Continuum Mechanics

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Continuum Mechanics George Backus Insitute for Geophysics and Planetary Physics University of California, San Diego S Samizdat Press Continuum Mechanics George Backus S Samizdat Press Golden White River Junction
Published by the Samizdat Press Center for Wave Phenomena Department of Geophysics Colorado School of Mines Golden, Colorado 80401 and New England Research 76 Olcott Drive White River Junction, Vermont 05001 c Samizdat Press, 1997  Samizdat Press publications are available via FTP from landau.mines.edu or 138.67.12.78. Or via the WWW from http://landau.mines.edu/~ samizdat Permission is given to freely copy these documents. Contents Dictionary D-1 Five Prerequisite Proofs PP-3 I Tensors over Euclidean Vector Spaces 1 1 Multilinear Mappings 3 1.1 1.2 1.3 1.4 Denition of multilinear mappings : : : : : : : Examples of multilinear mappings : : : : : : : Elementary properties of multilinear mappings Permuting a multilinear mapping : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2 Denition of Tensors over Euclidean Vector Spaces 3 4 6 7 11 3 Alternating Tensors, Determinants, Orientation, and n-dimensional Righthandedness 13 3.1 Structure of nV : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 3.2 Determinants of linear operators : : : : : : : : : : : : : : : : : : : : : : : 3.3 Orientation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 4 Tensor Products 4.1 Denition of a tensor product : : : : : : : : : : : : : : : : : : : : : : : : 4.2 Properties of tensor products : : : : : : : : : : : : : : : : : : : : : : : : : i 13 17 21 25 25 25 ii CONTENTS 4.3 Polyads : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5 Polyad Bases and Tensor Components 5.1 5.2 5.3 5.4 5.5 5.6 Polyad bases : : : : : : : : : : : : : : : : : : : : : : Components of a tensor relative to a basis sequence: Changing bases : : : : : : : : : : : : : : : : : : : : Properties of component arrays : : : : : : : : : : : Symmetries of component arrays : : : : : : : : : : Examples of component arrays : : : : : : : : : : : : ::::::: Denition : ::::::: ::::::: ::::::: ::::::: : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 28 31 31 33 35 36 37 38 6 The Lifting Theorem 41 7 Generalized Dot Products 47 7.1 7.2 7.3 7.4 Motivation and denition : : : : : : : : : : Components of P hqiR : : : : : : : : : : : Properties of the generalized dot product : Applications of the generalized dot product : : : : 8 How to Rotate Tensors (and why) 8.1 8.2 8.3 8.4 8.5 Tensor products of linear mappings : : : : : Applying rotations and reections to tensors Physical applications : : : : : : : : : : : : : Invariance groups : : : : : : : : : : : : : : : Isotropic and skew isotropic tensors : : : : : 9 Dierential Calculus of Tensors 9.1 9.2 9.3 9.4 9.5 Limits in Euclidean vector spaces : : : : : : Gradients: Denition and simple properties : Components of gradients : : : : : : : : : : : Gradients of dot products : : : : : : : : : : Curvilinear coordinates : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 47 50 51 57 67 67 71 73 76 80 89 89 95 102 105 108 CONTENTS iii 9.6 Multiple gradients and Taylor's formula : : : : : : : : : : : : : : : : : : : 113 9.7 Dierential identities : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 115 10 Integral Calculus of Tensors 10.1 Denition of the Integral : : : : : : : : 10.1.1 Mass distributions or measures 10.1.2 Integrals : : : : : : : : : : : : : 10.2 Integrals in terms of components : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 11 Integral Identities 11.1 11.2 11.3 11.4 11.5 11.6 11.7 Linear mapping of integrals : : : : : : : : : : : : : : Line integral of a gradient : : : : : : : : : : : : : : Gauss's theorem : : : : : : : : : : : : : : : : : : : : Stokes's theorem : : : : : : : : : : : : : : : : : : : Vanishing integral theorem : : : : : : : : : : : : : : Time derivative of an integral over a moving volume Change of variables of integration : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 119 119 119 120 123 127 127 128 130 131 132 134 135 II Elementary Continuum Mechanics 139 12 Eulerian and Lagrangian Descriptions of a Continuum 141 12.1 12.2 12.3 12.4 12.5 12.6 Discrete systems : : : : : : : : : : : : : : : : Continua : : : : : : : : : : : : : : : : : : : : Physical quantities : : : : : : : : : : : : : : Derivatives of physical quantities : : : : : : Rigid body motion : : : : : : : : : : : : : : Relating continuum models to real materials 13 Conservation Laws in a Continuum : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 141 142 144 147 151 159 163 13.1 Mass conservation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 164 13.1.1 Lagrangian form : : : : : : : : : : : : : : : : : : : : : : : : : : : 164 iv CONTENTS 13.2 13.3 13.4 13.5 13.6 13.7 13.1.2 Eulerian form of mass conservation : : : : : : : Conservation of momentum : : : : : : : : : : : : : : : 13.2.1 Eulerian form : : : : : : : : : : : : : : : : : : : 13.2.2 Shear stress, pressure and the stress deviator : : Lagrangian form of conservation of momentum : : : : : Conservation of angular momentum : : : : : : : : : : : 13.4.1 Eulerian version : : : : : : : : : : : : : : : : : : $ $ 13.4.2 Consequences of S T =S : : : : : : : : : : : : : : Lagrangian Form of Angular Momentum Conservation : Conservation of Energy : : : : : : : : : : : : : : : : : : 13.6.1 Eulerian form : : : : : : : : : : : : : : : : : : : 13.6.2 Lagrangian form of energy conservation : : : : : Stu : : : : : : : : : : : : : : : : : : : : : : : : : : : : 13.7.1 Boundary conditions : : : : : : : : : : : : : : : 13.7.2 Mass conservation : : : : : : : : : : : : : : : : : 13.7.3 Momentum conservation : : : : : : : : : : : : : 13.7.4 Energy conservation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 165 168 168 182 187 195 195 200 209 210 210 214 215 222 224 227 231 14 Strain and Deformation 235 15 Constitutive Relations 249 14.1 Finite deformation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 235 14.2 Innitesimal displacements : : : : : : : : : : : : : : : : : : : : : : : : : : 240 15.1 15.2 15.3 15.4 15.5 Introduction : : : : : : : : : : : : : : : : : : : Elastic materials : : : : : : : : : : : : : : : : : Small disturbances of an HTES : : : : : : : : Small disturbances in a perfectly elastic earth Viscous uids : : : : : : : : : : : : : : : : : : III Exercises : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 249 252 256 265 272 279 DICTIONARY D-1 Logic ) \implies" (e.g. A ) B where A and B are sentences.) , \implies and is implied by" (e.g. A , B where A and B are sentences.) i \if and only if" | same as , := \is dened as" 8 \for every" 9 \there exists at least one" 91 \there exists exactly one" 3 \such that" Sets (same as \Classes") (a1  : : :  a6 ) is an (ordered) sequence of objects. (a1 : : :  an) is also called an ordered n-tuple. Order is important, and (a1  a2 a3) 6= (a2 a1  a3). However, duplication, e.g. a1 = a2 , is permitted. fa1 : : :  a6 g is a set of objects. Order is irrelevant, and fa1  a2 a3g = fa2  a1 a3g = fa3 a1  a2g, etc. Duplication is not permitted. All a are dierent. (a1  : : :  6 a4  : : :  a6) := (a1  a2 a3  a5 a6). fa1 : : :  6 a4  : : :  a6 g := fa1  a2 a3 a5  a6g.  is the empty set, the set with no objects in it. R := set of all real numbers. R+ := set of all non-negative real numbers. R+ := set of all positive real numbers. D-2 DICTIONARY C := set of all complex numbers. f1 : : :  ng := set of integers from 1 to n inclusive. fx : p(x)g is the set of all objects x for which the statement p(x) is true. For example, fx : x = 2y for some integer yg is the set of even integers. It is the same as f2x : x is an integerg. 2 \is a member of." Thus, \a 2 A" is read \the object a is a member of the set A. The phrase \a 2 A" can also stand for \the object a which is a member of A" 62 \is not a member of" \is a subset of." A B means that A and B are sets and every member of A is a member of B . \is a proper subset of." A B means that A B and A 6= B (i.e., B has at least one member not in A). A \ B := the set of all objects in both A and B . (Read \A meet B .") A B := the set of all objects in A or B or both. (Read \A join B .") AnB := the set of all objects in A and not B . (Read \A minus B .") Cartesian A B := the set of all ordered pairs of (a b) with a 2 A and b 2 B . products 8 > A1 : : : An := the set of all ordered n-tuples (a1  : : :  an) with a1 2 A1, az 2 A2  : : :  an 2 > > > < An. > > > > : nA := the set of all ordered n tuples (a1 : : :  an) with a1  a2 : : :  an all members of A. Same as A : : : A (n times). DICTIONARY D-3 Functions Function is an ordered pair (Df  f ). Df is a set, called the \domain" of the function, and f is a rule which assigns to each d 2 Df an object f (d). This object is called the \value" of f at d. The function (Df  f ) is usually abbreviated simply as f . In the expression f (d), d is the \argument" of f . Range of function is a set Rf consisting of all objects which are values of f . Two equivalent denitions of Rf are Rf := fx : 9d 2 Df 3 x = f (d)g or Rf := ff (d) : d 2 Df g : f = g for functions means Df = Dg and for each d 2 D, f (d) = g(d). f (W ) for a function f and a set W is dened as ff (w) : w 2 W g. If W \ Df =6 0 then f (W ) =6 0. d j! f (d) is a shorthand way of describing a function. For example, if Df = R and f (x) = x2 for all x 2 R, we can describe f by saying simply x j! x2 under f . In particular, 2 j! 4 under f . f jW is the \restriction" of f to W . It is dened whenever W is a subset of Df . It is the function (W g) such that for each w 2 W , g(w) = f (w). Its domain is W . \Extension." If g is a restriction of f , then f is an extension of g to Df . f ( v0), f (u0 ). Suppose U and V are sets and f is a function whose domain is U V . Suppose u0 2 U and v0 2 V . Dene two functions g and h as follows: Dg = U and 8u 2 U g(u) = f (u v0) Dh = V and 8v 2 V h(v) = f (u0 v): D-4 DICTIONARY Df d f f -1 r Rf r = f(d ), -1 d = f (r ) Figure D-1: Then g is often written f ( v0), and h is often written f (u0 ). The dot shows where to put the argument of the function. P is dened when A Df and when f (A) consists of objects which can be added (e.g., real numbers or vectors). The symbol stands for the sum of all the values f (A) for a 2 A. The number of terms in the sum is the number of objects in A, so A must be nite unless some kind of convergence is assumed. a2A f (a) IU , the identity function on the set U , has domain U and eect u j ! u. That is, IU (u) = u, 8u 2 U . \Invertible." A function f is invertible if f (d) = f (d0) ) d = d0, 8d d0 2 Df . That is, if r 2 Rf , there is exactly one d 2 Df such that r = f (d). In other words, for each r 2 Rf , the equation f (d) = r has exactly one solution d 2 Df . \Inverse," f ;1. If f is invertible, its inverse f ;1 is dened to be the function whose domain is Rf and such that for each r 2 Rf , f ;1(r) is that unique d 2 Df such that DICTIONARY D-5 f (d) = r. That is, d = f ;1(r) is the unique solution of r = f (d). Note Df ;1 = Rf , Rf ;1 = Df . See Fig. D-1. For example, if the domain is R+ , both the functions x j! x2 and x j! x3 are invertible, and their inverses are x j! x1=2 and xj! x1=3 . However, if the domain is R, the function x j! x2 is not invertible. Note that if f is invertible, so is f ;1, and (f ;1);1 = f . Mappings A Mapping is an ordered triple (U V f ) in which U and V are sets, f is a function, Df = U , and Rf V . We say that \the function f maps U into V ." The mapping is often abbreviated simply as f if the context makes clear what U and V are. f : U ! V is the usual way of writing the mapping (U V f ) if one wants to note explicitly what U and V are. The symbol \f : U ! V " is read \the mapping f of U into V ." It can also stand for the sentence \f is a function which maps U into V ." In this latter usage it is equivalent to \f (U ) V ." F (U ! V ) := the set of all functions mapping U into V . Injective. If f is invertible, f : U ! V is an \injective" mapping or an \injection." Surjective. If Rf = V , f : U ! V is a \surjective" mapping or a \surjection." Bijection. If f : U ! V is both injective and surjective, it is \bijective," or a \bijection." Note: If f : U ! V is a bijection, so is f ;1 : V ! U . Composition g  f : Suppose f : U ! V and g : V ! W . Then g  f : U ! W is the mapping dened by requiring for each u 2 U that (g  f )(u) = gf (u)]. The function g  f is called the \composition" of g with f . Note that the order in which the functions are applied or evaluated runs backward, from right to left. D-6 DICTIONARY Note: If f : U ! V and g : V ! W and h : W ! X then h  (g  f ) = (h  g)  f: (D-1) Note: If f : U ! V and g : V ! W are bijections, so is g  f : U ! W , and (g  f );1 = f ;1  g;1: Note: If f : U ! V is a bijection, f ;1  f = IU and f  f ;1 = IV . Note: Suppose f : U ! V and g : V ! U . Then i) if g  f = IU , then f is an injection ii) if f  g = IV , then f is an surjection iii) if g  f = IU , and f  g = IV then f is a bijection and g = f ;1 . Note: If f : U ! V then f  IU = IV  f = f . Permutations A permutation is a bijection  : 1 : : :  n ! f1 : : :  ng. It is an \n-permutation" or a \permutation of degree n." Sn := the set of all n-permutations. It has n! members. The product  of two permutations  and  is dened to be their composition,   . If e = If1:::ng then e = e =  and ;1 = ;1 = e for all  2 Sn. These facts, and the above notes on pg. iv, make Sn a group under the multiplication dened by  =   . It is called the symmetric group of degree n. Its identity is e. (i j ) transposition or interchange. If i 6= j and 1  i j  n (i.e. fi j g f1 : : :  ng) then (i j ) stands for the permutation  2 Sn such that (i) = j , (j ) = i, and (k) = k if k 62 fi j g. This permutation is called the \transposition" or \interchange" of i and j . Note that (ij )  (ij ) = e, the identity permutation. Theorem 1 (See, e.g., Birkho and MacLane, A Survey of Modern Algebra.) If  2 Sn,  is the product of n ; 1 or fewer transpositions. There are many ways to DICTIONARY D-7 write  as a product of transpositions, but all have the same parity (i.e., all involve an even number of transpositions or all involve an odd number of transpositions.) even, odd. If  2 Sn,  is even or odd according as it can be written as the product of an even or an odd number of transpositions. sgn  := +1 if  is even, := ;1 if  is odd. Note: sgn (1  2 ) = (sgn 1 )(sgn 2 ), and sgn e = 1. sgn  is read \signum of ." Note: sgn  = sgn ;1 because   ;1 = e, and 1 = sgn e = sgn   ;1 = (sgn )(sgn ;1 ). But sgn  = 1. Arrays An array of order q is a function f whose domain is Df = f1 : : :  n1g : : : f1 : : :  nq g: The array is said to have dimension n1 n2 : : : nq . fi1 :::iq If i1 2 f1 : : :  n1 g i2 2 f1 : : :  n2g : : :  iq 2 f1 : : :  xq g, then f (i1  : : :  iq ) is often written fi1 i2:::iq or f i1:::iq or fi1 i2 i3 i4:::iq etc. There are 2q ways to write it. The object f (i1 : : :  iq ) is called the entry in the array at location (i1  : : :  iq ) or address (i1 : : :  iq ). The integers i1  : : :  iq are the \indices", and each can be either a subscript or a superscript. What counts is their right-left order. We will never use symbols like fklij . (n) is the n-dimensional Kronecker delta. It is an n n array with ij(n) = 0 if i 6= j , ij(n) = 1 if i = j . Usually the (n) is omitted, and it is written simply ij , or ij , or i j , or i j . "(n) z n factors }| { is the n-dimensional alternating symbol. It is an n n : : : n array, with nn entries, all 0 or +1 or ;1. It is dened as follows: D-8 DICTIONARY ) = 0 if any of i  : : :  i are equal i.e., if fi  : : :  i g 6= f1 : : :  ng. "i(1n:::i 1 n 1 n n If fi1  : : :  ing = f1 : : :  ng, then there is a unique  2 Sn 3 i1 = (1) : : :  in = ) = "(n) (n). In that case, "(i1n:::i (1):::(n) = sgn . n Usually the "(n) is written without the n as "i1:::in , or "i1 i2 :::in , or "i1 :::in , etc. Example: "11 = 0, "12 = 1, "21 = ;1, "22 = 0: Example (n = 3): (n = 2) "123 = "231 = "312 = 1 "213 = "132 = "321 = ;1 all other "ijk = 0. Thus, "111 = 0, "112 = 0, etc. Einstein index conventions. i) If an index appears once, without parentheses, in each term of an equation, the equation is true for all possible values of the index. Thus, \ij = ji" says that \ij = ji" for all i j 2 f1 : : :  ng. This is a true statement. \"ijk = "ijk " says that \"ijk = "jik for i j k"f1 2 3 g. This is a false statement. The correct statement is "ijk = ;"jik . ii) If an index appears twice in any expression but a sum or dierence, a sum over all possible values of that index is understood. Thus, if Aij is an n n array, Aii stands for Pni=1 Aii. If ui and vj are rst order arrays of dimension n, uivi stands for Pni=1 uivi , and ui=vi stands for Pni=1 ui=vi, but ui + vi and ui ; vi stand for the i'th entries in two arrays, not for Pi(ui + vi) or Pi(ui ; vi ). iii) If an index appears three or more times in an expression other than a sum or dierence, a mistake has been made. Thus, if Aijkl is an n n n n array, Aiiii is never written alone. The sum is written out as Pni=1 Aiiii. iv) Parentheses \protect" an index from the conventions. Thus, A(ii) refers to a particular element in the array, the one at location (ii). And v(i) = w(i) means equality only for the particular value of i under discussion, not for all possible values of i. The element Aiiii in an n n n n array is written A(iiii) to DICTIONARY D-9 make clear that no sum is intended. Strictly speaking, the denition of the Kronecker delta, using these conventions, reads (ij) = 0 if i 6= j , and (ii) = 1 for all i. An array A of order r is symmetric (antisymmetric) in its p'th and q'th indices if they have the same dimension, np = nq , and Ai1 : : : ip;1ipip+1 : : : iq;1 iq iq+1 : : : ir = Ai1 : : : ip;1 iq ip+1 : : : iq;1ipiq+1 : : : ir : (The + is for symmetric A, the ; is for antisymmetric A). The array is totally symmetric (antisymmetric) if it is symmetric (antisymmetric) in every pair of indices. Some properties of ij and "i1:::in are listed below. 1. ij = ji 2. ii(n) = n 3. ij Ajk1:::kp = Aik1:::kp for any array A of suitable dimension. ij Ak1 jk2:::kp = Ak1ik2 :::kp , etc. In particular, ij jk = ik . 4. "i1:::in is totally antisymmetric. z n factors }| { 5. If Ai1 :::in is an n n : : : n array which is totally antisymmetric, there is a constant  such that Ai1 :::in = "i1:::in : 6. "i1:::in "j1:::jn = P2Sn (sgn) i1j
(1) : : : i1 j
(n) (The Polar Identity). The two special cases of interest to us are n = 2: "ij "kl = ik jl ; il jk which implies "ij "kj = jk and "ij "ij = 2 n = 3: "ijk "lmn = il jmkn + im jnkl + injl km ;im jlkn ; il jnkm ; injmkl D-10 DICTIONARY 7. which implies "ijk "lmk = il jm ;im jl and "ijk "ljk = 2il and "ijk "ijk = 6. Some applications of "i1 :::ln are these. If Aij is a real or complex n n array whose determinant is det A, then Ai1j1 Ai2j2 : : : Ain jn "j1:::jn = (det A) "i1:::in : In real three-space, suppose x^1  x^2  x^3 is a right-handed triple of mutually perpendicular unit vectors (that is, x^1  (^x2 x^3 ) = 1 ). Suppose ~u = uix^i and ~v = vj x^j . Then ~u  ~v = uivi ~u ~v = x^i"ijk uj vk  or (~u ~v)i := x^i  (~u ~v) = "ijk uj vk : If ~r = rix^i is the position vector in real three-space, R3 , and w~ : R3 ! R3 is a vector eld, with w~ (~r ) = wi(~r )^xi then i r~  w~ = divergence of w~ = @w @ri k r~ w~ = curl of w~ = x^i "ijk @w @rj : A very important general property of arrays is the following: Remark 1 Suppose Aijk :::kp is antisymmetric in i and j , while Sijl :::lq is symmetric in 1 i and j . Then 1 Aijk1:::kP Sijl1:::lq = 0: Proof: The subscripts k and l are irrelevant, so we omit them. We have Aij Sij = ;AjiSji because Aij = ;Aji and Sij = Sji. Now replace the summation index DICTIONARY D-11 j by i and the index i by j . Then AjiSji = Aij Sij : Putting these two results together gives Aij Sij = ;Aij Sij , so Aij Sij = 0. Vector Space Facts Fields: A eld is a set F of objects which can be added, subtracted, multiplied and divided according to the ordinary rules of arithmetic. Examples are R, C , the set p of rational numbers, and the set of numbers a + b 2, where a and b are rational. We will use only the elds R and C , and usually F = R. Denition of vector spaces: A vector space is an ordered quadruple (V , F , ad, sc) with these properties: i) V is a non-empty set. Its members are called vectors. ii) F is a eld. Its members are called scalars. iii) ad : V V ! V . This function is called vector addition. iv) sc : F V ! V . This function is called multiplication by scalars. The remaining properties make more sense if we introduce a new notation. For any ~u~v 2 V and c 2 F we dene ~u + ~v : = ad(~u~v) c~v = ~vc := sc(c~v) The additional properties which make (V F ad sc) a vector space are properties of ad and sc as follows (letters with ! are vectors, those without ! are scalars): v) ~u + ~v = ~v + ~u vi) ~u + (~v + w~ ) = (~u + ~v) + w~ D-12 DICTIONARY vii) (a + b)~u = a~u + b~u viii) a(~u + ~v) = a~u + a~v ix) a(b~u) = (ab)~u x) 0~u = 0~v for any ~u~v 2 V xi) 1~u = ~u The vector 0~u is called the zero vector, written ~0. The vectors (;1)~v and ~u + (;1)~v are written ;~v and ~u ; ~v respectively. Example 1: F is a eld. V is the set of all ordered n-tuples of members of F , so a typical vector is ~u = (u1 : : : un) with u1 : : :  un 2 F: The addition and multiplication functions are ~u + ~v := ad(~u~v) := (u1 + v1  : : :  un + vn) c~v := sc(c~v) := (cv1  : : :  cvn) The zero vector is ~0 = (0 : : :  0). The space V is usually written F n. Example 2: F = R. V is the set of all ordered n tuples of members of R+ , so a typical vector is ~u = (u1 : : :  un) with u1 : : :  un 2 R+ : The addition and multiplication functions are ~u + ~v := ad(~u~v) := (u1v1  : : :  unvn) c~v := sc(c~v) := (v1c : : :  vnc ): The zero vector is ~0 = (1 1 : : :  1). Example 3: Let (V F ad sc) be any vector space and let S be any non-empty set. Dene a new vector space (VS  F adS  scS ) as follows: VS = F (S ! V ) the set of all functionsf : S ! V : DICTIONARY D-13 To dene adS and scS we note that if f and g are functions in VS and c 2 F then adS (f g) and scS (c f ) are supposed to be functions in VS , i.e., adS (f g) : S ! V and scS (c f ) : S ! V . To dene these functions, we must say what values they assign to each s 2 S . The denitions we adopt are these: adS (f g)] (s) = f (s) + g(s) scS (c f )] (s) = cf (s): (D-2) (D-3) Following the convention for vector spaces, we write adS (f g) as f + g and scS (c f ) as cf . Thus, f + g and cf are functions whose domains are S and whose ranges are subsets of V . The denitions (1) and (2) read thus: for any s 2 S (f + g)(s) = f (s) + g(s) (cf )(s) = cf (s): (D-4) (D-5) These are not \obvious facts." They are denitions of the vector-valued functions f + g and cf . In VS , the zero vector is the function which assigns to each s 2 S the zero vector in V. Example 4: F = real numbers = R. V = set of all continuous real-valued functions on the closed unit interval 0  x  1. ad and sc are dened as in Example 3, page D-12. (This is a vector space, because if f and g are continuous functions so is f + g and so is cf for any real c.) Example 5: F = R. V = set of all continuous positive real valued functions on 0  x  1. ad and sc as dened in Examples 3 and 4. This is not a vector space. There are scalars c 2 F and vectors f 2 V such that cf 6 2V . For example, if f 2 V then (;1)f 6 2V . In other words, the function sc in this case is not a mapping from F V to V . Notation: Usually the vector space (V F ad sc) will be called simply the vector space V . We will almost never use the notations ad(~u~v) or sc(c~v), but will use ~u + ~v D-14 DICTIONARY and c~v and ~vc. V will be called a vector space over F , or a real or complex vector space if F = R or C . Because of vector space rules v) and vi), there is no ambiguity about Pni=1 ~ui when ~u1 : : :  ~un 2 V . For example, if n = 4, all of ~u1 +u2 +(~u3 + u4)], (~u2 +~u4)+(~u3 +~u1), ~u1 + ~u2) + u~3] + u~4 ~u4 + (~u2 + ~u1)] + ~u3, etc., are the same. When ~u1 : : :  ~un 2 V and a1  : : :  an 2 F , we do not even need the P. We can use the index conventions to write Pni=1 ai~ui as ai~ui. Facts about ~0. If a 2 F , then a~0 = ~0 because a~0 = a(0~0) = (a0)~0 = 0~0 = ~0. Also, if a 2 F , ~v 2 V , and a 6= 0 , ~v 6= ~0, then a~v 6= ~0. For suppose a~v = ~0 and a 6= 0. Then a;1 2 F , and ~v = 1~v = (a;1a)~v = a;1 (a~v) = a;1~0 = ~0. Finally, ~0 + ~u = ~u because ~0 + ~u = 0~u + 1~u = (0 + 1)~u = 1~u = ~u. Linear mappings: Suppose V and W are both vector spaces over the same eld F , and L 2 F (V ! W ). We call L a \linear mapping" if for any ~u, ~v 2 V and c 2 F it is true that L(~u + ~v) = L(~u) + L(~v) L(c~u) = cL(~u): L(V ! W ) is the set of all linear mappings from V to W . 1 It is a vector space over F if it's ad and sc are dened as in example 3, page D-12. If L 2 L(V ! W ) is a bijection, then L;1 2 L(W ! V ). And then L is called an \isomorphism" and V and W are said to be isomorphic. Intuitively speaking, two isomorphic vector spaces are really the same space. Each member of V corresponds to exactly one member of W , and vice versa, and ad and sc can be applied in either V or W with the same result. For example, if F = R, then the two spaces in examples 1 and 2 on page D-12 are isomorphic. If V is the space in Example 2, 1 Note that if L 2 L(V ! W ) and M 2 L(U ! V ), then L  M 2 L(U ! W ). DICTIONARY D-15 then the isomorphism L : Rn ! V is given by L(u1  : : :  un) = (eu1  : : :  eun ) and L;1(v1  : : :  vn) = (ln v1  : : : ln vn): ~ sc~ ) with Subspaces: A subspace of vector space (V F ad sc) is a vector space (V~  F ad these properties: i) V~ V f = ad jVe Ve ii) ad iii) sfc = sc jF Ve .   If V~ is any subset of V , then (V~  F ad Ve Ve  sc  F Ve ) is a subspace of (V F ad sc) if and only if a) ad(Ve Ve ) Ve and also b) sc(F V~ ) Ve . Intuitively, subspaces are lines, planes and hyperplanes in V passing through the origin. (If Ve is a subspace of V , and ~v 2 Ve , then 0~v 2 Ve by iii), but 0~v = ~0. Thus the zero vector in V is the zero vector in every subspace of V .) Finite dimensional vector spaces. V is a vector space over eld F . U is an arbitrary non-empty subset of V . Linear combination: If ~u1 : : :  u~n 2 V and a1  : : :  an 2 F , the vector ai~ui is called a linear combination of f~u1 : : :~ung. Span: The set of all linear combinations of nite subsets of U is written sp U . It is called the span of U . It is a subspace of V , so it is called the subspace spanned or generated by U . If U is a subspace, U = sp U . D-16 DICTIONARY Finite dimensional. A vector space V is \nite dimensional" if V = sp U for some nite subset U V . Linear dependence. A set U V is \linearly dependent" if there are nitely many vectors ~u1 : : :  ~un 2 U and scalars a1  : : :  an such that i) at least one of a1  : : :  an is not 0, and also ii) ai~ui = ~0. It is an important theorem that if (~u1 : : :  ~un) is a sequence such that f~u1 : : :  ~ung is linearly dependent, then there is an m 2 f1 : : :  ng such that ~um 2 spf~u1 : : :  ~um;1g. Linear independence. A set U V is linearly independent if it is not linearly dependent. Linear independence of U means that whenever f~u1 : : :~ung U and ai~ui = ~0, then a1 = : : : = an = 0. Basis. Any linearly independent subset of vector space V which spans V is called a basis for V . The following are important theorems about bases. Theorem 2 If V is nite dimensional, it has a basis, and all its bases are nite and contain the same number of vectors. This number is called the dimension of V , written dim V . Theorem 3 If V is nite dimensional and W is a subspace of V , W is nite dimensional and dim W  dim V . Every basis for W is a subset of a basis for V. Theorem 4 If f~u1 : : :  ~ung spans V then dim V  n. If dim V = n, then f~u1 : : :  ~ung is linearly independent, hence a basis for V . Theorem 5 If V is nite dimensional and U is a linearly independent subset of V , U is nite, say U = f~u1 : : :  ~ung, and n  dim V . If n = dim V , U spans V and so is a basis for V . Theorem 6 Suppose B = f~b1  : : :~bn g is a basis for V . Then there are functions c1B  : : :  cnB 2 L(V ! F ) such that for any ~v 2 V ~v = cjB (~v)~bj : DICTIONARY D-17 The scalars v1 = c1B (~v) : : :  vn = cnB (~v), which are uniquely determined by ~v and B , are called the coordinates of ~v relative to the basis B . The linear functions c1B  : : :  cnB are called the coordinate functionals for the basis B . (A function whose values are scalars is called a \functional.") Clearly ~bi = i j~bj . Since the coordinates of any vector relative to B are unique, it follows that cjB (~bi ) = i j (D-6) Theorem 7 Suppose V and W are vector spaces over F , and B = f~b1 : : :  ~bng is a basis for V , and fw~ 1 : : :  w~ ng"W . Then 91L 2 L(V ! W ) 3 L(~bi ) = w~ i. (In other words, L is completely determined by LjB for one basis B , and LjB can be chosen arbitrarily.) (The L whose existence and uniqueness are asserted by the theorem is easy to nd. For any ~v 2 V , L(~v) = ciB (~v)w~ i.) This L is an injection if fw~ 1 : : :  w~ ng is linearly independent and a surjection if W = spfw~ 1 : : :  w~ ng. Linear operators: If L 2 L(V ! V ), L is called a \linear operator on V ." If B = f~b1  : : :  ~bn g is any basis for V , then the array Li j = cjB L(~bi )] is called the matrix of L relative to B . Clearly, L(~bi ) = Li j~bj . (The matrix of L relative to B is often dened as the transpose of our Li j . It is that matrix 3 L(~bi ) = ~bj Lj i . Our denition is the more convenient one for continuum mechanics.) The determinant of the matrix of L relative to B depends only on L, not on B , so it is called the determinant of L, written det L. A linear operator L is invertible i det L 6= 0. If it is invertible it is an isomorphism (a surjection as well as an injection). If L 2 L(V V ), ~v 2 V ,  2 F , and ~v 6= ~0, and if L(~v) = ~v (D-7) then  is an \eigenvalue" of L, and ~v is an \eigenvector"of L belonging to eigenvalue . The eigenvalues of L are the solution  2 F of the polynomial equation det(L ; IV ) = 0: (D-8) D-18 DICTIONARY Linear operators on a nite dimensional complex vector space always have at least one eigenvalue and eigenvector. On a real vector space this need not be so. If L and M 2 L(V V ), then L  M 2 L(V V ) and det(L  M ) = (det L)(det M ). Euclidean Vector Spaces Denition: A Euclidean vector space is an ordered quintuple (V R ad sc dp) with these properties: i) (V R ad sc) is a nite-dimensional (real) vector space ii) dp is a functional on V V which is symmetric, positive-denite and bilinear. In more detail, the requirements on dp are these: a) dp : V V ! R. (dp is a functional) b) dp(~u~v) = dp(~v ~u) (dp is symmetric) c) dp(~u ~u) > 0 if ~u 6= ~0. (dp is positive denite) d) for any ~u ~v 2 V , dp(~v ) and dp(~u ) 2 L(V ! R). (dp is bilinear). dp is called the dot-product functional, and dp(~u~v) is usually written ~u  ~v. In this notation, the requirements on the dot product are a)0 ~u  ~v 2 R (dp is a functional) b)0 ~u  ~v = ~v  ~u (dp is symmetric) c)0 ~u  ~u > 0 if ~u 6= ~0 (dp is positive denite) d)0 for any ~u ~u1 : : :  ~un ~v 2 V and a1  : : :  an 2 F (ai~ui)  ~v = ai (~ui  ~v) ~v  (ai~ui) = ai (~v  ~ui) 9 > = dp is bilinear >  Example 1: (V R ad sc) is as in Example 1, page D-12, and ~u  ~v = uivi. For n = 3 this is ordinary three-space. DICTIONARY D-19 Example 2: (V R ad sc) is as in Example 4, page D-13. If f and g are two functions in R V , their dot product is f  g = 01 dxf (x)g(x). Note that without continuity, f = 6 0 need not imply f  f > 0. For example, let f (x) = 0 in 0  x  1 except at x = 12 . There let f ( 21 ) = 1. Then f = 6 0 but f  f = 0. Length and angle. Schwarz and triangle inequalities. The length of a vector ~v in p a Euclidean space is dened as k~v k := ~v  ~v. All nonzero-vectors have positive length. From the denitions on page D-18, j~u  ~vj  k~uk k~vk (Schwarz inequality) j~u + ~vj  k~uk + k~uk (triangle inequality) If ~u 6= ~0 and ~v 6= ~0 then ~u  ~v=k~u kk~v k is a real number between ;1 and 1, so it is the cosine of exactly one angle between 0 and . This angle is called the angle between ~u and ~v, and is written 6 (~u~v). Thus, by the denition of 6 (~u~v), ~u  ~v = k~ukk~vk cos 6 (~u~v): In particular, ~u  ~v = 0 , 6 (~u~v) = 2 : If k~uk = 1, we will write u^ for ~u. Orthogonality: If V is a Euclidean vector space, and ~u~v 2 V , P V , Q V , then ~u?~v means ~u  ~v = 0: ~u?Q means ~u  ~q = 0 for all ~q 2 Q: P ?Q means ~p  ~q = 0 for all p~ 2 P and ~q 2 Q: Q? := f~x : ~x 2 V and ~x?Qg. This set is called \the orthogonal complement of Q." It is a subspace of V . Obviously Q?Q?. Slightly less obvious, but true in nite dimensional spaces, is (Q?)? = sp Q: n o n o Theorem 8 Q spans V , Q? = ~0 . In particular, if ~b1  : : :  ~bn is a basis for V and ~v  ~bi = 0, then ~v = ~0. Also, if ~v  ~x = 0 for all ~x 2 V , then ~v = ~0. Orthonormal sets and bases. An orthonormal set Q in V is any set of mutually perpendicular unit vectors. I.e., ~u 2 Q ) k~uk = 1 and ~u~v 2 Q, ~u 6= ~v ) ~u  ~v = 0. D-20 DICTIONARY An orthonormal set is linearly independent, so in a nite dimensional Euclidean space, any orthonormal set is nite, and can have no more than dim V members. An orthonormal basis in V is any basis which is an orthonormal set. If B = fx^1  : : :  x^ng is an orthonormal basis for V , the coordinate functionals ciB are given simply by ciB (~v ) = ~v  x^i 8~v 2 V: (D-9) That is, for any ~v, ~v = (~v  x^i )^xi . Theorem 9 Every Euclidean vector space has an orthonormal basis. If U is a subspace of Euclidean vector space V , U is itself a Euclidean vector space, and every orthonormal basis for U is a subset of an orthonormal basis for V . Linear functionals and dual bases: Let V be a Euclidean vector space. For any ~v 2 V we can dene a ~v 2 L(V ! R) by requiring simply ~v (~y) = ~v  ~y 8~y 2 V: (D-10) Theorem 10 Let V be a Euclidean vector space. The mapping ~v j! ~v dened by equation (9) is an isomorphism between V and L(V ! R): (Therefore we can think of the linear functionals on V , the members of L(V ! R), simply as vectors in V .) Proof: We must show that the mapping ~v j! ~v is an injection, a surjection, and linear. Linearity is easy to prove. We want to show ai~vi = ai ~vi . But for any ~y 2 V , ai~vi (~y) = (ai~vi)  ~y = ai (~vi  ~y) = ai ~vi (~y )] = (ai ~vi )(~y ): (See denition of ai ~vi in example 3, page D-12) Therefore the two functions ai~vi and ai~vi are equal. DICTIONARY D-21 To prove that ~v j! ~v is injective, suppose ~u = ~v . We want to prove ~u = ~v. For any ~y 2 V , we know ~u(~y) = ~v (~y), so ~u  ~y = ~v  ~y, or (~u ; ~v)  ~y = 0. In particular, setting ~y = ~u ; ~v we see (~u ; ~v)  (~u ; ~v) = 0, so (~u ; ~v) = ~0, ~u = ~v. To prove that ~v j! ~v is surjective, let  2 L(V ! R): We want to nd a ~v 2 V 3  = ~v . Let fx^1  : : :  x^n g be an orthonormal basis for V and dene ~v = (^xi)^xi : Then for any ~y 2 V , ~v (~y) = ~v  ~y = (^xi)(^xi  ~y) = (^xi )yi = (yix^i ) = (~y): Thus, ~v = . QED. Even in a Euclidean space, it is sometimes convenient to work with bases which are not orthonormal. The advantage of an orthonormal basis x^1  : : :  x^n is that for any ~v 2 V , ~v = (~v  x^i)^xi . An equation almost as convenient as this can be obtained for an arbitrary basis. We need some denitions. All vectors are in a Euclidean vector space V . Denition: Two sequences of vectors in V , (~u1 : : :  ~um) and (~v1  : : : ~vm ), are dual to each other if ~ui  ~vj = ij . Remark 2 If (~u1 : : :  ~um) and (~v1 : : : ~vm) are dual to each other, each is linearly independent. Proof: If ai~ui = ~0, then (ai~ui)  ~vj = 0, so ai(~ui  ~vj ) = 0, so aiij = 0, so aj = 0. Denition: An ordered basis for V is a sequence of vectors (~b1  : : :  ~bn) such that f~b1  : : :  ~bng is a basis for V . Remark 3 If two sequences of vectors are dual to each other and one is an ordered basis, so is the other. (Proof. If (~v1 : : : ~vm ) is a basis, m = dim V . Then, since (~u1 : : :  ~um) is linearly independent, it is also a basis. D-22 DICTIONARY Remark 4 If (~b1  : : :  ~bn) is an ordered basis for V , it has at most one dual sequence (~b1  : : :  ~bn). Proof: If (~b1  : : :  ~bn) and (~v1 : : : ~vn) are both dual sequences to (~b1  : : :  ~bn), then ~bi ~bj = ij , and ~vi  bj = ij , so (~bi ; ~vi ) ~bj = 0. Then (~bi ; ~vi)  (aj~bj ) = 0 for any ai : : :  an 2 R. But any ~u 2 V can be written ~u = aj~bj , so (~bi ; ~vi)  ~u = 0 for all ~u 2 V . Hence, ~bi ; ~vi = 0, ~bi = ~vi. Remark 5 If (~b1  : : :  ~bn ) is an ordered basis for V , it has at least one dual sequence (~b1  : : :  ~bn). Proof: Let B = f~b1  : : :~bn g, and let ciB be the coordinate functions for this basis. They are linear functionals on V , so according to theorem (10) there is exactly one ~bi 2 V such that ~bi = ciB . Then ~bi (~y) = ciB (~y ), 8~y 2 V , so ~bi  ~y = ciB (~y ), 8~y 2 V . In particular, ~bi ~bj = ciB (~bj ) = i j see (5)]. Thus (~b1  : : :  ~bn) is dual to (~b1  : : :  ~bn). From the foregoing, it is clear that each ordered basis B = (~bi  : : :  ~bn) for V has exactly one dual sequence B D = (~b1  : : :  ~bn ) , and that this dual sequence is also an ordered basis for V . It is called the dual basis for (~b1  : : :  ~bn). It is characterized and uniquely determined by ~bi  ~bj = i j : (D-11) There is an obvious symmetry: each of B and B D is the dual basis for the other. If ~v 2 V , and B = (~b1  : : :  ~bn) is any ordered basis, we can write ~v = vj~bj  (D-12) DICTIONARY D-23 where vj = cjB (~v) are the coordinates of ~v relative to B . Then ~bi  ~v = vj~bi  ~bj = vj ji = vi. Thus vi = ~bi  ~v: (D-13) Since (~b1 : : :  ~bn) = B D is also an ordered basis for V , we can write ~v = vj~bj : (D-14) Because duality is symmetrical, it follows immediately that vj = ~v  ~bj : (D-15) An orthonormal basis is its own dual basis it is self-dual. If ~bi = x^i then ~bi = x^i , and (11), (12) are the same as (13) and (14). We have vj = vj = x^j  ~v: In order to keep indices always up or down, it is sometimes convenient to write an orthonormal basis (^x1 : : :  x^n) as (^x1 : : :  x^n), with x^i = x^i. Then vj = vj = x^j  ~v = x^j  ~v: If (~b1  : : :  ~bn ) = B is an ordered basis for V , and B D = (~b1  : : :  ~bn) is its dual basis, and ~v 2 V , we can write ~v = vi~bi = vi~bi : (D-16) The vi are called the contravariant components of ~v relative to B , and the vi are the covariant components of ~v relative to B (and also the contravariant components of ~v relative to B D ). The contravariant components of ~v relative to B are the coordinates of ~v relative to B . The covariant components of ~v relative to B are the coordinates of ~v relative to B D . The covariant metric matrix of B D is gij = ~bi  ~bj : (D-17) D-24 DICTIONARY The contravariant metric matrix of B D is gij = ~bi  ~bj : (D-18) Clearly gij is the contravariant and gij the covariant metric matrix of B D . We have vi = ~bi  ~v = ~bi  (~bj vj ) = (~bi  ~bj )vj so vi = gij vj : (D-19) Similarly, or by the symmetry of duality, vi = gij vj : (D-20) The metric matrices can be used to raise or lower the indices on the components of ~v. Transposes: Suppose V and W are Euclidean vector spaces and L 2 L(V ! W ). Then there is exactly one LT 2 F (W ! V ) such that L(~v)  w~ = ~v  LT (w~ ) 8(~v w~ ) 2 V W: (D-21) This LT is linear, i.e. LT 2 L(W ! V ). Furthermore, (LT )T = L, and if a1  : : :  an 2 R and L1  : : :  Ln 2 L(V ! W ), then (aiLi)T = ai(LTi ) (D-22) The linear mapping LT is called the \transpose" of L (also sometimes the adjoint of L, but adjoint has a second, quite dierent, meaning for matrices, so we avoid the term). If L 2 L(V ! W ) and M 2 L(U ! V ), then L  M 2 L(U ! W ) and it is easy to verify that (L  M )T = M T  LT : (D-23) If L 2 L(V ! W ) is bijective, then L;1 2 L(W ! V ). It is easy to verify that LT : W ! V is also bijective and (L;1 )T = (LT );1 : (D-24) DICTIONARY D-25 Orthogonal operators or isometries: If V is a vector space, the members of L(V ! V ) are called \linear operators on V ." If V is Euclidean and L 2 L(V ! V ), then L is called \orthogonal" or an \isometry" if kL(~u) ; L(~v)k = k~u ; ~vk for all ~u and ~v in V . If L 2 L(V ! V ), then each of the following conditions is necessary and su"cient for L to be an isometry: a) kL~vk = k~vk 8 ~v 2 V b) L(~u)  L(~v) = ~u  ~v 8~u ~v 2 V c) LT  L = IV d) L  LT = IV e) LT = L;1 f) For every orthonormal basis B V , L(B ) is orthonormal. That is, if B = fx^1  : : :  x^ng is any orthonormal basis, then fL(^x1) : : :  L(^x)g is also an orthonormal basis. g) There is at least one orthonormal basis B 3 L(B ) is orthonormal. h) For every orthonormal basis B V , the matrix of L relative to B has orthonormal rows and columns (i.e., Li j Li k = jk , and Li j Lk j = ik ). i) There is at least one orthonormal basis B V relative to which the matrix of L has either orthonormal rows (Li j Lk j = ik ) or orthonormal columns (Li j Li k = ij ). If L is orthogonal, det L = 1. If det L = +1, L is called \proper." Otherwise L is \improper." #(V ) := set of all orthonormal operators on V . #+(V ) := set of all proper orthonormal operators on V . If L and M 2 #(V ) or #+(V ), the same is true of L  M and L;1 . Also IV 2 #(V ) and #+(V ). Therefore, #(V ) and #+(V ) are both groups if multiplication is dened D-26 DICTIONARY to mean composition. #(V ) is called the orthogonal group on V , while #+(V ) is the proper orthogonal group. If L 2 #+(V ), L is a \rotation." This implies that there is an orthonormal basis x^1  : : :  x^2m  x^2m+1 and angles #1 : : :  #m between 0 and  such that L(^x2i;1 ) = x^2i;1 cos i + x^2i sin i L(^x2i ) = ;x^2i;1 sin i + x^2i cos i L(^x2n+1 ) = x^2n+1 : 9 > > > = 8i 2 f1 : : :  mg > > >  The vector x^2n+1 is omitted in the above denition if dim V is even, and included if dim V is odd. If included, it is the \axis" of the rotation. If L 2 #(V ) is improper, L is the product of a rotation and a \reection." A reection is an L for which there is an orthonormal basis x^1  : : :  x^n such that L(^x1 ) = ;x^1 , L(^xn ) = x^k , if k  2. Symmetric operators: A symmetric operator on Euclidean space V is an L 2 L(V V ) 3 LT = L. (Antisymmetric means LT = ;L). Each of the following conditions is necessary and su"cient for a linear operator L to be symmetric. a) LT = L b) L(~u)  ~v = L(~v)  ~u 8~u ~v 2 V c) The matrix of L relative to every orthonormal basis for V is symmetric d) The matrix of L relative to at least one orthonormal basis for V is symmetric. e) There is an orthonormal basis for V consisting entirely of eigenvectors of L. That is, there is an orthonormal ordered basis (^x1 : : :  x^n) for V , and a sequence of real numbers (1 : : :  n) (not necessarily dierent) such that L(^x(i) ) = (i) x^(i)  8i 2 f1 : : :  ng: (D-25) Facts: The eigenvalues f1 : : :  ng are uniquely determined by L, but the eigenvectors are not. For any real  let  L ] denote the set of all ~v 2 V such that L(~v) = ~v. DICTIONARY D-27 Then  L ] is a subspace of V , and dim  L ] > 0 ,  is an eigenvalue of L. In that case,  L ] is called the eigenspace of L with eigenvalue . The eigenspaces are uniquely determined by L. If 1 6= 2 ,  L 1] ?  L 2 ] . Positive-denite (semidenite) symmetric operators. If L 2 L(V ! V ) is symmetric, L is said to be positive denite when L(~v)  ~v > 0 for all nonzero ~v 2 V (D-26) and positive semidenite when L(~v)  ~v  0 for all ~v 2 V: (D-27) A symmetric linear operator is positive denite (semidenite) i all its eigenvalues are positive (non-negative). Every symmetric positive (semi) denite operator L has a unique positive (semi) denite symmetric square root, i.e., a symmetric positive (semi) denite operator L1=2 such that L1=2  L1=2 = L (D-28) Polar Decomposition Theorem: (See Halmos, p. 169 ). Suppose V is Euclidean and L 2 L(V ! V ). Then there exist orthogonal operators O1 and O2 and positive semidenite symmetric operators S1 and S2 such that L = O1  S1 = S2  O2 : (D-29) Both S1 and S2 are uniquely determined by L (in fact S1 = (LT L)1=2 and S2 = (LLT )1=2 ). If L;1 exists, O1 and O2 are uniquely determined by L and are equal, and S1 and S2 are positive denite. If det L > 0 then O1 is proper. The existence part of the polar decomposition theorem can be restated as follows: for any L 2 L(V ! V ) there are orthonormal bases (^x1  : : :  x^n) and (^y1 : : :  y^n) in V and non-negative numbers 1  : : :  n such that D-28 DICTIONARY L(^x(i) ) = (i) y^(i)  i = 1 : : :  n: (D-30) If L is invertible, all (i) > 0. Stated this way, the theorem remains true for any L(V ! W ) if V and W are Euclidean spaces and dim V  dim W . If dim V > dim W , then (29) holds for i = 1 : : :  m, with m = dim W , while for i = m + 1 : : :  n (with n = dim V ) (29) is replaced by L(^xi ) = ~0: (D-31) Five Prerequisite Proofs PP-1 The Polar Identity Let Sn := group all permutations on f1 : : :  ng. "i1:::in := n dimensional alternating symbol := 0 if fi1  : : :  ing 6= f1 : : :  ng := sgn  if i1 = (1) : : :  in = (n)  2 Sn: Then the polar identity is X ( sgn )i1j
(1)    in j
(n) = "i1:::in "j1:::jn : (PP-1) 2Sn Lemma PP-1 Suppose Ai1:::in is an n : : : n dimensional n'th order array which is totally antisymmetric (i.e. changes sign where any two indices are interchanged). Then Ai1 :::in = (A12:::n) "i1:::in : (PP-2) Proof: If any two of fi1  : : :  ing are equal, A(i1 :::in ) = 0. If they are all dierent, there is a  2 Sn such that i1 = (1) : : :  in = (n). This  is a product of transpositions (interchanges), and by carrying out these interchanges in succession on A12:::n one obtains A(1):::(n). Each interchange produces a sign change, so the nal sign is + or ; according as  is even or odd. Thus, A(1):::(n) = ( sgn )A1:::n = "(1):::(n) A1:::n. Therefore, equation PP-2, holds both when fi1  : : :  ing contains a repeated index and when it does not. PP-3 PP-4 PROOFS Lemma PP-2 Suppose W is a vector space and f : Sn ! W and  2 Sn. Then X X ;1 X f () = f ( ) = f (): 2Sn 2Sn 2Sn Proof: (PP-3) Both the mapping  j! ;1 and the mapping  j!  are bijections of Sn to itself. Therefore, each of the sums in (PP-3) contains exactly the same terms. These terms are vectors in W , so they can be added in any order. Lemma PP-3 Dene Ai1 :::inj1:::jn = Then and for any  2 Sn, X ( sgn )i1 j
(1) : : : in j
(n) 2Sn Ai1 :::inj1:::jn = Aj1:::jni1:::in (PP-5) Ai1 :::inj(1):::j(n) = ( sgn )Ai1 :::in j1:::jn (PP-6) Proof: To prove (PP-5) note that by lemma 2, since sgn ;1 = sgn , Ai1 :::inj1:::jn = The set of n pairs is the same as the set because a pair (PP-4) X ( sgn )i1 j
;1(1) : : : inj
;1 (n): 2Sn 80 1 0 19 > > : ;1(1)  ;1(n) > 80 1 0 19 > B  : : :  @ A @ A > : 1  n > 0 B@ 1 iC A j PROOFS PP-5 is in the rst set of pairs i j = ;1 (i) and in the second set of pairs i i = (j ). Therefore i1 j
;1(1) : : : inj
;1(n) = i
(1) j1 : : : i
(n) jn and X ( sgn )i
(1)j1 : : : i
(n) jn 2Sn X = ( sgn )j1i
(1) : : : jn i
(n) 2Sn = Aj1:::jni1 :::in : Ai1:::in j1:::jn = This proves equation (PP-5). To prove equation (PP-6), let  2 Sn and let j1 = k(1)  : : :  jn = k(n) . Then from equation (PP-4), X ( sgn )i1 k
(1) : : : in k
(n) : 2Sn Now ( sgn )2 = 1 and so ( sgn ) = ( sgn )2 ( sgn ) = ( sgn ) ( sgn )( sgn )] = ( sgn )( sgn ), and X Ai1 :::in k(1):::k(n) = ( sgn ) ( sgn )i1 k
(1) : : : ink
(n) 2Sn X = ( sgn ) ( sgn )i1 k
(1) : : : in k
(n) by Lemma PP-2 2Sn = ( sgn )Ai1 :::in k1:::kn : Ai1 :::ink(1):::k(n) = This proves equation (PP-6). Now we can prove the polar identity. Let Ai1:::in j1:::jn be as in (PP-4). Then x i1  : : :  in. By lemma 1 and equation (PP-6), Ai1:::in j1:::jn = Ai1 :::in12:::n"j1:::jn : By equation (PP-5), interchanging the i's and j 's gives Ai1 :::in j1:::jn = A12:::n j1:::jn "i1 :::in : (PP-7) PP-6 PROOFS Set i1 = 1 : : :  in = n, and equate these two. The result is A12:::n j1:::jn "12:::n = A12:::n12:::n"j1:::jn so A12:::n j1:::jn = A12:::n 12:::n"j1:::jn : (PP-8) Substituting (PP-8) in (PP-7) gives Ai1 :::in j1:::jn = A12:::n 12:::n"i1:::in "j1:::jn : (PP-9) Then from the denition (PP-4), with ir = r, js = s A12:::n12:::n = X ( sgn )1(1) : : : n(n): 2Sn There are n! terms in this sum, and a term vanishes unless (1) = 1, (2) = 2 : : : (n) = n that is, unless  is the identity permutation e. But sgn e = 1, so only one of the n! terms in the sum is nonzero, and this term is 1. Therefore, A12:::n12:::n = 1, and equation (PP-9) becomes the polar identity (PP-1). Spectral Decomposition of a Symmetric Linear Operator S on a Euclidean Space V . Denition PP-1 S : V ! V is symmetric i S = S T , that is i ~u  S (~v) = S (~u)  ~v for all ~u, ~v 2 V . Theorem PP-11 Let S be a symmetric linear operator on Euclidean space V . Then V has an orthonormal basis consisting of eigenvectors of S . Lemma PP-4 Suppose S is symmetric, U is a subspace of V , and S (U ) U . Then S (U ?) U ? and S jU ? is symmetric as a linear operator on U ?. Proof: Suppose w~ 2 U ?. We want to show S (w~ ) 2 U ?. That is, for any ~u 2 U , we want to show ~u  S (w~ ) = 0. But ~u  S (w~ ) = S (~u)  w~ . By hypothesis, S (~u) 2 U and w~ 2 U ?, so S (~u)  w~ = 0. Thus, S (w~ ) 2 U ?. Therefore S (U ?) U ?. And if the denition (PP-1) above is satised for all ~u, ~v 2 V , it certainly holds for all ~u, ~v 2 U ?. Hence, S jU ? is symmetric. Lemma PP-5 Suppose S : V ! V is a symmetric linear operator on Euclidean space V . Then S has at least one eigenvalue  such that S (~v) = ~v, for some ~v = 6 ~0. Proof: PP-7 PP-8 PROOFS Let x^1 : : :  x^n be an orthonormal basis for V , and let Sij be the matrix of S relative to this basis, so S (^xi ) = Sij x^j . Then Sij = S (^xi )  x^j = x^i  S (^xj ) = S (^xj )  x^i = Sji so Sij is a symmetric matrix. A real number  is an eigenvalue of S i det(S ; I ) = 0, where I is the identity operator on V . (If the determinant is 0, then S ; I is singular, so there is a nonzero ~v 3 (S ; I )(~v) = ~0, or S (~v) ; ~v = ~0, or S (~v) = ~v.) The matrix of S ; I relative to the basis (^x1  : : :  x^n) is Sij ; ij , so we want to prove that the n'th degree polynomial in , det(Sij ; ij ), has at least one real zero. It certainly has a complex zero, , so there is a complex n-tuple (r1 : : :  rn) such that (Sij ; ij )rj = 0 or Sij rj = ri : Let  denote complex conjugation. Then riSij rj = (riri): (PP-10) Taking the complex conjugate of this equation gives (Sij is real) riSij rj =  (riri): Changing the summation indices on the left gives rj Sjiri =  (riri): (PP-11) But Sij = Sji, so the left hand sides of (PP-10) and (PP-11) are the same. Also, riri = riri 6= 0. Hence,  = . Thus  is real, and we have proved lemma (PP-5). Now we can prove theorem PP-1. Let  be any (real) eigenvalue of S . Let v^ be a unit eigenvector with this eigenvalue. Then S (spfv^g) spf~vg so by lemma PP-4, S (fv^g?) fv^g? and S jfv^g? is symmetric. We proceed by induction on dim V . Since dimfv^g? < dim V , we may assume theorem PP-4 true on fv^g?. Adjoining v^ to the orthonormal basis for fv^g? which consists of eigenvectors of S jfv^g? gives an orthonormal basis for V consisting of eigenvectors of S . PROOFS PP-9 Denition PP-2 Let S be a linear operator on Euclidean space V . A \spectral decom- position of S " is a collection of real numbers 1 < : : : < m and a collection of subspaces of V , U1  : : :  Um, such that i) S jU(i) = (i) I jU(i) for i = 1 : : :  m ii) V = U1  : : :  Um . (This means a) Ui ?Uj if i 6= j . That is, if ~ui 2 Ui and ~uj 2 Uj then u~i  ~uj = 0 b) every ~v 2 V can be written in exactly one way as ~v = ~u1 +    + ~um with ~ui 2 Ui: ) Corollary PP-1 If S is a symmetric linear operator on Euclidean space V , S has at least one spectral decomposition. Proof: Order the dierent eigenvalues of S which theorem PP-1 produces as 1 <    < m . Let Ui be the span of all the orthonormal basis vectors turned up by that theorem which have eigenvalue i. Corollary PP-2 S has at most one spectral decomposition. Proof: Let 10  <    < u0 and U10  : : : Uu0 be a second spectral decomposition. Let 0 be one of 10  : : :  u0 , and let ~v 0 be a corresponding nonzero eigenvector. Then S (~v 0) = 0~v 0. But ~v 0 = ~u1 +    + ~um where ~ui 2 Ui , and S (~v) 0 = 0~v 0 = 0~u1 + : : : + 0~um. Hence S (~v 0) = S (~u1) +    + S (~um) = 1~u1 +    + m~um. Hence, (0 ; 1 )~u1 +    + (0 ; m )~um = ~0: (PP-12) Since ~v 0 6= ~0, there is at least one ~uj 6= ~0. Then, dotting this ~uj into (PP-12) gives (0 ; (j) )~u(j)  ~u(j) = 0, so 0 ; j = 0, or 0 = j . Thus every i0 is PP-10 PROOFS included in f1  : : :  mg. By symmetry, every i is included in f10  : : :  u0 g. Therefore the two sets of eigenvalues are equal, and u = m. Since the eigenvalues are ordered by size, i0 = i for i = 1 : : :  m. Now suppose ~u 0 2 Ui0 . Then ~ui0 = ~u1 +    + ~um and S (~ui0) = (i)~u(0i) = i(~u1 +    + ~um) = S (~u1)+   + S (~um) = (1)~u1 +    + m~um so(i ; 1 )~u1 +    +(i ; m )~um = ~0. Dot in ~uj and we get (i ; j )~u(j)  ~u(j) = 0. Since i 6= j , ~u(j)  ~u(j) = 0, and ~uj = ~0. Thus ~u0i = ~ui. Thus, ~ui0 2 Ui . Thus Ui0 Ui. By symmetry, Ui Ui0 . Thus Ui = Ui0 . QED. Corollary PP-3 If a linear operator on a Euclidean space has a spectral decomposition, it is symmetric. Proof: Suppose S has spectral decomposition (1  U1)     (m Um ). Let ~u~v 2 V . We can write ~u = Pmi=1 ~ui, ~v = Pmi=1 ~vi with ~ui, ~vi 2 Ui . Then S (~u) = S (~v) = ~u  S (~v) = = S (~u)  ~v = = m X i=1 m X S (~ui) = m X i=1 m X i~ui S (~vi) = i~vi =1 1 m ! 0iX m X ~ui  @ j~vj A = j (~ui  ~vj ) i=1X m i=1 m X j =1 ij =1 i (~ui  ~vi ) 1 m ! 