Re: the basis of relativity



Ken S. Tucker wrote:
> Tom Roberts wrote:
> > Ken S. Tucker wrote:
> > > "the nonvanishing of the tensor R_abcd is the true
> > > expression of the presence of a gravitational field"
> >
> > And colloquially, the nonvanishing of R_abcd is called "having curvature".
>
> Ok, in view of Eq.(6.9.1), we have R_abcd =/=0 means
> gravitator(s) are present, and a curvature "K(a,b)"
> exists (non-zero, I presume), then we should study
> (6.10.1) giving the *tidal force*. Therein you will find
> the relativity of TWO geodesics separated by, delta x.
> My thinking is the curvature is now defined by TWO relative
> geodesics, and not at a point. So a real tidal effect
> requires a non-zero curvature. That's characteristic
> of a field that requires non-orthogonal CS's, wherein
> the curvature scalar R=/=0, is required for tidal
> effects.
>
> > > Let's go over to Eq.(6.9.1), and see K(a,b) aka Gaussian
> > > curvature relates to -R/2, and have a glance at (6.7.4),
> > > as Weinberg suggests, in the "Description in N Dimensions".
> > >
> > > It follows from the above that,
> > > R = g^uv R_uv is a weak field approximation,
> >
> > No, it does not. R = g^uv R_uv is the _DEFINITION_ of R, for ANY
> > MANIFOLD WHATSOEVER. Look in Weinberg, p142, where R is defined after
> > (6.6.7).
>
> Yes I did study that. In rebuttal the "g^uv" therein is
> defined at a point. I suggest a conflict of the pure
> Reimann tensor analysis designed for static problems when
> confronted with the dynamics of GR where two separate
> geodesics define "curvature" and the scalar "R".
> A question arises, which geodesic does the "g^uv" in
> "R = g^uv R_uv" apply to, in view of Eq.(6.10.1)?
>
> > > PS:IMO using G_uv=0 is a non-physical approximation,
> > > of the true G_uv=T_uv.
> >
> > No. Using G_uv=0 is fully and completely justified, and is no
> > approximation at all, for cases in which T_uv=0. This last holds, of
> > course, in any vacuum region (because T_uv=0 is what we mean by "vacuum"
> > in GR).
>
> No, the deflection of light and the Shapiro effect,
> in the neighbourhood of a gravitator are proof of a
> refraction of light. That refraction occurs is evidence
> the spacetime medium it not a complete and total vacuum.
> The velocity of light is perturbed, that's fact.
>
> > > For example, in the case of two
> > > binary relatively revolving gravitators, G_uv=T_uv
> > > would be required, since the particle within the
> > > field certainly cannot be regarded as negligable.
> >
> > If you have two small "gravitators", and nothing else in the manifold,
> > then everywhere outside the two gravitators T_uv=0.
>
> Again that depends upon how density is defined.
> I define density as measured by the refraction
> of light. If the velocity of c is perturbed, like by
> the presence of the sun then the density>0.
> We should argue that, it's germaine.
>
> > In that region my
> > above comment holds, and the field equation is G_uv=0 IN REGIONS OUTSIDE
> > THE GRAVITATORS. Of course one must solve the field equation throughout
> > the manifold, and not just in that region....
>
> see above...
>
> > > Trying to shoe-horn G_uv=0 into some physical law
> > > that makes fields, is likely a dead end considering
> > > the *Action* Weinberg describes on pg. 364.
> >
> > Sure. The field equation is G_uv = T_uv (in suitable units, with no
> > cosmological constant). But in regions with T_uv=0 it can be simplified
> > WITHOUT APPROXIMATION to G_uv = 0. This is just basic algebra.
>
> It's the zero in G_uv=0 that I think needs better definition.
>
> > Tom Roberts tjroberts@xxxxxxxxxx
>
> Regards
> Ken S. Tucker

Tucker adds, if we regard G_uv = T_uv =0 as a physical
law we would elevate both G_uv and T_uv to the status
of invariants, (because the RHS is zero and invariant).

IMO that's a dangerous decision.

I solved G_uv = T_uv by using,

G_00 = E(a)*E(b) = T_00,

(E(a) is Electric Field of charge "a"...).

Just as E=mc^2 really doesn't care about the
nature of energy, it seems reasonable to
convert the energy storage to electrical,

energy=ab/r.

Regards
Ken S. Tucker

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