Paradox 1 of 2.19.17 REVIEWED

Paradox 1 of 2.19.17 PhL 2.19.17 Notes added 2.24.17. This was quite a severe paradox and set me back probably 2 days. It is all clear now and it is written up at the end of Section 1.1 in frames doc (so I won't ever forget! ) It is subtle. This is a good one, at 8 AM I am totally 100% mystified by this Paradox! STATEMENT OF PARADOX 1. Suppose we know that (,,) = Rz(φ) (,,) (',',') = Rξ(θ) (,,) (G.5.7) (1) ok e'n = R-1 en e''n = S-1 e'n It then follows from the (1.1.1) and (1.1.2) theorem that = Rz(-φ) = Rξ(-θ) (G.5.9) (2) ok e'n = Rnm em e"n = S'nm e'm 2. Combine the transformations above to get (',',') = Rξ(θ) (,,) = [Rξ(θ)Rz(φ)] (,,) (G.5.7) (3) ok S-1 R-1 It then follows from the Basis Rule that = [Rξ(θ)Rz(φ)]-1 = Rz(-φ) Rξ(-θ) (G.5.10) (4) ok R S 3. On the other hand , from (2) it certainly seems that we can say = Rξ(-θ) = Rξ(-θ) { Rz(-φ) } = [ Rξ(-θ) Rz(-φ)] (5) e"n = S'nm e'm S'nm Rmk ek = [S' R]nm ek BUT (5) contradicts (4). Paradox! Which one is right and which one is wrong and why? Resolution: I have written up this case in Section 1 Concatenated Two Transformations near the end. One key fact is that S' = S" and you do not have S' = S. The equality of (4) and (5) follows from R S = S' R or S' = RSR-1 as noted on the very last line of this little subsection. I agree, this was a fairly subtle paradox and could not be cleared up basically until I wrote that section!!! Plan A Maybe it has to do with the fact that = Rz(-φ) is not really a normal "matrix equation". [no, this was not the problem ] It is these three vector equations = [Rz(-φ)]11 + [Rz(-φ)]12 + [Rz(-φ)]13 = [Rz(-φ)]21 + [Rz(-φ)]22 + [Rz(-φ)]23 = [Rz(-φ)]31 + [Rz(-φ)]32 + [Rz(-φ)]33 And we also then have ' = [ Rξ(-θ)]11 + [ Rξ(-θ)]12 + [ Rξ(-θ)]13 ' = [ Rξ(-θ)]21 + [ Rξ(-θ)]22 + [ Rξ(-θ)]23 ' = [ Rξ(-θ)]31 + [ Rξ(-θ)]32 + [ Rξ(-θ)]33 Now manually solve for ' ' = [ Rξ(-θ)]11 ( [Rz(-φ)]11 + [Rz(-φ)]12 + [Rz(-φ)]13 ) + [ Rξ(-θ)]12 ([Rz(-φ)]21 + [Rz(-φ)]22 + [Rz(-φ)]23 ) + [ Rξ(-θ)]13 ([Rz(-φ)]31 + [Rz(-φ)]32 + [Rz(-φ)]33 ) = { [ Rξ(-θ)]11 [Rz(-φ)]11 + Rξ(-θ)]12[Rz(-φ)]21 + [ Rξ(-θ)]13[Rz(-φ)]31 } + { [ Rξ(-θ)]11 [Rz(-φ)]12 + Rξ(-θ)]12[Rz(-φ)]22 + [ Rξ(-θ)]13[Rz(-φ)]23 } + { [ Rξ(-θ)]11 [Rz(-φ)]13 + Rξ(-θ)]12[Rz(-φ)]23 + [ Rξ(-θ)]13[Rz(-φ)]33 } This sure looks like; = [ Rξ(-θ) Rz(-φ)] Rξ(-θ) Rz(-φ) = The upper left element of the product matrix is 11 11 + 12 21 + 13 31 so we are going to get ' = [11 11 + 12 21 + 13 31 ] + ..... and that is exactly what I see above. So this argues that (5) is correct, so (4) must be wrong. But (4) is a trivial application of my theorem of Section 1.1 : (',',') = [Rξ(θ)Rz(φ)] (,,) = [Rξ(θ)Rz(φ)]-1 = Rz(-φ)Rξ(-θ) In other words e'n = R-1 en R-1 = [Rξ(θ)Rz(φ)] R = Rz(-φ)Rξ(-θ) e'n = Rnm em or = R So this theory does not explain the Paradox! [nice try, no cigar] **************************************************************************** Plan B. Go back to (1.1.1) and (1.1.2). Suppose en = R1 e'n and e'n = R2 e"n Then surely en = R1 e'n = R1 R2 e"n (1), I have no doubts about this Invert to get e"n = (R2)-1 (R1)-1 en (2) no doubts Now look at the claims of theorem (1.1.2) : e'n = R1nm em and e"n = R2'nm e'm (3) no doubts if THM is OK prime add later! but this line was my error! Then surely e"n = R2'nm e'm = R2'nm [ R1mk ek ] = [R2'R1]nk ek (4) no