new dAn fiddle

New dAn fiddling 1. Initial assumptions in x-space I am pretty sure these two expressions for dAn are correct: dAn = |det(Sab)| en Πi≠ndx'i = g'1/2 en Πi≠ndx'i dAn = σ (-1)n-1 Πxi≠nei Πi≠ndx'i = σ (-1)n-1 { εkabc...x (e1)a(e2)b ...... (eN)x } Πi≠ndx'i // n missing Correspondingly we have dx(n) = en dx'n dV = | dAn dx(n)| = |σ (-1)n-1 εkabc...x (e1)a(e2)b ...... (eN)x Πidx'i| = | σ det(S) Πidx'i | = |σ (g')1/2 Πidx'i| = (g')1/2 Πidx'i which results are consistent with Appendix B. The 3 quantities listed here have absolutely clean and unambiguous meaning in terms of the differential N-piped in x-space. It is useful to define the following "coordinate variation products" as follows, for later use: dL'n ≡ dx'n dA'n ≡ Πi≠ndx'i dV' ≡ Πidx'i = dA'n dL'n and then the above become dx(n) = en dL'n dAn = σ (-1)n-1 Πxi≠nei dA'n = g'1/2en dA'n dV = (g')1/2 dV' The quantities dL'n, dA'n, dV' may regarded as length, area and volume in the Cartesian-view of x'-space as discussed *****. As noted there, this is just a visualization tool. 2. Weights The weights of our three assumed tensor objects are dx(n) weight = 0 vector dAn weight = -1 vector density dV weight = -1 scalar density Throughout, we regard things like dx'n and Πi≠ndx'i and Πidx'i as positive scalar constants, and have been calling them the "variation products" of the curvilinear coordinates. The weight for dAn is obvious from the second formula above, given that the en have weight 0 and the true ε tensor has weight -1. This weight is NOT obvious from the formula d Πi≠ndx'i, although I have tried hard to extract the weight from this formula. It is an equation involving x-space objects en and dAn but includes the x'-space object g'1/2 and therefore does not fit the covariant mold. It might fit the "cross mold" as follows An = g'1/2 en A'n = g1/2 e'n // wrong! but that is pure conjecture. If these were true, we could apply WCT#2 with w=+1 and en having weight 0, but this would tell us that An had weight W+w = 0+1 = +1, and this is wrong. So forget this idea completely!! 3. The objects in x'-space. Assuming all three objects are tensors of weights shown, we may conclude that dx'(n) = R dx(n) dA'n = J R dAn = g'1/2 R dAn dV' = J dV = g'1/2 dV The first object may be written dx'(n) = R [en dx'n] = [Ren] dL'n = e'n dL'n so that [dx'(n)]i = (e'n)i dL'n = δni dL'n For the third object, using the expression dV = (g')1/2 dV above, one may write dV' = g'1/2 dV = g' dV' The remaining second object dA'n object may be calculated as follows: (dA'n)i = g'1/2 Rik (dAn)k = // g'1/2 = |J| = |J|-(-1), weight = -1 = g'1/2 Rik σ (-1)n-1 { εkabc...x (e1)a(e2)b ...... (eN)x } dA'n // n missing = g'1/2 Rik σ (-1)n-1 { εkabc...x Sa1 Sb2 ...... SxN } dA'n // n missing = g'1/2 Ski σ (-1)n-1 { εkabc...x Sa1 Sb2 ...... SxN } dA'n // n missing = g'1/2 σ { (-1)n-1εkabc...x Sa1 Sb2 .. .. Ski..... SxN } dA'n = g'1/2 σ {εabc..k..x Sa1 Sb2 .. .. Ski..... SxN } dA'n If i ≠k, then two S second indices must be the same and result is 0. If i =k, then {} = det(Sab) = σg'1/2. Therefore (dA'n)i = δni g' dA'n To summarize, the x'-space objects can be written in this manner [dx'(n)]i = δni dL'n length (dA'n)i = δni g' dA'n area dV' = g' dV' volume 4. Evaluation of x-space length and area magnitude. Consider from above dx(n) = en dL'n The magnitude squared is given by | dx(n)|2 = dx(n) dx(n) = (dL'n)2 en en = g'nn (dL'n)2 = h'n2(dL'n)2 so that dx(n) ≡ | dx(n)| = hn' dL'n Turning now to the area, from above dAn = g'1/2en dA'n so the magnitude squared is given by |dAn|2 = dAn dAn = (g'1/2 dA'n)2 en en = (dA'n)2 g' g'nn = (dA'n)2 g' (1/h'n)2 . Therefore one way to write the magnitude is this dAn ≡ |dAn| = (1/h'n) g'1/2 dA'n Another way to write this same dAn arises from the following fact: g' g'nn = cof(g'nn) which we now prove: (g'up)ab ≡ g'ab (g'dn)ab ≡ g'ab g'up = (g'dn)-1 = cof(g'dnT)/det(g'dn) = cof(g'dn)/det(g'dn) => (g'up)nn = cof[(g'dn)nn]/det(g'dn) or g'nn = cof[g'nn] / g' QED Therefore, one can write |dAn|2 = (dA'n)2 g' g'nn = (dA'n)2 cof(g'nn) and then one ends up with two forms for the area magnitude dAn ≡ |dAn| and one form for | dx(n)\ dAn = (1/h'n) g'1/2 dA'n dAn = dA'n dx(n) = h'n dL'n 5. Evaluation of x'-space length and area magnitude. dx'(n) = dx(n) since weight = 0 dA'(n) = g'1/2dA(n) since weight = -1 dV' = g'1/2 dV since weight = -1 Therefore dx'(n) = hn' dL'n dA'(n) = g'1/2 dA'n = (1/h'n) g' dA'n dV' = g' dV' 5. Summary of all results dL'n ≡ dx'n dA'n ≡ Πi≠ndx'i dV' ≡ Πidx'i = dA'n dL'n x-space x-space mag length dx(n) = en dL'n dx(n)= hn' dL'n area dAn = g'1/2en dA'n dAn = dA'n = (1/h'n) g'1/2 dA'n volume dV = (g')1/2 dV' dV = (g')1/2 dV' x'-space x'-space mag length (dx'(n))i = δni dL'n dx'(n) = hn' dL'n area (dA'n)i = δni g' dA'n dA'(n) = g'1/2 dA'n = (1/h'n) g' dA'n volume dV' = g' dV' dV' = g' dV' What is the famous volume result? dV = (g')1/2 dV' dV = J dV' and this agrees with Section 8.