appendix A covariance fiddling

Appendix A covariance fiddling (*) Covariance of the Ek formula of Appendix A. 1 Conclusion: 7 Previous notes on this subject, the conclusion is exactly the same, 7 Here are still other old notes, this is where I first tried the ω = -2 weight for Ek idea: 9 Most of this stuff is wrong, but I keep it anyway. I know that { εαabc...x (e1)a(e2)b ...... (eN)x } is a tensor density of weight -1, but I could never prove from this fact that (Ek)α is an ordinary vector because this thing did not fit my weight changing theorem form. But I know Ek is an ordinary vector from the other formula for Ek (*) Covariance of the Ek formula of Appendix A. First a review of this Appendix. The strange Ek formula is stated and rewritten with cross product notation finally in section (e). Dot products are computed in (f) and (g) and summarized in (h). Another notation is given in (i). Only in the very last section (j) is there mention of something in x'-space and that is the bit about rotations. So perhaps this new attack if successful will result in a replacement for section (j) and everything else can stand as is. Maybe I should delete section (j) and broach this topic as a section of Appendix D, where I am addressing the covariance of various objects and concepts in the rest of the doc. Appendix A describes the following object (Ek)α ≡ det(R) (-1)k-1{ εαabc...x (e1)a(e2)b ...... (eN)x } // κ(k) and (ek)κ are missing; N≥ 2 which is found to be equivalent, for Cartesian x-space (g=1), to the object Ek = g'kiei defined as such in Section 6 for an arbitrary x-space. The vectors Ek are the reciprocal base vectors and exist in x-space along with the tangent base vectors en. One wonders how the above equation might be "transformed into x'-space". The first step is to rewrite the above expression in standard notation, where det(R) = J -1 = σ|J|-1 = σ g'-1/2. In the curvilinear coordinates application, g' > 0 (s = +1) according to Section 5 (d), while σ = sign [ det(Sij)]. (Ek)α ≡ σ |J|-1 (-1)k-1{ εαabc...x (e1)a(e2)b ...... (eN)x } // κ(k) and (ek)κ are missing; N≥ 2 Use has been made of section (g) which shows that the cross product of contravariant vectors is a covariant vector density. It will now be assumed that Ek transforms as a vector density of some weight ω which we seek to determine. Part I. The plan is to make substitutions in the above equation. Start then with (E'k)i = |J|-ω Rij(Ek)j or Ek' = |J|-W R Ek Inverting the above gives (Ek)α = |J|+ω Sαj(Ek')j = |J|+ω Rjα (Ek')j (1) Meanwhile, from Section 3 en was defined as a contravariant vector so that (e'n)i = Rij(en)j which inverted gives (en)i = Sij(e'n)j = Rji (e'n)j // e.g., (e1)a = Ra'a (e'1)a' (2) Installing (1),(2) into the x-space equation for Ek then gives |J|+ω Rjα (Ek')j = σ |J|-1 (-1)k-1 { εαabc... Ra'a (e'1)a' Rb'b (e'2)b' ... Rx'x (e'N)x' } // k missing Applying Rnα to both sides and summing on α and using Rnα Rjα = δnj [ Section 7 (r) ] , |J|+ω (Ek')n = σ |J|-1 (-1)k-1 {εαabc... Ra'a (e'1)a' Rb'b (e'2)b' ..... Rnα....... Rx'x (e'N)x' } // k missing From Section 3 (a) one has (e'1)a' = δ1a' and so on, so then |J|+ω (Ek')n = σ |J|-1 (-1)k-1 { εαabc... R1aR2b ..... Rnα....... RNx } // k missing or (Ek')n = σ |J|-1-ω { (-1)k-1εαabc... R1aR2b ..... Rnα....... RNx } // k missing If n ≠ k, then n matches some other first index of an R factor, so {} = 0 by symmetry. If n = k, then Rnα = Rkα and in this case {..