christoffel v1

Christoffel Notes PhL 2.29.12 (3.8.12) This file documents how I learned about the parts of this subject, all of which I have written up in Appendix F and some in Appendix G. I will keep it just for fun. So red comments strewn below. Here is what I have to show: ξ S,R S',R' J J' [∂'aV'b – Γ 'kab V'k] = Rad Rbc [∂dVc – Γkcd Vk] where Γcab = ½ gcd( ∂agbd + ∂bgad – ∂dgab ) Γ'cab = ½ g'cd( ∂'ag'bd + ∂'bg'ad – ∂'dg'ab ) This looks like a lot of work. I turn then to Weinberg page 100. I go into my Picture D This was my first attack on this subject, it now appears in Appendix F (f) Problem 1: Define Γ' and Γ" and then find a relationship between them. and I use Weinberg's definition which makes the lower index symmetry manifestly true, Γλμν ≡ (∂xλ/∂ξα) ∂2ξα/∂xμ∂xν Γcab ≡ (∂xc/∂ξn) ∂2ξn/∂xa∂xb and I will now change so that ξ → x and x→x' so that] Γ'cab ≡ (∂x'c/∂xn) ∂2xn/∂x'a∂x'b = R'cn ∂'a (∂xn/∂x'b) = R'cn(∂'aS'nb) Now repeat this for going to x"-space Γ"cab = R"cn(∂"aS"nb) where as in my Section 5 we have R = R" S' => Rij = R"ikS'kj => R" = R R' S = R' S" Sij = R'ikS"kj => S" = S' S So here is we have so far: Γ'cab = R'cn (∂'aS'nb) => Γ'dki = R'dn (∂'kS'ni) Γ"cab = R"cn(∂"aS"nb) dx" = R dx' = R"S' dx' => ∂" = S ∂' => ∂"a = Rak∂'k Is there some way to relate Γ' to Γ" ? If you think of R' and R" as completely independent, then there can be no connection between them. But maybe we should think of R as fixed, and then R" = R R' and S" = S' S makes a connection between the " and ' worlds. Then the second line above would be Aside: I have a new source http://www.iopb.res.in/~phatak/relativity/node80.html now which says If I translate this second term to my notation I do x'→x" and x→x' so that second term is (∂'ρx"λ)∂"ν∂"μ x'ρ = Rλρ (∂"νSρμ) → Rci (∂"bSia) Γ"cab = R"cn(∂"aS"nb) = Rcd R'dn (∂"a {S'niSib}) = Rcd R'dn (∂"aS'ni)Sib + Rcd R'dn S'ni (∂"a Sib) Now only replace ∂"a = Rak∂'k in the first term = Rcd R'dn Rak( ∂'k S'ni)Sib + Rcd R'dn S'ni (∂"a Sib) = Rcd Rak Sib R'dn ( ∂'k S'ni) + Rcd R'dn S'ni (∂"a Sib) // reorder in 1st term = Rcd Rak Rbi Γ'dki + Rcd (R'dn S'ni) (∂"a Sib) = Rcd Rak Rbi Γ'dki + Rcd δdi (∂"a Sib) = Rcd Rak Rbi Γ'dki + Rci (∂"a Sib) and NOW we agree with our source, the first connection has been made! Thus Γ"cab = Rcd Rak Rbi Γ'dki + Rci (∂"aRbi) and finally I have a relationship between the two Γ objects! Only after doing all this work can we then convert the result to Picture A context which means x"→x' and x'→ x and R and S stay the same. And this gives result Γ'cab = Rcd Rak Rbi Γdki + Rci (∂'aRbi) // agrees Weinberg 4.5.2 I think In this context, we replace our x-space G frame with ξ and then we get Γcab = R'cn (∂aS'nb) Γ'cab = R"cn(∂'aS"nb) Wow, this is tricky stuff. These results above play no role afterwards! I think I like this idea of reusing my Picture D context then converting as shown above. And this was my first attack on this subject. Instead of proving this in tensor doc, I now prove the general case in Appendix F (g) for a covariant tensor of any rank. But here is a proof of this particular case. Problem 2: Show that [∂'aV'b – Γ 'kab V'k] = Rad Rbc [∂dVc – Γkcd Vk] NOW maybe I can show what I wanted to show which was this: [∂'aV'b – Γ 'kab V'k] = Rad Rbc [∂dVc – Γkcd Vk] I install on the left side LHS = [∂'aV'b – Γ 'cab V'c] = [∂'aV'b – { Rcd Rak Rbi Γdki + Rci (∂'aRbi) }V'c] = ∂'aV'b – Rcd Rak Rbi Γdki V'c – Rci (∂'aRbi) V'c Now we have to install a PL expansion for either ∂dVc or ∂'aV'b . I will do ∂'aV'b since it is sitting in my Business section and says (∂'aV'b) = (Rad∂d) (RbcVc) = Rad Rbc (∂dVc) + Rad(∂d Rbc)Vc So then our new LHS reads LHS = Rad Rbc (∂dVc) + Rad(∂d Rbc)Vc – Rcd