the Levi ee rule fiddling

Playing with the εε products seeking a covariant form I think this stuff is done OK in Appendix D now, but keep. Consider this well known property of ε εabcεab'c' = δbb'δcc' – δbc'δcb' So far I have only used this in Cartesian space where up and down don't matter. But suppose we have some general g operating in x-space. Is this still true? We want it to be a tensor equation, and the sum on a is not right, it has no tilt so it cannot be a tensor equation as it stands. Plan A: Here is the obvious hope : εabcεab'c' = δbb'δcc' – δbc'δcb' The hope is that each side is a mixed rank-4 tensor. But I have not even thought of δab as a tensor, so I am now back to that old question. If I think of it as gab then surely it is a tensor. So try this εabcεab'c' = gbb'gcc' – gbc'gcb' Now we for sure have a true tensor equation, but is it valid?. Let's try some tests. Consider now εabc... = det(gij) εabc... Then we can write det(gij) εabcεab'c' = gbb'gcc' – gbc'gcb' Is this true? Let b=2 and c=3 so this would then say det(gij) εa23εab'c' = g2b'g3c' – g2c'g3b' Now this forces a = 1 and we then have det(gij) ε123ε1b'c' = g2b'g3c' – g2c'g3b' Now let b'=2 and c'=3 as well det(gij) ε123ε123 = g22g33 – g23g32 det(gij) = g22g33 – g23g32 = 1 So that conclusion is wrong, something has malfunctioned. Maybe the correct equation is this: εabcεab'c' = det(gij) [ gbb'gcc' – gbc'gcb'] I have never seen anything like that ever. Plan B Let's start again with our hope in x-space εabcεab'c' = gbb'gcc' – gbc'gcb' What would this look like in x'-space? If covariant, we would say ε'abcε'ab'c' = g'bb'g'cc' – g'bc'g'cb' But then what happens if we fiddle the LHS? LHS = ε'abcε'ab'c' = det(g'ij) εabcεab'c Again, it seems that we really want this: εabcεab'c' = det(gij) [ gbb'gcc' – gbc'gcb'] ε'abcε'ab'c' = det(g'ij) [ g'bb'g'cc' – g'bc'g'cb'] The first line then can be written det(gij) εabcεab'c = det(gij) [ gbb'gcc' – gbc'gcb'] or εabcεab'c = [ gbb'gcc' – gbc'gcb'] This I know is correct, despite the bad up/down characteristics. Plan C as an aside. Let's try the Nikodem idea. He writes it as a true tensor δ'ik = RijRkm δjm Is this true? If so, what is the LHS object? δ'ik = RijRkm δjm = RijRkj = δik and that works So it would seem that δ and δ' are the same in both frames. Then δik= giaδak = gik So yes, it really is the same as g in its frame. ****************************************************************** I am back to my original yesterday mystery. Consider this statement of the Levi pdf,. This makes me realize: J = det(g') is wrong! This is only so if x-space is Cartesian. He is saying above that adding a power of det(g') fixes you up. So I am back to that issue, but without writing it as J. Plan D: Suppose Aa is a tensor density of weight W. Then we know that A'a = J-W RabAb Note that J2 = det(g')/det(g) Now consider this new object [ let det(g) always mean det(gij) ] Ba ≡ [det(g)]nAa Then how does B transform? Presumably the above implies that B'a = [det(g')]n A'a Then we find B'a = [det(g')]n A'a = ([det(g')]n / [det(g)]n) [det(g)]n [J-W RabAb ] = ([det(g')]n / [det(g)]n) J-W Rab [[det(g)]n Ab] = J2n J-W RabBb = J(2n-W) RabBb Then if you select n = W/2, you get your pure vector result! So THIS is the missing link. And THIS is why they always use that g symbol! Plan E. Let's back up a bit to Section 5 (k) where some fixes are needed. I get to the point where I have this J2 = det(') / det() = J2 How should I define the symbol g in a uniform manner? Quasi-Cartesian: Suppose det() = s = ± 1. Then s det() = 1 Suppose we go ahead and define g = det() Then J2 = J2 = g'/g J = | J | = / Then in special relativity we have s = -1 and then J = | J | = / It happens that in a free-fall x frame we have = 1 In curvilinear coordinates, we have J = | J | = / Now let's redo our theorem Plan D Revisited: Suppose Aa is a tensor density of weight W. Then we know that A'a = J-W RabAb J = | J | = / = (g'/g)1/2 J = σJ Now consider this new object [ let det(g) always mean det(gij) ] Ba ≡ gnAa Then how does B transform? Presumably the above implies that B'a = g'n A'a Then we find B'a = g'n A'a = (g'n /gn) gn [J-W RabAb ] = (g'n / gn) J-W Rab [gn Ab] = (g'/ g)n J-W Rab [gn Ab] = J2n J-W Rab [gn Ab] = J(2n-W) RabBb Then if you select n = W/2, you get your pure vector result! Theorem: If T is a tensor density of weight W, then gW/2 T is an ordinary tensor. The only problem here is what you do when g < 0? Let's try modifying Plan D again Plan D Revisited Again: Suppose Aa is a tensor density of weight W. Then we know that A'a = J-W RabAb J = | J | = / = (g'/g)1/2 J = σJ Now consider this new object [ let det(g) always mean det(gij) ] Ba ≡ (sg)nAa Then how does B transform? Presumably the above implies that B'a = (sg')n A'a Then we find, B'a = (sg)'n A'a = ((sg)'n /(sg)n) (sg)n [J-W RabAb ] = ((sg)'n / (sg)n) J-W Rab [(sg)n Ab] = (sg'/ sg)n J-W Rab [(sg)n Ab] = J2n J-W Rab [(sg)n Ab] = J(2n-W) RabBb Then if you select n = W/2, you get your pure vector result! Theorem: If T is a tensor density of weight W, then (sg)W/2T transforms as ordinary tensor. You can always add a constant and then k (sg)W/2T also transforms as an ordinary tensor. ****************************** Now lets go back to εabcεab'c' = gbb'gcc' – gbc'gcb' [ Theorem: If T is a tensor density of weight W, then (sg)W/2 T is an ordinary tensor. ] Realization: The object εabcεab'c' is the outer product of two tensors which has afterwards been contracted. thus, it has weight W = -2, not -1! Since the LHS is a tensor density of weight W = -2, I could make it a tensor by adding (sg)-1 , so here are my side by sides: (sg)-1 εabcεab'c' = gbb'gcc' – gbc'gcb' (sg')-1 ε'abcε'ab'c' = g'bb'g'cc' – g'bc'g'cb' Of course if x-space is Cartesian, then sg= 1 and you don't see that first factor. There can be no other constant factor on the LHS for this reason. Let's now mess with these to see if things are consistent. First, take this and assume g ≠ 1. (sg)-1 εabcεab'c' = gbb'gcc' – gbc'gcb' We know that εabc... = det(gij) εabc... = g εabc so install this to get (sg)-1 g εabc εab'c' = gbb'gcc' – gbc'gcb' sεabc εab'c' = gbb'gcc' – gbc'gcb' For special relativity, this seems to say - εabc εab'c' = gbb'gcc' – gbc'gcb' εabc εab'c' = gbb'gcc' – gbc'gcb' so that would be a good reason to use high and low to get a uniform result for any Quasi-Cartesian situation. Plan F: Let's conjecture that the following two equations are true (sg)-1 εabcεab'c' = [gbb'gcc' – gbc'gcb'] (sg')-1 ε'abcε'ab'c' = [g'bb'g'cc' – g'bc'g'cb'] Do these two equations for a covariant pair? They do "look the same" in both frames. What can one say about weights? Write it this way [gbb'gcc' – gbc'gcb'] = (sg)-1 εabcεab'c' Ubcb'c' = (sg)-1 Tbcb'c' Then our two equations are these Ubcb'c' = (sg)-1 Tbcb'c' U'bcb'c' = (sg')-1 T'bcb'c' We can apply weight changing theorem #1 with w = -2 and W = -2 to conclude that U is a tensor of the same type with weight W-w = -2 - (-2) = 0. So I think this is indeed a valid tensor equation. Now suppose we define Eabc ≡ (sg)-1/2εabc E'abc ≡ (sg')-1/2ε'abc Notice then that E123 = (sg)-1/2ε123 = (sg)-1/2 g = g1/2 if s = 1 Then I presume you also have E123 = (sg)-1/2ε123 = g-1/2 This agrees with page 31 of Hermann's translation of the 1900 Ricci Levi-Civita paper So I think I have nailed this subject pretty well and can add this to Appendix D later. Maybe the paper I quote above has the twiddle in the wrong place. The WCT #1 with w = -1 and W = -1 says that E has weight W-w = -1-(-1) = 0. So then EabcEab'c' = [gbb'gcc' – gbc'gcb'] has the same form in all coordinate systems. Now go back to the pair (sg)-1 εabcεab'c' = [gbb'gcc' – gbc'gcb'] (sg')-1 ε'abcε'ab'c' = [g'bb'g'cc' – g'bc'g'cb'] Both RHS's are of course the same. And we know that εab'c' = ε'ab'c'. So this would imply that (sg)-1 εabc = (sg')-1 ε'abc But we know that εabc... = det(gij) εabc... = g εabc... ε'abc... = det(g'ij) ε'abc... = g' ε'abc. and therefore (sg)-1 g εabc... = (sg')-1 g' ε'abc. (1/s) εabc.. = (1/s) ε'abc which is true. Conclusion: Here is the correct covariant form of this little theorem (sg)-1 εabcεab'c' = [gbb'gcc' – gbc'gcb'] You can raise or lower one or more indices on both sides and it remains true. I have never seen this stated in this way before.