is curl a vector

Resuming on the question of whether or not the curl is a tensorial vector. Part I: fiddling with rotations and the ε object Consider this result from Section 12 on the curl, where C ≡ [ curl B ] : C'n(x') = (1/ J(x')) εnab ∂'aB'b = [ curl B](x) Cn(x) = εnab ∂aBb = [ x B]n // Cartesian 1. I know from Appendix A (j) that [' x B']n = cof(Rni) [ x B]i so that, unless R is a rotation, this object does not transform as a contravariant vector. And if we have a rotation, J = 1, and C'n is a tensorial vector. BUT, the first line above does NOT say that C'n(x') = [' x B']n: [' x B']n = [(Σn ' ∂'n) x B']n = εnab (Σn ' ∂'n)a B'b = εnab ∂'a B'b where ' means what?? In x'-space, I would say ' is a unit vector and ( ')i = δn,i so I could think of these as the e'n . The factor (1/ J(x')) is missing! This is one reason the x notation is dangerous here. Maybe not. 2. What happens if we apply R to the second line? Rni Ci(x) = Rni εiab ∂aBb = Rni [ x B]i From Appendix A (j) I know that [' x B']n = cof(Rni) [ x B]i So this thing by itself is then not in general a tensorial vector, but maybe C'n(x') still is! 3. Could I still show that C'n(x') = Rni Ci(x)? That would require showing that (1/ J(x')) εnab ∂'aB'b = Rni εiab ∂aBb so we can start a sequence like this: εnab ∂'aB'b = J(x') Rni εiab ∂aBb ? εnab [Rad∂d] [RbcBc] = J(x') Rni εiab ∂aBb ? Suppose R is a linear transformation so then this is εnab RadRbc(∂dBc) = J(x') Rni εiab ∂aBb ? εnab RadRbc(∂dBc) = J(x') Rni εidc ∂dBc ? εnab RadRbc = J(x') Rni εidc ? Now use the result from 2 (f) of Matrix Addendum which says εna'b' Ra'dRb'c = εSdc cof(RnS) Then we can continue the chain above this way: εSdc cof(RnS) = J(x') Rni εidc ? εSdc cof(RnS) = J(x') εidc Sin ? Now apply Rnm to the right of both sides and sum on n: εSdc cof(RnS) Rnm = J(x') εidc Sin Rnm ? εSdc cof(RnS) Rnm = J(x') εidc (SR)im ? εSdc cof(RnS) Rnm = J(x') εidc δim ? εSdc cof(RnS) Rnm = J(x') εmdc ? | det(Rab)| εSdc cof(RnS) Rnm = εmdc ? Now use cof(RnS) = det(Rab) [ R-1]nS Then we have | det(Rab)| εSdc det(Rab) [ R-1]nS Rnm = εmdc ? Now suppose R-1 = RT . Is this valid for all positions of the R indices? Usually we have (R-1)ab = (RT)ab But we can jimmy indices any which way on such an equality, so therefore (R-1)ab = (RT)ab So replace R-1 by transpose | det(Rab)| εSdc det(Rab) [ RT]nS Rnm = εmdc ? | det(Rab)| εSdc det(Rab)RSn Rnm = εmdc ? σ εSdc RSn Rnm = εmdc ? Since σ = 1 for a rotation, yes this works out right. If d = c, both sides are 0 and equation is true. Consider this general fact cof(Rpq) = εiii...[i→q]...i R1i R2i ...... Rpi.......... RNi Specialize to N=3 cof(Rpq) = εi...[i→q]...i R1i ...... Rpi.......... R3i cof(R1q) = εqab R2a R3b and maybe we can translate this to say cof(R1q) = εqcd R2c R3d Then going back to εnab RadRbc = J(x') Rni εidc ? set n=1 for a test R2dR3c - R3dR2c = J(x') Rni εidc ? Nothing I do is going to generate that J(x') factor. ********************************************** Time to fiddle. Start with cof(R1q) = εqab R2a R3b Try writing this as cof(RPq) = (1/2) εPQRεqabRQaRRb Then cof(R1q) = (1/2)[ ε123εqab R2a R3b + ε132εqab R3a R2b } = (1/2)[ εqab R2a R3b –εqab R3a R2b } = (1/2)[ εqab R2a R3b –εqba R3b R2a } = (1/2)[ εqab R2a R3b –εqba R2a R3b } = (1/2)[ εqab R2a R3b +εqab R2a R3b } = εqab R2a R3b So at least for N=2 we seem to have cof(Rpq) = (1/2) εpabεqa'b'Raa'Rbb' This is totally new to me. How can I generalize this? Well, first lets' try to use the result. ************************** εSdc cof(RnS) = J(x') εidc Sin ? εSdc cof(RnS) = J(x') εidc (R-1)in ? Now I think we know that (R-1)in = cof(Rni)/ det(Rab) Then we have εSdc cof(RnS) = J(x') εidc cof(Rni)/ det(Rab) ? εidc cof(Rni) = J(x') εidc cof(Rni)/ det(Rab) ? Now we know that J(x') = 1/ |det(Rab)| = |det(Sab)| = |det(Rab)| Then we have εidc