old dAn fiddle

Fiddling with dAn PhL 11.18.11 My most recent belief is that dAn really does have weight -1 by Argument #2. Argument #1: weight comes out +1 Section 8 **** states that An = en = (sg')1/2 en A'n = e'n = (sg)1/2 e'n how justified?? // g = 1, s = 1 Taking a covariant component on both sides gives (An)i = (sg')1/2 (en)i (A'n)i = (sg)1/2 (e'n)i The reciprocal base vector en transforms as a regular tensorial vector (weight = 0). Weight Changing Theorem #2 from section (b) 7 can now be applied with w = +1 to show that (An) therefore transforms as a vector density of weight W + w = 0 + 1 = +1 Argument #2: weight comes out -1 // hard to deny this argument But now I am in trouble, because the other form dAn = σ (-1)n-1 Πxi≠nei seems to have weight -1: (An)i = σ (-1)n-1 εiabc..x (e1)a(e2)b.... (eN)x // en missing This is an outer product followed by contraction so things are additive. All the vectors are weight 0, and ε is weight -1, so the result must then be that A has weight -1. I think an implication of this claim is that (A'n)i = J Rij(An)j = Rij(An)j = σ (-1)n-1 Rij εjabc..x (e1)a(e2)b.... (eN)x = σ (-1)n-1 Rij εjabc..x Sa1Sb2....SxN = σ (-1)n-1 Sji εjabc..x Sa1Sb2....SxN where n is missing as usual. If i ≠n get 0 by usual argument. If i = n get (A'n)i = σ det(Sab) = σ σ = g' and we are forced then to have this result (A'n)i = g' δni I don't know how to interpret this result. The magnitude would be | A'n|2 = g'ij (A'n)i(A'n)j = g'2 g'ij δni δnj = g'2 g'nn = g'2 / h'n2 => | A'n| = g' / h'n Here is a little comparison: object in x-space object in x'-space magtrue magCartview en e'n g'nn 1 dAn dA'n g' (g'nn)1/2 1 dV dV' g' 1 Maybe here think of dV' = dA'n e'n = (dA'n)i (e'n)i = g' δniδni = g'. Back in x-space get An en = (An)i (en)i = σ (-1)n-1 εiabc..x (e1)a(e2)b.... (eN)x (en)i = σ det(S) = σ σ g'1/2 = g'1/2 If en has weight 0 and An has weight -1, then An en has weight -1 and we expect to find A'n e'n = |J|+1 An en = g'1/2 An en and that is exactly what we find! Here is the table corrected object in x-space object in x'-space magtrue magCartview endx'n e'ndx'n (g'nn)1/2dx'n dx'n ≡ dx'n dAn Πi≠ndx'i dA'n Πi≠ndx'i g' (g'nn)1/2 Πi≠ndx'i Πi≠ndx'i ≡ dA'n dV = g'1/2 Πidx'i dV' = g' Πidx'i g' Πidx'i Πidx'i ≡ dV' Notice with these meanings that we get dV' = g'1/2dV = J dV, which is just backwards from the usual statement, because these are not the meanings of dV and dV' in the usual statement. Argument #3: weight comes out +1 dAn = |det(Sab)| en Πi≠ndx'i = en Πi≠ndx'i (en)i = Rni dA'n = e'n Πi≠ndx'i // where from? (e'n)i = δni Then have (dAn)i = Rni Πi≠ndx'i (dA'n)i = δni Πi≠ndx'i What "theorem" applies here? Throw out the constant variations and say (An)i = Rni = Rmi δnm = Rmi(A'n)m Invert this to get (A'n)i = (1/) Rim (An)m This is the best yet. Since g = 1, can write this as (A'n)i = (g'/g)-1/2 Rim (An)m = |J|-1 Rim (An)m and we finally conclude that (An)m is a covariant vector density of weight +1. Argument #4 is in favor of weight = +1 // wobbly argument When I get to the div B think I will have div B = (1/dV) sum (BdA) and B is a regular vector, so I want dA/dV to be a regular vector. But we know dV' = J-1 dV so dV has weight +1 // agrees Weinberg p 99 Then I guess we can