Tensor-Analysis.pdf

Chapter 8 PHYSICAL LAWS must be independent of any particular coordinate systems used in describing them mathematically, if they are to be valid. A study of the consequences of this re-L quirement leads to tensor analysis, of great use in general relativity theory, differential ge...

Chapter 8 PHYSICAL LAWS must be independent of any particular coordinate systems used in describing them mathematically, if they are to be valid. A study of the consequences of this re-L quirement leads to tensor analysis, of great use in general relativity theory, differential geometry, mechanics, elasticity, hydrodynamics, electromagnetic theory and numerous other fields of science and engineering. SPACES OF N DIMENSIONS. In three dimensional space a point is a set of three numbers, called coordinates, determined by specifying a particular coordinate system or frame of reference. For example (x,y, z), (p, c,z), (r, 8, 55) are coordinates of a point in rectan- gular, cylindrical and spherical coordinate systems respectively. A point in N dimensional space is, by analogy, a set of N numbers denoted by (x1, x2,..., xN) where 1, 2,..., N are taken not as expo- nents but as superscripts, a policy which will prove useful. The fact that we cannot visualize points in spaces of dimension higher than three has of course nothing whatsoever to do with their existence. COORDINATE TRANSFORMATIONS. Let (x1, x2,..., xN) and (x1, x2,..., RN) be coordinates of a point in two different frames of reference. Suppose there exists N independent relations between the coordinates of the two systems having the form 1 _ -X'1 2 2 2 1 2 (1) xN = zN(x1, x2,..., xN) which we can indicate briefly by (2) xk = xk(x1, x2,..., xN) k = 1, 2,..., N where it is supposed that the functions involved are single-valued, continuous, and have continuous derivatives. Then conversely to each set of coordinates (x1, x2,..., xN) there will correspond a unique set (x1, x2,..., xN) given by (3) Xk = x,x,...,x) 1 2 N k = 1, 2,...,N The relations (2) or (3) define 4 transformation of coordinates from one frame of reference to another. 166 www.MathSchoolinternational.com  TENSOR ANALYSIS 167 THE SUMMATION CONVENTION. In writing an expression such as a1x1 + a2x2 +... + a1yx1 we can X use the short notation jZ1 xi. An even shorter notation is sim- ply to write it as ajxi, where we adopt the convention that whenever an index (subscript or super- script) is repeated in a given term we are to sum over that index from 1 to N unless otherwise spec- ified. This is called the summation convention. Clearly, instead of using the index j we could have used another letter, say p, and the sum could be written aoxO. Any index which is repeated in a giv- en term, so that the summation convention applies, is called a dummy index or umbral index. An index occurring only once in a given term is called a free index and can stand for any of the numbers 1, 2,..., N such as k in equation (2) or (3), each of which represents N equations. CONTRAVARIANT AND COVARIANT VECTORS. If N quantities A1, A2,..., AN in a coordinate sys- tem (x1, x2,..., x 1) are related to N other quantities A1, A2,..., ff in another coordinate system (x1, x2,..., xN) by the transformation equations A _ ax9 Aq p = 1, 2,..., N q=1 which by the conventions adopted can simply be written as A = axq axP Aq they are called components of a contravariant vector or contravariant tensor of the first rank or first order. To provide motivation for this and later transformations, see Problems 33 and 34 of Chapter 7. If N quantities A1i A2,..., AN in a coordinate system (x1, 12 ,..., x1) are related to N other quantities At, A2,..., Aff in another coordinate system (x1, x2,..., xN) by the transformation equations Ap = axp Aq p = 1, 2,..., N q=1 or AP axq A azp q they are called components of a covariant vector or covariant tensor of the first rank or first order. Note that a superscript is used to indicate contravariant components whereas a subscript is used to indicate covariant components; an exception occurs in the notation for coordinates. Instead of speaking of a tensor whose components are Ap or AP we shall often refer simply to the tensor AP or AP A. No confusion should arise from this. CONTRAVARIANT, COVARIANT AND MIXED TENSORS. If N2 quantities Aqs in a coordinate system _ (x 1,x2,..., x1) are related to N2 other quan- tities A in another coordinate system (x1, x2,..., x") by the transformation equations Air ax9 axs Aqs p,r = 1, 2,..., N S=1 q=1 x x or www.MathSchoolinternational.com 168 TENSOR ANALYSIS axq axrAgs axq axs by the adopted conventions, they are called contravariant components of a tensor of the second rank or rank two. The N2 quantities Aqs are called covariant components of a tensor of the second rank if axq axs A APr ax p -ay r qs Similarly the N2 quantities AS are called components o f a mixed tensor of the second rank if AP = ax P axs q As axq oxr THE KRONECKER DELTA, written 8k, is defined by J0 if jAk Sk 1 if j = k As its notation indicates, it is a mixed tensor of the second rank. TENSORS OF RANK GREATER THAN TWO are easily defined. For example, Akit are the compo- nents of a mixed tensor of rank 5, contravariant of order 3 and covariant of order 2, if they transform according to the relations A firm = axp axr azm axk ax 1. Agst ti axq