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Home , Christoffel symbols, Covariant transformation, Mixed tensor, Symmetric tensor, Tensor algebra

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Copyright © 2006, New Age International (P) Ltd., Publishers Published by New Age International (P) Ltd., Publishers

All rights reserved. No part of this ebook may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the publisher. All inquiries should be emailed to [email protected]

ISBN (13) : 978-81-224-2700-4

PUBLISHING FOR ONE WORLD NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS 4835/24, Ansari Road, Daryaganj, New Delhi - 110002 Visit us at www.newagepublishers.com To My parents This page intentionally left blank FOREWORD

It gives me great pleasure to write the foreword to Dr. Nazrul Islam’s book entitled “ and Their Applications. I know the author as a research scholar who has worked with me for several years. This book is a humble step of efforts made by him to prove him to be a dedicated and striving teacher who has worked relentlessly in this . This book fills the gap as methodology has been explained in a simple manner to enable students to understand easily. This book will prove to be a complete book for the students in this field.

Ram Nivas Professor, Department of and Astronomy, Lucknow University, Lucknow This page intentionally left blank PREFACE

‘Tensors’ were introduced by Professor Gregorio Ricci of University of Padua (Italy) in 1887 primarily as extension of vectors. A quantity having magnitude only is called and a quantity with magnitude and direction both, called Vector. But certain quantities are associated with two or more directions, such a quantity is called . The at a point of an elastic solid is an example of a Tensor which depends on two directions one normal to the area and other that of the force on it. Tensors have their applications to , Mechanics, Elasticity, Theory of Relativity, Electromagnetic Theory and many other disciplines of Science and Engineering. This book has been presented in such a clear and easy way that the students will have no difficulty in understanding it. The definitions, proofs of theorems, notes have been given in details. The subject is taught at graduate/postgraduate level in almost all universities. In the end, I wish to thank the publisher and the printer for their full co-operation in bringing out the book in the present nice form. Suggestions for further improvement of the book will be gratefully acknowledged.

Dr. Nazrul Islam This page intentionally left blank CONTENTS

Foreword ...... vii Preface ...... ix Chapter–1 Preliminaries...... 1-5 1.1. n-dimensional Space ...... 1 1.2. Superscript and Subscript ...... 1 1.3. The Einstein's Summation Convention ...... 1 1.4. Dummy Index ...... 1 1.5. Free Index ...... 2 1.6. Krönecker Delta ...... 2 Exercises ...... 5 Chapter–2 Tensor ...... 6-30 2.1. Introduction ...... 6 2.2. Transformation of Coordinates ...... 6 2.3. Covariant and Contravariant Vectors ...... 7 2.4. Contravariant Tensor of Rank Two ...... 9 2.5. Covariant Tensor of Rank Two ...... 9 2.6. of Rank Two...... 9 2.7. Tensor of Higher Order ...... 14 2.8. Scalar or Invariant ...... 15 2.9. Addition and Subtraction of Tensors ...... 15 2.10. Multiplication of Tensors ( of Tensors) ...... 16 2.11. Contraction of a Tensor ...... 18 2.12. Inner Product of Two Tensors ...... 18 2.13. Symmetric Tensors ...... 20 2.14. Skew- ...... 20 2.15. Quotient Law ...... 24 xii Tensors and Their Applications

2.16. Conjugate (or Reciprocal) Symmetric Tensor ...... 25 2.17. Relative Tensor ...... 26 Examples ...... 26 Exercises ...... 29 Chapter–3 and Riemannian Metric ...... 31-54 3.1. The Metric Tensor ...... 31 3.2. Conjugate Metric Tensor (Contravariant Tensor) ...... 34 3.3. Length of a Curve ...... 42 3.4. Associated Tensor ...... 43 3.5. Magnitude of Vector ...... 43 3.6. Scalar Product of Two Vectors ...... 44 3.7. Angle Between Two Vectors ...... 45 3.8. Angle Between Two Coordinate Curves ...... 47 3.9. Hypersurface ...... 48 3.10. Angle Between Two Coordinate Hyper ...... 48 3.11. n-Ply Orthogonal System of Hypersurfaces ...... 49 3.12. Congruence of Curves ...... 49 3.13. Orthogonal Ennuple ...... 49 Examples ...... 52 Exercises ...... 54 Chapter–4 Christoffel’s Symbols and Covariant Differentiation ...... 55-84 4.1. Christoffel’s Symbol...... 55 4.2. Transformtion of Christoffel’s Symbols ...... 64 4.3. Covariant Differentiation of a Covariant Vector ...... 67 4.4. Covariant Differentiation of a Contravariant Vector ...... 68 4.5. Covariant Differentiation of Tensors ...... 69 4.6. Ricci’s Theorem ...... 71 4.7. , and Curl ...... 75 4.8. The Laplacian Operator ...... 80 Exercises ...... 83 Chapter–5 Riemann-Christoffel Tensor ...... 85-110 5.1. Riemann-Christoffel Tensor ...... 85 5.2. Ricci Tensor ...... 88 5.3. Covariant Riemann-Christoffel Tensor ...... 89

5.4. Properties of Riemann-Christoffel Tensors of First Kind Rijkl ...... 91 5.5. Bianchi Identity ...... 94 5.6. ...... 95 5.7. Riemannian of ...... 96 Vn Contents xiii

5.8. Formula For Riemannian Curvature in Terms of Covariant Curvature Tensor of 98 Vn ...... 5.9. Schur’s Theorem ...... 100 5.10. Mean Curvature ...... 101 5.11. Ricci Principal Directions ...... 102 5.12. Einstein Space ...... 103 5.13. or Projective Curvature Tensor ...... 104 Examples ...... 106 Exercises ...... 109 Chapter–6 The e-systems and the Generalized Krönecker Deltas ...... 111-115 6.1. Completely Symmetric ...... 111 6.2. Completely Skew-symmetric ...... 111 6.3. e-system ...... 112 6.4. Generalized Krönecker Delta ...... 112 i jk 6.5. Contraction of ä áâã ...... 114 Exercises ...... 115 Chapter–7 Geometry ...... 116-141 7.1. Length of Arc ...... 116

7.2. in E3 ...... 120 7.3. Reciprocal Base System Covariant and Contravariant Vectors ...... 122 7.4. On The Meaning of Covariant Derivatives ...... 127 7.5. Intrinsic Differentiation ...... 131 7.6. Parallel Vector Fields ...... 134 7.7. Geometry of Space Curves ...... 134 7.8. Serret-Frenet Formulae ...... 138 7.9. Equations of A Straight Line ...... 140 Exercises ...... 141 Chapter–8 Analytical Mechanics ...... 142-169 8.1. Introduction ...... 142 8.2. Newtonian Laws ...... 142 8.3. of Particle ...... 143 8.4. Conservative Force Field ...... 144 8.5. Lagrangean Equation of Motion ...... 146 8.6. Applications of Lagrangean Equations ...... 152 8.7. Hamilton’s Principle ...... 153 8.8. Integral Energy ...... 155 8.9. Principle of Least Action ...... 156 xiv Tensors and Their Applications

8.10. Generalized Coordinates ...... 157 8.11. Lagrangean Equation of Generalized Coordinates ...... 158 8.12. Divergence Theorem, Green’s Theorem, Laplacian Operator and Stoke’s Theorem in Tensor Notation ...... 161 8.13. Gauss’s Theorem ...... 164 8.14. Poisson’s Equation ...... 166 8.15. Solution of Poisson’s Equation ...... 167 Exercises ...... 169 Chapter–9 Curvature of a Curve, ...... 170-187 9.1. Curvature of Curve, Principal Normal...... 170 9.2. ...... 171 9.3. Euler’s Condition ...... 171 9.4. Differential Equations of Geodesics ...... 173 9.5. Geodesic Coordinates ...... 175 9.6. Riemannian Coordinates ...... 177 9.7. Geodesic Form of a Line Element ...... 178 9.8. Geodesics in ...... 181 Examples ...... 182 Exercises ...... 186 Chapter–10 Parallelism of Vectors ...... 188-204 10.1. Parallelism of a Vector of Constant Magnitude (Levi-Civita’s Concept) ...... 188 10.2. Parallelism of a Vector of Variable Magnitude ...... 191 10.3. Subspace of Riemannian ...... 193 10.4. Parallelism in a Subspace ...... 196 10.5. Fundamental Theorem of Riemannian Geometry Statement ...... 199 Examples ...... 200 Exercises ...... 203 Chapter–11 Ricci’s Coefficients of Rotation and Congruence ...... 205-217 11.1. Ricci’s Coefficient of Rotation ...... 205 11.2. Reason for the Name “Coefficients of Rotation” ...... 206 11.3. Curvature of Congruence ...... 207 11.4. Geodesic Congruence ...... 208 11.5. Normal Congruence ...... 209 11.6. Curl of Congruence ...... 211 11.7. Canonical Congruence ...... 213 Examples ...... 215 Exercises ...... 217 Contents xv

Chapter–12 Hypersurfaces ...... 218-242 12.1. Introduction ...... 218 12.2. Generalized Covariant Differentiation ...... 219 12.3. Laws of Tensor Differentiation ...... 220 12.4. Gauss’s Formula ...... 222 12.5. Curvature of a Curve in a Hypersurface and Normal Curvature, Meunier’s Theorem, Dupin’s Theorem ...... 224 12.6. Definitions ...... 227 12.7. Euler’s Theorem ...... 228 12.8. Conjugate Directions and Asymptotic Directions in a Hypersurface...... 229 12.9. Tensor Derivative of Unit Normal...... 230 12.10. The Equation of Gauss and Codazzi ...... 233 12.11. Hypersurfaces with Indeterminate Lines of Curvature ...... 234 12.12. Central Quadratic Hypersurfaces ...... 235 12.13. Polar Hyperplane ...... 236 12.14. Evolute of a Hypersurface in an Euclidean Space ...... 237 12.15. Hypersphere ...... 238 Exercises ...... 241 Index ...... 243-245 This page intentionally left blank CHAPTER – 1

PRELIMINARIES

1.1 n-DIMENSIONAL SPACE In three dimensional rectangular space, the coordinates of a point are (x, y, z). It is convenient to write (x1, x2, x3) for (x, y, z). The coordinates of a point in four dimensional space are given by (x1, x2, x3, x4). In general, the coordinates of a point in n-dimensional space are given by (x1, x2, x3,...., xn) such n- dimensional space is denoted by Vn.

1.2 SUPERSCRIPT AND SUBSCRIPT ij In the symbol Akl , the indices i, j written in the upper position are called superscripts and k, l written in the lower position are called subscripts.

1.3 THE EINSTEIN'S SUMMATION CONVENTION

n 1 2 n i Consider the sum of the series S = a x + a x +...+ a x = S ai x . By using summation convention, 1 2 n i=1 drop the sigma sign and write convention as

n i i S aix = a x i=1 i This convention is called Einstein’s Summation Convention and stated as “If a suffix occurs twice in a term, once in the lower position and once in the upper position then that suffix implies sum over defined range.” If the range is not given, then assume that the range is from 1 to n.

1.4 DUMMY INDEX Any index which is repeated in a given term is called a dummy index or dummy suffix. This is also called Umbral or Dextral Index. i e.g. Consider the expression ai x where i is dummy index; then i 1 2 n ai x = a1x + a2x + ××× + anx 2 Tensors and Their Applications

j 1 2 n and ajx = a1x +a2x +×××+ anx These two equations prove that i j aix = aj x So, any dummy index can be replaced by any other index ranging the same numbers.

1.5 FREE INDEX Any index occurring only once in a given term is called a Free Index. j i e.g. Consider the expression ai x where j is free index.

1.6 KRÖNECKER DELTA

i The symbol d j , called Krönecker Delta (a German mathematician Leopold Krönecker, 1823-91 A.D.) is defined by

i ì1 if i = j d j = í î0 if i ¹ j ij Similarly dij and d are defined as ì1 if i = j ij = í d î0 if i ¹ j ì1 if i = j and dij = í î0 if i ¹ j Properties 1. If x1, x2, ... xn are independent coordinates, then ¶xi = 0 if i ¹ j ¶x j ¶xi = 1 if i = j ¶x j This implies that ¶xi = di ¶x j j ¶xi ¶xk It is also written as = d i . ¶xk ¶x j j i 1 2 3 n 2. di = d1 + d2 + d3 + ×× × + d n (by summation convention) i di = 1 +1+1 + ××× +1 i di = n ij j ik 3. a dk = a 3j j 31d1 + 32d2 + 33d3 + ××× + 3ndn Since a d 2 = a 2 a 2 a 2 a 2 (as j is dummy index) Preliminaries 3

32 1 3 n 2 = a (as ä 2 = ä 2 = ××× = ä2 = 0 and ä 2 = 1) In general,

ij j i1 1 i2 2 i3 3 ik k in n a d k = a dk + a dk + a dk + ××× + a dk + ×××+ a dk

ij j aik 1 2 n k a dk = (as dk = dk = × ×× = dk = 0 and dk =1)

i j i 4. d jdk = dk

i j i 1 i 2 i 3 i i i n d jdk = d1dk + d2dk + d3dk + ××× + didk + ××× + dndk = i (as äi = di = di = ××× = di = 0 and di =1) dk 1 2 3 n i

EXAMPLE 1 df ¶f dx1 ¶f dx2 ¶f dxn Write = + + ××× + using summation convention. dt ¶x1 dt ¶x 2 dt ¶x n dt Solution df ¶f dx1 ¶f dx2 ¶f dx n = + + ××× + dt ¶x1 dt ¶x2 dt ¶x n dt df ¶f dxi = dt ¶xi dt

EXAMPLE 2 i j Expand: (i) aij x x ; (ii) glm gmp Solution i j 1 j 2 j n j (i) aijx x = a1j x x + a2 j x x + ×××+ anjx x 1 1 2 2 n n = a11x x + a22x x + ×××+ annx x i j 1 2 2 2 n 2 aijx x = a11(x ) + a22(x ) + ×××+ ann (x ) (as i and j are dummy indices)

(ii) glmgmp = gl1g1p + gl2g 2p + ××× + glngnp , as m is dummy index.

EXAMPLE 3

If aij are constant and aij = aji, calculate: ¶ ¶ (i) (aijxix j ) (ii) (aijxix j) ¶xk ¶xk¶xl Solution ¶ ¶ (i) (aijxix j ) = aij (xi x j ) ¶xk ¶xk 4 Tensors and Their Applications

¶x j ¶xi = aijxi + aijx j ¶xk ¶xk

¶x j = aijxid jk + aij x jdik , as = d jk ¶xk

= (aijd jk )xi + (aijdik )x j

= aik xi + akj xj as aij djk = aik = aik xi + aki xi as j is dummy index

¶(aijxix j ) ¶ = 2aik xi as given aik = aki xk

¶(aijxix j ) (ii) = 2aikxi ¶xk

Differentiating it w.r.t. xl : ¶2(a x x ) ij i j ¶xi = 2aik ¶xk ¶xl ¶xl i = 2aikdl 2 ¶ (aijxi x j) i = 2alk as aikdl = alk. ¶xk ¶xl

EXAMPLE 4 i j If aij x x = 0 where aij are constant then show that aij + aji = 0 Solution Given i j aijx x = 0 l m Þ almx x = 0 since i and j are dummy indices Differentiating it w.r.t. xi partially,

¶ l m (almx x ) = 0 ¶xi

¶ l m alm (x x ) = 0 ¶xi l m ¶x m ¶x l alm x + alm x = 0 ¶xi ¶xi

l m ¶x m ¶x l = di Since = di and ¶xi ¶xi Preliminaries 5

l m m l almdi x + almdi x = 0 m l aimx + alix = 0 l m as almdi = aim and almdi = ali.

Differentiating it w.r.t. xj partially ¶xm ¶xl aim + ali = 0 ¶x j ¶x j m l aimd j + alid j = 0

aij + aji= 0 Proved.

EXERCISES

1. Write the following using the summation convention. (i) (x1)2 + (x2)2 + (x3)2 + . . . + (xn)2 2 1 2 2 2 . . . n 2 (ii) ds = g11 (dx ) + g22(dx ) + + gnn(dx ) 1 3 2 3 . . . n 3 (iii) a1x x + a2x x + + anx x 2. Expand the following:

¶ i (i) a xj (ii) ( ga ) (iii) Ak Bi ij ¶xi i 3. Evaluate: j i i j k i j (i) x d j (ii) d jdk dl (iii) d jdi

ij ij i 4. Express b y iy j in the terms of x variables where yi = cij xj and b cik = dk.

ANSWERS

i j 2 i j i 3 1. (i) x x (ii) ds = gij dx dx (iii) aix x . 1 2 3 . . . n 2. (i) ai1x + ai2x + ai3x + + ainx ¶ 1 ¶ 2 ¶ n (ii) 1 ( ga ) + 2 ( ga ) + ××× + ( ga ) ¶x ¶x ¶xn

k 1 k 2 k n (iii) A1 B + A2 B +... + An B

i i 3. (i) x (ii) dl (iii) n

4. Cij xi xj CHAPTER – 2

TENSOR ALGEBRA

2.1 INTRODUCTION A scalar (density, pressure, temperature, etc.) is a quantity whose specification (in any ) requires just one number. On the other hand, a vector (displacement, acceleration, force, etc.) is a quantity whose specification requires three numbers, namely its components with respect to some . Scalers and vectors are both special cases of a more general object called a tensor of order n whose specification in any coordinate system requires 3n numbers, called the components of tensor. In fact, scalars are tensors of order zero with 3° = 1 component. Vectors are tensors of order one with 31 = 3 components.

2.2 TRANSFORMATION OF COORDINATES In three dimensional rectangular space, the coordinates of a point are (x, y, z) where x, y, z are real numbers. It is convenient to write (x1, x2, x3) for (x, y, z) or simply xi where i = 1, 2, 3. Similarly in n- dimensional space, the coordinate of a point are n-independent variables (x1, x2,..., x n) in X-coordinate system. Let (x1, x 2,..., x n) be coordinate of the same point in Y-coordinate system. Let x1, x2,¼, x n be independent single valued function of x1, x2,...., xn, so that, 1 x = x1(x1, x2, ....,xn) 2 x = x 2(x1,x 2, ...., xn )

3 x = x 3(x1, x2, ...., x n)

M M xn = x n (x1,x 2, ..., x n) or xi = x i(x1, x2, ..., xn) ; i = 1, 2, …, n …(1) 7

Solving these equations and expressing xi as functions of x1,x 2,..., x n, so that

xi = xi (x1, x 2,...,xn ); i = 1, 2, ..., n The equations (1) and (2) are said to be a transformation of the coordinates from one coordinate system to another

2.3 COVARIANT AND CONTRAVARIANT VECTORS (TENSOR OF RANK ONE) Let (x1, x2, ..., xn) or xi be coordinates of a point in X-coordinate system and (x1, x 2,..., x n) or x i be coordinates of the same point in the Y-coordinate system. Let Ai, i = 1, 2, ..., n (or A1, A2, ..., An) be n functions of coordinates x1, x2, ..., xn in X-coordinate system. If the quantities Ai are transformed to Ai in Y-coordinate system then according to the law of transformation

¶xi ¶x j Ai = A j or A j = i ¶x j ¶x i A Then Ai are called components of contravariant vector. 1 2 n Let Ai, i = 1, 2,..., n (or A1, A2, …, An) be n functions of the coordinates x , x , ..., x in X-coordinate system. If the quantities Ai are transformed to Ai in Y-coordinate system then according to the law of transformation

¶x j ¶x i A = Aj or A = Ai i ¶xi j ¶x j Then Ai are called components of covariant vector. The contravariant (or covariant) vector is also called a contravariant (or covariant) tensor of rank one. Note: A superscript is always used to indicate contravariant component and a subscript is always used to indicate covariant component.

EXAMPLE 1 If xi be the coordinate of a point in n-dimensional space show that dxi are component of a contravariant vector. Solution 1 2 Let x1, x2, ..., xn or xi are coordinates in X-coordinate system and x ,x ,...,xn or xi are coordinates in Y-coordinate system. If xi = x i (x1, x2,...,xn)

i i i i ¶x ¶x ¶x dx = dx 1 + dx 2 + × × × + dx n ¶x1 ¶x 2 ¶xn 8 Tensors and Their Applications

¶xi d x i = dx j ¶x j It is law of transformation of contravariant vector. So, dx i are components of a contravariant vector.

EXAMPLE 2

¶f Show that is a covariant vector where f is a scalar function. ¶xi

Solution

1 2 i 1 2 Let x , x , ..., xn or x are coordinates in X-coordinate system and x , x , ..., x n or x i are coordinates in Y-coordinate system. Consider f(x1, x 2,...,x n ) = f(x1, x2,...,x n)

¶f ¶f ¶f ¶f = ¶x1 + ¶x2 + ××× + ¶xn ¶x1 ¶x2 ¶xn

¶f ¶f ¶x1 ¶f ¶x 2 ¶f ¶xn = + + ××× + ¶x i ¶x1 ¶x i ¶x2 ¶x i ¶xn ¶xi

¶f ¶f ¶x j = ¶x i ¶x j ¶xi

¶f ¶x j ¶f or = ¶x i ¶x i ¶x j ¶f It is law of transformation of component of covariant vector. So, is component of covariant ¶xi vector.

EXAMPLE 3 Show that the velocity of fluid at any point is a component of contravariant vector or Show that the component of tangent vector on the curve in n-dimensional space are component of contravariant vector. Solution dx1 dx2 dxn Let , , ¼, be the component of the tangent vector of the point (x1, x2,...,xn) i.e., dt dt dt dxi be the component of the tangent vector in X-coordinate system. Let the component of tangent dt Tensor Algebra 9

dx i vector of the point (x1,x 2,...,x n) in Y-coordinate system are . Then x1, x 2, ..., x n or x i being a dt function of x1, x2, ...,x n which is a function of t. So, d xi ¶x i dx1 ¶x i dx2 ¶x i dxn = + + ×× × + dt dt1 dt dx2 dt dxn dt

d xi ¶x i dx j = dt dx j dt dxi It is law of transformation of component of contravariant vector. So, is component of dt contravariant vector. i.e. the component of tangent vector on the curve in n-dimensional space are component of contravariant vector.

2.4 CONTRAVARIANT TENSOR OF RANK TWO Let Aij (i, j = 1, 2, ..., n) be n2 functions of coordinates x1, x2, ..., xn in X-coordinate system. If the quantities Aij are transformed to Aij in Y-coordinate system having coordinates x1, x 2, ..., x n . Then according to the law of transformation ¶x i ¶x j Aij = Akl ¶x k ¶xl Then Aij are called components of Contravariant Tensor of rank two.

2.5 COVARIANT TENSOR OF RANK TWO 2 1 2 n Let Aij (i, j = 1, 2, ..., n) be n functions of coordinates x , x , ..., x in X-coordinate system. If the 1 2 n quantities Aij are transformed to Aij in Y-coordinate system having coordinates x ,x ,...,x , then according to the law of transformation, ¶x k ¶xl A = A ij ¶x i ¶x j kl

Then Aij called components of covariant tensor of rank two.

2.6 MIXED TENSOR OF RANK TWO i 2 1 2 n Let Aj (i, j = 1, 2, ..., n) be n functions of coordinates x , x , ..., x in X-coordinate system. If the i i 1 2 n quantities Aj are transformed to Aj in Y-coordinate system having coordinates x , x ,...,x , then according to the law of transformation ¶xi ¶xl Ai = Ak j ¶xk ¶x l i Then Aj are called components of mixed tensor of rank two. 10 Tensors and Their Applications

Note: (i) The rank of the tensor is defined as the total number of indices per component. (ii) Instead of saying that “ Aij are the components of a tensor of rank two” we shall often say “ Aij is a tensor of rank two.” THEOREM 2.1 To show that the Krönecker delta is a mixed tensor of rank two. Solution Let X and Y be two coordinate systems. Let the component of in X-coordinate i i system d j and component of Krönecker delta in Y-coordinate be dj , then according to the law of transformation ¶x i ¶x i ¶xl ¶xk di = = j ¶x j ¶xk ¶x j ¶xl ¶x i ¶xl di = dk j ¶x k ¶x j l i This shows that Krönecker d j is mixed tensor of rank two.

EXAMPLE 4 ¶A If A is a covariant tensor, then prove that i do not form a tensor.. i ¶x j Solution

Let X and Y be two coordinate systems. As given Ai is a covariant tensor. Then ¶x k A = Ak i ¶x i Differentiating it w.r.t. x j

¶A ¶ æ ¶xk ö i = ç A ÷ j j ç i k ÷ ¶x ¶x è ¶x ø ¶A ¶x k ¶A ¶2xk i = k + A …(1) ¶x j ¶x i ¶x j k ¶x i¶x j

¶Ai It is not any law of transformation of tensor due to presence of second term. So, is not a ¶x j tensor.

i THEOREM 2.2 To show that d j is an invariant i.e., it has same components in every coordinate system.

i Proof: Since d j is a mixed tensor of rank two, then ¶xi ¶xl di = d k j ¶xk ¶x j l Tensor Algebra 11

¶xi æ ¶xl ö ç d k ÷ = k ç j l ÷ ¶x è ¶x ø ¶ xi ¶xk ¶xl ¶xk = , as d k = ¶xk ¶ x j ¶x j l ¶x j ¶x i ¶x i di = = di , as = di j ¶x j j ¶x j j i So, d j is an invariant.

THEOREM 2.3 Prove that the transformation of a contravariant vector is transitive. or Prove that the transformation of a contravariant vector form a . Proof: Let Ai be a contravariant vector in a coordinate system xi(i =1,2,...,n) . Let the coordinates xi be transformed to the coordinate system x i and x i be transformed to x i . When coordinate xi be transformed to x i , the law of transformation of a contravariant vector is ¶x p A p = Aq ... (1) ¶x q When coordinate x i be transformed to x i , the law of transformation of contravariant vector is ¶x i Ai = A p ¶x p ¶x i ¶x p i = Aq from (1) A ¶x i ¶xq i ¶x q i = A A ¶xq This shows that if we make direct transformation from xi to x i , we get same law of transformation. This property is called that transformation of contravariant vectors is transitive or form a group.

THEOREM 2.4 Prove that the transformation of a covariant vector is transitive. or Prove that the transformation of a covariant vector form a group.

i i Proof: Let Ai be a covariant vector in a coordinate system x (i = 1, 2, ..., n) . Let the coordinates x be i transformed to the coordinate system x and x i be transformed to x i . When coordinate xi be transformed to x i , the law of transformation of a covariant vector is ¶xq A = A ... (1) p ¶x p q When coordinate x i be transformed to x i , the law of transformation of a covariant vector is ¶x p A = Ap i ¶x i 12 Tensors and Their Applications

¶x p ¶xq A = Aq i ¶x i ¶x p ¶x q A = Aq i ¶x i This shows that if we make direct transformation from xi to x i , we get same law of transformation. This property is called that transformation of covariant vectors is transitive or form a group.

THEOREM 2.5 Prove that the transformations of tensors form a group or Prove that the equations of transformation a tensor (Mixed tensor) posses the group property. i i Proof: Let Aj be a mixed tensor of rank two in a coordinate system x (i =1, 2,...,n) . Let the coordinates xi be transformed to the coordinate system x i and x i be transformed to x i . When coordinate xi be transformed to x i , the transformation of a mixed tensor of rank two is ¶x p ¶xs A p = Ar ... (1) q ¶xr ¶x q s i When coordinate x be transformed to x i , the law of transformation of a mixed tensor of rank two is ¶x i ¶x q Ai = A p j ¶x p ¶x j q

i q p s ¶x ¶x ¶x ¶x r = p j r q As from (1) ¶x ¶x ¶x ¶x

¶x i ¶xs Ai = Ar j ¶xr ¶x j s

This shows that if we make direct transformation from xi to x i , we get same law of transformation. This property is called that transformation of tensors form a group. THEOREM 2.6 There is no distinction between contravariant and covariant vectors when we restrict ourselves to rectangular Cartesian transformation of coordinates. Proof: Let P(x, y) be a point with respect to the rectangular Cartesian axes X and Y. Let (x, y) be the coordinate of the same point P in another rectangular cartesian axes X and Y , Let (l1, m1) and (l2, m2) be the direction cosines of the axes X , Y respectively. Then the transformation relations are given by

x = l1x + m1y ü ý ...(1) y = l2x + m2 yþ and solving these equations, we have

x = l1x + l2y ü ý ...(2) y = m1x + m2yþ 2 2 put x = x1 , y = x , x = x1, y = x Tensor Algebra 13

Consider the contravariant transformation i ¶x j i = A ; j =1,2 A ¶x j i i ¶x 1 ¶x 2 i = A + A A ¶x1 ¶x2 for i =1,2 . 1 1 ¶x 1 ¶x 2 1 = A + A A ¶x1 ¶x2 2 2 ¶x 1 ¶x 2 2 = A + A A ¶x1 ¶x2 ¶x 2 2 From (1) = l , but x = x1 , y = x , x = x1 , y = x ¶x 1 Then ¶x ¶x1 = = l1 . ¶x ¶x1 Similarly, ¶x ¶x1 ü = m1 = ; ï ¶y ¶x2 ï ý ¶y ¶x 2 ¶y ¶x 2 ï ...(3) = l2 = ; = m2 = ¶x ¶x1 ¶y ¶x2 þï So, we have 1 1 2 ü A = l1A + m1A ï 2 1 2 ý ..(4) A = l2 A + m2A þï Consider the ¶x j A = Aj; j =1,2 i ¶x i ¶x1 ¶x2 A = A1 + A2 i ¶x i ¶x1 for i = 1, 2 . ¶x1 ¶x 2 A = A1 + A2 1 ¶x1 ¶x1 ¶x1 ¶x2 A = A1 + A2 2 ¶x 2 ¶x 2 From (3)

A1 = l1A1 + m1A2 ïü ý ...(5) A2 = l2A1 + m2A2þï 14 Tensors and Their Applications

So, from (4) and (5), we have 1 2 A = A1 and A = A2 Hence the theorem is proved.

2.7 TENSORS OF HIGHER ORDER (a) Contravariant tensor of rank r ... Let Ai1i2...ir be nr function of coordinates x1, x2, ..., xn in X-coordinates system. If the quantities Aiii2 ir are transformed to Ai1i2...ir in Y-coordinate system having coordinates x1, x 2, ..., x n . Then according to the law of transformation

¶xi1 ¶x i2 ¶x ir i1i2...ir p p ...p A = p p … p A 1 2 r ¶x 1 ¶x 2 ¶x r Then Aiii2...ir are called components of contravariant tensor of rank r. (b) Covariant tensor of rank s Let A be ns functions of coordinates 1, 2, ..., n in coordinate system. If the quantities j1 j2...js x x x X- A are transformed to A in coordinate system having coordinates x1,x 2 ,...,x n . Then j1 j2...js j1 j2...js Y- according to the law of transformation

¶xq1 ¶xq2 ¶xqs Aj j ...j = ××× Aq ,q ...,q 1 2 s ¶x j1 ¶x j2 ¶x js 1 2 s Then A are called the components of covariant tensor of rank s. j1 j2...js (c) Mixed tensor of rank r + s Let Ai1i2...ir be r+s functions of coordinates 1, 2, ..., n in coordinate system. If the quantities j1 j2...js n x x x X- Ai1i2...ir are transformed to Ai1i2...ir in coordinate system having coordinates 1 2 n . Then j1 j2...js j1 j2...js Y- x , x , ¼, x according to the law of transformation

¶ i1 ¶ i2 ¶ ir ¶ q1 ¶ q2 ¶ qs i1i2...ir x x x x x x p1p2 ...pr A ... = ××× A ... j1 j2 js ¶x p1 ¶x p2 ¶x pr ¶x j1 ¶x j2 ¶x js q1q2 qs Then Ai1i2...ir are called component of mixed tensor of rank . j1 j2...js r + s A tensor of type Aj1 j2 ...js is known as tensor of type (r,s ), In (r,s), the first component r i1i2...ir indicates the rank of contravariant tensor and the second component s indicates the rank of covariant tensor. ij i Thus the tensors Aij and A are type (0, 2) and (2, 0) respectively while tensor Aj is type (1, 1). Tensor Algebra 15

EXAMPLE

ijk Alm is a mixed tensor of type (3, 2) in which contravariant tensor of rank three and covariant tensor of rank two. Then according to the law of transformation

¶xi ¶x j ¶x k ¶x a ¶xb Aijk = Aabg lm ¶xa ¶xb ¶xg ¶x l ¶xm ab

2.8 SCALAR OR INVARIANT

A function f (x1, x2, ..., xn ) is called Scalar or an invariant if its original value does not change upon transformation of coordinates from x1, x2, ..., xn to x1, x 2, ..., x n . i.e. f(x1, x2,...,x n) = f(x1, x 2 ,...,x n) Scalar is also called tensor of rank zero. i For example, A Bi is scalar..

2.9 ADDITION AND SUBTRACTION OF TENSORS

THEOREM 2.7 The sum (or difference) of two tensors which have same number of covariant and the same contravariant indices is again a tensor of the same rank and type as the given tensors. Proof: Consider two tensors Ai1i2...ir and Bi1i2...ir of the same rank and type (i.e., covariant tensor of j1 j2...js j1 j2...js rank s and contravariant tensor of rank r. ). Then according to the law of transformation ¶ i1 ¶ i2 ¶ ir ¶ q1 ¶ q2 ¶ qs i1i2...ir x x x x x x p1p2 ...pr A ... = ××× ××× ××× A ... j1 j2 js ¶x p1 ¶x p2 ¶x pr ¶x j1 ¶x j2 ¶x js q1q2 qs and ¶ i1 ¶ i2 ¶ ir ¶ q1 ¶ q2 ¶ qs i1i2...ir x x x x x x p1p2 ...pr B ... = ××× × ××× ××× B ... j1 j2 js ¶x p1 ¶x p2 ¶x pr ¶x j1 ¶x j2 ¶x js q1q2 qs Then

¶xi1 ¶xi2 ¶xir ¶xq1 ¶xq2 ¶xqs i1i2 ...ir i1i2...ir ¼ p1 p2 ...pr p1 p2...pr A ... ± B ... = ××× (A ... ± B ... ) j1 j2 js j1 j2 js ¶x p1 ¶x p2 ¶x pr ¶x j1 ¶x j2 ¶x js q1q2 qs q1q2 qs If Ai1i2 ...ir ± B i1i2...ir = C i1i2...ir j1 j2 ...js j1 j2...js j1 j2...js and Ap1 p2...pr ± Bp1 p2...pr = C p1 p2...pr q1q2 ...qs q1q2 ...qs q1q2 ...qs So, ¶ i1 ¶ i2 ¶ ir ¶ q1 ¶ q2 ¶ qs i1i2...ir x x x x x x p1, p2,...,pr C ... = ××× ××× C , ,..., j1 j2 js ¶x p1 ¶x p2 ¶x pr ¶x j1 ¶x j2 ¶x js q1 q2 qs This is law of transformation of a mixed tensor of rank . So, C i1,i2 ,...,ir is a mixed tensor of r+s j1, j2 ,...,js rank r+s or of type (r, s). 16 Tensors and Their Applications

EXAMPLE 5

ij lm If Ak and Bn are tensors then their sum and difference are tensors of the same rank and type. Solution

ij ij As given Ak and Bk are tensors. Then according to the law of transformation ¶xi ¶x j ¶x r ij pq A = Ar k ¶x p ¶xq ¶x k and ¶xi ¶x j ¶x r ij pq B = Br k ¶x p ¶xq ¶x k then ¶xi ¶x j ¶x r ij ij pq pq A ± B = (Ar ± Br ) k k ¶x p ¶xq ¶x k If ij ij ij pq pq pq Ak ± Bk = Ck and Ar ± Br = Cr So, ¶xi ¶x j ¶x r ij pq C = Cr k ¶x p ¶xq ¶x k ij ij ij The shows that Ck is a tensor of same rank and type as Ak and Bk .

2.10 MULTIPLICATION OF TENSORS (OUTER PRODUCT OF TENSOR) THEOREM 2.8 The multiplication of two tensors is a tensor whose rank is the sum of the ranks of two tensors. Proof: Consider two tensors Ai1i2...ir (which is covariant tensor of rank and contravariant tensor of j1 j2...js s rank ) and Bk1k2 ...km (which is covariant tensor of rank and contravariant tensor of rank ). Then r l1l2...ln m n according to the law of transformation. ¶ i1 ¶ i2 ¶ ir ¶ q1 ¶ q2 ¶ qr i1i2...ir x x x x x x p1p2 ...pr A ... = ××× ××× A ... j1 j2 js ¶x p1 ¶x p2 ¶x pr ¶x j1 ¶x j2 ¶x js q1q2 qs and ¶ k1 ¶ k 2 ¶ km ¶ b1 ¶ b2 ¶ bn k1k 2...k m x x x x x x a1a2 ...am B ... = ××× ××× Bb b ...b l1l2 ln ¶xa1 ¶xa 2 ¶xa m ¶xl1 ¶xl2 ¶x ln 1 2 n Then their product is

¶xi1 ¶xir ¶x q1 ¶xqs ¶x k1 ¶x km ¶x b1 ¶xbn i1i2...ir k1k2...km ××× ××× ××× × ××× Aj j ...j Bl l ...l = a a 1 2 s 1 2 n ¶x p1 ¶x pr ¶x j1 ¶x js ¶x 1 ¶x m ¶xl1 ¶x lm

Ap1p2 ...pr Ba1a2 ...am q1q2...qs b1b2 ...b n Tensor Algebra 17

If ...... i1i2 ir k1 k2 km i i ...i k k ...k C 1 2 r 1 2 n j j ...j l l ... = A ... B ... 1 2 s 1 2 ln j1 j2 js l1l2 ln and C p1 p2...pr a1a 2...am = Ap1p2...,pr Ba1a2 ...am q1q2 ...qsb1b2 ...b n q1q2...,qs b1b2 ...b n So, ¶xi1 ¶x ir ¶x q1 ¶x qs ¶x k1 i1i2...ir k1k2 ...km ××× C ...... = ××× × a j1 j2 jsl1l2 ln ¶x p1 ¶x pr ¶x j1 ¶x js ¶x 1 ¶ k m ¶ b1 ¶ bn x x x p1 p2...pra1a 2 ...a h ××× × ××× C ... b b ...b ¶xa m ¶xl1 ¶x ln q1q2 qs 1 2 n i i ...i ...k This is law of transformation of a mixed tensor of rank + + + . . So, C 1 2 r k1k2 m is a r m s n j1 j2...jsl1l2...ln mixed tensor of rank r + m + s + n . or of type (r + m, s + n). Such product is called outer product or open proudct of two tensors.

i THEOREM 2.9 If A and Bj are the components of a contravariant and covariant tensors of rank one i then prove that A Bj are components of a mixed tensor of rank two. i Proof: As A is contravariant tensor of rank one and Bj is covariant tensor of rank one. Then according to the law of transformation ¶x i Ai = Ak ...(1) ¶xk and ¶xl B = B ...(2) j ¶x j l Multiply (1) and (2), we get ¶x i ¶xl AiB = Ak B j ¶x k ¶x j l i This is law of transformation of tensor of rank two. So, A B j are mixed tensor of rank two. Such product is called outer product of two tensors.

EXAMPLE 6

i kl Show that the product of two tensors Aj and Bm is a tensor of rank five.

Solution

i kl As Aj and Bm are tensors. Then by law of transformation ¶x i ¶xq ¶x k ¶x l ¶xt i p kl rs A = Aq and B = Bt j ¶x p ¶x j m ¶xr ¶xs ¶x m 18 Tensors and Their Applications

Multiplying these, we get

¶xi ¶xq ¶x k ¶xl ¶xt AiBkl = ApBrs j m ¶x p ¶x j ¶xr ¶xs ¶x m q t

i kl This is law of transformation of tensor of rank five. So, AjBm is a tensor of rank five.

2.11 CONTRACTION OF A TENSOR The process of getting a tensor of lower order (reduced by 2) by putting a covariant index equal to a contravariant index and performing the summation indicated is known as Contraction. In other words, if in a tensor we put one contravariant and one covariant indices equal, the process is called contraction of a tensor. ijk For example, consider a mixed tensor Alm of order five. Then by law of transformation, ¶xi ¶x j ¶x k ¶xs ¶xt A ijk = Apqr lm ¶x p ¶x q ¶xr ¶x l ¶x m st Put the covariant index l = contravariant index i, so that

¶xi ¶x j ¶x k ¶xs ¶xt Aijk = Apqr im ¶x p ¶x q ¶xr ¶x i ¶x m st

¶x j ¶x k ¶x s ¶xt = Apqr ¶xq ¶x r ¶x p ¶x m st

¶x j ¶x k ¶xt ¶x s = ds Apqr Since = ds ¶xq ¶x r ¶x m p st ¶x p p

¶x j ¶x k ¶xt Aijk = Apqr im ¶xq ¶x r ¶x m pt

ijk This is law of transformation of tensor of rank 3. So, Aim is a tensor of rank 3 and type (1, 2) ijk while Alm is a tensor of rank 5 and type (2, 3). It means that contraction reduces rank of tensor by two.

2.12 INNER PRODUCT OF TWO TENSORS ij l ij l Consider the tensors Ak and Bmn if we first form their outer product Ak Bmn and contract this by ij k putting l = k then the result is Ak Bmn which is also a tensor, called the inner product of the given tensors. Hence the inner product of two tensors is obtained by first taking outer product and then contracting it.

EXAMPLE 7 i If A and Bi are the components of a contravariant and covariant tensors of rank are respectively i then prove that A Bi is scalar or invariant. Tensor Algebra 19

Solution i As A and Bi are the components of a contravariant and covariant tensor of rank one respectively, then according to the law of the transformation

i q ¶x p ¶x i = A and B = Bq A ¶x p i ¶x i Multiplying these, we get i q ¶x ¶x p i i = A B A B ¶x p ¶x i q ¶xq ¶xq = ApB , since = dq ¶x p q ¶x p p q p = d p A Bq

i p A Bi = A Bp i This shows that A Bi is scalar or Invariant.

EXAMPLE 8

i kl i jl If Aj is mixed tensor of rank 2 and Bm is mixed tensor of rank 3. Prove that AjBm is a mixed tensor of rank 3. Solution

i kl As Aj is mixed tensor of rank 2 and Bm is mixed tensor of rank 3. Then by law of transformation ¶xi ¶x q ¶x k ¶x l ¶xt i p kl rs A = Aq and B = Bt ...(1) j ¶x p ¶x j m ¶xr ¶xs ¶x m Put k = j then ¶x j ¶x l ¶xt jl rs B = Bt ...(2) m ¶xr ¶xs ¶xm Multiplying (1) & (2) we get ¶xi ¶x q ¶x j ¶x l ¶xt AiB jl = Ap Brs j m ¶x p ¶x j ¶xr ¶xs ¶x m q t

i l t q j q ¶x ¶x ¶x q p rs ¶x ¶x ¶x = d A B since = = dq ¶x p ¶x s ¶x m r q t ¶x j ¶x r ¶xr r

¶xi ¶xl ¶xt i jl p qs q rs qs A B = Aq Bt since d B = B j m ¶x p ¶x s ¶x m r t t

i jl This is the law of transformation of a mixed tensor of rank three. Hence AjBm is a mixed tensor of rank three. 20 Tensors and Their Applications

2.13SYMMETRIC TENSORS A tensor is said to be symmetric with respect to two contravariant (or two covariant) indices if its components remain unchanged on an interchange of the two indices.

EXAMPLE (1) The tensor Aij is symmetric if Aij = A ji ijk ijk jik (2) The tensor Alm is symmetric if Alm = Alm

1 THEOREM 2.10 A symmetric tensor of rank two has only n(n +1) different components in n- 2 dimensional space. Proof: Let Aij be a symmetric tensor of rank two. So that Aij = Aji .

é 11 12 13 1n ù A A A L A ê 21 22 23 2nú êA A A L A ú The component of ij are ê A31 A32 A33 A3n ú A ê L ú ê M M M L M ú ê n1 n2 n3 nnú ë A A A L A û i.e., Aij will have n2 components. Out of these n2 components, n components A11, A22, A33, ..., Ann are different. Thus remaining components are (n2– n). In which A12 = A21, A23 = A32 etc. due to symmetry.

1 2 So, the remaining different components are (n - n) . Hence the total number of different 2 components 1 2 1 = n + (n - n) = n(n +1) 2 2

2.14SKEW-SYMMETRIC TENSOR A tensor is said to be skew-symmetric with respect to two contravariant (or two covariant) indices if its components change sign on interchange of the two indices.

EXAMPLE

(i) The tensor Aij is Skew-symmetric of Aij = -Aji ijk ijk jik (ii) The tensor Alm is Skew-symmetric if Alm = - Alm 1 THEOREM 2.11 A Skew symmetric tensor of second order has only n(n -1) different non-zeroo 2 components. Proof: Let Aij be a skew-symmetric tensor of order two. Then Aij = -A ji . Tensor Algebra 21

12 13 1n é 0 A A L A ù ê 21 23 2n ú ê A 0 A L A ú The components of Aij are ê A31 A32 0 A3n ú ê L ú ê M M M L M ú ê n1 n2 n3 ú ë A A A L 0 û

[Since Aii = -Aii Þ 2Aii = 0 Þ Aii = 0 Þ A11 = A22 = × ××Ann = 0] i.e., Aij will have n2 components. Out of these n2 components, n components A11, A22, A33, ..., Ann are zero. Omitting there, then the remaining components are n2 –n. In which A12 = – A21, A13 = – A31 1 2 etc. Ignoring the sign. Their remaining the different components are (n - n) . 2 1 Hence the total number of different non-zero components = n(n -1) 2 Note: Skew-symmetric tensor is also called anti-symmetric tensor.

THEOREM 2.12 A covariant or contravariant tensor of rank two say Aij can always be written as the sum of a symmetric and skew-symmetric tensor.

Proof: Consider a covariant tensor Aij. We can write Aij as 1 1 A = (A + A ) + (A - A ) ij 2 ij ji 2 ij ji

Aij = Sij + Tij 1 1 where S = ( A + A ) and T = (A - A ) ij 2 ij ji ij 2 ij ji Now, 1 S = (A + A ) ji 2 ji ij

S ji = Sij

So, Sij is symmetric tensor.. and 1 T = (A + A ) ij 2 ij ji 1 T = (A - A ) ji 2 ji ij 1 = - (A - A ) 2 ij ji

Tji = -Tij or Tij = -Tji

So, Tij is Skew-symmetric Tensor.. 22 Tensors and Their Applications

EXAMPLE 9

j k j k If f = a jk A A . Show that we can always write f = bjk A A where bjk is symmetric.

Solution As given

j k f = a jk A A ...(1) Interchange the indices i and j k j f = akj A A ...(2) Adding (1) and (2),

j k 2f = (a jk + akj )A A

1 j k f = (a + a )A A 2 jk kj j k f = bjk A A 1 where b = (a + a ) jk 2 jk kj

To show that bjk is symmetric.

Since 1 b = (a + a ) jk 2 jk kj 1 bkj = (a + a ) 2 kj jk 1 = (a + a ) 2 jk kj

bkj = bjk

So, bjk is Symmetric.

EXAMPLE 10

æ ¶T ¶Tj ö ç i - ÷ If T be the component of a covariant vector show that ç j i ÷ are component of a Skew- i è ¶x ¶x ø symmetric covariant tensor of rank two. Solution

As Ti is covariant vector. Then by the law of transformation ¶xk T = Tk i ¶x i Tensor Algebra 23

Differentiating it w.r.t. to x j partially,,

¶T ¶ æ ¶xk ö i ç T ÷ = j ç i k ÷ ¶x j ¶x è ¶x ø

¶2x k ¶x k ¶T = T + k ¶x j¶xi k ¶x i ¶x j

2 k k l ¶Tj ¶ x ¶x ¶x ¶T = T + k ...(1) ¶x j ¶x j¶x i k ¶xi ¶x j ¶x l Similarly, 2 k k l ¶Tj ¶ x ¶x ¶x ¶T = T + k ¶x i ¶x i¶x j k ¶x j ¶xi ¶xl Interchanging the dummy indices k & l

2 k k l ¶Tj ¶ x ¶x ¶x ¶T = T + l ...(2) ¶x i ¶x i¶x j k ¶x i ¶x j ¶xk Substituting (1) and (2), we get

k l ¶T ¶Tj ¶x ¶x æ ¶Tk ¶Tl ö i - = ç - ÷ ¶x j ¶xi ¶x i ¶x j è ¶xl ¶x k ø

¶T ¶T j This is law of transformation of covariant tensor of rank two. So, i - are component of ¶x j ¶xi a covariant tensor of rank two.

¶T ¶T j To show that i - is Skew-symmetric tensor.. ¶x j ¶xi Let

¶T ¶T j A = i - ij ¶x j ¶xi

¶Tj ¶T A = - i ji ¶xi ¶x j æ ¶T ¶T ö - ç i - j ÷ = ç j i ÷ è ¶x ¶x ø

Aji = - Aij

or Aij = -Aji

¶T ¶Tj So, A = i - is Skew-symmetric. ij ¶x j ¶xi

¶T ¶T j So, i - are component of a Skew-symmetric covariant tensor of rank two. ¶x j ¶xi 24 Tensors and Their Applications

2.15 QUOTIENT LAW By this law, we can test a given quantity is a tensor or not. Suppose given quantity be A and we do not know that A is a tensor or not. To test A, we take inner product of A with an arbitrary tensor, if this inner product is a tensor then A is also a tensor. Statement If the inner product of a set of functions with an atbitrary tensor is a tensor then these set of functions are the components of a tensor. The proof of this law is given by the following examples.

