temp2

7.9 More translations from developmental to standard notation (a) Translation of determinants of R and S. Chapter (f) showed that Rab → Rab, so one translates from old to new notation, det(R) = εabc...R1aR2b....RNx → det(Rij) = εabc...R1aR2b....RNx det(S) = εabc...S1aS2b.....SNx → det(Sij) = εabc...S1aS2b....SNx det(R) = εabc...Ra1Rb2....RxN → det(Rij) = εabc...Ra1Rb2....RxN det(S) = εabc...Sa1Sb2.....SxN → det(Sij) = εabc...Sa1Sb2....SxN (7.9.a.1) (b) Inverse of R and S. Again, section (f) showed that Rab → Rab. In the standard notation, imagine that there is some inverse R-1 defined by (R-1)caRab = δcb. The chain rule says that (∂xc/∂x'a) (∂x'a/∂xb) = δcb or Sca Rab = δcb , (7.9.b.1) and therefore it must be that (R-1)ca = Sca. A similar argument shows that (S-1)ca = Rca. Using the results of the next section which allow us to raise and lower indices on both sides of an equation, this relationships R-1 = S is valid for all four matrix position possibilities, (R-1)ik = Sik (R-1)ik = Sik (R-1)ik = Sik (R-1)ik = Sik (7.9.b.2) and of course the same is true for S-1 = R. Thus arise these translations from old to new notation: R-1 = S → (R-1)ik = Sik and all other index combinations S-1 = R → (S-1)ik = Rik and all other index combinations RR-1 = RS = 1 etc → Rik(R-1)ka = RikSka = δia etc (7.9.b.3) We have silently used a certain "down tilt" matrix multiplication here which will be explained later. (c) Tensor g raises and lowers any index. So far the following translation rules have been established: gab → gab ab → gab Rik → Rik Sik → Sik . (7.9.c.1) It was shown in developmental notation (5.6.3) how rank-2 contravariant and covariant tensors transform. Here then is how those statements translate to the new notation : M'ab = Raa'Rbb'Ma'b' → M'ab = Raa'Rbb'Ma'b' contravariant rank-2 tensor (7.9.c.2) 'ab = Sa'aSb'ba'b' → M'ab = Sa'aSb'bMa'b' . covariant rank-2 tensor Since g itself is such a rank-2 tensor, replace M by g to get g'ab = Raa'Rbb'ga'b' → g'ab = Raa'Rbb'ga'b' (7.9.c.3) 'ab = Sa'aSb'b a'b' → g'ab = Sa'aSb'b ga'b' . It was shown in section (d) that Va = gaa'Va' and Va = gaa' Va' so that gaa' lowers a vector index and gaa' raises a vector index. That is to say, gaa' converts a contravariant vector index into a covariant one, and gaa' does the reverse. What does gaa' do to the index of a rank-2 tensor? Consider the following definition: Mab ≡ gaa'Ma'b . (7.9.c.4) Since gab and gab are inverses, it follows that Mab = gaa' Ma'b . (7.9.c.5) How does this new object Mab transform? The claim is that it transforms as a mixed rank-2 tensor, which would mean that M'ab = Raa' Sb'b Ma'b' . (7.9.c.6) The upper index gets a factor Raa' and the lower index gets a factor Sb'b , consistent with (*) above. It is not hard to prove this claim: (7.32) (7.31) (7.30) (7.33) M'ab = g'aa'M'a'b = ( RacRa'd gcd ) ( Sea'SfbMef) = ( RacRa'd gcd ) ( Sea'Sfb gei Mif ) = [RacRa'd gcd Sea'Sfb gei] Mif = [Rac (Sea'Ra'd) gcd Sfb gei] Mif = [Rac (SR)ed gcd Sfb gei] Mif = [Rac δed gcd Sfb gei] Mif = [Rac gcd Sfb gdi] Mif (7.28) = [Rac (gcd gdi) Sfb] Mif = [Rac δci Sfb] Mif = [Rai Sfb] Mif inverses = Raa' Sb'b Ma'b' QED (7.9.c.7) Similarly one could define Mab ≡ gaa'Ma'b and one would find that M'ab = Sa'a Rbb' Ma'b' (7.9.c.8) so Mab is then another member of the family of rank-2 tensors. Finally were one to define Mab ≡ gbb'Mab' one would find that Mab transforms as in (7.30). To summarize the four transformation results M'ab = Raa' Rbb' Ma'b' M'ab = Raa' Sb'b Ma'b' M'ab = Sa'a Rbb' Ma'b' M'ab = Sa'a Sb'b Ma'b' . (7.9.c.9) One sees then a family of four tensors associated with M. One is contravariant, one is covariant, and the other two are mixed. [Later we will show that Sij = Rji and this allows one to write the above equations in a manner that is easier to remember. ] (d) Raising Lowering Rule: gaa' [----a'---] = [----a---] and gaa' [----a'---] = [----a---] (7.9.d.1) Here [----i---] represent a tensor with a certain contravariant index i and dashes indicate other indices which each could be up or down. Similarly [----i---] is another tensor in the same family where the index i that was up is now down. The notion of higher rank