doubts Write this as = [R2'R1] (5) no doubts as lin comb equation This does NOT say that e"n = [R2R1] en // this would contradict (2) above The previous is NOT a set of three component matrix equations! So beware that illusion!! Now where is my paradox here? I do have from (3) that = R1 and = R2' (6) no doubt as lin comb eqs It is tempting then to say that = R2' = R2' { R1 } = R2'R1 (7) agrees with (5) So far no paradox since this does not contradict anything yet written. Now go back to (1) en = R1 R2 e"n = [R1 R2] e"n ≡ R e"n where R = R1 R2 (8) no doubts Now use theorem (1.1.1) + (1.1.2) to say e"n = (R)nm em = (R1 R2)nm em (9) no doubts and we would write this as = [R1 R2] (10) no doubts as lincom eq The paradox is that this disagrees with (5) !! Resolution: R2'R1 = R1 R2 because R2' = R1 R2 (R1)-1. This is unfriendly notation but that is it! See frames doc clear writeup near end of Section 1.1. The red prime makes all the difference! ************************************************************* Plan C . Try to make the paradox presentation as brief as possible. Omit lin comb matrix notation. Write (1.1.2) for two transformations (these are lin comb equations) [ prime added in red ] e'n = Anm em and e"n = B'nm e'm A = R B = S (1) (e'm = Amk ek) Then e"n = B'nm e'm = B'nm [Amk ek ] = [B'A]nk ek (2) Now write the (1.1.1) versions of the two items in (1) (these are comp equations) en = A e'n and e'n = Be"n (3) Combine to get (this is a comp equation) en = [AB] e"n ≡ Re"n R ≡ AB (4) Apply the 1.1.1 and 1.1.2 theorem to (4) to get RS S'R e"n = Rnm em = [AB]nm em = [B'A]nk ek (5) This contradicts (2) ! Resolution: Same as previous, see end of frames doc Sec 1.1. ************************************************************* Plan D. When all else fails, activate Dirac notation! 1 = |en><en| Let's try to state Theorem 1.1.1 + 1.1.2 in Dirac notation en = R e'n e'n = Rnm em It will take a moment to convert this. Turn the gears: |en> = |R e'n> = R | e'n> = |e'm><e'm|R | e'n> = |e'm>R'mn = R'mn |e'm> = (R'-1)nm|e'm> Apply (R')kn to both sides (R')kn|en> = (R')kn(R'-1)nm|e'm> = δkm |e'm> = |e'k> or (R')nm|em>= |e'n> or |e'n> = (R')nm|em> So here is the comparison en = R e'n e'n = Rnm em |en> = |R e'n> |e'n> = (R')nm|em> So I learn (I guess for the first time), that the matrix which appears here (R')nm is in the prime basis! Now recall from "essay" that (e'n)i = Rni (en)i = δni Frame S components (en)'i = Rin (e'n)'i = δni Frame S' components Fact: (R')mn = <e'm|R | e'n> = <e'm| |ei><ei|R | |ej><ej|e'n> = <e'm|ei> Rij<ej|e'n> = (e'm )i Rij (e'n )j = Rmi Rij Rnj = Rmi Rij RTjn = Rmiδin = Rmn This shows that (R')mn = Rmn so the matrix in the no-prime basis is the same as in the prime basis. [ok] NOW let's do the paradox again: Repeat the above and just translate. ****************************************** [Here is where I found the resolution for the first time.] Write (1.1.2) for two transformations (these are lin comb equations) |e'n> = Σm A'nm |em> |e"n> = Σm B'nm |e'm> We know that in terms of matrices and their bases A'nm = Anm B"nm = B'nm [ I see the light! ] Note that the top right equation would be wrong if you wrote it as |e"n> = Σm Bnm |e'm> !!! Then |e"n> = B'nm |e'm> = B'nm [A'mk |ek> ] = [BA]'nk |ek> [ok] (2) Now write the (1.1.1) versions of the two items in (1) (these