cof(Rni) = |det(Rab)| εidc cof(Rni)/ det(Rab) ? But for rotation det(R) = 1 so this says εidc cof(Rni) = εidc cof(Rni) ? So I have finally showed that, if R is a rotation, THEN C'n is a contravariant vector, same conclusion as before. Part II. Definition of the Jacobian and dealing with Area transformation On page 98 Weinberg writes | ∂x/∂x'| where his notation means the determinant with no absolute value and he calls this the Jacobian. I look on the web and nobody but me is putting an absolute value on this definition. I know that if you just change the ordering of one pair of coordinates, the Jacobian changes sign, which does not seem reasonable. One pdf I just found says that absolute(J) is what appears in the change of variables, but the "Jacobian" itself does not include this absolute value. Why did I add the absolute value? I defined it that way in Section 5 (k), that is true. But where then do I use it? I think I had a problem in Appendix A which led to my doing this. I defined E there with det(R) as a leading factor. All is well through sections a,b,c,d,e. In section (f) I properly find det(R)det(S) = 1, no problem there. So nothing in this Appendix A even mentions Jacobian or uses abs value at all on any det. What about in the dAn business in Appendix B? Look in the 2-piped section there where I say An = |det(S)| En . Why do I have an abs value here? I put it there in my definition of An for the N-piped so that An and En point in the same direction, it is my definition of An. This definition then creates a σ in the second formula An = |det(S)| En = |det(S)| det(R) (-1)n-1 Πxi≠n ei = σ (-1)n-1 Πxi≠n ei σ ≡ sign(det(S)) = sign(det(R)) I cannot have |det(R)| in the Ek formula because then I won't get the right e E dot product! So I think this is all find, and I always use |det(S)| and don't mention J in this appendix. So nothing there would change if I removed the abs from J. Also, my rule about the direction out or ingoing of An or En is rock solid, depending only on the eE thing, so that would not change. What about Section 8? Here there are several occurrences of J which I now make red as I search and find them. There were a LOT of hits in this section where I would have to make changes. But I think they are all benign. I would have |J(x')| appearing in a lot of places. What about Section 9? Lots more abs values would have to appear. Suppose I define J as I do and just comment that the true Jacobian is the thing without the absolute values, and why I am doing this. That might be a good fix saving me from hundreds of edits. OK, this is what I will do, just remember to do it *********. Weinberg confirms abs of J in his 4.4.6 for coordinate change. Here is the official J: Jw(x') = det(S) = 1/det(R) → J(x') = det(Sij) = 1/det(Rij) = det(Rji) = det(Rij) so you have to keep the tilt in mind. Here is how he always writes it 1/ Jw(x') = det(Rij) = det(∂x'i/∂xj) = | ∂x'/∂x| Jw(x') = det(Rij) = det(Sji) = det(∂xi/∂x'j) = | ∂x/∂x'| Now while we're at it, I show that det(') = J2 det() g'(x') = J2 g(x) = JW2 g(x) Translating this to Weinberg one would have g'(x') = J2 g(x) = | ∂x/∂x'| 2 g(x) which agrees with his 4.4.2. Now consider Weinberg page 99 4.4.4 | ∂x'/∂x| means det(∂x'/∂x) // clearly stated on page 98 = det(∂x'i/∂xj) = det(Rij) = 1/Jw Now here then is the rule for tensor density: J'μν = | ∂x'/∂x|W Rμμ'Rνν'Jμ'ν' = (JW)-W Rμμ'Rνν'Jμ'ν' Something transforming in this manner is a tensor density of weight W. Above we had g'(x') = J2 g(x) = JW2 g(x) so this g(x) object is a scalar density of weight -2, just as he says. Now here is a point I missed, You can therefore write JW2 = g'/g So consider J'μν = | ∂x'/∂x|W Rμμ'Rνν'Jμ'ν' = (JW)-W Rμμ'Rνν'Jμ'ν' = (g'/g)-W/2 Rμμ'Rνν'Jμ'ν' Then just rewrite this way (g'W/2 J'μν) = Rμμ'Rνν (gW/2 Jμ'ν') This says