divT save old section

Save old divT section, eventually toss 3.6.12 Originally I had the divT calculation as the last subsection of the v appendix. I later decided that since the calculations are so parallel, it was clearer to put them in separate appendices with similar section labels. I then ran into a lot of trouble with divT . At first one term did not agree with my new calculation, but eventually I found that the old calculation was correct. There are still some "Plans" below in some added notes, all flailing. (g) Expansion of div(T) in general curvilinear coordinates The vector-like object A ≡ divT is defined in Cartesian coordinates by Ai = ∂jTij. As with ∂jVi, the object ∂jTij is not a tensor (ie, not a tensorial vector) except under rotations. Regarding ri as the ith row of the matrix Tij, this says that Ai = ∂j(ri)j = ri, so the components of divT are just the divergences of the rows of T. As in the v work done above, there is a fancy tensor method and a brute force method of finding how to express divT in curvilinear coordinates. In the fancy method, one can start with the following true rank-3 tensor, Tab;c ≡ Tab,c – ΓnacTnb – ΓnbcTan = Tab,c in x-space T'ab;c ≡ T'ab,c – Γ'nacT'nb – Γ'nbcT'an T'ab;c = RaαRbβRcγTαβ;γ Tab;c = RαaRβbRγc T'αβ;γ = RαaRβbRγc [T'αβ,γ – Γ'nαγT'nβ – Γ'nβγT'αn ] must be OK In Cartesian x-space Γ = 0, so one has simply ∂cTab = Tab,c = Tab;c = RαaRβbRγc [(∂'γT'αβ) – Γ'nαγT'nβ – Γ'nβγT'αn ] must be OK If only the first term were present on the right, ∂cTab would be transforming as a rank-3 tensor, so the presence of the other terms reminds us that ∂cTab is not a rank-3 tensor. In Cartesian space, up and down indices are the same, so ∂cTab = Tab,c = Tab;c = RαaRβbRγc [T'αβ,γ – Γ'nαγT'nβ – Γ'nβγT'αn ] danger #1 How about this different path ∂cTab = Tab,c = Tab;c = RαaRβbRγc [T'αβ,γ – Γ'nαγT'nβ – Γ'nβγT'αn ] [divT]a ≡ ∂bTab = Tab,b = Tab;b = RαaRβbRγb [T'αβ,γ – Γ'nαγT'nβ – Γ'nβγT'αn ] But this is just the same as below. This line above says that Tab;b and then [divT]a = ∂bTab = RαaRβbRγb [T'αβ,γ – Γ'nαγT'nβ – Γ'nβγT'αn ] Maybe this is going to work after all. For this vector-like object write divT = Σe [divT]'e ee [divT]'e = Ref[divT]f = Ref ∂bTfb but not a vector!!! but OK anyway!!!! = Ref RαfRβbRγb [T'αβ,γ – Γ'nαγT'nβ – Γ'nβγT'αn ] = g'eα g'βγ [T'αβ,γ – Γ'nαγT'nβ – Γ'nβγT'αn ] Now use Γ'cab = – Rbi (∂'aRci) Γ'nαγ = – Rγi (∂'αRni) Γ'nβγ = – Rγi (∂'βRni) Then [divT]'e = g'eα g'βγ[T'αβ,γ + Rγi(∂'αRni)T'nβ + Rγi(∂'βRni)T'αn ] = g'eα g'βγ[(∂'γT'αβ) + Rγi(∂'αRni)T'nβ + Rγi(∂'βRni)T'αn ] = g'eα g'βγ[Rγi(∂'αRni)T'nβ + Rγi(∂'βRni)T'αn + (∂'γT'αβ) ] = [g'eα g'βγ Rγi(∂'αRni)T'nβ + g'eα g'βγ Rγi(∂'βRni)T'αn + g'eα g'βγ (∂'γT'αβ) ] = [g'ea