Paradox2 v3

Paradox2 v3 PhL 3.9.17 I want to summarize how I got the right answer for (ω)' in doc paradox2 v2. Expression #1 : The first step is this: |(den(Φ))S'> ≡ |en(Φ+dΦ)> - |en(Φ)> = |e'i><e'i|en(Φ+dΦ)> - |e'i><e'i|en(Φ)> = |e'i> [R(Φ+dΦ)]'in - |e'i>[R(Φ)]'in = ( [R(Φ+dΦ)]'in -[R(Φ)]'in ) |e'i> = ( [R(Φ+dΦ)]in - [R(Φ)]in ) |e'i> // lincom ≡ (dR)in |e'i> dR = R(Φ+dΦ) - R(Φ) (matrices) (1) I show in a separate line item that you can remove the primes from either R : [R(Φ)]mn = <em(Φ)|R(Φ) |en(Φ)> = <em(Φ) |e'i><e'i| R(Φ) |e'j><e'j|en(Φ)> = [R(Φ)]'im [R(Φ)]'ij[R(Φ)]'jn = [ RT(Φ)R(Φ)R(Φ)]'mn = [R(Φ)]'mn (2) Notice right off the bat that everything is "lin comb" oriented here. There are no "operators". The second step to replace |e'i> as follows |e'i> = |ej(Φ)><ej(Φ) | e'i> = |ej(Φ)>[R(Φ)]ij (3) Inserting (3) into (1) then gives |(den(Φ))S'> = (dR)in [R(Φ)]ij |ej(Φ)> = [(dRT)R]nj |ej(Φ)> But (dRT)R = [ RT(Φ+dΦ) - RT(Φ)] R(Φ) = RT(Φ+dΦ)R(Φ) - 1 So at this point we have |(den(Φ))S'> = [ RT(Φ+dΦ)R(Φ) - 1]nj |ej(Φ)> Third step is to close this with <e'i | to get <e'i |(den(Φ))S'> = ( RT(Φ+dΦ)R(Φ) - 1 )nj <e'i |ej(Φ)> or [(den)S']'i = ( RT(Φ+dΦ)R(Φ) - 1 )nj R(Φ)ij = ( RT(Φ+dΦ)R(Φ) - 1 )nj R-1(Φ)ji = ( RT(Φ+dΦ) - RT(Φ) )ni = ( R(Φ+dΦ) - R(Φ) )in = (dR)in So we seem to have done a lot of work to get this simple result [(den/dt)S']'i = (dR/dt)in I could probably shorten that up a bit. Expression #2 : (den/dt)S' = – ω x en . (1.7.4) (G.7.1) and this is just a shorthand for [(den/dt)S']'i = - εijk(ω)'j(en)'k and this gives the correct ω in Frame S' or [(den/dt)S']'i = - εijk(ω)'jRkn Setting the two expressions equal - εijk(ω)'jRkn = (dR/dt)in Now multiply both sides by R-1nm and sum on n to get - εijk(ω)'jRkn R-1nm = (dR/dt)in R-1nm or - εijk(ω)'jδkm = [(dR/dt) R-1]im or - εijm(ω)'j = [(dR/dt) R-1]im or - εikm(ω)'k = [(dR/dt) R-1]im or - εikj(ω)'k = [(dR/dt) R-1]ij or εkij(ω)'k = [(dR/dt) R-1]ij ≡ Aij which I know is the correct answer. Go off and compute Aij I do this in Appendix G.7 and the result is Aij = - [ ( cosψ + sinθsinψ)1ij + ( sinψ - sinθcosψ)2ij+ ( + cosθ)3ij ] . (G.7.16) and this then yields the Goldstein results. ******************************************************************* NOW, instead of taking Frame S' components, try the above process but take Frame S