EXAMPLE 11 Show that the expression A(i,j,k) is a covariant tensor of rank three if A(i,j,k)Bk is covariant tensor of rank two and Bk is contravariant vector Solution Let X and Y be two coordinate systems. As given A (i, j, k)Bk is covariant tensor of rank two then p q ¶x ¶x r A(i, j,k )B k = A( p,q, r)B ...(1) ¶xi ¶x j

Since Bk is contravariant vector. Then ¶x k ¶xr Bk = Br or Br = B k ¶xr ¶x k So, from (1) p q r ¶x ¶x ¶x k A(i, j,k )B k = A(p,q,r) B ¶x i ¶x j ¶x k p q r ¶x ¶x ¶x k A(i, j,k )B k = A( p,q,r)B ¶xi ¶x j ¶x k ¶x p ¶xq ¶xr A(i, j,k) = A( p, q, r) ¶xi ¶x i ¶x k As Bk is arbitrary.. So, A(i, j,k) is covariant tensor of rank three.

EXAMPLE 12 i j i j If A (i, j, k)A B Ck is a scalar for arbitrary vectors A , B , Ck. Show that A(i, j, k) is a tensor of type (1, 2). Solution i j Let X and Y be two coordinate systems. As given A(i, j, k)A B Ck is scalar. Then i j p q A(i, j,k ) A B Ck = A( p,q,r )A B Cr ...(1) Tensor Algebra 25

i i Since A ,B and Ck are vectors. Then ¶xi ¶x p Ai = A p or Ap = Ai ¶x p ¶xi ¶x j ¶xq B j = Bq or Bq = B j ¶xq ¶x j ¶x k ¶xr C k = C r or C r = C k ¶xr ¶x k So, from (1) p q k i j ¶x ¶x ¶x i j A (i, j, k) A B Ck = A(p, q,r) A B Ck ¶xi ¶x j ¶xr i j As A , B ,Ck are arbitrary.. Then ¶x p ¶xq ¶xk A(i, j,k) = A( p, q,r) ¶xi ¶x j ¶xr So, A(i, j, k) is tensor of type (1, 2).

2.16 CONJUGATE (OR RECIPROCAL) SYMMETRIC TENSOR

Consider a covariant symmetric tensor Aij of rank two. Let d denote the determinant Aij with the elements Aij i.e., d = Aij and d ¹ 0 . Now, define Aij by

Cofactor of Aij is the determinant Aij ij = A d Aij is a contravariant symmetric tensor of rank two which is called conjugate (or Reciprocal) tensor of Aij .

ij THEOREM 2.13 If Bij is the cofactor of Aij in the determinant d = |Aij| ¹ 0 and A defined as

Bij Aij = d kj k Then prove that Aij A = di . Proof: From the properties of the determinants, we have two results.

(i) AijBij = d

Bij Þ A = 1 ij d

Bij A Aij = 1, given Aij = ij d 26 Tensors and Their Applications

(ii) AijBkj = 0

Bkj Aij = 0, ¹ 0 d d kj Aij A = 0 if i ¹ k from (i) & (ii) ì1 if i = k A Akj = í ij î0 if i ¹ k

kj k i.e., Aij A = di

2.17 RELATIVE TENSOR If the components of a tensor Ai1i2...ir transform according to the equation j1 j2...js

w ¶ ¶ k1 ¶ k2 ¶ k r ¶ j1 ¶ j2 ¶ js k1k 2...k r x i1i2...ir x x x x x x Al l ...l = A ... ××× × ××× 1 2 s ¶x j1 j2 js ¶xi1 ¶xi2 ¶xir ¶xlr ¶x l2 ¶x ls ¶x Hence Ai1i2...ir is called a relative tensor of weight w, where is the Jacobian of transformation. If j1 j2 ...jr ¶x w =1, the relative tensor is called a . If w = 0 then tensor is said to be absolute.

MISCELLANEOUS EXAMPLES 1. Show that there is no distinction between contravariant and covariant vectors when we restrict ourselves to transformation of the type

i i m i x = am x + b ; where a's and b's are constants such that i i r aram = dm Solution Given that i i m i x = am x + b ...(1) i m i i or am x = x - b ...(2)

i Multiplying both sides (2) by ar, we get i i m i i i i aram x = ar x - b ar dr x m ai xi - biai i i r m = r r as given aram = dm r ai xi - biai r m r x = r r as dm x = x s i i i i or x = asx - b as Differentiating Partially it w.r.t. to x i ¶x s = ai ..(3) ¶xi s Tensor Algebra 27

Now, from (1) i i s i x = asx + b i ¶x i = as ...(4) ¶x s The contravariant transformation is i ¶x s i s i = A = as A ...(5) A ¶x s The covariant transformation is s ¶x i A = As = as As ...(6) i ¶x i Thus from (5) and (6), it shows that there is no distinction between contravariant and covariant tensor law of transformation i i 2. If the tensors aij and gij are symmetric and u ,v are components of contravariant vectors satisfying the equations i (aij - kgij )u = 0, i, j =1,2,...,n i (aij - k'gij )v = 0, k ¹ k¢ . i j i j Prove that giju v = 0, aiju v = 0.

Solution The equations are i (aij - kgij )u = 0 ...(1) i (aij - k¢gij )v = 0 ...(2) Multiplying (1) and (2) by u j and v j respectively and subtracting, we get i j i j i j j i aiju v - aijv u - kgiju v + k¢giju v = 0 Interchanging i and j in the second and fourth terms,

i j j i i j i j aiju v - a jiv u - kgiju v + k ¢g jiu v = 0

As aij and gij is symmetric i.e., aij = a ji & gij = g ji

j i i j - kgijv u + k ¢giju v = 0 i j (k ¢ - k)giju v = 0 i j giju v = 0 since k ¹ k¢ Þ k - k¢ ¹ 0 Multiplying (1) by v j , we get j i i j aijv u - kgiju v = 0 i j i j aiju v = 0 as giju v = 0. Proved. 28 Tensors and Their Applications

3. If Aij is a Skew-Symmetric tensor prove that i k i k (d jdl + dld j )Aik = 0

Solution

Given Aij is a Skew-symmetric tensor then Aij = -Aji . Now, i k i k i k i k (d jdl + dld j )Aik = d jdl Aik + dld j Aik i i = d j Ail + dl Aij

= Ajl + Alj

i k i k (d jdl + dld j )Aik = 0 as Ajl = -Alj

4. If aij is symmetric tensor and bi is a vector and aijbk + a jkbi + akibj = 0 then prove that

aij = 0 or bk = 0 . Solution The equation is

aijbk + a jkbi + akib j = 0

Þ aijbk + a jkbi + akibj = 0 By tensor law of transformation, we have

¶x p ¶xq ¶xr ¶x p ¶xq ¶xr ¶x p ¶xq ¶xr a b + a b + a b = 0 pq ¶x i ¶x j r ¶x k pq ¶x j ¶x k r ¶x i pq ¶x k ¶x i r ¶x j

é¶x p ¶xq ¶xr ¶x p ¶xq ¶xr ¶x p ¶xq ¶xr ù + + a pqbr ê i j k j k i k i j ú = 0 ë ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x û or b = 0 Þ a pqbr = 0 Þ a pq = 0 r

Þ aij = 0 or bk = 0 m n m n a x x = b x x r 5. If mn mn for arbitrary values of x , show that a(mn) = b(mn) i.e.,

amn + anm = bmn + bnm a a = b If mn and bmn are symmetric tensors then further show the mn mn . Solution Given m n m n amnx x = bmnx x

m n (amn - bmn )x x = 0 Tensor Algebra 29

Differentiating w.r.t. xi partially

n m (ain - bin )x + (ami - bmi)x = 0 Differentiating again w.r.t. x j partially

(aij - bij) + (a ji - b ji) = 0

aij + a ji = bij + bji

or amn + anm = bmn + bnm or a(mn) = b(mn)

Also, since amn and bmn are symmetric then amn = anm , bmn = bnm . So,

2amn = 2bmn a mn = bmn

EXERCISES

1. Write down the law of transformation for the tensors

(i) Aij

ij (ii) Bk ijk (iii) Clm

pq s pq s 2. If Ar and Bt are tensors then prove that Ar Bt is also a tensor..

ij ij 3. If A is a contravariant tensor and Bi is covariant vector then prove that A Bk is a tensor of rank ij three and A Bj is a tensor of rank one. i i j 4. If A is an arbitrary contravariant vector and Cij A A is an invariant show that Cij + Cji is a covariant tensor of the second order. 5. Show that every tensor can be expressed in the terms of symmetric and skew-symmetric tensor. n n 6. Prove that in n-dimensional space, symmetric and skew-symmetric tensor have (n +1) and (n -1) 2 2 independent components respectively.

7. If Uij ¹ 0 are components of a tensor of the type (0, 2) and if the equation fUij + gU ji = 0 holds

w.r.t to a basis then prove that either f = g and Uij is skew-symmetric or f = –g and Uij is symmetric.

i k i k 8. If Aij is skew-symmetric then (BjBl + Bl Bj )Aik = 0 .

ij i 9. Explain the process of contraction of tensors. Show that aija = d j . 30 Tensors and Their Applications

pq pr 10. If Ar is a tensor of rank three. Show that Ar is a contravariant tensor of rank one.

aijl m g k l , m ,g k ij 11. If k i j is a scalar or invariant, i j are vectors then ak is a mixed tensor of type (2, 1).

h i h k i i 12. Show that if ahijkl m l m = 0 where l and m are components of two arbitrary vectors then

ahijk + ahkji + a jihk + a jkhi = 0

i j i j 13. Prove that AijB C is invariant if B and C are vector and Aij is tensor of rank two. 14. If A(r, s, t) be a function of the coordinates in n-dimensional space such that for an arbitrary vector Br of the type indicated by the index a A(r, s, t)Br is equal to the component Cst of a contravariant st tensor of order two. Prove that A(r, s, t) are the components of a tensor of the form Ar . ij 15. If A and Aij are components of symmetric relative tensors of weight w. show that

w-2 w+2 ij ij ¶x ¶x A = A and A = A ¶x ij ij ¶x 16. Prove that the scalar product of a relative covariant vector of weight w and a relative contravariant vector of weight w¢ is a relative scalar of weight w + w¢ . CHAPTER – 3

METRIC TENSOR AND RIEMANNIAN METRIC

3.1 THE METRIC TENSOR In rectangular cartesian coordinates, the distance between two neighbouring point are (x, y, z) and (x + dx, y + dy, z + dz) is given by ds 2 = dx2 + dy2 + dz2 .

In n-dimensional space, Riemann defined the distance ds between two neighbouring points xi and xi + dxi (i = 1,2,...n) by quadratic

2 1 2 1 2 1 n ds = g11(dx ) + g12dx dx + ×××+ g1ndx dx

2 1 2 2 2 n + g12(dx )dx + g22(dx ) + ×××+ g2ndx dx + ...... +

n 1 n 2 n 2 + gn1dx dx + gn2dx dx + ×× ×× + gnn(dx ) 2 i j ds = gijdx dx (i, j =1,2,...n) ...(1) using summation convention.

i Where gij are the functions of the coordinates x such that

g = gij ¹ 0 The quadratic differential form (1) is called the Riemannian Metric or Metric or line element for n- dimensional space and such n-dimensional space is called Riemannian space and denoted by Vn and

gij is called Metric Tensor or Fundamental tensor.. The geometry based on Riemannian Metric is called the Riemannian Geometry.

THEOREM 3.1 The Metric tensor gij is a covariant symmetry tensor of rank two. Proof: The metric is given by

2 i j ds = gijdx dx ...(1) 32 Tensors and Their Applications

Let xi be the coordinates in X-coordinate system and x i be the coordinates in Y-coordinate 2 i j 2 i j system. Then metric ds = gij dx dx transforms to ds = gijdx dx . Since distance being scalar quantity.

2 i j i j So, ds = gijdx dx = gijdx dx ...(2) The theorem will be proved in three steps. (i) To show that dxi is a contravariant vector. If x i = xi (x1, x2,...xn)

¶x i ¶xi ¶x i dxi = dx1 + dx2 + × ×× + dxn ¶xi ¶x2 ¶xn ¶x i dxi = dxk ¶xk It is law of transformation of contravariant vector. So, dx i is contravariant vector..

(ii) To show that gij is a covariant tensor of rank two. Since ¶x i ¶x j dxi = dxk and dx j = ¶xl ¶xk ¶xl from equation (2) ¶x i ¶x j g dxidx j = g dxk dxl ij ij ¶xk ¶xl ¶x i ¶x j g dxidx j = g ¶xk dxl ij ij ¶x k ¶xl ¶x i ¶x j k l k l g dx dx = gij dx dx kl ¶xk ¶xl

i j k l Since gijdx dx = gkldx dx (i, j are dummy indices).

æ ¶x i ¶x j ö ç g - g ÷dx kdxl = 0 ç kl ij k l ÷ è ¶x ¶x ø

¶x i ¶x j k l or gkl - gij = 0 as dx and dx are arbitrary.. ¶xk ¶xl ¶x j ¶x j g = g kl ij ¶xk ¶xl ¶x k ¶xl or g = g ij kl ¶x i ¶x j

So, gij is covariant tensor of rank two. Metric Tensor and Riemannian Metric 33

(iii) To show that gij is symmetric. Then gij can be written as 1 1 g = (g + g ) + ( gij - g ji) ij 2 ij ji 2

gij = Aij + Bij 1 where A = (gij + g ji ) = symmetric ij 2 1 B = (g - g ) = Skew-symmetric ij 2 ij ji i j i j Now, gijdx dx = (Aij + Bij)dx dx from (3)

i j i j (gij - Aij)dx dx = Bijdx dx (4)

i j Interchanging the dummy indices in Bij dx dx , we have

i j i j Bijdx dx = Bjidx dx i j i j Bijdx dx = - Bijdx dx

Since Bij is Skew-symmetric i.e., Bij = -Bji

i j i j Bijdx dx + Bijdx dx = 0

i j 2Bijdx dx = 0

i j Þ Bijdx dx = 0 So, from (4), i j (gij - Aij)dx dx = 0

i j Þ gij = Aij as dx ,dx are arbitrary..

So, gij is symmetric since Aij is symmetric. Hence gij is a covariant symmetric tensor of rank two. This is called fundamental Covariant Tensor.

i j THEOREM 3.2 To show that gijdx dx is an invariant.

Proof: Let xi be coordinates of a point in X-coordinate system and x i be coordinates of a same point in Y-coordinate system.

Since gij is a Covariant tensor of rank two.

¶x k ¶x1 Then, g = g ij kl ¶x i ¶x j 34 Tensors and Their Applications

¶xk ¶xl Þ g - g = 0 ij kl ¶x i ¶x j æ ¶xk ¶xl ö ç g - g ÷dxidx j ç ij kl i j ÷ = 0 è ¶x ¶x ø

¶xk ¶xl (g dxidx j ) = g dxidx j ij kl ¶x i ¶x j ¶x k ¶xl = g dxi dx j kl ¶x i ¶x j

i j k l gijdx dx = gkldx dx

i j So, gijdx dx is an ivariant.

3.2 CONJUGATE METRIC TENSOR: (CONTRAVARIANT TENSOR)

ij The conjugate Metric Tensor to gij , which is written as g , is defined by

Bij ij = (by Art.2.16, Chapter 2) g g

where Bij is the cofactor of gij in the determinant g = gij ¹ 0 . By theorem on page 26

kj k Aij A = di

kj k So, gijg = di ij Note (i) Tensors gij and g are Metric Tensor or Fundamental Tensors. ij (ii) gij is called first fundamental Tensor and g second fundamental Tensors.

EXAMPLE 1 Find the Metric and component of first and second fundamental tensor is cylindrical coordinates. Solution Let (x1, x2, x3) be the Cartesian coordinates and (x1, x2,x3) be the cylindrical coordinates of a point. The cylindrical coordinates are given by x = r cosq, y = r sin q, z = z So that x1 = x, x 2 = y, x3 = z and x1 = r, x 2 = q, x 3 = z ...(1)

Let gij and gij be the metric tensors in Cartesian coordinates and cylindrical coordinates respectively. Metric Tensor and Riemannian Metric 35

The metric in Cartesian coordinate is given by

2 2 2 ds 2 = dx + dy + dz

1 2 2 2 3 2 ds 2 = (dx) +(dx ) +(dx ) ...(2)

2 i j But ds = gijdx dx

1 2 1 2 1 3 2 1 = g11(dx ) +g12dxdx + g13dx dx + g21dx dx

2 2 2 3 3 1 + g22(dx ) + g23dx dx + g31dx dx

3 2 3 3 + g32dx dx +g33(dx ) ...(3) Comparing (2) and (3), we have

g11 = g22 = g33 =1 and g12 = g13 = g21 = g23 = g31 = g32 = 0 On transformation ¶xi ¶x j g = g , since g is Covariant Tensor of rank two. (i, j = 1, 2, 3) ij ij ¶xi ¶x j ij for i = j = 1. 2 2 æ ¶x1 ö æ ¶x2 ö æ ¶x3 ö g ç ÷ + g ç ÷ + g ç ÷ g11 = 11ç 1 ÷ 22ç 1 ÷ 33ç 1 ÷ è ¶x ø è ¶x ø è ¶x ø

since g12 = g13 = ××× = g32 = 0 . 2 2 2 æ ¶x ö æ ¶y ö æ ¶z ö g11 = g11ç ÷ + g22ç ÷ + g33ç ÷ è ¶r ø è ¶r ø è ¶r ø Since x = r cosq, y = r sin q, z = z ¶x ¶y ¶z = cosq, = sin q, = 0 ¶r ¶r ¶r and g11 = g22 = g33 =1.

2 2 g11 = cos q + sin q + 0

g11 = 1 Put i = j = 2.

2 2 2 æ ¶x1 ö æ ¶x2 ö æ ¶x3 ö g ç ÷ + g ç ÷ + g ç ÷ g22 = 11ç 2 ÷ 22ç 2 ÷ 33ç 2 ÷ è ¶x ø è ¶x ø è ¶x ø

2 2 2 æ ¶x ö æ ¶y ö æ ¶z ö g22 = g11ç ÷ + g22ç ÷ + g33ç ÷ è ¶q ø è ¶q ø è ¶q ø 36 Tensors and Their Applications

Since g11 = g22 = g33 =1 ¶x ¶y ¶z = -r sin q, = r cosq, = 0 ¶q ¶q ¶q

2 2 g22 = (- r sin q) + (r cos q) + 0

= r 2 sin 2 q + r 2 cos2 q

2 g22 = r Put i = j = 3. 2 æ ¶x1 ö æ ¶x 2 ö æ ¶x3 ö g ç ÷ + g ç ÷ + g ç ÷ g33 = 11ç 3 ÷ 22ç 3 ÷ 33ç 3 ÷ è ¶x ø è ¶x ø è ¶x ø 2 æ ¶x ö æ ¶y ö æ ¶z ö = g11ç ÷ + g22ç ÷ + g33ç ÷ è ¶z ø è ¶z ø è ¶z ø ¶x ¶y ¶z Since = 0, = 0, = 1. So, g =1 . ¶z ¶z ¶z 33 2 So, g11 = 1, g22 = r , g33 =1

and g12 = g13 = g21 = g23 = g31 = g32 = 0 (i) The metric in cylindrical coordinates

2 i j ds = gijdx dx i, j =1,2,3.

2 2 2 2 1 2 3 ds = g11(dx ) + g22 (dx ) + g33(dx )

since g12 = g13 = ××× = g32 = 0

2 2 2 2 ds 2 = dr + r (dq) + df (ii) The first fundamental tensor is

ég g g ù é1 0 0ù 11 12 13 ê ú ê ú = 0 2 0 gij = êg21 g22 g23ú ê r ú ê ú ê0 0 1 ú ëg31 g32 g33 û ë û

1 0 0 2 since g = gij = 0 r 0 0 0 1 g = r 2 Metric Tensor and Riemannian Metric 37

(iii) The cofactor of g are given by 2 2 B11 = r , B22 =1, B33 = r

and B12 = B21 = B13 = B23 = B31 = B32 = 0

Bij The second fundamental tensor or conjugate tensor is g ij = . g

cofactor of g in g g11 = 11 g

B r 2 g11 = 11 = =1 g r 2

B 1 g 22 = 12 = g r 2

B r 2 g 33 = 33 = =1 g r 2 and g12 = g13 = g 21 = g 23 = g 31 = g32 = 0

é1 0 0ù ê 1 ú Hence the second fundamental tensor in form is ê0 0ú . ê r 2 ú ê ú ë0 0 1û

EXAMPLE 2 Find the matrix and component of first and second fundamental tensors in spherical coordinates. Solution

Let (x1, x2, x3 ) be the cartesian coordinates and (x1, x 2, x 3) be the spherical coordinates of a point. The spherical coordinates are given by x = r sin q cosf, y = r sin q sin f, z = r cos q

So that x1 = x, x 2 = y, x3 = z and x1 = r, x 2 =q , x3 = f

Let gij and gij be the metric tensors in cartesian and spherical coordinates respectively.. The metric in cartesian coordinates is given by 2 2 2 ds 2 = dx + dy + dz

2 2 2 ds 2 = (dx1) + (dx2 ) + (dx3)

2 i j But ds = gijdx dx ; (i, j =1,2,3) 38 Tensors and Their Applications

Þ g11 = g22 = g33 =1 and g12 = g23 = g13 = g21 = g31 = g32 = 0 On transformation

¶xi ¶x j gij = gij ¶xi ¶x j

(since gij is covariant tensor of rank two) (where i, j = 1,2,3).

¶x1 ¶x1 ¶x 2 ¶x 2 ¶x3 ¶x3 g = g11 + g22 + g33 ij ¶x i ¶x j ¶x i ¶x j ¶x i ¶x1 since i, j are dummy indices. Put i = j = 1

2 2 2 æ ¶x1 ö æ ¶x 2 ö æ ¶x3 ö g ç ÷ + g ç ÷ + g ç ÷ g11 = 11ç 1 ÷ 22ç 1 ÷ 33ç 1 ÷ è ¶x ø è ¶x ø è ¶x ø

2 2 2 æ ¶x ö æ ¶y ö æ ¶z ö g = g11ç ÷ + g22ç ÷ + g33ç ÷ 11 è ¶r ø è ¶r ø è ¶r ø Since x = r sin qcos f, y = r sin q sin f, z = r cosq

¶x ¶y ¶z = sin qcos f, = sin qsin f, = cosq ¶r ¶r ¶r

and g11 = g22 = g33 =1. So, 2 2 2 g11 = (sin qcosf) + (sin q sin f) + cos q

g11 = 1 put i = j = 2

2 2 æ ¶x1 ö æ ¶x 2 ö æ ¶x3 ö g ç ÷ + g ç ÷ + g ç ÷ g22 = 11ç 2 ÷ 22ç 2 ÷ 33ç 2 ÷ è ¶x ø è ¶x ø è ¶x ø

2 2 2 æ ¶x ö æ ¶y ö æ ¶z ö g = g11ç ÷ + g22ç ÷ + g33ç ÷ 22 è ¶q ø è ¶q ø è ¶q ø

since g11 = g22 = g33 =1 ¶x ¶y ¶z = r cosq cosf, = r cosqsin f, = -r sin q ¶q ¶q ¶q

2 2 2 g22 = (r cosq cosf) + (r cosqsin f) + (- r sin q)

2 g22 = r Metric Tensor and Riemannian Metric 39

Put i = j = 3

2 2 2 æ ¶x1 ö æ ¶x 2 ö æ ¶x3 ö g ç ÷ + g ç ÷ + g ç ÷ g33 = 11ç 3 ÷ 22ç 3 ÷ 33ç 3 ÷ è ¶x ø è ¶x ø è ¶x ø

2 2 2 æ ¶x ö æ ¶y ö æ ¶z ö g = g11ç ÷ + g22ç ÷ + g33ç ÷ 33 è ¶f ø è ¶f ø è ¶f ø

since g11 = g22 = g33 =1 ¶x ¶y ¶z = -r sin q sin f, = r sin qcosf, = 0 and ¶f ¶f ¶f

2 2 g33 = (- r sin qsin f) + (r sin q cosf) + 0

2 2 g33 = r sin q So, we have

2 2 2 g11 = 1, g22 = r , g33 = r sin q

and g12 = g13 = g21 = g23 = g31 = g32 = 0 (i) The Metric in spherical coordinates is

2 i j ds = gijdx dx ; i, j =1,2,3

2 2 2 2 1 2 3 ds = g11(dx ) + g22(dx ) + g33(dx ) ds 2 = dr 2 + r 2dq2 + r 2 sin 2 qdf2 (ii) The Metric tensor or first fundamental tensor is 1 0 0 ég11 g12 g13 ù é ù ê ú ê 2 ú gij = êg21 g22 g23 ú = ê0 r 0 ú ê ú ê 2 2 ú ëg31 g32 g33û ë0 0 r sin qû and

1 0 0 2 4 2 g = gij = 0 r 0 = r sin q 0 0 r 2 sin 2 q 2 2 2 (iii) The cofactor of g are given by B11 =1, B22 = r , B33 = r sin q and B12 = B21 =

B31 = B13 = B23 = B32 = 0 B The second fundamental tensor or conjugate tensor is g ij = ij . g 40 Tensors and Their Applications

cofactor of g in g B g11 = 11 = 11 g g r 4 sin 2 q = r 4 sin 2 q g11 = 1 2 2 B22 r sin q g 22 = = g r 4 sin 2 q 1 g 22 = r 2 2 B33 r g 33 = = g r 4 sin 2 q 1 g 33 = r 2 sin 2 q and g12 = g13 = g 21 = g31 = g 32 = 0 Hence the fundamental tensor in matrix form is

é ù ê0 0 0 ú ég 11 g12 g 13ù ê 1 ú 0 0 ij ê 21 22 23ú ê 2 ú g = êg g g ú = r ê 31 32 33ú ê 1 ú ëg g g û ê0 0 ú ê r 2 sin 2 q ú ë û EXAMPLE 3 If the metric is given by

2 2 2 ds 2 = 5(dx1) + 3(dx 2) + 4(dx3) - 6dx1dx 2 + 4dx 2dx3

Evaluate (i) g and (ii) g ij .

Solution

2 i j The metric is ds = gijdx dx ; (i, j =1, 2, 3)

1 2 1 2 1 3 2 1 2 ds = g11(dx ) + g12dx dx + g13dx dx + g21dx dx 2 2 2 3 3 1 3 2 3 2 + g22(dx ) + g23dx dx + g31dx dx + g32dx dx + g33(dx )

Since gij is symmetric Þ gij = g ji

i.e., g12 = g21, g23 = g32, g13 = g31 Metric Tensor and Riemannian Metric 41

2 1 2 2 2 3 2 1 2 So, ds = g11(dx ) + g22(dx ) + g33(dx ) + 2g12dx dx

2 3 1 3 + 2g23dx dx + 2g13dx dx ...(1) Now, the given metric is

1 2 2 2 3 2 1 2 2 3 ds 2 = 5(dx ) + 3(dx ) + 4(dx ) - 6dx dx + 4dx dx ...(2) Comparing (1) and (2) we have

g11 = 5, g22 = 3, g33 = 4, 2g12 = -6 Þ g12 = -3 = g21

2g23 = 4 Þ g23 = 2 = g32, g13 = 0 = g31

g11 g12 g13 5 - 3 0

g = gij = g21 g22 g23 = - 3 3 2 = 4

g31 g32 g33 0 2 4

(ii) Let Bij be the cofactor of gij in g. Then 3 2 B11 = Cofactor of g = = 8 11 2 4 5 0 B22 = Cofactor of g = = 20 22 0 4

5 - 3 B = Cofactor of g = = 6 33 33 - 3 3 - 3 2 B = Cofactor of g = - = 12 = B 12 12 0 4 21

- 3 3 B = Cofactor of g = = -6 = B 13 13 0 2 31 5 - 3 B = Cofactor of g23 = – = -10 = B32 23 0 2

Bij Since gij = g We have

11 B11 8 22 33 3 12 21 13 31 3 23 32 5 g = = = 2; g = 5, g = , g = g = 3, g = g = - , g = g = - g 4 2 2 2 42 Tensors and Their Applications

Hence,

é 3 ù ê 2 3 - ú ê 2 ú ê 5 ú g ij = 3 5 - ê 2 ú ê ú 3 5 3 ê- - ú ëê 2 2 2ûú 3.3 LENGTH OF A CURVE i Consider a continuous curve in a Riemannian Vn i.e., a curve such that the coordinate x of any current point on it are expressible as functions of some parameter, say t. The equation of such curve can be expressed as xi = xi (t)

The length ds of the arc between the points whose coordinate s are xi and xi + dxi given by

2 i j ds = gijdx dx

If s be arc length of the curve between the points P1 and P2 on the curve which correspond to the two values t1 and t2 of the parameter t.

1 2 i j P2 t2 æ dx dx ö = ç ÷ s = ò ds ò ç gij ÷ dt P1 t1 è dt dt ø

NULL CURVE dxi dx j If g = 0 along a curve. Then s = 0. Then the points P1 and P2 are at zero distance, despite ij dt dt of the fact that they are not coincident. Such a curve is called minimal curve or null curve.

EXAMPLE 4 A curve is in spherical coordinate xi is given by

2 -1æ 1ö 3 2 x1 = t, x = sin ç ÷ and x = 2 t -1 è t ø Find length of arc 1 £ t £ 2. Solution In spherical coordinate, the metric is given by 1 2 1 2 2 2 1 2 2 3 2 ds 2 = (dx ) + (x ) (dx ) + (x sin x ) (dx ) Metric Tensor and Riemannian Metric 43

2 -1 1 3 2 given x1 = t, x = sin , x = 2 t -1 t

1 æ 1 ö 1 -1 2 dx1 = dt , dx 2 = ç - ÷dt, dx3 = 2 × (t2 -1) 2t dt 2 2 æ 1ö è t ø 2 1 - ç ÷ è t ø

dt 3 2t dx 2 = - , dx = dt t t 2 -1 t 2 -1

2 2 2 2 æ dt ö é æ 1 öù æ 2t ö 2 = (dt) + t 2ç - ÷ + tç sin sin -1 ÷ ç dt ÷ ds ç 2 ÷ ê ú ç 2 ÷ è t t -1 ø ë è t øû è t -1 ø

2 2 2 dt 4t 2 2 = dt + + (dt) ds t 2 -1 t 2 -1

2 5t 2 2 = dt ds t 2 -1 t ds = 5 dt t2 -1 Now, the length of arc, 1 £ t £ 2, is

2 2 é 2 ù t2 t 5 t - 1 ds = 5 dt = ê ú = units ò ò1 2 15 t1 -1 2 ê 1 2 ú t ë û1

3.4 ASSOCIATED TENSOR A tensor obtained by the process of inner product of any tensor Ai1i2...ir with either of the fundamental j1 j2... js ij tensor gij or g is called associated tensor of given tensor..

e.g. Consider a tensor Aijk and form the following inner product

ai a aj a ak a g Aijk = Ajk ; g Aijk = Aik; g Aijk = Aij

All these tensors are called Associated tensor of Aijk . Associated Vector ik k Consider a covariant vector Ai . Then g Ai = A is called associated vector of Ai . Consider a j j j contravariant vector A . Then g jk A = Ak is called associated vector of A . 44 Tensors and Their Applications

3.5 MAGNITUDE OF VECTOR The magnitude or length A of contravariant vector Ai. Then A is defined by

i j A = gij A A

2 i j or A = gijA A 2 j i Also, A = Aj A as gij A = Aj i.e., square of the magnitude is equal to scalar product of the vector and its associate.

The magnitude or length A of covariant vector Ai . Then A is defined by

ij A = g Ai Aj 2 ij or A = g Ai Aj A vector of magnitude one is called Unit vector. A vector of magnitude zero is called zero vector or Null vector.

3.6 SCALAR PRODUCT OF TWO VECTORS r r r r Let A and B be two vectors. Their scalar product is written as A× B and defined by r r i A× B = A Bi

r r i i j j Also, A× B = A Bi = gijA B since Bi = gijB

r r i ij i ij A× B = AiB = g AiBj since B = g B j Thus r r i i j 2 A× A = A Ai = gijA A = A

r i j i.e., A = A = gij A A Angle between two vectors r r Let A and B be two vectors. Then r r r r A× B = A B cosq

r r i j A × B gijA B Þ cosq = = r r i j i j A B gijA A gijB B

r i j r i j since A = gij A A ; B = gijB B This is required formula for cosq . Definition r i r The inner product of two contravariant vectors A (or A ) and B (or Bi) associated with a symmetric i j tensor gij is defined as gijA B . It is denoted by r r i j g(A,B) = gijA B Metric Tensor and Riemannian Metric 45

r r THEOREM 3.3 The necessary and sufficient condition that the two vectors A and B at 0 be orthogonal r r if g(A, B) = 0 r r Proof: Let q be angle between the vectors A and B then r r r r A× B = A B cosq r r or A× B = AB cosq

i j gijA B = AB cosq i j gij A B Þ cosq = ...(1) AB r r p If A and B are orthogonal then q = Þ cosq = 0 then from (1) 2 i j gijA B = 0

r r r r i j Þ g(A,B) = 0 since g(A,B) = gijA B i j Conversely if gij A B = 0 then from (1) p cosq = 0 Þ q = . 2 r r So, two vectors A & B are orthogonal. Proved. r r r r Note: (i) If A and B be unit vectors. Then A = B = 1. Then r r i j cosq = A× B = gijA B r r p p (ii) Two vectors A and B will be orthogonal if angle between them is i.e., q = then 2 2 p cosq = cosq = = 0 2

3.7 ANGLE BETWEEN TWO VECTORS

THEOREM 3.4 To show that the definition of the angle between two vectors is consistent with the requirement cos2q £ 1. OR To justify the definition of the angle between two vectors. OR To show that the angle between the contravariant vectors is real when the Riemannian Metric is positive definition. r r Proof: Let q be the angle between unit vectors A and B then i j j ij ij i cosq = gijA B = Aj B = AjB Bi = g AjBi = A Bi 46 Tensors and Their Applications

To show that q is real i.e., |cosq| £ 1. Consider the vector lAi + mB i when l and m are scalars. The square of the magnitude of

i i i i j j lA + mB = gij(lA + mB ) (lA + mB )

2 i j i j i j 2 i j = gijl A A + gijlmA B + m lg ij B A + m gijB B

= l 2 + 2lm cosq + m2 Since i j 2 i j 2 gijA A = A = 1; gijB B = B = 1. and

i j r r r 2 gijA B = cosq; as A & B are unit vector i.e., A = 1Þ A =1 . Since square of magnitude of any vector ³ 0. So, the square of the magnitude of lAi + mBi ³ 0. or l 2 + 2lm cosq + m2 ³ 0 (l + m cosq)2 + m2 - m2 cos2 q ³ 0 (l + m cosq)2 + m2(1- cos2 q) ³ 0 This inequality holds for the real values of l & m. if 1- cos2 q ³ 0 Þ cos2 q £ 1 cosq £ 1 Proved. THEOREM 3.5 The magnitude of two associated vectors are equal. i Proof: Let A and B be magnitudes of associate vectors A and Ai respectively. Then 2 i j A = gijA A ...(1) and 2 ij B = g Ai Aj ...(2) From equation (1) 2 i j A = (gij A )A 2 j A = Aj A ...(3)

i j since gijA = A (Associate vector) From equation (2) ij B2 = (g Ai )Aj

2 j B = A Aj ...(4) Metric Tensor and Riemannian Metric 47

ij j since g Ai = A from (3) and (4) 2 A2 = B Þ A = B i So, magnitude of Ai and A are equal.

3.8 ANGLE BETWEEN TWO COORDINATE CURVES i l Let a Vn referred to coordinate x , (i = 1, 2, ...n) . For a coordinate curve of parameter x , the coordinate xl alone varies. Thus the coordinate curve of parameter xl is defined as xi = ci , "i except i = l ...(1) where C i,s are constants. Differentiating it, we get dx i = 0, "i , except i = l and dxl ¹ 0 Let Ai and Bi be the tangent vectors to a coordinate curve of parameters x p and x q respectively.. Then Ai = dxi = (0,...0, x p, 0...0) ...(2)

Bi = dxi = (0,...0,x q,0...0) ...(3) If q is required angle then

i j gijA B cosq = i j i j gijA A gijB B

p q g pqA B = p p q q g pp A A gqqB B

p q g pq A B = p q g ppgqq A B

g pq cosq = ...(4) g ppgqq

which is required formula for q .

i j The angle wij between the coordinate curves of parameters x and x is given by

gij cos wij = giig jj 48 Tensors and Their Applications

If these curves are orthogonal then p cos w = cos = 0 ij 2 Þ gij = 0

i j Hence the x coordinate curve and x coordinate curve are orthogonal if gij = 0 .

3.9 HYPERSURFACE i i 1 1 The n equations x = x (u ) represent a subspace of Vn . If we eliminate the parameter u , we get (n –1) equations in xj,s which represent one dimensional curve. i i 1 2 Similarly the n equations x = x (u ,u ) represent two dimensional subspace of Vn. If we eliminating 1 2 i, the parameters u , u , we get n –2 equations in x s which represent two dimensional curve Vn. This two dimensional curve define a subspace, denoted by V2 of Vn. i i 1 2 n–1 Then n equations x = x (u , u , ... u ) represent n – 1 dimensional subspace Vn–1 of Vn. If we eliminating the parameters u1, u2, ...un–1, we get only one equation in xi,s which represent n –1 dimensional curve in Vn. This particular curve is called hypersurface of Vn. i Let f be a scalar function of coordinates xi. Then f(x ) = constant determines a family of hypersurface of Vn.

3.10 ANGLE BETWEEN TWO COORDINATE HYPERSURFACE Let f(xi ) = constant ...(1)

and y(xi) = constant ...(2) represents two families of hypersurfaces. Differentiating equation (1), we get

¶f i dx = 0 ...(3) ¶xi ¶f ¶f This shows that is orthogonal to dxi. Hence is normal to f = constant,since dx i is ¶xi ¶xi tangential to hypersurface (1). ¶y Similarly is normal to the hypersurface (2). If w is the angle between the hypersurface (1) ¶xi and (2) then w is also defined as the angle between their respective normals. Hence required angle w is given by ¶f ¶y gij ¶xi ¶x j cos w = ...(4) ¶f ¶f ¶y ¶y gij gij ¶xi ¶x j ¶xi ¶x j Metric Tensor and Riemannian Metric 49

If we take f = x p = constant ...(5) and y = xq = constant ...(6) The angle w between (5) and (6) is given by

¶x p ¶xq gij i j cos w = ¶x ¶x ¶x p ¶x p ¶xq ¶x q gij g ij ¶xi ¶x j ¶xi ¶x j

ij p q g di d j = ij p q ij q q g di d j g di d j

g pq cos w = ...(7) g ppg qq

i j The angle wij between the coordinate hypersurfaces of parameters x and x is given by

gij coswij = ...(8) giig jj

If the coordinate hypersurfaces of parameters xi and x j are orthogonal then p w = ij 2

Þ coswij = 0 from (8), we have g ij = 0 . Hence the coordinate hypersurfaces of parameters xi and xj are orthogonal if g ij = 0 .

3.11 n-PLY ORTHOGONAL SYSTEM OF HYPERSURFACES

If in a Vn there are n families of hypersurfaces such that, at every point, each hypersurface is orthogonal to the n -1 hypersurface of the other families which pass through that point, they are said to form as n-ply orthogonal system of hypersurfaces.

3.12 CONGRUENCE OF CURVES

A family of curves one of which passes through each point of Vn is called a congruence of curves.

3.13 ORTHOGONAL ENNUPLE

An orthogonal ennuple in a Riemannian Vn consists of n mutually orthogonal congruence of curves. 50 Tensors and Their Applications

ij THEOREM 3.6 To find the fundamental tensors gij and g in terms of the components of the unit tangent eh (h = 1, 2,... n) to an orthogonal ennuple. i Proof: Consider n unit tangents eh (h =1, 2,...n) to conguence eh (h = 1, 2,... n) of an orthogonal ennuple in a Riemannian Vn . The subscript h followed by an upright bar simply distinguishes one congruence from other. It does not denote tensor suffix.

The contravariant and covariant components of eh| are denoted by eh| and eh|i respectively..

Suppose any two congruences of orthogonal ennuple are eh| and ek| so that

i j h gijeh|ek| = dk ...(1)

ei e h h| k|i = dk from (1), i j gij eh| ek| = 0

i j and gijeh| eh| = 1 We define

cofactor of e in determinant e i h|i h|i eh| = eh|i Also, from the determinant property, we get n ei e i å h| h|j = d j ...(2) h=1 Multiplying by e jk

n ei e j k i jk å h| h| j g = d jg h=1

n ei ek ik or å h| h| = g ...(3) h=1

Again multiplying (2) by gik. n ei e g i å h| h|j ik = d jgik h=1

g or jk = å eh|keh| j ...(4) from (3) and (4)

n g e e ij = å h|i h| j ...(5) h=1 Metric Tensor and Riemannian Metric 51

n ij ei e j g = å h| h| ...(6) h=1 This is the required results.

Corollary: To find the magnitude of any vector u is zero if the projections of u on eh| are all zero. Proof: Let n i C ei u = å h h| ...(7) h=1 Then

n n i C ei e = C dh = C u ek|i = å h h| k|i å h k k h=1 h=1

i or Ck = u ek|i ...(8)

i i.e., Ck = projection of u on ek|i Using (8), equation (7) becomes

n i u je ei u = å h| j k| h=1 Now,

æ ö æ ö u2 uiu = ç C ei ÷ ç C e ÷ = i ç å h h| ÷ ç å k k|i ÷ from (7) è h ø è k ø

i = å ChCkeh|ek|i h,k

h = å ChCkdk h,k

C C = å h h h

n 2 C 2 u = å ( h ) h=1

2 This implies that u = 0 iff u = 0 iff Ch = 0 . Hence the magnitude of a vector u is zero iff all the projections of u (i.e. of ui) on n mutually

i orthogonal directions eh| are zero. 52 Tensors and Their Applications

Miscellaneous Examples 1. If p and q are orthogonal unit vectors, show that h i j k (ghjgik - ghk gij ) p q p q = 1 Solution Since p and q are orthogonal unit vectors. Then j 2 2 gij piq = 0, p = q = 1. Now, h i j k h j i k h k i j (ghjgik - ghk gij ) p q p q = ghjgik p p q q - ghk gij p q q p

h j i k j k i j = (ghi p p ) (gik q q ) - (ghk p q ) (gijq p ) = p2.q2 – 0.0 = 1 . 1 h j h k = 1 (since ghi p p =1& ghk p q = 0 ) 2. If q is the inclination of two vectors A and B show that

h j i k (ghigik - ghk gij)A A B B 2 sin q = h j j k ghj gik A A B B

Solution If q be the angle between the vectors A and B then

j i gij A B cosq = i j i k gij A A gikB B But sin 2 q =1 - cos2 q

i j h k 2 (gijB A ) (ghk A B ) sin q = 1- h j i k (ghjA A ) (gik B B )

h j i k (ghjgik - ghk gij )A A B B = h j i k ghjgik A A B B

3. If Xij are components of a symmetric covariant tensor and u, v are unit orthogonal to w and satisfying the relations i (X ij - agij)u + gw j = 0 i (Xij - bgij) v + dw j = 0 where a ¹ b prove that u and v are orthogonal and that Metric Tensor and Riemannian Metric 53

i j Xiju v = 0 Solution

i j i Suppose Xij is a symmetric tensor. Since u ,v are orthogonal to w then i u wi = 0 ...(1) i v wi = 0 ...(2)

i given (Xij - agij)u + gw j = 0 ...(3) i (Xij -bgij )v + dwj = 0 ...(4) where a ¹ b.

Multiply (3) & (4) by v j ,u j respectively and using (1) and (2), we have i j (Xij - agij )u v = 0 ...(5) i j (Xij - bgij)v u = 0 ...(6)

Interchanging the suffixes i & j in the equation (6) and since gij, X ij are symmetric, we get i j (Xij - agij )u v = 0 ... (7) Subtract (6) & (7) we get i j (b - a)giju v = 0 Since b ¹ a and b - a ¹ 0. Hence, i j giju v = 0 ...(8) So, u and v are orthogonal. Using (8) in equation (5) & (6), we get i i Xiju v = 0 Proved. 4. Prove the invariance of the expression gdx1dx 2...dx n for the element volume. Solution

Since gij is a symmetric tensor of rank two. Then ¶xk ¶xl g = g ij ¶x i ¶x j kl Taking determinant of both sides ¶x k ¶xl g = gkl ij ¶x i ¶x j ¶x Since = J (Jacobian) ¶x g = g gkl = g & ij 54 Tensors and Their Applications

So, g = gJ 2 or g J = g Now, the transformation of coordinates from xl to x i , we get

¶x 1 2 n 1 2... n = dx d x ... dx dx dx dx ¶x = Jd x1d x 2... d x n

g 1 2 n 1 2 n = d x d x ... d x dx dx ... dx g gdx1dx2...dxn = gd x1d x 2... d x n

So, the volume element dv = g dx1dx2... dxn is invariant.

EXERCISES

kl 1. For the Metric tensor gij defined g and prove that it is a contravariant tensor. ij 2. Calculate the quantities g for a V3 whose fundamental form in coordinates u, v, w, is adu2 + bdv2 + cdw2 + 2 fdvdw + 2gdwdu + 2hdudv 3. Show that for an orthogonal coordinate system 1 1 1 g11 = , g22 = , g33 = g11 g22 g33

4. For a V2 in which g11 = E, g12 = F, g21 = G prove that

g = EG - F2 , g11 = G g , g12 = - F g, g 22 = E g 1 5. Prove that the number of independent components of the metric g cannot exceed n(n+1) . ij 2 i i i ij i ij ij j i i i j ij 6. If vectors u , v are defined by u = g uj, v = g vj show that ui = g u , u vi = uiv and u giju = uig uj 7. Define magnitude of a unit vector. prove that the relation of a vector and its associate vector is reciprocal.

8. If q is the angle between the two vectors Ai and Bi at a point, prove that

h i j k (ghigik - ghkgij)A A B B 2 sin q = h i j k ghig jk A A B B 9. Show that the angle between two contravariant vectors is real when the Riemannian metric is positive definite. CHAPTER – 4

CHRISTOFFEL'S SYMBOLS AND COVARIANT DIFFERENTIATION

4.1 CHRISTOFFEL'S SYMBOLS The German Mathematician defined symbols

1 æ ¶g ¶g jk ¶gij ö ç ik + - ÷ [ij,k] = ç j i k ÷ , (i, j, k = 1,2,...n) ...(1) 2 è ¶x ¶x ¶x ø called Christoffel 3-index symbols of the first kind. ì k ü and í ý = g k l [ij, l] ...(2) îi jþ called Christoffel 3-index symbols of second kind, where g ij are the components of the metric Tensor or fundamental Tensor.

There are n distinct of each kind for each independent g ij . Since g ij is 1 symmetric tensor of rank two and has n (n +1) independent components. So, the number of 2 1 1 independent components of Christoffel’s symbols are n × n(n +1) = n2 (n +1). 2 2 ì k ü THEOREM 4.1 The Christoffel's symbols [ij,k] and í ý are symmetric with respect to the indices i îi jþ and j. Proof: By Christoffel’s symbols of first kind

1 æ ¶g ¶g jk ¶gij ö ç ik + ¶ - ÷ [ij,k] = ç j i k ÷ 2 è ¶x ¶x ¶x ø Interchanging i and j, we get

1 æ ¶g jk ¶g ¶g ji ö , = ç + ik - ÷ [ji k] ç i j k ÷ 2 è ¶x ¶x ¶x ø 56 Tensors and Their Applications

1 æ ¶g ¶g jk ¶g ij ö ç ik + - ÷ = = ç j i k ÷ since gij g ji 2 è ¶x ¶x ¶x ø [ ji, k] = [ij , k] Also, by Christoffel symbol of second kind ì k ü í ý = g k l [ij, l] îi jþ = g k l[ ji,l] since [ij,l]= [ji,l] ì k ü ì k ü í ý = í ý îi jþ î j iþ Proved. THEOREM 4.2 To prove that ïì k ïü (i) [ij,m] = g km í ý îïi j þï ¶gij (ii) [ik, j]+ [jk,i]= ¶xk ¶g ì i ü ì j ü ij - g jl - gim (iii) k = í ý í ý ¶x îl k þ îm kþ Proof: (i) By Christoffel’s symbol of second kind ì k ü í ý = g k l [ij,l] îi jþ

Multiplying this equation by gk m, we get ì k ü ï ï kl gk m í ý = gk m g [ij,l] îïi jþï l k l l = dm [ij,l] as gk m g = dm ì k ü gkm í ý = [ij,m] îi jþ (ii) By Christoffel’s symbol of first kind

1 æ ¶g k j ¶g ij ¶g ö ç + - ik ÷ [ik , j] = ç i k j ÷ ...(1) 2 è ¶x ¶x ¶x ø

1 æ ¶g ¶g ji ¶g jk ö and , = ç ki + - ÷ ...(2) [jk i] ç j k i ÷ 2 è ¶x ¶x ¶x ø adding (1) and (2), 1 æ ¶g ¶g ö ç ij + ji ÷ [ik, j]+ [jk,i] = ç k k ÷ since gij = g ji 2 è ¶x ¶x ø Christoffel's Symbols and Covariant Differentiation 57

1 ¶gij ¶gij [ik, j]+ [jk,i] = × 2 = 2 ¶xk ¶xk

ij i (iii) Since g glj = dl .

Differentiating it w.r.t. to x k , we get

ij ¶glj ¶g gij + g = 0 ¶xk lj ¶x k Multiplying this equation by g lm , we get

ij ¶glj ¶g g ijglm + glmg = 0 ¶xk lj ¶xk

ij ¶g ¶glj g lmg = - gijg lm lj ¶x k ¶xk

ij ¶g ¶glj dm = - gijg lm{ [lk, j]+ [ jk,l] } since = [lk, j]+ [jk,l]. j ¶xk ¶xk

im ¶g lm ij ij lm = - g {g [lk, j]}- g {g [ jk,l]} ¶x k

¶g im ì i ü ì m ü - glm - g ij k = í ý í ý ¶x îl k þ î j kþ Interchanging m and j, we get

ij ¶g ïì i ïü ïì j ïü - g lj - g im k = í ý í ý ¶x îïl k þï îïm k þï

¶g ij ì i ü ì j ü - g ij - g im lj jl or k = í ý í ý as g = g ¶x îl k þ îm k þ Proved.

