Grad of Tensor by the Lai method

Grad of Tensor by the Lai method PhL 10.13.12 ******************************************************************** How does Lai do it? Start with his (8A.12) which, converted to my notation, says di = Kijk dxj k His Γ I write as K so he is saying ∂ji = Kijkk or ∂jn = Knjkk In tensor doc App F I have (∂'jen) = Γ'kjn ek So write this last as ∂'j(h'nn) = Γ'kjnh'k k ∂'j(h'n) n + h'n∂'jn = Γ'kjnh'k k h'n∂'jn = Γ'kjnh'k k – ∂'j(h'n) n ∂'jn = [Γ'kjnh'k/ h'n] k – ∂'j(h'n)/ h'nn = [Γ'kjnh'k/ h'n – δn,k ∂'j(h'n)/ h'n] k Comparison then shows that Knjk = h'n-1 [h'kΓ'kjn–∂'j(h'n) δn,k] Kijk = h'i-1 [h'kΓ'kji– ∂'j(h'i) δi,k] so this is the gamma object he is using. Let's just see what the values come out to be. I did it in Maple, and the result is that my Kijk agrees with his Γijk for sphericals on page 503 top. Next, looking at page 501, I can associate Mijk = (T)'ijk M = T Lai page 503 (8A.15) which is good. How his claim on top of page 504 is that Mijmhm = ∂mTij + TqjΓqmi + TiqΓgmj Translated to my notation, this claim says (T)'ijmh'm = ∂mT'ij + T'qjKqmi + T'iqKgmj // no sum on m (T)'ijm = h'm-1∂mT'ij+ h'm-1T'qjKqmi + h'm-1T'iqKgmj (T)'ijk = h'k-1∂kT'ij+ h'k-1T'qjKqki + h'k-1T'iqKgkj = (h'i h'j h'k)-1 [h'i h'j (∂kT'ij)+ h'i h'j T'qjKqki + h'i h'j T'iqKgkj] My formula for this same object is: (T)'ijk = (h'i h'j h'k)-1 (T)'ijk where (T)'ijk = h'i h'j(∂'kT'ij) + ∂'k(h'i h'j) T'ij – Γ'nik(h'n h'jT'nj) – Γ'njk(h'i h'nT'in) So we are happy if we can show that [h'i h'j (∂kT'ij)+ h'i h'j T'qjKqki + h'i h'j T'iqKgkj] = h'i h'j(∂'kT'ij) + ∂'k(h'i h'j) T'ij – Γ'nik(h'n h'jT'nj) – Γ'njk(h'i h'nT'in) The first terms on each side match, so I have to show that h'i h'j T'qjKqki + h'i h'j T'iqKqkj = ∂'k(h'i h'j) T'ij – Γ'nik(h'n h'jT'nj) – Γ'njk(h'i h'nT'in) (*) We know that Kijk = h'i-1 [h'kΓ'kji– ∂'j(h'i) δi,k], KIJK = h'I-1 [h'KΓ'KJI– ∂'J(h'I) δI,K], Kqki = h'q-1 [h'iΓ'ikq– ∂'k(h'q) δq,i], Kqkj = h'q-1 [h'jΓ'jkq– ∂'k(h'q) δq,j], We can then evaluate the LHS of (*) above LHS = h'i h'j T'qjKqki + h'i h'j T'iqKgkj = h'i h'j T'qj h'q-1 [h'iΓ'ikq– ∂k(h'q) δq,i] + h'i h'j T'iq h'q-1 [h'jΓ'jkq– ∂k(h'q) δq,j] = h'i h'j T'nj h'n-1 [h'iΓ'ikn– ∂k(h'n) δn,i] + h'i h'j T'in h'n-1 [h'jΓ'jkn– ∂k(h'n) δn,j] = h'i2 h'j T'nj h'n-1 Γ'ikn + h'i h'j2 T'in h'n-1Γ'jkn – h'i h'j T'nj h'n-1∂k(h'n) δn,i – h'i h'j T'in h'n-1∂k(h'n) δn,j = h'i2 h'j h'n-1 Γ'ikn T'nj + h'i h'j2 h'n-1Γjkn T'in – h'i h'j T'ij h'i-1∂k(h'i) – h'i h'j T'ij h'j-1∂k(h'j) = h'i2 h'j h'n-1 Γ'ikn T'nj + h'i h'j2 h'n-1Γ'jkn T'in – h'j ∂k(h'i) T'ij – h'i∂k(h'j) T'ij = h'i2 h'j h'n-1 Γ'ikn T'nj + h'i h'j2 h'n-1Γ'jkn T'in – [ h'j ∂k(h'i) + h'i∂kh'j)] T'ij = h'i2 h'j h'n-1 Γ'ikn T'nj + h'i h'j2 h'n-1Γ'jkn T'in – ∂'k(h'i h'j) T'ij So I now want to show that ∂'k(h'i h'j) T'ij – Γ'nik(h'n h'jT'nj) – Γ'njk(h'i h'nT'in) = h'i2 