on 4, the Jacobian matrix J of ST 1 is the inverse of /. Thus, JJ = JJ = I, or дхг дх' дх дх' ' У [cf. Problem 2.42(a)]. It also follows that β = 11β. General Coordinate Systems In later developments it will be necessary to adopt coordinate systems that are not tied to rectangular coordinates in any way [via (3.1)] and to define distance in terms of an arc-length formula for arbitrary curves, with points represented abstractly by и-tuples (x1, x~, . . . , x"). Each such distance functional or metric will be invariant under admissible changes of coordinates, and admissible coordinate systems will exist for each separate metric. Under such metrics, R" will generally become non-Euclidean; e.g., the angle sum of a triangle will not invariably equal тт. Although the curvilinear coordinate systems presented above are explicitly associated with the Euclidean metric (since they are connected via (3.1) with rectangular coordinates and Euclidean space), those same systems could be formally adopted in a non-Euclidean space if some purpose were served by doing so. The point to be made is that the space metric and the coordinate system used to describe that metric are completely independent of each other, except in the single instance of rectangular coordinates, whose very definition (see Definition 1) involves the Euclidean metric. Usefulness of Coordinate Changes A primary concern in studying tensor analysis is the manner in which a change of coordinates affects the way geometrical objects or physical laws are described. For example, in rectangular coordinates the equation of a circle of radius a centered at the origin is quadratic, /-K2 , / -2ч2 _ 2 (x ) + (x ) = a but in polar coordinates, (3.2), that same circle has the simple linear equation xl = a. The reader is no doubt familiar with the sometimes dramatic change that takes place in a differential equation under a change of variables, which is nothing but a change of coordinates. This idea of changing the description of phenomena by changing coordinate systems lies at the heart of not only what a tensor means, but how it is used in practice. 3.2 FIRST-ORDER TENSORS Consider a vector field V=(V) defined on some subset У of R" [that is, for each i, the component V = V'(x) is a scalar field (real-valued function) as χ varies over У]. In each admissible coordinate system of a region ^L containing У, let the η components V1, V2, . . . , V" of V be expressible as η real-valued functions; say, as Γ1, T~, . . . , T" in the (xl)-system and f\ f2, . . . , f" in the (f')-system where (x) and (x1) are related by (3.1) and (3.6). Definition 2: The vector field V is a contravariant tensor of order one (or contravariant vector) provided its components (T') and (T") relative to the respective coordinate systems (x) and (£') obey the law of transformation τ — i contravariant vector T' = Tr ——r (lgi^n) (3-8)

CHAP. 3] GENERAL TENSORS 27 EXAMPLE 3.4 Let ^ be a curve given parametrically in the (x')-system by x' = x'{t) (aStSb) The tangent vector field Τ = (Tl) is defined by the usual differentiation formula Г' = — dt Under a change of coordinates (3.1), the same curve is given in the (f)-system by Г = x\t) = x\x\t), x\t)> · ■ ■ > *"(0) (aStSb) and the tangent vector for % in the (i')-system has components dx' But, by the chain rule, T' . dt dx dx1 dxr -, dx' ~r = τ~τ —r °r Τ = Τ -—-r at dx at dx proving that Τ is a contra variant vector. (Note that because Τ is denned only on the curve "#, we have Sf = <€ for this particular vector field.) We conclude in general that under a change of coordinates, the tangent vector of a smooth curve transforms as a contravariant tensor of order one. Remark 1: In some treatments of the subject, tensors are denned to possess certain weights, with (3.8) replaced by π —i weighted contravariant vector Tl — wTr ——r (l^jgn) (3-9) for some real-valued function w (the "weight of T")· In framing the next definition we (arbitrarily) shift to a subscript notation for the components of the vector field. Definition 3: The vector field V is a covariant tensor of order one (or covariant vector) provided its components (Τέ) and (Tt) relative to an arbitrary pair of coordinate systems (x') and (x'), respectively, obey the law of transformation π r covariant vector ft = Tr —— (1 Sr i: ^ n) (3.10) dX EXAMPLE 3.5 Let F(x) denote a differentiable scalar field defined in a coordinate system (x') of R". The gradient of F is defined as the vector field /^F dF_ £F_ ~ W ' dx1 '" ' -' dx'1 In a barred coordinate system, the gradient is given by VF= (dFldx), where F(x) = F°x(x). The chain rule for partial derivatives, together with the functional relations (3.6), gives dF _ 3F dxr dx1 ~ Sxr dx1 which is just (3.10) for Tt = dFldx1, Ti = dP/dx'. Thus, the gradient of an arbitrary differentiable function is a covariant vector. Remark 2: Tangent vectors and gradient vectors are really two different kinds of vectors. Tensor calculus is vitally concerned with the distinction between contravariance and covariance, and consistently employs upper indices to indicate the one and lower indices to indicate the other.

28 GENERAL TENSORS [CHAP. 3 Remark 3: From this point on, we shall frequently refer to first-order tensors, contravariant or covariant as the case may be, simply as "vectors"; they are, of course, actually vector fields, defined on R". This usage will coexist with our earlier employment of "vectors" to denote real и-tuples; i.e., elements of R". There is no conflict here insofar as the и-tuples make up the vector field corresponding to the identity mapping V'(x) = xl (i = 1, 2, ... , ή). But the vector (x) does not enjoy the transformation property of a tensor; so, to emphasize that fact, we shall sometimes refer to it as a position vector. 3.3 INVARIANTS Objects, functions, equations, or formulas that are independent of the coordinate system used to express them have intrinsic value and are of fundamental significance; they are called invariants. Roughly speaking, the product of a contravariant vector and a covariant vector always is an invariant. The following is a more precise statement of this fact. Theorem 3.1: Let S' and Ti be the components of a contravariant and covariant vector, respectively. If the inner product E = SrTr is defined in each coordinate system, then Ε is an invariant. EXAMPLE 3.6 In Examples 3.4 and 3.5 it was established that the tangent vector, (5") = (dx'/dt), to a curve 4i and the gradient of a function, (Tt~) = (dFldx'), are contravariant and covariant vectors, respectively. Let us verify Theorem 3.1 for these two vectors. Define ^ „rrr dF dxr Now, by the chain rule, dt so the assertion of Theorem 3.1 is that the value of dt И01 is independent of the particular coordinate system (x1) used to specify the curve. To visualize this, the reader should study Fig. 3-4, which shows how the composition F= F°(x'(t)) works out in R3. It is apparent here that the map F entirely bypasses the coordinate system (x1, x2, хъ). Thus, F— and with it, dF/dt— is an invariant with respect to coordinate changes. REAL LINE Fig. 3-4

CHAP. 3] GENERAL TENSORS 29 3.4 HIGHER-ORDER TENSORS Tensors of arbitrary order may be defined. Although most work does not involve tensors of order greater than 4, the general definition will be included here for completeness. We begin with the three types of second-order tensors. Second-Order Tensors Let V= (V) denote a matrix field; that is, (V) is an η x η matrix of scalar fields V'(x), all defined over the same region ^ = {x} in R". As before, it will be assumed that V has a representation (T'!) in (xl) and (V) in (£'), where (xl) and (£') are admissible coordinates related by (3.1) and (3.6). Definition 4: The matrix field V is a contravariant tensor of order two if its components (T'1) in (x1) and (T4) in (xl) obey the law of transformation contravariant tensor T1' = T" ——r ——s (1 = i, /= η) (3.11) Again going over to subscript notation for the components of the matrix field, we state Definition 5: The matrix field V is a covariant tensor of order two if its components (Ttj) in (xl) and (Ttj) in (xl) obey the law of transformation covariant tensor Τι;— Τ —— —— (1 = г, / 2= и) (3.12) ' r дх дх' Theorem 3.2: Suppose that (Tif) is a covariant tensor of order two. If the matrix [Г,.-]ии is invertible on °и, with inverse matrix [Г';]ии, then (T1') is a contravariant tensor of order two. Definition 6: The matrix field V is a mixed tensor of order two, contravariant of order one and covariant of order one, if its components (Т\) in (x!) and (T^) in (x1) obey the law of transformation mixed tensor f'; = T's ——r -^ (12= г, / ^ n) (3.13) OX· CfJC Tensors of Arbitrary Order Vector and matrix fields are inadequate for higher-order tensors. It is necessary to introduce a generalized vector field V, which is an ordered array of nm (m=p + q) scalar fields, (V/^'.'.'J ), defined over a region % in R"; let (T'^j'"'pj ) denote the set of component-functions in various coordinate systems which are defined on °U. Definition 7: The generalized vector field V is a tensor of order m- ρ + q, contravariant of order ρ and covariant of order q, if its components (T^j'"J ) in (xl) and (f-j'" ^ ) in (x1) obey the law of transformation ,, τ'Ί'2 ■■·',. Tr,r,...r dXh дх'2 дХ~1Р дх' дХ* dXS* ,„l/f4 general tensor Τ , ρ·. = Τ '/ / —-r у- ■ ■ = : г ■ ■ ■ —г- (3 14) S hl2...,q slS2...Sq дхг, дхг2 дхгр d-h d-h d-,q \J-it) with the obvious range for free indices. 3.5 THE STRESS TENSOR It was the concept of stress in mechanics that originally led to the invention of tensors (tenseur, that which exerts tension, stress). Suppose that the unit cube is in equilibrium under forces applied to three of its faces [Fig. 3-5(a)]. Since each face has unit area, each force vector represents the force

30 GENERAL TENSORS [CHAP. 3 FACE 3 σ31 σ33 FACE V σ32 σ21 'a12 - FACE 2 (A) Fig. 3-5 Φ) per unit area, or stress. Those forces are represented in the component form in Fig. 3-5(b). Using the standard basis e15 e2, e2, we have Is σ e„ 2.5 σ e. σ e. (stress on face 1) (stress on face 2) (stress on face 3) (3.15) Stress on a Cube Section The question arises: What stress F is transmitted to a planar cross section of the cube that has unit normal n? To answer this, refer to Fig. 3-6, which shows the tetrahedron formed by the cross section and the coordinate planes. Let A be the cross-sectional area. By the assumed equilibrium of the cube, the stresses on the хгх2-, xV-, and jt2jt3-bases of the tetrahedron are -v3, -v2, and -v15 respectively, as shown componentwise in Fig. 3-6. Hence, the forces on these same bases are .B^-Vj), Z?2(-v2), and β3(-ν1), respectively. For the tetrahedron itself to be in equilibrium, the resultant force on it must vanish: AF + Вг(- v3) + B2(~v2) + β3(-νι) = 0 or, solving for F, β, В, г=^ + ^+:Ь (3.16) But B3 is the projection of A in the x'x3-p\ane: B3 = Лпех or B3/A = nel. Similarly, B2IA = ne2 and Bl/A-ne3. Substituting these expressions and the expressions (3.15) into (3.16), we find that F = <r>erK (3.17) Contravariance of Stress Under Change of Coordinates An interesting formula results from (3.17) when we change the basis of R3 by a transformation of the form e, = α{ί. (with \a[\ #0). In terms of coordinates,

CHAP. 3] GENERAL TENSORS 31 Fig. 3-6 That is, we have a new coordinate system (*') that is related to (xl) via Note that here we have (3.18) dx! dx Substituting er = a'rfi into (3.17) yields the stress components (σ';) in the new coordinate system, as follows: with F = a»[n(a%)]{a>,ts) = а"а^аЦпЩ - ^(nf,)f;. дХ дх' дХГ dXS A comparison of (3.19) with the transformation law (3.11) leads to the conclusion that the stress components σ1' define a second-order contravariant tensor, at least for linear coordinate changes. -it rs ι rt} r, σ' — σ aras = σ (3.19) 3.6 CARTESIAN TENSORS Tensors corresponding to admissible linear coordinate changes, 3~ : χ - а';х' (\а'^0), are called affine tensors. If (a1.) is orthogonal (and ST is distance-preserving), the corresponding tensors are cartesian tensors. Now, an object that is a tensor with respect to all one-one linear transformations is necessarily a tensor with respect to all orthogonal linear transformations, but the converse is not true. Hence, affine tensors are special cartesian tensors. Likewise, affine invariants are particular cartesian invariants. Affine Tensors A transformation of the form ST : χ = α^χ' (|α'.| τ^Ο) takes a rectangular coordinate system (xl) into a system (xl) having oblique axes; thus affine tensors are defined on the class of all such

32 GENERAL TENSORS [CHAP. 3 oblique coordinate systems. Since the Jacobian matrices of ST and ST 1 are the transformation laws for affine tensors are: contravariant f' = a'rTr, f'' = a'ra'sT", f''k = a'ra'saktTrs\ ... covariant f, = b[Tr, f,; = b\ b) Trs, ^ fijk = b- b) b'k Trst, ... (3.21) mixed T\ = a\b) Ts, T'jk = a'rbsfb'kTrsn . . . Under the less stringent conditions (3.21), more objects can qualify as tensors than before; for instance, an ordinary position vector χ = (У) becomes an (affine) tensor (see Problem 3.9), and the partial derivatives of a tensor define an (affine) tensor (as implied by Problem 2.23). Cartesian Tensors When the above linear transformation ST is restricted to be orthogonal, then J'1 = JT, or b) = ai (l%i,j^n) so that the transformation laws for cartesian tensors are, from (3.21), contravariant T' = a\ Tr, f1' = dra's Trs, . . . covariant tt - a'rTr, Ttj = a\a's Trs, ... mixed f'j = dra'sTrs, A striking feature of these forms is that contravariant and covariant behaviors do not distinguish themselves. Consequently, all cartesian tensors are notated the same way—with subscripts: (3.22) Because an orthogonal transformation takes one rectangular coordinate system into another (having the same origin), cartesian tensors appertain to the rectangular (cartesian) coordinate systems. There are, of course, even more cartesian tensors than affine tensors. Note that JJT = / implies /2 = 1, or β = ±1. Objects that obey the tensor laws (3.22) when the allowable coordinate changes are such that allowable coordinate changes cartesian tensor laws xl = aijxj or xi = ajixj Ti = airTr, Т^ = а1га^Т„, k,.| = +i are called direct cartesian tensors. Solved Problems CHANGE OF COORDINATES 3.1 For the transformation of Example 3.2, (a) obtain the equations for 3"1; (b) compute /from (a), and compare with J"1. (a) Solving χ1 = xlx2, x2 = (x2)2 for x1 and x2, we find that \xl=xlx2 „__, \x 1---1--2 i^-jcVVF *2 = (*2)2 σ : U2 = VF {1)

CHAP. 3] GENERAL TENSORS 33 is a one-one mapping between the regions x2 > 0 and x2 > 0, and that = -хгЫТ χ1 = x'x2 (χΎ гг~г : -VF (2) is one-one between x2 < 0 and x2 > 0. Note that the two regions of the j'^-plane are separated by the line on which the Jacobian of 2F vanishes. (b) From Example 3.2, χ2 χ 0 2x Ά and so 1 2x2 0 "/] 2(x2)2 valid in both regions x2 > 0 and x2 < 0. Now, on *2 > 0, differentiation of the inverse transformation (1), followed by a change back to unbarred coordinates, yields J = "лс1 dx2 .dx1 Эх1 dx2 dx2 dx2. = (x2)-1'2 -, 0 χχ i(x ,χ2)'3'2 2 ч-1/2 = (χ2)-1 - 0 seen that on x2 > 0, / = J ' \ Similarly, from (2), with *2 <0, J= ~-(χ2) 0 -1/2 -К*2)'1'2. = -+(x2)- 0 -\x\x2T2' + HX2)-1 . -\x\x2T2 Ηχ2)-1 = j~l 3.2 For polar coordinates as defined by (3.2), (a) calculate the Jacobian matrix of 3~ and infer the region over which if is bijective; (b) calculate the Jacobian matrix of Э"1 for the region {(r, 0) | r>0, -я72<0<я72} i.e., the right half-plane, and verify that it is the inverse of the matrix of (a). («) ^ 1 2^1 2 г (χ COS Χ ) 5 (д: COS * ) ί*1 ν дх2У ' д ι 1 · 2ч <? / 1 · 2ч —7 (х sin д: ) —τ (* sin д: ) L<9* Лс . COS*2 -XlUnX2 Sin *2 Я1 COS Χ2 whence $-xx = r. Therefore, 2Г is bijective on the open set r>0, which is the entire plane punctured at the origin. (b) For ST'1 we have, over the right half-plane, dx_ 1-1 dx1 dX1 V(*')2 + (*2)2 *** V(^)2 + (^2)2 *Г if1 1 + (i2/*1)2 (*')2J (хГ + (Я2) dx' x1 dx2 (xlf + (x2)2 and so V(*-1)2 + (*-2)2 V(*_1)2 + (*2)2 -*2 (хУ + (хУ (хг)2 + {х2Г cos *2 sin x2 sin X2 COS X2 Now compute J 1: J-1 χ cos*2 x1 sin x2 - sin χ cos дг cos χ sin jc sin x2 cos jc2

34 GENERAL TENSORS [CHAP. 3 CONTRAVARIANT VECTORS 3.3 If V= (Γ') is a contravariant vector, show that the partial derivatives V- = dT'ldx', defined in each coordinate system, transform according to the rule Differentiate both sides of -,- „.,. дх' dXS m, д х' dXS Τ = Τ + Τ : ' s дх дх' dxrdxs эх' дх with respect to x', using the product rule if ' die' ■ дх' V дх дх' дТ дх г д (дх' By the chain rule for partial derivatives, (2.15), and dV __ dV dxs __ dxs дх' ~ dxs дх' ~ s дх' Substituting these expressions into (1) yields the desired formula. дх' дХг Ж' \дХг/ д (Sx'\_ die' \dxT)~~ д (дх'Х дх \дхг). дх дх (1) 3.4 Suppose that (Г!) is a contravariant vector on R" and that (Гг) = (χ', χ1) in the (x')-system. Calculate (T") in the (x')-system, under the change of coordinates -2 1 2 x —xx By definition of contra variance, дх дх дХ Note that the top row of the Jacobian matrix J is needed for the case i = 1, and the bottom row is needed for i =2. die1 дх дх dxi die1' dX2 die ~. 2 dX . "0 2 .X 2x2] xl . Thus, f1 = T1(0)+ 7'2(2jc2) = 2jcV which, in terms of barred coordinates, are :Τ=2ί2 f2 = T\x2) + TV) = (x2)2 + (x1)2 iW + Щ- 3.5 Show that a contravariant vector can be constructed the components of which take on a given set of values (a, b, c, . . .) in some particular coordinate system. (The prescribed values may be point functions.) Let (a, b, c, . . .) = (a) be the given values to be assigned in the coordinate system (x). Set V = a for the values in (У), and for any other admissible coordinate system (f), set V' = a'(dx'ldx). To show that (V) is a contravariant tensor, let (>'') and (yl) be any two admissible coordinate systems. Then, у =/'(x1, a2, .... x") and y' = g'(x\ x2, . . . , x"), and, by definition, the values of (V") in (y') and (y') are, respectively, V = ar(dy'ldxr) and f" = a(dyl!dxr). But, by the chain rule, Г = а dxr dy' dys ду" дх df_ df QED

CHAP. 3] GENERAL TENSORS 35 COVARIANT VECTORS 3.6 Calculate (T;) in the (*')-system if V= (Γ;) = (*2, χ1 +2x2) is a covariant vector under the coordinate transformation of Problem 3.4. To avoid radicals, compute J~l in terms of {x')\ Jl = -χ ι 2(x2)2 x2 h » By covariance, =, „ dX ^ OX ^ dx τ, = τ, — = т. — + тг — dx 1 дх г dx' 0' = 1,2) For Ί = 1, read off the partials from the first column of J \- f, = Tl(~xll2{x2)2) + T2(l/2x2) = ~хЧ2х2 + xV2x2 + 1 = 1 Similarly, for i = 2, use the second column of /_1: f2 = Тг(\1х2) + Г2(0) = x\llx2) = 1 Hence, (Г;) = (1,1) at all points in the (F)-system (x1 = 0 excluded). 3.7 Use the fact that V/ is a covariant vector (Example 3.5) to bring the partial differential equation df Bf χ — = у — dx y dy into simpler form by the change of variables x = xy, y- {yf; then solve. Write V/= (df/dx, df/dy)^(T.), (x\ x2) = (x, y), (x\ χ2) = (*, у), and f- Again calculating J first, then its inverse, we have JL = t — dXl ' dx' dx1 dx' = 7" у χ 0 2y 1 -x У 2(у)2 о I 2У . so that dx dXr die1 -L шТ = Τ ζ^— = Τ · - + Τ ·0 = liL У dx 1 df _ rf, _ T dxr _ —X ~dj~ г~ г~д?~ Г 2{yf + 2 2y 2(y)2 dx ' 2y iy χ df + ± d£ But, by (1), */ ίνϊ2 V ■*^ + y^-o 0) <?y 2(y)2 \ <?* ' <9y/ which implies that /= F(x), a function of χ alone; therefore, f—F(xy) is the general solution to (1).

36 GENERAL TENSORS [CHAP. 3 INVARIANTS 3.8 Prove Theorem 3.1. We must show that if (5") and (Γ,.) are tensors of the indicated types and order, then the quantity E^S'Ti is invariant with respect to coordinate changes; that is, Ё = Е, where Ε ~ S'fr But observe that S'^S r Эх' dXr and dx dx' so that, in view of (3.7), 3.9 Show that under linear coordinate changes of R", x' = a'jx' (|a^.|#0), the equation of a hyperplane Atx' = 1 is invariant provided the normal vector (At) is covariant. In view of Theorem 3.1, it suffices to show that (V) = (x') is a contravariant affine tensor. But this is immediate: Γι —j / / ι τ ι s χ = α,χ= a, I which is the transformation law (3.21). SECOND-ORDER CONTRAVARIANT TENSORS 3.10 Suppose that the components of a contravariant tensor Τ of order 2 in a coordinate system (*') of R2 are Tu = 1, T12 = 1, T21 = -1, and T22 = 2. (a) Find the components fiJ of Τ in the (*')-system, connected to the (x')-system via -2 1 2 X = X X (b) Compute the values of the f'1 at the point which corresponds to i1 = 1, j2 For economy of effort, the problem will be worked using matrices, (a) Writing dx dx' we have from (2.1b), t" =-T"—r —s = J1 Tn J'k Эх Эх ; That is, Τ = JTJ' = "2*1 0" 2 1 . X X . " 1 Γ .-1 2_ "2X1 . 0 (b) At the point (1, -2), fn=4(l) f2I=2(l] 2 = 4 (-2) -2 ;i)2 = x2' x\ = 4(x1)2 2^*2 + 2(*1)2' _2л:'л:2-2(л:1)2 2(jc')z + (χ2)2 _ = ~6 7'lz=2(l)(~2) + 2(l)z = ~2 f22 = 2(l)2 + (~2)z = 6 -2. 3.11 Show that if (5'') and (Γ'') are contravariant vectors on R", the matrix [U''] = [5,'r/J„„, defined in this manner for all coordinate systems, represents a contravariant tensor of order 2.

CHAP. 3] GENERAL TENSORS 37 Multiply s' = sr дх' dXr and OX dXS to obtain Эх Эх Эх Эх which is the tensor law. (The notion of the "outer product" of two tensors will be further developed in Chapter 4.) SECOND-ORDER COVARIANT TENSORS 3.12 Show that if Tl are the components of covariant vector T, then Si;. = Г,Т;- T.T, are the components of a skew-symmetric covariant tensor S. The skew-symmetry is obvious. From the transformation law for T, ,=,,=, - ,=, „ dxr „ dxs „ dxs „ dxr T.T. - Τ Τ = Τ : · Τ : - Τ 'Τ : ' дх 5 дх' s дх' ' дх' „, „, дхг дх" ^ ^ дхг дх" /rr, ^ ^ ^ „ дх' dxs = Γ Τ : ~ Τ Τ : ;= ΓΓ - Τ Τ ) : : ' s дх дх' s " дх дх' У r s s r) дх дх' or S = s — — "' " дх' dxJ which establishes the covariant tensor character of S. 3.13 If a symmetric array (Г,-) transforms according to - _ дх_ дх?_ дх_ дх_ Ц " dXS дх} дХ1 дХк show that it defines a second-order covariant tensor. TI; = Τ Л ι дх' дхк\ дх" дх' "\дхк дх" I дх1 дх дх*_ дх!_ _ дх' дх" " дх' дх' ,s JI' Jx4 \дХ^дХ^__т дх5 дх' I дх1 дх' ~ " s дх1 дх' 3.14 Let U = ([/,„) be a covariant tensor of order 2. Under the same coordinate change as in Problem 3.10, (a) calculate the components Όψ if Uu = x2, i/]2 = U2l = 0, U22 = x1; (b) verify that the quantity T''~UV] = Ε is an invariant, where the T'J and f"' are obtained from Problem 3.10. (a) In terms of the inverse Jacobian matrix, the covariant transformation law is Substituting UiS дх' dxs dxl " дх' 2X1 0 Lx2 x1 J'U JS = J"U J" ι rs j r rs I 2хг 2{x')2 or U- U = JTUJ x1 0 0 xx

38 GENERAL TENSORS [CHAP. 3 we find U- 1 2xl 0 x2 2(xlf 1 1 χ x2 0 0 xx 2xl 1 L 2{хУ x1 xxx2 + (x2f 4{xlf 2{x1)2 V)2 from which the Utj may be read off. (b) Continuing in the matrix approach, we note that Ε is the trace (sum of diagonal elements) of the matrix TUr. TU' = Ε = χ2 + 2xl 1 1 1 2. χ2 0 0 x1 and TUT- 4(У)2 2х1х2 + 2(х1)2 2x1x2~2(x1)2 2(x1f + (x2)2 2 1 X X _—x2 2x\ x'x' + jx2)2 2 2(хУ 2(x1f 1_ xx 0 1 2 3jc 2л:1 jc2 + 2xl E = x2 + 2xl = E 3.15 Prove Theorem 3.2. Observe first of all that if a covariant matrix (second-order tensor) U has inverse V in unbarred coordinates, then Ό has inverse V in barred coordinates; i.e., (U)'1 = U~l. Now, by Problem 3.14(a), U = /Γί/7 Inverting both sides of this matrix equation, applying Problem 2.13, and recalling that J J = /, we obtain W1 = j-'u~\jTy" = juxjt which is the contravariant law for U~l [see Problem 3.10(a)]. MIXED TENSORS 3.16 Compute the formulas for the tensor components (f'j) in polar coordinates in terms of (7^.) in rectangular coordinates, if the tensor is symmetric in rectangular coordinates. (In contrast to Section 3.1, it is now the curvilinear coordinates that are barred.) The general formula calls for the calculations f^Tr^-^- = — T — ' s dxr dxj 3xr s Эх1 Using (2.1b), this may be written in matrix form as where T= [7^]22 and where ; Г cos θ —rsinO I sin θ r cos θ . (Τ', = Τ>) (1)

CHAP. 3] GENERAL TENSORS 39 is the Jacobian matrix of the transformation from (r, 0) to (x, y). Thus, cos 0 sin 0 sin 0 cos 0 r r cos 0 sin 0 sin 0 cos 0 1 1 * 2 Tl Tl cos θ — r sin θ sin θ r cos θ Γ] cos θ + 7^ sin θ ~ϊΤ\ύηθ + rTlcosO' Tl cos θ + Tl sin θ - КГ' sin0 + КГ2 cos 0 The final matrix multiplication can be carried out routinely, simplifying by means of trigonometric identities: T\ cos2 0 + T\ sin 20 + T\ sin2 θ - ^ T\ sin 20 + /T' cos 20 + *- T\ sin 20 , j sin 20 j cos 20 2r + T + Ti sin 20 "27" Γ J sin2 0 - Tl sin 20 + T\ cos2 0 Observe that Τ does not share the symmetry of T: T\ = r 2T\. 3.17 Prove that the determinant of a mixed tensor of order two is invariant. By (1) of Problem 3.16, we have—whether or not Τ is symmetric— \Τ\ = \πΓ4 = \ι\\τ\\Γι\=β\τ\β-ι = \τ\ GENERAL TENSORS 3.18 Display the transformation law for a third-order tensor that is contravariant of order two and covariant of order one. Take ρ — 2 and q = 1 in Definition 7 and, to avoid unnecessary subscripts, write i, j, k, r, s, t in place of ι\, i2, jx, rit r2,si. Then (3.14) gives fy = Trs d* d*' 3x' k ' дх' дх5 dxk 3.19 Let Τ = (Т'[1т) denote a tensor of the order and type indicated by the indices. Prove that S = (Tk) = (T'lij) is a covariant vector. The transformation law (3.14) for Τ is ТУ 1 Mm дх1 дх' дх дх" дх" Jxr эх* Jx7* Ί>¥1 JF Set I = i, m—j and sum: дх1 дх"\! дх> дх"\ дх f = T> -T" — — — — — = Τ k" '"» дхт dxs дх" Эх1 дх' ,ии\дхт dx')\dxs dxj! дхк дх mv°r°s J^K дх' дхк дх' CARTESIAN TENSORS 3.20 Show that the permutation symbol (et-) defines a direct cartesian tensor over R2. Assume that etj is defined -the same way for all rectangular coordinate systems. If the coordinate change is f,- = ai;*,, where (al/)T(akl) = (δρ9) and

40 GENERAL TENSORS [CHAP. 3 we must establish the cartesian tensor law (3.22): ё„ = ersairais (n = 2) We examine separately the four possible cases: i=j=l ersawau = flnfl12 - β12βη = 0 = ёи i= 1, j = 2 ersaua2s = a11a22 - a12a21 = 1 = e12 i = 2,j=l e„a2rau = fl2,fl,2- fl22fln = -1 = ё21 i=j = 2 ersa2ra2s = a21a22 - a22a21 = 0 = <F22 3.21 Prove that (a) the coefficients ci;- of the quadratic form c-x'x' = 1 transform as an affine tensor and (b) the trace cH of (c;/) is a cartesian invariant. (a) If i' = α'μ' and x' = b^x', where (b1-) = (fl^)-1, the quadratic form goes over into 1 = сц(ЬХ)(Ь',х')1сг,х'х' with crs = b'rb'sctj. But this formula is just (3.21) for a covariant affine tensor of order two. (b) Assuming an orthogonal transformation, (b') = (fl,)T, we have — lis Hence, c,r = (b>y)ciy = 5^.· 3.22 Establish the identity between the permutation symbol and the Kronecker delta: eri!erkI - SikSjt - 8u8jk (3.23) The identity implies η - 3, so that there are potentially 34 - 8f separate cases to consider. However, this number can be quickly reduced to only 4 cases by the following reasoning: If either i = / or к = I, then both sides vanish. For example, if i =/', then on the left erij = 0, and on the right, Hence, we need only consider the cases in which both i¥-j and к Φ I. Upon writing out the sum on the left, two of the terms drop out, since 1Фу. eiijeikl """ e2ije2kl """ e3ije3kl ~ eV2'3,ei,kl V ~ ^ > / — ^ ) where (l'2'З') denotes some permutation of (123). Thus, there are left only two cases, each with two subcases. Case 1: e1,2,yerkl Φ 0 (with i = 2', / = 3'). Here, either k-2' and 1 = 3' or к = 3' and I = 2'. If the former, then the left member of (3.23) is +1, while the right member equals δ2.2'δ3.3. — δ2,3,δ3.2. = 1—0=1 If the latter, then both members equal -1, as can be easily verified. Case 2: er2,yerkl = 0 (with i = 2', j ~ 3'). Since к Φ /, either к = Г от 1= V. If Л = Г, then the right member of (3.23) equals ^Z'l'^3'/ ~~ ^2'f^3'l' = 0 — 0 = 0 If / = 1', we have d2.kdyi, - d2i,dyk = 0-0 = 0. This completes the examination of all cases, and the identity is established.

CHAP. 3] GENERAL TENSORS 41 Supplementary Problems 3.23 Suppose that the following transformation connects the (x:) and (x:) coordinate systems: . \xl = exp(x1 +x2) •J ■ 1 -2 /1 2ч [x — exp (x - χ ) (a) Calculate the Jacobian matrix J and the Jacobian β. Show that j? Φ 0 over all of R2. (b) Give equations for 3T~ \ (c) Calculate the Jacobian matrix J of ST 1 and compare with У \ 3.24 Prove that if (Г,) defines a covariant vector, and if the components Sif = Γ,Γ. + Γ;Γ, are defined in each coordinate system, then (S0) is a symmetric covariant tensor. (Compare Problem 3.12.) 3.25 Prove that if (Г;) defines a covariant vector and, in each coordinate system, we define Эх' όχ' ~ '' then (Γ ) is a skew-symmetric covariant tensor of the second order. [Hint: Model the proof on РгоЫеп/з.З.] 3.26 Convert the partial differential equation to polar form (making use of the fact that 3.27 Show that the quadratic form Q = g^x'x' is an affine invariant provided (g4) is a covariant affine tensor. [Converse of Problem 3.21(a).] 3.28 Prove that the partial derivatives of a contravariant vector (Tl) define a mixed affine tensor of order two. [Hint. Compare Problem 2.23.] 3.29 Prove that the Kronecker delta (δ^), uniformly defined in all coordinate systems, is a mixed tensor of order two. 3.30 Show that the permutation symbol (e;y) of order two, uniformly defined in all coordinate systems, is not—Problem 3.20 notwithstanding—covariant under arbitrary coordinate changes. [Hint: Use x1 = xlx2, χ- = χ2, at the point (*'') = (1,2).] 3.31 By use of (3.23), establish the familiar identity for the vector product of three vectors, u x (v x w) = (uw)v — (uv)w or, in coordinate form, eiik^j{ekrsvrws) = (μ,η',-Κ - (и^К 3.32 (α) Show that if (Tj) is a mixed tensor, then (Γ* + T[) is not generally a tensor, (b) Show that a mixed tensor of order two, symmetric in a given coordinate system, will transform as a symmetric tensor if the Jacobian matrix is orthogonal. 3.33 Prove: (a) If (T'j) is a mixed tensor of order two, T\ is an invariant; (b) if (S'lk) and (T') are tensors of the type and order indicated, SrjrT' is an invariant. df df дх ду V/is a covariant vector), and solve for f(x, y).

GENERAL TENSORS [CHAP. 3 If Тг(Т^) is a tensor, contravariant of order 3 and covariant of order 2, show that S = (T'lk) is a contravariant vector. Show that the derivative, dTldt, of the tangent vector T= (Tl) = (dx'/dt) to a curve x'! = x'(t) is a contravariant affine tensor. Is it a cartesian tensor? (a) Use the theory of tensors to prove that the scalar product uv = uivi of two vectors u = (и.) and v= (d;) is a cartesian invariant, (b) Is uv an affine invariant?

Chapter 4 Tensor Operations; Tests for Tensor Character 4.1 FUNDAMENTAL OPERATIONS From two given tensors, S = (^'.''\) Т = (Г^;;;;о (4.1) certain operations, to be described, will produce a third tensor. Sums, Linear Combinations Let p = r and q = s in (4.1). Since the transformation law (3.14) is linear in the tensor components, it is clear that S + Ύ ^(SYf1- +Τ\ή· ■■'<) (4.2a) 4 1лП ■ ■ ■ h HH ■ ■ ■ h' x ' is a tensor of the same type and order as the original two tensors. More generally, if T,, T2, . . . , Τμ are tensors of the same type and order and if λ,, λ2, . . . , λμ are invariant scalars, then A,T, + A2T2 + · · · + λμΎμ (4.2b) is a tensor of that same type and order. Outer Product The outer product of the tensors S and Τ of (4.1) is the tensor \sT]^(s';X;;)g-T^-^) (4.3) which is of order m^p + q + r + s (the sum of the orders of S and T), contravariant of order ρ + r and covariant of order q + s. Note that [ST] = [TS]. EXAMPLE 4.1 Given two tensors, S = (S)) and Τ = (Tk), the outer product [ST] = (S'fTk) = (Pljk) is a tensor because pi „if /„, дХ_ дХ*_\(т дХ?_\ = pr d±_ crf_ d±_ rik *jik V* dxr dx'/V» dxk' s" dxr dx>' dxk Inner Product To take the inner product of two tensors, one equates an upper (contravariant) index of one tensor to a lower (covariant) index of the other, and sums products of components over the repeated index. In effect, the contravariant and covariant behaviors cancel out, which lowers the total order of the two tensors. To state this more formally, set ia = u~ Ιβ in (4.1). Then the inner product corresponding to this pair of indices is ST-(^,;/.%?vr^^j {44) It is seen that there will exist ps + rq inner products ST and TS; in general, all of these will be distinct. Each will be a tensor of order m^p+q+r+s-2 43

44 TENSOR OPERATIONS; TESTS FOR TENSOR CHARACTER [CHAP. 4 EXAMPLE 4.2 From the tensors S = (S") and Τ = (Tklm), form the inner product U = (U{m)= (S"'Tkum). We have: w„. •(s"££)(t. = SrrT. dx dxq dx dx" dxr/\'sq' дхк ~дГ дхт a χ11 dxq\ dx' dxs dx' s<1,\dxp dx" =„PrT dx' dx5 dx' dx' dxk dx"' l'« ' dxr dx" dx"' _ vprT dx' dxs dx' _ dx' dxs dx' sp' ~d7 Jx~k ~df^ ~~ st ~d4 Jk~k Jr* which verifies that U is a tensor of order 3, contravariant of order 1 and covariant of order 2. EXAMPLE 4.3 With (Гу) and (TiJ) as in Theorem 3.2, As an inner product, the left side defines a second-order tensor that is contravariant of order one and covariant of order one. This constitutes a new proof (cf. Problem 3.29) of the tensor nature of the Kronecker delta. In the special case when S is a contravariant vector and Τ is a covariant vector, the inner product ST is of the form S'Tt, which is an invariant (Theorem 3.1). Because the tensor ST is of order m = p + q + r + s-2=1+0 + 0+ 1-2 = 0 an invariant is regarded as a tensor of order zero. Contraction Another order-reducing operation, like the inner product but applying to single tensors, is that of contracting a tensor on a pair of indices. In tensor S of (4.7) set ia = u = jp and sum on u; the resulting tensor (Problem 4.7), s' = (<-.::l:;:X) (4.5) is called a contraction of S, with contraction indices ia and }β. S' is contravariant of order ρ — 1 and covariant of order q - 1. Combined Operations It is clear that one may form new tensors from old in a variety of ways by performing a sequence of the tensor operations discussed above. For example, one might form the outer product of two tensors, then take an inner product of this with a third tensor; or contract on one or more pairs of indices, either before or after taking a product. It is noteworthy that an inner product of two tensors may be characterized as a contraction of their outer product: ST = [ST]'. See Fig. 4-1. S = (^), T = (T )- ■ST = (^"T ) = [ST]' OUTER PRODUCT [ST] = (^'^) Fig. 4-1 CONTRACTION

CHAP. 4] TENSOR OPERATIONS; TESTS FOR TENSOR CHARACTER 45 4.2 TESTS FOR TENSOR CHARACTER It is useful to have an alternative method for verifying tensor character that does not directly appeal to the tensor transformation laws. Roughly stated, the principle is this: If it can be shown that the inner product TV is a tensor for all vectors V, then Τ is a tensor. This idea is often referred to as the Quotient Rule for tensors; the official Quotient Theorem is our Theorem 4.2 below. The following statements are useful criteria or "tests" for tensor character; they may all be derived as special cases of the Quotient Theorem. (1) If Ty' = E is invariant for all contravariant vectors (V), then (Гг) is a covariant vector (tensor of order 1). (2) If TtjVl = Uj are components of a covariant vector for all contravariant vectors (V), then (Г,--) is a covariant tensor of order 2. (3) If T-U'V = E is invariant for all contravariant vectors (Ul) and (V), then (Гг/) is a covariant tensor of order 2. (4) If (Г-) is symmetric and T^V'V = Ε is invariant for all contravariant vectors (V), then (Γ;-) is a covariant tensor of order 2. EXAMPLE 4.4 Establishcriterion (1). Since Ε is invariant, Ε = Ε, or Τ у' = T;V. Substitute in this equation the transformation law for (V1) and change the dummy index on the right: flv* —\ = T.V or (T. - T, —)v> = 0 The latter equation must hold when (V) is any of the contravariant vectors represented in (*') by (δ!,), (δj),. . . ,(δ'η); their existence is guaranteed by Problem 3.5. Thus, for the kth of these vectors (lS/cSrc), —f I · 1 = О ОГ Γ = Τ, τ дхк/ к дхк (T*-f3h=o or Тк which is the law of transformation—from (*') to (x')—of a covariant vector. The method of Example 4.4 may be easily extended to establish the following result, which in turn implies the Quotient Theorem. Lemma 4.1: If Т'^ ' / ) U^ U^ ■ ■ ■ U^V'faV'fa · ■ ■ V{qq) = Ε is an invariant for arbitrary covariant vectors (U\a)) = U{a) (a = 1,2, . . . , p) and arbitrary contravariant vectors (ν'(ββ))=\{β) (β = 1,2, ... , q), then (TjJ2\\~'Fj ) is a tensor of the type indicated by its indices. Theorem 4.2 {Quotient Theorem): If T'lJ "' J kVk = S}^'.'.'Fj are components of a tensor for an arbitrary contravariant vector (V ), then (T^''' p . ) is a tensor of the type and order indicated. 4.3 TENSOR EQUATIONS Much of the importance of tensors in mathematical physics and engineering resides in the fact that if a tensor equation or identity is true in one coordinate system, then it is true in all coordinate systems. EXAMPLE 4.5 Suppose that in some special coordinate system, (У), the covariant tensor Τ = (Τ ) vanishes. The components of Τ in any other coordinate system, (x'), are given by =, π dxr dXs ,, ,, τ„· = Trs — ^7 = ° + ° + · ■ ■ + 0 = О " " дх дх' Therefore, Τ = 0 in every coordinate system.

46 TENSOR OPERATIONS; TESTS FOR TENSOR CHARACTER [CHAP. 4 EXAMPLE 4.6 Consider a putative equation RiJkUk = AWf'MjkI, (1) connecting six entities that may or may not be tensors. If it can be shown that (i) Τ = (Ttj) = (RtjliUk - AWf'M^U,) is a tensor, and (ii) a special coordinate system exists in which all Тц are zero, then (1) is valid in every coordinate system. EXAMPLE 4.7 A second-order covariant tensor, or a second-order contravariant tensor, that is known to be symmetric in one coordinate system must be symmetric in every coordinate system. (This statement does not extend to a second-order mixed tensor; see Problem 3.16.) Another application of the principle yields (Problem 4.15) a useful fact in tensor analysis, often taken for granted: Theorem 4.3: If (Ti;) is a covariant tensor of order two whose determinant vanishes in one particular coordinate system, then its determinant vanishes in all coordinate systems. Corollary 4.4: A covariant tensor of order two that is invertible in one coordinate system is invertible in all coordinate systems. Solved Problems TENSOR SUMS 4.1 Show that if λ and μ are invariants and S' and 7" are components of contravariant vectors, the vector defined in all coordinate systems by (AS' + μΤ') is a contravariant vector. Since λ = λ and μ — μ, as desired. 4.2 Prove that (a) the array defined in each coordinate system by (Γ;· - Γ-;), where (T;J) is a given covariant tensor, is a covariant tensor; (b) the array defined in each coordinate system by (T'j — T\), where (T'-) is a given mixed tensor, is not generally a tensor, but is a cartesian tensor. (a) By (4.2b), the array is a tensor if and only if (7*) = (7y,) is a covariant tensor. But the transformation law for (7(/) gives f _T S^_d^_ -* _ « dxs dx i! ~ " dx' dx1 °r "' sr dx' dxj " ■ which shows that (7*) is indeed a covariant tensor. (b) We give a second proof [recall Problem 3.32(a)], based on (4.2b). The question is whether (U') = (T{) is a tensor. From the transformation law for (7,), =y dx' dxs -■ dx' dx' * dx dx' ' r dx dxr Thus, ({/') does not obey a tensor law, unless, for all p, q, dx" dx" 1T i.e., unless the Jacobian matrix is orthogonal—as it is for orthogonal linear transformations (cartesian tensors).

CHAP. 4] TENSOR OPERATIONS; TESTS FOR TENSOR CHARACTER 47 OUTER PRODUCT 4.3 Show that the outer product of two contravariant vectors is a contravariant tensor of order two. ™-«£)(τ£)-*τ·££ With (5") and (Γ) as the given vectors, s't' = (s'^ \ dX , which is the correct transformation law for the outer product to be a contravariant tensor of order two, INNER PRODUCT 4.4 Prove that the inner product (TrUir) is a tensor if (T1) and (£//;) are tensors of the types indicated. With Vi = T'Ufi, *.-(r%){u.%%)-<rvM%-v.% which is the desired transformation law. 4.5 Prove that if g = (gi]) is a covariant tensor of order two, and U = (£/') and V=(V!) are contravariant vectors, then the double inner product gUV= g^-U'V is an invariant. The transformation laws are дх Эх" -ι , Эх' -j Эх' Multiply, and sum over i and /': iuv= gltu'v'- = grsu'v" %%%^ = 8„υ'ν»δ:δ: = gnu'v = guv CONTRACTION 4.6 Assuming that contraction of a tensor yields a tensor, how many tensors may be created by repeated contraction of the tensor Τ = (Γ1^)? Single contraction produces the four mixed tensors ra (T-jj (о (г-) and double contraction produces the two zero-order tensors (invariants) Τ""ΰ and T"v"u. Thus there are six tensors, in general all distinct. 4.7 Show that any contraction of the tensor Τ = {Tljk) results in a covariant vector. We may contract on either i = /' or i = k. For (Sk) = (T'tk), we have the transformation law с = τ' τ' дх Зх" дх = τ Rs дх τ' дх - дх \ -1* 1 „ дхг g-i g-k s,*r д~к ,, аЯЬ \ а-к and, for(i//)-(r;,), fj = г = Τ' д-^- **- — = "Г Ά1 — = Tr — = U — " 3xr дх' Эх' " r Эх1 sr дх' s дх' In either case, the transformation law is that of a covariant vector.

48 TENSOR OPERATIONS; TESTS FOR TENSOR CHARACTER [CHAP. 4 COMBINED OPERATIONS 4.8 Suppose that S = (S'l) and Τ = (Γ'.) are tensors from which a contravariant vector V= (V) is to be constructed using a combination of outer/inner products and contractions, (a) Show that there are six possibilities for V, which can all be distinct, (b) Verify that each possible V is obtainable as a contraction of an inner product ST. (a) Writing [ST]=U = (t/|'*), we obtain the contravariant vectors as the double contractions of U: (u:ik) (U'Zk) (UZ) (ut) (u'Z) (UZ) (b) The vector (UZk) = (S"u"T*) may be obtained by first taking the inner product (S*Tkv) and then contracting on i = и = /. Likewise for the other five vectors of (a). TESTS FOR TENSOR CHARACTER 4.9 Prove criterion (2) of Section 4.2 without invoking the Quotient Theorem. We are to verify that (Ttj) is a covariant tensor of order two if it is given that for every contravariant vector (V), TlfV = Uf are components of a covariant vector. Start out with the transformation law for (Uj) [from (*'') to (f')]: ϋ, = υ·^ or fi,V' = ТУ ^7 ' J dx' " ,s dx' Now substitute the transformation law for V' [from (x') to (χ*)]: ^ -t π /"- η дх \ Эх" т. ν = τ Α νρ — —: 11 "V дХР) дХ> Replace the dummy index i by ρ on the left and by r on the right: fyp=TV* **-K. or (f . - Τ ^^V=0 " dx' dx' \ " " 3xp dx'' The proof is concluded as in Example 4.4. 4.10 Prove criterion (3) of Section 4.2. Here we must show that (7\,.) is a covariant tensor, assuming that TtjU'V is invariant. Using criterion (1), we conclude that (f^U') is a covariant vector. Using criterion (2), it follows that since (Ul) is arbitrary, (T,) is a covariant tensor of order two, the desired conclusion. 4.11 Prove criterion (4) of Section 4.2. We wish to show that if (Tr) is a symmetric array such that T^V'V is an invariant for every contravariant vector (V), then (Γ.) is a (symmetric) covariant tensor of order two. Let (t/') and (V) denote arbitrary contravariant vectors and let (W) = (V + V). a contravariant vector by (4.2a). Then, Ti-W'W = 7V(t/' + V')(U'' + V) = τ,,.υ'υ' + τ,.ν'ί/'' + τ,,υ'ν' + τ.,ν'ν' ij i] ij i] = TlfU'U' + TtfV'V'+2T:jU'V' where the symmetry of (Tit) has been used in the last step. Now, by hypothesis, the left-hand side and the first two terms of the right-hand side of the above identity are invariants. Therefore, T^U'V must be an invariant, and the desired conclusion follows from criterion (3). 4.12 Use Lemma 4.1 to write a proof of the Quotient Theorem, Theorem 4.2.

CHAP. 4] TENSOR OPERATIONS; TESTS FOR TENSOR CHARACTER 49 In the notation of the theorem and lemma, S^'""' · U\l) i/<2) · · ■Ul")V'tiuV'2n ■ ■ ■ v'* is a tensor of order zero, or an invariant, for arbitrary U<a) and V(iJ); that is, Thh- 'p n(DfiP)...f|(p)y/i yh ...и'ч yk is an invariant, with (V*) also arbitrary. It then follows from Lemma 4.1 (with q replaced by q + 1) that (TV ,2 ? t) is a tensor, contravariant of order ρ and covariant of order q + 1. From the above method of proof, it is clear that the Quotient Theorem is equally valid when the "divisor" is an arbitrary covariant vector. This form of the theorem will be used in Problem 4.13. 4.13 Use the Quotient Theorem to prove Theorem 3.2. If U = (U') is a contravariant vector, the inner product ν=τυ=(Γ1/ί//) is a covariant vector. Moreover, because [7V]„„ has an inverse, it follows that as U runs through all contravariant vectors, V runs through all covariant vectors. Thus, υ = τ-1ν=(Γ''ν;.) is a tensor for an arbitrary (V,), making (7"') a contravariant tensor of order two. TENSOR EQUATIONS 4.14 Prove that if (T'jkl) is a tensor such that, in the (x')-system, T'jki = 3T'ljk, then T'jk, =3T',jk in all coordinate systems. We must prove that Т]к1=ЪТ\]к in (*''). But f" ~-\T = Tp ^— — ^- — - -\Tp ^- — — — '"*' "'* "' Эх" дх' дхк дх1 "' дх" Эх1 Эх1 дхк __ ρ дх' дхг Эх" дх' дх' дх' дх' dxs "' ~д~Р Л> IP ~di' ~ '" ~д~Р ~dF' 'дХ4 ~дХ~к ■ ^(ТР ЛТрЛ дЯ' дХ' дХ* дХ' -п yl"' iltrJ дхр дх'' дхк дх' as desired. 4.15 Prove Theorem 4.3. By Problem 3.14(a), the covariant transformation law has the matrix expression Г=/Г7У whence |7l=i2|7l Thus, |Г| =0 implies |Г| =0. 4.16 Prove that if a mixed tensor (Г1.) can be expressed as the outer product of contravariant and covariant vectors (U1) and (V-) in one coordinate system, then (Ггу) is the outer product of those vectors in general. We must prove that T"; = U'V- for any admissible coordinate system (x). But, by hypothesis,

50 TENSOR OPERATIONS; TESTS FOR TENSOR CHARACTER [CHAP. 4 Supplementary Problems 4.17 If (V) and (V) are contravariant vectors, verify that (2U' + ЗУ) is also a contravariant vector. 4.18 Verify that the outer product of a contravariant vector and a covariant vector is a mixed tensor of order two. 4.19 How many potentially different mixed tensors of order two can be defined by taking the outer productof S = (S'l) and Τ = (T'fk), then contracting twice? 4.20 Show that if T"kl are tensor components, T'/. is an invariant. 4.21 Prove that if T'jklU' = S'kl are components of a tensor for any contravariant vector {U'), then (T'jkl) is a tensor of the indicated type. [Hint: Apply the Quotient Theorem to (M'klj) = (T'jkl). More generally, the Quotient Theorem is valid for all choices of the inner product.] 4.22 Prove that if T'jklSkl = U't are tensor components for arbitrary contravariant tensors (Skl), then (T'jkl) is a tensor of the indicated type. [Hint: Follow Problem 4.9.] 4.23 Prove that if TllklUkU' = V'. are components of a tensor for an arbitrary contravariant vector (U:). and if (T'jkl) is symmetric in the last two lower indices in all coordinate systems, then <T'kl) is a tensor of the type indicated. 4.24 Show that Theorem 4.3 and Corollary 4.4 are equivalent. 4.25 Prove the assertion of Example 4.7. 4.26 Prove that if an invariant Ε can be expressed as the inner product of vectors ({/,) and (V) in one coordinate system, then Ε has that representation in any coordinate system.

Chapter 5 The Metric Tensor 5.1 INTRODUCTION The notion of distance (or metric) is fundamental in applied mathematics. Frequently, the distance concept most useful in a particular application is non-Euclidean (under which the Pythagorean relation for geodesic right triangles is not valid). Tensor calculus provides a natural tool for the investigation of general formulations of distance; it studies not only non-Euclidean metrics but also the forms assumed by the Euclidean metric in particular coordinate systems. Calculus texts often contain derivations of arc-length formulas for polar coordinates that apparently apply only to that one coordinate system; here we develop a concise method for obtaining the arc-length formula for any admissible coordinate system. The theory culminates in later chapters with a method for distinguishing between a metric that is genuinely non-Euclidean and one that is Euclidean but disguised by the peculiarities of a particular system of coordinates. 5.2 ARC LENGTH IN EUCLIDEAN SPACE The classical expressions from calculus for arc length in various coordinate systems lead to a general formula of the type f V dx' dx1 Si' ~dt ~dt dt (5.1a) where gtj = gt](xl, χ , . . . , χ") = gjt are functions of the coordinates and L gives the length of the arc a ^ t^ b of the curve xl = x\t) (lgiSn). EXAMPLE 5.1 The arc-length formula for Euclidean three-space in a rectangular coordinate system (x\ x2, x3) may be recalled: r ( (<Ь\ (<b\ (dx \ ι ( L dx' dx' _, This is (5.1a), with gtj - S;/. The formula in Example 5.1 has the equally informative differential form ds2 = (dx1)2 + (dx2)2 + (dx3)2 = 8tj dx1 dx' More generally, (5.1a) is equivalent to ± ds2 = gtj dx dx' (5.1 b) EXAMPLE 5.2 For convenient reference, formulas for the Euclidean metric in the nonrectangular coordinate systems heretofore considered are collected below. Polar coordinates: (x1, x2) = (r. Θ); Fig. 3-1. ds2 = (dx1)2 + (x'fidx2)2 (5.2) Cylindrical coordinates: (x\ x2, хъ) = (r, Θ, z)\ Fig. 3-2. ds1 = (dx1)2 + (x1f(dx1f + (ахъ)2 (5.3) Spherical coordinates: (χ1, χ2, χ3) = (ρ, φ, θ); Fig. 3-3. 51

52 THE METRIC TENSOR [CHAP. 5 ds2 = (dx1)2 + (x'fidx2)2 + (x1 sin x2f(dx3)2 (5.4) Affine coordinates: (see Fig. 5-1). ds2 = (dx1f + (dx2)2 + (dx3)2 + 2 cos a dx1 dx2 + 2 cos β dx1 dx3 + 2 cos γ dx2 dx3 (5.5) Formula (5.5) is derived in Problem 5.9. Note that the matrix (gtj) defining the Euclidean metric is nondiagonal in affine coordinates. Although formulated for Euclidean space, (5.1) is extended, in the section that follows, to provide the distance concept for non-Euclidean spaces as well. ζ t —<~y x2 5.3 GENERALIZED METRICS; THE METRIC TENSOR Assume that a matrix field g = (gtj) exists satisfying in all (admissible) coordinate systems (x) and in some (open) region of space: A. g is of differentiability class C2 (i.e., all second-order partial derivatives of the gtj exist and are continuous). B. g is symmetric (i. e., gtj = gjt). С g is nonsingular (i.e.. \gjj\ 5^0). D. The differential form (5.1b), and hence the distance concept generated by g, is invariant with respect to a change of coordinates. Sometimes, particularly in geometric applications of tensors, a property stt .ger than С above is assumed: C. g is positive definite [i.e., g,· i>V > 0 for all nonzero vectors ν = (у1, ν2, . . . , ν")]. Under property С, |giy| and gn, g21, . . . , gnn are all positive. Furthermore, the inverse matrix field g_1 is also positive definite.

