Re: Mathematical & Empirical Equivalence of Teleparallel & GR (was: Einstein was Precisely Backward)

"Ken S. Tucker" <dynamics@xxxxxxxxxxxx> wrote in message news:<1113385537.623220.189560@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>...
> whopkins@xxxxxxxxxxx wrote:
> > (So, anyway, Uncle Al, the foregoing means we're on the same page
> now.
> > Nix the other remarks about equivalence through redefinition of the
> > right-hand side. It may be possible, still, to effect that, but it's
> > awkward at best).
> >
> > If you're going to work out the general case, it helps to return to
> the
> > more basic definitions. If g_{mnp} denotes the components of the
> > covariant derivative of the metric: g_{mnp} = D_m g_{np}, and T_{mnp}
> > the index-lowered components of the torsion, then one can write (with
> > G_{mnp} denoting the index-lowered components of the connection):
> > G_{mnp} + G_{mpn} = d_m g_{np} - g_{mnp}
> > G_{mnp} - G_{nmp} = T_{mnp}.
> > Then it follows that
> > 2G_{mnp} =
> > T_{mnp} + d_n g_{mp} - g_{nmp}
> > + T_{pnm} - d_p g_{nm} + g_{pnm}
> > + T_{pmn} + d_m g_{pn} - g_{mpn}
> >
> > That's the correct ordering of indices, independent of whether the
> > metric is symmetric or not.
>
> Thank's again,
> That's a powerful and succint statement,
> and is about clear as you can get.
>
> 1)The T_{mnp} depends on a nonsymmetrical
> *connection*, but conventionally vanish.
>
> 2)The g_{mnp} conventionally vanish.
>
> If I understand Al correctly he's gunning for (2),
> (maybe (1) or both), ah, maybe Al could point to
> which one. I demo'd you can keep nonsymmetrical
> metrics and still retain (1) and (2) above if
> that helps.
>
> Is T_{mnp} supposed to be the sematic torsion?
>
> Regards
> Ken S. Tucker

And if he does it, I may just make a TV show about it.....

Simon Huntley

Tabitha Productions
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