archive old Section 7(g) and 7(h)

Archive of old Sections 7(g) and 7(h) PhL 4.5.15 (g) Matrix Multiplication in the Standard Notation. Although there are various forms of matrix multiplication, the most standard form is that obtained when all matrices have a "down-tilt" form. Consider this example, Cab = AacBcb (7.9.g.1) where it is assumed that all three objects are down-tilt rank-2 tensors. Although these are "split level" matrices, one can see that the index c has the correct "adjacency" property to justify matrix multiplication. A second requirement is that any matrix summed index should be a genuine contraction with one index up and the other down. The above tensor transformation rule can then be written in this more compact matrix notation ( SN means Standard Notation, dt means down-tilt) : C = AB // all down-tilt [C = AB]SN,dt . (7.9.g.2) By application of suitable g tensors to (7.9.g.1) (or by lowering index a and raising index b on both sides), one gets Cab = AacBcb . Application of the Tilt Reversal Rule on index c then gives Cab = AacBcb . (7.9.g.3) Again the adjacency and contraction rules are met, so this equation can also be represented by C = AB // all up-tilt [C = AB]SN,ut (7.9.g.4) where ut means up-tilt. It has just been shown that, [C = AB]SN,dt [C = AB]SN,ut . (7.9.g.5) The conclusion is that in Standard Notation, matrix notation can be used if all matrices in the equation being represented are either all down-tilt or all up-tilt. As will be shown later in section (u), such matrix equations are all "covariant" in that both sides of the equation have the same tensor transformation property, and this is due to the fact that the matrix summation index is a contraction. One might wonder about the conversion from developmental notation (DN) to standard notation (SN) of the following matrix equation, where A and B are contravariant rank-2 tensors, [ AB = C ]DN AacBcb = Cab → AacBcb = [???]ab . (7.9.g.6) It is not hard to show entirely in developmental notation that, unless STS = 1 (meaning S is a rotation), the object Cab is not a tensor. So for general underlying F, C is not a tensor, even though A and B are tensors. This fact is reflected in the LHS of the partially translated equation, where AacBcb is seen to not be a tensor because the c index is not a contraction (it is not tilted). The rules for translation from developmental to standard notation specify what to do with R, S, and with tensors. Since C is none of these objects, we cannot really "translate" the equation [ AB = C ]DN into standard notation. However, we could define Cab ≡ AacBcb and then write [ AB = C ]DN AacBcb = Cab → AacBcb = Cab [ AB = C ]SN,up (7.9.g.7) with the understanding that Cab is not a contravariant rank-2 tensor, and we now have yet a third kind of matrix multiplication in the Standard Notation, qualified by "up" meaning all indices are up on all objects. One could of course do the same thing with covariant matrices and write [ = ]DN accb = ab → AacBcb = Cab [ AB = C ]SN,dn (7.9.g.8) The situation can be summarized in this picture: (needs repair!! ) (7.9.g.9) In standard notation, there are four different kinds of "matrix multiplication" and only the down-tilt and up-tilt equations are the same equation, and only they are covariant The moral here is that if one chooses to use matrix multiplication notation in the Standard Notation, one needs to be very clear which multiplication form one is dealing with. What about the special case [ AB = 1 ]DN where A is a contravariant tensor? In this case, it must be that B = A-1 and then one has, [ AB = 1 ]DN Aac(A-1)cb = δab → Aac(A-1)cb = (AA-1)ab = (1)ab = δab that is to say: [ A(A-1) = 1 ]SN,up (7.9.g.10) and in this case all four Standard Notation equations are valid, the other three being: [ AB = 1 ]SM,dn Aac(A-1)cb = (AA-1)ab = (1)ab = δab [ AB = 1 ]SM,ut Aac(A-1)cb = (AA-1)ab = (1)ab = δab [ AB = 1 ]SM,dt Aac(A-1)cb = (AA-1)ab = (1)ab = δab (7.9.g.11) so that [ AB = 1 ]DN [ AB = 1 ]SN,up [ AB = 1 ]SN,dn [ AB = 1 ]SN,ut [ AB = 1 ]SN,dt (7.9.g.12) Now what about the special case [ RS = 1 ]DN ? Since R is not a contravariant tensor the above logic does not apply in the sequence stated. But we