I Correct Description of Black Hole Interior

[Onyx]

TL;DR Summary

Questions about a black hole’s interior

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. Also, is there a way to describe just a black hole’s with regular spherical coordinates?

 

[Nugatory]

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.

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.

Also, is there a way to describe just a black hole’s with regular spherical coordinates?

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.

 

[Onyx]

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.

I noticed the difference in terms: patch vs solution. By the interior “solution”, do you mean this?

 

[PeterDonis]

By the interior “solution”, do you mean this?

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.

 

[Onyx]

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. 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.”

 

[Nugatory]

I noticed the difference in terms: patch vs solution. By the interior “solution”, do you mean this?

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.

 

[PeterDonis]

In my OP I meant the interior coordinate patch of the Schwarzschild analytical extension.

That's what I thought. That is a vacuum solution, and as @Nugatory has posted, that's the solution he was talking about.

 

[Onyx]

But one thing that is still confusing to me is why so much emphasis is placed on the Einstein-Rosen

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.

Are r and t swapped?

 

[Onyx]

Finally, 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. Isn’t the “other universe” just a mathematical artifact?

 

[PeterDonis]

Are r and t swapped?

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.

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.

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.

Isn’t the “other universe” just a mathematical artifact?

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.

 

[Orodruin]

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.

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.

 

[martinbn]

A realistic black hole wil probably have less symmetries.

 

[PeterDonis]

A realistic black hole wil probably have less symmetries.

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]

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.

Yes, i meant the interior.

 

[cianfa72]

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.

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.

 

[Orodruin]

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.

Sure, but a coordinate chart nonetheless.

 

[cianfa72]

Sure, but a coordinate chart nonetheless.

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]

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) ?

It just means the chart does not cover the horizon itself.

 

[Ibix]

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.

Schwarzschild coordinates are similar. If you plot them on a Kruskal diagram, they even look like the above, although rotated 45°.

 

[DrGreg]

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) ?

It just means the chart does not cover the horizon itself.

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.

 

[pervect]

Andrew Hamilton has several paper on realistic black holes. Which if any of his ideas are accepted, I don't know, as has been mentioned, we can't experimentally test the interior solutions experimentally.

I've seen other similar papers by the same author, but the one google finds at the moment is:
https://arxiv.org/abs/1907.05292. The citation count is low, though, so it's hard to tell how influential and accepted this paper is.

Scanning the abstract, this is the first time that I recall seeing Hamilton mention the BKL solution. I know that Kip Thorne had at one time commented favorably (in a popularization, but not in a peer reviewed paper) on the BKL singularity.

https://en.wikipedia.org/wiki/BKL_singularity

It's likely that the Schwarzschild and Kerr solutions aren't stable, but I don't recall why I think that anymore. Anyway, there are some peer reviewed papers on the topic, it's hard to say if there is any consensus, I suspect the consensus opinion is that we don't know and can't find out for sure.

 

[PeterDonis]

It's likely that the Schwarzschild and Kerr solutions aren't stable, but I don't recall why I think that anymore.

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.)

 

[cianfa72]

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.

Sorry, the image of $\tan^{-1}(.)$ is $(-\pi/2, \pi/2)$. Therefore the above maps only two disconnected regions, I believe.

 

[Ibix]

Sorry, the image of $\tan^{-1}(.)$ is $(-\pi/2, \pi/2)$. Therefore the above maps only two disconnected regions, I believe.

$r$ can be negative in this case.

 

[cianfa72]

$r$ can be negative in this case.

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.