CHAP. 5] THE METRIC TENSOR 53 For later use we define the arc-length parameter for a curve 4 : χ = x'(t) (a^t^b): dx dx du du (5.6a) where ε = +1 or -1 according as ft; dx' dx' du du :0 or ft/ dx' dx' du du <0 The functional ε is called the indicator of the vector (dx'ldu) relative to the metric (&·■). One can, of course, use absolute value signs instead of the indicator, but the latter notation works better in algebraic manipulations. In terms of the arc-length parameter, the length of 4 is L = s(b). Differentiating (5.6a) and squaring yields the equivalent formula 2 ds dx' dx' Finally, introducing the differentials dt/ F'8i< dt dt dt (5.6b) the values of which are independent of the choice of curve parameter, we retrieve (5.1b) as ε ds2 = gr dx' dx' (5.6c) EXAMPLE 5.3 Suppose that on R3 a matrix field is given in (x') by (s,) (x1)2-! 1 0 1 (χ2)2 0 64 0 0 — where [(^)2-ip2)2#l (a) Show that, if extended to all admissible coordinate systems according to the transformation law for covariant tensors, this matrix field is a metric; i.e., it satisfies properties A—D above, (b) For this metric, compute the arc-length parameter and the length of the curve χ1 =2ί- 1 x2 - It2 (Oeiel) (a) Property A obtains since gt/ is a polynomial in x1 and x2 for each i, j. Since the matrix (g,.) is symmetric, property В holds. Since 64 j (л:1)2-! 1 9 ; 1 (x2f 64 {(*2Π(*)2-ι]-ι}^0 property С obtains. Property D follows from Problem 4.5. (b) It is convenient, here and later, to rewrite (5.6b) as the matrix product (ds eYdt Along the given curve, this becomes 1 ds dt [2 4i ЗГ] dx'\T/ / dx' (It - l)2 - 1 1 0 1 (2Г2)2 0 64 9. 2 0 0 = 64i6 + 64/4 + 16i2 = (8i3 + 4/) 2 At 3t2 (5.6 d)

54 THE METRIC TENSOR [CHAP. 5 Hence, ε = 1 and ■ s(t) = f (8ы3 + \ύ) du = [2ил + 2u2]'n = 2/4 + It1 from which L = 2(1)4 + 2(1)2 = 4. The properties postulated of g make it a tensor, the so-called fundamental or metric tensor. In fact, property D ensures that is an invariant for every contravariant vector (V) = (dx'ldt). (By solving an ordinary differential equation, one can exhibit the curve that possesses a given tangent vector.) Then, in view of property B, criterion (4) of Section 4.2 implies Theorem 5.1: The metric g = (g;-) is a covariant tensor of the second order. In Problem 3.14(a), the matrix equation U = / UJ was found for the transformation of a second-order covariant tensor U. If (xl) is a rectangular system and U = g is the Euclidean metric tensor, then in (x), U = G, and in (xl), U = G = /; thus we have proved Theorem 5.2: If the Jacobian matrix of the transformation from a given coordinate system (x) to a rectangular system (x1) is /= (дх'/дх1), then the matrix G = (gij) of the Euclidean metric tensor in the (x)-system is given by (5.7) fj Remark 1: Equation (5.7) illustrates the following well-known result of matrix theory: Any symmetric, positive definite matrix A has a nonsingular "square root" С such that A CTC. It should be emphasized that only the Euclidean metric admits of a representation of the form (5.7). For, by very definition, if g is non-Euclidean, there exists no coordinate system (xl) in which G = /. EXAMPLE 5.4 Cylindrical coordinates (x') and rectangular coordinates (x') are connected through : χ cos χ χ sin χ Thus 0 0 0 1 and the (Euclidean) metric for cylindrical coordinates is given by 1 · 2 n' -x sinx 0 COSJC sin x2 x1 cos x1 G^J'J- COSJC sin χ -x1 sinx2 x1 cosjc2 0 0 1_ cosjc2 sinx2 . 0 ~x smx xx cosx2 0 0 "1 0 0" 0 (x1)2 0 .0 0 1. or £n = £зз = 1' #22 = (x1)2· and gi,- =0 for i Φ;. These results verify (5.3). In spite of the apparent restriction to the Euclidean distance concept, in connection with such results as Theorem 5.2, the reader should keep in mind that one is free to choose as the metric tensor for R" any g that obeys properties A-D above. For instance, it can be shown by methods to be developed later that the metric chosen in Example 5.3 is non-Euclidean.

CHAP. 5] THE METRIC TENSOR 55 5.4 CONJUGATE METRIC TENSOR; RAISING AND LOWERING INDICES One of the fundamental concepts of tensor calculus resides in the "raising" or "lowering" of indices in tensors. If we are given a contravariant vector (T1) and if, for the moment, (g^) represents any covariant tensor of the second order, then we know (Problem 4.4) that the inner product (5,) = (g^T1) is a covariant vector. Now, if (gtj) is in fact the metric tensor whereby distance in R" is defined, it will prove useful in many contexts to consider (St) and (T1) as covariant and contravariant aspects of a single notion. Thus, we write Γ, instead of 5,: T^g.fT1 and say that taking the inner product with the metric tensor has lowered a contravariant index to a covariant index. Because the matrix (gtj) is invertible (property С of Section 5.3), the above relation is equivalent to Г = g'iTj where (gl') = (&;)"'; now we say that a covariant index has been raised to a contravariant index. Definition 1: The inverse of the fundamental matrix field (metric tensor), \g"L· = [gij] nn is called the conjugate metric tensor. Both metric tensors are freely applied to create new, more covariant (g) or more contravariant (g ) counterparts to given tensors. Thus, starting with the mixed tensor (T'[), j.ijk _ ir^jk 1 ik gir1 к and Τ = σ Ts = σ σ Tsr ijk ois ]k oisOjf- к ' 5.5 GENERALIZED INNER-PRODUCT SPACES Suppose that a metric g has been imposed on R" and that U and V are two vectors on the metric space. It is essential to the definition of a geometrically significant inner product UV that its value depend only on the vectors U and V, and not on the particular coordinate system used to specify these vectors. (There are other requirements on an inner product, but they are secondary.) This fact motivates Definition 2: To each pair of contravariant vectors U= (f/') and V= (V) is associated the real number uv= gyU'v' = υίνι = υ у1 (5.8) called the (generalized) inner product of U and V. In similar fashion, the inner product of two covariant vectors is defined as UV= giiU,Vj = U% = Uy' . (5.9) consistent with (5.8). We therefore have the rule: To obtain the inner product of two vectors of the same type, convert one vector to the opposite type and then take the tensor inner product. Remark 2: It follows from Problem 4.5—or, more fundamentally, from property D of g—that the inner product (5.8) or (5.9) is an invariant, as required.

56 THE METRIC TENSOR [CHAP. 5 According to (4.2), the set of all contravariant vectors on R" is a vector space, as is the set of all covariant vectors on R". With an inner product as defined above, these vector spaces become (generalized) inner-product spaces. 5.6 CONCEPTS OF LENGTH AND ANGLE Expressions (2.7) and (2.8) readily extend to a generalized inner-product space, provided the metric is positive definite. The norm (or length) of an arbitrary vector V= (V) or V= (V^) is the nonnegative real number |v||=Vv5 = Vvv^ (5.10) Remark 3: The norm of a vector—and thus the notion of a normed linear space—can be defined abstractly (see Problem 5.14), without reference to an inner product. EXAMPLE 5.5 Show that under the Euclidean metric (5.2) for polar coordinates, the vectors (!/'') = (3/5,4/5*') and (V1) = (-4/5, 3/5*1) are orthonormal. Using matrices, we have: №= </■!/, = *,!/'!/' = [§ ±] 1 0 0 (x1)2 ' 3 " 5 4 1.5x4 3 _4 5 5x τ] 3 ' 5 4xl 25 + 25*1 or ||U|| = 1; likewise, ||V|| = 1. Now we verify that the vectors are orthogonal: 3 4 JTV=gllUlV]- 5 5*1 3 J_ .5 5*1 1 0 0 (x1)2 5 3 1.5x4 _4 5 Зх1 25 + 25*1 Both normality and orthogonality depend, of course, on the metric alone, and not on the (polar) coordinate system. The angle θ between two non-null contravariant vectors U and V is defined by uv gipY' cos θ |u||||v|| y/gnU'VyfoxV* (0 ^ θ S 77) (5.11)

CHAP. 5] THE METRIC TENSOR 57 That θ is well-defined follows from the Cauchy-Schwarz inequality, which may be written in the form UV 1 υ ν 1 (see Problem 5.13). The tangent field to a family of smooth curves is a contravariant vector (Example 3.4), so that (5.11) yields the geometrical Theorem 5.3: In a general coordinate system, if (U1) and (V) are the tangent vectors to two families of curves, then the families are mutually orthogonal if and only if g^U'V' = 0. EXAMPLE 5.6 Show that each member of the family of curves given in polar coordinates by e1/r = a (sec θ + tan θ) (β&Ο) is orthogonal to each of the curves (limagons of Pascal) r = sin0 + с (с SO) (1) (2) (Figure 5-2 indicates the orthogonality of the curve a — 1 to the family of limagons.) In polar coordinates x1 = r, χ2 = Θ, and with curve parameter t, (1) becomes—after taking logarithms- = lna + In sec t + tan t\ χ = t With curve parameter u, (2) becomes χ = sin и + с (П (2') Fig. 5-2

58 THE METRIC TENSOR [CHAP. 5 Differentiation of (Γ) with respect to t yields 1 dx1 , dx 2 sec t = sec χ —— — 1 (χ1)2 Λ " dt so that the tangent vector to family (1') is (U\ U2) = (-(x1fsecx2,1) Similarly, the tangent vector to family (2') is (V\ F2) = (cos u, 1) = (cos x\ 1) Applying Theorem 5.3, with the Euclidean metric tensor in polar coordinates, g,,i/T' = gnt/V1+g22i/2F2 + 0 - (l)t-(x1)2 sec x2](cos x2) + (хУ(1)(1) -=-(хУ + (хУ=0 Observe that nonparametric forms of the tangent vectors are used in the orthogonality condition. This is necessary because the metric tensor at the intersection point (x\ x2) of a curve (1) and a curve (2) depends on neither the parameter t along (1) nor the parameter и along (2). Solved Problems ARC LENGTH 5.1 A curve is given in spherical coordinates (x) by χ = t χ = arcsin t i *3 = v7^T Find the length of the arc 1^ i^2. By (5.4), (l)4£)V>-(£)4w>'(£f so we first calculate the (dx'/dt)2: (dx'V Then and (5.1a) gives ux2\2 / -ι/t2 γ ι \dt) \y/i~(i/tfJ t2(t2-\) (dt) =1 + '\ν-ΐ)+('·7)·7 (dx"\2 /1 It \2 t2 \ dt 1 \2У/2-1/ t2~\ t2 2Г 2 -1 e -1 L-Ι -π= dt-W-l)] 2 = V6 1 5.2 Find the length of the curve 2 , (l^i^2) χ = ί if the metric is that of the hyperbolic plane (x > 0): 1 Si! = #22 = 7^2 #12= #21=0

CHAP. 5] THE METRIC TENSOR 59 Since (dx'ldi) = (0,1), (5.6d) yields (ε = 1) and f ) "Ю И " 1 ι 0 0 1 t\ 0 1 L -f1 Λ = In 2 GENERALIZED METRICS 5.3 Is the form ώτ + 3 dx dy + 4 φ2 + dz2 positive definite? It must be determined whether the polynomial Q = (ux)2 + Ъихи2 + 4(н2)2 + (и3)2 is positive unless η1 = и2 = и3 = 0. By completing the square, - 2n2 , ' / 2\2 ι / 3\2 . Q = (uxf + 3uxu2 + -4 (u2f + -4 (и2)2 + (и3)2 = [и1 + \ и2) + -4 (и2)2 + (и3)2 All terms are perfect squares with positive coefficients; hence the form is indeed positive definite. 5.4 Show that the formula (5.1a) for arc length does not depend on the particular parameterization of the curve. Given a curve % : x' = x'(t) (a^t^b), suppose that x'= x'(t) (a^l^b) is a different parameterization, where F= φ(ή, with φ'(ή>0 and a = ф(а), b = ф(Ь). Then, by the chain rule and substitution rule for integrals, L dx' dx' ?ii ~dt It dt /.V dx' dx' ?" ~dt~dJ -i С ι (t)dt = l_ yj dx^ dx^ "' dt dt ь ί (<PV)f dt dx' dx' " ~dt ~dJ dt = L TENSOR PROPERTY OF THE METRIC 5.5 Find the Euclidean metric tensor (in matrix form) for spherical coordinates, using Theorem 5.2. Since spherical coordinates (У) are connected to rectangular coordinates (x1) via -1 1-2 3 -2-1-2-3 -3 x = x sin χ cos χ χ = χ sin χ sin χ χ we have xl cos x2 j'j- sin χ cos χ sin χ sin χ xx cos x2 cos x3 xl cosjc2 sin x3 ~xx sinx2 sin χ xx sin x2 cos x3 COS X 1 · 2 -x sin χ 0 sin x2 cos χ χ cos χ cos χ -xx sin x2 sin x3 sin л^ sin x3 xx cos x2 sin χ xx sin x2 cos x3 COS* -x sin χ 0 Since G = 7r7 is known to be symmetric (see Problem 2.4), we need only compute the elements on or above the main diagonal: (sin x2)(l) +cos2 x2 (xx sinx2 cosx2)(l) — xx sin x2 cos x2 g21 ((;c1)2cos2jc2)(l) + (jc1)2sin2jc2 Sj, ι £32 513 ?23 1\2 „:_2 2 ((xI)2sin2x2)(l) where g13 = (x1 sin2 x2)(-sin χ cos x3 + cos x3 sin л: ) = 0 ^2з = ((x1)2 sin x2 cos x2)(—cosx3 sinx3 + sin x3 cos x3) = 0

60 THE METRIC TENSOR [CHAP. 5 Hence G 1 0 0 0 (xlf 0 0 0 (x1 sin x2)2 5.6 Find the components gtj of the Euclidean metric tensor in the special coordinate system (x) defined from rectangular coordinates (xl) by χ = χ , χ = exp (χ2 - χ ). We must compute J1 J, where J is the Jacobian matrix of the transformation x' = x'(x\ x2). Thus, we solve the above equations for the x': χ = x Hence χ = χ + In χ 1 0 1 (χ2)"1 1 1 0 (χ2)"1 and G or gu = 2, gl2 = g2l = (x2)"1, g22 = (x2) 1 0 ]_| 2 (x2) 1 (x2)"1]"1 /"2л"1 '--2^2 2r (хГ1 (χ2) 5.7 (a) Using the metric of Problem 5.6, calculate the length of the curve « : xl = 3i, x2 (0^ί^2) (ft) Interpret geometrically, (a) First calculate the dx'ldt: dx1 dt Then and = 2' d£_ dt dx dt 2-,-ii dxl\( dx + 2(^U)UJ+(^ dx dt ^2(9)+2e '(3)(e') + e-2'(e2')^25 L 5 dt = 10 (fo) From the transformation equations of Problem 5.6, the curve is described in rectangular coordinates by x2= 3X1; it is therefore a straight line joining the points which correspond to Z = 0 and i = 2, or (0,0) and (6,8). The distance from (0,0) to (6,8) is V62 + 82 = 10 as found in (a). 5.8 Making use of the Euclidean metric for cylindrical coordinates, (5.3), calculate the length of arc along the circular helix xl = a cos t χ = aunt x3 = bt with a and ft positive constants, from i = 0toi=c>0. See Fig. 5-3. In cylindrical coordinates (x'), where -11 2 -21-2 -33 x = x cos χ ;t=xsinx x=x the helical arc is represented by the linear equations 1 2 χ = α χ = t 1 - - -2 ~ - χ3 = bt (0S(Sc) [0 1 b] 1 0 0 0 2 α 0 0" 0 1_ "0" 1 _b_ [0 1 6] 0 2 1 /2

CHAP. 5] THE METRIC TENSOR 61 Fig. 5-3 whence L = f Va2 + b2 dt = cVa2 + b2 Jo 5.9 (Affine Coordinates in R ) Carpenters taking measurements in a room notice that at the corner they had used as reference point the angles were not true. If the actual measures of the angles are as given in Fig. 5-4, what correction in the usual metric formula, pxp2 = у Σ (*Ί - 4)2 should be made to compensate for the errors? We are asked, in effect, to display g = (gif) for three-dimensional affine coordinates (У). Instead of applying Theorem 5.2, it is much simpler to recall from Problem 3.9 that position vectors are contravariant affine vectors—in particular, the unit vectors u = (s;) v = (s'2) w = (s'3) along the oblique axes (Fig. 5-4). We can now use (5.11) in inverse fashion, to obtain: cos a : V^sf5?Vs„s^s2 Vg^Vgl a = 91° β = 90.5° γ = 89° ^^P^- A w Fig. 5-4

62 THE METRIC TENSOR [CHAP. 5 since, obviously, gn = g22 = g33 = 1 (±ds = ife1 for motion parallel to u; etc.). Likewise, cos β = 8гз cos У = en and the complete symmetric matrix is 1 -0.01745 -0.00873' G 1 cos a cos β cos a 1 cos γ cos β cos γ 1 It follows that the carpenters must use as the corrected distance formula -0.01745 1 0.01745 -0.00873 0.01745 1 РгРг = Vs*-(*i-*2)W-*D where the gy have the numerical values given above. RAISING AND LOWERING INDICES 5.10 Given that (V) is a contravariant vector on RJ, find its associated covariant vector (V,) in cylindrical coordinates (x') under the Euclidean metric. Since ш 1 0 0' 0 (x1)2 0 .0 0 1. and Vt — girV, we have in matrix form, 1 0 0 (X1)2 0 0 01 0 lj IVI V2 lv3\ = Γ ν1 Ί (x'fV2 L v3 J 5.11 Show that under orthogonal coordinate changes, starting with any particular system of rectangular coordinates, the raising and lowering of indices has no effect on tensors, consistent with the fact (Section 3.6) that there is no distinction between contravariant and covariant cartesian tensors. It suffices to show merely that gtj — δ^. = g" for any admissible coordinate system (x'), for then it will follow that Τ' = δ"'Τ r„* = e„7V Τ' t'j^Wi and so on. To that end, simply use formula (5.7), with J — (a'y) an orthogonal matrix. Because JT' — J l, we have G — J~XJ = I, or gtj — bip as desired. Since G1 = I1 = I, it is also the case that g" — Ъц. GENERALIZED NORM 5.12 Show that the length of any contravariant vector (V') equals the length of its associated covariant vector (Vj). By definition, ||(V')II = igijV'V and 11(^)11 = Vg%V, But, since V' = girVr and gtj = gM, s.v'V = gii(girvr)(g'svs) = g^g'Xv, = 8'}g"vrv. = g"vrv, and the two lengths are equal.

CHAP. 5] THE METRIC TENSOR 63 5.13 Assuming a positive definite metric, show that the basic properties of the cartesian inner product U ·ν are shared by the generalized inner product UV of contravariant vectors. (a) UV=VU (commutative property). Follows from symmetry of (g;;). (b) U(V + W) = UV + UW (distributive property). Follows from (1.2). (c) (AU)V=U(AV) = A(UV) (associative property). Follows from Ai/,.V" = U^W') = λί/,1Λ (d) U220 with equality only if U = 0 (positive-definiteness). Follows from the assumed positive- definiteness of (&,). (e) (UV)2 S (U2)(V2) (Cauchy-Schwarz inequality). This may be derived from the other properties, as follows. If U = 0, the inequality clearly holds. If U^fl, property (d) ensures that the quadratic polynomial Q( A) = ( AU + V)2 = U2A2 + 2UVA + V2 vanishes for at most one real value of A. Thus, the discriminant of Q cannot be positive: (UV)2-(U2)(V2)S0 and this is the desired inequality. 5.14 A generalized norm on a vector space is any real-valued functional φ[ ] that satisfies (i) φ[\] δ 0, with equality only if V= 0; (ii) ф[\\] = \\\ф[У\; (iii) φ[\] + V] ^ ф[И] + ф[\] (triangle inequality). Verify these conditions for <£[V] = ||V||, the inner-product norm under a positive definite metric. (i) and (ii) for ||V|| are evident. As for (iii), the Cauchy-Schwarz inequality gives ||U + V||2 = (U + V)2 = U2 + V2 + 2UV Ш ||U||2 + ||V||2 + 2 ||U|| ||V|| = (Ци|| + ||V||)2 from which (iii) follows at once. ANGLE BETWEEN CONTRAVARIANT VECTORS 5.15 Show that the angle between contravariant vectors is an invariant under a change of coordinate systems. The defining expression (5.1) involves only inner products, which are invariants. 5.16 In R2 the family of curves x2 = xl — с (parameterized as x1 = t, x2 = t — c), has as its system of tangent vectors the vector field U = (l, 1), constant throughout R2. If (x') represent polar coordinates, find the family of orthogonal trajectories, and interpret geometrically. The metric is given by "1 0 _0 (хУ so, by Theorem 5.3, the orthogonality condition becomes or, eliminating the differential du, dx1 + (x1)2 dx2 = 0

THE METRIC TENSOR [CHAP. 5 This is a variables-separable differential equation, whose solution is i_ 1 x2 + d The given family of curves in the usual polar-coordinate notation is r = θ + с, which is a family of concentric spirals (solid curves in Fig. 5-5). The orthogonal trajectories, 1 Г~ θ + d are also spirals, each having an asymptote parallel to the line θ = ~d; these are the dashed curves in Fig. 5-5. Note: To solve this problem in rectangular coordinates—that is, to find the orthogonal trajectories of the family tan 0jx2+y2~c) under the metric (gy) = (S,-)—would be difficult or impossible. Quite often, the complication of the metric involved in going over to a specialized curvilinear coordinate system is vastly outweighed by the degree to which the problem is simplified. Find the condition for two curves on a sphere of radius a to be orthogonal, if the curves are represented in spherical coordinates by % : 0 = /(φ) and % : θ = g{q>) The two curves can be parameterized in spherical coordinates (x') = (p, φ, θ) by % : \ Ψ = t % ; \ φ = и [θ=№ ie=g(") At an intersection point (α, φ0, θ0) the tangent vectors of %x and ^2 are, respectively, U = (0,1,/'(%)) and V=(0,l,g'(%))

CHAP. 5] THE METRIC TENSOR 65 These are orthogonal if and only if guU'V' = 0, or 0 = [0 1 /'(%)] 1 0 0 a2 0 0 0 1 0 0 (α sin <p0) = 0 + a2 + (a sin <p0)2/'(<p0)g'(<Po) = ("2 «n2 <p0)[csc2 «p0 + /'(%)«'(%)] Henee, the desired criterion is that/'(<p0)g'(<p0) = —esc2 <p0 at any intersection point (α, φ0, θ0). 5.18 Show that the contravariant vectors U = (0, 1, 2bx2) and V= (0, -2bx2, (*')2) are orthogonal under the Euclidean metric for cylindrical coordinates. Interpret geometrically along xl = a, 2 . 3 , .2 X = ί, Χ = Οί . g<,t/V = [0 1 2^2] 1 0 0 0 (χ1)2 0 0] 0 lj 0 Ί ~2bx2 L (χΎ J [0 1 2bx2] 0 -2ί)χ2(χ')2 (χ1)2 . = 0 - 2bx2(x1)2 + г^2^1)2 = 0 The geometric interpretation is that x1 = a, x2 = t, χ — bt2, for real t, represents a sort of variable-pitch helix on the right circular cylinder r- a, having tangent field U. Therefore, any solution of dxl du ν = ο dx du = VZ -2bx2 dr* du (1) or χ = a will represent a curve on that cylinder that is orthogonal to this pseudo-helix. See Problem 5.28. 5.19 Show that in any coordinate system (xl) the contravariant vector (recall Problem 3.5) V= (g'a) is normal to the surface xa = const, (a =1,2, . . . , h). Being "normal to a surface at the surface point P" means being orthogonal, at P, to the tangent vector of any curve lying in the surface and passing through P. Now, for the surface x" = const., any such tangent vector Τ has as its ath component dt We then have: ντ=gilv'r = gi.g'- r = gjigia r = s; r = r = о and the proof is complete. 5.20 Show that in any coordinate system (xl), the angle θ between the normals to the surfaces xa = const, and χβ = const, is given by cos θ - 8 (no sum) Ц) yg vr By Problem 5.19, U = (g'a) and V= (g"3) are the respective normals to xa — const, and χβ = const. Therefore, by the definition (5.11) δ~8'β _ 8"β cos θ UV Sijg 8 I^HIH ^8„q8Pa8qa \ΐ8,,8Ύβ VW^V^ Vg^V? In consequence of (1), orthogonal coordinates are defined as those coordinate systems (x') relative to which, at all points, giJ = 0 (tV/), or, equivalently, g,t =0 (tVy). Obviously, orthogonal coordinates need not be rectangular: witness polar, cylindrical, and spherical coordinates.

66 THE METRIC TENSOR [CHAP. 5 Supplementary Problems 5.21 Using the Euclidean metric for polar coordinates, compute the length of arc for the curve « : xL=2acost, x2 = t (OSfSir/2) and interpret geometrically. 5.22 Is the form Q(u, и2, и3) = 8(ί/)2 + (и2)2 - 6м V + (ы3)2 positive definite? 5.23 Using the metric "12 4 0 G= 4 1 1 _ 0 1 (x1)2 calculate the length of the curve given by x1 — 3 — t, x2 — Ы + 3, x3 = In t, where lSiSe. 5.24 A draftsman calculated several distances between points on his drawing using a set of vertical lines and his T-square. He obtained the distance from (1,2) to (4,6) the usual way: V(4-l)2 + (6-2)2 = 5 Then he noticed his T-square was out several degrees, throwing off all measurements. An accurate reading showed his T-square measured 95.8°. Find, to three decimal places, the error committed in his calculations for the answer 5 obtained above. [Hint: Use Problem 5.9 in the special case x\ = x\=Q, with a = 95.8°.] 5.25 In curvilinear coordinates (*'), show that the contravariant vectors U=(-*7*2,1,0) V=(l/x2,0,0) are an orthonormal pair, if (x') is related to rectangular coordinates (*') through ж ж ж ж ж ж ж where χ2 фЧ). 5.26 Express in (x) the covariant vectors associated with U and V of Problem 5.25. 5.27 Even though (g;.) may define a non-Euclidean metric, prove that the norm (5.10) still obeys the following "Euclidean" laws: (a) the law of cosines, (b) the Pythagorean theorem. 5.28 (a) Solve system (1) of Problem 5.18. (b) Does the solution found in (a) include all curves orthogonal to the given pseudo-helix? Explain. 5.29 Find the family of orthogonal trajectories in polar coordinates for the family of spirals x1 = ex2 (c = const.). [Hint: Parameterize the family as x1 = ce', x2 = e'.] 5.30 Find the condition for two curves, ζ =/(0) and ζ =g(0), on a right circular cylinder of radius a to be orthogonal. 5.31 Let (x') be any coordinate system and (gv) any positive definite metric tensor realized in that system. Define the coordinate axes as the curves %a : x' = t8'a (a = 1,2,. . . n). Show that the angle φ between the coordinate axes 4ia and %β satisfies the relation cos φ = —^=— (no sum) Vga„ V ξββ and is thus distinct, in general, from the angle θ of Problem 5.20.

CHAP. 5] THE METRIC TENSOR 67 5.32 Refer to Problems 5.20 and 5.31. (a) What property must the metric tensor (g,y) possess in (*') for the coordinate axis (βα to be normal to the surface xa = const, (in which case θ = φ)? (b) Show that the property of (a) is equivalent to the mutual orthogonality of the coordinate axes. 5.33 Under the metric „ _ \ 1 cos 2x „ -, „ , (2χ /π nonintegral) coszjc 1 J compute the norm of the vector V= (dx'/dt) evaluated along the curve x1 = — sin 2t, x2 — t, and use it to find the arc length between i = 0 and t= π/2. 5.34 Under the Euclidean metric for spherical coordinates, (5.4), determine a particular family of curves that intersect x1 = a x1 — bt x3 — t orthogonally. (Cf. Problem 5.28.)

Chapter 6 The Derivative of a Tensor 6.1 INADEQUACY OF ORDINARY DIFFERENTIATION Consider a contra variant tensor Τ = (Tl(\(t))) defined on the curve % : χ = x(t). Differentiating the transformation law Τ r dx' dx' with respect to t gives n 2 — i is д χ αχ dT_^dT_dx_ dt dt dxT dxsdxr dt which shows that the ordinary derivative of Τ along the curve is a contravariant tensor when and only when the x' are linear functions of the xr. Theorem 6.1: The derivative of a tensor is a tensor if and only if coordinate changes are restricted to linear transformations. EXAMPLE 6.1 With T= dx/dt the tangent field along <£ (under the choice t~s = arc length), the classical formula for the curvature of <£, dj dt will hold in affine coordinates but will fail to define an invariant in curvilinear coordinates, since dT/dt is not a general tensor. Clearly, to make the curvature of '€ an intrinsic property, we require a more general concept of tensor differentiation. This will entail the introduction of some complicated, nontensorial objects called Christoffel symbols. 6.2 CHRISTOFFEL SYMBOLS OF THE FIRST KIND Definition and Basic Properties The η functions ■ ijk (&*)+^(S*')~^(&y)J (6.1a) dx' w-"" дх' ч"л" dx" are the Christoffel symbols of the first kind. In order to simplify the notation here and elsewhere, we shall adopt the following convention: The partial derivative of a tensor with respect to xk will be indicated by a final subscript k. Thus, ■ ijk ■ 2 {-gijk + g.ki + gkti) (6.1b) EXAMPLE 6.2 Compute the Christoffel symbols corresponding to the Euclidean metric for spherical coordinates: G 1 0 0 (χ1)2 о о 0 0 (x1)2 sin2;c2 Here, g221 = 2x\ g311 = 2xl sin2 x2, g^2 = 2(x1)2 sin x2 cos x2, and all other gljk are zero. Hence, Г, unless the triplet ijk includes precisely two 2s (six cases) or precisely two 3s (six cases): 4jk

CHAP. 6] THE DERIVATIVE OF A TENSOR *221 9^ &221 &212 &122) X Γ223 = J (~Sl23 + #232 + #322 ) = 0 *212 2 ^ &212 &122 ^221J X * 232 О ^~^232 #322 ^223 J ~ ^ Γ122=2(- i + fei+fe)^' *322 О ^ &322 &223 &233J ^ and Г331 = 0 <~&>3i + £313 + Лзз) = -*' sin2 χ2 Г332 = J (~&32 + #323 + #23з) = ~(ΧΎ sil1 X COS X 1 2 Г323 = 2 (_^323 + #233 + £332) = 0 ) Sln X COS * Γΐ33 = 2 ^"gl33 + &1 + &>") = *' Sin2 ^ Γ233 = 2 (~&33 + #332 + #32з) = (ΧΎ ™ *' COS ^ (The nine nonzero symbols are marked with bullets.) The two basic properties of the Christoffel symbols of the first kind are: (i) Tijk - Tjik (symmetry in the first two indices) (ii) all Tijk vanish if all gtj are constant A useful formula results from simply permuting the subscripts in (6.1b) and summing: The converse of property (ii) follows at once from (6.2); thus: Lemma 6.2: In any particular coordinate system, the Christoffel symbols uniformly vanish if and only if the metric tensor has constant components in that system. Transformation Law The transformation law for the Tijk can be inferred from that for the gtj. By differentiation, 8цк д ( дх дх \ _ dg^ дх[ дх*_ д2х дх дх_ д2х дхк дх' дх' + 8rs дхкдх' дх1 + 8rs Jx1 дхкдх' iik. n — к \ ors n —i n —/' ' дХ V дХ дх' Use the chain rule on dgrsldxk dgrs dSrs dXl дх n — k Ί„/ n -k orst п —к дХ дХ дХ дХ Then rewrite the expression with subscripts permuted cyclically, sum the three expressions (arrows couple terms which cancel out), and divide by 2: ι—^ д ~Xr dXS _ _ дХ дх дХ дХ дХ дХ V дХ дХ дх д X дХ_ дхкдх' дх' о jki &sir dXS дх' дХГ дх' дХ дХ + 8гг д X дХ дХ дх' дХ + п 2 г п s д X дХ дх1дхк дх' _ _ дХ дХГ dXS Skij - g,rs JfkJ^J^] + gs д"х" дХГ д'ХГ dXS + ~—— — дх'дх дх дх'дХк дХ give - „ дХг дх" дх' Г = Г + σ *■ iik *■ rvl _ — i _ — i - — к δ, д 2Xr dXs ■ ijk -1 rst „ -i дХ дХ дХ rs 3ϊ'ίϊ< дХ дХ дХ (6.3)

70 THE DERIVATIVE OF A TENSOR [CHAP. 6 From the form of (6.3) it is clear that the set of Christoffel symbols is a third-order covariant affine tensor but is not a general tensor. Here again, conventional differentiation—this time, partial differentiation with respect to a coordinate—fails to produce more than an affine tensor (recall Problem 2.23). 6.3 CHRISTOFFEL SYMBOLS OF THE SECOND KIND Definition and Basic Properties The пъ functions !> = £%, (6.4) are the Christoffel symbols of the second kind. It should be noted that formula (6.4) is simply the result of raising the third subscript of the Christoffel symbol of the first kind, although here we are not dealing with tensors. EXAMPLE 6.3 Calculate the Christoffel symbols of the second kind for the Euclidean metric in polar coordinates. Since we have: G 1 0 0 (x1)2 * 111 _ 3 Sill ~ " 1 121 — I 211 _ 2 \ 8l21 """ #211 """ S\\2) ~ " 1 221 = 2 \ 8221 ~*~ 8212 ~*~ 8122) ~ ~x 1ц2 — Ί \ 8\\2~*~ 8\2\~*~ 821k) ~ " * * 122 _ * 212 — 2 \ 8l22 "*" &221 "*" 8212) ~ X * 222 — 2 ^222 — " To continue, СГ1 1 0 0 (x1)-2 From g12 = ϋ = g21, it follows that T'jk = g'rTjkr = g"Tjkl (no sum). Therefore, when i = 1, • Y\i = —xl Tljk = Q otherwise and when i = 2, • Γ?2 = Γ21 = 11 xx T2jk = 0 otherwise The basic properties of Yijk carry over to T'-k: (i) T'jlc = T'kj (symmetry in the lower indices) (ii) all T'jk vanish if all gtj are constant Furthermore, by Problem 6.25, Lemma 6.2 holds for both first and second kinds of Christoffel symbols. Transformation Law Starting with

CHAP. 6] THE DERIVATIVE OF A TENSOR 71 substitute for T-kr from (6.3) to obtain „ dx' dxr\(r dx11 dx" dxw\ ( st dx1 dxr\l д2хи dxL Y'ik \gS dx dx4Vuvw dx1 dxk dxr) + \8' dxs axlJ\om dx'dxk dx stT wdx_dx^dx^_ st dx d2Xu 8 *-uvw"t η s n-/' n-fc "' S guv" t n* „-/„-* OX dx' dX dX dX dX st dX dX dX st dX d X 8 i-uvt ~Zj ~7ч T=fc 8 gut dx' dx' dxK ' ° out dx dx'dxk Since gs'Tuut = Tsuv and gstgut= 8SU, after changing indices this becomes Г = Гг — — — + δ2χΓ — (f> <n ik st dxr dxJ dxk dx'dxk dxr l°"?J The transformation law (6.5) shows that, like (Tljk), (T'jk) is merely an affine tensor. An Important Formula d'x' _-s dXr dXS dx' JFJF = iy d~r ~ ■« 7F Ή'1 (6-6) See Problem 6.24. Needless to say, (6.6) holds when barred and unbarred coordinates are interchanged. 6.4 COVARIANT DIFFERENTIATION Of a Vector Partial differentiation of the transformation law ^ π дХГ T-=T : ι г п -ι dX of a covariant vector Τ = (Γ() yields df, dTr dxr „ d2xr + T, dxk dxk dx* r dxkdxi Using the chain rule on the first term on the right, and formula (6.6) on the second, results in the equations dtt = дТ\ dx^ d£_ /-, dx_ _ dx^_ dx!_ dXk dXS дх1 дхк Л1 <"* dXS st dX1 dXk = dTLdS_ irf_ ft f T, T dXr dXS dxs dx1 dxk lk ' " ' dx' dxk which rearrange to *T> t.fJ*L-r'T)**'** dxk ik ' V dxs " 4 dx1 dxk which is the defining law of a covariant tensor of order two. In other words, when the components of dT/dxk are corrected by subtracting certain linear combinations of the components of Τ itself, the result is a tensor (and not just an affine tensor). Definition 1: In any coordinate system (x'), the covariant derivative with respect to xk of a covariant vector Τ = (Tt) is the tensor

72 THE DERIVATIVE OF A TENSOR [CHAP. 6 Remark 1: The two covariant indices are notated i and ,k to emphasize that the second index arose from an operation with respect to the kth coordinate. Remark 2: From Lemma 6.2, the covariant derivative and the partial derivative coincide when the gtj are constants (as in a rectangular coordinate system). A similar manipulation (Problem 6.7) of the contravariant vector law leads to Definition 2: In any coordinate system (*'), the covariant derivative with respect to xk of a contravariant vector T = (T') is the tensor τ, = (ή)-(^ + ΓΛΓ' Of Any Tensor In the general definition, each covariant index (subscript) gives rise to a linear "correction term" of the form given in Definition 1, and each contravariant index (superscript) gives rise to a term of the form given in Definition 2. Definition 3: In any coordinate system (xl), the covariant derivative with respect to xk of a tensor Τ = (Thh 'f) is the tensor Τ k = (τ)) ''''' Λ, where I,I2 . . . lp _ I\l1 ■ ■ ■ Iq „ij i>2 ■ - > ri2 T»l> ■ ■ ■ 'p r'p i ;.;. ; L· — _ к "τ" l tk L i, /,.../ T l tk 1 i.i-, : "·" "·" 1 ik λ 1'2 ■ ilh--iq-k QXk tk hil--iq tk hh-'-lq '* l\h ■ ■ ■ Iq _ yt T/j!2 . . . ip _ -pi j,iji2 · ■ ■ 'p _ . . . _ pf T'i'2 · ■ ■ 'p /-j 7-, 1 Л*: 1tJ2---Jq J2k L h' ■■■ iq 1 iqk l hh ■■■' ^ ' > That Τ k actually is a tensor must, of course, be proved. This can be accomplished basically as in Problem 6.8, by use of Theorem 4.2 and an induction on the number of indices. Theorem 6.3: The covariant derivative of an arbitrary tensor is a tensor of which the covariant order exceeds that of the original tensor by exactly one. 6.5 ABSOLUTE DIFFERENTIATION ALONG A CURVE Because (Tl ) is a tensor, the inner product of (Г'у) with another tensor is also a tensor. Suppose that the other tensor is (dx'/dt), the tangent vector of the curve % : χ = x\t). Then the inner product „■ dx_s >r dt is a tensor of the same type and order as the original tensor (Γ'). This tensor is known as the absolute derivative of (T') along <#, with components written as ^)=(^+г»г^) "here r=T'«'» <*8> (see Problem 6.12). It is clear that, again, in coordinate systems in which the gVl are constant, absolute differentiation reduces to ordinary differentiation. The definition (6.8) is not an arbitrary one; in Problem 6.18 is proved Theorem 6.4 (Uniqueness of the Absolute Derivative): The only tensor derivable from a given tensor (T') that coincides with the ordinary derivative (dT'ldt) along some curve in a rectangular coordinate system is the absolute derivative of (T') along that curve.

CHAP. 6] THE DERIVATIVE OF A TENSOR 73 Remark 3: Theorem 6.4 concerns tensors with a given form in rectangular coordinates. Thus it presumes the Euclidean metric (see Section 3.1). Acceleration in General Coordinates In rectangular coordinates, the acceleration vector of a particle is the time derivative of its velocity vector, or the second time derivative of its position function χ = (x'(t)): . ,■ I d dx\ I d χ The (Euclidean) length of this vector at time t is the instantaneous acceleration of the particle: a = Vfy*'a; Since derivatives arc taken along the particle's trajectory, the natural generalization of — ( — I 8_ (d£\_chc_ dx_d^_ 1S 8t\dt)~ dt2 " dt dt Hence, in general coordinates, we take as the acceleration vector and the acceleration , /rfV , dx dx\ Ά = {α)Λ~ά7 +Τ"^Ίί) (6-9) a = ^J\gijaiai\ (6.10) Note that positive-definiteness of the metric is not assumed in (6.10). Curvature in General Coordinates In Euclidean geometry an important role is played by the curvature of a curve % : x' = x'(t), commonly defined as the norm of the second derivative of (x'(s)): I d2x d2x' where dsldt = \] 8^(ах11 dt)(dx! I dt) gives the arc-length parameter. The obvious way to extend this concept as an invariant is again to use absolute differentiation. Writing _δ_ ώΤ\ /rfV ■ <Ы_ dx" 8s ds )~\ ds2 + pq ds с where the arc-length parameter s = s(t) is given by (5.6), we have ^U^H^ + r^^) (6-n) K(s) = i\g~bbI\ (6.12) Geodesies An important application of (6.12) in curvilinear coordinates is the following. Suppose that we seek those curves for which к =0 (that is, the "straight" lines or geodesies). For positive definite metrics, this condition is equivalent to requiring that dV dxp dx4 к'4+ГЛ1=» (/ = 1.2......) (6ЛЗ) The solution of this system of second-order differential equations will define the geodesies x' = x'(s).

74 THE DERIVATIVE OF A TENSOR [CHAP. 6 EXAMPLE 6.4 In affine coordinates, where all g,.. are constant and all Christoffel symbols vanish, integration of {6.13) is immediate: χ' = α'ί + β' (i=l,2,...,n) where, s being arc length, g^a'a' = 1. Thus, from each point x= β of space there emanates a geodesic ray in every direction (unit vector) a. 6.6 RULES FOR TENSOR DIFFERENTIATION Confidence in the preceding differentiation formulas should be considerably improved when it is learned (see Problem 6.15) that the same basic rules for differentiation from calculus carry over to covariant and absolute differentiation of tensors. For arbitrary tensors Τ and S, we have: Rules for Covariant Differentiation sum (T + S)pfc = Tfc+Sit outer product [TS] fc = [T kS] + [TS k] inner product (TS) fc = Τ kS + TS k Since the absolute derivative along a curve is the inner product of the covariant derivative and the tangent vector, the above rules for differentiation repeat: Rules for Absolute Differentiation δ , _ δΤ 6S sum — (T + S) = —- + — δί ν ' δί δί с outer product ■— [TS] δί δΤ δί + δί inner product — (TS) = —— S + Τ —— δί δί δί Solved Problems CHRISTOFFEL SYMBOLS OF THE FIRST KIND 6.1 Verify that rfyfc = ryifc. By definition, Tijk = \ (-gljk + gjki + gklj) and Tjik = \ (-gjik + glkI + gk!l) But gijk = gjik, gjki = gkjl, and gkij = gikj, by symmetry of gij, and the result follows. 6.2 Show that if (g;>) is a diagonal matrix, then for all fixed subscripts a and β Φ a in the range 1, 2, . . . ,n, 0) Γααα = \ gaaa (not summed on a) Φ) ~Γααβ = Γαβα = Γβ*α = 2 8ααβ (not summed on a) (c) All remaining Christoffel symbols Tijk are zero.

CHAP. 6] THE DERIVATIVE OF A TENSOR 75 (a) By definition, Taaa = \{-gma + gaaa + gaaa) = \gaaa. (b) Since α Φ β, Ταβα = ΓΡαα = i {~8αβα + g„M + gMi,) = Η-0 + 0 + g^) = h gaaft (c) Let г, j, к be distinct subscripts. Then g;. = 0 and gi>Jt = 0, implying that Tijk = 0. 6.3 Is it true that if all Tijk vanish in some coordinate system, then the metric tensor has constant components in every coordinate system? By Lemma 6.2, the conclusion would he valid if the Ttjk vanished in every coordinate system. But (Г^) is not a tensor, and the conclusion is false. For instance, all fifk = 0 for the Euclidean metric in rectangular coordinates, but g22 = (xl) in spherical coordinates. CHRISTOFFEL SYMBOLS OF THE SECOND KIND 6.4 If (g;.) is a diagonal matrix, show that for all fixed indices (no summation) in the range — (- 1,2, ... ,n, («) Π, = r;„ = -^(bn|ga Φ) r;, = -^gf3f3a (α φ β) (с) All other Y'jk vanish. (a) Both (g/y.) and (g,-)-1 = (g") are diagonal, with nonzero diagonal elements. Thus, J_/i dg\_ а (l gaa \2 9χβ> 3χβ \2 Κβ = Ε"ταβί = §ααταβα = ^ = -^ (-1„ |ge 1/1 (b) Τ"ββ = 8ααΤββα = —{-2 8PP. (c) When i, j, к are distinct, T'jk = g"Tjkr = g"TJlcl = 0 (not summed on i). 6.5 Calculate the Christoffel symbols of the second kind for the Euclidean metric in spherical coordinates, using Problem 6.4. We have gn = 1, g22 = (я1)2, and g33 = (χ1)2 sin2 x2. Noting that gn is a constant and that all gaa are independent of x3, we obtain the following nonzero symbols from Problem 6.4(a): Γ" = Γζ3 = έ (^1η ((χ1 )2 si"2 χ2))= cot *' Similarly, from Problem 6.4(b), 2(1) Λκ1 ^ J τ-il jj- " и ls2 · 2 2n 1 ■ 2 2 33 = ~ 2(1) 7? ^X J Sln x ) = ~* sin x Гзз = , l4? —? ((x1 )2 sin2 x2 ) = — sin x2 cos jc2 33 20k1)2 dx~ vv y ;

THE DERIVATIVE OF A TENSOR [CHAP. 6 Use (6.6) to find the most general 3-dimensional transformation χ = χ'(χ) of coordinates such that (У) is rectangular and (*') is any other coordinate system for which the Christoffel symbols are fn = 1 f22 = 2 Гз3 = 3 all others = 0 Since T'sl = 0, (6.6) reduces to the system of linear partial differential equations with constant coefficients: <?V = pS dX_ дх'дх1 " dx K ' It is simplest first to solve the intermediate, first-order system ^ = i>-; (*: = §) (2) Since the systems (2) for r= 1,2,3 are the same, temporarily replace u's by us, and xr by a single variable x; thus, дй, 4 = г;л (з) For /' = 1, (3) becomes Эх ди1 -j _ -2 -3 —x = ГцМ, + Гпи2 + Гпи3 = ί/j <?W, - ( _ -2 __ -3 _ -^ = Г21и1 + Г2,и2 + Г2,и3 = 0 —| = Г зХ + Г з!И2 + Г з>3 = 0 Hence uj is a function of if1 alone, and the first differential equation integrates to give iij = bx exp xx (frj = constant) In the same way, we find for / = 2 and /' = 3: u2 = b2 exp2x2 (b2 = constant) ы3 = Ьъ exp 3i3 (fo3 = constant) Now we return to the equations dxldx' = ы( with the solutions just found for the ur dx , -ι dx , „_2 dx , 3 —7 = о.ехрл: -^ч-о7ехр2л: —-τ = о, exp 3jc (4) dx1 1 F ^jc 2 F Лс 3 F ; Integration of the first equation (4) yields x = blexpx1 + <р(х2,хг) and then the second and third equations give: дю _2 _2 / -3 ^ — b2 exp 2jc or φ = α2 exp 2jc + ψ(χ ) dx 1 -3 -3 b3 exp 3x or ф = α3 exp Здг + а άΦ _ «. _..._-,-з This means that, with αχ = £>15 χ = аг exp χ + a2exp2x + а3ехрЗх + α4 so that the general solution of (1) is хг = а[ехрх1 + a'2exp2x2 + а3ехрЗх3 + a'4 (5) for r=l, 2, 3. The constants u\ in (5) are unimportant; they merely allow any point in R3 to serve as the origin of the rectangular system (xr). The remaining constants may be chosen at will, subject to a single condition (see Problem 6.27).

CHAP. 6] THE DERIVATIVE OF A TENSOR 77 COVARIANT DIFFERENTIATION 6.7 Establish the tensor character of Τ k (Definition 2), where Τ is a contravariant vector. Beginning with the transformation law πι r^r dx' Γ = T —-r дх take the partial derivative with respect to xk and use the chain rule: 2-i ST' _ 3 I dic_\ д£_ _ дТ_ дх^_ дх^_ д х1 дх дхк дх" V дх') дхк дх3 дх* дхк дх*дхг дхк Now use (6.6), with barred and unbarred systems interchanged: дТ' дТ' дх' dxs mr/,„ дх' ;=,, дх" дх"\ дх ^к + Г Г' —г - Г дхК дх' дхг дх \ sr дх "" дх' дх 1 дх Since (дх"/дх')(дхУдхк) = 8"к and Т\дх"1дхг) = Т\ this becomes дТ_ = дТ^ д^_ дх^ , дх_ дх^ _ -, -„ ~ -к *> vs j „г ^ -к * sr * -,ι ~-к l kv * βχ ϋΧ ϋΧ дх дх дх or (by factoring and using the symmetry of the Christoffel symbols) дТ -, -, Ι дТ ~ „λ дх дх -t ^г дх' dxs 6.8 Show that (7%), as defined by (6.7), is a tensor, using the previously proven facts that T'k and TiJc are tensorial for all tensors (Tl) and (Γ,·). Let (У.) be any vector and set t/. = TrVT. The covariant derivative of the tensor (£A) is the tensor (Uj k), where dU; ■ 3 dTs. dV ,,k gxk ,k r gxk , r) ,k\ τ s) gxk s , dxk ,k r s But dVr „, dVr vrk = -ri- T'rkv, or -ri = vrk + rrkv, ,,k dyk rk s dyk r,k When the above expression for dVr/dxk is substituted into the preceding equation and the terms rearranged, the result is: ддф + ггкт) - rjkr)jvs = uIJc - rfvrik i.e., Tst kV, = tensor component It follows at once from the Quotient Theorem (Theorem 4.2) that (T'jk) is a tensor. 6.9 Extend the notion of covariant differentiation so that it will apply to invariants. First note that the partial derivative of an invariant is a tensor: ЭЁ _ ЭЕ _ дЕ дх" дх1 дх' дхг дх' Now, under any reasonable definition, (£ ■) must (1) be a tensor; (2) coincide with (дЕ/дх') in rectangular coordinates. The obvious choice is therefore ' ЭЕ\ <*■«> W

78 THE DERIVATIVE OF A TENSOR [CHAP. 6 6.10 Write the formula for the covariant derivative indicated by T'/.j. τι^^ + Γ,χί + τι,ή-Γ,χί 6.11 Prove that the metric tensor behaves like a constant under covariant differentiation; i.e., Sij,k = 0 for a11 hJ>k- By definition, since (gtj) is covariant of order 2, dg.. Sij.k ~ ~7~k ~ ^ ikSrj ~ * jkSir ~ Sijk ~ * ikj ~ * jki ~ " by (6.2). (In a similar manner, it follows that g''k = 0; see Problem 6.34.) Because of the above property of the metric tensor and its inverse, the operation of covariant differentiation commutes with those of raising and lowering indices. For example, ABSOLUTE DIFFERENTIATION 6.12 Prove that (6.8) is the result of forming the inner product of the covariant derivative (X'y) with the tangent vector (dx'ldt) of the curve. ■y dt \dx' rf I dt эх' dt r> dt dt " dt 6.13 A particle is in motion along the circular arc given parametrically in spherical coordinates by x1 = b, x2 = 77/4, χ3 = ωί (t = time). Find its acceleration using the formula (6.10) and compare with the result a = τω' from elementary mechanics. From Problem 6.5, we have along the circle Tl2i= -xx = -b Γ33 = -χ1 sin2 x2 = -fosin2 ^ = -- ,,11 , . , ■> . 17 17 1 Г., = Г,, = —г = γ Γ, = - sin χ cos x~ = - sin - cos — = - - 12 21 1 fr 33 4 4 2 X 11 „3 .,4 .2 .17 4 Tf, = Γ*, = — = г Г'3 = Г\2 = cot χ1 = cot ^ = 1 with all other symbols vanishing. The components of acceleration are, from (6.9), i_d1x1 , _! <fc' dxs _ 1 (dx2\2 ! /ifeV i/2*2 „·, <iir ife1 „ „„, dx1 dx2 - /ώ"\ „ / 1\, ,, ω2 Λ2 " ώ ί/r 13 ώ ώ 23 ώ ώ Together with the metric components along the circle, ?11 = 1 g22 = (*' )2 = fr2 g33 = (*' )2 «in2 x2 = the acceleration components give, via (6.10), a = \jgijaai = λ] (\)(-bw2llf + (b2)(-a>2!2)2 + 0 = Ьо2/л/2 Upon introducing the radius of the circle, using (3.4) with χ1 = χ = r and хъ = 0, 17 Ь r = b sin , - /- 4 V2 we obtain α = 7ω"

CHAP. 6] THE DERIVATIVE OF A TENSOR 79 6.14 Verify that x1 = asecx~ is a geodesic for the EucUdean metric in polar coordinates. [In rectangular coordinates (x, y), the curve is χ = a, a vertical line.] First choose a parameterization for the curve; say, xx = aseci 2= (~ττ!2<ί<ττΙ2) (1) Parameter t is related to the arc-length parameter s via ds j dx' dx' lidx1^2 dt Va/' dt dt -y(—j +(aseci)2( -j-) = Va2 sec2 t tan2 t + a2 sec2 t (a sec t)\ 1 + tan21= a sec2 Ζ Λ cos2t or so that for any function x(t) dx _ dx dt cos t dx ds dt ds a dt d2x d { cos2 t dx\ dt cos4 t d2x 2 sin t cos3 t dx ds2 dt\ a dt/ ds a2 dt2 a2 dt Now, taking the nonzero Christoffel symbols from Example 6.3, we can rewrite the geodesic equations (6.13) in terms of the independent variable t: _ d2xx j / dx2\2 _ cos4 t d2x1 2 sin t cos3 t dx1 г cos4 t ί dx2 0=Ί^+ 22\Ί*~) =~a^^F ? ^i + {~x)~^~\^t d2xl ,„ „ dx" Jdx2\ ,, ... ^--(2tan,) ^--4^) =0 (2) rfV 2 dx1 dx2 _ cos4/ ЛУ 2 sin/cos3 f rfx2 / 1 \/cos2A2 ^х1 rfx2 °~Ί^ + l2 ds ds ~ a2 dt2 a2 dt 2\xl)\ a ) dt dt d2x2 ,„ ч dx2 { 2\ dx1 dx2 n —=- - (2 tan t) — + -r — — = 0 (3) Л2 v > dt \xll dt dt y ' All that remains is to verify that the functions (1) satisfy the system (2)-(3). Substituting in (2): a(sec t + 2 sec t tan () - (2 tan ί)(α sec ttan t) - (a sec t)(\) = 0 Substituting in (3): 0- (2tani)(l) + ( )(a sec t tan t)(\) = 0 qed or and or DIFFERENTIATION RULES 6.15 Prove the rules for covariant differentiation stated in Section 6.6. (a) The sum rule obviously holds, as (6.7) is linear in the tensor components. (b) Let Τ = (Г.) and S = (S)) be two mixed tensors of order 2, with outer product U = (T^Sqs). Then, Kks: + TpTsik этрг . „_, w „ \ . /as \ I 3S ^X / \ OX

80 THE DERIVATIVE OF A TENSOR [CHAP. 6 dx dx I dUpJldxk = Upq u n,k and this proof of the outer-product rule extends to arbitrary Τ and S. (c) The inner-product rule follows from the outer-product rule and the following useful result: Contraction of indices and covariant differentiation commute. To prove this last, let R= (R'l). Then, / r)R'' \ Kflk, = {-jj + KK + rX - Κ,ιϊήδ] r)R'k = -^- + г;1л*+о = («*),, oed 6.16 Instead of Problem 6.15, why not: "Each rule is a tensor equation that is valid in rectangular coordinates, where covariant differentiation reduces to partial differentiation. Therefore, each rule holds in every coordinate system."? If the space metric is non-Euclidean, there is no way to transform to a rectangular coordinate system (in which the rules would indeed hold). 6.17 Infer the outer-product rule for absolute differentation from the corresponding rule for covariant differentiation. Let χ = x(i) be any curve, and T(x(i)) and S(x(i)) two tensors defined on the curve. Then, UNIQUENESS OF THE ABSOLUTE DERIVATIVE 6.18 Prove Theorem 6.4. Denote by ΔΤ/Δί any tensor that satisfies the hypothesis of the theorem. The tensor equation AT _ δΤ Δί ~ δί is valid in rectangular coordinates (*'), since, in (*'), both sides coincide with dT/dt. But then (Section 4.3) the equation holds in every coordinate system; i.e., AT _ δΤ ^δ7 = δί dxk dt Γ dxk Τ . S L -k dt 1 + dx' TS, — L ·* dt __ ΓδΤ Ί -г- S L δί J + Γ 8S1 T^~ L δί J Supplementary Problems 6.19 Find the general solution of the linear system a,. = const. d2x' = дх'дх" ~ a'k к with a'jk symmetric in the two lower subscripts. [Hint: Set y'k = dx'ldx - a'rkxr.]

