expansions of tensors

Expansions of tensors PhL 1.20.12 Motivation: Lai's continuum mechanics book. emphasized I will try to mimic what I have already written for vectors. V = V11 u1u1T + V12 u1u2T + ... = Σij Vij uiujT where for example (N=2) u1u2T= ( 0 1) = = a square matrix (u1u2T) u1 = ( 0 1) = u1 (u2u1) = u1 0 = 0 The object on the left, namely V, is an NxN matrix. So this is the generalization of V = V1 u1 + V2 u2 +... = ΣnVnun How can we find the Vij given V and the above expansion? V uk = (Σij Vij uiujT )uk = Σij Vij ui(uj uk) = Σi Vik ui Non-dyadic notation Let ei represent a generic set of N-dimensional basis vectors, possibly non-orthogonal. The notation shall be demonstrated for N=2 just to save space. The following objects may be constructed from the ei. (The transpose of a column vector v is a row vector vT. ) eiTek = ei ek = the usual dot product = ( a b) = ac+bd = a scalar eiekT = ( c d) = = a matrix (eiekT) en = [ ( c d)] = [( c d) ] = (cd+df) = a column vector = ei(ekT en) = ei (ek en) enT(eiekT) = (e f) [ ( c d)] = [(e f) ] ( c d)] = (ea+fb) (c d) = a row vector = (enTei)ekT = (en ei)ekT enT(eiekT) em = (enTei)(ekT em) = (en ei)(ek em) = a scalar Consider next an expansion of a matrix A onto the set of matrices eiTek , all of dimension N, A = Σik Aik (eiTek) Apply enT on the left and em on the right to get enT A em = enT [Σik Aik (eiTek)] em = Σik Aik enT(eiTek) em = Σik Aik (en ei)(ek em) At this point, we now further assume that the set of ei are in fact the tangent base vectors associated with a transformation F from x-space to x'-space. In this case we know that en ei = 'ni, the metric tensor, so Bnm ≡ enT A em = Σik Aik 'ni 'km = Σik 'ni Aik 'km = ( ' A ')nm B = ' A ' => A = g' B g' since ' g' = 1 Therefore one finds the coefficients in the above expansion to be A = g' B g' Aik = Σnm g'in Bnm g'mk = Σnm g'in (enT A em) g'mk = Σnmab g'in (en)a Aab (em)b g'mk = Σnmab g'in SanAabSbn g'mk = Σnmab g'in STnaAabSbn g'mk = (g' ST A S g')ik so A = g' ST A S g' = (R g RT) ST A S (R g RT) = R g (RTST) A (SR) g RT = R gAg RT Now if we further assume that x-space is Cartesian so that g = 1, then A = RART => Aik = Σab Ria Aab RTbk = Σab Ria Rkb Aab According to Section 5 (e), the coefficients Aik must in this case be the components of A in x'-space, which components are written A'ik. Thus Aik = A'ik and the following theorem has now been proven: Theorem: In Picture C, where x-space is Cartesian with g = 1, if one expands the contravariant rank-2 tensor A onto the basis matrices (eiTek), the expansion coefficients are the components of that same tensor A in x'-space: A = Σik A'ik (eiTek) Question: How would you generalize this to Aijk rank-3 tensor? You need a "basis cube" with a single 1 located in some position. We might define (cijk)abc = δiaδjbδkc Maybe direct product notation is better. Let's go back to the rank 2: A = Σik Aik (eiek) What does this mean? See matrix notes at end of matrix binder. (eiek)ab = (ei)a(ek)b Then we have Aab = Σik Aik (eiek)ab = Σik Aik (ei)a(ek)b = Σik Aik SaiSbk = Σik Sai Aik STkb But let's go to rank 3 right away Aabc = Σijk Aijk (eiejek)abc = Σijk Aijk (ei)a(ej)b(ek)c = Σijk Aijk SaiSbjSck = SaiSbjSck Aijk Since RS = 1, this can be inverted to give Aijk = RaiRbjRck Aijk But since Aijk is assumed to be a rank-3 tensor under transformation F, the LHS is A'ijk which are the contravariant components of tensor A in x'-space. That is to say, Aijk = A'ijk . Therefore one writes Aabc = Σijk A'ijk (eiejek)abc or A = Σijk A'ijk (eiejek) Here then are expansions for tensors of rank 1,2,3 and 4 onto direct products of tangent base vectors, Aa = Σi A'i (ei)a => A = Σi A'i ei Aab = Σij A'ij(eiej)ab => A = Σij A'ij(eiej) = Σij A'ij eiejT Aabc = Σijk A'ijk (eiejek)abc => A = Σijk A'ijk (eiejek) Aabcd = Σijkl A'ijkl (eiejekel)abcd => A= Σijkl A'ijkl (eiejekel) where