Lai version of DTijk v4

Lai's Version of (T)'ijk v3 PhL 4.24.15 I have never understood how they do this on page 503, but today I feel confident that I can translate their equations into my notation. 1. Assume this Picture !!!! The n vectors are in x-space, as usual. Assume the variations of n have this form dn(x') = [Γ()]kjndx'j k(x') I can rewrite this as ∂'jn(x') = [Γ()]kjnk(x') To simplify, lets define Gkjn ≡ [Γ()]kjn Then I have (∂'jn(x')) = Gkjn(x)k(x') dn = Gkjn(x') dx'j k(x') di = Gkji dx'j k dei = Γijk dxj ek Lai p 502 dev notation I think 2. Next consider ( Lai notation equations are in blue) Lai Me ei i dxj dx'j Γijk Gkji M = Mijk eiejej M = Σijk [M()]ijk ijk [M()]ijk = M'ijk T = Tij eiej T = Σij [T()]ij ij [T()]ij = T'ij Let's now install arguments as chosen above M(x') = Σijk [M()(x')]ijk i(x')j(x')k(x') M = rank 3 M(x') = Σijk M'ijk(x') i(x')j(x')k(x') T(x') = Σij [T()(x')]ij i(x')j(x') T = rank 2 T(x') = Σij T'ij(x') i(x')j(x') 3. Consider now, for Picture B above, T(x') = Σij T'ij(x') i(x')j(x') Then apply ∂'k to get ∂'kT(x') = (∂'kT'ij(x')) i(x')j(x') + T'ij(x')(∂'k i(x'))j(x') + T'ij(x')i(x')(∂kj(x')) Now use the relation above that (∂'jn(x')) = Grjn(x')r(x') so that (∂'k i(x')) = Grki(x')r(x') (∂'k j(x')) = Gskj(x')s(x') and then ∂'kT(x') = (∂'kT'ij(x')) i(x')j(x') + T'ij(x')(∂'k i(x'))j(x') + T'ij(x')i(x')(∂'kj(x')) = (∂'kT'ij(x')) i(x')j(x') + T'ij(x') Grki(x')r(x')j(x') + T'ij(x')i(x')Gskj(x')s(x') Now some index shuffling. In the middle term do i↔r and in the last term j↔s: = (∂'kT'ij(x')) i(x')j(x') + T'rj(x') Gikr(x')i(x')j(x') + T'is(x')i(x')Gjks(x')j(x') = [ ∂'kT'ij(x') + T'rj(x') Gikr(x') + T'is(x')Gjks(x')] i(x')j(x') and this verifies the result quoted above. 4. If we assume Picture B as above then (∂kT)rs = ∂kTrs = Mrsk and this is our first inkling of a connection between M and T. The index ordering above I take from this quote (T)ijk ≡ ∂kTij (J.1.3) with the idea that somehow M = T in the abstract. What does this look like in x'-space? That is a sticky wicket in this notation because ∂kTrs is not a tensor! I will assume however that Mrsk is a true tensor and what I really mean is this (T)rsk =Trs;k = Mrsk and in this notation we have in x'-space (T)'rsk =T'rs;k = M'rsk 5. Question 1: What tensor transformation rules apply to the above line going from x to x'-space? First, we know that M'rsk(x') =Rrb Rsc Rka Mbca(x) T'rs(x') = RrdRseTde(x) ∂'k = Rkd∂d 6. Going down a blind path, I could then evaluate, as in earlier appendices ∂'kT'rs(x') = ( Rkd∂d)(RrdRseTde) = Rkd [RrdRse(∂dTde) + (∂dRrd)RseTde + Rrd(∂dRse)Tde ] and this at least makes a connection between [ ∂'kT'rs(x')] and [∂dTde(x)] . 7. I don't really see any connection between this last item and the expansions in 2. Well there is this; Define Q'ijk(x') ≡ [∂'kT'ij(x') + T'aj(x') Gika(x') + T'ia(x')Gjka(x') ] I arbitrarily put Q in script with a prime but it means nothing right now, just a definition. Then from 3 above (∂'kT(x')) = Σij Q'ijk(x') i(x')j(x') dT(x') = (∂'kT(x'))dx'k = Σij Q'ijk(x')dx'k i(x')j(x') So OK, the first line above does make some kind of connection between an expansion and ∂'kT. However, I don't really know what (∂'kT(x)) means having no indices. It is a vector in direct product space. Probably I could dot both sides with ua(x')ub(x') and use ua(x') i(x') = [i(x')]a = [i(x')]a . Then maybe this is true ∂'kTab(x') = Σij Q'ijk(x') [i(x')]a [j(x')]b But the left side here is really unclear. Should there be a prime on the T? There is no prime when