} = det(Rij) = det(Sij) = J = σ|J|. Therefore, (Ek')n = δk,n σ |J|-1-ω σ|J|+1 = |J|-ω δk,n = g'-ω/2 δk,n To summarize Part I, if it is assumed that Ek is a vector density of weight ω, then Ek in x'-space comes out being the following (Ek')α = g'-ω/2 δk,α . Part II. Is there a value of ω which causes the following equation to be true: (E'k)α ≡ g-1/2 (-1)k-1{ ε'αabc...x (e'1)a(e'2)b ...... (e'N)x } // k missing If there were such an ω value, then the following two equations would be valid (Ek)α ≡ g'-1/2 (-1)k-1{ εαabc...x (e1)a (e2)b ...... (eN)x } (E'k)α ≡ g-1/2 (-1)k-1{ ε'αabc...x (e'1)a(e'2)b ...... (e'N)x } // k missing g = 1 This pair of equations has the form assumed in Weight Changing Theorem #2 of section (b) 8 with w=-1 and T = {...}. From section (f) it is known that weight (εαabc...x) = -1, and the vectors like (e1)a all have weight 0 since they are regular contravariant vectors. Therefore weight {....} = -1 which then plays the role of W in this theorem. The theorem says that Ek must have weight W + w = -1 - 1 = -2. Therefore, the candidate weight for Ek is ω = -2. Just as a check, does ω = -2 cause the second equation of the above pair to be true? (E'k)α ≡ g-1/2 (-1)k-1{ ε'αabc...x (e'1)a(e'2)b ...... (e'N)x } // k missing ? g'-ω/2 δk,α ≡ { (-1)k-1ε'α12...N } // k missing ? g'-ω/2 δk,α ≡ { ε'α12. α..N } // α is not in the hole where k was missing ? g'-ω/2 δk,α ≡ g' { ε'12. α..N } // from section (e) 2 ? g'-ω/2 δk,α ≡ g' { ε12. α..N } // from section (d) ? If α ≠ k, RHS = 0, in agreement with the LHS. If α = k, RHS = g', in agreement with the LHS when ω = - 2. The conclusions of Part II are that (E'k)i = |J|2 Rij(Ek)j and (Ek')α = g' δk,α so that Ek transforms under any underlying transformation F as a vector density of weight -2. Part III. Suppose now the vectors en and e'n of the Standard Notation are defined in this covariant manner en = g-1 En e'n = g'-1 E'n which fits the form of Weight Changing Theorem #1 of section (b) 7 with w = -2. The theorem says that the weight of en is then W - w where W= -2 is the weight of En, so weight(en) = -2 - (-2) = 0. Therefore vector en so defined is an ordinary contravariant vector (weight 0). Therefore (e'n)i= Rij (en)j and (e'n)i = g'-1 (E'n)i = g'-1 g' δn,i = δni => (e'n)i = g'ij(e'n)j = g'ij δnj = g'in= g'ni It follows from the first line above and from Section 3 (e'm)i = δmi that (e'n) (e'm) = (e'n)i (e'm)i = δniδmi = δnm and this is consistent with the covariance of the dot product (e'n) (e'm) = (en) (em) = δnm The other two dot products are easily verified: (e'n) (e'm) = g'ij (e'n)i(e'm)j = g'ij δniδmj = g'nm // = (en) (em) (e'n) (e'm) = g'ij (e'n)i(e'm)j = g'ijδniδmj = g'nm // = (en) (em) Just as was done in Appendix A (g), one may use the dot products above to conclude that En ≡ g'ni ei or (En)a ≡ g'ni (ei)a which was used as the definition of En in Section 6. Part IV. I don't think ω = -2 has any justification, what I did is just circular. So back to ω = 0. Start again with our old friend. (Ek)α ≡ σ |J|-1 (-1)k-1{ εαabc...x (e1)a(e2)b ...... (eN)x } // κ(k) and (ek)κ are missing; N≥ 2 Question: What does this look like if everything is expressed in terms of primed objects and ω = 0. The plan is to make substitutions in the above equation. Start then with (E'k)i = Rij(Ek)j or Ek' = R Ek // assumes ω = 0 Inverting the above gives (Ek)α = Sαj(Ek')j = Rjα (Ek')j (1) Meanwhile, from Section 3 en was defined as a contravariant vector so that (e'n)i = Rij(en)j which inverted gives (en)i = Sij(e'n)j = Rji (e'n)j // e.g., (e1)a = Ra'a (e'1)a' (2) Installing (1),(2) into the x-space equation for Ek then gives Rjα (Ek')j = σ |J|-1 (-1)k-1 { εαabc... Ra'a (e'1)a' Rb'b (e'2)b' ... Rx'x (e'N)x' } // k missing Applying Rnα to both sides and summing on α and using Rnα Rjα = δnj [ Section 7 (r) ] , (Ek')n = σ |J|-1 (-1)k-1 {εαabc... Ra'a (e'1)a' Rb'b (e'2)b' ..... Rnα....... Rx'x (e'N)x' } // k missing (Ek')n = σ |J|-1 (-1)k-1 Ra'a Rb'b .... Rnα ... Rx'x {εαabc... (e'1)a' (e'2)b' ....... (e'N)x' } Consider this internal item εαabc... Rnα Ra'a Rb'b ....... Rx'x = C εna'b'...x' based on the usual symmetry argument. To evaluate C, C ε123..N = εαabc... R1α R2a R3b ....... RNx = det(Rij) = det(Sab) = J = g'1/2 = C So we then know that εαabc... Rnα Ra'a Rb'b ....... Rx'x = g'1/2 εna'b'...x' and our equation above then becomes (Ek')n = σ |J|-1 (-1)k-1 εαabc...Ra'a Rb'b .... Rnα ... Rx'x { (e'1)a' (e'2)b' ....... (e'N)x' } = σ |J|-1 (-1)k-1 g'1/2 {εna'b'c'...x' (e'1)a' (e'2)b' ....... (e'N)x' } = σ (-1)k-1 {εnabc...x (e'1)a (e'2)b ....... (e'N)x } And now things do look somewhat similar (Ek)α ≡ g'-1/2 σ(-1)k-1{ εαabc...x (e1)a(e2)b ...... (eN)x } (Ek')α = σ (-1)k-1{ εαabc...x (e'1)a (e'2)b ....... (e'N)x } I am treating ε as just the permutation tensor, not going to ε' with it. So you asked: What happens (ω=0) if we express everything in terms of x'-space objects, and there is the answer staring you in the face. Part V. Suppose we think of εαabc...x as being "just a permutation bookkeeping object" so it is then NOT a tensor. It is NOT the Levi-Citiva tensor. Then what can be said about {...} ? Consider in this light Qα = εαabc...AaBbCc..... where A,B,C are contravariant vectors and ε is just the permutation object. What kind of object is Qα ? I think first of all it is not a tensor because it is contracted against ε which is not a tensor. That puts us in a pretty grim position! Q2 is just a sum of terms like A1B2C3..... . Part VI. OK, so let's go back to ε being the Levi tensor. We know that ε'abc... = g' εabc... so we can replace εαabc...x = g'-1 ε'αabc...x Then our two equations above are these: (Ek)α ≡ g'-1/2 σ(-1)k-1 { εαabc...x (e1)a(e2)b ...... (eN)x } (Ek')α = g'-1 σ (-1)k-1{ ε'αabc...x (e'1)a (e'2)b ....... (e'N)x } I ran into this pair (not sure if g' powers are correct) before, but those notes are now gone it seems. What can one say staring at this new pair of equations? (remember ω = 0 for Ek ) (1) the object {..} is a vector density of weight -1. (2) in either equation, you can reverse the tilt of any contracted index. (3) the object Ek is an ordinary vector (weight 0) This last item can be shown from the above TWO equations, but cannot be deduced from the first alone. The first line alone does not contain enough information to allow that conclusion! This is the conclusion I reached a few days ago. Here in fact is a quote from the that "failed tensor.." doc. Conclusion: When I consider (ek)α = (-1)k-1 J-1 εαabc.. (e1)a(e2)b... // k missing , it is true that Qα = εαabc.. (e1)a(e2)b... transforms as a vector density of weight -1. But as much as one wants to say that J-1 has weight +1 and "thus (ek)α has weight 0", that logic is just not valid. It happens that it is true for this very special vector Qα but it would never be true for some arbitrary vector Pα being a density of weight - 1. So I thought I had another example to put up, but I was wrong. *************************************************************************** Previous notes on this subject, the conclusion is exactly the same, I will store those notes here now: *************************************************************** I think it all boils down to studying this formula from Appendix A: (Ek)α ≡ det(R) (-1)k-1εαabc...x (e1)a(e2)b ...... (eN)x // κ(k) and (ek)κ are missing or in standard notation (ek)α ≡ det(R) (-1)k-1εαabc...x (e1)a(e2)b ...... (eN)x // κ(k) and (ek)κ are missing ek = g'ki ei OOOO. I claimed that this is "just a linear combination" and so ek is therefore a tensorial vector since ei is a