h'j h'n-1 Γ'ikn T'nj + h'i h'j2 h'n-1Γ'jkn T'in – ∂'k(h'i h'j) T'ij (**) Unfortunately, the Γ sum index is not in the same place on the two sides, so I have to do something about that. I do have this theorem: (∂'cg'ab) = – [g'an Γ' bcn + g'bn Γ'acn] For orthogonals, we have g'ab = δa,bh'a-2 (∂'cg'ab) = ∂'c[δa,bh'a-2] = δa,b∂'c (h'a-2) = -2 h'a-3δa,b(∂'ch'a) The theorem then reads +2 h'a-3δa,b(∂'ch'a) = g'an Γ' bcn + g'bn Γ'acn = h'a–2 Γ' bca + h'b–2 Γ'acb or +2 h'a-3δa,b(∂'ch'a) = h'a–2 Γ' bca + h'b–2 Γ'acb or +2 h'a-1δa,b(∂'ch'a) = Γ' bca + h'a2h'b–2 Γ'acb or Γ' bca = – h'a2h'b–2 Γ'acb + 2 h'a-1δa,b(∂'ch'a) or Γ' ikn = – h'n2h'i–2 Γ'nki + 2 h'n-1δn,i(∂'kh'n) Γ' jkn = – h'n2h'j–2 Γ'nkj + 2 h'n-1δn,j(∂'kh'n) If I use these in the RHS of (**), I get h'i2 h'j h'n-1 Γ'ikn T'nj + h'i h'j2 h'n-1Γ'jkn T'in – ∂'k(h'i h'j) T'ij = h'i2 h'j h'n-1 ********************************************************************** Go back here and try to rewrite: Kijk = h'i-1 [h'kΓ'kji– ∂'j(h'i) δi,k] Theorem said Γ' bca = – h'a2h'b–2 Γ'acb + 2 h'a-1δa,b(∂'ch'a) Γ' kji = – h'i2h'k–2 Γ'ijk + 2 h'i-1δi,k(∂'jh'i) Then Kijk = h'i-1h'kΓ'kji– h'i-1∂'j(h'i) δi,k = h'i-1h'k[– h'i2h'k–2 Γ'ijk + 2 h'i-1δi,k(∂'jh'i)] – h'i-1∂'j(h'i) δi,k = – h'ih'k-1 Γ'ijk + 2 h'i-2h'k δi,k(∂'jh'i) – h'i-1 δi,k (∂'jh'i) = – h'ih'k-1 Γ'ijk + 2 h'i-1 δi,k(∂'jh'i) – h'i-1 δi,k (∂'jh'i) = – h'ih'k-1 Γ'ijk + h'i-1 δi,k(∂'jh'i) = – h'ih'k-1 Γ'ijk + h'k-1 δi,k(∂'jh'i) = h'k-1[– h'i Γ'ijk + δi,k(∂'jh'i) ] So I now have an alternate form Kijk = h'k-1[– h'i Γ'ijk + δi,k(∂'jh'i) ] Now go back to the thing we wanted to show h'i h'j T'qjKqki + h'i h'j T'iqKqkj = ∂'k(h'i h'j) T'ij – Γ'nik(h'n h'jT'nj) – Γ'njk(h'i h'nT'in) (*) Now prepare to appropriate K objects Kijk = h'k-1[– h'i Γ'ijk + δi,k(∂'jh'i) ] KIJK = h'K-1[– h'I Γ'IJK + δI,K(∂'Jh'I) ] Kqki = h'i-1[– h'q Γ'qki + δq,i(∂'kh'q) ] Kqkj = h'j-1[– h'q Γ'qkj + δq,j(∂'kh'q) ] Then for LHS(*) we get h'i h'j T'qjKqki + h'i h'j T'iqKqkj = h'i h'j T'qj h'i-1[– h'q Γ'qki + δq,i(∂'kh'q) ] + h'i h'j T'iq h'j-1[– h'q Γ'qkj + δq,j(∂'kh'q) ] = h'j T'qj [– h'q Γ'qki + δq,i(∂'kh'q) ] + h'i T'iq [– h'q Γ'qkj + δq,j(∂'kh'q) ] = h'j T'nj [– h'n Γ'nki + δn,i(∂'kh'n) ] + h'i T'in [– h'n Γ'nkj + δn,j(∂'kh'n) ] = h'j T'nj [– h'n Γ'nki + δn,i(∂'kh'i) ] + h'i T'in [– h'n Γ'nkj + δn,j(∂'kh'j) ] = – h'n h'j Γ'nki T'nj – h'i h'n Γ'nkj T'in + h'j T'nj δn,i(∂'kh'i) + h'i T'in δn,j(∂'kh'j) = – h'n h'j Γ'nki T'nj – h'i h'n Γ'nkj T'in + h'j T'ij(∂'kh'i) + h'i T'ij (∂'kh'j) = – h'n h'j Γ'nki T'nj – h'i h'n Γ'nkj T'in + [ h'j (∂'kh'i) + h'i (∂'kh'j)] T'ij = – h'n h'j Γ'nki T'nj – h'i h'n Γ'nkj T'in + ∂'k(h'i h'j) T'ij and FINALLY we get agreement. Let's recheck the maple with this new form. Yes! So this is the way it works best Kijk = h'k-1[– h'i Γ'ijk + δi,k(∂'jh'i) ] or Γ(Lai)ijk = h'k-1[– h'i Γ'ijk + δi,k(∂'jh'i) ]