We don't really know, because nobody has gone into one to look and then come out to tell us. :)Onyx said:Does the interior coordinate patch of the Schwarzschild analytic extension really describe the interior of a black hole? After all, that portion would have mass.
The ##\theta## and ##\phi## coordinates are fine, but the Euclidean ##r## coordinate would not correspond to anything interesting, rather like ##x## and ##y## coordinates applied to a Mercator map of the cirved surface of the earth.Also, is there a way to describe just a black hole’s with regular spherical coordinates?
I noticed the difference in terms: patch vs solution. By the interior “solution”, do you mean this?Nugatory said:We don't really know, because nobody has gone into one to look and then come out to tell us. :)
But seriously, kidding aside...
There is no reason to think that the interior Schwarzschild solution is not valid in those regions where the magnitude of the curvature invariants are comparable to those outside the horizon. That region is vacuum and no more has mass than the vacuum outside the horizon. On the other hand, we do expect that the Schwarzschild solution will stop working when we're near enough to the singularity and the curvature exceeds anything that we've been able to study so far.
The ##\theta## and ##\phi## coordinates are fine, but the Euclidean ##r## coordinate would not correspond to anything interesting, rather like ##x## and ##y## coordinates applied to a Mercator map of the cirved surface of the earth.
No. In my OP I meant the interior coordinate patch of the Schwarzschild analytical extension. Also, in my second question, I forgot the word “interior” after “hole’s.”PeterDonis said:Is that what you meant in your OP? I don't think so. @Nugatory was talking about the solution you mentioned your OP. He wasn't introducing a new solution that you hadn't mentioned.
No of course not. I meant the Schwarzschild interior solution that you mentioned.Onyx said:I noticed the difference in terms: patch vs solution. By the interior “solution”, do you mean this?
That's what I thought. That is a vacuum solution, and as @Nugatory has posted, that's the solution he was talking about.Onyx said:In my OP I meant the interior coordinate patch of the Schwarzschild analytical extension.
Are r and t swapped?Nugatory said:No of course not. I meant the Schwarzschild interior solution that you mentioned.
However, that is also a different patch - there's a coordinate singularity at ##r=R_S## so the ##r## and ##t## instead the horizon are different coordinates than the ##r## and ##t## outside. It is an unfortunate accident of history that this was not recognized until long after the tradition of using the same letters to label coordinates on both sides of the horizon had taken root.
No. But their physical meanings, especially of ##t##, are different in the Schwarzschild coordinate patch inside the horizon. Note that the fact that ##r## is the "areal radius", i.e., that any given event with coordinate ##r## lies on a 2-sphere with surface area ##4 \pi r^2##, is still the same, and is a feature that is present in other coordinate charts as well. But in the Schwarzschild patch inside the horizon, ##r## is no longer spacelike, it's timelike, and ##t## is no longer timelike, it's spacelike. Those features make things work very differently from the way our intuitions would expect.Onyx said:Are r and t swapped?
The Einstein-Rosen bridge is a geometric feature of the maximum analytic extension; it is present independent of your choice of coordinates. Note also that there are coordinates, such as Kruskal, which cover the entire maximal analytic extension with a single patch.Onyx said:I don’t understand why there is so much emphasis placed on the Einstein-Rosen bridge aspect of the Schwarzschild analytic extension when we know that the coordinates break down after a certain point anyway.
It's a "geometric artifact" in the sense that it is not present in any actual model that is considered to be physically reasonable, because in any actual model of a black hole, the hole will have been formed by gravitational collapse of a massive object, and the only portions of the maximal analytic extension of the vacuum solution that will be present are the original exterior region and the black hole interior region outside of the collapsing matter. The Einstein-Rosen bridge, the "other universe" exterior region, and the white hole region will not be present.Onyx said:Isn’t the “other universe” just a mathematical artifact?
It can perfectly well be viewed as the same coordinate patch. What needs to be realized though is that that coordinate patch is not connected. All that is required from a coordinate patch is a smooth function from an open subset of ##\mathbb R^n## to the manifold. In the case of Schwarzschild coordinates ##(0,\infty)\setminus \{R_S\}## is indeed an open subset of ##\mathbb R## - although not connected.Nugatory said:No of course not. I meant the Schwarzschild interior solution that you mentioned.
However, that is also a different patch - there's a coordinate singularity at ##r=R_S## so the ##r## and ##t## instead the horizon are different coordinates than the ##r## and ##t## outside. It is an unfortunate accident of history that this was not recognized until long after the tradition of using the same letters to label coordinates on both sides of the horizon had taken root.