0X m X i~ui  @ ~vj A = i (~ui  ~vj ) i=1X m i=1 m X i=1 j =1 ij =1 i (~ui  ~vi ): Hence ~u  S (~v) = S (~u)  ~v. Corollary PP-4 Two linear operators with the same spectral decomposition are equal. Proof: PROOFS PP-11 Suppose S and S 0 have the same spectral decomposition (1  U1) : : :  (m  Um). Let ~v 2 V . Then ~v = Pmi=1 ~ui with ~ui 2 Ui . Then m m X X S (~ui) = i~ui = S 0(~ui) i=1 i=1 i=1 m ! X = S0 ~ui = S 0(~v): S (~v) = m X i=1 Since this is true for all ~v 2 V , S = S 0. PP-12 PROOFS Positive Denite Operators and Their Square Roots Denition PP-3 A symmetric linear operator S is positive denite if ~v  S (~v) > 0 for all ~v = 6 ~0. (S is positive semi-denite if ~v  S (~v)  0 for all ~v 2 V .) Corollary PP-5 A symmetric linear operator S is positive (semi) denite i all its eigenvalues are (non-negative) positive. Proof: ). Suppose S has an eigenvalue   0. Let ~v be a nonzero vector with S (~v) = ~v. Then ~v  S (~v) = (~v  ~v)  0, so S is not positive denite. (. Let (1  U1) : : :  (m  Um) be the spectral decomposition of S . Let (~v 2 V , ~v = 6 ~0. Then ~v = Pmi=1 ~ui with ~ui 2 Ui and at least one ~ui =6 ~0. Then S (~v) = Pmi=1 S (~ui) = Pmi=1 i~ui, so 0m 1 m X m ! X X ~v  S (~v) = ~ui  @ j uj A = j ~ui  ~uj = i=1 m X i=1 j =1 ij =1 i (~ui  ~ui) > 0: Note: The proofs require an obvious modication if S is positive semi-denite rather than positive denite. Theorem PP-12 Suppose S is a symmetric, positive (semi) denite linear operator in Euclidean space V . Then there is exactly one symmetric positive (semi) denite linear PP-13 PP-14 PROOFS operator Q : V ! V such that Q2 = S . (Q is called the positive square root of S , and is written S 1=2.) Proof: Existence. Let (1  U1 ) : : :  (m  Um ) be the spectral decomposition of S . Since i > 0, let i1=2 be the positive square root of i . Let Q be the linear operator on V whose spectral decomposition is (11=2  U1) : : :  (m1=2  Um). Then Q is symmetric by corollary PP-5, and positive denite by corollary PP-6. Let ~v 2 V . Then ~v = Pmi=1 ~ui with ~u 2 Ui . Then m X m X i1=2~ui i=1 i=1 m m X X Q2 (~v) = i1=2 Q(~ui) = i1=2 i1=2~ui i=1 i=1 m m X X = i~ui = S (~ui) = S (~v): Q(~v) = i=1 Q(~ui) = i=1 Hence, Q2 = S . Uniqueness. Suppose Q is symmetric, positive denite, and Q2 = S . Let (q1 U~1 ) : : :  (qu U~u) be the spectral decomposition of Q. Then if ~ui 2 U~i we have S (~ui) = Q2~ui = QQ(~ui) = Q(q(i)~u(i)) = q(2i)~u(i). Thus qi2 is an eigenvalue of S . Therefore (q12 U~1 ) : : :  (qu2  U~u) is a spectral decomposition for S . Therefore u = m, U~ = Ui, qi2 = i . Since qi > 0, qi = i1=2 . Therefore Q has the spectral decomposition (11=2 U1 ) : : :  (m1=2 Um ), and is the operator found in the existence part of the proof. The Polar Decomposition Theorem Theorem PP-13 Suppose V is a Euclidean vector space, L : V ! V is a linear operator, and det L > 0. Then there are unique linear mappings R1 , S1 , R2 , S2 with these properties: i) L = R1 S1 ii) L = S2 R2 iii) R1 and R2 are proper orthogonal (i.e. R;1 = RT and det R = +1) iv) S1 and S2 are positive denite and symmetric. Further, R1 = R2 , so S1 = R2T S2R2 . Before giving the proof we need some lemmas. Lemma PP-6 If det 6= 0, LT L and LLT are symmetric and positive denite. Proof: It su"ces to consider LT L, because LLT = (LT )T (LT ) and det LT = det L. First, (LT L)T = LT (LT )T = LT L, so LT L is symmetric. Second, if ~v 6= ~0, L(~v) 6= ~0 because L;1 exists. But then L(~v)  L(~v) > 0. And L(~v)  L(~v) = ~v  LT L(~v)] = ~v  (LT L)(~v)]. Thus, ~v  (LT L)(~v)] > 0, so LT L is positive denite. Lemma PP-7 If S is positive denite and symmetric, det S > 0. Proof: PP-15 PP-16 PROOFS Let x^1  : : : x^n be an orthonormal basis for V consisting of eigenvectors for S , namely s1 : : :  sn. (Some of these may be the same.) Relative to basis x^1  : : :  x^n, the matrix of S is diagonal, and its diagonal entries are s1  : : :  sn. Therefore det S = s1s2 : : : sn. But since S is positive denite, all its eigenvalues are positive. Therefore det S > 0. Lemma PP-8 If A and B are linear operators on V and AB = I then B ;1 exists and B ;1 = A. Hence, BA = I and A;1 exists and A;1 = B . Proof: If AB = I then (det A)(det B ) = det AB = det I = 1, so det B 6= 0. Hence, B ;1 exists. By its denition, B ;1 B = BB ;1 = I . And if we multiply AB = I on the right by B ;1 we have (AB )B ;1 = IB ;1 , or A(BB ;1) = B ;1, or AI = B ;1 , or A = B ;1. QED. Lemma PP-9 If L is a linear operator on Euclidean space and L;1 exists so does (LT );1, and (LT );1 = (L;1)T . Proof: Take transposes in the equation LL;1 = I . The result is (L;1 )T LT = I . By lemma PP-8, (LT );1 exists and equals (L;1 )T . Proof of Theorem PP-3: First we deal with R1 and S1. To prove these unique, we assume they exist and have the asserted properties. Then L = R1 S1 implies LT = (R1 S1)T = S1T R1T = S1R1;1 so LT L = S1R1;1 R1S1 = S1IS1 = S12 . Now LT L is a symmetric positive denite operator, and so is S1. Therefore, S1 must be the unique symmetric positive denite square root of LT L, i.e. S1 = (LT L)1=2 : (PP-13) PROOFS PP-17 Since det S1 > 0, S1;1 exists. Then L = R1 S1 implies R1 = LS1;1: (PP-14) Thus S1 and R1 are given explicitly and uniquely in terms of L. To prove the existence of an R1 and S1 with the required properties, we dene them by (PP-8) and (PP-9). Then S1 is symmetric positive denite as required, and we need show only that R1 is proper orthogonal, i.e. that det R1 = +1 and R1T = R1;1. But det R1 = det L det(S1;1) = det L= det S1 > 0 since det L > 0 and det S1 > 0. Therefore, if we can show R1T = R1;1, we automatically have det R1 = 1. By lemma PP-8 it su"ces to prove R1T R1 = I . But R1T = (LS1;1)T = (S1;1)T LT = (S1T );1LT = S1;1LT so R1T R1 = S1;1LT LS1;1 = S1;1S12S1;1 = I . Next we deal with R2 and S2 . To prove them unique, we suppose they exist. Then L = S2R2 so LT = (S2R2 )T = R2T S2T = R2;1S2 . But if det L > 0 then det LT = det L > 0, so there is only one way to write LT = R~1 S~1 with R~1 proper orthogonal and S~1 symmetric. If R2 is proper orthogonal, so is R2;1 , and we must have R2;1 = R~1 so R2 = R~1;1. Then R2 and S2 are uniquely determined by L. In fact, S2 = S~1 = (LT )T LT = LLT , and R2 = (R~1 );1 = (LT S2;1);1 = S2(LT );1. To prove existence of R2 and S2 with the required properties, we observe that LT = R~1 S~1 with R~1 proper orthogonal and S~1 symmetric. Then (LT )T = L = (R~1 S~1)T = S~1 T R~1T = S~1 R1;1. Therefore we can take S2 = S~1, R2 = R~1;1. Finally, to prove R1 = R2 , note that if L = S2R2 then L = R2R2;1 S2R2 = R2 (R2T S2R2 ). We claim that R2T S2R2 is symmetric and positive denite. If we can prove that, then L = R2(R2T S2 R2) is of the form L = R1S1 , and uniqueness of R1 and S1 requires R1 = R2 , S1 = R2T S2R2 . Symmetry is easy. We have (R2T S2 R2 )T = R2T S2T (R2T )T = R2T S2R2 . For positive deniteness let ~v 2 V , ~v 6= ~0. Then R2(~v) 6= 0 because R2;1 exists. Then R2 (~v)  S2 R2(~v)] > 0 because S2 is positive denite. Then ~v  R2T S2R2 (~v)] > 0, or ~v  (R2T S2R2 )(~v)] > 0. Thus R2T S2R2 is positive denite. We shall not need the PDT when det L < 0 or L;1 fails to exist. For these cases, see Halmos, p. 170. PP-18 PROOFS Representation Theorem for Orthogonal Operators Denition PP-4 A linear operator R on Euclidean vector space V is \orthogonal" i it preserves length that is, kR(~v)k = k~vk for every ~v 2 V . There are several other properties each of which completely characterizes orthogonal operators (i.e., R is orthogonal i it is linear and has one of these properties). We list them in lemmas. Lemma PP-10 R is orthogonal i R is linear and preserves all inner products that is, R(~u)  R(~v) = ~u  ~v for all ~u~v 2 V . Proof: Obviously if R preserves all inner products it preserves lengths. The interesting half is that if R preserves lengths then it preserves all inner products. To see this, we note that for any ~u and ~v 2 V , k~u + ~vk2 = (~u + ~v)  (~u + ~v) = ~u  ~u + ~u  ~v + ~v  ~u + ~v  ~v. That is k~u + ~vk2 = k~uk2 + 2~u  ~v + k~vk2 : (PP-15) Applying this result to R(~u) and R(~v), and using R(~u + ~v) = R(~u) + R(~v), gives kR(~u + ~v)k2 = kR(~u)k2 + 2R(~u)  R(~v) + kR(~v)k2 : PP-19 (PP-16) PP-20 PROOFS If R preserves all lengths, then kR(~u +~v)k = k~u +~vk, kR(~u)k = k~uk, kR(~v)k = k~vk, so subtracting (PP-15) from (PP-16) gives R(~u)  R(~v) = ~u  ~v: Lemma PP-11 R is orthogonal i RT = R;1. (Therefore if R is orthogonal R;1 exists.) Proof: If RT = R;1 then RT R = I . Then for any ~u~v 2 V , ~u  RT R(~v) = ~u  ~v so R(~u)  R(~v) = ~u  ~v. Hence R is orthogonal. Conversely, suppose R orthogonal. Then for any ~u~v 2 V , R(~u)  R(~v) = ~u  ~v, so ~u  RT R(~v) = ~u  ~v, so ~u  (RT R)(~v) ; ~v ] = 0. Fix ~v. Then we see that RT R(~v) ; ~v is orthogonal to every ~u 2 V , hence to itself. Hence it is ~0, so RT R(~v) = ~v. Since this is true for every ~v 2 V , RT R = I . Then by lemma PP-8, RT = R;1. Corollary PP-6 If either RT R = I or RRT = I , R is orthogonal. Proof: By lemma PP-8, either equation implies RT = R;1. Corollary PP-7 If R is orthogonal, so is R;1. Proof: By lemma PP-11, R;1 exists and is RT . But RT (RT )T = RT R = I , so RT satises the second equation in corollary PP-6. Hence RT is orthogonal. The easiest way to discover whether a linear R : V ! V is orthogonal is to study its matrix Rij relative to some orthonormal basis for V . Recall that we dene this matrix by R(^x)i = Rij x^j (PP-17) where (^x1  : : :  x^n ) is the orthonormal basis for V which we are using. The matrix Lij of an arbitrary linear operator L is dened in the same way, L(^x)i = Lij x^j : (PP-18) PROOFS PP-21 The matrix of LT relative to (^x1 : : :  x^n) is the transposed matrix Lji. That is LT (^xi ) = Lij x^j or (LT )ij = Lji: (PP-19) To see this, we note that x^k  L(^xi ) = Lij x^k  x^j = Lij kj = Lik , so Lij = L(^xi )  x^j : (PP-20) Therefore (LT )ij = LT (^xi )  x^j = x^i  L(^xj ) = Lji. Now we have Remark 6 If R is orthogonal, then its matrix relative to every orthonormal basis satises Rij Rik = jk and RjiRki = jk (\orthonormal columns") (PP-21) (\orthonormal rows") (PP-22) Proof: If R is orthogonal so is RT . Therefore, it su"ces to prove (PP-21). We have RT (^xj ) = (RT )jix^i = Rij x^i. Then RRT (^xj ) = Rij R(^xi ) = Rij Rik x^k . But if R is orthogonal, RRT = I , so RRT (^xj ) = jk x^k . Thus Rij Rik x^k = jk x^k , or (Rij Rik ; jk )^xk = 0. Fix j . Since x^1  : : :  x^n are linearly independent, we must have Rij Rik ; jk = 0. This is (PP-21). Remark 7 If R 2 L(V ! V ) and there is one orthonormal basis for V relative to which the matrix of R satises either (PP-21) or (PP-22), then R is orthogonal. Proof: Suppose that relative to the particular basis x^1  : : :  x^n, the matrix of R satises (PP-21). Multiply by x^k and add over k, using Rik x^k = R(^xi). The result is Rij R(^xi ) = x^j : PP-22 h i PROOFS By linearity of R, this means R(Rij x^i ) = x^j or R RT (^xj ) = x^j , or (RRT )(^xj ) = x^j . But then for any scalars v1 : : : ~vn we have (RRT )(vj x^j ) = vj RRT (^xj ) = vj x^j . Hence RRT (~v) = ~v for all ~v 2 V . Hence RRT = I . Hence R is orthogonal. Next suppose R satises (PP-22) rather than (PP-21). Then (RT )ij (RT )ik = jk , so the matrix of RT satises (PP-21). Hence, RT = R;1 is orthogonal. Hence so is (RT )T = (R;1 );1 = R. Corollary PP-8 If the rows of a square matrix are orthonormal, so are the columns. A matrix with this property is called orthogonal. What we have seen is that if R is orthogonal, so is its matrix relative to every orthonormal basis. And if the matrix of R relative to one orthonormal basis is orthogonal, R is orthogonal, so its matrix relative to every orthonormal basis is orthogonal. Now we discuss some properties of orthogonal operators which will eventually tell us what those operators look like. First, we introduce the notation in Denition PP-5 The set of all orthogonal operators on Euclidean space V is written #(V ). The set of all proper orthogonal operators (those with determinant + 1) is written #+(V ). Lemma PP-12 #(V ) and #+(V ) are groups if multiplication is dened as operator composition. Proof: All we need to prove is that both sets are closed under the operations of taking inverses and of multiplying. But if R 2 #(V ), we know R;1 2 #(V ) by remark PP-2. And det(R;1) = (det R);1 = 1 if det R = 1 so if R 2 #+(V ) then R;1 2 #+(V ). If R1 and R2 2 #(V ), then (R1 R2)T (R1 R2 ) = R2T R1T R1 R2 = R2T IR2 = R2T R2 = I , so R1R2 2 #(V ). And if R1 , R2 2 #+(V ) then det(R1R2 ) = (det R1)(det R2) = 1, so R1R2 2 #+(V ). PROOFS PP-23 Lemma PP-13 If R 2 #(V ), det R = +1 or ;1. Proof: RT R = I so (det RT )(det R) = det I . But det RT = det R and det I = 1. Thus (det R)2 = 1. Lemma PP-14 Suppose R 2 #(V ) and U is a subspace of V and R(U ) U . Then RjU 2 #(U ). Proof: By hypothesis, RjU is a linear operator on U . Since kR(~v)k = k~vk for all ~v 2 V , this is certainly true for all ~v 2 U . Hence, RjU satises denition PP-4 Lemma PP-15 Under the hypothesis of lemma PP-14, RT (U ) U . Proof: Suppose ~u 2 U . Then there is a unique u~ 0 2 U such that ~u = (RjU )(u~ 0) (because (RjU );1 exists as a linear operator on U ). But u~ 0 2 U , so (RjU )(u~ 0 ) = R(u~ 0). Thus ~u = R(u~ 0), and R;1(~u) = u~ 0. Then RT (~u) = u~ 0. Thus RT (~u) 2 U if ~u 2 U . QED Lemma PP-16 Under the hypotheses of lemma PP-14, R(U ?) U ?. Proof: Suppose w~ 2 U ?. We want to show R(w~ ) 2 U ?. That is, ~u  R(w~ ) = 0 for all ~u 2 U . But ~u  R(w~ ) = RT (~u)  w~ . By lemma PP-5, RT (~u) 2 U . By hypothesis w~ ?U . Hence, RT (~u)  w~ = 0, as required. Lemma PP-17 If R 2 #(V ) and  is an eigenvalue of R,  = 1. Proof: PP-24 PROOFS Suppose R(~v) = ~v and ~v 6= ~0. Then k~vk = kR(~v)k = k~vk =j  j k~vk so 1 =j  j. QED. Lemma PP-18 Let R 2 #(V ). Let x^1  : : : x^n be an orthonormal basis for V . Let Rij be the matrix of R relative to this basis. Suppose  is a complex zero of the n'th degree polynomial det(Rij ; ij ). Then j  j2= 1. Proof: There is a complex n-tuple (r1 : : :  rn) 6= (0 : : :  0) such that Rij rj = ri. Taking complex conjugates gives Rij rj = ri, or Rik rk =  ri. Multiplying and summing over i gives Rij Rik rj rk = riri: But Rij Rik = jk , so rj rj =j  j2 riri. Since riri > 0, we have j  j2= 1. Lemma PP-19 In lemma PP-18,  is also a zero of det(Rij ; ij ) = 0. Proof: Since (Rij ; ij )rj = 0 and (r1 : : :  rn ) 6= (0 : : : 0), det(Rij ; ij ) = 0. Lemma PP-20 Suppose R, Rij as in lemma PP-18. Suppose  is not real and (r1 : : :  rn) 6= (0 : : :  0) is a sequence of n complex numbers such that Rij rj = ri. Then ri ri = 0. Proof: Rik rk = ri so (Rij rj )(Rik rk ) = 2riri , or Rij Rik rj rk = 2riri, or jk rj rk = 2riri, or rj rj = 2riri , or (2 ; 1)riri = 0. Since  is not real, 2 ; 1 6= 0, so riri = 0. Remark 8 Suppose R 2 #(V ) and V contains no eigenvectors of R. Then V contains two mutually orthogonal unit vectors ~x and ~y such that for some in 0 < < , PROOFS 8 > > > < R(^x) = x^ cos # + y^ sin # > > > : R(^y) = x^ sin # + y^ cos #: PP-25 (PP-23) Proof: Choose an orthonormal basis z^1  : : :  z^n for V , and let Rij be the matrix of R relative to that basis. If 0 = det(Rij ; ij ) has a real root, , it is an eigenvalue of R, so R has an eigenvector. Hence no root of det(Rij ; ij ) is real. By lemma PP-18, we can write any root as ei where 0 < <  or  < < 2. By lemma PP-19, we can nd a root e;i with 0 < < . Then there is an n-tuple of complex numbers (r1 : : :  rn) 6= (0 : : :  0) such that Rjk rj = e;i rk : (PP-24) We may normalize (r1 : : :  rn) so that rjrj = 2 (PP-25) rj rj = 0: (PP-26) and by lemma PP-20 Now write rj = xj + iyj where xj and yj are real. Then separating the real and imaginary parts of (PP-24) gives Rjk xj = xk cos + yk sin Rjk yj = ;xk sin + yk cos : Separating the real and imaginary parts of (PP-26) gives xj xj ; yj yj = 0 xj yj = 0 PP-26 PROOFS and (PP-25) gives xj xj + yj yj = 2: From these equations, clearly xj xj = yj yj = 1 and xj yj = 0: Now take x^ = xj z^j and y^ = yj z^j , and we have the x^ y^, whose existence is asserted in the theorem. Corollary PP-9 dim V  2. Remark 9 Suppose R 2 #(V ) and V contains no eigenvectors of R. Then V has an orthonormal basis x^1  y^1 : : : x^m  y^m such that for each j 2 f1 : : :  mg there is an #j in 0 < j <  with 8 > > > < R(^xj ) = x^(j) cos (j) + y^(j) sin (j) : > > > : R(^yj ) = ;x^(j) sin (j) + y^(j) cos (j) (PP-27) Proof: By remark PP-4 we can nd mutually orthogonal unit vectors x^1  y^1 and 1 in 0 < 1 <  such that (PP-27) holds for j = 1. Then R(spfx^1 y^1g) spfx^1 y^1g. Hence, by lemma PP-16, R(fx^1 y^1g?) fx^1 y^1g?. Hence, by lemma PP-14, Rjfx^1  y^1g? 2 #(fx^1  y^1g?). Since R has no eigenvectors in V , it has none in fx^1  y^1g?. Thus Rjfx^1 y^1g? satises the hypotheses of remark PP-3, and we can repeat the argument, nding 2 in 0 < 2 <  an mutually orthogonal unit vectors x^2  y^2 in fx^1 y^1g? such that (PP-4) holds for j = 2. We can proceed in this way as long as the dimension of the remaining space is > 0. (The argument is by induction on dim V .) When the dimension of the remaining space is 0, we have an orthonormal basis for V with the required properties. PROOFS PP-27 Corollary PP-10 dim V is even. That is, if R 2 #(V ) and dim V is odd, R has an eigenvector. (This is obvious from another point of view. If dim V is odd, so is the degree of the real polynomial det(R ; I ). Hence, it has at least one real zero). Theorem PP-14 Suppose V is Euclidean vector space and R 2 #(V ). Then V has an orthonormal basis x^1  y^1 : : :  x^m  y^m z^1 : : : z^n w^1 : : :  w^p with these properties: i) R(w^j ) = ;w^j ii) R(^zj ) = z^j iii) for j 2 f1 : : :  mg, there is a j in 0 < j <  such that equation (PP-4) holds. Proof: Let Z and W be the eigenspaces of V with eigenvalues +1 and ;1. Let z^1  : : :  z^n be an orthonormal basis for Z and w^1 : : :  w^p an orthonormal basis for W . Then R(Z + W ) Z + W , so R(Z + W )?] (Z + W )?. Moreover, (Z + W )? contains no eigenvectors of R, since all their eigenvalues are +1 or ;1, and these are already in Z or W . Thus remark PP-4 applied to (Z + W )?. QED. Corollary PP-11 det R = (;1)p: Proof: Relative to the basis in theorem PP-15, the matrix of R looks thus: Each 2 2 diagonal block contributes a factor cos2 j + sin2 j = 1 to the determinant, whence the result. PP-28 PROOFS cos θ1 sin θ1 -sin θ1 cos θ1 cos θ2 sin θ2 -sin θ2 cos θ2 .. . cos θm sin θm -sin θm cos θm 1 .. { . n terms 1 -1 .. { . p terms Figure PP-1: Matrix of R. -1 PROOFS PP-29 Corollary PP-12 If R 2 #+(V ), then V has an orthonormal basis x^1  y^1 : : :  x^m  y^m z^1 : : :  z^n such that 8 > i) R(^zj ) = z^j for j 2 f1 : : :  ng: > > > > > > > > > < ii) For j 2 f1 : : :  mg there is a j in 0 < j   such that : > > > R(^xj ) = x^(j) cos (j) + y^(j) sin (j) > > > > > > > : R(^yj ) = ;x^(j) sin (j) + y^(j) cos (j) : (PP-28) Proof: Since det R = +1, in theorem PP-15, p is even. Therefore, we may divide w^1 : : :  w^p into pairs. For any such pair, we have (for example) R(w^1) = ;w^1 = w^1 cos  + w^2 sin  R(w^2) = ;w^2 = ;w^1 sin  + w^2 cos : If we adjoin these pairs to the pairs (^xj  y^j ) given by theorem PP-15, and take the corresponding angles as = , we have the result of the corollary. For obvious reasons, a linear operator on V which looks like (PP-27) in some orthonormal basis is called a \rotation." Corollary PP-12 says that every proper orthogonal operator is a rotation. If dim V = 3, the eigenvector z^1 is called the axis of the rotation, and the angle _1 in 0 < 1   is the angle of the rotation. This terminology breaks down only when m = 0 and n = 3 in (PP-27). In that case, R = I . The angle of rotation of I is 0, but its axis of rotation obviously cannot be dened. For dim V = 3, corollary PP-12 is called Euler's theorem. If R 2 L(V ! V ) and an orthonormal basis x^1 : : :  x^n for V can be found such that R(^x1 ) = ;x^1 , R(^xj ) = x^j for j  2 then R is said to be a reection along x^1 . Obviously det R = ;1 for a reection, PP-30 PROOFS so a reection is an \improper orthogonal" operator. Also, clearly a reection has the property R;1 = R, or R2 = I . If R1 is a reection and R2 is any other improper orthogonal operator, then R = R1R2 is orthogonal and det R = (det)R1)(det R2) = (;1)2 = 1, so R is proper orthogonal. But then R1 R = R12 R2 = R2, so R2 = R1 R. That is, every improper orthogonal operator is the product of a rotation and a reection. Part I Tensors over Euclidean Vector Spaces 1 Chapter 1 Multilinear Mappings 1.1 Denition of multilinear mappings M(V1 : : : Vq ! W ): V1  : : :  Vq  W are vector spaces over a single eld F , and M 2 F (V1 : : : Vq ! W ). If M satises any of the following three equivalent conditions, M is called a multilinear mapping from V1 : : : Vq to W . Denition 1.1.6 For any (~v1  : : : ~vq ) 2 V1 : : : Vq and any p 2 f1 : : :  qg, M (~v1  : : : ~vp;1 ~vp+1 : : : ~vq ) 2 L(Vp ! W ): (That is, M (~v1  : : : ~vq ) depends linearly on each of the separate vectors ~v1 : : :  v~q if the others are held xed.) Denition 1.1.7 If a 2 F , p 2 f1 : : :  qg, ~xp and ~yp 2 Vp, and (~v1  : : : ~vq ) 2 V1 : : : Vq then i) M (~v1  : : : ~vp;1 a~vp~vp+1 : : : ~vq ) = aM (~v1  : : : ~vp;1~vp~vp+n : : : ~vq ). Also ii) M (~v1  : : : ~vp;1 ~xp + ~yp~vp+1 : : : ~vq ) = M (~v1  : : : ~vp;1 ~xp~vp+1 : : : ~vq ) + M (~v1  : : : ~vp;1 ~yp~vP +1 : : : ~vq ) . 3 4 CHAPTER 1. MULTILINEAR MAPPINGS Denition 1.1.8 For each p 2 f1 : : :  qg, suppose a1(p)  : : :  an(pp) 2 F and ~v1(p) : : : ~vn(pp) 2    iq (q)  i1 : : : aiq M ~v (1)  : : : ~v (q) . Vp. Then M ai(1)1 ~vi(1)  : : :  a ~ v = a i1 iq (q) iq (1) (q) 1 Denition 1.1.9 M(V1 : : : Vq ! W ) denotes the set of all multilinear mappings from V1 : : : Vq to W . 1.2 Examples of multilinear mappings In these examples, V1  : : :  Vq are Euclidean vector spaces, and the eld F is R. The vector space W need not be Euclidean (i.e. need not have a dot product) but it must be real (i.e. its eld is R). Example 1.2.1 For each p between 1 and q, let ~up be a xed vector in Vp. Let w~ 0 2 W . For any (~v1  : : : ~vq ) 2 V1 : : : Vq , dene M (~v1  : : : ~vq ) = (~u1  ~v1 )    (~uq  ~vq )w~ 0: (1.2.1) To give the next example, some terminology is useful. Suppose that for each p 2   f1 : : :  qg Bp = ~b1(p)  : : :  ~bn(pp) is an ordered basis for Vp. Then the sequence of ordered bases (B1  : : :  Bq ) is called a \basis sequence" for V1 : : : Vq . Let BpD = (~b(1p)  : : :  ~b(npp) ) be the ordered basis for Vp which is dual to Bp. Then (B1D  : : :  BqD ) is \the basis sequence dual to (B1 : : :  Bq )." Example 1.2.2 Let (B1  : : :  Bq ) be a basis sequence for V1 : : : Vq . Let w~ i :::iq be any q'th order, n1 : : : nq array of vectors from W . For any (V~1 : : :  V~q ) 2 V1 : : : Vq 1 dene Then and also       M ~v(1)  : : : ~v(q) = ~b i(1)1  ~v (1) : : : ~b(iqq)  ~v(q) w~ i1 :::iq : (1.2.2) M 2 M (V1 : : : Vq ! W ) (1.2.3)   M ~bj(1) = w~ j1:::jq :  : : :~bj(1) q 1 (1.2.4) 1.2. EXAMPLES OF MULTILINEAR MAPPINGS 5 The multilinearity of M is almost obvious (1.2.2) is a sum of n1 n2 : : : nq terms like (1.2.1). Each of these terms is linear in each v~p if ~v1  : : : ~vp;1~vp+1 : : :~vq are xed. (q) To prove (1.2.4) we let ~vp = ~b(jpp) in the denition (1.2.2), obtaining M (~b(1) j1  : : :  ~bjq ) = i1 ~b (1) ) : : : (~b iq ~b (q) )w i1 iq ~ i :::i = w (~b(1) ~ j1:::jq . There are no other examples. j1 1 q (q) jq ~ i1 :::iq =  j1 : : :  jq w Every member of M(V1 : : : Vq ! W ) is like example (1.2.2). We have Remark 1.2.10 Suppose V1 : : :  Vq are Euclidean vector spaces and W is a real vector space. Suppose dim V1 = n1 : : :  dim Vq = nq . Suppose (B1 : : :  Bq ) is a basis sequence for V1 : : : Vq , and w~ i1:::iq is an n1 n2 : : : nq array of vectors from W . Then 91 M 2 M(V1 : : : Vq ! W ) 3 M (~bi(1)  : : :  ~bi(qq)) = w~ i1:::iq . 1 Proof: Example 1.2.2 shows at least one such M . To establish uniqueness suppose M does satisfy (1.2.3) and (1.2.4). Let (~b1(p)  : : :  ~b(npp) ) = BpD be the ordered basis for Vp which is dual to Bp = (~b1(p)  : : :  ~bn(pp)). For any ~v(p) 2 Vp, we have ~v(p) = v(ipp)~bi(pp) where ~v(ipp) = v(p)  ~bi(pp) . Then i1 ~b (1)  : : :  v iq ~b (q) ) (by denition 1.1.8) M (~v(1)  : : : ~v (q) ) = M (v(1) i1 (q) iq = bi(1)1 : : : v(iqq ) M (~bi(1)  : : :  ~bi(qq) ) (by hypothesis) 1 i1 : : : v i q w = v(1) (q) ~ i1 :::iq     1 = ~bi(1)  ~v(1) : : : ~b(iqq)  ~v(q) w~ i1 :::iq : In other words, M is the function dened by (1.2.2). It follows that every M 2 M(V1 : : : Vq ! W ) has the form (1.2.2). 6 CHAPTER 1. MULTILINEAR MAPPINGS 1.3 Elementary properties of multilinear mappings Remark 1.3.11 If one of ~v1  : : : ~vq is ~0 and if M 2 M(V1 : : : Vq ! W ) then M (~v1  : : : ~vq ) = ~0, the zero in W . Proof: We suppose ~v1 = ~0. Then M (~0~v2  : : : ~vq ) = M (0~0~v2 : : : ~vq ) = 0M (~0~v2  : : : ~vq ) = ~0, where we have used denition (1.1.7). Remark 1.3.12 M(V1 : : : Vq ! W ) is a vector space over F if we use the ad and sc dened in example 3, page D-12, or in equations (14.1), (14.2). Proof: Refer to \subspaces" on page D-15. We know from example 3, page D-12, that F (V1 : : : Vq ! W ) is a vector space over F . Obviously, M(V1 : : : Vq ! W ), is a subset of F (V1 : : : V q) ! W ). Therefore, it su"ces to prove that if a 2 F and M , N 2 M(V1 : : : V q ! W ), then aM and M + N 2 M(V1 : : : Vq ! W ). That is, if M and N are multilinear, so are aM and M + N . This is immediate from denition 1.1.6 because we know it for linear functions. Alternatively, we can use denition 1.1.8 and write everything out. We will do this for M + N . We have     i1 (1) iq (q)  iq (q)  = M ai(1)1 ~v(1)  : : : a ~ v + N a ~ v  : : :  a i1 (q) iq (1) i1 (q)~viq ( by denition of M + N )     i1 : : : a iq M ~v (1)  : : : ~v (q) + ai1 : : : aiq N ~v (1)  : : : ~v (q) = a(1) i1 iq i1 iq (q ) (1) (q) (because M and N are multilinear) h    i i1 : : : a iq M ~v (1)  : : : ~v (q) + N ~v (1)     ~v (q) = a(1) i1 iq i1 iq (a) because W is a vector space over F i1 ~v (1)  : : :  aiq ~v (q) (M + N ) a(1) i1 (q) iq 1.4. PERMUTING A MULTILINEAR MAPPING h  = ai(1)1 : : : ai(qq) (M + N ) ~vi(1)  : : : ~vi(qq) 1 7 i by denition of M + N: Comparing the two ends of this chain of equalities, we have a proof that M + N is multilinear if M and N are, and if M + N is dened by (14.1). 1.4 Permuting a multilinear mapping Denition 1.4.10 Suppose (~v1 : : : ~vq ) 2 V1 : : : Vq and 1  r < s  q. The q- tuple (~v1  : : : ~vr;1~vs~vr+1 : : : ~vs;1~vr ~vs+1 : : :~vq ) will be abbreviated (~v1  : : :~vq )(rs). It is simply (~v1  : : : ~vq ) with ~vr and ~vs interchanged. Note that if Vr = Vs, then (~v1  : : : ~vq )(rs) 2 V1 : : : Vq . Denition 1.4.11 Suppose M 2 M(V1 : : : Vq ! W ) and 1  r < s  q. Suppose Vr = Vs. Then we write (rs)M for the member of F (V1 : : : Vq ! W ) dened by requiring for any (~v1  : : : ~vq ) 2 V1 : : : Vq that (rs)M ] (~v1  : : : ~vq ) = M (~v1 : : : ~vq )(rs) : (1.4.1) That is (rs)M ] (~v1  : : : ~vr;1~vr ~vr+1 : : : ~vs;1~vs~ss+1 : : : ~vq ) = M (~v1 : : : ~vr;1~vs~vr+1 : : : ~vs;1~vr ~ss+1 : : : ~vq ) : Remark 1.4.13 (rs)M 2 M(V1 : : : Vq ! W ) if Vr = Vs. Proof: Obvious from (1.4.2) and denition (1.1.6). Remark 1.4.14 (rs) is a linear operator on M(V1 : : : Vq ! W ) if Vr = Vs. (1.4.2) 8 CHAPTER 1. MULTILINEAR MAPPINGS Proof: By remark (1.4.13), (rs) : M(V1 : : : Vq ! W ) ! M(V1 : : : Vq ! W ). It remains to prove that (rs) is linear. Suppose a1  : : :  aN are scalars and M1  : : : MN 2 M(V1 : : : Vq ! W ). We want to prove that (rs)(aj Mj ) = aj (rs)Mj ]. Let (~v1  : : : ~vq 2 V1 : : : Vq ). Then h (rs)(aj Mj )] (~v1 : : : ~vq ) = (aj Mj )(~v1  : : : ~vq )(rs) = aj Mj (~v1 : : : ~vq )(rs) h i   h i = aj (rs)Mj (~v1  : : : ~vq ) = aj (rs)Mj (~v1 : : : ~vq ): i Since this is true for any (~v1  : : : ~vq ) 2 V1 : : : Vq , we have the required result. Denition 1.4.12 Suppose M , r, s as in denition (1.4.2). If (rs)M = M , M is \symmetric" under (rs). If (rs)M = ;M , M is \antisymmetric" under (rs). Remark 1.4.15 If Vr = Vs, the members of M(V1 : : : Vq ! W ) which are symmetric under (rs) constitute a subspace of M(V1 : : : Vq ! W ). The same holds for the members antisymmetric under (rs). Proof: It su"ces to show that if a is any scalar and M , N are symmetric (or antisymmetric) under (rs), so are aM and M +N . We give the proof for antisymmetry. In that case, by hypothesis, (rs)M = ;M  (rs)N = ;N: But (rs) is linear, so (rs)aM ] = a(rs)M ] = a(;M ) = ;aM , and (rs)(M + N ) = (rs)M +(rs)N = ;M ; N = ;(M + N ). (Note that the set of members of M(V1 : : : Vq ! W ) which are symmetric (anti-symmetric) under (rs) is simply the eigenspace of (rs) with eigenvalue 1(;1). Of course, it is a subspace of M(V1 : : : Vq ! W ).) 1.4. PERMUTING A MULTILINEAR MAPPING 9 Denition 1.4.13 Suppose V1 =    = Vq = V . Suppose M 2 M( q V ! W ) and  2 Sq . Then M is dened as that member of F ( q V ! W ) such that for any ~v1  : : :~vq 2 V   (M )(~v1  : : : ~vq ) = M ~v(1)  : : : ~v(q) : Remark 1.4.16 If V1 =    = Vq = V then M 2 M( q V ! W ), and in fact  is a linear operator on M( q V ! W ). Moreover if ,  2 Sq and M 2 M( q V ! W ) then (M ) = ()M: (1.4.3) Proof: Except for (1.4.3), the proof is like that of remark (1.4.14). To prove equation (1.4.3), let ~v1 : : : ~vq 2 V and let w~ 1 = ~v(1)  : : :  w~q = ~v(q) . Then     (M ) (~v1 : : : ~vq ) = (M ) ~v(1)  : : : ~v(q)  = (M ) (w~ 1 : : :  w~ q ) = M w~ (1)  : : :  w~ (q)     = M ~v(1)]  : : : ~v(q)] = M ~v()(1)  : : : ~v()(q)  = ()M ] (~v1  : : : ~vq ) Since this is true for all ~v1  : : : ~vq 2 V , we have (1.4.3). Denition 1.4.14 Suppose V1 =    = Vq = V and M 2 M 2 ( q V ! W ). M is totally symmetric (antisymmetric) if M = M (M = ( sgn )M ) for every  2 Sq . 10 CHAPTER 1. MULTILINEAR MAPPINGS Remark 1.4.17 M is totally symmetric (antisymmetric) i (rs)M = M (= ;M ) for every transposition (rs). Proof: =) is obvious. For (=, recall that any  2 Sq is a product of transpositions. By remark 1.4.16 the eects of these transpositions on M can be calculated one after the other. If all leave M unchanged, so does . If all multiply M by ;1, then M is M multiplied by sgn  (= 1 for even/odd ). Remark 1.4.18 The totally symmetric (antisymmetric) members of M( q V ! W ) constitute a subspace of M( q V ! W ). Proof: We give the proof for the totally symmetric case. Suppose M1  : : :  MN are totally antisymmetric members of M( q V ! W ) and a1  : : :  aM are scalars. We want to show that aiMi is totally antisymmetric. But for any  2 Sq ,  is linear operator on M( q V ! W ), so (aiMi) = ai(Mi ) = ai ( sgn )Mi ] = ( sgn )(ai Mi). That is, aiMi is totally antisymmetric. Chapter 2 Denition of Tensors over Euclidean Vector Spaces If V1  : : :  Vq are Euclidean vector spaces, then M(V1 : : : Vq ) ! R is called the \tensor product of V1  : : :  Vq ". It is written V1  : : :  Vq . Thus V1  : : :  Vq := M(V1 : : : Vq ) ! R: (2.0.1) The multilinear functionals which are the members of V1  : : :  Vq are called the tensors of order q over V1 : : : Vq . z q times }| { If V1 = : : : = Vq = V , then V1  : : :  Vq = V  : : :  V , and this is abbreviated q V . Its members are called the tensors of order q over V . By convention, tensors of order zero are scalars that is  V = R: (2.0.2) For tensors of order 1, the notation requires some comment. When q = 1, multilinearity is simply linearity. That is M(V ! R) = L(V ! R). Thus the tensors of order 1 over a vector space V are simply the linear functional on V . This would not be a problem, except that for q = 1, (2.0.1) reduces to V1 := M(V1 ! R), which says V = L(V ! R): 11 (2.0.3) 12 CHAPTER 2. DEFINITION OF TENSORS OVER EUCLIDEAN VECTOR SPACES It is at this point that we require V to be a Euclidean vector space. If it is, then ~v 2 V can be safely confused with the linear functional ~v 2 L(V ! R) dened by (21.2). Tensors of order 1 over V are simply the vectors in V . When V is not a Euclidean vector space, (2.0.3) is false. (The situation is very subtle. If V is not Euclidean but is nite dimensional, every basis B for V generates an isomorphism between V and L(V ! R), namely ~bij! ciB . But each of these isomorphisms depends on the basis B chosen to produce it. There is no \natural", basis-independent isomorphism which can be used to identify vectors with linear functionals. See MacLane and Birkho, Algebra, MacMillan 1967, p. 237 for a more detailed discussion.) Tensors over vector spaces which are not Euclidean are constructed in a more complicated fashion than (2.0.1). We will not need this generality, and tensors over Euclidean spaces have some very useful properties for continuum mechanics. It is for this reason that we consider only tensors over Euclidean spaces. See M. Marcus's book, on reserve, for a treatment of tensors over arbitrary vector spaces. Chapter 3 Alternating Tensors, Determinants, Orientation, and n-dimensional Right-handedness In this section we discuss an application of tensors which some of you may have seen in your linear algebra courses. Let V be an n-dimensional Euclidean vector space. The set of totally antisymmetric tensors of order q over V is written q V . It is a subspace of q V . (See remark 1.4.18.) We want to study nV . The members of nV are called \alternating tensors over V ." 3.1 Structure of
nV We want to understand what the alternating tensors over V are like. It will turn out that there are not very many of them. In fact dim nV = 1. We want to prove this and to nd a basis \vector" in the vector space nV . First we note a technique for verifying that A 2 nV . We need to prove only that A 2 n V and that (rs)A = ;A for every transposition (rs) 2 Sn. If this latter fact is true, then clearly remark 2.0.1 tells us A 2 nV . A second property of an A 2 n V which is equivalent to membership in nV is that 13 14CHAPTER 3. ALTERNATING TENSORS, DETERMINANTS, ORIENTATION, AND N -DIMENSIONAL RIGHT-HANDEDNESS for any (~v1  : : : ~vn) 2 n V and any  2 Sn,   A ~v(1)  : : : ~v(n) = ( sgn )A (~v1  : : :~vn) : (3.1.1) This is obvious from the denition of A and nV . It implies the rst step in our attempt to understand alternating tensors: Lemma 3.1.21 If A 2 nV and two of ~v1  : : :~vn are equal, then A(~v1 : : : ~vn) = 0. Proof: If ~vr = ~vs, interchanging them will not change A(~v1  : : : ~vn). But sgn (rs) = ;1, so (3.1.1) says interchanging them changes the sign of A(~v1 : : : ~vn). Hence A(~v1 : : : ~vn) = 0. Equation (3.1.1) and lemma 3.1.21 are equivalent to the single equation A (~vi1  : : : ~vin ) = "i1 :::in A (v1 : : : ~vn) : (3.1.2) For if two of i1  : : :  in are equal, both sides of (3.1.2) vanish. If i1  : : :  in are all dierent, there is a permutation  2 Sn such that i1  = (1) : : :  in = (n). Then (3.1.2) reduces to equation (3.1.1). Lemma 3.1.21 can be extended to Remark 3.1.19 If A 2 nV and ~v1 : : : ~vn are linearly dependent, then A(~v1  : : : ~vn) = 0. Proof: By hypothesis, there are scalars c1 : : :  cn, not all zero, such that ci~vi = ~0. If ~v1 = ~0, then A(~v1  : : : ~vn) = 0 by remark 1.3.11. If ~v1 6= 0 then at least one of c2  : : :  cn is nonzero. Let cp be the last scalar which is nonzero. Then c1~v1 + : : : + cp~vp = ~0 and we can divide by cp to write pX ;1 ~vp = ai~vi i=1 3.1. STRUCTURE OF N V 15 where Then ai = ; cci : p 0 1 pX ;1 A (~v1  : : : ~vn) = A @~v1 : : : ~vp;1 ai~vi ~vp+1 : : : ~vnA i=1 pX ;1 = i=1 = 0 aiA (~v1 : : :~vp;1~vi~vp+1 : : : ~vn) by lemma 3.1.21: Remark 3.1.19 would be rather dull if A 2 nV implied A = 0. Therefore at this point we pause to construct a nonzero member of nV . Let B = (~b1  : : :  ~bn) be any basis for V . Let B D = (~b1  : : :  ~bn) be its dual basis. Dene AB 2 F ( nV ! R) by requiring     AB (~v1  : : : ~vn) = ~v1  ~bi1 : : : ~vn  ~bin "i1 :::in (3.1.3) for any (~v1 : : :~vn) 2 nV . From denition (1.1.6) it is clear that AB is multilinear,   so AB 2 n V , because the values of AB are scalars. Furthermore, AB ~b1  : : :  ~bn =   ~ ~i1  ~ ~in  b1  b : : : bn  b "i1 :::in = 1 i1 : : : n in "i1 :::in = "12:::n = 1. That is, if B = ~b1  : : :  ~bn ,   AB ~b1  : : : ~bn = 1 (3.1.4) when AB is dened by (3.1.3). Thus AB 6= 0. We claim also that AB 2 nV . To show this, is su"ces to show that (rs)AB = ;AB for any transposition (rs). The argument is the same for all (rs), so we consider (12). We have (12)AB ] (~v1 : : : ~vn) = AB (~v2 ~v1~v3 : : : ~vn)       = ~v2  ~bi1 ~v1  ~bi2 ~v3  ~bi3 : : : ~vn  ~bin "i1 :::in : If we relabel the summation index i1 as i2, and relabel i2 as i1 , this is       (12)AB ] (~v1 : : : ~vn) = ~v2  ~bi2 ~v1  ~bi1 ~v3  ~bi3 : : : ~vn  ~bin "i2 i1 i3 :::in 16CHAPTER 3. ALTERNATING TENSORS, DETERMINANTS, ORIENTATION, AND N -DIMENSIONAL RIGHT-HANDEDNESS     = ~v1  ~bi1 : : : ~vn  ~bin "i2 i1 i3 :::in     = ; ~v1  ~bi1 : : : ~vn  ~bin "i1 i2 i3 :::in = ;AB (~v1 : : : ~vn) : Since these equations hold for any (~v1 : : : ~vn) 2 nV , we have (12)AB = ;AB . QED Now for every ordered basis B in V we have managed to construct a nonzero member of nV , namely AB . It looks as if nV must be rather large. In fact it is not. We can now easily prove dim nV = 1 if n dim V: (3.1.5) To see this, let B = (~b1 : : :  ~bn) be a xed ordered basis for V . We will prove that if A 2 nV then A is a scalar multiple of AB . Thus, fAB g is a basis for nV . The argument goes thus. Let A 2 nV . For any (~v1  : : : ~vn) 2 nV we can write ~vp = vp i~bi where vp i = ~vp  ~bi, B D = (b~1  : : :  ~bn) being the basis dual to B . Then   A (~v1  : : : ~vn) = A v1 i1~bi1  : : :  vn in~bin  = v1 i1 : : : vn in A ~bi1  : : :  ~bin   = v1 i1 : : : vn in "i1 :::in A ~b1  : : :  ~bn        by (3:1:2)  = A ~b1  : : :  ~bn ~v1  ~bi1 : : : ~vn  ~bin "i1 :::in = A ~b1  : : :  ~bn AB (~v1  : : : ~vn) : Since this is true for all ~v1 : : : ~vn 2 V , we have for ordered basis B = (~b1  : : : ~bn) in V 3.2. DETERMINANTS OF LINEAR OPERATORS and for any A 2 nV     A = A ~b1  : : :  ~bn AB : 17 (3.1.6) Of course, A ~b1  : : :  ~bn 2 R, so (3.1.6) shows A 2 spfAB g. This proves (3.1.5). Equation 3.1.6 permits us to prove the following test for linear independence of a sequence of n vectors in an n-dimensional space. Remark 3.1.20 Suppose A 2 nV and A 6= 0, and dim V = n. Suppose ~v1  : : : ~vn 2 V . Then ~v1  : : : ~vn are linearly dependent if and only if A(~v1  : : : ~vn) = 0. Proof: =) is remark 3.1.2. To prove (=, we want to prove that if A(~v1  : : : ~vn) = 0 then ~v1  : : : ~vn are linearly dependent. If they are not, they are a basis B for V . Then (3.1.6) says A = A(~v1  : : : ~vn)AB , and since A 6= 0, we must have A(~v1  : : : ~vn) 6= 0, contrary to our hypothesis. 3.2 Determinants of linear operators The fact that dim nV = 1 makes possible a simple and elegant denition of the determinant det L of a linear operator L 2 L(V ! V ). Choose any nonzero A 2 nV . Dene AL] 2 F ( nV ! R) by AL] (~v1  : : : ~vn) = A L(~v1 ) : : :  L(~vn)]  8 (~v1 : : : ~vn) 2 nV: (3.2.1) Since L is linear and A is multilinear, denition (1.1.6) shows that AL] is multilinear. It is obvious that AL] is totally antisymmetric if A is. Thus AL] 2 nV . Since A 6= 0, A is a basis for nV , and there is a scalar kAL such that AL] = kALA: (3.2.2) From the denition (3.2.1), if a 2 R, (aA)L] = aAL]. Thus by (3.2.2), (aA)L] = kALaA. But, applying the general result (3.2.2) to aA instead of A, we have (aA)L] = kaAL(aA). Thus kaAL(aA) = (aA)L] = aAL] = akALA = kAL(aA): 18CHAPTER 3. ALTERNATING TENSORS, DETERMINANTS, ORIENTATION, AND N -DIMENSIONAL RIGHT-HANDEDNESS Since A 6= 0, kaAL = kAL. But every member of nV can be written aA for some a 2 R. Therefore kAL has the same value for all nonzero A 2 nV . It is independent of A, and depends only on L. Thus for any L 2 L(V ! V ) there is a real number kL such that for any nonzero A 2 nV , AL] = kLA: (3.2.3) This equation is obviously true also for A = 0, and hence for all A in nV . The determinant of L, det L, is dened to be kL. Thus, by the denition of det L, AL] = (det L) A (3.2.4) for any A 2 nV . Returning to the denition (3.2.1) of AL], we have A L (~v1 )  : : :  L (~vn)] = (det L) A (~v1  : : : ~vn) (3.2.5) for any A 2 nV and any (~v1  : : : ~vn) 2 nV . Now suppose (~v1  : : : ~vn) is any ordered basis for V , and Li j is the matrix of L relative to this basis. Then L(~vi) = Li j~vj so h i A L (~v1 )  : : :  L (~vn)] = A L1 j1~vj1  : : :  Ln jn ~vjn = L1 j1 : : : Ln jn A (~v1  : : : ~vn) because A is multilinear = L1 j1 : : : Ln jn "j1 : : : jnA (~vj1  : : : ~vjn ) because A is totally antisymmetric (3.1.1)   = det Li j A (~v1  : : : ~vn) where det Li j is the determinant of the n n matrix Li j . Comparing this with (3.2.5), we see that det L = det Li j : (3.2.6) 3.2. DETERMINANTS OF LINEAR OPERATORS 19 The determinant of a linear operator is the determinant of its matrix relative to any basis. In deducing (3.2.6) from (3.2.5) we have used the fact that A(~v1 : : : ~vn) 6= 0. This fact follows from remark (3.1.6). Some properties of det L are easy to deduce from (3.2.5). For example, if K and L are both linear operators on V , det (K  L) = (det K ) (det L) : (3.2.7) The proof is simple. From (3.2.5), A (K  L) (~v1)  : : :  (K  L) (vn)] = det (K  L) A (~v1 : : : ~vn) = A K (L(~v1 ))  : : :  K (L(~vn ))] = (det K ) A L (~v1 )  : : :  L (~vn)] = (det K ) (det L) A (~v1  : : : ~vn) : These equations are true for any A 2 nV and any f~v1 : : : ~vng V . If we choose A 6= 0 and (~v1  : : : ~vn) an ordered basis, then A(~v1  : : : ~vn) 6= 0, and we can cancel it, obtaining (3.2.7). This choice of A and (~v1  : : : ~vn) also shows that if we set L = IV in (3.2.5) (IV := identity mapping of V onto V ) then det IV = 1: (3.2.8) As another application of (3.2.5), let A 6= 0 and f~v1 : : : ~vng be a basis for V . Then, as remarked in theorem (18.2), L is an isomorphism () fL(~v1 ) : : :  L(~vn )g is linearly independent. From remark 3.1.20 this is true () A L(~v1 ) : : :  L(~vn )] 6= 0: From (3.2.5), since A(~v1     ~vn) 6= 0, AL(~v1 )     L(~vr )] 6= 0 , det L 6= 0. Thus a linear operator is invertible, and hence an isomorphism, i its determinant is nonzero. 20CHAPTER 3. ALTERNATING TENSORS, DETERMINANTS, ORIENTATION, AND N -DIMENSIONAL RIGHT-HANDEDNESS If det L 6= 0, then L;1 exists, and L  L;1 = IV , so (det L)(det L;1 ) = det(L  L;1 ) = det IV = 1. Thus   det L;1 = (det L);1 : (3.2.9) Finally we claim det LT = det L: (3.2.10) Recall that LT is the unique member of L(V ! V ) 3 L (~u)  w~ = LT (w~ )  ~u 8~u w~ 2 V: Let (^x1  : : :  x^n) be an ordered orthonormal basis for V . Relative to this basis B , the matrix of LT is  T j h  i h i L i = cjB LT ~bi = x^j  LT (^xi ) = x^i  L (^xj ) = ciB L (xj )] = Lj i   det LT = det LT i j and   = LT 1 j1 : : :  T  jn L n "j1:::jn = Lj1 1 : : : Ljn n"j1:::jn = Now for any  2 Sn, X L(1) 1 : : : L(n) n ( sgn ) : 2Sn n o Li j 2 L(1) 1 : : :  L(n) n n o () i = (j ) () j = ;1 (i) () Li j 2 L1 ; (1)  : : :  Ln ; (n) : Hence, n 1 o o n ;1 L(1) 1 : : :  L(n) n = L1 (1)  : : :  Ln ;1(n) 1 3.3. ORIENTATION 21 so, L(1) 1 : : : L(n) n = L1 ;1(1) : : : Ln ;1 (n) : Moreover, sgn  = sgn ;1. Hence   X ;1(1) det LT = L1  : : :  Ln ;1 (n) sgn  ;1 : (3.2.11) 2Sn Now suppose f : Sn ! W where W is any vector space. We claim X  ;1 X f  = f () : (3.2.12) 2Sn 2Sn The reason is that as  runs over Sn so does ;1. The mapping  ! ;1 is a bijection of Sn to itself. But from (3.2.11) and (3.2.12), X (1) det LT = L1 : : : Ln (n) sgn  2Sn = L1 i1 : : : Ln in "i1:::in = det L: If L 2 #(V ) (see page D-25) then L;1 = LT so L  LT = IV . Then (det L)(det LT ) = det IV = 1, so (det L)2 = 1. Thus det L = 1 if V 2 #(V ): 3.3 Orientation Two ordered orthonormal bases for V , (^x1  : : :  x^n) and (^x01  : : :  x^0n ), are said to have the same orientation i one basis can be obtained from the other by a continuous rigid motion. That is, there must be for each t in 0  t  1 an ordered orthonormal basis (^x1 (t) : : :  x^n(t)) with these properties: i) x^i(0) = x^i ii) x^i(1) = x^0i iii) x^i(t) depends continuously on t. 22CHAPTER 3. ALTERNATING TENSORS, DETERMINANTS, ORIENTATION, AND N -DIMENSIONAL RIGHT-HANDEDNESS Since (^x1 (t) : : :  x^n(t) is orthonormal, there is an Lt 2 #(V ) 3 x^i(t) = Lt (^xi ). From i) and iii) above, i)0 L0 = IV iii)0 Lt depends continuously on t (i.e., its components relative to any basis depend continously on t. For example, its components relative to (^x1  : : :  x^n) are Li j = x^j  Lt (^xi) = x^j  x^i (t), and these are continuous in t.) Then det L0 = 1 and det Lt depends continuously on t. But Lt 2 #(V ) for all t, so det Lt = 1 for all t. Hence det Lt = 1 and det L1 = 1. Thus, Remark 3.3.21 If two ordered o:n: bases for V have the same orientation, the orthogonal operators L which maps one basis onto the other has det L = 1. The converse of remark 3.3.21 is also true, as is clear from the comment on page D-26 and from Theorem PP-14. We will not prove that comment here. It is obvious when dim V = 2, and for dim V = 3 it is Euler's theorem. If two ordered orthonormal bases do not have the same orientation, we say they have opposite orientations. The orthogonal operator L which maps one basis onto the other has det L = ;1. Given any ordered orthonormal basis B = (^x1  : : :  x^n), if two other ordered orthonormal bases are oriented oppositely to B , they have the same orientation. Proof: Let the bases be B 0 and B 00 . Let L0 and L00 be the orthogonal operators which map B onto B 0 and B 00. Then L00 (L0 );1 is the orthogonal operator which maps B 0 onto B 00, and detL00  (L0);1 ] = (det L00 ) det ((L0 );1) = (det L00)(det L0);1 = (;1)(;1) = +1. Thus, we can divide the ordered orthonormal bases for V into two oppositely oriented classes. Within each class, all bases have the same orientation. We call these two classes the \orientation classes" for V . Choosing an orientation for V amounts to choosing one of the two orientation classes for V . It turns out that orientations are easily described in terms of alternating tensors. 3.3. ORIENTATION 23 Denition 3.3.15 We call A 2 nV \unimodular" if jA (^x1  : : :  x^n)j = 1 (3.3.1) for every ordered orthonormal basis (^x1  : : :  x^n) of V . If A and A0 are both unimodular, neither is 0, so there is a real nonzero a such that A0 = aA. Applying (3.3.1) to both A and A0 for one particular (^x1  : : :  x^n) shows a = 1. Thus if A and A0 are unimodular. Hence A0 = A (3.3.2) Remark 3.3.22 nV contains at most two unimodular members. Does it contain any? To see that the answer is yes, we prove Remark 3.3.23 Let B = (^x1 : : : x^n ) and B 0 = (^x01  : : : x^0n) be any two ordered orthonor- mal bases for V . Let AB be dened by (3.1.1). Then AB (B 0) (an abbreviation for AB (^x01  : : :  x^0n)) is +1 or ;1 according as B and B 0 have this same or opposite orientations. Therefore, AB and ;AB are unimodular. Proof: There is an L 2 #(V ) such that x^0i = L(^xi). Then AB (^x01  : : :  x^0n) = AB L(^x1 ): : :  L(^xn )] = (det L)AB (^x1  : : :  x^n). But det L = 1, and AB (B ) = +1 by (3.1.3). QED. If A and ;A are the two unimodular members of nV , we see from remark 3.3.23 that A is positive on one of the two orientation classes of V , and ;A is positive on the other orintation class. Choosing an orientation class for V amounts simply to choosing one of the two unimodular members of nV . Therefore we accept Denition 3.3.16 An oriented Euclidean vector space is an ordered pair (V A) in which V is Euclidean vector space and A is one of the two unimodular alternating tensors over V . A is called \the" alternating tensor of the oriented space. An ordered orthonormal basis (^x1  : : :  x^n) is \positively" or \negatively" oriented according as A(^x1 : : :  x^n) is +1 or ;1. 24CHAPTER 3. ALTERNATING TENSORS, DETERMINANTS, ORIENTATION, AND N -DIMENSIONAL RIGHT-HANDEDNESS Chapter 4 Tensor Products 4.1 Denition of a tensor product Denition 4.1.17 Suppose U1  : : :  Up V1 : : :  Vq are Euclidean vector spaces and P 2 U1  : : :  Up and Q 2 V1  : : :  Vq . Dene PQ as the member of F (U1 : : : Up V1 : : : Vq ! R) which assigns to any (p + q)-tuple ~u1 : : :  ~up~v1 : : :~vq ) 2 U1 : : : Up V1 : : : Vq the real number PQ (~u1 : : :  ~up~v1 : : : ~vq ) = P (~u1 : : :  ~up) Q (~v1  : : : ~vq ) : (4.1.1) By denition (1.1.6) it is clear that PQ is multilinear, so PQ 2 U1  : : :  Up  V1  : : :  Vq : (4.1.2) PQ is called the \tensor product" of P and Q. Note that if P were in M(U1 : : : Up ! W ) and Q were in M(V1 : : : Vq ! W 0) with W 6= R, W 0 6= R, then PQ could not be dened, because the product on the right of (4.1.1) would not be dened. Tensors can be multiplied by one another, but in general multilinear mappings cannot. The tensor product PQ is sometimes written P  Q. 4.2 Properties of tensor products 25 26 CHAPTER 4. TENSOR PRODUCTS Remark 4.2.24 Commutativity can fail. PQ need not equal QP . In fact, these two functions usually have dierent domains, namely U1 : : : Up V1 : : : Vq and V1 : : : Vq U1 : : : Up. Even if their domains are the same, it is unusual to have PQ = QP . For example, suppose p = q = 1 and U1 = V1 = V . Then P and Q are vectors in V , say P = ~u, Q = ~v. For any ~x and ~y 2 V we have P (~x) = ~u  ~x, Q(~y) = ~v  ~y so (PQ) (~x ~y) = P (~x) Q (~y) = (~u  ~x) (~v  ~y). Similarly (QP ) (~x ~y) = Q (~x) P (~y) = (~v  ~x) (~u  ~y) : If PQ = QP , then (PQ)(~x ~y) = (QP )(~x ~y) for all ~x ~y 2 V , so (~u  ~x) (~v  ~y) = (~v  ~x) (~u  ~y) : Then ~x  ~u (~v  ~y)] = ~x  ~v (~u  ~y)] so ~x  ~u (~v  ~y) ; ~v (~u  ~y)] = 0: (4.2.1) If we x ~y to be any particular vector in V , then (4.2.1) holds for all ~x 2 V . Therefore ~u (~v  ~y) ; ~v (~u  ~y) = ~0: (4.2.2) This holds for every ~y 2 V , so it holds for ~y = ~u. If ~u 6= ~0 then ~u~u 6= 0 so ~u(~v ~u);(~u~u)~v = ~0 shows that ~u and ~v are linearly dependent. And if ~u = ~0, then of course ~u and ~v are also linearly dependent. Thus PQ = QP implies ~u and ~v are linearly dependent. The converse is obvious. Remark 4.2.25 There are no divisors of zero. That is if P 2 U1      Up and Q 2 V1      Vp then PQ = 0 =) P = 0 or Q = 0: (4.2.3) To prove this, suppose PQ = 0 and Q 6= 0. We will show P = 0. Choose (~v1    ~vq ) 2 V1    Vq so that Q(~v1     ~vq ) 6= 0 (by hypothesis Q 6= 0 so this is possible). Then for any (~u1     ~up) 2 U1    Up, PQ = 0 implies P (~u1     ~up)Q(~v1     ~vq ) = 0: Cancelling Q(~v1    ~vq ), which is 6= 0, gives P (~u1     ~up) = 0. Since ~u1     ~up were arbitrary, P = 0. 4.2. PROPERTIES OF TENSOR PRODUCTS 27 Corollary 4.2.13 P1    Ps = 0 =) one of P1     Ps = 0. (Induction on s gives proof). Remark 4.2.26 Associative law. If P 2 U1      Up, Q 2 V1      Vq , and R 2 W1      Wr then (PQ) R = P (QR) : (4.2.4) Hence, we can write PQR without ambiguity. Proof: For any u~ = (~u1     ~up) 2 U1    Up , any v~ = (~v1     ~vq ) 2 V1    Vq , and any w~ = (w~ 1     w~ r ) 2 W1    Wr , we have (PQ)R] (~u v~ w~) := (PQ)(~u v~)] R(w~) := P (~u) (Qv~)R(w~)] (rule of arithmetic in R) := P (~u) (QR)(~v w~)] := P (QR)] (~u v~ w~): Remark 4.2.27 Right and left distributive laws. Suppose a1     am 2 R, P1      Pm 2 U1      Up, and Q 2 V1      Vq . Then (ai Pi)Q = ai (PiQ) and Q(ai Pi) = ai(QPi ): (4.2.5) Proof: For any u~ = (~u1     ~up) 2 U1    Up and any v~ = (~v1    ~vq ) 2 V1    Vq we have (aiPi)Q] (~u v~) = (aiPi)(~u)] Q(~v) = fai Pi(~u)]g Q(~v) = ai Pi(~u)Q(~v)] (real arithmetic) = ai (PiQ)(~u v~)] = ai (PiQ)] (~u v~): The second of equations (4.2.5) is proved in the same way. 28 CHAPTER 4. TENSOR PRODUCTS Remark 4.2.28 Multilinearity. For i 2 f1     sg, suppose Pi 2 U1(i)      Up(ii) . Then P1P2    Ps is separately linear in each of P1 P2     Ps. Proof: To prove linearity in Pr , write P = P1    Pr;1, Q = Pr , R = Pr+1    Ps. (If r = 1, P = 1 if r = s, R = 1.) Then we want to show that PQR depends linearly on Q. We take Q1      Qm 2 U1(r)      Up(rr) and a1      am 2 R. Then P (aiQi)R = (by 4.2.2) P (aiQi)R] (by 4.2.5) = P ai(QiR)] = (by 4.2.5) = ai P (QiR)] = (by remark 4.2.2) = ai(PQiR). 4.3 Polyads Denition 4.3.18 If (~v1     ~vq ) 2 V1    Vq , where V1      Vq are Euclidean spaces, then the tensor product ~v1~v2    ~vq is called a polyad of order q. It is a member of V1      Vq . It is a \dyad" if q = 2, a \triad" if q = 3, a \tetrad" if q = 4, a \pentad" if q = 5. By the denition 4.1.17 of a tensor product, for any (~x1      ~xq ) 2 V1    Vq we have (~v1    ~vq ) (~x1      ~xq ) = (~v1  ~x1 )    (~vq  ~xq ) : (4.3.1) By remark (4.2.28), ~v1~v2    ~vq depends linearly on each of ~v1     ~vq when the others are xed. We dene a mapping a) b) c)  : V1      Vq ! V1      Vq by requiring that for any (~v1     ~vq ) 2 V1    Vq we have  (~v1     ~vq ) = ~v1~v2    ~vq : Then  is multilinear. That is  2 M (V1    Vq ) ;! V1      Vq ) : (4.3.2) The tensor product PQ is sometimes written P  Q, and ~v1~v2    ~vq is written ~v1  ~v2      ~vq . This notation makes (4.3.2b) look thus: (~v1     ~vq ) = ~v1      ~vq . We will usually avoid these extra 's. 4.3. POLYADS 29 Remark 4.3.29 ~u1~u2    ~uq = 0 i at least one of ~u1 ~u2     ~uq is 0. Proof: (= follows from remark 1.3.11 and the multilinearity of  in (4.3.2). =) follows by induction on q from remark (4.2.25). Remark 4.3.30 Suppose for each p 2 f1     qg that ~up~vp 2 Vp and ap 2 R, and ~vp = a(p)~u(p). Suppose also a1    aq = 1. Then ~u1    ~uq = ~v1    ~vq . Proof: h Since  is multilinear, ~v1~v2    ~vq = (a1~u1)(a2~u2)    (a(q)~u(q)) = a1 a2    a(q) ~u1~u2    ~uq = ~u1 ~u2    ~uq . i Remark 4.3.31 Suppose that for each p 2 f1     qg, we have ~up~vp 2 Vp. Suppose also that ~u1    ~uq = ~v1    ~vq 6= 0. Then 9a1      aq 2 R such that a1 a2    aq = 1 and for each p 2 f1     qg we have ~vp = a(p)~u(p). Proof: By remark 4.3.29, no ~up or ~vp is ~0. By hypothesis, for any (~x1      ~xq ) 2 V1    Vq we have     (~u1  ~x1 )    ~u(q)  ~x(q) = (~v1  ~x1)    ~v(q)  ~x(q) : (4.3.3) Choose any p 2 f1    qg, and take ~xr = ~vr if r 6= p. Then for any ~xp 2 Vp (4.3.3) implies )    (~u(q)  ~v(q) ) ~v(p)  ~x(p) = (~u1  ~v(1~v)   ~v(~u) (p;1)(~v ~v(p;1) ~v)(~u(p))( ~~xv (p) )(~u(~vp+1)  ~)v(p+1) 1 1 (p;1) (p;1) (p+1) (p+1)  (~v(q)  ~v(q) ) or ~v(p)  ~x(p) = a(p)~u(p)  ~x(p) (4.3.4) where ~u(p+1)  ~v(p+1) )    (~u(q)  ~v(q) ) : a(p) = ((~u~v1  ~v~v1 ))    ((~u~v(p;1)  ~~vv(p;1) )( 1 1 (p;1) (p;1) )(~v(p+1)  ~v(p+1) )    (~v(q)  ~v(q) ) (4.3.5) 30 CHAPTER 4. TENSOR PRODUCTS Since (4.3.4) is true for all ~xp 2 Vp, it follows that ~vp = a(p)~u(p): Then ~v1    ~vq = (a1    aq )(~u1    ~uq ). But ~v1    ~vq = ~u1    ~uq by hypothesis, so (a1    aq ; 1)(~u1    ~uq ) = 0. Since ~u1    ~uq 6= 0, a1    aq ; 1 = 0, or a1    aq = 1 (see facts about ~0, p=s). Chapter 5 Polyad Bases and Tensor Components 5.1 Polyad bases It is true (see exercise 4) that some tensors of order  2 are not polyads. However, every tensor is a sum of polyads. We have Theorem 5.1.15 For p 2 f1 : : :  qg, suppose Vp is a Euclidean space and Bp = (~b(1p)      ~b(npp)) (2) (q) is an ordered basis for Vp. Then the n1 n2    nq polyads ~b(1) i1 ~bi2   ~biq are a basis for V1      Vq . Proof: First these polyads are linearly independent. Suppose that S i1 iq is an n1 n2    nq dimensional array of scalars such that (q ) S i1 iq~b(1) (5.1.1) i1   ~biq = 0: We want to prove S i1 iq = 0. Equation (5.1.1) is equivalent to the assertion that for any (~x1      ~xq ) 2 V1    Vq ,   ~(q)  S i1 iq ~b(1)  ~ x    bi  ~xq = 0: 1 i q 1 (5.1.2) Let BpD = (~b1(p)      ~bn(pp)) be the dual basis for Bp. Choose a particular q-tuple of integers (j1      jq ) and set ~x1 = ~bj(1)1      ~xq = ~bj(qq) in (5.1.2). The result is  ~j1  ~(q) ~jq  S i1 iq ~b(1) = bi  b = 0: i b 1 (1) q 31 (q) 32 CHAPTER 5. POLYAD BASES AND TENSOR COMPONENTS But for dual bases, (~bi  ~bj ) = i j , so S i1 iq i1 j1 = i jq = 0: Hence, S j1 iq = 0, as we hoped. (q) Second, the n1n2    nq polyads ~b(1) i1    ~biq span V1      Vq . For let T 2 V1      Vq . Dene   T i1 iq := T ~bi(1)1      ~bi(qq) (5.1.3) and (q) & := T i1 iq~b(1) (5.1.4) i1    ~biq : Clearly & is a linear combination of the polyads ~b(i1i)   ~b(iqq) . We will show that T = &. Since (B1D      BqD ) is a basis sequence for V1    Vq , by remark 1.2.10 it su"ces to show that     & ~bj(1)1      ~bj(qq) = T ~bj(1)1      ~bj(qq) : But  & ~bj(1)1      ~bj(qq)  h (q)   (j1 ) (q) i = T i1 iq ~b(1) i1    ~biq ~b(1)      ~biq  ~j1  ~(q) ~jq  i1 iq j1 = T i1 iq ~b(1) i1    iq jq i1  b(1)    biq  b(q) = T   = T j1 iq = T ~bj1      ~bjq : (1) (q ) QED. Corollary 5.1.14 If T 2 V1    Vq and p 2 f1     qg then T is a sum of n1    nq =np polyads, where n1 = dim V1     nq = dim Vq . Proof: From T = & and (5.1.4) we can write T = ~bi(1)   ~bi(pp;;11)~v i1 ip;1 ip+1 iq~bi(pp+1+1)   ~bi(qq) 1 where ~v(ip1) ip;1ip+1 iq = T i1 iq~bi(pp) 2 Vp: 5.2. COMPONENTS OF A TENSOR RELATIVE TO A BASIS SEQUENCE: DEFINITION 33 Corollary 5.1.15 Suppose V1     Vq are Euclidean spaces and W is a real vector space, not necessarily Euclidean. Suppose L : V1      Vq ! W and M : V1      Vq ! W are linear, and L(T ) = M (T ) for every polyad T . Then L = M . Proof: For an arbitrary T 2 V1      Vq , write T = PNj=1 Tj where the Tj are polyads. Then L(T ) = PNj=1 L(Tj ) = PNj=1 M (Tj ) = M (T ) because L and M are linear. 5.2 Components of a tensor relative to a basis sequence: Denition Denition 5.2.19 For each p 2 f1     qg, suppose Vp is a Euclidean space and Bp = (~b(1p)      ~b(npp)) is an ordered basis for Vp, with dual basis BpD = (~b1(p)      ~bn(pp) ). Suppose T 2 V1      Vq . Then T has 2q arrays of components relative to the basis sequence (B1     Bq ) for V1    Vq . Each array has dimension n1    nq . The arrays are as follows: 8   > i1 iq = T ~bi1      ~biq T > (1) ( q ) > > > > >   > i1 iq;1 i = T ~b(1) i1      ~b(q;1) iq;1  ~b (q) T > q i q > > > > <. .. > > > > >  (1) ~ q;1 ~ iq  > i q =T ~ > T bi1      biq;1  b(q) i1 iq;1 > > > > > > (q)  : Ti1 iq = T ~bi(1) ~      b iq 1 (5.2.1) An array is said to be \covariant" in its subscripts and \contravariant" in its superscripts. The rst array in (5.2.1) is the array of \contravariant components of T relative to (B1     Bq )." The last array is the array of \covariant components relative 34 CHAPTER 5. POLYAD BASES AND TENSOR COMPONENTS to (B1      Bq )." The other 2q ; 2 arrays are arrays of \mixed components relative to (B1      Bq )." Note that each array in the list (5.2.1) is the array of contravariant components of T relative to a suitable basis sequence. For example, T i1 iq;1 iq is the contravariant array relative to the basis sequence (B1      Bq;1 BqD ), and Ti1 iq is the contravariant array relative to (B1D  B2D      BqD ). Similarly, each array is the covariant array relative to some basis sequence. From (5.1.3) and (5.1.4) and the fact that T = & in those equations, we conclude (2) (q ) T = T i1 iq~b (1) i1 ~b i2   ~biq : (5.2.2) That is, if T 2 V1      Vq , then the coe"cients required to express T as a linear combination of the basis polyads from the basis sequence (B1     Bq ) are precisely the contravariant components of T relative to that basis sequence. By considering all 2q basis sequences, (B1     Bq ), (B1     Bq;1 BqD )     (B1D      BqD ) we obtain from (5.2.2) the 2q equations 8 > i1 iq~b (1)   ~b (q) T = T > i1 iq > > > > > > q;1 iq > T = T i1 iq;1 iq~b(1) i1    ~biq;1~b(q) > > > > > <. .. > > > > > > > T = Ti1 iq;1 iq = ~bi(1)1   ~b(iqq;;11)  ~bi(qq) > > > > > > i1    ~b iq  : T = Ti1 iq~b(1) (q) (5.2.3) (q ) i i Note 5.2.1 Suppose T = T~i iq~b(1) i    ~biq for some array T~ q . Since the polyads ~bi(1)   ~bi(qq) are a basis for V1      Vq , therefore T~i iq = T i iq . By using others of the 1 1 1 1 1 1 2q basis sequences one can reach the same conclusion for any of the 2q equations (5.2.3). Note 5.2.2 If S and T 2 V1      Vq and there is one basis sequence for V1    Vq relative to which one of the 2q component arrays of S is the same as the corresponding component array of T (e.g. Ti1 i2 i3 iq = Si i2 i3    iq ) then S = T . 5.3. CHANGING BASES 35 Note 5.2.3 The notation has been chosen to permit the restricted index conventions. A double index is summed only when it is a superscript at one appearance and a subscript at the other. The equation ai = bi holds for all possible values of i, but ai = bi should not occur unless we introduce orthonormal bases. Note 5.2.4 If each of the bases B1      Bq is orthonormal, then Bp = BpD , and all 2q component arrays (5.2.3) are identical, and all 2q equations (5.2.3) are the same. One uses the non-restricted index convention. 5.3 Changing bases Suppose T 2 V1    Vq . If we know the array of contravariant components of T relative to one basis sequence (B1     Bq ), we know T . Therefore, we ought to be able to calculate the contravariant components of T relative to any other basis sequence (B~1      B~q ). The procedure is this. Let ~ (p) ~ (p) ~ ~~ (p) ~~ (p)! Bp = b1      bnp  Bp = b1      bnp BpD 1 !  ~1 np ~ ~ np  ~ D ~ ~ ~ = b (p)     b (p)  Bp = b (p)      b (p) : The array of contravariant components of T relative to (B~1      B~q ) is  j1 jq T~j1 jq = T ~~b(1)      ~~b(q) !  i1 iq~ (1) ~ (q)  ~~ j1 jq ! ~ = T bi1    biq b(1)      ~b(q) !# "  ~~ j1 jq ~ (1) ( q ) i 1 iq ~ ~ ~ bi1    biq b(1)      b (q) = T   ~~ (q) ~~ jq ! i 1 iq ~ (1) ~ j1 = T bi1  b(1)    biq  b(q) : Thus  !  ! ~b j1    ~b (q)  ~~b jq : ~  T~j1 jq = T i1 iq ~bi(1) iq (q) (1) 1 (5.3.1) 36 CHAPTER 5. POLYAD BASES AND TENSOR COMPONENTS If we replace some of the bases in (B1     Bq ) and (B~1      B~q ) by their duals, we can immediately obtain (5.3.1) with some of the j 's as subscripts and some of the i's as subscripts on the components and superscript on the ~b's. For example, T~j1 j2 jq !  ! (1) j2 ~ ~ (1) (2) = T i1 iq;1 iq ~bi1  ~bj1 ~bi2  ~b(2)     ! ! jq;1 jq ~ ~ i ( q ; 1) q    ~biq;1  ~b(q;1) ~b(q)  ~b(q) (5.3.2) As an interesting special case of (5.3.1), we can take for (B~1      B~q ) the basis sequence obtained from (B1     Bq ) by replacing some of the Bp by BpD . For example, if B~1 = B1D and B~p = Bp for p  2, (5.3.1) becomes Tj1 j2 jq = T i1 j2 jq gi(1)1j1 (5.3.3) where gij(1) = ~bi(1)  ~bj(1) is the covariant metric matrix of B1. Similarly, if (B~1     B~q ) = (B1D      BqD ), (5.3.1) becomes Tj1 jq = T i1 iq gi(1)1 j1    gi(qqj)q : (5.3.4) Formulas like (5.3.3) and (5.3.4) (there are 4q such) are said to raise or lower the indices of the component array of T . 5.4 Properties of component arrays Remark 5.4.32 If S , T 2 V1    Vq and a b 2 R, then relative to any basis sequence for V1    Vq , (aS + bT )i1 iq = aS i1 iq + bT i1 iq : Proof: Immediate from denition of components, 5.2.19. 5.5. SYMMETRIES OF COMPONENT ARRAYS 37 Remark 5.4.33 Suppose P 2 U1     Up and Q 2 V1     Vq . Then relative to any basis sequence for U1    Up V1    Vq (PQ)i1 ipj1 jq = P i1 ip Qj1 jq : Proof: Immediate from denition of components, 5.2.19. Remark 5.4.34 Suppose T 2 V1      Vq and ~v(p) 2 Vp p = 1     q. Taking components relative to any basis sequence gives   T ~v(1)     ~v(q) = T i1 iq vi(1)1    vi(qq): Proof:  i1  ~v (p) = vi(pp)~bi(pp) so T (~v (1)     ~v (q) ) = T vi(1)1 ~b(1)      vi(qq)~b(iqq) (q) T ~bi1      ~b iq  = v (1)    v (q) T i1 iq . = vi(1)    v iq i1 iq (1) (q ) 1 5.5 Symmetries of component arrays If V1 =    = Vq = V , it is usual to consider only basis sequences (B1     Bq ) in which B1 =    = Bq = B . This is not necessary, but it does help if one wants to study the symmetries of T 2 V1      Vq = q V by looking at its arrays of components. If B1 =    = Bq = B , then one speaks of the arrays of components of T relative to the basis B . If we do have B1 =    = Bq = B = (~b1      ~bn), and if T 2 q V , then for any permutation  2 Sq we have    (T )i1 iq = (T ) ~b i1      ~b iq = T b i
(1)      ~b i
(q) = T i
(1) i