ì i ü ¶ log ( g ) THEOREM 4.3 To show that í ý = îi jþ ¶x j

Proof: The matrix form of gik is

ég11 g12 ... g1n ù ê ú g g ... G g = ê 21 22 2n ú ik ê ú ê M ú ëêgn1 gn2 ... gnn ûú 58 Tensors and Their Applications

g11 g12 ... g1n

g21 g22 ... g2n and g = gik = M

gn1 gn2 ... gnn il l But gik g = dk Take l = k.

ik k gik g = dk =1 G Þ g ik = [g ]-1 = ik (Theorem 2.13 , Pg 25) ik g g g where Gik is cofactor of ik in the determinant ik

Þ g = gikGik ...(1)

Differentiating w.r.t. gik partially

¶g ¶gik = Gik since =1 ¶gik ¶gik Now,

¶g ¶g ¶gik j = j ¶x ¶gik ¶x

¶gik = Gik ¶x j

ik But Gik = gg

¶g ik ¶g = gg ik ¶x j ¶x j

1 ¶g ik ¶gik = g g ¶x j ¶x j

1 ¶g ¶g = g ik{ [ jk,i] + [ij,k] } as ik = [ jk,i] + [ij,k ] g ¶x j ¶x j

= g ik [ jk,i] + g ik [ij,k]

ì k ü ì i ü = í ý + í ý î j kþ îi jþ Christoffel's Symbols and Covariant Differentiation 59

1 ¶g ì i ü ì i ü + j = í ý í ý as k is dummy indices g ¶x î j iþ î j iþ

1 ¶g ì i ü = 2 j í ý g ¶x î j iþ 1 ¶g ì i ü j = í ý 2g ¶x î j iþ

¶ log( g ) ïì i ïü j = í ý Proved. ¶x îï j i þï

EXAMPLE 1

If gij ¹ 0 show that

b ¶ ì b ü ¶ ïì ïü g j [ik ,a] - í ý ([bj,a]+ [aj,b] ) ab j í ý = ¶x îi kþ ¶x îïi k þï Solution By Christoffel’s symbol of second kind

ì b ü ba í ý = g [ik,a] îi k þ

Multiplying it by gab , we get

b ba ì ü [ ,a] gab í ý = gab g ik îi k þ b ïì ïü ba g ab í ý = [ik,a] as gab g = 1 îïi k þï

Differentiating it w.r.t. to x j partially

¶ é ì b üù ¶ g [ik,a] j ê ab í ýú = j ¶x ë îi k þû ¶x b b ¶ ïì ïü ïì ïü ¶g a b ¶ g + ik,a ab j í ý í ý j = j [ ] ¶x îïi k þï îïi k þï ¶x ¶x

¶gab since = [aj,b]+ [bj, a] ¶x j ¶ ì b ü ì b ü g + ([aj,b]+ [bj,a]) ¶ ab j í ý í ý = [ik,a] ¶x îi k þ îi k þ ¶x j 60 Tensors and Their Applications

¶ ì b ü ¶ ì b ü g [ik,a] - í ý ([ai,b] + [bj,a] ) ab j í ý = j Solved. ¶x îi kþ ¶x îi k þ

EXAMPLE 2 ì k ü Show that if gij = 0 for i ¹ j then (i) í ý = 0 whenever i, j and k are distinct. îi jþ ì i ü 1 ¶ log g ì i ü 1 ¶ log g ì i ü 1 ¶g = ii = ii = - jj (ii) í ý i (iii) í ý j (iv) í ý i îi iþ 2 ¶x îi jþ 2 ¶x î j jþ 2gii ¶x Solution The Christoffel’s symbols of first kind

1 æ ¶g jk ¶g ¶gij ö , = ç + ik - ÷ ...(1) [ij k] ç i j k ÷ 2 è ¶x ¶x ¶x ø (a) If i = j = k The equation (1) becomes 1 ¶g [ii, i] = ii 2 ¶xi (b) If i = j ¹ k The equation (1) becomes 1 é ¶g ¶g ¶g ù i k + ik – ii [ii, k] = ê i i k ú 2 ë ¶x ¶x ¶x û

Since gik = 0 as i ¹ k (given) 1 ¶g 1 ¶g [ii, k] = - ii or [ jj,i]= - jj 2 ¶xk 2 ¶xi (c) i = k ¹ j

1 é ¶g ji ¶g ¶gij ù + ii - [ij,i] = ê i j i ú 2 ë ¶x ¶x ¶x û 1 ¶g = ii , as g = 0, i ¹ j 2 ¶x j ij (d) i ¹ j ¹ k

[ij,k] = 0 as gij = 0 , g j k = 0 , i ¹ j ¹ k (i) as i, j, k are distinct i.e., i ¹ j ¹ k

ì k ü í ý = g kl [ij,l] since g kl = 0, k ¹ l îi jþ Christoffel's Symbols and Covariant Differentiation 61

ì k ü í ý = 0 îi jþ ì i ü (ii) í ý = gii[ii,i] îi iþ 1 ¶g = g ii × ii from (a) 2 ¶xi

1 ¶g ii 1 = ii as g = i g 2gii ¶x ii ì i ü 1 ¶ log g í ý = ii îi iþ 2 ¶xi (iii) i = k ¹ j ì i ü í ý = gii[ij,i] îi jþ 1 1 = [ij,i] as g ii = g ii gii

ì i ü 1 ¶ log gii í ý = j îi jþ 2 ¶x (iv) j = k ¹ i ì i ü í ý = gii[ jj,i] î j jþ

1 1 ¶g jj = - i from (b) gii 2 ¶x

ì i ü -1 ¶g jj í ý = i Solved. î j jþ 2gi i ¶x

EXAMPLE 3

If ds 2 = dr 2 + r 2dq2 + r 2 sin 2 qdf2, find the values of ì 1 ü ì 3 ü (i) [22, 1] and [13, 3], (ii) í ý and í ý î2 2þ î1 3þ Solution

The given metric is metric in spherical coordinates, x1 = r, x2 = q, x3 = f. Clearly, 2 2 2 = 0 g11 = 1, g22 = r , g33 = r sin q and gij for i ¹ j 62 Tensors and Their Applications

1 11 22 1 33 = Also, g = 1, g = , g 2 2 r 2 r sin q (See Ex. 2, Pg. 39, and gij = 0, for i ¹ j. ) (i) Christoffel Symbols of first kind are given by

1 é ¶g jk ¶g ¶gij ù + ik - , [ij,k] = ê i j k ú i, j,k =1,2,3 ...(1) 2 ë ¶x ¶x ¶x û Taking i = j = 2 and k = 1 in (1)

1 é ¶g21 ¶g21 ¶g22 ù [22,1] = ê 2 + 2 - 1 ú Since g21 = 0. 2 ë ¶x ¶x ¶x û

1 é ¶0 ¶0 ¶r 2 ù = ê 2 + 2 - 1 ú 2 ë ¶x ¶x ¶x û

1 ¶r 2 = - = -r 2 ¶r Taking i =1, j = k = 3 in (1)

1 é ¶g33 ¶g13 ¶g13 ù [13,3] = ê 1 + 3 - 3 ú 2 ë ¶x ¶x ¶x û

1 ¶r 2 sin 2 q = since g = 0 2 ¶r 13 [13, 3] = r sin 2 q (ii) Christoffel symbols of the second kind are given by ì k ü í ý = g kl [ij, l] = g k1[ij,1] + g k2[ ij,2] + g k3[ij,3] îi jþ Taking k = 1, i = j = 2.

ì 1 ü í ý = g11[22,1] + g12[22,2]+ g13[22,3] î2 2þ

ì 1 ü í ý = 1[22,1]+ 0[22,2]+ 0[22,3] Since g12 = g13 = 0 î2 2þ

ì 1 ü í ý = -r î2 2þ Christoffel's Symbols and Covariant Differentiation 63

and

ì 3 ü 31 32 33 í ý = g [13,1] + g [13, 2] + g [13,3] î1 3þ 1 = [13,3] Since g 31 = g32 = 0 r 2 sin 2 q

ì 3 ü 1 1 × sin 2 q = í ý = 2 2 r î1 3þ r sin q r

4.2 TRANSFORMATION OF CHRISTOFFEL'S SYMBOLS ij i The fundamental tensors gij and g are functions of coordinates x and [ij,k] is also function of

i -ij i coordinates x . Let gij, g and [ij,k ] in another coordinate system x . (i ) Law of Transformation of Christoffel's Symbol for First Kind Let [ij,k] is a function of coordinate xi and [ij,k ] in another coordinate system x i . Then

1 é ¶g ¶g ¶gij ù ik + ik - [ij,k ] = ê i j k ú ...(1) 2 ë ¶x ¶x ¶x û

Since gij is a covariant tensor of rank two. Then

¶x p ¶x2 g = g ...(2) ij ¶xi ¶x j pq

Differentiating it w.r.t. to x k , we get

¶g ¶ æ ¶x p ¶xq ö ij ç g ÷ k = k ç i j pq ÷ ¶x ¶x è ¶x ¶x ø

p q p q ¶ æ ¶x ¶x ö ¶x ¶x ¶g pq ç ÷g + = k ç i j ÷ pq i j k ¶x è ¶x ¶x ø ¶x ¶x ¶x

2 p q p 2 q p q r ¶g æ ¶ x ¶x ¶x ¶ x ö ¶x ¶x ¶g pq ¶x ij ç + ÷g + k = ç k i j i k j ÷ pq i j r k ...(3) ¶ x è ¶x ¶x ¶x ¶x ¶x ¶x ø ¶x ¶x ¶x ¶x Interchanging i, k and also interchanging p, r in the last term in equation (3)

2 p q p 2 q r q p ¶g æ ¶ x ¶x ¶x ¶ x ö ¶x ¶x ¶grq ¶x kj ç + ÷g + = ç i k j k i j ÷ pq k j p i ...(4) ¶x i è ¶x ¶x ¶x ¶x ¶x ¶x ø ¶x ¶x ¶x ¶x 64 Tensors and Their Applications and interchanging j, k and also interchange q, r in the last term of equation (3)

2 p q p 2 q p r q ¶g æ ¶ x ¶x ¶x ¶ x ö ¶x ¶x ¶x ¶g pr ik ç + ÷g + j = ç j i i i k j ÷ pq i k j q ...(5) ¶x è ¶x ¶x ¶x ¶x ¶x ¶x ø ¶x ¶x ¶x ¶x Substituting the values of equations (3), (4) and (5) in equation (1), we get

2 p q p q r 1 é ¶ x ¶x ¶x ¶x ¶x æ ¶grp ¶gqr ¶g pq öù 2 g + ç + - ÷ [ij, k ] = ê i j k pq i j k ç q p r ÷ú 2 ëê ¶x ¶x ¶x ¶x ¶x ¶x è ¶x ¶x ¶x øûú

2 p q p q r ¶ x ¶x ¶x ¶x ¶x 1 æ ¶grp ¶gqr ¶g pq ö = g + ç + - ÷ [ij, k ] i j k pq i i k ç q p r ÷ ¶x ¶x ¶x ¶x ¶x ¶x 2 è ¶x ¶x ¶x ø

¶2x p ¶xq ¶x p ¶x q ¶xr [ij, k ] = g pq + [pq, r] ...(6) ¶x i¶x j ¶x k ¶xi ¶x i ¶x k It is law of transformation of Christoffel's symbol of the first kind. But it is not the law of transformation of any tensor due to presence of the first term of equation (6). So, Christoffel's symbol of first kind is not a tensor. (ii) Law of Transformation of Christoffel's Symbol of the Second Kind

ïì k ïü ïì k ïü k l , = kl[ , ] = Let g [ij l] í ý is function of coordinates xi and g ij l í ý in another coordinate system îïi jþï îïi jþï x i . Then ¶2x p ¶xq ¶x p ¶xq ¶xr [ij, l] = g pq + [pq,r ] from (6) ¶x i¶x j ¶x l ¶x i ¶x j ¶x l As gkl is contravariant tensor of rank two.

¶x k ¶x l g kl = g st ¶x s ¶xt Now

¶x k ¶x l ¶2x p ¶x q ¶x k ¶x l ¶x p ¶xq ¶x r kl st st g [ij, l] = g g pq + g [pq, r] ¶x s ¶xt ¶x i x j ¶x l ¶xs ¶xt ¶xi ¶x j ¶xl

¶x k æ ¶x l ¶xq ö ¶ 2x p ¶x k æ ¶x l ¶x r ö ¶x p ¶xq ç ÷g st g + ç ÷ g st [pq,r ] = s ç t l ÷ i j pq s ç t l ÷ i j ¶x è ¶x ¶x ø ¶x x ¶x è ¶x ¶x ø ¶x ¶x

¶x k ¶2x p ¶x k ¶x p ¶xq ¶x l ¶xq = d2 g st g + dr g st[pq,r] as = dq ¶x s t ¶xi x j pq ¶x s t ¶x i ¶x j ¶xt ¶x l t

¶x k ¶2x p ¶x k ¶x p ¶xq sq sr r st sr = g g pq + g [pq, r] as d g = g ¶x s ¶x i¶x j ¶x s ¶xi ¶x j t Christoffel's Symbols and Covariant Differentiation 65

p 2 p k p 2 ¶x ¶ x s ¶x ¶x ¶x ì s ü kl d + g [ij, l] = s i j p s i j í ý ¶x ¶x ¶x ¶x ¶x ¶x î p qþ

sr ì s ü Since g sqg = ds and g [ pq,r ] = í ý pq p î p qþ

ïì k ïü ¶x k ¶2x s ¶x k ¶x p ¶x q ì s ü + í ý = s i j s i j í ý ....(7) îïi jþï ¶x ¶x ¶x ¶x ¶x ¶x îp qþ It is law of transformation of Christoffel’s symbol of the second kind. But it is not the law of transformation of any tensor. So, Christoffel’s symbol of the second kind is not a tensor.

¶xs Also, multiply (7) by , we get ¶x k

s k 2 s s k p q ¶x s ïì k ïü ¶x ¶x ¶ x ¶x ¶x ¶x ¶x ì s ü í ý = + í ý k k s i j k s i j p q dx îïi jþï ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x î þ

¶x s ¶x k Since = ds =1 ¶x k ¶xs s

2 s p q ¶x s ïì k ïü ¶ x ¶x ¶x ì s ü í ý = + í ý k i j i j p q ¶x îïi jþï ¶x ¶x ¶x ¶x î þ

2 s ¶xs ì ü ¶x p ¶xq ¶ x k - ì s ü = k í ý i j í ý ...(8) ¶x i¶x j ¶x îi jþ ¶x ¶x î p qþ

It is second derivative of xs with respect to x ’s in the terms of Christoffel’s symbol of second kind and first derivatives. THEOREM 4.4 Prove that the transformation of Christoffel’s Symbols form a group i.e., possess the transitive property. Proof: Let the coordinates xi be transformed to the coordinate system x i and x j be transformed to x i . When coordinate xi be transformed to x i , the law of transformation of Christoffel’s symbols of second kind (equation (7)) is

ì ü k 2 s k p q ï k ï ¶x ¶ x ¶x ¶x ¶x ïì s ïü + í ý = s i j s i j í ý ...(1) îïi jþï ¶x ¶x ¶x ¶x ¶x ¶x îï p q þï

When coordinate x i be transformed to x i . Then

ì r ü ì k ü ¶xi ¶x j ¶x r ¶2x k ¶x r ï ï + í ý = í ý u v k u v k îu vþ îïi jþï ¶x ¶x ¶x ¶x ¶x ¶x 66 Tensors and Their Applications

ì s ü ¶x k ¶x p ¶xq ¶x i ¶x j ¶x r ¶2x s ¶x k ¶xi ¶x j ¶x r = í ý + î p qþ ¶xs ¶x i ¶x j ¶x u ¶x v ¶x k ¶x i¶x j ¶xs ¶x u ¶x v ¶x k

¶2x k ¶x r + ¶x u¶x v ¶x k

2 ì r ü ì s ü ¶x p ¶x q ¶x r ¶ xk ¶x r í ý = í ý u v s + u v k + îu vþ î p qþ ¶x ¶x ¶x ¶x ¶x ¶x

¶ 2x s ¶x r ¶xi ¶x j ...(2) ¶x i¶x j ¶x s ¶x u ¶x v

¶x p ¶xi ¶x p as i u = u ¶x ¶x ¶x Since we know that

¶x s ¶x i ¶xs = ...(3) ¶xi ¶x u ¶x u

Differentiating (3) w.r.t. to x v , we get

¶ æ ¶xs ö ¶xi ¶xs ¶ æ ¶xi ö 2 s ç ÷ + ç ÷ ¶ x v ç i ÷ u i v ç u ÷ = ¶x è ¶x ø ¶x ¶x ¶x è ¶x ø ¶x u¶x v

¶2xs ¶x j ¶x i ¶xs ¶2x i ¶2xs + = ...(4) ¶x i¶x j ¶x v ¶x u ¶x i ¶x u¶x v ¶x u¶x v

¶x r Mutiply (5) by . ¶xs

¶2x s ¶x j ¶xi ¶x r ¶2xi ¶xs ¶x r ¶2xs ¶x r + = ¶x i¶x j ¶x v ¶x u ¶xs ¶x u¶x v ¶x s ¶x s ¶x u¶x v ¶xs Replace dummy index i by k in second term on L.H.S. ¶ 2xs ¶x j ¶xi ¶x r ¶2x k ¶x r ¶2xs ¶x r + = ...(5) ¶x i¶x j ¶x v ¶x u ¶x s ¶x u¶x v ¶x k ¶x u¶x v ¶xs Using (5) in equation (2), we get

ì r ü ì s ü ¶x p ¶x q ¶x r ¶ 2xs ¶x r + í ý = í ý u v s u v s ...(6) îu vþ î p qþ ¶x ¶x ¶x ¶x ¶x ¶x Christoffel's Symbols and Covariant Differentiation 67

The equation (6) is same as the equation (1). This shows that if we make direct transformation from xi to x i we get same law of transformation. This property is called that transformation of Christoffel's symbols form a group.

4.3 COVARIANT DIFFERENTIATION OF A COVARIANT VECT OR i i Let Ai and Ai be the components of a covariant vector in coordinate systems x and x respectively.. Then ¶x p A = Ap ...(1) i ¶x i

Differentiating (1) partially w.r.t. to x j ,

2 p p ¶A ¶ x ¶x ¶Ap i = A + ¶x j ¶x j¶x i p ¶xi ¶x j

2 p p q ¶A ¶ x ¶x ¶Ap ¶x i = A + ...(2) ¶x j ¶x j¶xi p ¶x i ¶xq ¶x j It is not a tensor due to presence of the first term on the R.H.S. of equation (2). Now, replace dummy index p by s in the first term on R.H.S. of (2), we have

2 s p q ¶A ¶ x ¶x ¶Ap ¶x i = A + ...(3) ¶x j ¶x j¶x i s ¶x i ¶x q ¶x j Since we know that from equation (8), page 65,

¶2xs ¶xs ïì k ïü ¶x p ¶xq ì s ü - i j = k í ý i j í ý ¶x ¶x ¶x îïi jþï ¶x ¶x î p qþ

¶2xs Substituting the value of in equation (3), we have ¶x i¶x j

æ s p q ö p q ¶A ¶x ïì k ïü ¶x ¶x ì s ü ¶x ¶Ap ¶x i = ç í ý - í ý÷ A + j ç k i j i j p q ÷ s i q j ¶x è ¶x îï þï ¶x ¶x î þø ¶x ¶x ¶x

ì ü s p q p q ï k ï ¶x ¶x ¶x ì s ü ¶x ¶x ¶Ap = í ý k As - i j í ýAs + i j q îïi jþï ¶x ¶x ¶x îp qþ ¶x ¶x ¶x

ì ü p q ¶A ï k ï ¶x ¶x æ ¶Ap ì s üö i A + ç - A ÷ j = í ý k i j ç q s í ý÷ ¶x îïi jþï ¶x ¶x è ¶x î p qþø 68 Tensors and Their Applications

p q ì ü ¶x ¶x æ ¶Ap ì s üö ¶Ai ï k ï ç ÷ - A í ý = - As í ý ...(4) j k i j ç q p q ÷ ¶x îïi jþï ¶x ¶x è ¶x î þø Now, we introduce the comma notation

¶A ì k ü A i - A i, j = j k í ý ...(5) ¶x îi jþ Using (5), the equation (4) can be expressed as

¶x p ¶x q Ai, j = A , ..(6) ¶xi ¶x j p q

It is law of transformation of a covariant tensor of rank two. Thus, Ai, j is a covariant tensor of rank two.

j So, Ai, j is called of Ai with respect to x .

4.4 COVARIANT DIFFERENTIATION OF A CONTRAVARIANT VECT OR Let Ai and Ai be the component of contravariant vector in coordinate systems xi and x i respectively.. Then i ¶x s Ai = A ¶x s s ¶x i or As = A ¶x i Differentiating it partially w.r.t. to x j , we get

¶As ¶2 xs ¶x s ¶Ai = Ai + ...(1) ¶x j ¶x j¶xi ¶x i ¶x j Since from equation (8) on page 65,

2 s ¶ x ¶x s ïì k ïü ¶x p ¶x q ïì s ïü A i - j i = k í ý i j í ý ¶x ¶x ¶x îïi jþï ¶x ¶x îï p q þï

¶2xs substituting the value of in the equation (1), we get ¶x j¶xi

s s ì ü p q s i ¶A ¶x ï k ï i ¶x ¶x ì s ü i ¶x ¶A j = k í ýA - i j í ýA + i j ¶A ¶x îïi jþï ¶x ¶x î p qþ ¶x ¶x

s q s ì ü p q s i ¶A ¶A ¶x ï k ï i ¶x i ¶x ì s ü ¶x ¶A q j = k í ý A - i A j í ý + i j ¶A ¶x ¶x îïi jþï ¶x ¶x îp qþ ¶x ¶x Christoffel's Symbols and Covariant Differentiation 69

¶x p Interchanging the dummy indices i and k in the first term on R.H.S. and put Ai = Ap we get ¶xi

s q s ì ü q s i ¶A ¶x ¶x ï i ï k ¶x p ì s ü ¶x ¶A q j = i í ýA - j A í ý + i j ¶x ¶x ¶x îïk jþï ¶x î p qþ ¶x ¶x

q æ s ì s üö s æ ì ü i ö ¶x ¶A p ¶x ï i ï k ¶A ç + A í ý÷ = ç í ýA + ÷ ¶x j ç ¶xq p q ÷ i ç k j j ÷ è î þø ¶x è îï þï ¶x ø

¶Ai ïì i ïü ¶xi ¶x q æ ¶As ì s üö + Ak ç + Ap ÷ j í ý = s j ç q í ý÷ ...(2) ¶x îïk jþï ¶x ¶x è ¶x îp qþø Now, we introduce the comma notation

¶A ì i ü i i + Ak A, j = j í ý ...(3) ¶x îk jþ Using (3), the equation (2) can be expressed as

p q i ¶x ¶x A = A , ...(4) , j ¶xi ¶x j p q It is law of transformation of a mixed tensor of rank two. Thus, Ai , j is a mixed tensor of rank two. Ai , j is called covariant derivative of Ai with respect to x j .

4.5 COVARIANT DIFFERENTIATION OF TENSORS Covariant derivative of a covariant tensor of rank two. i i Let Aij and Aij be the components of a covariant tensor of rank two in coordinate system x and x respectively then

¶x p ¶x q A = A ...(1) ij ¶xi ¶x j pq

Differentiating (1) partially w.r.t. to x k

p q p q ¶Aij ¶x ¶x ¶Apq ¶ æ ¶x ¶x ö + ç ÷ A k = i j k k ç i j ÷ pq ¶x ¶x ¶x ¶x ¶x è ¶x ¶x ø

p q r 2 p q p 2 q ¶Aij ¶x ¶x ¶Apq ¶x ¶ x ¶x ¶x ¶ x = + A + A ...(2) ¶x k ¶xi ¶x j ¶xr ¶x k ¶x k ¶xi ¶x j pq ¶xi ¶x k ¶x j pq

r ¶Apq ¶Apq ¶x as = (since A components in xi coordinate) ¶x k ¶xr ¶x k pq 70 Tensors and Their Applications

¶2x p ¶x q ¶2xl ¶xq Apq = A ¶x i¶x k ¶x j lq ¶x i¶x k ¶x j

¶2x p ¶x q q éì ü l p r ù A ¶x ï h ï ¶x ì l ü ¶x ¶x pq i k j = A êí ý - í ý ú ¶x ¶x ¶x lq j i k h p r i k ¶x ëêîï þï ¶x î þ ¶x ¶x ûú Since we know that from equation (8) on page 65.

¶2 xl ïì h ïü ¶xl ì l ü ¶x p ¶xr i k = í ý h - í ý i k ¶x ¶x îïi k þï ¶x î p r þ ¶x ¶x

2 p q ïì h ïü ¶x l ¶x 2 ì l ü ¶x p ¶x q ¶x r ¶ x ¶x A - A Apq = í ý lq h j í ý lq i j k ¶x i¶x k ¶x j îïi k þï ¶x ¶x î p r þ ¶x ¶x ¶x

2 p q ¶ x ¶x ïì h ïü ì l ü ¶x p ¶xq ¶xr Apq = í ýA - í ý A ...(3) ¶x i¶x k ¶x j hj lq i j k îïi kþï îp rþ ¶x ¶x ¶x

¶xl ¶xq as Ahj = A by equation (1) lq ¶x h ¶x j and

¶x p ¶2xq ¶x p ¶ 2xl Apq = A ¶xi ¶x j¶x k pl ¶x i ¶x j¶x k

¶x p éïì h ïü ¶xl ì l ü ¶xq ¶xr ù = A êí ý - í ý ú pl i j k h q r j k ¶x ëêîï þï ¶x î þ ¶x ¶x ûú

ïì h ïü ¶x p ¶xl ì l ü ¶xq ¶xr ¶x p = í ýApl i h - í ý j k i Apl îï j kþï ¶x ¶x îq rþ ¶x ¶x ¶x

¶x p ¶2xq ïì h ïü ì l ü ¶x p ¶xq ¶xr A pq i j k = í ýAih - í ý i j k Apl ...(4) ¶x ¶x ¶x îï j kþï îq r þ ¶x ¶x ¶x Substituting the value of equations (3) and (4) in equation (2) we get,

é¶A ù p q r ì ü ì ü ¶Aij pq ì l ü ì l ü ¶x ¶x ¶x ï h ï ï h ï = ê - Alqí ý - Apl í ýú + í ýAhj + í ý Aih k r p r q r i j k ¶x ëê ¶x î þ î þûú ¶x ¶x ¶x îïi k þï îï j kþï

ì ü ì ü p q r ¶Aij ï h ï ï h ï é¶Apq ì l ü ì l üù ¶x ¶x ¶x - A - A - A - A k í ý ih í ý hj = ê r lqí ý pl í ýú i j k ¶x îï j k þï îïi kþï ë ¶x î p r þ îq r þû ¶x ¶x ¶x Christoffel's Symbols and Covariant Differentiation 71

¶A ì h ü ì h ü A ij - A - A ij,k = k í ý ih í ý hj , then ¶x î j k þ îi kþ

¶x p ¶xq ¶xr Aij,k = A , pq r ¶x i ¶x j ¶x k

It is law of transformation of a covariant tensor of rank three. Thus, Aij,k is a covariant tensor of rank three.

k So, Aij,k is called covariant derivative of Aij w.r.t. to x .

k ij i Similarly we define the covariant derivation x of a tensors A and Aj by the formula ¶Aij ì i ü ì j ü ij + Alj + Ail A ,k = k í ý í ý ¶x îl k þ îl k þ ¶Ai ì i ü ì l ü i j + Al - Ai and Aj,k = k j í ý l í ý ¶x îl k þ î j kþ k ij...l In general, we define the covariant deriavative x of a mixed tensor Aab...c by the formula ¶Aij...l ì i ü ì j ü ì l ü ij...l ab...c + Apj...l + Aip...l + ×××+ Aij... p Aab...c,k = k ab...c í ý ab...c í ý ab...c í ý ¶x îp k þ î p k þ î p k þ

ij...l ì p ü ij...l ì p ü ij...l ì p ü - Apb...c í ý - Aap...c í ý - ××× - Aab... p í ý îa k þ îb k þ îc k þ

Note: Ai,k is also written as Ai,k = Ñk Ai.

4.6 RICCI'S THEOREM ij The covariant derivative of Kronecker delta and the fundamental tensors gij and g is zero. Proof: The covariant derivative xp of Kronecker delta is

¶di ì i ü ì l ü i j + dl - di d j,k = k j í ý l í ý ¶x îl k þ î j k þ

ì i ü ì i ü = 0 + í ý - í ý î j kþ î j k þ

¶di ì i ü ì i ü i j = 0; dl = d j,k = 0 as k j í ý í ý ¶x îl kþ î j k þ

Also, consider first the tensor gij and the covariant derivative of gij is

¶gij ì m ü ì m ü gij,k - g - g = k mjí ý im í ý ¶x îi k þ î j k þ 72 Tensors and Their Applications

¶gij ì m ü gij,k = -[ik, j]- [ jk,i] as gmj í ý = [ik, j] ¶xk îi kþ

¶gij But = [ik, j]+ [ jk,i] ¶xk

¶gij ¶gij So, gij,k = - ¶xk ¶xk

gij,k = 0 We can perform a similar calculation for the tensor gij. im i Since we know that g gmj = d j . Similarly taking covariant derivative, we get im im i g,k gmj + g gmj,k = d j,k i im = 0 as ¹ 0 But gmj,k = 0 and d j,k = 0. So, g,k gmj

EXAMPLE 4 Prove that if Aij is a symmetric tensor then ¶g j 1 ¶ j 1 jk jk A, = (Ai g )- A i j g ¶x j 2 ¶xi Solution

Given that Aij be a symmetric tensor. Then Aij = Aji ...(1)

We know that

¶A j ì j ü ì l ü j i + Al - A j Ai,k = k i í ý l í ý ¶x îl kþ îi k þ Put k = j, we get

¶A j ì j ü ì l ü j i + Al - A j Ai, j = j i í ý l í ý ...(2) ¶x îl jþ îi jþ

j ¶Ai l ¶(log g ) j hl = + Ai - A g [ij,h] ¶x j ¶xl l

¶A j Aj ¶ g = i + i - Ajh[ij, h] since Aij is symmetric. ¶x j g ¶x j

j j 1 ¶(Ai g ) jk A, = - A [ij,k ] ...(3) i j g ¶x j Christoffel's Symbols and Covariant Differentiation 73

But

1 æ ¶g jk ¶g ¶gij ö jk = A jk ç + ki - ÷ A [ij,k] ç i j k ÷ 2 è ¶x ¶x ¶x ø

1 æ ¶g jk ¶g ¶gij ö jk = ç A jk + A jk ki - Ajk ÷ A [ij,k] ç i j k ÷ 2 è ¶x ¶x ¶x ø

jk ¶gki ¶g ji A = Akj ...(4) ¶x j ¶xk On Interchanging the dummy indices j & k.

¶g ¶g ji Ajk ki = Ajk since Aij = Aji ¶x j ¶x k

jk ¶g jk ¶gij Þ A ki - A = 0 as g = g ...(5) ¶x j ¶xk ij ji Using (5), equation (4) becomes

1 jk ¶g jk Ajk [ij,k] = A 2 ¶xi

Put the value of Ajk [ij,k] in equation (3), we get

j j 1 ¶(Ai g ) 1 jk ¶g jk A, = - A Proved. i j g ¶x j 2 ¶xi

EXAMPLE 5 ì k ü ì k ü ì k ü ì k ü Prove that í ý - í ý are components of a tensor of rank three where í ý and í ý îi jþa îi jþb îi jþa îi jþb are the Christoffel symbols formed from the symmetric tensors aij and bij . Solution Since we know that from equation (8), page 65. ¶2xs ¶xs ïì k ïü ¶x p ¶xq ì s ü - i j = k í ý i j í ý ¶x ¶x ¶x îïi jþï ¶x ¶x î p qþ 2 s p q ¶x s ïì k ïü ¶ x ¶x ¶x ì s ü í ý = + í ý k ¶x i¶x j ¶xi ¶x j p q ¶x îïi jþï î þ

2 s p q k ïì k ïü é ¶ x ¶x ¶x ì s üù ¶x or í ý = ê + í ýú i j i j p q s îïi jþï ë¶x ¶x ¶x ¶x î þû ¶x 74 Tensors and Their Applications

Using this equation, we can write

2 s p q k ïì k ïü é ¶ x ¶x ¶x ì s ü ù ¶x = ê + í ý ú í ý i j i j p q s ïi jï ëê¶x ¶x ¶x ¶x î þaûú ¶x î þa

ïì k ïü é ¶ 2xs ¶x p ¶xq ì s ü ù ¶x k + and í ý = ê i j i j í ý ú s ïi jï ¶x ¶x ¶x ¶x p q ¶x î þb ëê î þb ûú Subtracting, we obtain

p q k ïì k ïü ïì k ïü éì s ü ì s ü ù ¶x ¶x ¶x - êí ý - í ý ú í ý í ý = p q p q i j s ïi jï ïi jï ëêî þa î þbûú ¶x ¶x ¶x î þa î þb Put

ì s ü ì s ü s í ý - í ý = Apq î p qþa î p qþb Then above equation can written as

¶x p ¶xq ¶x k A k = As ij pq ¶xi ¶x j ¶xs It is law of transformation of tensor of rank three. ì k ü ì k ü So, í ý - í ý are components of a tensor of rank three. îi jþa îi jþb

EXAMPLE 6 k If a specified point, the derivatives of gij w.r.t. to x are all zero. Prove that the components of covariant derivatives at that point are the same as ordinary derivatives. Solution Given that

¶gii = 0, " i, j, k at P ...(1) ¶xk 0 i Let Aj be tensor.. i i ¶Aj Now, we have to prove that A , = at P0. j k ¶xk ¶Ai ì i ü ì a ü i j + Aa - Ai Aj,k = k j í ý a í ý ...(2) ¶x îa k þ î j k þ Christoffel's Symbols and Covariant Differentiation 75

ì k ü ¶g Since and ij,.k both contain terms of the type ij and using equation (1) we get í ý [ ] k îi jþ ¶x ì k ü 0 = , í ý = [ij k] at P0 . îi jþ So, equation (2) becomes i ¶Aj Ai = at P j,k ¶xk 0

4.7 GRADIENT, DIVERGENCE AND CURL (a) Gradient If f be a scalar function of the coordinates, then the gradient of f is denoted by ¶f grad f = ¶xi which is a covariant vector. (b) Divergence The divergence of the contravariant vector Ai is defined by ¶Ai ì i ü div Ai + Ak = i í ý ¶x îk iþ i It is also written as A,i

The divergence of the covariant vector Ai is defined by ik div Ai = g Aik

EXAMPLE 7 1 ¶( g Ak ) Prove that div Ai = g ¶xk Solution:

If Ai be components of contravariant vector then ¶Ai ì i ü i Ai = + Ak div A = ,i i í ý ¶x îk iþ Since ì i ü ¶ 1 ¶ g í ý = (log g ) = îk iþ ¶x k g ¶xk So, i i ¶A 1 ¶ g k div A = + A ¶xi g ¶x k 76 Tensors and Their Applications

Since i is dummy index. Then put i = k, we get

¶Ak 1 ¶ g div Ai = + Ak ¶xk g ¶x k

1 ¶( g Ak ) div Ai = ...(1) g ¶xk Proved (c) Curl

Let Ai be a covariant vector then ¶A ì k ü A i - A i, j = j k í ý ¶x îi jþ ¶A ì k ü A j - A and j,i = i k í ý ¶x î j iþ are covariant tensor. ¶A ¶Ai j So, A, - A , = - is covariant tensor of second order, which is called curl of A . i j j i ¶x j ¶xi i Thus

curl Ai = Ai, j - Aj,i

Note: curl Ai is a skew-symmetric tensor. Since

Aj, i – A i,j = – (Ai,j – A j,i)

EXAMPLE 8

If Aij be a skew-symmetric tensor of rank two. Show that ¶A ¶A ij jk ¶Aki A , + A , + A , = + + ij k jk i ki j ¶xk ¶xi ¶x j Solution Since we know that ¶A ì l ü ì l ü ij - A - A Aij,k = k lj í ý il í ý ¶x îi kþ î j kþ

¶A ì h ü ì l ü jk - A - A Ajk,i = i lk í ý jl í ý ¶x î j iþ î j k þ

¶A ì l ü ì l ü ki - A - A Aki, j = j lií ý kl í ý ¶x îk jþ îi jþ Christoffel's Symbols and Covariant Differentiation 77

Adding these, we get

¶Aij ¶Ajk ¶A æ ì l ü ì l üö A + A + A + + ki - ç A + A ÷ ij,k jk,i ki, j = k i j ç lj í ý jl í ý÷ ¶x ¶x ¶x è îi kþ îk iþø æ ì l ü ì l üö æ ì l ü ì l üö - ç + ÷ - ç + ÷ ç Ali í ý Ail í ý÷ ç Akl í ý Alk í ý÷ è îk jþ î j kþø è îi jþ î j iþø ì l ü ì l ü ì l ü Since í ý is symmetric i.e., í ý = í ý etc. îi kþ îi kþ îk iþ

¶A ¶A ¶A ì l ü ij + jk + ki - (A + A ) = k i j í ý lj jl ¶x ¶x ¶x îi k þ

ïì l ïü ïì l ïü - í ý (Ali + Ail ) - í ý (Akl + Alk ) îïk jþï îïi jþï

Since Aij is skew-symmetric. Then Alj = -Ajl Þ Alj + Ajl = 0 . Similarly,,

Ali + Ail = 0 and Akl + Alk = 0 So, ¶A ¶A ij jk ¶Aki A , + A , + A , = + + ij k jk i ki j ¶xk ¶xi ¶x j THEOREM 4.5 A necessary and sufficient condition that the curl of a vanishes is that the vector field be gradient.

Proof: Suppose that the curl of a vector Ai vanish so that

curl Ai = Ai, j - Aj,i = 0 ...(1)

To prove that Ai = Ñf, f is scalar.. Since from (1),

Ai, j - Aj,i = 0

¶A ¶Aj Þ i - = 0 ¶x j ¶xi

¶A ¶Aj Þ i = ¶x j ¶xi

¶A ¶Aj j Þ i dx j = dx ¶x j ¶xi ¶ Þ dA = (A jdx j ) i ¶xi Integrating it we get

¶ j A = (Ajdx ) i ò ¶xi 78 Tensors and Their Applications

¶ j = Ajdx ¶xi ò ¶f A = , where f = A dx j i ¶xi ò j

or Ai = Ñf.

Conversely suppose that a vector Ai is such that

Ai = Ñf, f is scalar..

To prove curl Ai = 0 Now, ¶f A = Ñf = i ¶xi ¶A ¶2f i = ¶x j ¶x j¶xi 2 ¶Aj ¶ f and = ¶xi ¶xi¶x j

¶A ¶Aj So, i - = 0 ¶x j ¶xi ¶A ¶Ai j So, curl A = A, - A , = - = 0 i i j j i ¶x j ¶xi

So, curl Ai = 0 Proved.

THEOREM 4.6 Let f and y be scalar functions of coordinates xi. Let A be an arbitrary vector then (i) div (fA) = f div A + A ×Ñf (ii) Ñ(fy) = fÑy + yÑf

(iii) Ñ2(fy) = fÑ2y + yÑ2y + 2Ñf × Ñy

(iv) div (yÑf) = yÑ2f + Ñf ×Ñy

Proof: (i) Since we know that 1 ¶( g Ai ) div Ai = ...(1) g ¶xi

replace Ai by fAi , we get

1 ¶( g fAi ) div(fAi ) = g ¶xi Christoffel's Symbols and Covariant Differentiation 79

é i ù 1 ¶( g A ) i ¶f ê f + g A × ú = i i g ëê ¶x ¶x ûú i ¶f i 1 ¶( g A ) = × A + f ¶xi g ¶xi = Ñf × Ai + f div Ai Thus div(fA) = Ñf × A + f div A ...(2) (ii) By definition of gradient, ¶(fy) Ñ(fy) = ¶xi ¶y ¶f = f + y ¶xi ¶xi Thus Ñ(fy) = fÑy + yÑf ...(3) (iii) Taking divergence of both sides in equation (3), we get div (Ñfy) = div [fÑy + yÑf]

Ñ2(fy) = div (fÑy) + div (yÑf) = Ñf ×Ñy + f div (Ñy) + Ñy × Ñf + ydiv (Ñf)

Ñ2(fy) = fdiv (Ñy) + ydiv (Ñf) + 2Ñf × Ñy Thus, Ñ2(fy) = fÑ2y + yÑ2f + 2Ñf ×Ñy (iv) Replace A by Ñy in equation (3), we get div (fÑy) = Ñf × Ñy + f div (Ñy)

div (fÑy) = Ñf × Ñy + fÑ2y

THEOREM 4.7 Let Ai be a covariant vector and f a scalar function. Then (i) curl (fA) = A ´ Ñf + f curlA (ii) curl (yÑf) = Ñf ´ Ñy

Proof: (i) Let Ai be a covariant vector then

curl A = curl Ai = Ai, j - Aj,i

Replacing Ai by fAi , we get

curl (fAi ) = (fAi ), j - (fAj ),i 80 Tensors and Their Applications

= f, jAi + fAi, j - f,i Aj - fAj,i

= (Aif, j - Ajf,i) + f( Ai, j - Aj,i )

= Ai ´ Ñf + fcurl Ai So, curl (fA) = A´ Ñf + fcurlA ...(1) (ii) Replacing A by Ñy in equation (1), we get curl (fÑy) = fcurl (Ñy) + Ñy ´Ñf Interchange of f and y , we get curl (yÑf) = ycurl (Ñf) + Ñf ´ Ñy. Since curl (Ñf)= 0. So, curl (yÑf) = Ñf ´ Ñy. Proved.

4.8 THE LAPLACIAN OPERATOR The operator Ñ2 is called Laplacian operator read as "del square".

THEOREM 4.8 If f is a scalar function of coordinates xi then 1 ¶ æ ¶f ö Ñ2f = ç g gkr ÷ g ¶xk è ¶xr ø Proof: Since Ñ2f = div grad f ...(1) and ¶f grad f = , ¶x r which is covariant vector.

k But we know that any contravariant vector A associated with Ar (covariant vector) is k kr A = g Ar (Sec Art. 3.4, Pg 43) ¶f Now, the contravariant vector Ak associated with (Covariant vector) is ¶x r ¶f Ak = g kr ¶x r Since 1 ¶( g Ak ) div Ai = , (Sec Ex. 7, Pg 75) g ¶xk Christoffel's Symbols and Covariant Differentiation 81

So, from (1) æ ¶f ö ¶ç g g kr ÷ æ ¶f ö 1 è ¶xr ø Ñ2f = div ç g kr ÷ = Proved. è ¶x r ø g ¶xk

EXAMPLE 9 Show that, in the cylindrical coordinates, 1 ¶ æ ¶V ö 1 ¶2V ¶2V Ñ2V = ç r ÷ + + r ¶r è ¶r ø r 2 ¶q2 ¶z 2 by Tensor method Solution The cylindrical coordinates are (r,q, z). If V is a scalar function of (r,q, z). Now, Ñ2V = Ñ × ÑV Since ¶V ¶V ¶V ÑV = i + j + k ¶r ¶q ¶z Let ¶V ¶V ¶V A = , A = , A = ...(1) 1 ¶r 2 ¶q 3 ¶z

Then ÑV = iA1i + jA2 j + kA3k, since ÑV is covariant tensor. The metric in cylindrical coordinates is ds 2 = dr 2 + r 2dq2 + dz 2

here, x1 = r, x2 = q, x3 = z . 2 i j Since ds = gijdx dx g 2 11 = 1, g22 = r , g33 =1 and others are zero. 1 0 0 g11 g12 g13 2 2 g = gij = g21 g22 g23 = 0 r 0 = r 0 0 1 g31 g32 g33

g = r (See Pg. 34, Example 1) Now,

1 ¶( g Ak ) div (ÑV) = div Ai = g ¶xk 82 Tensors and Their Applications

1 2 3 1 ¶( g A ) 1 ¶( g A ) ¶( g A ) + + = 1 2 3 g ¶x g ¶x ¶x

1 é¶(rA1) ¶(rA2) ¶(rA3) ù div (ÑV) = ê + + ú ...(2) r ëê ¶r ¶q ¶z ûú We can write k kq A = g Aq (Associated tensor) k k1 k 2 k3 A = g A1 + g A2 + g A3 Put k = 1 1 11 12 13 A = g A1 + g A2 + g A3 1 11 12 13 A = g A1 as g = g = 0 Similarly, 2 22 A = g A2 3 33 A = g A3 and

Cofactor of g in g r 2 g11 = 11 = =1 g r 2 Cofactor of g in g 1 g 22 = 22 = g r 2

Cofactor of g in g r 2 g 33 = 33 = = 1 (See. Pg. 34, Ex.1) g r 2 So, 1 11 A = g A1 = A1 1 2 22 A = g A2 = A2 r 2 3 33 A = g A3 = A3.

or 1 = 2 1 3 A A1, A = A2, A = A . r 2 3 from (1), we get

¶V 2 1 ¶V 3 ¶V 1 = , A = , A = A ¶r r 2 ¶q ¶z from (2), Christoffel's Symbols and Covariant Differentiation 83

So,

i 1 é ¶ æ ¶V ö ¶ æ 1 ¶V ö ¶ æ ¶V öù 2 div A = r + r + r Ñ V = ê ç ÷ ç 2 ÷ ç ÷ú r ë ¶r è ¶r ø ¶q è r ¶q ø ¶z è ¶z øû

1 é ¶ æ ¶V ö ¶ æ 1 ¶V ö ¶ æ ¶V öù div (ÑV ) = ê ç r ÷ + ç ÷ + ç r ÷ú r ë ¶r è ¶r ø ¶q è r ¶q ø ¶z è ¶z øû

1 é ¶ æ ¶V ö 1 ¶2V ¶2V ù = ê ç r ÷ + 2 + r 2 ú r ë ¶r è ¶r ø r ¶q ¶z û

1 ¶ æ ¶V ö 1 ¶2V ¶2V div (ÑV ) = ç r ÷ + + r ¶r è ¶r ø r 2 ¶q2 ¶z 2

1 ¶ æ ¶V ö 1 ¶2V ¶2V Ñ2V = ç r ÷ + + r ¶r è ¶r ø r 2 ¶q2 ¶z 2

EXERCISES

1. Prove that the expressions are tensors ¶A ïìaïü ïìa ïü A ,l = ij - A - A (a) ij l í ý aj í ý ia ¶x îïilþï îï jl þï ¶Ar r i jk ì a ü r ì a ü r ì a ü r ì r ü a (b) A = - í ýA - í ýA - í ýA + í ýAijk i jk,l ¶xl îi lþ ajk î j lþ iak îk lþ ija îa lþ 2. Prove that j k j 1 ¶(Ai g) jì ü A = - Ak í ý i, j g ¶x j îi jþ

1 ¶ 3. If Aijk is a skew-symmetric tensor show that ( g Ai jk ) is a tensor.. g ¶xk 4. Prove that the necessary and sufficient condition that all the Christoffel symbols vanish at a point is that gij are constant. 5. Evaluate the Christoffel symbols in cylindrical coordinates.

