M&M chap5 support

M&M Chapters 4 & 5, support items PhL 9.1.08 1. Brute Force Proof that Vb,a is a tensor 1 Comments on Proof of Covariance of Vb,a 1 2. Proof that jac = 10 1. Brute Force Proof that Vb,a is a tensor I took these raw notes out of the main Chapter notes file. I finally got the proof done, it was not easy. I copied from the main file to here my summary of these raw notes, then come the raw notes: Comments on Proof of Covariance of Vb,a I did a brute force proof on my own to show that Vb,a transforms like a tensor (see M&M Chapter 5 "support" doc). The proof was very messy, but I just wanted to try it without looking at the M&M solution which is highly tuned and delicate. I learned many useful things along the way. Here is an outline of my proof; (1) By requiring the tensor transformation, I was quickly led to the need for the following to be true: {ab,c}'V'c = Tad(∂d Tbc)Vc – Tad Tbc{dc,e}Ve (2) I was disturbed by the fact that Tad(∂d Tbc)Vc did not appear symmetric in a,b (while the other two terms are), so this led me off on a digression where I found that I could rewrite this term in a manifestly symmetric form: Tad(∂d Tbc)Vc = –Tas Tbg Tfc Sfsg Vc where Sfsg = (∂sTfg) = = sym(s,g) Showing this required me to derive and then use a little "fact #2" based on ∂s(Tab Tcb) = 0, namely, (∂s Tce) = – Tae Tcb (∂sTab) = – Tae Tcb Sasb "Fact 2" This "fact" showed how to relate ∂s T** to that of ∂s T**. So the problem then became to show this to be true: {ab,c}'V'c = –Tas Tbg Tfc Sfsg Vc + Tad Tbc{dc,e}Ve LHS = RHS2 + RHS1 (3) I then expanded the LHS and ended up with two big terms LHS1 + LHS2 . LHS1 contained the three terms having factors of the form (∂*g**). I then compared this to RHS1, LHS1 = ½geh * Ve { Taa'Tbb' (∂a'gb'h) + Tba'Tab' (∂a'gb'h) –Tab'Tbc'(∂hgb'c') } RHS1 = ½ Tad Tbc ges ( ∂dgcs + ∂cgds – ∂sgdc )Ve and, by renaming indices, it was easy to show these terms were the same. So what remained to show was that LHS2 = RHS2. At this point then I had: RHS2 = –Tas Tbg Tfc Sfsg Vc LHS2 = + ½Tdh geh * Ve { Taa'Tbb'(∂a'Tdc')gb'c'+ Tba'Tab'(∂a'Tdc')gb'c' } + ½geh * Ve * { Taa'(∂a'Tbb') gb'h + Tba'(∂a'Tab') gb'h – (∂hTab')Tbc'gb'c' – Tab'(∂hTbc')gb'c'} where it took quite a bit of routine but detailed work to get this form for LHS2. There are 6 terms here, and each pair of terms shows the a,b symmetry. BUT, it was hard for me to see how this was going to equal RHS2 because I could see no cancellations of terms. (4) I then processed LHS2 into the following form, exposing the (∂sTdc) type factors to which I knew I could apply Fact #2. This gave: LHS2 = + ½Tdh Vh { TasTbc + TbsTac }(∂sTdc) (∂sTdc) = – Tfc Tdg Sfsg + ½ [Vc Tas – Vs Tac ] (∂sTbc) (∂sTbc) = – Tfc Tbg Sfsg + ½ [ Vc Tbs – Vs Tbc ] (∂sTac) (∂sTac) = – Tfc Tag Sfsg (5) I then installed Fact #2 three times on the right, and was able the express LHS2 as the sum of 6 terms: 1 = – ½Vg TasTbc Tfc Sfsg Tfc Sfsg = = not simple 2 = – ½Vg TbsTac Tfc Sfsg 3 = – ½ Vc Tas Tbg Tfc Sfsg 4 = + ½ Vs Tac Tbg Tfc Sfsg 5 = – ½ Vc Tbs Tag Tfc Sfsg 6 = + ½ Vs Tbc Tag Tfc Sfsg (6) At this point, I made use of the symmetry of Sfsg (main point!) to rewrite terms 4,5 and 6, then getting 1 = – ½Vg TasTbc Tfc Sfsg 2 = – ½Vg TbsTac Tfc Sfsg 3 = – ½ Vc Tas Tbg Tfc Sfsg 4 = + ½ Vg Tac Tbs Tfc Sfsg 5 = – ½ Vc Tbg Tas Tfc Sfsg 6 = + ½ Vg Tbc Tas Tfc Sfsg Now it became clear that terms 1 and 6 cancel, and also terms 2 and 4 cancel, and terms 3 and 5 are equal, so our total result is this: LHS2 = 3+5 = – Vc Tbg Tas Tfc Sfsg = RHS2 ! and this concluded the proof. (7) Among the "lessons learned" by doing the above, I realized that you cannot willy-nilly raise and lower indices on tensors that are acted upon by a derivative. For example, consider this simple example Psc = Ad (∂sTdc) ≠ Ad (∂sTdc) // where I am doing a "tilt change" In fact, calculation shows that Psc = Ad (∂sTdc) = Ad (∂sgdeTec) = Ad gde (∂sTec) + Ad(∂sgde) Tec = Ae (∂sTec) + Ad(∂sgde) Tec = Ad (∂sTdc) + extra term. Also, note that, just as ∂sVd is not a tensor ( our main proof!) , so also (∂sTdc) is not a tensor. Thus, even though T and A are tensors, Psc is not a tensor. Another more obvious example is this. Ac = (∂sgsc) ≠> Ac = (∂sgsc) If it did, we would conclude from the right equation that Ac = 0 since ∂s(δsc) = 0 and we would then think that Ac = gcdAd = 0, which is not true. Also, the fragment Ac is not a tensor! __________________________________________________________________________________ Mainline raw proof of covariance. Now suppose I just try my own brute force on this, where I copy paste and edit above for the first term: V'b,a = [( ∂'aV'b) – {ab,c}'V'c ] = [(Tad ∂d(Tbc Vc)) – {ab,c}'V'c ] = Tad Tbc ∂dVc + Tad(∂d Tbc)Vc – {ab,c}'V'c We want this to come out being the following Tad Tbc Vd,c = Tad Tbc [ ∂dVc – {dc,e}Ve ] = Tad Tbc ∂dVc – Tad Tbc {dc,e}Ve The first terms are the same, so we need to show this: Tad(∂d Tbc)Vc – {ab,c}'V'c = – Tad Tbc{dc,e}Ve Write the above as Tad(∂d Tbc)Vc = {ab,c}'V'c – Tad Tbc{dc,e}Ve ** STOP!!!!! Right here we have a "symmetry problem". I claim the RHS is symmetrical under a↔b, but the LHS does not seem to have this symmetry, so something is wrong right at this point, I have made some kind of mistake? The last term is symmetric as follows: Tad Tbc{dc,e}Ve = Tac Tbd{cd,e}Ve = Tbd Tac {dc,e}Ve So there is no point in my continuing this calculation until I resolve this confusion. Let's look more closely at the LHS of ** to see if it might be symmetric in a hidden way! Omit the Vc Qab = Tad(∂d Tbc) At the end of the previous section, we showed that (∂s Tce) = – Tge Tcf Sgsf or (∂d Tbc) = – Tgc Tbf Sgdf Thus we have that Qab = Tad(∂d Tbc) = – Tad Tgc Tbf Sgdf = – Tgc Tad Tbf Sgdf and now, finally, we see that in fact Qab really is symmetric! So we can continue our calculation. We had Tad(∂d Tbc)Vc = {ab,c}'V'c – Tad Tbc{dc,e}Ve which we