CHAP. 6] THE DERIVATIVE OF A TENSOR 81 6.20 A two-dimensional coordinate system (xl) is connected to a rectangular coordinate system (x') through -1 τ/ 1\2 r 2 -2 1 _r_ -, 2 x ~2(x ) + χ χ ■=■ —x +3x (a) Exhibit the metric tensor in (*'). (b) Calculate the Christoffel symbols of the first kind for (*') directly from the definition (6.1). 6.21 (a) Derive the formula __ d2Xr dxr l'k ~ дх'дх1 дхк when (χ') is rectangular and (x1) is any other coordinate system. [Hint: Interchange barred and unbarred coordinate systems in (6.3)]. (b) Derive the analogous formula __ d2Xr dX ik ~ дх''дхк ~dF when (f) is such that all gi; arc constant. [Hint: Interchange barred and unbarred coordinate systems in {6.5).] 6.22 Let the coordinate system (*') be connected to a system of rectangular coordinates (x') via x1 = exp (x1 + x2) x2 = exp(x — x~) Use Problem 6.21(b) to compute the nonzero Christoffel symbols of the second kind for (*'). 6.23 If x1 = — exp άΛχλ + exp d2x2 + exp аъхъ χ" = 2 exp ί/jjt ~- exp d2x + exp d3x χ = exp dxx — 2 exp d2x + 3 exp d3x and if all fjk = 0, find the T)k. 6.24 Derive (6.6) by solving (6.5) for the second derivative and then changing indices. 6.25 Prove that all Г^ vanish only if all g, are constant. 6.26 Calculate the nonzero Christoffel symbols of both kinds for the Euclidean metric in cylindrical coordinates, (5.3). 6.27 Express the condition that the transformation (5) of Problem 6.6 be bijective (Section 2.6). 6.28 Show that if l)jk are constant, then g . are linear in the variables (x); but that this is not necessarily true if Vjk are constant. (For a counterexample, use the metric gn — exp 2л:1, g12 = g21 = 0, g22 = 1.) 6.29 What is the most general two-dimensional transformation x' = x'(x) of coordinates such that (x1) are rectangular and the Christoffel symbols Г'^ in (x') are those for the metric of polar coordinates (Example 6.3)? 6.30 Is the covariant derivative of a tensor with constant components equal to zero as in ordinary differentiation? Explain your answer. 6.31 If T'jrs are tensor components, write out the components of the covariant derivative, T'jrs k. 6.32 Show that 8)k = 0 for all i, j, k. 6.33 For any tensor T, verify that (g*T) k = g*T k, where * denotes either an outer or inner product.

82 THE DERIVATIVE OF A TENSOR [CHAP. 6 6.34 Use Problem 6.32 and gjrg" = 8) to show that the covariant derivative of g"1 is zero. 6.35 Use the recursive method of Problem 6.8 to verify that (Ttj k) is a tensor. 6.36 Using tensor methods in polar coordinates, And the curvature of the circle x = b χ = t 6.37 If the metric for (x) is (V)2 οίο 1. (α) write the differential equations of the geodesies in terms of the dependent variables и = (χ1)2 and υ = χ2; (b) integrate these equations and eliminate the arc-length parameter from the solution. 6.38 Find the geodesies on the surface of a sphere of radius a by (a) writing the geodesic equations for the spherical coordinates x2 and x' (the j^-equation is trivial for x1 = a = const, and may be ignored); (b) exhibiting a particular solution of these two equations; and (c) generalizing on (b). Use Problem 6.5 for the Christoffel symbols.

Chapter 7 Riemannian Geometry of Curves 7.1 INTRODUCTION At this point, some new terminology is introduced, which commemorates the general formulation of и-dimensional geometry by Bernhard Riemann (1826-1866). Definition 1: A Riemannian space is the space R" coordinatized by (*'), together with a fundamental form or Riemannian metric, g^dx' dx', where g = (g,:) obeys conditions A-D of Section 5.3. Thus, in our preliminary treatment of angles, tangents, normals, and geodesic curves, in Chapters 5 and 6, we already entered Riemannian geometry—though largely restricted to familiar 3-dimensional coordinate systems and a positive definite (Euclidean) metric. The present chapter focuses on the theory of curves in a Riemannian space with an indefinite metric. It also takes up geodesies from a different viewpoint. 7.2 LENGTH AND ANGLE UNDER AN INDEFINITE METRIC Formulas (5.10) and (5.11) must be generalized to allow for changes in sign of the fundamental form. Definition 2: The norm of an arbitrary (contravariant or covariant) vector V is \\\\\^VeVi = ^eVyi (e = e(V)) where ε( ) is the indicator function (Section 5.3). Under this definition, ||V|| ^ 0, but it is possible that ||V|| = 0 for V# 0; such a vector is called a null vector. Moreover, the triangle inequality is not necessarily obeyed by this norm (see Problem 7.8). If V(i) is the tangent field to the curve x' = x\t) (a^t^b), then the length formula (5.1a) may be written as The angle between non-null contravariant vectors is still defined by (5.11), provided the new norm is understood: UV g-U'V' 11411141 yelgpqUpUq Ve2grsV V where ег = e(U) and ε2 = e(V). Because of the indefiniteness of the metric, we must distinguish two possibilities in the application of (7.2). Case 1: |UV|^||U|| ||V|| (the Cauchy Schwarz inequality holds for U and V). Then θ is uniquely determined as a real number in the interval [0, 77]. Case 2: |UV| > ||U|| ||V|| (the Cauchy-Schwarz does not hold). Then (7.2) takes the form cosd = к (|fc|>l) which has an infinite number of solutions for Θ, all of them complex. By convention, we always choose the solution 83

84 RIEMANNIAN GEOMETRY OF CURVES [CHAP. 7 f/lnf/c + VV-l) k>\ . {77 + iln(-k + Vk2-l) k<-l that exhibits the proper limiting behavior as k—> 1+ or k—>-l~. EXAMPLE 7.1 At the points of intersection, And the angles between the curves (i.e., between their tangents) % : (*ί) = (ί,0,0, t2) % : (4) = (к, 0,0, 2-и2) (t, и real), if the Rieiiiannian metric is ε ds2 = (ώ;1)2 + (<fc2)2 + ((it3)2 - (dx*)2 [This is the metric of Special Relativity, with x4 = (speed of light) x (time).] The curves meet in the two points P(l, 0, 0,1) and β(-1, 0, 0,1). At P, (where t = и = 1) the two tangent vectors are UP = (dx\ldt)p = (1, 0, 0, 2t)p = (1, 0, 0, 2) Vf = {ах\1аи)г = (1,0, 0, -2u)p = (1, 0, 0, -2) so that (7.2) gives cos θ = l(l)(l) + l(0)(0) + l(0)(0)-l(2)(-2) Vejl(l)2 + Ц0)2 + Ц0)2 ~ 1(2)2] У e2[l(l)2 + 1(0)2 + 1(0)2 - l(-2)2 5 43V+3 3 and θρ = i In [(5/3) + V(5/3)2-l] = i In3. Similarly we calculate (for —ί = и = 1) ϋβ = (1,0,0, -2)=\ρ νβ=(1,0,0,2) = υ, so that θβ = ΘΡ. 7.3 NULL CURVES If g is not required to be positive definite, a curve can have zero length. EXAMPLE 7.2 In R4, under the metric of Example 7.1, consider the curve x1 = 3 cos t x2 = 3 sin t хъ = 4ί χ4 = 5ί for 0 S ί S 1. Along the curve, 'dx'> . . (—3 sin t, 3 cos t, 4, 5) at I (ds\ dx' dx' and so the arc length is L = [ 0<ft = 0 Jo A curve is null if it or any of its subarcs has zero length. Here, a subarc is understood to be nontrivial; that is, it consists of more than one point and corresponds to an interval с ^ t^d, where с < d. A curve is null at a point if for some value of the parameter t the tangent vector is a null vector; i.e., dx dx' n *'/-л "л =0 The set of ί-values at which the curve is null is known as the null set of the curve.

CHAP. 7] RTEMANNIAN GEOMETRY OF CURVES 85 Under the above definitions, a curve can be null without having zero length (if there is a subarc with zero length); but a curve having zero length is necessarily null at every point, and hence a null curve. Example 7.2 gives such a curve. EXAMPLE 7.3 Under the Riemannian metric G- (χ1)2 -i -1 0 the curve (x\ χ2) = (ί, |ί3|/6) possesses a null subarc that renders the length of the curve much smaller than might be expected. In fact, because dx1/dt = 1 and dx2Jdt = δ ί2/2, where δ = ±1 and is positive if 13 0, ^ΊΤΛ -[(x1)^)2^^ ^] = ^(1)-2(1)(8Л2)]=е(г»-80 Since the quantity following the indicator is nonnegative, ε = +1 everywhere. But note that t2 - δ t1 = 0 if iSO. Hence, r-999 /*0 /.999 /*0 L=\ У/t2 - δ t2 dt = J V2? dt + 0i/i = V2j (-t)dt = -V2i2/2|0_j=V2/2« 0.707 The interpretation in rectangular coordinates (x\ x2) is queer: As a particle travels less than a millimeter along the curve, its "shadow" on the x'-axis travels a meter! Nonexistence of an Arc-Length Parameter For a positive definite metric, the arc-length parameter s is well-defined by (5.6) as a strictly increasing function of the curve parameter t. (Then it is also the case that t is a strictly increasing function of s.) This fact allowed us freely to convert between the two parameterizations in the Solved Problems to Chapter 6. However, it is clear that on a null curve, which possesses at least one interval ij < t< t2 of null points, it is impossible to define arc length s. Indeed, even isolated points of nullity pose analytical problems. For if s'(t0) = 0, then the chain rule, ds dt s'{t) K } breaks down at s0, the image of i0. When necessary, we get around the difficulty by restricting attention to curves that are regular. Definition 3: A curve is regular if it has no null points (i.e., ds/dt>u). It will be further assumed that all curves are of sufficiently high differentiability class to permit the theory considered; usually, this will require the assumption that curves are of class C2. 7.4 REGULAR CURVES: UNIT TANGENT VECTOR Let a regular curve r£ : x' = x'(s) be given in terms of the arc-length parameter; the tangent field is T = (dx'lds). By definition of arc length, = Jj|T(«)|| du and differentiation gives 1 = ||T(i)||, showing that Τ has unit length at each point of r-€. When it is inconvenient or impossible to convert to the arc-length parameter, we can, by (7.3),

86 RIEMANNIAN GEOMETRY OF CURVES [CHAP. 7 obtain Τ by normalizing the tangent vector U = (dx'/dt): T = 1 U: 1 и JU|| - s'(t) In Problem 7.20 is proved the useful Theorem 7.1: The absolute derivative δΤ/δί of the unit tangent vector Τ is orthogonal to T. (7.4) 7.5 REGULAR CURVES: UNIT PRINCIPAL NORMAL AND CURVATURE Also associated with a regular curve % is a vector orthogonal to the tangent vector. It may be introduced in two ways: (1) as the normalized δΎ/Ss, if it exists; (2) as any differentiable unit vector orthogonal to Τ and proportional to 8T/8s when ||δΤ7δί|| ¥=0. The latter definition is global in nature, and it applies to a larger class of curves than does the former. Analytical (Local) Approach At any point of % at which ||δΤ7δί|| #0, define the unit principal normal as the vector Nft δΤ 8s ι 8Ύ 8s The absolute curvature is the scale factor in (7.5): 8Ύ 8s 8T' 8T' ~~ Ve8,1 Ss 8s (7.5) (7.6) This notion of curvature was informally denned in [6.12). Calling this quantity "curvature" is suggestive of the fact that in rectangular coordinates ||δΤ/δί|| = ||dT7di|| measures the rate of change of the tangent vector with respect to distance, or how sharply <£ "bends" at each point. Substitution of (7.6) into (7.5) yields one of the Frenet equations: ST " K^O) (7.7) 8s k0N0 While this approach is simple and concise, it does not apply to many curves we want to consider; for instance, a geodesic—as defined by (6.13)—will not possess a local normal N0 at any point. Even if there is only one point of zero curvature and the metric is Euclidean, N0 can have an essential point of discontinuity there. EXAMPLE 7.4 The simple cubic у = xi has an inflection point at the origin, or s = 0 (by arrangement). As shown in Fig. 7-1, lim N„ = (0,-1) lim No = (0,1) To verify this analytically, make the parameterization χ = t, y = t, and calculate N0 as a function of t (j'(0 = Vl + 9f4). U = (*'(*),/(f)) = (l,3f2) 1 .. 1 s'(t) ds Vl + 9i 1 dT 6i (1.3П i'(0 dt (1+9Γ)' (~3i2,1) ds (l + 9i4)3'2

CHAP. 7] RIEMANNIAN GEOMETRY OF CURVES 87 Fig. 7-1 Nn = 1 dT tl\t\ (--3i2, 1) (i^O) 0 «o ds Vl + 9i4 The scalar factor tl\l\ accounts for the discontinuity in N0 at t — 0 (s = 0). Geometric (Global) Approach A unit principal normal to a regular curve % is any contravariant vector N = (N'(s)) such that, along r£, A. N' is continuously differentiable (class Cl) for each /; B. ||N|| = 1; C. N is orthogonal to the unit tangent vector T, and is a scalar multiple of STISs wherever ΐΐδτ/δίΐι#ο. The curvature under this development is defined as κ Ξ εΝ 17 = ε8ί>Ν' ^f (ε = ε(Ν)) If the metric is positive definite, the Frenet equation δΤ !7 = *N holds unrestrictedly along a regular curve (see Problem 7.13). (7.8) (7.9) EXAMPLE 7.5 For the curve of Example 7.4, conditions A, B, and С allow precisely two possibilities for N: N = + 1 vTTV (-3t2,1) N=-7=Li-(-3f2,l) for —oo<j<oo. Geoinetrically, these amount to reversing the normal arrows in either the left half or the right half of Fig. 7-1. The corresponding formulas for curvature are (ε = 1) 6t (\ + 9ty or 6/ On curves having everywhere a non-null δΤ/δί, either N =N„ (with к - кп) or N = —N„ (with k = —k0). Thus, the global concept applies to all curves covered by the local concept and, in addition, to all regular planar curves (see Problem 7.14) and all analytic curves (curves for which the χ are representable as convergent Taylor series in s).

RIEMANNIAN GEOMETRY OF CURVES [CHAP. 7 7.6 GEODESICS AS SHORTEST ARCS When the metric is positiye definite, a geodesic may be defined by the zero-curvature conditions (6.13), or, equivalently, by the condition that for any two of its points sufficiently close together, its length between the two points is least among all curves joining those points. The minimum-length development employs a variational argument. We need to assume that all curves under consideration are class C" (that is, the parametric functions which represent them have continuous second-order derivatives). Let χ ~ x'(t) represent a shortest curve (geodesic) passing through A = (x'(a)) and В = (x'(b)), where b - a is as small as necessary. Embed the geodesic in a one-parameter family of C2 curves passing through A and B: x' = X(t, u) = x\t) + (t - a)(b - 1)иф\г) where the multipliers φ'(ί) are arbitrary twice-differentiable functions. The length of a curve in this family is given by r. . Г аХ ах> . Г г-—_ А with ε = 1 for a positive-definite metric. Since X'(t, 0) = x'(t) (i = 1, 2, ... , n), the function L(u) must have a local minimum at и = 0. Standard calculus techniques yield the following expression of the necessary condition L'(0) = 0: rb Г .,„ де,= dxl dx' d ( .,„ dr'Y (t-a)(b-t)4>\t)dt = 0 (7.10) dgjj dx' dx' d ( _I/2 dx' ~д7 ~dt' ~dt ~ Jt\ 8ik~dt in which , „4 dx' dx' — w(,,0) = g,^-^ (7.11) Since (t — a)(b — t)>0 on (a, b) and φ (t) may be chosen arbitrarily, the bracketed expression in (7.10) must vanish identically over (a, b), for к = 1, 2, . . . , л; this leads to (Problem 7.21) d x' ^, dx1 dx 1 dw dx1 л „ ■7Т+Г*ТТ = ГТТ (i=l,2,...,n) (7J2) Л2 'k dt dt 2w dt dt v ' y v y System (7.12), with w defined by (7.11), are the differential equations for the geodesies of Riemannian space, in terms of the arbitrary curve parameter t. Assuming that these geodesies will be regular curves, we may choose t = s = arc length. Then HD'-d)1-^ - £=· so that (7.12) becomes d χ „,■ ώ; dxk , , which is precisely (6J3). It must be emphasized that L'(0) = 0 is only a necessary condition for minimum length, so that the geodesies are found among the solutions of (7.12) or (7.13). Null Geodesies Consider the case of indefinite metrics and class <€г curves which may have one or more null points. Since, at a null point, w = 0 in (7.11) the variational theory breaks down, because L(u) fails

CHAP. 7] RIEMANNIAN GEOMETRY OF CURVES 89 to be differentiable at such a point. Analogous to the zero-curvature approach, we consider the more general condition for geodesies d2x' ,„■ dx' dxk 8Ul ,. Λ „ ^ + i^^=¥^° 0 = 1,2,...,«) (7.14) where U = U' = (dx'ldt) is the tangent vector field. By properties of absolute differentiation, along a solution curve to (7.14); so w = const, along the curve. Since the curve has at least one point of nullity, w = 0 at all points, whence the curve is a null curve—called a null geodesic. In summary, the following system of η + 1 ordinary differential equations in the η unknown functions x'(t) will determine the null geodesies: d2x' ■ dx dx -άΤ+Υ"-ά7Ίΰ = * 0 = 1.2...-,π) dxr dxs *»ΊΓΊΓ=0 <7-i5> EXAMPLE 7.6 If the g;. are constants, (7.15) has the expected general solution x' = x'0 + a't with gt a'a' = 0 Imagining the χ to be rectangular coordinates, we interpret the null geodesies as a bundle of straight lines issuing from the arbitrary point x0; each line is in the direction of some null vector a. By elimination of the a' the equation of the bundle is found to be gij(xi-xi0)(xi~xi0) = Q tivity (gn = g22 = g33= - (S - x\f + (x* - x\f + (x> - xlf = (x* - xlf In particular, for the space of Special Relativity (gn = g22 = g33 = -g44 = 1, giy = 0 for iV/'), the null geodesies compose the 45° cone —see Fig. 7-2. Y- / / / I 45 _^ ^ -4 _ „4 Fig. 7-2

90 RIEMANNIAN GEOMETRY OF CURVES [CHAP. 7 Solved Problems LENGTH IN RIEMANNIAN SPACE 7.1 Determine the indicator of the tangent vector U to the curve 1 ^3 X = t 2 ,2 X = t X = t (-co</<co) if the fundamental form is (a) (dx1)2 + (x2f(dx2)2 + (xlf(dx3)2 - 6 dx1 dx3 + 2xlx2 dx2 dx3 (b) (dx1)2 + 2(dx2)2 + 3(dx3)2 (a) (3t2)2 + t\2tf + t\\f - 6(3ί2)(1) + 2(ί3)(ί2)(2ί)(1) = 9t6 + 9f - 1812 = 9t\t2 + 2)(i2 - 1) Since t2 + 2 is always positive, e(U) = · (b) e(U) = +1, because the form is positive definite. + 1 -1 + 1 -1 + 1 cai 0<ί<1 t = 0 ■KKO iS-1 7.2 Show that the following matrix defines a Riemannian metric on R : G = 2 Γ 1 2 (χ >0, -χ1 < χ < χ ) We must show that conditions A-D of Section 5.3 are satisfied. A. Since each g.. is linear in the x\ it is differentiable to any order. B. By observation, the matrix is symmetric. С \sA = C*2)2 ~ C*1)2 <^ over tne given domain. D. Extend the matrix to a tensor g by using the tensor transformation laws to define the g;. in terms of the g;.. This will then make the quadratic form g-y dx' dx1, hence the distance formula, an invariant. 7.3 Find the null set of the curve % : χ - (χ1) (χ1 >0) under the metric of Problem 7.2. Let 9? be parameterized by x1 = t, χ1 = ί2 (i>0). Then, along f<? dx' dx1 ''' dt dt [1 2i\ ·: 7]ti]-u *■ -ί + 2ί3 ί2(4ί2 - 3) which, for positive t, vanishes only at t = V3/2. 7.4 Find the arc length of the curve % in Problem 7.3 from x1 = 0 to xl = 1. Again using t = x1, observe that dx' dx' 8i' ~dt ~dt <0 for 0<i<V3/2 Hence, 3V3 + 1 ώ -^O-^r V5/2 -. о +Т1(4'2"3)3 V5/. 12 0.516

CHAP. 7] RIEMANNIAN GEOMETRY OF CURVES 91 7.5 7.6 Write g = det G for the determinant of a Riemannian metric. Prove that | g\ is a differentiable function of the coordinates. Applying the chain rule to \g\ = Vg , we have d\g\ _ g dg dx \g\ Эх' (1) Since dgldx1 exists (Property A) and \g\ Φϋ (Property C), the right-hand side of (1) is well-defined. Show that under the metric ε ds2 = (dx1)2 - (dx2)2 - (dx3)2 - (dx4)2 (another version of the metric for Special Relativity), the curve xl = A sinh t x2 = A cosh t x3 = Bt x4 = Ct (0 S ί S 1) with A2 = В + С2, is null at each of its points. dx dx' 8ii ~dt ~dl dx ~dt dx It dx dt dx^_ dt (A cosh tf - (A sinh tf - B2 - C2 /i2(cosh21 - sinh21) - B2 - C2 A2- B2 - C2~Q 7.7 At the point of intersection (0,0), find the angle between the curves a> . \xl = 2i-2 % ■ U2 = /2-i «, 4 1 и — 1 .x~ = 25w + 50м-75 if the Riemannian metric is given by ε ds2 = (dx1)2 — 2 dx1 dx2. At t = 1, Τ = (dx'ldt) = (2, 2); at к = 1, U= (dxlldu) = (4, 100). Hence, using matrices, TU=[2 2] 4 100 ] -200 e,P 2] and ε2[4 100] cos θ -ί "ί][ιο4ο]^^Κ-784)^784 -200 V4V784 25 7 This is Case 2 of Section 7.2; we have о^ + пм24 + Щ)2-1 ■π + i\n' 7.8 Verify that the vectors of Problem 7.7 do not obey the triangle inequality. As calculated, ||T|| + ||U|| = 2 + 28 = 30. But ||Т + и||2=г3[6 Ю2][_| "J][1J2] = e3(-1188) = 1188 whence ||T + U|| « 34.46 > ||T|| + ||U||. ARC-LENGTH PARAMETER, UNIT TANGENT VECTOR 7.9 Let r-€ : xl = xl(t) be any non-null curve, (a) Prove that arc length along ^ is defined as a strictly increasing function of t. (b) Exhibit the arc-length parameterization of <$.

92 RIEMANNIAN GEOMETRY OF CURVES [CHAP. 7 (a) For tx < t2, the Mean-Value Theorem of calculus gives 4'2) - ί(Ί) = (t2 - Qs'ir) (ί, < τ < t2) The right-hand side is nonnegative, so that s{t,) Ss(t2). But, in view of the identity s(t2) - sit,) = [s(Q - s(t3)] + [s(t3) - j(i,)] where t3 is any point in (tlt t2), the equality s^t^ = с = s(t2) would imply s{t3) = c; i.e., s{t) would be constant on \tlt t2], making s'(t) = 0 on (ί,, i2) and thus making <£ a null curve. We conclude that s{tt) < s(t2) whenever tx < t2 (b) The strictly increasing function s{t) will possess a strictly increasing inverse; denote it as t = 9{s). Then <€ admits the parameterization x' = x'{6{s)). 7.10 {a) In rectangular coordinates (χ1, χ ) but adopting the metric of Problem 7.7, find the null points of the parabola % : xl = t, χ - t1 (0 S t S §)■ (b) Show that the arc-length parameterization of 4Ϊ is differentiable to all orders except at the null points, (c) Find the length of <#. , ч ί ds\2 I ахг\2 „ dx1 dx so there is only one null point, at t— IIA. (b) s= VK1 ~4m) ^m Thus, for OS ig 1/4, ί = |o VT^7^ i/M = i [1 - (1 - 4i)3'2] and, for 1/4S/S1/2, s=\ V^_г4ΐί/м + Jl;4V4iΓгTί/м=^[l + (4ί-l)3,2] Inversion of these formulas gives 1 t = 0(ί)ί [1-(1-6ί) ] 0SiSl/6 I [1 + (6.5 - If'3] l/6iigl/3 (i) It is evident that θ(ί) is infinitely differentiable except at the null point s=l/6 (the image of t= 1/4); the same will be true of the functions xl = 9{s), x2 = 92{s). (c) Set t = 1/2 in the applicable expression for s: s=\[l + (2-lf'2]=\ 7.11 Find the arc length of the same curve 4 as in Problem 7.10, but with the normal Euclidean metric, ds = {dx ) + {dx2) . ds \( ахг\2 (dx2'*·2 so that L4',2V^TI^4^V^TI+iln(^ + V47TI)]1,^V2 + ln(1 + V2) -0,574 Jo L2 4 Jo 4 as compared to L —0.333 in Problem 7.10.

CHAP. 7] RIEMANNIAN GEOMETRY OF CURVES 93 7.12 Using the arc-length parameterization found for the curve 4Ϊ in Problem 7.10(b), compute the components T'(s) of the tangent vector and verify that this vector has unit length for all 5^1/6. We have (T') = (0', 200'), where 0 = 0(s) is the function defined by (1) in Problem 7.10(6). Hence, ||T||2 = ε(θ'2 - 400'2) = ε(\ - 40)0'2 But, by (1) of Problem 7.10(b), _( (l-6i)2'3 0ei§l/6 я^И1"6·*)-"3 0<s<l/6 (6s-l)2'3 1/6SJ51/3 [(6i-l)"1/3 1/6<ί<1/3 Therefore, ||T||2 = (ε)(±1) = +1, or ||T|| = 1 (s^M6). UNIT PRINCIPAL NORMAL, CURVATURE 7.13 Prove that the Frenet equation (7.9) holds at each point of a regular curve when the metric is positive definite. At a point where ||δΤ/δί|| 5^0, we have (from property С of N), N = A — (1) OS from some real A. Take the inner product with the vector N in (1); with ε — ε(Ν), εΝ2 = ελΝ —= λκ or l = λκ (2) OS Then A = 1/κ, and substitution into (1) yields (7.9). At a point where ||δΤ/δί|| = 0, both δΊ/ds = 0 (because the metric is positive definite) and к = 0 (by (7.8)); the Frenet equation then holds trivially. 7.14 For any regular two-dimensional curve % : x' = x'(s), define the contravariant vector n=(Ni)^(-T2/V\i\,Tl/VU[\) (7.16) where Τ = (Тг) is the unit tangent vector along <в and g = det (&.). Show that N is a global unit normal for %. We must show that the three properties of Section 7.5 are possessed by the given vector (except possibly at null points). A. Since <£ is regular, the V, and with them the Tt = gtjT', are in C1. The same is true of \g\ (Problem 7.5), which function is strictly positive. Therefore, the N' are also in C1. B. By (2.11), g" =gjg,g" = g21 = -gl2/g, and g22 = gjg. Hence, Ml2 = \8u(T22/\g\) + 2g12(-T1T2/\g\) + g22(TV\g\)\ = Y^\gg22Tl+2ggl2TlT2 + gguT\\ Ы — ' ° ' \ n,}T Τ I \T}T I — II Til2 \g\18 ' /M >]~m and so ||N|| = ||T|| = 1. C. N is orthogonal to T: Τ Τ N'T, = - -j^= Tl + -j= T2 = 0 Furthermore, when ||δΤ/δί|| 5^0, then N0 is defined and is also a vector orthogonal to Τ (by Theorem 7.1). In two dimensions this implies N= ±N0 = Α(δΤ/δί).

94 RIEMANNIAN GEOMETRY OF CURVES [CHAP. 7 7.15 For the curve and metric of Problem 7.10, determine the local normal N0 and, using Problem 7.14, a global normal N. Verify that the two stand in the proper relationship. We have gll = 1, g12 = g21 δΤ dT -1> #22 = ϋ (all constants), and T = (0', 200'); therefore, for i^ 1/6, 8s ds and Thus, With g = (0",20'2+200") Γ(2(1 - 6s)-413, (1 - 6y)-2'3 + (1 - 6i)-4'3) [ (-2(6* - 1)"4'\ (6s - IT2'3 - (6* - I)"4'3 δΤ 0<s<l/6 1/6<ί<1/3 Ν -** *° 8s/ 8s δΤ 2(1 - 6ί)~ 2(6ί-1)" 8s dT_ dV_ Veg" ds ds ((1-6J)"1'3,2(1-б*)1/3 + Н1-б*Г'3) (-(6s-iruM(6s-iYi3-{(6s-iy"3) -1, Problem 7.14 gives (s ^ 1/6): 7\ = gyT' =T" -T2 = 0'(1 ~ 2Θ) T2 = g2jT' = -Tl = ~e' \(\-6syV3 + \(\-6s)113 \(6s-\yll3-\{6s-\Y'3 (i-fo)' (6, - 1) -1/3 -1/3 N _ ,_ T T s _ { ((1 " 6^)-1/3, 2(1 - 6,)"1'3 + 1(1 - 6s)1'3) It is seen that, as expected, N= +N0 for s<l/6 and N= -N0 for s>l/6. neither N0 nor N is defined at the null point s = 1/6. For comparison, recall the situation in Examples 7.4 and 7.5: there the discontinuity in N0 occurred at a regular point (the cubic has no null points under the Euclidean metric), and N (either choice) was defined everywhere. 7.16 Under the metric of Special Relativity (Example 7.1), a regular curve r£ is given by 7 3S 3 4s X = 7 4 2 X — S for OSiSl. (a) Verify that s is arc length for <в, and show that the absolute derivative, δΤ/δί, of Τ is a null vector at every point of the curve (hence, a local principal normal N0 is nowhere defined on <β). Construct a global principal normal for c€ in such a manner that the corresponding curvature function is nonzero. Is more than one curvature function possible? (a) We have (Γ) = (2s, 3/5, 4/5,2s) and Ι&,.ΓΤ'Ι = \As2 +- (9/25) + (16/25) - 4s2| = 1 hence, s is an arc-length parameter. Also, since the g are constant, all Christoffcl symbols vanish and f = f=ft»·»·2) δ Τ 8s У|22 + 02 + 02-22| = 0 for all s. (b) Any differentiable unit vector orthogonal to Τ will do for N, which then determines the curvature through (7.8). In the orthonormality conditions 2SN1 + | N2 + | N3 - 2sN4 = 0 1 \2 , /лг2\2 , /лг3ч2 /лг4ч2 , , (ЛГ) + (Ν2)2 + (Ν3)1 - (Ν*) we may successively set N1-^ Ν4 = 0, Ν3 = Ν4 = 0, and Ν2 normals: Ν = 0 to obtain three candidate

CHAP. 7] RIEMANNIAN GEOMETRY OF CURVES 95 Νι = (θ,-|, |,θ) Ν 1 V(9/25)+4.y2 V 5 1 / 4 -ί,2л, 0,0 N3= . \—=,0,2s,0 V(16/25) + 4j2 V 5 The constant N, yields к1 = О; but N2 and N3 yield the distinct curvature functions -1 -1 2 vTTf? 3 Vi^W Note that the Frenet equation is invalid for all these normals. 7,17 Refer to Problems 7.10 and 7.15. Calculate the curvature functions к0 and к, and discuss the variability of к0 over the parabolic arcOSiil/3. Our previous results show that both curvatures are defined everywhere except s = 1/6, with к — к0 on 0Si<l/6 and к = ~к0 on 1/6 <i el/3 (cf. Problem 7.13). By Problem 7.15, δΤ \l-6s\ (s Φ 1 /6) whence K = T^fa (s^1/6> It is seen that κ0 has the same value, 2, at s = 0 (the vertex, or point of greatest Euclidean curvature) and the undistinguished point s=l/3. Moreover, near the ordinary (from the Euclidean viewpoint) point ί = 1/6, the absolute curvature becomes arbitrarily large. 7.18 (a) For any regular two-dimensional curve, derive the formula for absolute curvature *ο=νϊΐϊ ,i 8Тг 8T1 8s 8s (b) Use (7.17) to check Problem 7.17. (a) By (7.8) and the remarks made following Example 7.5, δΤ N. ' 8s лг δΤ' лг δΤ N,^— + N, ds ds Choosing the global normal (Nl) established in Problem 7.14, we have: Ni = g11N1 + g12N2^(gg22) N2^g2X + g22N2^(~gg12 - y (g1% + g12T2) = V\g\J U. Ш + (gg") VW V\i\ τ, (g2iT1 + g22T2)- V\g\ Γ1 ~vhV6 Μ,β ί2> m Substitution of these components in (1) yields (7.17). (b) For the metric of Problem 7.10, 1 "Ί ■1 OJ G (7.17) (1) V\i\ g = det G = -1 and absolute derivatives reduce to ordinary derivatives. Thus we can rewrite (7.17) in terms of the curve parameter t, as follows: Ti dT2 dt t2 dT1 dt dt ds dt ds 1 s\t) dt dT dt (τ1)2 s'(t) d_ /ту dt Vr1;

RIEMANNIAN GEOMETRY §F CURVES [CHAP. 7 Substitution of s'(t) = л/11 — 4ί| (ίΦ 1/4) and the components of the unit tangent vector, 1 dx1 1 , 1 <fe2 It Τ': gives: s'(t) dt s'(t) 2 2 О /„'/,ллЗ (,'(0)3 |i-4i| From Problem 7.12, |1-4ί| = |1-4β(ί)| = |1-6ί yielding exact agreement with Problem 7.17. s'(t) dt s'(t) Ш (ίΦΙ/4) 7.19 Compute the absolute curvature of the logarithmic curve <% : x1 = t, x2 = a In t, for \ ^ t< a, if the Riemannian metric is ε ds1 = (dx1)2 - (dx2)2 As the gif are constants (with g = — 1), we can proceed as in Problem 7.18(b). This time, the most convenient version of (7.17) is Substituting Ao = V s'(t T2 = ^- «o = VW (dx \ (dx \ \~dtl ~\~dt! dx1 t ) dt yV-i2 dx' a (τ2)2 j'(0 W dt\T 2 1-- i2 ;) = *'(0 л vV-^? -V7^?(^o) we find: к0 = at(a — t ) ' . 7.20 Prove Theorem 7.1. Along a regular curve we have ||Τ||2=εΤΤ=1 or TT= ε where the indicator ε is constant, |ε| = 1, on the curve. By the inner-product rule for absolute differentiation, and the fact that the absolute derivative of an invariant is the ordinary derivative, δΤ δΤ δΤ d δΤ T + T —=2T — = —(e) = 0 or T —=0 ds 8s ds ds GEODESICS 7.21 Establish (7.12). Start with the conditions 1/2 ^Sij_ d)c_ (bd_ _ d_ I _1/2 dx_ J7 dt ~di ~ dt\ W 8ik dt (1)

CHAP. 7] RIEMANNIAN GEOMETRY OF CURVES 97 By use of the product and chain rules, the expression on the right may be written 3/2 dw I dx'\ -irJd8ik dx'\dx _ll2 d2x -di\^^)+2w \JffF)fF + 2w 8ikfP Put this back in (1), multiply both sides by w1'2, and go over to the notation gijk = dg^/dx": which rearranges to dx' dx' _j dw dx' dx' dx' d x' 8i>k ~dt ~dt=~w 8ik4i~dt + 8ik> ~dt~dt+ 8ik ~d7 d2x' dx' dx' dx' dx' _ 1 dw dx' 8ik df ~ 8i/k ~dl ~dt + 8ik> ~dt ~dt ~ w 8ik ~dt ~dt Making use of the symmetry of gij, the third term on the left may be split into two similar terms, yielding d2xl dx1 dx[ dx[ dxJ_ dx^_ dxJ_ _ j_ dw dx1 8ik df ~ 8i,k It ~dt + s>ki ~dt dt + 8k'i ~dt dt ~ w 8ik ~dt ~dt Divide by 2, multiply by gpk, and sum on k: d2x pk dx' dx' _ 1 dw dx' d2xp dx' dxk _ 1 dw dxp ' ~άΨ ~ 8 ijk ~dt ~dl ~ 2w~ ' ~dl ~dt ОГ If + >k ft dT ~ 2w ~dt It which is (7.12). 7.22 In a Riemannian 2-space with fundamental form (dx1)2 - (x2)~2(dx2)2, determine (a) the regular geodesies, (b) the null geodesies. Here gu = 1, g12 = g21 = 0, g22 = ~(x2y2; Problem 6.4 gives Г d dx as the only nonvanishing Christoffel symbol. (a) The system (7.13) becomes \^(x2y2 2 X d χ „ d χ 1 / dx 0 -гт--^;\-^) =0 ds ds χ The first equation integrates to x1 = as + x^. In the second, let и = dx2lds: du 1 2 du dx' 2 и —ι j и = U or — = —γ or и = едг dx χ их from which /ίν2 2 ^2 J 2 2c cjc or —γ = с ds or jc = x„e ds As our notation indicates, an arbitrary point (x0, x2a) is the origin (s = 0) of a family of geodesies that seems to depend on two parameters, a and c. However, s must represent arc length, so that dX \ , 2ч-2( "* ±1 = Ш-^"\Л Hence, either a2 = c2 + 1 (the fundamental form is positive) or c2 = a2 + 1 (the fundamental form is negative). Both cases may be accounted for by a single parameter, A, if s is eliminated between the parametric equations for x1 and x2: regular geodesies x2 = x\ exp [A(x! - xl0)] (\ A| φ 1)

RIEMANNIAN GEOMETRY OF CURVES [CHAP. 7 λ = -0.1 Fig. 7-3 (b) System (7.14), in t, becomes iZ 1 a x о dx ~dt d χ 1 / dx 1 \ 2 /J 2-\2 It is clear that the solution may be found by formally replacing £ by t in part (a) and setting a2 = c2. Thus, the null geodesies through (x\, x2a) are given by null geodesies x2 = x\ exp [+(x1 - x\)\ and x2 = x\ cxp \r(xx - x\)\ Note that the null geodesies correspond to the exceptional values A = ±1 in part (a). Figure 7-3 is а sketch of the geodesies through the point (1, -1) in cartesian coordinates. 7.23 Without converting to arc length, verify that in spherical coordinates, under the Euclidean metric ds = (dx ) + (x1 dx2)2 + (x sin χ dx )2 any curve of the form % : x1 = a sec t, x2 = t + b, x3 = с (a, b, с constant) is a geodesic. (It should be apparent that % is a straight line.) The equations (7.12) must be verified. The Christoffel symbols l"jk for spherical coordinates are (Problem 6.5): i, Л. J. 0-7 л * J. oo 1·22 x sin χ

CHAP. 7] RIEMANNIAN GEOMETRY OF CURVES 99 I — ί 1 12 — 1 21 — , , 1 sin jc cos χ • __ ij Τ"1 Ι-1 Ι-1 Τ"1 + The derivatives of the x'(t) are: , a sec Ζ tan Z, —=r Λ: dt 2 .2 2 1, —~г = 0 and dt1 (α sec r)(tan2 ί + sec21) = 0 J 3 ,2 3 ax _ d χ ~dt ~~df With ε = 1, (7.ίί) gives and = (α sec Ζ tan Ζ)2 + (α sec Z)2( 1 f + 0 = a2 sec4 Ζ 1 ί/w _ (4a2_sec3 Z)(sec Ζ tan t) 2a2 sec4 f 2 tan Ζ 2w i/Z For convenience in the verification of (7.12), let LS denote the left side, and RS the right side, of the equation in question. We obtain: = (a sec Z)(tan2 Ζ + sec2 Z) - (a sec Z)(l)2 + 0 = 2a sec Ζ tan2 Ζ Λ:1 RS = (2 tan Z) — = (2 tan Z)(a sec Ζ tan Z) = 2« sec Ζ tan2 Ζ = LS i = 2 LS^ —r + 2Γ2 — — + Γ2 Λ Λ dt 3 <fe2 RS = (2 tan Z) — = 2 tan Ζ = LS "л" 0+ (asecZtanZ)(l) + 0 = 2tanZ a sec Ζ v /v 7 i = 3 LS ^ ,2 3 d χ ~df 13 dt dt 23 dt dt dx3 RS = (2 tan Z) — = 0 = LS Supplementary Problems 7.24 Determine the fundamental indicator ε(υ) if (Lri) = (2Z,-2Z, 1) at the point (*') = (Z2, - Ζ2 Ζ). The Riemannian metric is given by (8,,)- Ix1 χ2 Ο χ 2x 0 (4*V#(*3)2) 0 0 1 7.25 Find the null points of the curve % : χ1 = Ζ, χ2 = Ζ4 (Ζ real), if the metric is ε ds2 = 8(xl dx1)2 - 2 dx dx 7.26 Find the arc length of the curve in Problem 7.25 if 0^ ζШ2.

100 RIEMANNIAN GEOMETRY OF CURVES [CHAP. 7 7.27 Find the null points of the curve % : x1 = t3 + 1, x2 = t2, x2 = t, if the metric is eds2 = (dx1)2 - (dx2)2 - (x2 dx3)2 7.28 Find the arc length of the curve in Problem 7.27 if \ ^ t S 1. 7.29 Find the angle between the curves |V = 5i |V=m % : \x2=2 % : \x2 = 2 lx2=3t U3 = 3m2/25 at each of the points of intersection, if the fundamental form is (dx1)2 —(dx ) — (dx3)2. 7.30 If ε ds2 = (dx1 f - (dx2)2, (a) find the length L of the curve 41 : χ1 = \2t2, x2 = 8i3, for 0S t ё 2. (b) Find an arc-length parameterization, x' = x'(s), for %, with s = 0 corresponding to t = 1. (c) Show that the x'(s) are differentiable to all orders except at points of nullity. 7.31 Find the arc length of the curve of Problem.7.30, but with the Euclidean metric. 7.32 Compute Τ = (dx'lds) from the arc-length parameterization found in Problem 7.30 and verify that Τ has unit length at all points except s = 0. 7.33 Calculate the components N' of the unit principal normal of the curve of Problem 7.30, using (7.16) (Problem 7.14). 7.34 Calculate both the curvature к and the absolute curvature к0 for the curve of Problem 7.30. Discuss the numerical behavior of к0 along the curve. 7.35 Use the formula of Problem 7.18(b) to confirm the value of к0 found in Problem 7.34. 7.36 Compute к0 under the Euclidean metric for the curve of Problem 7.30; compare with the result obtained in Problem 7.34. For convenience, let t = 0 correspond to s = 8. 7.37 Without calculating an arc-length parameter, find the vectors Τ and N, and the curvature к, for the "parabola" x1 = t, x2 = t2 (0 ^ t ^ \) under the Riemannian metric sds2 = (dx1)2 - 2dxl dx2 7.38 Show that the first-quadrant portion (x' > 0) of the hypocycloid Ж of four cusps (x')ln + (x2)2/3 = a212 (a>0) may be parameterized as x1 = a cos31, x2 — a sin3 t, with 0 S t S π 12. Find the arc length under the two metrics (a) eds2 = (dx'f-(dx2)2- (b) ds2 = (dx1)2 + (dx2)2 (c) Without computing an arc-length parameter, find Τ and κ0 for Ж under both metrics. 7.39 (a) Determine the Christoffel symbols of the second kind for the Riemannian metric ε ds2 = xl(dx1)2 + x2(dx2f. (b) Without converting to an arc-length parameter, verify that all curves x1 = t2, x2 = (at3 + b)2'3, where a and b are arbitrary constants, are geodesies.

Chapter 8 Riemannian Curvature 8.1 THE RIEMANN TENSOR The Riemann tensor emerges from an analysis of a simple question. Starting with a covariant vector (V-;) and taking the covariant derivative with respect to x' and then with respect to χ produces the third-order tensor (Wbbs(vu) Does the order of differentiation matter, or does Vl]k = Vikj hold in general? Standard hypotheses concerning differentiability suffice to guarantee that the partial derivative of order two is order-independent, д2У, d2Vt дх'дхк дхк3х' but due to the presence of Christoffel symbols, such hypotheses do not extend to covariant differentiation. The following formula is established in Problem 8.1: where R)kl = -ф--±+ Г}1Ггк - Г)кГг1 (8.2) The Quotient Theorem (covariant form) immediately implies Theorem 8.1: The n" components defined by (8.2) are those of a fourth-order tensor, contravariant of order one, covariant of order three. (R'jki) is called the Riemann (or Riemann-Christoffel) tensor of the second kind; lowering the contravariant index produces Rtjki = 8i,Rrjki (8.3) the Riemann tensor of the first kind. In answer to our original question, we may now say that covariant differentiation is order- dependent unless the metric is such as to make the Riemann tensor (either kind) vanish. 8.2 PROPERTIES OF THE RIEMANN TENSOR Two Important Formulas The Riemann tensor of the first kind can be introduced independently via the following formula (see Problem 8.4): From (8.4) there follows dT,n dT,ki *№l =-^ --^ + rlfrrv rtt,rr, (8-4) 1 / d2gu , d% d2gik d2gjl j? — t \ °·ι _l °'K — blk 22L· j_ Г Гг — Г Гг (Я *\ iikl ~ 2 \ дх'дхк + эх1дх> дх'дх' дхдхк J + "Л ik l 'kA 'Ί ^) EXAMPLE 8.1 Calculate the components Rijkl of the Riemann tensor for the metric of Problem 7.22, ε ds2 = (dx1)2 - (x2y\dx2)2 101

102 RIEMANNIAN CURVATURE [CHAP. 8 The nonvanishing Christoffel symbols are Γ22 = -(х2уг and Г222 = £22Г22 = (x2)~2. The partial-derivative terms in (8.4) vanish unless all indices are 2; but then the two terms cancel. Likewise the Christoffel-symbol terms either vanish or cancel. We conclude that all sixteen components Rijkl = 0. Symmetry Properties Interchange of к and / in (8.2) shows that R'jkI = -R),k, whence Rljkl = -Rljlk. This and two other symmetry properties are easily established at this point; Bianchi's (first) identity will be demonstrated in Chapter 9. first skew symmetry Rljkt = —Rjikl (8.6) second skew symmetry Rijkl = —Rii[k block symmetry Rijkl = Rklij Bianchi's identity Rijkl + Riklj + Rtljk = 0 Number of Independent Components We shall count the separate types of potentially nonzero components, using the above symmetry properties. The first two properties imply that Raacd and Rabcc (not summed on a or c) are zero. In the following list, we agree not to sum on repeated indices. (A) Type Rabab, a<b: nA = nC2 = n(n - l)/2 (B) ТуреДв6вс>Ь<с: nB = 3 · „C3 = n(n - l)(n - 2)/2 (C) Type Rabcd or Racbd, a < b < с < d (for type Radbc, use Bianchi's identity): nc = 2 · „C4 = и(и-1)(и-2)(и-3)/12 In (A) the count is of combinations of η numbers two at a time (for a and b). In (B) one partitions the index strings into the three groups (with ИС3 in each group) for which a<b<c b<a<c b<c<a Either subtype of (C) has as many members as there are combinations of η numbers four at a time (for a, b, c, and d). Summing nA, nB, and nc, we prove Theorem 8.2: There are a total of n2(n2 — 1) /12 components of the Riemann tensor (Rijkl) that are not identically zero and that are independent from the rest. Corollary 8.3: In two-dimensional Riemannian space, the only components of the Riemann tensor not identically zero are R12U = ^2121 = "^1221 = —^2112- EXAMPLE 8.2 For the metric of spherical coordinates, ds2 = (dx1)2 + (x1 dx2f + (x1 sin*2 dx2)2 list and calculate the nonzero components Rijkl, if any. By Theorem 8.2 with n = 3, there are six potentially nonzero components: (A) Л1212, Л1313, Л2323 (B) п1213, л1232(= a2i23), n1323(— /ί3ι32) Because Rijkl = guR'lkl (diagonal metric tensor; no summation), we may instead compute the mixed components. From Problem 6.5, i = 1 Y\2=-χ1, Γ33 = -χ1 sin χ i = 2 T22 = Γ21 = — , Γ33 = -sin x~ cos x2

CHAP. 8] RIEMANNIAN CURVATURE 103 and (8.2) gives: 1 _ 01 22 _ °l 21 „, „I _r _! _ _ -ι _ rl |-.l _ rf pi _ « Λ212 - 1 -2 """ J 221 rl 1211,2 1 * 221 11 121122 " ■"313 = —; Г ~~Ί \~ * зз^ r\~ ι 3i^ гъ ~ —Sin X + J 33J + J 31J 33 = U η p2 л p2 лз2з= —Τ ψ + гззГг2 " Гз2Ггз = -cos2jc2 + ГззГ^2 - Гз2Г23 = -cos 2x2 - sin2 χ2 + cos2 x2 =0 Эх Эх pl ^23 _ ^21 pr pi _ pr pi рЗ pi _ р2 pl ,, "213 - 1 ι 3 -1 23 Π J 211 ГЗ J 231 31 * 21-1 23 " σΧ Эх . _ ό>Γ22 _ £Гзз _r _i _ .г „ι _ „ι „ι _ гз Γι _ п п232 , 3 - 2 "■" J 221 1-3 х 23J ,2 J 22J 13 J 23* 32 " Эх Эх pl _ 33 _ "^ 32 , pr pl _ pr pl _ _ry 1 ■ 2 2 , p2 pl _ p3 pl ^323 — - 2 - 3 ' ■* 33^ r2 * 22* r3 — ^X Sln X COS ■* ' * 331 22 * 32* 33 = -2л:1 sin x2 cos x2 + jc1 sin л:2 cos x2 + (cot х2)(хг sin2 л:2) = О Therefore, Rijkl = 0 for all i, j, k, I. 8.3 RIEMANNIAN CURVATURE The Riemannian (or sectional) curvature relative to a given metric (&.) is defined for each pair of (contravariant) vectors U = (CT), V= (Vf) as J?,.fc,i/'V;'i/V K = K(x;U,V)=- f r (G^^g^-g^,.) (S.7) This sort of curvature depends not only on position, but also on a pair of directions selected at each point (the vectors U and V). By contrast, the curvature к of a curve depends only on the points along the curve. Although it would seem desirable for К to depend only on the points of space, to demand this would impose severe and unrealistic restrictions, as will become apparent in Chapter 9. EXAMPLE 8.3 The numerator of (8.7) is an invariant, because (RiJkl) is a tensor. As for the denominator, the identity GpqrsVPWV\2)V[3)V'W = (V(1)V(3))(V(2)V(4)) - (V(1)V(4))(V(2)V(3)) (8.8) implies that the denominator is an invariant and proves (Lemma 4.1) that (Gijkl) is also a tensor. It follows that K(x; U, V) is an invariant, and thus it serves to generalize the Gaussian curvature of a surface to higher dimensions. Helpful in the calculation of К is the fact that the Gikl possess exactly the same symmetries as the Rijkl (see Problem 8.19). Moreover, if g= (gif) is diagonal, all the nonzero Gi/k, will be derivable from the type-Α terms Gabat = Saagbb (<* < Ъ\ ПО вШППШЮП) EXAMPLE 8.4 Evaluate the Riemannian curvature at any point (xl) of Riemannian 3-space in the directions (a) U =(1,0,0) and V= (0,1,1), and (b) U= (0,1,0) and V= (1,1, 0), if the metric is given by £n= 1 #22 = ^ £зз=2х2 g,y = 0 if i Ф) From Problem 6.4, the nonzero Christoffel symbols are: pi _ ._ ι F2 = Г2 = F2 = — F3 = F3 = 1 22 i * 12 J 21 r, 1 * 33 r. 1 * 23 * 32 r. 2 Since η = 3, only six (by Theorem 8.2) components of the Riemann tensor need be considered: Λ1212, Λ1313, Л2Ч23, Λ1213, Λ2ΐ23 and ^3i32- The metric being diagonal, we compute:

104 RIEMANNIAN CURVATURE [CHAP. 8 R1 /\212 <9Г„ дТ0 Эх 21 r pi «9 л:1 /?313 ~ ^ R which yield the three terms (A) R1212 = gnR212 = l/2x , R2323=822R 1 4* У r^r^ = 0-0 + 0 «213=0 Г2 Г1 = — 21 22 2/ «223=0 «132 - 4x1x2 2 323 l/2x2 (B) R2132 = g33R]32 =l/2x1 Theorem 8.2 also applies to the Gikl; but we may take the shortcut indicated in Example 8.3: (A) *J1212 8ΐ1§22 ^X ' "1313 £п£зЗ ^X > "2323 . :4jt л: 522633 Let us now give an expanded form of (8.7) in the case that type-C terms are absent. It is convenient to define the nA functions of two vectors U = ((/') and V= (V). Observe that the Wijkl possess all the symmetries of the Rijkl (or the Gijkl). Looking at the numerator of (8.7), we see that a given type-Α coefficient from the basic set generates, via its skew symmetries, the 4 terms Rabab(UaVbU"Vb - UbVUaVb - UaVbUbV + VbVVbVa) R W nabab " abab and these precisely exhaust the 2x2 = 4 terms in which the coefficient R;jkl involves the same distinct integers, a and b, in the first pair of indices as in the second pair. A given type-B coefficient from the basic set will generate, via its skew symmetries and block symmetry, the 8 terms "■abac\ ''oboe ~"~ "a cab) ~ ^"-abac" abac and these exactly correspond to the 22 x 2 = 8 ways of writing Rijkl such that the first and second index-pairs contain the common integer a but are otherwise composed of distinct integers b and c. Analyzing the denominator of (8.7) in the same fashion, we obtain as the desired formula: ^-* "ubub^abab + 2 Zj Rabac^abac }У11Л 4^1 {8J0) K = Σ GababWabab + 2 Σ GabacWab type A type В It is understood that the summation convention does not operate in (8.10); the indicated summations are over all the nonzero, independent Rijkl (Gjjkl), according to type. Now to the problem at hand. (a) For the data, (8.10) becomes R W + R W +2R W - _ *v1212 rr 1212 T Jv2323 "2323 τ ^М132 "3132 a w + a w + с w "1212 1212 T *Ji3i3r,i3i3 ^ "2323 "2З23 K = For we have W = W, 1 0 0 1 0 1 1 0 Ы1 ° °1 J ίθ 1 L 1 W = 1 "2323 0 1 = 0 w13 13 0 0 1 1 1 0 0 1 and so K = (b) For (\!2xl)(l) + (1 /2x2)(0) + 2(112xl)(0) (2x1)(l) + (2x2)(l) + (ix1x2)(0) [sh: 1 si =0 = 1 1 4x\x1 + x2)