the meaning of the direct product of R vectors is (ABCD ...)abcd... = AaBbCcDd..... Historically for N=2 the direct product has been called a "dyadic product", with this notation (VW)ab ≡ (VW)ab = VaWb and then the expansion of Aab above becomes Aab = Σij A'ij(eiej)ab => A = Σij A'ij(eiej) = Σij A'ij eiejT The section below now appears as Section 7 (w) in tensor doc, so do not edit anything here! (w) Notations for expansions of rank-n tensors: direct product, transpose, polyad; (v). Direct Product Notation. The key tool required here is the notion of a direct product of n tensorial vectors defined this simple way (ABC ...)abc... ≡ AaBbCc..... (ABC ...)abc... ≡ AaBbCc..... etc The tensor ABC ... is nothing more than the outer product of vectors A,B,C as discussed in section (a) for contravariant vectors, but later extended to any mixture of vector types. As noted in section (j), one can define a direct product of rank-2 tensors in this way (MN)ab,AB ≡ MaANbB (MN)ab,AB ≡ MaANbB etc and then the same idea can be applied to form a direct product of tensors of any rank, for example (MN)ab,AB,αβ = MaAαNbBβ etc // n = 3 The number of groups of indices equals the tensor rank n, and the number of indices within each group matches the number of tensors being direct-product-multiplied. In what follows, only the direct product of vectors shall be considered. Direct Product of Basis Vectors and a Theorem. In particular, two sets of x-space vectors are of interest: the axis-aligned unit vectors ui (Section 3 (d)) and the tangent base vectors ei (Section 3 (a)). For N dimensions, there are of course N such vectors of each type, whereas the rank of the tensor of interest is n. One can write for example (uiujuk ...)abc... = (ui)a (uj)b (uk)c ... = δiaδjbδkc ..... (eiejek ...)abc... = (ei)a (ej)b (ek)c ... = Sai Sbj Sck .... = Ria Rjb Rkc .... where (ei)a = Sai = Ria as shown in section (s). To be specific, we work below with N = 3. Consider this expansion of a contravariant rank 3 tensor Aabc, A = Σijk aijk (uiujuk) or Aabc = Σijk aijk (uiujuk)abc = Σijk aijk δiaδjbδkc = aabc In this case one finds of course that the coefficients aijk of the expansion are the x-space components Aijk of the tensor being expanded, so one can then write simply A = Σijk Aijk (uiujuk) or Aabc = Σijk Aijk (uiujuk)abc In the case of the tangent base vectors, consider A = Σijk Aijk (eiejek) or Aabc = Σijk Aijk (eiejek)abc = Σijk Aijk Sai Sbj Sck = Sai Sbj Sck Aijk Since RS = 1, this can be inverted to give Aabc = Rai Rbj Rck Aijk But since Aijk is a rank-3 contravariant tensor, the LHS here is A'ijk which are the x'-space components of the tensor A. Therefore the expansion can be written A = Σijk A'ijk (eiejek) or Aabc = Σijk A'ijk (eiejek)abc The following theorem has just been proven: Theorem: (a) When a tensor A of rank n with components Aijk... is expanded on the direct product of the axis-aligned unit vectors un, the coefficients of that expansion are the x-space components of A which are of course just Aijk.... (b) When a tensor A of rank n with components Aijk... is expanded on the direct product of the tangent base vectors en, the coefficients of that expansion are the x'-space components of A which are written A'ijk... . For N = 1 and A = V and this last claim becomes V = Σi V'i ei = V or Va = Σi V'i (ei)a which agrees with the vector expansion shown in section (s) above. Expansions of other tensor types can be obtained by raising and lowering indices in the obvious manner. For example Aabc = Σijk A'ijk (eiejek)abc Aabc = Σijk A'ijk (eiejek)abc Similarly, all the expansions of rank-1 tensors shown in section (s) can be generalized to the expansions of rank-n tensors. Polyadic Notation. Historically, a different notation was often used for the outer products of vectors. For example one had (AB)ij ≡ AiBj // = (AB)ij