obs old section 7 w

Old Section 7 (w) from tensor doc PhL 2.12.12 In an earlier tensor doc version, I had this very long subsection 7 (w). I decided to present it instead in two detailed appendices E and F, and then the new section 7 (w) is just a summary of the those appendices. So here is the "old" section 7 (w) : (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. 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 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 for dyadics. In the case n=2 the polyadic notation (dyadic) can be understood in the following manner that seems less abstract than what appears above: ab ≡ abT Here one knows that ab is a "dyadic" because there is no other meaning for two bolded column vectors abutting each other with no intervening operator, so no special notation like [ab] is needed to indicate that ab is a dyadic. The object abT on the other hand has a well-defined meaning in matrix algebra, abT = (b1 b2) = = a matrix and one sees that in fact (ab)ij = (abT)ij = aibj Superscript T means transpose, and the transpose of a column vector b is a row vector bT. This transpose notation can then be applied to the dyadic expansion of a 2x2 matrix A, A = Σnm Anm unum = Σnm Anm unumT = A11 u1 u1T + A12 u1 u2T + A21 u2 u1T + A22 u2 u2T = a matrix with A12 in the upper right corner where un is a column unit vector and unT is the corresponding row unit vector. For example, u1u2T= ( 0 1) = In our example here we happen to use dimension N = 2, but it could have been any N. Integer n on the other hand is the number of objects in the direct product which is n=2 above. For n > 2 this transpose of vector concept does not conveniently generalize. For n=3 the object uaubuc would be a cube of zeros with a single 1 located at coordinates a,b,c, and so on for n > 3. One cannot write this as uaubucT for example. The direct product notation seems clearest in this case. Sometimes a small-size dot • is used to indicate the action of a dyadic (matrix) on a vector. If A is a dyadic (matrix) then one defines: A• c ≡ Ac = a column vector (A• c)i = (Ac)i = Aijcj => A• c = ΣijAijcjui = Ac c • A ≡ cTA = a row vector (c • A)i = (cTA)i = cjAji => c • A = ΣijcjAji ui = cTA d • A• c = dTAc = a number = diAijcj It then follows that, for the particular dyadic A = ab , (ab) • c ≡ (ab) c = (abT)c = a(bTc) = a ( b c) = ( b c) a = a column vector c • [ab] ≡ cT [ab] = cT(abT) = (cTa) bT = (c a) bT = a row vector d • [ab] • c = dT [ab] c = dTabT c = (dTa)( bT c) = (d a)(b c) = a number Here is more detail on the first line showing a skeletal matrix structure, (ab) c = (a bT)c = abTc = a(bTc) = a(bc) { } = {(b1 b2)} = (b1 b2) = { (b1 b2) } = bc The same small dot is used to indicate the product of two dyadics, which is to say, matrix multiplication A• B ≡ AB Regarding this small size dot • : (1) from a matrix algebra point of view, it is completely superfluous except in the case c • A ≡ cTA ; (2) it is completely different from the dot used in bTc = bc . It is this larger dot which was the subject of Section 6 (b). Here are a few quotes from Morse and Feshbach (an = un and U is the dyadic) to illustrate some of the notation described above. Notice the impressive name "idemfactor" for the unit matrix. (The use of elaborate Gothic letters for dyadics certainly adds to the "archaic" feel. These are from pp 58-60 Vol I. ) Various other notations are used to indicate a dyadic, such as a double overbar or an over ↔ symbol. The reader is hopefully convinced that dyadic notation is just an older form of matrix notation and its associated linear algebra. Matrix Notation and Ambiguity. Tensors of rank N=2 are associated with matrices, and notational ambiguity can arise. Consider for the Cartesian unit basis vectors ui = ui , A = Aij (uiuj) = Aij uiujT => (ui)T A (uj) = Aij where use is made of the fact that (un)Tum = un um = δnm . No problem so far. Using now the ei basis one obtains by the same logic, since (en)Tem = en em = δnm , A = A'ij (eiej) = A'ij eiejT => (ei)T A (ej) = A'ij Again there is no ambiguity since, according to the Theorem above, A'ij are the x'-space components of tensor A under the transformation F which has tangent base vectors en . But suppose one uses some arbitrary complete set of basis vectors bi. As discussed in Section 6 (b), there will exist a unique set of dual vectors bi such that the relation (bn)Tbm = bn bm = δnm is true. Then one can write A = Aij (bibj) = Aij bibjT => (bi)T A (bj) = Aij The problem here is that one has to come up with a different font for every basis { bi} ! A possible solution would be indicate the basis as a superscript on the tensor, A = (A(b))ij (bibj) = (A(b))ij bibjT => (bi)T A (bj) = (A(b))ij The point of course is that the matrix elements Aij are really specific to a choice of basis. One can regard the tensor A as an operator Aop in an n dimensional Hilbert space En, and then one could write (bi)T A (bj) = (A(b))ij = < bi | Aop | bj > where | bj > are the abstract complete set of vectors which span the Hilbert Space (kets) and < bi | are the complete set of vectors which span the dual space (bras) as determined by < bi| bj > = δij . The matrix notational problem gets worse if one uses two different bases to define a matrix element, (ci)T A (bj) = (A(c,b))ij = < ci | Aop | bj > where perhaps A(c,b) could be used to describe such a matrix. Notice that there is no ambiguity in the notation Aop which describes the tensor A as a Hilbert Space operator, and for this reason many authors associate a rank-2 tensor or dyadic A with this operator Aop. For example, Lai et. al. use the bolded notation Aop = A. The main point of this subsection is that one must be aware of one's basis, and often this is not indicated by the notation Aij. Although contravariant indices were used in this discussion, the same conclusion applies to all other index positions. Comment: In quantum mechanics, observable (measureable) quantities are always matrix elements of a Hilbert Space operator which commutes with the Hamiltonian operator. Therefore, rank-2 tensors arise naturally in this subject. Mixed bases are used frequently, and the bra-ket notation mentioned above provides a clear, efficient and unambiguous notation. Both finite and infinite dimensional Hilbert Spaces occur. The Schrodinger wave function φE(x) = <x|φE> provides a simple example of a mixed basis matrix element of the unit operator 1, as in <x| 1op|φE>. The kets |φE> form a complete set of basis vectors of a Hilbert space (eigenkets of the Hamiltonian operator Hop with eigenenergies E), while kets |x> form a different complete basis of the same Hilbert Space, being the eigenkets of the position operator Xop. This Hilbert Space is infinite dimensional. In fact, <x'|1op|x> = 1(x) = δ(x'-x) where we use 1(x) to indicate the unit operator in the "x basis", known as the coordinate representation. Example of (v). The polyadic notation is regarded as "archaic" by many (eg, Wolfram) 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 an arbitrary 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. We now carry out this task as an illustration of the various notations discussed above. 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). Using the covariant vector rule Va = RbaV'b shown in section (p) one gets ∂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 [ not valid for non-linear F ] (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 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 v'n are covariant components of v 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 mentioned earlier, [ (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). The method outlined here of course can be used to obtain this matrix in any orthogonal coordinate system.