(q) :  (5.5.1) 38 CHAPTER 5. POLYAD BASES AND TENSOR COMPONENTS Here B D = (~b1      ~bn) is the basis dual to B . For mixed arrays of components, things are not quite so simple. For example, if (rs) is a transposition with r < s then   (rs)T ]i1 ir is;1 is is+1 iq = (rs)T ] ~b i1      ~b is;1  ~bis  ~b is+1    ~b iq   = T ~b i1      ~b ir;1  ~bis  ~b ir+1      ~b is;1  ~bir  ~bis+1    ~biq = T i1 ir;1 is ir+1 is;1ir is+1 iq : Whether T 2 q V is symmetric or antisymmetric under (rs) can be tested by looking at one array of components of T relative to one basis, as long as the r and s indices in that array are both covariant or both contravariant. If one index is covariant and the other is contravariant, two dierent component arrays must be compared to test for symmetry. For example, if T 2 V  V antisymmetric, then relative to any basis for V , T ij = ;T ji T i j = ;Tj i  Tij = ;Tji: (5.5.2) Proof: T (~bi ~bj ) = ;T (~bj  ~bi ) , T (~bi ~bj ) = ;T (~bj  ~bi) , T (~bj  ~bi) = ;T (~bj  ~bi) . Moreover, if any one of the three equations (5.5.2) is true relative to one basis B for V , then T is antisymmetric. For example, if T i j = ;Tj i then T and ;(12)T take the same values on the basis sequence (B D  B ) for V V . Therefore, by remark 1.2.10, T = ;(12)T . 5.6 Examples of component arrays Example 5.6.3 Let I be the identity tensor on Euclidean vector space V . By denition, I is that member of V  V such that for any ~u~v 2 V we have I (~u~v) = ~u  ~v: Let B = (~b1      ~bn) be any ordered basis for V , with dual basis B D = (~b1      ~bn ). Let gij = ~bi  ~bj and gij = ~bi  ~bj be the covariant and contravariant metric matrices relative to 5.6. EXAMPLES OF COMPONENT ARRAYS B . Then I ij Ii j Ii j Iij 39     = I ~bi ~bj = ~bi  ~bj = gij   = I ~bi ~bj = ~bi  ~bj = i j = I ~bi ~bj = ~bi  ~bj = i j   = I ~bi ~bj gij : (The reason for calling I the \identity tensor" will appear later.) Relative to an orthonormal basis, any component array of I is ij . From (5.2.3) it follows that I = gij~bi~bj = i j~bi~bj = i j~bi~bj = gij~bi~bj : Thus, I = gij~bi~bj = ~bi~bi = ~bi~bi = gij~bi~bj : (5.6.1) In particular, if (^x1      x^n) is an orthonormal basis for V , I = x^ix^i (5.6.2) Example 5.6.4 Let V , B , B D be as in example 5.6.3. Let A be any alternating tensor over V (i.e. any member of nV ). Then from (3.1.2)     Ai1 in = A ~b i1      ~b in = "i1 in A ~b 1     ~b n :     Ai1 in = A ~bi1    ~bin = "i1 in A ~b1    ~bn : It is not true that   Ai1 i2 in = "i1 i2 in A ~b1  ~b2      ~bn : (5.6.3) (5.6.4) If (V A) is an oriented Euclidean vector space, then A is unimodular. If B = (^x1      x^n) is an orthonormal basis for V then from (5.6.4) Ai1 in = "i1 in if (^x1      x^n) is positively oriented: Ai1 in = ;"i1 in if (^x1      x^n ) is negatively oriented: 40 CHAPTER 5. POLYAD BASES AND TENSOR COMPONENTS Chapter 6 The Lifting Theorem We must learn to perform a number of simple but useful operations on tensors of order q. Most of these operations will be easy to perform on polyads, so we express T 2 V1   Vq as a sum of polyads and perform the operations on the individual polyads, adding the results. The procedure works quite well once we have overcome a di"culty which is best understood by considering an example. Suppose 1  r < s  q and V1     Vq are Euclidean vector spaces and Vr = Vs. For any T 2 V1      Vq , we want to dene the \trace on indices r and s", written trrsT . First, suppose P is a polyad in V1      Vq , that is P = ~u1    ~ur    ~us    ~uq : (6.0.1) trrsP := (~ur  ~us) ~u1    6 ~ur    6 ~us    ~uq := (~ur  ~us) ~u1    ~ur;1~ur+1    ~us;1~us+1    ~uq : (6.0.2) Then we dene If T 2 V1      Vq , we write T= and we dene m X i=1 Pi Pi a polyad trrsT = m X i=1 trrsPi: There are two serious objections to this apparently reasonable procedure: 41 (6.0.3) (6.0.4) 42 CHAPTER 6. THE LIFTING THEOREM (i) Suppose P = ~u1    ~ur    ~us    ~uq = ~v1    ~vr    ~vs    ~vq . Is it true that (~ur  ~us)~u1    6 ~ur    6 ~us    ~uq = (~vr  ~vs)~v1    6 ~vr    6 ~vs    ~vq ? If not, trrsP is not uniquely dened by (6.0.2). (ii) Suppose T = Pmi=1 Pi = Pnj=1 Qj where Pi and Qj are polyads. Is it true that Pm tr P = Pn tr Q ? If not, tr T is not uniquely dened by (6.0.3). rs i=1 rs i j =1 rs j To attack this problem, we return to the very rst sloppiness, in (6.0.2). We hope that trrsP will be a unique polyad in V1    6 V r     6 V s     Vq , so that trrs maps the set of polyads in V1    Vq into the set of polyads in V1    6 V r    6 V s    Vq . But di"culty i) above leads us to look carefully at (6.0.1) and (6.0.2), and to recognize that until we have provided some theory all we have really done in (6.0.2) is to show how to take an ordered q-tuple (~u1     ~uq ) and assign to it a polyad of order q ; 2 in V1     6 V r     6 V s      Vq . The polyad assigned to (~u1     ~uq ) is M (~u1     ~uq ) = (~ur  ~us) ~u1    6 ~ur    6 ~us    ~uq : (6.0.5) This equation does clearly and unambiguously dene a function M 2 F (V1    Vq ! V1     6 V r     6 V s      Vq ). In fact, from denition (1.1.6), property d0 on page D-18 for dot products, and remark 4.2.28 for tensor products, it is clear that M is multilinear. Hence, M 2 M(V1    Vq ! V1     6 V r     6 V s      Vq ): (6.0.6) This M is all we really have. The process of constructing a linear mapping trrs : V1    Vq ! V1    6 V r    6 V s    Vq from the M of (6.0.5) is a very general one. That such a construction is possible and unique is the intent of the \lifting theorem." This theorem holds for all tensors, not just those over Euclidean spaces, and it is sometimes (as in Marcus, Multilinear Algebra) taken as the denition of V1      Vq . A formal statement of the lifting theorem for Euclidean vector spaces is as follows: Theorem 6.0.16 (Lifting theorem). Let V1     Vq be Euclidean vector spaces and let W be any real vector space. Let M 2 M(V1    Vq ! W ). Then there is exactly one 43 V1 x . . . x Vq M x W x V1 x . . . x Vq M Figure 6.1: M 2 L(V1      Vq ! W ) such that M = M   (6.0.7) where  2 M(V1    Vq ! V1      Vq ) is dened by (~v1     ~vq ) = ~v1    ~vq . Before proving the theorem, we discuss it and try to clarify its meaning. To make (6.0.7) less abstract we note that it is true () M (~v1     ~vq ) = M  (~v1    ~vq )]  8 (~v1     ~vq ) 2 V1    Vq : This is equivalent to M (~v1    ~vq ) = M (~v1    ~vq ) : (6.0.8) In other words, the lifting theorem asserts the existence of exactly one linear mapping M : V1      Vq ! W such that when M is applied to any polyad the result is the same as applying M to any q-tuple of vectors whose tensor product is that polyad. A diagram of the three functions may clarify matters: M maps V1    Vq into W , and  maps V1    Vq into V1      Vq . Both mappings are multilinear. The lifting theorem \lifts" M from V1    Vq to V1    Vq by producing a unique linear mapping M : V1      Vq ! W which satises (6.0.7) or (6.0.8). These equations mean that if we start at any (~v1     ~vq ) in (V~1    V~q ), and go to W either via M or via  and M , we will reach the same vector in W . Usually, when people draw the diagram as in Figure 6.2 they mean f : U ! W , g : U ! V , h : V ! W , and f = h  g. Proof of the lifting theorem: The proof is actually rather simple. Choose a basis sequence (B1      Bq ) for V1    Vq , with Bp = (~b(1p)      ~b(npp) ). Let BpD = (~b1(p)      ~bn(pp) ) be the basis for Vp dual to Bp. First 44 CHAPTER 6. THE LIFTING THEOREM h V W g f U Figure 6.2: we will prove that M is unique, i.e. there can be no more than one M 2 L(V1   Vq ! W ) which satises (6.0.7) or (6.0.8). Suppose M is such a mapping. Let T 2 V1   Vq . (q ) (1) (q) i i By (5.2.2), T = T i1 iq~b(1) i1    ~biq so M (T ) = T 1 q M (~bi1   ~biq ) because M is linear. Then M (T ) = T i1 iq M (~bi(1)      ~bi(qq) ) from (6.0.8) with ~v1 = ~bi(1)     ~vq = ~biq(q) . Thus 1 1 if such an M exists, then for any T 2 V1      Vq we must have    (q)  M (T ) = T ~bi(1)1      ~bi(qq) M ~b(1) i1      ~biq : (6.0.9) In other words, M is determined because M (T ) is known for all T 2 V1      Vq . It still remains to prove that there is an M 2 L(V1      Vq ! W ) which satises (6.0.7) and (6.0.8). Our uniqueness proof has given us an obvious candidate, namely the M : V1      Vq ! W dened by (6.0.9). This M is certainly a well dened function mapping V1      Vq into W . By the denitions of ad and sc in V1      Vq , M (T ) depends linearly on T , i.e. M 2 L(V1      Vq ! W ). For a formal proof, let T1     Tn 2 V1      Vq and a1      an 2 R. Then we want to show M (aj Tj ) = aj M (Tj ). From (6.0.9), and the denition of aj Tj , and the rules of vector arithmetic in W , given on page D-11, we compute   M aj Tj = = = =  j  ~ i1   (q)  ~ a Tj b(1)      ~b(iqq) M ~bi(1)      b iq 1 n j h ~ i1 io  (q)  ~      b a Tj b(1)      ~b(iqq) M ~bi(1) iq 1   i h  i1      ~b iq M ~b (1)      ~b (q) aj Tj ~b(1) iq i1 (q ) aj M (Tj ): QED.] 45 It remains only to prove that M satises (6.0.8). Let (~v1     ~vq ) 2 V1    Vq .  i1  i1 )    (~v  ~b iq ) = v i1    v iq , so Then if T = ~v1    ~vq we have T ~b(1)      ~b(iqq) = (~v1  ~b(1) q (q ) 1 (q) iq (1) ( q ) i 1 ~ ~ M (~v1    ~vq ) = v1    v(q) M (bi1      biq ). By hypothesis, M is multilinear, so M (~v1    ~vq ) = M (v1i1~bi(1)      vqiq~bi(qq) ). But this is just (6.0.8). QED. 1 As an application of the lifting theorem, let us return to trrsT . The M dened in (6.0.5) will be the M of the lifting theorem. The vector space W in that theorem will be W = V1     6 V r     6 V s      Vq . Then, M 2 L(V1      Vq ! V1     6 V r     6 V s      Vq ) is given to us by the lifting theorem, and we dene for any T 2 V1      Vq trrsT := M (T ): (6.0.10) If a T is a polyad, T = ~v1    ~vq , then trrs (~v1    ~vq ) = M (~v1    ~vq ) = M (~v1     ~vq ) = (~vr  ~vs) ~v1    6 ~vr    6 ~vs    ~vq : Thus the lifting theorem answers both the objections on page 41. We agree to dene trsr T := trrsT: (6.0.11) (6.0.12) As an application of this \machinery," let us nd the component arrays of tr12 T relative to the basis sequence (B3      Bq ) for V3    Vq . We have (3) (q) T = Ti1 i2i3 iq~bi(1)1 ~b(2) i2 ~bi3    ~biq : We must suppose V1 = V2 if tr12 T is to be dened. Then we take B1 = B2, so ~bi(1) = ~bi(2) . Since tr12 is a linear mapping, h i1 ~ (2)~ (3) ~ (q) i tr12 T = Ti1 i2 iq tr12 ~b(1) bi2 bi3    biq : (6.0.13) By (6.0.12), h (3)    ~b(q) i = ~bi1  ~b(2)  ~b(3)    ~b(q) : ~ tr12 ~bi(1)1 ~b(2) b i2 i3 iq iq (1) i2 i3 i We have agreed to take B1 = B2 , so ~bi(1)1  ~b(2) i2 =  1 i2 . Thus, h (q )  (3) (q) i  i1 (3) tr12 ~bi(1)1 ~b(2) i2 ~bi3    ~biq =  i2~bi3   ~biq : (6.0.14) 46 CHAPTER 6. THE LIFTING THEOREM Then, substituting this in (6.0.13), (q) tr12 T = Tj ji3 iq~b(3) i3    ~biq  so Similarly is proved from (tr12 T )i3 iq = Tj ji3 iq : (6.0.15) (tr12 T )i3 iq = T j j i3 iq (6.0.16) i2 ~b(3)   ~b (q) : ~b(2) T = T i1 i2 i3 iq~bi(1) i3 iq 1 Recall that for orthonormal bases we need not distinguish between superscripts and subscripts. Thus, if all bases are orthonormal and B1 = B2, (tr12 T )i3 iq = Tjji3 iq : (6.0.17) Chapter 7 Generalized Dot Products 7.1 Motivation and denition Suppose U , V , and W are Euclidean vector spaces, and ~u 2 U , ~x 2 V , ~y 2 V , w~ 2 W . Then ~u~x 2 U  V and ~yw~ 2 V  W . What could be more natural than to dene (following Willard Gibbs) (~u~x)  (~yw~ ) := ~u (~x  ~y) w~  (7.1.1) which is the same as (~u~x)  (~yw~ ) := (~x  ~y) ~uw~ ? Then the denition can be extended so that if P 2 U  V and R 2 V  W we dene P  R by writing P= m X i=1 Pi R= n X j =1 Rj (7.1.2) where Pi and Rj are dyads then we dene P  R := n m X X i=1 j =1 Pi  Rj (7.1.3) where Pi  Rj is dened by (7.1.1). It is in U  W , as is P  R. Does all this work? If ~u~x = ~u 0~x 0 and ~yw~ = ~y 0 w~ 0, is it true that (~x  ~y) ~uw~ = 0 Rj0 , is it true (~x 0  ~y 0 ) ~u 0w~ 0? And if P = Pmi=1 Pi = Pmi=10 Pi0 and R = Pnj=1 Rj = Pnj=1 0 that Pmi=1 Pnj=1 Pi  Rj = Pmi=10 Pnj=1 Pi0  Rj0 ? The lifting theorem will show that all of this works, and that it can even be generalized as follows: Suppose U1     Up, V1      Vq , W1     Wr are Euclidean spaces, and ~ui 2 Ui , 47 48 CHAPTER 7. GENERALIZED DOT PRODUCTS ~xj and ~yj 2 Vj , w~ r 2 Wr . We would like to dene a generalized dot product of order q as (~u1    ~up~x1    ~xq ) hqi (~y1    ~yq w~ 1    w~ r ) := (~u1    ~up) (~x1  ~y1)    (~xq  ~yq ) (w~ 1    w~ r ) = (~x1  ~y1)    (~xq  ~yq ) (~u1    ~upw~ 1    w~ r ) : (7.1.4) Then for any P 2 U1    Up  V1    Vq and any R 2 V1    Vq  W1    Wr we would like to dene P hqiR 2 U1      Up  W1      Wr by writing (7.1.2) and P hqiR := m X n X i=1 j =1 PihqiRj : (7.1.5) Incidentally, note the mnemonic device that (~u1    ~up~x1    ~x0q ) hqi (1~y1    ~y0q w~ 1  1 w~ r ) := ~u1    ~up B ~x1 C    B ~y1 C BB CC BB  CC @ A ~xq    BB CC BB  CC : @ A ~yq w~ 1    w~ r (7.1.6) Some authors dene (~u1    ~up~x1    ~xq ) hqi (~yq    ~y1    w~ 1    w~ r ) (~xq  ~yq ) ... ... := (~x2  ~y2) ~u1    ~up (~x1  ~y1 ) w~ 1    w~ r = (~xq  ~yq )    (~x1  ~y1) ~u1    ~upw~ 1    w~ r : De gustabus non disputandum est, but I think the rst denition has more convenient properties.] To prove that these denitions are unambiguous, we dene M 2 M(U1    Up V1    Vq V1    Vq W1    Wr ! U1      Up  W1      Wr ) 7.1. MOTIVATION AND DEFINITION 49 by requiring M (~u1     ~up ~x1     ~xq  ~y1    ~yq  w~ 1     w~ r ) := = (~x1  ~y1)    (~xq  ~yq ) ~u1    ~upw~ 1    w~ r : (7.1.7) Since M is multilinear, the lifting theorem provides a unique M 2 L(U1      Up  V1      Vq  V1      Vq  W1      W ! U1      Up  W1      Wr ) satisfying M (~u1      ~up  ~x1      ~xq  ~y1      ~yq  w~ 1      w~ r ) = M (~u1     ~up ~x1     ~xq  ~y1     ~yq  w~ 1     w~ r ) = (~x1  ~y1)    (~xq  ~yq ) ~u1    ~upw~ 1    w~ r (7.1.8) for all (~u1     ~up ~x1      ~xq  ~y1     ~yq  w~ 1     w~ r  ) 2 U1    Up V1    Vq V1    Vq V1    Vq W1    Wr : For any P 2 U1     Up  V1     Vq and any R 2 V1     Vq  W1     Wr , we dene P hqiR := M (PR): (7.1.9) Then if P and R are polyads, (7.1.8) says we do have (7.1.4). We can calculate P hqiR for any tensors and P and R by writing them as sums of polyads and doing the arithmetic justied by Remark 7.1.35 Suppose P1     Pm 2 U1      Up  V1      Vq and R1      Rn 2 V1      Vq  W1      Wr and a1     am b1      bn 2 R. Then (aiPi) hqi (bj Rj ) = aibj (Pi  Rj ) Proof: (7.1.10) 50 CHAPTER 7. GENERALIZED DOT PRODUCTS (ai Pi) hqi (bj Rj ) = M (aiPi) (bj Rj )] by denition of hqi = M aibj (PiRj )] by multilinearity of the tensor product = aibj M (PiRj ) because M is linear = aibj (PihqiRj ) by denition of hqi: Equation (7.1.10) is true for any Pi and Rj , but when Pi and Rj are polyads, (7.1.10) and (7.1.4) give a convenient way to calculate P hqiR for any tensors P and R. 7.2 Components of P hqiR Suppose P 2 U1    Up  V1    Vq and R 2 V1    Vq  W1    Wr . Suppose (j ) (i) (j ) (~b(1) 1      ~bmi ) is an ordered basis for Ui , and (~1     ~nj ) is an ordered basis for Vj , and (~6(1R)     ~6(PRk) ) is an ordered basis for Wk . We would like to compute the components of P hqiR relative to the basis sequence for U1    Up W1    Wr . We have P = P i1 ipj1 jq~bi(1)   ~bi(pp) ~j(1)1    ~j(qq) 1 k1    ~ kq ~6 (1)    ~6 (r) R = Rk1 kq l1 lr ~(1) lr (q) l1 (~(kpp) are the dual basis vectors to ~k(pp) ). Then (7.1.4) gives     k1    ~ (q)  ~ kq ~b (1)    ~b (p)~6 (1)    6(r) ~   P hqiR = P i1 ipj1 jq Rk1 kq l1 lr ~j(1) jq ip l1 lr (1) (q) i1 1 (r) = P i1 ipj1 jq Rk1 kq l1 lr jk11    jkqq i(1)   ~bi(pp~) 6(1) l1    ~6lr 1 But Rk1 kq l1 lr j1 k1    jq kq = Rj1 jq l1 lr so   P hqiR = P i1 ipj1 jq Rj1 jq l1 lr ~bi(1)   ~bi(pp)~6l(1)    ~6l(rr) : 1 1 Therefore, by note (5.2.1), (P hqiR)i1 ipl1 lr = P i1 ipj1 jq Rj1 jq l1 lr : (7.2.1) 7.3. PROPERTIES OF THE GENERALIZED DOT PRODUCT 51 It is important to remember that for (7.2.1) to hold, the same bases for V1     Vq (and their dual bases) must be used to calculate the components of P and R. Covariant or contravariant metric matrices can be used to raise or lower any index i or l or pair j in (7.2.1). The same collection of 2p+q+r formulas can be obtained immediately from (7.2.1) by choosing to regard certain of the original bases for the Ui, Vj or Wk as dual bases, the duals becoming the original bases. For example, (P hqiR)i1 i2 ipl1 lr;1 lr = = Pi1 i2 ipj1 j2 jg Rj1 j2 jq l1 lr;1 lr All the 2p+q+r variants of (7.2.1) collapse into a single formula when all the bases are orthonormal. 7.3 Properties of the generalized dot product Property 7.3.1 Bilinearity. For i 2 f1     mg and j 2 f1     ng, suppose Pi 2 U1      Up  V1      Vq and Rj 2 V1      Vq  W1      Wr and ai bj 2 R. Then (aiPi)hqi(bj Rj ) = ai bj (PihqiRj ). This has already been proved as remark (7.1.35). Property 7.3.2 Suppose P 2 U1      Up  V1      Vq and R 2 V1      Vq  W1      Wq : Then if P = 0 or R = 0, P hqiR = 0. Proof: Suppose P = 0. Then 0P = P so P hqiR = (0P )hqiR = 0(P  R) = 0: Note that the converse is false. We can have P hqiR = 0 even though P 6= 0 and R 6= 0. For example ~uh1i~v = ~u  ~v and this can vanish even though ~u 6= ~0 and ~v 6= ~0. Denition 7.3.20 Since ~uh1i~v = ~u  ~v for vectors, it is usual to write P h1iR as P  R, P h2iR as P : R, P h3iR as P   R and P h4iR as P : : R. The notation P : : R for P h5iR seems not to be used. 52 CHAPTER 7. GENERALIZED DOT PRODUCTS Property 7.3.3 P h0iR = PR. Proof: If P and R are polyads, this is obvious from (7.1.4) with q = 0. Both P h0iR and PR depend linearly on R, so if R = P Rj where the Rj are polyads, and if P is a polyad, we have P h0iR = P h0i(Pj Rj ) = Pj (P h0iRj ) = Pj (PRj ) = P (Pj Rj ) = PR. Therefore P h0iR = PR if P is a polyad and R is any tensor. If P = Pi Pi where the Pi are polyads, then for any tensor R, P 0R = (Pi Pi)h0iR = Pi (Pih0iR) = Pi(PiR) = (Pi Pi)R = PR. Thus P h0iR = PR for any tensors P and R. Property 7.3.4 P hqiR and RhqiP need not be the same even if both are dened. This is obvious from property 7.3.3 and the corresponding result for tensor products. An example with q = 1 is this. Let ~u~v w~  ~x 2 V . Let P = ~u~v, R = w~ ~x. Then P  R = (~v  w~ )~u~x and R  P = (~x  ~u)w~ ~v. If these two dyads are equal and not 0, remark (4.3.30) shows that ~u must be a multiple of w~ and ~x must be a multiple of ~v. Property 7.3.5 (Associativity). Suppose P 2 U1      Up  V1      Vq R2 V1      Vq  W1      Wr  X1      Xs T2 X1      Xs  Y1      Yt: Then P hqi (RhsiT ) = (P hqiR) hsiT: (7.3.1) Proof: First we prove (7.3.1) when P R, T are polyads. In that case (7.3.1) follows from (7.1.4) by simple computation, which we leave to the reader. That (7.3.1) holds for polyads can also be seen immediately from the mnemonic diagram (7.1.6). The generalized dot products hqi and hsi are formed by 7.3. PROPERTIES OF THE GENERALIZED DOT PRODUCT 53 P {{ R p q s { { { T t r Figure 7.1: dotting vertical pairs of vectors in the regions of overlap of P , R, T . It is clearly irrelevant to the outcome, whether we dot rst the q vectors in P with those below them in R or the s vectors in R with those below them in T . Next, suppose P , R and T1      Tn are polyads and T = Pi Ti. Then P hqi(RhsiT ) = P hqi Rhsi(Pi Ti )] = P hqi Pi(RhsiTi)] = Pi P hqi(RhsiTi)] = Pi P hqi(RhsiTi)] = (P hqiR) hsi (P Ti ) = (P hqiR) hsiT: This proves (7.3.1) when P and R are polyads and T is any tensor. Next suppose P , R1     Rn are polyads and T is any tensor and R = Pi Ri. Then P hqi(RhsiT ) = P hqi (Pi Ri)hsiT ] = P hqi Pi(RihsiT )] = Pi P hqi(RihsiT )] = P (P hqiR )hsiT ] = P (P hqiR )] hsiT = P hqi(E R )] hsiT = (P hqiR)hsiT . i i i i i i Thus (7.3.1) holds when P is a polyad and R and T are any tensors. Finally, suppose P1     Pn are polyads and R and T are any tensors, and P = Pi Pi, Then P hqi(RhsiT ) = (Pi Pi)hqi(RhsiT ) = Pi Pihqi(RhsiT )] = 54 CHAPTER 7. GENERALIZED DOT PRODUCTS P R T Figure 7.2: P (P hqiR)hsiT ] = P (P hqiR)] hsiT = (P P )hqiR] hsiT = (P hqiR)hsiT . i i i i i i This proves (7.3.1) for any tensors P , R, T . Figure (7.1) also shows how associativity can fail. If P and T overlap as in gure (7.2), there is trouble. The tensors P hqi(RhsiT ) and (P hqiR) hsiT may be dierent even if both are dened (they may not be). As an example, suppose all tensors are over a single space V . Let ~u~v w~  ~x ~y 2 V . Let P = ~u~v, R = w~ , T = ~x~y. Then we have the diagram P = 22 R = 2 T = 22 which exhibits the fatal P ; T overlap shown in Figure (7.2). We have P  (R  T ) = ~u~v  (w~  ~x)~y] = (w~  ~x) (~u~v)  ~y] = (w~  ~x)(~v  ~y)~u and (P  R)  T = ~u(~v  w~ )]  (~x~y) = (~v  w~ ) ~u  (~x~y)] = (~v  w~ )(~u  ~x)~y. If we choose ~x and ~y to be mutually perpendicular unit vectors, and take ~v = ~y, ~u = w~ = ~x, we have P  (R  T ) = ~u = ~x and (P  R)  T = ~0. Property 7.3.6 Suppose P , R 2 V1      Vq . Then P hqiR = RhqiP: (7.3.2) 7.3. PROPERTIES OF THE GENERALIZED DOT PRODUCT 55 v=y u =w=x Figure 7.3: Proof: If P = ~x1~x2    ~xq and R = ~y1~y2    ~yq then P hqiR = (~x1  ~y1)    (~xq  ~yq ) = (~y1  ~x1)    (~yq  ~xq ) = RhqiP , so (7.3.2) is true of P and R are polyads. But both RhqiP and P hqiR are bilinear in P and R, so the proof of (7.3.2) for general tensors P and R in V1      Vq is like the proof of (7.3.1). Property 7.3.7 Suppose P 2 V1      Vq and P 6= 0. Then P hqiP > 0: (7.3.3) Proof: For i 2 f1     qg, let (^x(1i)      x^(nii) ) be an orthonormal basis for Vi. By (7.2.1), P hqiP = Pi1 iq Pi1 iq : (7.3.4) This is a sum of squares of real numbers. It is > 0 unless every term is 0, i.e. Pi1 iq = 0. In that case P = 0. Theorem 7.3.17 On the vector space V1      Vq , dene the dot product dp(P R) = P hqiR for all P , R 2 V1      Vq . With this dot product, V1      Vq is a Euclidean vector space. Proof: 56 CHAPTER 7. GENERALIZED DOT PRODUCTS Properties (7.2.1), (7.3.2), (7.3.3) show that dp satises a) b), c), d) on page D18. Corollary 7.3.16 Let (^x(1i)      x^(nii) ) be an orthonormal basis for Vi, i = 1     q. Then (2) (1) the n1 n2    nq polyads x^(1) i x^i    x^iq are an orthonormal basis for V1      Vq . 1 2 Proof: By theorem 5.1.15 they are a basis. To prove them orthonormal, we note (q ) (q) (1) (^x(1) x(1) x(1) x(iqq)  x^(jqq) ) = i1 j1 i2 j2    iq jq : i1    x^iq )hq i(^ j1    x^jq ) = (^ i1  x^j1 )    (^ Denition 7.3.21 If P 2 V1      Vq , dene the \length" of P , kP k, in the way one usually does for vectors in a Euclidean space. That is, kP k is the non-negative square root of P hqiP . Remark 7.3.36 Suppose P 2 U1      Up  V1      Vq and R 2 V1      Vq  W1      Wr . Then kP hqiRk  kP kkRk: (7.3.5) Proof: Introduce an orthonormal basis for each space Ui , Vj and Wk . Take components relative to these bases. Abbreviate the p-tuple of integers (i1     ip) as I , the q-tuple (j1      jq ) as J , the r-tuple (l1    lr ) as L. Then (7.2.1) can be abbreviated X (P hqiR)IL = PIJ RJL : J By the ordinary Schwarz inequality for real numbers, X J !2 X PIJ RJL  J PIJ2 ! X K 2 RKL ! where K stands for the q-tuple (k1     kq ). (We have simply changed the summation index from J to K .) Summing over I and L gives X X IL J !2 0X 1 0X 1 2 A PIJ RJL  @ PIJ2 A @ RKL IJ KL 7.4. APPLICATIONS OF THE GENERALIZED DOT PRODUCT 57 u1 . . . u p T v1 . . . v q Figure 7.4: or X IL 0 10 1 X X 2 A: (P hqiR)2IL  @ PIJ2 A @ RKL IJ By equation (7.3.4), this last inequality is KL kP hqiRk2  kP k2kRk2: Taking square roots gives (7.3.5). 7.4 Applications of the generalized dot product Application 1: The value of a tensor. Suppose T 2 U1      Up  V1      Vq and (~u1     ~up~v1    ~vq ) 2 U1    Up V1    Vq . Then T (~u1     ~up~v1     ~vq  ) = (~u1~u2    ~up) hpiT hqi (~v1~v2    ~vq ) : (7.4.1) Note that (7.3.1) applies, so no parentheses are needed to say whether hpi or hqi is calculated rst. There is none of the overlap shown in gure 7.2. The diagram for (7.4.1) is shown in Figure 7.4. Therefore, at least (7.4.1) is unambiguous. To prove it true, suppose rst that T is a polyad, T = ~x1    ~xp~y1    ~yq . Then (~u1    ~up) hpi T hqi (~v1    ~vq ) = (~u1    ~up) hpiT ] hqi (~v1    ~vq ) 58 CHAPTER 7. GENERALIZED DOT PRODUCTS = (~u1  ~x1)    (~up  ~xp) ~y1    ~yq ] hqi (~v1    ~vq ) = (~u1  ~x1)    (~up  ~xp) (~y1    ~yq ) hqi (~v1    ~vq )] = (~x1  ~u1)    (~xp  ~up) (~y1  ~v1 )    (~yq  ~vq ) = T (~u     ~up~v1    ~vq ) : This proves (7.4.1) when T is a polyad. But both sides of (7.4.1) depend linearly on T , so if (7.4.1) is true for polyads T , it is true when T is a sum of polyads. That is, (7.4.1) is true for any tensor T 2 U1      Up  V1      Vq . Corollary 7.4.17 Suppose P is a xed tensor in U1      Up  V1      Vq , with the property that P hqiR = 0 for every R 2 V1    Vq  W1    Wr . Then P = 0. (Also, if R is a xed tensor in V1      Vq  W1      Wr with the property that P hqiR = 0 for every P 2 U1      Up  V1     Vq then R = 0 ). Proof: Choose a xed nonzero T 2 W1    Wr . For any (~v1     ~vq ) 2 V1    Vq , we know by hypothesis that P hqi (~v1    ~vq )T ] = 0. Figure (7.3) works, so P hqi(~v1    ~vq )]T = 0. By remark (4.2.25), P hqi(~v1    ~vq ) = 0. Then for any (~u1     ~up) 2 U1    Up we have (~u1    ~up)hpiP hqi(~v1    ~vq )] = 0. By equation (7.4.1), P (~u1     ~up~v1    ~vq ) = 0. Since (~u1     ~up~v1     ~vq ) is an arbitrary member of U1    Up V1    Vq , therefore P = 0. (The other half of the corollary, in parentheses, is proved similarly. Corollary 7.4.18 Suppose P , P 0 2 U1      Up  V1      Vq and for every R 2 V1      Vq  W1      Wr we have P hqiR = P 0hqiR. Then P = P 0. (Also, if R, R0 2 V1      Vq  W1      Wr and for every P 2 U1      Up  V1      Vq we have P hqiR = P hqiR0, then R = R0.) Application 2: Vector cross product. Suppose (V A) is an oriented three dimensional Euclidean space whose alternating tensor is A. For any ~u, ~v 2 V , we dene their A-cross product as A ~u ~v = Ah2i (~u~v) : (7.4.2) 7.4. APPLICATIONS OF THE GENERALIZED DOT PRODUCT 59 Usually we omit the A and write simply ~u ~v, but it should always be remembered that there are two oriented 3-spaces, (V A) and (V ;A), and so two ways to dene ~u ~v. We will now show that our cross product is the usual one. First, if ai, bj 2 R and ~ui, ~vj 2 V for i = 1     m and j = 1     n (ai~ui) (bj~vj ) = ai bj (~ui ~vj ) : (7.4.3) For (ai~ui) (bj~vj ) = Ah2i (ai~ui)(bj~vj )] = Ah2i (aibj ~ui~vj )] = aibj Ah2i(~ui~vj )] = aibj (~ui ~vj ). Second, w~  (~u ~v) = w~  Ah2i(~u~v)] = A(w~  ~u~v) = A(~u~v w~ ) so (~u ~v)  w~ = A (~u~v w~ ) : (7.4.4) Therefore, since A is totally antisymmetric, 8 > < (~u ~v)  w~ = (~v w~ )  ~u = (w~ ~u)  ~v > : = ; (~v ~u)  w~ = ; (~u w~ )  ~v = ; (w~ ~v)  ~u: (7.4.5) In particular, (~u ~v)  w~ = ;(~v ~u)  w~ for all w~ , so ~v ~u = ;~u ~v: (7.4.6) Therefore, ~u ~u = ~0. Setting w~ = ~u in (~u ~v)  w~ = (w~ ~u)  ~v gives (~u ~v)  ~u = (~u ~u)  ~v = 0. Thus, using (7.4.6), (~u ~v)  ~u = (~u ~v)  ~v = 0: (7.4.7) Relative to any ordered basis (~b1  ~b2 ~b3 ) for V we have   (~u ~v)i = Aijk uj vk = A ~b1  ~b2 ~b3 "ijk uj vk : If the ordered basis is positively oriented and orthonormal, A(~b1  ~b2 ~b3 ) = 1, so (~u ~v)i = "ijk uj vk : (7.4.8) If the ordered basis is negatively oriented and orthonormal, A(~b1  ~b2  ~b3 ) = ;1 so (~u ~v)i = ;"ijk uj vk : (7.4.9) 60 CHAPTER 7. GENERALIZED DOT PRODUCTS The notorious formula ~u (~v w~ ) = (~u  w~ ) ~v ; (~u  ~v) w~ (7.4.10) can be derived from (7.4.8) as follows: relative to any positively oriented orthonormal basis ~u (~v w~ )]i = "ijkuj (~v w~ )k = "ijkuj "klmvl wm = "ijk"klmuj vl wm = (il jm ; imjl ) uj vl wm see page D-9 = uj wj vi ; uj vj wi = (~u  w~ ) vi ; (~u  ~v) wi = (~u  w~ ) ~v ; (~u  ~v) w~ ]i : Since the vector on the left of (7.4.5) has the same components as the vector on the right, those two vectors are equal. From (7.4.5) and (7.4.10) we have (~u ~v)(~u ~v ) = ~u~v (~u ~v)] = ~uk~vk2~u ; (~v  ~u)~v] = k~uk2 k~vk2 ; (~u  ~v)2 , so k~u ~vk2 = k~uk2k~vk2 ; (~u ~v)2: (7.4.11) We have dened the angle between ~u and ~v as the in 0    such that ~u  ~v = k~ukk~vk cos . With this denition of , (7.4.11) implies k~u ~vk = k~ukk~vk sin : (7.4.12) Equations (7.4.7) and (7.4.12) determine that ~u ~v is ?~u and ~v and has length k~ukk~vk sin . This leaves two possibilities for a nonzero ~u ~v. The correct one is determined by the fact that A (~u~v ~u ~v) > 0 if ~u ~v 6= 0: (7.4.13) To prove (7.4.3), we note that A(~u~v ~u ~v) = A(~u ~v ~u~v) = (~u ~v)  Ah2i(~u~v)] = (~u ~v)  (~u ~v). Inequality (7.4.3) is usually paraphrased by saying that \~u~v ~u ~v is a positively oriented ordered triple." If we orient the space we live in by choosing A so 7.4. APPLICATIONS OF THE GENERALIZED DOT PRODUCT 61 Figure 7.5: that our right thumb, index nger and middle nger are positively oriented ordered triple of vectors when extended as in Figure (7.5), we will obtain the usual denition of ~u ~v in terms of the right hand screw rule. Application 3: Let U1      Up, V1     Vq be Euclidean spaces. Then U1      Up  V1      Vq = (U1      Up)  (V1      Vq ) : (7.4.14) Proof: What is meant is that there is a natural isomorphism between the spaces on the two sides of (7.4.14). Before proving this, we note that the tensor product of U1      Up and V1      Vq on the right in (7.4.14) is dened, because U1      Up and V1      Vq are Euclidean spaces. Their dot products are hpi and hqi. To construct the isomorphism whose existence is asserted by (7.4.14), let T 2 U1      Up  V1      Vq and dene Te : (U1    Up ) (V1    Vq ) ! R by requiring for any P 2 U1    Up and any Q 2 V1     Vq that Te(P Q) = P hpiT hqiQ: (7.4.15) 62 CHAPTER 7. GENERALIZED DOT PRODUCTS Then Te is bilinear, so Te 2 (U1      Up)  (V1      Vq ). From (7.4.15), (ag j Tj ) = aj Tej , so T 7! Te is a linear mapping of U1      Up  V1      Vq into (U1    Up )  (V1    Vq ). We claim this mapping is bijective. To see that it is injective, suppose T1 , T2 2 U1    Up  V1    Vq and Te1 = Te2 . Then for every P 2 U1    Up and Q 2 V1    Vq we have P hpiT1hqiQ = P hpiT2hqiQ, so P hpi(T1hqiQ ; T2hqiQ = 0 so P hpi(T1 ; T2 )hqiQ] = 0. Hence, (T1 ; T2 )hqiQ = 0 for all Q 2 V1      Vq . Hence, T1 ; T2 = 0. To see that T 7! Te is a surjection, suppose & 2 (U1      Up)  (V1      Vq ). Dene T 2 U1    Up  V1    Vq by requiring T (~u1     ~up~v1    ~vq ) = &(~u1    ~up~v1    ~vq ): Then P hpiT hqiQ = &(P Q) if P and Q are polyads. Hence P hpiT hqiQ = &(P Q) for all P 2 U1     Up and Q 2 V1     Vq . By (7.4.15), & = Te . Thus T 7! &, and the mapping T 7! Te is a surjection of U1      Up  V1      Vq onto (U1      Up)  (V1      Vq ). Application 4: Linear mappings as second order tensors. Denition 7.4.22 Suppose U and V are Euclidean vector spaces and L 2 L(U ! V ). $ Dene L2 F (U V ! R) by $ (7.4.16) L (~u~v) = L(~u)  ~v: $ $ $ Then L (~v) and L (~u ) are linear, so L2 M(U V ! R): That is, $ L2 U  V: Thus, (7.4.16) can be written in the equivalent form $ ~u L ~v = L(~u)  ~v: (7.4.17) (7.4.18) For a xed ~u 2 U , (7.4.18) is true for all ~v 2 V . Hence, since ~u  L~ and L(~u) are both in V , we have $ ~u L= L(~u): (7.4.19) $ $ Let  : L(U ! V ) ! U  V be dened by (L) =L. If we know L= (L), we can recover L from (7.4.19), so  is an injection. We claim  is also a surjection. That is, 7.4. APPLICATIONS OF THE GENERALIZED DOT PRODUCT 63 every T 2 U  V is (L) for at least one L 2 L(U ! V ). Suppose T is given and dene L $ $ $ by L(~u) = ~u  T . Then L (~u~v) = ~u L ~v = L(~u)  ~v = ~u  T  ~v = T (~u~v), so T =L= (L). We claim that  : L(U ! V ) ! U  V is also linear. If L1      Ln 2 L(U ! V ) and a1      an 2 R, we claim ! $ (aiLi )= ai Li : (7.4.20) The proof is this: for any ~u 2 U and ~v 2 V , ! ~u (aiLi ) ~v = (aiLi)(~u)]  ~v = fai Li(~u)]g  ~v $ = ai Li(~u)  ~v] = ai (~u Li ~v)  $ = ~u  ai Li  ~v: ! $ Thus (aiLi )(~u~v) = (ai L)(~u~v) for all ~u 2 U , ~v 2 V . Hence (7.4.20). We have now shown that  is a linear bijection between the two vector spaces L (U ! V ) and U  V . We can use  to regard any linear mapping L : U ! V as a tensor $ L2 U  V and vice-versa. The process of taking linear combinations can be done either to the tensors or to the linear mappings, so confusing tensors with linear mappings does no harm to linear combinations. Linear mappings can also be \multiplied" by composition. If K 2 L(U ! V ) and L 2 L(V ! W ) then L  K 2 L(U ! W ). Our identication of tensors with linear mappings almost preserves this multiplication. We have ! $ $ L  K =K  L : (7.4.21) The proof is simple. For any ~u 2 U , ! ~u (L  K ) = (L  K )(~u) = L K (~u)] $ $ = K (~u) L= (~u  K~ ) L $ = ~u  (K~  L) by (7.3.1): Since this is true for all ~u 2 U , corollary (7.4.18) gives (7.4.21). If IU : U ! U is the identity operator on U (that is, IU (~u) = ~u for all ~u 2 U ) then for any ~u1 ~u2 2 U , $ (7.4.22) I U (~u1 ~u2) = IU (~u1)  ~u2 = ~u1  ~u2: 64 CHAPTER 7. GENERALIZED DOT PRODUCTS $ That is, I U is what we have already called the identity tensor on U  see example 5.6.3. The origin of that name is now clear. And (7.4.22) can also be written $ ~u1 I U ~u2 = ~u1  ~u2: (7.4.23) This is true for all ~u1 and ~u2 2 U , so we have for any ~u 2 U $ $ ~u I U = I U ~u = ~u: (7.4.24) If L : U ! V then L  IU = IV  L = L. It follows then from (7.4.20) that $ $ $ $ $ I U  L=L  I V =L : (7.4.25) Actually, a more general result is true. If U , V , W1 ,    , Wq are Euclidean spaces and T 2 U  W1      Wq  V then $ $ (7.4.26) I U T = T  I V = T: The proof follows the usual lines. Since all three expressions in (7.4.26) are linear in T , it su"ces to prove (7.4.26) for polyads T . But for T a polyad, (7.4.26) follows immediately from associativity and (7.4.24). $ Thus, I U acts like a multiplicative identity. Then some second order tensors can have inverses. 7.4.23 If L 2 L(U ! V ) has an inverse L;1 2 L(V ! U ), then we dene Denition  $ $ ;1 L to be L;1 . Thus, using (7.4.20), we conclude $ $ $ $ $ $ L;1  L= I V  L  L;1 = I U : (7.4.27) Since the tensors in U  V and the linear mappings in L(U ! V ) can be thought of as essentially the same objects, any concept dened for one can be dened immediately for the other. Denition 7.4.23 is an example of this process. Other examples are below. Denition 7.4.24 If $L2 V  V then det $L:= det L. 7.4. APPLICATIONS OF THE GENERALIZED DOT PRODUCT 65 $ $ From denition 7.4.23 and 7.4.24 it is apparent that for any L2 V  V , L;1 exists i $ det L6= 0. $T $ $ $ Denition 7.4.25 If L2 U  V then L :=LT . Thus LT 2 V  U . Here LT is the of L : U ! V , as dened on page D-24. From the denition $transpose T $ $ of LT we have L (~v ~u) =LT (~v ~u) = LT (~v)  ~u = ~v  L(~u) = L(~u)  ~v =L (~u~v): Thus $ $ LT = (12) L : (7.4.28) $T $T $ $ It is also true that for any ( ~ u ~ v ) 2 U V , ~ v  L  ~ u = L ( ~ v  ~ u ) = ( ~ u ~ v ) = ~ u  L L ~v. $  $ Thus ~v  LT ~u ; ~u L = 0. This is true for that for any ~v 2 V so for all ~u 2 U $ $ LT ~u = ~u L : (7.4.29) $T $ ~v L =L ~v: (7.4.30)  $ $  Similarly, ~v LT ; L ~v  ~u = 0 for all ~u 2 U so if ~v 2 V Application 5: Linear mappings as higher order tensors. Suppose U1     Up V1    Vq are Euclidean vector spaces and L 2 L(U1     Up ! V1     Vq ). Both U1      Up and V1      Vq are Euclidean vectors spaces, with $ dot products hpi and hqi. Therefore, L corresponds to a second order tensor L in (U1      Up)  (V1      Vq ), exactly as in (7.4.16). That is if P 2 U1      Up and Q 2 V1      Vq then $ L (P Q) = L(P )hqiQ or $ P hqi L hqiQ = L(P )hqiQ: (7.4.31) Here hpi and hqi are to be interpreted as the dot products on U1    Up and V1    Vq , $ and L is a second order tensor in the space (U1      Up)  (V1     Vq ), which is the tensor product of two Euclidean spaces. 66 CHAPTER 7. GENERALIZED DOT PRODUCTS $ However, as we saw on page 61, L can also be interpreted as a tensor of order p + q in U1      Up  V1      Vq . Equation (7.4.31) continues to hold, but now hpi and hqi are interpreted as generalized dot products, and $L2 U1      Up  V1      Vq . Therefore, the linear mappings L 2 L(U1      Up ! V1      Vq ) can be viewed as $ $ tensors L2 U1    Up  V1    Vq . If we know L, we nd L from (7.4.31) by taking $ P and Q to be arbitrary polyads. If we know L2 U1      Up  V1      Vq , then we know it in (U1      Up)  (V1      Vq ) so we nd L by appealing to (7.4.19). That is, for any P 2 U1      Up, $ L(P ) = P hpi L : (7.4.32) The generalization from application 4 to application 5 is particularly useful in continuum mechanics. Chapter 8 How to Rotate Tensors (and why) 8.1 Tensor products of linear mappings For j 2 f1     qg, suppose Vj and Wj are Euclidean vector spaces and Lj 2 L(Vj ! Wj ). We would like to dene a linear mapping L1      Lq : V1      Vq ! W1      Wq by requiring that for any polyad ~v1    ~vq 2 V1      Vq we have (L1      Lq ) (~v1    ~vq ) = L1 (~v1)    Lq (~vq ) : (8.1.1) The lifting theorem will show that there is exactly one such L1      Lq . To nd it, we invoke that theorem. We dene M : V1    Vq ! W1      Wq by requiring for any (~v1    ~vq ) 2 V1    Vq that M (~v1     ~vq ) = L1 (~v1)    Lq (~vq ): Since L1     Lq are linear, M is multilinear, so we can use the lifting theorem with M as above and W = W1    Wq . The picture is shown in 8.1. Then we dene L1    Lq to be the M provided by the lifting theorem. With this choice, L1      Lq is linear and satises (8.1.1). 67 68 CHAPTER 8. HOW TO ROTATE TENSORS (AND WHY) V1 x . . . x Vq M x W1 x . . . x Wq x V1 x . . . x Vq M Figure 8.1:   (q) Remark 8.1.37 Suppose ~b(1j)      ~b(njj) is an ordered basis for Vj and Q = Qi iq~b(1) i    ~biq 2 V1      Vq . Then 1 h    i (q ) (L1      Lq ) (Q) = Qi1 iq L1 ~b(1) i1    Lq ~biq : 1 (8.1.2) Proof: h  (q) i L1   Lq is linear, so (L1   Lq )(Q) = Qi1 iq (L1      Lq ) ~b(1) i1    ~biq : Now use (8.1.1). Remark 8.1.38 Suppose for j 2 f1     qg that Uj , Vj , Wj are Euclidean and Kj 2 L(Vj ! Wj ) and Lj 2 L(Uj ! Vj ). Then (K1      Kq )  (L1      Lq ) = (K1  L1 )      (Kq  Lq ) : (8.1.3) Proof: We want to show for every P 2 U1      Uq that (K1      Kq )  (L1      Lq )] (P ) = (K1  L1 )      (Kq  Lq )] (P ): By linearity it su"ces to prove this when P is any polyad ~u1    ~uq . But (K1      Kq )  (L1      Lq )] (~u1    ~uq ) = (K1      Kq ) (L1      Lq )(~u1    ~uq )] = (K1      Kq ) L1(~u1)    Lq (~uq )] = K1 L1 (u1)] K2 L2 (u2)]    Kq Lq (uq )] (K1  L1 )(~u)] (K2  L2 )(~u2)]    (Kq  Lq )(~uq )] = (K1  L1 )      (Kq  Lq )] (~u1    ~uq ): 8.1. TENSOR PRODUCTS OF LINEAR MAPPINGS 69 Another way to try to understand L1      Lq is to exhibit explicitly for any P 2 V1      Vq the tensor (L1      Lq )(P ) 2 W1      Wq by evaluating it at any (w~ 1     w~ q ) 2 W1    Wq . We have Remark 8.1.39 Suppose that for j 2 f1     qg we have Lj 2 L(Vj ! Wj ) for Euclidean spaces Vj and Wj . Suppose P 2 V1      Vq and (w~ 1     w~ q ) 2 W1    Wq . Then h i (L1      Lq )(P )] (w~ 1     w~ q ) = P LT1 (w~ 1)     LTq (w~ q ) : (8.1.4) Proof: Both sides of this equation are linear in P , so it su"ces to prove the equation when P is a polyad, P = ~v1    ~vq . In that case, (L1      Lq )(P )] (w~ 1      w~ q ) = L1(~v1 )    Lq (~vq )] (w~ 1      w~ q ) = L1(~v1 )  w~ 1]    Lq (~vq )  w~ q ] h i h i = ~v1  LT1 (w~ 1)    ~vq  LTq (w~ q ) h i = P LT1 (w~ 1)     LTq (w~ q ) : Finally, it will be useful to understand how permutations aect tensor products of mappings. To do this we need Lemma 8.1.22 Suppose V1      Vq are Euclidean spaces, ~vj 2 Vj for j 2 f1     qg, and  2 Sq . Then  (~v1    ~vq ) = ~v;    ~v;q : (8.1.5) 1 (1) 1 ( ) Proof: From the denition (1.4.10),    (~v1    ~vq )] w~ ;1(1)      w~ ;1(q) = (~v1    ~vq ) (w~ 1     w~ q ) = (~v1  w~ 1)    (~vq  w~ q )        = ~v;1 (1)    ~v;1(q) w~ ;1(1)      w~ ;1(q) = ~v;1 (1)  w~ ;1(1)    ~v;1 (q)  w~ ;1(q) for any (w~ 1     w~ q ) 2 V1    Vq . QED. 70 CHAPTER 8. HOW TO ROTATE TENSORS (AND WHY) Now we can prove Remark 8.1.40 For j 2 f1     qg, suppose Lj 2 L(Vj ! Wj ), and suppose  2 Sq . Then    (L1      Lq ) = L;1(1)      L;1(q) : (8.1.6) Proof: The domain of L1      Lq is V1      Vq . We want to prove that if P 2 V1    Vq then (L1     Lq )](P ) = (L;1(1)     L;1 (q) )](P ), i.e., (L1      Lq )](P ) = (L;1 (1)      L;1 (q) )(P ). Both sides of the last equation are linear in P , so it su"ces to prove that equation when P is a polyad, say P = ~v1    ~vq . In this case  (L1      Lq ) (~v1    ~vq )] =  L1 (~v1 )    Lq (~vq )]     = L;1 (1) ~v;1 (1)    L;1 (q) ~v;1 (q) by 8.1.5      = L;1 (1)      L;1 (q)  (~v1    ~vq )] by 8.1.5 = L;1 (1)      L;1 (q) ~v;1(1)    ~v;1 (q) QED. Special Case: Suppose V1 =    = Vq = V and W1 =    = Wq = W and L1 =    = Lq = L. Then we write L1      Lq as q L. Then q L 2 L (q V ! q W )  (8.1.7) and q L is dened by requiring for all (~v1     ~vq ) 2 xq V that q L(~v1    ~vq ) = L(~v1 )    L(~vq ): (8.1.8) If P 2 q V , (8.1.8) leads to the gure of speech that (q L)(P ) is obtained by applying L to P . (Really q L is applied to P ). 8.2. APPLYING ROTATIONS AND REFLECTIONS TO TENSORS 71 If K 2 L(V ! W ) and L 2 L(U ! V ) then from (8.1.3) (q K )  (q L) = q (K  L) : (8.1.9) If L 2 L(V ! W ) and P 2 q V and (w~ 1     w~ q ) 2 xq W then from (8.1.4) h i (q L)(P )] (w~ 1     w~ q ) = P LT (w~ 1)     LT (w~ q ) : (8.1.10) If L 2 L(V ! W ) and  2 Sq then from (8.1.6) (q L) = (q L): (8.1.11) 8.2 Applying rotations and reections to tensors Recall that L 2 L(V ! V ) is \orthogonal" if LT = L;1 , and \proper orthogonal" if also det L = +1 rather than ;1. Recall the denitions on page D-25: #(V ) = set of all orthogonal operators on V #+(V ) = set of all proper orthogonal operators on V: Recall that according to corollary PP-12, every proper orthogonal operator is a rotation. As remarked on page PP-30, if L is improper orthogonal (det L = ;1) then L is the product of a rotation and a reection. If L is an orthogonal operator on V and Q 2 q V , then two more facts about (q L)(Q) make it even more reasonable to think of that tensor as the result of applying L to Q. These facts are remark 8.2.41 and its corollary. Remark 8.2.41 If L 2 #(V ), applying L to tensors over V preserves their generalized dot products. More precisely suppose p q r are any non-negative integers. Suppose P 2 p+q V , R 2 q+r V , and L 2 #(V ). Then h i h i (p+r L)(P hqiR) = (p+q L)(P ) hqi (q+r L)(R) : Proof: (8.2.1) 72 CHAPTER 8. HOW TO ROTATE TENSORS (AND WHY) Both sides of this equation are linear in P and R, so it su"ces to prove the equation when P and R are polyads, say P = ~u1    ~up~x1    ~xq and R = ~y1    ~yq~v1    ~vr . Then (p+r L)(P hqiR) = (p+r L) (~x1  ~y1)    (~xq  ~yq )~u1    ~up~v1    ~vr ] = (~x1  ~y1)    (~xq  ~yq )(p+r L)(~u1    ~up~v1    ~vr ) = (~x1  ~y1)    (~xq  ~xq ) L(~u1)    L(~up)L(~v1 )    L(v~r )]. So far we have not used L 2 #(V ). Now we use it to set ~x  ~y = L(~x)  L(~y). Therefore the above polyad is (p+r L)(P hqiR) = = L(~x1 )  L(~y1)]    L(~xq )  L(~yq )] L(~u1)    L(~up)L(~v1 )    L(~vr )] = L(~u1)    L(~up)L(~x1 )    L(~xq )] hqi L(~y1)    L(~yq )L(~v1 )    L(~vr )] h i h i = (p+q L)(~u1    ~up~x1    ~xq ) hqi (q+r L)(~y1    ~yq~v1    ~vr ) h i h i = (p+q L)(P ) hqi (q+r L)(R) : QED Corollary 8.2.19 The value of a rotated q-tensor at a rotated q-tuple of vectors is the value of the original tensor at the original q-tuple of vectors. The same is true for reections. in general, if L 2 #(V ) and Q 2 q V and (~v1    ~vq ) 2 xq V then (q L) (Q)] L (~v1)      L (~vq )] = Q (~v1     ~vq ) : Proof: (q L) (Q)] L (~v1 )      L (~vq )] = (q L) (Q)] hqi L (~v1)  L (~v2 )    L (~vq )] = (q L) (Q)] hqi (q L) (~v1~v2    ~vq )] = Qhqi (~v1    ~vq ) by remark 8.2.41 = Q (~v1     ~vq ) : 8.3. PHYSICAL APPLICATIONS 73 8.3 Physical applications Some physical properties of physical systems are tensors. For example, Ohm's law in a crystal says that if an electric eld E~ is applied, the resulting current density J~ depends $ $ linearly on E~ . This means that there is a second-order tensor K such that J~ = E~  K . $ ~ where K is a This K is the \conductivity tensor" of the crystal. In a liquid, J~ = EK $ $ scalar, so K = K I , but in a crystal all that can be said in general is that if there is $ $T $ no magnetic eld, K is symmetric (K =K ). This is Onsager's theorem for dissipative thermodynamics. A second example is an elastic crystal. As we will see later, its local state of deformation is described by a symmetric second order tensor $ , the strain tensor. Its local state $ of stress is described by another symmetric second order tensor S , the stress tensor. For $ small $ (i.e. k $ k << 10;2) S is a linear function of $ , so there is a tensor of fourth $ order over ordinary 3-space, the elasticity tensor E , such that S =$ h2iE . Thermodynamics assures that (12)E = (34)E = (13)(24)E = E , but E has no other symmetries in general so the dimenson of the space of all possible elasticity tensors is 21. (We will discuss all this in detail later.) $ A third example is the second moment tensor M of a collection of mass points. If there is a mass mv at position ~rv for v = 1     N , then N $ X (8.3.1) M := mv~rv~rv : v=1 This is the underlying tensor from which the inertia tensor and gravitational second degree harmonic tensor of a solid body are calculated. If V is real 3-space and L 2 L(V ! V ), then \subjecting a physical system to the mapping L" means that we move all the mass points (atoms) which make up the system, $ so that a mass point which was originally at ~rv is moved to the new location L(~rv ) = ~rv  L. If L 2 #+(V ), L subjects the body to a rigid rotation. If L 2 #(V ) but det L = ;1, L subjects the body to a rigid rotation and also a reection in some plane. If L 2 #(V ) but det L > 0 then L is the product of a rigid rotation and a symmetric linear operator. The latter stretches the body by dierent amounts in dierent directions. If det L < 0, a 74 CHAPTER 8. HOW TO ROTATE TENSORS (AND WHY) H H F Cl F Br Br Cl Left Handed Right Handed Figure 8.2: reection also occurs. It is actually possible to imagine subjecting a real body to a reection if we allow ourselves to disassemble and reassemble the body atom-by-atom. For example, iceland spar (calcite) crystals come in right-handed or left-handed version. Liquids can also be reected. For most, reection has no eect because the distribution of molecules in the liquid has no preferred directions and the molecules either have no right- or left-handedness, or the right- and left-handed versions are present in equal amounts. A liquid which is aected by reection is bromochlorouoromethane, C H F Cl Br. At one atmosphere of pressure, it is liquid between ;115C and 36C. the molecule is a tetrahedron with H, F, Cl, Br at the four vertices and C at the center. It can come in two forms, left handed or right handed. To reect a pot of left-handed C H F Cl Br means to replace it by a pot of right-handed C H F Cl Br. If we subject a system of mass points mv at positions ~rv to the mapping L 2 L(V ! $ V ), where V is real 3-space, then the second moment tensor M of that system (see (8.3.1)) $ changes to 2 L(M ). To see this, note that the second moment tensor of the deformed 8.3. PHYSICAL APPLICATIONS 75 system is N N X X $ = m L ( ~ r ) L ( ~ r ) = mv 2 L (~rv~rv ) ML v v v v=1 ! v=1   $  N  2  X = L mv~rv~rv = 2 L M : v=1 $ If we subject a homogeneously stressed crystal to the mapping L, the stress tensor S in $ that crystal will not change to 2 L(S ). Rather it will be determined by the elasticity tensor E of the crystal. These examples show that it is not always obvious how to calculate the eect of subjecting a physical system to a linear mapping L. Some tensor properties of that system simply have L applied to them, and others do not. The situation is much simpler if L 2 #(V ). To see this, suppose ~b1 , ~b2 , ~b3 is any basis for real 3-space V . Suppose Q 2 q V is a measurable tensor property of a particular physical system. Then (8.3.2) Q = Qi1 iq~bi1~bi2   ~biq : \Measuring" Q means measuring the real numbers Qi1 iq . Suppose we apply to the physical system a mapping L 2 #(V ). The new physical system will have instead of Q the tensor QL to describe that particular physical property. Suppose we apply L to the apparatus we used to measure Qi1 iq , and we use this mapped apparatus to measure the contravariant components of QL relative to the basis L(~b1 ), L(~b2 ), L(~b3 ). We assume either that no gravity or electromagnetic eld is present or, if they are, that they are also mapped by L. Then the experiment on QL is identical to that on Q, except that it has been rotated and possibly reected relative to the universe. As far as we know, the univese does not care the laws of nature are invariant under reection and rigid rotation. The universe seems to be isotropic and without handedness. (As a matter of fact, a small preference for one orientation or handedness exists in the weak nuclear force governing  -decay Yang, Lee and Wu won Nobel prizes for this discovery it has no measurable eects at the macroscopic level of continuum mechanics in an old universe.) Therefore, the numbers we read on our dials (or digital voltmeters) will be the same in the original and mapped systems. That is, the contravariant components of QL relative to L(~b1 ), 76 CHAPTER 8. HOW TO ROTATE TENSORS (AND WHY) L(~b2 ), L(~b3 ) will be the same as the contravariant components of Q relative to ~b1 , ~b2, ~b3 . Thus QL = Qi1 iq L(~bi1 ) L(~bi2 )    L(~biq ) = Qi1 iq (q L)(~bi1~bi2   ~biq ) h i = (q L) Qi1 iq~bi1~bi2   ~biq : Therefore, if L 2 #(V ) and the physical system is mapped by L, the new tensor property QL is obtained by applying L to the original Q QL = (q L)(Q): (8.3.3) A tensor property of a physical system rotates with that system, and is reected if the system if reected. 8.4 Invariance groups Sometimes, subjecting a physical system to certain rotations or reections will not aect any of its measurable physical properties. For example, suppose I work all morning on a cubic crystal of Na Cl. While I am at lunch, you rotate it in such a way as simply to permute its three crystal axes. When I return from lunch, I will be unable to detect that anything has happened to the crystal. The same is true if you manange to work so fast as to disassemble and reassemble the crystal so as to subject it to a reection along one of its crystal axes while I am at lunch. If I was working not on Na Cl but on a pot of left-handed CH F CL Br before lunch, you can subject it to any rotation whatever. I will not be able to detect the rotation when I return. However, if you reect it, I can detect that now I have a pot of right-handed CH F Cl Br. For example, it will rotate the plane of polarized light oppositely before and after lunch. Let S be any physical system in real 3-space V . For any L 2 #(V ), let LS be the physical system obtained by subjecting S to L. Let G (S ) be the set of all members of #(V ) which, when applied to S , change none of the measurable properties of S . That is, L 2 (V ) is a member of G (S ) if no measurement can distinguish between S and LS . 8.4. INVARIANCE GROUPS 77 These denitions have some simple but useful implications. Obviously I 2 G (S ) (8.4.1) where I is the identity operator on V . And if L1, L2 2 G (S ), then L2 S is indistinguishable from S , so L1 (L2 S ) is indistinguishable from S . Therefore Finally, we claim if L1  L2 2 G (S ) then L1  L2 2 G (S ): (8.4.2) if L 2 G (S ) then L;1 2 G (S ): (8.4.3) For LS and S are indistinguishable. Hence so are L;1(LS ) and L;1 S . But L;1 (LS ) is S , so S and L;1S are indistinguishable. Properties (8.4.1), (8.4.2), (8.4.3) show that G (S ), is a subgroup of #(V ). It is called the invariance group of the physical system S . Examples of invariance groups are these: Example 8.4.5 If S is water, G (S ) = #(V ). Example 8.4.6 If S is liquid, left-handed C H F Cl Br , G (S ) = #+(V ). Example 8.4.7 If S is polycrystalline calcite, with the microcrystals randomly oriented, and with equal numbers of right and left-handed microcrystals, and if we make only measurements involving lengths > > microcrystal diameter (e.g. study only seismic waves very long compared to the size of the microcrystals), G (S ) = #(V ). Example 8.4.8 If in example 8.4.7, 30% of the microcrystals are left handed and 70% are right handed, G (S ) = #+(V ). Example 8.4.9 If S is an Na Cl crystal, G (S ) is the \cubic group". It is the group of orthogonal operators which send into itself a cube centered on ~0. If x^1 x^2  x^3 are unit vectors parallel to the edges of the cube, G (S ) consists of those L 2 #(V ) such that L(^xi ) = (i) x^(i) for i = 1 2 3, where  2 S3 and 1  2 3 can be +1 or ;1 independently. 78 CHAPTER 8. HOW TO ROTATE TENSORS (AND WHY) If Q 2 q V describes some measurable property of physical system S , then clearly (q L)(Q) = Q for all L 2 G (S ). This fact limits the possible forms which Q can take, and is therefore of physical interest. It leads us to introduce Denition 8.4.26 Let V be a Euclidean vector space. Let G be a subgroup of #(V ). Then I q (G ) is the set of all Q 2 q V such that (q L) (Q) = Q for all L 2 G : (8.4.4) The set I q (G ) is called the space of q'th order tensors invariant under G . The tensors in I q (#(V )) are unchanged by any orthogonal operator. They are called \isotropic tensors". The tensors in I q (#+(V )) are unchanged by any rotation, but may be changed by reections. They are called \skew isotropic tensors." If G is the cubic group, the tensors in I q (G ) are unchanged by any rotation or reection which is physically undetectable in an Na Cl crystal. The following are simple consequences of denition 8.4.26. Corollary 8.4.20 I q (G ) is a subspace of q V . Proof: Clearly I q (G ) q V , so all we need prove is that any linear combination of members of I q (G ) is a member. If Q1     QN 2 I q (G ) and a1     aN 2 R and L 2 G , then (q L)(Qi ) = Qi, so (q L)(ai Qi ) = ai(q L)(Qi ) = ai Qi because q L is linear. Corollary 8.4.21 If G1 G2 then I q (G2 ) I q (G1 ). Proof: If Q 2 I q (G2), Q is unchanged by any L 2 G2 , and then certainly by any L 2 G1 . Hence Q 2 I q (G1). (The bigger the group G , the harder it is to be unchanged by all its members.) 8.4. INVARIANCE GROUPS 79 Corollary 8.4.22 If Q 2 I q (G ), and  2 Sq then Q 2 I q (G ). Proof: For any L 2 L(V ! V ), equation 8.1.11 gives (q L)   =   (q L). Then (q L)(Q) =  (q L)(Q)]. If Q 2 I q (G ) then for any L 2 G , (q L)(Q) = Q, so (q L)(Q) = Q. Hence Q 2 I q (G ). Corollary 8.4.23 If P 2 I p+q (G ) and R 2 I q+r (G ) then P hqiR 2 I p+r (G ). Proof: For any L 2 G , L 2 #(V ), so (p+r L)(P hqiR) = (p+q L)(P )] hqi (q+r L)(Q)] = P hqiR . Corollary 8.4.24 If Q 2 I q (G ) then the component array of Q is the same relative to any two ordered orthonormal bases (^x1      x^n ) and (^x01      x^0n) such that x^0i L(^xi ) for some L 2 G . Proof: For any Q 2 q V , Q = Qi1 iq x^i1    x^iq = Q0i1 iq x^0i1    x^0iq : For any L 2 L(V ! V ), (8.4.5)   (q L) (Q) = Qi1 iq L (^xi1 )    L x^iq : (8.4.6) If Q 2 I q (G ) and x^0i = L(^xi ) for some L 2 G , then (q L)(Q) = Q and comparing (8.4.5) with (8.4.6) gives Qi1 iq = Q0i i . 1 q Corollary 8.4.25 Suppose Q 2 q V . Suppose V has one ordered orthonormal basis (^x1      x^n) such that whenever L 2 G and x^0i = L(^xi ), then Q has the same component array relative to (^x1      x^n) and (^x01      x^0n). Then Q 2 I q (G ). Proof: Suppose L 2 G . Let x^0i = L(^xi ). By hypothesis Q0i1 iq = Qi1 iq . Comparing (8.4.5), and (8.4.6) gives (q L)(Q) = Q. Since this is true for every L 2 G , Q 2 I q (G ). 80 CHAPTER 8. HOW TO ROTATE TENSORS (AND WHY) 8.5 Isotropic and skew isotropic tensors Denition 8.5.27 Let V be a Euclidean vector space. The tensors in I q (#(V )) are called \isotropic". The tensors in I q (#+(V )) are called \skew isotropic". Any tensor describing a measurable property of water, or of polycrystalline calcite with randomly oriented microcrystals and equal numbers of left- and right-handed microcrystals, is isotropic. Any tensor describing a property of left-handed liquid C H F Cl Br , or of polycrystalline calcite with randomly oriented microcrystals and more of one handedness than the other, is skew isotropic. Since #+(V ) #(V ), corollary (8.4.21) implies I q (#(V )) I q (#+(V )): (8.5.1) That is, every isotropic tensor is skew isotropic. By corollaries 8.4.24 and 8.4.25, a tensor is isotropic i it has the same array of components relative to every orthonormal basis for V . Suppose (V A) is an oriented Euclidean space. By corollary 8.4.25, a tensor is skew isotropic if it has the same array of components relative to every positively oriented orthonormal basis for V . Also, it is skew isotropic if it has the same array of components relative to every negatively oriented orthonormal basis for V . If a tensor is skew isotropic, by corollary 8.4.25 it has the same array of components relative to every positively oriented orthonormal basis, and the same array relative to every negatively oriented orthonormal basis. The two arrays need not be the same (indeed, if they are, the tensor is isotropic). Example 8.5.10 The identity tensor $I is isotropic. For, relative to any orthonormal $ basis (^x1      x^n) for V , Iij = I (^xi x^j ) = x^i  x^j = ij . Example 8.5.11 If A is a unimodular alternating tensor, A is skew isotropic but not isotropic. For as we saw on page 39, A has the same component array relative to every positively oriented orthonormal basis, but its component array changes sign for negatively oriented bases. 