tensors is coming soon, but we just want to establish the general idea that ANY index on ANY tensor can be raised or lowered by an appropriate g tensor. For the rank-2 tensors this was demonstrated explicitly above, and section (d) showed it was valid for rank-1 tensors (vectors), gaa'Va' = Va gaa' Va' = Va . (7.9.d.2) Comment: Notice that in every equation shown above, the summed indices always occur in the contracted form discussed in section (e) above, which is to say, one index is up and the other is down. (e) Contraction Tilt Reversal Rule: [-----a---------a----] = [-----a---------a----] (7.9.e.1) This is proved in section (k) below, but since we are going to need it right now, here is a preview of that proof: ( note that gab gac = gba gac = δbc ) [-----a---------a----] = gab gac [-----b---------c----] = δbc [-----b---------c----] = [-----b---------b----] The upshot is that one can always "reverse the tilt" on any pair of contracted indices. (f) The Diagonal g Rule: gab = δab and gab = δab // and same for g' (7.9.f.1) This is proved in section (m) below, and here a preview: gab = gaa' ga'b // gaa'raises the first index of tensor ga'b = δab // because gij and gij are inverses of each other (g) Covariance and Matrix Multiplication Before continuing the process of translation from developmental to standard notation, we digress momentarily to consider the notion of covariance in developmental notation. As we shall discuss in more detail below in Section **, an equation is said to be covariant under the transformation x' = F(x) if it has "the same form" in both x-space and x'-space. The "same form" means that the equation looks the same but everything is primed in x'-space. Example 1: Newton's Law F = ma is covariant under rotations (x' = F(x) = Rx), and in x'-space this law takes the form F' = m'a' which has the same form as the equation in x-space F = ma. Once we know that F and a are contravariant vectors and m is a scalar, this conclusion is automatic from (2.3.2), F = ma F' = m'a' proof: F' = RF = R(ma) = m Ra = m a' = m' a' (7.9.g.1) where m = m' follows since mass is a scalar under rotation. Thus, Newton's Law has the same form when it is examined in two frames of reference related by a rotation. It is covariant. Example 2: Consider the equation A B = π where A and B are contravariant vectors and is the covariant dot product defined in (5.10.1), A B ≡ abAaBb . It was shown in (5.10.2) that the quantity A B transforms as a scalar under general transformation x' = F(x) so that A' B' = A B. Since the number π is also a scalar under any transformation (it is a constant), one could say that π' = π (it is the same number 3.14 in x'-space and x-space), so A B = π A' B' = π' , equation is covariant. (7.9.g.2) What we see here is that an equation is covariant IFF both sides of the equation transform as the same tensorial tensor type under the transformation of interest. In Example 1, both sides of F = ma transform as contravariant vectors under rotations, and in Example 2 both sides of A B = π transform as scalars under a general transformation. Example 3: Consider the outer product equation (7.1.1) Tab = UaVb where U and V are contravariant vectors. We show in (7.1.2) that Tab transforms as a contravariant rank-2 tensor. Both sides of this equation transform in this way, so in x'-space the equation becomes T'ab = U'aV'b. The equation is therefore covariant under the transformation x' = F(x) with dx' = Rdx. Approaching this example in a slightly different manner, suppose we define Tab ≡ UaVb where U and V are contravariant vectors. We then ask: Is Tab a contravariant rank-2 tensor? Line (7.1.2) shows that the answer is yes, T 'ab = U'aV'b = (Raa'Ua') (Rbb'Vb') = Raa' Rbb' Ua'Vb' = Raa' Rbb' Ta'b' (7.9.g.3) which matches the transformation rule as stated in (5.6.3). Example 4: Suppose A and B are tensorial contravariant rank-2 tensors. Is the equation AB = C covariant? If it were, we would have to show that in x'-space we have A'B' = C' where C is a contravariant rank-2 tensor. To investigate, we use the rule (5.7.1) which states how a contravariant rank-2 tensor transforms in terms of Picture A shown in (5.7.2) : A'B' = (RART)(RBRT) = RA(RTR)BRT // since A and B are contra rank-2 tensors (7.9.g.4) C' = RCRT = RABRT // assuming C is also a contra rank-2 tensor and AB = C If it were true that RTR = 1, one would find from the first line above that A'B' = RABRT = RCRT = C' and the answer would be yes, the equation AB = C is covariant. However, for a general Picture A transformation with metric tensor g in x-space and g' in x'-space, what we know about R comes from (5.7.6) : g' = R g RT . Even if g = 1 so x-space is Cartesian, this says g' = RRT, but this tells us nothing about RTR. So for a general transformation, we have RTR ≠ 1 and so the equation AB = C is NOT covariant. [In the special case that R is a rotation, so RT = R-1 (real orthogonal), then RTR= R-1R = 1.] As with Example 3, we can reformulate the current example in a different manner. Suppose we define C ≡ AB and specify that both A and B are contravariant rank-2 tensors. In this case, is C a contravariant rank-2 tensor? If it were, we would have to have (A'B') = R(AB)RT from (5.7.1). But we showed above that, since RTR ≠ 1, we end up with (A'B') ≠ R(AB)RT . Therefore, C ≡ AB is not a contravariant rank-2 tensor. Could C be a covariant rank-2 tensor? If it were, we would need to have (A'B') = ST(AB)S from (5.7.1). But above we show that A'B' = RA(RTR)BRT and this is completely different from (A'B') = ST(AB)S. Thus, C is not a covariant rank-2 tensor. Since C has two indices, the only way it could be a tensorial tensor is if it is either a contravariant or a covariant rank-2 tensor, but we have just ruled out both these possibilities. Therefore C ≡ AB is not a tensorial tensor of any kind whatsoever, even though A and B are tensorial tensors. Matrix Rule #1. In developmental notation, if A and B are contravariant rank-2 tensors, the matrix product AB is (in general) not a rank-2 tensor and is in fact not any kind of tensor. The equation C = AB is not covariant. Mimicking the above discussion, the reader can show that the same conclusion applies to C = B, C = A and C = : in none of these cases is C a tensor of any kind, and all these equations are non-covariant. Similarly, the Rule applies to X = ABC or X = ABCD and so on. (7.9.g.5) For this reason, we shall never ask how to transform an equation like X = ABC... from developmental to standard notation. Equations which are non-covariant are simply of no interest, and can never describe a physical relationship as we explain below in Section ***. The attentive reader might ask: What about the equation g' = R g RT which has the form X = ABC. And if g = 1, what about g' = RRT whose form is X = AB? In both these cases, the left hand side is a contravariant rank-2 tensor. These equations do not violate the Matrix Rule #1 above because the matrices R and RT are not tensors of any kind, as noted above in ***. Furthermore, one does not ask whether g' = R g RT is covariant or not because it is an equation relating objects in different spaces and not all objects in the equation are tensors. We now consider the notion of matrix multiplication using mixed rank-2 tensors. Since we never introduced such mixed tensors in our developmental notation, we have this discussion entirely in the Standard Notation. Consider Cij = AikBkj . // implied sum on k (7.9.g.6) The indices k have the right adjacency so one could think of this as being a matrix equation C = AB where all three objects are "down-tilt rank-2 mixed tensors". Down-tilt just means the two indices are tilting down like ij. We can ask again our questions of Example 4. If A and B are rank-2 tensors, is C = AB covariant? And if we define C ≡ AB, is C a rank-2 tensor? The answer to both questions is yes. To show that AB = C is covariant, we start with (7.9.c.9) applied to A and B: M'ab = Raa' Sb'b Ma'b' (7.9.c.9) so A'ikB'kj = (Ria' Sb'k Aa'b') (Rka" Sb"j Ba"b") = Ria' Sb'k Rka" Sb"j Aa'b'Ba"b" = Ria' (Sb'k Rka") Sb"j Aa'b'Ba"b" = Ria' (SR)b'a" Sb"j Aa'b'Ba"b" = Ria' δ b'a" Sb"j Aa'b'Ba"b" = Ria' Sb"j Aa'b'Bb'b" = Ria' Sb"j (AB)a'b" = Ria' Sb"j Ca'b" = C'ij . // using (7.9.c.9) a third time with M = C (7.9.g.7) Thus we have shown that AB = C A'B' = C' so our down-tilt