invert both sides (1/dV') = J (1/dV) and conclude that 1/dV is a scalar density of weight - 1. Then in the divergence thing we get div B : W = -1 + 0 + 1 = 0 true scalar Possible div B argument in favor of weight = -1 Once again, we have div B = (1/dV) Σn ∫ (BdAn) [div B] ' = (1/dV') Σn ∫ (BdA') In this second line I think of dA' as a true thing in x'-space, and dV' as the true volume there. The table above shows that dV' = g'1/2dV = J dV. Since dA'n = J dAn as well, we get cancellation of the J's and the two RHS's above are the same, and this shows why div B is a tensorial scalar of weight 0 ! *************************************************************** First discovery that (A'3)i = δ3i g' OK, stick with N=3 only, and consider this vector only (assume σ = 1) A3 = e1 x e2 The is pretty simple. A component would be (A3)i = [e1 x e2]i = εijk (e1)j(e2)k This seems to say weight A3 is -1. Can I verify that? If true, then we would have (A'3 )i = J Rij (A3)j I guess the primed world would say (A'3)i = [e'1 x e'2]i = ε'ijk (e'1)j(e'2)k = ε'ijk δ1jδ2k = ε'i12 = det(g'ij) εi12 = δ3i det(g'ij) = δ3i g' This seems a bit strange, because I had been thinking that A'3 was just an area in x'-space which would have magnitude 1 in the current context. So maybe I am onto something. Then we should find that (A'3)i = J Rij (A3)j Filling in, this says δ3i g' = g'1/2 Rij εjab (e1)a(e2)b = g'1/2 Rij εjab Sa1Sb2 = g'1/2 Sji εjab Sa1Sb2 = g'1/2 εjab Sji Sa1Sb2 If i = 3 this says g' = g'1/2 det(Sij) = g'1/2 g'1/2 = g' so things are consistent. How do I interpret this claim: (A'3)i = δ3i g' Try to interpret (A'3)i = δ3i g' in spherical coordinates. Take spherical coordinates to be specific, where g' = det(gij) = r4sin2θ. So dA3 = e1 x e2 dx'1dx'2 dAφ = er x eθ dr dθ = r x θ hrhθ dr dθ = φ dr rdθ and this seems quite reasonable. Now what does this mean: (A'3)i = δ3i g' = (e'3)i g' (A'φ)i = (e'φ)i r4sin2θ ??? (e'φ)i = (0,0,1) in x'-space Add the differentials now (dA'φ)i = (e'φ)i r4sin2θ dr dθ | dA'φ |2 = g'ij (A'φ)i(A'φ)j = g'ij δ3i g' δ3j g' = g'33 g'2 = h'3-2 g'2 | dA'φ | = h3'-1 g' drdθ = (rsinθ)-1 r4sin2θ drdθ = r3sinθ drdθ huh? But since dAφ = φ dr rdθ we get in x-space |dAφ| = r drdθ But I though we are supposed to have the same? but that is only if dA has weight 0 ! Question: what is the magnitude of a vector density? Suppose A and B are vector densities of weight w1 and w2. Then consider this object Dij = AiBj This object is an outer product, and thus has weight w1+w2. If we contract, it is still true. So AaBa = scalar density of weight w1 + w2. AaAa = scalar density of weight 2*w1. Let's define |Aa|2 ≡ AaAa Then |Aa| is a scalar density of weight w1. Attempt to understand | dA'φ | What happens in x'-space? I presume we make this definition there |A'a|2 ≡ A'aA'a Then it seems clear that it all works. |dA'| = J-w1|dA| // OK That is the new fact among us! Let's take her out for a spin: | dA'φ | = r3sinθ drdθ | dAφ | = r drdθ w1 = -1 J-w1 = J = r2sinθ Then should get |dA'| = J-w1|dA| = J |dA| r3sinθ drdθ = r2sinθ (r drdθ) and our test drive is OK. This object |dA'| is not very intuitive! Recall for volume that we got dV = J dV' which is backwards from this area result. But this is a different dV' !!