CHAP. 8] RIEMANNIAN CURVATURE 105 we have W = "1212 W = "3132 0 1 1 1 2 0 Oj 0 1 = 1 0 1 0 1 w = "2323 whence K = ;0 W1313 = (l/2*')(l) + 0 + 0 1 1 1 0 1 0 0 0 0 = 0 = 0 (2jc1)(1) + 0 + 0 \(xf Observations on the Curvature Formula I. If η = 2, (8.7) reduces to K = "1212 _ "1212 (8.11) 8^1822 812 8 (see Problem 8.7). Thus, at a given point in Riemannian 2-space, the curvature is determined by the gtj and their derivatives, and is independent of the directions U and V. II. The extension of (8.10) to include type-C terms is as follows: K = Σ Rababwabab + 2 Σ Rahacwahac + 2 Σ RaUwabcd - wadbc) + 2 Σ RachAWachd-wadcb) type A type В type С type С Σ GababWabab + 2 Σ GabacWabac + 2 Σ Gahcd(Wabcd-Wadbc) + 2 Σ Gacbd(Wacbd-Wadcb) type A type В type С type С (8.12) (see Problem 8.9). III. If linearly independent U and V are replaced by independent linear combinations of themselves, the curvature is unaffected; i.e., K(x; AU + vW, μ,υ + ω\) = K(x; U, V) (8.13) Therefore, at a given point x, the curvature will have a value, not for each pair of vectors U and V, but for each 2-flat passing through x. Isotropic Points If the Riemannian curvature at χ does not change with the orientation of a 2-flat through x, then χ is called isotropic. From (8.11), we have Theorem 8.4: All points of a two-dimensional Riemannian space are isotropic. It is not immediately clear whether any metric (gtj) could lead to isotropic points in R", η =ёЗ. But such is the case. Indeed, as is shown in Problem 8.12, R3 under a hyperbolic metric is isotropic at any point. 8.4 THE RICCI TENSOR A brief look will be given a tensor that is of importance in Relativity. The Ricci tensor of the first kind is denned as a contraction of the Riemann tensor of the second kind: ρ _ ηί i_k_ lJ_ , pr т^к pr j^k Kij-Kijk~ ϋχϊ dxk + l ikl rj l ijl rk Raising an index yields the Ricci tensor of the second kind: (8.14) (8.15)

106 RIEMANNIAN CURVATURE [CHAP. 8 By use of the following simple consequence of Laplace's expansion (2.5): Lemma 8.5: Let A = [atj(x)]nn be a nonsingular matrix of multivariate functions, with inverse B = [btj(x)]nn.Then -^(Ш|с1е1Л|)--^г-^(с1е1Л) = ^ да" dxl v ' u det Л дх' У ' sr дх1 the definition (8.14) may be put into a form (Problem 8.14) that makes evident the symmetry of the Ru «Γ ,. η-!. 1 *<>= ТТЛ <1п^ " Т7П ^7 (VH П) + rxy (SJ6) дхдХ \\g\ "X Here, as always, g = det G. Theorem 8.6: The Ricci tensor is symmetric. After raising a subscript to define the Ricci tensor of the second kind, R' - g'sRsj, and then contracting on the remaining pair of indices, the important invariant R = R\ results, called the Ricci (or scalar) curvature. By (8.16), R-g" d" ,. n—r, 1 S дХ дХ ι (inVfif) - 77r=f jrr ЫШП) + г;л. VI δ I (8.17) Solved Problems THE RIEMANN TENSOR 8.1 Prove (8.1). By definition of the covariant derivative, Substitute dV ν:. = —' - tv in (1), carry out the differentiation, and remove parentheses: d2V, ЭТ], dV, ,. dV dV. ^ V = V — Г —- - V —- + Г Г V — V —- + Γ ГУ (?\ У'·'" дхкдх> дхк Vs 1 '" дхк 'к дх' ik "s ik дхг +1 '" »V- {J) Interchanging / and к yields d2V. dTs dV dV dV '■k' дх'дхк дх' ' ik дх' u дх" " :ks 1"' dxr +1 k' ,rVs {J) Subtracting (3) from (2), one sees that the first, third, fourth, sixth, and seventh terms on the right of (2) cancel with the first, fourth, third, sixth, and seventh terms on the right of (3), leaving 3YS /9FJ V -V = V + ΓΓ Г1 V + —- V -FT* V

CHAP. 8] RIEMANNIAN CURVATURE 107 8.2 Show that at any point where the Christoffel symbols vanish, R'jkl + R'kl, + R'lfk-Q In this case the expression for R'jkl reduces to just дТ'п1дхк — dT'jkldx'. Therefore, ST' ЭГ' 3T' дТ' дТ' ЭТ\, ρ" -i- T> + R = — — -ι — — -ι — - '*' k" IJk дхк дх' дх1 дх> dxJ дхк As all the terms cancel, the desired relationship is proved. 8.3 Prove that for an arbitrary second-order covariant tensor (Ttj) (The general formula, L ij,kl * ij.lk Rikl * sj + " jkl *■ is 1\xi2 . . . ip,kl Tixh...ip,lk ^RiqklTi1...iq.1siq + 1...ip (8.18) which is credited to Ricci, is similarly established.) A direct approach would be quite tedious; instead, first establish that VtJk-V,M=-R\lkV' (1) for any contravariant vector (V) (see Problem 8.16). Now observe that (VTiq) is a covariant vector, to which (8.1) applies. Thus, (VqTiq)^-(VTiq)Jk = Rsik,V'Tsq (2) By the inner-product rule for covariant differentiation, (3) (VTiq),k, = V%Tiq + VqkTiqJ + VJ^k + ГГ,,И Interchange к and /: (VTlq)Jk = VJkTiq + у'Г,,., + V\TiqJ + VTtqJk (4) Subtraction of (4) from (3) will cancel the middle two terms on the right-hand sides, leaving KuVqTsq = (VI, - V%)Tiq + (Tiq,kl - TiqJk)V< (5) Now use (1) in the right member of (5): KklVqTsq = -RsqklVTiS + (Tiqrk, - TiqM)V which may be rearranged into [(T,q,kl ~ TiqJk) - (R°iklTsq + RsqklTis)]V4 = 0 But (V) is arbitrary, so the bracketed expression must vanish. qed PROPERTIES OF THE RIEMANN TENSOR 8.4 Establish (8.4). By definition, R = e Rs = e Ί _ e Ά + ζ TrTs -гГТ lv/y'JH Sislx jkl Sis - * Sis - I ' Sis1 jl1 rk Sis1 jk1 rl д(8иГц) _H^Vs _ d(Eis^Sjk) 3g^ _ dxk dxk μ дх1 вх1 ''■*"*"'"''*' 1/*1*

108 RIEMANNIAN CURVATURE [CHAP. 8 дхк дх' А эх' r") Лдхк irki T„ Recall from (6.2) that for arbitrary index /, Hi. Г = Г dX By substitution, Эх" Эх ρ — 'Jl '— -ι- Γ V — Γ Гг 8.5 Establish the first skew-symmetry property, Ri]kl = ~Rjikl. To save writing, let Gi - \ (-^гЧ + -Φί) and HI - Г, Г„ *' 2 \дхкдх' дхдх'1 ki "r kl Note the obvious symmetry properties G'ii = G'kl = G'fk and R''kl = Яkl = Η)* Also, it is clear that G'kl = G*'; furthermore, я£ = (g„rt))rkl = r;/ &rrw) = ιχ.Γ41ϊ = я*' Now, by (8.5), Rim = Glljk-Gik + H%-H'k and Rlikl = GH- Gik + НЦ- Hi: = Gik - G!;k + Wk - H% = -R ijkl 8.6 List the independent, potentially nonzero components of Rijkl for η = 5 and verify the formula of Theorem 8.2 in this case. Type Type Type A: B: C: ■"1212' "2323' "3434' ^4545 ^1213' ■"2123' ρ Λ3132> ■"4142' "5152' "l234' ■"1324' "l313' "2424' "3535 ■"1214' "2124' ■"3134' ■"4143' ■"5153' ■"1235' ■"1325' ■"1414' "2525 ρ "l215' ■"2125' η "3135' Λ4145' Л5154' ρ ■"1245' 'Μ 425' ■"1515 ■"1314' "2324' "з234' ■"4243' "5253' Л1345> ■"1435' ■"1315' "2325' ■"3235' ■"4245' "5254> "2345 ■"2435 "1415 ^2425 Я 3435 ■"4345 "5354 There are 10 components of types A and С each, and 30 of type B; or 50 altogether. From the formula, n2(n2-l) = 52(52-l) (25)(24) 12 12 12 RIEMANNIAN CURVATURE 8.7 Prove (8.11). By Corollary 8.3 and the corresponding result for the Gllkt, RiJklU'V'UkV> _ R1212[(U>)2(yy - Zt/Wv1 + (U2f(V1)2] _ R1212 _ *1212 K = G^U'V'irV G1212[(U1)2(V2)2-2U1V2U2V1+(U2)2(V1)2] G1212 gug2, 2 Ϊ12

CHAP. 8] RIEMANNIAN CURVATURE 109 8.8 Calculate К for the Riemannian metric ε ds = (x ) (dx ) - (x ) (dx ) , using the result of Problem 8.7. We have only to calculate /?1212 = guR\12· The nonvanishing Christoffel symbols are, by Problem 6.4, Consequently, Γ1 = 1 11 I 1 pi ^ p2 p2 ^_ 1 22 1 J 12 J 21 1 pi _ "*Γ22 _ "T21 j _ , _ 1 _n,rlrl - Γ2 Γ1 K212 ~ - 1 ,2 Г J 221 Μ 1211r2_/ 1x2 иТ122Д11 121122 1 1 1 1 \/ 1 1 and (Χ1)2 Χ1 V X1/ V xV\ χ1/ (χ1)2 τν- _ 8\\"-2\2 _ 4-211 _ \X J _ ι 811822 822 ~\x ) 8.9 Derive the form (8.12) of the curvature equation. We need only establish the summations over the type-C terms in the numerator; the rest of the work was done in Example 8.4. First of all, let us verify that all Rijkl with ijkl a permutation of abed, where a<b< c<d are distinct integers, are generated by the skew and block symmetries of the three components Rabcd, Racbd> and Radbc. Examination of Table 8-1, which uses an obvious notation for the symmetry operators, shows that all 4! = 24 permutations are accounted for. Consequently, the type-C part of the numerator of (8.12) is [cf. the equation preceding (8.10)] 22ZR„ si* W + ? У R W +2У Л Wj • ^-1 L^abcd y'abcd τ - ^-J ^acbd y4acbd ' ■i *-l Iyadbc '' adbc (1) Symmetry Operator I s. s2 ^1^2 2 1 В BS} = S2B BS^S^ BSjS2 = SjS2B Table 8-1 Subscript Chain abed abed bacd abdc bade cdab cdba dcab dcba acbd acbd cabd aedb cadb bdac bdea abac dbca adbc adbc dabc adcb dacb bead bcda cbad cbda The first summation is over all a < b< с < d that yield a nonzero Rabcd in the basic set; similarly for the second summation. The third summation does not involve the basic set, but the symmetries of Rijkl (shared by WjJkl) allow its absorption in the first two summations. Thus, by Bianchi's identity, 2RadbcWadbc = 2(~Rabcd - Racdb)Wadbc = -2RabidWadhr - 2RachdWadcb (2) and substitution of (2) in (1) produces the expression given in (8.12).

110 RIEMANNIAN CURVATURE [CHAP. 8 8.10 Prove (8.13). We have Wijkl(XU + v\, μυ+ωΥ): AU' + ν V μϋ' + wV λ ν μ ω 2 υ' ν χυ i + vV μϋ' + ων' U' V и" и1 Vk V1 XUk + vVk XU' + vV1 μί/* + ων* μυ' + wV' (λω-νμ)2ΨίιΊζ,(ν,\) so that the quantity (λω - νμ)2 factors out of all terms in (8.12) for K(x; AU + vY, μ\] + ων), leaving K(x; U, V). 8.11 Find the isotropic points in the Riemannian space R with metric «u = l «22 = йз = (*1)2 + 1 &, = ° О'^У) and calculate the curvature К at those points. Follow Example 8.4. By Problem 6.4, the nonzero Christoffel symbols are Г1 Г2 = Г2 = 1 12 J 21 / 1 (хГ + 1 Then: pi /\212 dTX dx 8Y\ ii+f г - Г Г = -1 - 1 ' J 22J 11 J 21J 22 ί pi ._ 33 , pi τ-Ί _ рЗ pi nrl 331 11 J 311 33 -Ί T-i2 _ Л323 l ЗЗ1 12 χ1 ~ (χΎ + ι (*Ύ г3 =г3 = 1 13 1 31 (-*') = "ТТ (*')2 + ι 1 (-*> (*Υ+1 1 V)2 + i ь <У)2 + 1 OO' + l pi — р2 — Р3 ™ О п213 Л123 л132 " which give (A) #i2i2 = guRZ12 = ~[(Х ) + 1] > ^1313 = #11^313 = "~ [(* ) +1] 7 ^2323 = #22^323 = ~ С* ) The corresponding terms for the denominator of (8.10) are (A) Gi2i2=gii^22 = (x1)2+1, G1313 = gng33 = (jr1)2 + 1, G2323=g22g33 = [(^)2+1]2 Thus K = -[(x1)2 + 1]-'W1212 - [(x1)2 + 1ГЧУ»» - (^)2^23 [(X1)2 + I]^i2i2 + K*1)2 + l]Wi3i3 + [(Χ1)' + 1]2W23 W1212 + Wlm + (x1)2^1)2 + 1]W2323 ^1212 + ^1313 + [(^)2 + 1]WM23 -[(^)2+i]- If К is to be independent of the Wijkl (which vary with Lne direction of the 2-flat), then (x1)2 = 1, or xx = ±1. Therefore, the isotropic points compose two surfaces, on which the curvature has the value K= -[1 + 1] 2-1 = -1/4. 8.12 Show that every point of R3 is isotropic for the metric ds2 = (xl)'2(dxlf + (xly2(dx2)2 + (х1у2(ахъ)2 Problem 6.4 gives as the nonvanishing Christoffel symbols: Г1 = - — Г1 =— Г1 = — xx 22 xx ъъ xx Г2 = Г2 = - — 1 12 J 21 1 Г3 1 13 г3 = 1 31 1

CHAP. 8] RIEMANNIAN CURVATURE 111 As in earlier problems, we proceed to calculate a basic set of Rijkt, via Rijkl = gitR)kl (no sum). , __ α ι 22 ^ ci 21 j ^ j ^ 1 _ η 4- Γ1 Γ1 — Γ2 Γ1 С**1 ijC2 22 rl 211 r2 ("r14)2 22 И 121122 1 J_ /_ 1_ (х1)2 + хЛ хх 1 \ 1 1 (хУ Similarly, R\1,= -I/(x1)2. For the remainder, the partial-derivative terms all drop out, yielding p2 _ r" r2 _ r" F2 = Γ1 F2 — Π =τ П323 J 33J r2 * 321 ГЪ J 331 12 " /,.1\2 (*T pi =p2 =P3 =0 iv213 -"123 Λ132 " Our basic set is thus (A) R12l2 ~ "l313 = R-2 ■ii(xy and, by Example 8.3, (A) G1212=G1313=G232, = 1/U1)4 is a basic set of Gijkl. Formula (8.10) or (8.12) now gives K _ Ri2l2Wl2l2 + JJ,31iW,il3 + Д232з^232з _ [-(л'ГЧОУ,^ + W„„ + W2323), G1212W1212 + G1313W1313 + G2323W2323 [(i,)-,](Wm2 + W1313 + lf2ra) It is seen that this Riemannian space is more than just isotropic; it is a space of constant curvature. THE RICCI TENSOR 8.13 For the metric of Example 8.4, calculate (a) Rijt (b) Rl}, (c) R. (a) From R(j = Rkijk = R]n + R2ij2 + R3lj3 and the fact that giy. = 0 for iΦ/', it follows that Я,7 = £ПЯ1У1+£22Я2<,2 + £33Яз„-з where gn = 1, g22 = 1 /2.x1, g33 = l/2x2. Now, a basic set of the Rijkt was computed as 1 1 *M221\ _*M212,J — — ~ Γ **2332\ ~~^2323/ — _ r. 2 "З12з(~~ "-21321 and the only other nonzero components of the form Raija generated by these are R =--L R =-J- Я =-J- Λ2112 2»-1 Λ3223 ~ 2 П3213 г. \ Hence, the nonzero Rtj may be read off from (1) as 1 А(хУ 1 1 if = σ22ρ л11 6 П2112 ^22 8 *M221 "*" S ^32 1 ρ _ „22p _ n33 5 n2332 /? ™ σ33/? — л12 5 n3123 \xxx2 1 2л:1 4(x2)2 „ззр __ ρ 6 Л3213 Л21 (b) (с) R) = g'*^, = g"Rit- (no summation on г) R = R\ + R22 + Rl = gn«n + g22«22 + g33«33 2л:1 ■A-w 4(xy 2хг 2xl 4(x2)2 2x2 \xxx2 Xх + 2(x2)2 (г*1*2)2 (1)

112 RIEMANNIAN CURVATURE [CHAP. 8 8.14 Derive (8.16) from (8.14). Formula (8.14) involves two summations of the form Γ*,. By (6.4) and (6.1b), 1 1 1 1 * is О * isr 9 О I 6(5r osn' oris) ~ 9 о 6sir 9ft Osri 96 ο,τι/ = i „"„ . = i „" ^» = JL (in -JUT) where Lemma 8.5 was used in the last step. Now substitute in (8.14): _<92(in^lH) эг a(invTu) dxldx> 3xs " ri iJ dx" ,^2(lnVTU) / 1 ,/ы^ 4. 1 S(^\g\) \ ~ дх'дх' ^Ш дх Ш Зх° "' ' d2(\nj\g\) 1 a Λ/ΓΓ™. , ΓΓΓ, Supplementary Problems 8.15 The absolute partial derivatives of a tensor Τ = (Τ" ) defined on a 2-manifold Μ : χ' — x'(u, υ) are defined as 5T=/„,... дх*\ δΤ 8u ~V>- <k Эй) аПй δυ ~V' ■■■■* ^ Since (dx'ldu) and (dx'ldv) are vectors, the inner products produce a pair of tensors of the same type and order as T; thus the operation of absolute partial differentiation may be repeated indefinitely. Prove that if (V) is any contravariant vector defined on M, 8u \ δν 1 δυ\ Su J ш Эй ди [Hint: Expand the left side and use Problem 8.16.] 8.16 Prove that for any vector (V), V'kl - Vllk = ~R'sklVs. 8.17 For an arbitrary second-order contravariant tensor (T"). show that ,kl 1 ,lk - "Kskl1 Kskl1 [Hint: Lower superscripts and use Problem 8.3.] 8.18 For an arbitrary mixed tensor (T'j) show that 1 j.kl 1 j.lk~ ~Kskll i+ Klkl< s 8.19 Verify the symmetry properties (8.6) for the Gi/kl [see (8.7)] and for the Wijkl [see (8.9)]. 8.20 Derive (8.5) from (8.4). [Hint: It is helpful to adopt the notation gtjkl for d2gij/3xk3x1.] 8.21 List the independent (nonzero) components of Rijkt when л = 4 and verify Theorem 8.2 for this case. 8.22 Calculate the Riemannian curvature К for the metric ε ds2 = (dx1)2 — 2x1(dx2)2.

CHAP. 8] RIEMANNIAN CURVATURE 113 8.23 Confirm that K = 0 for the Euclidean metric of polar coordinates, ds2 = (dx1)2 + (x1 dx2)2 (a) by a calculation; (b) by noting that К is an invariant. 8.24 Rework Example 8.4 for the pairs (a) U(1) = (1, 0,1), V(1) = (1,1,1) and (b) U(2) = (0,1, 0), V(2) = (2,1,2). (c) Explain why the answers should be the same for (a) and (b). 8.25 Let the surface of the 3-sphere of radius a be metrized by setting x1 = a in spherical coordinates and then allowing x\ x2 to replace x2, Xs, respectively: ds2 = a2(dx1)2 + (a sin x)\dx2)2 Determine К for this non-Euclidean R2. 8.26 If the metric for Riemanhian R3 is given by 8n = ί(χ2) Si2 = g(*2) £зз = Kx2) and git = 0 for iV/, write explicit formulas for (a) K(x2; U, V), (b) R. 8.27 Specialize the results of Problem 8.26 to the case f(x2) = g(x2) = h(x2). 8.28 Find the isotropic points for the Riemannian metric ds2 = (In χ2)(άχγ)2 + (In x2)(dx2)2 + (In x2)(dx2)2 (x2 > 1) and find the curvature К at those points. [Hint: Use Problem 8.27.] 8.29 Show that R3 under the metric has constant Riemannian curvature with all points isotropic, and find that curvature. 8.30 Show that in a Riemannian 2-space [for which (8.11) holds]: (a) Rtl = -giyK, (b) R) = -δ'-Κ, and (c) R = ~2K. 8.31 Calculate the Ricci tensor Rl} for Problem 8.13 using (8.16), and compare your answers with those obtained earlier. 8.32 Use Problem 8.30 to calculate the Ricci tensors of both kinds and the curvature invariant for the spherical metric of Problem 8.25. 8.33 Calculate the Ricci tensors of both kinds and the curvature invariant for the (hyperbolic) metric of Problem 8.12. [Hint: Problem 8.27 can be used to good advantage here.] 8.34 Prove that for any tensor (T"), symmetric or not, Т'Ц=Т')Г [Hint: Use Problem 8.17 and the symmetry of the Ricci tensor.] 8.35 Is identical vanishing equivalent for the Riemannian curvature and the Ricci curvature invariant? Can you And an example where one is zero everywhere but not the other? 8.36 Is constancy in space equivalent for the two curvatures К and R?

Chapter 9 Spaces of Constant Curvature; Normal Coordinates 9.1 ZERO CURVATURE AND THE EUCLIDEAN METRIC A fundamental question has run unanswered through preceding chapters: How can one tell whether a given metrization of R" is Euclidean or not? To be sure that the meaning of "Euclidean" is clear, let us make the formal Definition 1: A Riemannian metric g = (&,), specified in a coordinate system (x), is the Euclidean metric if, under some permissible coordinate transformation (3.1), g = (S;). Now, a coordinate system (x') in which gi} = 8if is (by Definition 1 of Chapter 3) a rectangular system. Hence our question amounts to: Does a given Riemannian space admit rectangular coordinates or does it not? Suppose that a rectangular system (xl) does exist. Then K=0, since all Christoffel symbols vanish in (x1). But Riemannian curvature is an invariant, so that K = 0 in the original coordinates (x) as well. Moreover, by invariance, Thus, the necessity part of the following theorem is immediate. Theorem 9.1: A Riemannian metric (&.) is the Euclidean metric if and only if the Riemannian curvature К is zero at all points and the metric is positive definite. To prove the sufficiency portion, we set up a system of first-order partial differential equations for η rectangular coordinates χ as functions of the given coordinates x' (j = 1, 2,. . . , n). The system that immediately comes to mind (Theorem 5.2) is G = JTJ, or дхк дхк ,12 „4 ,o ,x дХ дх' But (9.1) is generally intractable because of its nonlinearity. Instead, we select the linear system that results when barred and unbarred coordinates are interchanged in (6.6) and then the T'jk are equated to zero: ^ I»4| (9.2) ахах1 чК' ах' Setting w = xk and ui = dxkldx yields the desired first-order system dW : = U- dX (9.3) : = 1 U дх' " r EXAMPLE 9.1 It is proved in Problems 9.7 and 9.8 that when К = 0, (9.3) is solvable for a coordinate system (xk) for which all gf. are constants (i.e., all Г^ = 0); from these coordinates, rectangular coordinates can be reached, provided (gr) is positive definite. To make these results plausible, consider the two-dimensional metric #11 = 1 #12 = #21=0 #22 = (X2)2 This metric is obviously positive definite and, because the only nonvanishing Christoffel symbol is Тггг = 11 χ2, it has Λ1212 = 0 = К. It is possible to solve (9.1) directly for the corresponding cartesian coordinates, and then to verify that that solution is contained in the general solution to (9.3). 114

CHAP. 9J SPACES OF CONSTANT CURVATURE; NORMAL COORDINATES 115 Introduce the notation дХ1 „ дХ1 fl dx1 h dx2 h~ dx1 h~ dx2 {1) whereby [9.1) becomes the algebraic system Л/2+/з/4 = 0 (2) System (2) can be solved for three of the ft in terms of the fourth—say, /,: Now (7) becomes two simple first-order systems in x1 alone and x2 alone: T: \ dXX dx1 ^ and 11: A=*2vwl {dx dx1 ν л The unknown function /j is determined by the requirements that the two equations I and the two equations II both be compatible: ψ-г = -A (*2Vwb and -^ (-Vw!) = А (*2/,) The only function satisfying these two compatibility conditions is /i = const. = cos φ and I and II immediately integrate to give xx = хл cos φ + - (χ2)2 sin φ + с χ2 = -χ1 sin φ + - (χ2)2 cos φ + d (4) We are, of course, free to set φ = с = d = 0 in (4). Turning to (9.3), we have to solve ._ <9vv <9w (1) tt="i, Т^ = ы2 du, du, , <9ы, <?M, , И, (2) —j = 0, -4 = 0 (3) —τ = 0, —§ = ы2Г22 = 4 v ; dx dx ' dx dx χ Note that these equations include their own compatibility conditions! For instance, the second equation (2) and the first equation (3) ensure the compatibility of the two equations (1). The fact that system (9.3) is automatically compatible whenever К = 0 is crucial to the proof of Theorem 9.1. Integrating the above equations in the order (3)-(2)-(l), we get: w = fljjc1 + a2(x2)2 + a3 (flj, α2, α3 = const.) or, replacing the index k, χ = axx + a2(x )2 + a\ (of = const.) (5) As announced, (5) includes (4). For subsequent use, the following compatibility theorem for quasilinear systems [which include linear systems such as (9.3)] is stated here, without proof: Theorem 9.2: The quasilinear first-order system T7 =Fa,(mo> u,;. . . ,um,x\x2, . . . ,x") (A = 0,1, ... ,m; / = 1,2, ... , n)

116 SPACES OF CONSTANT CURVATURE; NORMAL COORDINATES [CHAP. 9 where the functions Fxj are of differentiability class С , has a nontrivial solution for the wA, bounded over some region of R", if and only if dFV г л. dF*i dF^ г л. dF^ μ πι ™ 1 <=-·//< л Ί^^ + Ί^ = Ί^^ + Ί^ (A = o,i,...,i»;is,<^„) [The v-summations run from 0 to т.] 9.2 FLAT RIEMANNIAN SPACES A Riemannian space, or the determining metric, is termed flat if there is a transformation of coordinates χ = x'(x) that puts the metric into the standard form ε ds2 = ελ(άχ1)2 + ε2(άχ2)2 + ■■■ + εη(άχη)2 (9.4) where εί = ±1 for each i. This condition generalizes the concept of the Euclidean metric. The essential distinction between the two concepts revolves about positive-definiteness; the analogue to Theorem 9.1 with positive-definiteness removed is: Theorem 9.3: A Riemannian space is flat if and only if K = 0 at all points. Corollary 9.4: If К = 0, then R = 0. Proof: If К = 0, then by Theorem 9.3 thespace is flat, and hence the gtj are constant forsome coordinate system (*''). It follows that all Tijk, T'jk, R)kl, Rtj, and R) vanish. Therefore, R = R\ = 0, and since Ricci curvature is invariant, R = 0. Remark 1: Problem 8.35 shows that the converse of Corollary 9.4 does not hold. EXAMPLE 9.2 Consider the Riemannian metric eds2 = (dx1 f + 4(x2)2(dx2)2 + 4(x*)2(dx")2 - 4(x*)2(dx<)2 (a) Calculate the Riemannian curvature, (b) Find a solution of system (9.3) from which it may be inferred that the space is flat. (a) Using Problem 6.4, we find as the nonvanishing Christoffel symbols Γ2 = — Γ3 = — Γ4 = — x2 33 x3 4* x4 Because T'jk = 0 unless i = j = k, the partial-derivative terms drop out of (8.2), leaving R'j*. = W* ~ ВД, = ВД - Γ„Γ„ = О (not summed) which in turn implies that Rijkl = 0 and К = 0. (b) For the above-calculated Christoffel symbols dux du2 u2 du2 иъ duA uA ~Z Г ^ "^ I 2 ~Z 3 3 "^ 4 ~ ~4 dX dx X dX X dX X with dutldXj = 0 for i Φ j. Integrating, £ / 2 3 4ч 2r I 1 3 <4\ 3/· / 1 2 4ч 4r / 1 2 3\ ul ~ h(X ' X ' X ) U2 = X f2(x ,X ,X ) U3 = X f,(X , X , X ) H4 = X f4(x ,Χ,Χ) for arbitrary functions /-. But the remaining equations (9.3), dwldx' — ы;, give rise to the compatibility relations du; _ dUj dx1 dx' which are satisfied only if fi = c. = const. Therefore, w = α,χ1 + a2(x2)2 + a,(x3)2 + α4(χ*)2 + a5

CHAP. 9] SPACES OF CONSTANT CURVATURE; NORMAL COORDINATES 117 and the transformation must be of the general form -к к \ . к / 2ч2 . k/ Зч2 . k/ 4\2 . к /к * * \ x = fljjc + a2(x ) + аг(х ) + a4(jc ) + a5 (ai constants) Wc wish to specialize the constants so that the covariant law G = JTGJ will hold, with G corresponding to (9.4). As a preliminary guess, set b, 0 0 0 0 0 b2 0 0 0 0 0 b3 0 0 0 0 0 b., 0 so that the covariant law becomes "1 4(*2)2 4(*3)2 -4(*<)\ By inspection, the choice Ьг - = b 'b, 2b2x2 2b3x3 2b4x\ - b, = fr4 = 1 will render ε "ει ε2 ε3 £4. 1 c2 C3 C4 "bi 2fo2jc2 2fo3jc3 . = 1. _ 2&„*4_ In connection with (9.4) there is an interesting theorem (Sylvester's law of inertia). Define as the signature of a flat metric (gi}) the ordered и-tuple (sgn e,,sgn ε2, . . . , sgn εη) composed of the signs of the coefficients in the standard form (i.e., the signs of gu, . . . , gnn). Theorem 9.5: The signature of a flat metric is uniquely determined up to order. 9.3 NORMAL COORDINATES It is possible to introduce local, quasirectangular coordinates in Riemannian space the use of which greatly simplifies the proofs of certain complicated tensor identities. Let О denote an arbitrary point of R", and p = (p') an arbitrary direction (unit vector) at O. Assuming a positive-definite metric, consider the differential equations for geodesies, dx' dxk «г ία χ ds [cf. (7.13)], along with initial conditions >k ds ds dx1 ds (9.6) Here the arc-length parameter is chosen to make s = 0 at O. Remark 2: Under an indefinite metric, there could exist directions at О in which arc length could not be defined; see, e.g., Problem 7.22. There would then be no hope of satisfying (9.6) with (p') arbitrary. It can be shown that for a given p, the system (9.5)~(9.6) has a unique solution; moreover, for each point Ρ in some neighborhood Μ of O, there is a unique choice of direction ρ at О such that the solution curve xl = x'(s) (a geodesic) passes through P. Accordingly, for each Ρ in Л, take as the coordinates of Ρ y'-sp' ■ (9.7) where s is the distance along the geodesic from О to P. The numbers (/) are called the normal coordinates (or geodesic or Riemannian coordinates) of P. EXAMPLE 9.3 Show that if the Riemannian metric ds2 = gtj dx dx' for R2 is Euclidean and there is a point О

118 SPACES OF CONSTANT CURVATURE; NORMAL COORDINATES [CHAP. 9 at which g12 = 0, then normal coordinates (/) with origin О arc constant multiples of (z'), for some rectangular coordinate system (z'). Because g12 = 0 at point O, the vectors Τ = (HYg^l, 0) and S = (0, HVgTi) aiey at O, an orthonormal pair. The space, being Euclidean, admits a rectangular coordinate system; in particular, a system (z') with origin О and unit vectors Τ and S (Fig. 9-1). Again because the space is Euclidean, the straight line segment OP is the unique geodesic connecting О with the arbitrary point P. With s = OP and ρ the direction vector of OP, we have the vector equation or componentwise, QED. = sp ζ'Ύ + ζ S = sp and sp =y Fig. 9-1 The chief value of Riemannian coordinates resides in the following theorem (Problem 9.10). Theorem 9.6: If the metric tensor (g-) is positive definite, then, at the origin of a Riemannian coordinate system (/), all dgijldyk, dg'!ldyk, Yijk, and Y'jk are zero. Remark 3: Recall that neither the partial derivatives of the metric tensor nor the Christoffel symbols are tensorial. Thus, their (y1)-representations can vanish at О without their ix')-representations doing so. For instance, because the transformation between (x') and (y:) has / = / at O, (6.5) gives: г'*(ж)|0- -.2 i д у дх'дх The right side is generally nonzero, unless the coordinate transformation happens to be linear. EXAMPLE 9.4 Prove Bianchi's first identity, Rijkl + Riklj + Riljk = 0. Theorem 9.6 implies that at O, the origin of normal coordinates, R ЭТт <?Г jki ijkl dyk dyl

CHAP. 9] SPACES OF CONSTANT CURVATURE; NORMAL COORDINATES 119 If we use the notation Tijkt for дУцк1ду\ for arbitrary i, j, k, I, then R,jkl Z Rikij - Rm- 1 /lik = r 1 kjtt 1 tkij -Г 1 jkil 1 ktij -т„* On summing these three relations and observing the cancellations which take place, we see that the desired identity holds at О in the coordinates (/). This tensor identity must therefore remain valid at О in the alias coordinates (*'). But О is any point of R", and the proof is complete. EXAMPLE 9.5 Prove Bianchi's second identity, Я,,*,,»+ «,■„»,*+ ««,-Ht.,= 0 (9.8) Working with the Riemann tensor of the second kind, we have, at the origin О of normal coordinates, *'■"■" ду" дуи\дук ду' >'' гк '* " = Г"' — Г"' 1 jlku J jklu since terms like (dYr]lldyll)Ylrk vanish along with the T'rk at O. From this, permutation of subscripts yields R'jkl.u + Rjlu.k + R'juk.l =0 at O, and the validity of (9.8) at О follows from the fact that covariant differentiation commutes with the lowering of a superscript (Problem 6.11). We conclude, as in Example 9.4, that (9.8) holds generally in (*'). A positive definite metric has been tacitly assumed, both here and in Example 9.4. The assumption can be dropped; see Problem 9.13. 9.4 SCHUR'S THEOREM From Chapter 8 it is known that although every point of Riemannian two-space is isotropic, the curvature (= R1212lg) can still vary from one isotropic point to the next. However, Problems 8.11, 8.12, 8.28, and 8.29 suggest that a different situation prevails in R3. To prove the general theorem, known as Schur's theorem, it is necessary to establish a preliminary result, a generalization of (8.11). Lemma 9.7: At an isotropic point of R" the Riemannian curvature is given by τζ „ abed __ abed /q q\ oac&bd oad&bc abed for any specific subscript string such that Gabcd Φ 0. [If Gabcd = 0, then Rabcd = 0 also.] For a proof, see Problem 9.8. Theorem 9.8 (Schur's Theorem): If all points in some neighborhood Л in a Riemannian R" are isotropic and n^3, then К is constant throughout that neighborhood. For a proof, see Problem 9.14. 9.5 THE EINSTEIN TENSOR The Einstein tensor is defined in terms of the Ricci tensor Rif and the curvature invariant R (Section 8.4): G'^R'j-^S'jR (9.10) It is clear that (G';) is in fact a mixed tensor of order two. As a direct generalization of the notion of the divergence of a vector field V= (V) relative to

120 SPACES OF CONSTANT CURVATURE; NORMAL COORDINATES [CHAP. 9 rectangular coordinates (xl), dVl dV2 dV dV -div V= —г + —τ + ■■■ + -τ~ΊΓ ^ ττ дх1 дх2 дхп дх we define the divergence of the general tensor T = (Г^'Д'/_''*■''''") with respect to its kth con- travariant index to be the tensor άίνΤ-(Γ£?"/ν·Λ) (9.77) In Problem 9.15 is proved Theorem 9.9: For any Riemannian metric, the divergence of the Einstein tensor is zero at all points. Solved Problems ZERO CURVATURE AND THE EUCLIDEAN METRIC 9.1 Test the compatibility conditions (Theorem 9.2) for the system —Τ - "Τ Τ^ - 2* и0 (1) дХ X дХ If it is compatible, solve the system. In the notation of Theorem 9.2, there is only the condition corresponding to λ = 0, / = 1, к = 2 to be satisfied. dFm p . dfoi ? dF02 dF02 ди0 Р°2 дх2 ~ ди0 m дх1 д (U0\ г д (иЛ д 2 и0 д 2 _^.2хио+_у1_(2хио)._ + _(2х1|0) 2xzu0 __ 2х2и0 Therefore, the system is compatible. The first equation (1) integrates to u0 = χ1 φ(χ2); the second equation then gives хгф' — 2χ2χιφ whence φ = с exp (χ2)2 Hence the solution of (1) is u0 = ex1 exp (x2)2. 9.2 Show that R3 under the metric ds2 = [(x1)2 + (x2)2](dx1)2 + [(x1)2 + (x2)2](dx2)2 + (dx3)2 is Euclidean. This metric has g33 = const., and gn and g22 independent of x2. Problem 6.4 then shows that V'jk = 0 whenever i, j, or к equals 3; consequently, of the six independent components of the Riemann tensor, only Л1212 is possibly nonzero. But (from Problem 6.4), with z = (x1)2 + (x2)2, 11 ζ v2--x- ζ Γ1 = 1 12 Γ2 - 1 12 1 21 -Г2 1 21 = = 2 X ζ 1 χ ζ Γ1 = 1 22 Γ2 = 1 22 Ζ 2 Χ ζ

9] SPACES OF CONSTANT CURVATURE; NORMAL COORDINATES 121 so that r> 1 22 21^ , y-il y-il , jn2 y-il y-il y-il тп2 тп Λ2Ι2 — - 1 ,2 "*" ■* 22-1 11 "■" * 22-1 21 * 21 * 12 ■* 21-* 2 1-Ί 22 <9л: <9л: Consequently, Ri2U = 0 = K. As the metric is clearly positive definite, Theorem 9.1 implies that the space is Euclidean. For the Euclidean space of Problem 9.2, exhibit a transformation from the given coordinate system (x) to a rectangular system (x'). Using the Christoffel symbols as calculated in Problem 9.2, we obtain from (9.3) the following system for the «,-: dxx 1 dux x Mj + χ дх2 ζ ди2 —χ иг H- дх2 ζ ди, дх и2 х2и2 ди. —ϊ =0 5х3 (1) (2) ди2 х ил + χ и2 дх1 ~ ζ ди, „ „., —, Н^=0 "4=0 "4=0 (3) дх1 дх2 дхъ ν ; Thus ил and u2 are functions of χ1, χ2 alone, and u3 = const. Since the gtj are all polynomials of degree 2 in x1, x2, use the method of undetermined coefficients, assuming polynomial forms u, = a^x1)2 + b/x2 + ct(x2)2 + d.x1 + e^2 + f, (i =1,2) The (compatibility) relation дил1дх2 = ди21дх1 implied by the second equation (1) and the first equation (2) requires frj =2a2 2c, = fr2 gj = ίί2 Similarly, dujdx1 = — ди21дх2 implies 2flj = ~fo2 frj = —2c2 i/j =■ —e2 Using the first equation (1), or г(Эи1/Эх1) ^jt1^ — х2ы2, we get: β!=0 α2=0 c1=fo2 2^-c, d^ = -e2 /1=0=-/2 It follows that frj = fr2 = Cj = c, = 0, and therefore (renotating dt and ег) ы, = αχλ + frjt2 ы2 = bx1 — ax2 и, = с {Note: This solution of (1 )-(2)-(3) may be obtained by the method of characteristics, without any prior assumptions.] The first equations (9.3), dW 1 . , 2 OW г 2 dW —r = ax -V bx —2 =bx -ax —τ — с дх1 Эх дхъ may now be integrated to give w = 2 (χ1ΐ + *>χ1χ2 ~ J (χ2Ϋ + cx3 + d or, replacing χ and corresponding superscripts, and with d =0, xk = у (χ1)2 + Ь W - у (χ2)2 + c\x2)2

122 SPACES OF CONSTANT CURVATURE; NORMAL COORDINATES [CHAP. 9 It is clear that we may take c1 = c2 = 0 = a3 = b3 and c3 = 1: -Λ _ 1 1/..1 1 jc = - a (x ) -V ο χ χ — - a (x ) x2 = i a2(x1)2 + b2x'x2 - i fl2(x2)2 -3 3 The Jacobian matrix is flV + fr1*2 fcV-flV 0 α У + &У b2xl - a2x2 0 0 0 1. Since JTJ = G, we must have (a1)2 + (fl2)2=l αΨ + a2b2=Q (b1)2 + (b2)2 = 1 so take a1 =0, a2 = 1, £r =0, bx -\. The transformation is, finally, -2 A f/ 12 / 2n2t -3 3 FLAT RIEMANNIAN SPACES 9.4 Determine whether the following metric is flat and/or Euclidean: eds2 = (dxlf-(x2)2(dx2)2 (n=2) Since the metric is not positive definite, it cannot be Euclidean. To determine flatness, it suffices to examine R12l2 - guRln- But Problem 6.4 shows that R\12 =0; hence the space is flat. 9.5 Show that if the metric tensor is constant, the space is flat and the coordinate transformation x = Ax, where Л is a rank-и matrix of eigenvectors of G = (g;i), diagonalizes the metric (i.e., gif = 0 if i*j). Since all partial derivatives of gtj are zero, all Christoffel symbols will vanish and all RIJkl = 0, making К = 0. Thus, by Theorem 9.3, the space is flat. By Chapters 2 and 3, if χ = Ax, then J = A and G = JTGJ = A'GA However, since G is real and symmetric, its eigenvectors form an orthogonal matrix which we now choose as A, with AGA~l - AGAT = D (diagonal matrix of eigenvalues of G) Hence, G-AGAT=D QED. 9.6 Find the signature of the flat metric eds2 = 4(dx1)2 + 5(dx2)2 - 2(dx3)2 + 2(dx4)2 - 4 dx2 dx3 - 4 dx2 dxA - 10 dx3 dx4 In view of Problem 9.5, it suffices to find the eigenvalues A of G = (gif). The characteristic equation is IG-A/I = 4-A 0 0 0 0 5-A -2 -2 0 -2 -2-A -5 0 -2 -5 2-λ (4-A) 5 - A -2 -2 -2 -2-A -5 -2 -5 2-A -(4-A) 5-A 2 0 -2 2 +A -3 + A -2 5 7-A = -(4 - A)[(5 - λ)(29 - A2) + 8(5 - A)] = -(4 - A)(5 - A)(37 - A2) = 0

CHAP. 9] SPACES OF CONSTANT CURVATURE; NORMAL COORDINATES 123 from which the eigenvalues are A = +4, +5, + V37, - V37. This means that there is a transformation which changes the metric into the form e ds2 = \(dxf + 5(dx2)2 + V3l(dx')2 - VJ7(dx4)2 = (dx1)2 + (dx2)2 + (dx3)2 - (dx'f with the obvious change of coordinates. Hence, the signature is (+ + Η—), or some permutation thereof (Theorem 9.5). 9.7 Show that the conditions Rljkl =0 are sufficient for the compatibility of (9.3). In the notation of Theorem 9.2, (9.3) takes the form (with m= n) λ>° S = F^"^-(X) The corresponding compatibility conditions are λ = 0 8]игТ1к = 8"кигТгч or urTrfk = игГк., which holds trivially, and r λ^ s *k r дхк г л* s .ι r dxi which rearranges to V dxk dx j ' * Aj1 чк x xkx sj lur Thus, RrXkj = 0 forces R\kj = 0 and compatibility. 9.8 Prove Lemma 9.7. As (Rtjkt) and (G kl) are tensors [see Example 8.3] and К is an invariant, is a tensor of the same type and order. It must be proved that all TiJkl = 0 at an isotropic point P. Since К is independent of direction at P, so are the Τ kl; and (8.7) gives TV, u'v'uW = о (T,jkl = Tijkl(P)) (i) If we define the second-order tensor (Sik) = (TijklV'Vl), we find that Sik = Ski, and by (1), SikU'Uk = 0 at Ρ for any (U'). It follows that all Sik =0 at P. Now set V =8'a. Then, at F, 0 = Sik = Tijkl8'a8la = T-aka for arbitrary (fixed) index a. Next set V' = 8'a + 8'b for arbitrary fixed indices a and fr: о = TiJklvvl = ri>w(«i + δίχβΐ + δ'6) = т1ака + rieti, + г,.6Ао + Tibkb or ^<a*b + Tlbka = 0. Therefore, since TIJkl obeys the same symmetry laws as Rijkl and Gi/JH, TlJU-Tm=0 (2) Tijkl + Tiklj+Tilik = 0 ■ (3) Adding (2) and (3), 2Γ„., + Γ/Α, = 0 (4) But, from (2), Γ,^ = Tljkl, so that (4) implies TIJkl = 0, as desired.

124 SPACES OF CONSTANT CURVATURE; NORMAL COORDINATES [CHAP. 9 9.9 Prove Theorems 9.1 and 9.3. We already know that if the space is either Euclidean or flat, K = 0. Suppose, conversely, that K = 0; then every point is isotropic, and Lemma 9.7 implies that all Rijkl vanish. It then follows from Problem 9.7 that there exists a coordinate system (i") for which t'jk = 0 or gtj = const. By Problem 9.5. there exists another coordinate system, (>''), in which the metric takes the form (for real constants a;) ε ds2 = erfidy1)2 + e2a22(dy2Y + ■■■ + ena2n(dy"Y The transformation y1 = axyx, y2 = a2y2,..., y" = any" now reduces the metric to ε ds2 = e^dy1)2 + e2(dy2)2 + ■ ■ ■ + en(dy"f (1) and the space is flat. This proves Theorem 9.3. If the given metric is positive definite, then in (1), ε, = 1 for each i. In this case the metric is Euclidean, proving Theorem 9.1. NORMAL COORDINATES 9.10 Prove Theorem 9.6. If (yl) are normal coordinates, then the geodesic through О and any point Ρ in some neighborhood Л of О has the parametric form y'= sp' (p'- const.) This geodesic thus obeys the differential equations dy' i , d2y' ~r = P and —-r- = 0 ds r ds But it must also satisfy (9.5), 3T/3s = 0, in the coordinates (y')\ d2y' r dy' dyk ds2 >k ds ds Thus, by substitution, T'kp'p = 0 for all directions (p') at O. But T'jk is symmetric for each i; hence, r'.fc = 0 at О foi all i, j, k. Also, Tijk = g^ = 0; hence, dgijldyk = Q at O, by (6.2). Finally, since g''gjr = 8'r, the product rule for differentiation yields dg''ldyk =0 at O. 9.11 Prove that at the origin of a Riemannian coordinate system (y1), rl Τ ' jFl —j- = —j (all / and k; summed on i) dy dy Since Г'уд. and dg''ldyk all vanish at the origin О of the Riemannian coordinate system, дГп а , ir„ , ,, а ι ι r)Vk r)Vk lir' & 2vk 9 &jir~T~Sirj'Srji) τ & ^ Sjirk ' 8irjk~*~ Srjik) \* ) at O, with gijkl = d2gijldykdyl. But, since g" = g", ir /·/ i> ir о о jirk о о jirk О о jrik О orjik and (1) becomes ^ 2 δ Sir^ 2 δ Sir*> iy' 9.12 Prove the identity Ri]kKu + Riljk^ = Rikulj + Rikjul. Covariant differentiation of Bianchi's first identity, (8.6), gives Rijk, u + Rikl u + Ra „ = 0. Then the second identity, (9.8), yields ^Ijkl.u ' "-iljk.u ~ ~Riklj,u ~ Rikju.l "■" KikulJ

CHAP. 9] SPACES OF CONSTANT CURVATURE; NORMAL COORDINATES 125 9.13 Show that Bianchi's identities remain valid under an indefinite metric. One can appeal to the topological fact that, at a given point Ρ of R", the directions for which a given metric (g,7) is indefinite span, at worst, a hyperplane. Hence, normal coordinates are possible along geodesies whose tangent vectors (p') at Ρ do not lie in the hyperplane; Problem 9.10 gives T'jkp'pk = 0 for these directions. But the T'jk are continuous, and any direction in the hyperplane is the limit of a sequence of directions not in the hyperplane. It follows that T'jkp'pJ = 0 for all (p'), yielding Theorem 9.6 and the Bianchi identities. SCHUR'S THEOREM 9.14 Prove Schur's theorem (Theorem 9.8). By Lemma 9.7, Rijk, = Gijk,K throughout N. Take the covariant derivative of both sides with respect to x", then permute indices (Gijklu = 0 because gtju = 0 in general): Rijki.u - GlJklK u Rijiu.k = GljluK k R.jukj - Gtjuk&j Add the three equations and apply (9.8): G,.A,Ku + G^K, + G,„,K;=0 (1) Multiply both sides of (1) by g'kg'1 and sum. Since gV<V, = g'Yig.kSj, ~ 8n8,k) = Bkk8'- 8f8'k = n2-n ё'кё'%ы = g'YW,. - gt»g„) = W - Ks\ = 8*u- пьки ^ g'kg"GVuk = gkg'\giugjk - gikgJU) = SkuS[ - 8kk8lu = 8lu - n8lu that summation yields the relation 0= (n2 - И)К B + (8ku - n8k)Kk + (δ'„ - «δ'„)Κ,, = (η2 - И)К „ + (1 - л)К „ + (1 - И)К „ = (и - 2)(и - 1)К „ For и S3, К „ = Ж/дх" = 0. Since и was arbitrary, К must be constant over N. qed THE EINSTEIN TENSOR 9.15 Prove Theorem 9.9. We must prove that Grir = 0. Multiply both sides of (9.8) by g''g'k and sum: 0 = g'lg'kRijkKu - g''g'kRiju,,k ~ g''g'kRjiuk,i - o'kRl - o'kRl _ e'lnk = g'kR - g'kR - g"R = Rk,u ~ Ru.k ~ Ru,l =2\2 R" ~ R"-k) or, changing и to i and к to r, \8rtR r - R[r = 0. But, by Problem 6.32, 8ritj = 0 for all i, j, r; hence, [r-,-18-^)^ = 0 or g;„=o 9.16 Show that Gtj, the associated Einstein tensor obtained by lowering the index i in G'y·, is symmetric. By definition, 1 _,_\ _ 1 G 4 gikGk = gik[Rk-^8kR) = Ril--gllR which is obviously symmetric (by symmetry of the Ricci tensor).

126 SPACES OF CONSTANT CURVATURE; NORMAL COORDINATES [CHAP. 9 Supplementary Problems 9.17 Solve, if compatible, the'system dujdx' = FA., with (a) Fm = х212иа F02 = xll2u0 \L/\ 01 Ι*ηΑ 02 1 11 Μ-ΓΐΛ 19 14-* Λ 9.18 Verify that ώ2 = (dx1)2 + (x1)2^2)2 represents the Euclidean metric (in polar coordinates). 9.19 Consider the metric ε ds2 = (dx1)2 - (x1 dx2)2 - (x1 dx3)2. Show that R1212 = 2 and that, therefore, the space is not flat. 9.20 Determine whether the following metric is flat and/or Euclidean: ε ds2 = (dx1)2 - (x'Yidx2)2 (n =2) 9.21 Determine whether the following metric is flat and/or Euclidean: ds2 = (dx1)2 + (Xy(dx2)2 + (dx1)2 9.22 Find the signature of the metric for R3 given by eds2 = 2(dx1)2 + 2(dx2)2 + 5(dx3)2 - 8 dx1 dx2 - 4 dx1 dx3 - dx2 dx3 9.23 Prove that R\jk = 0. [Hint: Use the first of (8.6).] 9.24 Use Problem 9.11 to obtain a simplified proof for Problem 8.34. 9.25 Show that the Einstein invariant, G = G\, vanishes if the space is flat. [Hint: Use Corollary 9.4.] 9.26 In the general theory of relativity one encounters the Schwarzschild metric, eds2 = e^dx1)2 + (x1)2^*2)2 + (sin2 x2)(dx3)2] - e\dx*f where both φ and ψ are functions of x1 and x" only. Calculate the nonzero components of the Einstein tensor.

