Paradox 4_30_15

Paradox of the Day 4_30_15 PhL 4.30.15 1. On the one hand, I compute g;j as follows: (g has a weight of -2) I call upon my own results for the covariant derivative of a scalar density of weight W, B;α = ∂αB + W Γκκα B (F.9.23) Γaan = (1/2) gad ∂ngad = (1/2)(1/g)∂ng = (1/) ∂n() . (F.4.2) Comment: In the sum for B;α , neither term separately is a vector density of weight W, but the sum is. I then find that g;j = ∂jg + W Γκκα g = ∂jg - 2 Γκκα g = ∂jg - 2(1/2)(1/g)∂jg g = ∂jg - (1/g)∂jg g = 0 . We get a cancelling minus sign here because W is negative! 2. On the other hand, I call upon these results, det(Mij) = (1/g) (1/N!) εab..x εAB...X MAa MBb ...MXx = true scalar (D.12.15) det(Mij) = g det(Mij) // scalar density of weight -2 g = det(gij) (D.12.18) which I apply to gij det(gij) = (1/g) (1/N!) εab..x εAB...X gAa gBb ...gXx = (1/g) (1/N!) εab..x εab..x = (1/g) (1/N!) εab..x g εab..x = 1 which certainly makes sense since det(gij) = det(1) = 1. Then we have det(gij) = g det(gij) = g which also makes sense. Just running a few checks on the above equations. Now write directly det(Mij) = g (1/g) (1/N!) εab..x εAB...X MAa MBb ...MXx = (1/N!) εab..x εAB...X MAa MBb ...MXx = scalar density weight -2 Write this for g g = det(gij) = (1/N!) εab..x εAB...X gAa gBb ...gXx We know that gab;c = 0 so this gab;c = 0 as well. One reason we know this last fact is that raising and lowering commutes with ; as I show somewhere and as some other author also pointed out. Since gab;c is a true tensor, you can always raise and lower indices on both sides. Apply Leibnitz g;j = (1/N!) [ εab..x εAB...X];j gAa gBb ...gXx because all the other expansion terms have something like gAa;j which vanishes. Now do the implied sums on AB...X to get g;j = (1/N!) [ εab..x εab...x];j New comment: I could have done this [ εab..x εab...x];j = [ N!];j = 0 but I did not realize this the first time through, so no paradox. Now assume this theorem is valid εabc...x;j = (1/) ∂j()εabc...x Theorem is wrong, see Plan C below! Since both sides are a tensor, you can also say that εabc...x;j = (1/) ∂j()εabc...x We then have ( I think my error is in these steps following) [ εab..x εab...x];j = εab..x [ εab...x];j + [ εab..x];j εab...x Since everything is a tensor, tilt indices in the second term to get [ εab..x εab...x];j = εab..x [ εab...x];j + [ εab..x];j εab...x = 2 [ εab..x];j εab...x Now use the assumed theorem εabc...x;j = (1/) ∂j()εabc...x and we end up with [ εab..x εab...x];j = 2 (1/) ∂j()εabc...x εab...x = 2 (1/) ∂j()εabc...x εab...x Now use this result, εabc... = det(gij) εabc... = g εabc... // relating all down to all up (D.5.10) to write then [ εab..x εab...x];j = 2 (1/) ∂j() g εabc...xεabc...x = 2 ∂j() N! and I have then shown that g;j = (1/N!) [ εab..x εab...x];j = 2 ∂j() I don't see where any minus signs can come from in this method which would cause a cancellation of terms as happened in item 1. Comments: I have confusion here about which tilt gives N! and I need to clarify that in tensor doc, I need to say when ε is perm tensor and when it is LC tensor. But this is just a factor of g, so the above paradox survives this confusion. 3. The Paradox is this: Method 1 says: g;j = 0 Method 2 says: g;j = 2 ∂j() These two results disagree, and that is The Paradox Of The Day. Plan A. I am reminded of the Weinberg minus sign page 39 (2.5.15) which I did not "adopt". That might play a role here. Weinberg on page 99 says εabcd = gaa'gbb'gcc'gdd'εa'b'c'd' assuming ε works like any tensor. He then argues that gaa'gbb'gcc'gdd'εa'b'c'd' = K εabcd He then evaluates K εabcd = gaa'gbb'gcc'gdd'εa'b'c'd' K ε1234 = g1a'g2b'g3c'g4d'εa'b'c'd' = det(gij) = g and so K = g and then εabcd = K εabcd = gεabcd and this is both what I get and what Weinberg gets on his page 39 (2.5.15) because he has g = det(g) = -1! So I don't think we have any disagreement after all. Plan B. Go back to my "assumed theorem" εabc...x;j = [(1/) ∂j()] εabc...x I am confused about the tensor nature of this claimed equation. I derived this for εabc only and had εabc;α = Γiiα εabc = Cα εabc I know that Γcab is NOT a true tensor, and in fact I show in (F.6.1) that Γ 'cab = Rcd Raα Rbβ Γdαβ + Rcα (∂'aRbα) // Weinberg p 100 (4.5.2) (F.6.1) and the extra term shows WHY it is not a tensor. So it seems likely that contracting this thing is ALSO not a