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1 total messages Started by brian a m stuckless Fri, 17 Mar 2006 11:06
Re: Solution to Einstein's Field Equations where T^uv not= 0?.
#1350
Author: brian a m stuckless
Date: Fri, 17 Mar 2006 11:06
49 lines
3571 bytes
Jay R. Yablon wrote: > 
> Many of the widely-studied solutions to Einstein's field
> equations are taken in vacuo, that is, at events where the
> energy momentum tensor T^uv=0.  This includes Schwarzchild
> and Kerr geometries, for example.
>
> Have there been many exact solutions found where T^uv not= 0?
$$                              No.
>  I am speaking of analytical solutions where the differential
> equations are solved exactly, *not* numerical approximations.
$$                        YABsolutely no.

>          Maxwell's energy tensor of electrodynamics
>   T^u_v = (1/4pi) [F^ut F_vt - (1/4) lambda^u_v F^st F_st].
>
> -=-  ..interested in solutions where F^uv_u=0 (free space)
> -=-    ..and where F^uv_u=J^v (space with current sources).
> Conditions of interest include static spherical symmetry in
> the nature of Schwarzchild, and rotation with spherical
> symmetry about the z-axis in the nature of Kerr.
                                              --Jay R Yablon.
> To be clear, I am *not* looking for solutions where the metric
> is assumed to be a Minkowski metric.  Lots of analyses assume
> a flat-space background for electrodynamics.
>
> Rather, I am looking for *exact* solutions, to the extent that
> such solutions are known, which derive a curved spacetime
> metric from the electromagnetic field strength tensor, that is,
> which derive g_uv = g_uv(F^uv) via the Maxwell tensor T^u_v,
> whereby T^u_v(g_uv, F_uv) simply becomes T^u_v(F_uv) once the
> g_uv(F^uv) are found.

$$ Maxwell used REAL "flat" plates in air to derive his equations.
$$ This is why GR is only "approximately" flat, at-great-distance.
$$ Even a dot has extreme "curvature", so you can imagine a point.
$$ GR is a "point-SURFACE manifold" at the end of it's WORLD-line.
$$ This is why GR is NOT a "local" theory (where it's all Newton).
$$ This is why GR is a "far-field" theory (where it's all Newton).
$$ [ The "SURFACE" of a GR-"POiNT" is "FLAT-at-a-GREAT-distance ].
$$
$$ Tom R ought derive a set from lab work ..using "CURVED" plates.
$$ [Just let the PLATEs be M1 and m1 and the air as the "AEther"].
$$
$$ Hope this helps,         ```Brian A M Stuckless, Ph.T (Tivity).
GR CUT OFF it's own WORLD-line, having DECLARED no PRiOR geometry.
p.s. A GR-"geodesic" is *NOT* Uncle Al's "OTHER LONGER way round".

> Thanks.> > Jay R. Yablon > > Email: jyablon@nycap.rr.com
Re: Solution to Einstein's Field Equations where T^uv not= 0?.
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