axs axt axi ax9 ki SCALARS OR INVARIANTS. Suppose 0 is a function of the coordinates xk, and let denote the functional value under a transformation to a new set of coordinates xk Then cb is called a scalar or invariant with respect to the coordinate transformation if _. A scalar or invariant is also called a tensor of rank zero. TENSOR FIELDS. If to each point of a region in N dimensional space there corresponds a definite tensor, we say that a tensor field has been defined. This is a vector field or a scalar field according as the tensor is of rank one or zero. It should be noted that a tensor or tensor field is not just the set of its components in one special coordinate system but all the possi- ble sets under any transformation of coordinates. SYMMETRIC AND SKEW-SYMMETRIC TENSORS. A tensor is called symmetric with respect to two contravariant or two covariant indices if its com- ponents remain unaltered upon interchange of the indices. Thus if Aqs r = AQS r the tensor is sym- metric in m and p. If a tensor is symmetric with respect to any two contravariant and any two co- variant indices, it is called symmetric. A tensor is called skew-symmetric with respect to two contravariant or two covariant indices if its components change sign upon interchange of the indices. Thus if Aqs r= -Aqs r the tensor is www.MathSchoolinternational.com  TENSOR ANALYSIS 169 skew-symmetric in m and p. If a tensor is skew-symmetric with respect to any two contravariant and any two covariant indices it is called skew-symmetric. FUNDAMENTAL OPERATIONS WITH TENSORS. 1. Addition. The sum of two or more tensors of the same rank and type (i.e. same number of contra- variant indices and same number of covariant indices) is also a tensor of the same rank and type. Thus if AQ 0 and Bq 0 are tensors, then CO = Aq 0 + Br is also a tensor. Addition of tensors is commutative and associative. 2. Subtraction. The difference of two tensors of the same rank and type is also a tensor of the same rank and type. Thus if Aq0 and Br are tensors, then Dr = AqO - Bq 0 is also a tensor. 3. Outer Multiplication. The product of two tensors is a tensor whose rank is the sum of the ranks of the given tensors. This product which involves ordinary multiplication of the components of the tensor is called the outer product. For example, Aqr BS = Cqs' is the outer product of Alir and BS. However, note that not every tensor can be written as a product of two tensors of lower rank. For this reason division of tensors is not always possible. 4. Contraction. If one contravariant and one covariant index of a tensor are set equal, the result in- dicates that a summation over the equal indices is to be taken according to the summation con- vention. This resulting sum is a tensor of rank two less than that of the original tensor. The process is called contraction. For example, in the tensor of rank 5, AgPr, set r=s to obtain Agrr = Bq " a tensor of rank 3. Further, by setting p = q we obtain 80 = C2 a tensor of rank 1. 5. Inner Multiplication. By the process of outer multiplication of two tensors followed by a contrac- tion, we obtain a new tensor called an inner product of the given tensors. The process is called inner multiplication. For example, given the tensors A' O and Bst, the outer product is Aq1 Bst* r Letting q = r, we obtain the inner product Ark B. Letting q = r and p = s, another inner product Ar1' Br is obtained. Inner and outer multiplication of tensors is commutative and associative. 6. Quotient Law. Suppose it is not known whether a quantity X is a tensor or not. If an inner prod- uct of X with an arbitrary tensor is itself a tensor, then X is also a tensor. This is called the quotient law. MATRICES. A matrix of order m by n is an array of quantities apq, called elements, arranged in m rows and n columns and generally denoted by all a12... aln all a12... asn a21 a22... a2 n an a22 a2n or a.41 ann a,ns ani2... ainn or in abbreviated form by (a1,q) or [apq] p = 1,..., m; q= 1,..., n. If m=n the matrix is a square matrix of order m by m or simply m; if m = 1 it is a row matrix or row vector; if n = 1 it is a column matrix or column vector. The diagonal of a square matrix containing the elements ass, ate,..., ann is called the princi- pal or main dia ogT 1. A square matrix whose elements are equal to one in the principal diagonal and zero else h is called a unit matrix and is denoted by 1. A null matrix, denoted by 0, is a matrix all of whose elements are zero. www.MathSchoolinternational.com 170 TENSOR ANALYSIS MATRIX ALGEBRA. If A = (apq) and B = (bpq) are matrices having the same order (m by n) then 1. A = B if and only if ap q = b1, q. 2. The sum S and difference D are the matrices defined by S = A + B = (aj,q + bpq) , D = A - B = (apq- bpq) 3. The product P = AB is defined only when the number n of columns in A equals the number of rows in B and is then given by P = AB = (apq) (bpq) = (apr brq) n where al,r brq = apr brq by the summation convention. Matrices whose product is defined r.1 are called conformable. In general, multiplication of matrices is not commutative, i.e. AB A BA. However the asso- ciative law for multiplication of matrices holds, i.e.