Chapter 10 Tensors in Euclidean Geometry 10.1 INTRODUCTION There exists a startling correlation between formulas of differential geometry, developed to answer questions about curves and surfaces in Euclidean 3-space, and tensor identities previously introduced to handle changes of coordinate systems. Differential geometry was used to great advantage by Einstein in his development of relativity. The metric will be assumed to be the Euclidean metric, and to emphasize this fact we shall designate the space by E3, which means R3 with the metric ds2 = {dxxf + (dx2)2 + (dx3)2 Moreover, we shall use the familiar notation (x, v, z) in place of (χ , χ , χ ). 10.2 CURVE THEORY; THE MOVING FRAME A curve ^ in E3 is the image of a class C3 mapping, r, from an interval .9 of real numbers into Ε , as indicated in Fig. 10-1. The image of the real number t in 3 will be denoted r(t) = (x(t),y(t),z(t)) {10.1) a vector field of class С . ζ ]—- Regular Curves The tangent vector of % is given by dr_._(dx dy dz\ <€ is said to be regular if r(i) Φ 0 for each t in 3. Remark 1: This corresponds to the definition of regularity, given in Section 7.3, in the case of a positive definite metric. 127 ■ь-^ Fig. 10-1

128 TENSORS IN EUCLIDEAN GEOMETRY [CHAP. 10 y2ll 1 in EXAMPLE 10.1 An elliptical helix (Fig. 10-2) is a helix lying on an elliptical cylinder x2/a2 + xyz-space; it is given by <ё : χ = a cos t, у = b sin t, z = ct, with $■ the entire real line. The pitch is defined as the number c. If a = b, the helix is called circular, with radius a. Fig. 10-2 EXAMPLE 10.2 The space curve % : χ = t, у = at2, ζ = bt3 (^=R) captures the salient local features of all curves; it is known as the twisted cubic. As indicated in Fig. 10-3, the projection of 41 in the ry-plane is a parabola, у = ax2; its projection in the xz-plane is a standard cubic curve, ζ : semicubical parabola (yldf — (z/b) . bx ; in the у ζ- plane, the = a\2lb2 Fig. 10-3 Arc Length Since the Euclidean metric is positive definite, every regular curve has an arc-length parameterization r = r(s), such that

CHAP. 10] TENSORS IN EUCLIDEAN GEOMETRY 129 f dx_ du du or ds Jt = r (10.3) (The dot, as in r, is used to denote differentiation with respect to t, and a prime, as in r', denotes differentiation with respect to s.) The mapping t—>s defined by (10.3) has the inverse relation ί—>ί given explicitly by t = 9(5), where ψ is also differentiable: dt ds <pXs): 1 (10.4) The Moving Frame Three vectors of fundamental importance to curve theory will now be discussed. Two of them were introduced in Chapter 7: the unit tangent vector— the (unique) vector τ = τ> = (^ί ^l i*i\ \ds ' ds ' ds I —and the unit principal normal—any unit, class С vector N that is orthogonal to Τ and is parallel to T' wherever Τ' Φ 0. The binormal vector associated with a curve is the unit vector Β = Τ x N [for the cross product, see (2.10)]; В is uniquely determined once N has been chosen. Not all regular curves have a principal normal vector (see Problem 10.1). However, it was proved in Problem 7.14 that all planar curves possess a principal normal, of the form N = (- sin Θ, cos Θ, 0) (plane ζ = 0) if Τ = (cos Θ, sin Θ, 0). The following result provides further information. Theorem 10.1: Every planar curve has a principal normal vector. If a space curve has a principal normal vector, that vector lies in the plane of the curve for any nonstraight planar segment of the curve. Along any straight-line segment, the principal normal can be chosen as any class C1 vector orthogonal to the unit tangent vector. At each point of % where N can be defined, the mutually orthogonal triplet of unit vectors Τ, Ν, В constitutes a right-handed system of basis elements for E3. This triad, which changes continuously along % (Fig. 10-4), is often called the moving frame or moving triad; the plane of Τ and N is known as the osculating plane. Fig. 10-4

130 TENSORS IN EUCLIDEAN GEOMETRY [CHAP. 10 The moving frame has been defined for the arc-length parameterization. When it is necessary to use the original parameter t instead, the following expressions may be established (Problem 10.4) for any point at which rΦ0 and rxf^O: \\r\\ \\r\\ \\r X r|| ||r X r|| Here, ε — ±1, the choice of sign depending on the choice of N as a class С vector. 10.3 CURVATURE AND TORSION Two important numbers, or more accurately, scalar fields, are associated with space curves. Definition 1: The curvature к and torsion τ of a curve r-€ : r = r(s) in Ε arc, respectively, the real numbers κ^ΝΤ' and τ^-ΝΒ' (10.6) The sign of к will depend on that chosen for N; however, since В and B' change in sign together with Ν, τ is uniquely determined. It follows (cf. Problem 7.13) that the absolute values of curvature and torsion are given by κ0 = |κ| = ||Τ'|| and τη = |τ| = ||Β'|| (70.7) Thus, κ0 measures the absolute rate of change of the unit tangent vector and the amount of "bending" a curve possesses at any given point, while τ0 measures the absolute rate of change of the binormal and the tendency of the curve to "twist" out of its osculating plane at each point. The significance of negative values for к and τ will become apparent later. Remark 2: It can be shown that the two functions к = k(s) and τ = t(s) determine the curve r-€ up to a rigid motion in E3. In the ί-parameterization of <£, wc have (Problem 10.7): ellrxfll detirf'rl κ= '' „3 " and τ= „ ,,2 (10.8) \\r\\ llrXr1l where ε = ±1 and [rf r] represents the 3 x 3 matrix having as row vectors r, f, and f. [Recall the identity a · (b x c) = det [a b c] for the triple scalar product of three vectors.] Serret-Frenet Formulas The derivatives of the vectors composing the moving triad are given by T' = κΝ ΓΤΊ' Γ 0 κ °ΊΓΤ" Ν'=-κΤ+τΒ or Ν = -к 0 τ Ν (10.9) Β'=-τΝ LbJ ί 0 -ι OJLb. Note the skew-symmetry of the coefficient matrix. The first of these formulas was established in Problem 7.13; the other two are derived in Problem 10.8. 10.4 REGULAR SURFACES Surfaces are generally encountered in the calculus in the form ζ = F(x, v); that is, as graphs of two-variable functions in three-dimensional space. Here, however, it is more convenient to adopt the

CHAP. 10] TENSORS IN EUCLIDEAN GEOMETRY 131 ?2 · Definition 2: A surface if in Ε is the image of а С vector function, r(x\ x2) = (f(x\ x2), g(x\ x1), h(x\ x2)) which maps some region Υ of E2 into E3. (See Fig. 10-5; in general, primes will designate objects in the parameter plane (У) corresponding to those on the surface in *>>z-space.) The coordinate breakdown of the mapping r, x=f{x\x2) y = g(x\x2) z = h(x\x2) (10.10) is called the Gaussian form or representation of if. Mapping r Point Ρ is a regular point of if if Г 2 дх дх Fig. 10-5 i J к df_ dg_ dh_ дх дх дх df dg dh дх дх дх 7^0 (10.11) at P'\ otherwise, Ρ is a singular point. If every point of if is a regular point, then if is a regular surface. Remark 3: Condition (10.11) is tantamount to the linear independence of the two vectors (drldx1)P and (drldx )P. Equivalently, and of more geometrical interest, the condition ensures that every curve in if through Ρ which we take to be the image under r of a regular curve in Ύ through P', is, in a neighborhood of P, regular in the sense of Section 10.2. EXAMPLE 10.3 For a C3 function F, show that the graph ζ = F(x, y) is a regular surface. The surface has the Gaussian representation and thus dr dr Γ Χ 2 дхх дх2 y=-x i J к 1 0 dFldx1 0 1 dFldx2 ζ — F(xl, χ ) dF dx' dF dx 2 ,lj*0 at an arbitrary surface point P. (This would be true if F were merely class C\)

132 TENSORS IN EUCLIDEAN GEOMETRY [CHAP. 10 Subscript Notation for Partial Derivatives From now on, write dr dr = r, „2 d Г 1 Ί. 1 dx1 ' dx- " dx dx so that, e.g., (10.11) takes the compact form гг Xr2^0. etc. 10.5 PARAMETRIC LINES; TANGENT SPACE Let (*') be taken as coordinates—for the moment, rectangular coordinates—in the parameter plane Ε , yielding two (orthogonal) families of coordinate lines: .1 _ - f . ι χ = t [x2 = d and \X2~C If (c, d) runs over V (the pre-image of surface Sf), then the images under r of these two families are the two sets of parametric lines (or coordinate curves) on У: r = r(t, d) = ρ(ί) г = r(c, σ) = q(a) χ -curves χ -curves Figure 10-6 suggests that the net of parametric lines is orthogonal also. This is not, of course, true in general. In fact, since the tangent fields to the ^-curves and the ^-curves are respectively dp/dt = r1 and dqlda = r2, the net is orthogonal if and only if rjr2 = 0 at every point of У. Tangent plane at Ρ The ^ mapping r χ — const, Fig. 10-6 For surface curves in general, the tangent vector of a curve passing through r(c, d) is a linear combination of the vectors гг and r2, as the following analysis shows. Let the curve be given in the parameter plane as c€' : xl = xx(t), x2 = x2(t); then the corresponding curve on the surface is « : r = r(*1(/),*2(i))^r(i) with tangent vector дГ dx дГ dx ! 2 i (10.12) Here, и = dx Idt, и = dx Idt, so that the vector (u1) in the parameter plane is the tangent to c€' at P' (see Fig. 10-6).

CHAP. 10] TENSORS IN EUCLIDEAN GEOMETRY 133 Definition 3: The collection of linear combinations of the vectors r^P) and r2(P) is called the tangent space of У at P. The unit surface normal is the unit vector η in the direction of n=|(r1xr2) (£HkiXr2||>0) (10.13) The geometric realization of the tangent space is obviously the tangent plane at P, and the surface normal can be identified with a line segment through Ρ perpendicular to this tangent plane; that is, orthogonal to the surface at P, as indicated in Fig. 10-6. To summarize this whole affair, the linearly independent (by regularity) triad of vectors r1? r2, η forms a moving frame for the surface, as shown in Fig. 10-7, much in the manner that a moving triad exists for a regular curve having a principal normal. Fig. 10-7 10.6 FIRST FUNDAMENTAL FORM Consider a curve on the regular surface if : r = r(x1. x2) given by % : r = r(x1(t),x2(t)) = r(t), with pre-image c€' : x' = x'(t) in the parameter plane. Using (10.12) and recalling that the (Euclidean) inner product is distributive over linear combinations of vectors, arc length along % is calculated as jt) = \\r\\2 = rr=(uir,)(u'rJ)^glJuiui (10.14a) in which g„. = r,r,. (lS/,/S2) (10.15) and, as above, и = dx'ldt. In the equivalent differential form, ds2 = gijdx'dx' = l (10.14b) the arc-length formula is known as the First Fundamental Form (abbreviated FFF) of the surface if. In view of (10.12) and the regularity of if, ||r|| = 0 if and only if ul = u = 0; this proves Lemma 10.2: The FFF of a regular surface is positive definite. Lemma 10.2 implies that g = det (gtj) > 0; in fact, we can use Lagrange's identity, (Γ>ΧΓ2)2 = (ή(Γ22)-(ΓΛ)2

134 TENSORS IN EUCLIDEAN GEOMETRY [CHAP. 10 to establish that cf. (10.13). g=E (10.16) EXAMPLE 10.4 Compute the FFF for the right helicoid (Fig. 10-8), r = (x1 cos x2, x1 sin x2, ax2) We have: whence and i-j = (cos x2, sin χ , 0) r2 — (—x1 sin x2, xl cos x2, a) Sn = r2 = con2 x2 + sin2 л:2 + О2 = 1 8i2 = rir2 = (cos x2)(~xl sin jc2) + (sin x2)(xi cos x2) = О g21 = r2 = (-χ1)2^2 x2) + (x'^cos2 x2) + a2 = (χ1)2 + α2 I = (djc1)2 + [(jc1)2 + a2](iijc2)2 Along with the FFF, tensor calculus enters the picture. For the intrinsic properties of a particular surface У in E3 (the properties defined by measurements of distance on the surface) are all implicit in (10.14b), which can be interpreted as a particular Riemannian metrization of the parameter plane. Thus, the study of intrinsic properties of surfaces becomes the tensor analysis of Riemannian metrics in R2—and this may be conducted without any reference to E3 whatever. Observe that the metrics under consideration will all be positive definite (Lemma 10.2) but not necessarily Euclidean (see Theorem 9.1). Accordingly, we shall drop the designation Ε for the parameter plane, which shall henceforth be referred to general coordinates (*'). EXAMPLE 10.5 The metric for R' corresponding to the right helicoid (Example 10.4) is non-Euclidean, as is demonstrated in Problem 10.27. Now the parameters xl and x2, which arc actual polar coordinates in the лу-planc of Ε (see Fig. 10-8), formally keep that significance when the plane is considered abstractly as parameter space. This is an instance of the formal use of a familiar coordinate system in a non-Euclidean space, as mentioned in Section 3.1. Unit Tangent Vector If r-€ : r = r(x\t),x\t)) is any curve on У, then by (10.12) and (10.14a), U Г: ygjku'uk (10.17)

CHAP. 10] TENSORS IN EUCLIDEAN GEOMETRY 135 Angle Between Two Curves Let c€1 and %2 be two intersecting curves on if that correspond to χ1 = φ'(ί) and χ' = ψ'(σ) (i = 1, 2) in the parameter plane. Writing и = άφ'/dt and v' = άψ'/άσ, we have for the angle θ between Tj of ^ and T2 of ^2: их-, υ'τ: gi-.uv' cosg=T1T7= . ' · , ; = . ; , (10.18) Compare (5.11). Theorem 10.3: The angle between the two parametric lines through a surface point is cosfl= _Ьг (10.19) Corollary 10.4: The two families of parametric lines form an orthogonal net if and only if g12 = 0 at every point of if. 10.7 GEODESICS ON A SURFACE A further link with tensors is provided by the concept of geodesies for regular surfaces. One can intuitively imagine stretching a string between two points on a surface, and pulling it tight: on a sphere this would lead to a great circular arc, and on a right circular cylinder, a helical arc. Since from our point of view the surface is disregarded and (g!;) is taken as a metric for the parameter plane, the problem has already been worked out (Section 7.6). Relative to the FFF of if, define the Christoffel symbols through formulas (6.1) and (6.4), η = 2. [Problem 10.48 gives an equivalent "extrinsic" definition, in terms of the vector г.] Then a geodesic on if is any curve r = r(x1(t), x2(t)) in the surface whose pre-image in the parameter R satisfies the system of differential equations (7.11 )—(7.12); if t = s = arc length, the governing system is (7.13). [Remember that the (non-Euclidean) distance measured by s in R is the Euclidean distance along the geodesic as a curve in E3.] Similarly, harking back to Section 6.5, the intrinsic curvature of a curve r£ in if is the function t(s) = \fg~bb] (10.20) —cf. (6.12)—where the intrinsic curvature vector (b1) (in R2) is given by (6.11). Remark 4: Intrinsic curvature can be shown to be the instantaneous rate of change of the angle between the tangent vector of r£ and another vector in the tangent space that is "transported parallelly" along the curve. Here, the term "parallel" refers to a certain generalization of Euclidean parallelism (see Problem 10.22). Theorem 10.5: A curve on a surface is a geodesic if and only if its intrinsic curvature к is identically zero. In contrast to the above intrinsic characterization of geodesies, there is an interesting and useful extrinsic characterization, proved as Problem 10.18. It adds a visual dimension that often allows the immediate identification of a geodesic. Theorem 10.6: A curve on a regular surface is a geodesic if and only if a principal normal N of the curve can be chosen that coincides with the surface normal η at all points along the curve.

136 TENSORS IN EUCLIDEAN GEOMETRY [CHAP. 10 10.8 SECOND FUNDAMENTAL FORM By taking the dot product of the surface normal wittuthe second partial derivatives of r with respect to x1 and χ , /,-nr, (10.21) we generate the coefficients of the Second Fundamental Form (SFF) of a surface: //;. dx dx1 - II (10.22) Curvature of a Normal Section If 2F is a plane containing the surface normal n at some point Ρ of У (Fig. 10-9), the curvature of the normal section of У (the curve of intersection of if and 9<), denoted %9, is given at point Ρ by the formula к 1 τ gklu и 1 (10.23) where (и1) =(dx'/dt) gives the direction, at P', of the curve corresponding to C9 in the parameter plane; see Problem 10.23. ~—~~ η 1 p\ 5 CiF Fig. 10-9 As & rotates about n, the curvature κφ of ^ at Ρ is periodic and will reach an absolute maximum and an absolute minimum: let max Kcs = κ min κ,. (10.24) The two section curves having these two extremal curvatures are called principal curves, and their directions are the principal directions. If кг = к2 at Ρ, all the normal sections at Ρ have the same curvature and unique principal directions do not exist. (In this case, Ρ is called an umbilical point of the surface.) Surface Curvature Two measures of the curvature of a surface Sf are commonly used. Definition 4: The Gaussian curvature of У at point Ρ is the number K= κ1κ2; the mean curvature is the number Η = кг + к2.

CHAP. 10] TENSORS IN EUCLIDEAN GEOMETRY 137 It will be proved in Problem 10.25 that the extreme curvatures Kj and κ2 are the roots of the following quadratic equation in λ: (gng22 "gi2)A2 -(/llg22 + /22g„ "2/12gl2)A + (/n/22 -f212) = 0 (10.25) The relations between the roots and the coefficients of a polynomial equation then give: i^ _ /ll /22 ~ / 12 τ τ _ JI1822 "*" /2281I ~ ^J 12§12 (1П?6\ 811822 ~ 812 811822 ~ 8l2 10.9 STRUCTURE FORMULAS FOR SURFACES Two fundamental sets of relationships involve the parts of the moving triad of a surface, 0Ί,Γ2,η). Equations of Weingarten Since n2 = l, d(j\2)ldx =2nn. = 0 (г = 1,2). Hence, for each г", n; lies in the tangent space: П; = u]r1 + u2r2, for certain scalars «f. Similarly, from orthogonality, 0 = (nr,.); = η;Γ/ + nr,; or n;r; = -ftj It follows (Problem 10.28) that ^ = -8%** (10.27a) for i = 1, 2. From the explicit form of the inverse metric matrix (g''), (10.27a) may be spelled out as follows: 812/12 ~ 822/n , 812/n ~~ 811/12 ni = ri + r2 8 8 (10.27b) 812/22 — 822/12 . 812/12 ~~ 811/22 n2 = rx + r2 8 8 Equations of Gauss Since (rl7r2,n) is a basis for E3, we can write r!; = «^Tj + «f;r2 + и3;п. Evaluation of the coefficients (Problem 10.29) leads to г;; = Г>,+/,п (10.28) An Identity Between FFF and SFF Since rijk =rikj, (10.28) implies (Г>, + fvn)k = (Fikrs + fikn)jt or (T'ti)kr, + Цггк + fijkn + f,nk = (Г?Дт, + Γ>,. + fikja + /Λη,. Dot both sides with r, and use the definition r;rs = gls and the relations r;rsfc = Гм (Problem 10.48) and r;n = 0: (rjOtg,/ + rijrski + fankri = (ПД&г + Vs/i + fik*jri Now substitute for the п; from (10.27a) and use rtr, = gti and gs'gti = 8\ to simplify the result: (J VS ri Г* ~~! ~T~k + ^ iVt^ r/ ~ "I £/-*■ r* Finally, introducing the Riemann tensor via (8.2) and (8.3), we obtain Rm=fikfn-fnf,k (Ю-29)

138 TENSORS IN EUCLIDEAN GEOMETRY [CHAP. 10 The left member of (10.29) depends only on the coefficients of I together with their first and second derivatives; the right member depends only on the coefficients of II. This essential compatibility relation between the two fundamental forms must hold at every point of a regular surface. The 'Most Excellent Theorem' of Gauss By (10.26) and (70.29), K= /11/22-/12 = gm2 {10J0) 811822 812 8 Thus, the numerator of К can be derived entirely from the FFF. Since the denominator is also obviously from the FFF, we have: Theorem 10.7 (Theorema Egregium): The Gaussian curvature is an intrinsic property of a surface, depending only on the First Fundamental Form and its derivatives. Remark 5: The motive for the definition (8.7) of Riemannian curvature is now apparent. 10.10 ISOMETRIES The practical question of whether inhabitants of a fog-enshrouded planet could, solely by measuring distances on the surface of the planet, determine its curvature, is answered in the affirmative by Theorem 10.7. A further important conclusion can be drawn. Suppose that two surfaces, У(1) : r(1) =r(1)(x\ x2) and Sf(2) : r(2) = r(2)(x\ x2), are defined over the same region V of the plane and that the First Fundamental Forms agree on V. This will obviously set up a correspondence between 5^(1) and 5^(2) in E3 that is bijective between small patches (induced by the neighborhoods of V over which both r(l) are bijective) of the two surfaces. This correspondence is called a local isometry between 5^(1) and 5^(2) because the two surfaces are, patch for patch, metrically identical. But then (Theorem 10.7) the Gaussian curvatures K(1) and K(2) must be equal at corresponding points. Theorem 10.8: If two surfaces are locally isometric, their Gaussian curvatures are identical. In the case of constant Gaussian curvature K, Beltrami's theorem tells us that there is a parameterization for if for which the FFF takes on the form: ds2 = a2(dx1)2 + (a2 sinh2 д:1)^*2)2 if K= -1/a2 <0 ds2 = (dx1)2 + (dx2)2 ifK = 0 ds2 = a2(dx1)2 + (a2 sin2 xl)(dx2)2 if K= 1/a2 > 0 EXAMPLE 10.6 The plane and the sphere are surfaces of constant zero curvature and constant positive curvature, respectively. For a surface of constant negative curvature, see Problem 10.49. Beltrami's theorem implies a partial converse of Theorem 10.8: Theorem 10.9 (Minding's Theorem): If two surfaces are of the same constant Gaussian curvature, they are locally isometric. Remark 6: A proof of Theorem 10.9 for zero curvature was given in Problem 9.9.

CHAP. 10] TENSORS IN EUCLIDEAN GEOMETRY 139 Solved Problems CURVE THEORY; THE MOVING FRAME 10.1 The curve (i<0) (ii?0) lies partly in the xy-plane and partly in the *z-plane (Fig. 10-10). Show that it is regular of class С , but that it possesses no principal normal vector. limN f-++0 Fig. 10-10 10.2 The component functions for r(/) are *(/) = / y(t) = When t < 0, y{t) = 4/3. As / -» 0, ,. У(?)-у(0) .. г4 hm -^^—7^ = hm — f—-о t-Q г—-о / /<0 z(0; 0 lim y(0-y(Q) /-0 /<0 ίίΟ 0 lim - = 0 r^ + 0 / f<0 /SO hence, y(t) is differentiable at / = 0. Clearly, y(t) = 0 for / > 0. A similar analysis applies to z(t). Hence: which are continuous functions. Continuing the analysis up to the third derivatives (24/ /<0 ·■·/,) =Л° г<0 Ίθ ifeO Z[) 124/ /a0 Я0 : Hence, x(t) being differentiable to all orders, r(/) is of class C3. Furthermore, because x(t) = 1, f(/) Φ 0 for all / and <£ is regular. However, the principal normal, which exists for the separate parts of <£ (lying in the xy-plane for / < 0 and in the xz-plane for / > 0), cannot possibly be continuous at / = 0, let alone differentiable. Hence, % does not possess a principal normal. (a) Describe the curve r = (cos t, sin t, tan1 i), where 0^ ί and where the principal value of the arctangent is understood. (£>) Find the arc length between the points r(0) and r(l).

140 TENSORS IN EUCLIDEAN GEOMETRY [CHAP. 10 (a) This is a form of the circular helix, except that the pitch decreases with increasing /. The curve lies on the right circular cylinder x2 + y2 = 1; beginning at (1, 0, 0), it winds around the cylinder and approaches the circle x2 + y2 = 1, ζ = π/2 asymptotically as <—»<». (*) r = I -sin /, cos /, /2 + l as It or — = vsin21 + cos2 t + 1 {t2 + iY A numerical method of integration is required. Using Simpson's rule on a programmable calculator, one obtains f1 V/4+2/2 Jo t2 + 1 + 2 dt« 1.27797806 10.3 Find the moving frame for the curve <g 3 - 3i3 4 + 4i3 , 3ί Ι (ί real) 5 ' 5 Show that the binormal vector В is constant, so that the curve is actually planar. Making the calculations required in (10.5): -9t2 Ylt2 and and ,3 /81 144 „.„ \\ — /4+ — i4 + 9 = 3\V+T 5 ' / " " V25 25 (-9Г/5,12г2/5,3) _ (-3t2,4t2,5) 3V/4 + 1 5V/4 + l 18/ 24/ 0 r xr rxr 5 ' 5 i J к 9 2 12 2 „ -- / — / 3 5 5 18/ 24/ 5 5 18|/ 72/ 54/ ΊΓ '" 5 ,0 18/ (4,3,0) V42 + 32 + 02 = 181/1 /..χ.. /„4 „Ύ 18 24 \ / 162 5 162 216 5 216 ,' (rr)r=(9/4+9)(-T/,T/,0)^-—/'- — /, —/5 + — /,0 (rr)r = 9-18 25 /J + 24-12 25 /3 + 0 12 ЛЗ 162 5 216 5 cA 3 -=- / , -z- t , 54/3 ,..,.. ,...,. , 162 216 rjJ\ 10/ 9 12 . (rr)r - (rr)r =(-—/, — /, -54/3j = 18/(^- - , -j , -3/· Ν=ε -18/(9/5,-12/5, 3/2) __£/ (3,-4,5/2) (3V/4 + l)(18|/|) ~ I'l 5V7T7 Now choose ε = +1 when /<0 and —1 otherwise, making N = (3,-4,5/2) 5V/4 + l Β=ε (-18//5)(4,3,0) 18 1/1 st /4 3 -e/ /4 1/1 V5 ' M-» 10.4 Establish the general formulas (10.5) for the moving frame of the curve 4ό : r = r(i), with an arbitrary parameter t.

CHAP. 10] TENSORS IN EUCLIDEAN GEOMETRY 141 By definition, ds It Krr)1 or ώ ds and we have at once for the unit tangent vector „ . ■ dt T = r' = r -r ds To obtain a principal normal, first calculate dt ^)1,2=^)"1,2^+*)-ш (Note the general formula </||u||/</< = uu/||u||.) Hence, and dt ^ dt . d ,, .. T = f-+f7 r ds dt " " dt IHI3 r (rf)r _ ||г||2У — (rf)r . A = (rr)r - (rf)r = _ r x (r χ f) ds ||rir luir where the last step used the vector identify u χ (v x w) = (uw)v - (uv)w. It follows that N can be constructed by normalizing the vector N* = -r x (r x r) Since r and r χ r are orthogonal, ||N*|| = ||r|| ||r χ f || and so, provided rxf^O, N=£ (rxr)xr _£ (rr)f-(rr)r ||r χ f || ||r|| ||r|| ||r χ r| Finally, for the binormal vector, with v=r xr^O, Г V X Г B = TxN= -77T77 x ε m ν llrl (rr)v - (rv)r _ ||r||2v-(0)r = jv_ llr||2ltvll llrll2llvll llvll CURVATURE AND TORSION 10.5 Find the curvature and the torsion of the circular helix S . S bs\ , . l—l Г2ч r = \ a cos - , asm - , — | (c = Va + b ) с с where s is arc length. By differentiation with respect to arc length, T = r' Normalizing T', choose and, correspondingly, a . s a s b - - sin - , - cos - , — с с с ее Τ' cos a . s — sin - , U N = s . s cos - , —sin - , 0 с с B = TxN a . s a so -sin - - cos - — с с с ее s s -cos - —sin - 0 с с о . s b s a — sin - , cos - , - с с с ее

142 TENSORS IN EUCLIDEAN GEOMETRY [CHAP. 10 B' —5 cos - , —* sin с с с in» Then, by (10.6), 1 - . » . ζ - . ni - b 7 s b . , s -, b κ = -τ cos - + ^sin-+0=-^ τ=-τ cos - + -^ sin - + 0 = -7 С С С С С С С с С С [If we introduce the "time" parameter t = cs, we then have: rfz b_ = dt~c2~T i.e. the rate at which the helix rises out of the xy-plane (its osculating plane at t = 0) is given by its (constant) torsion.] 10.6 Find the curvature and torsion of the curve r = (i2 + iV2, t2 - fV2, 2ί3/3) (ί real). Use the formulas (10.8): Hence f = (2/ + V2, It - V2,2r) ||r|| = \j(2t + V2)2 + (It - V2)2 + 4/4 = 2(/2 + 1) r = (2,2,4i) f=(0,0,4) rxr ί J к 2f + V2 2/-V2 2/2 2 2 4/ ^ 4(/2 - л/2, -(г2 + r/2), V2) (r χ f )2 = 16[(/2 - Л/2)2 + (г2 + Л/2)2 + 2] = 32(/2 + l)2 det [r r r] = r · (r χ f) = (0, 0, 4) · 4(/2 - л/2, - г2 - л/2, л/2) = 16V2 _ ε\/32(Γ + l)z _ e 16V2 _ 1 K~ 8(/2 + l)3 ~V2(f2+l)2 T~ 32(/2 + l)2 ~ V2(/2 + l)2 10.7 Prove (iO.S). Using the results of Problem 10.4, we have κ.-ΝΤ'-ί.· (*xf>x*V/' гх(гх?)У ||гх(гхг)|| Г ||f||||rxf||/ V ||f||4 1 ||f||5||rxf|| ||r||2||rxf||2 sin2 (it/2) _ ||rxr|| ||r||5 |r χ r|| ||r||3 The torsion requires the computation of B'. By (10.5), rxr ν εΒ rxr whence d I v r= — I — 1 = — · + A (—\ ν (vv)v_ ||v|| v-(vv)v εα dt\\M\) ΙΙνίΓ Λ\||ν||/ν llvll llvll3 llvll3 But ν = d(r χ t)ldl = (r χ if) + (f χ F) = r χ r; hence, εΒ , εΒ ||r xr||2(r xr')-[(r xr)(r χ r)](r xr) r rxr Dot this with εΝ-- (rr)f-(rf)r ||f II ||f χ f||

CHAP. 10] TENSORS IN EUCLIDEAN GEOMETRY 143 from (10.5), and use u · (u x w) = 0: (rr)||rxr||2[f-(rx?)] -0-0 + 0 ||r||2(-det [frif]) NB' г г χ r llrii г χ г det [r if if] T ~ IT- ^M2" rxr 10.8 Prove (α) Ν' = -кТ + тВ, (b) Β' = -τΝ. (α) Since NN = 1, Ν' is orthogonal to Ν, which puts it in the plane of Τ and B. Therefore, for certain real A and μ, Ν' = λΤ+μΒ (1) Dot both sides by T, then by B, and use TN = 0, к = NT', and τ = -NB': TN' = AT2 + μ ΤΒ = λ or λ = -Τ'Ν=-κ BN' = τ = АВТ + μ.Β2 = μ Substitution for A and μ in (7) then yields the desired results. (b) From Β = Τ χ N and part (α), В' = Τ' x Ν + Τ χ Ν' = (κΝ) Χ Ν + Τ Χ (-κΤ + τΒ) = 0 + 0 + τ(Τ Χ Β) = τ(-Ν) = -τΝ 10.9 Prove that if a curve has к' = 0 at some point, then N" is orthogonal to Τ at that point. From N'=-kT + tB, it follows that N" = -κ'Τ - κΤ + τ'Β + тВ'. but к' = 0; and from the Serret-Frenet formulas for T' and B' we obtain Ν" = -κ(κΝ) + τ'Β + τ(-τΝ) = (-κ2 - τ2)Ν + τ'Β As N" is in the plane ol N and B, it is orthogonal to T. SURFACES IN EUCLIDEAN SPACE 10.10 Show that a surface of revolution is regular and exhibit the unit surface normal. The Gaussian form of a surface of revolution about the z-axis (Fig. 10-11) is r = (fix1) cos x2, fix1) sin x\ gix1)) i fix1) > 0) so Tj = (/'(x1 )cosx2, fix1) sin x2, g'{x1)) r2 = i-fixl)s\nx2, fix1) cos x2, 0) and r,xr2 = (,—fg' cosx2, —fg' sinx2, ff (cos2 x2 + sin2 x2)), with norm Γ Λ /ΐ2 ,2 2 2 . 72 J2 ^ 2 2 < 75772 г л I г, 2 , Γ2 E = \f g' cos χ + f g' sm χ + ff =/V/ +g Now/ = /(jc1)t^0; further, the generating curve is regular, which means that, with / = x1, the tangent vector of that curve, (|.*2)-<Λ0.,Ί is non-null and f'2 + g'2 Φ 0. Therefore, E^Q and the surface is regular. The unit surface normal is

144 TENSORS IN EUCLIDEAN GEOMETRY [CHAP. 10 Generating curve *=/©=/(*') ζ = g{t) = g(xl) OQ = Д*1) Fig. 10-11 10.11 Identify the xl- and *2-curves for the right helicoid (Example 10.4) and describe the behavior of the unit surface normal along an ^-curve. The x'-curves (x2 = const.) are given by r = (0, 0, ax2) + ^(cos x2, sin x2, 0) (ί'έΟ) thus, they are rays parallel to the xy-plane. The x2-curves (x1 — const.) are given by i.e., circular helices of radii x1. We have: \Jx2 +y2 = x1 z = ax2 гг — (cos x2, sin x2, 0) r2 = (—x1 sin x2, x1 cos x2, a) *~y Fig. 10-12

CHAP. 10] TENSORS IN EUCLIDEAN GEOMETRY 145 cos χ sin jc* 0 ~xl sinx2 xl cosx2 a ■ (a sin χ , — flcosx2, x1) and a sin χ - a cos χ lkixr2|| V Vfl2 + (x1)2 ' V«2 + (л:1)2 ' V«2 + (л:1)2 = (cos ω)ιι + (sin ω)ν where ω = tan _1 (xVa), u = (sin x2, —cosjc , 0), v = (0, 0,1). On an χ'-ray, u and ν are fixed unit vectors, while ω increases from 0 to a π/2 as x1 increases from 0 to °°. Thus, η traces out a quarter-circle as the ray is described (see Fig. 10-12). 10.12 Find the FFF for any surface of revolution, and specialize to a right circular cone. With i-j and r2 as obtained in Problem 10.10, g„ = γιΓι = (/' cos*2)2 + (/' sin x2)2 + (g1)2 = f2 + g'2 811 = 821 =r1r2 = -/'/cosx2sinx2 + /'/sinx2cos.x2 + (g')(0) = 0 g22 = r2r2 = (-/ sin x2)2 + (/ cos x2)2 + 02 = f and i = (/'2 + ^2)(^)2 + /2(^2)2 For a right circular cone (Fig. 10-13), /= x1 and g = ax1; hence, 1 = (I + a^idx1)2 + (x^Xdx2)2 (1) (2) Ζ = ax z. 1^- a P. / A/ // I* С * x^e - χ = a cosh xx catenary: j Fig. 10-13 Fig. 10-14 10.13 Find the FFF for the catenoid (Fig. 10-14) and compute the length of the curve given by x1 = t,x2 = t (0^i^ln(l + V2)). Here fix1) = a cosh x1, g(x1)= ax1, and (7 ) of Problem 10.12 gives, along the curve, ds\ dx' Jdx2 and ) = (a2 cosh2 x'^^j-) + (a2 cosh2 -Х^^Ч ' = 2α2 cosh2 t rla (l + x/2) L = aV2 I cosh t dt = aV2 sinh [In (1 + V2)] = aVl

146 TENSORS IN EUCLIDEAN GEOMETRY [CHAP. 10 10.14 Let <6l and Чг be two curves on the right circular cone r = (я1 cos x~, x1 sin χ ,2л:1) whose pre-images in the parameter plane are σ At the point of intersection, find the angle between (€1 and <β2, and show that orthogonality in the xlx -plane does not carry over to the cone. The intersection point P' of the two pre-image curves is determined by the simultaneous equations t , 3 - t = σ and - = σ which give t — 2, τ = 1, and P' = (1, 1). Thus, the two tangent vectors at P' are: <"'· ν Λ -,-<-·»> С)-(г = (1,2) Considered in the Euclidean sense, (u) and (υ') are orthogonal. To express the angle between tangents at the image of P\ we adopt the metric (2) of Problem 10.12 (with я = 2) and apply (10.18) for xl = 1, x2 = 1: (l + 22)(-l)(l) + (l)2G)(2) -4 cos θ = j; = — ^ о Therefore, the curves are not orthogonal at the image of P'. 10.15 Prove Theorem 10.3 and verify Corollary 10.4 geometrically for the right helicoid (Example 10.4) and for any surface of revolution (Problem 10.12). The proof consists merely in taking (и')= (1,0) and (v') = (0,1) in (10.18). (Compare Problem 5.31.) As is clear from Problem 10.11, the right helicoid is a ruled surface, generated by a half-line (an x'-curve), pivoted on the z-axis, that rotates parallel to the xy-plane while the pivot point travels up the z-axis. A given point Ρ of the generator thus describes a helical χ -curve (Fig. 10-8), which is necessarily everywhere orthogonal to the generator (i.e., to the x'-curves). As for surfaces of revolution, it is clear that the parameter lines that match the revolved planar curve (xl-curves, or meridians) and the circles traced by individual points of the planar curve (x2-curves, or parallels of latitude) are mutually orthogonal. By previous computations, g12 = 0 for both the helicoid and for the general surface of revolution. 10.16 Show that under a change of coordinates χ =x (x\x~), x~ = x2(xl, x~) in the plane, the surface metric (gtj) transforms as a second-order covariant tensor. We have by substitution r^1, χ2) = τ(χ\χι, χ2), χ2(χ\ x2)) = r(x\ x2), the latter being the "new" parameterization for Sf. To compute the metric under this parameterization, write (by the chain rule for partial derivatives and the bilinearity of the inner product) - _-- _дГ дГ ( dt dXP\ ί дГ dXq\_ s,i = Γ,Γ;'= lix7'~дх~1 ~ Vdx^ дгУ\дх" dx'I дхр дх" _ дх" Эх" ~ГрГ" дх' ~dx4~Spq дх' dl1 which is the correct formula for tensor character. GEODESICS ШУ / Эх" \q дх1 10.17 (a) Find the Christoffel symbols of the second kind for the sphere of radius a. (b) Verify that the great circles passing through the north and south poles (i.e., the ^-curves) are geodesies.

CHAP. 10] TENSORS IN EUCLIDEAN GEOMETRY 147 (a) The FFF for the sphere of radius a may be calculated from Problem 10.12: £ll = fl £,2=0= #21 The formulas from Problem 6.4 can be used, since (g;) is diagonal; the nonzero Christoffel symbols are found to be: ΓΙ ·1 Ι Γ*2 ι-ι2 . 1 22 = -sin χ cos χ I 12 = 1 21 = cot jc (b) We want to show that the family of curves x1 = t, x1 = d = const, are integral curves of the differential system (7.11)-(7.12), which may be conveniently written as d x' dx' dx 1 dx' ~^tT+'k~dT~dt~ = 2~dT or, for the given metric, ά2χΛ . . , ,Jdx2\2 1 dx1 ι = 1 —-, (sin χ cos χ ) -—- I = - -—- dt V dt I 2 dt d χ , ,, dx1 dx 1 dx i-2 —γ- + (2cot χ) — — = - —r- dt dt dt 2 dt d I dx' dx ~dt \8>* ~dJ ~dt ln^( —) +(fl2sin2x1)l In I (dx2 dt \ \ dt I v '\ dt d dt Since dxlldt — 1 and dx2/dt = 0, both equations reduce to 0 = 0, and the verification is complete. 10.18 Prove Theorem 10.6: A curve on a regular surface is a geodesic if and only if, by proper choice of the principal normal, N = n. Let any curve on the surface be given by % : r = r(x\s), x2(s)), where s = arc length, Then, dx' and the first formula (10.9) gives ,T _,, d2x' dx' (dr. dx'\ d2x' dx' dx' .„4 KN = T=^r' + ^te^)^^r' + ^^r'> (i) Dot both sides of (1) by the vector rk and use the result of Problem 10.48: d x' , dx' dx' d x' , dx' dx' kl ΚΓλΝ= "ТУ rirk + ~5 J" rurk = ~TT Sik + ~1 J" Г,-/А (2) ds ds ds '' " ds ds ds ,Jk v ' Multiply both sides of (2) by g and sum on k: "i XT d^x' st' j. dx' dx' *'r ά2χ' , -pi dx' dx' Now if Я?? is a geodesic, the right side of (3) vanishes, and this implies that κγαΝ = 0 for к = 1,2. If к Ф0, then r1N = 0 = r2N; so that N is orthogonal to both r: and r2 (thus to the tangent plane). Therefore, but for orientation, N = n. If к = 0 at some point Ρ and there is a sequence of points along the curve approaching Ρ for which κ Φ 0, then, by continuity, N = n. Otherwise, к = 0 on an interval and the curve is a straight line on that interval, in which case its principal normal N can be chosen to agree with n. Conversely, if the curve has the property that Ν = η at all points, then rAN = rAn = 0 and the left side of (3) vanishes, showing that the pre-image of <€ satisfies the differential equations for a geodesic. 10.19 Apply Theorem 10.6 to the plane sections of a torus. In Fig. 10-15 is shown a torus and various examples of plane sections. In (a), an elliptical-shaped vertical section, the section cannot be a geodesic because the surface normal at Ρ does not lie in the plane of the curve (which contains the curve normal). In (b), a horizontal section that is a circle, again the surface normals do not lie in the plane, and this section is not a geodesic. The circles shown in (c) and (d) are geodesies, since the surface normal will coincide with a correctly chosen principal normal of the curve.

148 TENSORS IN EUCLIDEAN GEOMETRY [CHAP. 10 Geodesies Fig. 10-15 10.20 Demonstrate that the intrinsic curvature к of a curve on a surface can be different from its curvature к as a curve in E3. One example is a circle on a sphere of radius a. If the circle also has radius a, it is a great circle and, hence, a geodesic with zero intrinsic curvature. But its curvature as a (planar) curve in E3 is 1/a. Another example is the circular helix: its curvature is nonzero as a curve in E3, but as a geodesic on a circular cylinder its intrinsic curvature is zero. SECOND FUNDAMENTAL FORM 10.21 (a) Find the SFF for the right circular cone of Problem 10.12. (b) At the point P(l, 0, a), calculate the curvature of the normal section having the direction j at Ρ (see Fig. 10-16). ζ It /1 1 a. \ X Fig. 10-16

CHAP. 10] TENSORS IN EUCLIDEAN GEOMETRY 149 (a) From r = (x1 cos x2, x1 sin χ2, αχΛ) (χ1 > 0), we obtain: rt = (cos x2, sin x2, a) r2 = (-x1 sin x2, x1 cos x2, 0) ru = (0,0,0) r12 = r21 = (-sin x2, cosx2,0) r22 = (-хл cos χ2, -χλ sin χ2,0) and by Problem 10.10, n= (a2 + l)~1/2(-acos x2, -a sin x2,1). The coefficients in II are thus /n Ξ nrn = °> /,2 = /21 = nr12 = 0, /22 = nr22 = (a2 + iy1,2ax\ (b) The direction j at Ρ corresponds to the direction (и1, и2) at Ρ' = (1,0) in the parameter plane, j=M1r1(P) + M2r2(P) (0,1,0) = и'(1,0, α) + κ2(0,1,0) (0, 1, 0) = (и , и , аи ) Thus, и1 = 0 and и2 = 1. Appropriating I from Problem 10.12, we have _ IV, u2) _ /22(P)(m2)2 _ Д Ky i(M>2) (fl2 + i)(M1)2 + (i)V)2 /22^ VZTT 10.22 Develop geometrically a notion of "parallel transport" of a vector along a curve <6 on a regular surface if. Imagine $, and with it Sf, as being rolled without slipping onto a fixed plane &, in such fashion that the point of contact is aways on 4i and the tangent plane to У at the point of contacts always coincides with W. This maps <€ to a (planar) curve % * in cF that has the same arc-length parameter and the same tangent vector. Then, any vector in 9 that is attached to the point of contact and that remains parallel to itself (in the ordinary Euclidean sense) as the contact point describes ζ€* may—under the inverse mapping сё*—> <£—be considered as undergoing parallel transport along Ш. In general, parallel transport of a given vector around a closed curve on a surface does not reproduce the initial vector. 10.23 Prove (10.23). Start with the formula for the unit tangent vector of any curve 'eon У: и'г; / . dx' и ν**,«ν ν dt Then T = Л Ь=^)Г' + W=Ti r; = Q *, + 77=1=7 'uu' V ) where Q' is an abbreviation for the scalar coefficient of r,. Now the Frenet formula gives κΝ=Γ = Ϊ/Vgkluku\ together with (1), this yields: Q, ^ u'u' \lgklukul ' gk,uku Dot both sides of (2) with η (the surface normal) and use the fact that r;n = 0 for each i: κΝ=77===Γ. + 7Γ7^^ (2) кnN = —, r,7n = ^ Д. (3 ) If ^ is a normal section Ч& at Ρ, and κ, Ν, and the right side of (3) are all evaluated at P, then к = κ^, nN = η2 = 1, and (3) becomes the desired expression

150 TENSORS IN EUCLIDEAN GEOMETRY [CHAP. 10 CURVATURE OF SURFACES 10.24 Show that the maximum and minimum values of the function a^u'u' uTAu F(u) = ,, .t ι = ~τ— bkluu uBu where A = [а^]22, В = [Ь^]22, u = (и1, и2) Φ (0, 0), with В positive definite, are the two roots of the quadratic equation in λ 11 _ ^Ьц flX2 — ^Ь12 ■y, Ad^, Qti Ко ^^ det (A - λΒ) - (1) (hence, eigenvalues of Β ιΑ), and that the extreme values of F occur for vectors u satisfying (A - xB)u — 0, where χ takes on the two eigenvalues of B~ A (hence, eigenvectors of Β ~lA). We may assume without loss of generality that A and В are symmetric. Let S be any simple closed curve in the м1м2-р1апе having the origin in its interior. The Weierstrass theorem guarantees that F(u) asumes a largest value on S; say, F(w) = M. Because F is constant on rays emanating from the origin (F(Au) = F(u) for any λ Φ 0), the absolute maximum on <§ is both an absolute and a relative maximum in the ii'ii2-plane; hence, the gradient of F must vanish at w. We have: dF(u) (рк1ики1){1аИи') ~ (а,и'и')(2Ьр1и1) 2 ~^T~ = 77—ΓΤα = 7—η [apiW - F(u)(b ,u ) or VF(u) = -^— [Au - F(u) Bu] Therefore, ^4w - Μ 5w = 0, which shows (i) that Μ is an eigenvalue of В'гА and thus is a root of the characteristic equation (1); (ii) that w is an eigenvector belonging to M. A like consideration of the minimum value, m, of F on S leads to the other eigenvalue and associated eigenvector. 10.25 Prove that the extreme normal curvatures κλ, κ2 are the two roots of the quadratic equation (10.25). In Problem 10.24, take ац = fv and bkl = gkl; expand (1) to obtain (10.25). 10.26 Prove that the two normal section curves through Ρ on У giving rise to max κ^ = κγ and min κφ = κ2 are orthogonal when κ, Φ κ2 (that is, when Ρ is not an umbilical point on if). Let us prove the general result, in the notation of Problem 10.24. We have: Aw-MBw=0 Ay - m5v = 0 With the inner product of column vectors defined as ρ · q = pT5q, multiply the first equation by v' and the second by wr, and subtract: (m - M)v w = 0 Hence, if m φ Μ, ν and w are orthogonal. 10.27 Calculate К and Η for the right helicoid. Show that as xx-^>°°, К tends to zero (the surface becomes "flatter" as the distance from its axis increases without bound). From Problem 10.11 and Example 10.4, 1 (a sin x1, -a cos x2, x1) VOO2 + a rn = (0,0, 0) r12 = r21 = (-sinx2, cosjc2, 0) r22 = (—x1 cos x2, — x1 sin x2,0)

CHAP. 10] TENSORS IN EUCLIDEAN GEOMETRY 151 so that /n = nrn=0 Г12=Г21 = -а/У(х1)г + аг /22 = 0 and K. Мгг-fn 0~а2/[(хГ + а2} .() ,_!_ 511Й22 SI -g\2 [(x'f+a2] [(x'f + a2]2 n _ Λΐ?22 + /22g„ ~ 2/12g12 _ 0 + 0 - 2(0) _ 0 511622 612 STRUCTURE FORMULAS; ISOMETRIES 10.28 Complete the proof of (10.27a). The relations n; = щгк and nyr, = -ftj imply -fa = u)8ki Multiply both sides by g"' and sum over i, obtaining и, = —g'sfir Hence, ^ = <rk=-gl%rk=-g'%rk 10.29 Prove (10.28). Let the equation r-; = u]jr1 + и2р2 +- и^-п be rewritten as r;/ = uljrs + и^п. Dot with n, obtaining ul = f„\ therefore, «V = <rS + fijn i1 ) Dot (1) with rk and use Problem 10.48: Solve for u], Substitute back into (J): ГЦГк = U\jrsrk + 0 ОГ Г/,* = 4&* 8 rijk=uifg gslc or rij = uii8s=ulj г^Г^ + Д-п Supplementary Problems 10.30 (a) Describe geometrically the curve whose parametric vector equation is r = (cos i, sin i, (l-ί)'1) (0S(<1) What happens as t—> 1? (b) Use a programmable calculator and Simpson's rule to find the arc length for 0^ t^ 1/2 accurate to 6 places. 10.31 Find the exact length of the space curve r = (i2 + л/2, t2 - л/2, 2ί3/3) (-ISiSI). 10.32 (α) Using the arc-length parameterization of the right circular helix, r = I a cos - , a sin с с с find the coordinate equations of the tangent line to the helix at any point P = r(s). (b) Show that the tangent line intersects the xy-plane at a point Q = r*(s) such that PQ = s. (c) By thinking of a string wound along the helix, interpret the result of (b).

152 TENSORS IN EUCLIDEAN GEOMETRY [CHAP. 10 10.33 Show that for the curve у = χ5 in the xy-plane, parameterized as r = (i, t\ 0), the vector TV||T'|| has an essential point of discontinuity at t = 0. 10.34 Find the curvature and the torsion of the curve r = (i, t5 + a, t5 - a). 10.35 Prove that a curve is planar if and only if its torsion vanishes. 10.36 Prove that a planar curve with constant nonzero curvature к is a circle. [Hint: T= (cos Θ, sin Θ, 0) and N = (-sin Θ, cos Θ, 0) imply к = θ' or θ = κ.τ + a; show that the radius is Ι/κ.] 10.37 Verify the Serret-Frenet formulas for the circular helix. 10.38 Show that if included in the range of the map r(x\ x2), the vertex of the right circular cone is a singular point. 10.39 Calculate the unit normal for the catenoid as parameterized in Fig. 10-14 and show that the surface is regular. 10.40 Find the length of the curve on the right helicoid (Example 10.4) given by χΛ = t2, χ2 = In t, with 1 S t S 2, in the special case when the parameter a = 1. 10.41 Find the two possible directions for a curve <€' in the parameter plane whose image on the paraboloid of Fig. 10-17 meets the circle x2 + y2 = 4, ζ =4 at P(0, 2,4) at an angle of π/3. U= (χ1) (u1 иг) = (0, 1J ♦-У Fig. 10-17 10.42 Use Theorem 10.6 to show that an elliptical helix is not in general a geodesic on an elliptical cylinder. 10.43 Calculate the Christoffel symbols of the second kind for the right helicoid (Example 10.4). Show that circular helices on the surface are geodesies. 10.44 Exhibit the SFF for the general surface of revolution (Problem 10.10).

CHAP. 10] TENSORS IN EUCLIDEAN GEOMETRY 153 10.45 Establish the formulas below for any surface of revolution, with G = g'lf (see Problems 10.12 and 10.44): K= GG' and π /G'+g'd + G2) Use these formulas to verify that a sphere of radius a has Gaussian curvature 1/a2 and mean curvature -21a. 10.46 (a) Calculate К and Η for two different parameterizations of the paraboloid ζ = a(x2 + y2): (i) as the surface of revolution for which /= x\ g = a(x1)2; (ii) as the surface r = (x\ x2, а(хг)2 + α(χ2)2). (b) Interpret the results of (a). 10.47 Infer from Problem 10.45 that Η = 0 for any catenoid. [A surface with Η = 0 at all points is called a minimal surf ace. Among minimal surfaces are those that solve "soap-bubble" problems, which require a minimum in surface area.] 10.48 Prove that Гц4 = r#/rt. [Hint: (γ,γ,), = r,,ry + г/Г.,.] 10.49 Surfaces for which the Gaussian curvature is a negative constant are very rare. One such surface can be constructed as follows, (a) A tractrix is the involute of a catenary (see Problem 10.32(c)). Write the vector equation for the involute of the catenary r = (a coshx1, 0, ax1) (see Fig. 10-14). (b) Using Problem 10.45, show that К = —На2 for the tractroid generated by revolving the tractrix of (a) about the z-axis. 10.50 Prove that the catenoid, I = (a2 cosh2 x^idx1)2 + (a2 cosh2 x^dx2)2 and the right helicoid, 1 = (dx1)2 + [(x1)2 + a2](dx2)2 are locally isometric.

Chapter 11 Tensors in Classical Mechanics 11.1 INTRODUCTION Classical mechanics originated with the work of Galileo and was developed extensively by Newton (it is often called Newtonian mechanics). It deals with the motion of particles in a fixed frame of reference (rectangular coordinate system). The basic premise of Newtonian mechanics is the concept of absolute time measurement between two reference frames at constant velocity relative to each other (called Galilean frames). Within those frames, other coordinate systems may be used so long as the metric remains Euclidean. This means that some of the theory of tensors can be brought to bear on this study. 11.2 PARTICLE KINEMATICS IN RECTANGULAR COORDINATES Let Ρ be a particle whose path in E3 is given by <*? : x=(*'(i)) (11.1) where t represents time. The velocity vector of Ρ is defined as dx (dx1 dx2 dx3] x .2 .3 and the (instantaneous) velocity or speed as the scalar i^llvll^VCiT + C^ + Ot3)2 Ш-3) Further, define the acceleration vector as dv (d2xx d2x2 d2x3\ , „2 ..3 and the acceleration as а = Ы=У(х^)2 + (х2)2 + (х3)2 (11.5) If v= (υ1) and a = (a1), the preceding formulas have the component forms dx , r~r~i ι d χ V — —~ ν — \ ν ν и — ,τ dt dt In terms of the geometry of the curve <# we have v= vT, with ν = dsldt. Hence, d , ■ . dT ds „ -, a = — (υΎ) = νΎ + υΎ = υΎ + υ -у- — = υΎ + υ2Τ dty ' ds dt But Τ' = κΝ, and so Via the Pythagorean theorem, (11.7) implies (11.6) a=uT+Ki;2N (11.7) α = λ/ύ2+ kV (11.8) EXAMPLE 11.1 Formula (11.7) serves to define the tangential and normal accelerations of Ρ as i2 2 ύ = —rj and κυ2 = — (p = radius of curvature) dt Ρ 154

CHAP. 11] TENSORS IN CLASSICAL MECHANICS 155 respectively. For a particle with constant velocity, a = \\κν2Ν\\ = \κ\ ν2; i.e., the acceleration is proportional to the absolute curvature. 11.3 PARTICLE KINEMATICS IN CURVILINEAR COORDINATES There emerges the problem of expressing the preceding formulas in nonrectangular coordinate systems. This is not merely an academic consideration, for there are important situations in which the differential equations of motion can be solved only in polar or spherical coordinates (cf. Example 11.3), not to mention applications in relativistic mechanics. Let us start with the definitions of the velocity and acceleration vectors in a barred rectangular system: _,- dx' ■ dvl ν = —r- and a = —r- dt dt Because the tangent field of % is a tensor, the velocity components in an arbitrary coordinate system (xl) are just v' = dx'Idt. However, as we found in Chapter 6, the acceleration components must be written as absolute derivatives along <# : a = Sv'ISt. Hence, in an arbitrary coordinate system (xl), with (gf) representing the Euclidean metric, we have: i dx' i dv' ,-ri rs d2xl dxr dxs ν = —r- a = -г- + Tv ν = —y + Γ„ — — (НУ) dt dt " dt dt dt v ' and the speed and acceleration scalars are the invariants ν = y/gijvV « = Vft^ (11-Ю) EXAMPLE 11.2 Formulas (ί 1.9) give the contravariant components of velocity and acceleration. These are not the components used in classical physics and vector analysis. There, the metric for an orthogonal curvilinear coordinate system is written as ds2 = h\(dxl)2 + h\(dx2Y + h23(dx3f and one defines the physical components of the velocity vector as dxl dx2 dx} Thus, the physical components are related to the contravariant components via v(a) = yfg^~ava (a = 1,2,3; no summation) (1) Likewise for acceleration and force vectors. (See Problem 11.24.) Let us illustrate the distinction by calculating the components of acceleration in cylindrical coordinates, (x\ x2, хъ) = (хг,х2,хъ)= (r, θ, ζ), for which gn = 1, g22 = (χ1)2, g33 = 1. By Problem 6.26, Γ22 = —χ Γ12 = Г21 = I/xl (all others zero) so that (11.9) gives for the contravariant components ! d2r (άθ\2 2 d26 2 dr άθ , d2z a = —τ — r\ — I a = —τ + - — —- a = —τ (2) dt2 \dt/ dt2 r dt dt dt2 V ; The physical components are then obtained from (ί) as d2r /de\2 d26 „ dr άθ d2z Only the 0-component differs between (2) and (3); but the difference is significant. For instance, the coriolis acceleration of a particle is 2'rh, as in (3).