I just write the expansion T = expansion. It is really (∂'kT(x'))ab . This object is fuzzy and I cannot really continue to use it to mean anything. But OK, let's assume it has no prime somehow because we dotted with the u vectors, and let's see what happens. Then I have connected Q'ijk with something which at least has a meaning. Suppose I then multiply both sides of this by [r(x')]a [s(x')]b and sum on a and b. I end up then with things like Σa [i(x')]a [r(x')]a = i(x') r(x') These vectors are in x-space of Picture B. Therefore i(x') h'i = ei(x'). Σa [i(x')]a [r(x')]a = i(x') r(x') = (hi'hr;)-1 ei(x') er(x') = (h'ih'r)-1 g'ir(x) But we know that g'ir(x) = (h'i)2 δi,r so then (h'ih'r)-1 g'ir(x) = (h'ih'r)-1 (h'i)2 δi,r = δi,r So carrying out this process we get [r(x')]a [s(x')]b ∂'kTab(x') = Σij Q'ijk(x') δi,r δj,s = Q'krs(x') so we have now an expression for Q'krs, for what it is worth. At least I now know this: ∂'kTab(x') = M'abk = Σij Q'ijk [i(x)]a [j(x)]b This certainly looks fishy since we have some kind of mixed world. But carry on anyway. 8. What I am really interested in is M'abk = h'ah'bh'k M'abk. So I can write M'abk = h'ah'bh'k M'abk // Picture A,B notation M'abk = h'ah'bh'k RaARbB RkK MABK = h'ah'bh'k RaARbB RkK ∂KTAB But x-space is Cartesian, so MABK = ∂KTAB = ∂KTAB. From above I know that ∂'kTab(x') = Σij Q'ijk(x') [i(x')]a [j(x')]b . Now we know that ∂'k = RkK∂K and we can think of Tab(x') = Tab(x'(x) ) as a function of x. Then RkK(∂KTab) = Σij Q'ijk [i]a [j]b x' = x'(x) everywhere Now use the inversion rule to write this as (∂KTab) = RkK Q'ijk [i]a [j]b or (∂KTAB) = RsK Q'ijs [i]A [j]B so I then have M'abk = h'ah'bh'k RaARbB RkK MABK = h'ah'bh'k RaARbB RkK ∂KTAB = h'ah'bh'k RaARbB RkK {RsK Q'ijs [i]A [j]B } = h'ah'bh'k RaARbB RkKRsK [i]A [j]B Q'ijs Do I know anything about RkKRsK ?? Not an orthogonality rule. Could write as RkKRsK = g'ka gKb RabRsK = g'ka RaKRsK = g'ka δas = g'ks (as I suspected, should be doc'd) And for orthogs we get g'ks = δk,s (h's)-2 Then we have showed that RkKRsK = (h'k)-2δk,s I might as well install this right now to get M'abk = h'ah'bh'k RaARbB (h'k)-2 [i]A [j]B Q'ijk or M'abk = h'k-1 h'ah'bRaARbB [i]A [j]B Q'ijk Now this last statement is valuable. It relates [M()]abk = M'abk to QijK . So now the question is this: Can this last line be simplified? I first quote Mab ≡ h'a Rab (E.8.7) and then I can write M'abk = h'k-1 (h'aRaA)(h'bRbB) [i]A [j]B Q'ijk or M'abk = h'k-1 MaAMbB [i]A [j]B Q'ijk I quote again. Nin = h'n-1Rni = h'n-1(en)i = (n)i , // (7.13.1) (E.8.12) Then I can further simplify M'abk = h'k-1 MaAMbB [i]A [j]B Q'ijk NAi NBj M'abk = h'k-1 MaAMbBNAiNBj Q'ijk = h'k-1 (MaANAi)(MbBNBj) Q'ijk = h'k-1δaiδbj Q'ijk = h'k-1Q'abk which then says h'kM'abk = Q'abk or h'kM'abk = [∂'kT'ab(x') + T'qb(x') Gakq(x') + T'aq(x')Gbkq(x') ] 9. Now how would I convert this to Lai notation? From above I already have Mijk = M'ijk Tij = T'ij Γijk = Gkji hk = h'k Lai's equations are for Picture C1 so I replace x' → x everywhere. So start with h'kM'abk = [∂'kT'ab(x') + T'qb(x') Gakq(x') + T'aq(x')Gbkq(x') ] and make the translations to get hkMabk = [∂kTab + TqbGakq(x') + Taq Gbkq(x') ] Γqka Γqkb or Mabkhk = [∂kTab + Tqb Γqka + Taq Γqkb ] or Mijmhm = [∂mTij + Tqj Γqmi + Tiq Γqmj ] Finally, this agrees with (8A.26), this only took me two days!! 10. Now that I have finally gotten the right answer, I have to do it all again in some efficient manner. The secret was to get the Pictures right!!! // This is now written up as Section J.9 !!