A realistic black hole interior probably will, yes. But a realistic black hole exterior still has to be one of the Kerr-Newman family of solutions.martinbn said:A realistic black hole wil probably have less symmetries.
Yes, i meant the interior.PeterDonis said:A realistic black hole interior probably will, yes. But a realistic black hole exterior still has to be one of the Kerr-Newman family of solutions.
Neglecting ##\theta, \phi## coordinates for a moment, the target of the disconnected coordinate patch is ##(0,\infty)\setminus \{R_S\} \times \mathbb R##. It is a disconnected open set of ##\mathbb R^2##.Orodruin said:It can perfectly well be viewed as the same coordinate patch. What needs to be realized though is that that coordinate patch is not connected. All that is required from a coordinate patch is a smooth function from an open subset of ##\mathbb R^n## to the manifold. In the case of Schwarzschild coordinates ##(0,\infty)\setminus \{R_S\}## is indeed an open subset of ##\mathbb R## - although not connected.
Sure, but a coordinate chart nonetheless.cianfa72 said:Neglecting ##\theta, \phi## coordinates for a moment, the target of the disconnected coordinate patch is ##(0,\infty)\setminus \{R_S\} \times \mathbb R##. It is a disconnected open set of ##\mathbb R^2##.
By the way, since charts of a manifold's atlas are homeomorphisms then in the above case that means the coordinate patch (for that Schwarzschild spacetime region) is disconnected as well.
Sorry, does it mean that Schwarzschild spacetime at horizon is actually disconnected or does it just mean that we picked a disconnected patch around the horizon (not including it) ?Orodruin said:Sure, but a coordinate chart nonetheless.
It just means the chart does not cover the horizon itself.cianfa72 said:Sorry, does it mean that Schwarzschild spacetime at horizon is actually disconnected or does it just mean that we picked a disconnected patch around the horizon (not including it) ?
On a Euclidean plane define ##r=xy## and ##\theta=\tan^{-1}(y/x)##. That gives you a coordinate grid with lines of constant ##r## being ##y\propto 1/x## (hyperbolae centered on the origin) and lines of constant ##\theta## being radial lines. This coordinate system is defined everywhere except on the ##x## and ##y## axes, so covers four disconnected regions on a trivial geometry.cianfa72 said:Sorry, does it mean that Schwarzschild spacetime at horizon is actually disconnected or does it just mean that we picked a disconnected patch around the horizon (not including it) ?
cianfa72 said:Sorry, does it mean that Schwarzschild spacetime at horizon is actually disconnected or does it just mean that we picked a disconnected patch around the horizon (not including it) ?
There are other charts (e.g. Kruskal) on the same spacetime which cover the horizon as well as the whole of the disconnnected region covered by the Schwarzschild chart.Orodruin said:It just means the chart does not cover the horizon itself.
All of the Kerr-Newman family of black hole solutions are known to be unstable in their interiors against small perturbations--in other words, anything that breaks the exact symmetry of the solution, like a piece of matter falling in from some particular direction. The BKL singularity is stable against small perturbations and is, IIRC, one of the primary candidates for what a realistic black hole interior would look like. The fact that it is still spacelike, like the Schwarzschild singularity, suggests that the deep interiors of solutions like Kerr and Reissner-Nordstrom, i.e., the region near and inside the inner horizons, are unphysical and are not present in actual black holes whose exteriors are Kerr or Reissner-Nordstrom. (The "infinite blueshift" effect at the inner horizons is another reason for believing this.)pervect said:It's likely that the Schwarzschild and Kerr solutions aren't stable, but I don't recall why I think that anymore.
Sorry, the image of ##\tan^{-1}(.)## is ##(-\pi/2, \pi/2)##. Therefore the above maps only two disconnected regions, I believe.Ibix said:On a Euclidean plane define ##r=xy## and ##\theta=\tan^{-1}(y/x)##. That gives you a coordinate grid with lines of constant ##r## being ##y\propto 1/x## (hyperbolae centered on the origin) and lines of constant ##\theta## being radial lines. This coordinate system is defined everywhere except on the ##x## and ##y## axes, so covers four disconnected regions on a trivial geometry.
##r## can be negative in this case.cianfa72 said:Sorry, the image of ##\tan^{-1}(.)## is ##(-\pi/2, \pi/2)##. Therefore the above maps only two disconnected regions, I believe.
Sorry, ##r>0## covers I and III quadrants while ##r<0## II and IV. To get a one-to-one map the inverse function of ##\text{tan} (\theta)## must be extended to the range ##(-\pi, \pi)##. In other words one must specify which quadrant.Ibix said:##r## can be negative in this case.