Rak Rbi Γdki V'c – Rci (∂'aRbi) V'c Right off the bat then we see that the ∂dVc terms match on our two sides, so all we have to show is that this is true Rad(∂d Rbc)Vc – Rcd Rak Rbi Γdki V'c – Rci (∂'aRbi) V'c = – Rad Rbc Γkcd Vk Now replace V'c = RceVe to get Rad(∂d Rbc)Vc – Rcd Rak Rbi Γdki RceVe – Rci (∂'aRbi) RceVe = – Rad Rbc Γkcd Vk Rad(∂d Rbc)Vc – (Rcd Rce)Rak Rbi Γdki Ve – (RciRce) (∂'aRbi) Ve = – Rad Rbc Γkcd Vk Rad(∂d Rbc)Vc – δdeRak Rbi Γdki Ve – δie (∂'aRbi) Ve = – Rad Rbc Γkcd Vk Rad(∂d Rbc)Vc – Rak Rbi Γeki Ve – (∂'aRbe) Ve = – Rad Rbc Γkcd Vk Now try to get all Vc summation variables to be the same. Rad(∂d Rbe)Ve – Rak Rbi Γeki Ve – (∂'aRbe) Ve = – Rad Rbc Γecd Ve Rad(∂d Rbe)Ve – Rak Rbi Γeki Ve – (∂'aRbe) Ve = – Rak Rbi Γeik Ve Now if we know Γ has symmetry, we get cancellation of our two Γ terms!! So we then still have to show that Rad(∂d Rbe) = (∂'aRbe) We know that ∂'a = Rak∂k so we have to show that Rad(∂d Rbe) = (Rak∂k Rbe) Rak(∂k Rbe) = (Rak∂k Rbe) QED. This is the first time ever (in my current era after writing old tensor doc). I see a Christoffel Appendix coming soon! Now for the first time I am realizing the connection to the object (∂aen), and this now appears in Appendix F (a) as the very starting point. Review of a web site. Try http://www.mth.uct.ac.za/omei/gr/chap6/node2.html Write V = Vnen ∂aV = (∂aVn)en + Vn(∂aen) Now write ∂βen as a linear combination of the en, (∂aen) = Σm Γmna em So this is the definition from this source! Very good. This is also the missing piece in my Lai discussion, I could never figure out this derivative of en. Now apply eb to both sides, eb (∂aen) = Σm Γmna em eb = Σm Γmna δmb = Γbna So I then get this definition Γbna = eb (∂aen) or Γcab = ec (∂aeb) = ec (∂bea) and I can write this as Γcab = (ec)i (∂bea)i = (ec)i ∂b(ea)i = Rci(∂bSia) So in terms of my Picture D, what we really have is this Γ'cab = R'cn(∂'aS'nb) = 'ec (∂'b 'ea) Γ"cab = R"cn(∂"aS"nb) = "ec (∂"b "ea) Note that in our usual Picture A context, we refer to 'ea as just ea, a tangent base vector. Now how can we translate the above two lines to Picture A? Well, S' → S etc, so we have Γcab = Rcn(∂aSnb) = ec (∂b ea) Γ"cab = R"cn(∂"aS"nb) = "ec (∂"b "ea) The omei site seems to associate this Γ object with S and R so I could apply it to Picture A It would seem to be a property of the transformation, not of the individual spaces. Γbna = Rbi(∂aSin) except the deriviative is with respect to the x-space, not the x'-space coordinate. So the big quesion now is: what is the object Γ'bna ??? Hold on that for a moment and continue the above equation sequence ∂aV = (∂aVn)en + Vn(∂aen) (∂aen) = Σm Γmna em => ∂aV = (∂aVn)en + Vn Γmna em ∂aV = (∂aVk)ek + Vn Γkna ek ∂aV = [ (∂aVk) + Vn Γkna] ek ∂aV = Vk;a ek Vk;a ≡ (∂aVk) + Vn Γkna At this point the author suddenly talks about some a vector which he does not define. I guess it is just an arbitrary vector, that is fine. I think it is the dual vector, because later I see n em = δnm and I know it is unique. Good, I can then continue. This author does not show the dots of dot products. So n = en and his equation (2) is just what I have above Γbna = eb (∂aen) Now comes a LOT of algebra and now the typos are appearing. He should have primes on the V's on the left or at least some way to say it is in x'-space. But this is an excellent start. He never shows a Γ' object. This author drops the ball on the very first line after his equation (20). Maybe I can construct it. I cannot come up with an object V'k;a without an object Γ'kna. Let's look for another source, though this one gave me some help. ___________________________________________________________________________ Here I am