Lai version of DTijk v2

Lai's Version of (T)'ijk v2 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 dn = [Γ()]kjndxj k [Γ()]kjn ≡ hn-1 [hk Γkjn – (∂jhn) δk,n] I can rewrite this as ∂jn = [Γ()]kjnk To simplify, lets define Gkjn ≡ [Γ()]kjn Then I have (∂jn) = Gkjnk 2. Next consider ( Lai notation equations are in blue) Lai Me M = Mijk eiejej M = Σijk [M()]ijk ijk [M()]ijk = M'ijk T = Tij eiej T = Σij [T()]ij ij [T()]ij = T'ij 3. So how are these M and T related? In Cartesian space I know that (T)ijk ≡ ∂kTij ≡ Mijk assuming the idea that M = T (may not be true) If I write this as (T)ijk ≡ Tij;k = Mijk then the above is a true tensor equation, and we assume orthogs, so we have this covariance triplet, (T)ijk = Tij;k = Mijk (T)'ijk = T'ij;k = M'ijk (T)'ijk = T'ij;k = M'ijk I have another expansion which from tensor doc says (J.3.1), T = Σijk { [(T)()]ijk (hihjhk)-1} eiejek = Σijk {(T)ijk (hihjhk)-1} eiejek So if I can compute Lai's M'ijk I am done! I know from (E.8.20) that M" ijk = M'ijk = Mii'Mjj'Mkk' M i'j'k' // overloaded M Now the usual rule for going to covariant is this, M" ijk = M'ijk = Mii'Mjj'Mkk'M i'j'k' or M'ijk = MiaMjbMkcMabc But we know these two facts : Mik = Mik = Nki = Nki // g = 1, g' = diagonal (E.9.9) Nin = (n)i (E.8.12) Then write Mik = Nki = (i)k so then Mia = Nai = (i)a and then M'ijk = (i)a(j)b(k)cMabc (*) I am trying my best to "compute M'ijk". I found in v1 doc that T = Σij T'ij ij (∂kT) = Σij [(∂kTij) + Taj Gika + TiaGjka ] ij (∂cTab) = Σij [(∂cTij) + Taj Gica + TiaGjca ] [ ij]ab = Σαβ [(∂cTαβ) + Taβ Gαca + TαaGβca ] (α)a(β)b I could insert this in (*) above to get M'ijk = (i)a(j)b(k)c(∂cTab) = (i)a(j)b(k)c Σαβ [(∂kTαj) + Taj Gαka + TαaGjka ] (α)a(β)b Replace (x)y = Nyx everywhere M'ijk = NaiNbjNck Σαβ[(∂kTαβ) + Taβ Gαka + TαaGβka ] NaαNbβ Then NaiNaα = NiaNαa = δi,α using the rotation rule: RikRjk = δi,j or Rja = Rja // "rotation" . (7.9.6) OK, then we have M'ijk = Σαβ [(∂cTαβ) + Taβ Gαca + TαaGβca ] NaαNbβNaiNbjNck = Σαβ [(∂cTαβ) + Taβ Gαca + TαaGβca ] (NaαNai)(NbβNbj)Nck = Σαβ [(∂cTαβ) + Taβ Gαca + TαaGβca ] δiαδjβNck = Σαβ[(∂cTij) + Tij Gαca + TiaGjca ] Nck So at least this is an answer, though probably not the right answer, but it looks close. Write as M'ijk = Nck [(∂cTij) + Tij Gαca + TiaGjca ] Nck ≡ Sck h'k-1 = h'k-1 Rkc M'ijk = h'k-1 Rkc [(∂cTij) + Tij Gαca + TiaGjca ] hk M'ijk = Rkc [(∂cTij) + Tij Gαca + TiaGjca ] If that Rkc were not there, this would look close to page 504 top! *************************** T = Σij T'ij ij = Σij Tij uiuj Now apply ∂k just to the two equal expansions Σij [(∂kTij) + Taj Gika + TiaGjka ] ij = Σij (∂kTij) uiuj Then I avoid dealing with the strange object ∂kT. I could dot this with i'j' to get Σij [(∂kTij) + Taj Gika + TiaGjka ] δii'δjj' = Σij (∂kTij) uiuj i'j' or [(∂kTi'j') + Taj' Gi'ka + Ti'aGj'ka ] = Σij (∂kTij) (ui i')(uj j') But then (ui i') = Σc(ui)c (i')c = (i')i to get this strange result [(∂kTi'j') + Taj' Gi'ka + Ti'aGj'ka ] = Σij (∂kTij) (i')i (j')j or [(∂kTij) + Taj Gika + TiaGjka ] = Σnm (∂kTnm) (i)n (j)m Nin = (n)i maybe this says Nin = (n)i so Nni = (i)n Then [(∂kTij) + Taj Gika + TiaGjka ] = Σnm (∂kTnm) Nni Nmj = Σnm Nni(∂kTnm)Nmj and again this looks like a Hilbert Space rotation. Maybe somehow you get Σnm Nni(∂kTnm)Nmj = (∂'kT'nm) = h'k (∂kT'nm) and then somehow you get (∂kT'nm)h'k = [(∂kTij) + Taj Gika + TiaGjka ] and this does look like the Lai result. But my steps are all fuzzy here. End of this day.