8.5. ISOTROPIC AND SKEW ISOTROPIC TENSORS 81 Example 8.5.12 By corollary 8.4.23, $I $I is isotropic. So is $I $I $I , $I $I $I $I , etc. By corollary 8.4.22 , every permutation of these tensors is isotropic. By corollary 8.4.20, every linear combination of such permuted tensors is isotropic. Example 8.5.13 All the tensors in example 8.5.12 are skew isotropic. By corollary $ $$ $$$ 8.4.23, so are I A, I I A, I I I A, etc. By corollary 8.4.22, so are all permutations of these tensors. By corollary 8.4.20, so is every linear combination of such permuted tensors. We gain nothing new by using more than one factor A. The polar identity (10.2) shows that $$ $ if n = dim V then AA is a linear combination of n! permutations of I I    I (with n $ factors I ). On p. 64 of his book \The Classical Groups" (Princeton, 1946), Hermann Weyl shows that there are no isotropic tensors except those listed in example 8.5.12, and no skew isotropic tensors except those listed in example 8.5.13. Note the corollary that all nonzero isotropic tensors are of even order. We cannot consider Weyl's general proof, but we will discuss in detail the tensors of orders q = 0 1 2 3 4, which are of particular interest in continuum mechanics. We assume dim V  2. q = 0 Tensors of order 0 are scalars. They are unaected by any orthogonal transformation, so are isotropic and therefore also skew isotropic. q = 1 If dim V  2, the only skew isotropic tensor of order 1 is ~0. Therefore the only isotropic tensor of order 1 is ~0. Proof: Tensors of order 1 are vectors. We are claiming that if dim V  2 then the only vector which is unaected by all rotations is the zero vector. A formal proof of this intuitively obvious fact is as follows. Suppose ~v 2 I 1 (#+(V )). Let x^ and y^ be any two unit vectors. Each is the rst vector of a positively oriented ordered orthonormal basis (pooob), so by corollary 8.4.24, ~v(^x) = ~v(^y). That is ~v  x^ = ~v  y^ for any two unit vectors 82 CHAPTER 8. HOW TO ROTATE TENSORS (AND WHY) x^ and y^. If dim V  2, we can always nd a y^ such that ~v  y^ = 0. Then ~v  x^ = 0 for all x^, so ~v = ~0. q=2 I 2 (#(V )) = spf$I g I 2 (#+(V )) = spf$I g if dim V  3 I 2 (#+(V )) = spf$I  Ag if dim V = 2 and A is any nonzero member of 2V: (8.5.2) (8.5.3) (8.5.4) Proof: The containments  have all been established in examples 8.5.12 and 8.5.13, so we need prove only in (8.5.2, 8.5.3, 8.5.4). In all these cases, suppose T 2 I 2 (#+(V )). If x^ and y^ are any unit vectors, each is the rst vector of an ordered orthonormal basis (oob), and by changing the sign of one later basis vector, if necessary, we can assume that these oob's have the same orientation. Therefore T (^x x^) = T (^y y^) because T has the same component array relative to all oobs with the same orientation. Thus there is a scalar aT such that T (^x x^) = aT for every unit vector x^: (8.5.5) If fx^ y^g is orthonormal, (^x y^) is the rst pair of vectors in an oob. Replacing (^x y^) by (^y ;x^) in that oob gives a second oob with the same orientation. Hence, T (^x y^) = T (^y ;x^) = ;T (^y x^). T (^x y^) = ;T (^y x^) if fx^ y^g is orthonormal : (8.5.6) When dim V = 2, choose an oob (^x1  x^2 ) and let a = aT , b = T (^x1 x^2 ) $ =A(^x1  x^2). Then relative to this oob, Tij = aij + bA(^x1  x^2 )"ij = (a I $ +bA)ij , so T = a I +bA, and (8.5.4) is proved. To treat (8.5.2, 8.5.3), let 8.5. ISOTROPIC AND SKEW ISOTROPIC TENSORS 83 (^x y^) be orthonormal. Both (^x y^) and (^y x^) are the rst pair of vectors in an oob, so if T 2 I 2 (#(V )), T (^x y^) = T (^y x^) if fx^ y^g is orthonormal: (8.5.7) If dim V  3, both (^x y^) and (^y x^) are rst pairs in oobs with the same orientation, namely (^x y^ x^3     x^n) and (^y x^ ;x^3  x^4     x^n). If T 2 I (#+ (V )), we again have 8.5.7. In either case we have (8.5.6) and (8.5.7) so T (^x y^) = 0 if fx^ y^g is orthonormal: (8.5.8) Now let (^x1      x^n ) be an oob. By (8.5.5) and (8.5.8), relative to this $ oob we have Tij = aT ij , so T = aT I . QED q=3 I 3 (#(V )) = f0g I 3 (# + (V )) = f0g if dim V 6= 3 I 3 (#+(V )) = spfAg if dim V = 3 and A 6= 0 A 2 3V: (8.5.9) (8.5.10) (8.5.11) Proof: If dim V is even, let (^x1      x^n) be an oob for V . Then (;x^1      ;x^n) is another oob with the same orientation. Then if T 2 I 3 (#+(V )), T (^xi  x^j  x^k ) = T (;x^i  ;x^j  ;x^k ) = ;T (^xi x^j  x^k ) = 0 so Tijk = 0 . If dim V is odd and dim V  5, let (^x y^ z^) be any ordered orthonormal sequence. It is the rst triple of an oob (^x y^ z^ x^4     x^n) and (;x^ y^ z^ ;x^4  x^5    x^n) is an oob with the same orientation, so T (^x y^ z^) = T (;x^ y^ z^) = ;T (^x y^ z^) = 0. Similarly, T (^x y^  y^ ) = ;T (^x y^  y^ ) = 0 and T (^x x^  x^ ) == ;T (^x x^  x^ ) = 0. Thus if x^1  x^2  x^3 are orthonormal, T (x^i x^j  x^k ) = 0 for fi j kg f1 2 3 g. This comment applies to any three vectors from an arbitrary oob for V , so Tijk = 0 84 CHAPTER 8. HOW TO ROTATE TENSORS (AND WHY) for fi j kg f1     ng relative to any oob, and T = 0. Finally, suppose dim V = 3. If fx^ y^g is orthonormal, (^x y^) is the rst pair of an oob (^x y^ z^), and the oob (;x^ y^ ;z^) has the same orientation. Thus T (^x y^ y^) = T (;x^ y^ y^) = ;T (^x y^ y^) = 0. Similarly T (^y x^ y^) = 0 and T (^y y^ x^) = 0. Also T (^x x^ x^) = T (;x^ ;x^ ;x^) = ;T (^x x^ x^) = 0. Thus if T 2 I 3 (#+(V )) T (^x x^ x^) = 0 for every unit vector x^ (8.5.12) T (^x y^ y^) = T (^y x^ y^) = T (^y y^ x^) if fx^ y^g is orthonormal: (8.5.13) Finally, if (^x y^ z^) is an oob, then the following are oobs with the same orientation: (^y z^ x^) , (^z x^ y^) , (^y x^ ;z^) , (^x ;z^ y^) , (;z^ y^ x^) . Then if T 2 I 3 (#+(V )) T (^x y^ z^) = T (^y z^ x^) = T (^z x^ y^) = ;T (^y x^ z^) = ;T (^x z^ y^) = ;T (^z  y^ x^) (8.5.14) if fx^ y^ z^g is orthonormal. Let (^x1  x^2  x^3) be an oob for V . Let c = T (^x1  x^2  x^3)=A(^x1 x^2  x^3 ). Then (8.5.12 , 8.5.13 , 8.5.14) imply Tijk = cA(^x1  x^2  x^3 )"ijk = cAijk so T = cA. QED. Note that in all the foregoing arguments, only mutually ? vectors were used. These arguments remain valid if G is the cubic group and T 2 I q (G ) with q = 1 2 3 and we work only with unit vectors and oobs in the crystal axis directions. Therefore the conclusions apply to tensors of orders 1 2 3 known only to be invariant under $ the cubic group of Na Cl. For example, it follows that the conductivity tensor K $ $ of Na Cl must have the form K I . In Na Cl, Ohm's law J~ = E~  K reduces to the ~ , even though an Na Cl crystal is not isotropic. isotropic form J~ = EK q=4 I 4 (#(V )) $$ $$ $$ = sp I I  (23) I I  (24) I I (8.5.15) 8.5. ISOTROPIC AND SKEW ISOTROPIC TENSORS 85 I 4 (#+(V )) = 00if dim V 62 f2 4g $$ $$ $$  4 + I (# (V )) = sp I I  (23) I I  (24) I I  A (8.5.16) if dim V = 4 (8.5.17) I 4 (#+(V )) = spf$I $I  (23) $I $I  (24) $I $I  $ $ $ I A (23) I A (24) I A $ $ $ AI  (23)AI  (24)AI g if dim V = 2: (8.5.18) Here A is any nonzero alternating tensor, A 2 nV with n = dim V . Proof: The containments  have all been established in examples 8.5.12 and 8.5.13, so we need prove only . We consider only (8.5.15, 8.5.17), and leave (8.5.17, 8.5.18) to the reader with Sitzeisch. Therefore we assume T 2 I 4 (#+(V )), and if dim V 2 f2 4g we assume T 2 I 4 (#(V )). If dim V  5, then any four orthonormal vectors (w ^ x^ y^ z^) are the rst four vectors of and oob for V , say (w ^ x^ y^ z^ x^5     x^n). The vectors (;w ^ x^ y^ z^) are the rst four vectors of an oob with the same orientation, namely (;w ^ x^ y^ z^ ;x^5  x^6      x^n). Since T 2 I 4 (#+(V )), T has the same component array relative to both oob's, so T (w ^ x^ y^ z^) = T (;w ^ x^ y^ z^) = ;T (w ^ x^ y^ z^). Thus for dim V  5, T (w ^ x^ y^ z^) = 0 if fw ^ x^ y^ z^)g is orthonormal: (8.5.19) This result is also vacuously true for dim V = 3, because then we cannot nd four orthonormal vectors in V . Next, suppose dim V  3 and fx^ y^ z^g is orthonormal. Then fx^ y^ z^g and (;x^ y^ ;z^) are both the rst triple in oobs with the same orientation, so T (^x y^ z^ z^) = T (;x^ y^ ;z^ ;z^) = ;T (^x y^ z^ z^) = 0. Similarly, if fx^ y^ z^g is orthonormal T (^x y^ z^ z^) = T (^x z^ y^ z^) = T (^x z^ z^ y^) = T (^z x^ y^ z^) = T (^z  x^ z^ y^) = T (^z  z^ x^ y^) = 0: (8.5.20) 86 CHAPTER 8. HOW TO ROTATE TENSORS (AND WHY) Equations (8.5.20) are vacuously true if dim V = 2 because then we cannot nd three orthonormal vectors in V . Next, suppose fx^ y^g is orthonormal. Then (^x y^) and (;x^ y^) are the rst pairs in oobs, and if dim V  3 we may assume these oobs to have the same orientation, say (^x y^ x^3     x^n) and (;x^ y^ ;x^3  x^4    x^n). Then T (^x y^ y^ y^) = T (;x^ y^ y^ y^) = ;T (^x y^ y^ y^) = 0. Similarly T (^x y^ y^ y^) = T (^y x^ y^ y^) = T (^y y^ x^ y^) = T (^y y^ y^ x^) = 0 if fx^ y^g orthonormal:(8.5.21) Next, suppose that fx^ y^g and fx^0  y^0g are both orthonormal. Each pair (^x y^) and fx^0  y^0g is the rst pair in an oob, so if T 2 I 4 (#(V )) then T (^x x^ y^ y^) = T (^x0 x^0  y^0 y^0). If dim V  3, then (^x y^) and (^x0  y^0) are rst pairs in oobs with the same orientation, say (^x y^ x^3     x^n) and (^x0  y^0 x^3  x^4      x^n): Then in that case also, T (^x x^ y^ y^) = T (^x0 x^0  y^0 y^0). In either case, there is a scalar T depending only on T such that T (^x x^ y^ y^) = T if fx^ y^g is orthonormal: (8.5.22) Similarly, there are scalars uT and vT such that T (^x y^ x^ y^) = uT if fx^ y^g is orthonormal (8.5.23) T (^x y^ y^ x^) = vT if fx^ y^g is orthonormal: (8.5.24) Finally, if x^ and y^ are unit vectors, each is the rst member of an oob, and the two oobs can have the same orientation, so T (^x x^ x^ x^) = T (^y y^ y^ y^). Therefore, there is a scalar T depending only on T such that T (^x x^ x^ x^) = T for every unit vector x^: (8.5.25) $$ Then tensor I I satises (8.5.19)-(8.5.25) with  = 1,  = 1, u = 0, v = 0. $$ The tensor (23) I I satises (8.5.19)-(8.5.25) with  = 1 ,  = 0 , u = 1, 8.5. ISOTROPIC AND SKEW ISOTROPIC TENSORS $$ v = 0. The tensor (24) I I satises (8.5.19)-(8.5.25) with  = 1 ,  = 0 , u = 0, v = 1. Therefore, the tensor $$ $$ $$ S = T ; T I I ;uT (23) I I ;vT (24) I I (8.5.26) satises (8.5.19)-(8.5.25) with S = T ; T ; uT ; vT and S = uS = vS = 0. Now let (^x y^) be orthonormal in V and let (^x y^ x^3     x^n) be an oob for V . Then (^x0  y^0 x^3     x^n) is an oob with the same orientation if we take x^0 = x^ cos + y^ sin  y^0 = ;x^ sin + y^ cos : To see this, let A 2 nV , A 6= 0. Then A(^x0  y^0 x^3    x^n) = A(^x cos + y^ sin  ;x^ sin + y^ cos  x^3      x^n) = ; cos sin A(^x x^ x^3     x^n) + cos2 A(^x y^ x^3      x^n ) ; sin2 A(^y x^ x^3     x^n) + sin cos A(^y y^ x^3      x^n)   = cos2 + sin2 A(^x y^ x^3     x^n) = A(^x y^ x^3      x^n ): But if (^x y^ x^3      x^n ) and (^x0  y^0 x^3      x^n) have the same orientation and S 2 I 4 (#+(V )), the S (^x x^ x^ x^) = S (^x0 x^0  x^0 x^0 ). The S of (8.5.26) is a linear combination of members of I 4 (#+(V )), so it is in I 4 (#+(V )). In addition, S satises (8.5.19)-(8.5.25) with S = uS = vS = 0. Therefore S (^x x^ x^ x^) = S (^x0  x^0 x^0 x^0 ) = S (^x cos + y^ sin  x^ cos + y^ sin  x^ cos + y^ sin  x^ cos + y^ sin ) = cos4 S (^x x^ x^ x^) + sin4 S (^y y^ y^ y^) (8.5.27) since all other 14 terms in the multilinear expansion of S vanish. Thus S = (cos4 + sin4 )S . It is possible to choose so that cos4 + sin4 6= 1. Therefore S = 0. Therefore S = 0. Therefore $$ $$ $$ T = T I I +uT (23) I I +vT (24) I I : (8.5.28) 87 88 CHAPTER 8. HOW TO ROTATE TENSORS (AND WHY) This completes the proofs of (8.5.15) and (8.5.17). Notice that the proof of (8.5.26) works equally well if all the unit vectors are parallel to the crystal axes of an Na Cl crystal. Only (8.5.27) fails. But if x^ y^ z^ are crystal axes, then S = S (^xx^x^x^ + y^y^y^y^ + z^z^z^z^). Therefore, we have proved that if T is invariant under the cubic group we can write T = S (^xx^x^x^ + y^y^y^y^ + z^z^z^z^) $$ $$ $$ + T I I +uT (23) I I +vT (24) I I  where S = T ; T ; uT ; vT . (8.5.29) Chapter 9 Dierential Calculus of Tensors 9.1 Limits in Euclidean vector spaces Denition 9.1.28 Let U be a Euclidean vector space. Let ~u 2 U . Let t 2 R, t > 0. We dene three sets: i) B (~u t) := f~u0 : ~u0 2 U and k~u0 ; ~uk < tg ii) @B (~u t) := f~u0 : ~u0 2 U and k~u0 ; ~uk = tg iii) B( (~u t) := f~u0 : ~u0 2 U and k~u0 ; ~uk  tg . B (~u t) , @B (~u t) and B( (~u t) are called, respectively, the open ball, the sphere, and the closed ball centered at ~u with radius t. See gure 9.1 Denition 9.1.29 A subset D of Euclidean space U is \open" if every point in D is the center of a ball lying entirely in D. That is, if ~u 2 D then 9" > 0 3 B (~u ") D. See gure 9.2. Denition 9.1.30 Suppose U and V are Euclidean vector spaces, D U , f~ : D ! V , ~u 2 D and ~v 2 V . We say that lim~u!~u f~(~u) = ~v , or, equivalently f~(~u) ! ~v, as ~u ! ~u if for every " > 0 there is a (") > 0 such that whenever ~u 2 D and 0 < k~u ; ~uk < (") then kf (~u) ; ~v0 k < ": 89 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS t 90 u B(u, t) ∂ B(u, t) Figure 9.1: 9.1. LIMITS IN EUCLIDEAN VECTOR SPACES 91 . . D . Figure 9.2: 92 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS Denition 9.1.31 If f~(~u) = ~v in denition 9.1.30 then f~ is \continuous at ~u". Denition 9.1.32 If x and y are real numbers, denote the algebraically smaller of them by x ^ y (read \x meet y") and the algebraically larger by x _ y (read \x join y"). Thus 1 ^ 6 = 1, 1 _ 6 = 6, (;1) ^ 0 = ;1, (;1) _ 0 = 0. Corollary 9.1.26 Suppose U and V are Euclidean vector spaces, D U and f : D ! V . Suppose ~u0 2 D, ~v0 2 V and f~(~u) ! ~v0 as ~u ! ~u0. Then if ~u 2 D and k~u ; ~u0k < (1), we have kf (~u)k < 1 + k~v0 k: Proof: If ~u 2 D and k~u ; ~u0 k < (1), then kf (~u) ; ~v0 k < 1 so kf~(~u)k = kf~(~u) ; ~v0 + ~v0k  kf (~u) ; ~v0k + k~v0 k < 1 + k~v0k. Corollary 9.1.27 Suppose U and V are Euclidean vector spaces, D U , f~ : D ! V , f~ 0 : D ! V , ~u0 2 D, ~v0 and ~v00 2 V . Suppose that f~(~u) ! ~v0 and f~ 0(~u) ! ~v00 as ~u ! ~u0. Then (f~ ; f~ 0)(~u) ! ~v0 ; ~v00 as ~u ! ~u0. Proof: For any given " > 0 we must nd a 00(") > 0 such that ~u 2 D and k~u ; ~u0 k < 00(") implies k(f + f 0)(~u) ; ~v0 ; ~v00 k < ". By hypothesis, there are ("=2) > 0 and 0("=2) > 0 such that if ~u 2 D then kf~(~u) ; ~v0 k < "=2 if k~u ; ~u0k < ("=2) kf~0(~u) ; ~v00 k < "=2 if k~u ; ~u0k <  0 ("=2): (9.1.1) (9.1.2) Let 00(") = ("=2) ^ 0("=2). If k~u ; ~u0k < 00("), then both (9.1.1) and (9.1.2) are true, so k(f~ + f~ 0)(~u) ; ~v0 ; ~v00 k = kf~(~u) ; ~v0 + f~0 (~u) ; ~v00 k  QED. kf~(~u) ; ~v0k + kf~0(~u) ; ~v00 k  "=2 + "=2 = ". 9.1. LIMITS IN EUCLIDEAN VECTOR SPACES 93 Corollary 9.1.28 Suppose U , V , W , X are Euclidean vector spaces and D U and R : D ! V  W and T : D ! W  X . Dene R  T : D ! V  X by requiring (R  T )(~u) = R(~u)  T (~u) for each ~u 2 D. Suppose ~u0 2 D and as ~u ! ~u0 we have R(~u) ! R0 and T (~u) ! T0. Then (R  T )(~u) ! R0  T0 . Proof: Choose " > 0. We must nd (") > 0 such that if ~u 2 D and k~u ; ~u0k < (") then k(R  T )(~u) ; R0  T0k < ". We note (R  T )(~u) ; R0  T0 = R(~u)  T (~u) ; R0  T0 = R(~u)  T (~u) ; R(~u)  T0 + R(~u)  T0 ; R0  T0 = R(~u)  T (~u) ; T0] + R(~u) ; R0]  T0: Thus k(R  T )(~u) ; R0  T0k  kR(~u)  T (~u) ; T0 ] k + k R(~u) ; R0 ]  T0 k: The generalized Schwarz inequality, remark 7.3.36, gives kR(~u)  T (~u) ; T0] k  kR(~u)kkT (~u) ; T0 k k(R(~u) ; R0)  T0 k  kR(~u) ; R0 kkT0k Thus if ~u 2 D k(R  T )(~u) ; R0  T0k  kR(~u)kkT (~u) ; T0k + kR(~u) ; R0 kkT0k: (9.1.3) Now if ~u 2 D and 8 > if k~u ; ~u0k < R (1) then kR(~u)k < 1 + kR0 k (see corollary 9.1.26) > > > <  "  " if k ~ u ; ~ u k <  0 T 2(1+kR0 k) then kT (~u) ; T0 k < 2(1+kR0 k) > > > > : if k~u ; ~u0k < R  "  then kR(~u) ; R0 k < " : 2(1+kT0 k) 2(1+kT0 k) (9.1.4) 94   CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS   Take (") = R (1) ^ T 2(1+k"R0 k) ^ R 2(1+"kT0 k) . Then if k~u ; ~u0k < ("), we have all three of (9.1.4), so by (9.1.3), k(R  T )(~u) ; R0  T0k < (1 + kR0 k) 2(1 +"kR k) + 2(1k+T0kkT" k) < ": 0 0 QED. Corollary 9.1.29 Suppose U and V are Euclidean vector spaces, and D U , and ~u0 2 D and ~v0 2 V . Suppose f~ : D ! V is dened by f~(~u) = ~v0 for all ~u 2 D. Then f~(~u) ! ~v0 as ~u ! ~u0. Proof: Obvious from denition 9.1.30. Corollary 9.1.30 Suppose U and V are Euclidean vector spaces, and D U , and ~u0 2 D and ~v0 2 V . Suppose f~ : D ! V . i) If f~(~u) ! ~v0 as ~u ! ~u0, then ~bi  f~(~u) ! ~bi  ~v0 as ~u ! ~u0, for any basis (~b1      ~bn) of V , with dual basis (~b1      ~bn). ii) If there is one basis for V , (~b1      ~bn), such that ~bi  f~(~u) ! ~bi  ~v0 as ~u ! ~u0, then f~(~u) ! ~v0 as ~u ! ~u0. Proof: Regard ~bi as a function whose domain is D and whose value is ~bi for each ~u 2 D. Then i) follows from corollaries 9.1.28 and 9.1.29. And ii) follows from corollaries 9.1.29 , 9.1.28, and 9.1.27. Notice that the ranges of our functions can lie in tensor spaces since these are Euclidean vector spaces. The corollaries all apply verbatim. The dot products are, of course, the generalized dot products used to make tensor spaces Euclidean. The only corollary which perhaps deserves comment is 9.1.28. It now reads thus: Suppose U , V1     V W1      Ws, X1     Xt are Euclidean vector spaces, and D U and R : D ! V1  9.2. GRADIENTS: DEFINITION AND SIMPLE PROPERTIES 95     Vr  W1      Ws and T : D ! W1      Ws  X1      Xt. Dene RhsiT : D ! V1      Vr  X1      Xt by requiring (RhsiT )(~u) = R(~u)hsiT (~u) for each ~u 2 D. Suppose ~u0 2 D and as ~u ! ~u0 we have R(~u) ! R0 and T (~u) ! T0. Then (RhsiT )(~u) ! R0 hsiT0. 9.2 Gradients: Denition and simple properties Denition 9.2.33 We dene the \dot product" of two real numbers to be their ordinary arithmetic product. With this denition, R is a one-dimensional Euclidean vector space, and it has two orthonormal bases, f1g and f;1g, which we write as ^1 and ;^1 when thinking of R as a Euclidean space. When we think of a real number as a vector, we write it with an arrow, like ~2. Thus ~2 = 2^1, and 2 = ~2  ^1 = ^1  ~2. Denition 9.2.34 Suppose U and V are Euclidean vector spaces, D is an open subset of U , f~ : D ! V , and ~u 2 D. A linear mapping L : U ! V is a \gradient mapping for f~ at ~u " if there is a function R~ : U ! V (which may depend on f~ and ~u and L ) such that a) b) lim R~ (~h) = ~0 and f~(~u + ~h) = f~(~u) + L(~h) + k~hkR~ (~h) for any ~h 2 U such that ~u + ~h 2 D: ~h!~0 (9.2.1) Remark 9.2.42 f~ has at most one gradient mapping at ~u. Proof: If L1 and L2 are both gradient mappings for f~ at ~u then whenever ~u + ~h 2 D we have f~(~u + ~h) = = f~(~u) + L1 (~h) + k~hkR~ 1(~h) f~(~u) + L2 (~h) + k~hkR~ 2(~h): Therefore if ~u + ~h 2 D, L1 (~h) + k~hkR~ 1 (~h) = L2 (~h) + k~hkR~ 2 (~h): (9.2.2) 96 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS Now D is open, so there is an " > 0 such that if k~hk < " then ~u + ~h 2 D. Let ~k be any vector in U , and let t be any nonnegative real number. Then kt~kk < " as long as 0  t < "=k~kk. Therefore, if t is in this range, 9:2:2 is true for ~h = t~k. That is L1 (t~k) + kt~kkR~ 1 (t~k) = L2 (t~k) + ktkkR2 (t~k): But Li (t~k) = tLi (k) and kt~kk = tk~kk, so if t > 0 L1 (~k) + k~kkR~ 1 (t~k) = L2 (~k) + k~kkR~ 2(t~k): (9.2.3) This is true for all t in 0 < t < "=k~kk. Let t ! 0, and both R~ 1(t~k) ! ~0 and R~ 2 (t~k) ! ~0. Therefore L1 (~k) = L2 (~k). Since this is true for all k 2 U , we have L1 = L2 . QED Denition 9.2.35 In denition 9.2.34 if f~ : D ! V has a gradient mapping L at ~u, then f~ is \dierentiable at ~u ", and R~ is the \remainder function for f~ at ~u ". The tensor $ L2 U  V is called the \gradient tensor of f at ~u " or the \gradient of f~ at ~u ". It is ~ f~(~u). Since L(~h) = ~h $L, we can rewrite 9.2.33 as written r lim R~ (~h) = ~0 and h~ ~ i ~ ~ ~ f~(~u + ~h) = f~(~u) + ~h  r f (~u) + khkR(h) if ~u + ~h 2 D: a) ~h!~0 b) (9.2.4) Remark 9.2.43 Suppose U and V are Euclidean vector spaces, D is an open subset of $ U , ~u 2 D, and f~ : D ! V . Suppose there is a tensor L2 U  V and a function R~ : D ! V with these properties: a) b) lim R~ (~h) = ~0 and f~(~u + ~h) = f~(~u) + ~h  L + k~hkR~ (~h) if ~u + ~h 2 D: ~h!~0 (9.2.5) $ Then f~ is dierentiable at ~u, its gradient tensor at ~u is L, and its remainder function at ~u is R~ . Proof: 9.2. GRADIENTS: DEFINITION AND SIMPLE PROPERTIES 97 ~h 7! ~h $L is a linear mapping L, so (9.2.5) , (9.2.1). Therefore f~ is dier$ entiable. By Remark 9.2.42, L is the gradient mapping for f~ at ~u, so L is the gradient tensor. Then the R~ 's in (9.2.1b) and (9.2.5b) must be the same. Example 9.2.14 Suppose f~ : D ! V is constant. Then f~ is dierentiable at every ~ f~(~u) =$0 . ~u 2 D and r Proof: $ $ $ The hypotheses of remark 9.2.43 are satised with L= 0 , R (~h) = ~0: Example 9.2.15 Suppose f~ : D ! U is the identity mapping on D, i.e. f~(~u) = ~u for ~ f~(~u) =$I U , all ~u 2 D. Then f~ is dierentiable everywhere in D, and for each ~u 2 D, r ~ ~u =$I U .) the identity tensor on U . (This result is often abbreviated r Proof: $ $ The hypotheses of remark 9.2.43 are satised with L= I U , R~ (~h) = ~0. Example 9.2.16 Suppose U and V are Euclidean spaces, D is an open subset of U and $ $ T is a xed tensor in U  V . Suppose f~ : D ! V is dened by f~(~u) = ~u T for every ~ f~(~u) =T$. (Often abbreviated ~u 2 D. Then at every ~u 2 D, f~ is dierentiable and r r~ (~u T$) =T$.) Proof: $ $ The hypotheses of remark 9.2.43 are satised with L=T , R~ (~h) = ~0. Example 9.2.17 Suppose that for i 2 f1     ng we have ai 2 R and f~i : D ! V . ~ (aif~i)(~u) = Suppose all the f~i are dierentiable at ~u 2 D. Then so is ai f~i : D ! V , and r ~ f~i )(~u). ai(r Proof: Each f~i satises equation (9.2.4), so the hypotheses of remark 9.2.43 are sat$ ~ f~i(~u)] and R~ (~h) = ai R~ i(~h). ised for aif~i with L= ai r 98 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS Example 9.2.18 Suppose that D is an open subset of Euclidean space U and f : D ! R. In elementary calculus classes, f is dened to be dierentiable at ~u 2 D if there is a vector L~ 2 U and a function R : U ! R such that a) b) lim R(~h) = 0 and f (~u + ~h) = f (~u) + ~h  L~ + k~hkR(~h) if ~u + ~h 2 D: ~h!~0 (9.2.6) ~ f (~u). If we think of R as a oneThe vector L~ is called the gradient of f at ~u, written r dimensional Euclidean space, then we can think of f as a vector-valued function f~ = f ^1 (the vector is one dimensional). The scalar valued function f is dierentiable in the sense of equations (9:2:6) i f~ = f ^1 is dierentiable in the sense of equation (9.2.4), and then r~ f~(~u) = r~ f (~u)^1 r~ f (~u) = r~ f~(~u)  ^1: (9.2.7) Proof: To obtain (9.2.4) from (9.2.6), multiply all terms in (9.2.6) on the right by ^1. To obtain (9.2.6) from (9.2.4) dot ^1 on the right in all terms in (9.2.4). Here $ ~ f (~u)^1. clearly L= L~ ^1 = r Example 9.2.19 Suppose D is an open subset of R, regarded as a one-dimensional Euclidean space. Suppose f~ : D ! V has an ordinary derivative at u 2 D (we will write this derivative as @u f~(u) rather than df~=du, for reasons made clear later). Then f~ is dierentiable at ~u and r~ f~(u) = ^1@uf~(u): (9.2.8) Proof: By hypothesis, limh!0 for h 6= 0 then ~ f (u+h);f~(u) h  = @u f~(u). Dene R~ (h) = 0 if h = 0 and 8 9 < f~(u + h) ; f~(u) = h ~ ; @ f ( u ) R~ (h) = jhj : u : h 9.2. GRADIENTS: DEFINITION AND SIMPLE PROPERTIES 99 Then limh!0 R~ (h) = ~0, and f~(u + h) = f~(u) + h@uf~(u) + jhjR~ (h): (9.2.9) But h = ~h  ^1 so h@uf~(~u) = ~h  ^1@u f~(u). Also jhj = k~hk. Therefore (9.2.9) is $ (9.2.5) with L= ^1@uf~(u). Example 9.2.20 Suppose f : D ! R and ~g : D ! V are dierentiable at ~u 2 D. Then so is the product function f~g : D ! V , and h i h i r~ (f~g)(~u) = r~ f (~u) ~g(~u) + f (~u) r~ ~g(~u) : (9.2.10) Proof: By hypothesis, if ~u + ~h 2 D then ~ f (~u) + k~hkRf (~h f (~u + ~h) = f (~u) + ~h  r ~ ~g(~u) + k~hkRg (~h): ~g(~u + ~h) = ~g(~u) + ~h  r Multiplying these two equations gives h ~ i ~ f )~g + k~hkR~ fg (~h) (9.2.11) f (~u + ~h)~g(~u + ~h) = f (~u)~g(~u) + ~h  f (r ~g) + (r where, by denition, h i ~ ~g(~u) R~ fg (~h) = Rf (~h) ~g(~u) + ~h  r h i ~ f (~u) + k~hkRf (~h)R~ g (~h) +R~ g (~h) f (~u) + ~h  r h ~ i h~ ~ i ~ + ~h  r f (~u) h  r~g(~u) =khk: As ~h ! ~0, the three terms in R~ fg , which involve Rf or R~ g , obviously ! 0. So does the last term, because by Schwarz's inequality its length is  k~hkkr~ f (~u)kkr~ ~g(~u)k. Therefore remark 9.2.43 can be applied to (9.2.11). QED. 100 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS Example 9.2.21 The Chain Rule. Suppose U , V , W are Euclidean vector spaces Df is an open subset of U , Dg is an open subset of V , and f : Df ! Dg and g : Dg ! W . Suppose ~u 2 Df and ~v = f~(~u). Suppose f~ is dierentiable at ~u and ~g is dierentiable at ~v. Then their composition, ~g  f~ : Df ! W , is dierentiable at ~u and r~ (~g  f~)(~u) = r~ f~(~u)  r~ ~g(~v): (9.2.12) Proof: $ ~~ $ ~ Let Lf = r f (~u) and Lg = r ~g(~v). Let R~ f and R~ g be the remainder functions for f~ at ~u and ~g at ~v. Then R~ f (~h) ! ~0 as ~h ! ~0, and R~ g (~k) ! ~0 as ~k ! ~0, and $ f~(~u + ~h) = f~(~u) + ~h Lf +k~hkR~ f (~h) if ~u + ~h 2 Df  (9.2.13) $ ~g(~v + ~k) = ~g(~v) + ~k Lg +k~kkR~ g (~k) if ~v + ~k 2 Dg : (9.2.14) We hope to nd R~ gf : U ! W such that R~ gf (~h) ! ~0 as ~h ! ~0 and $ $ (~g  f~ )(~u + ~h) = (~g  f~ )(~u) + ~h  Lf  Lg + k~hkR~ gf (~h) whenever ~u + ~h 2 Df . This equation is the same as h i h i $ $ ~g f~(~u) + ~h) = ~g f~(~u) + (~h Lf ) Lg +k~hkR~ gf (~h): (9.2.15) To obtain (9.2.15), choose any ~h 2 U such that ~u + ~h 2 Df and dene ~k = f~(~u + ~h) ; f~(~u) = f~(~u + ~h) ; ~v: (9.2.16) Then ~k 2 V and ~v + ~k = f~(~u + ~h) 2 Dg , so (9.2.14) holds for the ~k of (9.2.16). Moreover, from (9.2.13) ~k = ~h $Lf +k~hkR~ f (~h): Substituting (9.2.17) in (9.2.14) gives (9.2.17) h ~ ~ i h ~ i ~ $ ~ ~ ~  $ ~ ~ ~ ~g f (~u + h) = ~g f (~u) + h Lf +khkRf (h)  Lg +kkkRg (k): 9.2. GRADIENTS: DEFINITION AND SIMPLE PROPERTIES 101 This is exactly (9.2.15) if we dene R~ gf (~0) = 0 and, for ~h 6= ~0, dene $ ~ R~ gf (~h) = R~ f (~h) Lg + k~kk R~ g (~k): (9.2.18) khk Applying to (9.2.17) the triangle and Schwarz inequalities gives   k~kk  k~hk k $Lf k + kR~ f (~h)k : (9.2.19) Therefore, as ~h ! ~0, we have ~k ! ~0, and k~kk=k~hk remains bounded. Hence R~ f g (~h) ! ~0 as h ! 0. QED. Example 9.2.22 Suppose Df is an open subset of Euclidean space U , and Dg is an open subset of R, and f : Df ! Dg and g : Dg ! R. Suppose f is dierentiable at ~u 2 Df and that g has a derivative, @v g(v), at v = f (~u). Then g  f : Df ! R is dierentiable at ~u ! and r~ (g  f )(~u) = @v g(v)] r~ f (~u) v = f (~u): (9.2.20) Proof: This fact is presumably well known to the reader. It is mentioned here only to show that it follows from the chain rule, example (9.2.21). If we regard R as a one-dimensional Euclidean space and consider f~ = f ^1 and ~g = g^1, then f~ is dierentiable at ~u, ~g is dierentiable at ~v = v^1, so ~g  f~ is dierentiable ~ f~ = r ~ f )^1, and r ~ (~g  f~) = r ~ (g  f )^1, and at ~u and satises 9.2.12. But r r~ ~g = ^1@v~g = ^1^1@v g. Thus 9.2.12 is r~ (g  f )^1 = (r~ f^)  ^1^1)@v g = (@v g)r~ f ^1. Dot ^1 on the right to obtain 9.2.20. Example 9.2.23 Suppose Df is an open subset of R, Dg is an open subset of V , and f~ : Df ! Dg , ~g : Dg ! W . Suppose f~ has a derivative, @uf~(u), at u 2 Df , and suppose ~g is dierentiable at ~v = f~(u). Then ~g  f~ has a derivative at u and ~ ~g(~v) ~v = f~(u): @u(~g  f~)(u) = @uf~(u)  r Proof: (9.2.21) 102 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS ~ f~(~u) = ^1@uf~(u). Therefore, by the chain f~ is dierentiable at ~u = u^1, with r ~ (~g  f~) = ^1@u (~g  f~). rule, ~g  f~ is dierentiable. Therefore @u (~g  f~) exists and r Substitute these expressions in (9.2.12) and dot ^1 on the left to obtain (9.2.21). Example 9.2.24 Suppose f~ and ~g are inverse to one another, and f~ is dierentiable at ~ f~(~u) and r ~ ~g(~v) are inverse to one another. ~u and ~g is dierentiable at ~v = f~(~u). Then r Proof: Let Df be the domain of f~, an open subset of Euclidean space U . Let Dg be the domain of ~g, an open subset of Euclidean space V . By hypothesis, f~  ~g = IDg ~ f~(~u)  r ~ ~g(~v) = r ~ IDf (~u) =$I U and and ~g  f~ = IDf . By the chain rule, r r~ ~g(~v)  r~ f~(~u) = r~ IDg (~v) =$I V . QED. 9.3 Components of gradients Throughout this chapter, U and V are Euclidean vector spaces, (~b1      ~bm) is a xed basis for U with dual basis (~b1      ~bm), and (~1     ~n) is a xed basis for V , with dual basis (~ 1     ~ n). Df is an open subset of U , and f~ : Df ! V . Denition 9.3.36 The i'th covariant component function of f~ relative to ~1      ~n is fi : Df ! R where, for every ~u 2 Df , h i fi (~u) = f~(~u) i = f~(~u)  ~i : (9.3.1) The ith contravariant component function of f~ relative to ~1     ~n is the i'th covariant component function of f~ relative to ~ 1    ~ n. That is, it is f i : Df ! R where, for any ~u 2 Df , h ii f i(~u) = f~(~u) = f~(~u)  ~ i: Evidently f~ = f i~i = fi~ i: (9.3.2) 9.3. COMPONENTS OF GRADIENTS 103 Theorem 9.3.18 Let ~1      ~n be any basis for V . Let Df be an open subset of U . Then f~ : Df ! V is dierentiable at ~u 2 Df i all n covariant component functions fi : Df ! R are dierentiable at ~u. If they are, then h i r~ fi (~u) = r~ f~(~u) i = r~ f~(~u)  ~i h i r~ f~(~u) = r~ f~(~u) i ~ i = r~ f~i(~u)~ i: (9.3.3) (9.3.4) Proof: ) Suppose f~ is dierentiable at ~u. Then it satises (9.2.4). Dotting ~i on the right in each term of (9.2.4) produces lim R (~h) = 0 ~h!~0 i h~ ~ ~ i ~ ~ fi(~u + ~h) = fi(~u) + ~h  r f (~u)  i + khkRi(h) where Ri = R~  ~i. By remark 9.2.43, fi is dierentiable at ~u, and its gradient is (9.3.3). Then (9.3.4) follows from (D-14) and (9.3.3). ( Suppose all the fi are dierentiable at ~u. We can regard ~ i as a constant function on Df , so it is dierentiable at ~u. Then by example 9.2.20 the product f(i) ~ (i) is dierentiable at ~u. By example 9.2.17 so is the sum fi~ i = f~. QED. Denition 9.3.37 Let ~b1     ~bm be any basis for U . Let Df be an open subset of U and suppose f~ : Df ! V . Then f~(uj~bj ) is a V -valued function of the real variables u1     un. If the partial derivative @ f~(uj~bj )=@ui exists at ~u = uj~bj , we abbreviate it as @ui f~(~u) or @i f~(~u) . Let ~b1     ~bm be the basis dual to ~b1      ~bm . If the partial derivative @ f~(uj~bj )=@ui exists at ~u = uj~bj , abbreviate it as @ui f~(~u) or @ i f (~u) . In summary ~(uj~bj ) @ f ~ ~ @if (~u) := @ui f (~u) := @ui (9.3.5) ~(uj~bj ) : @ i f~(~u) := @ui f~(~u) := @ f @u (9.3.6) i 104 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS Theorem 9.3.19 Suppose ~b1    ~bm is any basis for U , and Df is an open subset of U , and f~ : Df ! V . If f~ is dierentiable at ~u 2 Df then the partial derivatives @if~(~u) and @ i f~(~u) all exist, and ~ f~(~u) @if~(~u) = ~bi  r ~ f~(~u) @ i f~(~u) = ~bi  r r~ f~(~u) = ~bi @i f~(~u) = ~bi @ i f~(~u): (9.3.7) (9.3.8) (9.3.9) Proof: By example 9.2.23, if f~ is dierentiable at ~u = uj~bj then the partial derivative of f~(uj~bj ) with respect to ui exists (hold all uj xed except for ui) and is h i ~~ equal to @i(uj~bj )  r f (~u). But obviously, by the denition of @i, @iuj = i j so @i (uj~bj ) = ~bi : Thus we have (9.3.7). 9.3.8 is obtained similarly. To get (9.3.9) use the fact $ ~ f~(~u) =$I U r ~ f~(~u) = ~bi~bi  r ~ f~(~u) = ~bi @i f~(~u) and that I U = ~bi~bi = ~bi~bi. Then r r~ f~(~u) =$LU r~ f~(~u) = ~bi~bi  r~ f~(~u) = ~bi @ i f~(~u). Note: If dim U = 1, the converse of theorem 9.3.19 is true. See example 9.2.19. However, if dim U  2, that converse is false. A famous example is this. Let dim U = 2, V = R. Let x^, y^ be an orthonormal basis for U and dene f : U ! R by f (~0) = 0, p f (xx^ + yy^) = xy= x2 + y2 if x2 + y2 = 6 0. Then @xf (~0) = 0, @y f (~0) = 0, so if f is ~ f (~0) = 0^x + 0^y = ~0. Then by example 9.2.23, @t f (tx^ + ty^) = dierentiable at ~0 then r ~ f (~0) = 0 at t = 0. But f (tx^ + ty^) = t, so @t f (tx^ + ty^) = 1 at t = 0. This @t (tx^ + ty^)]  r contradiction shows that f is not dierentiable at ~0. Theorem 9.3.20 Suppose (~b1     ~bm ) and (~1      ~n) are bases for U and V respectively, with dual bases (~b1      ~bm) and (~ 1     ~ n). Suppose Df is an open subset of U and f~ : Df ! V and ~u 2 Df . Suppose f is dierentiable at ~u. Then the partial 9.4. GRADIENTS OF DOT PRODUCTS 105 ~ fj (~u). Moreover derivatives @if~(~u) and @ifj (~u) exist, and so do the gradients r h ~ i @i f (~u) j = @ifj (~u) h~ i rfj (~u) i = @ifj (~u) h~ ~ i rf (~u) ij = @ifj (~u): (9.3.10) (9.3.11) (9.3.12) Proof: ~ fj (~u) exists. By theorem By theorem 9.3.18, fj is dierentiable at ~u so r 9.3.19, @if~(~u) and @ifj (~u) exist because f~ and fj are dierentiable at ~u. By 9.3.4, h i r~ f~(~u) = r~ fj (~u) ~ j : (9.3.13) ~ f~(~u) = @i f~(~u) and ~bi  r ~ fj (~u) = @ifj (~u) . Therefore dotting By (9.3.7), ~bi  r ~bi on the left throughout (9.3.13) gives @if~(~u) = @ifj (~u)~ j : (9.3.14) This is equivalent to (9.3.10). Equation (9.3.9) applied to fj gives r~ fj (~u) = ~bi @ifj (~u): (9.3.15) This is equivalent to (9.3.11). Substituting it in (9.3.13) gives r~ f~(~u) = ~bi @i fj (~u)~ j : (9.3.16) This is equivalent to (9.3.12). 9.4 Gradients of dot products In this chapter we suppose that U , V , W , X are Euclidean spaces, that D is an open $ $ $ $ subset of U , and that f : D ! V  W and g : D ! W  X . We suppose that f and g $ $ are dierentiable at ~u 2 D. We want to show that f  g : D ! V  X is dierentiable at ~ ( $f  $g )(~u). ~u, and we want to calculate r 106 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS Method 1: No Bases. By hypothesis, $ $ $ ~ ~ $f (~u) + k~hk R f (~u + ~h) = f (~u) + ~h  r f (h) $ ~ $g $ ~ $g (~u) + k~hk R (~u + ~h) = g (~u) + ~h  r g (h): Dotting the rst equation into the second gives   $ $ $ $ $ ~ $ ( f  g )(~u + ~h) = ( f  g )(~u) + ~h  r f (~u)  g (~u) h ~$ i ~ $ ~ $ g (~u) + khk Rf g (h) + f (~u)  ~h  r (9.4.1) where $ $ ~ ~ ~ $  h~ ~ $ i ~ $ ~ Rf g (h) = f (~u) Rg (h) + h  r f (~u)  h  r g (~u) =khk  $  $ ~ $ ~ $ ~ ~ + h  r f (~u)  Rg (h)+ Rf (h) g (~u) h ~$ i ~ $ ~ $ ~ $ g (~u) + khk Rf (h) Rg (h): + Rf (~h)  ~h  r $ $ An application of Schwarz's inequality proves that Rf g (~h) ! 0 as ~h ! ~0. In (9.4.1) the expression   $ h ~$ i $ ~h  r ~ $ g (~u) f (~u)  g (~u)+ f (~u)  ~h  r (9.4.2) $ $ ~ ( $f  $g depends linearly on ~h, so (9.4.1) shows that f  g is dierentiable at ~u and that r ~ ( $f  $g )(~u) is equal to (9.4.2) for all ~h 2 U . )(~u) is that tensor in U  V  X such that ~h  r ~ ( $f  $g )(~u)? We need But what is r Lemma 9.4.23 If P 2 V  W , ~h 2 U , and Q 2 U  W  X then h~ i ~  h i P  h  Q = h  (12) P  (12)Q : Proof: Since both sides of this equation are linear in P and Q, it su"ces to prove the equation when P and Q are polyads, say P = ~vw~ , Q = ~uw~ 0~x. Then 9.4. GRADIENTS OF DOT PRODUCTS 107 h i P~  ~h  Q = (~h  ~u)(w~  w~ 0)~v~x , and ~h  f(12) P  (12)Q]g = = = = ~h  f(12) P  w~ 0~u~x]g ~h f(12) ~v(w~  w~ 0)~u~x]g (w~  w~ 0)~h  (12) ~v~u~x] (w~  w~ 0)~h  ~u~v~x] = (w~  w~ 0)(~h  ~u)~v~x: From lemma (9.4.23) it follows that we can write (9.4.2) as  $ $  $g  ~h  r ~ f (~u) g (~u) + (12) $ ~ f (~u)  12r (~u) : Therefore  h ~ $ i ~r( $f  $g )(~u) = r ~ $f (~u) $g (~u) + (12) $f (~u)  (12)r g (~u) : (9.4.3) Method 2: Introduce bases in U , V , W , X and take components. This is the procedure likely to be used in practical calculations, and the results are generally easier to use than (9.4.3). $ $ Since f and g are dierentiable at ~u, so are their component functions fjk and gk l . By example (9.2.20), so are the products of fj(k)g(k) l . By example (9.2.17) so are the sums fjk gk l , and from examples 9.2.20 and 9.2.17 r~ (fjk gk l ) = (r~ fjk )gk l + fjk (r~ gk l ): (9.4.4) Then (9.4.4) and (9.3.11) imply @i (fjk gk l ) = (@ifjk )gk l + fjk (@i gk l ): $ $ Now fjk gk l = ( f  g )jl and so (9.4.5) and (9.3.10) imply  $ $   $ h $ik @i ( f  g ) = @i f gk l + fjk @i g l  jl therefore jk  $ $   $ $  $  $ @i ( f  g ) = @i f  g + f  @i g  jl jl jl (9.4.5) 108 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS and  $  $ h $ i $ $ $ @i( f  g )(~u) = @i f (~u)  g (~u)+ f (~u)  @i g (~u) : (9.4.6) If we multiply (9.4.6) on the right by ~bi, where ~b1    ~bn is the dual to the basis we have introduced in U , we obtain   ~r( $f  $g ) = (r ~ $f ) $g +~bi $f (@i $g ) : (9.4.7) The last terms in (9.4.3) and (9.4.7) can be shown to be equal by lemma 9.4.23. In fact, neither (9.4.7) nor (9.4.3) is very useful. One usually works with (9.4.6) or (9.2.10). ~ ( $f  $g ) = r ~ ( $f ) $g +( $f )  (r ~ $g ), is generally false. It is The hoped for formula, r ~ or $ true if one of r f , is one-dimensional. If U is one dimensional, (9.4.6) reduces to $ $ $ $ $ $ @u ( f  g )(u) = @u f (u) g (u)+ f (u)  @u g (u): (9.4.8) $ If f is one dimensional, (9.2.10) is r~ (f $g )(u) = r~ f (~u) $g (u) + f (u)r~ $g (~u): (9.4.9) 9.5 Curvilinear coordinates The set Rn of real n-tuples ~u = (u1     un) is a vector space if, for ~u and ~v = (v1     vn) 2 R we dene ~u + ~v = (u1 + v1     un + vn) and, for ~u 2 Rn and a 2 R we dene a~u = (au1     aun). Rn becomes a Euclidean vector space if we dene ~u ~v = uivi. An orthonormal basis for Rn is e^1 = (1 0     0), e^2 = (0 1 0     0)    e^n = (0     0 1). Denition 9.5.38 Suppopse DV is an open subset of Euclidean space V . A \coordinate system" on DV is an open subset DU of Rn, together with an everywhere dierentiable bijection V~ : DU ! DV whose inverse mapping C~ : DV ! DU is also everywhere dierentiable. The real number i = ci(~v) = e^i  C~(~v) is the \value of the i'th coordinate at the point ~v, and ~v is uniquely determined by the values of its n coordinates u1     un. 9.5. CURVILINEAR COORDINATES 109 We have ~v(u1     un) = V~ (uie^i ) = V~ (~u) ui = ci (~v) = e^i  C~(~v): Denition 9.5.39 Suppose W is a Euclidean space and ~g : DU ! W is dierentiable. Then we will write @i~g for the partial derivative with respect to ui. That is j e^ ) @~ g ( u @i~g(~u) = @ui j : (9.5.1) If f~ : DV ! W is dierentiable, then so is f~  V~ : DU ! W (by the chain rule). We will abbreviate @i(f~  V ) as @if~. Thus n ~~ 1 (9.5.2) @if~(~v) = @ f V (u@u i   u )] if ~v = V~ (u1     un). Note that by theorem 9.3.18, the coordinate functions ci : DV ! R are dierentiable at every ~v 2 DV . And by theorem 9.3.19 the partial derivatives @i V~ exist at every ~u 2 DU . The key to understanding vector and tensor elds in curvilinear coordinates is Theorem 9.5.21 Suppose V~ : DU ! DV is a curvilinear coordinate system on open subset DV of Euclidean vector space V . Suppose n = dim U is the number of coordinates. Then dim V = n. For any ~v 2 DV , let ~u = C (~v). Then the two n-tuples of vectors in V , ~ c1 (~v)     r ~ cn(~v)), are dual bases for V . (@1 V~ )(~u)     @nV~ (~u) and (r Proof: By the denition of a coordinate system, C~  V~ = IU jDU . By the chain rule, r~ (C~  V~ )(~u) = r~ V~ (~u)  r~ C~(~v). By example 9.2.15, r~ (IU jDU ) =$I U . Hence, r~ V~ (~u)  r~ C~(~v) =$I U : (9.5.3) Similarly, V~  C~ = IV jDV , so r~ C~(~v)  r~ V~ (~u) =$I V : (9.5.4) 110 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS ~ V~ (~u) 2 U  V so r ~ V~ (~u) =$L for some L 2 L(U ! V ), and r ~ C~(~v) 2 Now r $ ~ C~(~v) =M V  U , so r for some M 2 L(V ! U ). According to (9.5.3) and $ $ $ (7.4.20), L  M = I U so M  L = IU : (9.5.5) $ $ $ According to (9.5.4) and (7.4.20), M  L= I V so L  M = IV : (9.5.6) According to (D-1), (9.5.5) and (9.5.6) together imply that L and M are bijections and M = L;1 . Therefore U and V are isomorphic. Hence they have the same dimension. Lemma 9.5.24 If U and V are nite dimensional vector spaces and L : U ! V is a linear bijection, and ~u1     ~un is a basis for U , then L(~u1)     L(~un) is a basis for V . Proof of lemma: Suppose ai L(~ui) = ~0 for a1      an 2 R. Then L(ai~ui) = ~0. Since L is a bijection, ai~ui = ~0. Since ~u1     ~un are linearly independent, ai =    = an = 0. Hence L(~u1)     L(~un) are linearly independent. Next, let ~v 2 V . There is a ~u 2 U such that ~v = L(~u). There are a1     an 2 R such that ~u = ai~ui. Then ~v = L(~u) = L(ai~ui) = aiL(~ui). Hence L(~u1)     L(~un) span V . ~ V~ (~u) = @i V~ (~u). By (9.3.3) , r ~ C~(~v)  e^j = r ~ cj (~v). And e^i $I U Now by (9.3.8), e^i  r e^j = e^i  e^j = i j . Therefore if we dot e^i on the left and e^j on the right on each side of (9.5.3) we obtain ~ cj (~v) = i j : (9.5.7) @i V~ (~u)  r ~ c1(~v)     r ~ cn(~v)) are dual Thus, the two n-tuples of vectors (@1 V~ (~u)     @nV~ (~u)) and (r to one-another. Hence they are linearly independent. Since dim V = n, they span V . Hence they are bases for V , and each is the basis dual to the other.1 A more intuitive way to put the proof of (9.5.7) is as follows. uj = cj (V~ (u1      un )) so i j = @i uj = @i cj (V~ (u1      un ))] = @i V~  r~ cj . 1 9.5. CURVILINEAR COORDINATES 111 Corollary 9.5.31 If f~ : DV ! W is dierentiable (W being a Euclidean space) then r~ f~(~v) = r~ ci(~v)@if~(~v): (9.5.8) Proof: ~ ci(~v). (9.5.8) is (9.3.9) with ~bi = r A function f~ : DV ! W is often called a vector eld on DV . If W is a tensor product of other vector spaces, f~ is called a tensor eld. It is often convenient to express f~ in terms of a basis for W which varies from point to point in DV . Suppose that for i 2 f1     N g, i : DV ! W is dierentiable everywhere in DV . Suppose also that at each ~v 2 DV , ~1 (~v)     ~N (~v) is a basis for W , with dual basis ~ 1(~v)     ~ N (~v). Then ~ i : DV ! W is also dierentiable in DV . At each ~v 2 DV we can write f~(~v) = f j (~v)~j (~v) = fj (~v)~ j (~v): Then, by (9.5.8), h i r~ f~(~v) = (r~ ci)@i f j ~j ~ ci)(@i f j )~j + f j (r ~ ci)(@i ~j ): = (r Now @i ~ j (~v) is a vector in W , so there are coe"cients ;i j k (~v) 2 R such that Then @i ~k (~v) = ;i j k (~v)~ )j (~v): (9.5.9) h i r~ f~(~v) = @i f j + ;i j k f k (r~ ci)~j (~v): (9.5.10) Equation (9.5.9) is called the \connection formula" for the basis ~ 1(~v)     ~ N (~v) in the coordinate system V~ : DU ! V . The scalars ;i j k (~v) are the \Christoel symbols" at ~v. The expression @i f j + ;i j k f k is often abbreviated Dif j and, in the older literature, is called the covariant derivative of the vector eld f~ : DV ! W . 112 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS If W = q V , then (~ 1     ~ N ) is often taken to be a q'th order polyad ~ c1      r ~ cn) and (@1 V~     @nV~ ). For example, if basis constructed from (r W = V  V and f~ : DV ! V  V , we might write $ ij ~ ~ f = f @i V @j V : (9.5.11) The Christoel symbols ;i j k for @i V~      @nV~ are dened by @i (@k V~ ) = ;i j k (@j V~ ): (9.5.12) If we use (@k V~ )(@l V~ ) as the basis ~1     ~N for V  V , we have from (9.5.12) h i @i (@k V~ )(@l V~ ) = ;i j k (@j V~ )(@l V~ ) + ;i j l (@k V~ )(@j V~ ): This is the connection formula (9.5.9) for (@k V~ )(@l V~ ). It leads to $ r~ f = (Dif jk )(r~ ci)(@j V~ )(@k V~ ) where Dif jk := @i f jk + ;i j l f lk + ;i k l f jl: Note that (9.5.12) implies ;i j k = ;k j i : (9.5.13) (9.5.14) (9.5.15) (9.5.16) In general this will not be true for ;i j k in (9.5.9). Instead of (9.5.11), we might prefer to write $ ~ ci)(r ~ cj ): f = fij (r (9.5.17) ~ ck )(r ~ cl ) analogous to (9.5.13). Then we need the connection formula for (r ~ ck . The Christoel symbols Therefore we need the analogue of (9.5.12) for r ~ cj  @k V~ = j k . introduced by that analogue will not be new. From (9.5.7), r ~ cj )  @k V~ + r ~ cj  @i (@k V~ ) = 0. From (9.5.12), Then @i (r 9.6. MULTIPLE GRADIENTS AND TAYLOR'S FORMULA 113 ~ cj )  @k V~ = ;;i l k r ~ cj  @l V~ = ;;i l k j l = ;;i j k . Thus ;;i j k r ~ ck = @i (r ~ cj )  (@k V~ )(r ~ ck ) = @i (r ~ cj ) $I V = @i (r ~ cj ). Hence, @i (r ~ cj ) = ;;i j k (r ~ ck ): @i (r (9.5.18) Therefore, where $ r~ f (~u) = (Difjk )(r~ ci)(r~ cj )(r~ ck ) (9.5.19) Difjk = @ifjk ; ;i l j flk ; ;i l k fjl : (9.5.20) 9.6 Multiple gradients and Taylor's formula Let U and V be Euclidean spaces with xed bases, (~b1      ~bm) and (~1      ~n) respectively. Take all components with respect to these bases. Let D be an open subset of U ~ f~(~u) 2 U  V for each and suppose f~ : D ! V is dierentiable at each ~u 2 D. Then r ~u 2 U , so we can dene a function r~ f~ : D ! U  V: (9.6.1) This function may itself be dierentiable everywhere in V , in which case we have another function. r~ r~ f~ : D ! U  U  V: (9.6.2) ~ pf~ : D ! (pU )  V exists, we say f~ is p times dierenThe process can continue. If r tiable in D. Now suppose f~ : D ! V is P times dierentiable in D. For any ~u 2 D write ~u = ui~bi . Let @i = @=@ i . Then r~ f~ = ~bi @i f~ h i h i r~ r~ f~ = r~ ~bi @if~ = ~bj @j ~bi @if~ = ~bj~bi @j @i f~ ... r~ pf~ = ~bi1   ~bip @i1    @ip f~: 114 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS Thus, for any ~h 2 U , if we write (p factors ) we have ~hp = ~h~h    ~h ~hphpir ~ pf~ = hi1    hip @i1    @ip f~: Fix ~h 2 U so that ~u + ~h 2 D. Let ~g(t) = f~(~u + th^ ) where h^ = ~h=k~hk: By the ordinary Taylor formula in one independent variable ~g(t) = P tp X @gp (0) + jtjP R~ P (T ) p ! p=0 where R~ P (t) ! 0 as t ! 0. But @t~g(t) = h^  r^ f~(~u + th^ ) = h^ i@i f~(~u + th^ ) h i ~ ) h^ i@i f~(~u + th^ ) @t2~g(t) = (h^  r h i = h^ j @j ^hi@i f~(~u + th^ ) = h^ j h^ i@j @i f~(~u + th^ ) ~ 2f~(~u + th^ ): = (h^ )2h2ir ~ pf~(~u + th^ ) so Similarly @tp~g(t) = (h^ )phpir ~ pf~(~u) @tp~g(0) = (h^ )phpir and ~ pf~(~u): tp@tp~g(0) = (th^ )phpir Thus, ~g(t) = P 1 X ~ pf~(~u) + jtjpR~ p(t): (th^ )phpir p ! p=0 (9.6.3) 9.7. DIFFERENTIAL IDENTITIES 115 Setting t = k~hk in this formula gives P 1 X ~ )pf~(~u) + k~hkpR~ p(t) f~(~u + ~h) = (~h)phpi(r p ! p=0 P 1 X i1    hip @    @ f~(~u) + k~hkpR ~ p(~h) = h i1 ip p ! p=0 (9.6.4) with R~ p(~h) ! 0 as ~h ! ~0. 9.7 Dierential identities Denition 9.7.40 Suppose U and V are Euclidean spaces, D is an open subset of U , $ ~ T$ (~u) 2 U  U  V . We dene and T : D ! U  V is dierentiable at ~u 2 D. Then r $ the divergence of T at ~u to be r~  T$ (~u) =$I U h2ir~ T$ (~u): (9.7.1) Remark 9.7.44 Suppose ~b1      ~bm is a xed basis for U and ~1     ~n is a xed basis $ for V . Write ~u = ui~bi , @i = @u@ i , T = T ik~bi ~k . Then r~  T$ (~u) = @i T ik ~k or, equivalently,  $ k r~  T (~u) = @iT ik : (9.7.2) (9.7.3) Proof: $ In theorem 9.3.20, replace V by U  V and f~ : D ! V by T : D ! U  V . ~ T$ (~u)@iT jk~bi~bj ~k . Then For U  V use the basis ~bj ~k . Then by (9.3.16) r   h i ~r T$ (~u) =$I U h2ir ~ T$ (~u) =$I U h2i @iT jk~bi~bj ~n = @i T jk $I U h2i~bi~bj ~k = @i T jk(~bi  ~bj )~k  = @iT jk i j ~k = @i T ik ~k . Corollary 9.7.32 Suppose f : D ! U and ~g : D ! V . Then r~  (f~~g) = (r~  f~)~g + f~  (r~ ~g): (9.7.4) 116 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS ^ z ^y ^x Figure 9.3: Proof: Use components. Then, because of (9.7.2), (9.7.4) is equivalent to @i (f igk ) = (@if i)gk + f i(@i gk ). In this form it is obvious. Denition 9.7.41 Suppose (U A) is an oriented three-dimensional Euclidean space and $ V is another Euclidean space. Suppose D is an open subset of U and T : D ! U  V $ ~ A T$ (~u), or simply is dierentiable at ~u 2 D. Then the A-curl of T at ~u, written r r~ T$ (~u), is dened to be r~ T$ (~u) = Ah2ir~ T$ (~u): (9.7.5) Note 9.7.5 There are two curls for T$: D ! U  V , one for each of the two unimodular alternating tensors over U . We will always choose the right handed A dened on page 61 so Figure 9.3 is positively oriented. The other curl is the negative of the one we use, because the two A's dier only in sign. Remark 9.7.45 Suppose (~b1      ~bm) is a basis for U and (~1     ~n) is a basis for V $ and T = Tkl~bk ~ l . Then or equivalently, h i r~ T$= Aijk @j Tkl ~bi~ l  (9.7.6) i ~r T$ l = Aijk @j Tkl: (9.7.7)  9.7. DIFFERENTIAL IDENTITIES 117 Proof: ~ T$)jkl = @j Tkl and (Ah2ir ~ T$)i l = Aijk (r ~ T$)jkl. (r Corollary 9.7.33 If f~ : D ! V is twice continuously dierentiable on D (i.e. r~ r~ f~ : D ! U  U  V exists and is continuous at every ~u 2 D) then r~ r~ f~ =$0 : (9.7.8) Proof: ~ r ~ f~)i l = Aijk @j (r ~ f~)kl = Aijk @j @k fl . Because of the continuity assump(r tion, @j @k = @k @j so Aijk @j @k fl = Aijk @k @j fl = Aikj @j @k fl = ;Aijk @j @k fl = 0: Corollary 9.7.34 If T$: D ! U  V is twice continuously dierentiable on D then r~  (r~ T$) = 0: Proof:  (9.7.9)  r~  r~ T$ l = @i(r~ T$)i l = @i Aijk @j Tkl = Aijk @i @j Tkl. Again this = 0 because @i@j = @j @i and Aijk = ;Aikj . Corollary 9.7.35 If T$: D ! U  V is twice continuously dierentiable then r~ (r~ T$) ~ (r ~  T$) ; r ~  (r ~ T$). (N.B. r ~  (r ~ T$) is often written r2 T$ :) =r 118 CHAPTER 9. DIFFERENTIAL CALCULUS OF TENSORS Proof: Let (~b1 ~b2  ~b3 ) be orthonormal and positively oriented, so (9.7.7) reads ~ T$)il = "ijk @j Tkl. Then (r   r~ (r~ T$) il = "ijk @j (r~ T$)kl = "ijk @j "kmn@m Tnl ] = "ijk "kmn@j @m Tnl = (im jn ; injm) @j @m Tnl = @j @i Tjl ; @j @j Til = @i(@j Tjl ) ; @j (@j Til ) ~  T$)l ; @j (r ~ T$)jil = @i (r  $  ~ ~ $  ~ ~ = r(r T ) ; r  (r T ) : il     ~ (r ~ T$) = r ~ (r ~  T$) ; r ~  (r ~ T$) : Thus, r il il il QED. Chapter 10 Integral Calculus of Tensors 10.1 Denition of the Integral 10.1.1 Mass distributions or measures Let D be any subset of Euclidean space U . A function m which assigns real numbers to certain subsets of D is called a mass distribution, or measure, on D if it has these properties: (if S is a subset of D to which m does assign a mass, that mass is written m(S ), and S is called \measurable-m") i) If S is measurable-m, then m(S )  0, ii) D and the empty set are measurable-m. iii) If S1 S2    is any nite or innite sequence of sets which are measurable-m , then S1 \ S2 , S1= S2 and S1 S2    are measurable-m  and m(S1 S2   ) = P m(S ) if no Si and Sj have common points (i.e., if S1  S2    are mutually disjoint.) The general theory of mass distributions is discussed in Halmos, \Measure Theory," Van Nostrand, 1950. An example of a \mass distribution" is (dim U )- dimensional volume. Some subsets of D are so irregular that a volume cannot be dened for them. All bounded open sets D do have volumes. As with volume, any mass distribution can be extended 119 120 CHAPTER 10. INTEGRAL CALCULUS OF TENSORS to permit innite masses. The reader will be presumed to be familiar with the following mass distributions (but not perhaps with the general theory). Example 10.1.25 D is an open subset of U . (~u) is the density of mass per unit of (dim U )-dimensional volume at ~u, and the mass in any measurable set S is m(S ) = R dV (~u)(~u) where dV (~u) is the innitesimal element of (dim U )-dimensional volume in S U. Example 10.1.26 D is a (dim U ) ; 1 dimensional \surface" in U . (~u) is the density of mass per unit of (dim U ) ; 1 dimensional \area" at ~u, dA(~u) is the innitesimal element R of such area at ~u, and the mass in any measurable set S is m(S ) = S dA(~u)(~u). Example 10.1.27 D is a curve in U , and  (~u) is the density of mass per unit of length at ~u, while dl(~u) is the innitesimal element of such length at ~u. Then the mass in any R measurable subset S of D is m(S ) = S dl(~u) (~u). Example 10.1.28 D is a nite set of points, ~u1     ~uN , and these points have masses m1      mN . The mass of any subset S of D is the sum of the masses M for which ~u 2 S . 10.1.2 Integrals Let F : D ! R be a real-valued function on D. The reader is presumed to know how to calculate the integral of f on D with respect to a mass distribution m. We will denote R this integral by D dm(~u)f (~u). In the foregoing examples, this integral is as follows: Example 10.1.25 Z D Example 10.1.26 Z D dm(~u)f (~u) = Z D Z dV (~u)(~u)f (~u): dm(~u)f (~u) = dA(~u)(~u)f (~u): S 10.1. DEFINITION OF THE INTEGRAL Example 10.1.27 121 Z D Example 10.1.28 dm(~u)f (~u) = Z D Z dl(~u) (~u)f (~u): C dm(~u)f (~u) = N X v=1 mv f (~uv ): R It is possible to invent functions f : D ! R so discontinuous that D dm(~u)f (~u) R cannot be dened in examples 10.1.25 , 10.1.26, 10.1.27. (Clearly D dm(~u)f (~u) is always dened in example 10.1.28.) If the integral can be dened, f is said to be integrable with respect to the mass distribution m, or integrable-dm. We will not need the general theory of mass distributions. All we need are the following properties of integrals: Remark 10.1.46 Suppose D1      DN are subsets of U , with mass distributions m1      mN . On D = D1    DN , dene a mass distribution m by m(S ) = PNv=1 mv (S \Dv ). Suppose f : D ! R and for each v, f jDv is integrable-dmv . Then f is integrable-dm and Z D dm(~u)f (~u) = N Z X v=1 Dv dmv (~u)(f jDv )(~u): (10.1.1) Remark 10.1.47 Suppose m is a mass distribution on D and for each v 2 f1     N g, cv 2 R and fv : D ! R is measurable-m. Then cv fv : D ! R is integrable-dm and Z D dm(~u)(cv fv )(~u) = cv Z D dm(~u)fv (~u): (10.1.2) Remark 10.1.48 Suppose m is a mass distribution on D, and c 2 R, and S is an mmeasurable subset of D. Suppose f : D ! R is dened by f (~u) = c if ~u 2 S , f (~u) = 0 if ~u 62 S . Then f is integrable-dm and Z D dm(~u)f (~u) = cm(S ): (10.1.3) Remark 10.1.49 Suppose m is a mass distribution on D and f : D ! R and g : D ! R are both integrable-dm and f (~u)  g(~u) for every ~u 2 U . Then Z D dm(~u)f (~u))  Z D dm(~u)g(~u): (10.1.4) 122 CHAPTER 10. INTEGRAL CALCULUS OF TENSORS Remark 10.1.50 Suppose m is a mass distribution on D and f : D ! R is integrabledm. Then so is jf j : D ! R, and (by 10.1.4) Z  Z  dm(~u)f (~u)  dm(~u) jf (~u)j : (10.1.5) D D Amended version of remark 10.1.50: Remark 10.1.51 Suppose f : D ! R is measurable-m. Then so is jf j, and f is integrable-dm i jf j is integrable-dm. If f is integrable-dm then Z  Z  dm(~u)f (~u)  dm(~u) jf (~u)j : D D Remark 10.1.52 Suppose f : D ! R and g : D ! R are measurable-m, and g is integrable-dm, and jf j  g. Then f is integrable-dm. Our goal is simply to extend the idea of integrating a function f with respect to a mass distribution so that we can integrate vector-valued and tensor-valued function. Since V1     Vq is a Euclidean vector space when V1     Vq are, it su"ces to consider vector-valued functions. Denition 10.1.42 Let U and V be Euclidean vector spaces. Let D be a subset of U , and let m be a mass distribution on D. Let f~ : D ! V . We say that f~ is integrable-dm if for every xed ~v 2 V the function (~v  f~) : D ! R is integrable-dm. (Obviously we dene ~v  f~ by requiring (~v  f~)(~u) = ~v  f~(~u) for all ~u 2 D.) We dene Imf~ : V ! R by requiring for each ~v 2 V Z Imf~(~v) = dm(~u)(~v  f~)(~u): (10.1.6) D Corollary 10.1.36 If f~ : D ! V is integrable-dm, then Imf~ 2 L(V ! R). Proof: Suppose ~v1     ~vN 2 V and c1     cN 2 R. Then cv~vv 2 V , so (cv~vv )  f~ is integrable-dm. Moreover, (cv~vv )  f~ = cv (~vv  f~), so by remark (10.1.47) Z h i Imf~(cv~vv ) = dm(~u) (cv~vv )  f~ (~u) ZD h i dm(~u) cv (~vv  f~) (~u) DZ = cv dm(~u)(~vv  f~ )(~u) = cv Imf~(~vv ): = D 10.2. INTEGRALS IN TERMS OF COMPONENTS 123 Corollary 10.1.37 If f~ : D ! V is integrable-dm, there is a unique vector V~ (m f~ ) 2 V such that for all ~v 2 V , Imf~(~v) = ~v  V~ (m f~ ). Proof: Immediate from theorem 10 and corollary 10.1.36. Denition 10.1.43 If f~ : D ! V is integrable-dm, we denote the vector V~ (m f~ ) by R dm(~u)f~(~u) and call it the integral of f~ with respect to m. From this denition and D (10.1.6), it follows that for every ~v 2 V Z  Z dm(~u)f~(~u)  ~v = dm(~u)(~v  f~)(~u) (10.1.7) D D R and that the vector dm(~u)f~(~u) is uniquely determined by the demand that (10.1.7) hold for all ~v 2 V . D Note 10.1.6 If V = V1      Vq , then the dot product on V is the generalized dot product hqi. Then T : D ! V1      Vq is integrable-dm if for every Q 2 V1      Vq , R the function QhqiT : D ! R is integrable-dm. If it is, then D dm(~u)T (~u) is the unique tensor in V1      Vq such that for every Q 2 V1      Vq Z D  dm(~u)T (~u) hqiQ = Z D dm(~u)(QhqiT )(~u): (10.1.8) Here (QhqiT )(~u) := QhqiT (~u)]. If Q = ~v1~v2    ~vq then (QhqiT )(~u) = T (~u)]hqiQ = T (~u)](~v1     ~vq ) so (10.1.8) implies the special case Z  D dm(~u)T (~u) (~v1    ~vq ) = Z D dm(~u)f T (~u)] (~v1     ~vq )g (10.1.9) for any (~v1     ~vq ) 2 V1    Vq . 10.2 Integrals in terms of components Remark 10.2.53 Suppose U and V are Euclidean spaces, D U , and m is a mass distribution on D. Suppose f~ : D ! V is integrable-dm. Suppose (~1     ~n) is a basis 124 CHAPTER 10. INTEGRAL CALCULUS OF TENSORS for V , with dual basis (~ 1     ~ n). Dene f i := ~ i  f~. Then f i : D ! R is integrable-dm and Z i Z i dm(~u)f~(~u) = dm(~u)f~ (~u): (10.2.1) D D Proof: This is a special case of denitions (10.1.42, 10.1.43) with ~v = ~ i in (10.1.43). Lemma 10.2.25 Suppose V1     Vq are Euclidean spaces and, for p 2 f1     qg, (~1(p)      ~n(pp)) is a basis for Vp, with dual basis (~(1p)      ~(npp)). Then the polyad bases f~i(1)    ~i(qq)g and i    ~ iq g are dual to one another. f~(1) (q ) 1 1 Proof:  ~ (1) ~ (q) ~ j1 ~ jq  i1    iq hqi (1)    (q) = i1 j1    iq jq . Corollary 10.2.38 Suppose U , V1     Vq are Euclidean vector spaces, D, m is a mass distribution on D, and T : D ! V1      Vq is integrable-dm. For p 2 f1     qg, suppose (~1(p)     ~n(pp)) is a basis for Vp, with dual basis (~(1p)      ~(npp)). Suppose T i iq = i    ~ iq )hq iT . Then T i iq : D ! R is integrable-dm and (~(1) (q ) 1 1 1 Z D dm(~u)T i1 iq (~u) = Z D dm(~u)T (~u) i1 iq : (10.2.2) Proof: Immediate from lemma 10.2.25 and remark 10.2.53 by setting, in remark i1    ~ iq g. 10.2.53 , f~1     ~ng = f~i(1)    ~i(1)q g and f~ 1     ~ ng = f~(1) (q) 1 Theorem 10.2.22 Suppose U and V are Euclidean vector spaces, D is a subset of U m is a mass distribution on D, and f~ : D ! V . Suppose (~1      ~n) is a basis for V and f i = ~ i  f~. Suppose that for each i 2 f1     ng, f i : D ! R is integrable-dm. Then f~ is integrable-dm. 10.2. INTEGRALS IN TERMS OF COMPONENTS 125 Proof: Let ~v 2 V , and vi = ~i  ~v. Then ~v  f~ = vif i. Each f i is integrable-dm, so vif i is integrable-dm by remark 10.1.47. Thus, ~v  f~ is integrable-dm. Since this is true for any ~v 2 V , denition 10.1.42 is satised, and f~ is integrable-dm. Corollary 10.2.39 Suppose U , V1      Vq are Euclidean spaces, D U , m is a mass distribution on D, and T : D ! V1      Vq . Suppose (~1(p)     ~nq p ) is a basis for i    ~ iq )hq iT and that each Vp with dual basis ((1p)      ~(npp) ). Suppose T i iq := ((1) (q) i i q T : D ! R is integrable-dm. Then T is integrable-dm. 1 1 1 Proof: Immediate from theorem 10.2.22 and lemma 10.2.25. Now it becomes obvious that remarks 10.1.46, 10.1.47, 10.1.48 are true if f~ : D ! V instead of f : D ! R, and if in remark 10.1.48, ~c 2 V . We simply apply those remarks to the component functions of f~ relative to a basis for V . Remark 10.1.49 has no analogue for vector-valued functions, because f~ < ~g is not dened for vectors. However, (10.1.5) does have a vector analogue. Suppose f~ : D ! V , and suppose f~ and also kf~k : D ! R are integrable-dm. Then Z Z k D dm(~u)f~(~u)k  D dm(~u)kf~(~u)k: (10.2.3) Proof: For any xed ~v 2 V , ~v  f~ is integrable-dm and so is k~vk kf~k. Also, j~v  f~j  k~vk kf~k. Therefore, using (10.1.4),  Z  Z  Z ~v  dm(~u)f~(~u) =  dm(~u)(~v  f~)(~u)  dm(~u)k~vkkf~(~u)k D D DZ = k~vk dm(~u)kf~(~u)k: D R If we set ~v = D dm(~u)f~(~u) in this inequality, we obtain Z   k~vk2  k~vk D dm(~u) f~(~u) : Cancelling one factor k~vk gives (10.2.3). 126 CHAPTER 10. INTEGRAL CALCULUS OF TENSORS We also have from remark 10.2.53, and theorem 10.2.22, Remark 10.2.54 Suppose U , V , W , X , Y are Euclidean spaces, D U , and m is a mass distribution on D. Suppose P 2 V  V  W , Q 2 X  Y , and T : D ! W  X is integrable-dm. Then so are P  T : D ! V  X and T  Q : D ! W  Y . Moreover Z Z dm(~u)(P  T )(~u) = P  dm(~u)T (~u) D D Z  Z  dm(~u)(T  Q)(~u) = dm(~u)T (~u)  Q: D D (10.2.4) (10.2.5) Proof: We consider P  T . The proof for T  Q is similar. Choose bases for V , W , X , Y and take components with respect to these bases. Then (P  T )(~u) = P  T (~u)] so (P  T )ik (~u) = (P  T )(~u)]ik = P  T (~u)]ik = P i j T jk (~u): By hypothesis, T is integrable-dm. Hence so is T jk. Hence so is P i j T jk . Hence so is (P  T )ik . Hence so is P  T . And Z = = ik Z dm(~u)(P  T )(~u) = dm(~u)(P  T ik )(~u) DZ dm(~u)PjiT jk (~u) = P i j dm(~u)T jk (~u) D Z jk  D Z ik P i j dm(~u)T (~u) = P  dm(~u)T (~u) : D D DZ Corollary 10.2.40 In remark 10.2.54, take W = W1    Wp and X = X1    Xq . Then Z Z dm(~u)(P hpiT )(~u) = P hpi dm(~u)T (~u) D Z  Z dm(~u)(T hqiQ)(~u) = dm(~u)T (~u) hqiQ: D D D (10.2.6) (10.2.7) Chapter 11 Integral Identities 11.1 Linear mapping of integrals Remark 11.1.55 Suppose U , V , W are Euclidean spaces, D U , m is a mass distribution on D, L 2 L(V ! W ), and f~ : D ! V is integrable-dm. Then L  f~ : D ! W is integrable-dm and Z  ~ ~ dm(~u)(*L  f )(~u) = L dm(~u)f (~u) : (11.1.1) D D R In other words, L can be applied to D dm(~u)f~(~u) by applying it to the integrand. Z Proof: $ Let L2 V  W be the tensor corresponding to L. Then (L  f~)(~u) = Lf~(~u)] = $ $ $ f~(~u) L= (f~ L)(~u) so L  f~ = f~ L, and remark 10.2.54 says this is integrable and that Z D Z $ dm(~u)(f~ L)(~u)  $ ZD = dm(~u)f~(~u)  L D  Z = L dm(~u)f~(~u) : dm(~u)(L  f~)(~u) = D QED. 127 128 CHAPTER 11. INTEGRAL IDENTITIES . u1 D . C u v . u2 τ( u ) Figure 11.1: Corollary 11.1.41 Suppose U , V , W , D, L and m as in remark 11.1.55. Suppose T : D ! q V is integrable-dm. Then (q L)  T : D ! q W is integral-dm and Z D dm(~u) (q L)  T ] (~u) = (q L) Z D  dm(~u)T (~u) : (11.1.2) Proof: A special case of remark 11.1.55, with V , W , L replaced by q V , q W and  q L. 11.2 Line integral of a gradient Theorem 11.2.23 Suppose U and V are Euclidean spaces, D is an open subset of U , and f~ : D ! V is continuously dierentiable at each ~u 2 D. Suppose C is a smooth curve in D whose two endpoints are ~u1 and ~u2. Suppose dl(~u) is the innitesimal element 11.2. LINE INTEGRAL OF A GRADIENT 129 of arc length on C , and ^(~u) is the unit vector tangent to C at ~u and pointing along C in the direction from ~u1 to ~u2. Then Z C ~ f~)(~u) = f~(~u2) ; f~(~u1): dl(~u)(^  r (11.2.1) Proof: We assume that the reader knows this theorem when V = R, and f~ is a realvalued function. Let (~1      ~n) be a basis for V , with dual basis (~ 1     ~ n), ~ f i(~u) and dene f i := ~ i  f~. Since f~ is dierentiable at each ~u, so is f i. Thus r ~ ~i =$0 , so by equation (9.4.9), r ~ f i(~u) = r ~ (f~  ~i)(~u) = exists. Moreover, r r~ f~(~u)  ~i . Both r~ f~ and ~i are continuous, and then so is r~ f~  ~i. Thus f i : D ! R is continously dierentiable. By the known scalar version of (11.2.1), Z ~ f i(~u) = f i(~u2) ; f i(~u1): dl(~u)^  r C If we multiply this equation by ~i and sum over i we obtain  ~ f i(~u) ~i dl(~u)^  r ZC h i ~ f i)~i (~u) by 10.2.7 with q = 0 = dl(~u) (^  r ZC ~ f~)(~u) because = dl(~u)(^  r Ch i ~ f i)~i = ^  (r ~ f i )~i = ^  r ~ (f i~i) = ^  r ~ f~: (^  r f~(~u2) ; f~(~u1) = Z QED. Corollary 11.2.42 Suppose D is open and arcwise connected (any two points in D can be joined by a smooth curve lying wholly in D). Suppose f~ : D ! V is continuously ~ f~(~u) =$0 for all ~u 2 D. Then f~ is a constant function. dierentiable and r Proof: For any ~u1 ~u2 2 D, f~(~u1) = f~(~u2) by theorem 11.2.23. 130 CHAPTER 11. INTEGRAL IDENTITIES 11.3 Gauss's theorem Theorem 11.3.24 Suppose U and V are Euclidean spaces, D is an open subset of U , @D is the boundary of D, and n^ (~u) is the unit outward normal at ~u 2 @D. Let D( = D @D = $ $ closure of D. Suppose f : D( ! U  V is continuous and f jD is continuously dierentiable at each ~u 2 D. Let dV (~u) and dA(~u) be the innitesimal volume element in D and the innitesimal surface area element on @D respectively ( dV (~u) has dimension dim U , and dA(~u) has dimension one less). Then Z Z $ $ dA(~u)(^n f )(~u) = dV (~u)(@~ f )(~u): (11.3.1) @D D Proof: We assume that the reader knows the corresponding theorem for vector-valued functions, f~ : D( ! U . Let (~1      ~n) be a basis for V , with dual basis $ i i $ (~ 1     ~ n). Dene f~ : D( ! U by f~ := f ~ i, so f~i(~u) = f (~u)~ i. Note that i i f~ (~u) is not a component but a vector in U . By corollary 9.1.28, f~ : D( ! U i is continuous, and by section 9.4, f~ is dierentiable. By corollary 9.1.28, i i i @~f~ is continuous. Thus, f~ : D( ! U is continuous and f~ jD is continuously i dierentiable. Therefore, f~ obeys Gauss's theorem: Z @D i dA(~u)(^n  f~ )(~u) = Z D ~  f~ i)(~u): dV (~u)(r If we multiply on the right by ~i and sum over i, (10.2.7) with q = 0 gives    Z i  ~  f~ i)~i (~u): dA(~u) (^n  f~ )~i (~u) = dV (~u) (r (11.3.2) @D D $ $ $ $ $ i i i But f~ ~i = f (~ i~i) = f (~ i~i ) = ( f  I V ) = f so (^nf~ )~i = n^ (f~ ~i) = n^ f~, and by section 9.4  i    ~ f~ ~i ) ~  f~ i)~i = $I U h2ir ~ f~ i ~i =$I U h2i (r (r h ~ ~i ~ i $ ~ $ ~ $ $ = I U h2i r (f i) = I U h2ir f = r f : Z Substituting these expressions in (11.3.2) gives 11.3.1. 