and then the object AB is a matrix known as the dyadic product of two vectors [ see e.g. p 58 of Morse and Feshbach where one sees U = AB and U is referred to as a "dyadic" ]. The expansion of a rank-2 tensor might then appear in these various somewhat mysterious (but valid) notations, A = Σnm Anm unum = A11 u1 u2 + A12 u1 u2 + ... = u1 A11 u2 + u1 A12 u2 + ... = Σnm Anm n m = A11 1 1 + A12 1 2 + ... = 1 A11 1 + 1 A12 2 + ... To add to the confusion, sometimes the Cartesian unit vectors are given different names u1 = 1 = = = i u2 = 2 = = = j u3 = 3 = = = k and then one might see for example (M&F p 59) A = A11 i i + A12 i j + A13 i k + ... This dyadic (dyad) notation is generalized to n > 2 so one can have triads and in general polyads ( see for example p 28 of Backus). The n=3 case would have A = Σnmp Anmp unumup = A111 1 1 1 + A112 1 1 2 + ... Notice that in the direct product definition (ABC ...)abc... = AaBbCc..... one cannot willy-nilly shuffle the order of A,B and C appearing in ABC without also shuffling the indices. This rule is less clear (but still valid) in the notation having an object like 1 3 2. In its favor, the polyadic notation is compact and just provides a shorthand to reduce the need for extra symbols, 1 3 2 ≡ (u1u3u2) Transpose notation. In the case n=2 the polyadic notation (dyadic) can be understood in the following manner that seems less abstract than what appears above, A = Σnm Anm unumT = A11 u1 u1T + A12 u1 u2T + ... = a matrix with A12 in the upper right corner where un is a column vector and unT is the corresponding row vector. Then one has, for example, u1u2T= ( 0 1) = Thus one could interpret any dyad AB as really being the matrix ABT. For n > 2 this transpose of vector concept does not conveniently generalize. For n=3 the object uaubuc would be a cube with a single 1 located at coordinates a,b,c, and so on for n > 3. One cannot write this as uaubucT for example. Example of (v). The polyadic notation is regarded as "archaic" by many (eg, Wolfram website) but it is quite embedded into the fields of continuum mechanics and hydrodynamics (even in current textbooks such as Lai et. al.) where there are many rank-2 tensors floating around. In these areas, one sometimes sees (A)ij ≡ ∂jAi where the indices are the reverse of the normal dyad definition because this makes certain equations look simpler. An example is the so-called material derivative of a vector field a in the Eulerian or spatial "view" of the motion of a blob of continuous matter ( eg, Lai (3.4.3) and (3.4.8) ), Dai/Dt = ∂tai + ai v = ∂tai + (∂jai) vj = ∂tai + (a)ij vj = ∂tai + [(a) v]i => Da/Dt = ∂ta + (a) v The matrix a is not a differential operator since the derivative does not act on the vector standing to the right of a, but one is still often interested in expressing a in curvilinear coordinates as is now outlined. Consider the matrix (v) where v is some arbitrary tensorial vector field (not necessarily the velocity appearing the previous equation). The first step is to write the Cartesian matrix equation using the transpose form of the dyadic notation discussed above, (v) = Σcd(v)dc uducT = Σcd(∂cvd) uducT . One can express ∂cvd in terms of x'-space coordinates and objects in exactly the Christoffel manner of section (v), so that ∂cvd = Σij (Ric∂'i)( Rjdv'j) = Σij Ric [(∂'iRjd) v'j + Rjd (∂'iv'j)] (*) Comment: As noted in section (v), for a non-linear transformation F the object ∂cvd is not a rank-2 tensor. If it were a rank 2 tensor, as in the case of linear F, it would be very easy to express ∂cvd in terms of x'-space objects and coordinates. One would just use the Theorem above to write A = Σij A'ij ei ejT A'ij = RiaRjbAab (v) = Σij (v)'ij ei ejT (v)'ij = RiaRjb(v)ab = RiaRjb∂bva = Rjb∂b Riava = (∂'jv'i) so that (v) = Σij (∂'jv'i) ei ejT and we are all done. When F is non-linear, as is the