6. Define covariant differentiation of a tensor w.r. to the fundamental tensor gij. Show that the covariant differentiation of sums and products of tensors obey the same result as ordinary differentiation. i 7. Let contravariant and covariant components of the same vector A be A and Ai respectively then prove that i div A = div Ai 84 Tensors and Their Applications

8. If Aij is the curl of a covariant vector. Prove that

Aij,k + A jk,i + Aki, j = 0

9. To prove that u ×Ñu = – ucurl u if u a vector of constant magnitude. 10. A necessary and sufficient condition that the covariant derivative vector be symmetric is that the vector must be gradient. 11. Show that, in spherical coordinates

2 2 1 ¶ æ 2 ¶V ö 1 ¶ æ ¶V ö 1 ¶ V Ñ V = 2 ç r ÷ + 2 çsin q ÷ + 2 2 2 r ¶r è ¶r ø r sin q ¶ qè ¶ q ø r sin q ¶ f by tensor method. CHAPTER – 5

RIEMANN-CHRISTOFFEL TENSOR

5.1 RIEMANN-CHRISTOFFEL TENSOR j If Ai is a covariant tensor then the covariant x derivative of Ai is given by ¶A ì a ü i - A Ai,j = j í ý a ...(1) ¶x îi jþ Differentiating covariantly the equation (1) w.r. to xk , we get a a ¶Ai, j ïì ïü ïì ïü - A - A Ai, jk = k í ý a, j í ý i,a ¶x îïi kþï îï j kþï ¶ æ ¶A ïì a ïü ö ïì a ïü æ ¶A ïì b ïü ö ç i - A ÷ - ç a - A ÷ = k ç j í ý a ÷ í ý ç j í ý b ÷ ¶x è ¶x îïi jþï ø îïi k þï è ¶x îïa jþï ø ïì a ïü æ ¶A ïì g ïü ö - ç i - A ÷ í ý ç a í ý g ÷ îïi k þï è ¶x îïi kþï ø

ïì a ïü ¶ í ý ¶ 2 A ïi jï ïì a ïü ¶A ïì a ïü ¶A i - î þ - a - a Ai, jk = k j k Aa í ý k í ý j ¶x ¶x ¶x îïi k þï ¶x îïi k þï ¶x ì a ü ì b ü ì a ü ì a ü ì g ü ï ï ï ï ï ï ¶Ai ï ï ï ï + í ý í ýAb - í ý a + í ý í ýAg ...(2) îïi kþï îïa jþï îï j kþï ¶x îï j k þï îïi aþï Interchanging j and k in equation (2), we get ì a ü ¶í ý ¶2 A i k ì a ü ¶A ì a ü ¶A i - î þ A - a - a Ai,kj = j k j a í ý j í ý k ¶x ¶x ¶x îi k þ ¶x îi jþ ¶x 86 Tensors and Their Applications

ì a ü ì b ü ì a ü ì a ü ì g ü ï ï ï ï ï ï ¶Ai ï ï ï ï ` + í ý í ýAb - í ý a + í ý í ýAg ...(3) îïi jþï îïa k þï îïk iþï ¶x îïk jþï îïi aþï Subtract equation (3) from (2), we get a ïì ïü ì a ü ¶í ý ¶ ì a ü ì b ü ï ï ì a ü ì b ü í ý A - A = ï ï ï ï îi jþ ï ï ï ï îi k þ i, jk i,kj í ý í ýAb - Aa - í ý í ýAb + A k j a îïi kþï îïa jþï ¶x îïi jþï îïa kþï ¶x Interchanging of a and b in the first and third term of above equation, we get

é ù ê ïì a ïü ïì a ïü ú ê¶í ý ¶í ý ú ì b ü ì a ü ì b ü ì a ü ê îïi k þï - îïi jþï ï ï ï ï ï ï ï ïú Aa A - A = + í ý í ý - í ý í ý ...(4) i, jk i,kj ê ¶x j k ú ê ¶x îïi k þï îïb jþï îïi jþï îïb k þïú ë û

a Ai, jk - Ai,kj = AaRi jk ...(5) where

ïì a ïü ïì a ïü ¶ í ý ¶í ý ïi k ï ïi j ï ïì b ïü ïì a ïü ïì b ïü ïì a ïü a î þ - î þ + - Ri jk = j k í ý í ý í ý í ý ...(6) ¶ x ¶ x îïi k þï îïb jþï îïi j þï îïb k þï

Since Ai is an arbitrary covariant tensor of rank one and difference of two tensors Ai, jk – Ai, kj is a a covariant tensor of rank three. Hence it follows from quotient law that Ri jk is a mixed tensor of rank a i j four. The tensor Ri jk is called Riemann Christoffel tensor or Curvature tensor for the metric gijdx dx . a The symbol Ri jk is called Riemann’s symbol of second kind. Now, if the left hand side of equation (4) is to vanish i.e., if the order of covariant differentiation is to be immaterial then a Ri jk = 0 a Since Aa is arbitrary. In general Ri jk ¹ 0, so that the order of covariant differentiation is not immaterial, It is clear from the equation (4) that “a necessary and sufficient condition for the validity of a inversion of the order of covariant differentiation is that the tensor Ri jk vanishes identically.. Remark The tensor

¶ ¶ ì i ü ì i ü í ý í ý ¶x k ¶xl îa k þ îa lþ Ri = + ...(7) ikl ì i ü ì i ü ì a ü ì a ü í ý í ý í ý í ý î j k þ î j lþ î j k þ î j lþ Riemann-Christoffel Tensor 87

a THEOREM 5.1 Curvature tensor Ri jk is anti symmetric w.r.t. indices j and k. Proof: We know that from (7) curvature tensor

¶ ¶ ïì a ïü ïì a ïü j k í ý í ý ¶x ¶x îïb j þï îïb k þï Ra = + i jk ïì a ïü ïì a ïü ì b ü ì bü í ý í ý ï ï ï ï ïi j ï ïi k ï í ý í ý î þ î þ îïi j þï îïi k þï

ïì a ïü ïì a ïü ¶í ý ¶í ý a b a b a îïi k þï îïi jþï ïì ïü ïì ïü ïì ïü ïì ïü Ri jk = j - k + í ý í ý - í ý í ý ¶x ¶x îïb jþï îïi kþï îïb k þï îïi jþï Interchanging j and k, we get

ì a ü ì a ü ¶í ý ¶í ý a îi jþ îi kþ ì a ü ì b ü ì a ü ì b ü R = - + í ý í ý - í ý í ý ikj ¶xk ¶x j îb k þ îi jþ îb jþ îi k þ

é ïì a ïü ïì a ïü ù ê¶í ý ¶í ý ú ê ïi k ï ïi jï ïì a ïü ïì bïü ïì a ïü ïì b ïüú - î þ - î þ + - = ê j k í ý í ý í ý í ýú ê ¶x ¶x îïb jþï îïi k þï îïb kþï îïi jþïú ê ú ë û

a a Rik j = – Ri jk

a So, Ri jk antisymmetric w.r.t. indices j and k.

Theorem 5.2 To prove that

a a a Ri jk + Rjk i + Rki j = 0 Proof: Since we know that

¶ ¶ ïì a ïü ïì a ïü j k í ý í ý ¶ x ¶x îïb jþï îïb k þï a + R = ïì a ïü ïì a ïü ì b ü ì bü ijk í ý í ý ï ï ï ï ïi j ï ïi k ï í ý í ý î þ î þ îïi jþï îïi k þï

ïì a ïü ïì aïü ¶í ý ¶í ý a îïi k þï îïi jþï ïì a ïü ïì b ïü ïì b ïü ïì a ïü Rijk = ...(1) j - k + í ý í ý - í ý í ý ¶x ¶x îïb jþï îïi k þï îïi jþï îïb k þï 88 Tensors and Their Applications

Similarly

ì a ü ì a ü ¶í ý ¶í ý a î j iþ î j k þ ì a ü ì b ü ì b ü ì aü Rj ki = - + í ý í ý - í ý í ý ...(2) ¶x k ¶xi îb k þ î j iþ î j k þ îb iþ and

ïì a ïü ïì a ïü ¶í ý ¶í ý ï ï ï ï ì a ü ì b ü ì b ü ì a ü a îk jþ îk iþ ï ï ï ï ï ï ï ï Rki j = i - j + í ý í ý - í ý í ý ...(3) ¶x ¶ x îïb iþï îïk jþï îïk iþï îïb jþï On adding (1), (2) and (3), we get

a a a Rijk + Rjki + Rk ij = 0 This is called cyclic property.

5.2 RICCI TENSOR

a The curvature tensor Ri jk can be contracted in three ways with respect to the index a and any one of its lower indices

a a a Rajk, Riak, Rija

Now, from equation (7), art. 5.1, ¶ ¶ ïì a ïü ïì a ïü j k í ý í ý ¶x ¶x îïb jþï îïb k þï + Ra = ïì a ïü ïì a ïü ì b ü ì b ü ajk í ý í ý ï ï ï ï ïa jï ïa kï í ý í ý î þ î þ îïa jþï îïa k þï ïì a ïü ïì a ïü ¶í ý ¶í ý îïa k þï îïa jþï ïì a ïü ïì b ïü ïì a ïü ïì b ïü = j - k + í ý í ý - í ý í ý ¶x ¶x îïb jþï îïa k þï îïb k þï îïa jþï

¶2 log g ¶2 log g = – ¶x j¶xk ¶xk¶x j ì a ü ¶log g = a Since í ý k and a and b are free indices R = 0 . îa k þ ¶x ajk a Also for Rija .

a Write Rij for Rija Riemann-Christoffel Tensor 89

¶ ¶ ì a ü ì a ü j a í ý í ý a ¶x ¶x îb jþ îb aþ Rij = Rija = + ì a ü ì a ü ì b ü ì b ü í ý í ý í ý í ý îi jþ îi aþ îi jþ îi aþ

ì a ü ì a ü ¶í ý ¶í ý îi aþ îi jþ ì a ü ì b ü ì a ü ì b ü Rij = - + í ý í ý - í ý í ý ¶x j ¶xa îb jþ îi aþ îb aþ îi jþ

ì a ü 2 ¶í ý ¶ log g îi jþ ì a üì b ü ì a üì b ü Rij = - + í ýí ý - í ýí ý ...(1) ¶x j¶xi ¶xa îb jþîi aþ îb aþîi jþ Interchanging the indices i and j we get ì a ü ì a ü ¶í ý ¶í ý î j aþ î j iþ ì aü ì b ü ì a ü ì b ü R = - + í ý í ý - í ý í ý ji ¶xi ¶xa îb iþ î j aþ îb aþ î j iþ

ì a ü 2 ¶í ý ¶ log g îi jþ ì b ü ì a ü ì a ü ì b ü R = - + í ý í ý - í ý í ý ...(2) ji ¶xi¶x j ¶xa îi aþ îb jþ îb aþ îi j þ (Since a and b are dummy indices in third term). Comparing (1) and (2), we get

Rij = Rji

Thus Rij is a symmetric Tensor and is called Ricci Tensor.. For a Riak : a a Riak = - Rika = -Rik 5.3 COVARIANT RIEMANN-CHRISTOFFEL TENSOR The associated tensor a Rij kl = gia Rjkl ...(1) is known as the covariant Riemann-Christoffel tensor or the Riemann-Christoffel tensor of the first kind.

Expression for Rijkl a Rijkl = giaRjkl 90 Tensors and Their Applications

æ ¶ ì a ü ¶ ì a ü ì b ü ì a ü ì b ü ì a üö = gia ç í ý - í ý + í ý í ý - í ý í ý÷ …(2) è ¶xk î j lþ ¶xl î j kþ î j lþ îb kþ î j kþ îb lþø Now, ¶ ì a ü ¶ é a ù a ¶g g ì ü - ì ü ia ia k í ý = k ê gia í ýú í ý k ¶x î j lþ ¶x ë î j lþû î j lþ ¶x

¶[ jl,i] ì a ü ¶g - ia = k í ý k ...(3) ¶x î j lþ ¶x Similarly, ¶ ì a ü ¶[ jk,i] ì a ü ¶g g - ia ia l í ý = l í ý l ...(4) ¶x î j k þ ¶x î j kþ ¶x é udv duv vduù êBy the formula = - ú ë dx dx dx û Using (3) and (4) in equation (2), we get

¶[jl,i] ïì a ïü ¶g ¶[ jk,i] ïì a ïü ¶g ïì b ïü ïì a ïü ïì b ïü ïì a ïü - ia - + ia + g - g Rijkl = k í ý k l í ý l ia í ý í ý ia í ý í ý ¶x îï j lþï ¶x ¶x îï j kþï ¶x îï j lþï îïb k þï îï j kþï îïb lþï

¶[ jl,i] ¶[ jk,i ] ì a ü ¶gia ì a ü ¶gia ì b ü ì a ü ì b ü ì a ü = - + í ý - í ý + g í ý í ý - g í ý í ý ¶xk ¶xl î j k þ ¶xl î j lþ ¶xk ia î j lþ îb kþ ia î j kþ îb lþ

¶[ jl,i] ¶[ jk,i] ì a ü ì a ü R - + il,a + al,i - ik,a + ak,i ijkl = k l í ý( [ ] [ ] ) í ý( [ ] [ ] ) ¶x ¶x î j k þ î j lþ

ì b ü ì b ü + í ý[bk,i] - í ý[bl,i] î j lþ î j kþ

¶[ jl,i] ¶[ jk,i] ì a ü ì a ü - + il,a - ik ,a Rijkl = k l í ý[ ] í ý[ ] ...(5) ¶x ¶x î j k þ î j lþ It is also written as

¶ ¶ ì a ü ì a ü k k í ý í ý Rijkl = ¶x ¶x + î j k þ î j lþ ...(6) [ jk,i] [ jl,i] [ik ,a] [il,a] But we know that

1 æ ¶g ¶g ji ¶g jl ö , = ç li + - ÷ ...(7) [jl i] ç j l i ÷ 2 è ¶x ¶x ¶x ø Riemann-Christoffel Tensor 91

1 æ ¶g ¶g ji ¶g ö and , = ç ki + - ik ÷ ...(8) [jk i] ç j k i ÷ 2 è ¶x ¶x ¶x ø Using (7) and (8), equation (5) becomes

1 ¶ æ ¶g ¶g ji ¶g jl ö 1 ¶ æ ¶g ¶g ji ¶g jk ö ç li + - ÷ - ç ki + - ÷ Rijkl = k ç j l i ÷ l ç j k i ÷ 2 ¶x è ¶x ¶x ¶x ø 2 ¶x è ¶x ¶x ¶x ø ì a ü ì a ü + í ý[il,a]- í ý[ik ,a] î j k þ î j lþ

æ 2 2 2 2 ö 1 ¶ g ¶ g jk ¶ g ¶ g jl ç il + - ik - ÷ + Rijkl = 2 ç j k i l j l i k ÷ è ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x ø ì a ü ì a ü í ý[il,a] - í ý[ik,a] ...(9) î j kþ î j lþ Since

ì a ü ì a ü ab í ý = g ab[ jk,b] and í ý = g [ jl,b] î j kþ î j lþ

æ 2 2 2 2 ö 1 ¶ g ¶ g jk ¶ g ¶ g jl R = ç il + - ik - ÷ ijkl ç j k i l j l i l ÷ 2 è ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x ø

+ gab [ jk ,b] [il,a] - gab [ jl,b] [ik,a] ...(10)

This is expression for Rijkl . The equation (9) can also be written as 2 2 1 æ ¶2g ¶ g ¶2g ¶ g ö ç il + jk - ik - jl ÷ Rijkl = ç j k i l j l i k ÷ 2 è ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x ø ïì a ïü ïì b ïü ïì a ïü ïì b ïü + g ab í ý í ý - g ab í ý í ý îï j k þï îïi lþï îï j lþï îïi k þï

5.4 PROPERTIES OF RIEMANN-CHRISTOFFEL TENSORS OF FIRST KIND Rijkl

(i) Rjikl = -Rijkl

(ii) Rijlk = -Rijkl

(iii) Rklij = Rijkl

(iv) Rijkl + Riklj + Riljk = 0 92 Tensors and Their Applications

Proof: We know that from equation (9), Pg. 91.

æ 2 ¶2g 2 ¶ 2g ö 1 ç ¶ gil jk ¶ gik jl ÷ Rijkl = + - - 2 ç j k i l 2 j l i k ÷ è ¶x ¶x ¶x ¶x ¶ x ¶x ¶x ¶x ø

ì a ü ì a ü + í ý[il,a]- í ý[ik ,a] ...(1) î j k þ î j lþ (i) Interchanging of i and j in (1), we get

æ ¶2g ¶2 ¶2g ¶2 ö 1 ç jl gik jk gil ÷ ì a ü ì a ü R = = - - + í ý [ jl,a]- í ý [ jk,a] jikl 2 ç i k j l i l j k ÷ i k i l è ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x ø î þ î þ

æ 2 2 2 2 ö a a 1 ¶ g ¶ g jk ¶ g jl ¶ g ïì ïü ïì ïü = - ç il + - - ik ÷ + í ý [ jl , a]- í ý [ jk , a] ç j k i l i k j l ÷ 2 è ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x ø îïi k þï îïi l þï

Rjikl = -Rijkl

or Rijkl = -Rjikl (ii) Interchange l and k in equation (1) and Proceed as in (i) (iii) Interchange i and k in (1), we get

æ 2 2 2 2 ö 1 ¶ g ¶ g ji ¶ g ¶ g jl ì a ü ì a ü R = ç kl + - ki - ÷ + [kl,a] - [ki,a] kjil ç j i k l j l k i ÷ í ý í ý 2 è ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x ø î j iþ î j lþ Now interchange j and l, we get æ 2 2 2 2 ö 1 ¶ gkj ¶ g ¶ g ¶ glj ì a ü ì a ü R ç + li - ki - ÷ + [kj,a] - [ki,a] klij = ç l i k j l j k i ÷ í ý í ý 2 è ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x ø îl iþ îl jþ For ì aü ì b ü ì a ü ï ï ab ï ï í ý[kj,a] = í ýg í ý îl iþ îïl i þï îïk jþï

ì b ü ab ì a ü = í ýg í ý îk jþ îl iþ

ì b ü = í ý[li,b] îk jþ

ì a ü ì a ü í ý[li,b] = í ý[li,a] îl iþ îk jþ Riemann-Christoffel Tensor 93

So,

æ ¶2g 2 2 ¶2g ö ì a ü ì a ü 1 ç kj ¶ gli ¶ gki lj ÷ Rklij = + - - + í ý [li,a] - í ý[ki,a] ç l i k j l j k i ÷ k j l j 2 è ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x ø î þ î þ

Rklij = Rijkl, from (1) from equation (1)

æ ¶2 ¶2g ¶2g ¶2g ö 1 ç gil jk jk jl ÷ ì a ü ì a ü R + - - + í ý[il,a]- í ý[ik,a] ijkl = 2 ç j k i l j l j k ÷ j k j l è ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x ø î þ î þ

æ 2 2 2 2 ö 1 ¶ gij ¶ g ¶ g ¶ gkj ì a ü ì a ü R ç + kl - il - ÷ + [ij,a] - [il,a] iklj = ç k l i j k j i l ÷ í ý í ý 2 è ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x ø îk lþ îk jþ

æ ¶2 ¶2g ¶2 g ¶2 ö 1 ç gik lj ij glk ÷ ì a ü ì a ü R = + - - + í ý[ik ,a]- í ý[ij,a] iljk 2 ç l j i k l k i j ÷ l j l k è ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x ø î þ î þ On adding these equations, we get

Rijkl + Riklj + Riljk = 0

This property of Rijkl is called cyclic property..

Theorem 5.3 Show that the number of not necessarily independent components of curvature tensor 1 2 2 does not exceed n (n -1) . 12 Or Show that number of distinct non-vanishing components of curvature tensor does not exceed 1 2 2 n (n -1) . 12

Proof: The distinct non-vanishing components of Rijkl of three types.

(i) Symbols with two distinct indices i.e., Rijij . In this case total number of distinct non-vanishing 1 components of R are n(n -1) . ijkl 2

(ii) Symbols with three distinct indices i.e., Rijik . In this case, total number of distinct non- 1 vanishing components R are n(n -1) (n - 2) . ijkl 2

(iii) Symbols Rijkl with four distinct indices. In this case, total number of distinct non-vanishing 2 2 n (n -1) components of R is . ijkl 12 94 Tensors and Their Applications

Hence the number of distinct non-vanishing components of the curvature tensor Rijkl does not 1 2 2 exceed n (n -1) . 12 Remark

When Rijkl is of the form Riiii i.e., all indices are same

In this case, Riiii has no components.

5.5 BIANCHI IDENTITY It states that i i i Rjkl,m + R jlm,k + Rjmk,l = 0

and Rhjkl,m + Rhjlm,k + Rhjmk,l = 0 1 Proof: Introducing geodesic coordinate in which Christoffel symbols are constant with the pole at P0 . Since we know that

¶ ¶ ì i ü ì i ü í ý í ý ¶x k ¶xl îmkþ îmlþ Ri = + jkl ì i ü ì i ü ì mü ì mü í ý í ý í ý í ý î jk þ î jlþ î jk þ î jl þ ì i ü ì i ü ¶í ý ¶í ý jl jk ì i üì mü ì m üì i ü i î þ - î þ + - Rjkl = k l í ýí ý í ýí ý ¶x ¶x îmk þî jl þ î jk þîmlþ

ì i ü ì i ü ¶2í ý ¶2í ý i î j lþ î j k þ R , = - ...(1) jkl m ¶xm ¶x k ¶xm¶xl ì i ü ì mü Since í ý, í ý etc. are constant at pole. î j kþ î j l þ So, their derivatives are zero.

¶ ¶ ïì i ïü ïì i ïü l m í ý í ý ¶x ¶x îïa lþï îïa mþï + Ri = ïì i ïü ïì i ïü ì a ü ì a ü jlm í ý í ý ï ï ï ï ï j lï ï j mï í ý í ý î þ î þ îï j lþï îï j mþï

1 Details of geodesic coordinate given in chapter curvature in curve. Geodesic. Riemann-Christoffel Tensor 95

ïì i ïü ïì i ïü ¶ í ý ¶í ý ï ï ï ï ì i ü ì a ü ì i ü ì a ü = î j mþ î j lþ ï ï ï ï ï ï ï ï l - m + í ý í ý - í ý í ý ¶x ¶x îïa lþï îï j mþï îïa mþï îï j lþï

ì i ü ì i ü ¶2í ý ¶2í ý i î j mþ î j lþ Rjlm,k = - ...(2) ¶x k¶xl ¶x k¶xm and

¶ ¶ ì i ü ì i ü m k í ý í ý i ¶x ¶x îb mþ îb k þ Rjmk = + ì i ü ì i ü ì b ü ì b ü í ý í ý í ý í ý î j mþ î j kþ î j mþ î j k þ

ì i ü ì i ü ¶í ý ¶í ý j k j m ì i ü ì b ü ì i ü ì a ü i î þ - î þ + - Rjmk = m k í ý í ý í ý í ý ¶x ¶x îb mþ î j k þ îb kþ î j mþ

ì i ü ì i ü ¶2í ý ¶ 2í ý i î j k þ î j mþ Rjmk = - ...(3) ¶x m¶xl ¶xk ¶xl On adding (1), (2) and (3), we get i i i Rjkl,m + R jlm,k + Rjmk,l = 0 ...(4)

i Multiplying Rjkl,m by ghi i.e., g Ri hi jkl,m = Rhjkl,m Then equation (4) becomes

Rhjkl,m + Rhjlm,k + Rhjmk,l = 0 ...(5) Since every term of equation (4) and (5) is a tensor. So, equation (4) and (5) are tensor equations and therefore hold in every coordinate system. Further, P0 is an arbitrary point of Vn . Thus there hold throughout Vn . Hence equation (4) or (5) is called Bianchi identity..

5.6 EINSTEIN TENSOR

i 1 i Theorem 5.4 To prove the tensor Rj - d jR is divergence free. 2 Proof: We know that from equation (5) 96 Tensors and Their Applications

Rhjkl,m + Rhjlm,k + Rhjmk,l = 0 Multiply it by g hlg jk , we get hl jk hl jk hl jk g g Rhjkl,m + g g Rhjlm,k + g g Rhjmk,l = 0

jk hl jk hl jk g R jk,m + g g (-Rhjml,k ) + g g (-Rjhmk,l ) = 0

Since Rhjlm = -Rhjml & Rhjmk = -Rjhmk

jk jk hl g R jk,m - g Rjm,k - g Rhm,l = 0

k l jk R,m - Rm,k - Rm,l = 0 Since g R jk = R

k k R,m - Rm,k - Rm,k = 0

k R,m - 2Rm,k = 0

k 1 R , - R, m = 0 m k 2 1 k k k Rm,k - dm R,k = 0 since R,m = d R,k 2 m

æ k 1 k ö ç Rm - dm R÷ = 0 è 2 ø, k

k 1 k R - d R is divergence free. m 2 m

k 1 k k i 1 i The tensor R - d R = G or Rj - d jR is known as Einstein Tensor.. m 2 m m 2

5.7 RIEMANN CURVATURE OF A Vn i i Consider two unit vectors p and q at a point P0 of Vn. These vectors at P0 determine a pencil of directions deferred by ti = api + bqi. a and b being parameters. One and only one geodesic will pass i through P0 in the direction of p . Similarly one and only one geodesic will pass through in the direction i i i q . These two geodesics through P0 determined by the orientation of the unit vectors p and q . Let this surface is denoted by S.

The of S at P0 is defined to be the Riemannian Curvature of Vn at P0 for the orientation determined by pi and qi. i Let the coordinates y of Vn are Riemannian coordinates with origin at P0. The equation of surfaces S in given by yi = ( pia + qib)s ...(1) i.e., yi = piu1 + qiu2 ...(2) Riemann-Christoffel Tensor 97

where as = u1 and bs = u2, three parameters namely a, b, s can be reduced to two parameters 1 u and u2 . Here u1 and u2 are coordinates of any current point on S. 2 a b Let ds = babdu du be the metric for the surface S. where ¶y i ¶y j bab = g (a, b = 1, 2) ij ¶ua ¶ub

ì k ü ì g ü Let í ý and í ý be the Christoffel symbols corresponding to the coordinates yi and ua . a b îi jþg î þ a b i j Let Rabgd and Rhijk be curvature tensor corresponding to the metrices bab du du and gijdy dy . Since the Greek letters a,b,g,d take values 1,2 and so that the number of independent non- 1 1 vanishing components of R is n2 (n2 -1) for n = 2, i.e., they are ×22 (22 -1) = 1. Let us abgd 12 12 a a transform the coordinate system u to u¢ and suppose that the corresponding value of R1212 are ' R1212 . Then ¶ a ¶ b ¶ g ¶ d ' u u u u R = Rabgd 1212 ¶u¢1 ¶u¢2 ¶u¢1 ¶u¢2

¶u1 ¶ub ¶u g ¶ud ¶u2 ¶ub ¶ug ¶ud = R1bgd + R2bgd ¶u¢1 ¶u¢2 ¶u¢1 ¶u¢2 ¶u¢1 ¶u¢2 ¶u¢1 ¶u¢2

¶u1 ¶u2 ¶ug ¶ud ¶u 2 ¶u1 ¶ug ¶ud = R12gd + R21gd ¶u¢1 ¶u¢2 ¶u¢1 ¶u¢2 ¶u¢1 ¶u¢2 ¶u¢1 ¶u¢2

¶u1 ¶u2 ¶u1 ¶u2 ¶u1 ¶u2 ¶u2 ¶u1 = R + R1221 1212 ¶u¢1 ¶u¢2 ¶u¢1 ¶u¢2 ¶u¢1 ¶u¢2 ¶u¢1 ¶u¢2

¶u2 ¶u1 ¶u1 ¶u2 ¶u2 ¶u1 ¶u2 ¶u1 + R2112 + R2121 ¶u11 ¶u¢2 ¶u¢2 ¶u¢2 ¶u¢1 ¶u¢2 ¶u¢1 ¶u¢2

2 é 1 2 1 2 2 2 1 ù ¶u ¶u ¶u ¶u ¶u ¶u¢ æ ¶u ¶u ö R2112ê - 2 + ç ÷ ú = ê ¶u¢1 ¶u¢ 2 ¶u¢1 ¶u¢2 ¶u¢2 ¶u¢2 ç ¶u¢1 ¶u¢2 ÷ ú ë è ø û

¶u1 ¶u1 ¶ ¢1 ¶ ¢2 2 u u ¶u = R1212J where J = = ¶u2 ¶u2 ¶u¢ 1 2 so, ¶u¢ ¶u¢ 2 R1212¢ = R1212J ...(3) 98 Tensors and Their Applications

Again ¶u g ¶ud bab¢ = bab ¶u¢a ¶u¢b ¶u g ¶u d Þ bab¢ = bab ¶u¢a ¶u¢b

or b¢ = bJ 2 ...(4) from (3) and (4), we get R¢ R 1212 = 1212 = K (say) ...(5) b¢ b This shows that the quantity K is an invariant for transformation of coordinates. The invariant K is defined to be the Guasian curvature of S. Hence K is the Riemannian Curvature of S at P0 . i Since Riemannian Coordinates y with the origin at P0 . We have as geodesic coordinates with the pole at P0 . Therefore

ì k ü ì g ü í ý , í ý = 0 at P0 i j a b î þg î þb Then æ ¶[ij,h] ¶[ik,h] ö R = ç - g + g ÷ at P . hijk ç k j ÷ 0 è ¶y ¶y ø and

æ ¶[bg, a]b ¶[bd, a]b ö R = ç - + ÷ at P . abgd è ¶ud ¶u g ø 0

¶[21,1] b ¶[22,1]b R = - + at P ...(6) 1212 ¶u2 ¶u1 0 from (5) we get

1 æ ¶[21, 1] ¶[22, 1] ö ç - b + b ÷ K = ç 2 1 ÷ ...(7) b è ¶u ¶u ø

This is required expression for Riemannian curvature at P0 .

5.8 FORMULA FOR RIEMANNIAN CURVATURE IN THE TERMS OF COVARIANT CURVATURE TENSOR OF Vn

ì k ü ì g ü a b Let í ý and í ý be the Christoffel symbols of second kind relative to the metrices babdu du i j a b î þg î þ Riemann-Christoffel Tensor 99

i j and gijdy dy respectively. We have ¶y i ¶y j ¶yk [ab, g] = [ij, k] ...(8) b ¶u a ¶ub ¶ug g

¶yi ¶y j ¶yk [21,1] = [ij,k ]g b ¶u 2 ¶u1 ¶u1

i j k = q p p [ij,k]g using (2) Now,

¶[21, 1]b ¶[ij, k]g = qi p j p k ¶u2 ¶u2

h i j k ¶[ij,k ]g ¶y = q p p ¶y h ¶u 2

i j k h ¶[ij,k ]g = q p p q ¶y h Interchanging h and k, we get ¶ ij,h ¶[21, 1]b i k j h [ ]g = q q p p ...(9) ¶u 2 ¶yk Similarly, ¶ ij,h ¶[22, 1]b i k j h [ ]g = q q p p ...(10) ¶u1 ¶yk Using (9) and (10), equation (6) becomes

é ¶[ij,h] ¶[ik ,h] ù phqi p jqk - g + g at P R = ê k j ú 0 1212 ë ¶y ¶y û

h i j k R1212 = p q p q Rhijk at P0 ...(11) Since

¶yi ¶y j bab = g ij ¶ua ¶ub

¶y i ¶y j b = g = g pi p j 11 ij ¶u1 ¶u1 ij h j b11 = ghj p p 100 Tensors and Their Applications

Similarly

i j i k b22 = gijq q = gikq q

i j j i h k b12 = gij p q = g ji p q = ghk p q

b11 b12 = - b = b11b22 b12b21 b21 b22

h i j k b = p q p q [ghigik - gij ghk ] ...(12) Dividing (11) and (12), we get h i k R p q q Rhijk 1212 = K = h i j k ...(13) b p q p q (gik ghj - gijghk ) i This is formula for Riemannian Curvature of Vn at P0 determined by the orientation of Unit vectors p i and q at P0 .

5.9 SCHUR’S THEOREM If at each point, the Riemannian curvature of a space is independent of the orientation choosen then it is constant throughout the space. i Proof: If K is the Riemannian curvature of Vn at P for the orientation determined by unit vectors p and qi then it is given by

h i j k p q p q Rhijk K = h i j k ...(1) (gik ghj - gijghk ) p q p q Let K be independent of the orientation choosen. Then equation (1) becomes R K = hijk gik ghj - gijg hk

Rhijk = K (gik ghj - gijghk ) ...(2)

We have to prove that K is constant throughout the space Vn .

If N = 2, the orientation is the same at every point. So, consider the case of Vn when n > 2.

Since gij are constants with respect to covariant differentation, therefore covariant differentation of (2) gives

Rhijk,l = (gik ghj - gijghk )K,l ...(3)

where K,l is the of K. Taking the sum of (3) and two similar equations obtained by cyclic of the suffices, j, k and l. Riemann-Christoffel Tensor 101

R + R + R (ghjgik - ghk gij)K ,i +(ghk gil - ghl gik )K , j +(ghlgij - ghjgil )K ,k = hijk,l hikl, j hilj,k ...(4) Here n > 2 therefore three or more distinct values to indices j, k, m can be given . hj hj j Multiplying (4) by g and using g ghl = dl , we get

h j j h (ngik -di ghk )k,l +(gildk - gikdl )K, j +(d i ghl - ngil )K ,k = 0 or, i (n -1)gikK,l + 0 + (1- n)gilK,k = 0 for d j = 0,i ¹ j

gik K,l - gilK,k = 0 Multiplying by g ik and using

ik ik k g dik = n, gilg = dl , we get k nK,l -dl K,k = 0 or (n -1)K,l = 0

or K,k = 0 as (n -1)K ,l = 0 ¶K or = 0. ¶x l integrating it, we get K = constant. This proves that the partial derivatives of K w.r.t. to x’s are all zero.

Consequently K is constant at P. But P is an arbitrary point of Vn . Hence K is constant throughout Vn .

5.10 MEAN CURVATURE

The sum of mean of a Vn for a mutually orthogonal directions at a point, is independent of the ennuple choosen. Obtain the value of this sum. Or

Prove that the mean curvature (or Riccian Curvature) in the direction ei at a point of a Vn is the sum of n – 1 Riemmanian curvatures along the direction pairs consisting of the direction and n – 1 other directions forming with this directions an orthogonal frame. i e i Proof: Let e h be the components of unit vector in a given direction at a point P of a Vn . Let k be the components of unit vector forming an orthogonal ennuple. i i ¹ Let the Riemannian curvature at P of Vn for the orientation determined by l h and e k (h k ) be denoted by Khk and given by p q lr ls eh ek k k R pqrs K = hk p q r s ( - ) eh ek eh ek g pr qqs g ps g qr

p q r s eh1ek1eh1ek1R pqrs = p r q s p s q r ...(1) (eh eh g pr ) (ek ek g qs) - (eh ek g ps ) (ek eh gqr ) 102 Tensors and Their Applications

i i Since unit vectors eh , l k are orthogonal. Therefore p r eh eh g pr = 1 p s and eh eh g ps = 0 etc. Using these in equation (1), then equation (1) becomes

p q r s eh ek eh ek R pqrs Khk = 1´ 1 - 0 ´ 0 p q r s Khk = eh ek eh ek R pqrs ...(2)

n n K = e peq er e s R å hk å h k h k pqrs k=1 k =1 n K = M . Put å hk h Then k=1 n e pe r eq e s R Mh = h h å k k pqrs k =1

p r qs = eh eh g R pqrs

p r qs = - eh eh g Rqprs

- e pe r R Mh = h h pr ...(3)

This shows that Mh is independent of (n – 1) orthogonal direction choosen to complete an orthogonal p ennuple. Here Mh is defined as mean curvature or Riccian curvature of Vn for the direction eh1 . Summing the equation (1) from h =1 to h = n, we get n M p r å h = - eh|eh|Rpr h=1 pr = - g Rpr = –R n M pr or å h = - g Rpr == -R h=1 This proves that the sum of mean curvatures for n mutual orthogonal directions is independent of the directions chosen to complete an orthogonal ennuple and has the value -R.

5.11 RICCI’S PRINCIPAL DIRECTIONS

i Let eh1 is not a unit vector and the mean curvature Mh is given by i j Rij eh eh Mh = i j gij eh eh Riemann-Christoffel Tensor 103

i j Þ ( Rij + M h gij)eh eh = 0

i Differentiating it w.r. to eh1, we get ¶M h g e j e k + 2( R + Mg )e j i jk h h ij ij h = 0 ...(1) ¶eh

For maximum and minimum value of Mh .

¶M h i = 0 ¶eh Then equation (1) becomes

i (Rij + Mgij)eh = 0 These are called Ricci's Principal direction of the space as they are principal directions of Ricci tensor Rij.

5.12 EINSTEIN SPACE

A space, which is homogeneous relative to the Ricci tensor Rij is called Einstein space. If space is homogeneous then we have

Rij = lgij ...(1)

Inner multiplication by gij , we get

ij ij R = ln since Rijg = R and g gij = n 1 Þ = R l n from (1) R R = g ij n ij R Hence a space is an Einstein space if R = g at every point of the space. ij n ij

Theorem 5.5 To show that a space of constant Curvature is an Einstein space. i i Proof: Let the Riemannian curvature K at P of Vn for the orientation determined by p and q , is given by

h i j k p q p q Rhijk K = h i j k p q p q (gik ghj - gijghk ) Since K is constant and independent of the orientaion. 104 Tensors and Their Applications

Rhijk K = (gik ghj - gijghk )

Rhijk = K (gik ghj - gijghk ) Multiplying by g hk

hk hk Kg (gik ghj - gijghk ) = g Rhijk

h K (di ghj - ngij) = Rij

hk h hk hk Since g gik = di ; g ghk = n and g Rhi jk = Rij

K (gij - ngij) = Rij

K (1 - n)gij = Rij ...(1) Multiplying by gij , we get ij Kn(1- n) = R as g Rij = R ...(2) from (1) & (2) R 1 R = (1- n)gij × = Rgij ij n(1 - n) n R Þ R = gij ij n

This is necessary and sufficient condition for the space Vn to be Einstein space.

5.13 WEYL TENSOR OR PROJECTIVE CURVATURE TENSOR

Weyl Tensor denoted as Whijk and defined by 1 W = R + (g R - g g ) hijk hijk 1 - n ki hj kh ij

Theorem 5.6 A necessary and sufficient condition for a Riemannian Vn(n > 3) to be of constant curvature to that the Weyl tensor vanishes identically throughout Vn . Proof: Necessary Condition:

Let K be Riemannian Curvature of Vn . Let K = constant.

We have to prove that Whijk = 0 Since we know that

h i j k p q p q Rhijk K = h i j k = constant (ghjgik - gijghk ) p q p q Riemann-Christoffel Tensor 105

Since K be independent of the orientation determined by the vector pi and qi . Then

Rhijk K = ...(1) ghj gik - gijg hk Multiplying by g hk , we get

hk hk g Rhijk = g K (ghjgik - gijghk )

k Rij = K (d j gik - ngij)

Rij = K (1 - n)gij ...(2)

Multiplying by g ij again, we get

ij ij g Rij = K (1 - n)gijg R = K (1 - n)n ...(3) Putting the value of K from (3) in (2), we get R R = g ...(4) ij n ij The equation (3) shows that R is constant since K is constant. Now, the W tensor is given by 1 W = R + [g R - g g ] hijk hijk 1 - n ik hj hk ij from (5), we get 1 é R R ù + - Whijk = Rhijk êgik ghj ghk gij ú 1 - n ë n n û R = Rhijk + [gik ghj - ghk gij ] n(1 - n)

R Rhijk = Rhijk + × , by. eqn. (1) n(1 - n) K

Rhijk W = R + K , by equation (3) hijk hijk K

Whijk = 2Rhijk

Since K is constant. The equation (5) shows that Whijk = 0 . This proves necessary condition. 106 Tensors and Their Applications

Sufficient Condition

Let Whijk = 0. Then we have to prove that K is constant.

Now, Whijk = 0. 1 Þ R + [g R - g R ] = 0 hijk 1 - n ik hj hk ij Multiplying by g hk , we get 1 R + [g hk g R - g hk g R ] = 0 ij 1- n ik hj hk ij 1 R + [dh R - nR ] = 0 ij 1- n i hj ij

Rij R + (1- n) = 0 ij 1- n

Þ 2Rij = 0

Þ Rij = 0 hk Since Rij = 0 Þ g Rhijk = 0. hk Þ g = 0 or Rhijk = 0

If Rhijk = 0, then clearly K = 0. So, K is constant. If g hk = 0 then

h i j k Rhijk p q p q K = h i j k (ghjgik - 0gij) p q p q

h i j k h i j k Rhijk p q p q Rhijk p q p q K = h j i k = 2 2 (p p ghj ) (q q gik ) p ×q

h i j k 2 2 K = Rhijk p q p q since p =1,q = 1

K = constant as Rij = 0 This proves sufficient condition.

EXAMPLE 1

For a V2 referred to an orthogonal system of parametric curves (g12 = 0) show that

R12 = 0, R11g22 = g22g11 = R1221

ij 2R1221 R = g Rij = g11g22 Consequently 1 Rij = Rg . 2 ij Riemann-Christoffel Tensor 107

Solution

12 Given that g12 = 0 so that g = 0. Also,

1 11 1 22 1 g ij = Þ g = & g = . gij g11 g22

The metric of V2 is given by

2 i j ds = gijdx dx ; (i, j = 1, 2)

2 1 2 2 2 ds = g11(dx ) + g22(dx ) Since g12 = 0.

hk We know that g1 Rhijk = Rij.

g11 g12 g11 0 and g = gij = = = g11g22 g21 g22 0 g22

(i) To prove R12 = 0 g hk R = gh1R R12 = h12k h 2 as Rh122 = 0 21 = g R2121 as R1121 = 0

R12 = 0

(ii) To prove R11g22 = R22g11 = R1221 hk 2k R11 = g Rh11k = g R211k

22 R2112 g R2112 = R11 = g22 So,

R2112 R11 = ...(1) g22 hk 1k and R22 = g Rh22k = g R122k

11 R1221 = g R1221 = g11

R1221 So, R22 = ...(2) g11 from (1) and (2)

R11g22 = R1221=R22g11 … (3) (iii) To prove

2R1221 R = g11g22 ij i1 i2 R = g Rij = g Ri1 + g Ri2 108 Tensors and Their Applications

11 22 12 = g R11 + g R22 for g = 0 R R = 11 + 22 g11 g22

R1221 R1221 = + [∵ by eqn. (3)] g22g11 g11g22

2R1221 R = g11g22 1 (iv) To prove R = Rg ij 2 ij 2R1221 R = g11g22

2R1221 R = as g = g11g22 g 1 R = Rg 1221 2 The eqn (3) expressed as 1 R g = Rg = R g . 11 22 2 22 11 it becomes

Rg Rg11g22 Rg11 R11 = = = 2g22 2g22 2 Rg Rg g Rg = 11 22 = 22 R22 = 2g11 2g11 2 So, 1 1 R = Rg11 & R22 = g22 11 2 2 1 R = g as R = 0 = g 12 2 12 12 12 1 This prove that R = R . ij 2 ij EXAMPLE 2 2 2 2 2 2 2 The metric of the V2 formed by the surface of a sphere of radius r is ds = r dq + r sin qdf 1 in spherical polar coordinates. Show that the surface of a sphere is a surface of constant curvature . r 2 Solution Given that ds 2 = r 2dq2 + r 2 sin 2 qdf2 Since r is radius of curvature then r is constant. Riemann-Christoffel Tensor 109

2 2 2 4 2 g11 = r , g22 = r sin q, g12 = 0, g = g11g22 = r sin q. We can prove 2 2 R1221 = r sin q

Now, the Riemannian curvature K of Vn is given by phqi p jq k K = h i j k . p q p q (ghjgik - ghk gij )

At any point of V2 there exists only two independent vectors.

Consider two vectors whose components are (1, 0) and (0, 1) respectively in V2 . Then

R1212 R1212 K = = g11g22 g

r 2 sin 2 q 1 K = = = constant. r 4 sin 2 q r 2

EXERCISES

1. Show that ¶[ jl, i] ¶[ jk , i] ïì a ïü ïì a ïü - + [il ,a]- [ik , a] Rijkl = k l í ý í ý ¶x ¶x îï j kþï îï j lþï 2. Show that

æ 2 ¶2 2 ¶2 ö 1ç ¶ gil g jk ¶ gik gjl ÷ ab Rijkl = + - - + g ([ jk ,b][il ,a] -[ jl ,b][ik ,a] ) . 2ç j k i l j l i k ÷ è ¶x ¶x ¶x ¶x ¶x ¶x ¶x ¶x ø 3. Using the formula of the problem 2. Show that

Rijkl = -R jikl = -Rijlk = Rklij and Rijkl + Riklj + Riljk = 0

4. Show that the curvature tensor of a four dimensional Riemannian space has at the most 20 distinct non-vanishing components. h 5. (a) If prove that the process of contraction applied to the tensor Rijk generates only one new tensor

Rij which is symmetric in i and j.

1 2 1 2 2 2 3 2 R = R , (b) If ds = g11(dx ) + g22(dx ) + g33(dx ) . Prove that ij ihhj (h, i, j being unequal). ghh 6. Show that when in aV 3 the coordinates can be chosen so that the components of a tensor gij are zero when i, j, k are unequal then 1 (i) Rhj = Rhiij gii 110 Tensors and Their Applications

1 1 (ii) Rhh = Rhiih + Rhjjh gii g jj 7. Prove that if

a 1 ¶R a aj R = R = g R then i,a i i ij 2 ¶x and hence deduce that when n > 2 then of an Einstein space is constant.

8. If the Riemannian curvature K of Vn at every point of a neighbourhood U of Vn is independent of the direction chosen, show that K is constant throughout the neighbourhood U. Provided n > 2.

9. Show that a space of constant curvature K0 is an Einstein space and that R = K0n(1 –n).

10. Show that the necessary and sufficient condition that Vn be locally flat in the neighbourhood of 0 is that Riemannian Christoffel tensor is zero.

11. Show that every V2 is an Einstein space. 12. For two dimensional manifold prove that R K = - 2 13. Show that if Riemann-Christoffel curvature tensor vanishes then order of covariant differentiation is commutative. CHAPTER – 6

THE e-SYSTEMS AND THE GENERALIZED KRÖNECKER DELTAS

The concept of symmetry and skew-symmetry with respect to pairs of indices can be extended to cover to pairs of indices can be extended to cover the sets of quantities that are symmetric or skew- symmetric with respect to more than two indices. Now, consider the sets of quantities A i... i k or A depending on indices written as subscripts or superscripts, although the quantities may not il ... ik k A represent tensor.

6.1 COMPLETELY SYMMETRIC

i1 ...ik The system of quantities A (or Ai1... ik ) depending on k indices, is said to be completely symmetric if the value of the symbol A is unchanged by any permutation of the indices.

6.2 COMPLETELY SKEW-SYMMETRIC

i1... ik A The systems A or ( i1... ik ) depending on k indices, is said to be completely skew-symmetric if the value of the symbol A is unchanged by any even permutation of the indices and A merely changes the sign after an odd permutation of the indices. Any permutation of n distinct objects say a permutation of n distinct , can be accomplished by a finite number of interchanges of pairs of these objects and that the number of interchanges required to bring about a given permutation form a perscribed order is always even or always odd. In any skew-symmetric system, the term containing two like indices is necessarily zero. Thus if one has a skew-symmetric system of quantities Aijk where i, j, k assume value 1, 2, 3. Then

A122 = A112 = 0

A123 = - A213, A312 = A123 etc.

In general, the components Aijk of a skew-symmetric system satisfy the relations.

Aijk = - Aikj = -Ajik

Aijk = Ajki = Akij 112 Tensors and Their Applications

6.3 e-SYSTEM

e (or e ) Consider a skew-symmetric system of quantities i1... in i1,..., in in which the indices i1... in assume

e i1... in values 1,2,...n. The system i1 ... in (or e ) is said to be the e-system if

ì= +1; when i1,i2,..., in ï ï an even permutation of number 1, 2, ..., n ï ï (or i1... in ) = -1; when , ,..., ei1... in e í i1 i2 in ï ï an odd permutation of number 1, 2, ..., n ï ï î= 0 in all other cases

EXAMPLE 1

Find the components of system eij when i, j takes the value 1,2. Solution

The components of system eij are

e11, e12, e21, e22. By definition of e-system, we have

e11 = 0, indices are same

e12 = 1, since i j has even permutation of 12

e21 = - e12 = -1 since i j has odd permutation of 12

e22 = 0, indices are same

EXAMPLE 2

Find the components of the system ei jk .

Solution By the definition of e-system,

e123 = e231 = e321 =1

e213 = e132 = e321 = -1

eijk = 0 if any two indices are same.

6.4 GENERALISED KRÖNECKER DELTA

A symbol di1... ik depending on k superscripts and k subscripts each of which take values from 1 to n, j1... jk is called a generalised Krönecker delta provided that (a) it is completely skew-symmetric in superscripts and subscripts The e-Systems and the Generalized Krönecker Deltas 113

(b) if the superscripts are distinct from each other and the subscripts are the same set of numbers as the superscripts. The value of symbol é=1; an even number of transposition is required to arrange the ê ê superscripts in the same order as subscripts. ... i1 ik ê d ... = -1; where odd number of transpositions arrange the superscripts j1 jk ê ê in the same order as subscripts ëê= 0, in all other cases the value of the symbol is zero

EXAMPLE 3

ij Find the values of dkl. Solution

ij By definition of generalised Kronecker Delta, dkl = 0 if i = j or k = l or if the set. ij is not the set kl.

11 22 23 i.e., d pq = d pq = d13 = ××× = 0 ij dkl = 1 if kl is an even permutation of ij 12 21 13 31 23 i.e., d12 = d21 = d13 = d31 = d23 = ××× =1

ij and dkl = -1 if kl is an odd permutation of ij. 12 31 13 21 i.e., d21 = d13 = d31 = d12 = ××× = -1

ii ...i 1 2 n i1...in Theorem 6.1 To prove that the direct product e e ... of two systems e and e is the j1 j2 jn j1j2... jn generalized Krönecker delta. ii ...i Proof: By definition of generalized Krönecker delta, the product e 1 2 ne has the following values. j1 j2... jn (i) Zero if two or more subscripts or superscripts are same.

(ii) +1, if the difference in the number of transpositions of i1,i2,...,in and j1, j2 ,...,jn from 1,2,...n is an even number.

(iii) –1, if the difference in the number of transpositions of i1, i2,...,in and j1, j2, ...jn from 1, 2,..n an odd number. Thus we can write

i i ...i e i1i 2 ... in e = d 1 2 n j1 j2 ... j n j1 j2... jn

THEOREM 6.2 To prove that

i1i2...in (i) e ... = d i1i 2 in j1 j2... jn

ii ...i (ii) e = d 1 2 n i1i2...in j1j2... jn 114 Tensors and Their Applications

Proof: By Definition of system, i1i2 ... i n (or e ) has the following values. e- e i1i2 ... in

(i) +1; if i1,i2,...,in is an even permutation of numbers 1,2,...n.

(ii) -1; if i1,i2,...,in is an odd permutation of numbers 1,2,...n (iii) 0; in all other cases Hence by Definition of generalized krönecker delta, we can write

i1i2...in i1i2...in (1) e = d12...n and 1 2... (2) e = d n i1i2...in i1i2...in

i jk 6.5 CONTRACTION OF äáâã

i jk Let us contract dabg on k and g . For n = 3, the result is

ijk ij1 ij2 ij3 ij dabg = dab1 + dab2 + dab3 = dab This expression vanishes if i and j are equal or if a and b are equal.

