Lai versin of DTijk v1

Lai's Version of (T)'ijk 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 Me M = Mijk eiejej M = Σijk [M()]ijk ijk Mijk = [M()]ijk = Mijk T = Tij eiej T = Σij [T()]ij ij Tij = [T()]ij = Tij T = Σij Tij ij 3. So how are these M and T related? In Cartesian space I know that (T)ijk ≡ ∂kTij = Mijk idea that M = T . If I write this as (T)ijk ≡ Tij;k = Mijk then (T)ijk = Mijk is a true tensor equation. It would then follow that (T)ijk = Mijk ok to here 4. Now go back to T = Σij Tij ij Practice taking a component of this tensor, TIJ = Σij Tij [ ij]IJ = Σij Tij [ ij] [uI uJ] = Σij Tij (i uI)( j uJ) = Σij Tij hihj (ei uI)( ej uJ) = Σij Tij hihj(ei)I (ej)J = Σij Tij hihj RiI RjJ which is a sum of vectors in double direct product space. Everything is a function of x. I am certainly allowed to apply ∂k to both sides. As shown in the Footnote below, this results in (∂kT) = Σij [(∂kTij) + Taj Gika + TiaGjka ] ij ≡ Σij Qijk ( ij) so this is also a vector in double direct product space. Again, try to take a component (∂kT)IJ = Σij Qijk hihj(ei)I (ej)J = Σij Qijk (i)I (j)J But maybe (∂kT)IJ = ∂kTIJ = (T)IJk. Then the above line reads (T)IJk = Σij Qijk (i)I (j)J = MIJk and for the first time I have some connection between T and Qijk , Maybe I can use my Hilbert Space stuff here, just a change of basis? Look at (E.7.11), A(b) = BABT Apply this for bn = n. Then Bni ≡ [bn]i = ( n)i and I suppose Bni ≡ [bn]i = ( n)i Then maybe Σij Qijk (i)I (j)J = Σij Qijk BiI BjJ = Σij Qijk (BT)Ii BjJ = Σij (BT)Ii Qijk BjJ = [BTQkB]IJ = [Q(e^)]IJ stop , flailing getting nowhere. How can I relate this to something called (T) ? (T) uk = (∂kT) ?? Well then I have (T) uk = Σij Qijk ( ij) Now maybe insert (T) = Σijk (T)ijk ijk Now practice a certain dot product on a generic tensor: A = Σij Aij ij A (rs) = Σij Aij (ij) (rs)) = Σij Aij δirδjs = Ars So this dot product results in a scripted tensor! So try this idea where A = ∂kT and we think of k as a fixed value, Then (∂kT) = Σij Qijk ( ij) (∂kT) (rs) = Qrsk (∂kT) (rs) = (∂kT)rs = [(∂kT)()]rs I can then identify Qrsn = [(∂nT)()]rs = [(T;n)()]rs // Cartesian How can I connect this with M? Think of this as a vector with index k. In general, ∂xf = f and then ∂nf = f Now dot this into (rs) to get (∂kT) (rs) = Σij Qijk ( ij) (rs) = Σij Qijk δirδjs = Qrsk Perhaps I can claim that (∂kT) (rs) = (∂kT)rs = ∂kTrs for each fixed index k. On the other hand, I know that T = Σijk [(T)()]ijk ijk = Σijk' (T)ijk' ijk' I could consider Qij to be a vector and write Qij = Σk Qijkk It would seem that Qijk = (∂kT) ij 4. Now consider this idea all in my notation, dT(x) = (∂kT)dxk which seems reasonable at the tensor level, where (∂kT) = ∂k[ Σij Tij ij ] = Σij(∂kTij) ij + ΣijTij (∂ki) j + ΣijTij i(∂kj) So at least I have an expression for and meaning for dT dT = (∂kT)dxk = [ Σij(∂kTij) ij + ΣijTij (∂ki) j + ΣijTij i(∂kj)] dxk Now before going further, lets replace (∂ki) = Gskis (∂kj) = Gskjs Then the bracket [...] above becomes [ Σij(∂kTij) ij + ΣijTij (∂ki) j + ΣijTij i(∂kj)] = [ Σij(∂kTij) ij + ΣijsTij Gskis j + ΣijsTij iGskjs ] = [ Σij(∂kTij) ij + ΣajiTaj Gikai j + ΣiajTiaGjka ij ] = [ Σij(∂kTij) + ΣajiTaj Gika + ΣiajTiaGjka ] ij = Σij [(∂kTij) + Taj Gika + TiaGjka ] ij So I have now shown that (∂kT) = Σij [(∂kTij) + Taj Gika + TiaGjka ] ij dT = Σij [(∂kTij) + Taj Gika + TiaGjka ] ij dxk = Σij (Q dx) ij Let's take the a,b component of the above equation (∂kT)rs = Σij [(∂kTij) + Taj Gika + TiaGjka ] [ij]rs = Σij [(∂kTij) + Taj Gika + TiaGjka ] δir δjs = [(∂kTrs) + Tas Grka + TraGska ] . But we can say (∂kT)rs = ∂kTrs = Mkrs in Cartesian space. Now in (8A.25) I think Lai makes this claim in my notation [(∂kTij) + Taj Gika + TiaGjka ] = Mijk and if this is true, then it must also be true that (∂kT) = ΣijMijk ij for fixed k So this looks like I am expanding a general tensor named (∂kT), and I would normally write (∂kT) = Σij[(∂kT)()]ij ij k = 1,2,3 Maybe rewrite this like so, (T)k = Σij[(T)k()]ij ij But meanwhile you can expand the (T) = Σk (T)k k M = Σijk [M()]ijk ijk (T) = Σijk [(T)()]ijk ijk This is my first use of a thing like dT, but it does make some sense I think. Now suppose we have dei = (ΓLai)ijk dxj ek So far M and T are unrelated objects in what I have written above. But I want this to be true M = (T) [M()]ijk = [(ΔT)()]ijk Mijk = (T)ijk = ∂k Tij J.1.3 Cartesian