that the object (gW/2 Jμν) then transforms as a simple tensor! Aha! Maybe I should have two different J's, one with and one without abs. Here is another example: g'(x') = J2 g(x) = JW2 g(x) = (g'/g) g(x) just says 1 = 1. Now here is another example dV' = | JW-1| dV 4.4.6 // agrees with my dV = J(x') dV' This says that the volume element is a scalar density of weight -1, just as he says. Then dV' = JW-1 dV = (g'/g)-1/2 dV => g'1/2 dV' = g1/2dV = scalar! and this also agrees with his statement. I think I can make good use of this stuff! Observation: I have often commented that, when an object has |det(S)| as part of its expression, that object is not a tensorial tensor. But that statement is wrong, because if the rest of the expression is a tensor density of the right weight, then the whole object IS in fact a tensor! Review of some Section 8 items. There I say dAn = |det(Sab)| en Πi≠ndx'i // = dAn , see comment above What is the corresponding dA'n in x'-space? I seem to avoid saying this, but in that Cartesian-view x'-space I think one would say dA'n = 'n Πi≠ndx'i for example dA'3 = '3 dx'1dx'2 = top of my cube drawing. dA'3 = dx'1dx'2 Now what can I say about the vector nature of dAn ? This takes me back to the old problem that these two dA objects don't transform into each other. I guess I could write 'n = e'n / h'n Then dA'n = (e'n / h'n )Πi≠ndx'i S dA'n = (en / h'n )Πi≠ndx'i = g'nm em (1/ h'n ) Πi≠ndx'i = g'nm(1/ h'n ) [ em Πi≠ndx'i] = g'nm(1/ h'n ) dAm / |det(Sab)| = g'nm(1/ h'n ) dAm |det(Rab)| So we then get this strange transformation rule dA'n = R { g'nm(1/ h'n ) dAm |det(Rab)| } dA'n = { g'nm(1/ h'n ) |det(Rab)| } R dAm dA'n = { g'nm(1/ h'n )J-1} R dAm Now is this some kind of tensor density thing? I don't think so. We don't just have a power of the Jacobian sitting in there. Even if orthogonal we would have dA'n = { h'n J-1} R dAm which again is not a vector density. It would be if I did not use the hat at the start. Observation: certainly for a non-orthogonal system, the object dAn = |det(Sab)| en Πi≠ndx'i is neither a tensorial vector, nor a tensorial vector density. [ wrong! it is a tensorial vector density.] Try again. Start again with dAn = |det(Sab)| ( Πi≠ndx'i) en What happens when I map this object into x'-space? dA'n = R dAn = |det(Sab)| ( Πi≠ndx'i) R en But what is R en ? [R en]i = Rij (en)j = Rij gja Rna = Rij Rna gja = g'in On the other hand, I would have thought that this was true [R en]i = (e'n)i = g'in R en = e'n and on the right I have looked it up in the new table and it agrees! So the conclusion is then that dA'n = R dAn = |det(Sab)| ( Πi≠ndx'i) R en = |det(Sab)| ( Πi≠ndx'i) e'n So we find then that dA'n = |det(Sab)| ( Πi≠ndx'i) e'n dAn = |det(Sab)| ( Πi≠ndx'i) en Somehow it seems that, since dA'n = R dAn , I have forced dAn to contravariant by definition! What are the magnitudes? enen = g'nn = (1/h'n2) => | en | = 1/h'n. So this would be the same as | e'n |. Then I would say that dA'n = dAn // magnitude comparison which seems to disagree with my famous cofactor gizmo. Perhaps here I have tried for force dAn to be a contravariant vector, and that is leading to a contradiction. I want this to be in x'-space (Curvilinear view) dA'n = ( Πi≠ndx'i) e'n Then one could say dA'n = |det(Sab)|-1 R dAn and then dAn is transforming as a vector density! If true, this is a new discovery for me. But now where is that cofactor magnitude thing going to come from??? I seem to get now: dA'n = ( Πi≠ndx'i) | e'n | = ( Πi≠ndx'i)( 1/h'n) dAn = |det(Sab)| ( Πi≠ndx'i) ( 1/h'n) This then suggests that dAn = |det(Sab)| dA'n and my cofactor has vanished! I end up with the volume rule ?? Well I showed in tensor that dAn = |det(Sab)| ( Πi≠ndx'i) en => dAn = ( Πi≠ndx'i) So it is true that |det(Sab)| ( 1/h'n) = To see why this is true, square