- C) _ (AB)C provided the matrices are conformable. Also the distributive laws hold, i.e. A(B+C) = AB + AC, (A +B) C = AC + BC. 4. The determinant of a square matrix A = (a,q) is denoted by I A I, det A, I I or det(ajq). If P=AB then IPI = IAI B. 5. The inverse of a square matrix A is a matrix A-1 such that AA-1 = 1, where I is the unit matrix. A necessary and sufficient condition that A-1 exist is that det A 0. If det A = 0, A is called singular. 6. The product of a scalar ?. by a matrix A denoted by X A, is the matrix (Xa pq) where each element of A is multiplied by X. 7. The transpose of a matrix A is a matrix AT which is formed from A by interchanging its rows and columns. Thus if A = (apq), then AT = (aqp). The transpose of A is also denoted by A. THE LINE ELEMENT AND METRIC TENSOR. In rectangular coordinates (x,y,z) the differential are length ds is obtained from By transforming to general curvilinear coordinates (see Problem 17, Chapter 7) this becomes ds 3 3 E I goq dupduq. Such spaces are called three dimensional Euclidean spaces. P=1 q=1 A generalization to N dimensional space with coordinates (x1, x2,..., xN) is immediate. We de- fine the line element ds in this space to be given by the quadratic form, called the metric form or metric, In the special case where there exists a transformation of coordinates from xI to xk such that www.MathSchoolinternational.com  TENSOR ANALYSIS 171 the metric form is transformed into (dz 1)2 + (d x2)2 +... + (d xN)2 or d x kd x k, then the space is call- ed N dimensional Euclidean space. In the general case, however, the space is called Riemannian. The quantities gpq are the components of a covariant tensor of rank two called the metric tensor or fundamental tensor. We can and always will choose this tensor to be symmetric (see Prob- lem 29). CONJUGATE OR RECIPROCAL TENSORS. Let g = g q denote the determinant with elements g and supp e g A 0. Define g pq by- pq gpq cofactor of gpq g Then gpq is a symmetric contravariant tensor of rank two called the conjugate or reciprocal tensor of gpq (see Problem 34). It can be shown (Problem 33) that sp gpq grq r ASSOCIATED TENSORS. Given a tensor, we can derive other tensors by raising or lowering indices. For example, given the tensor A pq we obtain by raising the index p, the, tensor A.q , the dot indicating the original position of the moved index. By raising the index q also we obtain.4'? A. Where no confusion can arise we shall often omit the dots; thus Apq can be written Apq. These derived tensors can be obtained by forming inner products of the given tensor with the metric tensor g pq or its conjugate gpq. Thus, for example p rp Apq A.q = g Arq, = grp gsq Ars A rs = grq A-ps q%ntk _ gpk g grm A A These become clear if we interpret multiplication by grp as meaning: let r= p (or p=r) in whatever follows and raise this index. Similarly we interpret multiplication by grq as meaning: let r= q (or q = r) in whatever follows and lower this index. All tensors obtained from a given tensor by forming inner products with the metric tensor and its conjugate are called associated tensors of the given tensor. For example A'4 and A. are asso- ciated tensors, the first are contravariant and the second covariant components. The relation be- tween them is given by AP = g pq Aq or AP = gpq Aq For rectangular coordinates g pq = 1 if p = q , and 0 if pA q , so that Ap = Ap, which explains why no distinction was made between contravariant and covariant components of a vector in earlier chap- ters. LENGTH OF A VECTOR, ANGLE BETWEEN VECTORS. The quantity APBP , which is the inner product of AP and Bq , is a scalar anal- ogous to the scalar product in rectangular coordinates. We define the length L of the vector AO or AP as given by www.MathSchoolinternational.com 172 TENSOR ANALYSIS L2 = AP AP = g1gA1'Aq = g1gAPAq We can define the angle 6 between AP and B1' as given by A1B1 cos 6 = (A1A1') (B1B1') THE PHYSICAL COMPONENTS of a vector A1' or A1' , denoted by Au, AV , and A. are the projec- tions of the vector on the tangents to the coordinate curves and are given in the case of orthogonal coordinates by Au = v Al = Al Av = 22 A2 = Aw = , s As = A V _9_1 1 922 933 Similarly the physical components of a tensor 12 A12 13 A 13 A A g A Auw = g1g A = etc. V'9 11922 -11-933 V19 CHRISTOFFEL'S SYMBOLS. The symbols are called the Christoffel symbols of the first and second kind respectively. Other symbols used in- stead o and 1 q. The latter symbol suggests however a tensor character, which is not trut-tf eneral. TRANSFORMATION LAWS OF CHRISTOFFEL'S SYMBOLS. If we denote by a bar a symbol in a co- ordinate system x k, then [ jk m [pq, r] a0 axq axr + ax1' a2 xq gpq axk axk a:x' ax's axj axk n s axn ax1 axq azn a2xq 1k pq axs axq axk axq ax3azk are the laws of transformation of the Christoffel symbols showing that they are not tensors unless the second terms on the right are zero. GEODESICS. The distance s between two points t1 and t2 on a curve xr= x'^(t) in a Riemannian space is given by s = J;1t2/;pq 2dx dt t at www.MathSchoolinternational.com  TENSOR ANALYSIS 173 That curve in the space which makes the distance a minimum is called a geodesic of the space. By use of the calculus of variations (see Problems 50 and 51) the geodesics are found from the differen- tial equation d2 xr + r dxp dxq = 0 ds2 pq ds ds where s is the are length parameter. As examples, the geodesics on a plane are straight lines where- as the geodesics on a sphere