123 If i = 1, and j = 2, we get dab3. Hence

é+ 1; if ab is an even permutation of 12 ê 12 ê- 1; if ab is an odd permutation of 12 dab = ê ëê 0; if abis not permutation of 12

Similarly results hold for all values of a and b selected from the set of numbers 1, 2, 3. Hence

é+ 1; if ij is an even permutation of ab ê ê- 1; if ij is an odd permutation of ab ê ij dab = ê 0; if two of the subscripts or superscripts are equal or when the ê ëê subscripts and superscripts are not formed from the same numbers.

1 If we contract dij . To contract dij first contract it and the multiply the result by . We obtain ab ab 2 a system depending on two indices 1 1 i ij i1 i2 i3 d = daj = (da1 + da2 + da3) a 2 2 The e-Systems and the Generalized Krönecker Deltas 115

1 i 1 12 13 It i = 1 in then we get da = (da2 + da3) da 2 1 This vanishes unless a =1 and if a = 1 then d1 =1.

i Similar result can be obtained by setting i = 2 or i = 3. Thus da has the values. (i) 0 if i ¹ a, (a,i =1,2,3,) (ii) 1 if i = a . By counting the number of terms appearing in the sums. In general we have 1 i ij ij d = daj and d = n(n -1) ...(1) a n -1 ij We can also deduce that

(n - k)! i1i2... irir-1... ik i1i2... ir d d = j1 j2... j j -1... j ...(2) j1 j2... jr (n - r )! r r k and n! di1i2... ir = n(n -1) (n - 2) ××× (n - r +1) = ...(3) j1 j2... jr n - r! or

i1i 2 ... in = ! ...(4) e ei1i2 ... in n and from (2) we deduce the relation

i1i2... irir+1... in = ! ...(5) e ej1 j2 ... jr jr +1... in n

EXERCISE

1. Expand for n = 3 i a 12 i j ab i j ij (a) dad j (b) dij x x (c) dij x y (d) dij 2. Expand for n = 2 ij 1 2 ij 2 1 ij i j ij (a) e ai aj (b) e ai aj (c) e aaab = e a.

ijk 3. Show that dijk = 3 ! if i, j, k = 1, 2, 3.

4. If a set of quantities Ai1i2...ik is skew-symmetric in the subscripts (K in number) then

i1...ik d A ... = k! Aj ... j j1... jk i1 ik 1 k

5. Prove that eijk g is a covariant tensor of rank three where where eijk is the usual permutation symbol. CHAPTER – 7

GEOMETRY

7.1 LENGTH OF ARC Consider the n-dimensional space R be covered by a coordinate system X and a curve C so that C : xi = xi (t), (i = 1,2...,n) ...(1) which is one-dimensional subspace of R. Where t is a real parameter varying continuously in the interval t1 £ t £ t2. The one dimensional manifold C is called arc of a curve. æ 1 2 n ö ç 1 2 n dx dx ,...,dx ÷ Let F ç x , x ,...,x , ÷ be a continuous function in the interval t £ t £ t . Wee è dt dt dt ø 1 2 æ dx ö dxi assume that F ç x, ÷ > 0, unless every = 0 and that for every positive number k è dt ø dt æ 1 2 n ö æ 1 2 n ö ç 1 2 n dx dx dx ÷ ç 1 2 n dx dx dx ÷ F ç x , x ,...,x ,k ,k ,..., k ÷ = kF ç x , x ,...,x , , ,... ÷ . è dt dt dt ø è dt dt dt ø The integral

t 2 æ dx ö s = F ç x, ÷ dt ...(2) òt1 è dt ø is called the length of C and the space R is said to be metrized by equation (2). æ dx ö Different choices of functions F ç x, ÷ lead to different metric geometrices. è dt ø If one chooses to define the length of arc by the formula

p q t 2 dx dx s = g pq(x) dt, (p,q = 1, 2, ..., n) ...(3) òt1 dt dt dx p dxq dx p where g (x) is a positive definite quadratic form in the variable , then the resulting pq dt dt dt geometry is the Riemannian geometry and space R metrized in this way is the Riemannian n-dimensional space Rn. Geometry 117

Consider the coordinate transformation T : x i = x i(x1,...,xn ) such that the square of the element of arc ds,

2 p q ds = g pqdx dx ...(4) can be reduced to the form ds 2 = dx idxi ...(5)

Then the Rn is said to reduce to an n-dimensional Euclidean manifold En.

The Y-coordinate system in which the element of arc of C in En is given by the equation (5) is called an orthogonal cartesian coordinate system. Obviously, En is a generalization of the Euclidean plane determined by the totality of pairs of real values (x1, x 2). If these values (x1, x 2) are associated with the points of the plane referred to a pair of orthogonal Cartesian axes then the square of the element of arc ds assumes the familiar form ds 2 = (d x1)2 + (d x2)2. æ dx ö æ dx ö æ dx ö THEOREM 7.1 A function F ç x, ÷ satisfying the condition F ç x,k ÷ = kF ç x, ÷ for every è dt ø è dt ø è dt ø k > 0. This condition is both necessary and sufficient to ensure independence of the value of the

t2 æ dx ö integral s = F ç x, ÷ dt of a particular mode of parametrization of C. Thus if t in C : xi = xi (t) òt1 è dt ø is replaced by some function t = f(s) and we denote xi[f(s )] by xi(l) . so that xi (t) xi(s) we have equality

t2 æ dx ö s2 F ç x, ÷ dt = F (x,x¢) ds òt1 è dt ø òs1 ds i where x¢i = and t = f (s ) and t = f (s ). ds 1 1 2 2 Proof: Suppose that k is an arbitary positive number and put t = ks so that t1=ks1 and t2=ks2. Then C : xi = xi (t) becomes C : xi (ks) = xi(t) dxi(ks) dxi (ks) and x¢i (s) = = k ds dt t2 æ dx ö Substituting these values in s = Fç x, ÷ dt we get òt1 è dt ø

s2 é dx(ks)ù s = F êx(ks), úkds òs1 ë dt û

s2 or s = F [x(s),x¢(s)]ds òs1 æ dx ö æ dx ö We must have the relation F(x,x¢) = F ç x,k ÷ = kF ç x, ÷. Conversely, if this relation is true è dt ø è dt ø 118 Tensors and Their Applications for every line element of C and each k > 0 then the equality of integrals is assumed for every choice of 1 parameter t = f (s),f (s) > 0,s1 £ s £ s2 with and t2 = f (s2). i i dx Note: (i) Here take those curves for which ( ) and are continuous functions in t1 £ t £ t2. x t dt æ dx ö æ dx ö æ dx ö (ii) A function F ç x, ÷ satisfying the condition F ç x, k ÷ = kF ç x, ÷ for every k > 0 is called è dt ø è dt ø è dt ø i dx positively homogeneous of degree 1 in the . dt

EXAMPLE 1 What is meant, consider a sphere S of radius a, immersed in a three-dimensional Euclidean 1 2 3 manifold E3, with centre at the origin (0, 0, 0) of the set of orthogonal cartesian axes O - X X X . Solution Let T be a plane tangent to S at (0, 0,–a) and the points of this plane be referred to a set of orthogonal cartesian axes O¢ –Y1Y2 as shown in figure. If we draw from O (0, 0, 0) a radial line OP, interesting the sphere S at P(x1, x2 ,x3) and plane T at Q(x1, x 2,-a) then the points P on the lower half of the sphere S are in one-to-one correspondence with points (x1, x 2) of the tangent plane T..

3 X

X2 O P2 P1 C 2 O´ Y 1 X

Q2

1 K Y Q1

Fig. 7.1

If P(x1, x2 ,x3) is any point on the radial line OP,, then symmetric equations of this line is x1 - 0 x2 - 0 x3 - 0 = = = l x1 - 0 x 2 - 0 - a - 0 or x1 = lx1, x2 = lx 2, x3 = -la ...(6) Since the images Q of points P lying on S, the variables xi satisfy the equation of S, (x1)2 + (x2 )2 + (x3)2 = a2

2 é 1 2 2 2 2 ù l (x ) + (x ) + a 2 or ëê ûú = a Geometry 119

Solving for l and substituting in equation (6), we get

1 ax ax 2 1 = , x2 = x 1 2 2 2 2 (x ) + (x ) + a (x1)2 + (x 2 )2 + (x 3)2

- a2 and x3 = ...(7) (x1)2 + (x 2)2 + (x3)2 These are the equations giving the analytical one-to-one correspondence of the points Q on T and points P on the portion of S under consideration. 1 2 3 1 1 2 2 3 3 Let P1(x , x , x ) and P2(x + dx ,x + dx , x + dx ) be two close points on some curve C lying on S. The Euclidean distance P1P2 along C, is given by the formula ds 2 = dxidxi, (i = 1, 2, 3) ...(8) Since i ¶x p i = dx , (P =1, 2) dx ¶x p Thus equation (8) becomes ¶xi ¶ xi ds 2 = d x pd x q ¶x p ¶x q p q = g pq(x)dx dx , (p,q = 1, 2) ¶xi ¶xi i . where g pq (x) are functions of x and g pq = ¶ x p ¶x q If the image K of C on T is given by the equations 1 1 ïìx = x (t) K : í 2 2 îïx = x (t), t1 £ t £ t2 then the length of C can be computed from the integral

p q t2 d x d x s = g pq dt òt1 dt dt

A straight forward calculation gives

1 (d x1)2 + (d x 2)2 + (x1d x2 - x 2dx1)2 a2 2 ds2 = é 1 ù ...(9) 1+ {( x1)2 + ( x2)2} ëê a2 ûú and 120 Tensors and Their Applications

2 2 2 æ d x1ö æ d x 2ö 1 æ d x 2 dx1 ö ç ÷ + ç ÷ + ç x1 - x 2 ÷ ç ÷ ç ÷ 2 ç ÷ t2 è dt ø è dt ø a è dt dt ø s = dt òt 1 1 1 + {(x1)2 + (x 2)2} a2 So, the resulting formulas refer to a two-dimensional manifold determined by the variables (x1, x 2) in the cartesian plane T and that the geometry of the surface of the sphere imbeded in a three-dimensional

Euclidean manifold can be visualized on a two-dimensional manifold R2 with metric given by equation (9). 1 If the radius of S is very large then in equation (9) the terms involving can be neglected. Then a2 equation (9) becomes 1 2 2 2 ds 2 = (dx ) + (d x ) . ...(10) Thus for large values of a, metric properties of the sphere S are indistinguishable from those of the Euclidean plane. The chief point of this example is to indicate that the geometry of sphere imbedded in a Euclidean 3-space, with the element of arc in the form equation (8), is indistinguishable from the Riemannian geometry of a two-dimensional manifold R2 with metric (9). the latter manifold, although referred to a cartesian coordinate system Y, is not Euclidean since equation (9) cannot be reduced by an admissible transformation to equation (10).

7.2 CURVILINEAR COORDINATES IN E3

Let P(x ) be the point, in an Euclidean 3-space E3 , referred to a set of orthogonal Cartesian coordinates Y.. Consider a coordinate transformation T : xi = xi (x1, x2 ,x 3), (i = 1, 2,3) ¶xi Such that J = j ¹ 0 in some region R of E . The inverse coordinate transformation ¶x 3

1 2 3 T -1 : x i = x i(x ,x , x ), (i =1,2,3) will be single values and the transformations T and T -1 establish one-to-one correspondence between the sets of values (x1, x2 ,x3) and (x1, x 2, x3). The triplets of numbers (x1, x2 ,x3) is called curvilinear coordinates of the points P in R. If one of the coordinates x1,x 2,x3 is held fixed and the other two allowed to vary then the point P traces out a surface, called coordinate surface. If we set x1 = constant in T then

x1(x1, x 2, x3) = constant ...(1) defines a surface. If constant is allowed to assume different values, we get a one-parameter family of surfaces. Similarly, x2 (x1,x 2, x 3) = constant and x3(x1, x 2, x3) = constant define two families of surfaces. The surfaces Geometry 121

1 2 3 x = c1, x = c2, x = c3 … (2)

3 3 Y X 2 X

1 P X

2 Y

1 Y

Fig. 7.2 intersect in one and only one point. The surfaces defined by equation (2) the coordinate surfaces and intersection of coordinate surface pair-by pair are the coordinate lines. Thus the line of intersection 1 2 3 3 of x = c1 and x = c2 is the x - coordinate line because along this the line the variable x is the only one that is changing.

EXAMPLE 2 Consider a coordinate system defined by the transformation

Y3

X1 X2

2 O Y

X3

Y 1 Fig. 7.3

x1 = x1 sin x 2 cos x3 ...(3)

x 2 = x1 sin x 2 sin x3 ...(4) x 3 = x1 cos x 2 ...(5) 122 Tensors and Their Applications

The surfaces x1 = constant are spheres, x2 = constant are circluar cones and x3 = constant are planes passing through the Y3-axis (Fig. 7.3). The squaring and adding equations (3), (4) and (5) we get, (x1)2 + (x 2 )2 + (x 3)2 = (x1 sin x 2 cos x3)2 + (x1 sin x2 sin x3)2 + (x1 cos x2)2 On solving

(x1)2 + (x 2 )2 + (x 3)2 = (x1)2

1 2 2 2 3 2 x1 = (x ) + (x ) + (x ) ...(6) Now, squaring and adding equations (3) and (4), we get (x1)2 + (x 2 )2 = (x1 sin x 2 sin x3 )2 + (x1 sin x2 sin x3)2

(x1)2 + (x 2 )2 = (x1)2 (sin x2)2

1 2 2 2 x1 sin x2 = (x ) + (x ) ...(7) Divide (7) and (5), we get (x1)2 +(x2)2 tan x 2 = x3 æ 1 2 2 2 ö -1ç (x ) + (x ) ÷ 2 tan or x = ç x3 ÷ ...(8) è ø Divide (3) and (4), we get

x 2 = tan x3 x1

2 -1æ x ö 3 tan ç ÷ Þ x = ç 1 ÷ ...(9) è x ø So, the inverse transformation is given by the equations (6), (8) and (9). If x1 > 0, 0 < x2 < p,0 £ x3 < 2p. This is the familiar spherical coordinate system.

7.3 RECIPROCAL BASE SYSTEMS Covariant and Contravariant Vectors r r r Let a cartesian coordinate system be determined by a set of orthogonal base vectors b1, b2,b3 then the position vector r of any point P(x1, x 2,x 3) can be expressed as r r i r = bix (i = 1, 2, 3) ...(1) Geometry 123

r 1 2 3 Since the base vectors bi are independent of the position of the point P(x , x ,x ). Then from (1), r r i dr = bidx ...(2) If P(x1, x 2,x 3) and Q(x1 + dx1, x 2 + d x2, x 3 + d x3) be two closed point. The square of the element of arc ds between two points is ds 2 = dr × dr from equation (2),

2 r i r j ds = bidx × bjdx

r r i j = bi ×b jd x d x

2 i j r r ds = dijd x d x ; since bi ×b j = dij

3 3 Y X 2 X a 3 a 2

a 1 1 X P r b 3

O b2 2 Y b1

1 Y

Fig. 7.4. r r r r r r r (b1,b2,b3 are orthogonal base vector i.e., b1 ×b1 =1& b1 ×b2 = 0) .

as äij =1, i = j ds 2 = i i d x d x ; = 0, i ¹ j a familiar expression for the square of element of arc in orthogonal cartesian coordinates. Consider the coordinate transformation 1 2 3 xi = xi (x , x , x ), (i = 1, 2, 3)

define a curvilinear coordinate system X. The position vector r is a function of coordinates xi. i.e., r = r (xi ), (i = 1, 2, 3) Then ¶r dr = dxi ...(3) ¶xi 124 Tensors and Their Applications

and 2 ds = d r × dr r r ¶r ¶r i j = × d x dx ¶xi ¶ x j

2 i j ds = gijdx dx where ¶r ¶r g = × ...(4) ij ¶ xi ¶ x j ¶r The vector is a base vector directed tangentially to X i - coordinate curve. ¶xi Put ¶r = ar ...(5) ¶xi i Then from (3) and (4) r r i g = ar × ar dr = aid x and ij i j ...(6) Now, from equations (2) and (6), we get

r j r i a jdx = bidx

r ¶x i ar dx j = b dx j j i ¶ x j

i r r ¶ x j Þ a = b , as d x are arbitrary j i ¶x j r So, the base vectors a j transform according to the law for transformation of components of covariant vectors. r The components of base vectors ai, when referred to X-coordinate system, are r r r a1 : (a1,0,0), a2 : (0,a2,0), a3 :(0,0,a3). ...(7) and they are not necessarily unit vectors. In general, r r = r × r ¹ 1, g = ar × ar ¹ 1. g11 = a1 × a1 ¹ 1, g22 a2 a2 33 3 3 ...(8) If the curvilinear coordinate system X is orthogonal. Then r r r r gij = ai × a j = ai a j cosqij = 0, if i ¹ j. ...(9) r r Any vector A are can be written in the form A = kd r where k is a scalar.. Geometry 125

r r ¶r i Since d r = d x we have ¶ xi r r ¶r i A = (kdx ) ¶xi r r i A = ai A i i r where A = kdx . The numbers Ai are the contravariant components of the vector A Consider three non-coplanar vectors r r r r r r a ´ a 2 a3 ´ a1 3 a1 ´ a2 1 2 3 ar = , ar = ar = r r r , r r r r r r ...(10) [a1a2a3] [a1a2a3] [a1a2a3 ] r r r r r r r where a2 ´ a3, etc. denote the vector product of a2 and a3 and [a1a2a3 ] is the triple scalar r r r prduct a1 × a2 ´ a3. Now, r r r r r r a2 ´ a3 × a1 [a1a2a3] r1 r = = 1. a ×a1 = r r r r r r [a1a2a3] [a1a2a3] r r r r r r a2 ´ a3 × a2 [a2a2a3] r1 r = = 0. a ×a2 = r r r r r r [a1a2a3] [a1a2a3] r r r Since [a2a2a3]= 0. Similarly, r1 r r2 r a × a3 = a × a1 = ××× = 0 r2 r r3 r a × a2 = 1, a ×a3 =1 Then we can write ri r i a × a j = d j

EXAMPLE 3

r r r r1r2 r3 1 g To show that [a1a2a3]= g and [a a a ] = where g = ij . g Solution

The components of base vectors ai are r r r a1 : (a1,0,0), a2 : (0,a2,0) and a3 : (0,0,a3) Then

a1 0 0 r r r 0 0 = [a1a2a3 ] = a2 a1a2a3 ...(11) 0 0 a3 126 Tensors and Their Applications

and

g11 g12 g13

g= gij = g21 g 22 g23

g31 g32 g33 from equations (8) and (9), we have r r 2 g11 = a1 × a1 Þ a1 = g11 Similarly 2 , 2 a2 = g22 a3 = g33 and r r r r g12 = a1 × a2 = 0, g13 = a1 × a3 = 0 etc. So,

2 a1 0 0 2 2 2 2 g = 0 a2 0 = a1 a2 a3 2 0 0 a3

g = a1a2a3 ...(12) from eqn. (11) and (12), we have r r r [a1a2a3 ] = g 1 Since the triple products [ar1ar2ar3] = . Moreover,, g

ar2 ´ ar3 r3 r1 r1 r2 r r a ´ a r a ´ a a = , a2 = , a3 = 1 [ar1ar2ar3] [ar1ar2ar3] [ar1ar2ar3 ]

The system of vectors ar1,ar2,ar3 is called the reciprocal base system.

Hence if the vectors ar1,ar2,ar3 are unit vectors associated with an orthogonal cartesian coordinates then the reciprocal system of vector defines the same system of coordinates. Solved. r r ri The differential of a vector r in the reciprocal base system is d r = a d xi. r where dxi are the components of dr . Then ds 2 = dr × dr ri r j = (a dxi ) ×(a dx j ) ri r j = a × a dxidx j

2 ij ds = g dxidx j where g ij = ari × ar j = g ji ...(13) Geometry 127

The system of base vectors determined by equation (10) can be used to represent an arbitrary r ri r vector A in the form A = a Ai , where Ai are the covariant components of A. ri ar , Taking scalar product of vector Aia with the base vector j we get ri r i ri r i Aia ×a j = Aid j = Aj as a × a j = d j.

7.4 ON THE MEANING OF COVARIANT DERIVATIVES r r ¶A a r THEOREM 7.2 If A is a vector along the curve in E3 . Prove that = A, aa ¶x j j i i ¶A i r Also, prove that A, = . Where A are component of A. j ¶x j r r Proof: A vector A can be expressed in the terms of base vectors ai as r i r A = A ai r r ¶r i r where ai = and A are components of A. ¶xi r j The partial derivative of A with respect to x is r ¶A ¶Ai ¶ar = ar + Ai i ...(1) ¶x j ¶x j i ¶x j r r Since gij = ai × a j. Differentiating partially it w.r.t. xk , we have r r ¶gij ¶a ¶a j = i ×ar + × ar ¶xk ¶x k j ¶x k i Similarly, r ¶g jk ¶a ¶ar = j × ar + k × ar ¶xi ¶xi k ¶xi j r r ¶gik ¶ai r ¶ak r and = × ak + ×ai ¶x j ¶x j ¶x j r Since A can be written as r r i ri A = aiA = a Ai r Taking scalar product with a j, we have r r i ri r ai × a j A = a × a j Ai i j Þ gijA = di Ai = Aj

r r ri r i i j As ai × a j = gij,a × a j = d j and d j Ai = A . i We see that the vector obtained by lowering the index in A is precisely the covariant vector Ai. 128 Tensors and Their Applications

i r The two sets of quantities A and Ai are represent the same vector A referred to two different base systems.

EXAMPLE 4 ja j Show that gia g = di . Solution Since we know that g r r ja r j ra ia = ai × aa and g = a × a Then ja r r r j ra giag = (ai × aa )(a × a )

r r j r ra = (ai × a ) (aa × a )

j a r r j j = di da as ai × a = di

ja j a giag = di as da =1. But ¶r ar = i ¶xi r 2r 2r r ¶ai ¶ r ¶ r ¶a j = = = ¶x j ¶x jdxi ¶xi¶x j ¶xi So, r r ¶ai ¶a j = ¶x j ¶xi Now,

1 æ ¶g ¶g jk ¶gij ö ç ik + - ÷ [ij,k ] = ç j i k ÷ , Christoffel’s symbol 2 è ¶x ¶x ¶x ø ¶g ¶g ¶g Substituting the value of ik , jk and ij , we get ¶x j ¶xi ¶x k r 1 ¶ai r [ij,k] = × 2 ×ak 2 ¶x j r ¶ai r [ij,k] = × ak ¶x j r ¶a 1 i [ , ]rk , = rk or j = ij k a as r a ¶x ak Hence ¶ar i ×ara = [ij,k ]ark × ara ¶x j Geometry 129

a = [ij,k ]g k , Since g ka = ark ×ara

¶ar ì a ü ka ì a ü i . ara = í ý, as [i j,k ]g = í ý ¶x j îi jþ îi jþ

r ïì a ïü ¶ai r = í ý aa ...(2) ¶x j îïi jþï ¶ar Substituting the values of i in equation (1), we get ¶x j

r i ¶A ¶A ì a ü ar + Aiar = j i í ý a ¶x j ¶x îi jþ ¶Aa ì a ü ar + Aiar = j a í ý a ¶x îi jþ r a ¶A é¶A ì a ü ù + Ai ar j = ê j í ý ú a ¶x ë ¶x îi jþ û r ¶A ¶Aa ì a ü = A,a ar since Aa = + Ai j j a , j j í ý ¶x ¶x îi jþ

a a Thus, the covariant derivative A,j of the vector A is a vector whose components are the r ¶A components of referred to the base system ar . ¶x j i ì a ü ¶ar If the Christoffel symbols vanish identically i.e., í ý = 0 the i = 0, from (2). îi jþ ¶x j Substituting this value in equation (1), we get r ¶A ¶Ai = ar ¶x j ¶x j i ¶Ai ì i ü i + Ab But A, j = j í ý ¶x îb jþ i ì i ü i ¶A = 0. A, = as í ý j ¶x j îb jþ Proved.

r r j THEOREM 7.3 If A is a vector along the curve in E3 . Prove that Aj,k a wheree Aj are components r of A.. r Proof: If A can be expressed in the form r ri A = Aia 130 Tensors and Their Applications

r where Ai are components of A. r The partial derivative of A with respect to x k is r ¶A ¶A ¶ari = i ari + A ...(1) ¶x k ¶x k i ¶xk

ri r i Since a × a j = d j , we have,

Differentiating it partially w.r. to xk , we get

ri r ¶a ¶a j ×ar + ari × = 0 ¶x k j ¶xk

i r ¶ar ¶a j ×ar = - ari × ¶x k j ¶x k r a ¶a j ì a ü ri r ì ü = ar = - a ×aa í ý, Since k í ý a î j k þ ¶x î j kþ ri r i But a × aa = da . Then

¶ari ì a ü ×ar i k j = - da í ý ¶x î j k þ ¶ari ì i ü ×ar k j = - í ý ¶x î j k þ ri 1 ¶a ì i ü r j r j = - í ýa , as r = a ¶x k î j k þ a j

¶ari substituting the value of in equation (1), we get ¶x k r ¶ ¶A ì i ü A i ari - A ar j = k ií ý ¶x k ¶x î j k þ ¶A ì i ü i ar j - A ar j = k i í ý ¶x î j kþ r ¶A é ¶A ì i üù i - A ar j k = ê k i í ýú ¶x ë¶x î j k þû r ¶A ¶Aj ì i ü r j A = - A = A a , Since j,k k i í ý ¶x k j,k ¶x î j k þ Proved. Geometry 131

7.5 INTRINSIC DIFFERENTIATION r Let a vector field A(x) and

i i C : x = x (t), t1 £ t £ t2 r be a curve in some region of E3. The vector A(x) depend on the parameter t and if A(x) is a differentiable vector then r r dA ¶A dx j = × dt ¶x j dt r dA dx j = Aa ar dt , j a dt Since we know

r é a ì a üù ¶A a r ¶A ï ï r A,j aa = ê + í ýúaa j = j (See Pg. 127, Theo. 7.2) ¶x ëê ¶x îïi jþïûú So,

r é¶Aa ì a ü ù dx j dA + Ai ar = ê j í ý ú a dt ë ¶x îi jþ û dt r é a ì a ü j ù dA dA i dx r = ê + í ý A úaa dt ë dt îi jþ dt û

a j dA ì a ü i dx The formula + í ýA is called the absolute or Intrinsic derivative of Aa with respect dt îi jþ dt dAa to parameter t and denoted by . dt

a a j dA dA ì a ü i dx dA dA So, = + í ýA is contravariant vector. If A is a scalar then, obviously, = . dt dt îi jþ dt dt dt Some Results

(i) If Ai be covariant vector a b dAi dAi ì ü dx = - í ý Aa dt dt îi bþ dt ij ij b b dA dA ì i ü aj dx ì j ü ia dx (ii) = + í ýA + í ýA dt dt îa bþ dt îa bþ dt i i b b dAj dAj ì i ü a dx ì a ü i dx (iii) = + í ýAj - í ýAa dt dt îa bþ dt îi bþ dt 132 Tensors and Their Applications

i i b b b dAjk dAjk ì i ü a dx ì a ü i dx ì a ü i dx (iv) = + í ýAjk - í ýAak - í ýAja dt dt îa bþ dt î j bþ dt îk bþ dt

EXAMPLE 5

dgij If g be components of metric tensor, show that = 0. ij dt Solution

The intrinsic derivative of gij is

b b dgij dgij ì a ü dx ì a ü dx = - í ýgaj - í ýgia dt dt îi bþ dt îb jþ dt

¶g dxb ì a ü dxb ì a ü dxb ij - - = b í ýgaj í ýgia ¶x dt îi bþ dt îb jþ dt é a ù b ¶gij ïì ïü ì a ü dx ê - í ýga – í ýg a ú = b j b j i dt ëê ¶x îïi bþï î þ ûú

b dg ì ¶gij ü dx ij - b, - b , = í b [i j] [ j i]ý dt î ¶x þ dt

ì a ü ïì a ïü as í ýgaj = [ib, j ] and í ý gia = [bj , i]. îi bþ îïb j þï

¶gij But = [ib, j]+ [ jb,i]. ¶xb

dgij So, = 0. dt

EXAMPLE 6 Prove that d(g Ai A j ) dAi ij = 2g Ai dt ij dt Solution i j Since gij A a is scalar.. Then i j i j d(gijA A ) d(gijA A ) = dt dt Geometry 133

d(Ai Aj ) = g , since g is independent of t. ij dt ij i j édA j i dA ù = gij ê A + A ú ë dt dt û Interchange i and j in first term, we get

i j j j d(gijA A ) é i dA i dA ù = gij êA + A ú as gij is symmetric. dt ë dt dt û

i j j d(gijA A ) dA = 2g Ai dt ij dt Proved.

EXAMPLE 7 Prove that if A is the magnitude of Ai then i Ai, j A A, j = A Solution

Given that A is magnitude of Ai . Then Since

i k i i gik A A = A A

i k 2 gik A A = A Taking covariant derivative w.r. to x j , we get i k i k gik A, j A + gik A A, j = 2 AA, j Interchange the dummy index in first term, we get

k i i k gki A, j A + gik A A, j = 2 AA, j

i k 2gik A A, j = 2 AA, j

i k gik A A, j = AA, j

i k A (gik A, j ) = AA, j

Ai A k i, j = AA, j since gik A, j = Ai, j

i Ai, j A A, j = A Proved. 134 Tensors and Their Applications

7.6 PARALLEL VECTOR FIELDS Consider a curve i i C : x = x (t), t1 £ t £ t2, (i = 1, 2, 3) r in some region of E3 and a vector A localized at point P of C. If we construct at every point of C a vector equal to A in magnitude and parallel to it in direction, we obtain a parallel field of vector along the curve C. 3 3 X Y 2 X C 1 X P

O 2 Y

1 Y

r r if A is a parallel field along C then the vector A do not change along the curve and we can write r dA r = 0. It follows that the components Ai of A satisfy a set of simultaneous differential equations dt ¶Ai = 0 or ¶t i b dA ì i ü a dx + í ýA = 0 dt îa bþ dt This is required condition for the vector field Ai is parallel.

7.7 GEOMETRY OF SPACE CURVES

Let the parametric equations of the curve C in E3 be i i C : x = x (t), t1 £ t £ t2 (i = 1, 2, 3). The square of the length of an element of C is given by

2 i j ds = gijdx dx ...(1) and the length of arc s of C is defined by the integral

i j t2 dx dx s = gij dt ...(2) òt1 dt dt from (1), we have dxi dx j g = 1 ...(3) ij ds ds Geometry 135

dxi Put = li. Then equation (3) becomes ds i j gijl l = 1 ...(4) r i r The vector l , with components l , is a unit vector. Moreover,, l is tangent to C, since its dxi components li , when the curve C is referred to a rectangular Cartesian coordinate Y,, becomes li = . ds These are precisely the direction cosines of the tangent vector to the curve C. r i Consider a pair of unit vectors l and mr (with components li and m respectively) at any point r P of C. Let l is tangent to C at P Fig. (7.6).

l l + dl

Px() (Q(x + dx) m C r + dr r

O Fig. 7.6 r r The cosine of the angle q between l and m is given by the formula i j cosq = gijl l ...(5) r r and if l and m are orthogonal, then equation (5) becomes

i j gijl m = 0 ...(6) Any vector mr satisfying equation (6) is said to be normal to C at P.. Now, differentiating intrinsically, with respect to the are parameter s, equation (4), we get dli dlj g lj + g li = 0 …(7) ij ds ij ds

as gij is constant with respect to s. Interchange indices i and j in second term of equation (7) we get dli dli g lj + g lj = 0 ij ds ij ds

Since gij is symmetric. Then

dlj 2g li = 0 ij ds 136 Tensors and Their Applications

j i dl Þ g l = 0 ij ds dlj we see that the vector either vanishes or is normal to C and if does not vanish ds dlj we denote the unit vector co-directional with by m j and write ds 1 dlj dlj m j = , K = ...(8) K ds ds

where K > 0 is so chosen as to make m j a unit vector.. The vector m j is called the Principal normal vector to the curve C at the point P and K is the curvature of C. r The plane determined by the tangent vector l and the principal normal vector mr is called the osculating plane to the curve C at P. Since mr is unit vector

i j gijm m = 1 ...(9) Also, differentiating intrinsically with respect to s to equation (6), we get dlj dlj g m j + g li = 0 ij ds ij ds or

dm j dli g li = - g m j ij ds ij ds dli = - Kg mim j Since = Kmi ij ds dm j g li = -K , since g mim j = 1. ij ds ij dm j g li + K = 0 ij ds dm j g li + g liljK = 0 as g lilj = 1 ij ds ij ij æ dm j ö iç j ÷ gijl ç + Kl ÷ = 0 è ds ø

j dm j This shows that the vector + Kl is orthogonal to i ds l . Geometry 137

r j Now, we define a unit vector v, with components v , by the formula

1 æ dm j ö i ç j ÷ v = ç + Kl ÷ ...(10) t è ds ø i dm i r r where t = + Kl the vector vr will be orthogonal to both l and m . ds To choose the sign of t in such a way that i j k g eijkl m v = 1 ...(11) r r so that the triad of unit vectors l , m and nr forms a right handed system of axes.

2 ¶ xi Since e is a relative tensor of weight and g = it follows that e = ge is an ijk -1 ¶ x j ijk ijk obsolute tensor and hence left hand side of equation (11) is an invariant vk in equation (11) is determined by the formula

k ijk v = e lim j ...(12)

a ma ijk 1 ijk where l and m j are the associated vectors g al and gia and e = e is an absolute tensor.. i i g The number t appearing in equation (10) is called the torsion of C at P and the vector vr is the binormal. We have already proved that in Theorem 7.2, Pg. 127. r ¶A = Aaar ¶xi ,i a r if the vector field A is defined along C, we can write r i i ¶A ¶x a ¶x r = A, aa ...(13) ¶xi ¶s i ¶s Using definition of intrinsic derivative, dAa dxi = A,a ds i ds Then equation (13) becomes r a r i r dA dA r ¶A dx dA = aa ×××; as = ...(14) ds ds ¶xi ds ds r r Let r be the position vector of the point P on C then the tangent vector l is determined by r dr r = liar = l ds i from equation (14), we get

2r r a d r dl dl r r Þ = = aa = c ...(15) ds2 ds ds 138 Tensors and Their Applications

r where cr is a vector perpendicular to l. With each point P of C we can associate a constant K , such that cr K = mr is a unit vector.. Since cr = mr K 1 dla mr = ar , from (15) K ds a from equation (8), we get a r a r a 1 dl m = m aa , since m = K ds

7.8 SERRET-FRENET FORMULA The serret-frenet formulas are given by i i i 1 dl dl i dl (i) mi = or = Km , K > 0 where K = K ds ds ds 1 æ dmi ö i dmi ç + Kli ÷ dm + Kli (ii) ni = ç ÷ or = tni – Kli where t = t è ds ø ds ds dnk (iii) = – tmk ds First two formulas have already been derived in article (7.7), equation (8) and (10). Proof of (iii) From equation (12), article (7.7), we have

ijk k e lim j = n

where li ,mi ,nk are mutually orthogonal. Taking intrinsic derivative with respect to s, we get dm k ijk dli ijk j dn e m j + e li = ds ds ds From formulas (i) and (ii), we get

dnk eijk Km m + eijkl (tn – Kl ) = i j i j j ds dnk eijkl (tn – Kl ) = , Since eijkm m = 0 i j j ds i j dnk eijkl n t – Keijkl l = i j j i ds Geometry 139

ijk Since e lil j = 0 dnk eijkl n t = i j ds ijk k ijk Since e lin j = m , but e are skew-symmetric. Then ijk k e lin j = – m So, dnk – mkt = ds dnk or = – tmk ds dni Þ = – tmi ds This is the proof of third Serret-Frenet Formula Expanded form of Serret-Frenet Formula.

dli ì i ü dxk d 2xi ì i ü dx j dxk + lj i + i (i) í ý = Km or 2 í ý = Km ds î j k þ ds ds î j k þ ds ds

dmi ì i ü dxk + m j (ii) í ý = tni – Kli ds î jkþ ds i k dv ì i ü j dx (iii) + í ýn = – tmi ds î jkþ ds

EXAMPLE 8 Consider a curve defined in cylindrical coordinates by equation ìx1 = a ï 2 íx = q(s) ï 3 îx = 0 This curve is a circle of radius a. The square of the element of arc in cylindrical coordinates is 1 2 1 2 2 2 3 2 ds 2 = (dx ) + (x ) (dx ) + (dx ) 1 2 so that g11 = 1, g22 = (x ) , g33 = 1, gij = 0, i ¹ j It is easy to verify that the non-vanishing Christoffel symbols are (see Example 3, Page 61) ì 1 ü ì 2 ü ì 2 ü 1 1 í ý = x , í ý = í ý = 1 . î22þ î12þ î21þ x 140 Tensors and Their Applications

dxi dq The components of the tangent vector l to the circle C are li = so that l1 = 0, l2 = , ds ds l3 = 0. i j Since l is a unit vector, gijl l = 1 at all points of C and this requires that 2 2 1 2æ dq ö 2æ dq ö (x ) ç ÷ = a ç ÷ = 1 è ds ø è ds ø

2 æ dq ö 1 So, ç ÷ = and by Serret-Frenet first formula (expanded form), we get è ds ø a2 1 k 2 dl ì 1 ü j dx ì 1 ü 2 dx 1 Km1 = + í ýl = í ýl = – ds î j k þ ds î2 2þ ds a

2 k 1 dl ì 2 ü j dx ì 2 ü 2 dx Km2 = + í ýl = í ýl = 0 ds î j k þ ds î2 1þ ds

3 k dl ì 3 ü j dx Km3 = + í ýl = 0 ds î j k þ ds

1 2 3 Since µ is unit vector, g mim j = 1 and it follows that K = , m1= –1, m = 0, m = 0 ij a

Similarly we can shows that t = 0 and n1 = 0,n2 = 0,n3 =1.

7.9 EQUATIONS OF A STRAIGHT LINE i Let A be a vector field defined along a curve C in E3 such that i i C : x = x (s) . s1 £ s £ s2, (i = 1, 2, 3). s being the arc parameter. If the vector field Ai is parallel then from article 7.6 we have

dAi = 0 ds

i b dA ì i ü a dx or + í ýA = 0 (1) ds îa bþ ds We shall make use of equation (1) to obtain the equations of a straight line in general curvilinear r coordinates. The characteristic property of straight lines is the tangent vector l to a straight line is r directed along the straight line. So that the totality of the tangent vectors l forms a parallel vector field. Geometry 141

dxi Thus the field of tangent vector li = must satisfy equation (1), we have d s dli d 2xi ì i ü dxa dxb = + í ý = 0 ds ds2 îa bþ ds ds

d 2xi ì i ü d xa dxb The equation + í ý = 0 is the differential equation of the straight line. ds2 îa bþ ds ds

EXERCISE

i j i i d(gijA B ) dA dB 1. Show that = g B j + g Ai dt ij dt ij dt

¶A ¶Aj 2. Show that A – A = i – i, j j,i ¶x j ¶xi

j a 3. If Ai = gijA show that Ai,k = gia A,k

i j d(gij A B ) i i 4. Show that = Ai,kB + Bi,k A dxk 5. Show that

d2li dK = mi – K(tni – Kli ) ds2 ds

2 d mi dt dK = ni – (K2 + t2)mi – li ds2 ds ds d2ni dt = t(Kli – tvi) – mi ds2 ds 6. Find the curvature and tension at any point of the circular helix C whose equations in cylindrical coordinates are 1 2 3 C : x = a, x = q, x = q

Show that the tangent vector l at every point of C makes a constant angle with the direction of X 3 - axis. Consider C also in the form y1 = a cosq, y2 = a sinq, y3 = q. Where the coordinates yi are rectangular Cartesion. CHAPTER – 8

ANALYTICAL MECHANICS

8.1 INTRODUCTION Analytical mechanics is concerned with a mathematical description of motion of material bodies subjected to the action of forces. A material body is assumed to consist of a large number of minute bits of matter connected in some way with one another. The attention is first focused on a single particle, which is assumed to be free of constraints and its behaviour is analyzed when it is subjected to the action of external forces. The resulting body of knowledge constitutes the mechanics of a particle. To pass from mechanics of a single particle to mechanics of aggregates of particles composing a material body, one introduces the principle of superposition of effects and makes specific assumptions concerning the nature of constraining forces, depending on whether the body under consideration is rigid, elastic, plastic, fluid and so on.

8.2 NEWTONIAN LAWS 1. Every body continues in its state of rest or of uniform motion in a straight line, except in so far as it is compelled by impressed forces to change that state. 2. The change of motion is proportional to the impressed motive force and takes place in the direction of the straight line in which that force is impressed. 3. To every action there is always an equal and contrary reaction; or the mutual actions of two bodies are always and oppositely directed along the same straight line. The first law depends for its meaning upon the dynamical concept of force and on the kinematical idea of uniform rectilinear motion. The second law of motion intorduces the kinematical concept of motion and the dynamical idea of force. To understand its meaning it should be noted that Newton uses the term motion in the sense of momentum, i.e., the product of mass by velocity, this, "change of motion" means the time of change of momentum. In vector notation, the second law can be stated as d(mvr) r = … (1) F dt Analytical Mechanics 143

If we postulate the invariance of mass then equation (1) can be written as r F = mar … (2) d(mvr) from (1) if r = 0 then = 0. F dt So that mvr = constant. hence vr is constant vector.. Thus the first law is a consequence of the second. The third law of motion states that accelerations always occur in pairs. In term of force we may say that if a force acts on a given body, the body itself exerts an equal and oppositely directed force on some other body. Newton called the two aspects of the force of action and reaction.

8.3 EQUATIONS OF MOTION OF A PARTICLE THEOREM 8.1 The work done in displacing a particle along its trajectory is equal to the change in the kinetic energy of particle. Proof: Let the equation of path C of the particle in E3 be C : xi = xi (t) … (1) and the curve C the trajectory of the particle. Let at time t, particle is at P {xi (t)}. If vi be the component of velocity of moving particle then dxi vi = … (2) dt and if ai be the component of acceleration of moving particle then

i i k dv dv ì i ü j d x ai = = + í ýv dt dt î j kþ dt

d 2xi ì i ü dx j dxk i + a = 2 í ý … (3) dt î j k þ dt dt dvi ì i ü where is the intrinsic derivative and the í ý are the Christoffel symbols calculated from the dt î j k þ metric tensor gij If m be the mass of particle. Then by Newton’s second law of motion dvi Fi = m = mai … (4) dt r We define the element of work done by the force F in producing a displacement dr by invariant r dW = F.dr . r i i Since the components of F and dr are F and dx respectively.. 144 Tensors and Their Applications

Then i j dW = gij F dx … (5) j i = Fj dx where Fj = gij F The work done in displacing a particle along the trajectory C, joining a pair of points P1 and P2, is line integral P 2 i W = Fi dx … (6) òP1 using equation (4) then equation (6) becomes

P i 2 dv j W = m gij dx òP1 dt

i j P2 dv dx = m gij dt òP1 dt dt

P i 2 dv j W = m gij v dt … (7) òP1 dt i j Since gij v v is an invariant then i j d (gijv v ) d i j = (gijv v ) dt dt i d i j dv or (g v v ) = 2 g v j dt ij ij dt i dv j 1 d i j Þ g v = (gijv v ) ij dt 2 dt using this result in equation (7), we get

P 2 m d i j W= (gijv v ) dt òP 1 2 dt

m i j P2 W= [gijv v ] 2 P1

Let T2 and T1 is kinetic energy at P2 and P1 respectively. W = T2 – T1 m m v2 where T = g viv j = is kinetic energy of particle. 2 ij 2

We have the result that the work done by force Fi in displacing the particle from the point P1 to 1 the point P is equal to the difference of the values of the quantity T = mv2 at the end and the 2 2 beginning of the displacement.

8.4 CONSERVATIVE FORCE FIELD

P2 F dxi The force field Fi is such that the integral W = i is independent of the path. òP1 i Therefore the integrand Fi dx is an exact differential i dW = Fi dx … (8) Analytical Mechanics 145

of the work function W. The negative of the work function W is called the force potential or potential energy. We conclude from equation (8) that ¶V F = – … (9) i ¶xi where potential energy V is a function of coordinates xi. Hence, the fields of force are called conservative ¶V if F = – . i ¶xi

THEOREM 8.2 A necessary and sufficient condition that a force field Fi, defined in a simply connected region, be conservative is that Fi,j = Fj,i. ¶V Proof: Suppose that F conservative. Then F = – i i ¶xi Now, ¶F ì k ü i - F F = j í ý k i,j ¶x îi jþ æ ¶V ö ¶ç - ÷ ¶ i ì k ü è x ø - F Fi,j = j í ý k ¶x îi jþ

¶2V ì k ü = - - í ýFk … (1) ¶x j¶xi î j iþ and ¶F ì k ü j - F F = i í ý k j,i ¶x î j iþ Similarly, ¶2V ì k ü - F Fj,i = – i j í ý k … (2) ¶x ¶x î jiþ From equation (1) and (2), we get

Fi,j = Fj,i conversely, suppose that Fi,j = F j,i Then ¶F ì k ü ¶F ì k ü i - F j - F j í ý k = i í ý k ¶x îi jþ ¶x î j iþ Þ

¶F ¶Fj ì k ü i as = i í ý due to symmetry.. ¶x j ¶x îi jþ 146 Tensors and Their Applications

¶V Take F = - i ¶xi Then 2 ¶Fi ¶ V = - ¶x j ¶x j¶xi ¶ 2V = - ¶xi¶x j ¶ æ ¶V ö = ç - ÷ ¶xi è ¶x j ø

¶Fi ¶Fj = ¶x j ¶xi ¶V So, we can take F = - . i ¶xi

Hence, Fi is conservative.

8.5 LAGRANGEAN EQUATIONS OF MOTION Consider a particle moving on the curve C : xi = xi (t) At time t, let particle is at point P (xi). 1 The kinetic energy T = mv2 can be written as 2 1 i j T = m gijx x 2 & & i Since x& = vi.

1 j k or T = m g jk x x … (1) 2 & & i Differentiating it with respect to x& , we get ¶T 1 æ ¶x j ¶xk ö m g ç & xk + x j & ÷ i = jk ç i & & i ÷ ¶x& 2 è ¶x& ¶x& ø

1 j k k j = m g (d x + d x ) 2 jk i & i &

1 j k 1 k j = m g d x + m g d x 2 jk i & 2 jk i &

1 k j = m (g x + g x ) 2 ik & ji &

1 j j = m(g x + g x ) as g = g 2 ij & ij & ij ji Analytical Mechanics 147

¶T j i = m gijx& ¶x& ¶T or = k …(2) i m gik x& ¶x& Differentiating equation (2) with respect to t, we get

d æ ¶T ö d k ç i ÷ = m (gik x& ) dt è ¶x& ø dt

é d k k ù = m ê gik x& + gik &x& ú ë dt û

é j ù ¶gik dx k k = m ê j x& + gik&x& ú ë ¶x dt û

d ¶T ¶gik j k k i = m j x& x& + m gik &x& … (3) dt ¶x& ¶x

1 j k Since T = m g jk x x 2 & & Differentiating it with respect to xi, we get

¶T 1 ¶g jk j k = m x& x& … (4) ¶xi 2 ¶xi Now,

d æ ¶T ö ¶T ¶g 1 ¶g ç ÷ - ik j k k jk j k i i = m j x& x& + m gik&x& - m i x& x& dt è ¶x& ø ¶x ¶x 2 ¶x ¶g k 1 gik j k 1 j k 1 jk j k = m gik&x& + m x& x& + m gik x& x& - m x& x& 2 ¶x j 2 2 ¶xi ¶g ¶g k 1 gik j k 1 ij k j 1 jk j k = m gik&x& + m x& x& + m x& x& - m x& x& 2 ¶x j 2 ¶xk 2 ¶xi

1 é ¶g ¶gij ¶g jk ù m g xk + m ik + - x jx k = ik&& ê j k i ú & & 2 ë ¶x ¶x ¶x û

= k j k m gik&x& + m[ jk ,i]x& x& = k il j k m gik &x& + m g gil[ jk,i]x& x&

l il j k = m gil&x& + m g gil[ jk,i]x& x& 148 Tensors and Their Applications

l il j k = m gil[&x& + g [ jk ,i]x& x& ]

é ì l ü ù il ì l ü = m g xl + x j xk , as g [ jk,i] = í ý il ê&& í ý& & ú j k ë î j k þ û î þ

d æ ¶T ö ¶T l l ì l ü j k ç ÷ - = m g al , Since a = &x& + í ýx& x& i i il j k dt è ¶x& ø ¶x î þ where al is component of acceleration d æ ¶T ö ¶T ç ÷ - or i i = m ai dt è ¶x& ø ¶x

d æ ¶T ö ¶T = F … (5) ç i ÷ - i i dt è ¶x& ø ¶x where Fi = m ai, component of force field. The equation (5) is Lagrangean equation of Motion. ¶V For a conservative system, F = – . Then equation (5) becomes i ¶xi d æ ¶T ö ¶T ¶V - - ç i ÷ i = i dt è ¶x& ø ¶x ¶x d æ ¶T ö ¶(T -V ) or ç i ÷ - i = 0 … (6) dt è ¶x& ø ¶x Since the potential V is a function of the coordinates xi alone. If we introduce the Lagrangean function L = T – V Then equation (6) becomes d æ ¶L ö ¶L ç i ÷ - i = 0 … (7) dt è ¶x& ø ¶x

EXAMPLE 1 Show that the covariant components of the acceleration vector in a spherical coordinate system with ds2 = (dx1)2 + (x1dx2 )2 + (x1)2 sin 2 x2(dx3)2 are

1 1 2 2 1 3 2 2 a1 = &x& - x (x& ) - x (x& sin x )

d 1 2 2 1 2 2 2 3 2 a = [(x ) x ]- (x ) sin x cos x (x ) 2 dt & &

d 1 2 2 3 and a = [(x sin x ) x ] 3 dt & Analytical Mechanics 149

Solution In spherical coordinate system, the metric is given by ds2 = (dx1)2 + (x1dx2 )2 + (x1)2 sin 2 x2(dx3)2 If v is velocity of the paiticle then 2 2 2 2 æ ds ö æ dx1 ö æ dx2 ö æ dx3 ö 2 ç ÷ 1 2ç ÷ 1 2 2ç ÷ v = ç ÷ = ç ÷ + (x ) ç ÷ + (x sin x ) ç ÷ è dt ø è dt ø è dt ø è dt ø 2 1 2 1 2 2 2 1 2 2 3 2 v = (x& ) + (x ) (x& ) + (x sin x ) (x& ) If T be kinetic energy then

1 2 T = mv 2

1 1 2 1 2 2 2 1 2 2 3 2 T = m[ (x ) + (x ) (x ) + (x sin x ) (x ) ] … (1) 2 & & & By Lagrangean equation of motion d æ ¶T ö ¶T ç ÷ - i i = Fi and m ai = Fi dt è ¶x& ø ¶x where Fi and ai are covariant component of force field and acceleration vector respectively. So, d æ ¶T ö ¶T ç ÷ - i i = m ai … (2) dt è ¶x& ø ¶x Take i = 1, d æ ¶T ö ¶T ç ÷ - m a1 = 1 1 dt è ¶x& ø ¶x from (1), we get

1 d 1 m 1 2 2 1 2 2 3 2 m a = m (2x ) - [2x (x ) + 2x (sin x ) (x ) ] 1 2 dt & 2 & &

1 dx& 1 2 2 1 2 2 3 2 a = - [x (x& ) + x (sin x ) (x& ) ] 1 dt

1 1 2 2 1 3 2 2 a1 = &x& - x (x& ) - x (x& sin x ) Take i = 2, d æ ¶T ö ¶T ç ÷ - m a2 = 2 2 dt è ¶x& ø ¶x

1 d 1 2 2 1 1 2 2 2 3 2 m a = m [(x ) 2x ] - m 2(x ) sin x cos x (x ) 2 2 dt & 2 &

d 1 2 2 1 2 2 2 3 2 a = [(x ) x ]- (x ) sin x cos x (x ) 2 dt & & 150 Tensors and Their Applications

Take i = 3 d æ ¶T ö ¶T m a3 = ç 3 ÷ - 3 dt è ¶x& ø ¶x

1 d 3 1 2 2 m a = m [2(x )(x sin x ) ]- 0 3 2 dt &

d 3 1 2 2 2 a = [x (x ) (sin x ) ] 3 dt &

EXAMPLE 2 Use Lagrangean equations to show that, if a particle is not subjected to the action of forces then its trajectory is given by yi = ait + bi where ai and bi are constants and the yi are orthogonal cartesian coordinates. Solution If v is the velocity of particle. Then we know that,

2 i j v = gij y& y& where yi are orthogonal cartesian coordinates. Since

gij = 0, i ¹ j

gij = 1, i = j So, 2 i 2 v = (y& ) But,

1 2 T = mv , T is kinetic energy.. 2

1 i 2 T = m( y ) 2 & The Lagrangean equation of motion is

d æ ¶T ö ¶T ç ÷ - ç i ÷ i = Fi dtè ¶y& ø ¶y Since particle is not subjected to the action of forces.