usual case for curvilinear coordinate transformations, the extra work shown below or its equivalent is required. The equation (*) above is of course valid for both orthogonal and non-orthogonal curvilinear coordinates. Just to simplify things from here on, we assume orthogonal coordinates which allows one to write ud = Σe h'e-1 Red e h'e-1 Red = Qed = a rotation matrix Varying the previous discussion, here we shall expand on the unit tangent base vectors n ≡ en/h'n . One may verify the above ud equation by applying ei to both sides and using the fact that un and en are both vectors in Cartesian x-space (g = 1, something assumed throughout this discussion), ud ej = (ud)i(ej)i = δdi Sij = Sdj = Rjd e ej = h'e-1 ee ej = h'e-1 g'ej = h'eδej => Rjd = Σe Qed h'eδej = Qjd h'j => Qjd = h'j-1 Rjd Transposing the ud equation and changing labels and indices, one can write ucT = Σf h'f-1 Rfc fT so that uducT = Σef h'e-1 Red h'f-1 Rfc e fT Installing into the starting equation, one finds after some algebra that (v) = Σcd(v)dc uducT = Σcd(∂cvd) uducT (v)dc = ∂cvd = Σij { [(∂'jv'i) + ΣdkRid (∂'jRkd) v'k] h'i-1 h'j-1 } i jT = Σij Pij i jT where Pij = [(∂'jv'i) + Rid (∂'jRkd) v'k] h'i-1 h'j-1 In this expression Pij the vectors v'n are covariant components in x'-space. Since the expansion is done on the unit vectors n , it is appropriate to replace the v'n by h'nv'n where v'n are the coefficients of vector v expanded on the n, as proven here: v = Σi v'iei = Σi (Σn g'inv'n) ei = Σi g'ii v'i ei = Σi h'i-2 v'i ei = Σi h'i-1 v'i i = Σi v'i i => h'i-1 v'i = v'i => v'n = h'n v'n It is often convenient to use components like v'n since they all have the same dimensions, whereas the covariant components v'n generally do not have the same dimensions. Therefore, (v) = Σij Pij i jT Pij = [(∂'jh'i) v'i + h'i (∂'jv'i) + Rid (∂'jRkd) h'k v'k] h'i-1 h'j-1 One might wonder about the utility of expanding (v) on the matrices ijT. The utility arises when the matrix (v) is applied to a vector b which itself is expanded on these i basis vectors: (v) b = (Σij Pij i jT) (Σn bn n) = Σijn Pij bn ( i jT) n where ( i jT) is a matrix applied to the vector n . One has [ i jT] n = i [jT n] = i [j n] = i δjn which is perhaps more obvious in skeletal matrix notation [ (x x) ] = [(x x) ] = (j n) = δjn Then, supposing one has the equation (v) b = c , (v) b = Σijn Pij bn ( i jT) n = Σijn Pij bn i δjn = Σin Pin bn i = Σi cii => ci = Σn Pin bn => bi = Σn P-1in cn and the equation is thus solved for vector b at the cost of inverting the matrix P. One might write (v)(x')ij = Pij to indicate that Pij is the object v expressed in terms of x' -space objects and coordinates. In fact, using the same matrix manipulations shown above, one can write (v) = Σij Pij i jT => iT (v) j = Pij (x x) = Pij In quantum mechanical bra-ket notation one would write Pij = iT (v) j = < i | v | j> = (v)(x')ij so that Pij is the matrix element of operator v in the j basis. Similarly pij = uiT (v) uj = < ui | v | uj> = (v)ij = ∂jvi is the matrix element of the same operator v in the Cartesian basis. It is not too difficult to have a computer algebra program compute the Pij from the expression given above, (v)(x')ij = Pij = [(∂'jh'i) v'i + h'i (∂'jv'i) + Rid (∂'jRkd) h'k v'k] h'i-1 h'j-1 For polar coordinates with 12 = rθ (the reverse of the ordering we have used earlier!), one finds that (v)(x')ij is given by (Maple did this) where the notation suggested in Example 1 of Section 14 is used, vr ≡ v'1 vθ ≡ v'2 and we apologize for Maple displaying v as ν . For spherical coordinates with 123 = rθφ, one finds similarly that (v)(x')ij is given by Both these results agree with Lai et al (2.33.23) and (2.25.25). M. Lai, E. Krempl and D.Ruben, Introduction to Continuum Mechanics, 4th Ed. (Elsevier, Amsterdam, 2009). G. Backus, Continuum Mechanics (Samizdat Press, Golden Colo., 1997).