two curls and their weight

Here I am reviewing my Section 13 Vec Lap work. The second term V is little more complicated. First, from Section 12 (d), C = curl B = ε'nab [(1/) ∂'a{ B'b} ] en = J-1C'n en J = (1) Confusion. The above can be written C = curl B = J-1 ε'nab [ ∂'a{ B'b} ] en = J-1C'n en J = and then I get C'n = ε'nab∂'aB'b C = curl B = J-1C'n en = g'-1/2 ε'nab∂'aB'b en = g'-1/2 ε'nab B'b;a en = g'-1/2 C'n en C = curl B= Cn un = εnab Bb;a un Looking at this last equation, I would say that Bb;a was a tensor with weight 0, and εnab was a tensor of weight -1, so writing Cn = εnab Bb;a, I would say Cn was a vector of weight -1. Then we know C'n = J RnmCm Now since Cn = εnab Bb;a is a "valid tensor density equation" my section 7 (u) or whatever says that in x'-space this equation should take the covariant form C'n = ε'nab B'b;a and that IS consistent with what I have written above. So I guess there is no confusion here. (2) Now suppose it happened that Bn were components of a weight - 1 vector. Then the logic of my derivation in Section 13 says B'b → J-1B'b and I claim then that C = curl B = J-1 ε'nab [ ∂'a{ J-1B'b} ] en Now here is how B transforms in this case, weight -1. B'n = J RnmBm The Big Question is: how do you now expand C. Suppose we assume C'n = J-W RnmCm C = JW C'n en and we seek to determine W. Then the above equation says JW C'n = J-1 ε'nab [ ∂'a{ J-1B'b} ] In Cartesian space we have C = curl B = Cn un = εnab Bb;a un Cn = εnab Bb;a Theorem: if Bb has weight W, then Bb;a has the same weight. Did I claim that somewhere? Maybe this is a wrong theorem! It is pretty important right now! I think this is wrong! No, this claim is right. I updated tensor doc to include covariant derivatives of tensor densities! Correct Theorem: If Bb has weight W, then (gW/2Bb);a is a tensor of weight 0. But can you say anything about Bb;a ?? I think this is NOT a tensor density!!! [ yes it is ] Theorem: If Bb is a vector density with weight -1, then (g-1/2Bb);a is a rank 2 tensor of weight 0. Now go back to the above analysis. The Cartesian claim is of course still valid as written. But now write it this way C = curl B = Cn un = εnab Bb,a un = εnab (g-1/2Bb),a un At this point, we can consider (g-1/2Bb);a = (g-1/2Bb),a – g-1/2 Γnba 2Bn and then the symmetry kills the second term just as in the weightless case, so we get C = curl B = Cn un = εnab Bb,a un = εnab (g-1/2Bb),a un = εnab (g-1/2Bb);a un Cn = εnab (g-1/2Bb);a NOW, we ask about the nature of Cn . The object (g-1/2Bb);a transforms as a zero weight rank 2 tensor! Therefore, vector Cn has weight -1 (and not weight -2, say ). So I think finally I have justified, V = curl C = ε'ncd [(1/) ∂'c{ J-1C'd} ] en = J-1V'n en (*) Review: If B has weight 0, we write Cn = εnab Bb;a and C then has weight -1. If B has weight -1, we write Cn = εnab (g-1/2Bb);a and C still has weight -1. In Section 15 we end up adding this g-1/2 factor in Cartesian space just to get this result for each cross product so the second term will end up being a regular vector.