for the first time making the connection between my (v) work and the Γ object. This all appears now in my new Appendix G on this subject. Now maybe I can just go to picture A Then I would rewrite that as Γ'cab(x') ≡ Rcn(∂'aSnb(x')) = Rcn(∂'bSna(x')) and you notice that the ab symmetry gets less obvious on these RHS objects! I can isolate the object on the right by applying Rcn' to both sides and using Rcn' Rcn = δn'n and thus (after n'→n) Rcn Γ'cab(x') = (∂'bSna) Now let's start off doing things as I did in my v section, ∂cvd = Σij (Ric∂'i)( Rjdv'j) = Σij Ric [(∂'iRjd) v'j + Rjd (∂'iv'j)] Now let's raise and lower a few things ∂cvd = Σij (Ric∂'i)( Rjdv'j) = Σij Ric [(∂'iRjd) v'j + Rjd (∂'iv'j)] = Σij Ric [(∂'iRjd) v'j + Rjd (∂'iv'j)] = Σij Ric [(∂'iSdj) v'j + Rjd (∂'iv'j)] Now consider Rcn Γ'cab(x') = (∂'bSna) Red Γ'eji(x') = (∂'iSdj) Installing this gives ∂cvd = Σij Ric [Red Γ'eij(x') v'j + Rjd (∂'iv'j)] and I have successfully gotten rid of my R derivative in favor of the Γ object. More fiddling 3.2.12. The Γ thing I remembered was not curl , it was v ! From here http://en.wikipedia.org/wiki/Curvilinear_coordinatesσ OK I think I see. Write (v)ij = ∂jvi = vi;j in Cartesian x-space Then to expand this I would write v = Σij (v)'ij ei ej where (v)'ij = R R v = v'i;j = [ (∂'jv'i) – Γcij v'c ] Then I would write v = Σij [ (∂'jv'i) – Γ'cij v'c ] ei ej This is in my Picture A . If I go to picture C or C1 then this would say v = Σij [ (∂jvi) – Γcij vc ] ei ej v = Σik [ (∂kvi) – Γcik vc ] ei ek and this DOES AGREE with the wiki page as I quote it justy above. So here is a possible formula one could use (v)ab = Σik [ (∂kvi) – Γcik vc ] (ei)a (ek)b = Σik [ (∂kvi) – Γcik vc ]RiaRkb and this is perhaps a nicer version of my formula, but you have to have Γcik sitting somewhere! That might be a pretty messy object if you use Γdab = (1/2) gdc [ ∂agbc + ∂bgca – ∂cgab] gab = RakRbk = (1/2) RdnRcn [∂a(RbkRck) + ∂b(RckRak) – ∂c(RakRbk) ] = (1/2) RdnRcn * [Rbk(∂aRck) + Rck(∂aRbk) + Rck(∂bRak) + Rak(∂bRck) – Rak(∂cRbk) – Rbk(∂cRak) ] 1 2 3 4 5 6 Now I know that (∂aRbn) = (∂bRan) . This makes 4 and 5 cancel. It makes 1 and 6 cancel. Then 2 and 3 add (I thought I just did this somewhere) Γdab = RdnRcn Rck(∂aRbk) = Rdnδnk(∂aRbk) = Rdn (∂aRbn) Γcab ≡ = Rcn(∂aRbn) Γcik ≡ = Rcn(∂iRkn) which of course is my original definition, duh. So this is what I expect to get (v)ab = Σik [ (∂kvi) – Γcik vc ] (ei)a (ek)b = Σij [ (∂'jv'i) – Rcn(∂'iRjn)v'c ] (ei)a (ej)b (v) = Σij [ (∂'jv'i) – Rcn(∂'iRjn)v'c ] ei ejT The result I got in the Appendix was this: (v) = Σij { Σab g'ja [Σd Rid (∂'aRbd) v'b + g'ib (∂'av'b)] } ei ejT = Σij { Σab δja [Σd Rid (∂'aRbd) v'b + δib (∂'av'b)] } ei ejT = { [Σd Rid (∂'jRbd) v'b + (∂'jv'i)] } ei ejT = [(∂'jv'i) + Σd Rid (∂'jRbd) v'b] ei ejT OK, pause for now. Something is wrong with the d tilt here. I think my v appendix should get rewritten and placed after the Christoffel one! But suppose I use my reversal theorem (∂'aRdn) = – Ren Rdm (∂'aRem) (∂'jRbd) = – Red Rbm (∂'jRem) My result then becomes = [(∂'jv'i) – Σd Rid Red Rbm (∂'jRem)v'b] ei ejT = [(∂'jv'i) – Σd Rid Red Rcn (∂'jRen)v'c] ei ejT If my d were up I would get = [(∂'jv'i) – Σd Rcn (∂'jRin)v'c] ei ejT Then I can do (∂'jRin) = (∂'iRjn) and we agree. So I have found an actual ERROR in my appendix thanks to doing the calculation another way. TBC. Idea: Next to a context picutre in the Christoffel appendix, put equations next to pictures like Γcik ≡ = Rcn(∂iRkn) These two identifies now appear in my Appendix F (c). Verification of the Γ theorems: http://www.mth.uct.ac.za/omei/gr/chap6/node3.html And here are the pair of them constrained-geodesic.com/manifold/manifold_pp_48_64.pdf