So, Fi = 0

d æ 1 i ö Then ç m2y& ÷ – 0 = 0 dt è 2 ø

dyi m & = 0 dt Analytical Mechanics 151

dyi or & = 0 dt

i i Þ y& = a Þ yi = ait + bi where ai and bi are constant.

EXAMPLE 3 Prove that if a particle moves so that its velocity is constant in magnitude then its acceleration vector is either orthogonal to the velocity or it is zero. Solution If vi be the component of velocity of moving particle then dxi vi = or vi = i dt x& given |v| = constant. Since i j 2 gijv v = | v | = constant Taking intrinsic derivative with respect to t, we get

d i j (g v v ) = 0 dt ij

æ d i d j ö ç v j i v ÷ gijç v + v ÷ = 0 è dt dt ø

dvi dv j g v j + g vi = 0 ij dt ij dt

dvi dvi g v j + g v j = 0, (Interchange dummy index i and j in second term) ij dt ji dt

i dv j 2g v = 0 as g = g ij dt ij ji

dvi g v j = 0 ij dt dvi dvi This shows that acceleration vector is either orthogonal to vi or zero i.e., = 0. dt dt 152 Tensors and Their Applications

8.6 APPLICATIONS OF LAGRANGEAN EQUATIONS (i) Free-Moving Particle If a particle is not subjected to the action of forces, the right hand side of equation (5), 148, vanishes. Then we have d æ ¶T ö ¶T ç i ÷ - i = 0 … (1) dt è ¶x& ø ¶x

1 i i If xi be rectangular coordinate system, then T = m y y . 2 & & i i i i Hence, the equation (1) becomes m &y& = 0. Integrating it we get y = a t + b , which represents a straight line. (ii) Simple Pendulum Let a pendulum bob of mass m be supported by an extensible string. In spherical coordinates, the metric is given by ds2 = dr 2 + r 2d f2 + r 2 sin 2 f d q2 If T be the kinetic energy, then

1 2 1 2 2 2 2 2 2 T = mv = m(r + r f& + r sin f q& ) … (1) 2 2 &

O Y 2

r

f R mg cos f 1 Y q P mg sin f Y 3 mg

Fig. 8.1. from Lagrangean equation of motion d æ ¶T ö ¶T ç ÷ - i i = Fi i = 1, 2, 3 dt è ¶x& ø ¶x

x1 = r ,x 2 = f,x3 = q.

So, take x1 = r Analytical Mechanics 153

d æ ¶T ö ¶T ç ÷ - = mg cos f - R dt è ¶r& ø ¶r from (1), we have R - f 2 - sin 2 fq 2 = g cos f - … (2) &r& r & r & m Take x2 = f, we have

2 r&f& + 2r&f& - r sin f cos fq& = – g sin f … (3) and take x3 = q, we have

d 2 2 (r q& sin f) = 0 … (4) dt If the motion is in one plane, we obtain from equations (2), (3), and (4), by taking q& = 0. R - f2 = g cos f - &r& r & m

r&f& + 2r&f& = – g sin f æ g ö If r& = 0, we get, &f& = -ç ÷ sinf which is equation of simple pendulum supported by an è r ø inextensible string. For small angles of oscillation the vibration is simple harmonic. For large vibration the solution is given in the term of elliptic functions.

8.7 HAMILTON’S PRINCIPLE

If a particle is at the point P1 at the time t1 and at the point P2 at the time t2, then the motion of the particle takes place in such a away that

t 2 i (dT + Fidx ) dt = 0 òt1 where xi = xi (t) are the coordinates of the particle along the trajectory and xi + dxi are the coordinates along a varied path beginning at P1 at time t1 and ending at P2 at time t2. Proof: Consider a particle moving on the curve

i i C : x = x (t), t1 £ t £ t2 At time t, let particle is at P(xi). If T is kinetic energy. Then

1 i j T = m gijx x 2 & &

or = i i , is a function of i and i . Let ' be another curve, joining and close T T (x , x& ) i.e T x x& C t1 t2 to be C is C' : x i (e,t) = xi (t) + dxi (t) 154 Tensors and Their Applications

At t1 and t2 xi = x i = xi + e d xi

i i Þ dx (t1) = 0 and dx (t2) = 0

i i But T = T (x , x& ). If dT be small variation in T.. Then ¶T ¶T dxi + dxi dT = i & i ¶x& ¶x Now,

t t2 ¶T ¶T 2 i æ i i i ö ç dx + dx + Fidx ÷ dt {(dT + Fi )dx }dt = i i & ò òt1 t1 è ¶x ¶x& ø

t2 ¶T t2 ¶T t2 dxidt + dxidt + F dxi dt = ò i ò i & ò i t1 ¶x t1 ¶x& t1 ¶T Integrating second term by taking i as 1st term ¶x&

t2 t2 ¶T é ¶T ù t2 d æ ¶T ö t2 d i + d i - d i + d i = i x dt ê i x ú ç i ÷ x dt Fi x dt òt1 ¶ ¶ òt1 dt ¶ òt1 x ë x& û t1 è x& ø

i i Since dx (t1) = 0,dx (t2) = 0.

t2 æ ¶T i ö then ç dx ÷ = 0. ¶ i è x& ø t1 So,

t t t 2 i 2 ¶T i 2 d æ ¶T ö i (dT + F dx )dt = dx dt - ç ÷dx dt ò i òt i òt ç i ÷ t1 1 ¶x 1 dt è ¶x& ø

t 2 i + Fidx dt òt1

t t2 2 i é ¶T d æ ¶T ö ù i (dT + F dx )dt = ê - ç ÷ + Fi údx dt ò i òt i i t1 1 ë¶x dtè ¶x& ø û since particle satisfies the Lagrangean equation of motion. Then d æ ¶T ö ¶T ç i ÷ - i = Fi dt è ¶x& ø ¶x ¶T d æ ¶T ö - or i ç i ÷ = – Fi ¶x dt è ¶x& ø Analytical Mechanics 155

So,

t t 2 i 2 i (dT + Fidx ) dt = [-Fi + Fi]dx dt òt1 òt1

t 2 i (dT + Fidx ) dt = 0 Proved. òt1

8.8 INTEGRAL OF ENERGY THEOREM 8.3 The motion of a particle in a conservative field of force is such that the sum of its kinetic and potential energies is a constant. Proof: Consider a particle moving on the curve i i C : x = x (t), t1 £ t £ t2 At time t, let particle is at P (xi). If T is kinetic energy. Then

1 i j T = m gijx x 2 & &

1 i j or T = m gijv v 2 As T is invariant. Then Taking intrinsic derivative with respect to t, we get dT dT = dt dt

d æ 1 i j ö = ç m gijv v ÷ dt è 2 ø

1 æ dvi dv j ö ç j + i ÷ = m gijç v v ÷ 2 è dt dt ø

1 æ dvi dv j ö ç j + i ÷ = mç gij v gij v ÷ 2 è dt dt ø 1 æ d i d i ö ç v j v j ÷ = m ç gij v + g ji v ÷ , Since i and j are dummy indices. 2 è dt dt ø

1 dvi = m 2g v j as g = g 2 ij dt ij ji

dT dvi = m g v j dt ij dt dT dv j or = m g vi dt ij dt 156 Tensors and Their Applications

dv j = m g a jvi as = a j ij dt dT = m a vi, since g aj = a dt i ij i dT = F vi, dt i

Since Fi = m ai is a covariant component of force field.

But given Fi is conservative, then ¶V F = – , where V is potential energy.. i ¶xi So, dT ¶V = – vi dt ¶xi ¶V dxi = - ¶xi dt dT dV = - dt dt dT dV d + = 0 Þ (T + V ) = 0 dt dt dt Þ T + V = h, where h is constant. Proved.

8.9 PRINCIPLE OF LEAST ACTION Let us consider the integral

P2 A = mv.ds (1) òp1

evaluated over the path

i i C : x = x (t), t1 £ t £ t2 where C is the trajectory of the particle of mass m moving in a conservative field of force. In the three dimensional space with curvilinear coordinates, the integral (1) can be written as

P i 2 dx j A = m gij dx òp1 dt

i j t(P2 ) dx dx = m gij dt ò t(P1) dt dt Analytical Mechanics 157

1 dxi dx j Since T = m g , we have 2 ij dt dt

t(P2 ) A = 2T dt ò t(P1) This integral has a physical meaning only when evaluated over the trajectory C, but its value can be computed along any varied path joining the points P1 and P2. Let us consider a particular set of admissible paths C' along which the function T + V, for each value of parameter t, has the same constant value h. The integral A is called the action integral.

The principle of least action stated as “of all curves C' passing through P1 and P2 in the neighbourhood of the trajectory C, which are traversed at a rate such that, for each C', for every value of t, T + V = h, that one for which the action integral A is stationary is the trajectory of the particle.”

8.10 GENERALIZED COORDINATES In the solution of most of the mechanical problems it is more convenient to use some other set of coordinates instead of cartesian coordinates. For example, in the case of a particle moving on the surface of a sphere, the correct coordinates are spherical coordinates r, q,f where q and f are only two variable quantities. Let there be a particle or system of n particles moving under possible constraints. For example, a point mass of the simple pendulum or a rigid body moving along an inclined plane. Then there will be a minimum number of independent coordinates required to specify the motion of particle or system of particles. The set of independent coordinates sufficient in number to specify unambiguously the system configuration is called generalized coordinates and are denoted by q1, q2,...qn where n is the total number of generalized coordinates or degree of freedom. i Let there be N particles composing a system and let x(a) ,(i =1,2,3),(a =1,2,...N ) be the positional coordinates of these particles referred to some convenient reference frame in E3. The system of N free particles is described by 3N parameters. If the particles are constrained in some way, there will be i certain relations among the coordinates x(a) and suppose that there are r such independent relations,

i 1 2 3 1 2 3 1 2 3 f (x(1) , x(1) ,x(1); x(2) , x(2), x(2) ;...x( N) x(N ) x(N ) ) = 0, (i = 1, 2, ..., r) … (1) By using these r equations of constraints (1), we can solve for some r coordinates in terms of the remaining 3N – r coordinates and regard the latter as the independent generalized coordinates qi. It is more convenient to assume that each of the 3N coordinates is expressed in terms of 3N – r = n independent variables qi and write 3N equations.

i i 1 2 n x(a) = x(a) (q ,q ,...,q ,t) … (2) where we introduced the time parameter t which may enter in the problem explicitly if one deals with moving constraints. If t does not enter explicitly in equation (2), the dynamical system is called a natural system. The velocity of the particles are given by differentiating equations (2) with respect to time. Thus

i i ¶x(a) j ¶x(a) xi = q& + … (3) &(a) ¶q j ¶t 158 Tensors and Their Applications

i i The time derivatives q& of generalized coordinates q the generalized velocities. For symmetry reasons, it is desirable to introduced a number of superfluous coordinates qi and describe the system with the aid of k > n coordinates q1, q2,..., qk. In this event there will exist certain relations of the form f j (q1....,qk ,t) = 0 … (4) Differentiating it we get j j ¶f i ¶f q& + = 0 … (5) ¶qi ¶t It is clear that they are integrable, so that one can deduce from them equations (4) and use them to eliminate the superfluous coordinates . In some problems, functional relations of the type j 1 2 k 1 k F (q ,q ,...,q ;q& ,...,q& ,t) = 0, ( j = 1, 2, 3, ..., m) … (6) arise which are non-integrable. If non-integrable relations (6) occurs in the problems we shall say that the given system has k – m degrees of freedom, where m is the number or independent non-integrable relations (6) and k is the number of independent coordinates. The dynamical systems involving non- integrable relations (6) are called non-holonomic to distinguish them from holonomic systems in which the number of degrees of freedom is equal to the number of independent generalized coordinates. In other words, a holonomic system is one in which there are no non-integrable relations involving the generalized velocities.

8.11 LAGRANGEAN EQUATIONS IN GENERALIZED COORDINATES Let there be a system of particle which requires n independent generalized coordinates or degree of freedom to specify the states of its particle. The position vectors xr are expressed as the function of generalized coordinates qi ,(i =1,2,...,n) and the time t i.e., 1 2 n xr = xr (q ,q ,...,q ,t); (r =1,2,3,)

r The velocity x& of any point of the body is given by ¶xr dq j ¶xr r = + x& ¶q j dt ¶t

r r ¶x j ¶x = q& + , ( j = 1, 2, ..., n) ¶q j ¶t

j where q& are the generalized velocities. Consider the relation, with n degree of freedom, 1 2 n xr = xr (q ,q ,...,q ) … (1)

i r involve n independent parameters q . The velocities x& in this case are given by r ¶x j r = q& , (r = 1, 2, 3; j = 1, 2, ..., n) …(2) x& ¶q j Analytical Mechanics 159

j where q& transform under any admissible transformation, q k = q k (q1,...,qn ), (k = 1, 2, ..., n) … (3) in accordance with the contravariant law. The kinetic energy of the system is given by the expression of the form

1 r s T = S m grs x x , (r,s = 1,2,3,) … (4) 2 & &

r where m is the mass of the particle located at the point x . The grs are the components of the metric tensor. r Substituting the value of x& from equation (2), then equation (4) becomes r s 1 ¶x ¶x i j T = S m grs q& q& 2 ¶qi ¶q j

1 i j T = aijq q … (5) 2 & & ¶xr ¶r s where a = S m grs , (r, s = 1, 2, 3), (i, j = 1, ..., n) ij ¶qi ¶q j

1 i j Since T = aijq q is an invariant and the quantities a are symmetric, we conclude that the a 2 & & ij ij are components of a covariant tensor of rank two with respect to the transformations (3) of generalized coordinates. i Since the kinetic energy T is a positive definite form in the velocities q& ,| aij |> 0. Then we construct the reciprocal tensor aij . Now, from art. 8.5, Pg. 146, by using the expression for the kinetic energy in the form (5), we obtain the formula,

d æ ¶T ö ¶T æ ì l ü ö ç ÷ - a ç ql + q jqk ÷ ç i ÷ i = ilç && í ý & & ÷ (6) dt è ¶q& ø ¶q è î j k þ ø ì l ü where the Christoffel symbol í ý are constructed from the tensor a . î j k þ kl Put

l ì l ü j k q&& + í ýq& q& = Ql î j k þ so, the equation (6), becomes

d æ ¶T ö ¶T ç ÷ - = l ç i ÷ i ail Q dt è ¶q& ø ¶q

= Qi (i = 1, 2, ..., n) … (7) 160 Tensors and Their Applications

¶xr ¶xr ¶x r ¶2xr ¶xr d æ ¶xr ö & = , & = q j & = ç ÷ Now, from the realtions j j i i j & and i ç i ÷ and using equations ¶q& ¶q ¶q ¶x ¶q ¶q dt è ¶q ø (2) and (4). Then by straightforward calculation, left hand member of equation (7) becomes

d æ ¶T ö ¶T ¶xr ç ÷ - m a ç i ÷ i = å r i … (8) dt è ¶q& ø ¶q ¶q i in which aj = gij a is acceleration of the point P. Also, Newton's second law gives

m ar = Fr … (9) 's where Fr are the components of force F acting on the particle located at the point P.. From the equation (9), we have

¶xr ¶x r m a = Fr å r ¶qi å ¶qi and equation (8) can be written as

r d æ ¶T ö ¶T ¶x ç ÷ - = Fr … (10) ç i ÷ i å i dt è ¶q& ø ¶q ¶q comparing (7) with (8), we conclude that

¶xr Q = Fr i å ¶qi where vector Qi is called generalized force. The equations

d æ ¶T ö ¶T ç ÷ - = … (11) ç i ÷ i Qi dt è ¶q& ø ¶q are known as Lagrangean equations in generalized coordinates. They give a system of n second order ordinary differential equations for the generalized coordinates qi. The solutions of these equations in the form C : qi = qi (t) Represent the dynamical trajectory of the system. If there exists a functions V (q1,q 2,...,q") such that the system is said to be conservative and for such systems, equation (11) assume the form

d æ ¶L ö ¶L ç ÷ - = 0 … (12) ç i ÷ i dt è ¶q& ø ¶q where L = T – V is the kinetic potential. Analytical Mechanics 161

Since L (q, q& ) is a function of both the generalized coordinates and velocities. ¶L ¶L dL i + i = i &q& i q& … (13) dt ¶q& ¶q

¶L d æ ¶L ö ç ÷. from (12), we have i = ç i ÷ ¶q dt è ¶q& ø Then equation (13), becomes dL ¶L d æ ¶L ö qi + ç ÷ qi = i && ç i ÷ & dt ¶q& dt è ¶q& ø d æ ¶L ö ç qi ÷ = ç i & ÷ … (14) dt è ¶q& ø

i since L = T – V but the potential energy V is not a function of the q& ¶L ¶T qi qi = 2T i & = i & ¶q& ¶q&

1 i j since T= aijq q . 2 & & Thus, the equation (14) can be written in the form d(L - 2T ) d(T +V ) = = 0 dt dt which implies that T + V = h (constant). Thus, along the dynamical trajectory, the sum of the kinetic and potential energies is a constant.

8.12 DIVERGENCE THEOREM, GREEN'S THEOREM, LAPLACIAN OPERATOR AND STOKE'S THEOREM IN TENSOR NOTATION (i) Divergence Theorem r Let F be a vector point function in a closed region V bounded by the regular surface S. Then r r ˆ òdiv F = ò F ×n ds … (1) V S where nˆ is outward unit normal to S. Briefly the theorem states that the integral with subscript V is evaluated over the volume V while r the integral in the right hand side of (1) measures the flux of the vector quantity F over the surface S. r In orthogonal cartesian coordinates, the divergence of F is given by the formula 1 2 3 r ¶F ¶F ¶F div F = + + … (2) ¶x1 ¶x 2 ¶x 3 162 Tensors and Their Applications

r If the components of F relative to an arbitrary curvilinear coordinate system X are denoted by Fi then the covariant derivative of F i is

i ¶F ì i ü k F,i = + í ýF j ¶x j îk jþ i r The invariant F, j in cartesian coordinates represents the divergence of the vector field F . Also,

r i j i j F×nˆ = gijF n = F ni since gijn = ni Hence we can rewrite equation (1) in the form FidV Fin dS ò ,i = ò i … (3) V S (ii) Symmetrical form of Green's Theorem 1 2 3 1 2 3 Let f (x , x , x ) and y (x , x , x ) be two scalar function in V.. Let fi and yi be the of f and y respectively, so that ¶f ¶y Ñf = fi = and Ñy = yi = ¶xi ¶xi

Put Fi = fyi and from the divergence of we get j ij ij F,i = g Fi, j = g (fyi, j + yif j ) Substituting this in equation (3), we get gij (fy + y f )dV fy ni dS ò i, j i j = ò i … (4) V S

Since Ñy = yi , then ij 2 g yi, j = Ñ y … (5) ij Also, the inner product g yif j can be written as

ij g yif j = Ñf.Ñy where Ñ denote the gradient and Ñ2 denote the Laplacian operator.. Hence the formula (4) can be written in the form (gijfy + g ijy .f )dV f nˆ × ÑydS ò i, j i J = ò V S (fÑ2y + Ñf × Ñy)dV f nˆ × ÑydS ò = ò V S

f Ñ2y dV ò = òf nˆ ×Ñy - òÑf × ÑydV … (6) V S V

i ¶y where nˆ ×Ñy = y n = . i ¶n Analytical Mechanics 163

Interchanging f and y in equation (5), we get

yÑ2f dV ò = òynˆ ×Ñf - òÑy ×Ñf dV … (7) V S V Subtracting equation (5) from equation (6), we get

2 2 æ ¶y ¶f ö (fÑ y - yÑ f)dV ç f - y ÷ dS ò = ò ¶n ¶n … (8) V S è ø This result is called a symmetric form of Green's theorem. (iii) Expansion form of the Laplacian Operator The Laplacian of y is given by

2 ij Ñ y = g yi, j from (5) when written in the terms of the christoffel symbols associated with the curvilinear coordinates i x covering E3, æ ¶2y ì k ü ¶y ö 2 g ijç - ÷ Ñ y = ç i j í ý k ÷ … (9) è ¶x ¶x îi jþ ¶x ø and the divergence of the vector F i is ¶F i ì i ü i + F j F = i í ý …(10) ,i ¶x î j iþ ì i ü ¶ But we know that í ý = log g î j iþ ¶x j The equation (10) becomes

i ¶F æ ¶ ö j F i = + ç log g ÷ F ,i ¶xi è ¶x j ø

1 ¶( g F i) or F i = … (11) ,i g ¶xi ¶y i ij ij If putting F = g = g y j in equation (11), we get ¶x j æ ij ¶y ö ¶ç g g j ÷ ij 1 è ¶x ø g y j,i = … (12) g ¶xi But from equation (5), we know that 2 ij Ñ y = g y j,i 164 Tensors and Their Applications

Hence equation (12) becomes

æ ij ¶y ö ¶ç g g j ÷ 2 ij 1 è ¶x ø Ñ y = g y j,i = g ¶xi It is expansion form of Laplacian operator. (iv) Stoke's Theorem r Let a portion of regular surface S be bounded by a closed regular curve C and let F be any vector point function defined on S and on C. The theorem of Stokes states that r r ˆ ònˆ.curlF ds = ò F.l ds … (13) S C r where l is the unit tangent vector to C and curl F is the vector whose components in orthogonal cartesian coordinates are determined from

e1 e2 e3 ¶ ¶ ¶ r 1 2 3 r curl F = ¶ x ¶x ¶x = Ñ ´ F … (14) F1 F 2 F3 where ei being the unit base vectors in a cartesian frame. We consider the covariant derivative Fi,j of the vector Fi and form a contravariant vector i ijk G = – e Fj,k … (15) r we define the vector G to be the curl of F . dxi r i ijk Since ˆ . curl = n G = -e F , n and the components of the unit tangent vector l and . n F i j k i ds Then equation (13) may be written as i - eijk F n ds dx ò j,k i = Fi ds … (16) ò ds S C i r The integral ò Fidx is called the circulation of F along the contour C. c

8.13 GAUSS'S THEOREM The integral of the normal component of the gravitational flux computed over a regular surface S containing gravitating masses within it is equal to 4pm where m is the total mass enclosed by S. Proof: According to Newton's Law of , a particle P of mass m exerts on a particle Q of unit m mass located at a distancer r from P. Then a force of magnitude F = . r 2 Consider a closed regular surface S drawn around the point P and let q be the angle between the unit outward normal to nˆ to S and the axis of a cone with its vertex at P.. This cone subtends an element of surface dS. Analytical Mechanics 165

The flux of the produced by m is

r mcosq r 2dw F.nˆ dS = 2 ò ò r cos q S S r 2dw where dS = and dw is the solid angle subtended by dS. cos q Thus, we have, r F.nˆ dS m dw = 4pm ò = ò … (1) S S

n F q

dS

r

dw

P m

S

Fig. 8.2.

If there are n discrete particles of masses mi located within S, then n m cosq r i i F.nˆ = å 2 i=1 ri and total flux is

n r 4p m F.nˆ dS = å i … (2) ò i=1 S The result (2) can be easily generalized to continuous distributions of matter whenever such distribution no where melt the surface S. The contribution to the flux integration from the mass element r dV contained within V, is r cos qr dV F.nˆ dS 2 dS ò = ò r S S and the contribution from all masses contained easily within S is æ cosq rdV ö r ç ÷ dS F.nˆ dS = ò ç ò r 2 ÷ … (3) ò S è V ø S 166 Tensors and Their Applications where ò denotes the volume integral over all bodies interior to S. Since all masses are assumed to be V interior to S,r never vanishes. So that the integrand in equation (3) is continuous and one can interchange to order of integration to obtain æ cos qdS ö r r ç ÷ dV F.nˆ dS = ò òs r 2 … (4) ò V è ø S cos q dS But = 4p. Since it represents the flux due to a unit mass contained within S. ò r 2 S Hence

r 4p rdV = 4pm F.nˆ dS = ò … (5) ò V S where m denotes the total mass contained within S. Proved. Gauss's theorem may be extended to cases where the regular surface S cuts the masses, provided that the density S is piecewise continuous. Let S cut some masses. Let S' and S" be two nearby surfaces, the first of which lies wholly within S and the other envelopes S. Now apply Gauss's theorem to calculate the total flux over S" produced by the distribution of masses enclosed by S since S" does not intersect them. We have r ( .ˆ) ò F n i dS = 4pm S" r where the subscript i on F × nˆ refers to the flux due to the masses located inside S and m is the total mass within S. On the other hand, the net flux over S' due to the masses outside S, by Gauss's theorem is r ( × ˆ) ò F n o ds = 0 S' r where the subscript o on F × nˆ refers to the flux due to the masses located outside S. Now if we S' and S" approach S, we obtain the same formula (5) because the contribution to the r ( × ˆ) total flux from the integral ò F n o dS is zero. S' 8.14 POISSON'S EQUATION By divergence theorem, we have r r ò F.nˆ ds = òdiv F dV S V and by Gauss's Theorem, r 4p rdV F.nˆ ds = ò ò V from these, we have S Analytical Mechanics 167

r ò(div F - 4pr )dV = 0 v Since this relation is true for an arbitrary V and the integrand is piecewise continuous, then r div F = 4pr By the definition of potential function V, we have r F = -ÑV and div ÑV = Ñ2V So, r div F = 4pr div ( -ÑV ) = 4pr Ñ2V = – 4pr which is equation of poisson. If the point P is not occupied by the mass, then r = 0. Hence at all points of space free of matter the potential function V satisfies Laplace's equation Ñ2V = 0

8.15 SOLUTION OF POISSON’S EQUATION We find the solution of Poisson’s Equation by using Green’s symmetrical formula. We know that Green's symmetrical formula æ ¶y ¶f ö 2 2 ç f - y ÷ dS (fÑ y - yÑ f)dV = ò ¶n ¶n … (1) ò S è ø V where V is volume enclosed by S and f and y are scalar point functions. 1 Put f = where r is the distance between the points ( 1, 2, 3) and ( 1, 2, 3) and V is r P x x x Q y y y the gravitational potential. n

Qy()

r S Px()

n

Fig. 8.3. 168 Tensors and Their Applications

1 Since has a discontinuity at i = i , delete the point P(x) from region of integration by r x y surrounding it with a sphere of radius e and volume V'. Apply Green’s symmetrical formula to the 1 region V – V' within which and V possess the desired properties of continuity.. r

2 2 1 In region V -V',Ñ f = Ñ = 0. r Then equation (1) becomes

1 1 1 2 æ 1 ¶y ¶ ö æ 1 ¶y ¶ ö Ñ y dV ç - y r ÷ dS + ç - y r ÷ ds ò r = ò ç r ¶n ¶n ÷ ò ç r ¶n ¶n ÷ … (2) V -V' S è ø S' è ø where nˆ is the unit outward normal to the surface S + S' bounding V – V' . S' being the surface of the ¶ ¶ sphere of radius e and = - . ¶n ¶r Now

æ 1 ¶y ¶ 1 ö æ 1 ¶y ¶ 1 ö ç - y r ÷ dS ç - - y r ÷ dS ò ç r ¶n ¶n ÷ = ò ç r ¶r ¶r ÷ S' è ø S' è ø

æ 1 ¶y y ö ç - - ÷ r 2 dw = ò r ¶r r 2 S' è ø

æ ¶y ö - ç r + y÷ dw = ò ¶r S' è ø

æ 1 ¶y ¶(1 ) ö æ ¶y ö ç - y r ÷ - e ç ÷ dw - 4py ò ç r ¶n ¶n ÷ = ò è ¶r ø … (3) S' è ø S' r=e where y is the mean value of V over the sphere S' and w denote the solid angle.

Let y(x1,x 2, x3) = y(P) as r ® 0 then as e ® 0 from (3), we have

æ 1 ¶y ¶ 1 ö ç - y r ÷ ò ç r ¶n ¶n ÷ = -4py (P) S' è ø Then equation (2) becomes

1 1 2 æ 1 ¶y ¶ ö Ñ ydV ç - y r ÷ dS - 4py (P) ò r = ò ç r ¶n ¶n ÷ V S è ø

1 2 Since e ® 0 then Ñ y dV = 0. ò r V' Analytical Mechanics 169

1 1 1 2 1 1 ¶y 1 ¶ r y( ) = - Ñ ydV + dS - y dS … (4) P 4p ò r 4p ò r ¶n 4p ò ¶n V S S This gives the solution of Poisson's equation at the origin. If y is regular at infinity, i.e., for sufficiently large value of r, y is such that m ¶y m (y) £ and £ … (5) r ¶r r 2 where m is constant. If integration in equation (4) is extended over all space, so that r ® ¥ . Then, using equation (5), equation (4) becomes

1 Ñ2y y(P) = - dV … (6) 4p ò ¥ r

But y is a potential function satisfying the Poisson's equation i.e. Ñ2y = -4pr . Hence, from (6), we get

dV r y(P) = ò r ¥ This solution is Unique.

EXERCISES

1. Find, with aid of Lagrangian equations, the trajectory of a particle moving in a uniform gravitational field. 2. A particle is constrained to move under gravity along the line yi = cis (i = 1, 2, 3). Discuss the motion. 3. Deduce from Newtonian equations the equation of energy T + V = h, where h is constant. 4. Prove that

y nidS Ñ2y dV ò i = ò S V ¶y where yi = . ¶xi 5. Prove that the curl of a gradient vector vanishes identically. CHAPTER – 9

CURVATURE OF CURVE, GEODESIC

9.1 CURVATURE OF CURVE: PRINCIPAL NORMAL

i Let C be a curve in a given Vn and let the coordinates x of the current point on the curve expressed as functions of the arc length s. Then the unit tangent t to the curve the contravariant components dxi i = ...(1) t ds The intrinsic derivative (or desired vector) of ti along the curve C is called the first curvaturee r pr vector of curve C relative to Vn and is denoted by p . The magnitude of curvature vector is called first curvature of C relative Vn and is denoted by K: So,

i j K = gij p p r where Pi are contravariant components of p so that dx j i = ti, j P ds

æ ¶ti ì i üö dxi ç + ta ÷ = ç j í ý÷ è ¶x îa jþø ds

i j j ¶t dx a dx ì i ü = j + t í ý ¶x ds ds îa jþ

dti dxa dx j ì i ü = + í ý ds ds ds îa jþ

d 2xi dx j dxk ì i ü = 2 + í ý, Replacing dummy index a by k ds ds ds îk jþ Curvature of Curve, Geodesic 171

d 2xi dx j dxk ì i ü ì i ü ì i ü i p = 2 + í ý as í ý = í ý ds ds ds î j kþ î j kþ îk jþ r If nˆ is a unit vector in the direction of p, then we have pr = k nˆ The vector nˆ is called the Unit principal normal.

9.2 GEODESICS Geodesics on a surface in Euclidean three dimensional space may be defined as the curve along which lies the shortest distance measured along the surface between any two points in its plane. But when the problem of find the shortest distance between any two given points on a surface is treated properly, it becomes very complicated and therefore we define the geodesics in V3 as follows: (i) Geodesic in a surface is defined as the curve of stationary length on a surface between any two points in its plane.

(ii) In V3 geodesic is also defined as the curve whose curvature relative to the surface is everywhere zero.

By generalising these definitions we can define geodesic in Riemannian Vn as

(i) Geodesic in a Riemannian Vn is defined as the curve of minimum (or maximum) length joining two points on it.

(ii) Geodesic is the curve whose first curvature relative to Vn is zero at all points.

9.3 EULER'S CONDITION THEOREM 9.1 The Euler condition for the integral t 1 i i ò f (x , x& )dt to to be staionary are ¶f d æ ¶f ö - i ç i ÷ = 0 ¶x dt è ¶x& ø dxi where i = i = 1, 2, 3,... x& dt i Proof: Let C be a curve in a Vn and A, B two fixed points on it. The coordinates x of the current point

P on C are functions of a single parameter t. Let t0 and t1 be the values of the parameter for the points A and B respectively. To find the condition for the integral t 1 i i ò f (x , x& )dt ...(1) t0 to be stationary. Let the curve suffer an infinitesimal deformation to C¢, the points A and B remaining fixed while the current points P(xi) is displaced to P' (xi + hi) such that hi = 0 at A and B both. 172 Tensors and Their Applications

P' C’

A B

P C

Fig. 9.1

In this case the value of integral (1) becomes I ¢ So,

t 1 i i i i I ¢ = F(x + h , x& + h& )dt ò t0 By Taylor's theorem

æ ¶f ¶f ö F( x + h, y + k) = f (x, y) + ç h + k ÷ + × ××× è ¶x ¶y ø Then

t1é æ ¶F ¶F ö ù F(xi, xi) + hi + hi + ××× dt I ¢ = ê & ç i i & ÷ ú ò t0 ë è ¶ x ¶x ø û

t1 t1 æ ¶F ¶F ö F (xi, xi) dt + ç hi + hi ÷ dt I ¢ = ò & ò i i & t0 t0 è ¶x ¶x& ø

(Neglecting higher order terms in small quantities hi )

t1æ ¶F ¶f ö = I + ç hi + hi ÷ dt I ¢ ò i i & t0 è ¶x ¶ x& ø

t1æ ¶F ¶F ö I¢ - I = ç hi + hi ÷ dt dI = ò i i & ...(2) t0 è ¶x ¶ x& ø

i i ¶z j where h& = x& ¶x j Now,

t 1 t t1 ¶F é ¶F i ù 1 d æ ¶F ö i hidt h – ç ÷h dt i & = ê i ú i ò t ë¶x û òt0 dt è ¶x ø 0 ¶x& & t0 &

t1 d æ ¶F ö - ç ÷hidt = ò i ...(3) t0 dt è ¶x& ø Curvature of Curve, Geodesic 173

t é é ¶F ù 1 ù êsince hi (t) = 0, hi(t ) = hi (t ) = 0ú ê ëê¶xi ûú 1 0 ú ë & t0 û Then equation (2) becomes

ti é ¶F d æ ¶F öù - ç ÷ hidt dI = ò ê i i ú ...(4) t0 ë ¶ x dt è ¶ x& øû The integral I is stationary if dI = 0.

t1é ¶F d æ ¶F öù - ç ÷ hidt i.e., if ò ê i i ú = 0 t0 ë ¶x dt è ¶x& øû Since hi are arbitrary and hence the integrand of the last integral vanishes, so that

¶F d æ ¶F ö i - ç i ÷ = 0, (i = 1, 2,..., n) … (5) ¶x dt è ¶x& ø Hence the necessary and sufficient condition for the integral (1) to be stationary are

¶F d æ ¶F ö – ç ÷ = 0, (i = 1, 2, …, n) ¶xi dt è dxi ø These are called Euler's conditions for the integral I to be stationary.

9.4 DIFFERENTIAL EQUATIONS OF GEODESICS

To obtain the differential equations of a geodesic in a Vn, using the property that it is a path of minimum (or maximum) length joining two points A and B on it. i Proof: Consider a curve C in Vn joining two fixed points A and B on it and x (t) be the coordinates of point P on it. The length of curve C is

B dxi d x j s = gij dt ...(1) ò A dt dt ds dxi d x j = g dt ij dt dt Put

ds d xi d x j = g = F (say) ...(2) dt ij dt dt i j = or s& = gijx& x& F Then equation (1) becomes B s = F dt ...(3) ò A 174 Tensors and Their Applications

Since curve C is geodesic, then the integral (3) should be stationary, we have from Euler's condition

¶F d æ ¶F ö - i ç i ÷ = 0 ...(4) ¶x dt è ¶x& ø k k Differentiating equation (2) with respect to x and x& we get, ¶F 1 ¶g ij xi x j k = k & & ¶x 2s& ¶x

¶F æ 1 ö 1 2 g xi = g xi and k = ç ik & ÷ ik & ¶x& è 2s& ø s&

d æ ¶F ö 1 1 ¶g 1 - s g xi + ik x j xi + g xi ç k ÷ = 2 && ik & j & & ik&& dt è ¶x& ø s& s& ¶x s& Putting these values in equation (4), we get

1 ¶gij i j é 1 i 1 ¶gik j i 1 i ù x& x& - ê- &s&gik x& + x& x& + gik&x& ú = 0 2s& ¶xk ë s2 s& ¶x j s& û

s æ ¶g 1 ¶g ö g xi - && g xi + ç ik - ij ÷xix j ik && ik & ç j k ÷ & & = 0 s& è ¶x 2 ¶x ø

i &s& i i j gik&x& - gik x& + [k,ij]x& x& = 0 s& multiplying it by gkm , we get

km i &s& km i km i j g gik &x& - g gikx& + g [k,ij]x& x& = 0 s&

km ì m ü But g kmg = dm and g [k, ij]= í ý ik i îi jþ

ì mü m &s& m ï ï i j &x& - x& + í ý x& x& = 0 s& îïi jþï

d 2xm s dxm ì m ü dx j dxk æ Replacing dummy ö - && + ç ÷ 2 í ý = 0 ç ÷ ...(5) dt s& dt î j kþ dt dt èindex i by k ø This is the differential equation for the geodesic in parameter t. Taking s = t, s& =1, &s& = 0. Then equation (5) becomes

d 2xm ì m ü dx j dxk 2 + í ý = 0 ...(6) ds î j k þ ds ds Curvature of Curve, Geodesic 175

which may also written as

k æ m ö dx ç dx ÷ ds ç ds ÷ = 0 è ø,k Then the intrinsic derivative (or derived vector) of the unit tangent to a geodesic in the direction of the curve is everywhere zero. In otherwords, a geodesic of Vn is a line whose first curvature relative to

Vn is identically zero. THEOREM 9.2 To prove that one and only one geodesic passes through two specified points lying in a neighbourhood of a point O of a Vn. OR

To prove that one and only one geodesic passes through a specified point O of Vn in a prescribed direction.

Proof: The differential equations of a geodesic curve in a Vn are d 2xm ì m ü dx j dxk 2 + í ý = 0 ds î j k þ ds ds These equations are n differential equations of the second order. Their complete integral involves 2n arbitrary constants. These may be determined by the n coordinates of a point P on the curve and the n components of the unit vector in the direction of the curve at P. Thus, in general, one and only one geodesic passes through a given point in a given direction.

9.5 GEODESIC COORDINATES A cartesian coordinate system is one relative to which the coefficients of the fundamental form are constants. Coordinates of this nature do not exists for an arbitrary Riemannian Vn.. It is, however, possible to choose a coordinate system relative to which the quantities gij are locally constant in the neighbourhood of an arbitrary point P0 of Vn. Such a cartesian coordinate system is known as geodesic coordinate system with the pole at P0.

The quantities gij are said to be locally constants in the neighbourhood of a point P0 if

¶gij = 0 at P ¶xk 0

¶gij and ¹ 0 elsewhere ¶xk ì k ü This shows that [ij, k] = 0, í ý = 0 at P . îi jþ 0 k Since the covariant derivative of Aij with respect to x is written as

¶A ïì h ïü ïì h ïü ij - A - A Aij,k = k í ý ih í ý hj , see pg. 71 ¶x îï j k þï îïi kþï 176 Tensors and Their Applications

k The covariant derivative of Aij at P0 with respect to x reduces to the corresponding ordinary derivatives. Hence

¶Aij Aij,k = at P ¶xk 0 THEOREM 9.3 The necessary and sufficient condition that a system of coordintes be geodesic with pole at P0 are that their second covariant derivatives with respect to the metric of the space all vanish at P0. Proof: We know that (equation 8, Pg. 65)

2 s ¶xs ì k ü ¶x p ¶xq ì s ü ¶ x - = k í ý i j í ý ¶x j¶x j ¶x îi jþ ¶x ¶x î p qþ

s p q ¶ æ ¶xs ö ¶x ì k ü ¶x ¶x ì s ü ç ÷ í ý - í ý or j ç i ÷ = k i j ...(1) ¶x è ¶x ø ¶x îi jþ ¶x ¶x î p qþ

Interchanging the coordinate system xi and xi in equation (1), we get

¶ æ ¶x s ö ¶x s ì k ü ¶x p ¶x q ì s ü ç ÷ - j ç i ÷ = k í ý i j í ý ¶x è ¶x ø ¶x îi jþ ¶x ¶x î p qþ

p q ¶x ¶x ì s ü ¶ æ ¶x s ö ¶x s ì k ü - í ý ç ÷ - i j = j ç i ÷ k í ý ¶x ¶x î p qþ ¶x è ¶x ø ¶x îi jþ

s ¶ s s ì k ü ¶x s = j (x,i )- x,k í ý since k = x,k at P0 ¶x îi jþ ¶x k s ì ü = (x, ) since í ý = 0 at P0 i , j îi jþ Thus,

¶x p ¶x q ì s ü s - x,ij = i j í ý ...(2) ¶x ¶x î p qþ Necessary Condition s Let x be a geodesic coordinate system with the pole at P0 so that ì s ü í ý = 0 at P0 î p qþ Hence from (2), we have s x,ij = 0 at P0 Sufficient Condition s Conversely suppose that x,ij = 0 at P0 . Curvature of Curve, Geodesic 177

Then equation (2) becomes ì s ü ¶x p ¶x q í ý i j = 0 î p qþ ¶x ¶x

ì s ü ¶x p ¶x q Þ í ý = 0 at P , as ¹ 0 and ¹ 0 at P0 î p qþ 0 ¶xi ¶x j s So, x is a geodesic coordinate system with the pole at P0.

9.6 RIEMANNIAN COORDINATES A particular type of geodesic coordinates introduced by Riemann and known as Riemannian coordinates. i Let C be any geodesic through a given point P0 , s the length of the curve measured from P0 and x the quantities defined by æ dxi ö i ç ÷ x = ç ds ÷ ...(1) è øo i the subscript zero indicating as usual that the function is to be evaluated at P0 . The quantities x i i represents that only one geodesic will pass through P0 in the direction of x in Vn. Let y be the coordinates of a point P on the geodesic C such that yi = sxi ...(2) i where s is the arc length of the curve from P0 to P . The coordinates y are called Riemannian coordinates. i The differential equation of geodesic C in terms of coordinates y relative to Vn is given by d 2 yi ì i ü dyi dy j + 2 í ý = 0 ...(3) ds î j kþ ds ds ì i ü where í ý is a christoffel symbol relative to the coordinates yi. î j kþ The differential equation (3) will be satisfied by (2), we have,

ì i ü i ï ï i j dy 0 + í ý x x = 0 since = xi îï j k þï ds

ì i ü i j or í ýx x = 0 ...(4) î j k þ using equation (2), equation (4) becomes

i j ì i ü y y yi í ý = 0 as = xi î j k þ s s s

ì i ü i j or í ý y y = 0 ...(5) î j kþ

The equation (5) hold throughout the Riemannian Vn . 178 Tensors and Their Applications

Since yi ¹ 0, y j ¹ 0, from (5) we get ì i ü í ý = 0 at P0 î j kþ

Hence the Riemannian coordinates are geodesic coordinate with the pole at P0 . THEOREM 9.4 The necessary and sufficient condition that the coordinates yi be Riemannian coordinates

ì i ü i j is that í ý y y = 0 hold throughout the Riemannian V . î j kþ n ì i ü ï ï i j Proof: If yi are Riemannian coordinates then the condition í ýy y = 0 (from equation 5) throughout îï j k þï the Riemannian Vn .

2 i j ì i ü d y ïì i ïü dy dy i j + i i Conversely if í ý y y = 0 hold then 2 í ý = 0 are saitsfied by y = sx . î j k þ ds îï j kþï ds ds Hence yi are Riemannian coordinates.

9.7 GEODESIC FORM OF A LINE ELEMENT Let f be a scalar invariant whose gradiant is not zero. Let the hypersurface f = 0 be taken as coordinates hypersurface x1 = 0 and the geodesics which cut this hypersurface orthogonally as the coordinate lines of parameter x1, this parameter measuring the length of arc along a geodesic from the hypersurface x1 = 0 . i Since dx1 is the length of the vector m is given by 2 i j u = gij u u

1 2 1 1 i.e., (dx ) = g11dx dx Þ g11 = 1 ...(1) Now, if vi is the tangent vector to the hypersurface x1 = 0 then we have vi = (0, dx2, dx3,..., dxn ) since the vectors ui and vi are orthogonal vectors. Then, i j giju v = 0

1 j i Þ g1ju v = 0, [u = 0, i = 2, 3, ..., n]

j 1 Þ g1jv = 0, as u ¹ 0.

Þ g1j = 0, for j = 2, 3, …, n. ...(2) Again the coordinate curves of parameter x1 are geodesics. Then s = x1. If ti is unit tangent vector to a geodesic at any point then t1 = 1 and ti = 0, for i = 2, 3, …, n. Curvature of Curve, Geodesic 179

Now,

d xi d xi ti = = ds dx1 1 dxi dx = 0 Þ = 1 and 1 for i ¹ 1 dx1 dx d 2xi d 2xi and = = 0 for i = 1, 2,..., n ds2 dx12 Also, the differential equation of geodesic is

d 2xi ì i ü dx j dx k + 2 í ý = 0 ds î j k þ ds ds using above results, we have

ì i ü dx1 dx1 í ý = 0 î11þ ds ds ì i ü Þ í ý = 0 î11þ Þ gij[11, j] = 0 ij Þ [11, j] = 0 as g ¹ 0

1 æ 2¶g1 j ¶g ö ¶g ç - 11 ÷ = Þ 11 = 0 ç i j ÷ = 0, since g11 i 2 è ¶x ¶x ø ¶x j So,

¶g1j = 0 for j ¹1 ...(3) ¶x1

from equations (1), (2), and (3), we have

¶g1j g11 = 1, g = 0; ( j = 2, 3, …, n), = 0, ( j = 2, 3, …, n) 1j ¶x1 The line element is given by

2 i j ds = gij dx dx

2 1 1 j k ds = g11dx dx + g jk dx dx 2 1 2 j k ds = (dx ) + g jkdx dx ; ( j = 2, 3, …, n, k = 2, 3, …, n) ...(4) The line element (4) is called geodesic form of the line element. Note 1: We note that the coordinate curves with parameter x1 are orthogonal to the coordinate curve xi = ci (i = 1, 2, ..., n) at all points and hence to the hypersurfaces x1 = c at each point. Note 2: The existence of geodesic form of the line element proves that the hypersurfaces f = x1 = constant form a system of parallels i.e., the hypersurfaces f = x1 = constant are geodesically parallel hypersurfaces. 180 Tensors and Their Applications

THEOREM 9.5 The necessary and sufficient condition that the hypersurfaces f = constant form a system of parallel is that (Ñf)2 = 1. Proof: Necessary Condition Suppose that hypersurface f = constant form a system of parallels then prove that (Ñf2) = 1. Let us take the hypersurface f = 0 as the coordinate hypersurface x1 = 0. Let the geodesics cutting this hypersurface orthogonally, be taken as coordinate lines of parameter x1. Then the parameters x1 measures are length along these geodesics from the hypersurface x1 = 0. This implies the existence of geodesic form of the line element namely

2 1 2 i j ds = (dx ) + gijdx dx ...(1) where i, j = 2, 3,..., n. From (1), we have

g11 = 1, g1i = 0 for i ¹ 1. from these values, it follows that g11 = 1, g1i = 0, for i ¹1 Now,

ij ¶f ¶f (Ñf)2 = Ñf × Ñf = g ¶xi dx j

¶x1 ¶x1 = gij = gijd1d1 ¶xi ¶x j i j so (Ñf)2 = g11 =1 (Ñf)2 = 1 Sufficient Condition Suppose that (Ñf)2 = 1 then prove that the hypersurface f = constant from a system of parallels. Let us taken f = x1 and orthogonal trajectories of the hypersurfaces f = x1 constant as the coordinate lines of parameter x1. Then the hypersurfaces x1 = constant xi = constant (i ¹ 1) are orthogonal to each other. The condition for this g1i = 0 for i ¹ 1. Now, given that (Ñf)2 = 1

ij ¶f ¶f Þ g = 1 ¶xi ¶x j

¶x1 ¶x1 Þ gij = 1 ¶xi ¶x j

ij 1 1 Þ g di d j = 1 Þ g11 = 1 Curvature of Curve, Geodesic 181

Thus g11 = 1 and g1i = 0 for i ¹1. Consequently g 11 = 1, g1i = 0, for i ¹ 1. Therefore, the line element 2 i j ds = gijdx dx is given by

2 1 2 i k ds = (dx ) + gikdx dx ; (i, k = 2, 3, …, n) which is geodesic form of the line element. It means that the hypersurfaces f = x1 = constant form a system of parallels.

