Markus Hanke

And how do you decide that they are simultaneous, given the finite speed of light and non-uniform and non-stationary curvature of space-time between here and there ?

Because, in the general case, you cannot cover the entirety of a Lorentzian 4-manifold with a single coordinate chart. What that basically means is that both space and time are local phenomena only; observers at different points on the space-time manifold may thus not agree on clock readings and ruler measurements. In fact, even the very notion of comparing such readings in different frames of reference is at best pretty problematic since there is no universal coordinate chart.

That's true, but if you are not in a vacuum then [math]\displaystyle{T_{\mu \nu }\neq 0}[/math], and the Ricci tensor would not vanish. All I was trying to say is that a vanishing Ricci tensor does not automatically imply a flat space-time; it is a necessary, but not a sufficient condition since, as mentioned earlier, the Riemann curvature tensor is not completely determined by its contractions in 4 dimensions.

In four dimensions the Ricci tensor does not fully determine Riemann curvature, so [math]\displaystyle{R_{\mu \nu }=0}[/math] does not necessarily mean that we are deal with a globally flat space-time. All it means that we are considering a vaccum, and that space-time can locally be considered to be Minkowskian. To determine the global geometry one would need additional constraints on the system, as for example the Schwarzschild metric does by demanding that it vanishes at infinity, and that it reduces to Newtonian gravity for weak fields.

In the above sentence, swap "due to" with "is" - that will give a more accurate statement. Consider for a moment elfmotat's avatar, and imagine two observers standing on the equator of that sphere. Now let our observers start walking north, along the lines as drawn in the avatar picture. What happens ? The further north they get, the more they approach each other. At the north pole, they meet. This happens not because of any force between them, but simply because of the geometry of the surface they are moving on. Likewise GR - two bodies will approach one another not because there are any forces between them, but because of the geometry of space-time itself. Gravity ceases to be a phenomenon external to a body, and becomes a geometric property of space-time.

On what grounds do think it is permissible to use SR when analysing a binary pulsar system ? Surely you are aware that SR requires inertial reference frames.

Gravitation by Thorne/Misner/Wheeler is a very in-depth treatment of the subject, and considered the "gold standard" by many. A good understanding of exterior calculus / differential forms is also helpful in understanding GR, but not necessarily required.