components. Expression #1 We just copy the previous result |(den(Φ))S'> = [ RT(Φ+dΦ)R(Φ) - 1]nj |ej(Φ)> and this time we close with <ei | top get <ei |(den(Φ))S'> = ( RT(Φ+dΦ)R(Φ) - 1 )nj <ei |ej(Φ)> or [(den)S']i = ( RT(Φ+dΦ)R(Φ) - 1 )nj δij or [(den)S']i = ( RT(Φ+dΦ)R(Φ) - 1 )ni or [(den)S']i = [ ( RT(Φ+dΦ)R(Φ)R-1(Φ) - R-1(Φ) )R(Φ) ]ni or [(den)S']i = [ ( RT(Φ+dΦ) - R-1(Φ) )R(Φ) ]ni or [(den)S']i = [ ( RT(Φ+dΦ) - RT(Φ) )R(Φ) ]ni or [(den)S']i = [ RT(Φ)( R(Φ+dΦ) - R(Φ) ) ]Tni or [(den)S']i = [ RT(Φ)(dR) ]in or [(den/dt)S']i = [ RT(Φ)(dR/dt) ]in = [(dR/dt)T R(Φ)]ni Expression #2 Now I am not exactly sure what to do. Here is the naive approach: (den/dt)S' = – ω x en so [(den/dt)S']i = -εijk(ω)j(en)k or [(den/dt)S']i = -εijk(ω)jδnk or [(den/dt)S']i = - εijn(ω)j Setting the two expressions equal - εijn(ω)j = [ RT(Φ)(dR/dt) ]in = RT(Φ)im (dR/dt)ma R-1(Φ)ab R(Φ)bn = [ RT(Φ)(dR/dt)R-1(Φ)R(Φ)]in = [ R-1(Φ)A R(Φ)]in or - εikn(ω)k = [ R-1(Φ)A R(Φ)]in or - εikj(ω)k = [ R-1(Φ)A R(Φ)]ij or εkij(ω)k = [ (R-1ij Jj]ij ≡ Bij Well, this does give a different result, though I don't know if it is the right result. Go off and compute Bij We know that A = - [ ( cosψ + sinθsinψ)(iJ1) + ( sinψ - sinθcosψ)(iJ2)+ ( + cosθ)(iJ3) ] Idea: Write A = akJk where we know the ai as above. Then using Theorem 1, Theorem 1: R-1JR = RJ (G.4.1) we find, B = R-1(Φ)A R(Φ) = akR-1(Φ)Jk R(Φ) = ak Rkj Jj = akQk where Qk ≡ iRkj Jj We compute the vector Qk as follows Meanwhile, the vector ak is a1 = - i( cosψ + sinθsinψ) a2 = - i( sinψ - sinθcosψ) a3 = - i( + cosθ) I enter these quantities into Maple Then I compute B = akQk = A Q : We then transcribe the last result B = (- sinθsinφ - cosφ)(iJ1) + (sinφ - sinθcosφ)(iJ2) + (-cosθ - )(iJ3) so Bij = (- sinθsinφ-cosφ)ε1ij + (sinφ - sinθcosφ)ε2ij + (-cosθ - )ε3ij and B23 = - sinθsinφ-cosφ B31 = sinφ - sinθcosφ B12 = -cosθ - I then have from above that εkij(ω)k = Bij so then ε123 ω1 = B23 ω1 = - sinθsinφ-cosφ ε231 ω2 = B31 ω2 = sinφ - sinθcosφ ε312 ω3 = B12 ω3 = -cosθ - If I now negate the three angles the results are ω1 = sinθsinφ+cosφ = cosφ + sinθsinφ ω2 = sinφ - sinθcosφ = sinφ - sinθcosφ ω3 = cosθ + = + cosθ My result computed in Appendix G was (after sign changes) (ω)x = cosφ + sinθsinφ (ω)y = sinφ - sinθcosφ (ω)z = + cosθ . (G.6.16) (G.7.20) and hurray, exact agreement!!!!!