Paradox of the Day 5_1_15

Paradox of the Day 5_1_15 PhL 5.1.15 Comment: I think Part 1 is the one that is wrong, but I cannot figure out the logical reason why. 1. On the one hand, consider my (15.2.5) analysis: {...}1a = {...}2a when both sides are evaluated in Cartesian x-space . (15.2.5) (Bj;j);a = Bj;a;j // realization of the above! Mjja = Mjaj When both sides are evaluated in Cartesian space, each ; becomes a , and claim is true since (Bj;j);a = (Bj,j),a = ∂a(∂jBj) = ∂a∂jBj = ∂a∂jBj Bj;a;j = Bj,a,j = ∂j(∂aBj) = ∂j∂aBj = ∂j∂aBj = ∂a∂jBj . Continuing: (a) Yes, the two objects (Bj;j);a and Bj;a;j are the same in Cartesian x-space, this has to be true. "One can then transform both objects to x'-space in the usual manner, {...}'1a = Rab{...}1b and {...}'2a = Rab{...}2b . " (15.2.6) (B'j;j);a = Rab(Bj;j);b and B'j;a;j = Rab Bj;b;j Why do I claim these two equations on the line just above? Left: Bj;j is a scalar, call it S. Then S;a is a vector. A vector transforms as S';a = RabS;b. This then says that (B'j;j);a = Rab(Bj;j);b which is the left equation above. Right: Bj;a;j is a vector, call it Va. A vector transforms as V'a = RabVb. This then says that B'j;a;j = Rab Bj;b;j which is the right equation above. I continue with my Section 15.2 proof: "Therefore, since {...}1a = {...}2a in x-space, one must have {...}'1a = {...}'2a in x'-space. " Well, I would write in x-space that (B'j;j);a = Rab(Bj,j),b B'j;a;j = Rab Bj,b,j (*) (b) Yes, the two right sides of these equations are equal, this is true. And it is true the that right sides are indeed the same since in Cartesian space (Bj,j),b = ∂b(∂jBj) = ∂b∂jBj Bj,b,j = ∂j(∂bBj) = ∂j∂bBj Since the right sides of (*) are exactly the same the left sides must also be equal and we get (B'j;j);a = B'j;a;j (c) If A = B and C = D and A = D, then A = C, so this is true. So Part 1 claims that these two objects are equal. 2. On the other hand, Let us consider the validity of (Bj;j);n = Bj;n;j. (*) First, process the left side of (*) (Bj;j);n = Σj(Bj;j);n = Σj(ΣiδjiBi;j);n = Σij δji(Bi;j;n) = Σij δji(Σm gjmBi;m;n) = Σijmδjigjm (Bi;m;n) = Σijmδjigjm (ΣsgnsBi;ms) = Σijms δjigjmgns (Bi;ms) = Σjms gjmgns (Bj;ms) Now process the right side of (*) ΣjBj;n;j = Σjmsδnm δjs Bj;m;s then (*) becomes Σjms gjmgns (Bj;ms) = Σjmsδnm δjs Bj;m;s ? or Σjms[ gjmgns - δnm δjs] Bj;m;s = 0? For this to be true for an arbitrary tensor B, we would need to have gjmgns = δnm δjs (**) But this is NOT TRUE. Therefore (Bj;j);n ≠ Bj;n;j Specific Example. Suppose you had orthogonal coordinates. Then gjm = hj2 δjm while gns = hn-2 δns and in that special case you would have gjmgns = hj2 δjm hn-2 δns = (hj/hn)2 δjmδns . Now examine (hj/hn)2 δjmδns = δnm δjs ?? Suppose for example that j = m = 2 and n = s = 3. Then you have (h2/h3)2 = 0 which is obviously not true. So this gives a clear case where (**) is false. Plan A. Work with some simpler case, "have you