11.4. STOKES'S THEOREM 131 Figure 11.2: 11.4 Stokes's theorem Theorem 11.4.25 Suppose (U A) is an oriented Euclidean 3-space, and A is used to dene cross products and curls. Suppose D is an open subset of U , and S is a smooth two-sided surface in D, with boundary curve @S . Suppose n^ : S ! U is one of the two continuous functions such that n^(~u) is a unit vector in U perpendicular to S at ~u 2 S . If ~u 2 @S , let ^(~u) be the unit vector tangent to @S at ~u and pointing so that n^(~u) ^(~u) $ points into S . Suppose that f : D ! U  V is continuously dierentiable everywhere in D. Suppose that dA(~u) is the element of area on S and dl(~u) is the element of arc length on @S . Then  Z Z $ $ ~ dl(~u)(^  f )(~u) = dA(~u) n^  (r f ) (~u): (11.4.1) @S S Proof: Again, we assume the reader knows this theorem when V = R, i.e. for vector elds f~ : D ! U . Let (~1     ~n) be a basis for V , with dual basis i (~ 1     ~ n), and dene f~i = f~  ~ i. Then we can use Stoke's theorem for f~ , $ i i.e., replace f by f~ in (11.4.1). If we multiply both sides of the result on the 132 CHAPTER 11. INTEGRAL IDENTITIES right by ~i, sum over i, and use (10.2.7) with q = 0, we get     Z i  ~ f~ i) ~i (~u): dl(~u) (^  f~ )~i (~u) = dA(~u) n^  (r @S S $ $ i i i As in the proof of Gauss's theorem, f~ ~i = f , so (^  f~ )~i = ^  (f~ ~i) = ^ f . ~ f~i)]~i = n^  (r ~ f~ i)~i]. Finally And ^n  (r     r~ f~ i ~i = Ah2ir~ f~ i ~i  i  ~ f~ )~i = Ah2i (r  i   $ ~ (f~ ~i ) = Ah2i r ~ f = Ah2i r ~ $ = r f: Z This nishes the proof. 11.5 Vanishing integral theorem Lemma 11.5.26 Suppose D is an open subset of Euclidean space U , f : D ! R is R continuous, and D0 dV (~u)f (~u) = 0 for every open subset D0 of D. Here dV (~u) is the volume element in U . Then f (~u) = 0 for all ~u 2 D. Proof: Suppose ~u0 2 D and f (~u0) 6= 0. If f satises the hypotheses so does ;f , so if f (~u)0) < 0 we may replace f by ;f . Therefore, we may assume f (~u0) > 0. Then there is a  > 0 such that if j~u ; ~u0 j <  then ~u 2 D and jf (~u) ; f (~u0 )j < f (~u0)=2. But then ;f (~u0)=2 < f (~u) ; f (~u0) < f (~u0)=2, so f (~u) > f (~u0)=2 if ~u 2 B (~u0 ), the open ball of radius  centered at ~u0. Then Z Z dV (~u)f (~u) > dV (~u) f (2~u0) = jB (u~0 )j f (2~u0)  B(~u0 ) B(~u0 ) where jB (~u0 )j = volume of B (~u0 ). But jB (~u0 )jf (~u0)=2 > 0, and R B (~u0 ) is an open subset of D, so we must have B(~u0 ) dV (~u)f (~u) = 0. This contradiction shows f (~u0) = 0. 11.6. TIME DERIVATIVE OF AN INTEGRAL OVER A MOVING VOLUME x x 133 Wδ t x x x V I . { u n x III x II x x ∂K(t) ∂K(t+ δ t) Figure 11.3: Theorem 11.5.26 Suppose U and V are Euclidean spaces, D is an open subset of U , and f~ : D ! V is continuous. Suppose that for every open subset D0 of D, we have R dV (~u)f~(~u) = ~0 . Then f~(~u) = ~0 for all ~u 2 D. V V D0 Proof: Let (~1     ~n) be a basis for V , with dual basis (~ 1     ~ n). Let f i = ~ i  f~. Then f i : D ! R is continuous, and for every open subset D0 of D, R dV (~u)f i(~u) = 0. Hence, f i = 0 by lemma 11.5.26. Hence f~ = f i~ = ~0 . i V D0 134 CHAPTER 11. INTEGRAL IDENTITIES 11.6 Time derivative of an integral over a moving volume Suppose that for each instant t we are given an open set K (t) in Euclidean vector space U , with boundary surface @K (t) and outward unit normal n^. Suppose that K (t) changes with time su"ciently smoothly that its surface @K (t) has a well dened speed normal to itself. At ~u 2 @K (t), this speed is W (~u t), and it is > 0 if @K (t) is moving outward from K (t), < 0 if inward. In gure 11.3, W > 0 on the part of @K (t) marked with x's and W < 0 on the part of @K (t) marked with o's. Suppose that all the sets K (t) are subsets (K (t) = K (t) @K (t) = closure of K (t)) of an open set D, and f~ : D R ! V , where V is another Euclidean space. Thus for each ~u 2 D and t 2 R, f~(~u t) is a vector in V . We suppose this vector depends continuously on ~u and is continuously dierentiable in t (i.e. f~( t) is continuous for each t 2 R, and f~(~u ) is continuously dierentiable for each ~u 2 D ). Then d Z dV (~u)f~(~u t) = Z dV (~u)@ f~(~u t) t dt K (t) KZ(t) + dA(~u)(W f~)(~u t): @K (t) (11.6.1) Here @t f~(~u t) stands for @ f~(~u )](t), or @ f~(~u t)=@t. Proof: Let @K (t)+ stand for the part of @K (t) marked with x's in gure 11.3, while @K (t); stands for the part marked with o's. Let II be the set K (t) \ K (t + t), while I is K (t)nK (t + t), and III is K (t + t)nK (t). See gure 11.3. Let F~ (t) = Then Z K (t) dV (~u)f~(~u t): Z Z F~ (t) = dV (~u)f~(~u t) + dV (~u)f~(~u t): II Z Z I ~F (t + t) = dV (~u)f~(~u t + t) + dV (~u)f~(~u t + t): II III 11.7. CHANGE OF VARIABLES OF INTEGRATION 135 Correct to rst order in t, f~(~u t + t) = f~(~u t) + (t) @t f~(~u t) in D dV (~u) = (t)W (~u t)dA(~u) in III where W > 0 dV (~u) = ;(t)W (~u t)dA(~u) in I where W < 0: Thus, correct to rst order in t, Z dV (~u)f~(~u t) = ;(t) I Z Z ~ dV (~u)f (~u t) = (t) III Z @K (t); @K (t)+ dA(~u)W f~(~u t) dA(~u)W f~(~u t): Therefore, Z III Z Z dV (~u)f~(~u t) ; dV (~u)f~(~u t) = (t) Also, I @K (t) dA(~u)W f~(~u t): (11.6.2) Z Z h i dV (~u) f~(~u t + t) ; f~(~u t) = (t) dV (~u)@t f~(~u t) II Z II Z = (t) dV (~u)@t f~(~u t) ; (t) dV (~u)@t f~(~u t): K (t) I R R But I dV (~u)@t f~ = ;(t) @K (t); dA(~u)W@tf , correct to rst order in t. Therefore, correct to rst order in t, Z Z h~ i ~ dV (~u) f (~u t + t) ; f (~u t) = (t) dV (~u)@t f~(~u t): (11.6.3) II K (t) Combining (11.6.2, 11.6.3), we have, correct to rst order in t, F~ (t + t) ; F~ (t) = t "Z @K (t) dA(~u)W f~(~u t) + Z K (t) # dV (~u)@t f~(~u t) : Dividing by t and letting t ! 0 gives (11.6.1). 11.7 Change of variables of integration Suppose (U AU ) and (V AV ) are two oriented n-dimensional Euclidean spaces. Suppose L 2 L(U ! V ). For any ~u1     ~un 2 U , dene A(~u1     ~un) = AV L(~u1)     L(~un)]. 136 CHAPTER 11. INTEGRAL IDENTITIES Clearly A 2 nU , so there is a scalar kL such that A = kLAU . We denote this scalar by det L and call it the determinant of L. Then AV L(~u1 )     L(~un)] = (det L)AU (~u1     ~un) (11.7.1) for all (~u1     ~un) 2 X nU . Note that det L changes sign if we replace either AU or AV by its negative. Thus det L depends not only on L but on how U and V are oriented. For a given L, det L can have two values, one the negative of the other, and which value it has depends on which of the two unimodular alternating tensors AU and AV are used to orient U and V . $ If (^x1     x^n) is a pooob in U and (^y1     y^n) is a pooob in V and Lij = x^i  L y^j , then1 L(^xi ) = Lij y^j and so det L = (det L)AU (^x1      x^n) = AV L(^x1 )     L(^xn )] = AV Lij1 y^j1      Lnjn y^jn ] = L1j1    Lnjn AV (^yj1     y^jn ): Therefore det L = L1j1  Lnjn "i1 in = det(Lij ): (11.7.2) Now suppose H and K are open subsets of U and V respectively and ~r : H ! K is a continuously dierentiable bijection of H onto K . For ~x 2 H write ~x = xi x^i and dene ~ ~r )ij = x^i  r ~ ~r ]  y^j so the determinent of rj = y^j  ~r . At any ~x 2 H we have @irj = (r ~ ~r (~x), as we see from (11.7.2). the matrix @i rj (^x) is det r Finally, suppose f : K ! R is integrable with respect to volume in V . Then f  r : H ! R is integrable with respect to volume in U and Z   @ r j  f (rj y^j )dr1    drn = f  ~r (xk x^k )] det @x  dx1    dxn: K H i Z (11.7.3) The reader is presumed to know this formula for changing variables in a multiple integral. The determinant det(@rj =@xi ) is the Jacobian of the coordinate transformation. We note that @rj =@xi = @i rj , so by (11.7.2), at ~x 2 H ! @r j ~ det @xi = det r ~r (~x):  1 $ $ Proof: Lij y^j = xi  L y^j y^j = x^i  L (11.7.4) 11.7. CHANGE OF VARIABLES OF INTEGRATION 137 The local volume elements in U and V are dVU (~x) = dx1    dxn and dVV (~r) = dr1    drn, so (11.7.3) can be written Z K dVV (~r)f (~r) = Z H   ~ ~r (~x) (f~  ~r )(~x): dVU (~x) det r (11.7.5) Now suppose W is another Euclidean space, with basis (~1     ~p) and dual basis (~ 1     ~ p). Suppose f~ : K ! W is integrable. Then so is fi = ~i  f~, and (11.7.5) holds for each fi. If we multiply the resulting equations by ~ i and sum over i we obtain Z K dVV (~r)f~(~r) = Z H   ~ ~r (~x) (f~  ~r )(~x): dVU (~x) det r (11.7.6) 138 CHAPTER 11. INTEGRAL IDENTITIES Part II Elementary Continuum Mechanics 139 Chapter 12 Eulerian and Lagrangian Descriptions of a Continuum 12.1 Discrete systems For some purposes, some physical systems (for example, the solar system) can be regarded as consisting of a nite number N of mass points. A complete description of the motion of such a system consists simply of N vector-valued functions of time the value of the  'th function at time t is the position of the  'th particle at time t. If the values of the positions ~r1(t)    ~rN (t) of all particles are known for all times, nothing more can be said about the system. To predict the motion of such a \discrete" system, one must usually add other properties to the model. In the case of the solar system, one ascribes to each particle a mass, m being the mass of the  'th particle. Then, if all the velocities and positions are known at some time t0, all the functions ~r (t) can be calculated from Newton's laws of motion and gravitation. If the system of particles is a classical (pre-quantum) atom, one needs not only their masses but their charges in order to calculate their motion. And, of course, for some purposes the discrete model of the solar system is too crude. It cannot model or predict the angular acceleration of the earth's moon due to the tidal 141 142 CHAPTER 12. EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM bulges on the earth. A point mass has no tidal bulges. Nevertheless, the discrete model of the solar system is very useful. No one would suggest abandoning it because the solar system has properties not described by such a model. 12.2 Continua A parcel of air, water or rock consists of such a large number of particles that a discrete model for it would be hopelessly complicated. A dierent sort of model has been developed over the last three centuries to describe such physical systems. This model, called a \continuum", exploits the fact that in air, water and rock nearby particles behave similarly. The continuum model for a lump of material regards it as consisting of innitely many points, in fact too many to count by labeling them even with all the integers. (There are too many real numbers between 0 and 1 to label them as x1  x2     . Any innite sequence of real numbers between 0 and 1 will omit most of the numbers in that interval. This theorem was proved by Georg Cantor in about 1880.) Nearby points are given nearby labels. One way to label the points in the continuum is to give the Cartesian coordinates of their positions at some particular instant t0 , relative to some Cartesian coordinate system. Then each particle is a point to which is attached a number triple (x1  x2 x3 ) giving the Cartesian coordinates of that particle at the \labelling time" t0 . This number triple is a label which moves with the particle and remains attached to it. Nearby particles have nearly equal labels, and particles with nearly equal labels are close to one another. The collection of all particles required to describe the motion will usually ll up an open set in ordinary real 3-space. Of course there are many ways to label the particles in a continuum. The label attached to a particle could be the number triple (r  ) giving the values of its radius, colatitude and longitude in a system of polar spherical coordinates at the labelling time t0. Another label would be simply the position vector ~x at time t0 of the particle relative to some xed origin and axes which are unaccelerated and non-rotating (so that Newton's laws can be used). In fact, labelling by number triples can also be thought of as labelling 12.2. CONTINUA 143 by vectors, since a number triple is a vector in the Euclidean vector space R3 . (Rn is the set of real n-tuples, with addition and multiplication by scalars dened coordinate-wise and with the dot product of (u1     un) and (v1      vn) dened to be uivi.) In order to leave ourselves freedom to choose dierent ways of labelling, we will not specify one particular scheme. We will simply assume that there is a three-dimensional oriented real Euclidean space, (L AL ), from which the labels are chosen. The labels describing a continuum will be the vectors ~x in a certain open and subset H of label space L. We will denote real physical space by P , and we will orient it in the usual way, denoting the \right handed" unimodular alternating tensor by AP , or simply A. We use this notation: ~r L(~x t) := position at time t of the particle labelled ~x: (12.2.1) The label ~x is in the open subset H of label space L, and the particle position ~r L(~x t) is in P . One special case which is easy to visualize is to take L = P , and to choose some xed time t0, and to label the particles by their positions at t0 . With this labelling scheme, called \t0 -position labelling," we have r L(~x t0) = ~x for t0 -position labelling: (12.2.2) The motion of a continuum is completely described by giving the position of every particle at every instant, i.e. by giving ~r L(~x t) for all ~x 2 H and all t 2 R. This amounts to knowing the function r L : H R ! P: (12.2.3) This function is called the Lagrangian description of the motion of the continuum. We denote by K (t) the set in position space P occupied by the particles of the continuum at time t. We assume that particles neither ssion nor coalesce each particle retains its identity for all time. Therefore we need never give two particles the same label ~x, and we do not do so. But then if ~r 2 K (t) there is exactly one particle at position ~r at time t. We denote its label by ~x E (~r t). Thus x E (~r t) := label of particle which is at position ~r at time t: (12.2.4) 144 CHAPTER 12. EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM From the denitions, it is clear that if ~r 2 K (t) and ~x 2 H then ~r = ~r L(~x t) , ~x = ~x E (~r t): (12.2.5) But (12.2.5) says that ~r L( t) : H ! K (t) is a bijection, whose inverse is ~x E ( t) : K (t) ! H . If we know the function ~x E ( t), we can nd the function ~r L( t). For any ~x 2 H , ~r L (~x t) is the unique solution ~r of the equation ~x = ~x E (~r t). Therefore, knowing the label function ~x E is equivalent to knowing ~r L. Either function is a complete description of the motion of the continuum, and either can be found from the other. For any xed t, ~r L( t);1 = ~x E ( t): (12.2.6) In order to enforce that nearby particles have nearby labels we will assume not only that ~r L and ~x E are continuous, but that they are continuously dierentiable as many times as needed to make our arguments work. (Usually, twice continuously dierentiable will su"ce.) Since (12.2.6) implies ~r L( t)  ~x E ( t) = IP jK (t), the identity function on P , restricted to K (t), and also ~x E ( t)  ~r L( t) = ILjH , the identity function on L, restricted to H , therefore the chain rule implies that if ~r = ~r L(~x t) or, equivalently, ~x = ~x E (~r t), then 8 > L (~x t)  r E (~r t) = $ ~ ~ > ~ ~ r r x IL > < (12.2.7) > > $ > ~ ~x E (~r t)  r ~ ~r L(~x t) = I P : :r 12.3 Physical quantities If the motion of a continuum is to be predicted from physical laws, more properties of the continuum must be considered than simply the positions of its particles at various times. Most of the useful properties are local, i.e. they have a value at each particle at each instant. Examples are temperature, mass density, charge density, electric current density, entropy density, magnetic polarization, the electrical conductivity tensor and, in a uid, the pressure. 12.3. PHYSICAL QUANTITIES 145 A local physical property f can be described in two ways: at any instant t, we can say what is the value of f at the particle labelled ~x, or we can say what is the value of f at the particle whose position is ~r at time t. The two values are the same, but they are given by dierent functions, which we denote by f L and f E . We call these the Lagrangian and the Eulerian descriptions of f . They are dened thus: f L(~x t) = value of physical quantity f at time t at the particle labelled ~x (12.3.1) f E (~r t) = value of physical quantity f at time t at the particle whose position is ~r at time t: (12.3.2) The physical quantity f can be a scalar, vector or tensor. From the denitions, obviously 8 > L E > > < if ~r = ~r (~x t) or ~x = ~x (~r t) then > > > : f L(~x t) = f E (~r t): (12.3.3) This is equivalent to asserting 8 h i > L(~x t) = f E ~r L (~x t) t and > f > < > h i > > : f E (~r t) = f L ~x E (~r t) t : (12.3.4) These two assertions are equivalent in turn to 8 > L L E > > < f ( t) = f ( t)  ~r ( t) > > > f E ( t) = f L( t)  ~x E ( t): : (12.3.5) If we have the Lagrangian description of the motion, the function ~r E , then we can nd the labelling function ~x E . Then using (12.3.4) or (12.3.5), we can nd both f E and f L if we know either one of them. So far, our discussion of \physical quantities" has been somewhat intuitive. We have found some mathematical rules which are satised by certain entities that we feel comfortable to call physical quantities. In the spirit of mathematical model building, it is 146 CHAPTER 12. EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM useful to give a precise denition to what will qualify as a \physical quantity". Because of what we have learned above, we introduce Denition 12.3.44 Given a continuum whose motion is described by ~r L : H R ! P , and whose labelling function is ~x E , with ~x E ( t) : K (t) ! H , a physical quantity f is an ordered pair of functions (f L f E ) with these properties: i) The domain of f L is H R ii) The domain of f E is a subset of P R, and the domain of f E ( t) is K (t) iii) If ~r L(~x t) = ~r , or ~x E (~r t) = ~x, then f L(~x t) = f E (~r t) . The function f L is the Lagrangian description of the physical quantity f , and f E is the Eulerian description of f . Property iii) is equivalent to 8 h i > L (~x t) = f E ~r L (~x t) t or to > f > < > > > f E (~r t) = f L h ~x E (~r t) ti for all t 2 R: : (12.3.6) These in turn are equivalent to 8 > L E L > > < f ( t) = f ( t)  r ( t) or to > > > f E ( t) = f L( t)  ~x ( t) for all t 2 R: : (12.3.7) Temperature, mass density, and the other items enumerated beginning on page 144 are physical quantities in the sense of denition 12.3.44. The advantage of having such a formal denition is that it enables us to introduce new physical quantities to suit our convenience. We can dene a new physical quantity f by giving either its Eulerian description f E or its Lagrangian description f L. The missing description can be obtained from (12.3.6) or (12.3.7), assuming that we have ~r L, the Lagrangian description of the motion of the continuum. A somewhat trivial example is the f such that f L(~x t) = 7 for all ~x t. Clearly f E (~r t) = 7, and we can regard 7 (or any other constant) as a physical quantity. 12.4. DERIVATIVES OF PHYSICAL QUANTITIES 147 A more interesting example is a physical quantity which we call \particle position", and denote by ~r . It is dened by requiring that its Eulerian description be ~r E (~r t) = ~r for all t 2 R and ~r 2 K (r): (12.3.8) From (12.3.6), its Lagrangian description is the ~r L of (12.2.1), the Lagrangian description of the motion. Another interesting example is a physical quantity we call \particle label", and denote by ~x . It is dened by requiring that its Lagrangian description be ~x L(~x t) = ~x for all (~x t) 2 H R: (12.3.9) From (12.3.6), its Eulerian description is ~x E  the label function. The fact that the physical quantity ~x is conserved at each particle turns out to be very useful in some variational formulations of the equations of motion of a uid, although at rst glance the fact seems rather trivial. 12.4 Derivatives of physical quantities It will be useful to introduce the following notation for the derivatives of a physical quantity f = (f L f E ). f L(~x t +  ) ; f L(~x t) Dt f L(~x t) = lim !0  = time derivative of f L(~x t) at the particle label ~x: (12.4.1) h i ~ L(~x t) = r ~ f L( t) (~x) = gradient tensor of the function f L( t) Df at the particle label ~x: (12.4.2) f E (~r t +  ) ; f E (~r t) @t f E (~r t) = lim !0  = time derivative of f E (~r t) at spatial position ~r: (12.4.3) h i ~ f E ( t) (~r) = gradient tensor of the function f E ( t) @~f E (~r t) = r at the position ~r in P: (12.4.4) 148 CHAPTER 12. EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM From these denitions, (12.3.7), and the chain rule, it is clear that if ~r = ~r L(~x t) (or ~x = ~x E (~r t)) then ~ L(~x t) = D~ ~ r L(~x t)  @~f E (~r t) Df (12.4.5) ~ L(~x t): @~f E (~r t) = @~ ~x E (~r t)  Df (12.4.6) It is also useful to relate Dtf L and @t f E . We will calculate the former from the latter. We assume that ~r and ~x are chosen so ~r = ~r L(~x t): Then we dene ~h( ) for any real  by ~h( ) = ~r L(~x t +  ) ; ~r L(~x t): We have 8 > L ~ ~ > > < h( ) = Dt ~r (~x t) +  R1 ( ) where > > > : R~ 1 ( ) ! 0 as  ! 0: (12.4.7) We also have h i h i f L(~x t +  ) = f E ~r L(~x t +  ) t +  = f E ~r + ~h( ) t +  h i h i = f E ~r + ~h( ) t + @t f E ~r + ~h( ) t + R2 ( ) where R2( ) ! 0 as  ! 0. Dene h i R3 ( ) = R2 ( ) + @t f E ~r + ~h( ) t ; @t f E (~r t) : Since ~h( ) ! ~0 as  ! 0 therefore 8 > > R3 ( ) ! 0 as  ! 0 and > < > h i > > : f L(~x t +  ) = f E ~r + ~h( ) t + @t f E (~r t) + R3 ( ): (12.4.8) 12.4. DERIVATIVES OF PHYSICAL QUANTITIES Also 8 > E E ~ ~ ~ E ~ ~ > > < f (~r + h t) = f (~r t) + h  @ f (~r t) + khkR4(h) > > > : where R4 (~h) ! 0 as ~h ! ~0: 149 (12.4.9) Substituting (12.4.7) in (12.4.9) gives f E (~r + ~h t) = f E (~r t) + Dt ~r L(~x t)  @~f E (~r t) +R5 ( ) (12.4.10) where R5 ( ) = R~ 1 ( )  @~f E (~r t)   h i + Dt ~r L (~x t) + R~ 1 ( ) R4 ~h( ) : Thus R5 ( ) ! 0 as  ! 0. Moreover, f E (~r t) = f L(~x t), so substituting (12.4.10) in (12.4.8) gives h i f L(~x t +  ) = f L(~x t) +  @t f E (~r t) + Dt ~r L(~x t)  @~f E (~r t) + R3 ( ) + R5( )] (12.4.11) where R3( ) + R5( ) ! 0 as  ! 0. From 12.4.11 it follows immediately that 8 > L L E ~ E > > < Dt f (~x t) = @t f (~r t) + Dt ~r (~x t)  @ f (~r t) > > > : where ~r = ~r L(~x t) ( or ~x = ~x E (~r t)): (12.4.12) The formal mathematical relationship between labels and positions is symmetrical. That is, we can interpret ~r as label and ~x as position. Therefore, we can infer immediately from (12.3.8) that ~ L(~x t) @t f E (~r t) = Dt f L(~x t) + @t ~x E (~r t)  Df where r = ~r L(~x t) (or ~x = ~x E (~r t)): (12.4.13) 150 CHAPTER 12. EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM Alternatively, (12.4.13) can be computed in the same way as was (12.4.12). Note the consequence of (12.4.12, 12.4.13) that for any physical quantity f , ~ L(~x t) = 0: Dt ~r L(~x t)  @~f E (~r t) + @t ~x E (~r t)  Df (12.4.14) The notation has become somewhat cumbersome, and can be greatly simplied by introducing some new physical quantities. Denition 12.4.45 Let ~r be the physical quantity \particle position" (see bottom of ~ , @~f , page 146) and let f be any physical quantity. The physical quantities ~v, ~a , Df Dt f , @t f are dened as follows: ~vL(~x t) ~aL (~x t) ~ )L(~x t) (Df (@~f )E (~r t) (Dt f )L(~x t) (@t f )E (~r t) = Dt ~r L (~x t) (12.4.15) = Dt~vL(~x t) ~ L(~x t) = Df = @~f E (~r t) (12.4.16) = Dtf L(~x t) (12.4.17) (12.4.18) (12.4.19) = @t f E (~r t): (12.4.20) The names of these physical quantities are as follows: ~v = particle velocity ~a = particle acceleration ~ = label gradient of f Df @~f = spatial gradient of f Dtf = substantial time derivative of f = material time derivative of f = time derivative moving with the material. @t f = partial time derivative of f . 12.4. DERIVATIVES OF PHYSICAL QUANTITIES 151 With these denitions, the relations among the derivatives of a physical quantity become ~v ~a ~ Df Dtf @~f @t f = Dt ~r = Dt~v ~ r  @~f = D~ = @t f + ~v  @~f ~ = @~ ~x  Df ~ = Dt f + @t ~x  Df (12.4.21) (12.4.22) (12.4.23) (12.4.24) (12.4.25) (12.4.26) The function ~vE is called the Eulerian description of the motion of the continuum.1 If the Lagrangian description is known, the Eulerian description can be obtained from (12.4.21). The converse is true, in the following sense. Remark 12.4.56 Suppose the Eulerian description ~vE of a continuum is given. Suppose that at one instant t0 , the function ~r L( t0) is known that is, for each ~x, the position at time t0 of the particle labelled ~x is known. Then the Lagrangian description of the motion can be calculated. That is, the position ~r L(~x t) of the particle labelled ~x can be calculated for every time t. Proof: By denition of ~v, we have for ~r = ~r L(~x t) the relation Dt ~r L(~x t) = ~vE (~r t) or h i Dt ~r L(~x t) = ~vE ~r L(~x t) t : (12.4.27) But (12.4.27) is an ordinary dierential equation in t if ~x is xed. Therefore, for any xed particle label ~x, we can solve (12.4.27) for ~r L(~x t) if we know ~r L(~x t0 ) for one t0 . QED Corollary 12.4.43 If ~vE is known, then the Lagrangian description of the motion can be calculated without further information if the particles are \t0 -position labelled", i.e. if their labels are their positions at some tme t0 . 1 If @t~vE = ~0, the motion is called \steady". 152 CHAPTER 12. EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM Proof: With t0-position labelling ~r L (~x t0) = ~x, so ~r L(~x t0) is known for all particle labels ~x. 12.5 Rigid body motion For this discussion we need a lemma about linearity: Lemma 12.5.27 Suppose V and W are real vector spaces and f : V ! W . Suppose f has the following properties, for any nonzero ~x ~y ~z 2 V : i) f (~0) = ~0 ii) f (;~x) = ;f (~x) iii) f (a~x) = af (~x) for any positive real a (i.e., a 2 R a > 0). iv) If ~x + ~y + ~z = ~0 then f (~x) + f (~y) + f (~z) = ~0. In that case, f is linear. Proof: a) For any c 2 R and ~v 2 V , f (c~v) = cf (~v) a1) If c~v = ~0, either c = 0 or ~v = ~0, so f (c~v) = ~0 = cf (~v). a2) If c~v 6= ~0 and c > 0, use iii) above. a3) If c~v = 6 ~0 and c < 0, then f ((;c)~v) = (;c)f (~v) by iii) above. But f ((;c)~v) = f (;c~v) = ;f (c~v) by ii) above. Hence ;f (c~v) = ;cf (~v). b) For any ~u~v 2 V , f (~u + ~v) = f (~u) + f (~v). b1) If ~u or ~v is ~0, one of f (~u) and f (~v) is ~0 by i) above. E.g., if ~u = ~0, f (~u + ~v) = f (~v) = f (~u) + f (~v). 12.5. RIGID BODY MOTION 153 b2) If ~u and ~v are nonzero but ~u + ~v = ~0, then ~v = ;~u so f (~u + ~v) = f (~0) = ~0 = f (~u) ; f (~u) = f (~u) + f (;~u) = f (~u) + f (~v). b3) If ~u, ~v and ~u +~v are nonzero, let w~ = ~u +~v. Then ~u +~v +(;w~ ) = ~0 so by iv), f (~u)+ f (~v)+ f (;w~ ) = ~0. By ii) this is f (~u)+ f (~v) ; f (w~ ) = ~0, or f (w~ ) = f (~u) + f (~v). QED The motion of a continuum is called \rigid body motion" if the distance separating every pair of particles in the continuum is independent of time. We want to study rigid body motions. Step 1: Choose one particular particle in the continuum, and call it the pivot particle. Step 2: Choose t0 2 R. Choose a reference frame for real physical space P so that its origin is at the position of the pivot particle at time t0 . Step 3: Introduce t0 -position labelling to give the Lagrangian description of the motion, ~r L. Let H denote the open subset of L = P consisting of the particle labels. The pivot particle is labelled ~0, so ~0 2 H , and ~r L(~x t0 ) = ~x for all ~x 2 H: Step 4: For each t 2 R, dene a mapping ~rt : H ! P by ~rt (~x) = ~r L(~x t) ; ~r L(~0 t) and dene R~ : R ! P as R~ (t) = ~r L (~0 t). Thus ~r L(~x t) = R~ (t) + ~rt (~x) (12.5.1) for all ~x 2 H and t 2 R. Moreover, for any t 2 R and any particle labels ~x, ~y 2 H , ~r L(~x t) ; ~r L(~y t) = ~rt (~x) ; ~rt(~y), so, by the denition of a rigid body, k~rt (~x) ; ~rt(~y)k = k~x ; ~yk for all t 2 R and ~x ~y 2 H: (12.5.2) 154 CHAPTER 12. EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM Also, from the denition of R~ (t) and (12.5.1), ~rt (~0) = ~0 for all t 2 R: (12.5.3) The main result implied by the above is contained in Theorem 12.5.27 Suppose f : H ! P where H is an open subset of real Euclidean vector space P . Suppose f (~0) = ~0 and for any ~x ~y 2 H , kf (~x) ; f (~y)k = k~x ; ~yk. Then there is a unique L 2 #(P ) (i.e., L is linear and orthogonal) such that for all ~x 2 H $ f (~x) = ~x L= L(~x): (12.5.4) Proof: Lemma 1: kf (~x)k = k~xk for any ~x 2 H . Proof: kf (~x)k = kf (~x) ; ~0k = kf (~x) ; f (~0)k = k~x ; ~0k = k~xk . Lemma 2: f (~x)  f (~y) = ~x  ~y for all ~x ~y 2 H . Proof: kf (~x) ; f (~y)k2 = k~x ; ~yk2 so kf (~x)k2 ; 2f (~x)  f (~y) + kf (~y)k2 = k~xk2 ; 2~x  ~y + k~yk2 : But by lemma 1, kf (~x)k2 = k~xk2 and kf (~y)k2 = k~yk2 . Lemma 3: Suppose a1      aN 2 R, ~x1      ~xN 2 H , and ai~xi 2 H . Then f (ai~xi) = ai f (~xi). Proof: kf (ai~xi);ai f (~xi)k2 = kf (ai~xi)k2;2aj f (~xj )f (ai~xi) +kaif (~xi)k2 = kai~xi k2 ; 2aj f (~xj )  f (ai~xi)] +aiaj f (~xi)  f (~xj ). By lemma 2, this is kai~xi k2 ; 2aj xj  (ai~xi )] + aiaj (~xi  ~xj ) = kai~xi k2 ; aiaj ~xi  ~xj = 0. Lemma 4: Choose  > 0 so small that if k~xk <  then ~x 2 H . Dene L : P ! P as follows: i) L(~0) = ~0 ii) L(~x) = (k~xk=)f (~x=k~xk) if ~x 6= ~0. 12.5. RIGID BODY MOTION 155 Then L(~x) = f (~x) for all ~x 2 H , and L satises conditions i)-iv) of lemma 12.5.27. Proof: First, if ~x 2 H then in lemma 3 take N = 1, a0 = =k~xk, ~x1 = ~x. Lemma 3 then implies f (~x=k~xk) = =k~xkf (~x), and from ii) above, obviously L(~x) = f (~x). Now for i)-iv) of lemma 12.5.27. i) is obvious ii) L(;~x) = (k; ~xk=) f (;~x=k; ~xk) = (k~xk=)f (;~x=k~xk): In lemma 3 above, take N = 1, a0 = ;1, ~x1 = ~x=k~xk. Then f (;~x=k~xk) = ;f (~x=k~xk): Thus L(;~x) = ;L(~x). iii) L(a~x) = (ka~xk=)f (a~x=ka~xk) = a(k~xk=)f (a~x=ak~xk) = a(k~xk=)f (~x=k~xk) = aL(~x) if a > 0 and ~x 6= ~0. iv) Suppose ~x + ~y + ~z = ~0 and ~x ~y ~z nonzero. Choose  so small ( 2 R  > 0) that ~x + ~y + ~z 2 H . Then ~x + ~y + ~z = ~0, so, using lemma 3 , ~0 = f (~0) = f (~x + ~y + ~z) = f (~x)+ f (~y)+ f (~z). Since ~x 2 H , f (~x) = L(~x), and similarly for ~y ~z. Thus ~0 = L(~x) + L(~y) + L(~z). By iii) above, L(~x) = L(~x), etc., so ~0 = L(~x) + L(~y) + L(~z)]. Thus L(~x) + L(~y) + L(~z) = ~0. It follows from lemma 4 above and lemma 12.5.27 that L : P ! P is linear. For any ~x 6= ~0 in P , choose  so small ( 2 R,  > 0) that ~x 2 H . Then (kL(~x)k ; k~xk) = kL(~x)k ; k~xk = kL(~x)k ; k~xk = kf (~x)k ; k~xk = k~xk ; k~xk = 0. Thus kL(~x)k = k~xk for all ~x 2 P . Therefore L 2 #(P ). QED $ By theorem (12.5.27), for each t 2 R there is an orthogonal tensor L (t) 2 P  P such that in equation (12.5.1) we can write $ ~r L(~x t) = R~ (t) + ~x L (t) for all ~x 2 H: (12.5.5) Then clearly ~ r L(~x t) =$L (t) D~ for all ~x 2 H: (12.5.6) 156 CHAPTER 12. EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM ~ r L(~x t) is continuous in ~x and t, so $L (t) is a continuous function We are assuming that D~ $ of t. At the labelling time we have R~ (t0 ) = ~r L(~0 t0) = ~0 and so ~x L (t0) = ~r L (~x t0) = ~x. Thus $ $ R~ (t0) = ~0 (12.5.7) L (t0 ) = I P $ and det L (t0) = +1. But det L(t) depends continuously on t since L(t) does, and det L(t) is always +1 or ;1 since L(t) is orthogonal. Therefore det L(t) = +1 L(t) 2 #+(P ) for all t 2 R: (12.5.8) Every rigid motion can be written as in (12.5.5) with (12.5.7) and (12.5.8). Now let us nd the Eulerian description of the motion whose Lagrangian description is (12.5.5). We have $ ~vL(~x t) = Dt ~r L (~x t) = t R~ (t) + ~x  t L (t): (12.5.9) $ Note: Since R~ and L are functions of t alone, we write their time derivatives as t R~ and $ t L, so as not to confuse them with @t or Dt.] To obtain the Eulerian description from (12.5.9) we must express ~x in terms of ~r, where as usual ~r = ~r L(~x t) or $ ~r = R~ (t) + ~x L (t): (12.5.10) $ $T $;1 Since L2 #+(P ), L =L . Therefore the solution of (12.5.10) is h i $T ~x = ~r ; R~ (t)  L : (12.5.11) $ Note: It follows that x E (~r t) = ~r ; R~ (t)] L (t)T .] Substituting this (12.5.9) and using ~vL(~x t) = ~vE (~r t) gives h i $ $ ~vE (~r t) = t R~ (t) + ~r ; R~ (t)  L (t)T  t L (t): We introduce these abbreviations: V~ (t) = t R~ (t) (12.5.12) 12.5. RIGID BODY MOTION $ $ $ + (t) =L (t)T  t L (t): Then the Eulerian description of the motion is h i $ ~vE (~r t) = V~ (t) + ~r ; R~ (t)  + (t): Remark 12.5.57 t $L (t)]T = t LT (t). 157 (12.5.13) (12.5.14) Proof: $ $ $ $ For any ~x 2 P , ~x L (t) =LT (t)  ~x, so ~x  t L (t)] = t LT (t)]  ~x. But $ $ $ $ ~x  t L] = (t L)T  ~x, so t L (t)]T  ~x = t LT (t)]  ~x for all ~x 2 P . Hence, we have the remark Remark 12.5.58 $+ (t) = ; $+ (t)T . I.e., $+ (t) is antisymmetric. Proof: $ $ $ $ $ $ $ Since L (t) is orthogonal, LT (t) L (t) = I P for all t. Hence, (t LT )  L + LT $ $ T (t $L) = t $I P =$0 for all t. But $+ = $L  t $L and +T = (t $L)T  (LT )T = $ $ $ $ $ (t LT )  (L). Thus +T + += 0 . QED Now dene $ $ 1 (12.5.15) + (t) = 2 Ah2i + (t) where A is the unimodular alternating tensor we have chosen to orient real physical space P . By exercise 6, $ (12.5.16) + (t) = A  +~ (t): We claim that for any vector ~x 2 P we have $ ~x + (t) = +~ (t) ~x: (12.5.17) Proof: Take components relative to a pooob. Then  $   ~x + = ~x  A  +~ j = xi "ijk+k j     = "kij +k xi = Ah2i+~ ~x j = +~ ~x j : 158 CHAPTER 12. EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM Thus (12.5.14) can be written h i ~vE (~r t) = V~ (t) + +~ (t) ~r ; R~ (t) ~  V = tR~ : (12.5.18) The vector +~ (t) is called the angular velocity of the motion relative to the pivot particle. Obviously V~ (t) is the velocity of the pivot particle, and R~ (t) is its position. An obvious question not answered by the foregoing discussion is this: if we are given two arbitrary dierentiable functions R~ : R ! P , +~ : R ! P , and dene V~ = t R~ , then (12.5.18) gives the Eulerian description of a continuum motion. Is this always a rigid body motion, or do rigid body motions produce particular kinds of functions R~ : R ! P and +~ : R ! P ? In fact, any choice of R~ and +~ will make (12.5.18) the Eulerian description of a $ $ rigid body motion. To see this, use the given + (t) as in (12.5.16). + (t) is, of course, $ antisymmetric. Now dene L (t) as follows: $ $ $ t L = L  + : (12.5.19) $ $ $ (This is obtained from 12.5.13 as if LT =L;1, but we don't assume that.) We obtain L (t) by solving (12.5.19) with the initial condition $ $ (12.5.20) L (t0 ) = I P : Remark 12.5.59 $L (t) obtained from (12.5.19,12.5.20) is proper orthogonal. Proof: $ $T  $ $T $  $T t L  L = t L  L + L  t L $ $ $T $ $ $T = L+L +L L+ $ $ $T $ $T $T = L+L +L+ L $ $ $T  $T $ $ $T $ = L  + + +  L =L  O  L =O : $ $ $ Thus, L (t) L (t)T is independent of t. At t = t0, it is I P , so it is always $ $ $ I P . This proves L (t) 2 #(P ). And det L (t) is continuous in t and always 1 and +1 at t0, so it is always +1. Thus $L (t) 2 #+(P ) for all t. QED 12.6. RELATING CONTINUUM MODELS TO REAL MATERIALS 159 Now we can write (12.5.18) in the form (12.5.14). As we have seen in deriving (12.5.18), the motion whose Lagrangian description is (12.5.5) then has Eulerian description (12.5.14) or, equivalently (12.5.18). But as long as L(t) 2 #+(P ) for all t, (12.5.5) is a rigid-body $ motion. Indeed, for any ~x and ~y 2 P , k ~r L(~x t) ; ~r L(~y t)k = k(~x ; ~y) L (t)k = k~x ; ~yk if L(t) 2 #(P ). 12.6 Relating continuum models to real materials In a rock there is some jumping of atoms out of lattice positions, i.e., \self-diusion", but crystal structures stay mostly intact. In water and air, however, individual molecules move at random. After t seconds, two water molecules which were initially neighbors are separated, on average, by about (10;5t)1=2 cm and two air molecules at sea level are separated by (0:2t)1=2 cm. What is the physical meaning of a \particle" in water or air? We cannot mean a molecule, because then ~r L(~x t) would quickly become a very discontinuous function of ~x. In reality, continuum models are \two scale" approximations. To see what we mean, let us examine how to make a continuum model of a gas in which the individual molecules are moderately well modelled as randomly colliding very small hard spheres, with all their mass at the center. Let B (~r ) be the open ball of radius  with center at ~r. Let Mt B (~r )] be the sum of the masses of all the molecules whose centers are in B (~r ) at time t. Let jB (~r )j = 43=3 = volume of B (~r ). The average density in B (~r ) at time t is hi (~r t) = jB (~r1 )j Mt B (~r )] : (12.6.1) To use a continuum model, we must be able to choose  so large that the jumps in hi (~r t) as a function of ~r and t, which occur when individual molecules enter or leave B (~r ) as ~r and t vary, are very small fractions of hi (~r t). But at the same time,  must be so small that (12.6.1) does not sample very dierent physical conditions inside B (~r ). That is,  must be so small that k~rk can be many times  and yet hi (~r + ~r t) 160 CHAPTER 12. EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM λ r Figure 12.1: will dier from hi (~r t) by only a very small fraction. Therefore, the average physical conditions must not change appreciably over a length . For example, if we want to use a continuum approximation to study sound waves of wavelength  in air, we must use  <<  in (12.6.1), and yet we must have  >> average distance between nearest neighbors. Actually, to do all of continuum mechanics accurately (including momentum and heat) we must have  >> mean free path of air molecules. We cannot nd such a  to use in (12.6.1) unless  >> mean free path of air molecules. This mfp varies from 10;5 cm at sea level to 10 cm at 100 km altitude, and determines the shortest sound wave treatable by continuum mechanics. In the spirit of (9.2.3), we would dene the average momentum in B (~r ) at time t as (12.6.2) hp~i (~r t) = jB (~r1 )j P~t B (~r )] where P~t B (~r )] is the sum of the momenta m~v for all the molecules in B (~r ) at time t. Then we dene the average velocity at ~r t to be ~ h~vi (~r t) = hhp~ii ((~~rr tt)) = MPt BB((~r~r)])] : (12.6.3) t If  is the shortest length scale in our gas over which occur appreciable fractional changes in average physical properties, and if  >> mean free path, we can choose any  in mfp <<  << , and model our gas as a continuum whose Eulerian velocity function is ~vE (~r t) = h~vi (~r t): (12.6.4) 12.6. RELATING CONTINUUM MODELS TO REAL MATERIALS 161 The Lagrangian description of this continuum is obtained by choosing a labelling scheme and then integrating (12.4.27) with ~vE given by (12.6.4). Usually t0-position labelling is convenient. Evidently in a gas, the \particles" of the continuum model are ctitious. In a solid, they can be thought of as individual molecules, at least for times short enough that self diusion is unimportant. An example of the breakdown of the continuum model is a shock wave in air. They are considerably less thick than one mean free path. In fact, continuum models work better inside shocks than is at present understood. 162 CHAPTER 12. EULERIAN AND LAGRANGIAN DESCRIPTIONS OF A CONTINUUM Chapter 13 Conservation Laws in a Continuum In this whole chapter we will use the following notation. (L AL) is oriented label space and (P AP ) is oriented real physical space. H is the open subset of L consisting of the labels ~x of all the particles making up the continuum and H 0 is any open subset of H with piecewise smooth boundary @H 0 (i.e. @H 0 consists of a nite number of smooth pieces). The open set in P occupied by the particles at time t will be written K (t), and the open subset of K (t) consisting of the particles with labels in H 0 will be written K 0(t). As t varies, K (t) and K 0 (t) will change position in P , but will always contain the same particles. The sets K (t) and K 0 (t) are said to move with the material. The outward normal velocity W (~r t) of @K 0 (t) at ~r 2 @K 0 (t), to be used in (11.6.1), is W (~r t) = n^P (~r t)  ~vE (~r t), where ~v is the particle velocity and n^ P (~r t) is the unit outward normal to @K 0 (t) at ~r 2 @K 0 (t). We will use dVL(~x) to denote a very small open subset of H 0 containing the particle label ~x. We will denote the positive numerical volume in L of this small set by the same symbol, dVL(~x). We use dVP (~r) to denote a very small open subset of K 0(t) containing the particle position ~r we also use dVP (~r) for the positive numerical volume of this small set in P . We will always assume when using this notation for the sets that ~r = ~r L(~x t) and dVP (~r) =  ~r L( t)]dVL(~x)], so dVP (~r) consists of the positions at time t of the particles whose labels are in dVL(~x). It follows that, for the numerical volumes,   ~ r L(~x t) dVL(~x): dVP (~r) = det D~ 163 (13.0.1) 164 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM We assume that the reader knows this Jacobian formula from calculus. (We will accidentally prove it later.) Note that L means Lagrangian description and L means label space. 13.1 Mass conservation 13.1.1 Lagrangian form The mass of the collection of particles with labels in dVL(~x) is dm = E (~r t)dVP (~r) (13.1.1) where E (~r t) is the mass per unit of volume in physical space. This is our mathematical model of how mass is distributed in a continuum. Because of (13.0.1), we can also write dm = ~L(~x t)dVL(~x) (13.1.2) ~ r L (~x t)]. But E (~r t) = L (~x t) since we are assuming if we set ~L(~x t) = E (~r t)j det D~ ~ r L(~x t)j = (j det D~ ~ r j)L(~x t), so ~r = ~r L(~x t). Therefore ~L (~x t) = L(~x t)j det D~ ~ r j: ~ = j det D~ (13.1.3) The physical quantity ~ is the mass per unit of volume in label space. The material with labels in H 0 always consists of the same particles, so it cannot change its mass. Therefore, dm in (13.1.2) must be independent of time. Since dVL(~x) is dened in a way independent of time, it follows that ~L(~x t) must be independent of t. Therefore, for all ~x 2 H and all t, t0 2 R, ~L (~x t) = ~L (~x t0 ): (13.1.4) An equivalent equation is Dt ~L(~x t) = 0, or (Dt )L(~x t) = 0, so Dt ~ = 0: (13.1.5) 13.1. MASS CONSERVATION 165 Either equation (13.1.4) or (13.1.5) expresses the content of a physical law, the law of conservation of mass. The mathematical identity (13.1.3) makes this law useful. From (13.1.3) and (13.1.4) we deduce     ~ r L(~x t) = L(~x t0) det D~ ~ r L (~x t0) L (~x t) det D~ (13.1.6) for all ~x 2 H and all t, t0 2 R. If we use t0 -position labelling, then r~L(~x t0) = ~x, ~ r L (~x t0) =$I P and det D~ ~ r L(~x t0) = det I$p= det IP = 1. Thus, with t0-position so D~ labelling,   ~ r L(~x t) = L (~x t0): L(~x t) det D~ (13.1.7) 13.1.2 Eulerian form of mass conservation An informal physical argument like (13.1.1, 13.1.2) is possible here, but rather confusing. The bookkeeping becomes clearer if we account for real, nite masses and volumes. The total mass in the set K 0(t) in physical space P at time t is M E K 0 (t)] = Z K 0 (t) dVP (~r)E (~r t): (13.1.8) The physical law of mass conservation is d M E K 0 (t)] = 0 (13.1.9) dt because K 0(t) always consists of the same material. Using (11.6.1) with w = n^P ~vE (~r t) we have the mathematical identity d M E K 0(t)] = Z dV (~r)@ E (~r t) + Z dAP (~r)^nP  ~vPE (~r t) P t dt K 0 (t) @K 0 (t) where n^ P is the unit outward normal to @K 0 (t). An application of Gauss's theorem to the surface integral gives d M E K 0(t)] = Z dV (~r) h@  + @~  (~v)iE (~r t): (13.1.10) P t dt K 0(t) Comparing (13.1.9) and (13.1.10), we see that if K 0 = K 0 (t) then Z K h ~  (~v)iE (~r t) = 0: dV ( ~ r ) @  + @ P t 0 (13.1.11) 166 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM Now let K 0 be any open subset of K (t), with piecewise smooth boundary @K 0 . Let H 0 be the set of labels belonging to particles which are in K 0 at time t. Then for this set of particles, K 0(t) = K 0, so (13.1.11) holds. In other words, (13.1.11) holds for every open subset K 0 of K (t), as long as @K 0 is piecewise smooth. Therefore, by the vanishing integral theorem, @t  + @~  (~v)]E (~r t) = 0 for all t 2 R and all ~r 2 K (t). Therefore @t  + @~  (~v) = 0: (13.1.12) Equation (13.1.12) is the Eulerian form of the law of mass conservation. It is called the \continuity equation". Since @~  (~v) = (@~)  ~v + (@~  ~v) and Dt  = @t  + ~v  (@~), the continuity equation can also be written Dt  + (@~  ~v) = 0: (13.1.13) By the chain rule for ordinary dierentiation, Dt ln  = Dt=, so Dt (ln ) + @~  ~v = 0: (13.1.14) A motion of a continuum is called incompressible if it makes L(~x t) independent of t at each particle ~x. The density is constant at each particle, but can vary from particle to particle. A material is incompressible if it is incapable of any motion except incompressible motion. A motion is incompressible i Dt L(~x t) = 0 for all ~x 2 H and t 2 R, i.e. i Dt = 0. Comparing (13.1.13), the motion is incompressible i @~  ~v = 0: (13.1.15) An extremely useful consequence of the continuity equation follows from Lemma 13.1.28 Suppose U and V are Euclidean spaces, D is an open subset of U , and f~ : D ! U and ~g : D ! V are dierentiable. Then @~  (f~~g) = (@~  f~)~g + f~  (@~~g): (13.1.16) 13.1. MASS CONSERVATION 167 Proof: Take components relative to orthonormal bases in U and V . Then (13.1.16) is equivalent to @i (figj ) = (@i fi)gj + fi(@i gj ). But this is the elementary rule for the partial derivative of a product. Remark 13.1.60 Suppose f~ is any physical quantity taking values in a Euclidean space V . Suppose K 0(t) is any open set moving with the material in a continuum (i.e., always consisting of the same particles). Then d Z dV (~r)(f~)E (~r t) = Z dV (~r)(D f~)E (~r t): (13.1.17) P t dt K 0(t) P K 0 (t) Proof: By (11.6.1), we have d Z dV (~r) f~E (~r t) = Z dV (~r)@ (f~)E (~r t) p t dt K 0(tZ) P K 0 (t) + 0 dAP (~r)^nP  (~vf~)E (~r t) @K (t) where n^P is the unit outward normal to @K 0 (t) and dAP is an element of area. Applying Gauss's theorem to the surface integral gives d Z dV (~r)(f~)E (~r t) dt K 0(tZ) P h  iE = 0 dVP (~r) @t (f~) + @~  ~vf~ (~r t): K (t) h i The integrand is (@t )f~ + (@t f~) + @~  (~v) f~ + ~v  (@~f~) because of lemma (13.1.28). This integrand can therefore be written @t  + @~  (~v)]f~ + @t f~ + ~v  (@~f~)]. From (12.4.21-12.4.26) and (13.1.12) this is 0f~ + Dt f~, which proves (13.1.17). An alternative proof of (13.1.17) may be physically more enlightening. We use (13.0.1) to change the variable of integration in (13.1.17) to ~x instead of ~r. Then Z K 0 (t) dVP (~r)E (~r t)f~E (~r t) 168 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM Z   ~ r L(~x t) L (~x t)f~L(~x t) = 0 dVL(~x) det D~ H (use 13.1.3) Z = dV (~x)~L(~x)f~L(~x t): 0 L H Since H 0 and ~L(~x) do not vary with t, (dVL(~x)~L(~x) = dm) d Z (t)dV (~r)(f~)E (~r t) = Z dV (~x)~L (~x)D f~L(~x t) L t dt K 0 Z P H 0  ~ r L(~x t) L (~x t)(Dt f~)L (~x t) = 0 dVL(~x) det D~ H Z = dVP (~r)E (~r t)(Dtf~)E (~r t): K (t) QED 13.2 Conservation of momentum 13.2.1 Eulerian form We use the notation introduced on page 163. We denote by P~ E K 0 (t)] the total momentum at time t of the material in subset K 0(t) of P at time t. This material consists of the particles whose labels are in H 0 and whose positions at time t are in K 0 (t). The law of conservation of momentum is d P~ E K 0(t)] = F~ E K 0(t)]  dt (13.2.1) where F~ E K 0 (t)] is the total force at time t on the material in K 0 (t). As it stands, (13.2.1) is essentially a denition of F~ E K 0(t)], and our task is to calculate this force and then to extract from (13.2.1) a local form of the law of conservation of momentum in the same way that (13.1.12) was extracted from (13.1.9). To do this, we also need an expression for P~ E K 0(t)], but this is simple. The mass in dVP (~r) is dVP (~r)E (~r t) at time t, so its momentum is dVP (~r)E (~r t)~vE (~r t). Summing over all the volume elements dVP (~r) in K 0(t) gives Z P~ E K 0 (t)] = 0 dVP (~r)(~v)E (~r t) (13.2.2) K (t) 13.2. CONSERVATION OF MOMENTUM 169 If we apply to this equation both (13.1.17) and (12.4.21-12.4.26) we obtain the identity d P~ E K 0(t)] = Z dV (~r)(~a)E (~r t) (13.2.3) P dt K 0(t) where ~a = Dt~v = @t~v + ~v  @~~v = particle acceleration. Finding F~ E K 0(t)] is less simple. The expression for it is one of the major eighteenth century advances in mechanics. If , is the gravitation potential, so that ~g = ;@~, is the local acceleration of gravity, then the gravitational force on the matter in dVP (~r) at time t is dVP (~r)E (~r t)~gE (~r t). If E~ and B~ are the electric and magnetic elds, and Q and J~ are the electric charge density and electric current density then the electromagnetic force on the matter in dVP (~r) at time t is dVP (~r)(QE~ + J~ B~ )E (~r t). The total gravitational plus electromagnetic force on the matter in dVP (~r) is where dF~B (~r) = dVP (~r)f~E (~r t) (13.2.4) ~ f~ = ~g + QE~ + J~ B: (13.2.5) The total gravitational plus electromagnetic force on the matter in K 0 (t) is F~BE K 0 (t)] = Z K 0 (t) dVP (~r)f~E (~r t): (13.2.6) This is called the \body force" on K 0(t), and f~ is the body force density per unit of physical volume, the physical density of body force. If we accept F~BE as a good model for F~ E , the only force acting on a cubic centimeter of ocean or rock is ~g (assuming E~ = B~ = ~0), and yet neither is observed to fall at 981cm2=sec. Something important is still missing in our model of F~ E . The physical origin of our di"culty is clear. The forces in (13.2.5) are calculated from the average distribution of molecules and charge carriers, as in Figure 12.1. In addition to these long range average forces, we expect that the molecules just outside @K 0 (t), will exert forces on the molecules just inside @K 0 (t), and these contribute to F~ E K 0(t)]. Also, in gases and, to some extent, in liquids, individual molecules will cross @K 0 (t), and the entering molecules may have, on average, dierent momenta from the exiting molecules. 170 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM S ^ nP ( r ) r nt o Fr Ba ck dAP ( r ) Figure 13.1: This will result in a net contribution to the rate of change of the total momentum in K 0(t). Therefore, it is part of dP~ K 0 (t)]=dt and hence, by (13.2.1), part of F~ E K 0 (t)]. Both the intermolecular force and the momentum transfer by molecular motion can be modelled as follows: Fix time t and choose a xed ~r 2 K (t). Choose a very small nearby plane surface S in K (t) such that S passes through ~r. Choose an even smaller surface dAP (~r) which passes through ~r and lies in S . Arbitrarily designate one side of S as its \front", and let n^P (~r) be the unit normal to S extending in front of S . Assume that S is so small that the molecular statistics do not change appreciably across S , but that S is considerably larger in diameter than the intermolecular distance or the mean free path. Then the total force exerted by the molecules just in front of dAP (~r) on the molecules just behind dAP (~r) will be proportional1 to the area of dAP (~r). We write this area as dAP (~r). The proportionality constant is a vector, which we write S~force. It depends on ~r and t, and it may also depend on the orientation of the surface S , i.e. on the unit normal n^P (~r). If the material is a gas or liquid, there will also be a net transfer of momentum from front to back across dAP (~r) because molecules cross dAP (~r) and collide just after crossing, and the population of molecules one mean free path in front of dAP (~r) may be 1 if the linear dimension of dA is >> distance between molecules 13.2. CONSERVATION OF MOMENTUM 171 ^n P r dAP (r ) K’ ( t) Figure 13.2: statistically dierent from the population one mean free path behind dAP (~r). The net rate of momentum transfer from just in front of dA(~r) to just behind dAP (~r) will produce a time rate of change of momentum of the material just behind dAP (~r) that is, it will exert a net force on that material. This force will also be proportional to dAP (~r), and the proportionality constant is another vector, which we write S~mfp. This vector also depends on ~r,t and n^P (~r). The sum S~ (~r t n^P ) = S~force(~r t n^P ) + S~mfp (~r t n^P ) is called the stress on the surface (S n^P ). The total force exerted by the material just in front of dAP (~r) on the material just behind dAP (~r) is dF~S (~r) = dAP (~r)S~ (~r t n^ P ): (13.2.7) This is called the surface force on dAP (~r). Summing (13.2.7) over all the elements of area dAP (~r) on @K 0 (t) gives F~SE K 0(t)] = Z @K 0(t) dAP (~r)S~ ~r t n^ P (~r t)] (13.2.8) where n^P (~r t) is the unit outward normal to @K 0 (t) at ~r 2 @K 0 (t). The expression (13.2.8) is called the surface force on K 0 (t). It is the total force exerted on the material just inside @K 0 (t) by the material just outside @K 0 (t). The total force on K 0(t) is the sum 172 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM of the body and surface forces, F~ E K 0(t)] = F~BE K 0(t)] + F~SE K 0(t)], so F~ E K 0 (t)] = Z K 0 (t) dVP (~r)f~E (~r t) + Z @K 0(t) dAP (~r)S~ ~r t n^ P (~r t)] : (13.2.9) Combining the physical law (13.2.1) with the mathematical expressions (13.2.3) and (13.2.9) gives Z K dV (~r)(~a ; f~)E (~r t) = 0 P Z @K 0 dAP (~r)S~ ~r t n^ P (~r)] : (13.2.10) where K 0 = K 0 (t) and n^P (~r) is the unit outward normal to @K 0 at ~r 2 @K 0 . As with (13.1.11), (13.2.10) must hold for every open subset K 0 of K (t), as long as @K 0 is piecewise smooth. This conclusion is forced on us if we accept the mathematical models (13.2.9) for F~ E K 0(t)] and (13.2.2) for P E K 0(t)]. To convert (13.2.10) to a local equation, valid for all ~r 2 K (t) at all times t, (i.e., to \remove the integral signs") we would like to invoke the vanishing integral theorem, as we did in going from (13.1.11) to (13.1.12). The surface integral in (13.2.10) prevents this. Even worse, (13.2.10) makes our model look mathematically self-contradictory, or internally inconsistent. Suppose that K 0 shrinks to a point while preserving its shape. Let  be a typical linear dimension of K 0 . Then the left side of (13.2.10) seems to go to zero like 3, while the right side goes to zero like 2. How can they be equal for all  > 0? Cauchy resolved the apparent contradiction in 1827. He argued that the right side of (13.2.10) can be expanded in a power series in , and the validity of (13.2.10) for all  shows that the rst term in this power series, the 2 term, must vanish. In modern language, Cauchy showed that this can happened i at every instant t, at every ~r 2 K (t), $E there is a unique tensor S (~r t) 2 P  P such that for each unit vector n^ 2 P , $E S~ (~r t n^) = n^ S (~r t): (13.2.11) If equation (13.2.11) is true, we can substitute it in the surface integral in (13.2.10) and use Gauss's theorem to write Z Z E $E ~ $ dA ( ~ r )^ n ( ~ r )  dV ( ~ r ) @ ( ~ r  t ) = (~r t): S S P P P @K 0 K0 13.2. CONSERVATION OF MOMENTUM Then (13.2.10) becomes Z 173  $E ~ ~ dV (~r) ~a ; f ; @  S (~r t) = 0: K0 Since this is true for all open subsets K 0 of K (t) with piecewise smooth boundaries @K 0 , the vanishing integral theorem implies that the integrand vanishes for all ~r 2 K (t) if, as we shall assume, it is continuous. Therefore $ ~a = @~ S +f~: (13.2.12) $E This is the Eulerian form of the momentum equation. The tensor S (~r t) called the $ Cauchy stress tensor at (~r t). The physical quantity S is also called the Cauchy stress tensor. The argument which led Cauchy from (13.2.10) to (13.2.11) is fundamental to continuum mechanics, so we examine it in detail. In (13.2.10), t appears only as a parameter, so we will ignore it. Then (13.2.10) implies (13.2.11) because of Theorem 13.2.28 (Cauchy's Theorem) Suppose U and V are Euclidean spaces, K is an open subset of U , NU = set of all unit vectors n^ 2 U , f~ : K ! V is continuous, and S~ : K NU ! V is continuous. Suppose that for any open subset K 0 of K whose boundary @K 0 is piecewise smooth, we have Z K dV (~r)f~(~r) = 0 U Z @K 0 dAU (~r)S~ ~r n^U (~r)]  (13.2.13) where n^U (~r) is the unit outward normal to @K 0 at ~r 2 @K 0 . Then for each ~r 2 K there $ is a unique S (~r) 2 U  V such that for all n^ 2 NU , $ S~ (~r n^ ) = n^ S (~r): (13.2.14) This theorem is true for any value of dim U  2. We give the proof only for dim U = 3, the case of interest to us. For other values of dim U , a proof can be given in exactly the same way except that ~u1 ~u2 must be replaced by the vector in Exercise 7.] Proof: 174 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM K Figure 13.3: Two lemmas are required. Lemma 13.2.29 Suppose f~ and S~ satisfy the hypotheses of theorem 13.2.28. Let ~r0 be any point in K and let K 0 be any open bounded (i.e., there is a real M such that ~r 2 K 0 ) k~rk  M ) subset of U , with piecewise smooth boundary @K 0 . We don't need K 0 K . Then Z dAU (~r)S~ ~r0  n^ U (~r)] = ~0V (13.2.15) 0 @K if n^U (~r) is the unit outward normal to @K 0 at ~r 2 @K 0 and dAU (~r) is the element of area on @K 0 . Proof of Lemma 13.2.29: For any real  in 0 <  < 1, dene ~r : U ! U by ~r (~r) = ~r0 + (~r ; ~r0 ) for all ~r 2 U . Since ~r (~r) ; ~r0 = (~r ; ~r0 ), ~r shrinks U uniformly toward ~r0 by the factor . The diagram above is for   1=2. Dene K 0 = ~r (K 0) so @K 0 = ~r (@K 0 ). Choose ~r 2 @K 0 and let ~r = ~r (~r). Let dA(~r) denote a small nearly plane patch of surface in @K 0 , with ~r 2 dA(~r), and use dA(~r) both as the name of this set and as the numerical value of its area. Let the set dA (~r ) be dened as ~r dA(~r)], and denote its area also by dA (~r ). Then by 13.2. CONSERVATION OF MOMENTUM 175 geometric similarity dA (~r ) = 2dA(~r): (13.2.16) Let n^(~r) be the unit outward normal to @K 0 at ~r, and let n^ (~r ) be the unit outward normal to @K 0 at ~r . By similarity, n^(~r) and n^ (~r ) point in the same direction. Being unit vectors, they are equal: n^ (~r ) = n^(~r): Since ~r0 is xed, it follows that Z ~ ^ (~r )] = 2 0 dA (~r )S ~r0  n @K Z @K 0 dA(~r)S~ ~r0  n^ (~r)] : (13.2.17) (13.2.18) If  is small enough, K 0 K . Then, by hypothesis, we have (13.2.13) with K 0 and @K 0 replaced by K 0 and @K 0 . Therefore Z ~ 0 dV (~r)f (~r) = K Z n~ o ~ ~r0 n^ (~r )] dA ( ~ r ) S  ~ r  n ^ ( ~ r )] ; S K0 Z + 0 dA (~r )S~ ~r0 n^ (~r )] : @K From (13.2.18) it follows that Z Z ~ ~r0  n^ (~r)] = 12 dV (~r)f~(~r) dA ( ~ r ) S  K0 @K 0 Z + 12 0 dA (~r )  @K  ~S ~r0 n^ (~r )] ; S~ ~r  n^ (~r )] : (13.2.19) Let mS~ () = maximum value of kS~ (~r0 n^) ; S~ (~r n^)k for all ~r 2 @K 0 and all n^ 2 NU . Let mf~() = maximum value of jf~(~r)j for all ~r 2 K 0 . Let j@K 0 j = area of @K 0 , j@K 0 j = area of @K 0 . Let jK 0 j = volume of K 0 , jK 0j = volume of K 0. Then j@K 0 j = 2j@K 0 j and jK 0 j = 3jK 0 j, so (10.2.3) and (13.2.19) imply  Z  dA(~r)S~ ~r0 n^(~r)]  jK 0jmf~() + j@K 0 jmS~ (): @K 0 (13.2.20) 176 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM As  ! 0, mf~() remains bounded (in fact ! kf (~r0)k) and mS~ () ! 0 because S~ : K NU ! V is continuous. Therefore, as  ! 0, the right side of (13.2.20) ! 