4-vector dot A = invariant => A is a 4-vector?
in [Relativity]
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From: Dono. on 2 Mar 2010 14:34 On Mar 2, 7:11 am, Schoenfeld <schoenfeld.fore...(a)gmail.com> wrote: > > since subsitution of [5] into the RHS of [4] gives [1] > > [6] A'.B' = (TA).(T^-1 B) > = (T T^-1) A.B > = A.B > Matrix multiplication does not commute, so you can't write (TA)T^-1=(T T^-1) A
From: mpalenik on 2 Mar 2010 15:07 On Mar 2, 2:34 pm, "Dono." <sa...(a)comcast.net> wrote: > On Mar 2, 7:11 am, Schoenfeld <schoenfeld.fore...(a)gmail.com> wrote: > > > > > since subsitution of [5] into the RHS of [4] gives [1] > > > [6] A'.B' = (TA).(T^-1 B) > > = (T T^-1) A.B > > = A.B > > Matrix multiplication does not commute, so you can't write > > (TA)T^-1=(T T^-1) A This is why I should have used indices in my original response. We have: T^beta_alpha (T^-1)^gamma_beta A^alpha B_gamma but T^beta_alpha (T^-1)^gamma_beta = delta^gamma_alpha I hope I wrote that correctly. It's always hard for me to tell in ASCII.
From: Dono. on 2 Mar 2010 16:58 On Mar 2, 12:07 pm, mpalenik <markpale...(a)gmail.com> wrote: > On Mar 2, 2:34 pm, "Dono." <sa...(a)comcast.net> wrote: > > > On Mar 2, 7:11 am, Schoenfeld <schoenfeld.fore...(a)gmail.com> wrote: > > > > since subsitution of [5] into the RHS of [4] gives [1] > > > > [6] A'.B' = (TA).(T^-1 B) > > > = (T T^-1) A.B > > > = A.B > > > Matrix multiplication does not commute, so you can't write > > > (TA)T^-1=(T T^-1) A > > This is why I should have used indices in my original response. We > have: > > T^beta_alpha (T^-1)^gamma_beta A^alpha B_gamma > > but T^beta_alpha (T^-1)^gamma_beta = delta^gamma_alpha > > I hope I wrote that correctly. It's always hard for me to tell in > ASCII. You can also use an interesting property of the dot product in order to achieve the same result: <A_matrix*u_vector, B_matrix_v_vector>=<B_transposed*A*u_vector,v_matrix> Schoenfeld's derivation will hold only if T_transposed=T (i.e. T=symmetric). Otherwise, it doesn't.
From: Dono. on 2 Mar 2010 18:25 On Mar 2, 1:58 pm, "Dono." <sa...(a)comcast.net> wrote: > On Mar 2, 12:07 pm, mpalenik <markpale...(a)gmail.com> wrote: > > > > > > > On Mar 2, 2:34 pm, "Dono." <sa...(a)comcast.net> wrote: > > > > On Mar 2, 7:11 am, Schoenfeld <schoenfeld.fore...(a)gmail.com> wrote: > > > > > since subsitution of [5] into the RHS of [4] gives [1] > > > > > [6] A'.B' = (TA).(T^-1 B) > > > > = (T T^-1) A.B > > > > = A.B > > > > Matrix multiplication does not commute, so you can't write > > > > (TA)T^-1=(T T^-1) A > > > This is why I should have used indices in my original response. We > > have: > > > T^beta_alpha (T^-1)^gamma_beta A^alpha B_gamma > > > but T^beta_alpha (T^-1)^gamma_beta = delta^gamma_alpha > > > I hope I wrote that correctly. It's always hard for me to tell in > > ASCII. > > You can also use an interesting property of the dot product in order > to achieve the same result: > > <A_matrix*u_vector, > B_matrix_v_vector>=<B_transposed*A*u_vector,v_matrix> > > Schoenfeld's derivation will hold only if T_transposed=T (i.e. > T=symmetric). Otherwise, it doesn't.- Hide quoted text - > Turns out that the general form of Lorentz matrix IS indeed symmetric: http://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_form So, Schoenfeld's proof is now complete.
From: Schoenfeld on 2 Mar 2010 21:13
On Mar 3, 5:34 am, "Dono." <sa...(a)comcast.net> wrote: > On Mar 2, 7:11 am, Schoenfeld <schoenfeld.fore...(a)gmail.com> wrote: > > > > > since subsitution of [5] into the RHS of [4] gives [1] > > > [6] A'.B' = (TA).(T^-1 B) > > = (T T^-1) A.B > > = A.B > > Matrix multiplication does not commute, so you can't write > > (TA)T^-1=(T T^-1) A That step is valid using tensors [6] A'^r B'_r = A^i (@x'^r/@x^i) B_j (@x^j/@x'r) = A^i B_j (@x'^r/@x^i) (@x^j/@x'r) = A^i B_j KroneckaDelta_ij = A^r B_r |