rewrite as {ab,c}'V'c = Tad(∂d Tbc)Vc + Tad Tbc{dc,e}Ve = Qab Vc + Tad Tbc{dc,e}Ve or the following, where all terms are manifestly symmetric in a,b : {ab,c}'V'c = – Tgc Tad Tbf Sgdf Vc + Tad Tbc{dc,e}Ve LHS = RHS To see if this equation is true, write out the LHS: LHS = {ab,c}'V'c = g'cd [ab,d]' V'c = ½ g'cd ( ∂'ag'bd + ∂'bg'ad – ∂'dg'ab ) V'c = ½TcfTdh gfh * { Taa'∂a'(Tbb'Tdc'gb'c') + Tba'∂a'(Tab'Tdc'gb'c') – Tda'(∂a'Tab'Tbc'gb'c') } TceVe which is an amazing mess. We can contract the first and last T at least to get something a little simpler: LHS = ½Tdh geh * { Taa'∂a'(Tbb'Tdc'gb'c') + Tba'∂a'(Tab'Tdc'gb'c') – Tda'(∂a'Tab'Tbc'gb'c') } Ve Each of the derivatives creates 3 terms so we have a total of 9 terms. I can always get the rightmost two objects by permutations, so write the above as LHS = ½Tdh geh * { Taa'∂a'(Tbb'Tdc'gb'c') + P(abd→bad) – P(abd→dab } Ve // OK to here 10 AM So expand this first term in {} to get Taa'∂a'(Tbb'Tdc'gb'c') = Taa'(∂a'Tbb')Tdc'gb'c' + Taa'Tbb'(∂a'Tdc')gb'c') + Taa'Tbb'Tdc'(∂a'gb'c') So we now have LHS = ½Tdh geh * Ve { Taa'(∂a'Tbb')Tdc'gb'c' + Taa'Tbb'(∂a'Tdcc')gb'c') + Taa'Tbb'Tdc'(∂a'gb'c') + P(abd→bad) – P(abd→dab } Now let's write it all out: LHS = ½Tdh geh * Ve { Taa'(∂a'Tbb')Tdc'gb'c' + Taa'Tbb'(∂a'Tdc')gb'c') + Taa'Tbb'Tdc'(∂a'gb'c') + Tba'(∂a'Tab')Tdc'gb'c' + Tba'Tab'(∂a'Tdc')gb'c') + Tba'Tab'Tdc'(∂a'gb'c') – Tda'(∂a'Tab')Tbc'gb'c' – Tda'Tab'(∂a'Tbc')gb'c') – Tda'Tab'Tbc'(∂a'gb'c') } // OK to here 10 AM Now, we can contract the opening Tdh in 7 of the 9 terms to get LHS = ½Tdh geh * Ve { Taa'(∂a'Tbb')Tdc'gb'c' + Taa'Tbb'(∂a'Tdc')gb'c') + Taa'Tbb'Tdc'(∂a'gb'c') + Tba'(∂a'Tab')Tdc'gb'c' + Tba'Tab'(∂a'Tdc')gb'c') + Tba'Tab'Tdc'(∂a'gb'c') – Tda'(∂a'Tab')Tbc'gb'c' – Tda'Tab'(∂a'Tbc')gb'c') – Tda'Tab'Tbc'(∂a'gb'c') } = + ½Tdh geh * Ve { Taa'Tbb'(∂a'Tdc')gb'c'+ Tba'Tab'(∂a'Tdc')gb'c' } + ½geh * Ve { Taa'(∂a'Tbb') gb'h + Taa'Tbb' (∂a'gb'h) + Tba'(∂a'Tab') gb'h + Tba'Tab' (∂a'gb'h) – (∂hTab')Tbc'gb'c' –Tab'Tbc'(∂hgb'c') – Tab'(∂hTbc')gb'c'} // OK to here 10 AM Now just for the moment, let's keep only the (∂a'gb'h) type terms and call them LHS1 LHS1 = ½geh * Ve { Taa'Tbb' (∂a'gb'h) + Tba'Tab' (∂a'gb'h) – Tab'Tbc'(∂hgb'c') } Meanwhile, here is where we are trying to get to: RHS = – Tgc Tad Tbf Sgdf Vc + Tad Tbc{dc,e}Ve Now go off and do these things {ab,c} ≡ gcs [ab,s] => {dc,e} ≡ ges [dc,s] [ab,c] ≡ ½ ( ∂agbc + ∂bgac – ∂cgab ) => [dc,s] ≡ ½ ( ∂dgcs + ∂cgds – ∂sgdc ) => {dc,e} = ½ ges ( ∂dgcs + ∂cgds – ∂sgdc ) // OK to here 10 AM so we have RHS = – Tgc Tad Tbf Sgdf Vc + ½ Tad Tbc ges ( ∂dgcs + ∂cgds – ∂sgdc )Ve If this is to work out, it is pretty clear that the ∂dgcs terms have to match, so we need RHS1 = ½ Tad Tbc ges ( ∂dgcs + ∂cgds – ∂sgdc )Ve LHS1 = ½geh * { Taa'Tbb' (∂a'gb'h) + Tba'Tab' (∂a'gb'h) –Tab'Tbc'(∂hgb'c') } Ve At least we have the right number of factors everywhere. We now need to do some index renaming: LHS1 = ½geh * { TadTbc (∂dgch) + TbcTad (∂cgdh) –TadTbc(∂hgdc) } Ve which I just did live. Now all three T factors are the same so pull them out to get (and h→s) LHS1 = ½TadTbc ges [(∂dgcs) + (∂cgds) – (∂sgdc)] Ve RHS1 = +½ Tad Tbc ges ( ∂dgcs + ∂cgds – ∂sgdc )Ve and we have a match on this set of terms. // OK to here 10 AM __________________________________________________________________________ It then remains to show that all the other terms match. On the LHS we have these terms: LHS2 = + ½Tdh geh * Ve { Taa'Tbb'(∂a'Tdc')gb'c'+ Tba'Tab'(∂a'Tdc')gb'c' } + ½geh * Ve * { Taa'(∂a'Tbb') gb'h + Tba'(∂a'Tab') gb'h – (∂hTab')Tbc'gb'c' – Tab'(∂hTbc')gb'c'} and on the RHS we have these terms RHS2 = – Tgc Tad Tbf Sgdf Vc // OK to here 10 AM Now some live index renaming on the LHS terms: + ½Tdh geh * Ve { TasTbt(∂sTdc)gtc+ TbtTas(∂tTdc)gsc } + ½geh * Ve { Tas(∂sTbc) gch + Tbt(∂tTac) gch – (∂hTac)Tbtgct – Tas(∂hTbc)gsc} // OK to here 10 AM Separate out the last terms like this: + ½Tdh geh * Ve { TasTbt(∂sTdc)gtc+ TbtTas(∂tTdc)gsc } + ½geh * Ve { Tas(∂sTbc) gch + Tbt(∂tTac) gch } – ½geh * Ve { (∂hTac)Tbtgct + Tas(∂hTbc)gsc} Now rename summed indices to try and reduce the different forms of (∂sTdc): + ½Tdh geh * Ve { TasTbt(∂sTdc)gtc+ TbsTat(∂sTdc)gtc } + ½geh * Ve {Tas(∂sTbc) gch + Tbs(∂sTac) gch } – ½ges * Ve { (∂sTac)Tbtgct + Tat(∂sTbc)gtc} where now we have only three different "forms" like (∂sTdc). Group by these forms: + ½Tdh geh * Ve { TasTbt + TbsTat }(∂sTdc)gtc + ½geh * Ve {Tas(∂sTbc) gch – ½ges * Ve Tat(∂sTbc)gtc + ½geh * Ve Tbs(∂sTac) gch – ½ges * Ve (∂sTac)Tbtgct And write again factoring from last lines: + ½Tdh geh Ve { TasTbt + TbsTat }(∂sTdc)gtc + [ ½geh Ve Tasgch – ½ges Ve Tatgtc ] (∂sTbc) +[ ½geh Ve Tbsgch – ½ges Ve Tbtgct ] (∂sTac) Now use up as many gab as possible just to get less symbols showing: + ½Tdh Vh { TasTbc + TbsTac }(∂sTdc) + ½ [Vc Tas – Vs Tac ] (∂sTbc) + ½ [ Vc Tbs – Vs Tbc ] (∂sTac) and now all the gab are gone. Now, finally, we want to use our Fact 2 rule: (∂s Tce) = – Tae Tcb Sasb We have to install this in 3 locations in the above. (∂sTdc) = – Tfc Tdg Sfsg (∂sTbc) = – Tfc Tbg Sfsg (∂sTac) = – Tfc Tag Sfsg We then have: – ½Tdh Vh { TasTbc + TbsTac } Tfc Tdg Sfsg – ½ [Vc Tas – Vs Tac ] Tfc Tbg Sfsg – ½ [ Vc Tbs – Vs Tbc ] Tfc Tag Sfsg Now in the first line, write Tdh Tdg = δhg and rewrite all – ½Vg { TasTbc + TbsTac } Tfc Sfsg = 1 + 2 – ½ [Vc Tas – Vs Tac ] Tfc Tbg Sfsg = 3 + 4 – ½ [ Vc Tbs – Vs Tbc ] Tfc Tag Sfsg = 5 + 6 Recall that RHS2 = – Vc Tas Tbg Tfc Sfsg is where we are trying to arrive. Write the terms out: 1 = – ½Vg TasTbc Tfc Sfsg 2 = – ½Vg TbsTac Tfc Sfsg 3 = – ½ Vc Tas Tbg Tfc Sfsg 4 = + ½ Vs Tac Tbg Tfc Sfsg 5 = – ½ Vc Tbs Tag Tfc Sfsg 6 = + ½ Vs Tbc Tag Tfc Sfsg Now finally use symmetry of S to change some terms swapping s and g 1 = – ½Vg TasTbc Tfc Sfsg 2 = – ½Vg TbsTac Tfc Sfsg 3 = – ½ Vc Tas Tbg Tfc Sfsg 4 = + ½ Vg Tac Tbs Tfc Sfsg 5 = – ½ Vc Tbg Tas Tfc Sfsg 6 = + ½ Vg Tbc Tas Tfc Sfsg Now