are arcs of great circles. THE COVARIANT DERIVATIVE of a tensor Ap with repect to xq is denoted by Ap,q and is de- fined by _ aAp s _ As Ap,q axq pq a covariant tensor of rank two. The covariant derivative of a tensor Ap with respect to x9 is denoted by Apq and is defined by Ap q - aAp 1P1),AS axq qs a mixed tensor of rank two. For rectangular systems, the Christoffel symbols are zero and the covariant derivatives are the usual partial derivatives. Covariant derivatives of tensors are also tensors (see Problem 52). The above results can be extended to covariant derivatives of higher rank tensors. Thus APi... pn r i... n, q _ ri- aAPl nC axq Api... pX s 1,4 Sr2...rn - r2Sq Ap. s rap rn ri... rn_is iq sp..Pln + {PiAr2r + P2 Aplsp3...Pin A pi.. pin - i s n qs i... rn +... + Pin qs 1... rn pi... pin q is the covariant derivative of Ari rn with respect to x The rules of covariant differentiation for sums and products of tensors are the same as those for ordinary differentiation. In performing the differentiations, the tensors g pq , gpq and 80 maybe treated as constants since their covariant derivatives are zero (see Problem 54). Since covariant derivatives express rates of change of physical quantities independent of any frames of reference, they are of great importance in expressing physical laws. PERMUTATION SYMBOLS AND TENSORS. Define a pqr by the relations e123 =em1 =e312 =+1, e213=e132-= e321=-1, e pqr = 0 if two or more indices are equal epgr and define in the same manner. The symbols epgr and epgr are called permutation symbols in www.MathSchoolinternational.com 174 TENSOR ANALYSIS three dimensional space. Further, let us define It can be shown that E pqr and Epgr are covariant and contravariant tensors respectively, called permutation tensors in three dimensional space. Generalizations to higher dimensions are possible. TENSOR FORM OF GRADIENT, DIVERGENCE AND CURL. 1. Gradient. If , V2c , in (a) cylindrical coordinates, (b) spherical coordinates. (a) In cylindrical coordinates g11=1, g22 =1/P2, g =1 (see Problem 35 (a)). Then from Problem 58, v245 Vg-- axk (,rg- gkr a l aa P C aP cP aP) + ao c P a 1a aa cP az ) ] az 1a a(f) 1D2 A) P aP cP aP' + p2 22 + az2 (b) In spherical coordinates g11=1, g22=1/r2, g = 1/r2 sin28 (see Problem 35 (b)). Then v%) _ 7g axk''gkra 1 a 2 (r sin 8 M) ar) + ae (Sin 8 a ae) + V (sin 4 ] a 1 a4) r2 sin 8 -6r t a a r2 ar (2 r' r + 1 r2 sine ae a (sin B a8 ) + 1 2 sin2 8 a2 -42 www.MathSchoolinternational.com 202 TENSOR ANALYSIS INTRINSIC DERIVATIVES. 62. Calculate the intrinsic derivatives of each of the following tensors, assumed to be differentiable functions of t: (a) an invariant (l), (b) Ai, (c) Ak , (d) Al kn. _ ,q = a = d4) , th e or di nary der ivative. (a) b4 d q ax dtq t (b) 8AJ =A dxq aAi + 1 As dxq = aAi dxq + 1 As dxq 8t q dt axq qs dt axq dt qs dt dAI + 1 As dxq dt qs dt SAk (c ) St Aj k,q dxq dt {;q}4+{}A:) dt q dAkj Aj dz + As dxq dt kq s dt qs k dt jk jk ,. Almn , jk dxq l aAlnn s Akn S A jk lq mq lsn s Ajk + Ask + k Ajs dxq nq Ins {i} qs Inn qs lmn dt dAA dt lmn Is Ajk s1nn dt dxq _ s Ajk dxq s Ajk _ dx q lq lmqf lsn dt nq ins dt + j Ask dxq + £kAJS dxq qs lmn dt qs Inn dt 63. Prove the intrinsic derivatives of gjk , gjk and are zero. 89jk = (g ) dxq = 0, 8glk = gjk dxq = 0, S bk -8 dxq =0 by Problem 54. 8t k ,q dt St 'q dt bt k,q dt RELATIVE TENSORS. 64. If Aq and Bts are relative tensors of weights w1 and w2 respectively, show that their inner and outer products are relative tensors of weight w1 + w2- By hypothesis, A jw1 az axq A B lm _ jw2 13F-' azm axt Brs k axP axk q' n axr axs axn t jw, w2 axq axq axl axm axt The outer product is Ak Bnm Aq Btrs axp az k axr axs az n a relative tensor of weight w1 + w2. Any inner product, which is a contraction of the outer product, is also a relative tensor of weight w1 + w2. www.MathSchoolinternational.com  TENSOR ANALYSIS 203 65. Prove that Vg- is a relative tensor of weight one, i.e. a tensor density. -3.p -3..q The elements of determinant g given by g transform according to gj = g q k axe axk oq Taking determinants of both sides, g = axe axe g = J2g or JV , which shows azj -3-x 66. Prove that dV = Vrg- Al dx2... dx1 is an invariant. By Problem 65, dV = Vg- dx1 dx... dxy = vrg- J dx1 dx`2... dx'A _ vg ax dz1 dx... dxN = dx1 dx2... dx" = dV a) From this it follows that if is aninvariant, then f...fdv = f...fdv V V for any coordinate systems where the integration is performed over a volume in N dimensional space. A similar statement can be made for surface integrals. MISCELLANEOUS APPLICATIONS. 