matrix equation is covariant. If we define C ≡ AB where A and B are down-tilt mixed rank-2 tensors, then C will be a rank-2 down-tilt tensor providing we can show that (A'B')'ab = Raa' Sb'b (AB)a'b = Raa' Sb'b Ca'b . But this is just what was shown above (albeit with different indices), so yes, C is also a down-tilt mixed rank-2 tensor. One way to clarify the intention of C = AB is to write the matrix equation as Cdt = AdtBdt where the notation Adt means the down-tilt mixed rank-2 tensor having components Aij. It is easy to show that the conclusions reached above apply similarly to an all up-tilt matrix equation Cij = Aik Bkj . // implied sum on k (7.9.g.8) We thus arrive at: Matrix Rule #2. In Standard Notation, it is reasonable to use matrix notation in the following two situations involving mixed rank-2 tensors: Cij = AikBkj Cdt = AdtBdt dt = down-tilt Cij = AikBkj Cut = AutBut ut = up-tilt (7.9.g.9) In special relativity the down-tilt matrix form is most often used, and one just writes C = AB without bothering with the dt clarifying subscripts. This is consistent with the usual statement x'μ = Λμνxν to describe a Lorentz transformation acting on the contravariant vector xν (where Λ = our R ) (h) Matrix Inverse, Transpose and Determinant Matrix Inverses. Consider the matrix equation AB=1 where (assuming det(A) ≠ 1) we can write B = A-1. As demonstrated above, AB = 1 can only be a covariant equation if A and B are both down-tilt or both up-tilt mixed rank-2 tensors. Then we are talking about either AdtBdt = 1dt or AutBut = 1ut, and the corresponding Bdt = (A-1)dt and But = (A-1)ut, all these being matrix equations . Thus AdtBdt = 1dt Bdt = (A-1)dt AikBkj = δij Bij = (A-1)ij AutBut = 1ut But = (A-1)ut AikBkj = δij Bij = (A-1)ij (7.9.h.1) In this context, we have shown that if A is a mixed rank-2 tensor, then (A-1) is a mixed rank-2 tensor as well, assuming it exists. In Section (b) above we considered SR = 1 and reached the conclusions shown just above for the cases A = S and R = B. It happens that in this special case, S and R are not tensors, but the results are still valid. We shall see below that the matrix 1dt is really gdt, the down-tilt form of the metric tensor g, and similarly for 1ut : 1dt = gdt (1)ij = gij = δij = δi,j 1ut = gut (1)ij = gij = δij = δi,j (7.9.h.2) Thus first equation above can be written AdtBdt = gdt where all three matrices are down-tilt mixed rank-2 tensors. And this is also true for AutBut = gut . Transpose Matrices. In the developmental notation we have equations like (5.7.1) M' = R M RT and = RT ' R which involve transposes of matrices. Although it is possible to define a notion of "matrix transpose" in the Standard Notation, we have found that this leads to much confusion, and it is best to simply not allow such a notation. In converting an equation to standard notation, one should remove the transpose notation right at the start within the developmental notation, and then convert the equation to standard notation. For example we start in developmental notation, M' = R M RT => M'ad = RabMbc(RT)cd = RabMbcRdc = RabRdcMbc (7.9.h.3) Then we make the conversion using (7.6.2) M'ad = RabRdcMbc → (M')ad = RabRdcMbc (Standard Notation) (7.9.h.4) and this then is the Standard Notation rule for the way a contravariant rank-2 tensor transforms, as was shown in (7.9.c.9). Determinant of a Matrix. . In developmental notation one writes det(A) = εabc... A1aA2bA3c..... = εabc... Aa1 Ab2Ac3..... (7.9.h.5) where the Aij are components of the contravariant rank-2 tensor A and where ε is the permutation tensor discussed above in **. We have argued above that the notion of a rank-2 tensor being a matrix in Standard Notation is only viable for mixed rank-2 tensor of either the down-tilt or up-tilt variety. Thus, the matrix determinants of interest in Standard Notation would be these: det(Adt) = det(Aij) = εabc... A1a A2bA3c..... = εabc... Aa1 Ab2Ac3..... (7.9.h.6) det(Aut) = det(Aij) = εabc... A1a A2bA3c..... = εabc... Aa1 Ab2Ac3..... (7.9.h.7) These determinants for a rank-2 tensor A will never come up in this document, but they have come up for the non-tensor objects R and S as shown in ****.