we see that terms 1 and 6 cancel, and also terms 2 and 4 cancel, and terms 3 and 5 are equal, so our total result is this: 3+5 = – Vc Tbg Tas Tfc Sfsg RHS2 = – Vc Tas Tbg Tfc Sfsg so finally we have our proof!!! What a horrendous mess. 2. Proof that jac = Oddly, the Jacobian is not even mentioned in M&M's discussion of transformations. Here is some info from the Wolfram site: If I make the change x→ x' and y→ x in the above, we get jac = det( ) = det(Tab) and dV = jac dV' On page 195 we get from M&M that dV = dV' so it must be true that jac = which is what we shall prove here, hopefully. Meanwhile, we know these facts g = det(gij) gkm = Σi Tki Tmi = Σi (∂xi/∂ x'k) (∂xi/∂ x' m) // page 192 5-4a For use below write g1i = Σs T1s Tis The context here is that Tab transforms us from a Cartesian system xi with gij = δij to a new system x'i with some new g'ij. However, we shall refer to this simply as gij (instead of g'ij) and det = g (instead of g'). This new x' world is that of the curvilinear coordinates where q = x'. So we will really prove that jac2 = g. We can start writing things out: jac = εstuT1sT2tT3u = jac1 We can do this another way which is this (rows swap columns) jac = εijkTi1Tj2Tk3 = jac2 These will both come in very handy very soon! Meanwhile, we have to pick one way or the other for the g determinant g = εijk g1ig2jg3k = εijk T1s Tis T2t Tjt T3u Tku //6 sums and 6 T's which we can just reorder to get g =( εijk Tis Tjt Tku )T1s T2t T3u Now define a rank 3 tensor from the first terms of g Fstu ≡ εijk Tis Tjt Tku This Fstu is completely antisymmetric. For example, Fstu = εijk Tis Tjt Tku = εjik Tjs Tit Tku = εjik Tit Tjs Tku = – εijk Tit Tjs Tku = -Ftsu That fact means that Fstu is zero if any two indices are equal. Thus, although you might think this tensor had 3x3x3 = 27 terms, there are only 6 non-vanishing terms, and all those are equal to ± F123 according to εstu. Thus we have this pretty amazing and powerful claim Fstu = εstuF123 = εstu εijk Ti1 Tj2 Tk3 This allows us to rewrite g as follows: g = ( εijk Tis Tjt Tku )T1s T2t T3u = Fstu T1s T2t T3u = (εstu εijk Ti1 Tj2 Tk3) T1s T2t T3u = (εijk Ti1 Tj2 Tk3)( εstu T1s T2t T3u) = jac2 * jac1 = jac * jac = jac2 Very spiffy I would say, offhand. QED, Now add three more sums to j2 j2 = (εijkTisTjtTku) δs1 δt2 δs3 (εabcTasTbtTcu) Let's look at only the abc = 123 term in j2 " j2abc=123 = (εijkTisTjtTku) δs1 δt2 δs3 (T1sT2tT3u) = δs1 δt2 δs3 εijkTisTjtTku T1s T2t T3u g = εijk Tis TjtTku T1s T2t T3u Thus, our selected term from j2 matches one of 27 terms in the g sum, in terms of stu sums. Interesting. We need some symmetry help! Rewrite g = εijk Tis T1s Tjt T2t Tku T3u _________________________________ j2 = εijkT1iT2jT3k εabcT1aT2bT3c // 6 sums and 6 T's g = εijk T1s T2t T3u Tis Tjt Tku j2 = εijk εabc T1i T2j T3k T1a T2b T3c Let's just "look at" the ijk = 123 term in g g = ε123 T1s T2t T3u T1s T2t T3u j2 = ε123 εabc T1i T2j T3k T1a T2b T3c