0. Inequality (13.2.20) is true for all su"ciently small  > 0, and the left side is non-negative and independent of . Therefore the left side must be 0. This proves (13.2.15) and hence proves lemma 13.2.29. We also need Lemma 13.2.30 Suppose S~ : NU ! V . Suppose that for any open set K 0 with piecewise smooth boundary @K 0 , S~ satises Z @K 0 dA(~r)S~ ^n(~r)] = ~0V (13.2.21) where n^ (~r) is the unit outward normal to @K 0 as ~r 2 @K 0 . Suppose F : U ! V is dened as follows:  ! F (~0U ) = ~0V and if ~u 6= ~0U  F (~u) = k~ukS~ k~~uuk : (13.2.22) Then F is linear. Proof of Lemma 13.2.30: a) F (;~u) = ;F (~u) for all ~u 2 U . To prove this, it su"ces to prove S~ (;n^ ) = ;S~ (^n) for all n^ 2 NU : (13.2.23) Let K 0 be the at rectangular box shown at upper right. For this box, (13.2.21) gives h i L2 S~ (^n) + L2S~ (;n^ ) + "L S~ (^n1) + S~ (;n^1 ) + S~ (^n2 ) + S~ (;n^2 ) = ~0V : Hold L xed and let " ! 0. Then divide by L2 and (13.2.23) is the result. b) If c 2 R and ~u 2 U , F (c~u) = cF (~u). i) If c = 0 or ~u = ~0U , this is obvious from F (~0U ) = ~0V . ii) If c > 0 and ~u 6= ~0U , F (c~u) = kc~ukkS~ (c~u=kc~uk) = ck~ukS~ (c~u=ck~uk) = ck~ukS~ (~u=k~uk) = cF (~u) . 13.2. CONSERVATION OF MOMENTUM 177 ε L ^ n2 ^ -n1 ^ -n ^ n ^ n 1 ^ -n2 L Figure 13.4: 178 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM iii) If c < 0, F (c~u) = ;F (;c~u) by a) above. But ;c > 0 so F (;c~u) = (;c)F (~u) by ii). Then F (c~u) = ;(;c)F (~u) = cF (~u). c) F (~u1 + ~u2) = F (~u1) + F (~u2) for all ~u1, ~u2 2 U . i) If ~u1 = ~0U , F (~u1 + ~u2) = F (~u2) = ~0V + F (~u2) = F (~u1) + F (~u2). ii) If ~u1 6= ~0U and ~u2 = c~u1 then F (~u1 + ~u2) = F (1 + c)~u1] = (1 + c)F (~u1) = F (~u1) + cF (~u1) = F (~u1) + F (c~u1) = F (~u1) + F (~u2). iii) If f~u1 ~u2g is linearly independent, let ~u3 = ;~u1 ; ~u2. We want to prove F (;~u3) = F (~u1) + F (~u2), or ;F (~u3) = F (~u1) + F (~u2), or F (~u1) + F (~u2) + F (~u3) = ~0V : (13.2.24) To prove (13.2.24) note that since ~u1 , ~u2 are linearly independent, we can dene the unit vector ^ = (~u1 ~u2)=k~u1 ~u2k. We place the plane of this paper so that it contains ~u1 and ~u2, and ^ points out of the paper. The vectors ~u1, ~u2, ~u3 form the three sides of a nondegenerate triangle in the plane of the paper. ^ ~ui is obtained by rotating ~ui 90 counterclockwise. If we rotate the triangle with sides ~u1 , ~u2, ~u3 90 counterclockwise, we obtain a triangle with sides ^ ~u1, ^ ~u2, ^ ~u3. The length of side ^ ~ui is k^ ~uik = k~uik, and ~ui is perpendicular to that side and points out of the triangle. Let K 0 be the right cylinder whose base is the triangle with sides ^ ~ui and whose generators perpendicular to the base have length L. The base and top of the cylinder have area A = k~u1 ~u2k=2 and their unit outward normals are ~ and ;~ . The three rectangular faces of K 0 have areas Lk~uik and unit outward normals ~u(i) =k~u(i)k. Applying (13.2.21) to this K 0 gives 3 X AS~ (^ ) + AS~ (;^) + Lk~uikS~ (~ui=k~uik) = ~0V : i=1 But S~ (^ ) = ;S~ (;^) so dividing by L and using (13.2.22) gives (13.2.23). Corollary 13.2.44 (to Lemma 13.2.30.) Under the hypotheses of lemma 13.2.30, there $ is a unique S 2 U  V such that for all n^ 2 NU $ S~ (^n) = n^ S : (13.2.25) 13.2. CONSERVATION OF MOMENTUM 179 u3 ν^ x u1 3 . u ^ν x u1 ν^ u2 ^ν x u2 L u 1 ^ν x u 3 ν^ x −ν^ ^ν ν^ x u Figure 13.5: 2 180 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM Proof: Existence. Take $S =F$, the tensor in U  V corresponding to the F 2 L(U ! $ $ V ) dened by 13.2.22. Then n^  S = n^  F = F (^n) = S~ (^n). Uniqueness. If n^  $S 1= n^ $S 2 for all n^ 2 NU then(cn^ ) S$1 = (cn^) $S 2 for all n^ 2 NU and c 2 R. But every ~ 2 U is cn^ for some c 2 R and n^ 2 NU , so $ $ $ ~u S 2 for all ~u 2 U . Hence S 1=S 2 . Now we return to the proof of Cauchy's theorem (theorem 13.2.28). If ~r0 is any xed point in K1 then by lemma 13.2.29, the function S~ (~r0 ) : NU ! V satises the hypothesis of lemma 13.2.30. Therefore by corollary 13.2.44 there $ is a unique S (~r0 ) 2 U  V such that for any unit vector n^ 2 U , S~ (~r0 )](^n) = $ n^ S (~r0). But S~ (~r0 )](^n) = S~ (~r0 n^ ) so we have (13.2.14). Having completed the proof of Cauchy's theorem, we have proved that if $E (13.2.10) holds for all K 0 then Cauchy's stress tensor S (~r t) exists and satises (13.2.11). And as we have seen, (13.2.11) leads automatically to the Eulerian momentum equation (13.2.12). It is important to have a clear phys$E ical picture of what the existence of a Cauchy stress tensor S (~r t) means. If dAP is any small nearly plane surface in the material in physical space P at time t, and if ~r 2 dAP , choose one side of dAP to be its front, and let n^P be the unit normal extending in front of dAP . Then the force dF~S exerted by the material just in front of dA on the material just behind dA is, at time t, $E dF~S = dAP n^P  S (~r t): (13.2.26) The stress, or force per unit area, exerted on the material just behind dA by the material just in front is $E S~ (~r t n^ P ) = n^P  S (~r t): (13.2.27) If K 0 is any open subset of K (t), with piecewise smooth boundary @K 0 , then at time t the total surface force on the matter in K 0 is, according to (13.2.8), 13.2. CONSERVATION OF MOMENTUM 181 ^n P r . front dAP back Figure 13.6: and (13.2.27), F~SE K 0] = Z @K $E ~ dA ( r )^ n ( ~ r )  (~r t): S P P 0 (13.2.28) Using Gauss's theorem, we can write this as F~SE K 0 ] = Z K E ~ $ dV ( ~ r ) @ (~r t): S P 0 (13.2.29) The total force on the matter in K 0 is, according to (13.2.9) and (13.2.29),  E Z ~F E K 0] = dVP (~r) f~ + @~ $ (13.2.30) S (~r t): 0 K If the particles are all motionless, ~r L(~x t) is independent of time, so ~vL(~x t) = Dt ~r L(~x t) = ~0, and ~aL(~x t) = Dt~vL(~x t) = ~0, and the momentum equation (13.2.12) reduces to ~0 = @~ S$ +f~ (13.2.31) the \static equilibrium equation." In this case, we see from (13.2.30) that the surface force on the matter in K 0 exactly balances the body force, for every open subset K 0 of K (t). If we have both ~a = ~0 and f~ = ~0, then $ @~ S = ~0: (13.2.32) $ The physical quantity S is called a \static stress eld" if it satises (13.2.32). An example is an iron doughnut in orbit around the sun (to cancel gravity). 182 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM Figure 13.7: Cut out a slice, close the gap by pressure, weld, and remove the pressure, as in gure 13.7 The resulting doughnut will contain a nonzero static stress eld. The net force on every lump of iron in the doughnut vanishes because the body force vanishes and the surface force sums to ~0. 13.2.2 Shear stress, pressure and the stress deviator We want to discuss some of the elementary properties of stress and the Cauchy stress tensor. In gure (13.6), choosing which side of dA is the \front" is called \orienting dA", and the pair (dA n^) is an oriented surface. Two oriented surfaces can be produced from dA, namely (dA n^ ), and (dA ;n^ ) depending on which side of dA we call the front. We want to study the state of stress at one instant t at one position ~r 2 K (t), so we will omit ~r t and also the superscript E . We will work only with Eulerian descriptions at the single point ~r and time t. Thus the stress S~ (~r t n^ ) on small nearly plane oriented $E $ area (dA n^) will be written S~ (^n), and S (~r t) will be written S . Denition 13.2.46 $ Sn(^n) := n^  S~ (^n) = n^  S n^  $  ~SS (^n) := S~ (^n) ; n^Sn(^n) = n^  $ S ;n^ n^  S n^ $ pn(^n) := ;Sn (^n) = ;n^  S n^: (13.2.33) (13.2.34) (13.2.35) The vector n^Sn(^n) is called the normal stress acting on the oriented surface (dA n^), the vector S~S (^n) is called the tangential or shear stress acting on (dA n^), and pn(^n) is the 13.2. CONSERVATION OF MOMENTUM 183 pressure acting on (dA n^ ). Corollary 13.2.45 n^  S~S (^n) = 0 S~ (^n) = n^Sn(^n) + S~S (^n) = ;n^ pn(^n) + S~S (^n): (13.2.36) (13.2.37) The shear stress is always perpendicular to n^, parallel to dA. The normal stress acts oppositely to n^ if pn(^n) > 0. To visualize S~ as a function of n^, note that the set NP of unit vectors n^ 2 P is precisely @B (~0 1), the spherical surface of radius 1 centered on ~0 in P . Thus S~ attaches to each n^ on the unit sphere NP a vector S~ (^n) whose radial part is n^SN (^n) and whose tangential part is S~S (^n). This suggests Denition 13.2.47 The average value of pn(^n) for all n^ 2 NP is called the average or mean pressure at ~r t. It is written hpni(~r t). We do not write the (~r t) in this chapter. Thus Z 1 hpni = 4 N dA(^n)pn(^n) P where dA(^n) is the element of area on NP . Remark 13.2.61 hpni = ;1=3 tr S$ (13.2.38) Proof: Let y^1 y^2, y^3 be an orthonormal basis for P , and take components relative to this basis. Then $ pn(^n) = ;n^  S n^ = ;niSij nj = ;Sij ninj : By the symmetry of NP , Z dA(^n)ni nj = 0 if i 6= j and Z Z Z Z 2 2 2 dA(^n)n21 = dA(^n)n22 = dA(^n)n23 = dA(^n) (n1 + n32 + n3) = 43 : NP NP NP NP NP 184 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM Therefore, Z NP dA(^n)ni nj = 43 ij and Z hpni = ; 41 Sij N dA(^n)ni nj = ; 13 ij Sij P $ 1 = ; Sii = ; 1 tr S : 3 3 $ Only part of S produces the shear stress S~S (^n). To see this we need a brief discussion of tensors in U  U , where U is any Euclidean space. Lemma 13.2.31 tr $I U = dim U . Proof: $ $ Let y^1     y^n be an orthonormal basis for U . Then I U = y^iy^i and tr I U = y^i  y^i = n = dim U . Lemma 13.2.32 If $S 2 U  U , tr $S =$I U h2i S$. Proof: $ $ $ $ With y^i as above, tr S = Sii = ij Sij = ( I U )ij Sij = I U h2i S . Denition 13.2.48 $S 2 U  U is \traceless", or a \deviator" if tr $S = 0. Corollary 13.2.46 S$ is a deviator , S$ is orthogronal to all isotropic tensors in U  U . Proof: $ $ The isotropic tensors are the scalar multiples of I U . But S is orthogonal to $ $ $ ^ I U ,S h2i I U = 0. Now use lemma 13.2.32. Remark 13.2.62 If S$2 U  U , there is exactly one isotropic tensor  $I U and one $ deviator S D such that $ $ $ (13.2.39) S=  I U + SD : 13.2. CONSERVATION OF MOMENTUM 185 Proof: Uniqueness. Since tr $S D = 0, (13.2.39) implies tr $S =  tr $I U = (dim U ). $ Thus  is uniquely determined by S . Then, from (13.2.39), the same is true $ of S D . $ Existence. Choose  = (tr $S )= dim U and dene SD by (13.2.39). Then $ $ $ $ $ $ tr S = (tr I U ) + tr S D = tr S +tr S D so tr S D = 0. Denition 13.2.49 In (13.2.39),  $I U is the isotropic part of S$, and $S D is the deviatoric $ $ $ part of S . If U = P and S is a Cauchy stress tensor, S D is the stress deviator. Remark 13.2.63 Suppose  $I P and S$D are the isotropic and deviatoric parts of Cauchy $ stress tensor S . Then i) hpni = ;  $ $ $ $ $ ii) If S =  I P (i.e., S D = 0 ) then pn(^n) = ; for all n^ and S S (^n) = ~0 for all n^  $ $ iii) S and S D produce the same shear stress for all n^ (i.e., the shear stress is due entirely $ to the deviatoric part of S ). Proof: $ $ i) By (13.2.39) tr S = 3. By remark (13.2.61), tr S = ;3hpni. $ $ ii) Substitute  I P for S in (13.2.35) and (13.2.34). $ iii) From (13.2.34), if n^ is xed, S~S (^n) depends linearly on S . By (13.2.39), $ $ S~S (^n) is the sum of a contribution from  I P and one from S D . By ii) $ above, the contribution from  I P vanishes. Remark 13.2.64 If a Cauchy stress tensor produces no shear stress for any n^ , it is isotropic. Proof: 186 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM $ The hypothesis is, from (13.2.34), that n^  S = n^ Sn(^n) for all unit vectors n^. Let y^1, y^2, y^3 be an orthonormal basis for P . Then setting n^ = y^i gives $ $ $ $ $ $ y^i S = y^(i) Sn(^y(i)). But S = I P  S = (^yiy^i) S = y^i(^yi S ) = P y^iy^iSn(^yi). That is 3 $ X = y^iy^iSn(^yi): S i=1 $T $ $ $ But then S =S , so n^ S y^i = y^i S n^, and thus n^Sn(^n)  y^i = y^iSn(^yi)  n^ , or Sn(^n)(^n  y^i) = Sn(^yi)(^n  y^i). This is true for any n^ , so we may choose an n^ with n^  y^i 6= 0 for all of i = 1 2 3. Then Sn(^n) = Sn(^y1) = Sn(^y2) = Sn(^y3), $ $ so S = Sn(^n)^yiy^i = Sn(^n) I P . QED Fluid Motion: As an application of Cauchy stress tensors and the momentum equation, we consider certain elementary uid problems. Denition 13.2.50 A uid is a material which cannot support shear stresses when it is in static equilibrium. A non-viscous uid is a material which can never support shear stresses. Corollary 13.2.47 In a non-viscous uid, or in a uid in static equilibrium, the Cauchy $E $ stress tensor is isotropic, and S (~r t) = ;hpni(~r t) I P . Proof: $E $ From remark (13.2.64), S (~r t) = (~r t) I P . From remark 13.2.63, i),  = ;hpni. Denition 13.2.51 In a non-viscous uid, or in a uid in static equilibrium, hpni is called simply \the" pressure, written p. Thus $ $ S = ;p I P : Remark 13.2.65 @~  (p $I ) = @~p. (13.2.40) 13.2. CONSERVATION OF MOMENTUM 187 Proof: $ $ $ $ $ $ By exercise 11a, @~  (p I P ) = @~p I P +p@~ I P . But @~ I P = I P h2i@~ I P = 0 $ $ $ because I P is constant and @~ I P = 0. Also @~p I P = @~p. QED. Therefore, the Eulerian momentum equation in a non-viscous uid is ~a = ;@~p + f~: (13.2.41) Since ~a = Dt~u = @t~u + ~u  @~~u, this is often written    @t~u + ~u  @~~u = ;@~p + f~: (13.2.42) In a uid in static equilibrium, ~a = ~0, so @~p = f~: (13.2.43) Therefore, @~ f~ = ~0. We have Remark 13.2.66 Static equilibrium in a uid subjected to body force f~ is impossible unless @~ f~ = ~0. Remark 13.2.67 If a uid is in static equilibrium in a gravitational eld, then the pres- sure and density of the uid are constant on gravitational equipotential surfaces which are arcwise connected. Proof: Let , be the gravitational potential. Then f~ = ;@~,, so (13.2.43) shows that @~p is a scalar multiple of @~,. And @~ f~ = ;@~ @~,. Since @~ f~ = ~0, @~ is also a scalar multiple of @~,. Therefore, if C is any curve lying in a gravitational equipotential surface, and if ^(~r) is the unit tangent vector to C at ~r, then ~ (~r)  @~(~r) = ^(~r)  @~p(~r) = 0. Therefore, by theorem 11.2.23,  and p are the same at both ends of C . 188 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM 13.3 Lagrangian form of conservation of momentum We use the notation introduced on page 163. If ~x is a point in label space L, we use dVL(~x) to stand for a small set containing ~x and also for the volume of that set. The mass per unit volume in label space is ~L(~x), independent of time, so the total mass of the particles whose labels are in dVL(~x) is ~L(~x)dVL(~x). Their velocity at time t is ~vL(~x t), so their momentum is dVL(~x)~L(~x)~vL(~x t). The total momentum of the particles whose labels are in open subset H 0 of H is, at time t, P~tL H 0] = Z H0 dVL (~x)~L(~x)~vL(~x t) = Z H0 dVL(~x)(~~v)L(~x t): Let F~tLH 0] denote the total force exerted on those particles at time t. Then the law of conservation of momentum requires d P~ L H 0] = F~ L H 0] : (13.3.1) t dt t Since the set H 0 does not vary with time, we also have the mathematical identity d P~ L H 0] = Z dV (~x)~L(~x)D ~vL(~x t) L t dt t H0 or d P~ L H 0] = Z dV (~x) (~~a)L (~x t) : (13.3.2) L dt t H0 The particles whose labels are in H 0 occupy the set K 0 (t) in physical space P at time t. Therefore P~tLH 0] = P~ E K 0 (t)], and (13.3.2) should agree with (13.2.3). That it does can be seen by changing the variables of integration from ~r to ~x in (13.2.3) by means of (11.7.6) and then using (13.1.3). L H 0 ] denote the total body force on the particle with labels in H 0 . Then Let F~Bt L H 0 ] = F ~BE K 0(t)]. By changing the variables of integration in (13.2.6) from ~r to ~x F~Bt via (11.7.6), we nd L H 0 ] = F~Bt or Z H  det D~ ~ r L(~x t) f~L(~x t) dV ( ~ x ) L 0 L H 0 ] = F~Bt Z H ~f~L (~x t) dV ( ~ x ) L 0 (13.3.3) 13.3. LAGRANGIAN FORM OF CONSERVATION OF MOMENTUM 189 ^n L x . front dAL ( x ) back Figure 13.8: where  ~f~ = det D~ ~ r  f~: (13.3.4) The vector ~f~ is the body force density per unit of label-space volume. Formula (13.3.4) can be seen physically as follows. The body force on the particles in the small set dVP (~r) is dVP (~r)f~E (~r t) at time t. But these are the particles whose labels are in dVL(~x), so by L (13.0.1) this force is dVL(~x)~f~ (~x t) with ~f~ given by (13.3.4). The surface force at time t exerted by the particles whose labels are just outside @H 0 L H 0 ]. The particles on the particles whose labels are just inside @H 0 we will denote by F~St with labels just outside (inside) @H 0 are those whose positions at time t are just outside (inside) @K 0 (t), so F~StL H 0] = F~SE K 0 (t)] : (13.3.5) L H 0 ] + F L H 0 ]. It remains to study the surface force F L H 0 ]. ~St ~St Then F~tL H 0] = F~Bt Let dAL(~x) be any small nearly plane patch of surface in label space L containing the label ~x 2 H . Choose one side of dAL to be the front, and let n^L be the unit normal to dAL(~x) on its front side. Figure 13.8 is the same as gure (13.6) except for the labelling. Now, however, the picture is of a small patch of surface in label space, not physical space. The force dF~S exerted by the particles with labels just in front of dAL on the particles with labels just behind dAL is proportional to dAL, as long as the orientation of that small surface does not change, i.e., as long as n^ L is xed. We denote the vector \constant" of ~~ L. The vector S~~ will change if n^L changes, and it proportionality by S~~. Then dF~SL = SdA 190 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM ^n L ( x ) . x H′ ∂H ′ Figure 13.9: also will depend on ~x and t, so at time t at label ~x dF~SL = ~S~(~x t n^ L)dAL(~x): (13.3.6) (Compare (13.2.7.) The vector S~~ we call the \label stress". The total surface force on the particles in H 0 is then Z L H 0 ] = ~FSt dAL(~x)S~~ ~x t n^ L(~x)] (13.3.7) 0 @H where n^L(~x) is the unit outward normal to @H 0 at ~x 2 @H 0 . Now we can combine (13.3.3) and (13.3.7) to obtain F~tL H 0] = Z H dV (~x)f~L(~x t) + 0 L Z dAL(~x)S~~ ~x t n^L(~x)] : (13.3.8) dAL(~x)S~~ ~x t n^L(~x)] : (13.3.9) @H 0 Inserting this and (13.3.2) in (13.3.1) gives Z H  L Z ~ ~ dV ( ~ x )  ~ ~ a ; f ( ~ x  t ) = 0 L @H 0 Equation (13.3.9) is true for every open subset H 0 of H for which @H 0 is piecewise smooth. It is mathematically of the same form as (13.2.10) and the same application of Cauchy's $L theorem (13.2.28) shows that there is a unique tensor S~ (~x t) 2 L  P such that for every unit vector n^L 2 L, ~S~(~x t n^ L) = n^L S$L (~x t): (13.3.10) $L The tensor S~ (~x t) is the Piola-Kircho or body stress tensor. We would prefer to call it the label stress tensor, but history prevents us. Proceeding in the way which led from 13.3. LAGRANGIAN FORM OF CONSERVATION OF MOMENTUM 191 (13.2.10) to (13.2.12) we note that by Gauss's theorem Z Z $L ~  S$L (~x t) dA ( ~ x )^ n  ( ~ x  t ) = dV ( ~ x ) D S L L L 0 0 so @H H  $!L ~ ~ ~ dVL(~x) ~~a ; f ; D  S~ (~x t) = 0: H0 Since this is true for all open H 0 H with @H 0 piecewise smooth, the vanishing integral theorem gives Z $ ~ ~~a = ~f~ + D~  S: (13.3.11) This is the Lagrangian form of the momentum equation. Equation (13.3.6) can now be written $L L ~ dFS = dAL(~x)^nL  S~ (~x t) (13.3.12) where dF~SL, dAL(~x) and n^L refer to gure 13.8. Equations (13.3.11) and (13.2.12) refer to the same motion. The relation between ~ and  is (13.1.3). The relation between ~f~ and f~ is (13.3.4), which can also be written ~f~ f~ (13.3.13) ~ =   both sides of this equation being the force per unit mass on the material. But what is the $ $ relation between S~ and S ? Suppose that dAL(~x) is a small nearly plane surface in label space L, and that we have chosen one of its sides as the front. Let n^ L be the unit normal to dAL(~x) on its front side. At time t, let ~r = ~r L(~x t). Let dAP (~r) be the small nearly plane surface in physical space P consisting of all positions at time t of particles whose labels are in dAL(~x). Let dAL(~x) and dAP (~r) denote both the sets of points which make up the small surfaces and also the areas of those sets. Choose the front side of dAP (~r) so that particles are just in front of dAP (~r) if their labels are just in front of dAL(~x). Let n^ P be the unit normal to dAP (~r) on its front side. Then the force exerted on the particles just behind $E dAP (~r) by the particles just in front can be written either as dAP (~r)^nP  S (~r t) or as $L dAL(~x)^nL  S~ (~x t). Therefore, these two expressions must be equal: $L $E dAP (~r)^nP  S (~r t) = dAL(~x)^nL  S~ (~x t) (13.3.14) 192 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM ^n L ^n P front x . . front r dAP ( r ) dAL ( x ) back back Figure 13.10: if and ~r = ~r L(~x t) (13.3.15) h i dAP (~r) = ~r L ( t) dAL(~x)] : (13.3.16) Notice that in (13.3.14), dAP and dAL are positive real numbers, areas, while in (13.3.16) they stand for sets, the small surfaces whose areas appear in (13.3.14). Our program is to express dAP (~r)^nP in terms of dAL(^x)^nL in (13.3.14), to cancel $ $ dAL(^x)^nL , and thus to obtain the relation between S and S~. Equation 13.3.14 holds, whatever the shape of dAL(~x). It is convenient to take it to be a small parallelogram whose vertices are ~x, ~x + ~1, ~x + ~2, ~x + ~1 + ~2. Then dAP (~r) will be a very slightly distorted parallelogram with vertices ~r ~r + ~1 ~r + ~2 ~r + ~1 + ~2 = ~r L(~x t) = ~r L(~x + ~1 t) = ~r L(~x + ~2 t) = ~r L(~x + ~1 + ~2 t): Correct to rst order in the small length k~ik, we have ~ r L(~x t) ~r + ~i = ~r L(~x t) + ~i  D~ so ~ r L(~x t): ~i = ~i  D~ (13.3.17) 13.3. LAGRANGIAN FORM OF CONSERVATION OF MOMENTUM 193 ^n P ^n L front x +ξ 2 ξ2 x . x +ξ +ξ 1 2 . ρ1 dA L ( x ) . r r+ ρ 2 dAP ( r ) back ξ1 r+ ρ x +ξ 1 1 . r+ ρ +ρ 1 2 Figure 13.11: We naturally take ~1 and ~2 in dierent directions, so k~1 ~2 k 6= 0 (recall that L and P are oriented 3-spaces, so cross products are dened). We number ~1 and ~2 in the order which ensures that ~1 ~2 points into the region in front of dAL(~x), so ~1 ~2 = k~1 ~2kn^L. But k~1 ~2k = dAL(~x), so dAL(~x)^nL = ~1 ~2: (13.3.18) Also, clearly, k~1 ~2 k = dAP (~r), and ~1 ~2 is perpendicular to dAP (~r), so ~1 ~2 = k~1 ~2 kn^P . Thus dAP (~r)^nP = ~1 ~2: (13.3.19) But which sign is correct? We have to work out which is the front side of dAP (~r), because that is where we put n^ P . If ~3 2 L is small, ~x + ~3 is in front of dAL(~x) , ~3  n^L > 0, and hence ,   ~3  ~1 ~2 > 0: (13.3.20) The particle with label ~x + ~3 has position ~r + ~3 , with ~3 given by (13.3.17). Then 8 > ~3  (~1 ~2 ) = AP (~3 ~1  ~2) = AP (~1  ~2 ~3 ) > > > > > < ~ ~ L ~ ~ L ~ ~ L  = A 1  D~r  2  D~r  3  D~r P > > > > >  h ~ L i ~ ~ ~  i  > ~ r L(~x t) AL ~1 ~2 ~3 = det D~ : = hdet D~ r (~x t) 3  1 2 : (13.3.21) 194 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM Thus the position ~r + ~3 is in front of dAP (~r) , ~3  (~1 ~2) has the same sign as ~ r L(~x t). Let sgn c stand for the sign of the real number c. That is, sgn c = +1 if det D~ c > 0, = ;1 if c < 0, and = 0 if c = 0. Then position ~r + ~3 is in front of dAP (~r) , ~ r L(~x t): sgn ~3  (~1 ~2) = sgn det D~ (13.3.22) We have chosen the front of dAP (~r), and the direction of n^ P , so that ~r + "n^P is in front ~ r L (~x t). Thus of dAP (~r) when " > 0. Therefore sgn n^P  (~1 ~2 ) = sgn det D~ ~ r L(~x t) ~1 ~2 = k~1 ~2 kn^P sgn det D~ and ~ r L(~x t): dAP (~r)^nP = (~ ~2) sgn det D~ (13.3.23) Using (13.3.18) and (13.3.23), we hope to relate to dAP n^P and dALn^L , so we try to relate ~1 ~2 and ~1 ~2. The two ends of (13.3.21) give   ~ ~3  (~1 ~2) = ~3  ~1 ~2 det D~ r L(~x t) (13.3.24) for any ~1, ~2, ~3 2 L, if ~1 , ~2 , ~3 are given by (13.3.17). Substituting (13.3.18) and (13.3.23) in (13.3.24) and using c sgn c = jcj, we nd   ~ r L(~x t) : ~3  dAP (~r)^nP ] = ~3  dAL(~x)^nL ] det D~ ~ r L(~x t), so But ~3 = ~3  D~   ~ r L(~x t)  dAP (~r)^nP ] = ~3  dAL(~x)^nL] det D~ ~ r L(~x t) : ~3  D~ Since this is true for any ~3 2 L, we must have   ~ r :(~x t)  dAP (~r)^nP ] = dAL(~x)^nL det D~ ~ r L(~x t) D~ or, letting T stand for transpose,   h ~ L iT ~ r L(~x t) : dAP (~r)^nP  D~ r (~x t) = dAL(~x)^nL det D~ (13.3.25) 13.3. LAGRANGIAN FORM OF CONSERVATION OF MOMENTUM $L If we dot S~ (~x t) on the right of each side of (13.3.25), we obtain $L ~ r L(~x t)T  S~ (~x t) dAP (~r)^nP  D~ $L   ~ r L(~x t) : = dAL(~x)^nL  S~ (~x t) det D~ 195 (13.3.26) But (see 13.3.14) $L $E ~ dFS = dAL(~x)^nL  S~ (~x t) = dAP (~r)^nP  S (~r t) where ~r = ~r L (~x t). Thus $L $L dAL(~x)^nL  S~ (~x t) = dAP (~r)^nP  S (~x t): Substituting this on the right side of (13.3.26) gives " # $L L T ~ ~ dAP (~r)^nP ]  D~r (~x t)  S (~x t) $L    L ~ = dAP (~r)^nP ]  S (~x t) det D~r (~x t) : (13.3.27) We have proved (13.3.27) when dAP is a small parallelogram, as in Figure (13.3.14). But ~1 and ~2 can be arbitrary as long as they are small. Hence so can dAP (~r)^nP = ~1 ~2 . Thus (13.3.27) is true when dAP (~r)^nP is any small vector in P . Since (13.3.27) is linear in dAP (~r)^nP , it is true whatever the size of that vector. Therefore (13.3.27) implies $L   ~ r L(~x t)T  S~ (~x t) = S$L(~x t) det D~ ~ r L (~x t)  D~ or  ~ T $~ $  ~  D~r  S = S det D~r  : (13.3.28) $ $ ~ r L (~x t) has an inverse, @~ ~x E (~r t), This gives S in terms of S~. By (12.2.7, 13.3.24), D~ ~ r L(~x t)T has the inverse @~ ~x E (~r t)T . Dotting this on the left where ~r = ~r L (~x t), so D~ in (13.3.28) gives $  T $  ;1 S~ = @~ ~x  S det @~ ~x  (13.3.29) ~ r );1 = (det D~ ~ r);1 . where we use det(D~ 196 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM By appealing to (13.3.1), we can write (13.3.28) and (13.3.29) as 0$1 0 $1   S~ C ~ r T B @ S A = D~ @  ~ A 0$1 0 $1 B@ S~ CA = @~ ~x T  @ S A : ~  (13.3.30) (13.3.31) 13.4 Conservation of angular momentum 13.4.1 Eulerian version Refer again to the notation on page 163. All torques and angular momenta are calculated relative to the origin O~ P in physical space P . Let L~E K 0 (t)] denote the total angular ~ E K 0(t)] denotes the total torque on them. momentum of the particles in K 0 (t), while M The law of conservation of angular momentum is d L~E K 0 (t)] = M ~ E K 0 (t)] : (13.4.1) dt Our problem is to express this law locally. The momentum of the particles in dVP (~r) is dVP (~r)E (~r t)~vE (~r t) and this produces, relative to O~ P , the angular momentum h i ~r dVP (~r)(~v)E (~r t) = dVP (~r)(~r ~v)E (~r t): It is possible that the individual atoms in dVP (~r) might have angular momentum this might be from the orbits of electrons not in closed shells, from the spins of unpaired electrons, or from nuclear spin. Orbital angular momentum is quantized in integer multiples of h( , while spin angular momentum is in units of h( =2. Here h( = h=2 and h is Planck's constant, so h( = 1:0544 1;34 joule sec. If the atoms are not randomly aligned, there may be an intrinsic atomic angular momentum of ~lE (~r t) per kilogram at position ~r at time t. Then the intrinsic angular momentum in dVP (~r) is dVP (~r)(~l)E (~r t), and the total angular momentum in dVP (~r) is h  iE dL~E = dVP (~r)  ~r ~v + ~l (~r t): 13.4. CONSERVATION OF ANGULAR MOMENTUM 197 Figure 13.12: The total angular momentum in K 0(t) is LE K 0(t)] = From (13.1.12), Now so Z K 0 (t) h  iE dVP (~r)  ~r ~v + ~l (~r t): d LE K 0(t)] = Z dV (~r) hD ~r ~v + ~liE (~r t): P t dt K 0(t) Dt(~r ~v) = (Dt~r) ~v + ~r Dt~v = ~v ~v + ~r ~a = ~r ~a d LE K 0 (t)] = Z dV (~r) h ~r ~a + D ~l iE (~r t): (13.4.2) P t dt K 0 (t) To calculate the torque on the particles in K 0(t) we note that the torque exerted by the body force dVP (~r)f~E (~r t) is ~r dVP (~r)f~E (~r t), or dVP (~r)(~r f~)E (~r t). The torque exerted $E by the surface force dAP (~r)S~ (~r t n^ P ) is ~r dAP (~r)S~ = ;dAP (~r)S~ ~r = ;dAP (~r)^n S $ $ $ $ (~r t)] ~r. If Q2 P  P is a polyad, (^n Q) ~r = n^  Q ~r], so this is true for all Q. $ $ (Exercise 11b gives a denition of ~r Q, and Q ~r is dened similarly.) Thus  $ E ~ ~r dAP (~r)S (~r t n^P ) = ;dAP (~r)^nP  S ~r (~r t): 198 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM front ^n P r . back dAP ( r ) Figure 13.13: Therefore, the torque about O~ P which is exerted on the particles in K 0 (t) by the body and surface force acting on them is Z  $ E Z  E ~ dVP (~r) vecr f (~r t) ; 0 dAP (~r)^nP  S ~r (~r t): K 0(t) @K (t) By applying Gauss's theorem to the surface integral, we can write this torque as Z  $ E ~ ~ dVP (~r) ~r f ; @  S ~r (~r t): K 0 (t) (13.4.3) In addition to this torque, there may be a torque acting on each atom in dVP (~r). This would be true, for example, if the material were a solid bar of magnetized iron placed in a magnetic eld B~ . If the magnetization density was M~ , there would be a torque dVP (~r)M~ B~ acting on dVP (~r) and not included in (13.4.3). In general, we might want to allow for an intrinsic body torque of m ~ joules/meter3, so that (13.4.3) must be supplemented by a term Z dVP (~r)m ~ E (~r t): 0 K (t) Finally, in a magnetized iron bar, atoms just outside @K 0 (t) exert a torque on atoms just inside @K 0 (t), so there is a \torque stress" acting on @K 0 (t). The torque exerted on the material just behind dAP (~r) by the material just in front is proportional to dAP as long as the orientation of that small patch of surface, i.e., its unit normal, stays xed. We write the proportionality constant as M~ (~r t n^ P ). The torque exerted on the material $E just behind dAP by the material just in front, other than that due to n^P  S (~r t), is 13.4. CONSERVATION OF ANGULAR MOMENTUM 199 M~ (~r t n^P )dAP (~r), so we must also supplement (13.4.3) by a term Z dAP (~r)M~ (~r t n^ P (~r t))  0 @K (t) where n^ P is the outward unit normal to @K 0 (t). Thus we nally have    E Z M~ E K 0 (t)] = 0 dVP (~r) ~r f~ ; @~  S$ ~r + m~ (~r t) KZ(t) + @K 0(t) dAP (~r)M~ ~r t n^ P (~r t)] : If we substitute (13.4.2) and (13.4.4) in (13.4.1), we obtain E R dV (~r)  ~r ~a + D ~l ; ~r f~ + @~  $ ~r ; m ~ (~r t) S t K 0(t) P R = @K 0(t) dAP (~r)M~ ~r t n^ P (~r t)] : (13.4.4) (13.4.5) This equation has the same mathematical form as (13.2.10), and leads via Cauchy's the$ orem to the same conclusion. At ~r t there is a unique tensor M E (~r t) 2 P  P , which we call the (Cauchy) torque-stress tensor, such that for every unit vector n^P 2 P we have $ M~ (~r t n^P ) = n^P  M E (~r t ): (13.4.6) Substituting this on the right in (13.4.5), applying Gauss's theorem to the surface integral, and the vanishing integral theorem to the volume integral, gives $    ~  ~r ~a + Dt~l ; ~r f~ + @~  S ~r ; m ~ = @~  M: (13.4.7) Before we accept this as the Eulerian angular momentum equation, we note that it can be greatly simplied by using the momentum equation (13.2.12). We have (~r ~a) = ~r f~   $ = ~r ~a ; f~ = ~r (@~ S ), so (13.4.7) is  $ $  $ ~ ~ ~ Dt l + ~r @  S + @  S ~r = m ~ + @~ M : (13.4.8) This can be still further simplied. Let y^1 , y^2, y^3 be a poob in P , (^x1  x^2  x^3) a pooob in L, and take components relative to these bases. Then  $ $ ~ ~r (@  S ) = "ijk rj (@~ S )k = "ijk rj @l Slk : i 200 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM Also, for ~u , ~v, w~ 2 P , (~u~v) w~ = ~u(~v w~ ) (this is how we use the lifting theorem to $ $ dene Q w~ for Q2 P  P and w~ 2 P ) so (~u~v) w~ ]ij = ~u (~v w~ )]ij = ui (~v w~ )j = ui"jklk wl = (~u~v)ik "jklw: $ Therefore for any Q2 P  P , $  Q w~ = Qik "jklwl : ij Therefore $  S ~r li = Slk "ikj rj  $  $  @~  S ~r = @l S ~r = @l Slk "ikj rj ] i li = "ikj @l (Slk rj ) = "ikj (@l Slk ) rj + Slk @l rj ] : Referring to page 199 and using "ijk + "ijk = 0, we get    $  ~ ~ ~ ~r @  S + @  S ~r = "ikj Slk @l rj : i  $ But @l rj = lj , so this is "ikj Sjk = ;ijk Sjk = ; AP h2i S . Thus i  $ $  $ ~r @~ S + @~  S ~r = ;AP h2i S : Thus (13.4.8) is $ $ Dt~l = AP h2i S +m ~ + @~ M : (13.4.9) (13.4.10) This is the Eulerian form of the angular momentum equation. In non-magnetic materials $ it is believed that ~l, m ~ are M and all negligible, and that the essential content of (13.4.10) is $ AP h2i S = ~0: (13.4.11) 13.4. CONSERVATION OF ANGULAR MOMENTUM 201 That is, "ijk Sjk = 0. Multiply by "ilm and sum on i, and this becomes (see page D-9) (jlkm ; jmkl) Sjk = 0 or Slm ; Sml = 0 or $ $ S T =S : (13.4.12) The conservation of angular momentum requires that the Cauchy stress tensor be symmetric, if intrinsic angular momentum, body torque and torque stress are all negligible. $ 13.4.2 Consequences of S T =S$ In the absence of intrinsic angular momentum, torque and torque stress, the Cauchy stress tensor is symmetric. This greatly simplies visualizing the stress at a point ~r at time t. $E $ We will abbreviate S (~r t) as S in this chapter, because we are looking at one time t and one position ~r in physical space. We denote by S~ (^n) the stress on the oriented small area (dA n^ ) at ~r t (i.e. we write S~ (~r t n^ P ) as S~ (^n), and n^P as n^, dAP as dA). We denote by $ S the linear operator on P corresponding to the tensor S . Then for any unit vector n^ , $ S~ (^n) = n^ S = S (^n): (13.4.13) Since S : P ! P is a symmetric operator, P has an orthonormal basis x^ y^ z^ consisting of eigenvectors of S . That is, there are scalars a, b, c such that S (^x) = ax^ S (^y) = by^ S (^z ) = cz^: $ $ $ Then x^ S = ax^, y^ S = by^, z^ S = cz^ so $ $ $ $ S = I P  S = (^xx^ + y^y^ + z^z^)  S = x^ (ax^) + y^ (by^) + z^ (cz^) so $ S = ax^x^ + by^y^ + cz^z^: (13.4.14) (13.4.15) 202 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM The unit vectors x^, y^, z^ are called \principal axes of stress" and a, b, c are \principal stresses." By relabelling, if necessary, we can always assume that a  b  c. Also if we change the sign of any of x^ y^ z^ both (13.4.14) and (13.4.15) remain unchanged. Therefore we may assume that (^x y^ z^) is positively oriented. For any vector ~n 2 P we write ~n = xx^ + yy^ + zz^: (13.4.16) The unit sphere NP consists of those ~n 2 P for which x2 + y2 + z2 = 1: (13.4.17) $ The pressure pn(^n) acting on (dA n^) is pn(^n) = ;n^  S n^ = ;(ax2 + by2 + cz2 ). It can be thought of as a function of position n^ = xx^ + yy^ + zz^ on the spherical surface (13.4.17). In the earth, pn > 0, so it is convenient to introduce the principal pressures A, B , C , dened by A = ;a B = ;b C = ;c: (13.4.18) Then A  B  C . The cases where equality occurs are special and are left to the reader. We will examine the usual case, A < B < C: (13.4.19) We want to try to visualize pn(^n) and the shear stress S~S (^n) acting on (dA n^ ) as functioning of n^ , i.e., as functions on the spherical surface (13.4.17). We have pn(^n) = Ax2 + By2 + Cz2 (13.4.20) $ and, since S~ (^n) = n^  S , S~ (^n) = S~S (^n) ; pn(^n)^n (13.4.21) S~ (^n) = ;Axx^ ; Byy^ ; Czz^: (13.4.22) First consider pn. Its value is unchanged if we replace any of x y z by its negative in (13.4.16). Therefore pn is symmetric under reections in the coordinate planes x = 0, 13.4. CONSERVATION OF ANGULAR MOMENTUM 203 y = 0, z = 0. Therefore it su"ces to study pn(^n) when n^ is in the rst octant of (13.4.17), i.e., when x  0 y  0 z  0: (13.4.23) If p is any constant, then the level line pn(^n) = p on Np satises Hence, it satises Ax2 + By2 + Cz2 = p x2 + y2 + z2 = 1: (13.4.24) (A ; p) x2 + (B ; p) y2 + (C ; p) z2 = 0: (13.4.26) (13.4.25) This equation and (13.4.17) have no real solutions (x y z) unless A  p  C (13.4.27) so the values of pn(^n) all lie in the interval (13.4.27). The value p = A requires y = z = 0, x = 1, so pn(^n) = A only for n^ = x^. The value p = C requires x = y = 0, so pn(^n) = C p p only for n^ = z^. The value of p = B makes (13.4.25) into x B ; A = z C ; B , and these two planes intersect the sphere (13.4.17) in great circles, one of which lies in (13.4.23). For any value of p in A < p < C , (13.4.25) and (13.4.26) imply all three of the following: 8 > > (C ; A) x2 + (C ; B ) y2 = C ; p > < (13.4.28) (B ; A) x2 ; (C ; B ) z2 = B ; p > > > : (B ; A) y2 + (C ; A) z2 = p ; A: Therefore the projection of the level curve (13.4.25), (13.4.17) onto the xy or yz plane is part or all of an ellipse, and its projection on the xz plane is part of a hyperbola with p p asymptotes x B ; A = z C ; B . The level lines pn(^n) = B on Np, with p = B , p p are the great circles where Np intersects the planes x B ; A = z C ; B . The level lines of pn on NP are the dashed curves in gure 13.14. Next, consider the shear stress S~S (^n). At any n^ 2 NP , this is a vector tangent to NP . It is dened by (13.4.21), and from (13.4.22) S~ (^n) = ;1=2 @~pn(^n) where pn is to be considered as a function dened for all ~n 2 P by (13.4.20), even when k~nk 6= 1. At a point n^ 2 NP , let ^(^n) be the unit 204 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM ^z pn = C pn = B ^y pn = B ^x pn = A Figure 13.14: 13.4. CONSERVATION OF ANGULAR MOMENTUM 205 vector tangent to the level curve of pn passing through n^ . Then n^  ^(^n) = 0, and since ^(^n) is tangent to a curve lying in the level surface pn(~n) = constant in P , it follows that ^(^n)  @~pn(^n) = 0. Therefore, from (13.4.29), we have ^(^n)  S~ (^n) = 0 and ^(^n)  n^ = 0: Then from (13.4.21), ^(^n)  S~S (^n) = 0. Thus the shear stress S~S (^n) at n^ 2 NP is perpendicular to the level line of pn which passes through n^. The \lines of force" of the shear stress on NP are the curves which are everywhere tangent to S~S (^n). They are the solid curves in gure 13.14, and are everywhere perpendicular to the level lines of pn. These \lines of force" give the direction of S~S (^n) everywhere on NP . Finally, we would like to nd kS~S (^n)k on NP , which we abbreviate as (^n) = kS~S (^n)k: (13.4.29) Again we consider only the rst octant, x  0, y  0, z  0. The other octants are obtained by reection in the three coordinate planes x = 0, y = 0, z = 0. First consider the edges of the rst octant, where x = 0 or y = 0 or z = 0. On the quarter circle n^ = x^ cos + y^ sin , 0   =2, subjecting (13.4.20) and (13.4.21), (13.4.22) to a little algebra yields S~S (^n) = sin cos (^x cos ; y^ sin ) on n^ = x^ cos + y^ sin : (13.4.30) Therefore, from (13.4.29), (^n) = 12 (B ; A) sin 2 if n^ = x^ cos + y^ sin : (13.4.31) (^n) = 1=2(C ; A) sin 2 if n^ = x^ cos + z^ sin (13.4.32) (^n) = 1=2(C ; B ) sin 2 if n^ = y^ cos + z^ sin : (13.4.33) This reaches a maximum value of (B ; A)=2 at = =4, half way between x^ and y^ on the quarter circle. And (^x) = (^y) = 0. Similarly To see what happens to (^n) when x > 0, y > 0, z > 0, we appeal to 206 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM Lemma 13.4.33 On any level line of pn in the rst octant on NP , (^n) decreases monotonically as y increases. Proof: From (13.4.21), kS~ (^n)k2 = (^n)2 + pn(^n)2 so (^n)2 = kS~ (^n)k2 ; pn(^n)2: (13.4.34) We can parametrize the level curve pn(^n) = p on NP by the value of y at n^, namely y^  n^ = y. We want to show that d (^n) < 0 on the curve p (^n) = p on N 0 rst octant: (13.4.35) n P s dy On that level curve, of course pn(^n) is constant, so (13.4.34) shows that (13.4.35) is equivalent to d kS~ (^n)k2 < 0 on the curve p (^n) = p on N 0 rst octant: (13.4.36) n p s dy The vector dn^=dy on pn(^n) = p is tangent to that level curve. Hence n^  dn^=dy = 0 and @~p(^n)  dn^=dy = 0, so S~ (^n)  dn^=dy = 0. Hence, for some  dn^ = n^ S~ (^n): dy Now dn^=dy = x^(dx=dy) + y^ + z^(dz=dy) on the level curve. Hence y^  ddyn^ = 1 so 1 =  y^  n^ S~ (^n): But n^ = x^x + y^y + z^z, ;S~ (^n)Axx^ + Byy^ + Czz^, so    0 1 0    ;y^  n^ S~ (^n) =  x y z  = ;xz(C ; A):    Ax By Cz  Thus  = 1=xz(C ; A), and 1 dn^ = ~ dy xz(C ; A) n^ S (^n): (13.4.37) 13.4. CONSERVATION OF ANGULAR MOMENTUM 207 By the chain rule, dyd kS~ (^n)k2 = ddyn^  @~kS~ (^n)k2 . From (13.4.22), kS~ (~n)k2 = A2 x2 + B 2y2 + C 2z2  so @~kS~ (~n)k2 = 2A2xx^ + 2B 2 yy^ + 2C 2zz^: Therefore h ~ i  2  d kS~ (^n)k2 = 2 2 y y^ + C 2 z z^ n ^ S (^ n )  A x x ^ + B dy xz(C ; A)        y z   x  1 1 1  = xz(C2; A)  ;Ax ;By ;Cz  = ; (C 2;y A)  A B C   2   2 2 2   A x B 2 y C 2z  A B C  = ;2y(C ; B )(B ; A) < 0: This proves lemma 13.4.30 . QED. From (13.4.31, 13.4.32, 13.4.33) and lemma 13.4.33, we infer that the largest value of p (^n), the magnitude of the shear stress, is (C ; A)=2, and this occurs at n^ = (x^  z^)= 2, and for no other n^. It is of some interest to see which pairs (pn(^n) (^n)) are possible on NP . On the three quarter-circles in (13.4.31, 13.4.32, 13.4.33) we have 0 1 0 1 A+B ; B;A cos 2 p (^ n ) B CA if n^ = x^ cos + y^ sin (z = 0) @ n CA = B@ 2 B;A 2 n(^n) 2 sin 2 1 0 1 0 C ; A A + C ; cos 2 C if n^ = x^ cos + z^ sin (y = 0) p (^n) B @ n CA = B@ 2 C ;A 2 A sin 2 n (^n) 2 1 0 1 0 C ; B B + C ; cos 2 C if n^ = y^ cos + z^ sin : (x = 0) p (^n) B A @ n CA = B@ 2 C ;B 2 sin 2 n(^n) 2 (13.4.38) (13.4.39) (13.4.40) For z = 0, the possible pairs (pn(^n) (^n)) are given by (13.4.38) and trace out the seimicircle marked z = 0 in Figure 13.15. For y = 0, the possible pairs are given by 208 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM MOHR σ( ^n ) MOHR . . start stop y=0 z=0 x=0 A B p θ increases in the direction of the arrows. Figure 13.15: C p n ( ^n ) 13.4. CONSERVATION OF ANGULAR MOMENTUM 209 (13.4.39) and trace out the semicircle marked y = 0 in Figure 13.15. For x = 0, use the third semicircle in Figure 13.4.38. In gure 13.14, suppose we start at a point with y = 0 and move along the level line of pn till we hit either x = 0 or z = 0. Then in the ( pn) plane we will start at the point marked \start" in Figure 13.15, and we will remain on the vertical dashed line pn = p. According to lemma 13.4.33, as we increase y on the level curve pn ; p in Figure (13.14), we will decrease (^n), so we move down the dashed line in Figure 13.15 until we strike the point marked \stop." Therefore, the possible pairs (pn(^n) (^n)) which can occur on NP are precisely the points in the shaded region in Figure 13.15. Coulomb suggested that a rock will break when the maximum value of (^n) exceeds a critical value characteristic of that rock. Navier pointed out that when two plates are pressed together, the greater this pressure the harder it is to slide one over the other. Navier suggested that the fracture criterion is that there be an orientation n^ for which (^n) > c +pn(^n), where c and  are constants characteristic of the rock (when p(^n) = 0, Navier's fracture criterion reduces to Coulomb's). The constant  is a sort of internal coe"cient of friction. Mohr suggested that Coulomb and Navier had over-simplied the problem, and that the true fracture criterion is (n) > f p(^n)], where f is a function characteristic of the rock, which must be measured empirically, and about which one can say in advance only that f (p) > 0 df=dp > 0: (13.4.41) Even with such a general criterion, much can be said. The curve  = f (p) is marked \MOHR" in Figure 13.15. If A, B , C are moved so as to produce an overlap of the shaded region with the Mohr curve, the rock will break at rst contact of the curve and the shaded region. This rst contact will always occur on the plane y = 0, at a value of in 13.4.32 which is between 0 and =4, so the normal to the plane of fracture will lie in the xz plane, closer to x^ than to z^. The circles in Figure 13.15 are called Mohr circles. 210 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM 13.5 Lagrangian Form of Angular Momentum Conservation Refer again to the notation introduced on page 163. If L~Lt(H 0) is the angular momentum ~ Lt(H 0) is the total torque on them, of the collection of particles with labels in H 0, and M the law of conservation of angular momentum is d L~L(H 0) = ML(H 0): t dt t Using the notation introduced on page 195 and page 164, Z h i ~LLt(H 0) = dVL(~x) ~~l + ~~r ~v L (~x t) 0 (13.5.1) H so Moreover, d L~L (H 0) = Z dV (~x) h~ D ~l + ~r ~aiL (~x t) : L t dt t H0 MLt (H 0) = (13.5.2)  ~ L dV ( ~ x ) ~r f~ + m ~ (~x t) L H0 "  $L! # Z ~ + 0 dAL (~x) ~r n^L  S~ + M~ (~x t n^ L (~x)) Z H ~ is body torque per unit where n^L (~x) is the unit outward normal to @H 0 at ~x 2 @H 0 , m ~~ (~x t n^L(~x))dAL is the force exerted by the material with of volume in label space, and M labels just in front of (dAL n^ L) on the material with labels just behind (dAL n^ L). Again $ $ $ ~ ~ we have ~r (^nL  S ) = ;(^nL  S ) ~r = ;n^L  (S~ ~r), so using Gauss's theorem, " $ !#L Z ~ L 0 Mt (H ) = H 0 dVL (~x) ~r f~ + m~ ; D~  S~ ~r (~x t) (13.5.3) Z ~~ ~x t n^ L(~x)] : + dAL (~x) M (13.5.4) H0 Therefore $ !# " R dV (~x) ~ D ~l + ~r ~a ; ~r ~f~ ; m ~ + D~  S~ ~r t H0 L R ~~ ~x t n^ L(~x)] : = 0 dAL (~x) M @H (13.5.5) 13.6. CONSERVATION OF ENERGY 211 Applying Cauchy's theorem to (13.5.5) shows that for each (~x t) there is a unique $ M~ L (~x t) 2 L  P such that for any unit vector n^ 2 L, $ $L M~ ~x t n^] = n^  M~ (~x t) : (13.5.6) If we substitute this on the right in (13.5.5) and use Gauss's theorem and the vanishing integral theorem, we obtain $ ! $ ~ ~ ~ ~ ~ ~ ~Dtl + ~~r ~a ; ~r f ; m~ + D  S~ ~r = D~  M: Using the Lagrangian momentum equation (13.3.11), we can write this as  $! $ ! $ ~ ~ ~ ~ ~ ~ ~Dtl + ~r D  S + D  S ~r = D  M~ + m: ~ Finally, by the sort of argument which led to (13.4.9), we can show  $! $ ! " T $# ~ ~ ~ ~ ~ r  S~  ~r D  S + D  S ~r = ;AP h2i D~ so the Lagrangian form of the angular momentum equation is " T $# $ $ ~ r  S~ : ~Dt~l = D~  M~ + m~ + AP h2i D~ (13.5.7) $ $ $ $ The relation between M~ and M~ is the same as that between S~ and S . If the intrinsic $ angular momentum ~l, body torque m~ , and torque stress M~ are all negligible, (13.5.7) reduces to " T $# ~ r  S~ = 0: AP h2i D~ (13.5.8) A glance at (13.3.30) shows that (13.5.8) is the same as (13.4.11), so we learn nothing new from (13.5.7) in this case. 13.6 Conservation of Energy 13.6.1 Eulerian form The velocity ~vE (~r t) imparts to the matter in dVP (~r) a kinetic energy dVP (~r)E (~r t) 12 k ~vE (~r t)k2. In addition, that material has kinetic energy of molecular motion, and the 212 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM potential energy of the intermolecular forces. The sum of these latter is called the internal energy. We denote by U E (~r t) the internal energy per kilogram in the matter at ~r t. Then the total energy in dVP is The total energy in K 0(t) is  1 E 2 dVP (~r)  2 v + U (~r t) : E E K 0 (t)] = Z  1 E 2 dVP (~r)  2 v + U (~r t) : K 0 (t) (13.6.1) The law of conservation of energy is d E E K 0(t)] = W E K 0(t)] (13.6.2) dt where W E K 0 (t)] is the rate at which energy is being added to the matter in K 0(t), i.e., the rate at which work is being done on that matter. The body force f~ does work on dVP (~r) at the rate of dVP (~r)(~v  f~)E (~r t) watts. If $E ~r 2 @K 0 (t) then the surface stress n^P  S (~r t) does work on the particles just inside $E @K 0 (t) at the rate of dAP (~r)^nP  S (~r t)  ~vE (~r t) watts. Thus the purely mechanical component of W E K 0(t)] is  $ E Z Z  ~E dVP (~r) ~v  f (~r t) + 0 dAP (~r)^nP  S ~v (~r t) : 0 K (t) @K (t) Using Gauss's theorem, this can be written Z K 0 (t)  $ E dVP (~r) ~v  f~ + @~  S ~v (~r t) : (13.6.3) In addition to this mechanical method of adding energy to K 0 (t) there are nonmechanical methods. Examples are i) ohmic \heating" at a rate of dVP (~r)kJ~k2= watts, where J~ = electric current density and  = electrical conductivity. ii) \heating" by absorption of light (e.g., sunlight in muddy water). iii) radioactive \heating". 13.6. CONSERVATION OF ENERGY 213 front ^n P r . back dAP ( r ) Figure 13.16: All these \heating" mechanisms together will add energy to dVP (~r) at a total rate of dVP (~r)hE (~r t), where hE is the sum of the heating rates per unit volume. The resulting contribution to W E K 0(t)] is Z K 0 (t) dVP (~r) hE (~r t) watts: (13.6.4) Finally, energy can leak into K 0(t) through the boundary, @K 0 (t). The physical mechanisms for this are molecular collision and diusion, but that does not concern us. We are interested only in the possibility that if dAP (~r) is a small nearly plane surface containing ~r, with a designated front to which a unit normal n^P is attached, the energy can leak across dAP (~r) from front to back at a rate proportional to dAP as long as the orientation of dAP is not varied, i.e., as long as n^P is xed. We write the constant of proportionality as ;H (~r t n^ P ), so that energy ows from front to back across dA(~r) at a net rate of ;H (~r t n^ P )dAP (~r) watts. This \heat ow" contributes to W E K 0 (t)] the term ; Z @K 0 (t) dAP (~r) H (~r t n^ P ) watts: The minus sign in the denition of the proportionality constant is a convention established for two centuries, and we must accept it. Indeed, there is aesthetic reason to put a minus $ $ sign in the denition of S~ , to avoid the ;p I term. Our choice of signs for S~ and S was also dictated by history, and it is unfortunate that the same choice was not made in both cases. 214 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM The sum of all the foregoing rates at which energy is added to the matter in K 0 (t) gives W E K 0(t)] = Z KZ0 (t) ;  $  E dVP (~r) ~v  f~ + @~  S ~v + h (~r t) @K 0(t) dAP (~r) H (~r t n^P ) : From (13.6.1) and (13.1.12), " # dE E K 0 (t)] = Z dV (~r) D  1 v2 + U E (~r t) : P t dt 2 K 0 (t) (13.6.5) (13.6.6) (13.6.7) Substituting (13.6.6) and (13.6.7) in (13.6.2) gives Z  $  E  1 2 + U ; ~v  f~ ; @~  S v  ~v ; h (~r t) dV ( ~ r ) D P t 2 K 0 (t) Z = ; 0 dA (~r) H (~r t n^ ) : @K (t) (13.6.8) (13.6.9) Applying Cauchy's theorem 13.2.28 to (13.6.9) shows that at each (~r t) there is a unique vector H~ E (~r t) 2 P such that for all unit vectors n^ 2 P H (~r t n^ ) = n^  H~ E (~r t) : (13.6.10) The vector H~ E (~r t) is called the heat-ow vector. Heat ows from front to back across (dAP  n^P ) at the rate of ;dAP (~r)^nP  H~ E (~r t) watts. Inserting (13.6.10) in (13.6.9) and applying Gauss's theorem converts (13.6.9) to  $  E  1 2 + U ; ~v  f~ ; @~  S ~ ~ dV ( ~ r ) D v  ~ v ; h + @  H (~r t) : (13.6.11) P t 2 K 0 (t) Since K 0 (t) can be any open subset of K (t) with piecewise smooth boundary, the vanishing 0= Z integral theorem gives   $  Dt 12 v2 + U + @~  H~ = h + ~v  f~ + @~  S ~v : This equation can be simplied. We have   Dt v2 = Dt (~v  ~v) = 2~v  Dt~v = 2~v  ~a: (13.6.12) 13.6. CONSERVATION OF ENERGY 215 Also, taking components relative to an orthonormal basis,   ~@  $ S ~v = @i (Sij vj ) = (@i Sij ) vj + Sij @i vj  $ $ = @~ S  ~v+ S h2i@~~v: (13.6.13) (13.6.14) Therefore (13.6.12) is  $ $ ~a  ~v + Dt U + @~  H~ = h + f~  ~v + @~ S  ~v+ S h2i@~~v: $ But the momentum equation says ~a = f~ + @~ S , so those terms cancel, leaving only $ Dt U + @~  H~ = h+ S h2i@~~v: (13.6.15) This is the Eulerian form of the internal energy equation, or heat equation. 13.6.2 Lagrangian form of energy conservation The argument should be familiar by now. The total energy of the material with labels in H 0 is, at time t,   L Z EtL H 0] = 0 dVL (~x) ~ 12 v2 + U (~x t) : (13.6.16) H The rate at which energy is added is WtL H 0] " $ ! #L ~ = 0 dVL (~x) f~  ~v + D~  S~  ~v + h( (~x t) H Z ; @H 0 dAL (~x) 6 H (~x t n^ L(~x)) : Z (13.6.17) (13.6.18) The law of conservation of energy is d E L H 0] = W L H 0] (13.6.19) t dt t and from (13.6.16) we have the identity d E L H 0] = Z dV (~x) ~D  1 v2 + U L (~x t): (13.6.20) L t dt t 2 H0 Substituting (13.6.18) and (13.6.20) in (13.6.19) gives $ ! #L "   ~ Z 1 2 dVL (~x) ~Dt 2 v + U ; f~  ~v ; D~  S~  ~v ; h( (~x t) H0 Z (13.6.21) = ; 0 dAL(~x) 6 H (~x t n^ L(~x)) : @H 216 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM At any (~x t), Cauchy's theorem 13.2.28 assures us of the existence of a unique vector 6 H~ L(~x t) 2 L such that for any unit vector n^ 2 L, 6 H (~x t n^) = n^ 6 H^ L(~x t). Substituting this on the right in (13.6.21), and applying Gauss's theorem and the vanishing integral theorem gives 1  ~ $  2 ~ ~ ~ ~Dt 2 v + U ; f  ~v ; D  S ~v h( = ;D~  6 H: But Dt v2 = 2~v  Dt~v = @~v  ~a, and $ !  $ ! $ ~v D~  S~  ~v = D~  S~  ~v + S~h2iD~ so  $! $ ~ ~ ~ ~~a  ~v + ~Dt U = f~  ~v ; D~  S~  ~v; S h2iD~ v ; h( = ;D~  6 H: $ The Lagrangian momentum equation, ~~a ; ~f~ ; D~  S~ = ~0, permits some cancellation and leaves us with $ ~ v: (13.6.22) ~Dt U + D~  6 H~ = h( + S~h2iD~ This is the Lagrangian form of the internal energy equation. Using (13.3.25) and the identity ~ dAP n^  H~ = dALn^ L 6 H one proves  H~ = D~ ~ r T  6 H~  ~ in the same manner as (13.3.30) was proved. (13.6.23) 13.7 Stu Mass, momentum, angular momentum and energy are all examples of a general mathematical object which we call a \stu." The idea has produced some confusion, so we discuss it here. 13.7. STUFF 217 $E Denition 13.7.52 A \stu" is an ordered pair of functions (
~ E  F ) with these properties: V is a Euclidean space, and for each t 2 R, K (t) is an open subset of physical space P  and
~ E ( t) : K (t) ! V (13.7.1) $E (13.7.2) F ( t) : K (t) ! P  V: $E E The function
~ is called the spatial density of the stu F is the spatial current density or spatial ux density of the stu. And ~ E , the creation rate of the stu, is dened to be $E E ~ E = @t
~ + @~ F : (13.7.3) Denition 13.7.53 Suppose t 2 R, ~r 2 K (t), and dV (~r) is a small open subset of K (t) containing ~r dV (~r) will also denote the volume of that subset. Suppose dA(~r t) is a small nearly plane oriented surface containing ~r, and n^ is the unit normal on the front side of dA(~r t). We also use dA(~r t) to denote the area of the small surface. Suppose dA(~r t) moves with speed W normal to itself, with W > 0 when the motion is in the direction of n^. Then E i) dV (~r)
~ (~r t) is called the amount of stu in dV (~r) at time t $E E ii) dA(~r t)^n F (~r t) ; W
~ (~r t)] is called the rate at which the stu ows across dA(~r t) n^] from back to front. iii) dV (~r)~ E (~r t) is called the rate at which stu is created in dV (~r) at time t. Denition 13.7.54 Let K 0 (t) be any open subset of K (t), with piecewise continuous boundary @K 0 (t) and unit outward normal n^(~r) at ~r 2 @K 0 (t). Suppose K 0(t) moves so that the outward velocity of @K 0 (t) normal to itself at ~r 2 @K 0 (t) is W (~r t). Then i) R ~ K 0 (t) dV (~r)
-~ K 0 (t)] E (~r t) is called the amount of stu in K 0 (t) at time t. Denote if by " # $E R E ii) @K 0(t) dA(~r) n^ (~r) F (~r t) ; W (~r t)
~ (~r t) is called the rate at which stu ows out of K 0(t) across @K 0 (t) at time t. Denote it by F~ K 0 (t)]. 218 iii) CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM R is called the rate at which the stu is created in K 0 (t) at time t. Denote it by ~;K 0(t)]. K 0 (t) dV (~r)~ E (~r t) Remark 13.7.68 d=dt -~ K 0(t)] = ~;K 0 (t)] ; F~ K 0 (t)] for any open subset K 0(t) of K (t), moving in any way whatever. Proof: d -~ K 0 (t)] = Z dV (~r)@
~ E (~r t) + Z ~ E (~r t) dA ( ~ r ) W
t dt K 0 (t) @K 0(t) Z Z $E E F~ K 0(t)] = dV (~r)@~ F (~r t) ; dA(~r)W
~ (~r t): K 0 (t) @K 0 (t) Therefore,  ! E d -~ K 0(t)] + F~ K 0(t)] = Z dV (~r) @
~ E + @~ $ F t dt K 0 (t) Z = 0 dV (~r)~ E (~r t) = ~; K 0(t)] : K (t) Denition 13.7.55 Suppose the K (t) in denition 13.7.52 is the set of positions occu- pied by the particles of a certain continuum at time t. Suppose  and ~v are the density of mass and the particle velocity in the contiuum. Then dene E i) ~E (~r t) =
~ (~r t)=E (~r t) $E $E E ii) F (~r t) = F (~r t) ; ~vE (~r t)
~ (~r t). $ $ Let ~,
~ , F , F be, respectively the physical quantities whose Eulerian descriptions $E E $E $ E are ~E ,
~ , F , F . Then the denitions of ~ E and F imply ~ =
~ = or
~ = ~ (13.7.4) $ $ $
~ = F ;~v
~ or F =F +~v~: (13.7.5) The physical quantity ~ is called the density of stu per unit mass of continuum material (\material density of the stu," for short), while
~ is the density of stu per unit volume $ of physical space. The physical quantity F is the ux density of the stu relative to the $ material in the continuum, or the material ux density, while F is the ux density of the stu in space. 13.7. STUFF 219 $ Remark 13.7.69 Suppose the two pairs of functions (
~  F ) and ( F~ ) are related by (13.7.4). Then $ $ @t
~ + @~ F = Dt ~ + @~ F $ so the creation rate of the stu (
~  F ) is $ ~ = Dt ~ + @~ F : (13.7.6) (13.7.7) Proof:   @t
~ = (@t ) ~ +  @t ~ i $ h @~  F~ = @~ F + @~  (~v) ~ + (~v)  @~~ h i h  i $ @t
+ @~  F~ = @t  + @~  (~v) ~ +  @t ~ + ~v  @~~ + @~ F $ = 0~ + Dt ~ + @~ F : QED. Stus can be added if their densities and uxes have the same domain and same range. We have $ Denition 13.7.56 Suppose (
~  F ) for each v = 1     N is a stu, and that all these stus have the same K (t) and V in denition (13.7.1). Suppose c1     cN 2 R. $ $ Then PN =1 c (
~  F ) stands for the stu (PN =1 c
~ ) (PN =1 c F ). $ Corollary 13.7.48 The stu PN =1 c (
~  F ) has creation rate PN =1 c  . Proof: Obvious from (13.7.3). Corollary 13.7.49 In a continuum with particle velocity ~v and density , the material density and material ux density of the stu PN =1 c (
 F~ ) are, respectively PN =1 c  and PN =1 c F~ v. 220 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM Proof: Obvious from (13.7.4) and (13.7.5). Now we consider a number of stus which arise out of the conservation laws. The name of each stu is capitalized and underlined, and then its spatial density
~ , spatial $ $ ux F , material density ~, material ux F , and creation rate ~ are given. The stu is $ $ not dened by
~ or ~ alone. Both
~ and F , or both ~ and F must be given. Then  $ is calculated from (13.7.3). The choice of F is dictated by convenience in applications. Mass:
= , F~ = ~v,  = 1, F~ = ~0,  = @t
+ @~  F~ = @t  + @~  (~v) = 0. Mass is not created, and its material ux is zero. $ Momentum:
~ = ~v, F = ~v~v; $S , ~ = ~v, F$= ; $S , ~ = Dt +@~ F~ = Dt~v ;@~ $S = f~ . $ Intrinsic Angular Momentum:
~ = ~l , F = ~v~l; M$ , ~ = ~l, F$= ; M$ , so  = $ $ Dt~l ; @~ M = m ~ + AP h2i S . Kinetic Angular Momentum: $ $
~ = ( ~r ~v) F = ~v( ~r + ~v)+ S ~r  $ $ ~ = ~r ~v F =S ~r $   $ $ ~ = Dt ( ~r ~v) + @~  S ~r = ~r ~a ; ~r @~ S ; AP h2i S $ ~ = ~r f~ ; AP h2i S : Total Angular Momentum:   $   $ $
~ =  ~l + ~r ~v  F = ~v ~l + ~r ~v ; M + S ~r $ $ $  = ~l + ~r ~v F = ; M + S ~r : Since total angular momentum is the sum of intrinsic and kinetic angular momentum in the sense of denition (13.7.4), its creation rate is the sum of their creation rates, by corollary (13.7.48). Thus ~ = m ~ + ~r f~: 13.7. STUFF 221 Internal Energy:
= U F~ = ~vU + H~  = U F~ = H~  = Dt U + @~  H~ so $  = h+ S h2i@~~v: An alternative denition of internal energy suggests itself if h is entirely hnuc, the radioactive heating rate. Let Lnuc(~x t) be the nuclear energy in joules per kilogram of mass near particle ~x. As the radioactive nuclei near ~x decay, they emit  -rays and massive particles. We will assume that these lose most of their energy by collision with molecules so close to ~x that the macroscopic properties of the continuum are nearly constant over the region where the collisions occur. Then kinetic energy of molecular motion is added to the material near ~x at the rate ;Dt Lnuc(~x t) watts/kg. To convert to watts/meter 3 , we must multiply by the density, so hLnuc(~x t) = ;L (~x t)Dt Lnuc(~x t) or hnuc = ;Dt nuc: (13.7.8) We can regard U + nuc as total internal energy density, and dene a stu called TOTAL INTERNAL ENERGY with
=  (U + nuc)  F~ = ~v (U + nuc) + H~ ~  = U + nuc F~ = H: The creation rate of this energy is  = Dt  + @~  F~ = Dt U + @~  H~ + Dt nuc $ $ = h+ S h2i@~~v ; hnuc. If h = hnuc, then  =S h2i@~~v. This \total internal energy" is not very useful when the value of nuc is unaected by what happens to the material. That is the situation at the low temperatures and pressures inside the earth (with one possible small exception see Stacey, Physics of the Earth, p. 28). In such situations, hnuc= is a property of the material which is 222 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM known at any particle ~x at any time t, independent of the motion of the continuum. By contrast, the molecular U is not known at ~x t until we have solved the problem of nding how the continuum moves. In the deep interiors of stars, pressures and temperatures are high enough to aect nuclear reaction rates, so it is useful to include nuc with U . Kinetic Energy:  = = = =  =   ~F = ~v 1 v2 ; S$ ~v 2 $ F~ = ; S ~v $  1 2 ~ ~ ~ Dt  + @  F = 2 Dt v ; @  S ~v ~v  Dt~v ; @i (Sij vj ) relative to an orthonormal basis for P ~v  ~a ; (@i Sij ) vj ; Sij @ivj  $ $ ~a ; @~ S  ~v; S h2i@~~v $ f~  ~v; S h2i@~~v:
= 12 v2   = 12 v2 IK Energy: (Internal plus Kinetic Energy). This is the sum of internal energy and kinetic energy in the sense of denition 13.7.56, so it has     $
=  12 v2 + U  F~ = ~v 12 v2 + U + H~ ; S ~v $ F~ = H~ ; S ~v  = 21 v2 + U  = h + f~  ~v: Potential Energy: Suppose the body force density f~ is given by f~ = ;@~ where the potential function E (~r t) is independent of time, i.e., @t  = 0. Then Dt = @t  + ~v  @~ = ~v  @~ so Dt  = ;f~  ~v. Then the stu \potential energy" is dened by  = 
=  F~ = ~0 F~ = ~v 13.7. STUFF 223 so its creation rate is  = Dt  = ;f~  ~v   . IKP Energy: (Internal plus kinetic plus potential energy). If f~ = ;@~ with @t  = 0, then we dene the stu \IKP energy" as the sum of internal, kinetic and potential energies in the sense of denition (13.7.56). Then $  = 12 v2 + U +  F~ = H~ ; S ~v  1  1 $ 2 2 ~
=  2 v + U +   F = ~v 2 v + U +  + H~ ; S ~v: By corollary 13.7.48,  is the sum of the three  's for kinetic, internal and potential energy, so  = h. If we include nuclear potential energy in U and neglect other sources of h (e.g. solar heating), then  = 0. Usually, this is not done, and IKP energy is called \total energy." 13.7.1 Boundary conditions So far we have dealt only with physical quantities which were continuously dierentiable to the extent needed to justify using Gauss's theorem and the vanishing integral theorem. Now we analyze what happens when there is a surface S in the continuum, across which physical quantities may have jump discontinuities. That is, lim E (~r t) exists as ~r approaches ~r0 2 S from either side of S , but the two limits are dierent. Examples of such surfaces are the interface between the ocean and atmosphere, the ocean bottom, the core-mantle boundary, and a shock wave in air, water or rock. (Shock waves in rock are excited only by nuclear blasts and meteor impacts. Earthquakes are too weak to excite them.) We will assume that it is possible to choose a front side of S (M.obius strips are excluded) and that we have done so. We will denote the unit normal to S on its front, or positive, side by ^, and we will call its back side the negative side. S is a large surface, not nearly plane, but piecewise smooth, so ^ can vary with position ~r 2 S . Moreover, we will permit S to move, so that S (t) may vary with time t. Then so will ^(~r t), the normal to S (t) at ~r 2 S (t). We will write W (~r t) for the velocity with which S (t) moves 224 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM ^ ) ν( r S r . + front back Figure 13.17: normal to itself at point ~r 2 S (t), and we take W > 0 if the motion is in the direction of ^, W < 0 if opposite. If the surface is given in physical space P , we write it as S E (t). If it is given in label space L, we write it as S L(t). If  is any physical quantity with a jump discontinuity across S (t), we write E (~r0  t)+ for the limiting value of E (~r t) as ~r ! ~ro 2 S E (t) from in front of S E (t). We write E (~r0  t); for the limit of E (~r t) as ~r ! ~r0 from behind S E (t). And we dene h i+ E (~r0 t) ; = E (~r0 t)+ ; E (~r0 t);: (13.7.9) This quantity is called the jump in E across S E (t). We introduce a similar notation for label space, with hL i+  (~x0  t) ; = L (~x0  t)+ ; L (~x0  t); : We will need the surface version of the vanishing integral theorem. This says that if V is a Euclidean vector space, S is a surface in P or L, and f~ : S ! V is continuous, and R dA(~r)f~(~r) = O~ for every open patch S 0 on S , then f~(~r) = O~ for all ~r 2 S . Here an V V S0 \open patch" on S is any set S 0 = K 0 \ S , where K 0 is an open subset of P or L. The 13.7. STUFF 225 proof of the surface version of the vanishing integral theorem is essentially the same as the proof of the original theorem 11.5.26, so we omit it. We will consider only the Eulerian forms of the conservation laws, so we will omit the superscript E . The reader can easily work out the Lagrangian forms of the law. We use the notation introduced on page 163, but now we assume that the surface S (t) passes through K 0 (t). Therefore, we must augment the notation. We use K 0 (t)+ for the part (excluding S (t)) of K 0(t) which is in front of S (t), @ + K 0(t) for the part of @K 0 (t) in front of S (t), and @K 0 (t)+ for the whole boundary of K 0 (t)+, including K 0(t) \ S (t). If  is a function dened and continuous on the closed set K 0 (t) = K 0 (t) @K 0 (t), except for a jump discontinuity on S (t), we write + for the continuous function on K 0(t)+ dened as  except on K 0(t) \ S (t), where its value is the + of equation (13.7.9). We label objects behind S (t) in the same way, except that + is replaced by ;. The unit outward normal to @K 0 (t)+ will be written n^+ . The unit outward normal to @K 0 (t); will be written n^ ;. The unit outward normal to @K 0 (t) will be written n^ . Then n^+ = n^ on @ + K 0(t) and n^ + = ;^ on K 0 (t) \ S (t). And n^; = n^ on @ ; K 0(t), while n^; = ^ on K 0 (t) \ S (t). As on page 163, we choose K 0(t) so that it always consists of the same particles. Then the outward normal velocity of @K 0 (t)+ relative to itself is n^  ~ on @ + K 0(t) and ;W on K 0(t) \ S (t). Here ~v is the particle velocity in the continuum. The outward normal velocity of @K 0 (t); relative to itself is n^  ~v on @ ; K 0 (t) and W on K 0(t) \ S (t). 13.7.2 Mass conservation Let M K 0 (t)] denote the mass of the material in K 0(t). Since K 0(t) always consists of the same particles, the physical law of mass conservation is d M K 0(t)] = 0: (13.7.10) dt We need a mathematical identity like (13.1.9). We begin with Z 0 M K (t)] = dV (~r) (~r t) : (13.7.11) K 0 (t) This looks obvious until we recognize that there are physical situations which we might want to model by placing a mass per unit area on S (t), which would have to be added 226 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM Figure 13.18: 13.7. STUFF 227 to (13.7.11). Examples are S (t) = the soap lm in a soap bubble in air, or S (t) = the ice pack on a polar sea. We ignore these possibilities and assume mass on S (t) = 0: (13.7.12) Next, we try to dierentiate (13.7.10) using (11.6.1). The jump discontinuity on S (t) \ K 0(t) prevents this, so we break (13.7.11) into h i h i M K 0(t)] = M K 0 (t)+ + M K 0(t); with h i Z M K 0(t)+ = 0 + dV (~r)+(~r t) K (t) h 0 ;i Z M K (t) = 0 ; dV (~r); (~r t): (13.7.13) (13.7.14) (13.7.15) K (t) The two integrands in (13.7.14) and (13.7.15) are continuous, so we can apply (11.6.1) to obtain Z Z d M hK 0 (t)+i = Z ++ ++ ; dV ( ~ r ) @  dA n ^  ~ v dA W+: (13.7.16) t dt K 0 (t)+ @ + K 0 (t) K 0 (t)\S (t) We would like to use Gauss's theorem on the right in (13.7.16), but @ + K 0 (t) is not a closed surface. To close it we must add K 0(t) \ S (t). Then Z @ + K 0(t) dAn^  (~v)+ = = Z Z@K 0(t)+ dA n^ +  (~v)+ ; dV @~  (~v)+ + + K 0 (t) Z Z K 0 (t)\S (t) K 0(t)\S (t) dA n^ +  (~v)+ dA ^  (~v)+ Using this in (13.7.16) gives h i+ Z d M hK 0 (t)+i = Z ~ dV @  + @  ( ~ v ) + 0 dA  (^  v^ ; W )]+ : t dt K 0 (t)+ K (t)\S (t) But @t  + @~  (~v)]+ in K 0(t)+ is just the value of @t  + @~  (~v) there, so it vanishes, and d M hK 0 (t)+i = Z dA (^  ~v ; W )]+ : 0 dt K (t)\S (t) By an exactly similar argument, d M hK 0 (t);i = Z dA (^  ~v ; W )]; : dt K 0 (t)\S (t) (13.7.17) (13.7.18) 228 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM Therefore, d M K 0(t)] = Z dA (^  ~v ; W )]+; : dt K 0(t)\S (t) Combining (13.7.10) and (13.7.19) gives for K 0 = K 0(t) (13.7.19) Z dA (^  ~v ; W )]+; = 0: (13.7.20) Since K 0 can be any open subset of K (t), K 0 \ S (t) can be any open patch on S (t). Therefore, by the surface version of the vanishing integral theorem K 0 (t)\S (t)  (^  ~v ; W )]+; = 0 on S (t): (13.7.21) By analogy with denition 13.7.53 ii), we call either (^  ~v ; W )]+ or (^  ~v ; W )]; the ux of mass across S (t), and write it m(~r t) since it depends on ~r and t. Thus m = (^~v ; W )]+ = (^~v ; W )]; : (13.7.22) If m = 0, S (t) is a boundary between materials. It is then called a contact surface. It is not crossed by particles. Then ^  v~ + = ^  ^; = W: (13.7.23) If m 6= 0, S (t) is a shock wave or shock front. 13.7.3 Momentum conservation Again refer to gure 13.18. Let P~ K 0(t)] be the momentum in K 0 (t), while F~ K 0(t)] is the force on that material. Then mometum conservation requires d P~ K 0 (t)] = F~ K 0 (t)] : dt (13.7.24) We must nd mathematical identities for the two sides of (13.7.24). Because of (13.7.12), we have Z P~ K 0(t)] = 0 dV ~v: (13.7.25) K (t) 13.7. STUFF 229 Again we cannot appeal directly to (11.6.1) to dierentiate (13.7.25), because of the jump discontinuity on S (t). Therefore we imitate the procedure beginning on page 224. We have h i h i P~ K 0 (t)] = P~ K 0 (t)+ + P~ K 0 (t); (13.7.26) where h i Z ~P K 0 (t)+ = dV (~v)+ K 0 (t)+ h i Z P~ K 0(t); = 0 ; dV (~v);: K (t) Then, appealing to (11.6.1), we have d P~ hK 0 (t)+i = Z dV @t (~v)+ dt KZ0 (t)+ + + 0 dA n^  (~v~v )+ Z@ K (t) ; K 0(t)\S(t) dA W (~v)+ and Z @ K 0 (t) + dA n^  (~v~v)+ = Z @K 0 (t)+ ; = Z K 0 (t)\S (t) KZ0 (t)+ + Thus Z dA n^ +  (~v~v)+ dA n^ +  (~v~v)+ dV @~  (~v~v)+ K 0(T )\S (t) dA ^  (~v~v)+ : h d P~ hK 0 (t)+i = Z ~  (~v~v)i+ dV @ ( ~ v ) + @ t dt KZ0 (t)+ + 0 dA (^  ~v ; W ) ~v ]+ : K (t)\S (t) Now h i h i @t (~v) + @~  (~v~v) = @t  + @~  (~v) ~v +  @t~v + ~v  @~~v $ = Dt~v = ~a = @~ S +f~ (13.7.27) (13.7.28) 230 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM so, using Gauss's theorem $+ d P~ hK 0(t)+i = Z + S dA n ^ dt @K Z 0(t)+ Z ~ + 0 + dV f + 0 dA (^  ~v ; W ) ~v]+ Z K (t) Z K (t)\S(t) $ = 0 + dV f~ + + 0 dA n^  S K (t) @ K (t$) + Z + 0 dA ;^ S + ((^  ~v) ; W ) ~v : K (t)\S (t) Using 13.7.22, we nally get Z $ d P~ hK 0 (t)+i = Z ~++ dV f dA n ^  S + 0 dt K 0 (t)+  @ K$(t) + Z + 0 dA ;^ S +m~v : (13.7.29) K (t)\S (t) An exactly similar argument gives Z $ d P~ hK 0(t); i = Z ~+ dV f dA n ^  S dt K 0(t); @ ; K 0 (t)  Z $; ; K 0(t)\S(t) dA m~v ; ^ S : (13.7.30) Therefore, adding (13.7.29) and (13.7.30), $ d P~ K 0(t)] = Z dV f~ + Z dA n ^  S dt K 0 (t) @K 0 (t)  Z $+ + 0 dA m~v ; ^ S : (13.7.31) K (t)\S (t) ; We still need a mathematical identity for F~ K 0 (t)] in (13.7.24). This seems easy: Z Z $ ~ ~F K 0 (t)] = dV f + dA n ^  : (13.7.32) S 0 0 K (t) @K (t) But (13.7.32) does involve assumptions. If S (t) is an air-water interface holding an electric charge per unit area, there will be an electrostatic force per unit area on S (t) which must be added to (13.7.32). In addition, the surface integral on the right in (13.7.32) assumes that the stress on the material just inside @K 0 (t) by the material just outside @K 0 (t) $; $+ is n^  S behind S (t) and n^ S in front of S (t), all the way to S (t). In fact, if S (t) is an air-water interface, for example, the water molecules just behind S (t) and the air 13.7. STUFF 231 molecules just in front will be arranged dierently from those in the bulk of the air and water. Therefore, along the curve @K 0 (t) \ S (t) there will be an additional force per unit length exerted by the molecules just outside @K 0 (t) on the molecules just inside @K 0 (t). This force per unit length is called the surface stress in S (t). For an air-water interface it is tangent to S (t) and normal to @K 0 (t) \ S (t), and its magnitude is the surface tension. We assume there is no surface stress or surface force per unit area on S (t): (13.7.33) Then (13.7.32) is correct, and substituting (13.7.31) and (13.7.32) in (13.7.24) gives  Z $+ dA m~v ; ^ S = 0 (13.7.34) K 0\S (t) ; for any open subset K 0 of K (t). By the surface version of the vanishing integral theorem, it follows that  $+ m~v ; ^ S = 0 on S (t): (13.7.35) ; If S (t) is a boundary between two materials rather than a shock, then m = 0 so (13.7.35) reduces to  $+ ^ S = 0 on S (t): (13.7.36) ; At a boundary between two materials (a contact surface), the stress on the boundary (the \normal stress") is continuous across that boundary. Equation (13.7.35) has a simple physical interpretation. Consider a small patch dA on S (t). Mass mt dA crosses dA from back to front in time t, and gains momentum mt ~v]+; dA = m~v]+; tdA. Therefore, the patch requires momentum to be supplied to it at the rate m~v ]+;dA per second. Where does this come from? In the material the $ ux density of momentum is ; S , so momentum arrives at the back of dA at the rate + ; ;^ $S dA and leaves the front at the rate ;^ S$ dA. The net momentum accumulation $ $+ rate in dA, available to supply the required m~v]+;dA, is ;^  S ; dA + ^  S +dA $ = ^  S ]+;dA. Thus (13.7.35) simply says that the dierence in momentum ux into the two sides of dA supplies the momentum needed to accelerate the material crossing dA. If m = 0, this reduces to action = reaction, i.e. (13.7.36). 232 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM 13.7.4 Energy conservation The physical law of conservation of energy is d E K 0(t)] = W K 0(t)] dt (13.7.37) where E K 0(t)] is the total kinetic plus internal energy in K 0(t) and W K 0 (t)] is the rate at which energy is being added. We refer again to gure 13.18 and follow the by now well-trodden path to nd mathematical identities for both sides of (13.7.37). We have h i h i E K 0(t)] = E K 0 (t)+ + E K 0 (t); where   + h i Z E K 0 (t)+ = K 0(t) dV  12 v2 + U with a similar formula for E K 0(t); ]. Then (13.7.38) +  1 + d E hK 0(t)+ i = Z 2+U dV @  v t dt 2 K 0(t)+  1 + Z Z + + 0 dA n^  ~v 2 v2 + U ; @ K (t)  1 + 2+U dA W  v : 2 K 0(t)\S (t) Also R = Thus, R K 0(t)+ h  i h  i  ~v 12 v2 + U + = R@0 K 0(t) dA n^ +  ~v 21 v2 + U + i h  ; RK 0(t)\S(t) dA n^+  ~v 12 v2 + U + h  i+ R h  i+ dV @~  ~v 21 v2 + U + K 0(t)\S(t) dA ^  ~v 12 v2 + U : ^ @ K 0 (t) dA n + +   1   1 + d E hK 0(t)+i = Z 2 2 ~ dV @t  2 v + U + @  ~v 2 v + U dt K 0(t)+ +  1 Z 2 + 0 dA (^  ~v ; W )  2 v + U : K (t)\S (t) Now on page 221 we see that IK energy is a stu with 1 -= 2 v2 + U  1  $ 2 ~  F = ~v 2 v + U + H~ ; S ~v: 13.7. STUFF h i 233 Therefore @t - + @~  F~ =  = h + f~  ~v for this stu. Thus @t ( 21 v2 + U ) + @~  h  1 2 i ~ ~ ~ $ ~v 2 v + U + @  H ; @  S ~v = h + f~  ~v, and    + d E hK 0(t)+i = Z ~  H~ + @~  S$ ~v ; h + f~  ~v dV ; @ dt K 0(t)+ +  Z + 0 dA m 21 v2 + U  Z K (t)\S(t) + Z $ + + ~ ~ = 0 + dV h + f  ~v + 0 + dA n^  ;H + S ~v K (t) @K (t) + 1 Z 2 + 0 dA m 2 v + U K (t)\S (t)   ~ + Z $  d E hK 0 (t)+i = Z ~ dV h + f  ~v + + 0 dA n^  ;H + S ~v dt K 0 (t)+ @ K (t)  1   $ + Z 2 + 0 dA m 2 v + U + ^  H~ ; S ~v : (13.7.39) K (t)\S (t) Similarly,   ~ ; Z $  d E hK 0(t);i = Z ~ dV h + f  ~v + + 0 dA n^  ;H + S ~v dt K 0 (t); @ K (t)  1   $ ; Z 2 ; K 0(t)\S(t) dA m 2 v + U + ^  H~ ; S ~v : (13.7.40) Therefore   d E K 0(t)] = Z dV h + f~  ~v + Z ~ + S$ ~v dA n ^  ; H dt K 0 (t) @ + K 0(t)  1   $ + Z + 0 dA m 2 v2 + U + ^  H~ ; S ~v : (13.7.41) K (t)\S (t) ; Also, W K 0(t)] = Z  ~  Z  ~ ~  dV h + f  ~ v + dA n ^  ;H + S  ~v : K 0(t) @K 0(t) Therefore, (13.7.37) implies  1   $ + 2 dA m 2 v + U + ^  H~ ; S ~v = 0 (13.7.42) K 0 \S (t) ; for every open subset K 0 of K (t). By the surface version of the vanishing integral theorem,   1 $ + (13.7.43) m 2 v2 + U + ^  H~ ; ^ S ~v = 0 on S (t): ; Z 234 CHAPTER 13. CONSERVATION LAWS IN A CONTINUUM If the material is stationary, m = 0 and ~v = 0, so (13.7.43) reduces to the thermal boundary condition h ~ i+ ^  H ; = 0: (13.7.44) $ $ At a material boundary which is not a shock, m = 0, so ^  S ]+ = ^  S ];, and $ $ $ $ $ $ $ ^  S ~v]+; = ^  S ]+  ~v+ ;(^  S );  ~v; = (^  S )  ~v]+; where ^ S = ^  S ]+ = ^  S ];. Also, from (13.7.23), if ~v = v ^ + ~vS where vv = ^  ~v, then v + = v ;, so ~v]+; = ~vS ]+. Therefore (13.7.43) becomes h ~ i+  $ ^  H ; = ^ S  ~vS ]+; : (13.7.45) The quantity ^  H~ ]+; is the net heat ow per unit area per second from the boundary into the surrounding medium. Equation (13.7.45) says that heat is generated mechanically when the two materials slip past one another at the relative velocity ~vS ]+; and when the $ surface stress (^  S ) on the boundary has a component in the direction of the tangential relative slip velocity. Brune, Henyey and Roy (J. Geophys. Res. 74, 3821) used the absence of an observed heat ow anomaly along the San Andreas fault, together with an $ estimate of ~vS ]+; on the fault, to put an upper bound on the component of ^ S in the direction of slip. They used (13.7.45). Even when m 6= 0, (13.7.43) can be made to resemble 13.7.45. We have  1 + 1 m v~ + ; v~ ;  v~ + + v~ ; 2 mv = 2 2  ;  $+ $;  1 = 2 ^  S ; S  v~ + + v~ ; from 13.7.35 so 1 2 Dene $ mv2 ; ^ S ~v +  $+ $; + $+ ~; $; ~; 1 + = 2 ^  ; S v~ ; S v~ + S v + S v ; $+ $; = ;^  1 S + S  ~v]+; : 2 $ + 1 $+ $; hS i; = 2 S + S = average S$ on S: Then (13.7.43) can be written in general as h ~ i+ $ ^  H + m U ; = ^  hS i+;  ~v]+; : (13.7.46) (13.7.47) Chapter 14 Strain and Deformation 14.1 Finite deformation Consider a continuum which at time t occupies open subset K (t) of physical space P . Suppose K 0 (t0) is a very small open subset of K (t0 ). We would like to be able to visualize the particles in K 0(t0 ), and to see their spatial positions at a particular later time t. The time dierence t ; t0 is not small, but K 0 (t0) is a very small set, in a sense to be made precise shortly. We use t0 -position labelling. Then label space L is the same as physical space P , and ~r L(~x t) = position at time t of the particle which was at position ~x at time t0 . The particles in which we are interested have their labels in the set H 0 = K 0(t0 ). First choose a particular particle ~xA 2 K 0(t0 ). Let ~rA be its position at time t. Then ~rA = ~r L(~xA t). Consider any other particle ~x 2 K 0(t0 ), and let ~r be its position at time t. Then ~r = ~r L(~x t). Dene ~ = ~r ; ~rA, ~ = ~x ; ~xA. Then ~ = ~r L(~x t) ; ~r L (~xA t) = ~r L(~xA + ~ t) ; ~r L(~xA t) so  ~ r L (~xA  t) + k~kR~ ~ ~ = ~  D~ where R~ (~) is the remainder function for ~r L( t) at ~xA . 235 (14.1.1) 236 . . r L . CHAPTER 14. STRAIN AND DEFORMATION ( , t) x ρ . ξ . r rA x A K ′(t) H ′=K ′(t 0 ) Figure 14.1: Now we assume that K 0 (t0) has been chosen so small that for every ~x 2 K 0(t0 ),  ~ r L (~xA t) k: kR~ ~ k << kD~ (14.1.2) This is the sense in which K 0 (t0) must be a small set. If (14.1.2) is satised, then we can neglect the remainder term in (14.1.1) and write ~ r L (~xA t) : ~ = ~  D~ (14.1.3) If we measure initial positions in K 0 (t0) as displacement vectors from an origin at ~xA , and nal positions in K 0 (t) as displacement vectors from an origin ~rA, the mapping from initial to nal positions ~ 7;! ~ is linear, and is (14.1.3). Thus the eect of the motion on the particles in K 0(t0 ) is clear. The particle at ~xA is displaced to ~rA. The relative positions of nearby particles, relative to particle ~xA 's position, suer the linear mapping (14.1.3). ~ r L(~xA t0)j is never 0 for any t0 . But ~r L(x t0 ) = As we have noted (see 12.2.7), j det D~ ~ r L(~x t0 ) =$I P and det D~ ~ r L(~xA  t0) = 1. Then, since det D~ ~ r L(~xA  t0) depends ~x, so D~ continuously on t0 and is never 0, it cannot change sign, so it is always positive. Therefore ~ r L (~xA t) > 0: det D~ (14.1.4) It follows from the polar decomposition theorem (page D-27) that there are unique tensors $ $ C and R2 P  P with these properties: 14.1. FINITE DEFORMATION 237 $ i) C is symmetric and positive denite, $ ii) R is proper orthogonal (i.e., a rotation through some angle between 0 and  about some axis w^). iii) Thus (14.1.3) is $ ~ r L (~xA  t) =C$  R D~ : (14.1.5) $ $ ~ = ~ C  R : (14.1.6) Thus the eect of the motion on K 0 (t0) is to displace ~xA to ~rA then to subject the relative position vectors (relative to particle ~xA) to the symmetric operator C  and then to rotate everything through angle about an axis passing through ~rA in the direction of w^. $ The displacement ~xA 7;! ~rA and the rotation R are easy to visualize. The mapping $ $ $ C may be worth discussing. Since C T =C there is an orthonormal basis y^1, y^2, y^3 for P which consists of eigenvectors of C. That is, there are scalars c1, c2, c3 such that C (^yi) = c(i) y^(i)  i = 1 2 3: (14.1.7)  $ P $ $ $ $ $ Then y^i C = c(i) y^(i) so C = I P  C = (^yi y^i)  C = y^i y^i C = 3i=1 c(i) y^(i) y^(i) . Thus $ (14.1.8) C = c1 y^1y^1 + c2y^2y^2 + c3y^3y^3: $ Since C is positive denite, c1 and c2 and c3 must all be positive. Now ~ = 1y^1 +2y^2 +3y^3 $ so ~ C = c11y^1 + c2 2y^2 + c3 3y^3. If we imagine a cube with edges of unit length parallel $ to the axes y^1, y^2, y^3, the eect of C is to multiply the i'th edge of the cube by ci, transforming that cube to the rectangular parallelipiped in gure 14.2. Suppose that in K 0(t0 ) we imagine a small cube, with edges of length " parallel to the $ axes y^1, y^2, y^3 which are eigenvectors for C . We can use the cube on the left in 14.2 by shrinking its edges to " and placing the hidden vertex at ~xA. Then the eect of ~r L ( t) on this cube is i) to translate it so the hidden vertex is at ~rA 238 CHAPTER 14. STRAIN AND DEFORMATION ^y 3 c2 ^y 3 1 c3 1 ^y 2 1 ^y ^y 2 c1 ^y 1 1 Figure 14.2: 14.1. FINITE DEFORMATION 239 ii) to stretch or compress it by the factor ci in the direction y^i, as shown in Figure 14.2 iii) to rotate it through angle , with 0   , about an axis passing through ~rA in the direction w^. Now (14.1.5) implies $ $0 ~ r L (~xA  t) =R D~ C (14.1.9) where 3  $  $ $0 $;1 $ $ X C =R  C  R= ci y^i R y^i R : (14.1.10) i=1 $ Therefore we can perform the rotation R as step ii), and the stretching or compression as step iii). But then, from (14.1.10), the orthogonal axes along which stretching or $ $ $ compression occurs are the rotated axes y^1 R, y^2 R, y^3 R. All this motivates the following denition: Denition 14.1.57 Suppose a continuum is labelled with t0-positions, and ~r L : K (t0 ) R ! P is the resulting Lagrangian description of the motion. Suppose t 2 R. Then ~r L( t) : K (t0 ) ! K (t) is called the \deformation" of the continuum from t0 to t. For ~ r L(~xA  t) is the \deformation gradient" at particle ~xA . The rotation any ~xA 2 K (t0), D~ $ $ R in (14.1.5) and (14.1.9) is called the \local rotation at particle ~xA", and C in (14.1.5) $ is the \prerotation stretch tensor at ~xA ", while C 0 in (14.1.9,14.1.10) is the \postrotation $ $ L ~ r L(~xA  t) as G stretch tensor at ~xA ." We will write D~ t0 (~xA  t). For any t0 , Gt0 , is a physical quantity. $ $ Various tensors obtained from C or C 0 are called nite strain tensors. The commonest $ $ is probably C ; I P , and the most useful is probably 3 X (ln ci) y^iy^i: i=1 All these strain tensors reduce to the innitesimal strain tensor when the deformation is $ $ nearly the identity mapping (see next section), and to 0 when ~r L( t) = I P . In the foregoing discussion, the displacement of particle ~xA to position ~rA was so easy to visualize that we said very little about it. It is worth considering. If we x t0 , the 240 CHAPTER 14. STRAIN AND DEFORMATION displacement vector of any particle ~x at time t is ~sL (~x t) = ~r L (~x t) ; ~r L (~x t0 ) : (14.1.11) As the notation indicates, we will consider the physical quantity ~s, called displacement, whose Lagrangian description is the ~sL dened by (14.1.11). The denition requires that we x t0. If we change it, we change the denition of ~s.   For t0 -position labelling, ~r L(~x t0 ) = ~x = ~x L (~x t) so ~sL(~x t) = ~r L ; ~x L (~x t). Therefore ~s = ~r ; ~x for t0-position labelling: (14.1.12) Equation (14.1.12) makes it convenient to use t0 -position labelling when studying dis~ x =$I P so D~ ~x =$I P . placements. In particular, for such labelling, (D~ ~x )L(~x t) = D~ Thus, from (14.1.12) ~ r =$I P +D~ ~ s for t0-position labelling: D~ $ $ Also, (@~ ~r )E (~r t) = @~~r = I P so @~ ~r = I P . Then from (14.1.12) (14.1.13) $ @~ ~x = I P ;@~~s for t0 -position labelling: (14.1.14) ~ r ) =$I P , so from (14.1.13) and (14.1.14), ($I P ; $@ ~s) ($I P +D~ ~ s) =$I P , or But (@~ ~x )  (D~ $ ~ ~ ~ ~ $ I P +D~s ; @~s ; @~s  D~s = I P , or   ~ s = @~~s + @~~s  D~ ~ s = @~~s  $I P +D~ ~s : D~ (14.1.15) 14.2 Innitesimal displacements We use t0-position labelling throughout this section, and we discuss the displacement ~s dened by (14.1.11) or (14.1.12). Denition 14.2.58 A tensor T$2 P  P is \small" if k T$ k << 1. That is k T$ h2i T$ $ $ k << 1. A tensor Q2 P  P is \close to the identity" if Q ; $I P is small. 1 2 14.2. INFINITESIMAL DISPLACEMENTS 241 $ $ Remark 14.2.70 Suppose T$2 P  P and k T$ k < 1. Dene (T$)n =T n, so T 0=$I P , $ $ $ $ $ $ $ $ $ $ $ $ T 1=T , T 2=T  T , T 3=T  T  T , etc. Then ( I P + T );1 and limN !1 PNn=0(;1)n T n both exist, and they are equal. Proof: k N X n=M $ (;1)n T n k  N X $ n k T$ kM kT k  $ : 1;k T k n=M As M , N ! 1, this ! 0, so the Cauchy convergence criterion assures the $ existence of limN !1 PNn=0(;1)n T n. Then $ $ X N N N $ $n $n X X n n (;1) T + (;1)n T n+1 I P + T  (;1) T = n=0 n=0 n=0 N +1 $ $ X n $n NX $N +1 = (;1) T ; (;1)n T n= I P +(;1)N T : n=0 n=1 $N +1 $ But k T k  k T kN +1, so letting N ! 1 on both sides gives $ $ X 1 $ $ +  (;1)n T n= I P : IP T n=0 QED ~ sk << 1, then k@~~sk << 1, and to rst order in D~ ~ s, Corollary 14.2.50 If kD~ ~ s = @~~s: D~ (14.2.1) Proof:   $ ~ ;1 n ~ s n . Then By remark 14.1.14, ( I P +D~ s) exists and equals P1 n=0 (;1) D~ ~ s);1k  P1 ~ skn = 1=(1 ; kD~sk). By (14.1.15) , @~~s = D~ ~ s  ($I P k($I P +D~ n=0 kD~ ~ s );1, so +D~     ~ skk IP + D~ ~ s ;1 k  kD~ ~ sk= 1 ; kD~ ~ sk : k@~sk  kD~ 242 CHAPTER 14. STRAIN AND DEFORMATION Remark 14.2.71 If T$2 P  P and T$ is symmetric and k T$ k < 1, then $I P + T$ is positive denite and its positive denite square root is 0 1 $ $1=2 X 1 B 12 C $n IP + T = @ A T n=0 n where (14.2.2) 0 1 B@ N CA = N (N ; 1)    (N ; n + 1)=n! (= 1 if n = 0): n Proof: $ $ $ If k T k < 1, all T 's eigenvalues lie between ;1 and 1. Therefore I P + T has all itseigenvalues between 0 and 2. Therefore, it is positive denite. That  $ P1 1n=2 T n exists, is positive denite, and when squared gives $ + $ IP T n=0 can be proved in the same way as remark (14.2.70). Alternatively, the eect   1=2 $ of PNn=0 n T n can be studied on each of the vectors in an orthonormal $ basis for P consisting of eigenvectors of T . Then one simply uses the known $ $ validity of (14.2.2) when I P is replaced by 1 and T by any of its eigenvalues. $ Corollary 14.2.51 Suppose T T =T$ and k T$ k < 1. Then correct to rst order in k T$ k, $ $1=2 $ 1 $ I P + T =I P +2 T : Theorem 14.2.29 Use the notation of denition (14.1.9), and dene the displacement ~s ~ sk << 1. Then to rst order in D~ ~ s we have by (14.1.12). Suppose that kD~ ~s @~~s = D~ i $ $0 $ 1 h ~ ~ s (~xA t)T s (~xA  t) + D~ C =C = I P + 2 D~ i $ $ 1 h~ ~ s (~xA  t)T : s (~xA t) ; D~ R= I P + 2 D~ Proof: (14.2.3) (14.2.4) (14.2.5) 14.2. INFINITESIMAL DISPLACEMENTS 243 ~ s(~xA  t) as T$. We have already proved (14.2.3). For the rest, abbreviate D~ $ $ $T $T $ $ $;1 ~ r =C$  R ~ r )  (D~ ~ r )T =C$  R Since D~ we have (D~  R  C =C  R  R $ $ $ $ $2  C =C  I P  C =C , so    T 1=2 "$ $ $ $ !#1=2 $ ~ r  D~ ~r = IP + T  IP + TT C = D~ "$ $ $ $ $ #1=2 = IP + T + TT + T  TT ( $ $ $ $ ) $ 1  $ $T $ $T ! = I P + 2 T + T + T  T + O k T + T T + T  T T k2  !  $  $ 1 $ $T = I P + T + T + O k T k2 : 2 $ $0  ~ T  ~  $0T $T $ $0 $0 $;1 $ $0 ~ r =R Also, D~  C , so D~r  D~r =C  R  R  C = C  R  R  C $0 $ $0  $02 =C  I P  C = C so  T  1=2 "$ $ ! $ $#1=2 ~ r  D~ ~r C~ 0 = D~ = IP + TT  IP + T "$ $ $ $ $# = IP + TT + T + TT  T ( $ $ $ $ ) $ 1  $T $ $T $! = I P + 2 T + T + T  T + O k T T + T + T T + T k2  $  $ 1  $ $T  = I P + T + T + O k T k2 : 2   $ n o $ $  $ $ ~ r =C ;1  $I P + T$ and C ;1=$I P ; 12 T$ + T T +O kT~ k2 Finally, R=C ;1 D~ so $  $ $ ! n o! $ $ $ 1 R = I P ; 2 T + T T + O kT~ k2  I P + T n o $ 1  $ $T ! = I P + 2 T ; T + O kT~ k2 : QED. Denition 14.2.59 Use t0-position labelling so the displacement ~s is dened by (14.1.12). ~ sL(~x t)k << 1. Then the tensor Suppose kD~ i $ L (~x t) = 1 hD~ ~ sL(~x t) + D~ ~ sL(~x t)T 2 (14.2.6) 244 CHAPTER 14. STRAIN AND DEFORMATION is called the innitesimal strain tensor at particle ~x at time t, and $ h~ L i ~ sL(~x t)T s (~x t) ; D~ wL (~x t) = 12 D~ is the innitesimal rotation tensor. The vector w~ L (~x t) = 12 AP h2i w$L (~x t) is called the innitesimal rotation vector at particle ~x at time t. (14.2.7) (14.2.8)   Corollary 14.2.52 w~ L(~x t) can be written 1=2D~ ~sL(~x t) or 1=2 @~ ~s L (~x t) because ~ s, and, to rst order in D~ ~ sL (~x t) w~ L (~x t) = 12 AP h2iD~ (14.2.9)  L w~ L (~x t) = 12 AP h2i @~~s (~x t) : (14.2.10) Proof: (14.2.10) follows from (14.2.9) and (14.2.3). To prove (14.2.9) note that for $ any tensor T we have $ $ AP h2i T T = ;AP h2i T : (14.2.11) To see this, look at components relative to a pooob in P . Then (14.2.11) is equivalent to "ijk Tkj = ;"ijk Tjk . But by changing index names we get "ijk Tkj = "ijkTjk = ;"ijk Tjk . Applying (14.2.11) to (14.2.7, 14.2.8) with $ ~ L s (~x t) gives (14.2.9). QED T = D~ Corollary 14.2.53 Let y^1, y^2, y^3 be an orthonormal basis for P consisting of eigenvectors ~ sL(~xA  t), of $ (~xA  t), where ~xA is as in denition 14.1.9. Then to rst order in D~ 3 X $ $0 $ $  = = + ( ~ x  t ) = (1 + i ) y^iy^i (14.2.12) C C IP A i=1 where 1 , 2, 3 are the eigenvalues of $ (~xA t) belonging to the eigenvectors y^1 , y^2, y^3. $ And R is a rotation in the right-handed-screw sense through angle kw~ L(~xA  t)k about the axis through O~ P in the direction w~ L(~xA  t). 14.2. INFINITESIMAL DISPLACEMENTS 245 ξ δφ ω sin θ θ ω x ξ ξ OP Figure 14.3: Proof: We show $ (~xA  t) = P3i=1 i y^iy^i in the same way we showed (14.1.8). The $ rest of (14.2.12) follows from (14.2.4). To prove the results about R, we note from exercise 9c that w$L (~xA t) = AP  w~ L (~xA  t) : ~s Then from (14.2.5). to rst order in D~ $ $ R= I P +AP  w~ L(~xA t): $ If we apply R to a vector ~, the result is $ ~ R= ~ + ~  AP  w~ L (~xA  t) : We claim ~  AP  ~v = ~v ~. Looking at components relative to a pooob,   (~  Ap  ~v)i = j "jik vk = "ikj vk j ~v ~ i. Therefore $ ~ R= ~ + w~ L (~xA  t) ~: (14.2.13) Abbreviate w~ L (~xA t) by w~ , and examine gure 14.3 where ~  R~ = ~ + w~ ~. Since kw~ ~k = kw~ kk~k sin , clearly  = kw~ k. QED. 246 CHAPTER 14. STRAIN AND DEFORMATION Now suppose t = t0 + t (14.2.14) where t is small enough that it is a good approximation to write ~r L (~x t) = ~r L (~x t0 ) + tDt ~r L (~x t0) + O (t)2 : (14.2.15) Then, since Dt ~r = ~v and ~sL(~x t) = ~r L(~x t) ; ~r L(~x t0), we have ~sL (~x t) = t~vL (~x t0 ) + O (t)2 (14.2.16) Then, to rst order in t, i $ L (~x t) =$ L (~x t + t) = t 1 hD~ ~ vL (~x t0 ) + D~ ~ vL (~x t0 )T 0 2 h~ L i ~ vL (~x t0)T w$L (~x t) =w$L (~x t0 + t) = t 1 D~ v ( ~ x  t ) ; D~ 0 2 h~ L i w$L (~x t) =w$L (~x t0 + t) = t 1 Aph2i D~ v ( ~ x  t ) 0 2 = t 12 D~ ~vL (~x t0) : (14.2.17) (14.2.18) (14.2.19) Consider the material near particle ~xA at various times t0 + t, with small t. Relative to particle ~xA, the material stretches as in (14.2.12), the fractional stretching i being proporh~ L i ~ vL (~xA  t0)T , tional to t, i = _ it, where _ i is one of the eigenvalues of 1=2 D~ v (~xA  t0) + D~ and y^i in (14.2.12) is the corresponding eigenvector. And relative to particle ~xA , the material also rotates through angle tk 12 D~ ~vL (~xA t)k about an axis through ~xA and in direction 1~ L ~ L 2 D ~v (~x0  t). In other words, it rotates with angular velocity 1=2D ~v (~xA  t). To rst $ $ order in t C 0=C , so the rotation can occur before or after the stretching. We prefer to think of them as occurring simultaneously. The foregoing analysis began by choosing a time t0 at which to introduce t0-position labels. We can choose t0 to be any time we like the foregoing results are true for all t0 2 R. Therefore, we can dene physical quantities $_ , w$_ , w~_ by requiring for any ~x 2 P and any t 2 r that 0 i $_ E (~x t ) = 1 hD~ ~ vL (~x t0) + D~ ~ vL (~x t0)T 0 2 (14.2.20) 14.2. INFINITESIMAL DISPLACEMENTS 247 h~ L i E ~ vL (~x t0)T w$_ (~x t0 ) = 1 D~ v ( ~ x  t (14.2.21) 0 ) ; D~ 2 E w$_ (~x t0 ) = 1 D~ ~vL (~x t0) : (14.2.22) 2 Here it is understood that to evaluate the left side of (14.2.20, 14.2.21, 14.2.22) at any particular t0 , we use that t0 to establish t0 -position labels in establishing the Lagrangian descriptions needed on the right. Each t0 on the left calls for a new Lagrangian description on the right, the one using t0-position labels. ~ r L(~x t0) =$I P , so @~E f E (~x t0) = D~ Lf L(~x t0 ) for any But with t0-position labels, D~ ~ vL can be replaced by @~~vE on the right in 14.2.20. Since physical quantity f . Therefore D~ the Eulerian description of physical quantities are independent of the way particles are labelled, this replacement leads to the conclusions $_ = 1 @~~v + @~~vT  (14.2.23) 2    w$_ = 1 @~~v ; @~~v T (14.2.24) 2 (14.2.25) w~_ = 21 @~ ~v: The quantity 2w~_ is called the vorticity, and $_ is the strain-rate tensor. 248 CHAPTER 14. STRAIN AND DEFORMATION Chapter 15 Constitutive Relations 15.1 Introduction Let us examine the conservation laws as a system. They are Eulerian Form Lagrangian Form Mass Dt  + @~  ~v = 0 Dt ~ = 0 (15.1.1) $ $ Momentum Dt~v = @~ S +f~ ~Dt~v = D~  S~ +~f~ (15.1.2) $ $ ~ = h~ + S~ h2iD~ ~ v: Internal Energy Dt U + @~  H~ = h+ S h2i@~v ~DtU + D~  H (15.1.3) If we knew the initial values of , ~v and U , or ~, ~v and U , at some instant t0 , we might hope to integrate (15.1.3) forward in time to nd later values of these quantities. To do this, clearly we must know f~ and h, or ~f~ and h~ , and we will assume henceforth that these $ $ ~ are given to us. But there still remains the question of nding H~ and S , or H and S~. The equations which give their values when the state of the material is known are called the \constitutive equations" of the material. Ideally, we would like to be able to deduce these from the molecular pictures of matter, but theory and computational techniques are not yet adequate to this task for most materials. We must rely on experimental measurements. The outcomes of these experiments suggest idealized mathematical models for various classes of real materials. 249 250 CHAPTER 15. CONSTITUTIVE RELATIONS ^n K ∂K Figure 15.1: In modelling a real material, one usually begins by considering its homogeneous thermostatic equilibrium states (HTES). Then one studies small deviations from thermostatic equilibrium, by linearizing in those deviations. Large deviations from thermodynamic equilibrium are not fully understood, and we will not consider them. They make plasma physics di"cult, for example. A homogeneous thermostatic equilibrium state (HTES) is said to exist in a material when no macroscopically measurable property of that material changes with time, and when the macroscopic environments of any two particles are the same. The second law of thermodynamics then implies that there can be no heat ow in the material. The material can have a constant velocity, but if it does we study it from a laboratory moving with the same velocity, so we may assume ~v = ~0 in an HTES. Then DtU = @t U . But @t U = 0 $ so, from (15.1.3), h = 0. The Cauchy stress tensor S must be the same everywhere at all $ times in an HTES, so @~ S = ~0, and (15.1.3 momentum]) implies f~ = ~0. Therefore, in an HTES we have  = constant H~ = ~0 ~v = ~0: U = constant h = 0 $ S = constant f~ = ~0 If the material occupies a volume K with boundary @K and unit outward normal n^ , then in order to maintain the material in an HTES we must apply to the surface @K a 15.1. INTRODUCTION 251 $ stress n^ S , and we must prevent surface heat ow and volume heating. That is we need 8 $ > > surface stress applied to @K at ~ r is n ^ ( ~ r )  S > < (15.1.4) n^(~r)  H~ = 0 at every ~r 2 @K > > > : h = 0 at every ~r 2 K: Two HTES's are dierent if they dier in any macroscopically measurable quantity, including chemical composition. Most pure substances require only the values of a few of their physical properties in order to specify their HTES's. For example, two isotropically pure samples of H2O (no deuterium, only O16) liquid which have the same density and the same internal energy per unit mass will have the same values of pressure, temperature, entropy per gram, electrical resistivity, index of refraction for yellow light, neutron mean absorption length at 15 kev, etc. For water, the set of all possible HTES's is a twoparameter family. (We have ignored, and will ignore, electrical properties. It is possible to alter U for water by polarizing it with an electric eld.) Changes from one HTES to another are produced by violating some of the conditions (15.1.4). Changes produced by altering the surface stress applied to @K are said to \do work," while changes produced by letting h 6= 0 are said to involve \internal heating or cooling," and n^  H~ 6= 0 on @K involves \heat ow at @K ." For the materials in which we are interested, these changes can be described as follows. We pick one HTES as a reference state, and call it HTES0 We use t0 -position labelling, where t0 is any time when the material is in HTES0 . We carry out a change by violating (15.1.4), and wait until time t1 when the material has settled into a steady state, HTES1. The Lagrangian description of the motion is ~r L(~x t), and it is independent of t for t  t0 and for t  t1 . We write ~r 01 (~x) = ~r L (~x t1) : $ ~ r 01 (~x) : G01 (~x) = D~ (15.1.5) (15.1.6) Since the environments of all particles must be alike in HTES1 , they must have been subjected to the same stretch and the same rotation so $ (15.1.7) G01 is constant, independent of ~x: 252 CHAPTER 15. CONSTITUTIVE RELATIONS $ (G01 is called the deformation gradient of HTES1 relative to HTES0.) Then from (15.1.6), $ ~r 01 (~x) = ~x G01 + ~r 01 (~0). If ~r 01 (~0) 6= ~0, we may move the material with a uniform displacement so ~r 01 (~0) = ~0. Thus we may assume $ ~r 01 (~x) = ~x G01 : (15.1.8) $ Here ~x is particle position in HTES0 and ~x G01 is particle position in HTES1. If we want to study another HTES, say HTES2, we can adopt either HTES0 or HTES1 as the $ reference state. If we use HTES1 for refrence, then particle position in HTES2 is ~x1  G12 $ where ~x1 is particle position in HTES1 and G12 is calculated from (15.1.6) but with t1 $ position labelling. If a particle has position ~x in HTES0, then ~x1 = ~x G01 and so ~x2 = $ $ $ $ position in HTES2 = ~x1  G12 = ~x G01  G12 . But also ~x2 = ~x G02 , so $ $ $ (15.1.9) G02 =G01  G12 for any HTES0 , HTES1, HTES2. 15.2 Elastic materials The observed HTES's of many material are approximated well by the following idealization: Denition 15.2.60 An elastic material is one whose HTES's have these properties: $ $ i) If HTES0 is any HTES, there is an M0 > 0 such that for any G2 P  P with k G ; $I P k < M0 , there is an HTES1 whose deformation gradient relative to HTES0 is $ $ G10 =G  ii) Once a reference state HTES0 is chosen, any other equilibrium state HTES1 is com$ pletely determined by its deformation gradient G01 relative to HTES0 , and by its entropy per unit mass N1. iii) U , the internal energy per unit mass, is unchanged by rigid rotations. That is, if $ $ G2 P  P and R2 #+(P ) then $  $ $  U G N = U G  R  N : (15.2.1) 15.2. ELASTIC MATERIALS 253 Now consider two dierent HTES's of an elastic material, say HTES1 and HTES2 , neither of them the reference state HTES0 . Suppose ~r L : K0 R ! P is the Lagrangian description of the motion which carries the material from state 1 to state 2. Here K0 is the region occupied by the material in HTES0 , so the total mass of the material is M = jK0j~ (15.2.2) where jK0j = volume of K0 . The total internal energy of the material is U i = MUi = jK0 j~Ui in HTESi : (15.2.3) Let 412W be the work done on the material and 412 Q the heat added to it in going from HTES1 to HTES2 . Energy conservation requires 412 W + 412 Q = U 2 ; U 1 : (15.2.4) Now suppose HTES1 and HTES2 are not very dierent. Then we will write df = f2 ;f1 for any property f of the two states. Thus (15.2.4) is d U = 412W + 412Q. According to the second law of thermodynamics, correct to rst order in small changes, 412Q  d N = ( N 2 ; N 1) (15.2.5) where N i = jK0j~Ni = total entropy of state i, and is the smaller of 1 and 2 . It is possible to nd processes ~r L in which equality is approximated arbitrarily closely in (15.2.5). These are called \reversible processes." In such processes 412Q = jK0 j~dN: (15.2.6) In any transition from state 1 to state 2, reversible or not, $ ~ ~ ~ ~v ~DtU = ;D  H + h+ S~ h2iD~ so, integrating over K0 and using Gauss's theorem, ~ Z K0 dVL(~x)Dt U = Z K0 dVLh~ ; Z @K0 Z $ ~ ~v ~ dAL n^L  H + dVL S~ h2iD~ K0 254 CHAPTER 15. CONSTITUTIVE RELATIONS or d U = Q_ + Z dV $~ h2iD~ LS ~ v dt K0 where Q_ is the rate at which heat is added to the material. Thus, integrating from t1 to t2 Z t2 Z $ ~ v: U 2 ; U 1 = 412Q + dt dVL S~ h2iD~ t1 Comparing with (15.2.4) we see that K0 Z t2 Z $ ~ v: dt dVL S~ h2iD~ t1 K0 $ Now if the transformation is done slowly, S~ will be nearly constant, about equal to its 412W = value in HTES1 or HTES2 so $ 412W =S~ h2i Z K0 dVL Z t2 ~ v: dtD~ t1 ~r L(~x t) ~ v = DD ~ t ~r , (D~ ~ v)L(~x t) = DD ~ t ~ r L (~x t) But ~v = dt ~r so D~ = Dt D~ $ $ ~ v = Dt G = (Dt G)L(~x t). Thus, D~ , and $  $ Z $ ~ 412W =S h2i K dVL(~x) G (~x t2) ; G (~x t1) : 0 $ $ But G (~x t1 ) and G (~x t2 ) refer to HTES's. They are independent of ~x. Thus $ 412W = jK0j S~ h2id G$ : (15.2.7) $ $ From (15.2.4), (15.2.6) and (15.2.7), d U = jK0j S~ h2id G +jK0j ~ dN . Since d U = jK0j~dU , dividing by jK0j~ gives $ $ S~ dU = d G h2i ~ + dN (15.2.8) $ for small reversible changes between HTES's. Here and S~ refer to either the initial or the nal state, since they are close. $ But U is a function of G and N . Equation (15.2.8) says this function is dierentiable, and that $  @ $  (15.2.9) G N = @N U G N 15.2. ELASTIC MATERIALS 255 $ $ S~ (G N ) = @~$ U $ N   (15.2.10) G G ~ h i $ where @~$G U (G N ) is an abbreviation for @~U ( N ) (G~ ). Since P  P is a Euclidean space, the theory of such gradients is already worked out. In particular, if y^1, y^2, y^3 is a pooob for P , then the dyads y^iy^j are an orthonormal basis for P  P , and $ (15.2.11) G= Gij y^iy^j : Then so, from (15.2.10) $ $  @U ( G @$G U G N = y^iy^j @G N ) ij $ S~ij = @U (G N ) :  @Gij To obtain the Cauchy stress tensor we use (13.3.30), namely 0 $1 $ 0$ 1 @ S A =GT  B@ S~ CA :  ~ The equation $  U = U G N (15.2.12) (15.2.13) (15.2.14) is called the equation of state of the elastic material. It can be solved for N as a function $ of G and U , $  N = N G U : (15.2.15) This serves as the basis for Denition 15.2.61 A perfectly elastic material is one in which the small region near $ any particle ~x at any time t behaves as if it were in an HTES with G= D~ r~L(~x t) and U = U L (~x t). $ In a perfectly elastic material, we will have (see page 250) h = 0, H~ = ~0, and S~ will be given by (15.2.10), so the missing information needed to integrate (15.1.1, 15.1.2, 15.1.3) is supplied by (15.2.12), which is known if the equation of state is known. 256 CHAPTER 15. CONSTITUTIVE RELATIONS 15.3 Small disturbances of an HTES In seismology, the earth is at rest before an event. This resting state is used to label particles by their positions in the resting state. Then particle displacements are given by ~sL (~x t) = ~r L (~x t) ; ~x: (15.3.1) Usually it is a very good assumption in seismology that ~ sL (~x t) k << 1: kD~ (15.3.2) Expressed in terms of deformation gradients, we have $ ~ $ ~ r = I P +D~s G= D~ so where and $ $ $ G= I P + G (15.3.3) $ ~  G= D~ s k G$ k << 1: (15.3.4) We want to study HTES's which are close to the zeroth or initial state in the sense of (15.3.4). We use subscripts 0 to refer to the initial state. Then, since particles are labelled by their positions in HTES0 , $ $ (15.3.5) G0 = I P ~0 = 0 $ $ S~=S0 : (15.3.6) (15.3.7) Furthermore, if we write f = f1 ; f0 for any physical quantity, f1 being its value in HTES1 , (since f is constant throughout the material, we needn't distinguish between f L $ and f E ) then from (15.2.8), to rst order in N and  G we have $ $ U = N + 1 S 0 h2i G : 0 15.3. SMALL DISTURBANCES OF AN HTES 257 $ $ $ $ $ $ If  GT = ; G, then G= I P + G is a rotation to rst order in  G, and therefore N = 0 implies U = 0. Consequently, $ $ $ $ S 0 h2i G= 0 if  GT = ; G : $ $T! $T  $ $  $ $ But then for any  G, S 0 h2i  G ; G = 0, so S0 ; S0 h2i G= 0: Thus $ $ S0=S0T : (15.3.8) The Cauchy stress tensor is symmetric in every HTES of an elastic material. We can go further. Since HTES1 is close to HTES0 , all the changes f are small, so $ $ $0 we can linearize (15.2.12) and (15.2.13). The changes  = 1 ; 0,  S~=S~1 ; S~ are $ $ $ given in terms of N = N1 ; N0 and  G=G1 ; G0 by $  $  @ G0 N0 $  = N @N +  G h2i@~$G G0 N0 $~ $  @ S G0  N0 $ $ $~ $  $~ +  G h2i@G S G0  N0 :  S = N @N $$ $ We dene the following tensors, Q02 4 P , W 02 2 P , B0 2 R: $  $$ ~ $~ $  ~ ~ Q0 = @$G S G0  N0 = ~0 @$G @$G U G0  N0 $ @ $~ $  N  = ~ @ @~$ U $  N  = W0 @N S G0 0  @N G G0 0  $ $ = ~@~$G @ U G0 N0 = ~0@~$G @~G0 N0 @N $  @ 2  $  @ B0 = @N G0  N0 = @N 2 U G0  N0 : Then 0 1 $ @ W$ 0 A  = NB0 +  G h2i ~ 0 (Note: ~ = .) Also $ $ $ $$ ~  S = N W 0 + G h2i Q0 : (15.3.9) (15.3.10) (15.3.11) (15.3.12) (15.3.13) 258 CHAPTER 15. CONSTITUTIVE RELATIONS $ Furthermore, if y^1, y^2, y^3 is any pooob for P , and we write G as in (15.2.11) and write $$ Q0= Qijkly^iy^j y^k y^l then Qijkl = @ 2 U=@Gij @Gkl = @ 2 U=@Gkl @Gij so $$ $$ Qijkl = Qklij or (13)(24) Q0 =Q0 : (15.3.14) $ $ $ We would also like to calculate  S =S 1 ; S 0, the change in the Cauchy stress tensor. From (15.2.13) and (13.1.3), $  $ ;1 $T $~ S 1= det G01  G 01 S 1 : $ Now det G (t) depends continuously on t in going from HTES0 to HTES1 , and can never $ $ vanish. It is 1 in HTES0, so is always > 0. Hence j det G j = det G and $  $ ;1 $T $~ (15.3.15) S 1= det G01 G 01 S 1: $ $ $ $ Now G01 =G= I P + G and, taking components relative to y^1, y^2, y^3    1 + G11 G12 G13   $ det G =  G21 1 + G22 G23    G32 1 + G33   G31  $  $ $ = 1 + Gii + O k G k2 = 1 + tr  G +Ok  Gk2: Thus Then  $;1 $  $ 2 det G = 1 ; tr  G +O k G k : (15.3.16) $T ! $ $~  $  $ $ S 1 = 1 ; tr  G I P + G  S 0 + S  $ where we have used (15.3.15) and have linearized in  G. Keeping only terms up to rst $ order in  G, we have $ $ $~ $ $ $ S 1=S 0 ; tr (G)] S 0 + GT  S 0 + S  15.3. SMALL DISTURBANCES OF AN HTES or 259 $T $ $  $ $ $~  S =  S +  G  S 0 ; S 0 tr  G : (15.3.17) If we take components relative to the pooob, y^1, y^2, y^3 from P , and write $ S 0= Sij y^iy^j (15.3.18) then (15.3.17) becomes Skl = S~kl + Gik Sil ; Gjj Skl = (Gij ) Qijkl + NWkl + Gij (jk Sil ; ij Skl ) = (Gij ) (Qijkl + Sil kj ; ij Skl ) + NWkl : $$ We dene a tensor F 2 4 P by $$ F 0= (Qijkl + Sil kj ; ij Skl) y^iy^j y^k y^l  or, Then and $ $  $ $ $ $$ $ F 0=Q0 +(24) S 0 I P ; I P S 0 : (15.3.19) Fijkl = Qijkl + Sil kj ; ij Sk (15.3.20) $ $ $$ $  S =  G h2i F 0 +N W 0 : (15.3.21) $T $ W 0 =W 0 : (15.3.22) $$ $$ This equation is of exactly the same form as (15.3.13), but Q0 is replaced by F 0 . $ $ Now  S is the dierence between the Cauchy stress tensors in two HTES's, so  S T = $ $ $  S for any  G whatever and any N . First, take  G= 0. Then we must have Next, take N = 0. Then Skl = Gij Fijkl = Slk = Gij Fijlk  260 CHAPTER 15. CONSTITUTIVE RELATIONS so $ for every  G. Hence or Gij (Fijkl ; Fijlk ) = 0 Fijkl = Fijlk (15.3.23) $$ $$ (34) F 0=F 0 : (15.3.24) $ Now in general, G is a joint property of two HTES's, the zeroth or initial state used $ $$ $ $ $ to label particles, and the nal state. However, S 0, B0 , W 0 , Q0 and F 0 are properties $$ $ $ of the initial state alone,$HTES0 . As we have seen S 0 and W 0 are symmetric, and Q0 $ satises (15.3.14), while F 0 satises (15.3.24). If the initial HTES is isotropic, all these tensors must be isotropic. From this follows Remark 15.3.72 In an isotropic HTES of an elastic material, there are constants p, , ,  such that (,  are the Lam/e parameters of the material) 8$ $ > = ; p > S IP > > $ > < W = ; $I P $$ $ $ $ $ > > =  +  (23) + (24)] F I I I P I P or P P > > > Fijkl = ij kl +  (ik jl + il kj ) relative to any pooob for P: : (15.3.25) Proof: $$ In our discussion of I #+(P )] we proved all of this except that for F we proved only the existence of three scalars , ,  such that Fijkl = ij kl + ik jl + il kj : But by (15.3.23), ik jl + il kj = iljk + ik lj , or ( ;  )(ik jl ; il jk ) = 0. Setting i = $k = 1, j = l = 2 gives  ;  = 0, so  = , and we have the $ expression for F given in the remark 15.3.72. 15.3. SMALL DISTURBANCES OF AN HTES 261 Corollary 15.3.54 For HTES's close to an isotropic HTES0 , 8   > <  = (N ) B ; tr $ (=) (15.3.26) > :  S$=  tr $  $I P +2u $   ~ T  $ $T $ ~ where  = 1=2( G + G ) = 1=2 D~s + D~s = innitesimal strain tensor. In most cases, the HTES0 of the elastic material is not isotropic, either because the $$ $ material itself is anisotropic or because S 0 is anisotropic. Then the stiness tensor F 0 $ $ $ gives the response of S to small changes  G in the deformation gradient G away from $ $ G0= I P . The stiness tensor has two sets of symmetries (see 15.3.23): Fijkl = Fijlk : (15.3.27) The second symmetry is obtained (15.3.14) and (15.3.20). Let Rijkl := jk Sil ; ij Skl: (15.3.28) Then from (15.3.20, Qijkl = Fijkl ; Rijkl, so from (15.3.14) Fijkl = Fklij + Rijkl ; Rklij : (15.3.29) Since Fijkl have to be measured experimentally, and there are 81 such components, we would like to use (15.3.27) and (15.3.29) to see how many of these components are indpendent. Only those need be measured and recorded. To answer this question we need one more symmetry, deducible from (15.3.27) and (15.3.29). We have from those two equations Fijkl = Fklji + Rijkl ; Rklij : Interchange i and j in (15.3.29), so Fkljk = Fjikl ; Rjikl + Rklji: Then Fijkl = Fjikl + Rijkl ; Rklij + Rklji ; Rjikl: (15.3.30) 262 CHAPTER 15. CONSTITUTIVE RELATIONS $$ Now dene E 02 4 P by Eijkl = 41 Fijkl + Fjikl + Fklij + Flkij ] : (15.3.31) Then it is a trivial algebraic identity that Fijkl = Eijkl + 14 Fijkl ; Fijkl] + 41 Fijkl ; Fjikl] + 14 Fijkl ; Fklij ] + 41 Fijkl ; Flkij ] = Eijkl + 1 Fijkl ; Fjikl] + 1 Fijkl ; Fklij ] 4 4 1 + Fijkl ; Fklij + Fklij ; Flkij ] 4 = Eijkl + 41 Fijkl ; Fjikl] + 21 Fijkl ; Fklij ] + 14 Fklij ; Flkij ] : Using (15.3.30) once as written and once with the interchanges i $ k and j $ l, and using (15.3.29 once as written, we get Fijkl = Eijkl + 14 Rijkl ; Rklij + Rklji ; Rjikl] + 14 Rklij ; Rijkl + Rijlk ; Rlkij ] + 1 Rijkl ; Rklij ] 2 = Eijkl + 1 (Rijkl ; Rklij ) + 1 (Rijlk ; Rlkij ) 2 4 1 ; 4 (Rjikl ; Rklji) = Eijkl + 21 (jk Sil ; ij Skl ; li Skj + klSij ) + 14 (jlSik ; ij Slk ; kiSlj + lk Sij ) ; 41 (ik Sjl ; jiSkl ; lj Ski + kl Sji) : Therefore Fijkl = Eijkl + 12 kl Sij ; ij Skl + jlSik ; ik Sjl + jk Sil ; il Sjk ] (15.3.32) 15.3. SMALL DISTURBANCES OF AN HTES 263 where we have used ij = ji and Sij = Sji. Now we must measure the six independ$$ ent components of Sij and the components of Eijkl in order to know F 0 . How many $$ independent components does E have? From (15.3.31) Eijkl = Ejikl = Eijlk = Eklij : (15.3.33) Therefore we can think of Eijkl as a 6 6 matrix whose rows are counted by (ij ) = (11), (12), (13), (22), (23), (33) and whose columns are counted by (kl) = (11), (12), (13), (22), (23), (33). We need not consider (ij ) = (21), (31), or (32) because of the rst of equations (15.3.33). The fact that Eijkl = Eklij means that this 6 6 matrix is symmetric. Therefore it contains only 21 independent components. These can be chosen arbitrarily but not all choices describe stable materials. A tensor in 4 P with the symmetries (15.3.33) is called an \elastic tensor." The set of all such tensors in a subspace of 4P , which $we call E 4(P ). We have just proved $ that dim E 4(P ) = 21. We have also shown that if F 2 4 P has the$ symmetries (15.3.27, $ 4 15.3.28, 15.3.29) then it can be written in the form (15.3.32) with 2 E (P ). Conversely, E $$ $ $ $ $ $ $ if E 2 E 4(P ) and S T =S and F is given by (15.3.32), then F has the symmetries (15.3.27, 15.3.28, 15.3.29), a fact which is easily veried and is left to the reader. $$ Therefore it is not possible to reduce the problem further. In principle, to measure F 0 , it may be necessary $ to measure the six independent components of S 0 and the 21 independent components of $$ $ $ E 0. Actually, only ve of the components of S 0 need be measured. Suppose 40 is the $ stress deviator for S 0 , so $ $ $ $ (15.3.34) S 0 = ;p0 I P + 40 with ; 3p0 = tr S 0 : $ Then substituting (15.3.34) in (15.3.27) shows that the isotropic part of S 0 makes no $$ contribution to F 0 , and we can rewrite (15.3.32) as Fijkl = Eijkl + 12 kl 4ij ; ij 4kl + jl 4ik ; ik 4jl + jk 4il ; il 4jk] : $$ $$ $$ Therefore F 0 has 26 independent components, 21 in E 0 and 5 in 40 . (15.3.35) 264 CHAPTER 15. CONSTITUTIVE RELATIONS $$ $$ Suppose a material is originally$isotropic in an HTES with S 0= 0 . Suppose the mater$ $ ial is strongly compressed so that S 0= ;p0 I P where p0 is very large (say of the order $$ $$of mantle pressure, 0.1 to 1.35 megabar). The new HTES will still be isotropic, and F 0 =E 0 will have the form (15.3.25), which involves only two constants, the Lam/e parameters  $ and . Suppose then that a small (a few kilobars) stress deviator is$ added, say 40. Then $$ $$ $$ $ will change by a small amount  which depends linearly on 4 , and E0 $ $ E F 0 will change 0 $ $ $ $ $ $ $ by  F =  E + terms involving 40 in (15.3.35). Now  E = J0h2i 40 where J0 2 6P . This tensor is a property of the original highly compressed isotropic HTES. Therefore it is a member of I 6( +(V )), i.e. it is isotropic. By Weyl's theorem, it is a linear combination $ $ $ of all possible permutations of I P I P I P . These are as follows: ij kl mn ij kmln ij knlm ik jl mn ik jmln ik jnlm il jk mn il jmkn il jnkm im jk ln im jl kn imjnkl in jk lm in jlkm injmkl : Since 4mn = 4nm, we can combine two terms which dier only in interchanging m and n. Since 4mm = 0, we can discard terms with mn . Thus there are scalars ,  ,  , ",  such that 2Jijklmn = (ij kmln + ij knlm)+  (ik jm ln + ik jnlm) + (il jmkn + il jnkm) +(im jk ln + injk lm) +"(imjl kn + injl km) + (imjnkl + in jmkl). Moreover, Jijklmn must satisfy (15.3.33) for any xed m, n, so  =  =  = " and  =  . Therefore J0 involves only two independent constants, and 2Jijklmn = ij((kmln + kllm ) + kl (im jn + in jm) + ik (jmln + jnlm) + il (jmkn + jnkm) ) +jk (im ln + inlm ) + jl (im kn + in km) : Then Eijkl = Jijklmn4mn =  (ij 4kl + kl 4ij ) + (ik 4jl + il 4jk + jk 4il + jl4ik ) : 15.4. SMALL DISTURBANCES IN A PERFECTLY ELASTIC EARTH 265 Thus, nally, correct to rst order in 4ij , (15.3.35) gives Fijkl = ij kl + u (ik jl + il jk ) (15.3.36)  1  1 (15.3.37) +  ; 2 ij 4kl +  + 2 kl 4ij  1  1 +  ; (ik 4jl + il 4jk) +  + (jk 4il + jl 4ik ) : (15.3.38) 2 2 $$ If the anisotropy in F 0 arises solely because$ an isotropic material has been subjected to $ a small (anisotropic) deviatoric stress then F 0 involves only 9 independent constants, the $ $ ve independent components of 40 and the four constants , u, ,  . If the anisotropy was already present in the crystal structure of the material, either because it was a single $ crystal, or because its micro-crystals were not randomly oriented, then E 0 can involve 21 $$ and F 0 26 independent parameters. 