67. Express in tensor form (a) the velocity and (b) the acceleration of a particle. k (a) If the particle moves along a curve xk = xk(t) where t is the parameter time, then vk = is its ve- dt locity and is a contravariant tensor of rank one (see Problem 9). k 2k (b) The quantity dt = t2 is not in general a tensor and so cannot represent the physical quantity acceleration in all coordinate systems. We define the acceleration ak as the intrinsic derivative of the velocity, i.e. ak = Stk which is a contravariant tensor of rank one. 68. Write Newton's law in tensor form. Assume the mass M of the particle to be an invariant independent of time t. Then Mak = Fk a contravariant tensor of rank one is called the force on the particle. Thus Newton's law can be written S kk Fk = Mak = M www.MathSchoolinternational.com 204 TENSOR ANALYSIS 69. Prove that ak = 6vk = dx 2 k k dx0 dxq St dt2 pq dt dt ' Since vk is a contravariant tensor, we have by Problem 62 (b) 2 8vk = dvk + k vs dxq d xk + k dxq VP St dt qs dt dt2 qp dt 2 _ d xk + k dxp dxq d- i pq dt dt 70. Find the physical components of (a) the velocity and (b) the acceleration of a particle in cylin- drical coordinates. (a) From Problem 67(a), the contravariant components of the velocity are dx1 dp dx2 do dx3 dz and dt dt ' dt dt dt dt Then the physical components of the velocity are v911 dx1 dt = dp dt 922 dx2 dt ' dt do and v 933 dx3 dt = dz dt 911=1,g22=p2.g33=1. using (b) From Problems 69 and 49 (b) , the contravariant components of the acceleration are d2x1 + dx2 dx2 _ d2 p do al _ p( dt ) 2 dt2 22 dt dt dt2 a 2 d2 x2 2 dxl dx2 2 dx2 dxl do 2 dp do dt2 + 12 dt dt + 21 dt dt dt2 + p dt dt and a3 d2 x3 _ d2 z dt2 dt2 Then the physical components of the acceleration are 11 al = 0 -- p 22 a2 =p + 2pq5 and 33 a 3 =z where dots denote differentiations with respect to time. 71. If the kinetic energy T of a particle of constant mass M moving with velocity having magnitude v is given by T = 2Mv2 = 2Mg pq 0 xq, prove that d (aT) _ aT = Ma k dt axk axk where ak denotes the covariant components of the acceleration. Since T = 2Mgpq zp zq, we have www.MathSchoolinternational.com  TENSOR ANALYSIS 205 d ('3T ag j aT I agpq zp zq, aT = Mg zq and = zq + q,, kq M(gk 9 axk axk '31k dt ax ax Then d ( aT) - aT = M + agk q zjzq - 1 agpq zPzq dt azk axk (gkq zq axk 2 axk agkp M (gkq zq + 1 (agkq 2 + - agpq) xp xq axp axq axk M(gkq zq + [pq,k] xp xq) = Mg xr + r zj z q = Mg ar = Mak kr pq kr using Problem 69. The result can be used to express the acceleration in different coordinate systems. 72. Use Problem 71 to find the physical components of the acceleration of a particle in cylindrical coordinates. Since ds2 = dp2+,02dca2+ dz2, v2 = ( )2 = z2 and T = 2Mv2 = e(;;? + Z2) , 2+ From Problem 71 with x1 = p, x2 = 0, x3 = z we find a1 = P a2 = dt a3 z Then the physical components are given by a, a2 ag 1V,gj-l 22 33 since g11 = 1, g22 = p2 , g33 = 1. Compare with Problem 70. 73. If the covariant force acting on a particle is given by Fk = - a where V (x1..... xj') is the k potential energy, show that dt (aLk) - a Lk = 0 where L = T -- V. From L = T -- V, aL _ aT since V is independent of zk. Then from Problem 71, axk r axk d aT aT _ Mak Fk -- aV and d (aL) - aL = = 0 dt ( azk azk axk dt azk axk The function L is called the Lagrangean. The equations involving L, called Lagrange's equations, are important in mechanics. By Problem 50 it follows that the results of this problem are equivalent to the statement that a particle moves in such a way that f L dt is an extremum. This is called Hamilton's t1 principle. www.MathSchoolinternational.com 206 TENSOR ANALYSIS 74. Express the divergence theorem in tensor form. Let A k define a tensor field of rank one and let vk denote the outward drawn unit normal to any point of a closed surface S bounding a volume V. Then the divergence theorem states that fffA kk dV II A k vk dS V S For N dimensional space the triple integral is replaced by an N tuple integral, and the double integral by an N -1 tuple integral. The invariant A kk is the divergence of Ak (see Problem 57). The invariant Ak vk is the scalar product of Ak and vk, analogous to A n in the vector notation of Chapter 2. We have been able to express the theorem in tensor form; hence it is true for all coordinate systems since it is true for rectangular systems (see Chapter 6). Also see Problem 66. 75. Express Maxwell's equations (a) div B = 0, (b) div D = 47rp, (c) Vx E = - aB , (d) VxH = 4Z f in tensor form. Define the tensors Bk, Dk, Ek, Hk, 1 k and suppose that p and c are invariants. Then the equations can be written (a) Bkk0 (b) D kk = 47Tp (c) --- EJkq Ek,q 1 aBj or EJWq Ek.q 1 aBj C at J 47r1 (d) - Ejkq or Ejkq Hk , 4 c Hk, q c These equations form the basis for electromagnetic theory. 76. (a) Prove that Al,gr A0,rq = Rngr An where A0 is an arbitrary covariant tensor of rank one. (b) Prove that R qr is a tensor. (c) Prove that R' is a tensor. Rpgrs gns aAM (a) Ap, gr = (AO.q) r axr - 1 {Pr A. J.q - {'}A qr O9 J,_ (BA. aA i A. - i j k - {}Ak) - PS l Pq Pr qr axq P1 a2 A Aj k {;;iq} A..- -- + A Pq -ax, Pr axq Pr Iq k aAP l 1 + 1 A qr axq qr P1 By interchanging q and r and subtracting, we find www.MathSchoolinternational.com  TENSOR ANALYSIS 207 A p,qr _ i k Ak a j JA. - {pJq}{J}A k A p rq I r k pr jq axr P4 - p r kq Aj a ; A 9 _ kj pq kr Ai a xr Pq RpJgr = A I where Rqr = {:r}{q} _ - {:q}{Lr} + pr axr P4 axq i Replace j by n and the result follows. (b) Since Ap,gr - Ap,rq is a tensor, Rqr An is a tensor; and since An is an arbitrary tensor, Rqr is n n a tensor by the quotient law. This tensor is called the Riemann-Christoffel tensor, and is sometimes n written R, pqr, R pqr , or simply R pqr (c) Rpgrs = gns Rpgr is an associated tensor of Rpgr and thus is a tensor. It is called the covariant curvature tensor and is of fundamental importance in Einstein's general theory of relativity. SUPPLEMENTARY PROBLEMS Answers to the Supplementary Problems are given at the end of this Chapter. 