9.8 GEODESICS IN EUCLIDEAN SPACE i Consider an Euclidean space Sn for n-. Let y be the Euclidean coordinates. The differential equation of geodesics in Euclidean space is given by d 2 yi ì i ü dy j dy k + 2 í ý = 0 ...(1) ds î j kþ ds ds

In case of Euclidean coordinates the fundamental tensor gij is denoted by aij and

i ì1, if i = j gij = a = d = í ij j î0, if i ¹ j

¶gij ¶aij = = 0 ¶xk ¶xk

ì k ü This implies that í ý = 0, [ij,k ] = 0 relative to Sn. îi jþ Then equation (1) becomes d 2 yi = 0 ds2 Integrating it, we get dyi = ai, where ai is constant of integration. ds Again Integrating, we get yi = ais + bi, where bi is constant of integration ...(2) The equation (2) is of the form y = mx + c. Hence equation (2) represents a straight line. Since equation (2) is a solution of equation (1) and therefore the geodesic relative to Sn are given by equation (2). Hence geodesic curves in Euclidean space Sn are straight lines. 182 Tensors and Their Applications

i i THEOREM 9.6 Prove that the distance l between two points P (y ) and Q (y' ) in Sn is given by n i i 2 l = å (y¢ - y ) i=1

Proof: We know that geodesics in Sn are straight line. Then equation of straight line in Sn may be taken as yi = ais + bi ...(1) Let P (yi) and Q ( y' i) lie on equation (1). Then yi = ais + bi, y¢i = ais¢ + bi y¢i - yi = ai(s¢ - s) ...(2) Then equation (2) becomes i i y¢ - y = ail n n 2 i 2 i i 2 l å (a ) = å (y¢ - y ) i=1 i=1

But ai is the unit tangent vector to the geodesics. Then

n i 2 å (a ) = 1 i=1 So, n i i 2 2 (y¢ - y ) l = å i=1 n i i 2 l = å(y¢ - y ) i=1

EXAMPLE 1

Prove that Pythagoras theorem holds in Sn . Solution

Consider a triangle ABC right angled at A i.e., ÐBAC = 90 o .

i C (y)3

i i A (y)1 B (y)2 Fig. 9.2 Curvature of Curve, Geodesic 183

Then the lines AB and AC are orthogonal to each other. So, AB× AC = 0

i i i i or aij(y2 - y1) (y3 - y1) = 0

n (yi - yi ) (y i - y i) or å 2 1 3 1 = 0 ...(1) i=1 By distance formula, we have

n i i 2 2 (y - y ) (AB) = å 2 1 ...(2) i=1

n i i 2 2 (y - y ) (AC) = å 3 1 ...(3) i=1

n i i 2 2 (y - y ) (BC) = å 3 2 ...(4) i=1 Now, equation (4) can be written as

n i i i i 2 2 [(y - y ) + (y - y )] (BC) = å 3 1 1 2 i=1

n [(yi - yi )2 + (yi - yi )2 + 2(yi - yi ) ( yi - yi ) ] = å 3 1 1 2 3 1 1 2 i=1

n n (yi - yi )2 + (yi - yi )2 + 2 ´ 0, = å 3 1 å 1 2 [from (1)] i=1 i=1

n n (yi - yi )2 + ( yi - yi )2 = å 3 1 å 1 2 i=1 i=1

(BC)2 = (AC)2 +(AB)2

Hence Pythagoras theorem holds in Sn .

EXAMPLE 2 Prove that if q is any solution of the differential equation (Ñq)2 = f (q) then the hypersurfaces q = constant constitute a system of parallels. Solution Given that (Ñq)2 = f (q) ...(1) 184 Tensors and Their Applications

Then prove that the hypersurfaces q = constant form a system of parallel. Suppose dq dq f = , Then, df = ò f (q) f (q)

df 1 or = dq f (q)

¶f ¶f ¶q 1 Now, Ñf = = = Ñq ¶xi ¶q ¶xi f (q)

2 æ 1 ö 2 = ç Ñq÷ (Ñf) ç ÷ è f (q) ø 1 1 = (Ñq)2 = f (q); from (1) f (q) f (q) (Ñf)2 = 1 This proves that the hypersurfaces f = constant form a system of parallels and therefore the hypersurfaces q = constant.

EXAMPLE 3

Show that it is always possible to choose a geodesic coordinates system for any Vn with an arbitrary pole P0. Solution i Let P0 be an arbitrary pole in a Vn . Let us consider general coordinate system x . suppose the i i j value of x and P0 are denoted by x0. Now consider a new coordinate system x defined by the equation. 1 ì h ü j j m m j i l m m x = am ( x - x0 ) + ah í ý (x - x0) (x - x0 ) ...(1) 2 îl mþ j The coefficients am being constants and as such that their determinant do not vanish. Now we shall prove that this new system of coordinated x j defined by equation (1) is a geodesic j coordinate system with pole at P0 i.e., second covariant derivative of x vanishes at P.. Differentiating equation (1) with respect to xm , we get j ¶x j 1 jì h ü i l = am + ah í ý2 ( x - x0) ...(2) ¶xm 2 îl mþ

æ ¶x j ö ç ÷ = a j at P ...(3) ç ¶xm ÷ m 0 è ø0 Now, the Jacobian determinant Curvature of Curve, Geodesic 185

æ ¶ x j ö ç ÷ a j ¹ 0 ç ¶ m ÷ = m è x ø0

and therefore the transformation given by equation (1) is permissible in the neighbourhood of P0 . Differentiating equation (2) with respect to x j, we get

æ ¶2x j ö ç ÷ j ì h ü j m = a í ý ...(4) ç ¶x ¶x ÷ h l m è ø0 î þ0 But we know that

æ ¶ 2x j ö ì h ü æ ¶x j ö (x j ) = ç ÷ - í ý ç ÷ ,lm 0 ç ¶xl¶xm ÷ l m ç ¶xh ÷ è ø0 î þ0 è ø0

j ì h ü ì h ü j = a í ý - í ýa , (from (3) and (4)) h l m lm h î þ0 î þ

j (x,lm)0 = 0

Hence equation (1) is a geodesic coordinate system with pole at P0 .

EXAMPLE 4 If the coordinates xi of points on a geodesic are functions of arc lengths s and f is any scalar function of the x's show that d pf dxi dx j dxl = f,ij... l ×× × ...(1) dsp ds ds ds Solution

Since the coordinates xi lie on a geodesic. Then

d 2xi ì i ü dx j dxk 2 + í ý = 0 ...(2) ds î j kþ ds ds Here the number of suffices i j...l is p. We shall prove the theorem by mathematical induction method. Since x¢s are functions of s and f is a scalar function of x¢ s , we have

df ¶f dxi df dxi = or = f,i ...(3) ds ¶xi ds ds ds 2 j 2 i d f ¶f,i dx d x = + f,i ds2 ¶x j ds ds2 i j j k ¶f, dx dx ïì i ïü dx dx i - f , = j ,i í ý from (2) ¶x ds ds îï j k þï ds ds 186 Tensors and Their Applications

i j j k ¶f, dx dx ïì m ïü dx dx i - f, = i m í ý ¶x ds ds îï j kþï ds ds

i j i j ¶f,i dx dx ì mü dx dx = j - f,m í ý (adjusting the dummy index.) ¶x ds ds îi jþ ds ds æ ¶f ì m üö dxi dx j ç ,i - f ÷ = ç j ,m í ý÷ ...(4) è ¶x îi jþø ds ds Equations (3) and (4) imply that the equation (1) holds for p = 1 and p = 2.

Suppose that the equation (1) holds for p indices r1,r2,...,rp so that

r d pf dxr1 dxr1 dx p = f, ××× ...(5) dsp r1, r2 , ..., rp ds ds ds Differentiating the equation (5) with respect to s, we get

r r p+1 ¶f r1 p p+1 2 r1 r2 rp d f ,r1r2...rp dx dx dx d x dx dx = r ×× × + f,r r ... r ××× + × ×× dsp+1 ¶x p+1 ds ds ds 1 2 p ds 2 ds ds r dxr1 d 2x p + ×××f, ... × ×× ...(6) r1 r2 rp ds ds 2 d 2x r1 substituting value of etc. from (2) in (6) and adjusting dummy indices, we have ds2

p+1 æ ¶ f ìm ü ìm üö d f ç ,r1 r2 ... rp ï ï ï ï÷ - f,m r ... r í ý ×× ×f,r r ...m í ý = rp+1 2 p 1 2 p+1 ç ¶ r r r r ÷ ds è x îï 1 p+1þï îï p p+1þïø

dxr1 dxr2 dxrp+1 ××× ds ds ds r dxr1 d xr2 d x p+1 = f ×× × ,r1r2..rprp+1 ds ds ds This shows that the equation (1) holds for next values of p. But equation (1) holds for p = 1, 2, ... Hence equation (1) holds for all values of p.

EXERCISES

1. Prove that at the pole of a geodesic coordinate system, the components of first covariant derivatives are ordinary derivatives. 2. If xi are geodesic coordinates in the neighbourhood of a point if they are subjected to the transformation

i i 1 i j k l x = x + c jkl x x x 6 Curvature of Curve, Geodesic 187

i where Cs¢ are constants then show that x are geodesic coordinates in the neighbourhood of O. 3. Show that the principal normal vector vanishes identically when the given curve is geodesic. 4. Show that the coordinate system xi defined by

i 1 ì i ü j k xi = x + í ýx x 2 î j kþ is geodesic coordinate system with the pole at the origin. 5. Obtain the equations of geodesics for the metric ds2 = e-2kt (dx2 + dy2 + dz2) + dt2 6. Obtain the differential equations of geodesics for the metric

2 2 2 1 2 2 f ( x)dx + dy + dz + dt ds = f ( x)

2 é d2x 1 d æ dxö 1 d dt d2y d2z d2t d(log f ) dx dt ù ê ú Ans : 2 - (log f ) ç ÷ + 2 (log f ) = 0; 2 = 0; 2 = 0; 2 - = 0 ëê ds 2 dx è dsø 2 f dx ds ds ds ds dx ds ds ûú 7. Find the differential equations for the geodesics in a cylindrical and spherical coordinates. 8. Find the rate of divergence of a given curve C from the geodesic which touches it at a given point. CHAPTER – 10

PARALLELISM OF VECTORS

10.1 PARALLELISM OF A VECTOR OF CONSTANT MAGNITUDE (LEVI-CIVITA’S CONCEPT) Consider a vector field whose direction at any point is that of the Unit Vector ti. In ordinary space, the field is said to be parallel if the derivative of ti vanishes for all directions ui (say) and at every point of the field i.e., ¶t i u j = 0 ¶x j i Similarly in a Riemannian Vn the field is said to be parallel if the derived vector of t vanishes at i each point for every direction u at each point of Vn. i.e., i j t, j = u = 0

It can be shown that it is not possible for an arbitrary Vn. Consequently we define parallelism of vectors with respect to a given curve C in a Vn. i A vector u of constant magnitude is parallel with respect to Vn along the curve C if its derived vector in the direction of the curve is zero at all points of C i.e., dx j ui = 0 … (1) ds , j where s is arc-length of curve C. The equation (1) can be written in expansion form as

i j é ¶u m ì i üù dx ê j + u í ýú = 0 ë¶x îm jþû ds

i j j ¶u d x m ì i ü d x + u í ý = 0 ¶x j ds îm jþ ds i j du m ì i ü dx + u í ý = 0 … (2) ds îm jþ ds Parallelism of Vectors 189

This concept of parallelism is due to Levi-Civita. The vector ui is satisfying the equation (1) is said to a parallel displacement along the curve

Now, multiplying equation (1) by gil, we get æ j ö ç i dx ÷ gilç u, j ÷ = 0 è ds ø dx j (g ui ) = 0 il , j ds

dx j (g ui) = 0 il , j ds

dx j u l, j ds = 0

dx j or u = 0 i, j ds

é ¶u ìm üù dx j i - u ê j m í ýú = 0 ë¶x îi jþû ds

j dui ìm ü dx - um í ý = 0 … (3) ds îi jþ ds The equation (2) and (3) can be also written as

m ì i ü j dui = - u í ý dx … (4) îm jþ

ìm ü j and du = um í ý d x … (5) i îi jþ i The equation (4) and (5) give the increment in the components u and ui respectively due to displacement dxj along C. THEOREM 10.1 If two vectors of constant magnitudes undergo parallel displacements along a given curve then they are inclined at a constant angle. Proof: Let the vectors ui and vi be of constant magnitudes and undergo parallel displacement along a curve C, we have (from equation (1), Pg. 188.) j i d x ü u, = 0ï j ds ï j ý i d x ï … (1) v, = 0 j ds þï at each point of C.

Multiplying (1) by gil, we get dx j (g ui ) = 0 il , j ds 190 Tensors and Their Applications

dx j u = 0 l, j ds

dx j or u = 0 … (2) i, j ds Similarly, dx j v = 0 … (3) i, j ds Let f be the angle between ui and vi then i u .vi = uv cos q Differentiating it with respect to arc length s, we get d d(uiv ) (uv cos q) = i ds ds

dx j = (uiv ) i , j ds

dq dx j dx j - uvsin q = ui v + uiv … (4) ds , j ds i i, j ds Using equation (1) and (3), then equation (4) becomes dq - uvsin q = 0 ds dq Þ sin q = 0, as u ¹ 0,v ¹ 0 ds dq Þ Either sinq = 0 or = 0 ds

Þ Either q = 0 or q = constant. Þ q is constant. Since 0 is also a constant. THEOREM 10.2 A geodesic is an auto-parallel curve. Proof: The differential equation of the geodesic is given by (See Pg. 174, eqn. 6)

d 2xm ì m ü dx j dxk + 2 í ý = 0 ds î j k þ ds ds

æ m ö m j k d ç dx ÷ ì ü dx dx ç ÷ + í ý = 0 ds è ds ø î j k þ ds ds

¶ æ dxm ö d x j ì m ü d x j d xk ç ÷ + j ç ÷ í ý = 0 ¶x è ds ø ds î j kþ ds ds Parallelism of Vectors 191

é ¶ æ dxm ö ì m ü dx k ù dx j ç ÷ + ê j ç ÷ í ý ú = 0 ëê¶x è ds ø î j kþ ds ûú ds

m j æ d x ö dx dx j ç ÷ m = 0 ç ds ÷ ds = 0 or t, j è ø, j ds dxm This shows that the unit tangent vector suffer a parallel displacement along a geodesic curve.This ds confirms that geodesic is an auto-parallel curve. Proved.

10.2 PARALLELISM OF A VECTOR OF VARIABLE MAGNITUDE Two vectors at a point are said to be parallel or to have the same direction if their corresponding components are proportional. Consequently the vector vi will be parallel to ui at each point of curve C provided vi = fui … (1) where f is a function of arc length s. i If u is parallel with respect to Riemannian Vn along the curve C. Then, dx j u,i = 0 … (2) j ds i The equation (1) shows that v is of variable constant and parallel with respect to Riemannian Vn so that

d x j d x j v,i = (fui) j ds , j ds

d x j = (f ui + fui ) , j , j ds

dx j d x j = f ui + fui , j ds , j ds

d x j d x j = f ui Since ui = 0 , j ds , j ds

¶f d x j = ui ¶x j ds

j d x df i vi = u , j ds ds

d f vi = from (1) d s f 192 Tensors and Their Applications

i d(log f) = v ds dx j d(log f) vi = v i f (s) where f (s) = … (3) , j ds ds i Hence a vector v of variable magnitude will be parallel with respect to Vn if equation (3) is satisfied. Conversely suppose that a vector vi of variable magnitude such that dx j v,i = = vi f (s) j ds i to show that v is parallel, with respect to Vn . Take i i u = v y (s) … (5) Then d x j dx j u,i = (viy), j ds j ds dx j dx j = vi y + y, vi , j ds j ds ¶y dx j = vi f (s)y + vi ¶x j ds dx j é dyù i i y ( ) + u, j = v ê f s ú … (6) ds ë ds û dy Select y such that yf (s) + = 0. ds Then equation (6) becomes dx j ui = 0 , j ds This equation shows that the vector ui is of constant magnitude and suffers a parallel displacement along curve C. The equation (5) shows that vi is parallel along C. Hence necessary and sufficient condition that a vector vi of variable magnitude suffers a parallel displacement along a curve C is that dx j i i v, j = v f (s). ds EXAMPLE 1 Show that the vector vi of variable magnitude suffers a parallel displacement along a curve C if and only if dxk (vivi - vivl ) = 0, i = 1, 2, ..., n. ,k ,k ds Solution From equation (4), we have j i dx i v,j = v f (s) … (1) ds Parallelism of Vectors 193

Multiplying by v l, we get j l i dx l i v v, j = v v f (s) ds Interchange the indices l and i, we get dx j vivl = vi vl f (s) … (2) , j ds Subtract (1) and (2), we get dxk (vlui - vivl ) = 0 by interchanging dummy indices j and k. ,k ,k ds

10.3 SUBSPACES OF A RIEMANNIAN MANIFOLD

i Let Vn be Riemannian space of n dimensions referred to coordinates x and having the metric 2 i j a ds = gij dx dx . Let Vm be Riemannian space of m dimensions referred to coordinates y and having 2 a b, the metric ds = axb dy dy where m > n. Let Greek letters a , b, g take the values 1, 2, ..., m and Latin indices i, j, k ... take the values 1, 2, … n. i a If the n independent variables x are such that the coordinates ( y ) of points in Vm are expressed i as a function of x then Vn is immersed in Vm i.e. Vn is a subspace of Vm. Also Vm is called enveloping space of Vn. Since the length ds of the element of arc connecting the two points is the same with respect to Vn or Vm. it follows that i j a b gij dx dx = aabdy dy a b i j ¶y ¶y i j g dx dx = aab dx dx ij ¶xi ¶x j ¶ya ¶yb Þ g = aab … (1) ij ¶xi ¶x j As dxi and dxj are arbitrary.

This gives relation between gij and aab. THEOREM 10.3 To show that the angle between any two vectors is the same whether it is calculated with respect to Vm or Vn. i j Proof: Consider two vectors dx and dx defined at any point of Vn and suppose that the same vectors a a i i in Vm are represented by dy and dy respectively. If q is the angle between dx and dx then

i j gijdx dx cos q = … (1) i j i j gijdx dx gijdx dx If f is the angle between the vectors dya and dya then

a b aabdy dy cos f = a b a b aabdy dy aabdy dy 194 Tensors and Their Applications

a b ¶y i ¶y j aab i d x j dx = ¶x ¶x a b a b ¶y i ¶y j ¶y i ¶y j aab d x dx aab dx dx . ¶xi ¶x j ¶xi ¶x j

a b ¶y ¶y i j aab dx dx ¶xi ¶x j = a b a b ¶y ¶y i j ¶y ¶y i j aab dx dx aab dx dx . ¶xi ¶x j ¶xi ¶x j

i j gijdx dx cosf = … (2) i j i j gijdx dx gijdx dx

¶ya ¶yb Since g = aab (from equation (1), art. 10.3) ij ¶xi ¶x j from (1) & (2) cos q = cos f Þ q = f. Proved. a i THEOREM 10.4 If U and u denote the components of the same vector in Riemannian Vm and Vn respectively then to show that a i ¶y a = u U ¶xi

Proof: Let the given vector be unit vector at any point P of Vn. Let the component of the same vector x's and y's be ai and Aa respectively. Let C be curve passing through s in the direction of the given vector then a a i a ¶y ¶y dx ¶y i a = = = a … (1) A ¶s ¶xi ds ¶xi But the components of a vector of magnitude a are a times the corresponding components of the Unit vector in the same direction. Then a U a = aA , ui = aai … (2) Multiplying (1) by a, we get a ¶y i a = (aa ) aA ¶xi a ¶y i Or a = u , using (2) Proved. U ¶xi

THEOREM 10.5 To show that there are m-n linearly vector fields normal to a surface Vn immersed in a Riemannian Vm. a Proof: Since Vn is immersed in Vm, the coordinates y of points in Vm are expressible as functions of i coordinates x in Vn. Parallelism of Vectors 195

Now, dya ¶y a dxi = ...(1) ds ¶xi ds for the curve xi = s, we have dya ¶ya = ...(2) ds ¶xi dya ¶ya Since is a vector tangential to the curve in V . Then from equation (2) it follows that ds m ¶xi i a in Vm is tangential to the coordinate curve of transmeter x in Vn. Let the Unit vectors N in Vm be normal to each of the above vector fields of Vn then a ¶y b aab N = 0 ¶xi a b ¶y b a Þ (aab N ) = 0 Þ a N y = 0 ...(3) ¶xi ab ,i ¶ya Þ Na = 0, i = 1, 2, …, n & a = 1, 2, …, m ¶xi æ ¶ya ö a ç ÷ The equation (3) are n equations in m unknowns N (m > n). The coefficient matrix ç i ÷ is è ¶x ø of order m × n and the rank of this matrix is n.

It means that there will be only m – n linearly independent solution of N a . This shows that there are m – n linearly independent normals to Vn to Vm.

EXAMPLE 2 Show that ¶y a ¶yb ¶yg ¶2ya ¶xb [ij, k] = [ab,g] + aab ¶xi ¶x j ¶xk ¶xi¶x j ¶xk a b where [ab, g] and [i j,k ] are the Christoffel’s symbols of first kind relative to metrics aabdy dy and i j gijdx dx . Solution

Since relation between aab and gij is given by ¶ya ¶yb g = aab ij ¶xi ¶x j Differentiating it with respect to xk, we get

a b g 2 a g a 2 b ¶gij ¶aab ¶y ¶y ¶y ¶ y ¶y ¶y ¶ y = + a + a … (1) ¶xk ¶xg ¶xi ¶x j ¶xk ab ¶xi¶xk ¶x j ab ¶xi ¶x j¶xk 196 Tensors and Their Applications

Similarly ¶a a b g 2 a b a 2 b ¶g jk bg ¶y ¶y ¶y ¶ y ¶y ¶y ¶ y = a + aab + aab … (2) ¶xi ¶y ¶xi ¶x j ¶xk ¶x j¶xi ¶xk ¶x j ¶xk ¶xi and a b g 2 a b a 2 b ¶g ki ¶aga ¶y ¶y ¶y ¶ y ¶y ¶y ¶ y = + a + a … (3) ¶x j ¶xb ¶xi ¶x j ¶xk ab ¶xk ¶x j ¶xi ab ¶xk ¶xi¶x j But we know that

1 æ ¶g jk ¶g ¶gij ö [ , ] = ç + ki - ÷ … (4) ij k ç i j k ÷ 2 è ¶x ¶x ¶x ø 1 æ ¶g ¶g ¶g ö ç bg + ga - ab ÷ and, [ab, g] = ç a b g ÷ … (5) 2 è ¶y ¶y ¶y ø Substituting the value of (1), (2), (3) in equation (4) and using (5) we get, ¶y a ¶yb ¶y g ¶2ya ¶yb [ij,k ] = [ab,g] + aab . ¶xi ¶x j ¶xk ¶xi¶x j ¶xk

10.4 PARALLELISM IN A SUBSPACE

a i THEOREM 10.6 Let T and t be the components of the same vector t relative to Vm and Vn i a respectively. Let the vector t be defined along a curve C in Vn. If p and q are derived vectors of t along C relative to Vn and Vm respectively. Then ¶ya qa = p ¶xi i Proof: Since from equation (1), theorem, (10.4), Pg. 194 we have a i ¶y a = t … (1) T ¶xi dT a dti ¶y a ¶2 y a dx j = + ti … (2) ds ds ¶xi ¶xi¶x j ds Now, dx j i i p = t, j … (3) ds b a a dy q = T,b ds æ ¶T a ì a üö dyb ç + T g ÷ = ç b í ý÷ è ¶x îgbþø ds ¶T a dyb ì a ü dyb + g = b T í ý ¶x ds îgbþ ds a b dT g ì a ü dy qa = + T í ý … (4) ds îgbþ ds Parallelism of Vectors 197

dT a Putting the value of and g from (2) and (1) in equation (4), we get ds T dti ¶y a ¶2 ya dx j ¶yg dyb ì a ü a + ti + ti q = i i j i í ý ds ¶x ¶x ¶x ds dx ds îgbþ ¶t i ¶ya dx j ¶2 ya dx j ¶yg dy b ì a ü + t i + ti = j i i j i í ý ¶x ¶x ds ¶x ¶x ds ¶x ds îgbþ

i a j 2 a j g b j ¶t ¶y dx i ¶ y dx i ¶y ¶y dx ì a ü = + t + t í ý j i i j i j gb ¶x ¶x ds ¶x ¶x ds ¶x ¶ x. ds î þ

¶t i ¶y a d x j dx j æ ¶2 ya ¶yb ¶y g ì a üö a + t i + ç + ÷ q = j i ç i j j i í ý÷ … (5) ¶x ¶x ds ds è ¶x ¶x ¶x ¶x îgbþø But we know that

¶ya ¶yb ¶yg ¶2 ya ¶yg [ij,k ] = [ab,g] + aag ¶xi ¶x j ¶xk ¶xi¶x j ¶xk

¶yb ¶y g ¶yd ¶2 y a ¶y d [ ji,k ] = [bg,d] + aad ¶x j ¶xi ¶xk ¶xi¶x j ¶xk

ì a ü ¶yb ¶y g ¶yd ¶2 ya ¶yd a + a = í ý ad j i k ad i j k îbgþ ¶x ¶x ¶x ¶x ¶x ¶x

¶yd æ ì a ü ¶yb ¶y g ¶2y a ö a ç + ÷ = ad k ç í ý j i i j ÷ … (6) ¶x è îbgþ ¶x ¶x ¶x ¶x ø ¶y d Multiplying (5) by aad , we get, using (6), ¶xk d i j a d j ¶y a ¶t dx ¶y ¶y i dx aad q = aad + t [ij,k] ¶xk ¶x j ds ¶xi ¶x k ds

¶yd dx j æ ¶ti ì p ü ö ¶ya ¶yb q ç g + ti g ÷ g = a Or d k = ç j ik í ý pk ÷ , since ij ab i j ¶x ds è ¶x îi jþ ø ¶x ¶x

dx j æ ¶ti ì i ü ö ç g + t a g ÷ = ç j ik í ý ik ÷ ds è ¶x îa jþ ø dx j = g t,i ik ds j 198 Tensors and Their Applications

dx j = (g ti ), ik j ds dx j = t k, j ds ¶y a dx j qa = t , ¶xk k j ds ¶y a or qa = p …(7) ¶xk k

Proved.

Properties of Vm (i) If a curve C lies in a subspace Vn of Vm and a vector field in Vn is parallel along the curve C with regard to Vm, then to show that it is also a parallel with regard to Vn. a Proof: If a vector field t is a parallel along C with respect to Vm. then its derived vector q vanishes i.e., qa = 0, a = 1, 2,... m.

Hence equation (7) becomes pk = 0, k = 1, 2, ..., n. This shows that the vector field t is also parallel along C with respect to Vn. (ii) To show that if a curve C is a geodesic in a Vn it is a geodesic in any subspace Vn of Vm.

a ¶ya Proff : Science, q = p ¶xi i a Let t be unit tangent vector to the curve C them T is unit tangent vector to C relative to Vm and ti is unit tangent vector to C relative to Vn. Now, pi = 0 Þ curve C is a geodesic in Vn a q = 0 Þ curve C is a geodesic in Vm

Also, qa = 0, "a Þ pi = 0, "i This shows that if a curve C in Vn is a geodesic relative to Vm then the same curve is also a geodesic relative to Vn. (iii) A necessary and sufficient condition that a vector of constant magnitude be parallel with respect to Vn along C in that subspace, is that its derived vector relative to Vm for the direction of the curve be normal to Vn. Proof: Let t be a vector of constant magnitude. This vector t is parallel along C relative to Vn iff pi = 0, "i ¶y a iff qa = 0 ¶xi ¶ya But qa = 0 implies that q is normal to V . ¶xi a n a ¶y a i For = y, lying in V is tangential to a coordinate curve of parameter x in V . ¶ xi i m n Parallelism of Vectors 199

These statements prove that a necessary and sufficient condition that a vector of constant magnitude be parallel along C relative to Vn is that its derived vector i.e., qa along C relative to Vm be normal to Vn. (iv) A necessary and sufficient condition that a curve be a geodesic in Vn is that its principal subnormal relative to Vm the enveloping space be normal to Vn at all points of the curve.

Proof: In particular let the vector t be unit tangent vector to the curve C. In this case qa is

called principal normal the curve C. Also pi = 0, "i implies that the curve C is a geodesic relative to Vn. Using result (iii), we get at once the result (iv). (v) To prove that the tendency of a vector is the same whether it is calculated with respect to Vm. Proof: Since, we have ¶y a qa = p ¶xi i ¶ya dxi dxi qa = p ¶xi ds i ds dya dxi q = p a ds i ds dyb dya dx j dxi T = t a,b ds ds i, j ds ds

i.e., tendency of Ta along C = tendency of ti along C. i.e., tendency of t along C relative to Vm = tendency of t along C relative to Vn.

10.5 THE FUNDAMENTAL THEOREM OF RIEMANNIAN GEOMETRY STATEMENT With a given Riemannian metric (or fundamental tensor) of a Riemannian manifold there is associated a symmetric affine with the property that parallel displacement (or transport) preserves scalar product. i i Proof: Let C be a curve in Vn. Let p and q be two unit vectors defined along C. Suppose that the unit i i vectors p and q suffer parallel displacement along the curve C in Vn, then we have dx j pi = 0 … (1) , j ds dx j and q,i = 0 … (2) j ds

Let gij be the given fundamental tensor of a Riemannian manifold. Hence, the scalar product of i i i j vectors p and q is gij p q . Now,

dxk (g piq j ), = 0 ij k ds 200 Tensors and Their Applications

æ dxk ö æ dxk ö æ dxk ö ç i ÷ j + iç j ÷ + ç ÷ i j gijç p,k ÷ q gij p ç q,k ÷ ç gij,k ÷ P q = 0 … (3) è ds ø è ds ø è ds ø Using equation (1) and (2), the equation (3) becomes, æ k ö ç dx ÷ i j ç gij,k ÷ p q = 0 è ds ø gij, k = 0 dxk (Since pi and qi are unit vectors and ¹ 0) ds ¶g ìm ü ì m ü ij - g - g k mj í ý im í ý = 0 ¶x îi kþ î j kþ

¶gij - [ik, j]- [ jk,i] = 0 ¶xk

¶gij = [ik, j] + [ jk ,i] … (4) ¶xk Now, using equation (4), we have

¶g jk ¶g ¶gij + ki - = [ ji,k ]+ [ki, j]+ [kj,i] +[ij,k ] – ([ik,i] + [ jk ,i]) ¶xi ¶x j ¶xk since [ij,k ] = [ ji,k]. So,

¶g jk ¶g ¶gij + ki - = 2[ij,k ] ¶xi ¶x j ¶xk

1 æ ¶g jk ¶g ¶gij ö [ , ] = ç + ki - ÷ … (5) ij k ç i j k ÷ 2 è ¶x ¶x ¶x ø

ì k ü lk But we know that í ý = g [ij,l] îi jþ from (5), we have

ì k ü 1 æ ¶g jl ¶g ¶gij ö g lk ç + li - ÷ í ý = ç i j l ÷ Proved. îi jþ 2 è ¶x ¶x ¶x ø EXAMPLE 3 i a If t and T are contravariant components in x's and y's respectively, of n vector field in Vn immersed in Vm. Show that a a i a i T, j = y,ijt + y,i t, j Solution Since we know that a i ¶y a = t T ¶xi Parallelism of Vectors 201

i a a T = t y, j. Taking covariant differentiation of both sides, we get

a i a T, j = (t y,i ), j

i a i a = t, j y,i + t (y,i ), j

a i a i a T, j = t y,ij + t, j y,i EXAMPLE 4 dxi dx j Show that g remains constant along a geodesic. ij ds ds Solution dxi Let ti = . Then ds dxi dx j g = g tit j = t 2 ij ds ds ij Since we know that geodesics are autoparallel curves. Then dx j ti = 0 , j ds

i j or t, jt = 0 … (1) Now, 2 dt d i j d i = (g t t ) = (t t ) ds ds ij ds i dx j = (t t i) = (t ti ) t j i , j ds i , j dt2 = (t ti )ti + (t i ti)t = 0, from (1) ds i, j , j i Integrating it we get t2 = constant. dxi dx j So, g remains constant along a geodesic. ij ds ds

EXAMPLE 5 If ti are the contravariant components of the unit tangent vector to a congruence of geodesics. Show that i t (ti, j + t j,i ) = 0 and also show that | ti, j + t j,i |= 0. 202 Tensors and Their Applications

Solution Let ti denote unit tangent vector to a congruence of geodesic so that i j t, jt = 0 … (1) Since geodesics are auto-parallel curves. Then to prove that i (i) t (ti, j + t j,i ) = 0

(ii) | ti, j + t j,i |= 0 Since i 2 t ti = t = 1 i t ti = 1 Taking covariant derivative of both sides, we get i (t ti), j = 0 i i or t, jti + t ti, j = 0 Since t is a free index. Then we have i i i t, jt + ti, jt = 0 i i 2t, jt = 0 i Þ ti, jt = 0 … (2) from (1)

i j gikt, jt = 0 j or tk, jt = 0 i i i Þ tk,it = 0 Þ t j,it = 0 Þ t t j,i = 0 i Thus t t j,i = 0 … (3) Adding (2) and (3), we get

i t (ti, j + t j,i ) = 0 Also, since ti ¹ 0, "i. Taking determinants of both sides we get i | t (ti, j + t j,i) | = 0

Þ | ti, j + t j,i | = 0 Solved. EXAMPLE 6

When the coordinates of a V2 are chosen so that the fundamental forms is 1 2 2 2 (dx&') + 2g12dx dx + (dx ) , prove that the tangent to either family of coordinate curves suffers a parallel displacement along a curve of the other family. Parallelism of Vectors 203

Solution The metric is given by

2 i j ds = gijdx dx i,j = 1,2. Comparing it with

2 1 2 1 2 2 2 ds = (dx ) + 2g12dx dx + (dx ) (1) We have,

g11 = 1, g22 = 1. In this case, we have coordinate curves of parameters x1 and x2 respectively. The coordinate x1 curve is defined by xi = ci, "i, except i = 1. (2) and the coordinate x2 curve is defined by xi = di, "i, except i = 2 (3) where ci and di are constants. Let pi and qi be the components of tangents vectors to the curves (2) and (3) respectively. Then we have pi = dxi = 0, "i, except i = 1 and qi = dxi = 0, "i, except i = 2 So, pi = (dxi, 0) and qi = (0,dxi) Let t be the unit tangent vector to the curve (2). Hence i i dx i p = t = = (1,0) where p = dxi ds p So, we have dx j pi = 0 , j ds Hence the tangent to the family of coordinates curves (2) suffers a parallel displacement along a curve of the family of curves (3).

EXERCISES

1. Explain Levi-Civita's concept of parallelism of vectors and prove that any vector which undergoes a parallel displacement along a geodesic is inclined at a constant angle to the curve. 2. Show that the geodesics is a Riemannian space are given by d 2xm ìm ü dxi dxk + 2 í ý = 0 ds îi k þ ds ds Hence prove that geodesics are auto-parallel curves. 204 Tensors and Their Applications

3. Establish the equivalence of the following definitions of a geodesic. (i) It is an auto–parallel curve.

(ii) It is a line whose first curvature relative to Vn identically zero. (iii) It is the path extremum length between two points on it.

4. If u and v are orthogonal vector fields in a Vn, prove that the projection on u of the desired vector of u in its own direction is equal to minus the tendency of v in the direction of u. 5. If the derived vector of a vector ui is zero then to show that vector ui has a constant magnitude along curve. 6. Prove that any vector which undergoes a parallel displacement along a geodesic is inclined at a constant angle to the curve.

7. Prove that there are m – n linearly independent vector fields normal to a surface Vn immersed in a Riemannian Vm and they may be chosen in a multiply infinite number of ways. But there is only one vector field normal to the hyperface. 8. Show that the principal normal vector vanishes identically when the given curve in geodesic. 9. Show that if a curve is a geodesic of a space it is a geodesic of any subspace in which it lies.

10. Define parallelism in a subspace of Riemannian manifold. If a curve C lies in a subspace Vn of Vm and a vector field in Vn is parallel along C with respect to Vm. Then show that it is also parallel with respect to Vn. CHAPTER – 11

RICCI'S COEFFICIENTS OF ROTATION AND CONGRUENCE

11.1 RICCI'S COEFFICIENTS OF ROTATION i Let eh| (h=1, 2, …n) be the unit tangents to the n congruences eh , of an orthogonal ennuple in a e j Riemannian Vn,. The desired vector of el|i in the direction of ek| has components el|i, jek| and the i projection of this vector on eh| is a scalar invariant, denoted by glhk, so that

i j glhk = el|i, jeh|ek| …(1)

The invariants glhk are Ricci's Coefficients of Rotation. Since i being a dummy index, has freedom of movement. Then equation (1) may be written as

j glhk = el|i, jeh|iek| …(2)

The indices l, h and k are not tensor indices. But these indices in glhk are arranged in proper way, the first index l indicates the congruence whose unit tangent is considered, the second h indicates the direction of projection and the third k is used for differentiation. THEOREM 11.1 To prove thast the Riccie's coefficients of rotation are skew-symmetric in the first two indices i.e., glhk = –glhk i Proof: If eh| (h = 1, 2, …, n) be n unit tangents to n congruences eh| of an orthgonal ennuple in a Vn then i eh|iel| = 0 convariant differentiation with respect to xj, we get i eh|iel| ,j = 0

i i eh|i, jel| + el|, jeh|i = 0 206 Tensors and Their Applications

j multiplying by ek| and summing for j, we get

i i i j eh|i, jel|ek| + ei|, jeh|iek| = 0

ghlk + glhk = 0 or ghlk = –glhk …(1) Note: Put l = h in equation (1), we get

gllk = –gllk or 2gllk = 0 or gllk = 0 THEOREM 11.2 To prove that

g e j å lhk h|i = e e h l|i, j k| Proof: since we know that

i j glhk = el|i, jeh|ek|

Multiplying by eh|m and summing for h. g e i j å lhk h|m = åel|i,jeh|ek|eh|m h h

j i e | , e | (e e ) = l i j k å h| h| m h i e e j di e e i = l|i, j k| m since å h| h|m = dm h i j = (el|i, jd m ) ek|

glhkeh|m j å = el|m, jek| h Replacing m by i, we get

glhkei|m j å = e | , e | h l i j k

11.2 REASON FOR THE NAME "COEFFICIENTS OF ROTATION" i Let Cm be a definite curve of the congruence whose unit tangent is em| and P0 a fidxed point on it. Let 1 i u be a unit vector which coincides with the vector el| at P0 and undergoes a parallel displayment along the curve Cm. Thus i i u = el| at P0 … (1)

i j and u,jeh| = 0 … (2) Ricci's Coefficients of Rotation and Congruence 207

i i If q is the angle between the vectors u and eh| , we have i cos q = u eh|i

Differentiating it with respect to arc length sm along Cm, we get dq – sin q i j = (u eh|i ), j em| dsm

i i j = (u eh|i, j + u, jeh|i )em| dq – sin q i j i j = u eh|i, j em | + u , je …(3) dsm m | p at the point P ,q = , we have 0 2 dq – i j = u eh|i, jem| dsm i j = u eh|i, jem| i j = el|eh|i, jem|; from (1)

i j = eh|i, jel|em| dq – = ghlm dsm dq Þ = – ghlm … (4) dsm q d i In Eucliden space of three dimensions, is the arc-rate of rotation of the vector eh| about the dsm curve Cm. Hence the quantities ghlmare called coefficients of rotation of the ennuple. Since it was discovered by Ricci and hence it is called Ricci's coefficient of rotation.

11.3 CURVATURE OF CONGRUENCE i The first curvature vector pl| of curve of the congruence is the derived vector of eh| in its own i direction. Where eh| (h = 1, 2, …, n) be unit tangents to n congruence eh| of an orthogonal ennuple in a Vn. i If ph| be the contravariant component of first curvature vector pl|. Then by definition, we have i i j ph| = eh,jeh| from theorem (11.2), we have i i ghlhel| ph| = å … (1) l 208 Tensors and Their Applications

i The magnitude of ph| is called curvature of the curve of congruence eh| and denoted by Kh| . Now,

2 i j Kh| = gij ph| ph|

æ ö æ ö g ç g ei ÷ ç g ei ÷ = ijç å hlh l| ÷ ç å hmh m| ÷ è l ø è m ø

g ei ei g g = åå ij l| m| hlh hmh l m

l d g g , i i l = åå m hlm hmh Since gijel|em| = d l m m

= å ghmhghmh m

2 (g )2 Kh| = å hmh m

This is the required formula for Kh|.

11.4 GEODESIC CONGRUENCE If all the curves of a congruence are geodesics then the congruence is called a geodesic congruence. THEOREM 11.3 A necessary and sufficient condition that congruence C of an orthogonal ennuple be a geodesic congruence is that the tendencies of all the other congruences of the ennuple in the direction of C vanish indentically. Or To obtain necessary and sufficient conditions that a congruence be a geodesic congruence. Proof: From equation (1), Pg. 207, we have pi i h| = å ghlhel| l i i i Since el| ¹ 0 , "h and hence ph| = 0 iff ghlh = 0, "h . But pl| = 0 iff the congruence C is geodesic congruence.

Hence C is a geodesic congruence iff ghlh = 0, "h . But

ghlh = – glhh

So, C is a geodesic congruence iff – glhh = 0, –"h .

or C is a geodesic congruence iff glhh = 0. i Hence glhh = 0 are the necessary and sufficient conditions that congruence with unit tangents eh| i i be geodesic congruence. Again glhh is the tendency of the vector el| in the direction vector eh| . Ricci's Coefficients of Rotation and Congruence 209

Thus a congruence C of an orthogonal ennuple is a geodesic congruence iff the tendency of all other congruences in the direction of C vanish identically.

11.5NORMAL CONGRUENCE A normal congruence is one which intersects orthogonally a family of hypersurfaces.

THEOREM 11.4 Necessary and sufficient conditions that the congruence eh| of an orthogonal ennuple be normal is that

gnqp = gnpq 1 2 n Proof: Consider a congruence C of curves in a Vn. Let f(x , x ,¼, x ) = constant be a family of hypersurfaces. To determines a normal congruence whose tangent vector is grad f or Ñ f. Let ti be the convariant components of unit tangent vector to C. The congruence C is a normal congruence to a family of hypersurface f(x1, x2,¼, xn ) = constant if

f,1 f,2 f, = = ¼ n = y (say) … (1) t1 t2 tn In order that (n – 1) differential equations given by equation (1) admit a solution which is not constant, these must constitute a complete system. From (1)

f,i = y or f,i = yti ti ¶f Þ yt = i ¶xi Differentiating it with respect to x j , we get 2 ¶y ¶ti ¶ f ti + y = … (2) ¶x j ¶x j ¶xi¶x j Interchanging indices i and j, we get 2 ¶y ¶t j ¶ f t j + y = … (3) ¶xi ¶xi ¶x j¶xi

Subtracting (2) and (3) we get

æ ¶y ¶ti ö æ ¶y ¶ti ö ç ti + y ÷ – ç t j + y ÷ = 0 è dx j ¶x j ø è ¶xi dx j ø

æ ¶t ö ç ¶ti j ÷ æ ¶y ¶y ö yç – ÷ + çti – t j ÷ = 0 è ¶x j ¶xi ø è ¶x j ¶xi ø 210 Tensors and Their Applications

Multiplying by tk,

æ ¶t ¶t j ö ¶y ¶y ç i – ÷ + t t – i t ytk ç j i ÷ i k j k i i = 0 …(4) è dx ¶x ø ¶x dx By cyclic permutation of i, j, k in (4), we get

æ ¶t j ¶t ö ¶y ¶y yt ç – k ÷ + t t – i t = 0 … (5) i ç k j ÷ j i k k i j è ¶x ¶x ø ¶x ¶x and

æ ¶tk ¶ti ö ¶y ¶y yti ç – ÷ + tkt j – tit j = 0 … (6) è ¶xi ¶xk ø ¶xi ¶xk On adding (4) , (5) and (6) we get

é æ ¶t ¶t ö æ ¶t ¶t ö æ ¶t ¶t öù ç i – j ÷ + ç j – k ÷ + ç k – i ÷ yêtk ç j i ÷ ti ç k j ÷ t j i k ú = 0 ëê è ¶x ¶x ø è ¶x ¶x ø è ¶x ¶x øûú

or tk (ti, j – t j,i ) + ti (t j,k – tk, j ) + t j (tk,i – ti,k ) = 0 … (7)

as y ¹ 0 , where i, j, k = 1, 2, …, n. These are the necessary and sufficient conditions that the given congruence be a normal congruence.

Now suppose that the congruence is one of an orthogonal ennuple in Vn. Let en|i be the unit tangents of given congruence C so that ti = en|i Then equation (7) becomes

en|k (en| i,j – en| j,i) + en| i (en| j,k – en| k,j) + en| j (en|k,i – en|i,k) = 0 … (8) i R Now, multiplying equation (8) by e p|eq| , we get

i k i k i k en|k (en|i, j – en| j,i )e p|eq| + en|i(en| j,k - en|k, j )e p| eq + en| j (en|k,i – en|i,k )e p|eq| where p and q are two new indices chosen from 1, 2, …, n – 1. i.e., p,q and n are unequal.

k q But en|k eq| = = 0, q ¹ n dh k = p = 0, p ¹ n en|ie p| dh so, we have

i k en| j (en|k,i – en| j,k ) ep| eq| = 0

i k i k en| j (en|k,ieq|ep| – en| j,k ep|eq| ) = 0

en|j (gnqp – gnpq) = 0 i k Since en|k,i ep| eq| = gnpq Ricci's Coefficients of Rotation and Congruence 211

Þ gnqp – gnpq = 0, en|j ¹ 0 Þ gnpq = gnqp, … (9) Conversely if equation (9) in true then we get equation (8). Which implies that equation (7) are satisfied bty en|i Hence en| is a normal congruence. Thus necessary and sufficient conditions that the congruence en| of an orthogonal ennuple be a normal congruence are that

gnqp = gnpq (p, q = 1, 2, …, n – 1 such that p ¹ q) THEOREM 11.5 Necessary and sufficient conditions that all the congruences of an orthogonal ennuple be normal. Proof: If all the congruences of a orthogonal ennuple are normal. Then

gnqp = gnpq (p, q = 1, 2, …, n – 1 such that p ¹ q) If the indices h, k, l and unequal then

ghkl = ghlk … (1) But due to skew-symmetric property i.e., ghlk = – glhk. So,

ghkl = ghlk = – glhk = – glkh, from (1) = gklh (skew-symmetric property). = gkhl, from (1) ghkl = – ghkl, (skew-symmetric property). ghkl + ghkl = 0 Þ 2 ghkl = 0 Þ ghkl = 0 where (l, h, k = 1, 2, …, n such that h, k, kl are unequal).

