Any coordinate system in GR?
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From: LEJ Brouwer on 1 Sep 2006 23:08 Tom Roberts wrote: > The geodesically complete extension of the Schwarzschild charts is > unique, given by the Kruskal chart. You say that the maximal extension of the (exterior and interior) 'Schwarzschild' charts is unique. But could you possibly enlighten us also as to whether the geodesically complete extension of just the _exterior_ Schwarzschild chart is unique? If it is not unique, could you please list the possible extensions? Thank you. - Sabbir
From: Tom Roberts on 2 Sep 2006 07:54 LEJ Brouwer wrote: > Tom Roberts wrote: >> The geodesically complete extension of the Schwarzschild charts is >> unique, given by the Kruskal chart. > > You say that the maximal extension of the (exterior and interior) > 'Schwarzschild' charts is unique. But could you possibly enlighten us > also as to whether the geodesically complete extension of just the > _exterior_ Schwarzschild chart is unique? It is known to be unique. Tom Roberts
From: I.Vecchi on 4 Sep 2006 12:54 JanPB ha scritto: > I.Vecchi wrote: > > > > Consider the following alternative scenario. Take the complete Kruskal > > solution and paste together the horizons of the white and of the black > > hole. This is still a well-behaved extension. > > I don't see why the metric should remain smooth (at tne horizon) after > such regluing. I would say it does, also based on stuff I've read on the web, but I may be wrong. . > > > The infalling > > observers, raising his eyes after checking on his clock that he's > > passed the horizon of the black hole, would just find himself gushing > > out of a white hole horizon. After browsing around I think this is > > called an Einstein-Rosen bridge. > > The Einstein-Rosen bridge is a spacelike surface (really a 3D manifold) > corresponding to the lines T=const. (horizontal) on the K-S diagram. As > you can see from the K-S it can be traversed only if you move faster > than light. I'll try to work that out. From what I read, as well as from my wobbly intuition, the Einstein-Rosen solution is well-behaved at horizon. I would say that such a space-like surface (photons staying put on the horizon) is there in any case and it is spanned by the K-S charts (which , as far as I understand, get smoothly pasted together in the Einstein-Rosen case), but that may reflect my current inadequate understanding of the problem. > > > I surmise this may be the same as a > > blackhole were the singularity at r=0 is blown up to infinity > > (together with tau_0) through a coordinate change for the interior > > domain 2m<r<0. Repeating the above question, how do we know that this > > is not the "right" chart (i.e, the chart corresponding to space-time > > measurements of an infalling observer)? > > Not sure what you mean... Proper measurements don't change when > coordinates change. True, but I still have doubts about (or don't understand, as you prefer) the argument fixing an infalling observer's proper arrival time at the singularity at r=0 . That's Darryl's argument that "you can *compute* r from your proper time s", where it's still not clear to to me whether the proper time is uniquely defined (this is the doubt that has been motivating all my posts). So I still suspect that the Einstein-Rosen solution may actually be described by what is regarded as the interior solution, short of additional assumptions . Essentially, as far as I understand, you are saying that Einstein-Rosen solutions (aka wormholes) are not smooth at the horizon. I'll study this a bit more, gather references and then possibly state my point again. IV
From: LEJ Brouwer on 4 Sep 2006 14:52 Tom Roberts wrote: > LEJ Brouwer wrote: > > Tom Roberts wrote: > >> The geodesically complete extension of the Schwarzschild charts is > >> unique, given by the Kruskal chart. > > > > You say that the maximal extension of the (exterior and interior) > > 'Schwarzschild' charts is unique. But could you possibly enlighten us > > also as to whether the geodesically complete extension of just the > > _exterior_ Schwarzschild chart is unique? > > It is known to be unique. > > > Tom Roberts It is the unique _maximal_ extension (I would still appreciate a reference for that. Steve Carlip I think pointed out that the full Kruskal extension is mentioned to be the unique maximum extension for both the (interior+exterior) Schwarzschild solution in Hawking & Ellis, but that they merely state the result with giving a proof). Anyway, this is not the case I am interested in. I am interested specifically in whether 'non-maximal' extensions are possible of the _exterior_ Schwarzschild solution - usually it is extended into the sector corresponding to the black hole interior, but my question whether it is also possible to extend just to the sector representing the other exterior solution? This will require that light cones be orientated in the opposite direction in the other exterior solution - I drew a picture of the scenario I had in mind (i.e. the infinite cone) in another thread, which I think you must have read. Basically my question is whether this sewing of patches is consistent. I am not interested at this moment on whether the situation is physically well motivated or not. Thanks, Sabbir.
From: Tom Roberts on 4 Sep 2006 15:24
LEJ Brouwer wrote: > [The Kruskal extension] > is the unique _maximal_ extension (I would still appreciate a > reference for that. Hawking and Ellis is the best reference I know for this. > Anyway, > this is not the case I am interested in. I am interested specifically > in whether 'non-maximal' extensions are possible of the _exterior_ > Schwarzschild solution Given that the maximal extension is known, all other extensions are simply submanifolds of it. So take the Kruskal manifold and cut away any parts you wish, and you'll still have an extension of the exterior region (as long as you don't cut part of the exterior away (:-)). You seem to be thinking/hoping/dreaming that there is some other solution -- as long a smoothness and geodesic connectivity are required there isn't (though I'm not certain what all caveats apply). Cutting away parts of the complete manifold is, of course, unphysical -- it make mathematical sense and is well defined, but physically it makes no sense to model the universe with a manifold from which objects can simply disappear! So geodesic completeness for timelike geodesics is required on physical grounds; the only exception is at a curvature singularity where it is clear that theory (GR) breaks down. Tom Roberts |