have rules for translating R and S, so [ RS = 1 ]DN RacScb = δab → RacScb = δab [ RS = 1]SN,dt (7.9.g,13) and since the four SN forms are equivalent (S = R-1), we end up with the same conclusion as above, [ RS = 1 ]DN [ RS = 1 ]SN,up [ RS = 1 ]SN,dn [ RS = 1 ]SN,ut [ RS = 1 ]SN,dt (7.9.g.14) confirming what was stated in a preliminary way in item 2 above. Finally, what about the special case [ STS = 1 ]DN ? Again using the rule that Sij → Sij, [STS = 1 ]DN ScaScb = δab → ScaScb = δab ??? (7.9.g.15) The combination ScaScb does not have the proper adjacency to be matrix multiplication. In order to achieve adjacency, we must have some notion of "transpose" in the Standard Notation. That notion is the subject of the next section, and the result is going to be this: (AT)ab = Aba (AT)ab = Aba (AT)ab = Aba (AT)ab = Aba . (7.9.g.16) Therefore Sca = (ST)ac and we then have [STS = 1 ]DN ScaScb = δab → (ST)ac Scb = δab ??? (7.9.g.17) Although adjacency of the c index is now obtained, the form matches none of the Standard Notation forms for matrix multiplication defined above. Thus, no contact is being made with any of these standard forms by this translation. Since STS = 1, one must have S = (ST)-1 and therefore all four of our stated SN forms are equivalent. So we then have this interesting situation: [STS = 1 ]DN (ST)ac Scb = δab [STS = 1 ]SN,dt the other three SN forms (7.9.g.18) What this says is that [STS = 1 ]DN and [STS = 1 ]SN,dt are completely different equations. Specifically, [STS = 1 ]DN ScaScb = δab → (ST)ac Scb = δab or Sca Scb = δab (7.9.g.19) [STS = 1 ]SN,dt (ST)acScb = δab or Sca Scb = δab . (7.9.g.20) The very last equation on the right can be written (as will be seen below) Rac Rbc = δab and this is one of the standard orthogonality conditions on R and this is always true for any underlying transformation F. Therefore, [STS = 1 ]SN,dt is always true, and this means that in terms of Standard Notation down-tilt matrix multiplication, the matrix S is always real orthogonal. In contrast, in terms of Developmental Notation matrix multiplication, S is only real orthogonal if it is a rotation matrix! That is to say, in the general case: [ST = S-1 ]DN [ST = S-1 ]SN,dt (7.9.g.21) So the very meaning of the equation ST = S-1 is dependent on which notion of matrix multiplication is involved. The operator S-1 is really defined in terms of SS-1 = S-1S = 1 and this involves matrix multiplication. One final thing to note: (Sab)DN = (Sab)SN ≠ (Sab)SN = gac(Scb)SN // unless it happens that g = 1 (7.9.g.22) so clearly when one writes Sab or Rab one should be clear which notation is involved. (h) Transpose of a rank-2 tensor. If A is a contravariant rank-2 tensor, the translation mapping (AT)ab = Aba → (AT)ab = Aba (7.9.h.1) seems obvious, and the object AT therefore also transforms as a rank-2 tensor. Once (AT)ab = Aba is established in standard notation, one can apply the metric tensor g to lower either or both of the indices of this equation, to get (AT)ab = Aba , (AT)ab = Aba, and (AT)ab = Aba. Notice in all four equations that the indices on the two sides of the equation are reflected in a vertical axis passing between the indices. This causes a left index to become a right index and vice versa, as one would expect for transposing a matrix. Moreover, on each side of all four equations, each index has the same contravariant/covariant sense. This same argument also applies to R and S even though they are not tensors. The only difference is that the first index of Rba is lowered by g' while the second by g. For example, (RT)ab = Rba where b is a g' type index and a is a g type index, as will be elaborated in section (o) below. Similarly (ST)ab = Sba . To summarize the situation with transposes in Standard Notation: (AT)ab = Aba (RT)ab = Rba (ST)ab = Sba (AT)ab = Aba (RT)ab = Rba (ST)ab = Sba (AT)ab = Aba (RT)ab = Rba (ST)ab = Sba (AT)ab = Aba (RT)ab = Rba (ST)ab = Sba . (7.9.h.2) Notice that the rule is not (AT)ab = Aba which would be a straight swap of indices (and would result in AT not being a tensor). The straight swap idea works for the pure contravariant and pure covariant forms of A, but not for the tilted forms! As shown in Theorem 