11.6 CURL OF CONGRUENCE The curl of the unit tangent to a congruence of curves is called the curl of congruence.

If en| is a given congruence of curves then Curl en|i = en|i,j – en| j,i If curl en| i = 0 then congruence is irrotational. THEOREM 11.6 If a congruence of curves satisfy two of the following conditions it will also satisfy the third (a) that it be a satisfy the third (b) that it be a geodesic congruence (c) that it be irrotational. i Proof: Consider an orthogonal ennuple and eh| (h = 1, 2, …, n) be n unit tangent to n congruences of this orthogonal ennnuple. From theorem (11.2), we have g e j å lhk h|i = el|i, jek| h 212 Tensors and Their Applications

Putting l = n and j = m, we have n g e m å nhk h|i = en|i,m ek| h=1

Now, multiplying by ek| j and summing with respect to k from 1 to n, we have n m å gnhkeh|iek| j = en|i,mek| ek| j h,k=1 m m m = en|i,md j Since ek| ek| j = d j n g e e å nhk h|i k| j = en|i, j … (2) h,k =1 By definition of curl of congruence, we have

curl en|i = en| i,j – en| j,i n n = å gnhkeh|iek| j – å gnhkeh| jek|i; from (2) h,k=1 n,k=1

n n

g e | e | – g e | e | curl en| i = å nhk h i k j å nkh k j hi h,k =1 h,k=1

n

curl en|i = å (gnhk – gnkh) eh| iek| j … (3) h,k =1 This double sum may be separated into two sums as follows. (i) Let h and k take the values 1, 2, …, n – 1. (ii) Either h = n or k = n or h = k = n. Now, the equation (3) becomes

n-1 n–1

(g – g ) e | e | + (gnhn – gnnh)e | e | curl en| i = å nhk nkh hi k j å hi n j h,k =1 h=1

n (g – g )e e + (g – g )e e + å nnk nkn n|i n| j nnn nnn n|i n|j k=1

Since we know that gnnk = gnnh = gnnn= 0. So, n–1 n–1 n–1

(g nhk – g nkh )eh|i ek| j + g nhneh|i en| j – g nknen|i ek| j curl en|i = å å å h,k=1 h=1 k=1

n–1 n–1

(gnhk – gnkh) eh|i ek| j + gnhn(eh|ien| j – en|iek| j) curl en|i = å å … (4) h,k h=1 Ricci's Coefficients of Rotation and Congruence 213

The first term on R.H.S. of equation (4) vanishes

if gnhk – gnkh = 0 i.e.,

if gnhk = gnkh.

i.e., if the congruence en| is normal. Again the second term of R.H.S of equation (4) vanishes

if gnhn = 0 i.e., if ghnn = 0

i.e., if the congruence en| is a geodesic congruence. Further, if first and second term on right hand side of equation (4) both vanishes then

curl en|i = 0

Hence we have proved that if the congruence en| satisfies any two of the following conditions then it will also satisfy the third.

(a) en| is a normal congruence

(b) en| is irrotational (c) en| is a geodesic congruence

11.7 CANONICAL CONGRUENCE It has been shown that given a congruence of curves, it is possible to choose, in a multiply infinite number of ways, n – 1 other congruences forming with the given congruence an orthogonal ennuple. Consider the system of n – 1 congruence discovered by Ricci, and known as the system canonical with respect to the given congruence.

THEOREM 11.7 Necessary and sufficient conditions that the n – 1 congruences eh| of an orthogonal ennuple be canonical with respect to en| are gnhk + gnkh = 0; (h, k = 1, 2, …, n – 1, h ¹ k ). th Proof: Let the given congruence en| be regarded as n of the required ennuple. Let en|i be unit tangent to given congruence.

Xij = ½ (en| i,j + en| j,i) … (1) Let us find a quantity r and n quantities ei satisfying the n + 1 equations

i en|ie = 0ïü i ý i, j = 1, 2, …, n … (2) (X ij – wgije + rEn| j = 0þï where w is a scalar invariant. Writing equation (2) in expansion form, we have 1 2 n en|1e + en|2e + ¼¼ + en|ne = 0 … (3) and

i 2 n (X1j – wg1 j) e + (X 2j – wg2 j )e + ¼ ( Xnj – wgnj )e + ren| j = 0 214 Tensors and Their Applications

for j = 1, 2, …, n. 1 2 n (X11 – wg11) e + (X21 – wg21) e + ¼ + (X n1 – wgn1) e + ren|1 = 0 1 2 n (X12 – wg12) e +(X 22 – wg22) e + ¼+ (Xn2 – wgn2 )e + ren|2 = 0 M 1 2 n (X1n – wg1n) e + (X 2n – wg2n ) e + ¼+ (Xnn – wgnn ) e + ren|n = 0 from (3) 1 2 n en|1e + en|2e +¼en|ne + r.0 = 0 eliminating r and the quantities e1, e2, …en, we have the equation

(X11 – wg11) (X 21 – wg21) ¼ (X n1 – wgn1) en|1

(X12 – wg12) (X22 – wg22) ¼ (Xn2 – wgn2 ) en|2 M

(X1n – wg1n ) (X 2n – wg2n ) ¼ (X nn – wgnn) en|n

en|1 en|2 ¼ en|n 0

which is of degree n – 1 in w. Hence there will be n – 1 roots of w and these roots be w1, w2, …, i wn – 1. All roots are real. Let wh be one of these roots and let the corresponding values of r and e be i i denoted by rh and eh| respectively. Then rh and eh| will satisfy the equation (2), we have i en|ieh| = 0 … (4) i and (X ij – wgij) eh| + rhen|j = 0 … (5)

Similarly wk be another root of these roots, we have i en|iek| = 0 … (6) i (X ij – wgij) ek| + rhen| j = 0 … (7)

j j Multiplying (5) by ek| and (7) eh| and using (4) and (6), we get

i j (X ij – whgij) eh|ek| = 0 … (8)

i j (X ij – wk gij) ek|eh| = 0 … (9)

Since Xij and gij are symmetric tensor in i and j. Now, interchanging i and j in equation (9), we get i j (X ij – wk gij) eh|ek| = 0 … (10) Subtracting (10) and (8), we get i j (wh – wk ) gijeh|ek| = 0 … (11)

Since wk – wk ¹ 0 as wh ¹ wh. i j Þ gijeh|ek| = 0 … (12) Ricci's Coefficients of Rotation and Congruence 215

i j This shows that eh| and ek| unit vectors are orthogonal to each other and hence the congruence eh| and ek| (h ¹ k) are orthogonal to each other. Hence the n – 1 congruence eh| (h = 1, 2, …, n) thus determined form an orthogonal ennuple with en|. Using equation (12), equation (10) becomes i j Xij eh| ek| = 0 1 Since from (1), X = (e | , + e | , ). Then we have, ij 2 n i j n j i

1 i j (e | , + e | , e |e | ) = 0 2 n i j n j i h k i j i j en|i, jeh|ek| + en| j,iekeh = 0

gnhk + gnkh = 0; (h, k = 1, 2, …, n – 1 such that h ¹ k ) … (13)

Conversely if equation (13) is true then (n – 1) congruences eh| of the orthogonal ennuple and canonical with respect to en| . Hence necessary and sufficient condition that the n – 1 congruences eh| of an orthogonal ennuple be canonical with respect to en| are gnhk + gnkh = 0; (h, k = 1, 2, …, n such that h ¹ k )

THEOREM 11.8 Necessary and sufficient conditions that n –1 mutually orthogonal congruences eh| orthogonal to a normal congruence en| , be canconical with respect to the later are gnhk = 0 where k, h = 1, 2,…, n – 1 such that h ¹ k.

Proof: By theorem (11.7), Necessary and sufficient conditions that (n – 1) congruences eh| of an orthogonal ennuple be canonical with respect to en| are gnhk + gnkh = 0 (h, k = 1, 2, …, n – 1 such that h ¹ k).

If the congruence en| is normal. Then gnhk = gnkh

The given condition gnhk + gnkh = 0 becomes

gnhk + gnhk = 0 2gnhk = 0 gnhk = 0 Proved.

EXAMPLE 1 1 If e are the congruences canonical with respect to e prove that (i) w = g (ii) r = g n| n| h nhk h 2 hnn, (iii) If en| is a geodesic congruence, the congruences canonical with respect to it are given by i (Xij - wgij)e = 0 = 0 Solution

Suppose (n – 1) congruences eh| of an orthogonal ennable in a Vn are canonical with respect to the cougruence eh| then i en|ieh| = 0 … (1) 216 Tensors and Their Applications

i and (X ij – whgij) eh| + rhen| j = 0 … (2) where 1 X = (e + e ) … (3) ij 2 n| i,j n| j,i i Since en| unit tangents then i j gij eh|eh| = 0 … (4) i (i) Multiplying equation (2) by eh| , we get

i j j (X ij – whgij) eh| eh| + rh en| j eh| = 0

i j i j j Xij eh|eh| – whgijeh|eh| = 0 since en| jeh| = 0 (from (1))

i j i j Xijeh|eh| – wh = 0, since gijeh|eh| = 1

1 i j (e | , + e | , ) e |e | – w = 0; from (3) 2 ni j n j i h h h

1 i j i j (e | , e |e | + e | , e |e |) – w = 0 2 ni j h h n j i h h h 1 (g + g ) – w = 0 2 nhh nhh h wh = gnhh j (ii) Multiplying (2) by en| , we get

i j j (X ij – whgij) eh| en| + rh en| j en| = 0

i j i j X ijeh|en| – whgijeh|en| + rh ´1 = 0 from (1)

1 i j i j (en|i, j + en| j,i ) eh|eh| + rh = 0, since e e = 0 2 h| h|

1 i j i j (e | , e | e | + e | , e | e |) + r = 0 2 ni j h n n j i h h h 1 (g + g ) + r = 0 2 nhn nnh h 1 r = – g , since g = 0 h 2 nhn nnh 1 or r = g h 2 hnn

(iii) If en| is a geodesic congruence, then ghnn = 0 from the result (ii), we have 1 r = ´ 0 h 2 rh = 0 Ricci's Coefficients of Rotation and Congruence 217

from equation (2), we get i i (Xij –whgij) eh| + 0en| j = 0 or (X ij – whgij ) eh| = 0 i this gives (Xij – wgij ) e = 0 .

EXAMPLE 2 Prove that when a manifold admits an orthogonal systyem of n normal congruences then any of these in canonical with respect to each other congruence of the system. solution i Let eh| (h = 1, 2, …, n) be unit tangents to n normal congruences of an orthogonal ennuple in a Vn. So that glhk = 0, where l, h, k = 1, 2,...n such that l, h, k being unequal. It is required to show that a congruence eh| is canonical with respect to the congruence ek| (h, k = 1, 2, …, h ¹ k). We know that the n – 1 congruence en| of an orthogonal ennuple be canonical to en| iff gnhk + gnkh = 0 This condition is satisfied by virtue of equation (1). Hence (n – 1) congruences eh| of an orthogonal ennuple are canonical to en|. Similarly we can show that any n – 1 congruences are canonical to the remaining congruence. It follows from the above results that any one congruence is canonical with respect to each other congruence of the system.

EXERCISE

1. If f is a scalar invariant n i i å f,ij eh|eh| = D2f h=1

n – 1 2 2. The coefficient of r in the expansion of the determinant |f,ij – rgij| is equal to Ñ f.

3. If eh| are the unit tangents to n mutually orthogonal normal congruences and e11+ be21 is also a normal congruence then ae11 – be21 is a normal congruence. 4. Show that ¶ ¶f ¶ ¶Q ¶Q – (g – g ) = å lhk lkh ¶s ¶sh ¶sk ¶sk ¶sh l l

where Sh denotes the arc length of a curve through a point P of an ennuple and Q is a scalar invariant.

5. If the congruence eh| (h = 1, 2, … n – 1) of an orthogonal ennuple are normal, prove that they are canonical with respect to other congruence eh| . CHAPTER – 12

HYPERSURFACES

12.1 INTRODUCTION

We have already studied (Art 10.3 chapter 10) that if m > n then we call Vn to be a subspace of Vm and consequently Vm is called enveloping space of Vn. We also know that there are m – n linearly independent a normals N to Vn. (Art 10.3 Theorem 10.5, chapter 10). If we take m = n + 1 then Vn is said to be hypersurface of the enveloping space Vn + 1 i 2 Let Vn be Riemannian space of n dimensions referred to cordinates x and having the metric ds = i j a gij dx dx . Let Vn be Riemannian space of m dimensions referred to coordinates y and having the metric 2 a b ds = aab dy dy . Where m > n. Let Greek letters a, b, g … take the values 1, 2, …, m and latin indices i, j, k, … take the values 1, 2, …n. Then we have, the relation between aab and gij ¶ya ¶yb gij = aab … (1) ¶xi ¶x j a i Since the function y are invariants for transformations of the coordinates x in Vn, their first covariant derivatives with respect to the metric of Vn are the same as their ordinary derivatives with respect to the variables xi. ¶ya i.e., ya = ,i ¶xi Then equation (1) can be written as a b gij = aab y,i y, j … (2)

a The vector of Vn + i whose contravariant components are y,i is tangential to the curve of i a parameter x in Vn. Consequently if N are the contravariant components of the unit vector normal to Vn. Then we have b a aabN y,i = 0, (i = 1, 2, …, n) … (3) a b and aabN N = 1 … (4) Hypersurface 219

12.2GENERALISED COVARIANT DIFFERENTIATION

Let C be any curve in Vn and s its arc length. The along this curve the x's and the y's may be expressed b as function of s only. Let ua and n be the components in the y's of two unit vector fields which are i parallel along C with respect to Vm. Similarly w the components in x's of a unit vector field which is parallel along C with respect to Vn. Now, ua is parallel along C relative to Vm then we have, dyb ua,b = 0 ds é¶u ì g üù dyb a – ê b ug í ýú = 0 ë ¶y îabþû ds

¶ua dyb ì g ü dyb – b ug í ý = 0 ¶y ds îabþ ds

b dua ì g ü dy – ug í ý = 0 ds îabþ ds

ì a ü dyb dua = ug í ý … (1) ds îb gþ ds b similarly v is parallel along C relative to Vm then

b a dv g ì g ü dy = – v í ý … (2) ds îa bþ ds i and w is parallel along C relative to Vn then k dwi j ì i ü dx = – w í ý … (3) ds î j kþ ds

The Christoffel symbol with Greek indices being formed with respect to the aab and the y's and christoffel symbol with Latin indices with respect to the gij and the x's. Let Aa be a , defined along C, which is mixed tenser of the second order in the y's and bi a b i Abi a covariant vector in the x's. Then the product ua v w is scalar invariant and it is a function of s along c. Its derivative with respect to s is also a scalar invariant. a b i Abi Differentiating ua v w with respect to s, we have

a b i d b i a dA du dv dw ( ) b i bi a b i a i a b a uav w Abi = u v w + v w A + ua w Ab + uav Ab dt a ds ds bi ds i ds i

dAa d j b i bi dud b i d dv i a dw b a = uav w + v w Abi + uaw Adi + uav Ab ds ds ds ds j 220 Tensors and Their Applications

a g g b i dAbi d b i ì a ü dy a b i ì d ü dy = uav w + Abi.uav w í ý – Adi.uav w í ý ds îdg þ ds îbgþ ds k a b i ì j ü dx – Abj ua v w í ý , by equation (1), (2) & (3). îikþ ds

é a g g k ù b i dAbi d ì a ü dy a ì d ü dy a ì j ü dx = uav w ê + Abi.í ý – Adi í ý – Abj.í ý ú ëê ds îd gþ ds îbgþ ds îi k þ ds ûú = Scalar, along C b i Since the outer product uav w is a tensor and hence from quotient law that the expression within a a the bracket is tenser of the type Abi and this tensor is called intrinsic derivative of Abi with respect to s. The expression within the bracket can also be expressed as

k é a ù dx ¶Abi d ì a ü g a ì d ü g a ì j ü ê + Abií ý y – Adi í ý y – Abj í ýú ds ê k dg ,k bg ,k ik ú ë ¶x î þ î þ î þû

dxk Since is arbitrary.. ds So, by Quotient law the expression within the bracket is a tensor and is called tensor derivative of a k a Abi with respect to x . It is denoted by Abi;k . Then we have

¶Aa ì a ü ì d ü ì j ü a bi d g a g a Abi;k = + Abi í ýy,k – Adi í ýy,k – Abj í ý. dxk îdg þ îbg þ îikþ

a a k Abi;k is also defined as generalised covariant derivative Abi with respect to x . Note:– Semi-colon(;) is used to denote tensor differentiation.

12.3 LAWS OF TENSOR DIFFERENTIATION

THEOREM 12.1 Tensor differentiation of sums and products obeys the ordinary rules of differentiation. a a Proof: Suppose Ab , Bb and Bg are tensors in Vm. (i) To prove that a a a a (Ab + Bb );k = Ab;k + Bb;k a a a Let the sum Ab + Bb be denoted by the tensor Cb . Now,

¶C a ì a ü ì a ü a b a c a c C = + Cb í ý y,k –Ca í ý y,k b;k ¶xk îa cþ îbcþ Hypersurface 221

¶(Aa + Ba) ì a ü ì a ü a a b b + (Aa + Aa) yc (Aa + Ba ) y c \ (A + B ) = k b b í ý ,k – a a í ý ,k b b ;k dx îa cþ îb cþ

é a ù ¶Ab aì a ü c a ì a ü c ê + Ab í ý y,k – Aa í ýy,k ú = ê k ac bc ú + ë ¶x î þ î þ û

é a ù ¶Bb aì a ü c aì a ü c ê + Bb í ý y,k – Ba í ý y,k ú ê k ac bc ú ë ¶x î þ î þ û

a a a a (Ab + Bb );k = Ab;a + Bb;k Hence the result (i) (ii) Prove that

a a a (Ab + Bg ); k = Ab;k Bg + Ab Bg;k

a a a Let Ab Bg = Dbg Then Dbg is a tensor we have

a ¶Dbg ì a ü ì a ü ì a ü a + Da c a c a c D = k ;k í ý y – Dag í ýy,k – Dba í ýy,k bg;k ¶x îacþ îbcþ îgcþ

a ¶(Ab Bg ) ì a ü ì a ü ì a ü a + Aa B y c a c a c \ ( Ab Bg );k = k bg g í ý ,k – Aa Bg í ýy,k – Ab Ba í ýy,k dx îacþ îbcþ îbcþ

é¶Aa ì a ü ì ü ù b a c a a c aé ¶Bg ì a ü c ù ê + Ab í ý y,k – Aa í ýy,k úBg A ê – B y ú = ê k ac bc ú + b k aí ý ,k ë dx î þ î þ û ëê dx îgcþ ûú

a a a (Ab Bg);k = Ab;k Bg + Aa Bg ; k Hence the result (ii). Note:– a, b, g, a, c take values from 1 to m while k take values from 1 to n.

THEOREM 12.2 To show that aab; j = 0

or To prove that the metric tensor of the enveloping space is generalised covariant constant with respect to the Christoffel symbol of the subspace. Proof: We have (see pg. 220)

¶aab ì g ü d ì g ü d aab;i = – agbí ý y,i – aagí ýy,i dxi îabþ îbdþ

¶aab d = – ( [ad,b] + [bd,a] ) y, dxi i 222 Tensors and Their Applications

¶aab ¶aab d or a = – y,i ab;i dxi ¶yd

¶a ¶a ¶y ¶a ¶a ab – ab × d ab ab or aab; = d = – = 0 Proved. i dxi ¶y dxi ¶xi ¶xi

12.4 GAUSS'S FORMULA

At a point of a hypersurface Vn of a Riemannian space Vn + 1, the formula of Gauss are given by a a y;ij = Wij N Proof: Since ya is an invariant for transformation of the x's and its tensor derivative is the same as its covariant derivative with respect to the x’s, so that

dya ya = ya = … (1) ;i ,i dxi Again tenser derivative of equation (1) with respect to x’s is a ( a ) a y;ij = y;i ; j = ( y,i )

¶ a a ì l ü b g ì a ü = ( y,i ) – y,l í ý + y,i y, j í ý dx j îijþ îbgþ

¶ æ ¶ya ö ì l ü ì a ü = ç ÷ – ya + ybyg j ç i ÷ ,l í ý ,i , j í ý ¶x è ¶x ø îijþ îbg þ

2 a l a a ¶ y a ì ü b g ì ü y = – y,l í ý + y,i y, j í ý … (2) ;ij ¶xi¶x j îijþ îbgþ Interchanging j and i in (2), we have ¶ 2y a ì l ü ì a ü a – y a + yb y g y; = j i ,l í ý , j ,i í ý ji ¶x ¶x î jiþ îbgþ

¶ 2y a ìl ü ì a ü a – y a + yb y g y; = i j ,l í ý ,l , j í ý … (3) ji ¶x ¶x îijþ îbgþ

ì a ü ì a ü [On interchanging b and g in third term of R.H.S. of (3) and using í ý = í ý ]. îbg þ îgbþ On comparing equation (2) and (3), we have a a y;ij = y; ji

a So, y;ij is symmetrical with respect to indices i & j. Hypersurface 223

i j a b Let gijdx dx and aabdy dy be fundamental forms corresponding to Vn and Vn + 1 respectively.. Then

¶ya ¶yb g = aab ij ¶xi ¶x j

a b or gij = aab y;i y; j Taking tensor derivative of both sides with respect to xk a yayb + a ya yb + a ya yb gij;k = ab;k ;i ; j ab ;i ; jk ab ;ik ; j

But gij;k = 0 and aab;k = 0. we have,

a b a b 0 = aab y;i y; jk + aab y;ik y; j

a b a b or aab y;ik y, j + aab y,i y; jk = 0, using (1), Art. 12.4 … (4) By cyclic permutation on i, j, k in (4), we have

a b a b aab y;ji y,k + aab y, j y;ki = 0 … (5)

a b a b and aab y;kj y,i + aab y,k y;ij = 0 … (6) subtracting equation (4) from the sum of (5) and (6), we get.

a b 2aab y;ijy,k = 0

a b or aab y;ij y,k = 0 … (7)

a b b a This shows that y;ij is normal (orthogonal) to y,k . Since y,k is tangential to Vn and hence y;ij is normal to Vn. Then we can write a a y;ij = N Wij (8) a where N is unit vector normal to Vn and W ij is a symmetric covariant tensor of rank two. Since a y;ij is a function of x's the tenser Wij is also a function of x's. The equation (8) are called Gauss’s formula. From equation (8) a a y;ij = Wij N

a b b a y;ij aabN = Wij aab N N

2 = Wij N 224 Tensors and Their Applications

a b y;ij aab N = Wij, N = 1

or W a b ij = y;ijaabN . The quadratic differential form i j Wijdx dx is called the second fundamental form for the hypersurface Vn of Vn + 1. The components of tenser Wij are said to be coefficient of second fundamental form. i j Note: The quadratic differential form gijdx dx is called first fundamental form.

12.5 CURVATURE OF A CURVE IN A HYPERSURFACE AND NORMAL CURVATURE a i If U and u be the contravariant components of the vector u relative to Vn and Vn + 1 respectively then we have (from chapter 10, Theorem 10.4)

a ¶y i a = u = yaui … (1) U ¶xi ,i

Let the derived vector to vector u along C with respect to metric of Vn and Vn + 1 are denoted by p and q respectively. Then

j i i d x p = u , j ds b a a d y and q = U,b ds j a dx = U; , (from equation 1, Art 12.4) … (2) j ds Taking the tensor derivative of each side of equation (1) with respect to x's, we have a a i a i U; j = y;ij u + y,i u, j

a a By Gauss's formula, we have y;ij = WijN

a a i a i Then U; j = Wij N u + y,i u, j

a Putting the value of u; j in equation (2), we have

dx j a a i a i q = (Wij N u + y,i u, j) ds æ j ö ç i dx ÷ a a i or qa = ç Wij u ÷ N + y,i p … (3) è ds ø j i i dx where p = u, j ds Hypersurface 225

Now suppose that vector u is a the unit tangent t to the curve C. Then the derived vectors q and p are the curvature vectors of a C relatively to Vn + 1 and Vn respectively. Then equation (3) becomes. æ i j ö ç dx dx ÷ a a i qa = ç Wij ÷ N + y,i p … (4) è ds ds ø

dxi dx j Taking K = W , we get n ij ds ds

a a a i q = Kn N + y,i p ...(5) a Kn is called Normal curvature of Vn at any point P of the curve C and Kn N is called normal curvature vector of Vn + 1 in the direction of C. Meaunier's Theorem

If Ka and Kn are the first curvature of C relative to Vn + 1 and normal curvature of Vn respectively r r and w is the angle between N and C (C being the unit vector of Vn + 1 then the relation between Ka, Kn and w is given by

Kn = Ka cosw Proof: We know that a a i a q = KnN + p y,i ...(1) dxi dx j where K = W n ij d s d s Let Ka and Kg be the first curvatures of C with respect Vn + 1 and Vn respectively then

a b i j Ka = aabq q , Kg = gij p p r r Let w be the angle between N and C . r r Then N ×C = N × C cos w = cosw as |N| = |C| = 1 r r N.C = cosw ...(2) r r If b is a unit vector of Vn + 1 in the direction principal normal of C with respect to Vn then equation (1) becomes r r r KaC = K gb + Kn N … (3) r Taking scalar product of equation (3) with N , we have r r r r r r Ka N .C = K gN.b + KnN.N

Ka cosw = Kg × 0 + Kn ×1; from (2)

Kn = Ka cosw … (4) Proved. 226 Tensors and Their Applications

EXAMPLE 1 Show that the normal curvature is the difference of squares of geodesic curvatures. Solution We know that (from Meurier’s Theorem, equation 3) r r r KaC = Kgb + KnN Taking modulus of both sides, we get

2 2 2 Ka = Kg + Kn

2 2 2 or Kn = Ka – Kg .

Theorem 12.3 To show that the first curvature in Vn + 1 of a geodesic of the hypersurface Vn is the normal curvature of the hypersurface in the direction of the geodesic. Proof: From, example 1, we have

2 2 2 Ka = Kg + Kn i If C is a geodesic of Vn then p = 0 Þ Kg = 0 Then we have 2 2 Ka = Kn

Þ Ka = Kn. Proved.

Dupin's Theorem

The sum of normal curvatures of a hypersurface Vn for n mutually orthogonal directions is an ij invariant and equal to W ijg . i Proof: Let eh| (h = 1, 2, …, n) be unit tangents to n congruences of an orthogonal ennuple in a Vn. Let

Knh be normal curvature of the hypersurface Vn in the direction of the congruence eh| . Then i j Knh = Wij eh| eh| The sum of normal curvatures for n mutually orthogonal directions of an orthogonal ennuple is a Vn is

n n K W ei e j å nh = å ij h| h| h=1 h=1

n W ei e j = ij å h| h| h=1 ij = Wij g = Scalar invariant Proved. Hypersurface 227

12.6 DEFINITIONS

(a) First curvature (or mean curvature) of the hypersurface Vn at point P. It is defined as the sum of normal curvatures of a hypersurface Vn form mutually orthogonal directions at P and is denoted by M. Then ij M = Wij g (b) Minimal Hypersurface

The hypersurface Vn is said to be minimal if M = 0 ij i.e., Wij g = 0 (c) Principle normal curvatures

The maximum and minimum values of Kn are said to be the principle normal curvatures of Vn at P. Since these maximum and minimum values of Kn correspond to the principal directions of the symmetric tensor Wij . (d) Principal directions of the hypersurface at a point P.

The principal directions determined by the symmetric tensor Wij at P are said to be principal directions of the hypersurface at P.

(e) Line of curvature in Vn A line of curvature in a hypersurface Vn is a curve such that its direction at any point is a principal direction.

Hence we have n congruences of lines of curvature of a Vn. THEOREM 12.4 To show that the mean curvature of a hypersurface is equal to the negative of the divergence of the unit normal. or To show that the first curvature of a hypersurface is equal to the negative of the divergence of the unit normal. or To show that the normal curvature of a hypersurface for any direction is the negative of the tendency of the unit normal in that direction. Proof: Let N be the unit normal vector to the hypersurface Vn in y's and let Na be its covariant components. Let t be the unit tangent vector to N congruences eh| (h = 1, 2, …, n) of an orthogonal a ennuple in Vn and let Th| be the contravariant components in Vn + 1 of t. Since t is orthogonal to N, therefore a Th| Na = 0 … (1)

Taking covariant derivative of equation (1) with regard to yb provides

a a Th|,b Na + Th| Na,b = 0 … (2) b Multiplying equation (2) by Th , we get a b a b Th|,b Th| N a + Th| Th| Na,b = 0 228 Tensors and Their Applications

a b b a (Th|,bTh| )Na = – Na,b Th| Th| … (3)

Ta T b Tb a Now h|,b h| is the first curvature of the curve eh| and Na,b . h| . Th| is the tendency of Na is the direction of eh| . Hence equation (3) implies that the normal component of the first curvature of the eh| relative to Vn + 1 or the normal curvature of Vn in the direction of the curve eh|

= – tendency of N in the direction of the curve eh| … (4) Taking summation of both of (3) and (4) for h = 1, 2, …, n we have i.e., mean curvature or first curvature of a hypersurface = – divergence of the unit normal. Corollary: To prove that

M = – divn + 1N Proof: Since N is a vector of unit magnitude i.e., constant magnitude, its tendency is zero. also by definition, the divn + 1 N and divn N differ only by the tendency of the vector N in its direction. But the tendency of N is zero hence it follows that divn N = – div n + 1 N Hence M = – div n N = – div n + 1 N.

12.7 EULER'S THEOREM

Statement r The normal curvature Kn of Vn for any direction of a in Vn is given by n 2 Khcos ah Kn = å h=1 r where Kh are the and ah are the angles between direction of a and the congruence eh| Proof: The principal directions in Vn determined by the symmetric covariant tensor teser W ij are given by

i (Wij – Kh gij) ph| = 0 … (1)

where Kh are the roots of the equation

Wij – Kgij = 0 … (2)

i and ph| are the unit tangents to n congruences of lines of curvature.

The roots of the equation (2) are the maximum and minimum values of the quality Kn defined by i j Wij p p Kn = i i … (3) gij p p Hypersurface 229

i Multiplying equation (1) by p k| , (K ¹ h), we get i i (Wij – Kngij ) ph| pk| = 0 The principal directions satisfy the equation i j Wij ph| pk| = 0 (h ¹ k) … (4)

Let eh| be the unit tangents to the n congruences of lines of curvature. Then the principal curvatures are given by i i Kh = Wij eh| eh| (h = 1, 2, …, n) … (5) r Any other unit vector a in Vn is expressible in the form e | cosa ar = å h h h n i ei cosa or a = å h| h … (6) h=1 r r i where cosah| = a × eh| = a eh|i i r a being the contravariant components of a and ah being the inclination of vector a to eh|. The r normal curvature of Vn for the direction of a is given by i j Kn = Wija a from (6), we get æ ö æ ö W ç ei cos a ÷ ç e j cos a ÷ Kn = ij ç å h| h÷ çå k| h÷ è h ø è h ø

n i j = å (Wijeh|ek|) cosah cosak h,k =1

n (W ei e j )cos2 a = å ij h| h| h h=1 n 2 Khcos ah Kn = å … (7) h=1 This is a generalisation of Euler’s Theorem.

12.8 CONJUGATE DIRECTIONS AND ASYMPTOTIC DIRECTIONS IN A HYPERSURFACE r r The directions of two vector, a and b , at a point in Vn are said to be conjugate if i j Wija b = 0 … (1) and two congruences of curves in the hypersurface are said to be conjugate if the directions of the two curves through any point are conjugate. 230 Tensors and Their Applications

A direction in Vn which is self-conjugate is said to be asymptotic and the curves whose direction are along asymptotic directions are called asymptotic lines. r Therefore the direction the vector a at a point of Vn be asymptotic if i j Wij a a = 0 … (2) The asymptotic lines at a point of a hypersurface satisfies the differential equation

i j Wij dx dx = 0 … (3)

THEOREM 12.5 If a curve C in a hypersurface Vn has any two of the following properties it has the third

(i) it is a geodesic in the hypersurface Vn (ii) it is a geodesic in the enveloping space Vn + 1 (iii) it is an asymptotic line in the hypersurface Vn. Proof: Let C be a curve in the hypersurface Vn. The normal curvature Kn of the hypersurface Vn in the direction of C is given by

2 2 2 Ka = Kg + Kn … (1) where Ka and Kg are the first curvatures of C relative to enveloping space Vn + 1 and hypersurface Vn respectively.

Suppose C is a geodesic in the hypersurface Vn [i.e., (i) holds] then Kg = 0. If C is also a geodesic in the enveloping space Vn + 1 [i.e., (ii) holds] then Ka = 0. Now, using these values in equation (1), we have 2 Kn = 0 Þ Kn = 0

Implies that C is an asymptotic line is the hypersurface Vn i.e., (iii) holds. Hence we have proved that (i) and (ii) Þ (iii) Similarly we have proved that (ii) and (iii) Þ (i) and (i) and (iii) Þ (ii)

12.9 TENSOR DERIVATIVE OF THE UNIT NORMAL a i The function y are invariants for transformations of the coordinates x in Vn their first covariant derivatives with respect to the metric of Vn are the same as their ordinary derivatives with respect to the variables xi.

a a a ¶y i.e., y = y, = … (1) ;i i ¶xi a The unit Normal N be the contravariant vector in the y’s whose tensor derivative with respect to the x’s is ¶N a ì a ü a + N b yd N; = i í ý ,i … (2) i ¶x îbdþ Hypersurface 231

b a Since aabN N = 1 … (3) Tensor derivative of this equation with respect to xi gives a b a b aabN N;i + aabN;i N = 0 Interchanging a and b in Ist term, we get

b a a b or aba N N;i + aab N;i N = 0

b a a b aabN N;i + aab N;i N = 0

b a 2aab N N;i = 0

b a aab N N;i = 0 … (4)

a which shows that N;i is orthogonal to the normal and therefore tangential to the hypersurface. a a Thus N;i can be expressed in terms to tangential vectors y,k to Vn so that a a a N;i = Ai y,k … (5) k where Ai is a mixed tensor of second order in Vn to be determined. a a Since unit normal N is orthogonal to tangential vector y,i in Vn. Then a b aabN y,i = 0 Taking tensor derivative with respect to x j, we get

a b a b b b aab N;j y,i + aab N y,ij = 0 since y,ij = Wij N

a b a b or aab N;j y,i + aab N Wij N = 0

a b a b or aab N;j y,i + Wij(aab N N ) = 0

a b from equation (3), aab N N = 1

a b or aab N;j y,i + Wij = 0

k or gki Aj + Wij = 0

é ¶y a ¶yb ù since a ya ya = a = g ê ab ,k ,i ab k i ki ú ë ¶x ¶x û Multiplying this equation by gim, we get

im k im gki g Aj + Wij g = 0 232 Tensors and Their Applications

m k im dk Aj + Wijg = 0

m im Ai + Wij g = 0

m im Ai = – Wij g Substituting this value in equation (5), we get

a ik a jk a N;i = Wij g y,k = – W ji g y,k

a jk a N;i = – Wij g y,k … (6)

a This is the required expression for the tensor derivative of N . Theorem 12.6 The derived vector of the unit normal with respect to the enveloping space, along a curve provided it be a line of curvature of the hypersurface. a Proof: Since the tensor derivative of N is

a jk a N;i = – Wij g y,k … (1)

Consider a unit vector ei tangential to the curve C.

a i jk a i Then N;i e = – Wij g y,k e … (2)

a i i The direction of N;i e is identical with that of e a i a i Then N;i e = – ly,i e , (l is scalar constant) from (2), we have

jk a i a i Wij g y,ke = ly,i e

b Multiplying both sides of this equation by aaby,l

jk i a b i a b Wij g e (aab y,k y,l ) = le (aab y,i y,l )

jk i i a b Wijg e gkl = le gil since gil = aab y,i y,l

i i i Wijdle = le gil i i Wile – le gil = 0

i (Wil – lgil ) e = 0 (l = 1, 2, …, n) i This equation implies that the direction of e is a principal direction for the symmetric tensor W il i i.e., e is a principal direction for the hypersurface Vn. Hence by definition the curve C is a line of curvature Vn. Hypersurface 233

12.10 THE EQUATION OF GAUSS AND CODAZZI since we know that (pg. 86, equation 5) p Ai, jk – Ai,kj = ApRijk … (1) where Ai is a covariant tenser of rank one and difference of two tensers Ai, jk – Ai,kj is a covariant tenser of rank three.

a ' a It y,i are components of a covariant tensor of rank one in x s. Then replacing Ai by y,i in equation (1), we get

a a a p a ph p ph y,ijk – y,ikj = y,p Rijk = y,pg Rhijk , Since Rijk = g Rhijk … (2) where Rhijk are Riemann symbols for the tensor gij We know that a a y,ij = Wij N … (3)

a ik a and N,i = – Wij g y,k … (4) a Let Rgde are Riemann symbols for the tensor aab and evaluated at points of the hypersurface using equation (3) and (4), equation (2) becomes

a ph a a g d e y, p g [Rhijk – (Whj Wik – Whk Wij)] – N (Wij,k – Wik, j ) – Rgde y,i y, j y,k … (5)

b Multiplying equation (5) by aaby,l and summed with respect to a. Using the relations

a b aab y,i y,l = 0

b a and gab = N y,l = 0, we get b g d e Rlijk = (WljWik – WlkWij ) + Rbgde y,l y,i y, j y,k … (6) b Multiplying (6) by aabN and summing with respect to a. Using relations b a aab N y,l = 0

b b and aab N B = 0

b g d e we get Wij,k – Wik, j + Rbgde N y,i y, j y,k = 0 … (7) Hence, The equation (6) are generalisation of the Gauss Characteristic equation and equation (7) of the Mainardi-Codazzi equations. 234 Tensors and Their Applications

12.11 HYPERSURFACES WITH INDETERMINATE LINES OF CURVATURE A point of a hypersurface at which the lines of curvature are indeterminate is called an Umbilical Point. The lines of curvature may be indeterminate at every point of the hypersurface iff

Wij = wgij … (8) where w is an invariant The mean curvature M of such a hypersurface is given by

ij ij M = Wijg = w gijg = wn M Þ w = . n So that the conditions for indeterminate lines of curvature are expressible as M W = gij , from (i) … (9) ij n

If all the geodesics of a hypersurface Vn are also geodesics of an enveloping Vn + 1. They hypersurface Vn is called a totally geodesic hypersurface of the hypersurface Vn + 1. THEOREM 12.7 A totally geodesic hypersurface is a minimal hypersurface and its-lines of curvature are indeterminate. Proof: We know that 2 2 2 Ka = Kn + Kg … (1) and a hypersurface is said to be minimal if M = 0 … (2) and the lines of curvature are indeterminate if M W = g … (3) ij n ij

If a hypersurface Vn is totally geodesic then geodesics of Vn are also geodesics of Vn + 1. i.e., Ka = 0 = kg Now, from (1), we have

Kn = 0 But normal curvature Kn is zero for an asymptotic direction. Hence a hypersurface Vn is totally geodesic hypersurface iff the normal curvature Kn zero for all directions in Vn and hence

Wij = 0

ij M = Wij g = 0 i.e., equation (2) is satisfied. Hence, the totally geodesic hypersurface is minimal hypersurface. In this case equation (3) are satisfied hence the lines of curvature are indeterminate. Hypersurface 235

12.12 CENTRAL QUADRATIC HYPERSURFACE i Let x be the cartesian in Euclidean space Sn, so that the components gij of the fundamental tensor are constants. Let yi be the Riemmannian coordinates. If a fixed point O is taken as a pole and s the distance of any point P then Riemannian coordinates yi of P with pole O are given by yi = sx i … (1) xi is unit tangent in the direction of OP. ¢ Let aij be the components in the x s of a symmetric tensor of the rank two and evaluated at the pole O. Then the equation

i j y aijy = 1 … (2) represents a central quadratic hypersurface Substituting the value of equation (1) in equation (2), we get i i sx aijsx = 1 1 xia xi = … (3) ij s2 The equation (3) showing that the two values of s are equal in magnitude but opposite in sign. The positive value of s given by equation (3) is the length of the radius of the quadric (2) for the direction xi .

THEOREM 12.8 The sum of the inverse squares of the radii of the quadric for n mutually orthogonal ij directions at O is an invariant equal to aij g . i Proof: If eh| , (h = 1, 2, …n) are the contravariant components of the unit tangents at O to the curves i of an orthogonal ennuple in Sn. The radius Sh relative to the direction eh| is given by

i j 1 aij eh| eh| = 2 sh n h s –2 a ei e j or å ( h) = å ij h| h| h=1 h=1

n a ei e j = ijå h| h| h=1

n (s )–2 ij å h = aij g Proved. h=1 236 Tensors and Their Applications

THEOREM 12.9 The equation of hyperplane of contact of the tangent hypercone with vertex at the point Q(y ). Proof: Given (from equation 2, pg. 235) i j aij y y = 1 Differentiating it

i j i j aij dy y + aij y dy = 0

i j j i or aij dy y + aij y dy = 0

i j 2aij dy y = 0

i j aij dy y = 0 This shows that dyi is tangential to the quadric. Hence yj is normal to the quadric. The tangent hyperplane at the point P ( y j) is given by

i i j (Y – y )aij y = 0

i j i j aijY y = aij y y

i j i j aijY y = 1 since aij y y = 1 … (4) This equation represents the equation of tangent hyperplane at P ( y i) . If the tangent hyperplane P ( y j) passes through the point Q (y–i). Then we have

i j y aij y = 1 … (5) Thus all points of the hyperquadric, the tangent hyperplanes at which pass through Q lie on the hyperplane (5) on which yi is the current point. This is the hyperplane of contact of the tangent hypercone whose vertex is Q(y i) .

12.13 POLAR HYPERPLANE The polar hyperplane of the point R(~y i) with respect to quadric (2) is the locus of the vertices of the hypercones which touch the hyperquadric along its intersections with hyperplanes through R. If Q(y i) is the vertex of such tangent hypercone, then R lies on the hyperplane of contact of Q so that i ~ j y aij y = 1 Consequently for all positions of the hyperplane through R, Q lies on the hyperplane i ~ j y aij y = 1 … (6) This is required equations of the polar hyperplane of R and R is the pole of this hyperplane. Hypersurface 237

12.14 EVOLUTE OF A HYPERSURFACE IN EUCLIDEAN SPACE i Consider a hypersurface Vn of Euclidean space Sn + 1 and let x (i = 1, 2, … n) be coordinates of an arbitrary point P of Vn whose components relative to Sn + 1 are a y (a = 1,2,Ln +1) a a Let N be a unit normal vector at P relative to Sn + 1 so that tensor derivate N becomes covariant derivative. a a jk a So, N;i = N,i = – Wij g y,k … (1) a a a y;ij = y,ij = Wij N … (2) n+1 ya ya and gij = å ,i , j … (3) a=1 Let P(y a ) be a point on the unit normal Na such that distance of P from P is r in the direction of Na such that a a ya = y + rN … (4)

i a Suppose P undergoes a displacement dx in Vn then the corresponding displacement dy of P is given by

a a a i a d y = ( y,i + rN,i )dx + N dr … (5)

a a i a The vector ( y,i + rN,i ) dx is tangential to Vn whereas N dr is a normal vector. Therefore if the displacement of P(y a ) be along the normal to the hypersurface then we have

a a i ( y,i – rN,i 0) dx = 0 … (6) Using equation (1) in equation (6), we get

a jk a i ( y,i – rg y,k )dx = 0

a Multiplying it by y,i and summing with respect to a, we get, using equation (3), as

jk i (gil – r Wijg glk )d x = 0

j i (gil – rWijdl ) d x = 0

i (gil – rWil ) dx = 0

i (gij – rWij) d x = 0 … (7) This shows that the directions dxi given by equation (7) are principal directions of the hypersurface where the roots r of the equation |gij – rWij| = 0 are called principal radii of normal curvature. The a locus of P(y ) satisfying the condition (4) is called evolute of the hypersurface Vn of Sn + 1 where r is a root of (7). The evolute is also a hypersurface of Sn + 1. 238 Tensors and Their Applications

12.15 HYPERSPHERE

The locus of a point in Sn which moves in such way that it is always at a fixed distance R from a fixed point C(ba ) is called a hypersphere of radius R and centre C. Therefore the equation of such a hypersphere is given by

n a a 2 å (y – b ) = R2 … (1) a=1

THEOREM 12.10 The Riemannian curvatrure of a hypersphere of radius R is constant and equal to 1 . R2 Proof: Let the hypersurface be a Vn of Sn + 1 and let its centre be taken as origin of Euclidean coordinates in Sn + 1. Then the hypersphere is given by a 2 å (y ) = R2, (a = 1, 2, …, n + 1) … (2) a i For the point in Vn the y's are functions of the coordinates x on the hypersphere. Differentiating equation (2) with respect to xi, we get ya ya å ,i = 0 … (3) a and again differentiating it with respect to x j, we get. y a ya + ya ya å , j ,i å ,ij = 0 … (4) a a By Gauss formula, a a y,ij = Wij N … (5) Using (5), equation (4) becomes

a a gij + å y Wij N = 0 … (6) a a a a From equation (3) it follows that y,i is perpendicular to y . But y,i is tangential to Vm. Hence a a y is normal to Vn.. The equation (2) implies that the components of the unit vector N are given by

ya a = Þ ya = RNa … (7) N R Using (7), equation (6) becomes

g + R N aN a W ij å ij = 0 a

g + RW (N a )2 or ij ijå = 0 a

a 2 or gij + RWij = 0 since (N ) = 1

240 Tensors and Their Applications

2. To prove that for a space Vn with positive constant Riemannian curvature K these exists sets of n + 1 real coordinate ya satisfying the condition

n+1 a 1 1 å( y ) = where R2 = a=1 K K Proof: Using (9) in equation (1), we get (y a)2 1 å = Proved. a K

EXAMPLE 2 Show that the directions of two lines of curvature at a point of a hypersurface are conjugate. Solution

i The principal direction eh| are given by

i j Wij eh| eh| = 0 … (1) The shows that principal directions at a point of a hypersurface are conjugate. Thus we say that two congruences of lines of curvature are conjugate.

EXAMPLE 3

Show that the normal curvature of hyper surface Vn in an asymptotic direction vanishes. Solution

Let us consider a curve C in a Vn. If C is an asymptotic line then it satisfies the differential equation i j Wij dx dx = 0 dxi dx j i.e., W = 0 … (1) ij ds ds Now the normal curvature Kn of the hypersurface Vn in an asymptotic direction in a Vn is given by dxi dx j K = W n ij ds ds i.e., Kn = 0 from (1)

EXAMPLE 4 To prove that if the polar hyperplane of the point R passes through a point P then that of P passes through R. Solution

Let P(yi ) and R(yi ) be two points. The polar hyperplane of R(yi ) is

242 Tensors and Their Applications

11. Prove that the necessary and sufficient condition that system of hypersurface with unit normal N be isothermic is that r(N div N – N.ÑN ) = 0. 12. Show that the normal curvature of a subspace in an asymptotic direction is zero.

13. If straight line through a point P in Sn meets a hyperquadric in A and B and the polar hyperplane of P in Q prove that P, Q are harmonic conjugates to A, B.

14. What are evolutes of a hypersurface in an Euclidean space. Show that the varieties r1 = constant in

evolute are parallel, having the curves of parameter r1 as orthogonal geodesics of Vn. INDEX

A Covariant tensor of rank two 9 Covariant vector 7 Absolute 131 Curl 76 Addition and Subtraction of Tensors 15 Curl of congruence 211 Associated tensor 43 Curvature 136 Asymptotic directions 229 Curvature of Congruence 207 B Curvature tensor 86 Bianchi identity 94 D Binormal 137 Degree of freedom 157 C Dextral Index 1 Divergence 75 Canonical congruence 213 Divergence Theorem 161 Christoffel’s symbols 55 Dummy index 1 Completely skew-symmetric 111 Dupin’s Theorem 226 Completely symmetric 111 Concept 188 E Congruence of curves 49 Einstein space 103 Conjugate (or Reciprocal) Symmetric Tensor 25 Einstein tensor 95 Conjugate directions 229 Einstein’s Summation Convention 1 Conjugate metric tensor 34 Euler’s condition 171 Conservative force field 144 Euler’s Theorem 228 Contraction 18 Evolute 237 Contravariant tensor of rank r 14 Contravariant Tensor of rank two 9 F Contravariant vector 7 Covariant tensor of rank s 14 First curvature 227 First curvature vector of curve 170 244 Tensors and Their Applications

First fundamental tensor 34 M Free Index 2 Magnitude of vector 44 Fundamental tensor 31 Mainardi-Codazzi equations 233 Fundamental theorem 199 Mean curvature 101 G Meaunier’s Theorem 225 Metric Tensor 31 Gauss Characteristic equation 233 Minimal curve 42 Gauss’s formula 222 Minimal Hypersurface 227, 234 Gauss’s theorem 164 mixed tensor of rank r + s 14 Generalised KrÖnecker delta 112 mixed tensor of rank two 9 Generalized coordinates 157 Geodesic congruence 208 N Geodesic coordinate system 175 Newtonian Laws 142 Geodesics 171 Normal congruence 209 Gradient 75 n-ply orthogonal system of hypersurfaces 49 Green’s Theorem 162 Null curve 42

H O Hamilton’s principle 153 Orthogonal Cartesian coordinates 120 Hypersphere 238 Orthogonal ennuple 49 Hypersurface 48, 218 Osculating plane 136 I Outer Product of Tensor 16 Inner product of two tensors 18 P Integral of energy 155 Parallel vector fields 134 Intrinsic derivative 131 Parallelism 188 K Poisson’s equation 166 Polar hyperplane 236 Kinetic energy 144 Potential energy 145 Krönecker Delta 2 Principal directions of the hypersurface 227 Principle normal vector 136 L Principle normal curvatures 227 Lagrangean equation 148 Principle of least action 156 Lagrangean function 148 Projective curvature tensor 104 Laplace’s equation 167 Laplacian operator 80, 163 Q Length of a curve 42 Quotient Law 24 Levi-civita’s concept 188 Line element 31 R Line of curvature 227 Reciprocal base systems 122 Index 245

Relative Tensor 26 Stoke’s Theorem 164 Ricci tensor 88 Straight line 140 Ricci’s principal directions 102 Subscripts 1 Ricci’s coefficients of rotation 205 Superscripts 1 Ricci’s Theorem 71 Symmetric Tensors 20 Riemann curvature 96 Riemann’s symbol 86 T Riemann-Christoffel Tensor 85 Tensor density 26 Riemannian coordinates 177 Torsion 137 Riemannian Geometry 31, 116 Totally geodesic hypersurface 234 Riemannian Metric 31 Transformation of Coordinates 6 Riemannian space 31 U S Umbral 1 Scalar product of two vectors 44 Unit principal normal 171 Schur’s theorem 100 Second fundamental tensors 34 W Serret-Frenet formula 138 Weyl tensor 104 Simple Pendulum 152 Work function W 145 Skew-Symmetric Tensor 20