156 TENSORS IN CLASSICAL MECHANICS [CHAP. 11 11.4 NEWTON'S SECOND LAW IN CURVILINEAR COORDINATES The momentum vector of a particle of mass m is defined as Μ = m\. Relative to a rectangular coordinate system (with the property of being an inertial frame), Newton's second law of motion effectively defines the force vector acting on a particle as F = dMIdt. Accordingly, in curvilinear coordinates (x), the law reads: δΜ δν , F= —= w —=wa (11.11) assuming a constant mass. Therefore, the contravariant components of force are given by d xl „/ dx dx ' — ma — m\ and the covariant components by Р = ™' = т\-& + Г"-&-&) <1L12e> „ ,_,,. ( d xr _, dxr dxs\ *. = ^ = Чй'-^ + Г»<-Л Л") {1112Ь) By introducing a scalar invariant called the kinetic energy of the particle, τ 1 2 1 if T= -mv =2 m8av v one can (see Problem 11.5) put (11.12b) into the equivalent Lagrangian form d ( дТ\ дТ Fi=7 — j (11.13) 1 dt\dv> dx1 y ' The partial derivatives in (11.13) are taken with Τ considered as a function of six independent variables, the x' (via the g/y) and the v'. EXAMPLE 11.3 (Motion under a Central Force) (a) Obtain the differential equation for the trajectory of a particle acted on by a force that is always directed from (or toward) a fixed point O. (b) Solve the equation of (a) when the central force is gravitational, thus determining the orbit of a satellite. (a) By Problem 11.18, the motion will be confined to a plane through O. Take О as the origin of polar coordinates (x1, x2) = (r, Θ) in the plane; the force field then has the form F=(F\ 0). Taking the acceleration components from (2) of Example 11.2, we have as the equations of motion: Z7l 1 r = ma m dr Idd dt2 r\ dt )] 0 = ma2 = m The 0-cquation has the first integral d θ 2 dr άθλ = m d_ I 2 dQ dt2 r dt dt\~r2 dt V dt 2 άθ r — = q = const. dt ч (conservation of angular momentum), which can be used to change the parameter of the trajectory from / to Θ. Thus, writing и = \lr, we have: dr άθ du~l , 2ч/ -τ du\ du d и dt2 n do2 ^" ' qu de 2 - q~^i(qu ) = -qu 2 and the r-equation becomes ,2 -^ + u = g(u,e) (1) in which g(u, θ)= -F\u'\ &)!mq2u2

CHAP. 11] TENSORS IN CLASSICAL MECHANICS 157 (b) For the gravitational field, F1 = -k/r2 = -ku2 (k>0; attractive force), so that g(u, Θ) = Q = const, and the solution of (1) is и = Ρ cos θ + Q, or 1/Q 1 + e cos θ which is a conic having eccentricity e= PIQ and focus at O, the classical result. (2) 11.5 DIVERGENCE, LAPLACIAN, CURL The divergence of a contravariant vector u = (u) on Ε is defined by (9.11), with use of Problem 8.14: j- ι du ■ r du r д /л _ ди ,1 д , _ The product rule for partial differentiation yields the compact form diVU=W^(^H/) (11.14) Another notation for the divergence is V-u. The Laplacian of a scalar field / is given by V/=div(grad /). Since in general coordinates divergence is defined for contravariant tensors only, while grad / = (dfldx1) is a covariant tensor (Example 3.5), we first raise the subscript and then find the divergence by (11.14): vz/=divU siJL ι а ( dx'I Vg dxl [Vgg- dx' (11.15) The Laplacian figures importantly in electromagnetic theory via the scalar wave equation, (11.16a) dt1 д f ii —2 —k V / (k — const. = wave speed) In cartesian coordinates only, one defines the Laplacian of a vector field as V u = (V u), where V«' = uxx + и + uzz, and writes the vector wave equation, dt2 = k2 V2u (11.16b) as an abbreviation for three scalar wave equations. EXAMPLE 11.4 Write the Laplacian for cylindrical coordinates. Using gu = 1, g22 = llix1)2, g33 = 1, and g= (x1)2 in (11.15), v2/= 1 Idx' \ dx1/ dx \x dx1' dx V dx "/ll-"- /rl\2 J22 "*" J3,3, "τ" 1 /l (^>0) the last line employing subscript notation for the partial derivatives. The curl of a vector field u — (u)—symbolized curl u, V x u, or rot u—is given in a rectangular coordinate system (xl) by curl u = \e du ''^Tx1 (11.17a) where eijk is the permutation symbol (Chapter 3). The definition may be rewritten as a determinantal operator:

158 TENSORS IN CLASSICAL MECHANICS [CHAP. 11 curl u; ei d Ϊ7 1 e2 d дх 2 dx~ (П.17b) in which (e1,e2,e3) = (i, j, к) is the standard orthonormal basis. Unlike the gradient and the divergence, the curl cannot be extended to curvilinear coordinate systems by a tensor formula. Remark 1: Not everything in mathematical physics is a tensor. Problem 11.11 shows that (11.17) defines a direct cartesian tensor, but that is all. This is not to say that the curl operator cannot be formulated and used in curvilinear coordinates (see any text in vector analysis). It is only that the curl in spherical coordinates (say) and the curl in rectangular coordinates are not related tensorwise. Nonrelativistic Maxwell's Equations Let Ε = electric field strength D = electric displacement Η = magnetic field strength В = magnetic induction J = current density ρ = charge density e — dielectric constant μ = magnetic permeability с = velocity of light Then the famous Maxwell's equations may be written as follows: 1 dB curl Ε + - — = 0 с dt divB = 0 curl Η 1 (11.18) div D = ρ 1 JO С dt С From standard formulas in electromagnetic theory, D = eE, Β = μΗ, and J = pu, where u denotes the velocity field of the charge distribution; (11.18) becomes μ Ж с dt e dE curl Ε = curl Η -- + % С dl С If the charge distribution is in free space (e = e0 equations into the form 1 iH curl Ε = с dt μ divH = 0 div Ε = - μ0), a proper choice of units brings the div Η = 0 curl Η = 1 дЕ о + ί- u с dt с (11.19) div Ε = ρ Work with Maxwell's equations requires the vector identities listed below (see Problems 11.10 and 11.21). V-(Vxu) = 0 (for any u) (11.20) Vx(Vxu)=V(V-u)-V2u (11.21) — (V-u)=V· dt v ' o>u dt — (V xu) =V x — dtK ' dt (11.22) (11.23)

CHAP. 11] TENSORS IN CLASSICAL MECHANICS 159 Solved Problems VELOCITY AND ACCELERATION 11.1 Find the velocity and acceleration vectors and the scalars υ and a for a particle whose equation of motion (along a twisted cubic) is χ = (ί, ί , г3) (—1 = t = 1)· Determine the extreme values of ν and a, and where they are assumed. v=(l,2/,3/2) a =(0,2, 6/) and and u = Vl +4Г +9t4 a = V4 + 36i2 Hence υ and α have maxima at t = ±1, where ν = Vl4 and a = V40. They have minima at / = 0, where υ = 1 and α = 2. PARTICLE DYNAMICS 11.2 A particle travels at constant speed ν on a curve with positive curvature. Show that its acceleration is greatest where the curvature is greatest. By (11.8) with ν = 0, a = κυ2 or α/κ = const. 11.3 Compute the contravariant acceleration components in a coordinate system (x) connected to a rectangular coordinate system (xl) by x1 = (x1)2, x2 = x2, x3 = x3. Use (5.7): \2xl 0 0 G = 7r7= 0 10 _ 0 0 1 Hence, the Christoffel symbols are given by д Го ι 2x 0 L о 0 0 1 0 0 lj = A(xlf 0 1 0 1 0 . 0 0 1. Г1 and (11.9) gives - In \(x' )2 = —ϊ (all others zero) Z, J X ■Xх 1 (dxl\2 , d2x2 3 d2. NEWTON'S SECOND LAW 11.4 Show that Newton's second law is consistent with Newton's first law: A particle that is not acted upon by an outside force is at rest or is in motion along a straight line at constant velocity. Assume a rectangular coordinate system. F = 0 implies dvidt = 0, or ν = d (constant). Then, dx 1U = A or χ = /d + xn which is the parametric equation for a point (if d= 0) or for a straight line (if d^0), along which υ = lldll = const. 11.5 Prove the equivalence of (11.13) and (11.12b). For simplicity, take m = 1 in (11.13). By the chain rule and the symmetry of (gu),

160 TENSORS IN CLASSICAL MECHANICS [CHAP. 11 dvr ~dF 2VV) + J7-dFv ',·, dxs r dvr 1 '"' ~di ~ 2 ,ν ν + e.v ν dv \ , s 1 „ , 1 ?ir -^ - 2 8rSiv ν + 2 Ss,rv ν + - girsv ν The final expression is exactly the right-hand side of (11.12b) (for m = 1). 11.6 Solve (1) of Example 11.3 when the force field is of the form g(u, 0) = Au + h(d) where Л is a constant and h(6) is periodic of period 2π. With primes denoting θ-derivatives, we must solve u" + u = Au + h(e) or m" + (1-v4)m = A(0) The general solution to the homogeneous equation is \P cos (VI- Α θ + a) ιι = \αθ + β Α<ί Α=ί Qexp(VA-10) + Rexpi-VA^Je) A>\ A particular solution of the nonhomogeneous equation may be obtained in the form и = uHw, where uH is any particular solution of the homogeneous equation. In fact, substitution in the differential equation yields 2u'Hw' + uHw" = h or (u2Hw')' = uHh and this last equation can be solved by two quadratures: W'W= ~4^ L И*(Ф) *(Ф) άφ and w(0) = -η— I ия(ф)Н(ф) The integrals are easily evaluated when к(ф) is represented as a Fourier series. άφ 11.7 If /г(0) = О in Problem 11.6, identify the orbits corresponding to (a) A = 0, (b) A = 1, (c) Λ = 5/4. (α) The curve l/r= Pcos(0 + α), or rcos(0 + a) = 1/F, is а straight line (Fig. 11-1). (b) The curve 1/r = αθ + /3 is a hyperbolic spiral that degenerates into a circle for a = 0. (c) The curve 1/r = 2е"'2 + Re~e/2 is a complex spiral which, in the case Q = Q, R = 1, reduces to the simple logarithmic spiral r = e"/2. Fig. 11-1

CHAP. 11] TENSORS IN CLASSICAL MECHANICS 161 DIFFERENTIAL OPERATORS 11.8 Calculate the Laplacian for spherical coordinates by the tensor formula. (The calculation is very tedious by other methods.) We have G = ..1л4 · 2 2 10 0 0 (χ1)2 0 0 0 (л:1 sin*2)2 С 1 0 0 0 (χ1)'2 0 0 0 (jc^injc2)" and g = (χ1)4 sin2 x2, so that in (11.15), «*%-<>·***('·%+<·£+<■% Therefore, У df _ /,.1л2/„;„ ,.2λ tf «,*%-(,'П**% vj g" Я-, = (^("(sin ,!) -L -ίζ = (sin *!) Д <2л:' cU' 3j df . ι ч τ, . ,. 1 Vgg3i fj = (^)2(sin x2) -^ cU (JC SinjC ) cU .2\2 ,„ч (esc л: ) 3 and so д дх1 дх' -I с»* (xl)2(sinx2)-^ дх , ■ г df (sin *) -^j <9* дх (cscx2) —ч v ' дх3 -2x\smx2) "Й + (^(sin*2) -^ + (cos*2) -^ + (sinx2)-^ OX (dX ) dX {dX ) д f + (esc x2) / ^ Зч2 (дх2)2 In writing the final steps we convert to ρ = χ1, φ = χ2, and θ = χ3: 1 Vg дх' 1 ρ2 sin <p (2ρ sin <ρ) — + (ρ2 Sin <ρ) -4 + (cos φ) -f- + (sin φ) -\ + (CSC <ρ) -^ ν <9<ρ <9<ρ' ,202 ύ. + ι g2f { ι g2f + ΙΕ + cot<p */ <9ρ2 ρ2 <9<ρ2 ρ2 sin2 <ρ дв2 ρ dp p2 dip *(3)e3· 11.9 Calculate the divergence in spherical coordinates (ρ,φ,θ) of (a) a contravariant vector, u = (и1); (b) a vector specified by its physical components, u = w^ej + w(2)e2 + и( ° (α) We plug into the formula (11.14): a- Id, du' ,1 <9 div u = -7= —i (л/1»ы ) = —τ + и -η= —; (Vg) Vg дх v s ' дх' Vg дх' v &J dU' i 1 ί r, κι . 2i = -■ + U —т-Ъ 5 7 К* ) Sin Χ Ι <?*' (χ1)2 sin χ2 дх u ; J <2ы' , ./ 2 \ 2/cos x2\ 3.,,. = Τ"? + и ( - ) + и ( -—2 ) + и (0) с** \Х I ^SinjC / ди ди ди 2 , , Thus divu = -τ- + —— + -— + - u[ + (cot <р)ы ф <9φ ΰθ ρ

162 TENSORS IN CLASSICAL MECHANICS [CHAP. 11 (h) By Example 11.2, we apply (11.4) to the contravariant vector having components u. "(1) и = —г- и 1 χ л: sin jc Hence, from (a), д д /"(2,\ д ( «(з, \ /2\ ит = ^М + 1 ίίίίΣ) + 1 *"»> + 2 со^ φ ρ ify? ρ sin .ρ ^ ρ (ρ) ρ "<<ρ) It is in this last form that "the divergence in spherical coordinates" is generally encountered in reference books. 11.10 Establish in rectangular coordinates the identity Vx(VXu)=V(V-u)-V2u (1) ("curl curl equals grad div minus del-square"). Both sides of (1) are (cartesian) vectors; we shall show that they are componentwise equal. By (11.7), the i'th component of curl u is el]k(dukldx!). Therefore, the /th component of curl (curl u) is [use (3.23)]: д dxr ( duk^ Vs'k dxJ/ ' *'"*"* дх'дх' *'**"'* Я 2 k -/-(drvu)-W дх .,2 к д и дх'дх' д2и' дх'дх' „2 / ό и дх'дх' The first term on the right is recognized as the ith component of grad (divu), and the second term is (by definition) the /th component of the Laplacian of the vector u. qed 11.11 Prove that the array represented in rectangular coordinates (x1) by Ί к·, dU '"iik ~d~x> is a direct cartesian tensor. It suffices to show that (el]k) is a direct cartesian tensor, since (du'ldx1) is known to be a (direct) cartesian tensor. Therefore, given the orthogonal transformation χ' = α\χ\ with |e'.| = +l, define the 33 = 27 quantities дх' dxs дх' { ■ k T^e"'Ji^'Jik = e-a'a*a' We observe that: (0 Tijk =0 when two subscripts have the same value; e.g., T.-22 = ersta'ra2a2 = - e rtsara2sa2 = -ersta'ra2a2 = -ri22 (") ^23 = «„tfl»? = l<!=+l (n0 Tijk changes sign when any two subscripts are interchanged; e.g. But these three properties identify Tijk with eijk, and the proof is complete.

CHAP. 11] TENSORS IN CLASSICAL MECHANICS 163 11.12 Show that in a vacuum with zero charge (p =0), the electric field Ε satisfies the vector wave equation dt2 c2 V2E From Maxwell's equations (11.19), along with the identity (11.23), 1 д ,_ „, 1 /1 д2Е\ 1 д2Е Vx(VxE) =— — (VxH)- , ,,- 2 2 v ' с dty ' c\c at I с dt2 But V· Ε = 0 and Problem 11.10 imply V χ (V χ Ε) = -V2E, and the wave equation follows. Supplementary Problems 11.13 Show that if υ is constant, a particle describes equal lengths of arc in equal periods of time. 11.14 (β) Show that a particle whose path is given by χ = (cos t, sin t, cot /), for я74 S t < it 12, has velocity decreasing to VI as t-+irl2. (b) What is the behavior of the acceleration as ί-»π/2? (с) Find the extreme values of υ and α for this particle. 11.15 For what kind of motion, if any, is а = dvldtl 11.16 Develop a formula for a for a particle that has constant speed v. 11.17 Calculate the acceleration components (contravariant) in spherical coordinates (ρ, φ, θ). 11.18 Prove that motion under a central force is planar. 11.19 Calculate the Laplacian for cylindrical coordinates (r, Θ, z). 11.20 Show that V2/= g'%r [Hint: Write (11.15) at the origin of Riemannian coordinates.] 11.21 Prove (11.22) and (11.23). 11.22 Prove that curl (grad /) = 0 for any C2 scalar field /. 11.23 Show that in a charge-free vacuum, Η also satisfies the vector wave equation. 11.24 Show that, relative to an orthogonal curvilinear coordinate system (x1, x2, Xs), an arbitrary contravariant vector v= (υ') has the representation v=u(1)e1 + u(2)e2 + i>(3)e3 where u(a) is the physical component and ea is the unit normal to the surface x" = const. [Hint: Use Problems 5.19 and 5.20].

Chapter 12 Tensors in Special Relativity 12.1 INTRODUCTION If the motion of a light pulse were an ordinary phenomenon, its velocity с to one observer would appear to a second observer, moving at velocity ν relative to the first, to have the value c— v. This hypothetical property of light depends on the concept of absolute time measurement for all observers. However, beginning with the landmark Michelson-Morley experiment in 1880, all experimental data force us to abandon this reasonable hypothesis and to accept instead the now undisputed fact that the velocity of light, rather than the measurement of time, is an absolute of nature. Light is observed to have a single velocity, с = 2.9979 X Ю8 m/s, independent of the observer's motion away from or towards its source. This calls for adjustments to the equations of Newtonian mechanics which become major when high-velocity particles are involved. 12.2 EVENT SPACE It is first necessary to wed the concepts of time and space. Thus, each event (atomic collision, flash of lightning, etc.) is assigned four coordinates (t, x, y, z), where t is the time (in seconds) of the event and (x, y, z) is the location (in meters) of the event in ordinary rectangular coordinates. Such coordinates are called space-time coordinates. Definition 1: An event space is an R4 whose points are events, coordinated by (x') = (x , x\ x2, x3), where x° - ct is the temporal coordinate, and (χ1, χ , χ ) = (χ, у, ζ) the rectangular positional coordinates, of an event. Two events ^(xj and E2(x2) are identical if x\ = x2 for all i; simultaneous, if xx = x2', and copositional, if x\ — x'2 for i = 1, 2, 3. The spatial distance between E1 and E2 is the number d^iiAx1)2 + (Ax2)2 + (Ax')2 {12.1) where Ax = х'г — x\ for i = 1, 2, 3. Inertial Reference Frames The general setting for Einstein's Special Theory of Relativity (henceforth abbreviated SR) consists of two or more observers Ο,Ο,Ο, . . . , moving at constant velocities relative to each other, who set up space-time coordinate systems (x), (x), (x'),... to record events and make calculations for experiments they conduct. Such coordinate systems in uniform relative motion are called inertial frames provided Newton's first law is valid in each system. All the systems are assumed to have a common origin at some instant, which is taken as t = t = 1 = ■ · · = 0. Light Cone A flash of light at position (0, 0, 0) and time t = 0 sends out an expanding spherical wave front, with equation x2 + y1 + z2 = cV, or (^0)2-(^)2-(^2)2-(«3)2 = 0 (12.2) (12.2) is the equation of the light cone in event space, relative to the inertial frame (x). Figure 12-1 shows the projection of the light cone onto the hyperplane x3 = 0. In any other inertial frame, (x1), the equation of the light cone is exactly the same (since all observers measure the velocity of light as (x°)2-(xl)2^(x2f~(x3f=0 164

CHAP. 12] TENSORS IN SPECIAL RELATIVITY 165 Fig. 12-ί In Example 7.6 (using slightly different notation), we identified the light cone with the null geodesies of R under the metric of SR. Relativistic Length For an arbitrary event E(x), the quantity (x0)2 - (x1)2 — (x2)2 - (x3)2 may be positive, negative, or zero. The relativistic distance from E(x) to the origin £Ό(0) is the real number s^O such that 2 .- CK2 .- K2 / 2\2 , Ъ\1 , ,Λ\ ss = (χ ) - (χ ) - (χ ) - (χ ) (ε = ±1) More generally, the length of interval or relativistic distance between ^(Xj) and E2(x2) is the unique real number As ^ 0 such that s(As)2 = (Ax0)2 - (Ax1)2 - (Ax2)2 - (Αχ3)2 (ε = ±1) (12.3) where Ax' = x2 - x\ for i = 0,1, 2, 3. The chief significance of this length-concept is to be found in Theorem 12.1: Relativistic distance is an invariant across all inertial frames. For a proof, see Problem 12.6. Interval Types The interval between E^x^ and E2(x2) is (1) spacelike if (Ax')2 + (Ax2)2 + (Ax3)2 > (Ax0)2 (or ε = -1; predominance of distance over time); (2) lightlike if (Ax0)2 = (Ax1)2 + (Ax2)2 + (Ax3)2 (equality of time and distance); (3) timelike if (Ax°)2>(Ax'f + (Ax2)2 + (Δχ3)2 (or ε = 4-1; predominance of time over distance). By Theorem 12.1, the categorization is independent of the particular inertial frame.

166 TENSORS IN SPECIAL RELATIVITY [CHAP. 12 12.3 THE LORENTZ GROUP AND THE METRIC OF SR Imagine two observers, О and O, in uniform relative motion at speed v. They approach each other during negative time, coincide at zero time, and then recede from each other during positive time (Fig. 12-2(a)). Let О and О set up independent reference frames (x) and (x) by means of identical but separate clocks, with t = Ί = 0 when О - O, and identical metersticks. Newton's first law will be assumed to hold in both frames, making them inertial frames. Ф) Fig. 12-2 Common observation of events sets up a correspondence Э~ : x' = F\x\x\x2,x3) (12.4) that is bijective, since each event is assigned a unique set of coordinates. In most of what follows, it will be assumed that О and О perform a simplifying maneuver at the instant of coincidence, whereby their x-axes point in the same direction along the line of motion and their y-axes and their z-axes coincide. In the ensuing translation, the y-axes and the z-axes remain parallel (Fig. 12-2(b)). Postulates of SR (1) Principle of Relativity: The laws of physics are the same in all inertial frames. (2) Invariance of Uniform Motion: A particle with constant velocity in one inertial frame will have constant velocity in all inertial frames.

CHAP. 12] TENSORS IN SPECIAL RELATIVITY 167 (3) Invariance of Light Speed: The speed of light is invariant across all inertial frames. Postulate 2 requires that the bijective transformation (12.4) be such as to map straight lines into straight lines. Consequently, each F' must be a linear function. Since F'(0) = 0, constants a1, exist such that Τ : χ' = α)χ> (12.5) Lorentz Matrices and Transformations The invariance of the equation of the light cone (in consequence of Postulate 3) may be expressed as £,7*ν=0 = £,^ (12.6) where g00 = l, gn - g22 = g33 = -1, and giJ- = 0 for i^j. Substitution of (12.5) in (12.6) yields (Problem 12.4): or, in matrix form, or, written out. giM=g„ (12.7a) ATGA = G (12.7b) , CK2 , \\2 , 2\2 , 3\2 1 К) ~(flo) ~(flo) ~(ao) =1 (a))2-(a))2-(a2)2-(a))2 =-I (/ = 1,2,3) (12.7c) 0 0 11 2 2 33a ί · _l ·\ aiaj - aia.j - aiaj - aiaj = 0 (i¥=j) It is easy to see (Problem 12.8) that requiring gifx'x' = 0 to be invariant is tantamount to requiring the invariance of g^x'x'= q for every value of q. Thus, (12.7) is a criterion for the quadratic form g^x'x' to be invariant. Definition 2: Any 4X4 matrix (or corresponding linear transformation) that preserves the quadratic form χ Gx is called Lorentz. In Problem 12.10 it is shown that the set of Lorentz matrices constitutes a group (the Lorentz group) under matrix multiplication. Metric of SR If the terms gtj = gtj are denned for the (x')-system, then (12.7a) becomes grs = gifa'ra's, which makes (gtj) a covariant tensor of the second order under Lorentz transformations of coordinates. Accordingly, the metric for R4 is chosen as ε ds2 = gtj dx dx> = (dx0)2 - (dx1)2 - (dx2)2 - (dx3)2 (12.8) 12.4 SIMPLE LORENTZ MATRICES Let us suppose that О and О have performed an alignment of their ryz-axes. Then any right circular cylinder with axis along the line of relative motion must have the same equation in the two systems; i.e., (x1)2 + (x3)2 is invariant. It follows (see Problem 12.11) that the Lorentz transformation for this situation has the form f-0 U0. 0 1 _ O.il χ = α0χ + αλχ = ax + ox ST χ — a^x τ д.χ — dx ι ex (to q\ 2 2 X — X -3 _ 3

168 TENSORS IN SPECIAL RELATIVITY [CHAP. 12 By (12.7), d2 = l 1 ab - de = 0 (12.10) By considering the coordinates which О and О would assign to each other's origin (see Problem 12.12), we find that d = -(vlc)a = - βα and a = e (12.11) (The notation β = vie is standard in SR.) From (12.10) and the fact that a > 0 (since both clocks can be assumed to run in the same sense), it follows that α = (1-β2)χι2 = ε b Therefore, the coordinate transformation takes on the simplified form 0 „ 1 о x - βχ ST or X = X -3 3 ^X = X Any 4x4 matrix (linear transformation) of the form a b 0 0 A β(\-β2)-ιη = ά mplified form \ l ~β " Vi-j82 Vi-β2 ^ г1-, о VI - j82 VI - j82 0 0 1 L о oo 0 0 0 iJ b a 0 0 0 0 10 0 0 0 1. where a - b =1, will be termed simple Lorentz. The relative velocity in the physical situation modeled by A is recovered as β - —bla. (12.12) (12.13) EXAMPLE 12.1 By Problem 12.9, the inverse of a simple Lorentz matrix is a ~b 0 0" -b 0 0 a 0 0 0 0 1 0 0 1 which is itself a simple Lorentz matrix, corresponding to a reversal in sign of β. [If the velocity of О with respect to О is v, then the velocity of О with respect to О is —v.] A Decomposition Theorem The possibility of simplifying an arbitrary Lorentz matrix by a suitable rotation of axes can be expressed in purely mathematical terms. (See Problems 12.14 and 12.15.) Theorem 12.2: An arbitrary Lorentz matrix L = (a'y) has the representation where L* is a simple Lorentz matrix with parameters a = \аЦ = εαϋ0 and b -V(flo)2 ~ 1» an^ ^i and R2 are orthogonal Lorentz matrices defined by R^LR^L*)-1 and K2 = [e, r' t'V (12.14) Corollary 12.3: Here, ei = (l,0,0,0), r' = (slb)(0, a\, a°2, a\) = (0, r), s' = (0,s), t'=(0,t), with s and t chosen to complete a 3 x 3 orthogonal matrix [r s t]. If L = (a'j) connects two inertial frames, then the relative velocity between the frames is υ = cVl - (flo)'2 (12.15)

CHAP. 12] TENSORS IN SPECIAL RELATIVITY 169 12.5 PHYSICAL IMPLICATIONS OF THE SIMPLE LORENTZ TRANSFORMATION Length Contraction For any fixed χ , {12A3) gives Ax1--F=Ax1 or Αχ1 = λ/ΐ- β2 Axl<Ax' If the frame О is moving at a uniform velocity υ relative to O, distances in О appear to observer О to be foreshortened in the direction of the motion by the factor у 1 - β . Time Dilation For any fixed x\ (12.13), inverted, gives Ax°=-r=^=Ax° or Af= . l ;AJ>AJ Vl-β2 λί^β If the frame О is moving at a uniform velocity υ relative to O, the clock of observer О appears to observer О to run slow by the factor у 1 - β2 . Composition of Velocities If О has velocity иг relative to О and О has velocity u2 relative to O, then the Newtonian composition of velocities predicts that О has velocity v3 = vl + v2 relative to O. Although the error does not show up unless υγ and v2 are substantial fractions of the velocity of light, SR shows the Newtonian theory to be incorrect. The correct formula (Problem 12.20) is V, + V-, υ3= l 2-5 (12.16) 3 1 + vxv2lc2 У ' 12.6 RELATIVISTIC KINEMATICS 4-Vectors We begin with ordinary velocity and acceleration of a particle within a single inertial frame (x ). By introducing the concept of proper time, we shall be able to obtain velocity and acceleration as contravariant vectors with respect to Lorentz transformations (to be called 4-vectors from now on). In general, (V) is a 4-vector if it transforms according to the law V' = ay', where (a') is the Lorentz matrix of (12.5). It is customary to use the following notation for 4-vectors: (У) = (У°,У) where V°^Vr and V-(V\ V2, V3) = (Vx, Vy, Vz) V is referred to as the time component of the vector and (Vx, Vy, Vz) are the usual space components. All indices are understood to range over the values 0, 1, 2, 3, unless specifically noted otherwise. Nonrelativistic Velocity and Acceleration In the inertial frame (x) = (x, y, z) let a particle describe the class C2 curve Ж : (х') = (а,г(1))=(с1,х(1),у(1),г(1))' Then we have the classical formulas (".-) = (^Η*,ν) 02.17)

170 TENSORS IN SPECIAL RELATIVITY [CHAP. 12 where v — drldt and v = ||v|| = yu2 + υ2 + υ\\ («,) = 1^)-(0,а) (12.18) where a = ay I at and a = ||a|| = у a'x + a2y + a2z . As denned, neither the velocity nor the acceleration is a tensor under Lorentz transformations. In fact (Problem 12.22), if (ϋ;) and (at) are the like quantities in (xl), then (12.13) yields the relations ζ = , , - »*-» - »У v1-^ - ^VT^p^ u0 = c=u0 ux=- -5 uv=- -5- ux=- -5- (ί2.ίΡ) 1-ихи/с ' l-vxv!c l-vxv!c _ _ αχ(1-β2Ϋ'2 - К+(«уах-^ву)(«/с2)](1-)82) 0 * (Ι-ϋ,υ/c2)3 y (l-vxvlc2f [αζ+(νζαχ-νχαΣ)(ν1ο2)](1-β2) (l-vxvlc2f (12.20) The inverse relations can be quickly obtained by replacing υ by - υ and interchanging barred and unbarred terms throughout. For example, the second formula in (12.19) inverts to Vr + V 1 + υ vie2 which is just (12.16) as applied to v1 = vx and v2 = v. Proper Time; Velocity and Acceleration 4-Vectors Let us reparameterize the curve Ж, choosing now the quantity s 1 Γ dx dx' , άτ .г тт. ,..,■. „..ч τ = - = - Л eg;, -г- 1- du or Τ = V1 - и'/с2 (ί2.2ί ) с с Jf0 ν й1' au du at y ' where, as always, ν <c. The new parameter τ (a distance divided by a velocity) is known as the proper time for the particle; by Problem 12.23, a clock attached to the particle (and thus accelerating and decelerating along with it) reads т. When τ-derivatives replace t-derivatives, velocity and acceleration become tensorial; i.e., the components t dx ,: du1 ά2χ' /лп „ч "=-Г- Ь--Г = -ГТ (12.22) άτ άτ J- v are taken by (12.13) into _0 и - /Зм1 _! -βυ° + U1 -ζ ζ -j /tooi\ κ =—, κ =—, и —и и =и (12.23) ь« = »^Ж в·-=^4 *2"»' «'"»' <12·2") у ι-β γι-β2 The important identities utu' = с2 м;Ь; = О (12.25) are proved in Problem 12.24, and Problem 12.25 establishes the following connecting formulas between the numerical values of the relativistic and the nonrelativistic components: " ,, a; (va)u, b' = TIT! + г,л U 2Ч2 (12-26) rfV άτ2 -2 Μ = Ρ = 2 = Ь2 -3 и = б3-- 3 = и = b3 VwV 1 - ν21 с2 с2(1 - tfV)2

CHAP. 12] TENSORS IN SPECIAL RELATIVITY 171 Instantaneous Rest Frame At time t= txi the particle moving along Ж has instantaneous position Px =г(^) and instantaneous speed v(tx). An instantaneous rest frame for the particle is an inertial frame that translates at speed £(ij) along the tangent to Ж at Ρλ, in such manner that its origin coincides with P1att=t1. See Fig. 12-3. Fig. 12-3 We shall say that a particle's motion (with respect to frame O) is uniformly accelerated if its spatial acceleration relative to an instantaneous rest frame O, - 1 / -2 ι -2 . ^2 a = a = v^ + fly + fl2 does not vary along the trajectory Ж. EXAMPLE 12.2 An electron fired at right angles into a uniform magnetic field undergoes uniformly accelerated motion (in a circle). 12.7 RELATIVISTIC MASS, FORCE, AND ENERGY The appropriate SR version of Newton's second law depends on the concept of mass to be adopted. Rest Mass and Relativistic Mass The rest mass of a particle is its mass as measured or as inferred from Newtonian mechanics in any instantaneous rest frame for that particle. The relativistic mass (in О) of a particle with spatial velocity ν (with respect to О) is m m=vr^n {12-27) VI — и !c where m is the rest mass of the particle. As is shown in Problem 12.27, (12.27) is a necessary consequence of conservation of momentum.

172 TENSORS IN SPECIAL RELATIVITY [CHAP. 12 Relativistic Momentum and Force The 4-momentum of SR is defined by (р') = (р°,р) = (ти') = (т>,) (12.28) and the (nontensorial) Lorentz force (F0, F) is defined as the time derivative of the 4-momentum: _ dp _ d I mc \ __ d I my F°- dt dt\y/r^l?) F ^vvwv' {12'29) Like velocity, force becomes tensorial when proper time is introduced. Thus, the 4-force (Minkowski force) of SR is defined as (Лэ(|) (12-30) From (12.21), we have the connecting formula K'= 7f7=m (1231) VI - ν I c The following identities for the Lorentz and Minkowski forces are proved in Problem 12.29: Relativistic Energy According to the classical work-energy theorem, the rate at which work is performed on a particle, (vF), equals the rate of increase of the particle's kinetic energy. Thus, the last identity (12.32) suggests the definition for SR тс л 2 £= /ι ,г,г^тс (12Х) VI- ν I c for the energy of a particle moving at speed v. In the limit as v—*0, (12.33) becomes E = mc2 (12.34) This is Einstein's famous formula for the rest energy Ε of a particle with rest mass m. 12.8 MAXWELL'S EQUATIONS IN SR It is helpful to look briefly at the way the metric for Special Relativity affects the formulas for the divergence and the Laplacian, and to consider a new kind of matrix that will be useful in formulating Maxwell's equations. Vector Calculus and Lorentz Transformations For the metric of SR, all Christoffel symbols vanish, so that and div(grad f) = Uf-- divu ; £(' a2f (OX ) 1 d2f г dt2" du dx dx1' a2f (dx1)2 -v2/ g" d2f дх'дх' d2f (dx2)2 (12.35) *2f (dx'f (12.36)

CHAP. 12] TENSORS IN SPECIAL RELATIVITY 173 Note that in SR the Laplacian operator is notated D, with V2 reserved for its spatial part. It is verified in Problem 12.31 that D/is an invariant under Lorentz transformations, which means that the scalar wave equation has the same form, D/ = 0, in every inertial frame. If we introduce the differential operator d'=- g' dx' then we can express the equation of continuity for a vector (w;) = (w0, w) as (12.38) is equivalent to dwjdt = cdivw. (12.37) (12.38) Maxwell's Equations in Event Space First, introduce for any two 3-vectors U = (Ul) and V= (V) the two antisymmetric matrices [Л44 = 0 Vх V2 V3 -Vх 0 -и3 u2 -V2 и3 0 -и1 -ν -и и 0 L /ί / J 44 0 и1 и2 и3 и1 0 -У3 V2 и2 V3 0 -У1 и3' -V2 V1 0 . (12.39 а) The second matrix may be obtained from the first by making the replacements V—» —U and U—»V; because these replacements constitute an anti-involution—i.e., fv = —f—the two matrices are said to be dual to each other. In terms of individual components (e denotes the permutation symbol of order 3): f'i = — fi' fuq = —Vq fpq in which i, j ё 0 and p, q Ш 1. By their concoction, these matrices turn out to be tensors under Lorentz transformations (provided the row-divergences vanish in all inertial frames); a proof is given in Problem 12.32. Moreover, these tensors have the properties (Problem 12.33) (12.39b) 0; and a£l afi af[ ox' ' ox' ' ox' dx> curl U + -divV 1 ^V с dt ox' dx' divU dx! ' dx' curlV - 1 SV с dt (12.40) (12.41) We now show how Maxwell's equations in vacuum, (11.19), may be extended to space-time via dual tensors of this sort. The equations are: 1 ΟΉ с dt 1 Ш с dt div Η = 0 divE: curl Ε + curl Η — 0 Ρ (12.42) (12.43) in the last of which ν is the classical spatial velocity (12.17) of the charge-cloud p. Define per (12.39) the tensors &=[F!i] 44 ' 0 Иг я, я я, E3 0 Ε, -н; -E2 Ε: 0 £ = [П44- 0 Ег -Е2 -Еъ ■ Е, 0 -Я3 Я2 Е2 н3 0 ~Нг Е3 -н2 Нг 0 (12.44) (with U = Ε and V= Η). In view of the first equations (12.40) and (72.47), (12.42) may be written as

174 TENSORS IN SPECIAL RELATIVITY [CHAP. 12 dF'j -=0 (12.45 a) ox' Similarly, if we make a 4-vector out of ν and ρ by the prescription (O = (p,^v) (12.46) (see Problem 12.52), then the remaining Maxwell's equations, (12.43), are rendered tensorial as OF" dx> sl (12.45b) Equations (12.45) are the relativistic Maxwell's equations, valid in every inertial frame. Because F is antisymmetric, we have from (12.45b): dxl dx dx' dx y μ \ dx1) ' ' so that the covariant vector (i.) obeys the equation of continuity (12.38). Solved Problems EVENT SPACE 12.1 Calculate ε and /As for the event pairs: (a) £\(5, 1, -2,0) and £2(0,3,1,-3), (b) ЕД5, 1, 3, 3) and E2(2, -1,1, 1), (c) Ex(l, 2, 4, 4) and £2(4,1, 2, 6), (d) E1 = flash of light in Chicago at 7 p.m. and E2 = flash of light in St. Louis (400 miles away) at 7.000 000 61 p.m. (e) Determine the interval type in each case. (β) ε(Δί)2 = 52 - (-2)2 - (-3)2 - 32 = 25 - 4 - 9 - 9 = 3, or Δί = V3 and ε = 1. (b) ε(Δί)2 = 9-4-4-4= -3, or Δ* = V3 and ε = -\. (c) ε(Δί)2 = 9 - 1 - 4 - 4 = 0, or Δϊ = 0 and ε = 1. (d) With с = 186 300 mi/sec, ε(Δί)2 = (0.002 196cf - (400)2 = 7375 mi2, or As ~ 85.8 mi and ε = 1. (e) Time like, spacelike, lightlike, and timelike, respectively. 12.2 Show that (a) simultaneous events have a spacelike interval; (b) copositional events have a timelike interval; (c) the interval between two light flashes is lightlike if they are simultaneous to an observer who is present at the site of one of the flashes. (β) ε(Δί)2 = 02 - (Ax1 f - (Ax2 f - (Ax3 f < 0 (b) ε(Δί)ζ = (ΔΛ:ο)2-0>0 (c) Let the observer measure the proximate flash as ЕДО, 0,0,0). The distant flash E2(c At, Ax1, Ax2, Ax^) will be registered simultaneously, at x° = 0, if ΥίΔχ1)2 Η (Αχ2)2 + (Ax3)2 At= с But then s(As)2 = 0 and the interval is lightlike. (Note that the (negative) time coordinate of E2 is calculated, not measured.)

CHAP. 12] TENSORS IN SPECIAL RELATIVITY 175 THE LORENTZ GROUP 12.3 Prove the following lemma involving the metric of SR, gtj, as given by (12.6). Lemma 12.4: If С = (cfy) is a symmetric 4x4 matrix such that c^x'x1 = 0 for all (x) such that gijx'x' ~ 0, there exists a fixed real number λ for which c/; = \gtj (C — AG). Observe that the vector (1, ±1,0,0) satisfies gtjx'x' = 0. Hence, substituting these components into the equation ctx'x' = 0 yields cOo±coi±cio + cii=0 or coo + cii=0=coi = Cio (by symmetry of C). Similarly, using the vectors (1,0, ±1,0) and (1,0,0, ±1), we get c0o=-cn = -c22=-c33 = A cjy=0 (i=0or/ = 0) Finally, employing the vectors (V2,1,1, 0), (V2,1, 0,1), and (V2,0,1,1), we obtain c12 = c13 = c23 = 0. 12.4 Establish the transformation (12.7) between inertial frames under the postulates for SR. From (12.6) and (12.5), that is, g^x'x' = 0 = gilx'x' = g,t(aX)(a'sxs) = grsaia'jx'xi fi'fijXX1 =0 whenever g^x'x1 =0 (1) Now apply Lemma 12.4 to (2), with grsaria/ = cy, where C= (ctJ) = ATGA is symmetric. We obtain ^Χ< = λ&/ or ATGA = \G (2) It remains to show that A = l. Since G2=I, multiplication of (2) by the matrix A_1G gives (С(\~гАт)С)А = I, which shows that the inverse of A is B= ν GA'G л α0/Α —a^l Κ —α01λ —al/λ -α°/λ. α] /Α α J/Α α^/λ -α°/Α α2/λ α2/Α α2/Α -β°/λ β'/λ allk all Κ "lb% (3) In particular, b0 = α^ΙΚ. Now since observers О and О are receding from each other at constant velocity υ and are using identical measuring devices, it is clear that each views the other in the sam^ way. It follows that Aq = £>o and A = а°01Ь°0 = 1 (see Problem 12.5). 12.5 With reference to Problem 12.4, give a "thought-experiment" which leads to the conclusion that αϊ) = b°0. Consider the motion of О in (7's frame: Transform the point (ct, 0, 0, 0) under 1 to get -o - о . χ = ct = anct 7 0. f = flnf Thus, 1 second on O's clock is a° seconds on O's; reciprocally, 1 second on O's clock is b°0 seconds on O's. Thus, a„ = b"0.

176 TENSORS IN SPECIAL RELATIVITY [CHAP. 12 12.6 Prove Theorem 12.1 from (12.7). [Note that Problem 12.4 did not make use of Theorem 12.1, so the proof will be logically correct.] By (12.7), (glt) is a covariant tensor under Lorentz transformations, so that gtj Ax' Ax1 is an invariant (under Lorentz transformations). 12.7 Verify that the following matrix is Lorentz: 'V5 V2 V6 1 1 0 2 V6 2 0 0 1 2 1 2 V2 2 0 1 2 _1 2 V2 2 We verify directly the conditions (12.7c): (V3)2-l2-l2-02=3-2=l (^)2"(^)2-(^ 2 τ ο -О2 = 2---- 2 2 V6\ •'-(г)-H)-(*W- <^-<.,(£)-<.>(£)-.-о №(„)-(1,(1)-(1)(-1)-,о)(^).о (v,K„,-(^)(l)-(f)(-i)-(o,(ti^„ Ш) ♦(№(£)(£ ■°-ϊ-Η-° 12.8 Show that a matrix A which preserves xTGx = 0 necessarily preserves xTG\= q. This is really Problem 12.6 in another guise. By Problem 12.4, A must satisfy ATGA = G. But then (Ax)TG(Ax) = xT(ATGA)x = xTGx = q 12.9 (a) Exhibit the inverse, B, of a given Lorentz matrix, A. (b) If we define a matrix A to be pseudo-orthogonal when there exists a matrix J whose square is the identity and A J A - J, show that all Lorentz matrices are pseudo-orthogonal. (a) Set λ = 1 in (3) of Problem 12.4. (b) If A is a Lorentz matrix, then G clearly fills the role of / in the definition of pseudo-orthogonal matrix. 12.10 Prove that the Lorentz matrices compose a group under matrix multiplication. We are required to show that (a) the product of two Lorentz matrices is Lorentz, (b) the inverse of a Lorentz matrix is Lorentz. (a) (PQ)TG(PQ) = QT(PTGP)Q = QrGQ = G (b) Using Problem 12.4 with \ = 1, B = A'1 = GATG, and BTGB = (GATG)TGB = GAGZB = GAB = G

CHAP. 12] TENSORS IN SPECIAL RELATIVITY 177 SIMPLE LORENTZ MATRICES 12.11 Derive the simple form (12.9) of the transformation equations for SR by considering how observers О and О will view events occurring on a circular cylinder about their common x-axis. Fig. 12-4 At any time t, let Ex and £2 be two events taking place at the points of space (q, 1,0) and (q, 0,1), respectively, which lie on a unit cylinder about O's x-axis (Fig. 12-4). Thus, with ρ = ct, we have space-time coordinates Ег(р, q, 1, 0) and E2(p, q, 0,1). Since the axes of О are not turning with respect to O's, these two events will be viewed by observer О as E}(p, q, 1,0) and E2(p*, q*, 0,1), respectively. The transformation equations (12.5) give: (I) Observing just the last equation of (I) and the third equation of (II), we may, since ρ and q arc arbitrary, take ρ = q = 0, then ρ = 1, q = 0, and ρ = 0, q = l. It follows that all six of the coefficients aj = a\ = 0. Using the third equation of (I) and the last equation of (II), we хъ. Now to Ρ q 1 0 = a0p + aYq + a2 = al„p + a\q + a\ = a-lP + a\q + a\ = alp + a\q + a\ (Π) p* = a°0p + a\q + a° q* = flo/> + a\q + a\ 0 = alp + afg + 03 1 = alp + a\q + a\ vanish: a0= ax = a3 = a0 find that flj = a\ = 1, It follows that the last two equations of 5" reduce to x2 concentrate on the first two: x1 = 0, the instant when x1 - ■ хг and xz ■■ p = q = 0 implies ρ * = q * = 0 and a% ■- If ρ = q = 0, then event Ej is (0, 0,1, 0)—occurring when t = ί = 0 at 0. That is, /?= q =0, with the result a° = a] = 0. Similarly, using £2, = 0. 12.12 Consider event Ex, a lightning flash at the point (v,0, 0) at time ί = 1 s in O's frame, and event E2, a lightning flash at (-и, 0,0) at time i= 1 s in O's frame. By determining the corresponding events in the opposing frames of reference, deduce (12.11).

178 TENSORS IN SPECIAL RELATIVITY [CHAP. 12 Since at t = 1 observer О has reached the point (v, 0, 0), the lightning strikes O's origin at time t. Hence, Ex has coordinates (c, v,0, 0) in О and (ci, 0,0, 0) in O. Substituting these into ST we obtain ci = ac + £>l> 0= dc + ev The second equation gives d= -ββ. Since О has progressed backwards to the point (-v,0, 0) in О at the time ί = 1 at which E2 occurs, this event has coordinates (ci, 0, 0, 0) in О and (c, — υ, 0,0) in O. Substituting these into 5" yields c= act+ b(0) which upon division give d = — β a. Hence, a = e. dct + e(0) 12.13 Show that a 4 x 4 matrix is both Lorentz and orthogonal if and only if it has the form R 1 0 0 0 0 ri Г2 гз 0 *1 *2 s3 0 Ί h h (1) where the 3x3 matrix [r s t] is orthogonal. A Lorentz matrix A = (a'y) is also orthogonal if and only if its inverse B, as obtained in Problem 12.4 (with A = 1), is equal to AT and is itself orthogonal. This observation immediately yields the form (1). 12.14 Prove Theorem 12.2. Since \\r\\2 = b~2[(a01)2 + (a02)2 + (al)2] = b-2[(a°0)2-l] = l (using Problem 12.36), the matrix [r s t] is orthogonal and R2 has the form of the matrix in Problem 12.13, making it Lorentz and orthogonal. It follows that R2 is orthogonal (and Lorentz), with R~г = R2; hence, L = R1L*R2. Now, as the product of Lorentz matrices, R1 is Lorentz; to show it is orthogonal, consider LR^(L*)~\ which may be written as 0 '"l r-, 0 0 ь о o' a 0 0 0 1 0 0 0 1. Ь0гг + c0r2 + d0r3 b0sr + c0s2 + dgSj b0t^ + c0t2 + d0t3 a -b 0 0 -b a 0 0 0 1 0 0 0 1. [The omitted rows have the form (a„blr1 + ctr2 + dlri,bisx + ср2 + dis3,b!tl + ctt2 + а,1ъ), with i=l,2,3.] We first concentrate on proving that the top row and first column of this product are (±1, 0, 0, 0). The 00-element of the product is a0a + (Ь0гг + c0r2 + d0r3)(-b) = εα20 + | (b20 + c20 + d20)(-b) = ε(α20 - b\ - c20 - d20) = ε again using the fact that the transpose of a Lorentz matrix is Lorentz. The next element in the top row of the product is -a0b + (b0r1 + c0rz + d0r3)a= -a0b + - (r2)εα0 = -a0b + ba0 = 0 For the third and fourth elements, !>0s1 + c0s2 + d0s3 = - rs = 0 and b 0ij + c0t2 + dgt^ = - rt = 0

CHAP. 12] TENSORS IN SPECIAL RELATIVITY 179 Now for the first column of the product; its elements, beginning with the second, are (for ί = 1, 2, 3) α,α + (Ь,г, + ctr2 + d,r3)(- b) = εα,α0 - e{btb0 + c,c0 + d,d0) = 0 Hence, the product matrix becomes *i = ε 0 0 0 0 0 R .0 and the 3 X3 matrix R must be orthogonal, since R1 is Lorentz. 12.15 Apply Theorem 12.2 to the Lorentz matrix of Problem 12.7, and demonstrate the physical significance of this matrix by computing the velocity υ between the two observers involved. We proceed to calculate a, b, and the vectors r, s, and t: a = V3 ε = 1 i>=-V3_rT=-V2 Hence, we may take s = (0,1,0) and t = (0,0,1), yielding r = V2 (V2, 0,0) = (-1,0,0) «! = = Corolla "V3 1 1 . 0 "V3 1 1 . 0 V2 V6/2 V6/2 0 -V2 -V6/2 -V6/2 0 ry 12.3, 0 1/2 -1/2 -V2/2 0 1/2 -1/2 -V2/2 R о 1" 1/2 -1/2 V2/2J. 0 1/2 -1/2 V2/2 ι "1 0 0 -1 0 0 .0 0 1 0 0 0-10 0 0 1 0 0 0 TV3 V2 V2 V3 0 0 .L 0 0 0 0 1 0 0" 0 0 1. 0" 0 0 1. "V3 V2 V2 V3 0 0 . 0 0 0 0" 0 0 1 0 0 1. = ■Ί "1 0 0 .0 0 0" 0 0 1 0 0 1. 0 -V2/ -V2/ 0 0 2 1/2 2 -1/2 -V2/2 0 1/2 -1/2 V2/2. LENGTH CONTRACTION, TIME DILATION 12.16 A pole-vaulter runs at the rate (V3/2)c (in m/s) and carries a pole that is 20 m long in his reference frame [the rest length of the pole is 20 m]. He approaches a barn that is open at both ends and is 10 m long, as measured by a ground observer. To the ground observer, will the pole fit inside the barn? What is the pole-vaulter's conclusion? To the ground observer, the pole undergoes length contraction with the factor yl - β2, where β = V3/2. Hence, the length of the pole in the frame of the ground observer is 20Vl - (V3/2)2 = 10 m and so, for her, the pole exactly fits inside the barn (instantaneously). To the runner, however, the barn is 10(1/2) = 5 m long, so that the 20-m pole does not fit. This example shows that order relations are not preserved under the Lorentz transformation.

