Paradox another attempt v1

This is the Title PhL 1.11.15 Note that page numbering is turned on in this template and view is 125%, located in phil/roaming/microsoft/templates size about 219K. en(Φ) = R(Φ)e'n |en(Φ)> = |R(Φ)e'n> |en(Φ)> = R(Φ)|e'n> = |e'i><e'i|R(Φ)|e'n> = |e'i> [R(Φ)]'in = |e'i> [R(Φ)]in |en(Φ+dΦ)> = R(Φ+dΦ)|e'n> = |e'i><e'i|R(Φ+dΦ)|e'n> = |e'i> [R(Φ+dΦ)]'in = |e'i> [R(Φ+dΦ)]in |(den(Φ))S'> ≡ |en(Φ+dΦ)> - |en(Φ)> = |e'i> [R(Φ+dΦ)]in - |e'i> [R(Φ)]in = ( R(Φ+dΦ)]in - [R(Φ)]in ) |e'i> Then < e'j |(den(Φ))S'> = ( R(Φ+dΦ)]in - [R(Φ)]in ) < e'j |e'i> = ( R(Φ+dΦ)]jn - [R(Φ)]jn ) = [(den(Φ))S']'j So here suddenly I am getting this result [(den(Φ))S']'j = R(Φ+dΦ)]jn - [R(Φ)]jn where the matrices on the right are the unprimed matrices! The only facts I used above were <e'i|R(Φ)|e'n> = <e'i|ej><ej| R(Φ)|ek><ek| e'n> = R(Φ)ij R(Φ)jk R-1(Φ)kn = R(Φ)in [R(Φ+dΦ)]'jk = <e'i|R(Φ+dΦ)|e'n> = <e'i|ej><ej| R(Φ+dΦ)|ek><ek| e'n> = R(Φ)ij R(Φ+dΦ)jk R-1(Φ)kn = ???? This is the Catch 22! Now if you think of Back up, show time dependence t on things better. en(Φ(t)) = R(Φ(t))e'n |en(Φ(t))> = |R(Φ(t))e'n> 1 = |e'i><e'i| = |ei(Φ(t))><ei(Φ(t))| so you think of this completeness AT ANY TIME t. Then you get |en(Φ(t))> = |R(Φ(t))e'n> <e'm |en(Φ(t))> = <e'm |R(Φ(t))e'n> = <e'm |R(Φ(t))| e'n> = R'mn(Φ(t)) Then it follows that <e'm |en(Φ(t)+dΦ)> = R'mn(Φ(t) + dΦ) and maybe this fixes the Catch 22 !! Try this without the clutter of showing t : <e'm |en(Φ)> = R'mn(Φ) <e'm |en(Φ+dΦ)> = R'mn(Φ+dΦ) **************************************************************** Let's now try to present this to the Critic Listener. Consider that viewed from Frame S' the Frame S basis vectors are moving and so en(Φ) = R(Φ)e'n or |en(Φ)> = |R(Φ)e'n> . (*) Completeness can be written 1 = |e'i><e'i| in the fixed Frame S' basis 1 = |en(Φ)><en(Φ)| for any Φ in the moving Frame S basis For example, at a slightly later time when Φ has changed to Φ+dΦ we get 1 = | en(Φ+dΦ) > < en(Φ+dΦ) | Close (*) on the left with the non-moving <e'm | to get <e'm | en(Φ)> = <e'm | R(Φ)e'n> = <e'm | R(Φ)| e'n> = [R(Φ)]'mn Then at some slightly later time where Φ has changed to Φ+dΦ we can write <e'm | en(Φ+dΦ)> = <e'm | R(Φ+dΦ)e'n> = <e'm | R(Φ+dΦ)| e'n> = [R(Φ+dΦ)]'mn This all seems just fine. Now consider [R(Φ)]'mn = <e'm | R(Φ)| e'n> = <e'm |ei(Φ)><ei(Φ)| R(Φ) |ej(Φ)><ej(Φ)|e'n> = [R(Φ)]mi [R(Φ)]ij [R-1(Φ)]jn = [R(Φ)R(Φ)R-1(Φ)]mn = [R(Φ)]mn At a later time we replace Φ by Φ+dΦ everywhere in the above line to get [R(Φ+dΦ)]'mn = [R(Φ+dΦ)]mn Therefore we have shown that <e'm | en(Φ)> = [R(Φ)]'mn = [R(Φ)]mn <e'm | en(Φ+dΦ)> = [R(Φ+dΦ)]'mn = [R(Φ+dΦ)]mn Now consider |(den(Φ))S'> ≡ |en(Φ+dΦ)> - |en(Φ)> Close with <e'm | to get <e'm |(den(Φ))S'> ≡ <e'm |en(Φ+dΦ)> - <e'm |en(Φ)> = [R(Φ+dΦ)]mn - [R(Φ)]mn This says that [(den(Φ))S']'m = [R(Φ+dΦ)]mn - [R(Φ)]mn = (dR)mn Even though we are taking a Frame S' component on the left, we end up with the Frame S standard rotation matrices on the right! What happens if you instead take a Frame S component? <em(Φ) |(den(Φ))S'> ≡ <em(Φ) |en(Φ+dΦ)> - <em(Φ) |en(Φ)> = <em(Φ) |e'i><e'i|en(Φ+dΦ)> - <em(Φ) |e'i><e'i|en(Φ)> = [R(Φ)]im [R(Φ+dΦ)]in - [R(Φ)]im [R(Φ)]in = [R(Φ)]im { R(Φ+dΦ) - R(Φ) }in = [R(Φ)]im {dR}in So this is a different result !!!!!!!!!!!!! = [R(Φ+dΦ)]mn - [R(Φ)]mn