both sides |det(Sab)|2 g'nn = cof(g'nn) det(g'ab) g'nn = cof(g'nn) etc So the whole issue is just this ( 1/h'n) factor and that is back to the Cartesian view thing. Reconstruct the above pieces. (a) Assume that dAn transforms as a vector density, so that (ignoring the abs val issue) dA'n = |det(Sab)|-1 R dAn = |J|-1 R dAn => weight = -1 (b) In Cartesian x-space, we seem to know without ambiguity that dAn = |det(Sab)| ( Πi≠ndx'i) en In x-space, there is no distinction between covariant and Cartesian magnitudes, and we have dAn ≡ |dAn| = |det(Sab)| ( Πi≠ndx'i) (1/h'n) // = |dAn|C = { |det(Sab)| (1/h'n) } ( Πi≠ndx'i) = { } ( Πi≠ndx'i) (c) From the tensor density rule (a), this is compatible with this x'-space object dA'n = ( Πi≠ndx'i) e'n for which we know that |dA'n| = ( Πi≠ndx'i) (1/h'n) |dA'n|C = ( Πi≠ndx'i) ≡ dA'n (d) We find then that |dAn| / |dA'n|C = dAn/ dA'n = Part III: The Covariant Curl 1. Weinberg casually mentions that this is the covariant curl, page 106 Bμ;ν - Bν:μ = ∂νBμ - ∂μBν because the fancy cross terms cancel. Let's check on that right here: Bμ;ν = ∂νBμ – Γλμν Bλ Bν;μ = ∂μBν – Γλνμ Bλ But Γλμν written in any of its forms is symmetric in μ and ν, so QED. 2. Could I formulate the entire curl section in terms of the above object? I could first show the following fact (as I have already done) ( Bdx)3 = [∂'1B'2 – ∂'2B'1] dx'1 dx'2 But this entire line integral concept only works for N=3. In N=4 with a 4-cube, the areas are 3-cubes and what is your integral? You somehow might want to get the above for some line integral buried in that thing somewhere. For any N, eventually there are some loops you could integrate around: N=4 face = 3-cube sub-face = 2-cube = pgram = something you could int around N=5 face = 4 cube sub-face = 3 cube sub-sub-face = 2-cube You could just pick some particular sub sub...sub face and integrate around it and get something like the above. Perhaps you would do this Fij dx'idx'j ≡ ( Bdx)ij = [∂'iB'j – ∂'jB'i] dx'idx'j Then you could refer to Fij as "the curl tensor". Would the edges of this pgram be these? ei and ej two out of N ? I think so. Pick any pair, and they do define a p-gram no matter how large N is. So my integral evaluate goes through and I get the result exactly as shown above for the RHS. But then what? I have nowhere to go unless I come up with a vector area element. For an N-piped, what direction would such a 2-face point? In the cube, we use e1 x e2 to describe this face's normal arrow in 3-space. But this cross product is not defined for N=4. This piece of area is not an N-vector, it is a tensor which has no direction. Well, again, there must be some concept of this area in the sense of dAn of the N=3 case. The two vectors e1 and e2 do exist in x-space which is Cartesian N-space. The volume of the 2-piped we showed volume(2) = | e1 x e2 | = | εab (e1)a(e2)b | = | det [ e1, e2 ] | where I leave off the |det(S)| item. So at least we know the size of the area dA12 in terms of the ei vectors. But these formulas are no good because εab means nothing for general N. The det is not defined for a non-square matrix. Still, you would think there was some "area" associated with the face spanned by the vectors e1 and e2. They span a 2D surface I think. How about this idea for N=4 A12 = | εab34 (e1)a(e2)b | This is the idea used going from 2-piped to 3-piped and here we take it up one more. OUCH. I think I just found a bug in Appendix B. Yes it was a very large one having to do with face area, and it is now completely fixed after spending about an hour doing it! Resume. If you express the 4D e1 in the en expansion, only the 1st component is non-zero. And if you express e2, only the 2nd component is non-zero. A linear combination of these would be points on my area of interest, my "face" of 2 dimensions