4 below, this tilted transpose form has some interesting implications. Theorem 1: In the Standard Notation, S is a real-orthogonal matrix. This means all of the following: [S-1 = ST]SN [SST = 1]SN [STS = 1]SN (7.9.h.3) where [...]SN can mean any of the four Standard Notation matrix multiplication forms given in the previous section. This theorem does not apply to the matrix S in the developmental notation as also explained in the previous section. That is to say, the equations [SST = 1]DN and [SST = 1]SN are different. Proof of theorem: start with (3) above which says gij is a covariant rank-2 tensor, g'ab = Sa'aSb'b ga'b' // next, apply gb'b to both sides (or just raise index b on both sides) g'ab = Sa'aSb'b ga'b' // next, reverse the tilt of the b' index g'ab = Sa'aSb'b ga'b' // next, use the Diagonal g rule in two places δab = Sa'aSb'b δa'b' = Sa'aSa'b // next, use (ST)aa' = Sa'a δab = (ST)aa'Sa'b // next use matrix notation as per above [1 = STS]SN,ut => ST = S-1 => 1 = SST (7.9.h.4) Theorem 2: In the Standard Notation, R is a real-orthogonal matrix. This means all of the following: [R-1 = RT]SN [RRT = 1]SN [RTR= 1]SN (7.9.h.5) In other words, the previous theorem also applies to R. Again, this fact is not true of the developmental notation matrix Rab. The proof is very similar but just different enough to warrant showing it : Proof of theorem: start with (3) above which says gij is a covariant rank-2 tensor, g'ab = Raa'Rbb'ga'b' // next, apply gb'b to both sides (or just lower index b on both sides) g'ab = Raa'Rbb'ga'b' // next, reverse the tilt of the b' index g'ab = Raa'Rbb'ga'b' // next, use the Diagonal g rule in two places δab = Raa'Rbb'δa'b' = Raa'Rba' // next, use (RT)a'b = Rba' δab = Raa'(RT)a'b // next use matrix notation as per above [1 = RRT]SN,dt => RT = R-1 => 1 = RTR (7.9.h.6) Comment: In the developmental notation, neither R nor S is a real-orthogonal matrix, unless by accident, but in the standard notation R and S are always real-orthogonal. Theorem 3: (a) [S = RT]SN and [R = ST ]SN (b) Sab = Rba and Sab = Rba ( reflect indices in vertical line between them) (7.9.h.7) Proof of theorem: (a) As shown in the previous section, S = R-1 for any index positions in the standard notation. From Theorem 2, RT = R-1 in standard notation. Therefore S = RT (and so ST = RTT = R) . (b) From (a) , S = RT => Sab = (RT)ab = Rba S = RT => Sab = (RT)ab = Rba QED An implication of this theorem is that one can completely eliminate references matrix S in tensor analysis and that is what is usually done. This is like replacing S with R-1 in the developmental notation. Theorem 4: For a standard notation tilted matrix A, det(A) ≠ det(AT). (7.9.h.8) This surprising result points out a potential hazard of using tilted matrices in the standard notation, and perhaps is an indication of why people avoid matrix notation and just write out all the components. For a traditional matrix Aab the determinant is given by either of these forms (figure on rows or columns) det(A) = εab..A1aA2b.... = εab..Aa1Ab2 ... det(AT) = εab..AT1aAT2b.... = εab..Aa1Ab2...... = det(A) (7.9.h.9) In the tilted standard notation, however, one has det[Aij] = εab..A1aA1b.... = εab..Aa1Ab2 ... det[(AT)ij] = εab.. (AT)1a(AT)1b.... = εab...Aa1Ab2 (7.9.h.10) but this is not the same as either of the det[Aij] forms! Here is a simple example: det(A) = εab..A1aA1b... = = A11 A22 – A21 A22 det(AT) = εab...Aa1Ab2 = = A11 A22 – A21 A12 ≠ det(A) (7.9.h.11) The point is that the standard notation transpose rule (AT)ij = Aij doesn't just swap the indices, it also changes the tilt. That means that the columns of one determinant matrix are not the same as the rows of the other and that is why det(A) ≠ det(AT). Corollary: [RTR = 1]SN does not imply that det(R) = ± 1 for a tilted R matrix. (7.9.h.12) The usual proof goes that det(RTR) = det(RT)det(R) = det(R)det(R) = [ det(R) ]2 = 1, but of course the part saying det(RT) = det(R) is no longer valid for a tilted R matrix in standard notation. Therefore, [RTR = 1]SN does not lead us to the conclusion that the Jacobian of Chapter 5 (k) is forced to be ± 1 (it is always the tilted R and S matrix that appears in equations like 1/J = det(Rij) ).