Rci(∂'aRbi)T'bc + g'eb Raj(∂'aRcj)T'bc + g'eb g'ca (∂'aT'bc) ] error old 2nd term old 3rd term divT = Σe [divT]'e ee Plan A. According to Appendix F (h) the object Tij;k transforms as a rank-3 tensor, given that Tij is a rank-2 tensor. Contracting the j and k indices gives object Tij;j which transforms as a vector. This then is the starting point Ai ≡ Tij;j But I don't now how Tij;j expans in terms of Γ's. And this relies on unproven claims in my other appendix, so I don't like it Plan B. Do we know that the rows of a rank-2 tensor are vectors? T'ij = Rii'Rjj'Ti'j' [r'(i)]j = Rjj' [Rii' Ti'j'] = Rjj' Rii' [r(i')]j' The answer is NO. The other rows are "mixed in". So I guess A ≡ divT is then NOT a vector! And that agrees with Ai = ∂jTij because ∂kTij is not a rank-3 tensor. But I know I can expand a vector-like object from Appendix E. A = ΣiA'i ei where A'j = RjiAi but A' are not the x'-space components of A. Where does this lead us? A'j = RjiAi = Rji∂jTij But even this is not helpful . Plan C. Write Tab;c ≡ Tab,c – ΓnacTnb – ΓnbcTan = Tab,c in x-space T'ab;c ≡ T'ab,c – Γ'nacT'nb – Γ'nbcT'an T'ab;c = RaαRbβRcγTαβ;γ Tab;c = SaαSbβScγT'αβ;γ = SaαSbβScγ[T'αβ,γ – Γ'nαγT'nβ – Γ'nβγT'αn ] Now use this Γ'cij = Rck(∂'jRik) Γ'nαγ = Rnk(∂'γRαk) Γ'nβγ = Rnk(∂'γRβk) Insert the last two lines to get Tab,c = Tab;c = SaαSbβScγ[T'αβ,γ – Γ'nαγT'nβ – Γ'nβγT'αn ] = SaαSbβScγ[T'αβ,γ – Γ'nαγT'nβ – Γ'nβγT'αn ] = SaαSbβScγ[T'αβ,γ – Rnk(∂'γRαk)T'nβ – Rnk(∂'γRβk)T'αn ] To summarize Tab,c = SaαSbβScγ[T'αβ,γ – Rnk(∂'γRαk)T'nβ – Rnk(∂'γRβk)T'αn ] = [SaαSbβScγ T'αβ,γ – SaαSbβScγ Rnk(∂'γRαk)T'nβ – SaαSbβScγ Rnk(∂'γRβk)T'αn ] and this must be the same as (∂kTij) = (Rak∂'a)(RbiRcjT'bc) = [Rak (∂'aRbi)RcjT'bc + Rak Rbi(∂'aRcj)T'bc + Rak RbiRcj (∂'aT'bc) ] The last terms match but tilts are wrong on the inside terms. Changing them would certainly give the right answer. But my other method seems a lot simpler! Now expand in a triple as per Appendix E T = Σijk T'ij;k (eiejek) So that is pretty quick. But how to you do the contraction? ********************************************************* Plan D. Let's use our new information which says Tab;c ≡ (∂cTab) – ΓnacTnb – ΓnbcTan n Tab;c = (∂cTab) + Γacn Tnb – ΓnbcTan Now look at the second line in our two spaces Tab;c = (∂cTab) + Γacn Tnb – ΓnbcTan = (∂cTab) // x-space Cartesian T'ab;c = (∂'cT'ab) + Γ'acn T'nb – Γ'nbcT'an // x'-space Then write T'ab;c = RaαRbβRcγTαβ;γ Invert to get Tαβ;γ = RaαRbβRcγ T'ab;c = RaαRbβRcγ[(∂'cT'ab) + Γ'acn T'nb – Γ'nbcT'an] Now contract so γ = α: ∂αTαβ = Tαβ;σ = RaαRbβRcα[(∂'cT'ab) + Γ'acn T'nb – Γ'nbcT'an] But this is still not quite what I want. I need to go off and do a general index raising rule, then come back.