down from the space dimension. Therefore, I think points on this face would have 3rd and 4th coordinates zero. Then the A34 thing I show above might be right! IN that formula, e1 and e2 are 4-vectors, but each only has the first two components non zero. It then all boils down to the original pgram e1 x e2 for the area. Then maybe dA12 = | εab34 (e1)a(e2)b |dx'3dx'4 This I think "really is" the area of the thing I am interested in. now consider εab34 (e1)a(e2)b Is this perhaps a double cofactor? What you get crossing out two rows and two columns? M&M don't mention such things, perhaps it would be called a second order minor or some such? The matrix obtained by crossing out the 3rd row and 3rd column is this M = cof(S33) Mij = [cof(S33)]ij a 3x3 matrix where this matrix has only rows and columns 1,2 and 4. Now cross out row and column 4: N = cof(M44) = cof([cof(S33)]44) Nij = cof([cof(S33)]44)ij = cof2(S33,S44)ij // making up a notation Notation really only works if two elements are in different rows and different columns. So then I conjecture that εab34 (e1)a(e2)b = cof2(S33,S44)12 Well, this is going nowhere fast, just a strange idea, let it rest for now. Well suppose this were valid, where would you go with it? perhaps Fij dx'idx'j ≡ ( Bdx)ij = [∂'iB'j – ∂'jB'i] dx'idx'j = CijdAij and Cij is the curl tensor? I keep thinking dyadic where you do things like A = Σij Aij ei ej I used something like this somewhere, very dim. Part IV: The regular N=3 Curl. As things now stand, I get to this point with things: C dAn = ( Bdx)n C ≡ curl B ( Bdx)n = εnab ∂'aB'b ( Πi≠n dx'i) I then process the LHS using these two items, C = Σk=13 C'k ek dAn = J en ( Πi≠n dx'i) and doing this I find that C dAn = C'n J ( Πi≠n dx'i) I then compare the two sides to obtain C'n J = εnab ∂'aB'b At this point, I could "activate" the εnab object into a full tensor. NOTE: At this point I took a long digression on ε, and ended up writing Appendix D. If should be noted that some authors use W → -W for the weight. ******************************** Application. Thinking of the above as in some appendix. Now how would I USE the above data? Back to the curl. I end up with C'n(x') = (1/ J(x')) εnab ∂'aB'b = J-1 ε'nab ∂'aB'b What can I say about the vector nature of C', if anything? First, write it as C'n(x') = (1/ J(x')) ε'nab (1/2) [ ∂'aB'b – ∂'bB'a] = (1/ J(x')) ε'nab (1/2) [ B'b;a - B'a;b] with the prime now on ε', as per above. We also have Cn(x) = εnab ∂aBb ε'nab (1/2) [ ∂aBb – ∂bBa] = εnab (1/2) [ Bb;a - Ba;b] Let's now assume that Cn transforms as a vector density with some weight W: C'n(x') = J-W Rnn'Cn'(x) Installing the pieces above, (1/ J(x')) ε'nab (1/2) [ B'b;a - B'a;b] = J-W Rnn' εn'ab (1/2) [ Bb;a - Ba;b] (*) The LHS pieces may be written ε'nab = J Rnn' Raa'Rbb'εn'a'b' [ B'b;a - B'a;b] = Rac'Rbd' [ Bd';c' - Bc';d'] Then LHS = J-1 (1/2) J Rnn' Raa'Rbb'εn'a'b' Rac'Rbd' [ Bd';c' - Bc';d'] = (1/2) Rnn' (Raa'Rac')( Rbb'Rbd') εn'a'b'[ Bd';c' - Bc';d'] = (1/2) Rnn' δa'c'δb'd' εn'a'b'[ Bd';c' - Bc';d'] = (1/2) Rnn' εn'a'b'[ Bb';a' - Ba';b'] = Rnn' εn'ab(1/2) [ Bb;a - Ba;b] But this equals the RHS of (*) with W = 0. Therefore it has been shown that Cn(x) = [ curl B ]n(x) transforms as a contravariant vector under any F. This is the result I was hoping to find. How then does this relate to my Appendix A (j) where I claim cross products are NOT contravariant unless F = a rotation? Let's do it all from scratch Qa ≡ εabc...x BbCcDd.....Xx Then Q'a = ε'abc...x B'bC'cD'd.....X'x = ( J Raa'Rbb'Rcc'...Rxx' )εa'b'c'...x' Rbb" Rcc".... Rxx" Bb"Cc"... Xx" = J Raa'(Rbb'Rbb")( Rcc'Rcc") ....