4-vector dot A = invariant => A is a 4-vector?
in [Relativity]
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From: eric gisse on 2 Mar 2010 22:12 Dono. wrote: > On Mar 2, 6:31 pm, Schoenfeld <schoenfeld.fore...(a)gmail.com> wrote: >> On Mar 3, 12:13 pm, Schoenfeld <schoenfeld.fore...(a)gmail.com> wrote: >> >> >> >> > 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 >> >> TYPO: the above refers to KroneckaDelta^i_j >> >> NOTE: Standard vector algebra just isn't flexible enough and you >> actually "lose structure". Some people even say Maxwell's original >> Electromagnetism (written with Quaternions) has lost structure after >> Gibbs coverted it to standard vector algebra. I wonder what physical >> principles exist but remain unknown because they can't be expressed >> with Gibbs's representation of Maxwell's E&M. > > > > I got it working with standard vectors and matrices, there is no need > for tensors. > Mind you, I am not disputing the validity of the tensorial proof, I > just tightened the matrix-vector proof. This remark is valid for bot > you and mpalenik. > It is nice to have a sane converstaion, uninterupted by kooks once > in awhile :-) As long as you don't ask Schoenfeld about Apollo or 9/11.
From: Dono. on 2 Mar 2010 22:12
On Mar 2, 7:09 pm, mpalenik <markpale...(a)gmail.com> wrote: > On Mar 2, 10:05 pm, "Dono." <sa...(a)comcast.net> wrote: > > > > > On Mar 2, 6:45 pm, mpalenik <markpale...(a)gmail.com> wrote: > > > > On Mar 2, 6:25 pm, "Dono." <sa...(a)comcast.net> wrote: > > > > > 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.- Hide quoted text - > > > > > - Show quoted text - > > > > Also, (T^-1)T = T(T^-1) = I for any matrix and its inverse. > > > True but no germaine to the discussion, you really need T to be > > symmetric in order for the proof to work. Since the general Lorentz > > transform is symmetric, the proof worked out ok in the end.- Hide quoted text - > > > - Show quoted text - > > Yeah, my mistake. I was thinking for a second that T(T^-1) = (T^-1)(T > +), when it's actually ((T^-1)+)(T+) where + indicates the transpose. I hope you and Shoenfeld stick around for a long time, it is good to see some people that really know their stuff. |