tried n = 2" idea. Let S be a scalar. I know that S,a = S;a and both of these are true vectors. That says S',a = RabS,b and S';a = RabS;b This case is too simple and does not show my problem, so escalate up a little. The problem seems ONLY to arise when I have a divergence involved. Let S = scalar, and consider S;b;a;a and S;a;b;a which seem to be two different objects. How do these objects transform? In general we have S';x;y;z = RxARyBRzC S;A;B;C Then look at our two contracted cases S';b;a;a = RbARaBRaC S;A;B;C = RbAδBC S;A;B;C = RbAS;A;B;B = RbaS;a;s;s Now do the other case S';a;b;a = RaARbBRaC S;A;B;C = δACRbBS;A;B;C = RbBS;A;B;A = RbaS;s;a;s So this says that S';b;s;s = RbaS;a;s;s S';s;b;s = RbaS;s;a;s Now suppose x-space is Cartesian. then semicolons go to commas, and up and down don't matter so S';b;s;s = RbaS,a,s,s = Rba ∂a∂s2S S';s;b;s = RbaS,s,a,s = Rba ∂a∂s2S In this special case that x'-space is Cartesian, we find that in x'-space, these are equal S';b;s;s = S';s;b;s So this is the Part 1 approach for this example. Now let's try to find a conflicting Part 2 method. S';b;s;s = δsj S';b;j;s = δsj δbaS';a;j;s S';s;b;s = δsaS';a;b;s = δsaδbj S';a;j;s To see if the two tensors on the left are equal, I have to test δsj δbaS';a;j;s = δsaδbj S';a;j;s ? b is fixed [ δsj δba - δsaδbj] S';a;j;s = 0 ? (*) You MIGHT now argue : For an arbitrary complex scalar S'(x'), this can only be true if δsj δba = δsaδbj But suppose s = j = 2 and b = a = 3. Then get 1 = 0 So maybe this is a simpler case that captures my confusion. Perhaps (*) is valid for a different reason! Let's write it out in terms of x-space objects, [ δsj δba - δsaδbj] RaARjJRsS S;A;J;S = [ δsj δba - δsaδbj] RaARjJRsS ∂A∂J∂S (S) where we now expose a triple symmetry for ∂A∂J∂S (S). Call this MAJS which is totally symmetric. Then = [ δsj δba - δsaδbj] RaARjJRsSMAJS b is fixed = δsj δbaRaARjJRsSMAJS - δsaδbjRaARjJRsSMAJS = RbARsJRsSMAJS - RsARbJRsSMAJS Now A,J,S are all dummy indices. In the second term, do J↔A and rewrite = RbARsJRsSMAJS - RsJRbARsSMJAS Now in the second term do MJAS = MAJS to get = RbARsJRsSMAJS - RsJRbARsSMAJS = 0 So, we find that (*) is indeed true, but NOT because [ δsj δba - δsaδbj] = 0. To show it is true, you have to make use of the fact that x-space is Cartesian!! So perhaps it is in fact Part 2 that is wrong above, not Part ! So let's go through part 2 again: 2. On the other hand (version 2.0) Let us consider the validity of (Bj;j);n = Bj;n;j. (*) First, process the left side of (*) (Bj;j);n = Σj(Bj;j);n = Σj(ΣiδjiBi;j);n = Σij δji(Bi;j;n) = Σij δji(Σm gjmBi;m;n) = Σijmδjigjm (Bi;m;n) = Σijmδjigjm (ΣsgnsBi;ms) = Σijms δjigjmgns (Bi;ms) = Σjms gjmgns (Bj;ms) Now process the right side of (*) ΣjBj;n;j = Σjmsδnm δjs Bj;m;s then (*) becomes Σjms gjmgns (Bj;ms) = Σjmsδnm δjs Bj;m;s ? or Σjms[ gjmgns - δnm δjs] Bj;m;s = 0 ? (***) For this to be true for an arbitrary tensor