77. Write each of the following using the summation convention. (a) a1 x1x3 + a2x2x3 +... + a) xNx3 (c) Al B1 + A2 B2 + A3 B3 +... + AjV BN A21 + g23 g + g24g (b) B1 + A22 B2 + A' B3 + + By `42I (d) g21 g11 + g22 g21 31 41 8221 + 8222 (e) B111 + B 122 12 + 21 22 78. Write the terms in each of the following indicated sums. az,7 k , N=2 (a) axk(iAk), N=3 (b) Bp C (c) xk a' 79. What locus is represented by akxkxk = 1 where zk, k = 1, 2,..., N are rectangular coordinates, ak are positive constants and N = 2, 3 or 4 ? 80. If N = 2, write the system of equations represented by apq xq = b p. k 81. Write the law of transformation for the tensors (a) Ak , (b) B , (c) Can , (d) An. www.MathSchoolinternational.com 208 TENSOR ANALYSIS 82. Determine whether the quantities B(j,k,m) and C(j,k,m,n) which transform from a coordinate system xti to another xti according to the rules ax ax k axr B(j,k,m) axp axq axq axs C(j,k,m,n) (a) B(p.q,r) = (b) C(p,q,r.s) = axp axq axx axq az k azr axn are tensors. If so, write the tensors in suitable notation and give the rank and the covariant and contra- variant orders. 83. How many components does a tensor of rank 5 have in a space of 4 dimensions ? 84. Prove that if the components of a tensor are zero in one coordinate system they are zero in all coordinate systems. 85. Prove that if the components of two tensors are equal in one coordinate system they are equal in all co- ordinate systems. k k 86. Show that the velocity = vk of a fluid is a tensor, but that v is not a tensor. t 87. Find the covariant and contravariant components of a tensor in (a) cylindrical coordinates p, 0, z , (b) spherical coordinates r, 6, if its covariant components in rectangular coordinates are 2x --z, x2y, yz. 88. The contravariant components of a tensor in rectangular coordinates are yz, 3, 2x+y. Find its covariant components in parabolic cylindrical coordinates. q 8s 89. Evaluate (a) 8q Bas, (b) Sq Sr Aqs, (c) 8p 8 8s , (d) 8q 8r 8s. Arq r 90. If is a tensor, show that Ar is a contravariant tensor of rank one. 1 j=k 91. Show that is not a covariant tensor as the notation might indicate. 0 j#k 92. If A0 = a Aq prove that Aq = -axq AP 93. If A r = azp axs 'IS prove that As = axq axr Ar axq azr s s axp axs. 94. If (D is an invariant, determine whether a is a tensor. axp axq 95. If Aq and Br are tensors, prove that A0 Br and Aq Bq are tensors and determine the rank of each. Pq 96. Show that if Ars is a tensor, then + ASS is a symmetric tensor and Ars - Asr is a skew-symmetric tensor. 97. If Apq and Brs are skew-symmetric tensors, show that Cps = Ap4 Brs is symmetric. 98. If a tensor is symmetric (skew-symmetric), are repeated contractions of the tensor also symmetric (skew- symmetric) ? 99. Prove that Apq xp xq = 0 if Apq is a skew-symmetric tensor. www.MathSchoolinternational.com  TENSOR ANALYSIS 209 100. What is the largest number of different components which a symmetric contravariant tensor of rank two can have if (a) N = 4. (b) N = 6 ? What is the number for any value of N ? 101. How many distinct non-zero components, apart from a difference in sign, does a skew-symmetric covariant tensor of the third rank have 102. If AIrs is a tensor, prove that a double contraction yields an invariant. 103. Prove that a necessary and sufficient condition that a tensor of rank R become an invariant by repeated contraction is that R be even and that the number of covariant and contravariant indices be equal to R/2. Brs 104. If Apq and are tensors, show that the outer product is a tensor of rank four and that two inner prod- ucts can be formed of rank two and zero respectively. 105. If A(p, q) Bq = C where Bq is an arbitrary covariant tensor of rank one and C is a contravariant tensor of rank one, show that A(p,q) must be a contravariant tensor of rank two. 106. Let AP and Bq be arbitrary tensors. Show that if AP Rq C(p, q) is an invariant then C(p,q) is a tensor which can be written C C. 107. Find the sum S = A+B, difference D = A-B, and products P = AB and Q = BA, where A and B are the matrices (a) A = 3 -1 B 4 3 2 4 ' _ (2-1 - 2 0 1 -i 1 2 (b) A -1 -2 2 B= 3 2 -4 (_1 3 -1 -i -2 2 108. Find (3A-2B)(2A-B), where A and B are the matrices in the preceding problem. 109. (a) Verify that det (AB) = {det A } {det B } for the matrices in Problem 107. (b) Is det (AB) = det (BA) ? I -3 2 -1 1111. Let A = B= 1 3 -2 2 1 2 Show that (a) AB is defined and find it, (b) BA and A +B are not defined. 2 -1 3 x = 1 111. Find x, y and z such that 1 2 -4 y -.3 -1 3 -2 z 6 A'1 112. The inverse of a square matrix A, written is defined by the equation AA-1 = 1, where 1 is the unit matrix having ones down the main diagonal and zeros elsewhere. _ -1 1 Find A-' if (a) A = (_5 (b) A = (12 1 -1. 4 1 -1 2 A-1 Is A = 1 in these cases ? www.MathSchoolinternational.com 210 TENSOR ANALYSIS 2 1 -2 113. Prove that A = 1 -2 3 has no inverse. 4 -3 4 114. Prove that (AB) 71- = B-1 A-1, where A and B are non-singular square matrices. 115. Express in matrix notation the transformation equations for (a) a contravariant vector (b) a covariant tensor of rank two (c) a mixed tensor of rank two. 1 116. Determine the values of the constant X such that AX = INX, where A = -3 and X is an arbi- trary matrix. These values of X are called characteristic values or eigenvalues of the matrix A. 117. The equation F(X) = 0 of the previous problem for determining the characteristic values of a matrix A is called the characteristic equation for A. Show that F(A)=0, where F(A) is the matrix obtained by re- placing A. by A in the characteristic equation and where the constant term c is replaced by the matrix cl, and 0 is a matrix whose elements are zero (called the null matrix). The result is a special case of the Hamilton-Cayley theorem which states that a matrix satisfies its own characteristic equation. 