180 TENSORS IN SPECIAL RELATIVITY [CHAP. 12 12.17 (the Twin Paradox) One of a pair of twins embarks on a journey into outer space, taking one year (earth time) to accelerate to (3/4)c, then spends the next 20 years cruising to reach a galaxy 15 light-years away. An additional year is spend in decelerating in order to explore one of its solar systems. After one year of exploration (β =0), the twin returns to earth by the same schedule—one year of acceleration, 20 years of cruising, and one year of deceleration. Estimate the difference in the ages of the twins after the journey has ended. In order to apply SR, replace the four periods of acceleration or deceleration by four periods of uniform motion at speed (3/8)c (the time-average speed under constant acceleration). These account for 4 years by the earth clock; but to the space twin, who measures proper (shortest) time intervals, the time lapse is (β =3/8) 4\/T-(3/8)2 = 3.71 years Similarly, the 40 earth-years of cruising at β = 3/4 corresponds to a proper-time interval of 40\/l - (3/4)2 « 26.46 years Thus, the space twin has aged 3.71 + 26.46 +1 = 31 years while the earth twin has aged 4 + 40 + 1 = 45 years. The space twin returns biologically younger by some 14 years. While the accelerations and decelerations between the two twins were reciprocal, the forces in the situation acted on the space twin alone. 12.18 Prove the basic integrity of (12.16) by solving algebraically for v2 as a function of u, and v3 to verify that v2 follows the correct format for composition of velocities. Solving, -υ, + i>3 v-, ~ 1 - v1v3/c which is precisely (12.16) under the substitution (u1; v2, ι>3)—>(-ΐΊ, ι>3, v2). 12.19 A light source at О sends a spherical wavefront (Fig. 12-5(a)) advancing in all directions at velocity c; it reaches the ends of a diameter AB centered at О simultaneously, as determined by O. But as far as О is concerned, the spherical wave, centered at O, moves with him (invariance of the light cone) and therefore reaches point В before it reaches point A. Calculate the time difference on O's clock for these two events (light reaching В and light reaching A) if β = 1/2 and if AB = 6 m. Since AB = 6 m and О is the midpoint of segment AB, О assigns spatial coordinates B(3, 0, 0) and A(-3, 0, 0) to the endpoints. It takes 3/c seconds for light to reach A and B, so О calculates the time coordinate as x° = c(3/c) = 3 m. The space-time coordinates of the two events are thus EB(3,3,0,0) and EA(3, -3, 0, 0) Fig. 12-5

CHAP. 12] TENSORS IN SPECIAL RELATIVITY 181 Substitute_ these values and β = 1/2 into the first equation (12.13) to obtain tB = V3/c, tA =3V3/c. Hence, Δί~= l\fblc (in s), while Δί =0. It is seen that simultaneity is not an invariant of Lorentz transformations. 12.20 Derive the composition of velocities formula, (12.16). According to Section 12.4, we must have u; = -biclai for i = 1, 2, 3. Composing the simple Lorentz transformations, we have аг Ьг О О b, a, 0 0 0 0 10 L0 0 0 1. fl, b2 0 0 b-, a2 0 0 0 0 1 0 0] 0 0 lj = fljfl2 + b1b2 агЬ^ + a2bl 0 0 a2b1+a1b2 blb2 + a1a2 0 0 0 0 10 0 0 0 1. whence a3 = ara2 + ЬгЬ2, b3 = агЬ2 + а2Ьг, and axb2c a2bxc b2c (a1b2 + a2b1)c ага2 ага2 α. ν-, = -■ α,αΊ ЬгЬ2 1 + 1 + υΛυ2Ιε 12.21 A physicist wants to compose two equal velocities υ — υλ = υ2 to produce a resultant velocity that is 90% of the velocity of light. What velocity must he use? From (12.16), 0.90c = 2v 1 + v2lc2 or 0.90 = 2β_ 1 + β2 Solving the quadratic, β~ 0.627 (as compared to the Newtonian value 0.45). VELOCITY AND ACCELERATION IN RELATIVITY 12.22 Establish (12J9) and (12.20), the Lorentz transformations of velocity and acceleration, that define how О tracks the motion of a particle in O's frame. To simplify notation, let у = (1 - /32)"1'2. Then ST is ct = y(ct- βχ) χ = γ(-βα + χ) y = y z = z Differentiate the first equation with respect to Fand use the chain rule: dt dt 1 c = y(c- βυχ) dt or Now differentiate the last three equations: vx = y(~fic + vx) dt y(\ - vvjc ) y(-v + vx) vx-v dt dt dt y(l-vvjc2) l-vxv!c2 = v, Vl ~ β2 dt y(\ - vvjc2) 1 - vrv/c2 ν,λ/1-β2 - - dt_- υ*~υ* dt~ \-vxvlc2 By differentiation of the velocity components just found, _ dvx dt _ (ax - 0)(1 - vxv/c2) - (vx - v)(0 - axv/c2) 1 dt dt (l-vxv/c2)2 y(\ - vxv!c ) axvxv/c2 I vxaxv/c2 - axv2/c2 _ ax(\ - β2)3/2 7(1 - vxvlc2Y (1 - vxv!c2)3

182 TENSORS IN SPECIAL RELATIVITY [CHAP. 12 (1 - vxvlczf 1 - i^/c2 (1 - vjc2)3 { ** ' The formula for dz is derived as that for dy, with ζ replacing у throughout. 12.23 Show that if the curve of motion in O's frame is the path of О itself, the clock in O's frame (the clock moving with the particle) measures proper time. By (3) of Problem 12.4, 0 0-0 1-1 2-2 3-3 x = a0x — a0x - a0x — a0x xl = - a°x° + α]χα + a2 x2 + a3x3 (i = 1, 2, 3) Now the motion of б relative to itself is obviously χΊ = χ2 = χ2 = ϋ. Hence, U t) 10 2 0 3 0 χ = aucu χ = — fl,ci< χ = — a2cu χ = — аъси give the trajectory of О in O's frame, with parameter u= t. Therefore, the tangent field to the trajectory is "■* 1 / о о о 0ч — J = (a0c, -a,c, -a2c, - a3c) so that the proper time parameter for this curve is defined as r=-cj'o λ/\(α°0α)2 - (a\cf - (a"2c)2 - (а"3с)2\ du = УК^оУ - (a0,)2 - (a°2)2 - (a°3)2\ [ du Because the inverse transformation is Lorentz, the factor in front of the integral sign equals 1, and so T = t. 12.24 Derive the identities (12.25). By (12.21), i · „ ' У_/ 0ч2 / 1ч2 / 2Ч2 / Зч2 utu = gyU и = (и ) — (и ) — (и ) — (и ) = к-,)2 - (о2 - (vy)2 - (o2i(f )2=[-2 - *2] γ-^ and from this, 0=^-(с2) = 4~(щи') = 2и^ dr dr 12.25 Establish the formulas in (12.26). dx' dx' dt ν. dr dt dr Vl - v2/c du dr dr Lift \-Λ/ΐ-β2/ε2> _ д,(1 - v2lc2f12 - u,-(l/2)(l - v2/c2)-1,2(-2axvx~2ayvy - 2atvz)/c2 dt. 1 - u2/c2 dr a,(l ~ t;2/c2) + υ,(αχνχ + avvy + azvz)lc2 a. (va)u, /1 л2, 2Ч2 , ~2/ 2 "*" 2/1 ~2/ 2ч2 (1 - и /с ) 1 - υ 1с с (1 - υ /с ) 12.26 Derive the equation for uniformly accelerated motion along the x-axis of an inertial frame: 2 2 2 c JC - C( = —

CHAP. 12] TENSORS IN SPECIAL RELATIVITY 183 Let О be an instantaneous rest frame at some point i1; and let О be a given (stationary) frame in which the motion curve is traced. Since the motion is along the х-axis of O, u„ = υ, = a = a, = 0 and υ, = υΎ = αν = αΎ =0 whereby αΛ = a = const, (assuming ax >0). At ί = tr, υ = vt (the constant velocity of О is by definition equal to the instantaneous velocity of the particle); thus, from (12.20), 2, 243/2 ax(\ - v'lcy~ ax(\ - vxlc~) (1-^u/c) (l-vx/c) (l-vjc) Since ij is arbitrary, (1) must hold for all i. Writing i, χ for the derivatives of x(t), we have from (ί): 3·· / 2 ·243/2 /ТЧ с х- а(с - χ ) (2) Make the substitution у = χ and (2) becomes с dy _ Г ^з72 = J α 3 4" /2 2чЗ с ^- = а(с - у ) or к ■У ) Л Р) Standard techniques of integration yield the first integral cy v< 2 2 с -у = at (4) (where wc have taken the initial velocity to be zero). Solving (4) for у (assumed positive for positive i) and then integrating the equation χ = y(t), we obtain JC = cVc2 + a2t2/a or x2 - c2t2 = с I a1 (where we also take the initial position as zero). This is the desired equation, which represents a hyperbola in the xt plane. By contrast, the Newtonian equation is the parabola χ = \at. RELATIVISTIC MASS, FORCE, AND ENERGY 12.27 Show that the observed mass of a particle with rest mass m, moving at velocity υ, is m = m(\ - v2lc )~12, by considering the following experiment. Let each observer О and О carry a ball with rest mass m near his origin and so situated as to collide obliquely at t = t = 0 (when their origins coincide). See Fig. 12-6. Suppose this collision imparts reciprocal velocities of ε in the positive x-direction and negative x-direction. Calculate the momentum of the system before and after collision (which is preserved), and what each observer sees based on the equations of SR; then take the limit as ε^Ο. The velocity vectors Vj and v2 of balls Bt and B2 before collision are, as seen by O, (0,0,0) = 0 and (u,0,0) = in. Observer O calculates these vectors as У! = (-и,0,0) and v2 = (0,0,0) (either by Вг{т) BiH у У (a) Before impact (b) After impact Fig. 12-6

184 TENSORS IN SPECIAL RELATIVITY [CHAP. 12 reciprocity or by use of (12.19). After collision, observer О calculates the velocity of Вг as \г = (ε, δ, 0) = ε'ι + 5j, assuming Bt has the proper alignment with B2. Reciprocally, observer О calculates the velocity of B2 as v2 = (-ε, —δ, 0). To find v2, use the inverse of (12.19), with vx = — ε and ϋ = — δ: .1- ευ/c2/ V Ι-ευ/c2 Thus, observer О calculates the net momentum vector of the system as follows, using the rest mass m of B1 for m1 and the "perceived mass" m of B2 for m2: before impact m1v1 + m2v2 = тДО) + m2(vi) = mv'\ v2 = uvi + vj = { ^тт^ ji + \ — _ 2 )j α/ier impact mlv1 + m2v2 = m(ei + 6j) + m ε + υ V ί-δ\Ι i-β2 1 - ευ/c / V 1 — ευ/c l+ ^—Z^T- J - и - e \. / с л S\/l-/32\ mf + m ^ 7T li + m5 - m -j- j 1 - ευ/c / V 1 - eu/c / Since ϋ is using the universal laws of physics as they apply to his frame (Postulate 1 of SR), the two momentum vectors above must be the same. Hence, mv = me + m 2 and 0 = m - m r 1 — εν/c 1 — εν/c (after division by δ). Now take the limit as e—>0: mv = mv and 0 = m - m\ 1 - β2 The right-hand equation is the connection between m and m. 12.28 Show that the Minkowski force is a 4-vector. We must show that K' = a'-K', if x' = a^x', where (a'y) is any Lorentz matrix. Since τ is invariant and a'- = const, we may differentiate the coordinate transformation with respect to τ across the equal sign: d d dr(x^Jr (.*')=-jiVj*') or (proving that (и') is a 4-vector). Multiply both sides by m and differentiate again, using the fact that the rest mass of a particle is invariant: K' = — (тй1) = — (тй') = — (a\mu>) = a\ — (mu>) = a K' 12.29 Establish (12.32). Definition (12.30), К' = d(mu)Idr = mb\ along with the second identity (12.25), gives at once u.K' = 0. From кД' = 8ии'К* = u°K° - uqKq = 0 and the first formula (12.26), we have vaK° vqKq = 0 or cK° = vK Vl - v2/c2 Vl - v2lc By (12.31) and cK° = vK, 1 „ 1 vF = - vK Vl - ν 2/c2 = K° Vl - v2lc2 = Fn Using the first definition (12.29), vF = cF° *WwV

CHAP. 12] TENSORS IN SPECIAL RELATIVITY 185 The result is 12.30 Show that as ΰ—*0, Ё = тс' + \πιΰ2 + 0(v*lc2). Interpret this result. The expression for relativistic energy, E= mc2(l - u2/c2)-1'2 may be expanded by the binomial theorem: (1 +*)" = 1 + ax + ^^—'-x2 + -·- (—1 < л: < 1) - 2 I -2 Эти"4 Ε = тс + χ mv Η ,—l· · ■ ■ 2 8c" Thus, at low speeds, the total energy of particle is very nearly the sum of its rest energy (which includes all sorts of potential energy) and its classical kinetic energy. MAXWELL'S EQUATIONS IN SR 12.31 Prove that □/=□/. As the git are constants, D/= g"f tj = invariant. 12.32 Prove that if (F'1) is any matrix of functions of the 3-vectors U and V such that dF''/дх' = 0 (i = 0,1, 2, 3) for all inertial frames and Fu(0, 0) = 0 for all i, j, where 0 = (0, 0, 0), then (F'J) is a second-order contravariant tensor under Lorentz transformations. Let (m;.) be any constant, covariant vector under Lorentz transformations [hence, (iZ;) = (bkuk) is also constant]. Define S' =■ ukFkf S' = ukFki By the given conditions 3F"/dxJ = 0, ?n - - dH _ η _3S' dx' ~~ Uk dx1 ~~ ~~ dx' Suppose that at some point (x'0), S' = ti(S°, S\ S7, S3); then, dS' „ dht dS' dxk (,t dh'\ dS dx' 0 = ^^^ or [b'7e)7?=0 for an arbitrary matrix (dS'ldxk) having dS'ldx1 = 0. By a well-known lemma (Problem 12.57), there exists a real number A= A(5'0, S\ S2, S3) such that (1) (2) b'W = K3' Now differentiate both sides of (1) with respect to Sl: k d2h' ЭК k ' dS'dS' dS1 ' which is symmetrical in / and /; therefore, dS' '' dS'0' for all /, k, I. Let к = / Φ j in (3): ЭК дК ЭК dsk ds> or ds> Hence A is constant with respect to the 5" and (1) inverts to give Sh' = 0 (3) (4)

186 TENSORS IN SPECTAL RELATIVITY [CHAP. 12 Integrating (4), ti = S' = Ka'jS' + Г (5) For the special assignment U = V= 0, we have (since 0 = 0): S' = K)(0) + (и2)(0) + (ιι3)(0) + (и4)(0) = 0 S'= (щ)(0) + (й2)(0) + (й3)(0) + (й4)(0) =0 Together, (5) and (6) imply T' = 0 (г = 0,1, 2, 3); consequently, S' = \aijSi (7) Similarly, there exists a real number μ such that Sl = μ^β' (8) It follows that S' = ka^b'kSk = λμ.5', or λμ. = 1. But we can exploit the reciprocal relationship between observers О and O, as in Problem 12.4; to show that A = μ. Therefore, A = μ = 1 and (7) or (8) becomes the transformation law of a (contravariant) 4-vector. Finally, we conclude from the Quotient Theorem that if Fk'uk = 5" is a tensor for an arbitrary covariant vector (и,.), (Fu) is a second-order contravariant tensor. 12.33 Prove the relations (12.40) and (12.41). By (12.39) and the constancy of the gijr дх' and, for ρ = 1, 2, 3, df dfp0 df" дх' ~ Ί)Ι + дх" ~ df° df " dx" + dx4 df d . ттг, dVp dUr (1 d\ , \ The other two formulas arc derived from these by replacing U, V by V, —U. Supplementary Problems 12.34 Suppose two events consist of light signals, and an observer sends one of the signals himself. Classify the space-time interval between the events if the observer sees the distant light signal (a) before he sends his own signal, (b) after he sends his own signal, (c) at the same time he sends his own signal. 12.35 Assuming that any velocity less than с is attainable, suppose that a concert in Los Angeles begins at 8:0508 p.m. and one in New York City, 3000 miles away (consider this the accurate distance), begins at 8:0506 p.m. could a person physically attend both events (opening measures only)? Is the pair of events timelike or spacelike? 12.36 Show that the transpose of a Lorentz matrix is Lorentz. 12.37 Verify the expressions (12.12). 12.38 An event occurs at O's origin at some time t. (a) How does О view this event? (b) What is the significance of a° > 0? 12.39 Write out the simple Lorentz transformation connecting inertial frames О and О that move apart at 80% of the velocity of light.

CHAP. 12] TENSORS IN SPECIAL RELATIVITY 187 12.40 (a) Confirm that a photon (a particle with the velocity of light in some inertial frame) will be viewed as having the velocity of light in all other inertial frames, (b) What must be the rest mass of such a particle? 12.41 Show that the following matrix is Lorentz, and use Theorem 12.2 to find the matrices L*, Д1; and R2, and the velocity υ between the two observers. 5/4 1/2 1/4 -1/2" -3/4 -5/6 -5/12 5/6 0 2/3 2/15 11/15 . 0 -1/3 14/15 2/15. 12.42 Verify that the following matrix is Lorentz and calculate the velocity between the two observers without finding the simple Lorentz matrix L*. "3/V3 1/V3 2/V3 -1/vT 1 1 1 0 ^10 1 -1 0 1/V3 -1/V3 -1/V3. 12.43 Show that by definition the proper-time parameter τ is an invariant with respect to all Lorentz transformations. 12.44 Verify the formula for the composition of velocities by (i) multiplying the two simple Lorentz matrices below; (ii) calculating from (12.15) the velocities belonging to the two matrices and to their product; (iii) showing that the three velocities obey (12.16). 17/8 -15/8 0 0" -15/8 17/8 0 0 0 0 10 .0 0 0 1. 12.45 An electron gun shoots particles in opposite directions at one-half the velocity of light. At what relative velocity are the particles receding from each other? 12.46 Show that the composition of two velocities less than с is also less than с 12.47 How slow would your watch run relative to a stationary clock if you were moving at 2/3 the velocity of light? 12.48 At the age of 20, an astronaut left her twin brother on earth to go exploring in outer space. The first two years the spaceship gradually accelerated to a cruising speed 95 percent of the velocity of light. Traveling at that speed for 25 years, it reached a distant galaxy (23.75 light years away) and then decelerated for two years. Two years were spent exploring the galaxy before the journey back home, which followed the schedule of the trip outward. How old is the astronaut when she rejoins her 80-year-old brother? (Use an average rate for clock-retardation during the 8 years in acceleration/deceleration.) 12.49 How fast would a pole-vaulter have to run for his 20-foot pole to fit (instantaneously) inside a barn, in the judgment of a ground observer for whom the barn is 19 feet 11 inches long? 12.50 An alternate definition of uniformly accelerated motion is motion under a constant Lorentz force. Verify that the two definitions are equivalent for one-dimensional motion. 12.51 Show that ξ"α\α[ = giJ. 12.52 Prove that the array (12.46) is a 4-vector. "13/12 5/12 0 0" 5/12 13/12 0 0 0 0 10 . 0 0 0 1.

188 TENSORS IN SPECIAL RELATIVITY [CHAP. 12 12.53 Show that the matrices #and ^of (12.44) are connected via F'!: the matrix product GFG = P.] eukiSkrSbF"· Wint: First evaluate 2*ijkl 12.54 Define Faraday's two-form by Φ = G&G = 0 E, Ez Б, -^ 0 -H3 я, -E2 Я3 0 -я, -£з -н2 я, 0 = [Ф,Д or, inversely, #= G^G. Show that (α) SP is related to Φ through F'J = - \eljkl<&kl; (b) Maxwell's equations can be written in terms of the single matrix Φ as ЭФ дх дФ к дх' дФ у* дх' = 0 дФ дх1 12.55 The energy flux in an electromagnetic field is specified by the Poynting vector, p=ExH. By direct matrix multiplication or otherwise, derive the formula - ψ®- &&)■■ 0 0 * 0 * * * * 0 Ръ 0 * 0 -ρ Pi 0 (antisymmetric matrix) 12.56 Verify that for simple Lorentz matrices A (hence, no rotation of axes allowed): (a) iF(U, V) = G&(\, V)G; (b) &(V, V) = BT&(Y,V)B, where B = A~K, and (c) .#(U, V) = A3F(V, V)AT (thereby proving that (F1') is a contravariant tensor under simple Lorentz transformations). 12.57 Prove that if Λ = [Α,]„ satisfies А^ВЦ 0 for every B—[Blt]n„ that has zero trace (B;/ = 0), then A = A/, for some real A. [Hint: First take 5 as having all elements zero, except for one off-diagonal element. Then choose Baa = -Βββ = 1 (α Φ β; no summation), with all other BtJ zero.]

Chapter 13 Tensor Fields on Manifolds 13.1 INTRODUCTION The modern, noncoordinate approach to tensors will be introduced as an important alternative to the coordinate-component approach employed exclusively in the previous chapters. This will entail somewhat more sophisticated mathematics. 13.2 ABSTRACT VECTOR SPACES AND THE GROUP CONCEPT Linear algebra provides a means of systematically studying the algebraic interplay between real numbers (scalars) and a wide variety of different types of objects (vectors). Vectors can be matrices, π-tuples of real numbers, functions, differential operators, etc. In this chapter, we shall adopt the convention of using uppercase boldface characters for sets (of points, of real numbers, of elements of a group, etc.), and lowercase boldface for vectors (as in the preceding chapters). However, the latter will be gradually phased out in favor of light uppercase characters, not only for easier reading, but also in conformity with notation used in many standard textbooks. The concept of vector spaces requires a careful distinction between the scalars a,b, c,. . . , and the objects of study (vectors), u, v, w,. . . . We shall always identify the scalars with the field of real numbers, although any field could serve for the construction of an abstract vector space. Algebraic Properties of a Vector Space In terms of two binary operations, the axioms for a vector space are as follows. Addition Axioms 1. u + ν is always a vector 2. u + v = v + u 3. (u + v) + w = u + (v + w) 4. There is a vector 0 such that u + 0 = u. 5. For each u there is a vector -u such that u + (—u) =0. Scalar Multiplication Axioms 6. a · u = flu is always a vector 7. fl(u + v) = flu + flv 8. (a + b)u = au + bu 9. (ab)u = a(b\i) 10. lu = u EXAMPLE 13.1 We give notation for four familiar vector spaces. (a) R" = the n-tuples of reals under componentwise addition and scalar multiplication. (b) P" =the polynomials (in a variable t) of degree η or less. If p(t) = a/, q(t) = b(t\ let p(t) + q(t) = (a,. + bt)tl and r -p{t) = (ra,.)i''. (c) Cfc(R) = the continuously Λ-times differentiable functions (of i), / : R^-R (mapping the reals into the reals). To define + and ·, write/(i) + g(t) = (f + g)(t) and r-f(t) = (rf)(t). (d) M"(R) = the η χ η matrices over R. If A = (aif) and В = (Ь .), addition and scalar multiplication are defined by A + В = (acj + Ьи) and rA = (ra,y). 189

190 TENSOR FIELDS ON MANIFOLDS [CHAP. 13 Algebraic Properties of a Group Axioms 1-5 make a vector space an abelian (commutative) group under addition. In the general definition of a group, the binary operation is designated as "multiplication" and the commutative requirement is dropped. Multiplication Axioms 1. uv belongs to the group. 2. (uv)w = u(vw). 3. There is an identity element e such that eu = ue= u. 4. For each и there is an inverse element u~ such that uu~ = ulu = e. EXAMPLE 13.2 Some frequently encountered groups follow. (a) The reals R over ordinary addition; the reals over ordinary multiplication if 0 is removed from the set. (b) The cube roots of unity, C3 = {1, ω, ω2}, over ordinary multiplication of complex numbers, where ω = |(—1 + iV3). Groups of this type are called cyclic and are generally denoted by C* (the cyclic group of order к). С* is necessarily abelian. (c) The 4-group {e, u, s, b), under the rules u2 = s2 = b2 = e, b=us, and the associative law of multiplication. The 4-group is abelian, but it is not equivalent to the cyclic group on four elements, C4. (d) M"(R), under matrix addition. (e) GL(n, R) = the real, nonsingular η χ η matrices under matrix multiplication; this is the general linear group (nonabelian). GL(n, R) contains many very important smaller groups (called subgroups). Some of these are: SL(n, R) = the real η χ η matrices with determinant +1; SO(rc) — the η x η orthogonal matrices; and L(n) =the η x η Lorentz matrices [see the definition of L(4) in Section 12.3]. (/) GL(n, C)=the complex, nonsingular nxn matrices under matrix multiplication. An important subgroup is the unitary group, U(n), consisting of all η χ η Hermitian matrices (such that A'1 = AT, where the bar denotes complex conjugation). 13.3 IMPORTANT CONCEPTS FOR VECTOR SPACES Basis A basis for a vector space is a maximal, linearly independent set of vectors bl7 b2,. . . . If this set is finite, possessing η elements, the vector space is finite-dimensional, of dimension n. Otherwise, the space is said to be infinite-dimensional. EXAMPLE 13.3 (1) It is obvious that a basis for R" is the set of vectors βι = (1,0,0,...,0), e2=(0,l,0,...,0), ..., e„ = (0,0,...,0,l) called the standard basis. (2) P" is finite-dimensional, of dimension η + 1; one basis is {t'}, OSiSn. (3) The vector space of all polynomials is infinite-dimensional, as is the vector space C*(R). See Problem 13.4. Isomorphisms, Linear Mappings Two mathematical systems of the same type (such as two vector spaces or two groups) are called isomorphic if they are structurally identical and differ only in nomenclature. In the case of two vector spaces, an isomorphism is a one-to-one (bijective) linear mapping φ from one space to the other, where the term linear refers to the properties (for all vectors u, v, and scalars a): <p(u + v) = <p(u) + <p(v) and φ(α\ι) = fl<p(u) (13.1) For groups, an isomorphism would be a bijection ψ with the property ф(ии)= ф(и)ф(и), for all elements of the group. A more general mapping that is important for groups is a homomorphism, which merely requires that ф(ии) = ф(и)ф(и) for all и and υ, without necessarily requiring one-to- oneness.

CHAP. 13] TENSOR FIELDS ON MANIFOLDS 191 Product of Vector Spaces If U and V are any two vector spaces, the ordinary cartesian product U x V, the set of ordered pairs (u, v) with u in U and ν in V, can be made into a vector space by defining addition and scalar multiplication of pairs via (p, q) + (r, s) = (p + r, q + s) and a(p, q) = (ap, aq) Such a product space is denoted U® V; if U =V, write U® V as U2. More generally, the product of any number of vector spaces \l,\2, . . . ,\k may be easily defined as above; this product is denoted У!®У2®У3® ·· · ®\k. If \l =V2 = · · · =\k =V, the product space is written V*. (This notation is also often used for the tensor product of two vector spaces, a concept which will not be treated here.) 13.4 THE ALGEBRAIC DUAL OF A VECTOR SPACE If a vector space V be mapped linearly into the reals R, satisfying (13.1), the mapping is called a linear functional, or one-form. As in Example 13.1(c), we can make the set of all linear functional on V into a vector space itself, with the zero functional as that mapping which takes every vector in V into 0 in R. Definition 1: The algebraic dual of a vector space V is the set V* of all linear functional made into a vector space under ordinary pointwisc addition and scalar multiplication: (/ + g)(v) = /(v) + g(v) ( A/)(v) = A/(v) Since any linear functional on R" can be expressed as a linear function of the coordinates, ν = ν el + v'e2 + ■ ■ ■ + v"en —* f(v) = axv + a2v2 + ■ ■ ■ + αηυ" where at = /(e,) for each i, the functional is completely determined by the π-tuple (a1, a2, . . . , an). Differential Notation: One-Forms Thus, different functional correspond to different π-tuples, as f±*(a1?a2, .. . ,an) g±*(bl,b2, . . . ,b„) and it has become customary to represent linear functional by the compact notation of one-forms: ω = αλάχ + a2dx + · · · + an dx" σ = bl dx + b dx + · · · + bn dx" But why dx' for the coordinates? The motivation comes from differential geometry. Recall that any class C1 multivariate function F(x1, x1, . . . , x") on R" has the gradient VF = (dFldx1) and the directional derivative (in the direction (dx, dx2, . . . , dx")) dF = —7 dx + —j dx' + ■ ■ ■ + ——^ dx" dx dx dx which, at a specific point in space, is a one-form that defines a linear functional on R" (i.e., the set of all directions). Recall too that, just as in ordinary one-dimensional calculus, dx = Ax = an unspecified real number not necessarily small. EXAMPLE 13.4 (a) In R3, find the image of ν = (1, 3, 5) under the one-forms (linear functionals) ω = 4<ί*1 - dx2 σ = 2ώ' + 3 dx1 - dx% ω + σ = 6 dx1 + 2 dx2 - dx3 (b) What is the relationship among ω(ν), σ(ν), and (ω + <τ)(ν)?

192 TENSOR FIELDS ON MANIFOLDS [CHAP. 13 (a) ω(ν) = 4·1-1·3 + 0·5 = 4-3 + 0=1 σ(ν) = 2 · 1 + 3 · 3 - 1 ■ 5 = 2 + 9 - 5 = 6 (ω + σ)(ν) = 6·1 + 2·3-1·5 = 6+6-5 = 7 (b) ω(ν) + €τ(ν) = 1 + 6 = 7 = (ω + σ)(ν) For vector spaces different from R" we agree to use the procedure of Example 13.4 on the components of vectors relative to an arbitrary basis. That is, to evaluate the image of ν = t/bj + v2b2 + · · · + v"bn = t/b,- under the one-form ω = a; dx', simply write ω(ν) = <о(и*Ъ}) = (at dxl)(v''bj) = α,υ' (13.2) A dual reading of (13.2) gives a better understanding of the relationship between V and V* (between vectors and one-forms). If we regard the ai as fixed (tantamount to fixing a basis in V) while the v' vary—the "normal" situation—then a linear map from V to R is uniquely defined. If, on the other hand, the vector components v' are held fixed and the coefficients ai are allowed to vary (this amounts to fixing a basis in V*), a linear map from V* to R is denned (the latter map is actually an element of the space V**). The expression at ν is bilinear in the two vector variables ν and ω. Theorem 13.1: If V is a finite-dimensional vector space, then V* is finite-dimensional, of the same dimension, and is isomorphic to V. A proof is given in Problem 13.6. Dual Basis A basis bj, b2, . . . , b„ for V determines one for the dual space V* in a very natural way. Each ν in V has a representation ν = v'bj and thus defines a linear functional <p(v) = υι dx1 + υ2 dx2 + ■■■ + υ" dx" (13.3) Then the η linear functional (vectors in V*) defined by <p(b,.) = p; (i = l, 2,..., я) (13.4a) form a basis for V* (see Problem 13.6); we say that the basis {β1} in V* is the dual of the basis {b;} in V. The evaluation rule (13.2) provides a simpler characterization of the dual basis: β'(ν) = (<p(b;))(vJbj) = (0 - dx1 + 0 · dx2 + ■ ■ ■ + 1 · dx' + ■ ■ ■ + 0 · άχη)(υ%;) = ν! (13.4b) Thus, β' = dx1 is the linear functional that picks out the ith component relative to {bk} of any vector in V. A special application of (13.4b) gives β'0>;.) = δ;. (13.5) for all i, j. EXAMPLE 13.5 The standard basis e = {βι, e2,.. ., e„} for R" generates the standard basis for (R")*, given in terms of one-forms as P\e) = dx1 ft2(e) = dx2 ... βη(έ) = άχ" Suppose, then, that R3 is referred to the (nonstandard) basis ft, = (1,1,0) b2 = (1,0,1) ft3 = (0,1,1) This may be written in terms of the standard basis {ej through a formal matrix multiplication: V b2 b3 = "1 1 0" 1 0 1 .0 1 1. "еГ e2 _ез_ Express the dual basis {β'} for (R3)* in terms of its standard basis (dx') as a similar matrix product.

CHAP. 13] TENSOR FIELDS ON MANIFOLDS 193 Let β' = a'j dxx + a'2 dx2 + a3 ахъ\ we must solve for the components a'.. For i = 1, we have from (13.5): β'Ο^) = β\ΐ, 1, 0) = flj · 1 + a' · 1 + a\ ■ 0 = a\ + aj = χ + у = 1 β1(^) = β1(1,0, l)=rl + jO+2'l = i + 2 = 0 β'0>3) = β'(0, l,l) = x-0 + y-l + z-l = y + z = 0 (where χ = α\, у = α\, ζ = α\). Solving, χ= \= у, ζ = - j. A similar analysis may be used to determine the a2 and a3. The final result is β1 = \dxx + \dx2 - \ахъ β2= \dxx - \dx2 + \dx2 β3= -£<&' + 4й(д:2+ ^(£t3 or β1 β2 β3 1 2 1 2 1 2 I _1 2 1 1" 2 1 2 1 2 dxx dx2 dx\ Observe that the two basis-connecting matrices are formal inverses of each other. Change of Basis in V and V* The result of Example 13.5 can be generalized. Let {bj and {bj be two bases for V, and let {β1} and {β1} be the respective dual bases for V*. Then Ь; = Л/Ьу -> β; = Α^ with (А') = (А)У1 (13.6) the arrow denoting implication. (See Problem 13.7.) 13.5 TENSORS ON VECTOR SPACES The concept of a multilinear functional is needed: If /(v , v2, . . . , v'") represents a mapping of m vector variables into the reals such that the restricted mapping obtained by holding all but one of the variables fixed is a linear functional, then /is said to be multilinear in all its variables. Definition 2: A type-(q) tensor is any multilinear functional Τ : (V*)/'®V9- one-forms and q vectors into the reals; the real image is denoted ■R mapping ρ Τ(ω\ . ρ ι ,ν*) EXAMPLE 13.6 Let T represent a linear functional in what follows. A type-(J) tensor takes on real values Γ(ω) for all one-forms ω as argument. As we shall see later, such a tensor can be identified with a contravariant vector. A type-(5) tensor takes on real values Γ(ν) for all vectors ν as argument; it can be shown to correspond to a covariant vector. A type-( [) tensor takes on real values ί/(ω; ν) for all ordered pairs in V* ® V as argument, with U a bilinear functional. EXAMPLE 13.7 For л-dimensional vectors, the ordinary scalar product u · ν = uv defines a type-( °) tensor, in the form G(u, v) = uv, since the elementary properties of the scalar product make G a bilinear mapping of a vector pair into the reals. More generally, an inner product defined arbitrarily by the quadratic form G(u, v) = uTE\ where Ε is an η χ η matrix, defines a type-( %) tensor. Definition 3: A type-^) tensor G(u, v) is (i) symmetric if, for every u and v, G(u, v) = G(v, u) (ii) nonsingular if [G(u, v) = 0, identically in u] —> ν = 0 and (iii) positive definite if, for any nonzero vector u, · G(u,u)>0 A type-(2) tensor that is symmetric and nonsingular is called a metric tensor. (A positive definite tensor is necessarily nonsingular.)

194 TENSOR FIELDS ON MANIFOLDS [CHAP. 13 EXAMPLE 13.8 Let С = [Cy]„„ be a square matrix and let (a,.) and (V) be the respective components of ω and ν relative to the standard basis in R" and its dual. Then the matrix product T(<a; \) = <aCv = a^'jV' (a bilinear form) defines a type-( \) tensor over the vector space R". Tensor Components In the three types of tensors considered in Examples 13.6-13.8, we may define tensor components in the following manner, which may be generalized to arbitrary tensors in an obvious way. Let b,, . . . , b„ be a basis for V, and β1, . . . , β" its dual in V*. Then, for each i, write type (I) f=T(p) type (l) Г,= Г(Ь;) type(\) Γ; = ΐχβ'; b7.) EXAMPLE 13.9 Find the components, relative to the standard basis for V=R", of a type-(J) tensor on V constructed by the recipe of Example 13.8. By construction, Γ(ω; ν) = а^С'р', for all ω and v. Substituting ω = βρ = dxp and ν = b = e , we find r^r(^p;e9) = 5fc;5;=Cp Thus, the components of Τ are independent of those of the arguments, ω and v, and depend only on the components of the matrix C. Effect of Change of Basis on Tensor Components Under a change of basis, (13.6), type (I) T= Γ(β') = T(Alrp) = AlrT(p) = TA\ type φ f. = Т(Ъ;) = Т(А[ЪГ) = А]Т(ЪГ) = TrA\ type{\) Г>к = Γ(β', β'; Ък) = T(A'rp, А>#; А\Ъ,) = А\А[А\Т(р, β*; Ь,) = Т\°А\А\Ак EXAMPLE 13.10 If V is Euclidean R", the change of basis b; - Д'Ьу induces the change of coordinates χ = Α\χ', for which J=(*L) = A and so J=J~l = A \dx'l The above transformation formulas then reduce to the classical laws for affine tensors—compare (3.21). 13.6 THEORY OF MANIFOLDS A manifold is the natural extension of a surface to higher dimensions, and also to spaces more general than R". It is helpful at first to think of a manifold as just a hypersurface in R". By the term neighborhood of a point we shall understand either the set of all points in R" within some fixed distance from the given point, or any set containing these points. A neighborhood of ρ will be denoted U^. If the concept used for distance in Rm is Euclidean, then every neighborhood U contains a solid, spherical ball (or "hyperball," if η > 3), having some positive radius and centered at p. A neighborhood of a point ρ in a set is the intersection of a neighborhood U and the set. A set is open if each of its points has a neighborhood completely composed of points of the set. An open neighborhood is simply a neighborhood that is also an open set (in the case of a solid-ball neighborhood, the outer boundary of the ball would have to be removed in order to make it an open neighborhood).

CHAP. 13] TENSOR FIELDS ON MANIFOLDS 195 Descriptive Definition of a Manifold A manifold is a set which has the property that each point can serve as the origin of local coordinates that are valid in an open neighborhood of the point, which neighborhood is an exact "copy" of an open neighborhood of a point in R". Though such a definition allows the manifold to lie in a metric space, topological space, Banach space, or other abstract mathematical system, it is best that we begin with manifolds in a simpler space, like R"\ Accordingly: Definition 4: A manifold is any set Μ in Rm which has the property that for each point ρ in the manifold there exists an open neighborhood U in Μ and a mapping <pp which carries U into a neighborhood in R". The mapping is required to be a homeomorphism; i.e. (1) φρ is continuous. (2) ψ is bijective from \Jp onto its range, φρ(Χ}ρ). (3) φ~ι is continuous. See Fig. 13-1. Fig. 13-1 Coordinate Patches, Atlas The neighborhoods U for ρ in Μ provide a means for locally ascribing coordinates to Μ which have the correct dimension (e.g., a plane lying in 3-space is actually 2-dimensional and, as a manifold, has a coordinatization by pairs of reals instead of triples, as in Section 10.4). For any point ρ in M, the pair (Vp, φ ) is called a coordinate patch (also chart, or local coordinatization) for M, while any collection of such pairs for which the neighborhoods U together cover Μ is called an atlas for M. Since the coordinate patches make Μ и-dimensional at each point, Μ is sometimes referred to as an n-manifold. Often a finite number of charts is sufficient for an atlas (Example 13.11). It can be proved that if a manifold in Rm is closed, and bounded in terms of the distance in Rm, a finite number of charts will always be sufficient. EXAMPLE 13.11 (a) The 2-sphere, denoted S2, is the ordinary sphere in 3-dimensional space (y'), centered at (0,0,0) having radius a. It may be coordinatized by an atlas of only two charts, as follows. [Note that the usual spherical coordinates (φ, θ) fail to give a one-one mapping at the poles, where θ is indeterminate.] Τ 2 1 ι _ 2a χ У ~ (x1)2 + (x2)2 + a2 Τ 2 2 , 2a x {x ) +{x ) + a I 1ч2 . / 2Ч2 2 {χ ) + (χ ) - а &α / 1ч2 / 2ч2 . 2 {χ ) +(χ ) + а (е=±1)

196 TENSOR FIELDS ON MANIFOLDS [CHAP. 13 As illustrated in Fig. 13-2, the chart corresponding to ε = +1 has Up centered on the south pole (whose coordinates are x1 = x2 = 0) and including every point of the sphere except the north pole. The other chart (ε = -1) is the mirror image of the first chart in the equatorial plane. For a derivation of this atlas, see Problem 13.15. У1 Ρ = S(0, 0, -a) y1 S Fig. 13-2 (b) The η-sphere S" in R"+1 may be defined as the set of points (/) in R" + 1 such that iff+ (УУ+ {/? + ■■■+ {y'l+1f = a2 (centered at (0, 0, 0,... , 0), radius a). A coordinate patch for a neighborhood of (0, 0,... , a) is: yl=xx y2 = x2 ■·■ >■" = *" у"+1=^а2-(хУ-(х2)2 (χ"γ where the mapping is into the и-dimensional neighborhood (дс1)2 + ■ · ■ + (χ") < a2 (the interior of S""1). Establishing the analogous patches around the other "diametrically opposite" endpoints, we obtain an atlas of 2л + 2 charts. (A smaller atlas requires a more clever approach.) Differentiable Manifolds Inevitably, there will exist pairs (Up, <pp) and (U?, <pq) whose neighborhoods overlap in Μ (Fig. 13-3); so the common region U Π U = W, called an overlapping set, generates a map φ between the images of W under φρ and (pq. Explicitly (trace the circuit in Fig. 13-3), φ = φ ° φ~ρι. It is clear that φ and φ'1 are both continuous. If φ and φ~ι are of class Ck (have continuous partial derivatives of order к at each point) then the overlapping set W is said to be of class Ck. Fig. 13-3

CHAP. 13] TENSOR FIELDS ON MANIFOLDS 197 Definition 5: A differentiable manifold is a manifold which possesses an atlas such that all overlapping sets are of class C1. A Ck (C* or Сш) manifold has an atlas whose overlapping sets are of class Ck (C°° or C). Remark 1: Recall the distinction between infinitely differentiable (C°°) and analytic (Сш). One way to ensure that a manifold be Ck in the present context is to demand that each φ and φ~ι be class Ck. As a matter of convenience, we assume from now on that all manifolds are C°° manifolds. EXAMPLE 13.12 In the case of the spherical manifolds of Example 13.11, the mapping functions φ~* are cither rational with nonvanishing denominators or square roots of positive polynomials. These are certainly C°° manifolds (in fact, C°). To bring the notation closer to that of differential geometry (cf. Section 10.4), we now redesignate the maps <p~l linking Μ with coordinates (*'): let — 1/12 n\ /12 n\ ///12 n\\ φρ (X , X , . . . , X ) = x(x , X , . . . , X ) =(y\X , X , . . · , X )) for l^j^m. (See Fig. 13-4.) Fig. 13-4 13.7 TANGENT SPACE; VECTOR FIELDS ON MANIFOLDS Intuitively expressed, a vector field У on a manifold Μ is simply a tangent vector to Μ which varies in some continuous (and differentiable) manner from point to point (Fig. 13-5). More precisely, it is a rule that gives a tangent vector at every point of M. One way to obtain a vector field (if we are in RJ) is to take the variable normal vector η and cross it with some fixed vector a; thus V=nxa is a differentiable vector field. But this definition takes us outside the manifold (is extrinsic). We seek a way to remain on the manifold itself (which is immediately applicable to abstract manifolds not imbeddable in a familiar space); such endeavors are called intrinsic methods. The clue is to consider some curve on Μ and to define V as the tangent field of that curve. If the curve is defined through a coordinate patch by с = r(x\t), x\t), ..., x'\t)) = (y'(x\ x2,..., xn)) (1 Ш] ё m) then the chain rule gives dc ~dt dx __ dx dx' dt ~ dxi dt = V or ν=νν where the vectors r; = drldx1 are themselves tangent to M.

198 TENSOR FIELDS ON MANIFOLDS [CHAP. 13 Fig. 13-5 Definition 6: For any (differentible) manifold Μ having coordinate patch Up at any point p, the span of the vectors rt, r2, . . . , r„ [evaluated at φ (p)] is the tangent space at p, denoted Γ (Μ). The union of all tangent spaces Γ (Μ), for ail ρ in M, is called the tangent bundle of M, denoted Γ(Μ). Although each Tp(M) is a vector space, this does not guarantee that T(M) is a vector space. For example, the sum of a vector in Tp(M) and a vector in Tq(M) will not generally be tangent to M. Definition 7: A vector field V on a manifold Μ is any C" function that maps Μ to its tangent bundle T(M). That is, for each point ρ in M, the image \(p) = V is a vector belonging to the tangent space Tp(M) at p. Explicitly, for certain scalar functions V, V= v-r, = (у1^£) U =1,2,..., m) (13.7a) A fundamental theorem regarding vector fields on manifolds may be proved from the basic theory of systems of ordinary differential equations. Theorem 13.2: Every vector field on a manifold Μ possesses a system of flow curves or integral curves on M, defined as curves for which the tangent vector at each point coincides with the given vector field at that point. Notation To de-emphasize the particular choice of coordinatization map φ : Vp —> R", it is customary to omit the vector r from the above description of the basis for Tp(M), and to write д д д . , . дх дх дх —\ > —г . · · · > -^п ш place of —т , —-г ,. . . , —ц дХ дХ дХ дХ дХ дХ or, even more cursorily, дх, д2,. . . , дп. Many textbooks use this last notation exclusively, and write (13.7a) as ν=ν% (13.7b) Such shorthand is particularly convenient when the coordinate maps φ~ι for a particular manifold are unspecified (for example, when the manifold is defined by an equation F(y\ y2,. . . , ym) = 0 for some real-valued function F). In this situation, since r1( r2, . . . , xn are not explicitly defined, we use the notation Ex, E2,. . . , En to denote the coordinate frame on M, whose restriction to Τ (Μ), for each ρ in M, is a basis for Γρ(Μ). We make the identifications E-t = di (i = 1, 2,. . . , ή), giving V= ViEi (13.7 c)

CHAP. 13] TENSOR FIELDS ON MANIFOLDS 199 Extrinsic Representation of Vector Fields It is possible to represent a vector field У on a manifold without reference to coordinate patches (which often have the disadvantage of being complicated or difficult to construct); one can stay entirely in the (y!) system, which we assume to be rectangular. Suppose Μ is given by a single equation F(yl, y2,. . . , ym) — 0, for some С function F. A one-form σ = ωίάγ, where ω. = wi(y\ У2, ■ ■ ■ , ym), is said to be restricted to Μ if the point {y) is required to lie on M; that is, F(yl, y~, . . . , ym) = 0. As is well known from multidimensional calculus, the gradient VF = (dF/dy1) is normal to M, so that if we further require that the restriction of σ map VF into zero, ω, —: =0 dyl then V — σ is a vector field on Μ (having components dy'). EXAMPLE 13.13 Consider the paraboloid Ρ in R3 given by F(y\y2,/)-(ylY + (y2f~y3-0 Show that the restriction of σ = у1 у2 dy1 + (у2)2 dy2 + 2y2y2 dy2 to Ρ is a vector field on P. We must show that the scalar product of (y1)/2, (y2)2, 2y2>·3) and VF=(2y', 2y\ -1) is zero: (уУ)(2у1) + (у2)2(2у2) + (2уУ)(-1) = [(у1)2 + (у2)2-Л2)'2 = 0 13.8 TENSOR FIELDS ON MANIFOLDS Dual Tangent Bundle At each ρ in M, let Γ*(Μ) denote the dual of the vector space ГДМ), and denote by Г*(М) the union of all spaces Г*(М). The set Г*(М), called the dual tangent bundle of M, is not necessarily a vector space (just as T(M) was not). We need to make explicit certain elements of Г*(М). Differentials on Μ The differential of a function / : R" —> R is rigorously defined as a two-vector function df : (R") —> R which maps each pair (x, v)—where χ is a point in R" and ν = (dxl, dx2,. . . , dx") is a direction in R"—to the real number df(x, v) = -^r dx' + Дг dx2 + · ■ ■ + —n dxn = fi dx (13.8) дХ дХ дХ Here, the ft are evaluated at x. If / is any real-valued Ck function on M, then the differential of Л1 2 ri\ r/ 1/1 n\ 2/1 n\ m / I nw X , X , . . . , X ) = f(y (X , . . . , X ), у (X , . . . , X ),..., У (Χ , . . . , Χ )) called a differential field on Μ, is df=j-i(mx)))dxl = -f-kf1dx' = (yf-rl)dxi dx dy dx —a one-form. Thus, df may be thought of as a mapping from Μ to T*(M): we agree that the evaluation of df at ρ is the one-form (V/(p) · r^p)) dx in T*(M). Let us now compare the two kinds of fields on M. Mapping Restricted to U^ vector field V : M-» Г(М) У=У;£1 differential field df : M-» T*(M) <o = (Vf-El)dxi

200 TENSOR FIELDS ON MANIFOLDS [CHAP. 13 Definition 8: A tensor field of type (Ϊ) on a manifold Μ is a mapping Τ : [T*(M)]r ® [Γ(Μ)]*-»· Ck(Rm) taking r differential fields and s vector fields on Μ to real-valued Cfc-functions / on Rm. It is assumed that the evaluation of Γ at a point ρ on Μ is given by Τρ{ω\ . . . , ω'; V,, . . . , Vs) = Цс^, . . . , α/ρ; Vlp, . . . , Vsp)^f(p) and that each map Τ is multilinear. EXAMPLE 13.14 (a) At each fixed ρ on M, the mapping Tp is a tensor [on the vector space [T*(M)Y®[T (M)Y, of type (;)], per Definition 2. (b) Any vector field У on Μ can be interpreted as a type-(o) tensor field via a mapping Γ(ω)= ω(ν); compare Problem 13.20. Solved Problems ABSTRACT VECTOR SPACES AND THE GROUP CONCEPT 13.1 (a) Show that the set of polynomials Pi(0 = l + f Pi(t)=t + t2 p3(t) = t2 + t3 ρ4(ί) = '3-1 is a basis for the vector space P3 (polynomials of degree §3). (b) Find the components of the polynomial p(t) — t3 relative to this basis. (a) Since the dimension of P3 is 4 and there are 4 vectors, it suffices to show they are linearly independent. Suppose that, for all t, λ\\ + t) + \2{t + t2) + \\t2 + i3) + λV - 1) = 0 or (A1 - A4) · 1 + (A1 + k2)t + ( A2 + λ3)ί2 + (λ3 + Α4)ί3 = 0 Since this is an identity, we must have 0 = A1 - A4 = A1 + A2 = A2 + λ3 = λ3 + λ4 Thus A1 = A4 A1 = —A2 = A3; so the last equation gives A1 + A1 = 0 or A1 = 0, and all A' vanish, thus proving linear independence. (b) To find the linear combination yielding p(t) = i3, write X\l + t) + \2(t + t2) + A3(i2 + i3) + A4(i3 - 1) = i3 i.e. (A1 ~ A4) · 1 + (A1 + A2)i + (A2 + A3)i2 + (A3 + A4 - l)i3 = 0 or A^A4 A1 = -A2 = A3 A1 + A1-1 = 0 Hence, A1 = - A2 = A3 = A4 = 1 /2. 13.2 (a) Model the 4-group by manipulating an ordinary 85 by 11 sheet of paper, in the following way: Let s be the operation of turning the sheet over sideways (as in a book) and setting it on its original location; u, the operation of turning the sheet upside down (end-for-end); b, both operations (s followed by u, resulting in a 180° rotation of the page, face up); and e, doing nothing (identity). Interpret the group operation (multiplication) as one operation followed by another (thus, for example, by the above definitions, b = su, reading from left to right), (b) Show that the 4-group cannot be isomorphic to the cyclic group on four elements, C4.

CHAP. 13] TENSOR FIELDS ON MANIFOLDS 201 (a) This is one of those problems in mathematics that is best handled without formulas or equations. By simple observation, the operation us also results in a 180° rotation; hence, b = su = us. It is also clear that if we apply s twice, or и twice, the sheet is left in its original state; s2 = u2 = e. Next, observe that the associative law is valid, so long as we keep the order of the operations intact. It follows that b2 = (su)(us) = s(u)2s = s2 = e When we multiply all the group elements by b, we obtain: be = b bb = e bu = (su)u = su2 = s bs = (su)s = (us)s = и Hence, the multiplication table for this group may be displayed and may be seen to coincide with that for the 4-group: • e s и b e e s и b s s e b и и и b e s b ~T~ и s e (b) For the cyclic group, {e, z, z2, z3}, with z4 = e for some z, we could not have z2 characteristic property of all elements of the 4-group. : e, which is the 13.3 The simple Lorentz group can be studied by compressing the 4 x 4 matrices down to 2 x 2 matrices: 'a b 0 0" a b~ a b 0 0" b a 0 0 0 0 10 0 0 0 1. ~* b a ] (a2-b2 = l) Show explicitly that all real 2x2 matrices of the above form constitute an abelian group (the group L(2)) under matrix multiplication, and that L(2) is a subgroup of the following two larger groups: GL(2, R) : matrices of the form , , ad¥=bc SU(2) : matrices of the form Since for a matrix in L(2), ad — be = a — , dl' ad — be = 1 1 all such matrices belong to SU(2), which, in turn, is a subgroup of GL(2,R). Now verify the group properties: (1) uv belongs to the group for all u, v. If then AB a b A ЬЛ a. = a _b \c d] Id ci b~ a. B = с d .d с ac + bd ad + be _ be + ad bd + a χ y~ 1У x. and x2 - y2 = det AB = (det A)(det B) = (1)(1) = 1 (2) (uv)w = u(vw). Yes: matrix multiplication is associative. (3) For some e and all u, eu = ue = u. Yes: the identity matrix has l2 - 02 = 1, so is a member of L(2). (4) Given и, и~ги = ии~г = e for some и l. a bV b aJ 2 1.2 a — b a -b -b a а -ЬЛ -b a\ which is in L(2).

202 TENSOR FIELDS ON MANIFOLDS [CHAP. 13 (5) uv = vu (abelian group). В А с d d с a b' .b a. ca + db cb + da .da + cb db + ca. χ у у χ AB VECTOR SPACE CONCEPTS 13.4 (a) Show that the space Ρ of all real-valued polynomials in a real variable χ is infinite- dimensional, (b) Conclude that C*(R) is infinite-dimensional. 13.5 (a) Suppose that Ρ had the finite basis {рг, р2, exist constants a1,. . . ,an such that alPl(x) + a2(x) + ■■■ + anPn(x) = p(x) Write (1) for the η + 1 values хг < x2 < ■ · · < xn + 1 as a matrix equation pn}. Then, for any real polynomial p(x), there (1) Pl(*l) Pi(x2) Рл(Хп+л) + a. />z(*i) p2(x2) + a. Pn(*l) Рп(Хг) Ра(Ха + л) P(Xl) p(x2) P(x„+i) (2) The column vectors on the left are elements of R"+1, and as there are η of them, they do not span R"+1 (see Problem 13.5). To finish the proof, we have only to choose a vector on the right of (2) that is not in the span of those on the left—say, (zt, z2, . . ., z„ + 1)—and then to exhibit a polynomial ρ that takes on those values at x1,x2,. . . ,xn+1. The polynomial provided by Lagrange's interpolation formula does the job. (b) For any k, the vector space C*(R) contains the infinite-dimensional subspace P; thus, it too is infinite-dimensional. The set S of all linear combinations of a fixed set of η vectors, {bj, b2, . . . , b„}, is called the span of the given vectors; it is obviously a vector space. Prove that this space has dimension mg n, with equality if and only if the given vectors are linearly independent. First we show that any n + \ vectors in S are linearly dependent. Suppose, on the contrary, that {u,,u2,. . . ,un + 1} are linearly independent. Then, because the sequence of vectors ui bi b2 . . . b„ is necessarily dependent, the well-known exchange lemma tells us that a sequence ui bi b;-i b/ + i b„ also spans S. Repeating the argument η — 1 times, we arrive at the result that the vectors span S, making un+1 dependent on them—a contradiction. If, therefore, {b,} is linearly independent, it constitutes a basis for S, and m= n. On the other hand, if only m < η of the bf are linearly independent, the above argument shows that any basis consists of exactly m vectors. DUAL SPACE 13.6 Prove Theorem 13.1. It is almost trivial that any two vector spaces of dimension η are isomorphic [if {b,(1)} and {bj2)} are bases, set up the correspondence ί/b,-1' •w-n'b-2']. Thus it is necessary to prove only that V* is л-dimensional if V is; in other words, to prove that the set of vectors {β'} defined by (13.4) (i) is linearly independent and (ii) has V* as its span. (Problem 13.5 will then immediately yield Theorem 13.1.)

CHAP. 13] TENSOR FIELDS ON MANIFOLDS 203 Proof of (f): By (13.5), for / = 1, 2,. .. , n, λ,β'(ν) = 0 -> λ,β'(^) = 0 -> Лг5/ = 0 -> λ,=0 Proof of (»): If β(ν) is an arbitrary element of V*, then, by (13.4b), β(ν)=β(υ>.)=β(Μ«''=β(^)β'(ν) that is, β is a linear combination of the β'. 13.7 Prove the inverse relation between the matrices A and A of (13.6). By definition b, = А{ъ, and β; = A'k$k, so look at (13.5): ji;(b,) = δ?. By the algebra of mappings and the fact that each β; and β*" is linear, we have δ> = β'(*,.) = (А№)(ЪЛ = А'Ж(Ъ,) = Α№(ΑΧ) ^А{А^к(Ъг)=А'кАгХ = А'^ that is, AA = I. TENSORS ON VECTOR SPACES 13.8 Which of the following represent linear mappings of (R )* (taking the one-forms on R into the reals), and so constitute (contravariant) tensors of type (J)? (a) T(axdx + a2dx + a3 dx )г=ала1аъ (b) T(aidx') = ax — a3 (c) T(aidxL)-\ (d) T(aidx) = 0 (b) and (d)—the only linear mappings. 13.9 Associated with a particular basis {bj of a vector space of dimension n, we are given some set of numbers {C'k; i, j, к = 1, . . . , η}. Then we define another set of numbers (and assume a similar definition for all changes of bases), {C'k; i, j, к - 1, . . . , η}, such that с'к-А\А'Лс; and call these numbers the components of the "tensor" С on the new basis {bj. Show that this "tensor" is indeed a tensor per Definition 2. Wc have only to define the functional ΓΚ, ω2; ν) = Τ(α,β; bfl'; v%k) = afiftv* which, by inspection, is a type (i) tensor. We have: T'i = Γ(β', β'; Ък) = Τ(δ[β\ 8$'; дкЪс) = 8'г8',С78'к= С'[ f'i = Γ(β', β';bk) = ЦА',р, Α'β*; А\Ъ,) = Α',Α{Α[Τ(β\ β°; bt) = Α^Α',σ/ = C\ which show that Τ and С coincide in all coordinate systems. 13.10 In terms of the components gtj of a metric tensor G(u,v), show that: (a) G is symmetric if and only if gt] - gjt for all i, j. (b) G is nonsingular if and only if |g,;| Φ 0. (c) G is positive definite if, for all vectors (u) Φ 0, g^uu' Φ 0 and gu > 0. By Section 13.5, g,y = G(b,, b,) where {b,} is some basis for V. Then, if u = и'Ь,. and ν = v'bl are any two vectors in V, G(u,y)=u'viG(bi,bi) = giju'vi

204 TENSOR FIELDS ON MANIFOLDS [CHAP. 13 (a) G(u, v) = G(v, u), for all u, v, if and only if gu^v'= g^v'u'= g^u'v' °r (glt-g,t)u'v' = Q for all real u, υ', which is true if and only if gtj = gjr (b) In matrix form, the nonsingularity criterion reads: [MrGi;=0, for all и] -> υ=0 But uTGv vanishes for all и if and only if Gv is the zero vector. Hence the criterion takes the form Gv = 0 -> υ = 0 which defines G as a nonsingular matrix (a matrix with nonvanishing determinant). (c) For each fixed u and a scalar parameter A, we have g-iu1 + ЛЬ', )(и' + Ab() = glJ(u' + λδ'ΟΚ + \δ[) = G(u, u) + b\ + gllA2 - P( A) where fr = (giy + gn)u'. If u is not in the span of b1? the quadratic form is, by hypothesis, nonzero. Hence, the discriminant of F(A) is negative: b2 b2~4guG(u,u)<0 or G(u,u)>7—SO It only remains to note that if u = кЪ1 (к^О), then G(u, u) = «2g11, which is again positive. 13.11 Show that positive-definiteness of a type-^) tensor G implies its nonsingularity. If G(u,v)=0 for all u and some v, then G(v, v)=0; and so, by positive-definiteness, v = 0. 13.12 A covariant tensor A(u, v) is antisymmetric if and only if Л(и, ν) = - Л (ν, u), for all u, v. Show that a criterion for antisymmetry is: A(u, u) = 0 (all u) By bilinearity, A(u + v, u + v) = A(u, u) + A(u, v) + A(y, u) + A(\, v) Thus, if v4(u,u) = 0 for all u, 0 = 0+ A(u,y) + A(\,u) + 0 or A(u,\) = -A(y,u) Conversely, suppose A(u, v) = ~A(\, u), for all u and v. Then, with u = v, we have A(u, u) = ~^4(u, u), or v4(u, u) = 0. MANIFOLDS 13.13 (a) Show that the 1-sphere S1 (a circle in R2) can be made into a C°° 1-manifold by constructing an atlas with two charts, (b) Show that a one-chart atlas does not exist [thus, a circle is not homeomorphic to a line or interval]. (a) The standard parameterization of the circle, _l iy1 = flCOS0 ... л „ . i) : ζ · a (0<θ<2ιτ) ψ {у = a sin θ ν ; is insufficient, since the inverse map φ is discontinuous at point ρ (Fig. 13-6). But if we define _i |y1 = flCosjt:1 ,ι,χ -ι |y1=flCosjt:1 ... ι _ , <P„ : 2 . ι (-ir<x <ττ) ψη : у г -ι (Q<x <2ττ) then (S1 — q, φρ) and (S1 —p, <pq) will constitute an atlas. Since there are no 'singular' points involved, it is clear that φ , φ4 and their inverses are C°°.

