zonde said:
Where did you get that Einstein considered SR defective?
SR certainly is not defective. It has limited applicability. And GR is meant to overcome this limit. But that means one important thing - GR should reduce to SR at appropriate limits.
Einstein was philosophically attracted to Mach's ideas. Ideally, he felt there should be no physical significance to anything except relative motion. The idea of distinguishable accelerated motion in an empty universe was abhorrent to him. He hoped that general covariance and his GR program would show that inertial resistance to acceleration arose from motion relative the distant mass of the universe. SR's preference for inertial frames bothered him and was one of several major motivators for his development of GR. [Einstein later realized GR failed in his Machian objective, but still succeeded, in his view, of displacing or, at least weakening, any special position for inertial frames.]
Of course GR reduces SR sufficiently locally everywhere. That is built into the mathematical structure of pseudo-riemannian geometry in the same way local Euclidean geometry is built into Riemannian geometry.
zonde said:
And that in turn means that if two different coordinate systems when reduced (under appropriate limits) to SR give coordinate systems that are not related by Lorentz transform then they describe two different physical situations.
No, this is not a correct statement of the way GR contains SR. The relationship is local not global. In general, a GR solution has no global coordinates that resemble Minkowski coordinates at all. In a limited sense you can say that for asymptotically flat spacetimes (which, by the way, does
not include our universe), there are coordinates systems that approach Minkowski at infinity. However, not only are the 'too many' of them, they are not generally related by Lorentz transforms. This whole statement of your is pretty much a complete misunderstanding of the relationship between SR and GR.
As an aside, you should be aware that the GP coordinates used in the river model reduce Minkowski coordinates at infinity, just like SC ones do. In fact they reduce to exactly the same coordinates at infinity because they share the same center of symmetry and both make explicit the asymptotic flatness of SC geometry.
zonde said:
So we can try to compare coordinate systems in that sense to find out if they are equal.
Would you still say that this somehow goes against Einstein's view?
Yes. The only place for Lorentz transforms in GR is local (in the limit of a small region of spacetime where curvature= tidal gravity can be ignored).
zonde said:
What argumentation you can provide for this statement?
It is actually mathematically obvious. The metric for GP coordinates as given in the paper in post #1 of this thread is:
ds^2 = - dt^2 + (dr + βdt^2)^2 + r^2(dθ^2 + sin^2θd\varphi^2)
For a slice of constant t, you have dt=0. Then you have, for the spatial geometry of the slice:
ds^2 = dr^2 + r^2 (dθ^2 + sin^2θd\varphi^2)
which is just the flat Euclidean metric in polar coordinates. Interesting, wouldn't you say, for coordinates that have no horizon coordinate singularity and go smoothly through the horizon to the singularity? The horizon itself is there as a physical phenomenon, but there is no coordinate singularity there, and no infinite coordinate time there. Those are artifacts of Schwarzschild coordinates.
Meanwhile, for t=constant slice for Scwharzschild coordinates, using e.g. the form given in:
http://en.wikipedia.org/wiki/Schwarzschild_metric
you get:
ds^2 = (1/(1-R/r)) dr^2 + r^2 (dθ^2 + sin^2θd\varphi^2)
(where I am using R for SC radius). This is non-Euclidean spatial geometry.