118. Prove that (AB) = BT A_T. 119. Determine the metric tensor and conjugate metric tensor in (a) parabolic cylindrical and (b) elliptic cylindrical coordinates. 120. Prove that under the affine transformation -'r = as xp + br, where ap and br are constants such that apaq = bq , there is no distinction between the covariant and contravariant components of a tensor. In the special case where the transformations are from one rectangular coordinate system to another, the tensors are called cartesian tensors. 121. Find g and gjk corresponding to ds2 = 3 (dx1)2 + 2 (dx2)2 + 4 (dx3)2 - 6 dx1 dx3. 122. If Ak = gikAi , show that AJ = g Ak and conversely. J,k 123. Express the relationship between the associated tensors (a) Apq and q, (b) A qr and Aj ' ql, (c) Apgr and A..,, 124. Show that (a) APq B.s = A1'gBprs , (b) B7 r = AAgr Bpr = B. Hence demonstrate the gen- eral result that a dummy symbol in a term may be lowered from its upper position and raised from its lower position without changing the value of the term. qr 125. Show that if A B Cr A; = B; q a free index in a tensor equation may be raised or lowered without affecting the validity of the equa- tion. 126. Show that the tensors g gpq and 89 are associated tensors. pq' gpq axe 127. Prove (a) Ejk ax _ g pq axQk (b) gdk axp , = axp ax ax ax 128. If AP is a vector field, find the corresponding unit vector. www.MathSchoolinternational.com  TENSOR ANALYSIS 211 129. Show that the cosines of the angles which the 3 dimensional unit vector Uti make with the coordinate U1 U2 U3 curves are given by. g11 y g22 933 130. Determine the Christoffel symbols of the first kind in (a) rectangular, (b) cylindrical, and (c) spherical coordinates. 131. Determine the Christoffel symbols of the first and second kinds in (a) parabolic cylindrical, (b) elliptic cylindrical coordinates. 132. Find differential equations for the geodesics in (a) cylindrical, (b) spherical coordinates. 133. Show that the geodesics on a plane are straight lines. 134. Show that the geodesics on a sphere are arcs of great circles. 135. Write the Christoffel symbols of the second kind for the metric ds2 = (dx1)2 + [(x2)2 - (x1)2] (dx2)2 and the corresponding geodesic equations. 136. Write the covariant derivative with respect to xq of each of the following tensors: AXkl' (a) Aik, (b) Alm' (c) Ak1X, (d) (e) Ainn j - jk 1 138. Use the relation A - g Ak to obtain the covariant derivative of A from the covariant derivative of Ak. 139. If 4> is an invariant, prove that ,pq= i.e. the order of covariant differentiation of an invariant is immaterial. 140. Show that Ejgr and Epgr are covariant and contravariant tensors respectively. 141. Express the divergence of a vector AP in terms of its physical components for (a) parabolic cylindrical, (b) paraboloidal coordinates. 142. Find the physical components of grad in (a) parabolic cylindrical, (b) elliptic cylindrical coordinates. 2 143. Find V 4) in parabolic cylindrical coordinates. 144. Using the tensor notation, show that (a) div curl Ar = 0, (b) curl grad = 0. 145. Calculate the intrinsic derivatives of each of the following tensor fields, assumed to be differentiable functions of t : , k (a) Ak, (b) Al , (c) Aj Bk, (d) OAk where 0 is an invariant. k 146. Find the intrinsic derivative of (a) gjk A , (b) 8k Aj A. (c) gjk r A. 9 9 r 8A 147. Prove dt (gpq A A A q) = 29Pq A q 8t www.MathSchoolinternational.com 212 TENSOR ANALYSIS 148. Show that if no external force acts, a moving particle of constant mass travels along a geodesic given by p as(ds) = a. 149. Prove that the sum and difference of two relative tensors of the same weight and type is also a relative tensor of the same weight and type. 150. if Apq is a relative tensor of weight w, prove that g-V/2 Apq. is an absolute tensor. 151. If A(p,q) Bqs = where Br is an arbitrary relative tensor of weight wl and Cpr is a known relative tensor of weight w2, prove that A(p,q) is a relative tensor of weight w2---w1. This is an example of the quotient law for relative tensors. 152. Show that the quantity G(j,k) of Solved Problem 31 is a relative tensor of weight two. 153. Find the physical components of (a) the velocity and (b) the acceleration of a particle in spherical co- ordinates. 154. Let Ar and Br be two vectors in three dimensional space. Show that if ,\ and,i are constants,then Cr = X Ar+ LRr is a vector lying in the plane of Ar and Br. What is the interpretation in higher dimen- sional space ? 155. Show that a vector normal to the surface 0 (xi, x2, x3) = constant is given by AO =9a. Find the corresponding unit normal. as 156. The equation of continuity is given by V (0- V) + = 0 where cr is the density and v is the velocity of ac a fluid. Express the equation in tensor form. 157. Express the continuity equation in (a) cylindrical and (b) spherical coordinates. 158. Express Stokes' theorem in tensor form. 159. Prove that the covariant curvature tensor Rpqrs is skew-symmetric in (a) p and q, (b) r ands , (c) q ands. 160. Prove Rpqrs = Rrsjiq 161. Prove (a) Rj,grs + Rpsqr + Rjirsq = 0, (b) R¢grs + Rrgps + Rrspq + R¢srq = 0. 