CHAP. 13] TENSOR FIELDS ON MANIFOLDS 205 Fig. 13-6 (b) Suppose (U, φ) covered S1 (U = S1) and φ mapped S1 to the real line (x1), with both φ and φ'1 continuous. It is not too difficult to see that φ maps the circle to a closed interval I: for continuous maps take bounded, closed sets to bounded, closed sets, and connected sets to connected sets; and the only bounded, closed, connected subsets of the real line are closed finite intervals. For any point Ρ on the circle let P' be its diametrically opposite point. The map g(t) = φ[(φ~\ή)'] takes a real number t in I, maps it to a unique point Ρ on S1, goes to the (unique) diametrically opposite point P', and returns to a unique real number t' in I; it is thus a continuous map from I to I. As such, it must (by a familiar theorem of analysis) have a fixed point: £('о) = lo f°r some t0 ш I But this means that φ sends some pair of diametrically opposite points on S1 to the same real number, denying one-oneness of φ. 13.14 A manifold in R is defined by the charts (k ι '(*) y' = χ' cos χ cos x3 2 1 2.3 у = χ cos χ sin χ 3 1-2 у = χ sin χ ly4 = a(x2 + x3) -2,-ί,0,ί,2,--.;χ1>0) (к~1)~<х3<(к + 1)~ (α) Show that on each coordinate patch, the mapping r(fc) is one-to-one; hence, <p(fc) = r(k) '■ U(k)~^R3 exists. (£>) Show that both <p(fc) and φ^\ are continuous, (c) Show that the manifold is generated by a line in R4 moving along an axis orthogonal to it, with the axis, in turn, orthogonal to the hyperplane y4~0 (use vector geometry in R4). Verify that the parameter χ measures the distance from a given point on the manifold to the axis, (d) Show that the parametric section x3 =0 is a right helicoid (Example 10.4), lying in the hyperplane 0 (R coordinatized by у1, у3, у4). Assume that r(/t)(jt') = r(/t)(ii'); we want to show that (x1, x2, x3) = (и1, и2, и3). Now, χ cos χ cos χ χ1 cos χ2 sin x3 и cos и cos и 1 2-3 и cos и sin и tan x~ = tan u~ But, for U(/t), the argument of the tangent function is restricted to a range of тт units; so x3 = и3. It follows that /2. 3\ /2.3ч ^ 2 2 a(x + χ ) = a(u + и ) —> χ = и Finally, from χ1 sin χ2 = и1 sin и2, we obtain χ1 = и1.

206 TENSOR FIELDS ON MANIFOLDS [CHAP. 13 (b) From the form of φ^ sr(fc), this function is C". To solve for (У) in terms of (/) (to find <p(/t)), write (У1)2 + (У2)2 + (гУ = (^(cos2 x2)(cos2 x3 + sin2 x3) + (^/(sin2 x2) = (x1)2 or x1 = νυΨ+ΪΤΐ+ΪΤ? (since x1 > 0). Then sin*2 = 7 = V(>Wy2)2 + (y3): or, for a suitable branch of the function sin-1 , 3 У V(/)2 + (y2)2 + (/)2 χ =sin It is seen that ί^ = ν(/)2 + (/)2 + (/)2 «"(*) ■■ jc =sin V(/)2 + (/)2 + (/): 3 jc3 = sin , . νν(/)2 + (/)2 + (/)2 is continuous (in fact, C). (c) The axis orthogonal to y4 = 0 is the vector e4 in R4. At the point y4 = a(x2 + x*) = const, on the manifold, we have (with x2, x3 constants and x1 = t) y1 = t cos x2 cos x3 у1 = t cos x2 sin jc3 y3 = ί sin x2 y4 = const. —a straight line with direction vector orthogonal to e4. A previous calculation gives the distance from (y\ y2, y\ y4) on Μ to (0, 0, 0, a(x2 + x3)) as V(/)2 + (/)2 + (y3)2 = x1 (d) Set χΊ = 0 and the map reduces to 11 1 2/Λ 31-2 4 1 у = χ cos χ у = 0 у = * sin * у = ax 13.15 Derive the charts of Example 13.11(a), using stereographic projection (Fig. 13-7). As Ρ is a "convex" combination of Q and N, (/, Г, /) = λ(*\ χ\ 0) + (1 - A)(0, 0, a) (1) Fig. 13-7

CHAP. 13] TENSOR FIELDS ON MANIFOLDS 207 To determine λ (λ>0), write я2 = (У1)2 + (уУ + (уУ = (Ах1)2 + (Ах2)2 + [(1 ~ А)я]2 and solve, obtaining 2а2 (2) (χ Υ + (Ху + а [Note that A is less than or greater than 1 according as Ρ lies in the northern or southern hemisphere; λ Φ 0, so this patch omits the north pole.] Together, (1) and (2) yield the chart ε = +1; the chart ε = —1 is obtained by changing α to -a in the above (stereographic projection from the south pole). VECTOR FIELDS ON MANIFOLDS 13.16 The hypcrboloid of one sheet 4(j/')2 + 4(у2)2 - (y3)2 = 16 is a C°° 2-manifold M, by the coordinatization (k - 1, 2) Iyl - 2 cos xl cosh x2 y2 = 2 sin x1 cosh x2 ((fe - 2)77 < x1 < fetr) y3 = 4 sinh x2 with U(1) - ир and ρ = (2, 0, 0), U(2) = U? and q> = (-2, 0, 0). Represent the vector field on Μ given by (V"') = (4 sinh jc2, 4 cosh x2) in terms of (a) a vector basis for the tangent space Г (М), and (£>) extrinsically. (c) Describe this field geometrically. (a) By the usual tools of surface theory (Section 10.5): r = (2 cos хл cosh x2, 2 sin χ1 cosh x2, 4 sinh x2) Ел = r-j = (—2 sin x1 cosh jc2, 2 cos x1 cosh χ2, 0) E2 = r2 = (2 cos χ1 sinh дг, 2 sin χ1 sinh x2, 4 cosh x2) V= VEi = (—8 sin x1 sinh jc2 cosh jc2, 8 cos x1 sinh x2 cosh x2, 0) + (8 cos хл sinh x2 cosh x2, 8 sin x1 sinh x2 cosh x2, 16 cosh2 x2) = (4(cos*1 — sin x1) sinh 2x24(cosx1 + sinхл) sinh 2x2, 16cosh2 jc2) (fr) From the equations for y\ у2, у3 we may calculate: cosh x2 = i VC/V' + C)'2)2 sinh *2 =!/ ι У1 ■ г У2 cos χ = ,——— sinx - V(?)2 + (/)2 V(/)a + o2)a so that £, = (-/,/,0) ,1 .3 2 3 ^шттт-'штгт-2^177^' (V') = (/,2\V)2 + (/)2) V^V%^(-y2y\yy,0) + (yy,yy,4(y1f+4(y2)2) -(у\уг~у2),у\уг + y2),(y3f +16) (using the equation of the hyperboloid). Hence, in terms of the coordinates (y'), V=^ = y3(y1-/)dy1 + y\yl+f)dy2 + [(y3)2 + l6\dy3

208 TENSOR FIELDS ON MANIFOLDS [CHAP. 13 (c) See Fig. 13-8 and note that the first component is zero in the plane y1 = y2 Hence, along the curve of intersection, the field is always parallel to the y2y 3-plane. Similarly, along y1 = -yz, the field is parallel to the y1y3-plane. On the circle y3 =0 the field is (0,0, 16), or vertical. Since the third component is S16, there is always a vertical component. Fig. 13-8 13.17 Show that the restrictions of (a) σ-j = yl dy2 - у dy1 and (b) σ2 = (у -у3) dyl ~ (у1 + у3) dy2 + (у1 + у2) ауъ to the sphere (у1)2 + (у2)2 + (у3)2 = α2 are vector fields. (See Fig. 13-9 for a graph of selected values of σΡ) By the well-known "Hairy-Ball Theorem" (every head of hair has a cowlick), every continuous vector field on S2 (and also on S", for all even integers ή) is zero at some point on the sphere. In fact, the field must vanish at some point of an arbitrarily selected, open hemisphere, (c) Find the zero points explicitly. -*r

CHAP. 13] TENSOR FIELDS ON MANIFOLDS 209 (a) The normal vector to S is ω = 2y dy1 + 2y2 dy2 + 2уъ dy3 and 1 2 σ, · ^ ω = (-/)(/) + (/)(/) + (0)(y3) = 0 1 (ft) σ2· ϊ ω = (/ - y3)( у1) - (/ +y3)(y2) + (у1 + y2)(f) (с) If ^=0, -у2**/ =0 and 02=02 + (>>3)2 = a2, or у3 = ±α. Thus, the zero points are (0,0, ±a). For σ2 = 0, /-уз = у+уз = /+/=0 ^ ^^„y and (y1)2 +(y2)2 + (У)2 = а2 = 3(У)2, or / = ±a/Vb. Hence, the zero points are ±(аЛ/3, ~fl/V3, -a/V3). 13.18 Consider a manifold whose coordinatization is not easily determined (SO(rc), of Example 13.2(e), is such a manifold in R" ) for which, therefore, base vectors r, = Et for ^(M) are unavailable. Develop a reasonable definition of ГДМ) in this situation, which possesses the salient properties of a "tangent space" at point p. To get an idea of what may be desirable, examine the case when the vectors r15 r2,. . . , rn are available. Each tangent vector has the form V= V'r,., and when V is the tangent vector of a curve <€ on Μ—the image of <€' : x' = x'(t) in the coordinate space R"—then dx' ■ dx V= -г- r. or V = —r dt ' dt Thus (V') is a direction vector. Recall that r = r(x\ . . . , x")^r(y\x\ . . .,x"), y\x\ . . . ,x"),..., y"\x\ . . .,x")) dy1 dy2 dy" whence ., , , , ,,·.., ; \ Эх Эх дХ Thus when we write V'r, we are actually indicating the m directional derivatives y'SsV/'y (!=/ =m) with each y'(x\ x2, . . . , x") a C°° real-valued function (defined on Μ if we identify the points of Μ with their coordinates (x') in R"). It is customary to let the directional derivative of a function/ : R"^R in the direction V be denoted V(/)»V/-V Thus, each vector V maps a differentiable, real-valued function f to its directional derivative in the direction V. The properties of this mapping are immediate: If / and g denote any two differentiable functions from Μ to R, with fg denoting the ordinary product of two functions, and if a and b are two scalar constants, then linearity V(af +bg) = aV(f) + bV(g) Leibniz' rule V(fg) =V(f)g + fV(g) With this in mind, and armed with the knowledge that the directional derivatives of all functions on Μ would be enough information to construct the basis {r,} when r is known, we frame Definition 9: By C°°(p) will be understood the real-valued C°° functions on U such that any two functions that agree on some neighborhood of ρ are identified. Definition 10: The tangent space Tp(M) at ρ is the set of all mappings Vp : C*°(p)^>R that satisfy for all a, b in R and /, g in C"(p) the two conditions

210 TENSOR FIELDS ON MANIFOLDS [CHAP. 13 (i) Vp(af+bg) = aVp(f) + bVp(g) (ϋ) VfUs) = VpU)E + fVp(g) with the vector-space operations in ГДМ) defined by (Up+Vp){f)^Up(f) + Vp(f) (aVp)(f)^aVp(f) Any У in Γ (Μ) will be called a tangent vector to Μ at p. This definition has the advantage not only of dispensing with coordinates, but of enabling one to extend naturally a mapping F : M^N (from one manifold to another) to a mapping F* : Tp(M)^> Tp.(S) at each point ρ in M, where p' = F(p). Such an extension cannot be accomplished using the more elementary definition. Remark 2: The vectors of Τ (Μ) as originally defined, if regarded as mappings on C"(p), are members of the abstract Tp(M) (Definition 10). In more advanced treatments it is shown that the reverse is true and that dim rp(M) = dimM= n. Hence, the two approaches to tangent spaces are equivalent. TENSOR FIELDS 13.19 Show that tensor fields always have the property of being bilinear with respect to scalar functions (as well as to scalar constants), unlike differential operators. We must show that for any scalar function / on Μ and any tensor Τ of type (°r), Г(У,,. . . , fVt, . . . , Vr) =/ Г(У1; . . . , Vn . . . , Vr) This is true, since it is true at each point ρ of M: W, . . . , fVn . . . , Vr) = T(Vlp, ..., f(p)Vlp, ..., Vrp) = f(p) T(Vlp, ...,Vtp,..., Vrp) = /rp(y1,...,y,.,...,yj 13.20 Show how to interpret the tangent vector to a curve on a surface S as a (contravariant) tensor of type (o). Let с = c(i) be a given curve on Μ = S, with C*{) dt эх' dt Define for any one-form ω = at dz the linear mapping from Γ*(Μ) to R: dx I dx T(a>) = at^^a>Kdt Under the standard basis {dz\ dz2,. . . , dz") of T*(M), and with ω = dz' sjj. dz, ^ „, , 1ч „г dx' dx' T^T(dZ') = S'- = - (We saw earlier that the dx'ldt were contravariant components.) 13.21 Show how to interpret the gradient of a function as a tensor of type (?). Let /have gradient V/= (dfldx'). Define the linear mapping Г(У) = у<4 (4 fixed v ' dx' \dxl Use the basis {£,, E2,. . . , EJ for ГДМ); then with V= £,. = 5f£y, Г1.= Г(£|) = в; Д = 4 v "; ' dx' dx

CHAP. 13] TENSOR FIELDS ON MANIFOLDS 211 Supplementary Problems 13.22 The set of all 2 x 2 matrices of the form Γ±1 0" L 0 ±1_ where all possible combinations of signs are taken, forms a four-element subset of GL(2, R). Is it a subgroup? 13.23 Prove that SU(n), the set of all η x η matrices over the complex numbers having determinant +1, is a subgroup of GL(n, C). [Hint: det AB = (det v4)(det B) holds for complex matrices.] 13.24 Show that the operator L(f) = J^ f(x) dx is a linear functional over the set of continuous, real-valued functions on [0,1]. 13.25 In terms of the standard basis {dx'} of (R3)*, a new basis is defined by β1 = dx1 - 2 dx3 β2 = 2 dx1 + dx2 β3 = dx1 + dx3 Find the corresponding dual basis {b,.} for R3 in terms of (ег), using (13.6). Check your answer by making several calculations of the form ω(υ) = ώ(ϋ) (a change of basis does not affect the value a linear functional assigns to a vector), 13.26 Consider a tensor Γ(ω; ν) over a vector space of dimension η and its dual, with components T'r (a) Show that the trace τ(Τ) = T\ is invariant under changes of bases, (b) Find τ(Τ) for the tensor defined by Γ(ω; ν) = ω (ν). 13.27 Show that every metric tensor G induces a one-to-one mapping (which is an isomorphism, since it is linear) G : V—>V* from a vector space to its dual, under the definition: For each fixed u in V, let G(u) be the linear functional G(u)(v) = G(u, v), for all ν in V. This proves for vector spaces of arbitrary dimension: Theorem 13.3: If V possesses a metric tensor, then V is isomorphic to its dual V*. 13.28 Find a convenient atlas showing that the set in R4 given by the equation / 1\2 . / 2x2 ι / 3\2 / 4x2 2 (У ) +(У ) +(У ) ~(y ) =a can be made into a C°° 3-manifold. [Hint: Use radicals, as in Example 13.11(b); here, 6 charts will suffice.] 13.29 Show that the restriction of а = уг dy2 - y2 dy1 + y3 dy4 - y* dy3 on R4 to the sphere S3 is a nonzero vector field on S". 13.30 Extend Problem 13.29 to the sphere S2Ar_1 (fee2). 13.31 Show that if there are only two points, pt andp2, on S2 where a vector field is zero, those points must be antipodal (endpoints of a diameter). 13.32 Show, by geometric reasoning, that there exists a continuous, nonzero vector field on the torus. 13.33 Show that the restrictions of the following one-forms to S4 (y1)2 + · · · + (>5)2 = 1, are vector fields on S4, and find the points where they are zero: (α) σ = >'2 dy1 - y1 dy2 + y4 dy3 — y3 dy4 (b) a^(y2-y3-y4)dyl + (y3-yl)dy2 + (y1 ~ y2 + y5) dy* + /dy4 - y3dy5

212 TENSOR FIELDS ON MANIFOLDS [CHAP. 13 ■3L34 Although no nonvanishing continuous vector field exists on the 2-sphere S2, there are three, mutually orthogonal, unit vector fields on S3 CR4. These are, in the extrinsic representation of S3, σλ = -у1 dy1 + у2 dy2 + / dy3 - у3 dyA 3jl 4 j 2 , 1 j 3 , 2 τ 4 σ2=-y dy — у dy + у dy + у dy σ3 = -у4 dy1 + у3 dy2 - у2 dy3 + у1 dy* Show this. [Note: Manifolds with such vector-field bases are called parallelizable. The manifolds S\ S3, S7— and no other n-spheres—and the torus are examples.] 13.35 Without resorting to coordinate patches, express extrinsically the collection of tangent spaces Г(М), if Μ is the hyperboloid of one sheet (y1)2 - 4(y2)2 + 4(y3)2 =4. 13.36 For the manifold Μ of Problem 13.35, consider the coordinate patch /=*' y2=x2 / = Vl~(*V2)2+(x2)2 valid for y3 >0. Find an expression for an arbitrary vector in Tp(M). 13.37 One way to show that two surfaces meet at right angles is to show that along the curve of intersection the normal vector to one lies in the tangent space of the other. Illustrate this idea for the sphere (У1)2 + (У2)2 + (У3)2 = 16 and the cone (уУ = %У1)2 + %y2f> the latter coordinatized by 11 2 2 3 -,, Г/ 1ч2 , / 2ч2 у =х у =х у =ЗУ(дг ) +(х )

Answers to Supplementary Problems CHAPTER 1 1.15 Ajfrj + агЬг + a3b3 + a4b4 + a5fr5 + аьЬь 1.16 R\kl + R2jk2 + Rjk3 + Rjki. The index i is a dummy index, while / and к are free indices; there are summations. 1.17 Xj 1.18 (a) n; (b) δ,,δ,, = δ„ = η; (с) diftI = с„ = cn + с22 + с33 + · · · + с„„ 1.19 ai3ba (n=3) 1.20 α,.^,.^. (η = 3) 1.21 у,. = с/Л (η = 2) 1.22 alt (fc = 2,3) 1.23 ±-^ацх^ац^{х^а>Ъ^а1к 1.24 alk[(Xt)2 + 2jcl.jc/t] [not summed on fe] 1.25 (a(j/ + fl,v/ + a,-,-,)*,*,- 1.26 aw + a;/t 1.27 (a) b)T^ (b) atlbjrx,\ (с) ^ААА^Л 1.28 (с) α;/(χ. + Xj) = a,.,(£/x, + ε,χ,.) = β,,.ε,.*, + β,,.ε,.*,. = α,.,ε,.χ, + β,-,ε,·*,. = 2fli/£,.*,. CHAPTER 2 2.24 (α) and (b) 2.25 11 12 13 14 15 и и и и и 21 22 23 24 25 и и и и и 31 32 33 34 35 («О 11 12 13 5" 0 5. (с) и 21 и 31 и 41 и 51 _и (Ь) 1 .2 и 22 и 32 Μ 42 Μ 52 Μ 2 2 и 23 ы 33 43 и 53 и -4" -2_ (<*) "10 0 0 0 0 0 10 0 0 0 .001000 2.29 (а) 17; (&) 0; (с·) -1 2.30 (а) -а12а21а33а44 + а12а21а34а43 + а12а23а31а44 - a12fl23fl34fl41 - fl12fl24a31fl43 + a12fl24fl33fl41 a„ a„ a,, а12Л12 21 "23 "24 ^31 33 fl:<4 «41 fl43 fl44 213

214 ANSWERS TO SUPPLEMENTARY PROBLEMS 2.32 И (b) 1 3 1 -4 .3 2 2.33 One need verify only that (i) interchanging a pair of consecutive indices changes the sign of a single factor in the product; (ii) ΠΡί=Πι = ι р>я \p-q\ 2.34 2тт73 2.35 One pair are (2, 3, 0) and (-3, -2, 5). 2.36 2.37 Q = xx + 2x2 — хъ + 8xlx2 + 6*j*3 -3 -k ~\ 3" -\ -1 0 0 2.38 Л ^ 0 1 0 3 0 0 0. 2.39 c, = crbrl, where (fr„) = (я,)"1. 2.40 g11 = 13/49Jg12 = g21=4/49,g22=5/49 2.41 «/(Ϊ, y) = 3 = d(x, y) CHAPTER 3 3.23 (a) / = ~2exp(2jc1)<0 \х1 = ±1п(х1х2) <*> *-1 : l**-Sin(*V) <*·* >0> (с) / = 1/2* 1/2* 1/2* -1/2* J exp (χ1 + χ ) exp (x1 + x2) exp(x* ~ x2) —exp (x1 — x2) 3.26 -1 = 0, so that /(*, у) = Дг) = /(V^+y2) = S(x2 + У*)- 3"29 8^^ = ^iF = 8J = 8> 3.30 The inverse Jacobian matrix at (1,2) is χ χ 0 1 By Problem 3.14(a), covariance of the matrix would imply the matrix equation or which is patently false. 0 1 1 0 "2 01 .1 1_ 0 1" -1 oJ " 0 1 .-1 0 " 0 2 _-2 0 2 1 .0 1 ■?J] ;]

ANSWERS TO SUPPLEMENTARY PROBLEMS 215 3.32 (a) (7" + T') represents a tensor if and only if Эх' Эх" Эх' дх> Эх1 дх K s r) 3xr эх1 5 3xr Эх'' ' дх дх' which requires that JTJ = J TJT. This last relation, in turn, generally requires that J — JT; i.e., J must be an orthogonal matrix. (b) f = JTJ, so that fT = Τ if J = JT. 3.35 As Τ is a tensor (Example 3.4), it is an affine tensor: f' = drTr. Thus, dT _ , dT dt ~a' dt showing dTldt also to be an affine tensor. Any affine tensor is a fortiori a cartesian tensor. 3.36 (a) (7,(7, = (airur)(alsu.s) = airalsurus = drsurus = urur (b) No, because distance and angles are not preserved under arbitrary linear transformations. Specifically, consider χl = Зх1, χ2 = χ + χ2. A scalar product in (x') is uivi = (3hj, «j + u2) · (3i)j, υι + v2) = IOhjUj + u1v2 + u2v1 + u2vz This clearly will not coincide with u1v1 + u2v2. CHAPTER 4 4.19 Write [ST] = (U','*m). There are (\) ways of choosing locations for the contraction indices и and υ among the contravariant indices, and, for each of these, (2) ways of choosing locations among the covariant indices. A given quartet of locations can be filled in 2 inequivalent ways. Thus, the desired number is 4.23 First, use the device of Problem 4.11 to establish that T'Jk!UkV' are tensor components for all (t/') and (V); then apply the Quotient Theorem twice. CHAPTER 5 5.21 L = απ; semicircle of radius a. 5.22 No: Q(1,0, 3) =-1. 5.23 L = 2 + e 5.24 The true distance formula, PlP2 = У/(х\ - x\f + (x\ - x\f - 0.2021125(^5 - x\)(x\ - x\), yields 4.751. for an error of +0.249. 5.25 G ■■ 5.26 (t/,) - (0,1, 0), (V:) = (jc2, jc1, 0) (x2f xLx2 0 xlx2 1 + (x1)2 0 0 0 1 5.27 (fl) ||U+V||2 = (U + V)2=U2 + V2+2UV= ||U||2 + ||V2|| + 2 ||U|| ||V|| cos θ (b) Take θ = π/2 in (α). 5.28 (α) χ2 = Cexp (~2bx3/a ) (a one-parameter family of spirals on the cylinder x1 = a)

216 ANSWERS TO SUPPLEMENTARY PROBLEMS (b) No: the curves of (a) have tangent field V all along their length; but, for orthogonality, it is necessary only that the tangent at intersections with the pseudo-helix be V. For example, the curve x2 — x3 on xx - a is also orthogonal to the pseudo-helix at the point x2 — -a2/2b, л3 = a4/4b. 5.29 xl = аещ>(-(х2)212) (d = const.) 5.30 f'(0o)g'(da) = —a2 at intersection points. 5.32 (a) g'a = λ.(α)δ'α, which is tantamount to g'1 = gtj — 0 for i Φ j. 5.33 ||V|| = 1, L=tt/2 5.34 χ =α, x3 — b cot x2 + с (c~ const.) CHAPTER 6 6.19 x' = - a'rsxrxs + b'X + c' (the b) and c1 constants) ΙόζΛ:1)2 + 1 4л:1-3' Ax1 - 3 10 6.20 (a) G (b) Гш = 16л:1, ГИ2 = 4, all others 0 6.22 The values, in (x'), of the dx'ldx' are easiest found by inverting J'= (dx'ldx1). Final results are: Г1 =Г1 =Г2 =Г2 =1 1 11 J" 22 J 12 J 21 * 6.23 From Problem 6.21(b), T'jk — 0 for /' Φ к; while, for /' = к — a (no summation on a), "" дха \дх"1 дх \ " Эх I дх " " 6.26 1 221 — 1 212 — Fi22 ■* > ^ 21 ^ 12 ί 'X , 1 22 Х ajexpi1 2α1 ехр 2л ЗазехрЗл" 6.27 β= α2ехрл1 2а2. схр 2л За3ехрЗл" 3 -1 ι 3 л -2 -, 3 „ -ь -3 ^схрл 2а2ехр2л За3ехрЗл Hence the condition is det (а'^ФО. 6.29 x' = A'x^inx2 + B'x1 cosx2 + С (ί = 1,2), with A1 B1 A2 B2 -- [6 ехр (л1 + 2x2 + Зл3)] det (а))Ф0 Φ0 for a bijection. 6.30 No, because of the presence of Christoffel symbols in (6.7). 6.31 Τ дТ\ jrs.h дХ к -1 uA /лг /A wrs rfc ^ ;'i/5 .?/: jru 6.36 κ=1/ί> 6.37 (a) d и d ν = 0 (fr) л2 ^^(x1)2 + q (a two-parameter family of "parabolas")

ANSWERS TO SUPPLEMENTARY PROBLEMS 217 6.38 (a) dj2 ,__._2 2Jdx3^ ds2 ι ■ 2 2ч/ ах \ (sinjc cosjc)I-t-I =0 . ± dx2 dx3 + (2cotx )-t--7-=0 ds2 ds ds (b) x2 = -s x3=0 a (c) The solution (b) represents an arc of a particular great circle (x2 + z2 = a2, in the usual cartesian coordinates) on the sphere. By symmetry of the sphere, all great-circular arcs, and only these, will be geodesies. CHAPTER 7 '+1 0<|i|Sl/2 7·24 ε~' -1 M>l/2 7.25 / = 0,1 7.26 L = 8V2/3 7.27 / = V5/3 [z = 0, which makes r = |g,v|=0, is disallowed] 7.28 L = (64 + 11VTT) /216 «0.465 7.29 0 = iln2 at (0,2,0); θ = cos-1 (7/4VTT) at (5,2,3) 7.30 (a) L = 8(1 +3V3)« 49.57 (b) xl = 3(as2'3 + 4), x2 = (σί2/3 + 4)3/2, where i+1 -8Ss>0 σ~ί-1 0£sS24V3 (c) The null points are Ζ = 0 (s = —8) and / = 1 (s = 0). 7.31 L = 8(5V5-1)« 81.44 7.32 For s 5^0, T = (2|i|"1/3,Vir + 4i-2/3) and ||T||2 = \-σ\ = +1 7.33 Ν1 = Γ2, Λί2 = Τ1 7.34 For s^-8,0, 3s\W,3+4 7-36 Ko = |k|=3|J|(J2'3-4)1'2 (^8) At the null point (0, 0), both Euclidean and Riemannian absolute curvatures become infinite; but at the null point (12,8), only the Riemannian curvature becomes infinite. 7.37 T= 1 — 4/| 1/2(1,2/), N=|l-4z|"1/2(l,l-2z), к =2 |1-4z|"3/2 7.38 (a) L = a. (b) L = 3a/2. (c) Riemannian: T= |cos2z|~1/2(-cos Z, sin ζ), κ0 = (2/3a)(csc2z)|cos2z|"3/2; Euclidean: Τ = (-cos Z, sin Ζ), κ0 = (2/За) esc 2ί 7.39 (a) Tln = \!2x\ Г22 = 1/2*2 others zero

218 ANSWERS TO SUPPLEMENTARY PROBLEMS CHAPTER 8 8.16 By Problem 6.34 and (8.1), П, ~ V\lk = g'\Vrkl - VrJk) = girR'rklVs = 8"(8«R'rn)V = (g"R,rkl)V = -R'MV 8.22 K = 1/4(jc1)2 x'+x1 8.24 (a) and (b) К = 4(*1)V(l+2*2) (c) ϋ(2) = -ϋ(1)+ν(1),ν(2)=ϋ(1)+ν(1) 8.25 Κ=1/α2 8.26 Basic sets of nonvanishing terms are: r,2 \S\) K1212 — 4 \^7 r J> «1313 ~ 4 > «2323 _ 4\ / and (A) Gi2i2=/g, G1313=/A, G2323=g/i so that ρ ту _|_D W _|_D W / ч τ/γ 2 |т -уч _ «1212 "1212 τ «1313 " 1313 τ «2323 ''2323 {) **x>v,v)- fgWi2i2 + βψ^ + f AWr23a , -2(ln|/|)"(W1212 + W„,3)-(ln|/|y2W14„ 8.27 (a) K(x2;U,V)= lAiL^izi? 2зму__ ν щ; тз Τ/(."1212 """ "1313 """ "232з) (Ь) д= 2/3 8.28 Isotropic points compose the surface χ = е~зп, over which K = 2e3/27. 8.29 К =-1/4 8.32 Rn=-1, K12=K21=0, R22=sin2jc1; «J =-1/a2 = Я2, Я2=0 = Я2; Д =-2 8.33 Дп=Д22 = Д33=2/(х1)2 others 0; «*| = A*2 = Й3 = 2, others 0; A* =6 8.35 gi} = (x1 )\ has Д = 0, К φ 0 (use Problem 8.27). 8.36 No implication either way. CHAPTER 9 9.17 (a) u0 = ±Vjc'jc2 + α (α = const.); (fr) incompatible 9.20 flat, non-Euclidean 9.21 Euclidean 9.22 (+ + -)

ANSWERS TO SUPPLEMENTARY PROBLEMS 219 9.26 With the notation / = dfldx1, for any function /: Φ, 1 xx (χ1)2 n2_-v( Фи Φΐ.ΨΐΦΐ,Ψι ΦΛ,-*(<Ρ44,<Ρ4 <РА\ пЪ 1 (χ ) + e -J Pi 1 Ix1 (x1)2 G\ = -<pAe->lxl G\ = <pAe-*lxl CHAPTER 10 10.30 (a) The curve lies on a right circular cylinder of unit radius, beginning at the point (1,0,1) and rising in helix fashion, approaching °° asymptotic to the vertical line χ = cos 1, у = sin 1, as i—» 1. (b) 10.31 16/3 L = — Ц— dt« 1.13209039 Jo 1-i) 10.32 (α) Τ = (—(ale) sin (sic), (ale) cos (sic), blc). Hence the tangent line, r(t) = r + iT, has the coordinate equations s at . s . s at s bs bt χ — a cos - — — sin - у = a sin - + — cos - ζ = 1 с с с с с с ее (b) Q corresponds to t = —s and PQ — || — iT|| = s. (c) The interpretation is that Q can be thought of as the free end of the taut string as it is unwound from the helix. [The locus of Q, r* = r(s) — sr'(s), is called an involute of the helix.] 10.33 10.34 tl\t\ τ ||T'|| ~ (1 + 25ί8) = 20f3V2 K"(l+50i8)3''2 171 (~5 Λ 1,0) 7=0 10.35 Let the curve r = r(s) lie in the plane br = const., where b = const, and ||b|| = 1. Differentiate twice with respect to s: bT = 0 and bT' = 0; hence, b(«N) = 0 or bN = 0. It follows that b = B, the binormal vector, so that B' = 0 and τ = -B'N = 0. Conversely, if τ — 0 for a curve r = r(s), then B' = ~rN = 0 and В is a constant unit vector. Define the function β(^) = Β· (r(i) ~ r(0)); we have (2' = Br' = BT=0 whence Q — const. = Q(Q) = 0. Therefore, the curve lies in the plane Br = Br(0) = const. 10.38 £ = |jc1| Va2 + 1=0 atx^O. 1 10.39 £ = a coshx^O, n = cosh χ γ (~cos x2, —sin x2, sinh x1) 10.40 f V5i4 + 1 1 9-V6 + ln^ (V6 + 1)

220 ANSWERS TO SUPPLEMENTARY PROBLEMS 10.44 II =ff. f'i (dx1)2 + {g (ώ:2)2 Ю.46 (e) (o K^[1+4flV)2]2, н=[^?Ш гл „ ^ ,r 4fl[l+2fl2(f')2+2fl2(f2)2] W K [l+4fl2(f,)2+4a2(f2)2]2' И [l+4fl2(f1)2+4fl2(f2)2]3/2 (b) Consistent with the invariance of К and H, the change of parameters χг = xl cos χ2, χ2 = χ1 sin jc2—i.e., the transformation from polar to rectangular coordinates in the parameter plane— takes the forms (i) into the forms (ii). 10.49 (a) r* = (a sech x\ 0, ax1 - a tanh xl) 10.50 The two FFFs correspond under the mapping χί = a sinh xl, x2 = x2. CHAPTER 11 11.14 (a) D=Vl + csc4i -> V2; (b) a = Vl + 4 esc4 t cot2 t-> 1; (c) maxu=V5, max α = VT7 (no minima) 11.15 Rectilinear motion [use (11.8) to show that к must vanish]. 11.16 From (11.7) and (10.9), a= -kVT + kd2N + κτν3Β. „ ,„ ! d2p . 2 4/d0\2 /ί/<ρ\2 2 ά2ψ 2 dp άφ , . Ύ^\2 11.17 α = —-у - (ρ sm <p) ~r I ~ Pi ~r I . fl = —rr + ;—τ — (sin φ cos <p) — . dt2 κμ ψ'\άί) H\dt/' dt ρ dt dt v ψ Ύ'\άί/ ■ α3 = —- + - ~ — + (2 cot <ρ) — "τ ί/ί2 ρ dt dt v ry </ί </ί 11.18 Let the center of force be the origin of rectangular coordinates for E3, with the particle's path given by r = r(i). By Newton's second law, /r = mi, so that d / -4 " if 7 ГХГ =ГХГ = ГХ -Г dty ' \m and rXr = p = const. It follows that ρ·r = 0. 11.19 vy-^4+4 + ^ <9r r2 8Θ dz2 r dr CHAPTER 12 12.34 (a) timelike; (b) spacelike; (c) lightlike 12.35 Yes: travel at 4167 mi/sec <s c. Timelike interval. 12.36 Premultiply ATGA = G by AG, and postmultiply by A~lG. 12.38 (a) t = a0t, χ = ~a\ct, χ = ~a20ct, x3 = —a\ci. (b) a°0 > 0 if ί and ί have the same sign; that is, if the clocks of the two observers are both turning clockwise or both counterclockwise. !-> in -» 0 1 -1 0 , 3 1 -2 2 -3 3 12.39 χ = - χ ~ - χ χ =-^ χ +- χ χ ~χ χ =x

ANSWERS TO SUPPLEMENTARY PROBLEMS 221 12.40 (b) zero 12.41 5/4 -3/4 0 0 -3/4 0 0' 5/4 0 0 0 1 0 0 0 1. «1 1 0 0 0 0 1 0 0 0 0 4/5 3/5 0 0 -3/5 4/5 «, 1 0 0 0 0 -2/3 1/3 -2/3 0 -1/3 2/3 2/3 0 " 2/3 2/3 -1/3 J 12.42 12.45 12.47 12.48 12.49 12.50 12.52 12.54 V d = (3/5)c 1. d = (4/5)c Approximately 25% slow. About 45 years old. «17 000 mi/sec For constants F and a = Flm, and with F = (F, 0, 0) and \ = (vx, 0,0), (12.29) becomes identical with (1) of Problem 12.26. Since Ss'/Sx' — 0— ds'ldx' (the equation of continuity), (sl) may be identified with the vector (5") of Problem 12.32. (a) By analogy with the evaluation oi{ejjklPkl in Problem 12.53, Hl -H2 -#3 ... ^ ^ ,.0 F — F fig /„φ \1 = " ^03 -*-U2 = V D3 D2 =[M 0 * * * φ ^23 0 * * Φ» -φ03 0 * ~Φΐ2 Φ ^U2 -Φ 0 . 0 * * * ~Η, 0 * * -Η2 Ε3 0 * -н3 ~Ε2 Я, 0 (b) Let (abed) denote a permutation of (0 1 2 3). Then ФаЬ eabcdFc (no summation) and *Ф. дхс дФс, дхь дФь дха dFcd гаЬЫ дхс idFc '"ьы\ дхс dFbd dFa Эх" дха bcai дх« dFid Эх' ■Q The second set of equations is derivable directly from (12.45b), the definition of Ф, and the fact that the gtJ are constants. CHAPTER 13 13.22 Yes; it is isomorphic to the 4-group. 13.26 (a) By (13.6), t; = T(p, ь,) = τ(Λ'β', А*ък) = А'^цр*, bj=β*η = г; (b) т(Г) = л 13.27 Suppose that о(и:) = G(u2). Then G(iij, v) = G(u2, v), or by symmetry, G(v, uj = G(v, u2) for all ν By nonsingularity, Uj = u2.

222 ANSWERS TO SUPPLEMENTARY PROBLEMS 13.28 -1 у^Уа2-(х>)2-(х2)2 + (х3)2 У =x y-=x \y -=x /=*' Vp:yl>0 p = (fl,0,0,0) / = ±ν«2-(^)2-(^)2 + (^)2 у =x 4 3 у = x u„ u_ <7 = 9 ' = (0 y2>0 >'2<0 a, 0,0) Ψ±τ 1 л / 2 / К 2 / 2~ч2 , / 3x2 у = -Л/а - (х ) - (* ) + (л ) 2 1 1 1 2 2 и, и_ г = г = (0 у3>0 у3<0 Ο,ω,Ο) >·3 = ±ν«2-(^)2-(^2)2 + (χ3)2 [у 13.30 (7 = /i/y1 - y'^+Zdy3 - у3 dy' + y6dy5 - y5dy6 + ■■■ + y2kdy2k-1 - y2k-1 dy2k 13.31 If p1 and p2 are not antipodal, there exists a closed hemisphere containing neither one, on which the given (continuous) vector field is nonzero—an impossibility by Problem 13.17(b). 13.32 As shown in Fig. 13-10, let a unit tangent vector be constructed to a generating circle; as the circle is revolved to generate the torus, the tangent vector is obviously propagated continuously to all points of the torus. Fig. 13-10 13.33 Zero points are: (a) (0,0,0, 0, ±a); (b) ±(0, a/V3, 0, a/V3, alVb). 13.35 With ω =2(y1 dyl - \y2 dy2 + Ay3 dy3), σ = f dyl + g dy2 + h dy3 must be orthogonal to ω, for C" functions /, g, h. Hence, ylf-*y2g + *y3h=Q Replace /, g by 4y3F, y3G, and solve for h. Similarly, replace g by y1G and h by у:Н and solve for/; etc. All possible tangent vectors are given by one of three distinct types (F, G, Η denote arbitrary C" functions of y\ у2, у3): (1) a = 4y3Fdy1 + y3G dy2 + (y2G~y1F)dy3 (2) a = (4y2G-4y3H)dyl + ylG dy2 + ylH dy3 (3) v=\y2Fdy' + {ylF+y3H)dy2 + y2Hdy3 13.36 U = СГг, = (2i/V4 - (л:1)2 + 4(x2)2, 2U2\'4 - (x1)2 + Цх2)\ -χ'ϋ1 + 4x2U2) for any two C°° functions U1, U2 on (x1, x2). 13.37 The normal vector to the sphere is represented by а = уг dy1 + y2 dy2 + y3 dy3; the tangent space of the cone is given by Set Ul=yl and U2=y2 (t/,, U2, (З^и1 + 2>x2U2)({xf + (χ2)2)-1'")

Index Absolute derivative, of a tensor, 72 uniqueness of, 72, 80 partial derivative on 2-manifolds, 112 Acceleration, in curvilinear coordinates, 73 nonrelativistic, 169 in rectangular coordinates, 154 relativistic 4-vector, 170 Admissible change of coordinates, 25 Affine coordinates, 12, 23 Affine invariants, 31 Affine tensors, 31 Angle, between curves on a surface, 135 between vectors, 10, 56 Antisymmetry, 9, 204 Arclcngth, of curves in space, 51, 53, 129 of curves on a surface, 133 Arc-length parameter, 53, 128, 129 Atlas, coordinate patches of, 195 for a sphere, 195, 196 Associated metric {see Conjugate metric tensor) Associated tensor {see Raising, lowering indices) Barred, unbarred coordinates, 11 Basis, change of, 30, 193 dual, 192 of R3, 10 standard, 10, 61, 190 (for R") of a vector space, 190 Beltrami's theorem, 138 Bianchi identities: first, 102, 108 second, 119 Bijection, 12 Bilinear functional, 193 Binormal vector, 129 Cartesian coordinates (rectangular), 13, 23 Cartesian invariants, 31 Cartesian tensor, 31, 32 Catenoid, 145, 153 Cauchy-Schwarz inequality, 57, 63, 83 Central force, 156 gravitational, 157 Chain rule for partial derivatives, 13 Chart, 195 Christoffel symbols, definition, 68, 70 for cylindrical coordinates, 81 for polar coordinates, 70 for spherical coordinates, 69, 75 transformation laws, 69, 71 Circular helix, 141 Compatibility conditions, for partial differentia equations, 115, 116 for zero curvature, 123 Components: of vectors, 23 of tensors, 26 physical, 161, 163 Composition of velocities (relativistic), 169 Cone: circular, 145 light, 164, 165 Conjugate metric tensor, 55 Constant curvature, 138, 153 {See also Zero curvature) Continuity equation, 173 Contraction, of indices (tensors), 173 of length (relativity), 169 Contra variant; tensor, 26, 29 vector, 26, 28 Coordinate axes (general), 66 Coordinate patch (frame), common notation for 198 for a manifold, 195 for a sphere, 195, 196 Coordinate system: admissible change of, 25 affine, 13, 23 Cartesian (rectangular), 13, 23 curvilinear, 13, 23 cylindrical, 24 general, 26 normal (geodesic), 117 polar, 23 spherical, 24 transformation of, 25 Coordinates {see Coordinate system) Coriolis force, 155 Co variant, derivative of a tensor, 71, 72 Covariant tensor, 27, 29 Covariant vector, 27, 28 Cubic curve (twisted cubic), 128 Curl, 157, 158 Curvature: of a curve, 68, 73, 86, 87, 130 (formula) Gaussian, 136, 137 (formula) geodesic, 73 intrinsic (for curve on a surface), 135 invariant (Ricci), 106 and isometry, 138 mean, 136, 137 (formula) of normal section, 136 Riemannian, 103^ 105 (formula) zero (Riemannian), 114, 116

224 INDEX Сшгге: definition, 127 length of, 51, 53, 85 parameterization by arc length, 53, 128, 129 null, 84 planar, 129 regular, 85 space, 27, 28, 51, 127 on a surface, 132, 135, 136 Curvilinear coordinates, 13, 23 axis of, 66 Cylindrical coordinates, 24, 51 Christoffel symbols for, 81 definition, 24 metric tensor, 51 Cyclic group, 190 Derivative: absolute, 72 absolute partial derivative on 2-manifolds, 112 covariant, 71, 72 of a determinant, 106 of metric tensor, 78 of tensors, 68, 71, 72, 74 (product rule) Determinant, of metric tensor, 91 of a square matrix, 9 Differentiability class of a function, 23 Differentiable manifold, 197 Differential fields on a manifold, 199 Differential forms (as linear functionals), 191 Differential geometry, 127 Differentiation (see Derivative) Dilation of time (relativity), 169 Dimension of space, 190 Directional derivatives (manifolds), 199 Distance formula, 12 (See also Metric, Length) Divergence, in spherical coordinates, 161 of a tensor, 120, 157 Dot product (see Scalar product) Dual tangent bundle, 199 Dual tensors (see Raising, lowering indices) Dual of vector space, 191, 192 (basis) Dummy index, 1 Dynamics (see Kinematics of a particle) Eigenvalues and eigenvectors, 122, 150 Einstein: invariant, 126 and Special Relativity, 164 summation convention, 1 tensor, 119, 120 Elliptical helix, 128 Energy: Einstein's equation for, 172 Newtonian concept, 156 Energy (continued): relativistic, 185 Equation of continuity, 173 Equations of Gauss, 137 Euclidean space, 54, 114 Event space, 164 Faraday's 2-form, 188 Field (see Matrix field; Vector field) First curvature (see Curvature) First Fundamental Form, of catenoid, 145 of general surface, 133 of helicoid, 134 of right circular cone, 145 of surface of revolution, 145 Flat space, 116 Flow curves (manifold), 198 Force: central, 156, 157 gravitational, 157 Lagrangian formula for, 156 Lorentz (relativistic), 172 Minkowski (relativistic 4-vector), 172 (See also Newton's second law) Forms, differential, 191 Four-group, 190 Four-vector (relativity), 169 Frame (of reference): coordinate, of manifolds, 195 Galilean, 154 inertial, 164 moving, of a curve, 129, 130 (formulas) moving, of a surface, 133 of observer, in relativity, 164, 166 Free indices, 1, 2 Frenet equations, 86, 87, 130 Functional (linear), 191 Fundamental indicator, 53, 83 Fundamental form or tensor, 54, 83 (See also Metric) Galilean frame, 154 Gauss: equation (formula) of, 137 theorem of, 138 Gaussian curvature, 136 Gaussian representation of surfaces, 131 General linear group, 190 Geodesic: as curve of zero curvature, 73 null, 73, 88, 89 as shortest arc, 88 on a surface, 135 Geodesic coordinates (see Normal coordinates) Geodesic curvature, 73 Gradient, 157

INDEX 225 Group: cyclic, 190 definition of, 190 four-group, 190 general linear, 190 Lorentz, 166, 176 unitary, 190 Kinematics of a particle, in curvilinear coordinates, 155 in rectangular coordinates, 154 in Special Relativity, 169, 170 Kinetic energy, 156, 172, 185 Kronecker delta, definition, 3 generalized (permutation symbol), 9 Hairy-ball theorem, 208 Helicoid, 134 Helix, 128, 141 Hermitian matrices, 190 Homeomorphism, 190 Hyperbolic motion (relativity), 183 Identity element of group, 190 Identity matrix, 9 Index (indices): for matrices, 8 notation, 1 range of, 2 tensor notation, 8, 23, 26 Indefinite metric: Cauchy-Schwarz inequality, 83 definition, 83 indicator of, 53, 83 Indicator, 53, 83 Inertial frames, 164 Inner product, as a contraction, 44 of tensors, 43 of vectors, 55 Inner-product space, 55 Instantaneous rest frame, 171 Intervals in space-time, 165 Integral curves (see Flow curves) Intrinsic curvature (curves), 135 Intrinsic properties (surfaces), 134 Invariant: relativistic distance, 165 Ricci curvature, 106 Riemannian curvature, 103 tensor symmetry, 46 zero-order tensor, 44 Inverse, of Jacobian matrix, 26 of a matrix, 9, 10 of a transformation, 25 Involute (of catenary), 153 Isomorphism, 190 Isometry: between catenoid and helicoid, 153 definition, 138 (See also Curvature) Isotropic point, 105, 119 Jacobian, 25 Jacobian matrix, 25 Lagrange multiplier method, application of, 150, 175, 185, 188 Lagrange's identity, 133 Lagrangian form of second law of motion, 156 Laplacian: in cartesian coordinates, 157 in cylindrical coordinates, 163 relativistic, 173 in spherical coordinates, 161 Length: of a curve, 51, 53, 85 of interval (relativity), 165 of a vector, 9, 56 (See also Arc length; Contraction) Light cone, 164, 165 Light velocity, 164 Lightlike interval, 165 Linear: combination (tensors), 43 combination (vectors), 202 functional, 191 group, 190 independence, 15, 190 transformation, 11 Lorentz force, 172 Lorentz group, 166, 176 Lorentz matrix, 167 simple, 168 Lowering, raising indices, 55 Magnitude of vector (see Length) Manifold, definition of, 194, 195 differentiable, 197 parallelizable, 212 Mass, relativistic, 171 Matrices, basic definitions involving, 9, 10 Matrix field, 29 Maxwell's equations, in Newtonian mechanics, 158 in Special Relativity, 173, 185 Mean curvature, 136, 137 Metric (metric tensor): for affine coordinates, 52 for cylindrical coordinates, 51 determinant of, 91 Euclidean, 26, 51, 54, 114 general, 26, 52, 54, 193 non-Euclidean, 26, 114 for polar coordinates, 51

226 INDEX Metric (metric tensor) (continued): for Riemannian geometry, 83 for Special Relativity, 84, 89, 165 for spherical coordinates, 51 Michelson-Morley experiment, 164 Minding's theorem, 138 Minimal surfaces, 153 Minkowski force (4-vector), 172 Mixed tensor, 29 Momentum: Newtonian, 156 relativistic, 172 Motion equations (see Kinematics of a particle) Moving frame (on a curve) 129, 130 (formulas) Multilinear functional, 193 Natural basis (see Standard basis) Newtonian mechanics, 154 Newton's second law, in curvilinear coordinates, 156 in Special Relativity, 172 Non-Euclidean metric, 26, 52, 54 Norm of a vector: Euclidean length, 9 generalized (Riemannian metric), 56, 63 Normal coordinates, 117 Normal vector, principal, 86, 87, 129 Normed linear space, 56 Null: curve, 84 geodesic, 88, 89 point of curve, 84 vector, 83 One-form, 191 Order, of differentiation, 101 of a tensor, 29 Orthogonal: coordinates, 65 families of curves, 57 matrix, 9 trajectories, 57 transformation, 31 vectors, 9 Osculating plane, 129 Outer product, 43 Parallel transport, 135, 149 Parallelizable manifolds, 212 Parametric equations, for curves, 127, 128 for surfaces, 131 Parametric lines, 132 Permutation symbol, 9 Perpendicularity (see Orthogonal) Physical components, 155, 161, 163 Pitch of helix, 128 Planar curve, 129 Polar coordinates, Christoffel symbols for, 70 definition, 23 metric for, 51 Position vector, 28 Positive-definiteness, of bilinear functionals, 193 of inner product, 63 of metric tensor, 52 Principal curvatures, 136 Principal normal (curve), 86, 87, 129 Product: inner, 43 outer, 43 scalar, 43 of vectors, 10 of vector spaces, 191 Product rule, 74 Proper time, 170 Pseudo-orthogonality (Lorentz matrices), 176 Pythagorean theorem, 66 Quadratic forms, 10 Quotient theorem, 45 Raising, lowering indices, 55 Range of free indices, 2 Rectangular coordinates, 13, 23 Regular curves, 85, 127 Regular surfaces, 131 Relativistic: acceleration, 170 energy, 172 force, 172 mass, 171 motion, 170 velocity, 170 Relativity, special theory, 164 postulates for, 166 Rest mass, 171 Ricci curvature invariant, 106 Ricci tensor, 106 Riemann, Bernhard, 83 Riemann tensor, 101 Riemann-Christoffel tensor, definition, 101 symmetry properties, 102 Riemannian: coordinates (see Normal coordinates) curvature, 103, 105 (formula) space, 83 Right helicoid, 134 Scalar multiplication, 8, 189 Scalar product (vectors), 9 Schur's theorem, 119 Schwarz inequality (see Cauchy-Schwarz inequality)

INDEX 227 Schwarzschild metric, 126 Second Fundamental Form, of a cone, 148, 149 definition, 136 of surface of revolution, 152 Second curvature (see Torsion of a curve) Second law of motion (see Newton's second law) Serret-Frenet formulas (see Frenet equations) Signature of flat metric, 117 Singular point (surface), 131 Simultaneity (relativity), 164, 181 Skew-symmetry (see Antisymmetry) Space: of constant curvature, 138 event space (relativity), 169 Euclidean, 54, 114 flat, 116 R", 23 Riemannian, 83 Space components (relativity), 169 Space curve, 127 Space-time, 164 Spacelike interval (relativity), 165 Special theory of relativity (see Relativistic; Relativity) Sphere, coordinate patches of, 195, 196 Spherical coordinates, Christoffel symbols for, 75 definition, 24 metric for, 51 Standard basis, 10, 18, 30, 61, 190 Stcreographic projection, 206 Stress tensor, 30, 31 Subgroup, 190 Subscripts, superscripts (see Index) Suffix (see Index) Sum of tensors, 43 Summation convention, 1, 2, 3 Surface, definition of, 131 Surface of revolution, curvature of, 153 First Fundamental Form for, 145 Gaussian form of, 143, 144 Sylvester's theorem, 117 Symmetry: of bilinear functionals, 193 of matrices, 9 of tensors, 46 Tangent bundle (manifold), 197, 198, 209 (generalized) dual, 199 Tangent plane, 133 Tangent vector (curve), 85, 127, 129 as a tensor, 27 Tensor: cartesian, 31. 32 components of, 26 Tensor (continued): contravariant, 26, 29 covariant, 27, 29 differentiation of, 68, 71, 74 metric (fundamental), 52, 54 mixed, 29 noncoordinate definition, 193, 200 order of, 29 symmetry of, 106 on vector space, 193 (See also Einstein tensor; Ricci tensor; Riemann tensor) Tensor field (manifold), 200 Tests for tensor character, 45 Theorema egregium, 138 Time, absolute, 164 Time component (relativity), 169 Time dilation (relativity), 169 Timelike interval (relativity), 165 Torsion of a curve, 130 Torus, 147 Trace of a matrix, 38 Transformation: affine, 31 of Christoffel symbols, 69, 71 of coordinates, 23 linear, 11 Lorcntz, 167 orthogonal, 9 Tractrix and tractroid, 153 Transpose of a matrix, 9 Triad (see Moving frame) Triple scalar product, 130 Triple vector product, 41 Twin paradox (relativity), 180 Twisted cubic, 128 Umbilic point (surface), 136 Uniform acceleration (relativity), 171 Unitary group, 190 Vector: contravariant, 26, 28 covariant, 27, 28 length of, 9, 56 in R", 8 (See also Position vector) Vector field, 26 on a manifold, 197, 198 Vector product in R3, 10 Vector space, 8 axioms, 189 Velocity: composition of two velocities. 169 of light, 164 nonrelativistic, 169

228 INDEX Velocity (continued): Weingarten's formulas, 137 of a particle, 154 Work-energy equation, 172 relativistic, 170 Wave equation, 157 Zero curvature, 114, 115 Weighted tensor, 27 vanishing Ricci curvature, 116

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