tensor. In fact Q'b ≡ Γ 'aab = Rad Raα Rbβ Γdαβ + Raα (∂'aRbα) = δdα Rbβ Γdαβ + Raα (∂'aRbα) = Rbβ Γbαβ + Raα (∂'aRbα) and this definitely looks NON vectorish. Question. Suppose tensor Mabcd is antisymmetric on a.b.c. Does that mean you can write Mabcd = εabcQd ??? I am not sure! For d = 1 it seems reasonable. Assume answer is yes. Then my assumed theorem is OK. But if it is OK, then Γiiα must transform as a vector density! But that just seems impossible [(1/) ∂'a()] = Rab [(1/) ∂b()] ?? This says ∂'a() = (g'/g)1/2 Rab ∂b() ∂'a() = J Rab ∂b() Well, that would be correct if ∂b() were a vector density of weight -1. δ(cd...x; c',d'...x') ******************************88 Suppose in N = 2 we know that Mabc is antisymmetric on a and b (a true tensor). Then εabMabc ≡ Qc = a true vector Then what? Can you claim that any such Mabc can be written in the form εabVc where Vc is a vector? Let's hypothesize that you can do this. Then what is Vc? Mabc = εabVc εabMabc = εab εabVc = 2Vc => Vc = (1/2) εabMabc So this shows that IF Mabc can be factored as shown, THEN Vc must be give as shown. OK, I hand you some Mabc AS on ab. You can of course go off and compute Vc = (1/2) εabMabc But how would you "solve this" for Mabc? You cannot do so because it is contracted on a,b. So you cannot conclude that Mabc = εabVc. I really need a simple counterexample to get happy here. How about this: Mabc = AaBbc - AbBac Is this really of the form Mabc = εabVc Well, Try Vc = (1/2) ΣabεabMabc and see what you get Mabc = εabVc = (1/2) εab ΣijεijMijc But now use my specific form Σab εabMabc = Σab εab[AaBbc - AbBac] So I then get Mabc = εabVc = (1/2) εab Σij εijMijc = (1/2) εab Σij εij[AiBjc - AjBic] = (1/2)Σijεabεij [AiBjc - AjBic] = (1/2)Σij(δaiδbj - δajδbi) [AiBjc - AjBic] = Σijδaiδbj[AiBjc - AjBic] = [AaBbc - AbBac] So for that example it works. It will probably work for any example that you try which is factored as in my case. Resume manana. Plan C. Let's look at my assumed theorem a little, In the edit log I have shown this theorem. Theorem 1: If rank-4 tensor Mabcd is totally antisymmetric on the first three indices, then Mabcd can be written as Mabcd = εabcVd where Vd = (1/3!)εijkMijkc It just then also be true that Mabcd = εabcVd where Vd = (1/3!)εijkMijkd Now apply this as follows: εabc;d = εabcVd where Vd = (1/3!)εijkMijkd = (1/3!)εijk εijk;d Now how did Γ get into this picture? Oh yes: εabc;α = Γanαεnbc + Γbnαεnca + Γcnαεnab = Cαεabc // suspected step!! Aha! Since ε has weight, I have omitted a term here!!! I even suspected it without knowing. The correct expansion is εabc;α = Γanαεnbc + Γbnαεnca + Γcnαεnab + (-1)Γκκαεabc = Γanαεnbc + Γbnαεnca + Γcnαεnab + (-1)Γκκαεabc Cαεabc = εabc;α = Γanαεnbc + Γbnαεnca + Γcnαεnab + (-1)Γκκαεabc To find Cα do standard order, Cαε123 = ε123;α = Γ1nαεn23 + Γ2nαεn31 + Γ3nαεn12 + (-1)Γκκαε123 = Γ11αε123 + Γ22αε231 + Γ33αε312 + (-1)Γκκαε123 = Γ11αε123 + Γ22αε123 + Γ33αε123 + (-1)Γκκαε123 = Γ11α + Γ22α + Γ33α + (-1)Γκκα = Γκκα + (-1)Γκκα = 0 So therefore Cα = 0 and I have shown that εabc;α = Cαεabc = 0 Conclusions. I now have lots of interesting facts. (a) g;j = 0 (b) Theorem 1: If rank-4 tensor Mabcd is totally antisymmetric on the first three indices, then Mabcd can be written as Mabcd = εabcVd where Vd = (1/3!)εijkMijkc (c) εabc;α = 0 This last is consistent with [constant];α= 0. Now maybe I can show this [g2];j = 0 Write this as [gg];j = g;g + gg: = 0+0 = 0 Thus I know that [gn];j = 0 for n = 1,2,3.... Next the object g-1/2. We know g has weight -2, so g-1/2 has weight +1. Therefore (g-1/2);j = ∂j(g-1/2) + 1 * Γκκj (g-1/2) = ∂j(g-1/2) + 1 * (2g)-1∂jg (g-1/2) = (-1/2) g-3/2(∂jg) + 1 * (1/2) g-1 ∂jg (g-1/2) = (-1/2) g-3/2(∂jg) + (1/2) g-3/2 ∂jg (g-1/2) = 0 Then consider (g+1/2);j = (g g-1/2);j = 0 by Leibnitz So I then know this as well (gn-1/2);j = 0 n = 0,1,2..... All fascinating new facts! But how about this: (g-3/2);j = Well how about we just start with gα for arbitrary α. The weight of gα must be -2α since' g' = J2 g g has weight - 2 g'α = J2α gα gα has weight -2α Then (gα);j = ∂j(gα) + (-2α) * Γκκj (gα) = α gα-1(∂jg) - 2α [ (1/2)(1/g)(∂jg) ] (gα) // (F.4.2) = α gα-1(∂jg) - α (∂jg) (gα-1) = 0 This proof is valid for any real value of α !!!! This needs to get into tensor doc. One implication is this [gα stuff]; = gα [stuff]; It comes out just like a constant.