B, we would need to have WRONG! gjmgns = δnm δjs (**) But this is NOT TRUE. Therefore (Bj;j);n ≠ Bj;n;j Specific Example. Suppose you had orthogonal coordinates. Then gjm = hj2 δjm while gns = hn-2 δns and in that special case you would have gjmgns = hj2 δjm hn-2 δns = (hj/hn)2 δjmδns . Now examine (hj/hn)2 δjmδns = δnm δjs ?? Suppose for example that j = m = 2 and n = s = 3. Then you have (h2/h3)2 = 0 which is obviously not true. So this gives a clear case where (**) is false. WRONG because (***) is 0 for a different reason! Go back and show all the primes now [ g'jmg'ns - δnm δjs] B'j;m;s = 0 ? n is fixed Write as [ g'jmg'ns - δnm δjs] RjJRmMRsS BJ;M;S = [ g'jmg'ns - δnm δjs] RjJRmMRsS ∂M∂S(BJ) = g'jmg'ns RjJRmMRsS ∂M∂S(BJ) - δnm δjs RjJRmMRsS ∂M∂S(BJ) n is fixed = RjJ RjM RnS ∂M∂S(BJ) - RjJRnMRjS ∂M∂S(BJ) Now in the second term do M↔S and then reorder the two derivatives = RjJ RjM RnS ∂M∂S(BJ) - RjJRnSRjM ∂M∂S(BJ) = δJM RnS ∂M∂S(BJ) - δJM RnS∂M∂S(BJ) = 0 hurray!!!! Again, you have to make use of the fact that x-space is Cartesian, then shows that (Bj;j);n = Bj;n;j if you make the assumption that x-space is Cartesian. Then Part 1 and Part 2 give the same result and the paradox is removed. Fact: It is not IN GENERAL true that (B'j;j);n = B'j;n;j. It is only true if x-space is Cartesian and you get to x'-space from x-space and that transformation defines your tensor world. Appendix A: a possible third method which does not pan out (Bj;j);n = Bj;n;j. (*) Process these two expressions explicitly using Appendix F expansions Left side: (Bj;j);n = gnc Bj;j;c LHS = RHS But Ba;b;α = ∂α Ba;b + Γaαn Bn;b – ΓnbαBa;n (F.9.21) so Bj;j;c = ∂c Bj;j + Γjcn Bn;j – ΓnjcBj;n (F.9.21) Then LHS = gnc[(∂c Bj;j) + Γjcn Bn;j – ΓnjcBj;n ] Now do right side: Bj;n;j. But now use Ba;b;α ≡ ∂α Ba;b + Γaαn Bn;b + Γbαn Ba;n . (F.9.9) with ; added so Bj;n;j ≡ ∂j Bj;b + Γjjn Bn;b + Γbjn Bj;n Now want to get second indices down, so write RHS = Bj;n;j ≡ ∂j (gbcBj;c) + Γjjn gbcBn;c + Γbjn gncBj;c So the question is then whether or not these two are equal: LHS = RHS ? gnc(∂c Bj;j) + Γjcn gncBn;j – ΓnjcgncBj;n = ∂j (gbcBj;c) + Γjjn gbcBn;c + Γbjn gncBj;c ? Wow, nothing obvious at all here! It seems unlikely certainly. Have to install lots of dummy indices., LHS1 = gnc(∂c Bj;j) = gnd(∂d Bj;j) = δjcgnd(∂dBj;c) LHS2 = Γcdn gndBn;c = δnjΓcdn gndBj;c LHS3 = ΓcjdgndBj;c = RHS2 = Γddn gbjBn;j Now the equation of interest is this: δjcgnd(∂dBj;c) + δnjΓcdn gndBj;c + ΓcjdgndBj;c = ∂j(gbcBj;c) + Γddn gbjBn;j + Γbjn gncBj;c Now compute ∂j(gbcBj;c) = gbc(∂jBj;c) + (∂jgbc)Bj;c so then our interest is δjcgnd(∂dBj;c) + δnjΓcdn gndBj;c + ΓcjdgndBj;c = gbc(∂jBj;c) + (∂jgbc)Bj;c + Γddn gbjBn;j + Γbjn gncBj;c or δjcgnd(∂dBj;c) + δnjΓcdn gndBj;c + ΓcjdgndBj;c = gbc(∂jBj;c) + (∂jgbc)Bj;c + Γddn gbjBn;j + Γbjn gncBj;c Ii give up on this pathway. You really have to expand everything all the way down to get an answer.