obs old section 7 i

(i) A few S,R matrix facts in new notation It was noted in the Introduction that the translation from developmental to standard notation is not trivially obvious especially when it comes to matrix objects with both up and down indices. This section addresses some of the issues that arise. Inverse. Section (g) 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 and therefore it must be that (R-1)ca = Sca. A similar argument applies to S-1 and this then justifies the first two lines in the little table below. So in standard notation one has R-1 = S, just as in the developmental notation. Transpose. If A is a rank-2 tensor, the translation mapping (AT)ab = Aba → (AT)ab = Aba 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 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 explained in section (o) below. Similarly (ST)ab = Sba . Generally one does not see transposed objects appearing much in standard notation equations, but with the above rules, which are replicated in the table below, there is no reason one cannot use them. S and R are real orthogonal but only in standard notation. Below in section (q) it will be shown that Rab = Sba which shows this exact same reflection of indices in the vertical axis between the indices. Therefore, one has (ST)ab = Sba = Rab which says ST = R (but only in the standard notation). In the previous section on inverses it was shown that R-1 = S (in both notations) so of course S-1 = R. Since S-1 = R and ST = R, it must be that S-1 = ST so that, in standard notation, S (and R as well) are real orthogonal matrices : 1 = STS = SST and 1 = RTR = RRT. In the developmental notation, these two equations appear as ' = ST S and g' = R g RT, see Section 5 (f) and Example 2 below. The reason for this strangeness is that in the developmental notation (AT)ab = Aba and one just swaps the indices, whereas in standard notation (AT)ab = Aba one does not just swap indices, that is, (AT)ab ≠ Aba. Example 1: (standard notation, using result of section (q)) 1 = STS => δac = (STS)ac = (ST)abSbc = SbaSbc = RabSbc = (RS)ac = (1)ac = δac Example 2: (standard notation) g'ab = Raa'Rbb'ga'b' // translation to standard notation of g'ab = Raa' ga'b' (RT)b'b g'ab = Raa'Rbb'ga'b' δab = Raa'Rbb'δa'b' δab = Raa'Rba' δab = Raa'(RT)a'b = (RRT)ab => 1 = RRT Here then is a table showing the translation of some equations from developmental to standard notation. In general, one avoids using "matrix notation" (such as A=MB) in the standard notation, and instead writes out the index structure. This is perhaps a slight disadvantage of the notation, but it is only a small disadvantage, since one cannot use matrix notation for tensors of rank > 2. R-1 = S → (R-1)ik = Sik // or any other index combo S-1 = R → (S-1)ik = Rik RR-1 = RS = 1 etc → Rik(R-1)ka = RikSka = δia etc (ST)ab = Sba → (ST)ab = Sba // see explanation above (RT)ab = Rba → (RT)ab = Sba det(R) = εabc...R1aR2b....RNx → det(Rij) = εabc...R1aR2b....RNx det(S) = εabc...S1aS2b.....SNx → det(Sij) = εabc...S1aS2b....SNx g' = R g RT or g'ab = Raa'Rbb'ga'b' → g'ab = Raa'Rbb'ga'b' ' = ST S or 'ab = STaa'STbb'a'b' → g'ab = Sa'aSb'b ga'b' The right sides of the last two equations show that gab transforms as a contravariant rank-2 tensor, while gab transforms as a covariant rank-2 tensor. We proved these facts during development in Section 5 (f). One can invert these right side equations as follows g'ab = Raa'Rbb'ga'b' => gab = (R-1)aa'(R-1)bb'g' a'b' = Saa' Sbb' g' a'b' g'ab = Sa'a Sb'b ga'b' => gab = (S-1)a'a (S-1)b'b g'a'b' = Ra'a Rb'b g'a'b' and here then is a summary, g'ab = Raa'Rbb' ga'b' gab = Saa'Sbb' g'a'b' g'ab = Sa'a Sb'b ga'b' gab = Ra'a Rb'b g'a'b' If x-space is Cartesian with g = 1, the first column above simplifies to g'ab = RacRbc // g = 1 g'ab = Sca Scb = Rac Rbc ( see section (q) below) // g = 1 There are times when one wants to think in terms of matrices, and the following "trick" can be used (gdn)ab ≡ gab (Rdt)ab = Rab etc (gup)ab ≡ gab (Rut)ab = Rab where dn means all indices are down in the real object, up means all indices are up, dt means "down tilt" and ut means "up tilt".