162. Prove that covariant differentiation in a Euclidean space is commutative. Thus show that the Riemann- Christoffel tensor and curvature tensor are zero in a Euclidean space. 163. Let T 0 = dsP be the tangent vector to curve C whose equation is xP = x '(s) where s is the arc length. q (a) Show that g,g TP Tq - 1. (b) Prove that gig TO T = 0 and thus show that Nq = K &sq is a unit normal to C for suitable K. (c) Prove that Nq is orthogonal to Nq as 164. With the notation of the previous problem, prove: TO S N q (SNq (a) gig T' Nq = 0 (b) gig = - K or gpq T + K Tq) = 0. r Hence show that Br = I (6N + K Tr) is a unit vector for suitable T orthogonal to both and Nq. www.MathSchoolinternational.com  TENSOR ANALYSIS 213 165. Prove the Frenet-Serret formulas $s K Np, S= TB1'- KT 1', sB = - TN1' where T N1' and BP are the unit tangent, unit normal and unit binormal vectors to C, and K and T are the curvature and torsion of C. 166. Show that ds2 = c2(dx4)2 - dxk dxk (N=3) is invariant under the linear (affine) transformation x2 x3 18 x1 = y(x1- vx4) , = x2 , = x3 , z4 = y(x4 -- x1) 2 where 'y,,8, c and v are constants, 8 = v/c and y = (1-,8 )71/2 This is the Lorentz transformation of special relativity. Physically, an observer at the origin of the xi system sees an event occurring at position x1,x2,x3 at time x4 while an observer at the origin of the 'xi system sees the same event occur- ring at position 3F1,`x2,`x3 at time z4. It is assumed that (1) the two systems have the x1 and Z1 axes coincident, (2) the positive x2 and x3 axes are parallel respectively to the positive x2 and x3 axes, (3) the xi system moves with velocity v relative to the xi system, and (4) the velocity of light c is a constant. 167. Show that to an observer fixed in the xi ((i) system, a rod fixed in the xi (xi) system lying parallel to the x1 (xi) axis and of length L in this system appears to have the reduced length LA - This phenomena is called the Lorentz-Fitzgerald contraction. ANSWERS TO SUPPLEMENTARY PROBLEMS. 77. (a) akxkx3 (b) A23B (c) Ak Bk (d) g2q gq1 , N=4 (e) Bar, N= 2 78. (a) 2 1 (vg A1) + ax2 (v g A2) 2x3 (V -g A3) (c) azj ax 1 + axi axe ax ax A + +... + (b) All BP C, + A2' B1' C2 + A12 B2 C, + A22 B2 C2 axi ax-'IR ax2 azm -axl -a--n 79. Ellipse for N= 2, ellipsoid for Nj= 3, hyperellipsoid for N= 4. a11x1 + a12x2 = b1 80. a21x1 + a22x2 = b2 ij 81. (a) A-pq r - ax-p axq ax k Ak (c) C axx axn Cxn axi -ax, avr q ax1' axq (b) B-1'gr = axp azq ax ax's Bijk axi All S (d) A axi axi axk aTs = axi 82. (a) B(/, k, m) is a tensor of rank three and is covariant of order two and contravariant of order one. It can be written (b) C(j, k, m, n) is not a tensor. 83. 45 = 1024 87. (a) 2p cos2 c - z cos 0 + p3 sin2 0 cos2o, -- 2p2 sin 0 cos 0 + pz sin 0 + p4 sin 0 cos3 pz sin 0. www.MathSchoolinternational.com 214 TENSOR ANALYSIS (b) 2 r sin2 6 cos2 0 - r sin 6 cos 6 cos (p + r3 sin46 sin2 0 cos2 0 + r2 sin 6 cos2 6 sink, 2r 2 sin 6 cos 6 cos2 0 - r2 cos2 6 cos 4 + r4 sin3 6 cos (9 sin2 O cos2 O - r3 sin26 cos 6 sink, - 2r2 sin2 6 sin cos ) + r2 sine cos 6 sin + r4 sin4 6 sino cos3 o 88. u2vz + 3v , 3u - uv2z , u2 + uv - v2 89. (a) Bq r,s (b) P (c) bs , (d) N 94. It is not a tensor. 95. Rank 3 and rank 1 respectively. 98. Yes. 100. (a) 10 , (b) 21 , (c) N(N+ 1)/2 101. N(N- 1) (N- 2)/6 107. (a) S 7 2 D -1 -4 P 14 10 18 8 0 3 ' 4 5 0 2 Q -8 -2 3 -1 3 1 1 -1 1 -4 6 1 8 -3 (b) S = 2 0 -2 , D= -4 -4 6 , p= -9 -7 10 , Q= 8 -16 11 -2 1 1 0 5 -3 9 9 -16 -2 10 -7 3 -16 20 108. (a) 52 104 -86 ) (b) 9 163 -136 110. -4 5 3 104 17 -2 1 -61 -135 132 1/3 1/3 0 111. x=-1, y=3, z=2 112. (a) 2 5/2 3/2 1 (b) - 5/3 1/3 1. Yes -1 0 1 x1 ax3 -ax' ax1 ax1 ax1 A12 A13 ax1 aaG2 az3 (b) ax2 ax2 ax2 A 22 A 23 ax 1 ax- 2 - az3 ax1 ax2 ax3 ax3 '3X3 A31 A32 A33 A 32 A 33 ax ax ax1 ax2 ax3 ax1 ax1 ax1 ax2 (c) ax ax2 ax1 ax2 Al 3 A2 3 A3 3 ax3 ax ax1 ax2 u 2 +v2 0 0 U'2 2+v2 0 0 2 + v2 0 ! \ 116.\.=4,-1 119. (a) 0 11 , 0 u2 +v 1 2 0 0 0 1 0 0 1 www.MathSchoolinternational.com  TENSOR ANALYSIS 215 a2(sinh2u + sin2v) 0 0 1 0 0 (a2(sinh2u + sin2v) (b) 0 a2(sinh2u + sin2v) 0 0 1 0 a2(sinh2u + sin2v) 0 0 1 0 0 1 4/3 0 1 121. g = 6, 0 1/2 0 1 0 1 Ap9 _ gPi A 9, (b) A. q r..r rl Jk 123. (a) Ap q A..1 (c) = gpj gqk g A A 128. or AY , gig AP A q 130. (a) They are all zero. (b) [22,1] =-p, [12,2] _ [21,2] = p. All others are zero. (c) [22,1] _ -r, [33.1] r sin28, [33,2] _ - r2 sin 8 cos 8 [21,2] [12,2] =r, [31,3] = [13,3] = r sin28 [ 32,31 _ [ 23,3] = r2 sin 8 cos e. All others are zero. 131. (a) [ii,1] = u, [22,2] = v, [11,2] _ - v, [ 22,1] -u, [12,1]= [21,1]=v, [21,2]= [12,2]=u. 1 _ u 2 v 1 -u 2 -v 11 u2 + v2 ' 22 u2 + v2 ' 22 u2 + v2 11 u2 + v2 1 _ 1 v 2 1 2 u All others are zero. 21 12 u2 + v2 21 12 u2 + v2 (b) [11,1] = 2a2 sinh u Cosh u , [22,2] = 2a2 sine cosv , [11,2] _ -2a2 sinv cosv [22.1] =-2a2sinh u cosh u, [12,1] _ [21,1] = 2a2 sinv cosv, [21,2]= [12,2]= 2a2sinhu coshu 51 sinh u coshu 2 _ sinv cosv 1 - sinh u cosh u 11 sinh2u + sin2v ' 22 sinh2u + sin2v ' 122 5 - sinh2u + sin2v 52 -sin v cosv 1 1 sinv cos v 1 11 sinh2u + sin2v 21 12 sinh2u + sin2v ' j2 = 2 sinh u cosh u. All others are zero. 21 12 y sinh2u + sin2v d dsP d2 0 + 2 do dq d2 0, ds2z 132. (a) =0 - p( d2 0. ds2 p ds ds d'r d