Re: the basis of relativity

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".


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).


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).



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. 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....


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.


Tom Roberts	tjroberts@xxxxxxxxxx
.