Lai version of DTijk v3

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. First, here is something I have shown. Here I choose to make x be the arg of en. This may be the wrong choice, but at least I am making a choice and displaying it. The other choice is en(ξ) The whole idea here is that x-space is NOT Cartesian, it is the curvilinear coordinates! dn = [Γ()]kjndxj k [Γ()]kjn ≡ hn-1 [hk Γkjn – (∂jhn) δk,n] 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 dxj k di = Gkji dxj 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 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. Now look again at the dT and related objects: Here T is an abstract rank-2 tensor which I expand in a particular basis, assumed all functions of x. (∂kT(x)) = Σij [(∂kT'ij(x)) + T'aj(x) Gika(x) + T'ia(x)Gjka(x) ] i(x)j(x) dT(x) = (∂kT)dxk = Σij [(∂kT'ij(x)) + T'aj(x) Gika(x) + T'ia(x)Gjka(x) ] i(x)j(x) dxk At this point T and M have no connection at all. Now I derived the above in some earlier document, so let's rederive right here to make sure it is done correctly. Start with 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 that in Picture C1 above the x-space is Cartesian, we may say (∂kT)rs = ∂kTrs = Mrsk and this is our first inkling of a connection between M and T. We are later going to invoke Picture C1' where x' is going to be non-Cartesian. 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. 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 ≡ [(∂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 i(x)j(x) dT(x) = (∂kT)dxk = Σij Q'ijkdxk 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 [i(x)]a [j(x)]b and then I have connected Qkij 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) Now comes a confusion. These vectors are in ξ-space of Picture C1. Therefore i(x) hi = ei(x) where there is no prime on the hi. Then continue the above Σa [i(x)]a [r(x)]a = i(x) r(x) = (hihr)-1 ei(x) er(x) = (hihr)-1 gir(x) where g is for x-space in Picture C1. For Cartesian x-space, we know that hi = 1 and g = δ, wo then Σa [i(x)]a [r(x)]a = (hihr)-1 gir(x) = δi,r which is what your intuition suggests in Cartesian space. So carrrying out this process we get [r(x)]a [s(x)]b ∂kTab(x) = Σij Q'ijk δi,r δj,s = Q'krs(x) so we have now an expression for Qkrs, for what it is worth. At least I now know this: ∂kTab(x) = Mabk = Σij Q'ijk [i(x)]a [j(x)]b 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 ∂KMAB But x-space is Cartesian, so ∂KMAB = ∂KMAB and I know that ∂KTAB(x) = Σij Q'ijK [i(x)]A [j(x)]B so I then have M'abk = [ h'kh'ah'b RkKRaARbB [i(x)]A [j(x)]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 = [ MaAMbB h'kRkK [i(x)]A [j(x)]B ] QijK 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 = [ MaAMbB h'kRkK [i(x)]A [j(x)]B ] Q'ijK NAi NBj or M'abk = [ MaAMbBNAiNBjh'kRkK] Q'ijK = [ MaANAiMbBNBjh'kRkK] Q'ijK = [ (MaANAi)(MbBNBj)h'kRkK] Q'ijK = [ δaiδbj)h'kRkK] Q'ijK = h'kRkK Q'abK = h'kRkK [(∂KT'ab(x)) + T'rb(x) GaKr(x) + T'as(x)GbKs(x) ] This is similar to something I found before, and it seems very close to the answer I seek which is shown on Lai page 504 top. 9. Now how would I convert this to Lai notation? From above I already have Mijk = M'ijk Tij = T'ij Γijk = Gkji Then in Lai notation I have shown that Mijk = h'kRkK [(∂KTij(x)) + Trj GiKr(x) + TisGjKs(x) ] ΓrKi ΓsKj or Mijk = h'kRkK [∂KTij + TrjΓrKi + TisΓsKj ] or Mijk = h'kRkK [∂KTij + TqjΓqKi + TiqΓqKj ] or But here is what appears in