Markus Hanke

Let me attempt to show that within the axioms of SR it is actually impossible to arrive at any paradoxes; SR is an internally self-consistent system, rendering any alleged "paradoxes" physically meaningless. Fundamental Postulates of Special Relativity (1) The substance existing at any world point can always be conceived to be at rest, if time and space are interpreted suitably. In other words - locally, all events can be considered to be in an inertial frame (2) Between all inertial frames the same laws of physics apply, regardless of their states of relative motion (3) Space-time can be considered isotropic and homogenous (4) Rulers and clocks function independently of their past history Self-Consistency Condition (5) In order for (1)-(4) to hold, the line element measuring the distance between two events in space-time must be the same for all observers, i.e. must not vary if going from one frame to another, regardless of their states of relative motion. Going from one frame to another, and then back to original frame, will yield the same event in space-time. Mathematical Proof The distance between two events in space-time can be defined as a line element of the form [latex]\displaystyle{ds^2=d\mathbf{R}\cdot d\mathbf{R}=g_{\mu \nu }dx^{\mu }dx^{\nu }}[/latex] wherein [latex]g_{\mu \nu }[/latex] shall be called the metric tensor, and can be thought of as a 4x4 matrix which transforms according to certain rules. The mathematical description of going from one inertial frame into another is realised by introducing a linear transformation between two vectors x' and x of the form [latex]\displaystyle{{x}'^{\mu }=L{^{\mu }}_{\nu }x^{\nu }+a^{\mu }}[/latex] wherein L is a general transformation matrix which represents an as-per-yet unspecified boost and rotation in arbitrary directions, and the 4-vector a represents a shift of origin. We now demand the following restriction to hold : [latex]\displaystyle{L^{T}gL=g}[/latex] which corresponds to the simple observation that, when performing a rotation and its inverse, you always arrive at the original vector, i.e. a rotation and its inverse chained together will yield the unity matrix. In tensor language this corresponds to [latex]\displaystyle{g_{\mu \nu }L{^{\mu }}_{\rho }L{^{\nu }}_{\sigma }=g_{\rho \sigma }}[/latex] In order to prove (5) one now only needs to show that such a transformation L leaves the space-time line element ds invariant, meaning that the distance between two events in space-time is the same for all inertial observers : [latex]\displaystyle{g_{\mu \nu }d{x}'^{\mu }d{x}'^{\nu }=g_{\mu \nu }L{^{\mu }}_{\rho }L{^{\nu }}_{\sigma }dx^{\rho }dx^{\sigma }=g_{\mu \nu }dx^{\mu }dx^{\nu }}[/latex] Quod erad demonstrandum. What this means is that, because above line element is invariant under said transformation, all inertial observers experience the same laws of physics. References Minkowski, Hermann (1908/9). "Raum und Zeit". Jahresberichte der Deutschen Mathematiker-Vereinigung: 75–88. English translation: Space and Time. In: The Principle of Relativity (1920), Calcutta: University Press, 70-88 Address the above, then.

No, it is valid in general for any Lorentzian 4-manifold. You cannot, in general, cover such a manifold with just one coordinate chart; physically that means that there is no universal frame of reference. This would hold both in a "Steady State" as well as an expanding universe. Also, as other posters have quite correctly pointed out, there is no "centre point" to such a manifold.

Exactly, that is part of a wider problem when talking about simultaneity of two spacially separated events on a general Lorentzian 4-manifold. There just isn't any "universal" frame of reference against which we can decide simultaneity, and attempting to do so in the local coordinate charts will lead to problems.

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.