15.4 Small disturbances in a perfectly elastic earth Neglecting dissipation, we will treat the earth as a perfectly elastic body in studying its response to earthquakes. Before an earthquake, we assume the earth is at rest in static equilibrium. We label the particles by their equilibrium positions, so we use t0-position labelling where t0 is any time before the earthquake. When the earth is at rest, we have 8 L > < ~r (~x t) = ~x (15.4.1) > : G$L (~x t) = ( D ~r )L (~x t) =$I P so the Eulerian and Lagrangian descriptions of any physical quantity are the same, and the two kinds of density and two kinds of stress tensor are the same. Using subscript 0 to denote the equilibrium state, 0 (~x) = ~0 (~x) $ $ S 0 (~x) = S~0 (~x) : The body force is f~ (~x) = f~~0 (~x) = 0(~x)~g0(~x) where ~g is the acceleration of gravity, and ~g0 is its equilibrium value. The conservation equations reduce to ~0 = @~ S$0 ;0~g0 = D~  S$~ 0 ; ~0~g0: (15.4.2) 266 CHAPTER 15. CONSTITUTIVE RELATIONS Now suppose an earthquake occurs, after which with ~r L (~x t) = ~x + ~sL (~x t)  (15.4.3) ~ sL (~x t) k << 1 for all ~x t: kD~ (15.4.4) Then $ $ GL (~x t) = D~ r~L (~x t) = I P +D~ s~L (~x t) : (15.4.5) ~ s is quite small, and it is almost always possible to For earthquakes, ~s as well as D~ replace the exact conservation laws and constitutive relations by linearized versions in ~ s higher than the rst are omitted. In these problems it is which all powers of ~s or D~ useful to be able to convert easily between Eulerian and Lagrangian representations. For example, gravity is easy to calculate in the Eulerian representation, and the constitutive relations are simple in the Lagrangian representation. We discuss this conversion now. For any physical quantity f , we have f L (~x t) = f E (~r t) where ~r = ~r L (~x t) : If ~r L(~x t) = ~x + ~s L(~x t) then to rst order in ~s we have   f L (~x t) = f E ~x + ~sL (~x t)  t = f E (~x t) +~s L (~x t)  @~f E (~x t) : (15.4.6) If f is small of rst order in ~s, we have f L (~x t) = f E (~x t) or f L = f E if f = Ok~s k: (15.4.7) If f is small of order ~s, there is no need to distinguish between its Eulerian and Lagrangian representations, so we can write both f E (~x t) and f L (~x t) simply as f (~x t). That is f (~x t) := f E (~x t) = f L (~x t) if f = Ok~s k: (15.4.8) This is true in particular for f = ~s. If f is of order ~s, we can use f not merely for the pair of functions (f L f E ) but for either function alone, since to order ~s they are equal. Even 15.4. SMALL DISTURBANCES IN A PERFECTLY ELASTIC EARTH 267 when f is not small of order ~s, f E (~x t) ; f0 (~x) is, so to rst order in ~s, ~sL(~x t)  @~f E (~x t) = ~s(~x t)  @~f0(~x) and (15.4.6) for any f is f L(~x t) = f E (~x t) + ~s(~x t)  @~f0(~x) (15.4.9) or, as a relationship between functions, f L = f E + ~s  @~f0  (15.4.10) for any physical quantity f . Relations between derivatives are also needed. For any physical quantity f , Dt f = @t f + ~v  @~f . But ~v = Dt~s = Ok~sk, and f ; f0 is of order ~s, so correct to rst order in ~s, ~v  @~f = ~v  @~f0 , and Dt f = @t + ~v  @~f0 : (15.4.11) If f is Ok~sk then ~v  @~f0 is Ok~sk2, so correct to Ok~s k Dt f = @t f if f = Ok~sk: (15.4.12) In particular, correct to Ok~sk ~v = Dt~s = @t~s ~a = Dt~s = @t2~s: (15.4.13) ~ = (D~ ~ r )  @~f = ($I P +D~ ~ s)  @~f = @~f + D~ ~ s  @~f . If If f is any physical quantity, Df f = Ok~s k then correct to Ok~s k ~ = @~f if f = Ok~sk: Df (15.4.14) ~ s = @~~s. Therefore, correct to Ok~sk, D~ ~ s  @~f = @~~s  @~f In particular, correct to Ok~sk, D~ = @~~s  @~f0 . Thus for any f ~ = @~f + @~~s  @~f0 correct to Ok~sk: Df (15.4.15) For any physical quantity f we dene Lf and E f , the Lagrangian and Eulerian derivatives of f from its equilibrium value. Both are physical quantities, and they are dened by  L L  f (~x t) = f L (~x t) ; f0 (~x) (15.4.16) 268 CHAPTER 15. CONSTITUTIVE RELATIONS  E E f (~r t) = f E (~r t) ; f0 (~r) : (15.4.17) Both Lf and E f are small of order k~sk, so we can apply to them the convention (15.4.8) even when f is not small of order ~s. Thus, for the functions, Lf = f L ; f0 (15.4.18) E f = f E ; f0: (15.4.19) Then Lf ; E f = f L ; f E = ~s  @~f0 from (15.4.10), so L f = E f + ~s  @~f0 to rst order in ~s: $ $ $ $ From (15.4.1), G0 = I P so ~s  @~ G0 = 0 . Therefore $ $ L G= E G= @~~s to rst order in ~s: (15.4.20) (15.4.21) With the help of this machinery, the Eulerian and Lagrangian conservation equations and the constitutive relations can be linearized. They produce either Eulerian or Lagrangian equations of motion, and it is conventional to work with the former, even in seismology. We will give the derivation of these equations, based on the Lagrangian constitutive relations and Eulerian conservation equations. The Eulerian mass conservation equation is @t E (~r t) + @~  E (~r t) ~vE (~r t) = 0: We have E = 0 + E  and @t 0 = 0, so, correct to rst order in ~s,   @t E  + @~  (0~v) = 0: But ~v = Dt~s = @t~s so, correct to rst order in ~s, h i @t E  + @~  (0~s) = 0: For t = t0 (before the earthquake) E  = 0 and ~s = 0. Therefore E  = ;@~  (0~s) : (15.4.22) 15.4. SMALL DISTURBANCES IN A PERFECTLY ELASTIC EARTH 269 The Eulerian momentum conservation equation is  2 E $E Dt ~s (~r t) = @~ S (~r t) ; E (~r t) ~gE (~r t) : $E $ $ Here S =S 0 +E S , E = 0 + E , ~gE = ~g0 + E~g, so correct to rst order in ~s,     $ ~  E $ 2 ~ 0 @t ~s = @  S 0 +@   S ; 0~g0 ; 0 E~g ; E  ~g0: But then we can apply (15.4.2) to obtain  $     2 ~ 0@t ~s = @  E S ; 0 E~g ; E  ~g0: (15.4.23) We nd E  from (15.4.22). We nd E~g as E (;@~) = ;@~E + @~0 = ;@~(E ; 0), so   E~g = ;@~ E  (15.4.24) where  is the gravitational potential. It is obtained from @ 2  = 4; where ; = Newton's gravitational constant. Subtracting @ 2 0 = 40; gives @ 2 E  = 4;E : (15.4.25) $ In order to solve (15.4.23), it remains to express E S in terms of ~s. We approximate $ the neighborhood of each particle in the earth as a HTES, and we assume that S can be calculated everywhere at all times as if the local material were in such a state. Thus at each particle ~x we can use (15.4.20):  $ $ L $ $$ L  S =  G h2i F 0 (~x) + LN W 0 (~x) : (15.4.26) We must use the Lagrangian representation here, because (15.3.21) applies to a particular lump of matter, the matter close to particle ~x. The initial HTES0 at particle ~x is the HTES appropriate for the region near that particle before Thus this HTES0 will $$ the earthquake. $ vary from particle to particle, and the tensors F 0 and W 0 really will depend on ~x. The heat conductivity of the earth is so low that during the passage of earthquake waves almost no heat has time to ow, so there is no change in entropy per gram at each particular particle. That is LN = 0 everywhere at all times. (15.4.27) 270 CHAPTER 15. CONSTITUTIVE RELATIONS $ $ Thus equation (15.4.26) gives L S in terms of ~s, since L G= @~~s. And then (15.4.20) $ gives E S as $ $ $ E S = L S ;~s  @~!S 0 $$ $ $ E ~  S = @~sh2i F 0 ; ~s  @~ S 0 : Thus (15.4.23) nally becomes, to rst order in ~s, "  $$ #  $ 2 ~ ~ ~ 0@t ~s = @  @~s h2i F 0 (~x) ; @  ~s  @~ S 0   h i ;0 E~g + @~  (0~s) ~g0 (15.4.28) where E~g is obtained from ~s via (15.4.22), (15.4.24) and (15.4.25). $ For waves short compared to the radius of the earth, it is safe to neglect ~g and @~ S 0 . If we assume that the undisturbed state is the same everywhere and is isotropic, then Fijkl = ij kl +  (ik jl + il jk ) so "  $$ #   @~s h2i F 0 =  @~  ~s kl +  (@k sl + @l sk ) kl ~  $$  $ h i @~s h2i F 0 =  @~  ~s I P + @~ ~s + (@~s)T : If we assume that there is no ~x dependence in the equilibrium state (homogeneous earth) then  and  are independent of ~x, and (15.4.28) becomes "  $$ #  ~   2 ~ 0 @t sl = @k @~s h2i F 0 = @l @  ~s +  @k @k sl + @k @l sk kl   0 @t2~s = ( + ) @~ @~  ~s + @ 2~s   h  i = ( + 2) @~ @~  ~s +  @ 2~s ; @~ @~  ~s     = ( + 2) @~ @~  ~s ; @ @~ ~s : Divide this equation by 0 and dene c2p =  + 2  c2s =  : 0 0 (15.4.29) 15.4. SMALL DISTURBANCES IN A PERFECTLY ELASTIC EARTH Then     @t2~s = c2p@~ @~  ~s ; c2s @~ @~ ~s : 271 (15.4.30) Since we are assuming 0 , ,  independent of ~x, the same is true of c2p and c2s . Taking the divergence of (15.4.30) gives     @t2 @~  ~s = c2p@ 2 @~  ~s : (15.4.31) The scalar @~  ~s propagates in waves with velocity cp. These are called P -waves (primary   waves) or compression waves. Taking the curl of (15.4.30) and using @~ @~ ~v = @~ @~  ~v ; @ 2~v gives     @t2 @~ ~s = c2S @ 2 @~ ~s : (15.4.32) The vector @~ ~s propagates in wave with velocity cs. These are called S -waves (secondary waves) or shear waves. There is an interesting inequality involving cs and cp which we can derive by studying $$ (15.3.21) when N = 0 and F 0 is isotropic. In that case $ $  $ " $  $T #  S =  I P tr  G +   G +  G : $ $ For a pure shear, ~s(~r) = (~r  y^3)^y1, we have  G= y^3y^1 so  S = (^y1y^3 + y^3y^1) and $ y^3   S = y^1. If   0, the material either does not resist shear or aids it once it has begun, so we must have  > 0: (15.4.33) $ $ $ $ For a pure dilatation ~s(~r) = "~r, we have  G= " I P , so  S = "(3 + 2) I P . If " > 0, the material expands slightly. If this does not produce a drop in pressure, the material will explode. Therefore 3 + 2 > 0: (15.4.34) Combining these two relations gives 3( + 2) > 4 > 0, so 2 0 < cs2 < 3  0 < cs < 0:866: (15.4.35) cp 4 cp Therefore the P wave always travels faster than the S wave, and arrives rst. Hence it is the \primary" wave. 272 CHAPTER 15. CONSTITUTIVE RELATIONS 15.5 Viscous uids A uid is a material whose HTES's support no shear, so the Cauchy stress tensor is always isotropic. We need now a more precise denition. A uid supports no shear in an HTES, and each HTES is uniquely determined by the density and entropy per unit mass, or by volume per unit mass,  , and entropy per unit mass, N . Clearly  = 1=, and by our assumption U = U ( N ) : (15.5.1) From elementary thermodynamics, it is well known that dU = ;pd + dN so (15.5.2) p ( N ) = ; @U (@ N )  (15.5.3) ( N )  ( N ) = @U@N (15.5.4) $ $ S = ;p I P : (15.5.5) In an HTES, the Cauchy stress tensor is Now suppose the uid deviates slightly from an HTES. At each point in space, ~r, and each instant t, there will be a well dened density E (~r t) and a well dened energy per unit mass, U E (~r t). We set  = 1=E (~r t), U = U E (~r t) and use (15.5.1) to calculate N E (~r t) as if the uid were in an HTES at (~r t). Then with N = N E (~r t) and  = 1=E (~r t) we use (15.5.3) to calculate pE (~r t) as if the uid were in an HTES at (~r t). We call this pressure the thermostatic equilibrium pressure at ~r t and write it pEHTES(~r t). The uid at (~r t) is not in fact in the HTES dened by U E (~r t) and E (~r t), so the actual $ Cauchy stress tensor S E (~r t) is not given by (15.5.5) but by $ $ $ S E (~r t) = ;pEHTES (~r t) I P + V E (~r t) (15.5.6) $ $ where (15.5.6) denes V E . The tensor V E (~r t) is called the viscous stress tensor at (~r t). It vanishes in HTES's. 15.5. VISCOUS FLUIDS 273 The deviation from an HTES can be measured by the failure of to be constant and $ by the failure of ~v to vanish. Thus V E (~r t) can be expected to depend on the functions $ E and ~vE in the whole uid. We expect V E to vanish when E is constant for all ~r t. In fact, ~vE = constant for all ~r t is simply an HTES moving with constant velocity, so we expect $ $ V E (~r t) = 0 if E and v~E are constant for all ~r and t: $ For ordinary uids, we would also expect V E (~r t) to be aected only by the behavior of the functions E and ~vE near ~r t. This behavior is determined by the derivatives, so we $ would expect V E (~r t) to be determined by the values at (~r t) of 8 > E ~E > < @  @t  > > : @~v~E  @t v~E  @~@~ E  @~t @~ E  @t2 E     @~@~v~E  @~t @~v~E  @t2 v~E     (15.5.7) We will assume that the deviation from an HTES is small. In that case, V~ E (~r t) can be expanded in Taylor series in the quantities (15.5.7), and we can drop all but the linear terms. We will also assume that the inuence of distant uid on (~r t) drops o very rapidly with distance in either space or time, so that only rst derivatives need be considered in (15.5.7). In fact, physical arguments suggest that a term @~m @tn E or @~m @tn v~E in (15.5.7) $ contributes to V (~r t) in the ratio (=L)m (t=T )n where  and t are mean free path and mean time between collisions for a molecule and L and T are the macroscopic wavelength and time period of the disturbance being studied. $ On the basis of the foregoing approximations, we expect that V E (~r t) will depend linearly on @~ E (~r t), @t E (~r t), @~~vE (~r t) and @t~vE (~r t). $$ We$are thus led to a model in which at any point (~r t) there are tensors F E (~r t) 2 $ $ $ $ 4 P , J E (~r t) and K E (~r t) 2 3 P , and H E (~r t) 2 2 P such that !$ $  ~  $$ ~  !$ $ = @ ~ v h 2 i + @  + ( @ ~ v )  (15.5.8) V F J K + (@t ) H : t 274 CHAPTER 15. CONSTITUTIVE RELATIONS It seems$reasonable to assume that the uid is skew isotropic. Then the same must be ! ! $ $ $ $ true of F , J , K , and H . Therefore at each (~r t) there are scalars , ,  , ,  ,  such that relative to any pooob for P Fijkl Jijk Kijk Hij = ij kl + ik jl + il jk = "ijk = "ijk = ij: It is observed experimentally that @~~v, @~ , @t~v and @t can be adjusted independently. $ Therefore we can take all but @~ to be 0. Then Vij = "ijk @k . This V is antisymmetric. If we assume that the intrinsic angular momentum, body torque and torque stress in the uid are negligible, then Sij = Sji, so Vij = Vji. Therefore we must have  = 0. By the same argument,  = 0 and  =  . Thus we conclude that   Vij = ij @~  ~v +  (@i vj + @j vi) + ij @t : (15.5.9) If ~vE = ;"~r, the uid is being compressed at a uniform rate. If @t = 0 then (15.5.9) gives Vij = ;ij (3 + 2)". An extra pressure, over and above pHTES , is required to conpress the uid at a nite rate. The quantity  =  + 2=3 is therefore called the bulk viscosity:  =  + 2=3: (15.5.10) Then      Vij = ij @~  ~v +  @ivj + @j vi ; 23 ij @~  ~v + ij @t : (15.5.11) The quantity  is called the shear viscosity because in a pure shear ow, ~vE (~r) = w~r  y^3y^1, $ $ $ V = w(^y3y^1 + y^1y^3) and y^3 V = y^1w. The tangential stress y^3 V is  times the rate of shear, w. In the ocean, =  100. However, except in sound waves, @~  ~v  0, so the large value of  in the ocean is not noticed except in the damping of short sound waves. The second law of thermodynamics gives more information about (15.5.11), namely  = 0   0   0: (15.5.12) 15.5. VISCOUS FLUIDS 275 The argument is as follows. At a particle ~x at time t, U and  are measurable physical quantities even though the material is not in an HTES. If the local values of U and  are used in (15.5.1), a local value of N can be found for particle ~x at time t. Then (15.5.3) and (15.5.4) give local values of p and at particle ~x at time t. These local values of N , p, are not measured. They are calculated from the measured  and U as if the material L (~x t), pL (~x t), near ~x at time t were in an HTES. Therefore we denote them by NHTES HTES $ L (~x t). Note that pHTES is not ;1=3 tr S  the latter is a measurable quantity. If we HTES follow particle ~x, (15.5.2) implies Dt U = ;pHTES Dt  + HTES Dt NHTES: (15.5.13) There are no other candidates for and N , So we drop the subscripts HTES on and N . $ We do not do so on pHTES, because ;1=3 tr S is another candidate for the pressure and we do not want to confuse the two. If we multiply (15.5.9) by  and use (13.6.15) we obtain ;@~  H~ + h+ S$ h2i@~~v = ;pHTES Dt  + Dt N: (15.5.14) But Dt  =  ;1 Dt = Dt ln  = ;Dt ln  = @~  ~v. Therefore (15.5.14) is   Dt N + @  H = h + S h2i@~v + pHTES @~  ~v : (15.5.15) $ $ $ $ But pHTES(@~  ~v) = pHTES I P h2i@~~v so S h2i@~~v + pHTES(@~  ~v) = (S +pHTES I P )h2i(@~~v) $ ~ )= =V h2i@~~v from (15.5.6). If we divide (15.5.15) by and use the identity @~  (H= (@~  H~ )= + H  @~(1= ), (15.5.15) becomes 0 1   ~ $ DtN + @~  @ H A = h + 1 V h2i@~~v + H~  @~ 1 : (15.5.16) If we dene the stu \entropy" as the stu whose density per unit mass is N and ~ , then (15.5.16) shows that its creation rate  is whose material ux density is H=   $  = h + H~  @~ 1 + 1 V h2i@~~v: (15.5.17) Now h can be positive (e.g. ohmic heating) or negative (e.g. heat radiation by a transparent liquid into empty space if the liquid is a blob in interstellar space). However, 276 CHAPTER 15. CONSTITUTIVE RELATIONS  ; h= is the creation rate of entropy by local events in the uid. The second law of thermodynamics is generalized to motions near HTES by requiring that everywhere at all times  ; h  0: (15.5.18) $ To use (15.5.18), we must calculate V h2i@~~v in (15.5.17). Since Vij = Vij , we have $ 1 1 1 V h2i@~~v = Vij @ivj = 2 (Vij + Vji) @i vj = 2 Vij @ivj + 2 Vji@vj = 1 Vij @i vj + 1 Vij @j vi = Vij 1 (@i vj + @j vi) 2 2 2 $ 1 ~ ~ T  = V h2i @~v + @~v : 2 Now suppose we write (15.5.10) as   ~   2 ~ Vij = ij @  ~v + @t +  @i vj + @j vi ; 3 ij @  ~v (15.5.19) and write 1 (@ v + @ v ) =   1 @~  ~v + 1 @ v + @ v ; 2  @~  ~v : (15.5.20) ij 2 ij j i 3 2 i j j i 3 ij $ Then we have expressed V and 1=2@~~v +(@~~v)T ] as the sum of their isotropic and deviatoric parts. Since an isotropic and a deviatoric tensor are orthogonal, there are no cross terms $ when we calculate V h2i 21 @~~v + (@~~v)T ] from (15.5.19) and (15.5.20). Therefore      T 2 $ $ V h2i@~~v = @~  ~v @~  ~v + @t + 2 k@~~v + @~~v ; 3 I P @~  ~vk2 : Thus for a viscous uid (15.5.18) requires  1   2   ~ T 2 $ ~ 2 ~  ~ ~ ~ ~  H  @ +  @  ~v + 2 @~v + @~v ; 3 I P @  ~v  +  @  ~v @t  0: (15.5.21) We can arrange experiments in which @~ = ~0, and @t = 0, and ~vE (~r t) = "~r +  (~r  y^3)^y1. Then @~~v = "Ip +  y^3y^1, @~  ~v = 3", and (15.5.21) becomes 9"2 + 2 2  0. Since this inequality must hold for " = 0,  6= 0 and for " 6= 0,  = 0, we must have   0,   0. We can also arrange experiments with ~vE (~r t) = "~r, @~ = ~0, and @t of any value. In such an experiment, (15.5.21) becomes 9"2 + 3"@t  0: (15.5.22) 15.5. VISCOUS FLUIDS 277 If  6= 0, take " > 0 and @t < ;3". Then (15.5.22) is violated. Therefore we must have  = 0. The fact that (15.5.12) are observed experimentally is one of the arguments for (15.5.18) as an extension of the second law of thermodynamics. 278 CHAPTER 15. CONSTITUTIVE RELATIONS Part III Exercises 279 Set One Exercise 1 Let (~b1      ~bn) be an ordered basis for Euclidean vector space V . Let (~b1      ~bn ) be its dual basis. Let gij and gij be its covariant and contravariant metric matrices. Prove a) ~bi = gij~bj b) ~bi = gij~bj c) gij gjk = i k d) gij gjk = i k Note that either c) or d) says that the two n n matrices gij and gij are inverse to each other. Thus we have a way to construct (~b1     ~bn) when (~b1      ~bn ) are given. This is the procedure: Step 1. Compute gij = ~bi  ~bj . Step 2. Compute the inverse matrix to gij , the matrix gij such that gij gjk = i k or gij gjk = i k (either equation implies the other). Step 3. ~bi = gij~bj . Exercise 2 V is ordinary Euclidean 3-space, with the usual dot and cross products. The function M is dened by one of the following six equations, where ~v1, ~v2 ~v3, ~v4 are arbitrary vectors 281 282 in V . In each case, state whether M is multilinear, whether M is a tensor, and whether M is totally symmetric or totally antisymmetric. State why you think so. i) M (~v1 ~v2) = ~v1  ~v2 . ii) M (~v1 ~v2 ) = ~v1 ~v2 . iii) M (~v1 ~v2 ) = (~v1  ~v2 )~v2. iv) M (~v1 ~v2) = (~v1 ~v2 )  ~v2. v) M (~v1 ~v2~v3) = (~v1 ~v2)  ~v3 . vi) M (~v1 ~v2~v3~v4) = (~v1  ~v2 )(~v3  ~v4 ). Solutions 1.a) For any vector ~v 2 V , we know that   ~v = ~v  ~bj ~bj See equations (D-14) and (D-15). Let i 2 f1     ng, and apply this result to the vector ~v = ~bi . Thus we obtain ~bi = ~bi  ~bj ~bj  or ~bi = gij~bj : 1.b) For any ~v 2 V , ~v = (~v  ~bj )~bj (see equations (D-12) and (D-13). Apply this result to ~v = ~bi . Thus ~bi = (~bi  ~bj )~bj , or ~bi = gij~bj . Note that part b) can also be obtained immediately by regarding (~b1      ~bn) as the original basis and (~b1     ~bn) as the dual basis. 1.c) Dot ~bk into a) and use ~bi  ~bk = i k . 1d. Dot ~bk into b) and use ~bi  ~bk = i k . 283 2.i) M is multilinear because a dot product is linear in each factor. M is a tensor because its values are real numbers (scalars). (It is of order 2.) M is totally symmetric because ~v1  ~v2 = ~v2  ~v1 . 2.ii) M is multilinear because a cross product is linear in each factor. M is not a tensor because its values are vectors in V , not scalars. M is totally antisymmetric because ~v1 ~v2 = ;~v2 ~v1. 2.iii) M is not multilinear because M (~v1  2~v2) = 4M (~v1 ~v2 ). M is not a tensor and is neither totally symmetric nor totally antisymmetric because M (~v1 ~v2) is generally 6= M (~v2 ~v1) and 6= ;M (~v2 ~v1). 2.iv) M (~v1 ~v2 ) = 0, so M is multilinear. M is a tensor because its values are real numbers. (It is of order 2.) M is both totally symmetric and totally antisymmetric because M (~v1 ~v2) = M (~v2 ~v1) = ;M (~v2 ~v1) for all ~v1 , ~v2 2 V . 2.v) M is multilinear because it is linear in each of ~v1, ~v2 , ~v3 when the other two are xed. M is a tensor because its values are scalars. (It is of order 3.) M is totally antisymmetric. To show this, we must show that (12)M = (13)M = (23)M = ;M: We have M (~v2~v1 ~v3) = (~v2 ~v1 )  ~v3 = ; (~v1 ~v2 )  ~v3 = ;M (~v1 ~v2~v3) so (12)M = ;M . And we have M (~v3~v2 ~v1) = (~v3 ~v2)  ~v1 = (~v2 ~v1 )  ~v3 = ;M (~v1 ~v2~v3) 284 so (13)M = ;M . Finally M (~v1~v3 ~v2) = (~v1 ~v3)  ~v2 = (~v2 ~v1 )  ~v3 = ;M (~v1 ~v2~v3) so (23)M = ;M . 2.vi) M is multilinear because it is linear in each of ~v1 , ~v2 , ~v3 , ~v4 when the other three are xed. M is a tensor because its values are scalars. (It is of order 4.) M is neither totally symmetric nor totally antisymmetric. It is true that (12)M = M and (34)M = M . However, (23)M is neither M nor ;M . To see this, we must nd ~u1, ~u2, ~u3, ~u4 2 V such that M (~u1  ~u3 ~u2 ~u4) 6= M (~u1  ~u2 ~u3 ~u4), and we must nd ~v1~v2 ~v3~v4 2 V such that M = (~v1~v3 ~v2~v4) 6= ;M = (~v1 ~v2~v3~v4). Let x^ y^ z^ be an orthonormal basis for V and let ~u1 = ~u2 = ~v1 = ~v2 = x^, ~u3 = ~u4 = ~v3 = ~v4 = y^. Then M (~u1  ~u2 ~u3 ~u4) = 1 and M (~u1  ~u2 ~u3 ~u4) = 0. Note: One might be tempted to say that (~v1  ~v2 ) (~v3  ~v4) is \obviously" dierent from (~v1  ~v3)(~v2  ~v4 ). Yet if dim V = 1, we do have (~v1  ~v2)(~v3  ~v4) = (~v1  ~v3)(~v2  ~v4), and M is totally symmetric. Extra Problems In what follows, V is an n-dimensional Euclidean vector space and L 2 L(V ! V ). 1E. Recall that l 2 R is an \eigenvalue" of L i there is a nonzero ~v 2 V such that L(~v) = l~v. a) Show that if l is an eigenvalue of L and p is any positive integer, then lp is an eigenvalue of Lp. b) If  2 Sq (i.e.,  : f1     qg ! f1     qg is a bijection ), show that there is a positive integer p  q! such that p = e, the identity permutation. c) Regard  2 Sq as a linear operator on q V . Suppose c 2 R is an eigenvalue of  : q V ! q V . Show that c = 1. 285 2E. a) For any c 2 R, show that det(cL) = cn det L. b) Recall that L;1 exists i the only solution ~u of L(~u) = ~0 is ~u = ~0. Use this fact to give a pedantically complete proof that l 2 R is an eigenvalue of L i det(L ; lIV ) = 0, where IV is the identity operator on V . 3E. a) Show that det(L ; lIV ) is a polynomial in l with degree  n. That is, det(L ; lIV ) = c0 ln + c1ln;1 +    + cn. Hint: Choose a basis B = (~b1     ~bn) for V . Let Li j be the matrix of L relative to B , i.e., L(~bi ) = Li j~bj . Show that relative to B the matrix of L ; lIV is Li j ; li j . Finally, use det L = L1 i1    Ln in "i1 in with L ; lIV replacing L.] b) Show that c0 = (;1)n and cn = det L. (Thus the degree of det(L ; lIV ) is n.) c) (;1)n;1c1 is called the trace of L, written tr L. Calculate tr L in terms of the matrix Li j of L relative to B . Solutions to Extra Problems 1E. a) By hypothesis, ~v 6= ~0 and L(~v) = l~v. Then L2 (~v) = LL(~v)] = L(l~v) = lL(~v) = l2~v. And L3(~v) = LL2 (~v)] = L(l2~v) = l2 L(~v) = l3~v. Etc. (or use mathematical induction on p). b) Sq has q! members. Therefore, among the q!+1 permutations e  2 3     q!, two must be the same, say a = b a < b. Then ;a a = e = ;ab = b;a . Take p = b ; a. c) If c is an eigenvalue of , and p is as in b) above, then cp is an eigenvalue of p, so there is a nonzero M 2 q V such that eM = cpM . But eM = M , so cp = 1. Since c 2 R, c = 1. 286 2E. a) Let A 2 nV , A 6= 0. Let (~b1      ~bn) = B be a basis for V . Then A(cL)(~b1 )     (cL)(~bn)] = det(cL)A(~b1      ~bn). But A(cL)(~b1 )     (cL)(~bn)] = AcL(~b1 )     cL(~bn)] = cnAL(~b1 )     L(~bn )] = cn(det L)A(~b1      ~bn). Since A(~b1      ~bn) 6= 0, det(cL) = cn(det L). b) l is an eigenvalue of L , 9~v 2 V 3 ~v 6= ~0 and L(~v) = l~v. (Denition of eigenvalue.) , L(~v) ; lIV (~v) = ~0 ( def. of IV ) , (L ; lIV )(~v) = ~0 ( def. of L ; lIV ): But 9~v 2 V 3 ~v = 6 ~0 and (L ; lIV )(~v) = ~0 , det(L ; lIV ) = 0. 3E. a) (L ; lIV )(~bi ) = L(~bi ) ; lIV (~bi) = Li j~bj = l~bi = (Li j ; li j )~bj , so the matrix of L ; lIV relative to B is Li j ; li j . * Therefore det(L ; lIV ) = (Ll i1 ; ll i1 )    (Ln in ; ln in )"i1 in . This is a sum of n! terms. Each term is a product of n monomials in l, whose leading coe"cients are 0 or ;1. Therefore each term is a polynomial in l of degree  n. Hence so is the sum of all n! terms. b) cn is the value of the polynomial when l = 0. Hence it is the value of det(L ; lIV ) when l = 0, i.e., it is det L. b and c) To nd c0 and c1 , note that (*) in the solution 3E.a) is a sum of nn terms, of which n! have a nonzero factor "i1 in . Among these nonzero terms, one has i1 = 1, i2 = 2     in = n. It is (y)  1    L1 ; ll l    L(n) (n) ; l(n) (n) "1 n      = L1 1 ; l L2 2 ; l    L(n) (n) ; l : In all other nonzero terms, there must be at least two failures among i1 = 1, i2 = 2     in = n, because fi1      ing = f1     ng. Therefore, in each 287 nonzero term in (*) except (y) above, at least two factors have no l, so the total degree as a polynomial in l is  n ; 2. Therefore, all the terms in ln and ln;1 in det(L ; lIV ) come from (y). But (y) is h i (;1)nln + (;1)n;l L1 1 + L2 2 +    + L(n) (n) ln;1 +    : Thus c0 = (;1)n and c1 = (;1)n;1Li i. Hence tr L = Li i. 288 Set Two Exercise 3 Let (^x1  x^2      x^n) = B be an ordered orthonormal basis for Euclidean space V . Let B 0 = (^x2 x^1  x^3      x^n) (i.e., just interchange x^1 and x^2 ). Show that B and B 0 are oppositely oriented. Exercise 4 If V is Euclidean vector space, the \identity tensor" on V is dened to be that I 2 V  V such that for any ~x ~y 2 V I (~x ~y) = ~x  ~y: Suppose that ~u1    ~un, ~v1    ~vn are xed vectors in V such that I= n X i=1 ~ui~vi : a) Prove that n  dim V . (Hence, if dim V  2, I is not a dyad ). b) Prove that if n = dim V then (~u1     ~un) is the basis of V dual to (~v1    ~vn). Solutions 3. Let A be one of the two unimodular alternating tensors over V . Let  2 Sn. Then A(^x(1)      x^(n) ) = ( sgn )A(^x1      x^n). Hence (^x1      x^n) and (^x(1)      x^(n) ) have the same or opposite orientation according as sgn  = +1 or ;1. 289 290 4. Change notation so as to use our restricted summation convention. Write I = ~ui~vi. Then for any ~x ~y 2 V we have h  i ~x  ~y = I(~x ~y) = (~ui  ~x)(~vi  ~y) = ~x  ~ui ~vi  ~y : (23) h   i ~x  ~y ; ~y  ~vi ~ui = 0: (24) Thus For any xed ~y 2 V , (24) is true for all ~x 2 V . Thus ~y ; (~y  ~vi)~ui = ~0, or   ~y = ~y  ~vi ~ui for any ~y 2 V: (25) a) (25) shows that f~u1     ~ung spans V . Hence, n  dim V . b) If n = dim V , then (~u1     ~un) is an ordered basis for V . Let (~u1     ~un) be its dual basis. Choose xed j k 2 f1     ng and set ~x = ~uj , ~y = ~uk in (23). Then    ~uj  ~uk = ~ui  ~uj ~vi  ~uk or   j k = i j ~vi  ~uk = ~vj  ~uk : Thus (~v1    ~vn) is a sequence dual to (~u1     ~un). Since (~u1     ~un) is a basis for V , there is exactly one such dual sequence, namely (~u1     ~un). Hence ~vi = ~ui. Exercise 5 Let V be an n-dimensional Euclidean space. Let A be any nonzero alternating tensor over V (that is, A 2 nV and A 6= 0). Let (~b1      ~bn) be an ordered basis for V , with dual basis (~b1      ~bn). Let 4 = A(~b1      ~bn). For each i 2 f1     ng dene   ~vi = 1 (;1)i;1Ahn ; 1i ~b    6~b   ~b : 4 1 i n Show that ~vi = ~bi . When n = 3, express (~b1  ~b2 ~b3 ) explicitly in terms of (~b1  ~b2 ~b3 ), using only the dot and cross products, not A. 291 Exercise 6 Suppose L 2 L(V ! V ). a) Show that there are unique mappings S , K 2 L(V ! V ) such that S is symmetric (S T = S ) K is antisymmetric (K T = ;K ) and L = S +K . (Hint: then LT = S ;K .) b) Suppose dim V = 3 and A is one of the two unimodular alternating tensors over V . Suppose K 2 L(V ! V ) is antisymmetric. Show that there is a unique vector ~k 2 V such that K$ = A  ~k. Show that ~k = 12 Ah2i K$ and that K$ = ~k  A. Hint: Take components relative to a positively-oriented ordered orthonormal basis. Use (3.1.2) and page D-9.] Exercise 7 Let (~b1      ~bn) be a basis for Euclidean vector space V , its dual basis being (~b1      ~bn). Let L 2 L(V ! V ). $ a) Show that L= ~bi L(~bi) = ~biL(~bi ). b If L is symmetric, show that V has an orthonormal basis x^1      x^n with the property $ that there are real numbers l1      ln such that L= Pnv=1 l(v) x^v x^v . Hint: Let l(v) be the eigenvalues of L. See page D-27.] Solutions 5.a) We show that ~bj  v~ i = j i, and then appeal to the fact that the dual sequence of an ordered basis is unique. We have ~bj  v~ i = 1 (;1)i;1 ~bj  Ahn ; 1i ~b1     6~bi   ~bn  4 i;1   = (;41) A ~bj  ~b1      ~bi;1  ~bi+1     ~bn : 292 It takes i ; 1 interchanges to change (~bj  ~b1     ~bi;1 ~bi+1      ~bn) into (~b1      ~bi;1  ~bj  ~bi+1      ~bn), and each changes the sign of A, so ~bj  v~ i = 1 A ~b1      ~bi;1 ~bj  ~bi+1     ~bn : 4 If j 6= i, ~bj appears twice, so A(~b1      ~bi;1 ~bj  ~bi+1      ~bn) = 0 and ~bj  v~ i = 0. If j = i, then A(~b1      ~bi;1 ~bi  ~bi+1     ~bn) is 4, so ~bj  v~ i = 1. QED. 5.b) When n = 3, 4 = A(~b1  ~b2  ~b3) = ~b1  Ah2i(~b2~b3 ) = ~b1  (~b2 ~b3 ), and ~b 1 = 1 Ah2i(~b ~b ) = 1 ~b ~b 2 3 4 42 3 ~b 2 = ; 1 Ah2i(~b2~b3 ) = ; 1 ~b1 ~b3 = 1 ~b3 ~b1 4 4 4 ~b3 = 1 Ah2i(~b1~b2 ) = 1 ~b1 ~b2 : 4 4 6.a) Uniqueness: If L = S + K with S = S T and K T = ;K then LT = S ; K so S = 1=2(L + LT ) and K = 1=2(L ; LT ). Existence: dene S = 1=2(L + LT ) and K = 1=2(L ; LT ). Then S T = S , K T = ;K , and L = S + K . $ 6.b) Uniqueness: Suppose K = A  ~k. Relative to any pooob we have Kij = "ijkkk so $ Kij "ijl = "ijl"ijk kk = 2l k kk = 2kl . Thus, 2~k =K h2iA. Also 2kl = "lij Kij , so $ 2~k = Ah2i K . Existence: Given K$ antisymmetric, dene ~k = 1=2Ah2i K$ . We have relative to any pooob  ~ A  k ij = "ijk kk = 12 "klmKlm   = 12 i l j m ; i mj l Klm = 12 Kij ; 21 Kji = Kij : $ $ Hence, K = A  ~k. Also Kij = "ijk kk = kk "kij so K = ~k  A. $ $ $ 7.a) ~bi  L= L(~bi ). Multiply by ~bi and sum over i to get ~bi L(~bi ) = ~bi~bi  L] = (~bi~bi ) L $ $ $ = I V  L=L. By viewing ~bi as the original and ~bi as the dual basis, one then also $ nds ~biL(~bi ) =L. 293 7.b) From page D-27, V has an orthonormal basis x^1      x^n consisting of eigenvectors of L. That is, there are numbers l1     ln 2 R such that L(^x1 ) = l1 x^1      L(^xn ) = lnx^n. By 7a) n n n X X $ X = x ^ L (^ x ) = x ^ ( l x ^ ) = lv (^xv x^v ) : L v v v v v v=1 v=1 v=1 Extra problems 4E. Suppose f~b1      ~bm g is a linearly independent subset of Euclidean space V (so m  dim V  m < dim V is possible!). Let (~u1     ~um) and (~v1    ~vm) be any m-tuples of vectors in V . Show that ~ui~bi = ~vi~bi implies ~ui = ~vi. 5E. Suppose V is a Euclidean vector space and a, b, c, d, e are non-negative integers. Suppose A 2 a V , C 2 cV , E 2 eV . We want to study whether it is true that Ahbi (C hdiE ) = (AhbiC ) hdiE: (26) i) Using p 54 of the notes, nd conditions on a, b, c, d, e which imply (26). ii) If dim V  2, show that whenever a, b, c, d, e violate the conditions found in i), it is possible to have A 2 a V , C 2 cV and E 2 l V which violate (26). (Hint: use polyads.) 6E. In oriented Euclidean 3-space (V A), show that ~u, ~v, w~ can be chosen so that ~u (~v w~ ) 6= (~u ~v) w~ . Solutions to Extra Problems 4E. Let U = spf~b1      ~bm g. Then U is a subspace of V and hence a Euclidean vector space under V 's dot product. Also, (~b1      ~bm) is an ordered basis for U , so in U it has a dual basis (~b1      ~bm). If u~ i~bi = v~ i~bi then (u~ i~bi )  ~bj = (v~ i~bi )  ~bj , so u~ i(~bi  ~bj ) = v~ i(~bi  ~bj ), so u~ i i j = v~ ii j . Then u~ j = ~vj . QED. 294 a A p q b c C q ss r d E A t e Figure Ex-1: Exercise 5E.i 5E.i) The picture that makes (26) work is in gure Ex-1 There must be p q r s t  0 such that a = p + q, b = q, c = q + r + s, d = s, e = s + t. solving for p q r s t gives p = a ; b, q = b, r = c ; b ; d, s = d, t = e ; d. The conditions that p q r s t a b c d e  0 reduce to 0  b  a 0  d  e b + d  c: 5E.ii) The question is, is it possible to violate the inequalities in i) in such a way that Ahbi(C hdiE ) and (AhbiC )hdiE are both dened. Let hki(P ) = order of tensor P . Thus hki(A) = a, hki(AhbiC ) = a + c ; b, etc. A product P hqiR is dened only if hki(P )  q and hki(R)  q. If Ahbi(C hdiE ) is dened we must have a  b, hki(C hdiE ) = c + e ; 2d  b and c  d, e  d. If (AhbiC )hdiE is dened we must have d  e and hki(AhbiC )  d, or a + c ; 2b  d, and also b  a, b  c. If both triple products are dened we must have b  a, b  c, d  c, d  e, b + 2d  c + e and 2b + d  a + c. The only way i) can fail if all these hold is to have b + d > c. Choose ~ 1      ~ a , ~1    ~c, ~"1    ~"e 2 V and let A = ~ 1    ~ a , C = ~1    ~c, E = ~"1    ~"e. Then we have gure Ex-5ii And AhbiC = (~ a;b+1  ~1)    (~ a  295 a A b C c d E e Figure Ex-2: Figure Ex-5.ii ~b)~ 1    ~ a;b~b+1    ~c (AhbiC )hdiE = (~ a;b+1  ~1)    (~ a  ~b)(~ a+c;2b;d+1  ~"1)    (~ a;b  ~"d+b;c) (~b+1  ~"d+b;c+1)    (~c  ~"d) ~ 1    ~ a+c;2b;d ~"d+1    ~"e. Similarly C hdiE = (~c;dn  ~"1)    (~c  ~"d)~1    ~c;d~"d+1    ~"e and Ahbi(C hdiE ) = (~ a;b+1 ~1)    (~ a;b+c;d ~c;d)(~ a;b+c;d+1  ~"d+1)    (~ a ~b+2d;c) (~c;d+1 ~1)    (~c  ~d )~ 1    ~ a;b~b+2d;c+1    ~"e. Choose all ~ , ~ , ~" so the dot products are 6= 0. Now a ; b > a + c ; 2b ; d, so (AhbiC )hdiE = Ahbi(C hdiE ) implies that ~ a;b and ~"2d+b;c are linearly dependent. By choosing them linearly independent we have (AhbiC )hdiE 6= Ahbi(C hdiE ). 6E. ~u (~v w~ ) = (~u  w~ )~v ; (~u  ~v)w~ (~u ~v) w~ = w~ (~v ~u) = (w~  ~u)~v ; (w~  ~v)~u: These are equal i (~u  ~v)w~ = (~v  w~ )~u. Choose ~u and w~ mutually ? and of unit length, with ~v = ~u + w~ , and this equation will be violated. 296 Exercise 8 $ Let U be a Euclidean vector space. Let D be an open subset of U . Let T 2 U  U . For $ each ~u 2 U , let f~(~u) =T ~u and g(~u) = ~u  ~u. Show that at every ~u 2 D, f~ : D ! U ~ f~(~u), r ~ g(~u), and the remainder and g : D ! R are dierentiable. Do so by nding r functions, and by showing that R(~h) ! 0 as ~h ! ~0. Exercise 9 Suppose that Df and Dg are open subsets of Euclidean vector spaces U and V respectively. Suppose that f~ : Df ! Dg and ~g : Dg ! Df are inverse to one another. Suppose there is a ~u 2 Df where f~ is dierentiable and that ~g is dierentiable at ~v = f~(~u). Show that dim U = dim V . Hint: See page 108.] Exercise 10 Suppose V is a Euclidean vector space, (U A) is an oriented real three-dimensional Euclidean space, D is an open subset of U , and f~ : D ! U and ~g : D ! V are both dierentiable at ~u 2 D. i) Which proposition in the notes shows that f~~g : D ! U  V is dierentiable at ~u? ii) Show that at ~u     r~ f~~g = r~ f~ ~g ; f~ r~ ~g: Note: In this equation, one of the expressions is undened. Dene it in the obvious way and then prove the equation.] Solutions 8a.    $ $ $  f~ ~u + ~h = T  ~u + ~h =T ~u+ T ~h 297 $T = f~(~u) + ~h T : ~ f~(~u) =T$T and R~ (~h) = ~0 for all ~h. Therefore r 8b.   g ~u + ~h = = =  ~  ~ ~u + h  ~u + h ~u  ~u + 2~h  ~u + ~h  ~h    g (~u) + ~h  2~u] + ~h ~h :  ~ g(~u) = 2~u, and R(~h) = ~h. As ~h ! 0, R(~h) ! 0 because ~h ! ~0 Therefore r  means ~h ! 0. $ ~ $ $ $ $ ~~ $ $ $ 9. Dene P = r f (~u) and Q= r ~g(~v). By example 9.2.24, P  Q= I U and Q  P = I V . Therefore, by 7.4.20, the linear mappings P : U ! V and Q : V ! U satisfy Q  P = IU and P  Q = IV . By iii) on page D-6, P and Q are bijections. Since they are linear, they are isomorphisms of U to V and V to U . By preliminary exercise 3, dim U = dim V . 10.i) Method 1 on page 105 or method 2 on page 107, applied to f~ ^1 : D ! U  R1 and ^1~g : D ! R1  V .  $ $ $ 10.ii) For ~u 2 U and P 2 U  V we dene ~u P := Ah2i ~u P . Then, relative to any $ basis ~b1 , ~b2 , ~b3 for U and ~1     ~n for V , we have (~u P )i l = Aijk j Pkl . Then h ~  ~ ii   r f ~g l = Aijk @j f~~g kl = Aijk @j (fk gl) = Aijk (@j fk ) gl + fk @j gl ]  ~ ~ i = r f gl ; Aikj fk @j gl h ~ ~ ii   = r f ~g l ; Aijk fj r~ ~g kl h ~ ~  ii  ~ ~ i = r f ~g l ; f r~g l h ~ ~  ~ ~ ii = r f ~g ; f r~g l : QED: 298 Extra Problems $ 7E. Let U be a Euclidean vector space. Let T 2 U  U . Let D be the set of all ~u 2 U $ $ such that ~u T ~u > 0. If ~u 2 D, dene f (~u) = ln(~u T ~u). i) Show that D is open. ~ f (~u). ii) Show that f : D ! R is dierentiable at every point ~u 2 D, and nd r 8E. Let V be the set of all innite sequences of real numbers, ~x = (x1  x2   ), such that P1 x2 converges. If ~y = (y  y    ) is also in V and a b 2 R, dene a~x + b~y = 1 2 n=1 n (ax1 + by1  ax2 + by2    ). Dene ~x  ~y = P1 n=1 xn yn . The sum converges by the Cauchy test because, for any M and N , Schwarz's inequality implies j PNn=m xn ynj  (PNn=m x2n )1=2 (PNn=m yn2 )1=2 . Furthermore, j~x  ~yj2 = j limN !1 PNn=1 xn ynj2 =< limN !1 (PNn=1 x2n )(Pnn=1 yn2 ) = k~xk2k~yk2, so Schwarz's inequality works in V .] V is a real dot-product space but not a Euclidean space, because it is not nite dimensional. If ~v0  ~v1~v2     is a sequence of vectors in V , dene limn!1 ~vn = ~v0 by the obvious extension of denition 9.1.30, i.e. limn!1 k~v0 ; ~vn k = 0. i) Suppose limn!1 ~vn = ~v0 . Show that for any ~x 2 V , limn!1 ~x  ~vn = ~x  ~v0 . ii) Construct a sequence ~v1 ~v2    in V such that limn!1 ~x  ~vn = 0 for every ~x 2 V , but limn!1 ~vn does not exist. (For dierentiability, V has two dierent denitions, leading to Frechet and Gateau derivation respectively based on norms and on components). Solutions $ $ $ 7E. If T = 0 , D is empty and i) and ii) are trivial. Therefore assume k T k > 0. Dene $ g(~u) = ~u T ~u. 299 $ $ $ i) g(~u + ~h) ; g(~u) = ~h T ~u + ~u T ~h T ~h. Therefore, by the triangle and generalized Schwarz inequalities,   ~    $  g ~u + h ; g (~u)  k~hkk T k 2k~uk + k~hk : Assume ~u 2 D so g(~u) > 0. Then ~u 6= ~0. Let " be the smaller of k~uk and $ $ g(~u)=(3k~ukk T k). If k~hk < " then jg(~u + ~h) ; g(~u)j < "k T k(2k~uk + k~hk) $ < 3"k T kk~uk  g(~u). Hence g(~u + ~h) ; g(~u) > ;g(~u), so g(~u + ~h) > 0. That $ is, if ~u 2 D and " = smaller of ~u and g(~u)=(3k~ukk T k) then the open ball B (~u ") D. Hence D is open, $ $T ii) By i) above, g(~u + ~h) = g(~u) + ~h  (T + T )  ~u + k~hkR(~h) where R(~h) = $ (~h T ~h)=k~hk for ~h 6= ~0 and = ~0 for ~h = ~0. Thus g is dierentiable at ~u, with T r~ g(~u) = (T$ + T$ ) ~u. By the chain rule, f  g is dierentiable, with f (v) = lnv, ~ g(~u) where v = g(~u). This is v;1r ~ g(~u), so and its gradient is @v f (v)r  $ $T    $  T + T  ~u ~rln ~u T ~u (~u) = : $ ~u T ~u 8E. i) j(~x  ~vn ; ~x  ~v0 )j = j~x  (~vn ; ~v0 )j  k~xkk~vn ; ~v0 k ! 0 if k~vn ; ~v0 k ! 0. 8E. ii) Let ~vn = (0     0 1 0   ), all 0 except for a 1 in the n'th place. If ~x 2 V , 2 ~x  ~vn = xn ! 0 as n ! 1 because P1 1 xn converges. But if k~vn ; ~v k ! 0 for some ~v 2 V then there is an N so large that if n > N then k~vn ; ~vk < 1=2. If m, n > N p then k~vm ; ~vn k = k~vm ; ~v + ~v ; ~vn k  k~vm ; ~vk + k~vn ; ~vk = 1. But k~vm ; ~vn k = 2. Exercise 11 A mass distribution m occupies a subset D of ordinary Euclidean 3-space U . Let n^ be any unit vector in U and denote by n^R the straight line through U 's origin ~0 in the direction 300 of n^. Let w~ (~r) be the perpendicular distance of the point ~r from the axis n^R, so that R the moment of inertia of the mass distribution about that axis is J (^n) = D dm(~r)w~(~r)2. $ $ $ Dene T : D ! U  U by requiring for each ~r 2 D that T (~r) = r2 I U ;~r~r. Here r2 = ~r  ~r $ $ R $ and I U is the identity tensor in U  U . The tensor J = D dm(~r) T (~r) is called the inertia $ R $ tensor of the mass distribution, and is usually written J = D dm(~r)r2 I U ;~r~r ]. $ a) Show that J is symmetric. $ b) Show that J (^n) = n^  J n^. $ c) Show that to calculate J it is necessary to calculate only six particular integrals of real-valued functions on D with respect to the mass distribution m. (Once these six functions are calculated, the moment of inertia of the body about any axis is found from 11b.) Exercise 12 In problem 11, suppose the mass distribution rotates rigidly about the axis n^ R with angular velocity +~ = +^n, so that the mass at position ~r has velocity ~v(~r) = +~ r. The R angular momentum of the mass distribution about ~0 is dened as L~ = D dm(~r)~r ~v(~r)], R and the kinetic energy of the mass distribution is dened as K = 1=2 D dm(~r)v(~r)2, where v2 means ~v  ~v. Show the following: $ a) L~ =J +~ $ b) K = 1=2 +~  J +~ c) L~ need not be J (^n)+~ . Exercise 13 $ The second moment tensor of the mass distribution in problem 11 is dened as M = R dm(~r) ~r ~r. D 301 $ R a) Show that tr M = D dm(~r)r2. $ $ $ $ b) Express J in terms of M , tr M and I U . $ $ $ $ $ c) Express M in terms of J , tr J and I U . (J can be observed by noting the reaction $ R of the mass distribution to torques. Then 13c) gives M and 13a) gives D dm(~r)r2 from such observations). d) If the mass distribution is a cube of uniform density , side 2a and center at ~0, nd J (^n) when n^ R is the axis through one of the corners of the cube. Exercise 14 Let y^1, y^2, y^3 be an orthonormal basis for physical space P . Let c be a xed scalar. For any ~r 2 P , write ~r = riy^i. Consider a steady uid motion whose Eulerian description ~vE is given by ~vE (~r t) = r1y^1 + c r2y^2 = ~r  (^y1y^1 + cy^2y^2) for all ~r 2 P and all t 2 R. a) Find the Lagrangian description of this motion, using t0-position labels. b) Show that the paths of the individual particles lie in planes r3 = constant, and are hyperbolas if c = ;1 and straight lines if c = +1. c) Find the label function ~x E : P R ! P . (It will depend on which xed t0 is used to establish t0-position labels.) d) Find the Eulerian and Lagrangian descriptions of the particle acceleration ~a. Solutions $ $ 11 a) The permutation operator (12): P  P ! P  P is linear, and (12) T (~r) =T (~r), so Z Z Z $ $ $ (12) dm(~r ) T (~r ) = dm(~r) (12) T (~r ) = dm(~r ) T (~r ): D D D 302 . ~ w( r) . n^ r ^n r . Figure Ex-3: b) we (~r )2 = r2 ; (^n  ~r )2 $ = r2n^ I P n^ ; n^  r^r^  n^  $  = n^  r2 I P ;~r ~r  n^ : Hence Z h i dm(~r)^n  r2IP ; ~r ~r  n^ D Z  $  $ 2 = n^  dm (~r) r I P ;~r ~r  n^ = n^ J n^: J (^n) = D (27) (28) $ $T $ $ c) Choose a pooob, x^1 , x^2 , x^3 . Then J = Jij x^i x^j and since J =J , Jij = Jji. Thus J is known if we know J11, J22, J33 , J12 , J23, J31 relative to one pooob. But Jij = 12 a) D h i dm (^r) r2ij ; rirj : i Z h  i dm (~r) ~r + ~r = dm (~r) ~r +~ ; ~r ~r  +~  $ D ZD dm (~r ) r2 I P ;~r ~r  +~ = D  $  Z $ 2 = dm (~r ) r I P ;~r~r  +~ =J +~ : L~ = Z Z h~ D 303 b) K = = = = = = 1 Z dm (~r ) +~ ~r  +~ ~r  2 ZD 1 dm (~r) +  h~r +~ ~ri 2 ZD 1 dm (~r ) +~  hr2+~ ; ~r ~r  +~ i 2 ZD 1 dm (~r) +~  r2 $ ;~r ~r   +~  IP 2 D Z 1 +~  dm (~r) r2 $ ;~r~r  +~ IP 2 D 1 +~  $ +~ : 2 J c) Let the mass distribution consist of a single mass point m at position ~r. Then  $  $ 2 J = m r I P ;~r ~r   ~L = J$ +~ = m+ r2 $I P ;~r ~r  n^ h i = m+ r2n^ ; ~r (~r  n^)  $  h i ~ J (^n) + = n^  J n^ n^+ = m+ r2 ; (^n  ~r )2 n^: If L~ = J (^n)+~ then r2 ; (^n  ~r)2 ]^n = r2n^ ; ~r(~r  n^ ), so (~r  n^)2 n^ = (~r  n^)~r. To violate this condition choose n^ so it is neither parallel nor perpendicular to ~r. 13 a) tr : P  P ! R is linear, so Z Z $ tr M = tr dm (~r) ~r ~r = dm (~r) tr ~r~r = Z D D dm (~r) r2: D  $  R $ R $ R 2 b) J = dm(~r) r I P ;~r ~r =  D dm (~r) r2] I P ; D dm (~r) ~r ~r $ $  $ $  J = I P tr M ; M : c) From the above equation, since tr : P  P ! R is linear, $ $ $ $ $  $   $ tr J = tr I P tr M ; tr M = 3 tr M ;tr M = 2 tr M : $ $ $ $ $ $ Hence, tr M = 1=2 tr J and M = 1=2 I P (tr J ); J . 304 d) Let x^1 , x^2 , x^3 be a pooob parallel to the edges of the cube. Then Mij = Z dx1 dx2 dx3 xixj  Za Za =  dx1 dx2 dx3 xi xj : ;a ;a ;a R Thus Mij = 0 if i 6= j and M11 = M22 = M33 = 4a2 ;aa x2 = 8a5 =3. Therefore $ $ $ M = 8a5 =3 I P and tr M = 8a5 so from () above $ 16a5 $ J= 3 I P : $ Then for any n^ , n^ J n^ = 16a5=3. 14 a) cube Za dr1 = r1 dr2 = cr2 dr3 = 0 so dt dt dt 1 1 ( t ; t ) 2 2 r = A e 0  r = A ec(t;t0 )  r3 = A3: The label ~x = xi y^i belongs to the particle which was at that position at t = t0. For this particle when t = t0 , ri = xi , so xi = Ai. Thus () r1 = x1 e(t;t0 ) r2 = x2 ec(t;t0 )  r3 = x3 : ~r = riy^i = e(t;t0 ) x1 y^1 + ec(t;t0 ) x2 y^2 + x3 y^3 h i = ~x  y^1y^1e(t;t0 ) + y^2y^2ec(t;t0 ) + y^3y^3 : h i ~r L (~x t) = ~x  y^1y^1e(t;t0 ) + y^2y^2ec(t;t0 ) + y^3y^3 : b) y^3  ~r is independent of time for each particle, so each particle lies in a plane y^3  ~r = constant. And for a particular particle, r1r2 = x1 x2 e(1+c)(t;t0 ) = x1 x2 if c = ;1, which is a hyperbola. And r2=r1 = e(c;1)(t;t0 ) x2=x1 = x2 =x1 , if c = +1. Then this is a straight line through the y^3 axis. c) From () above, x1 = r1e;(t;t0 ) , x2 = r22e;c(t;t0 ) , x3 = r3, so ~x = xi y^i = r1 e;(t;t0 )y^1 + r2e;c(t;t0 ) + r3y^3 h i = ~r  y^1y^1e;(t;t0 ) + y^2y^2 e;c(t;t0 ) + y^3y^3 : h i ~x E (~r t) = ~r  y^1y^1 e;(t;t0 ) + y^2y^2 e;c(t;t0 ) + y^3y^3 : 305 d) h i ~aL (~x t) = Dt2 ~r L (~x t) = ~x  y^1y^1 e(t;t0 ) + y^2y^2 ec(t;t0 ) h i ~aE (~r t) = ~x E (~r t)  y^1y^1 e(t;t0 ) + c2y^2y^2 ec(t;t0 ) h i ~aE (~r t) = ~r  y^1y^1 e;(t;t0 ) + y^2y^2 e;c(t;t0 ) + y^3y^3 h i  y^1y^1 e(t;t0 ) + c2 y^2y^2 ec(t;t0 ) h i = ~r  y^1y^1 + c2y^2y^2 : Another way to get the same result is to observe that ~vE (~r t) = ~r  (^y1y^1 + c y^2y^2) has no explicit t dependence, so @t~vE (~r t) = ~0. Also @~~vE (~r t) = y^1y^1 + c y^2y^2, so ~vE  @~~vE (~r t) = ~r  (^y1y^1 + c2 y^2y^2). But ~aE (~r t) = = = Dt~vE (~r t) @t~vE (~r t) + ~vE  @~~vE (~r t)   ~0 + ~r  y^1y^1 + c2y^2y^2 : Exercise 15 a) Given a continuum and a function f : R ! W which depends on t alone (W is a Euclidean vector space), invent a physical quantity which can reasonably be called f , and show that (Dtf )L(t) = (@t f )E (t) = t f (t). b) A continuum undergoes rigid body motion. At time t the position, velocity and acceleration of the pivot particle are R~ (t), V~ (t) = t R~ (t) and A~ (t) = t V~ (t), and the angular velocity relative to the pivot particle is +~ (t). Find the Eulerian description of particle acceleration in the material. 306 Exercise 16 Several decades ago, before the big bang was accepted, Fred Hoyle suggested that the expanding universe be explained by the continuous creation of matter. Then mass is not quite conserved. Suppose that a certain region of space is occupied by a continuum, and that at position ~r at time t new matter is being created at the rate of  E (~r t) kilograms per cubic meter per second. Make the necessary changes in the derivation of the Eulerian form of the law of conservation of mass and in equations (13.1.12) and (13.1.17). Exercise 17 $ At time t0 in a certain material, the Cauchy stress tensor S is symmetric. a) Let K 0 be an open set in the region occupied by the material, and suppose its boundary @K 0 is piecewise smooth. Show that the total force F~ and the total torque L~ about ~0 exerted on K 0 by the stress on @K 0 are the same as those exerted by a ctitious $ body force with density @~ S newtons / m3 acting in K 0. $ $ $(0) b) Write the Taylor series expansion of S about ~0 in physical space P as S (~r) =S $(0) $(1) $(2) +~r  S (1) + 1=2(~r ~r)h2iS (2) +    where S , S , S ,    are constant tensors in P  P , 3 P , 4P ,    and (12)S (2) = S (2) . Neglect all terms except those involving $(0) $(1) $(2) S , S , and S , and calculate F~ and L~ of a) for K 0 = B (~0 c), the solid ball of radius c centered on ~0. Hint 1: Do all calculations relative to an orthonormal basis in P . R Hint 2: Use the symmetry of B (~0 c) to evaluate B dV (~r)rirj . Solutions 15 a) The physical quantity has as its Lagrangian description f L(~x t) = f (t). Then its Eulerian description is f E (~r t) = f (t). Since @~f~ = 0, Dtf = @t f + ~v  @~f~ = @t f . 307 b) h i ~vE (~r t) = V~ (t) + +~ (t) ~r ; R~ (t) h i $ = V~ (t) + ~r ; R~ (t)  + (t): $ @~~vE =+ so h  i $ ~vE  @~~vE = ~vE  += +~ ~vE = +~ V~ + +~ +~ ~r ; R~ i $ $ h @t~vE = A~ (t) ; V~ (t) + + ~r ; R~  @t +   = A~ ; +~ V~ + @t +~ ~r ; R~ :  E   h  i (Dt~v)E = ~aE = @t~v + ~v  @~~v = A~ + @t +~ ~r ; R~ + +~ +~ ~r ; R~ h ~ i  $ $ $ E ~ ~a (~r t) = A(t) + ~r ; R(t)  @t + + +  + : 16. The amount of mass in K 0(t) is M K 0 (t)] = Z K 0 (t) dVP (~r)E (~r t): It is a mathematical identity (derivative over moving volume) that  E E d M K 0(t)] = Z dV @ E + Z dA n ^  p p ~v  0 (t) P t 0 (t) dt K @K Z h iE = 0 dVp @t  + @~  (~v) (~r t) : K (t) Hoyle's new physical law is that d M K 0(t)] = Z dV  E (~r t): p dt K 0(t) Therefore Z h iE ~ dV @  + @  ( ~ v ) ;  (~r t) = 0: p t K 0 (t) Since this is true for every open set K 0 (t) with piecewise smooth boundary, the vanishing integral theorem gives @t  + @~  (~v ) = : 308 Then for any other physical quantity f~, d Z dV f~ E = Z dV h(@ ) f~ +  @ f~ iE P t t dt K 0(t) P KZ0 (t) h  iE + 0 dAP n^  ~v f~ Z K (t) h    iE = 0 dVP (@t ) f~ +  @t f~ + @~  ~vf~ ZK (t) n   h i oE = 0 dVP (@t ) f~ +  @t f~ + @~  (~v ) f~ + ~v  @~f ZK (t) n   h i oE = 0 dVP  @t f~ + ~v  @t f~ + @t  + @~  (~v) f~ ZK (t) h iE = 0 dVP Dt f~ +  f~ K (t) 17 a)  $ $ Z dAn^ S = 0 dV @~ S @K 0 K Z  $ Z $  dA~r n^ S = ; 0 dA~n S ~r 0 Z @K = = = K i  $  Z ~  S ~r = ; dV @j (Sjk rl "kli) dV @ K0Z K0 i ;"ikl K 0 dV (rl @j Sjk + jl Sjk ) Z ;"ikj K 0 dV (rj @l Slk + jl Slk ) Z "ijk 0 dV (rj @l Slk + Sjk ) Z K $ dV ~r @~ S : = ; = i Z K0 i b) (1) + 1 rk rl S (2) Sij = Sij(0) + rk Skij klij 2 (1) + 1  k rl +  l rk  S (2) @i Sij = Siij i klij 2 i (1) + 1 rl S (2) + rk S (2)  = Siij ilij kiij 2 (1) + rk S (2) = v + rk w @i Sij = Siij j kj kiij ( denes vj and wkj ): 309 $ @~ S $ ~r @~ S  $ ~ ~r @  S i Z $ dV @~ S B Z dV rj = 0 and B Ah2itr23 S (2) . = ~v + ~r w$ :   = ~r ~v + ~r ~r w$ = (~r ~v)i + "ijk rj rlwlk Z  Z $ = jB j~v + dV ~r  w~ : dV ~r = 0 so B B 4  4  F~ = 3 c3~v = 3 c3 tr12 S (1) Z Z ~L = dV ~r @~ $ S  Li = dV "ijk (rj vk + rj rl wlk ) : B B R dV r r = 4=15 c5  so L = 4=15 c5 " w , L~ = 4=15 c5 j l jl i ijk jk B Exercise 18 a) The angular momentum of a single dysprosium atom is 15h=4 where h is Planck's constant. The density of dysprosium is 8:54gm=cm3, and its atomic weight is 162.50. Compute the angular momentum of a stationary ball of solid dysprosium with radius r cm if all the dysprosium atoms are aligned in the same direction. If the dysprosium atoms had no angular momentum, how rapidly would the ball have to rotate as a rigid body in order to have the same angular momentum? b) Let y^1, y^2, y^3 be a positively oriented ordered orthonormal basis for physical space P . A permanent alnico magnet has magnetization density M~ = M y^1 where M = 106 amp/meter (close to the upper limit for permanent magnets). The magnet is held at rest in a uniform magnetic eld B~ = B y^2 with B = 1 tesla (about the upper limit for commercial electromagnets). Then the volume torque on the magnet is m ~ = M~ B~ joules /meter3 . Suppose that the magnet's intrinsic angular momentum density ~l $ $ does not change with time, and that its torque stress tensor M vanishes. Let S be $ $T the Cauchy stress tensor in the magnet. Find 1=2(S ; S ), the antisymmetric part $ of S , in atmospheres of pressure (one atmosphere = 1:013 105 newtons/meter2 = $ $ $T 1:013 106 dynes /cm2 ). If the symmetric part of S , 1=2(S + S ), vanishes, sketch 310 1 meter ^n 5 ^n 1 ^z S5 S1 ^n 4 S4 ^y S2 ^n 2 ^x 1 meter ^n 3 S3 Figure Ex-4: the stresses acting on the surface of a small spherical ball of material in the magnet. Exercise 19 a) Give the obvious denition of the stu \y component of momentum". Give its (- F~ ), its ( F~ ), and its creation rate  . b Show that in a material at rest with no body forces, the material ux density F~ of y momentum satises @~  F~ = 0. (If F~ were the velocity of a uid, the uid would be incompressible.) c) An aluminum casting 10 cm thick and 1 meter on each side is shaped as in gure Ex-4. A steel spring is compressed and placed between the jaws of the casting as shown. The spring and casting are at rest and gravity is to be neglected. Roughly sketch 311 the eld lines of F~ (the material ux density of y^ momentum) in the casting. (If F~ were a force eld, the eld lines would be the lines of force if F~ were a velocity eld, they would be the streamlines.) Hint for c): Estimate qualitatively the sign of the y^ component of the force exerted by the aluminum just in front of (Si n^ i) on the aluminum just behind it for the surfaces i = 1 2 3 4 5 sketched in the gure. Note for c): The x^ axis points out of the paper, the y^ axis is parallel to the axis of the spring, and the z^ axis points up, as shown in the gure. Solutions 18 a) There are 6:025 1023=162:50 = 3:7 1021 atoms in a gram of Dysprosium. Each has angular momentum (15=4)(6:625 10;27) erg sec. If all are aligned, one gram of Dy has angular momentum l = 2:932 10;5 erg sec/gm. A Dy ball of radius r cm has intrinsic angular momentum (4=3)r3l. If this angular momentum were not intrinsic, but due to an angular velocity ! of rigid rotation, the angular momentum would be (2=5)r2(4=3)r3!. Thus l = (2=5) r2!, or ! = (5=2) l=r2 = 7:33 10;5=r2 radians/sec, with r in cm. b) ~0 0 0 0 $ = m ~ + Ah2i S  so relative to the pooob y^1 y^2 y^3 = mi + "ijk Sjk : Then = mi"ilm + "ilm"ijk Sjk = mi"ilm + (lj mk ; lk mj ) Sjk = mi"ilm + Slm ; Sml : () 12 (Slm ; Sml ) = ; 21 "lmimi : In our problem, m ~ = MB y^3 so 1 (S ; S ) = ; 1 MB 1 (S ; S ) = 1 (S ; S ) = 0 and 2 12 21 2 2 13 31 2 23 32 312 m .. .. Figure Ex-5: 1 $ ; $T  = 1 $ ; $T  y^ y^  so 2 S S  2 S S ij i j 1 $ ; $T = 1 MB (^y y^ ; y^ y^ ) : 2 1 1 2 2 S S 2 1 MB = 106 joules/meter3 = 106 newtons/meter2 2 2 2 6 10 = 2 1:013 105 atm = 4:94 atmospheres: $ $ $ $T $ $T $ $T $ $ For any S , S = 1=2(S + S ) +1=2(S ; S ). If 1=2(S + S ) = 0, then S = 1=2(S T ; $S ). From (*), $S = ;1=2 A  m~ . The stress S~ (^n) is n^  S$= ;1=2^n  A  m~ so Sj (^n) = ;1=2ni"ijk mk = 1=2"jiknimk = 1=2(^n m ~ )j . Thus S~ (^n) = 21 n^ m ~: The stress is directed along lines of latitude about the north pole m ~ , points west, and has magnitude equal to the sine of the colatitude of n^ relative to m ~. 19 a) - = ~v  ~y   = ~v  y^ $ F~ = ~v~v; S  y^ F~ = ; $S y^  $   $ ~ ~  = Dt  + @  ; S y^ = Dt~v ; @  S  y^ = f~  y^: 313 n^5 ^n 1 ^z n^ 4 ^y ^x ^n 2 ^n 3 Figure Ex-6: $ $ $ b) If f~ = 0 and ~v = ~0 then @~ S = ~0 so (@~ S )  y^ = @~  (S y^) = 0. $ $ c) F~ = ; S y^. The double arrows show n^i  S on the ve surface sections. Then  $   $ $ ~ on S1 n^1  F = y^  ; S y^ = ;y^ S y^ = ; n^1  S  y^ > 0  $   $  $ ~ on S2  n^ 2  F = ;z^  ; S y^ = ; ;z^ S  y^ = ; n^2  S  y^ > 0  $   $ > 0 above ~ on S3  n^3  F = ;y^  ; S y^ = ; ;n^3  S  y^ < 0 below  $   $ ~ on S4 n^4  F = z^  ; S y^ = ; n^4  S  y^ > 0  $   $ on S5 n^5  F~ = y^  ; S y^ = ; n^5  S  y^ > 0: Thus the ow lines are as in gure Ex-7. 314 Figure Ex-7: Exercise 20 Geological evidence indicates that over the last 5 million years, the west side of the San Andreas fault has moved north relative to the east side at an average rate of about 6 cm/year. On a traverse across the fault, Brune, Henyey and Roy (1969) (J.G.R. 74, 3821) found no detectable extra geothermal heat ow due to the fault. They estimate that they would have detected any anomalous heat ow larger than 1:3 10;2 watts/meter2 . Use (13.7.45) to obtain an upper bound on the northward component of the average stress exerted by the material just west of the fault on the material just east of the fault. Assume that the fault extends vertically down from the surface to a depth D, and that the shape of the heat ow anomaly due to the fault is a tent function whose width is D on each side of the fault. 315 up American Plate North Pacific Plate east D San Andreas Fault Extra heat flow due to fault, watts /m 2 . D D Figure Ex-8: . east 316 Solutions 20. We assume that the heat ow is steady. Since we are interested only in averages, we $ may assume that S ]+; is constant on the fault, as is ^  H~ ]+; . D is probably < 20 km (see Brune et al. 1969) so we may treat the fault as innitely long. Then H~ lies in east-west vertical planes. Consider a length L along the fault. The amount of heat produced by this section of the fault in LD^  H~ ]+;. It all ows out of the surface of the earth in a rectangle of length L along the fault and width 2D across the fault, the prole being the triangular one sketched as the bottom gure in the exercise. The area under that triangle is Dh where h is the triangle's height, so the total heat ow out in the length L and width 2D is LDh. This must equal LD^  H^ ]+; in the steady state. Thus h = ^  H^ ]+;. Since no anomaly was detected, h  1:3 10;2w=m2. $ Thus ^  H^ ]+; = ^  S ]+;  ~v]+;  1:3 10;2w=m2. Let x^ be a unit vector pointing north, and let y^ = ^ point west. Then ~v]+; = 6^x cm=yr = 6 10;2x^ meters/yr = 2 10;9x^ meters/sec. The heat production must be positive, so $+   0  ^  S  x^ 2 10;9  1:3 10;2 ; or  $+ 0  ^ S  x^  65 105 pascals. ; The material just west of the fault exerts on the material just east of the fault, a $ $ stress ^  S ]+;. The northward component of this stress is ^  S ]+;  x^, and it lies between 0 and 65 bars (105 pascal ). $ See Brune et al., for a discussion of energy loss from ~v  S ]+;  ~v]+; by seismic radiation. This exercise works only for the parts of the fault where most energy is not radiated, i.e., the parts of the fault which creep. Exercise 21 Below are the Eulerian descriptions of three motions of continua. In these descriptions, fx^ y^ z^g is a right-handed orthonormal basis for physical space P ,  is a constnat real 317 number, and +~ is a constant vector. For each motion, nd the local strain rate tensor $ , ! , and the its eigenvectors, and its eigenvalues. Also nd the local rotation rate tensor $ local angular velocity w~ . a) ~vE (~r t) = ~r  (^xx^ ; y^y^) = (xx^ ; yy^) b) ~vE (~r t) = ~r  (^yx^) = yx^ c) ~vE (~r t) = +~ ~r . Exercise 22 !_ be the Eulerian descriptions of the velocity, local strain rate tensor and local Let ~v, $_ , $ rotation rate tensor in a certain continuum. Let x^1 , x^2 x^3 be a right-handed orthonormal basis in physical space P . Let position vectors be written ~r = rix^i . Let @i denote the partial derivative with respect to ri. Take tensor components relative to x^1  x^2  x^3. a) Show that @ ! = @  ;@  i jk j ki k ij b) Show that @i @j  kl ;@j @k  li +@k @l  ij ;@l @i  jk = 0. $ c) Show that if  : P ! P  P is any tensor-valued function of ~r such that everywhere  = and b) holds, then a function $! : P ! P  P can be found which satises ij ji ! = ; ! and also a) above. Show that this function is unique up to an additive ij ji $ constant tensor + which is antisymmetric. d) Show that if $ is the function $ in c), there is a velocity function ~v : P ! P whose local strain rate tensor is  . Show that any two such functions ~v and ~v dier by the 1 2 velocity eld of a rigid motion that is, there are xed vectors +~ and V~ such that for all ~r 2 P , ~v2(~r) ; ~v1(~r) = V~ + +~ ~r. Hint for c) and d): Suppose gi : P ! R for i = 1 2 3. You know from elementary vector calculus that there is a \potential function" f : P ! R such that gi = @if i R @i gj = @j gi everywhere and if this is so, then f (~r) = f (~0) + 0~r drigi (line integral). 318 Exercise 23 Use the notation beginning on page 256 for discussing small disturbances of an HTES of an elastic material whose reference state is HTES0 . $$ $ $ a) Using only 0 , S 0, B0, W 0 , 0 , Q0, explicitly express U ; U0 as a function of N and $ $ $ $  G, correct to second order (i.e., including the terms (N )2 , N G and  G  G.) $$ b) The tensor F 0, is called the isentropic or adiabatic stiness tensor because if N = 0 (no change in entropy) then $ $ $$  S =  G h2i F : $$ Find the isothermal stiness tensor F00 , the tensor such that if  = 0 then $ $ $$0  S =  G h2i F0 : $ $0 $$ $ Express F0 in terms of F 0 , W 0 B0, 0 . Exercise 24 Using the notation on page 260, give the results of 23 a), b) when HTES0 is isotropic. Solutions 21 a)  T @~~vE =  (^xx^ ; y^y^) = @~~vE  so $ =  (^xx^ ; y^y^)  $ ! = 0 , so ~!= ~0, eigenvectors and eigenvalues being (^x ), (^y ;), (^z  ;0). $ 319 b)  T @~~vE = y^x^ @~~vE = x^y^ $ =  (^xy^ + y^x^)  $! =  (^yx^ ; x^y^) 2 2 eigenvectors and eigenvalues of $ are  !  ! x^p+ y^    x^p+ y^  ;   (^z  0) : 2 2 2 2 $ c) Let += A  +~ . Then w~ = 12 Ah2i $! = 21 2 (^y x^ ; x^ y^) = ; 12 z^ ~vE = Ah2i+~ ~r = ;+~  A  ~r $ $ = ; + ~r = ~r + : $  T $ $ Then @~~vE =+, @~~vE = ; +, so $ = 0 (so all eigenvalues are 0 and any vector is an eigenvector) $! =$ ~! = +~ : + 22. a) !_ jk = 12 (@j vk ; @k vj ) _ ij = 12 (@ivj + @j vi ) @k _ ij = 12 (@k @i vj + @k @j vi) Replace (ijk) by (kij ) and this becomes @j _ ki = 12 (@j @k vi + @j @ivk ) : @i !_ jk = 12 (@i @j vk ; @i @k vj ) : Comparing these three equations gives a). 320 b) @i @l !_ jk = @l @i!_ jk so, by a), @i (@j _ kl ; @k _ lj ) = @l (@j _ ki ; @k _ ij ) or, since _ kl = _ lk and @i @j = @j @i , @j @i _ kl ; @i @k _ lj + @k @i_ li ; @l @j _ ik = 0: Now make the replacements j ! i, i ! j , and one proves b). c) Fix i and j . a) can be written () @k !_ ij = @i _ jk ; @j _ ki: Then b) implies @k @l !_ ij = @l @k !_ ij , which are exactly the integrability conditions for (*). They insure that the line integral Z ~r ~0 drk (@i _ jk ; @j _ ki) will be independent of path. Thus the solution of (*) exists, and has the form !_ ij (~r) = ;!_ ij (~0) + Z ~r ~0 drk (@i _ jk ; @j _ ik ) : The integral is antisymetric in ij . If !_ ij (~0) is chosen antisymmetric in ij , !_ ij (~r) will be antisymmetric for all ~r. d) Let !_ ij (~r) be given as in c). We want to try to solve @i vj = _ ij + !_ ij : This will be possible, and the solution will be vj (~r) = Vj + Z ~r ~0 dri (_ ij + !_ ij )  as long as for each xed j the integrability condition @k @i vj = @i@k vj is satised, i.e., (y) @k (_ ij + !_ ij ) = @i (_ kj + !_ kj ) : But !_ ij was obtained by solving (*) on the preceding page. Substituting (*) in (y) makes (y) an identity. 321 23. a) $ U ; U0 = N@N U +  G h2i@~$G U  + 1 (N )2 @N2 U + N G h2i@~$G @N U 2 $ $ ~$ ~$ 1 + 2  G  G h4i@G @G U + third order terms $ S$0 1 = N 0 +  G h2i  + 2 (N )2 B0 0 $$ $   $ 0 + 1  $  $ h4i Q0 + third order term: +N G h2i W  2 G G  0 b) 0 $ W$ 0  = NB0 +  G h2i  0 $ $ $ $$  S = N W 0 + G h2i F 0 : $ $ If  = 0, then N = ; G h2i W0 B00 , so 1 0 $ $ @ $$ W$ 0W$ 0 A  S =  G h2i F 0 ;  B : 0 0 Therefore $$ $$ $ $ F00 =F0 ; W 0BW 0 : 0 0 24. In an isotropic material, $ S0 $ W0 Fijkl Qijkl $ = ;p0 I $ = ; I = ij kl +  (ik jl + il kj ) = Fijkl + ij Skl(0) ; jk Sil(0) = Fijkl + p0 (jk il ; ij kl ) Qijkl = ( ; p0) ij kl +  ik jl + ( + p0 ) jk il 322 a)  $ 1 p 0 U ; U0 = N 0 ;  tr  G + 2 (N )2 B0 0  $ 1 n  $2  ;  (N ) tr  G + 2 ( ; p0) tr  G +   G$ h2i G$ 0 0 $ $T o + ( + p0)  G h2i G + third order terms: b) 0ij kl + 0 (ik jl + il jk ) = ij kl +  (ik jl + il jk ) ; 0B2 0 ij kl: Thus 0 = , 0 =  ; 2=0 B0.