are comp equations) |en> = |A e'n> = A | e'n> and |e'n> = |B e"n> = B | e"n> (3) Combine to get (this is a comp equation) |en> = AB | e"n> ≡ R | e"n> R ≡ AB (4) Apply the 1.1.1 and 1.1.2 theorem to (4) to get |e"n> = Rnm |em> = [AB]nm |em> = [AB]nk |ek> (5) The choices of basis here are no-prime or double prime, and single prime would be wrong! So we end up with |e"n> = B'nm A'mk |ek> = [BA]'nk |ek> (2) [ok] |e"n> = Anm Bmk |ek> = [AB] nk ek (5) Now maybe we no longer have a contradiction due to the basis difference! Recall A'nm = Anm B"nm = B'nm Maybe I can show that [BA]'nk = [AB] nk which would require <e'n|BA|e'k> = <en|AB|ek> ? <e'n|B|e'i><e'i|A|e'k> = <en|A|ei><ei|B|ek> ? B'niA'ik = Ani Bik ? B'A = AB ? B' = ABA-1 looks like tensor but what exactly is it? How are Frame S and Frame S' related? en = A e'n where A is our rotation matrix. Therefore for any tensor we have T' = ATA-1 Recall G.4.10 that RTR-1 = T' So THAT is the resolution! I need to get this clarified EARLY in frames doc. perhaps Section 1! Perhaps another appendix to encapsulate this ? [ I worked it into the end of Section 1.1 ] What is my rule for a combination of two transformations? Start with the individual transformations : en = R e'n e'n = Rnm em Rij = R'ij e'n = S e"n e"n = S'nm e'm S'ij = S"ij So there are the two single rules. We can combine the above to get e"n = S'nmR'nk ek = (SR)'nk ek What then is the combined rule? en = (RS)e"n e"n = (RS)nm em (RS)nm = (RS)"nm We then end up with two seemingly different statements e"n = (SR)'nk ek e"n = (RS)nk ek One must be very careful with the basis of a matrix, here indicated by no-prime, ' or ". The two statements above which look very different are in fact the same because (these equations are for matrices) (SR)' = RS ? S'R' = RS ? S'R = RS ? // since R = R' S' = RSR-1 yes! This is how a rank-2 tensor transforms under R. Let's do a vector notation now = (SR)' = (RS) // use either one! ******************** OK, so in the original paradox problem, which one was wrong and which was right? I will just copy it down from above and make changes as needed 1. Suppose we know that e'n = R-1 en e'n = R-1 en (,,) = Rz(φ) (,,) (',',') = Rξ(θ) (,,) (1) Lets denote the three coordinate systems this way (,,) = Frame S (,,) = Frame S^ (',',') = Frame S~ (',',') = Frame S' It then follows from the (1.1.1) and (1.1.2) theorem that for lincoms, = Rz(-φ) = R^ξ(-θ) (2) where Rz(-φ) = R^z(-φ) R^ξ(-θ) = R~ξ(-θ) 2. Combine the transformations above to get (',',') = Rξ(θ) (,,) = [Rξ(θ)Rz(φ)] (,,) (3) It then follows from the (1.1.1) and (1.1.2) theorem that = [Rξ(θ)Rz(φ)]-1 = Rz(-φ) Rξ(-θ) (4) where Rξ(θ) = R~ξ(θ) and Rz(φ) = R~z(φ) 3. On the other hand , from (2) it certainly seems that we can say = R^ξ(-θ) = R^ξ(-θ) { R^z(-φ) } = [ Rξ(-θ) Rz(-φ)]^ (5) Here we have R^ξ(-θ) = R~ξ(-θ) and Rz(-φ) = R^z(-φ) . So we have these two equations = Rz(-φ) Rξ(-θ) (1) = R^ξ(-θ) R^z(-φ) (2) If we want to use the standard Frame S matrix, we must choose (1) !!!! In order to always use rule (1) with regular matrices in Frame S, you must make a link all the way from your coordinates of interest over to the Frame S basis vectors. THAT was a pretty good paradox (!!) and will have update implications. [ update is that I wrote this all up at the end of Section 1.1 AND added lots of Dirac stuff ]