( Rxx' Rxx") εa'b'c'...x' Bb"Cc"... Xx" = J Raa' εa'b'c'...x' Bb'Cc'... Xx' = J Raa'Qa' and we conclude that Qa transforms as a contravariant vector density of weight -1. Now look how this plays out here Ek ≡ det(R) (-1)k-1 e1 x e2 x ......x eN // ek missing; N > 2 = J-1 (-1)k-1 e1 x e2 x ......x eN // ek missing; N > 2 Here the cross product has weight -1, the factor J-1 has weight + 1, so Ek comes out pure vector! Hurray, lots of pieces are falling out now. Another piece: that M = cof(R) thing has also fallen into place, all consistent. What about dAn ? dAn = |det(Sab)| en Πi≠ndx'i = J en Πi≠ndx'i [dAn]α = |det(Sab)| (en)α Πi≠ndx'i = J (en)α Πi≠ndx'i This looks like a contravariant vector density of weight -1 because of the J factor. The other form: dAn = σ (-1)n-1 Πxi≠nei Πi≠ndx'i If we think of the ei as covariant vectors, then [Πxi≠nei]α = εαab..... (e1)a(e2)b .... = contravariant vector density of weight -1. Then [dAn]α = σ (-1)n-1 [Πxi≠nei]α Πi≠ndx'i = contravariant vector density with weight -1 So both expressions give the same weight! No they don't agree. The first seems to say the weight is the same whether I put α up or down! Continue Mon Nov 14. Problem #1: Consider Cn = J Bn where Bn is a contravariant vector. What is Cn ? We know that B'n = RnaBa , but what are we supposed to do with that J sitting there? In x'-space is it still J ? In my curl case, I was able to find an expression for C'n by other means, and then I also knew Cn. I guess you could do this: C = C'nen but that might not be true for a density C! Probably not. Do we have some examples? You need to know "how it transforms" to get your answer. Well, suppose I just apply R to both sides of Cn = J Bn . But if Cn is not a vector, I don't think I can just say that RC = C' in the usual way. So maybe there is no solution to this problem without further information? But Cn is 100% defined, what more could you want? Is J itself a tensor density? The answer is a resounding NO. I think this problem does not have a solution and is in fact ill-posed. On the other hand, I have a real-world example above which is this: (setting all dx'n = 1) dA1 = |det(Sab)| e1 dA1 ~ Πxi≠nei // Cartesian only ~ e2 x e3 In the second case I write (dA1)α = [e2 x e3] = εαab(e2)a(e3)b Since everything on the right side has a known transformation rule, I know (one rank-3 tensor density and two covariant vectors), I know that (dA1)α is a vector density with a certain weight. But how do I arrive at this same conclusion from the equation dAn = |det(Sab)| en ?? Ah, I do know that dA'n = e'n I think. I am back now to "x'-space mysteries". In N=3, dA'3 = e'3 dx'1dx'2 is "what I want to say". But where does this come from? dA'3 = (0,0, dx'1dx'2) since e'3 = (0,0,1). So I need a formula for dA that works for any g? Look at the App C picture What do you MEAN by area in x'-space, how would you even define the concept? Maybe this is a topic I need to formally address in tensor.doc. Operationally, we always mean dρdθ as "the area". OK, let's stay with the existing plan. Then we have dA'n = e'n dAn = |det(Sab)| en = J en Are en and e'n related in some way? Well en = g'nmem = g'nm Se'm // this last from Section 3 (a) So this seems to indicate that dAn = J g'nm Se'm = J g'nm S dA'm = J S dA'n Now suddenly the position on the dAn label is starting to matter! This last result is then dA'n = J-1 R dAn => dAn has weight +1 . Now go back to the "other method": (dA1)α = [e2 x e3] = εαab(e2)a(e3)b Here since εαab has weight -1, I get a different answer! Well how about just playing with the e vectors and forget dA for a while. Appendix A: e1 = det(R) e2 x e3 = J-1 e2 x e3 (e1)α = J-1 εαab(e2)a(e3)b I think I know that εαab(e2)a(e3)b "does", given that en are covariant vectors. So I am back to my original "product question". I would like to say that εαab(e2)a(e3)b has weight -1 (agrees Wein) and J-1 has weight +1, so the product is then a pure vector. But I think that argument is wrong. What I do know is this e'1 = g'1a e'a (e'1)α = g'1a (e'a)α = g'1a δaα = g'1α // agrees with my table So maybe now I can deduce a transform rule, given that (e'1)α = g'1α (e1)α = J-1 εαab(e2)a(e3)b // = g'1i ei But I also know that (I think I know) e'1 = R e1 e1 is a true vector e1 = J-1 e2 x e3 = g'1n en Using this set of equations one can then say e'1 = R e1 = J-1 R e2 x e3 Getting nowhere fast. Go back to Fα ≡ εαab(e2)a(e3)b F'α ≡ ε'αab(e'2)a(e'3)b = (J Rαα'Raa'Rbb' εα'a'b') (Rac(e2)c) ( Rbd(e3)d) = (J Rαα'εα'a'b') (e2)a' (e3)b' = J Rαα' εα'ab(e2)a(e3)b = J Rαα'Fα' = J RαaFa and this says that object F has weight -1. I am just repeating something I did earlier. e1 = J-1F e'1 = J-1F' = RF = J R e1 wrong! ***************** C ≡ J B B' = RB JB' = JRB = R(JB) = RC Conjecture that C' = J-1 RC Then C' = J-1 JB' = B' I need another source on tensor densities, Weinberg is not sufficient to answer my extremely simple questions. So another week on this subject I guess. Wiki on tensor density makes a point that I did not appreciate. Fact: If x-space is Cartesian (and can generalize to Quasi-Cartesian), then J2 = g'(x') = det('(x')) and in this case you can regard J as something that exists in x'-space and then you can talk about how J transforms! In general relativity, you would say "if x-space is a locally inertial coordinate system" and g there is the Minkowski metric (-1,1,1,1). Now with this in mind, let's as the question again. Suppose x-space is Cartesian, and consider: C ≡ J B = B = J(x') Consider this as an equation in either space. Then we have C = B // in Cartesian x-space C' = B' // in x'-space Now it follows that C' = RB = R C = J R C and this last shows that C' transforms in such a way that it is a vector density of weight -1. This is the result I have been wanting to see. Here is how the wiki site states it more generally in the relativity context: So I will state a theorem for my own use: Theorem 1: If x-space is Cartesian, then consider the product of two objects, P = J-W Q J = where Q is an ordinary tensor, and where J-W is a function. Just for clarification, we can write this equation as follows: P(x) = Q(x) // x-space, since g = 1 there, so J = 1 P'(x') = J'-WQ'(x') // x'-space with g = g' so J' = J'(x') = The theorem then claims that object P is a tensor density of weight W. Proof: P'(x') = J'-WQ'(x') = J'-W [ R R R...R Q(x) ] = J'-W [ R R R...R P(x) ] and this last line is the definition of a tensor density of weight W (Weinberg convention). Theorem 2: In the previous theorem, if Q is a tensor density of weight W1 , then P as defined there will be a tensor density of W1 + W. This just follows from the direct product rule. Now more playing in the sandbox before the next theorem. Suppose Cα is a contravariant vector of weight W. That tells us that C'α = J-W RαaCa What can we say about the covariant vector Cα ? In the usual way, applying g to both sides gives C'α = J-W RαaCa and I believe this says that Cα also has weight W. This agrees with Weinberg's statement page 100 in item (C): "raising and lowering indices does not change the weight of a tensor density". But I thought that the following was true: εabc... = tensor of weight -1 (says Weinberg page 99) εabc... = tensor of weight -1 (says Weinberg page 100) So he is consistent. But I thought someone said differently somewhere. Our same tensor density site says this: (he/she may use the negative of the Weinberg convention, yes! ) Now consider this strange paragraph: But this does not agree at all with what I do because he has that extra factor of g sitting there. I don't like this one bit, because then ε is not a tensor object that I am used to, density or otherwise. I Book by Michael Spivak mentioned. Application of Theorem 2 Recall e1 = det(R) e2 x e3 = J-1 e2 x e3 where x-space is Cartesian! Consider now Qα = εαab(e2)a(e3)b