tensorial vector. But the coefficients are not just numbers, they are g' components! But if I declare both ek and ek to be contravariant vectors by definition, I guess it is OK. Continuing along: (ek)α ≡ det(R) (-1)k-1εαabc...x (e1)a(e2)b ...... (eN)x // κ(k) and (ek)κ are missing Now det(R) = det(Rij) = 1/J = σ (g/g')1/2 = σ g'-1/2 so the above can be stated as (ek)α ≡ σ g'-1/2 (-1)k-1εαabc...x (e1)a(e2)b ...... (eN)x // κ(k) and (ek)κ are missing Now just for fun, how are things related in x'-space. I suspect this is true (no prime on ε ! ) (e'k)α = (-1)k-1εαabc...x (e'1)a(e'2)b ...... (e'N)x // k missing [ Inject: From ek = g'kiei apply R to get (weight 0 ass) e'k = g'kie'i =>( e'k)j = g'ki(e'i)j = = g'kiδij = g'kj. Then (e'k)α = g'kα = δkα so is correct. ] which would mean that δkα = (-1)k-1εαabc...x δ1aδ2b..... δNx // k missing = (-1)k-1εα123...N // k missing = ε123.. α...N // put α into the missing k slot This is then TRUE. So I seem to have these two equations (all OK on check) (ek)α ≡ σ g'-1/2 (-1)k-1εαabc...x (e1)a(e2)b ...... (eN)x // k missing (e'k)α = (-1)k-1εαabc...x (e'1)a(e'2)b ...... (e'N)x // k missing I could if I wanted make this replacement in the second equation ε'abc... = |J|2 εabc... = (g'/g) εabc... = g' εabc... => εαabc...x = g'-1 ε'αabc...x Then the two equations would read (ek)α ≡ σ g'-1/2 (-1)k-1εαabc...x (e1)a(e2)b ...... (eN)x // k missing (e'k)α = g'-1 (-1)k-1ε'αabc...x (e'1)a(e'2)b ...... (e'N)x // k missing I stare at this for 30 minutes and I see nothing usable. (agreed) Go back to the earlier (ek)α ≡ σ g'-1/2 (-1)k-1εαabc...x (e1)a(e2)b ...... (eN)x // k missing (e'k)α = (-1)k-1εαabc...x (e'1)a(e'2)b ...... (e'N)x // k missing where in both cases ε is the permutation ε. Can I at least show that ek is a true vector? Can I at least show that (e'k)α = Rαj (ek)j // def of contravariant vector (e'k)α = Rαj (ek)j // I think this is OK, δkα = Rαj { det(Rij) (-1)k-1εαabc...x (e1)a(e2)b ...... (eN)x } ? δkα = det(Rij) (-1)k-1Rαj {εαabc...x (e1)a(e2)b ...... (eN)x } ? (e1)a = Sa1 = R1a and k missing on the right δkα = det(Rij) (-1)k-1Rαj {εαabc...x R1a R2b ...... RNx } ? If k ≠α, then k = one of the other values and both sides 0. If k=α, then 1 = det(Rij) (-1)k-1 {εαabc...x R1a R2b .. Rαj.... RNx } = det(Rij) det(Rij) = det(Rij) det(Sji) = det(R)det(S) = 1 So it DOES work out in this sense. But it defies density analysis! ****************************************************************** Here are still other old notes, this is where I first tried the ω = -2 weight for Ek idea: ***************************************************************** (ek)α ≡ det(R) (-1)k-1εαabc...x (e1)a(e2)b ...... (eN)x // κ(k) and (ek)κ are missing (ek)α ≡ g'-1/2 (-1)k-1 εαabc...x (e1)a(e2)b ...... (eN)x // κ(k) and (ek)κ are missing In x'-space, I have (e'k)α ≡ (-1)k-1ε'αabc...x (e'1)a(e'2)b ...... (e'N)x // κ(k) and (ek)κ are missing which I can try to verify in this manner g'kα = (-1)k-1εαabc...x g1a g1b .... gNx // k missing This is too hard, I don't know how to deal with this RHS. So start over with the other tilt: (ek)α ≡ g'-1/2 (-1)k-1εαabc...x (e1)a(e2)b ...... (eN)x // κ(k) and (ek)κ are missing (e'k)α ≡ (-1)k-1ε'αabc...x (e'1)a(e'2)b ...... (e'N)x // κ(k) and (ek)κ are missing where I just conjecture the second line. But then try to verify second line δkα = (-1)k-1 ε'α123...N Need 1 = ε'123...N = g' and it does not work! To make it work we need to have (e'k)α = g' δkα but this is NOT what I have been using. I need some other symbol for the bookkeeping permutation tensor maybe. (ek)α ≡ g'-1/2 (-1)k-1pαabc...x (e1)a(e2)b ...... (eN)x // κ(k) and (ek)κ are missing (e'k)α ≡ (-1)k-1 pαabc...x (e'1)a(e'2)b ... (e'N)x // κ(k) and (ek)κ are missing In this case things work right in my little condo world. But now p is not a real tensor. I can then say nothing about weights. Plan A Idea. Imagine going through appendix A and these facts are correct as derived there (Ek)α ≡ det(R) (-1)k-1{ εαabc...x (e1)a(e2)b ...... (eN)x } // κ(k) and (ek)κ are missing; N≥ 2 Ek = g'ki ei (*) all in developmental notation, all in x-space only. I have already assumed by definition that en is a contravariant vector, and that seems OK. But what if Ek were a vector density of some weight W. That would mean that (Ek')i = J-W Rij(Ek)j or Ek' = J-W R(Ek) Now apply J-W R to equation (*) above to find that Ek' = g'ki J-W R ei = g'ki J-W e'i = g'ki g'-W/2 e'i Earlier I have assumed W = 0, but let's see if this leads anywhere useful. Now go back to (Ek)α ≡ det(R) (-1)k-1{ εαabc...x (e1)a(e2)b ...... (eN)x } // κ(k) and (ek)κ are missing; N≥ 2 and write this in covariant notation as (Ek)α ≡ g'-1/2 (-1)k-1{ εαabc...x (e1)a(e2)b ...... (eN)x } // κ(k) and (ek)κ are missing; N≥ 2 Now we know from above that (Ek')i = J-W Rij(Ek)j => (Ek)i = J+W Sij(Ek')j (en)a = Rab(e'n)b // weight is 0 here Insert these results into our known good equation to get J+W Sαj(Ek')j = g'-1/2 (-1)k-1{ εαabc...x (e1)a(e2)b ...... (eN)x } // k missing = g'-1/2 (-1)k-1{ εαabc...x Sa1 Sb2...... SxN } // k missing Our goal is to invert this to find out what Ek' might be. Apply Rβα to both sides and sum on α J+W Rβα Sαj(Ek')j = g'-1/2 (-1)k-1 Rβα { εαabc...x Sa1 Sb2...... SxN } (Ek')β = J-W g'-1/2 (-1)k-1 Rβα { εαabc...x Sa1 Sb2...... SNx } k missing (Ek')β = g'-W/2 g'-1/2 (-1)k-1 { εαabc...x Sa1 Sb2........ Sαβ......... SxN } If β ≠ k, then {} = 0 and so LHS must be 0. If β = k, then (-1)k-1{} = det(Sij) = g'1/2. Therefore we have found that (Ek')β = g'-W/2 g'-1/2 g'1/2 δk,β = g'-W/2 δk,β and we have at least answered the question that was asked. Now I would like to find a value of W which makes this be true: (g=1) (E'k)α ≡ g-1/2 (-1)k-1{ ε'αabc...x (e'1)a(e'2)b ...... (e'N)x } // k missing Just plugging in for both sides gives g'-W/2 δk,α = (-1)k-1{ ε'α123...N } // k missing = δk,α g' It would appear that W = -2 makes this work. So let's now assume W = -2, and write our two equations (Ek)α ≡ g'-1/2 (-1)k-1{ εαabc...x (e1)a (e2)b ...... (eN)x } (E'k)α ≡ g-1/2 (-1)k-1{ ε'αabc...x (e'1)a(e'2)b ...... (e'N)x } // k missing Now what does Weight Changing Theorem #2 tell us? In this application w = -1. The object in {} we know has weight W' = -1. The theorem then says the thing on the left Ek has weight W'+w = -1 -1 = -2 which amazingly is consistent!!! Tentative conclusion: the object (Ek)α transforms with weight -2. Therefore (Ek')β = g'-W/2 δk,β = g' δk,b and this then is "something new" but I think reflects the spherical coords example above. The next question: how should we define ek ? If we take ek ≡ Ek, then all the dot products come out right as Appendix A showed. But then we seem to have a weight anomaly here: Ek en = δkn Is this a tensor equation? What does it say in x'-space? E'k e'n = (Ek')β(e'n)β = g' δk,β δnβ = g' δk,n Perhaps then the two equations are really these: Ek en = g δk,n E'k e'n = g' δk,n I think this would be consistent with section (2) 7, since the equations are "covariant". To be compatible with Section 7 one should then define ek ≡ Ek / g ek' = E'k / g' Now consider then these equations ek = g-1 Ek and covariant What does section (2) 6 now say? The weight of ek should be W - w where W = -2 for Ek and where g-1 = gw/2 => w = -2, so we get ek having weight -2 - (-2) = 0 which is correct! Wow! Maybe just maybe this is going somewhere.