Vectors and parity in tensor doc world

Vectors and parity in tensor doc world PhL 2.25.12 Sent here from "use of abs value in tensor density...". A tensor density theorem is used by me to argue that C = curl B is a true tensorial vector. The question arises: what is the connection between this claim and "parity"? Overview. I first show below how my former claimed vector density rule C' = |J| R C leads to a contradiction -- the contradiction is that C' = -C when R = -I is a parity transformation for C = A x B. When I fixed this problem, lots of things started working right. The fix is C' = J R C . Second, I realized that covariance applies as well to any tensor density equation! Third, I show that in general C = curl B is a tensor density of weight -1 just as is C = A x B. 1. Consider first C = A x B where A and B are tensorial vectors. I consider this in App D (g). I argue that since ε has weight -1, that C is a vector density with weight -1 and therefore, I say, C'a = |J| RabCb Now just imagine A and B sitting in x-space for N=3. Under parity, A and B invert but C does not invert. Here parity is acting on the vectors in x-space, it is not changing x-space the coordinate system. The picture here I can steal vectors and parity.vsd In both cases cross product points toward viewer, so C has not changed. That is to say, C' = C. So think of R or S as just diag(-1's). Then our equation above says C'a = |J| RabCb = RabCb = -Ca => C' = - C So we have a conflict. NOTE: The desired correct result is C' = C under parity. Here is a parity transformation, a linear one in fact x' = F(x) = -x F = F a matrix = -I x' = F x F = -I So in this situation, we have S = -I and R = -I as well. If I take g = 1 then g' = RRT = 1 as well. But we have J = det(S) = -1. Consider the claim made above, that C' = |J| R C => C' = |-1| (-I) C = - C // gives the wrong result! I think I would then argue that C' = |J| R C is not a valid transformation rule for this particular R. It is well known that C does not change sign, and such a C is called a pseudovector, a term I don't use in tensor doc. How might I fix this up? Perhaps this is the correct transformation: C'a = J RabCb C' = J R C Then we get the correct result for this example. And this would then agree with Weinberg's tensor density definition and usage. [ yes this is the right thing to do! ] Covariance question. Now can we add primes to the equation C = A x B ? If we did we would get C' = A' x B' = (-A) x (-B) = A x B = C which is right! Therefore, the equation C = A x B is covariant for R = parity. 2. Where is σ and sign of detR first mentioned in tensor doc? Section 6 (i) on Handedness. Go read it. // Read it, sounds good, no changes required. Next σ occurrence is in Section 8 and I ignore all those σ's since Section 8 is pending rewrite right now. Next is App B regarding the area, which I have just examined in " should sigma...". So σ is not used in very many places. 3. Let's then have a look at our covariance section. This is section 7 (u) which I now read. // OK, have read it all, it all seems OK, parity is never mentioned. No assumption I can see is made about the sign of detR, so I cannot argue to add any ad hoc σ factors to the statement of covariance! If someone hands you an equation A = B + C, you would not say A' = σ(B'+C') since there is one free index! So I see no edits to be made to this section! 4. I think it has something to do with the ε tensor. This is what causes pseudo vectors to appear. [ no, nothing new was needed for the ε tensor! ] 5. I have been editing Appendix D, replacing |J| with J everywhere, carefully reading all the arguments. In the cross product section, I now end up with this Q'a = J RabQb which applied to the above situation gives C'a = J RabCb and this then does provide the "fix" suggested above. 6. OK, lets get back to "the covariance problem" with C = A x B. This is NOT a tensor equation because neither side is a tensorial vector, so I don't expect to just "prime everything". My argument for priming everything in section 7 (u) only applied to true tensor valid equations. So I have to go compute things to see what this looks like with things primed. So C = A x B Ci = εijkAjBk C'a = J RaiCi = J Rai εijkAjBk But replace Aj = RjbA'b C'a = J RaiCi = J Rai εijk(RjbA'b)( RkcB'c) = J RaiRjbRkcεijk(A'b)( B'c) = J εabcdet(R) (A'b)( B'c) = εabc (A'b)( B'c) => C' = A' x B' And this is the correct result. This result does not depend on the sign of det(R), by the way. Claim: The covariance rule of section 7 (u) probably applies to tensor density matching equations as well as tensor matching equations. [ I later proved this and updated that section to include it. ] 7. Is C = curl B a vector density of weight -1 ? Written as x B in Cartesian coordinates, it looks a lot like A x B so I am inclined to say yes. If we write out components Cn(x) = εnab∂aBb(x) I have to quickly use the Appendix D trick to rewrite this as Cn = εnab Bb;a Now I think Bb;a is a true tensor, that being the whole point of the Christoffel stuff, and it has weight 0. Then ε has weight -1, and so then yes, Cn is a vector density of weight -1. This seems to be a viable proof. Therefore, the correct Cn expansion on the en is this C = curl B = JW ΣnC'nen = J-1ΣnC'nen and this means I have to make major edits to all later Sections involving curls!!! I just did a full edit of Section 12 on C = curlB and everything went fine. Yes, the curl for N=3 is a vector density of weight -1 ! I really believe it now. There is no need to use that Weight Change Theorem (which only I have "heard of"). I used this theorem to prove that curl is a vector density, but I don't need that theorem any more, I know it a priori.! I am going to have problems, however in my fancy Eabc... section without the use of this theorem. I will deal with that in my doc on Appendix D rewrite.