If no singularity, what’s inside a big black hole?

[Bernie G]

I'm not terribly happy about the possible calculation above for the size of a theoretical radiation ball inside the Schwarzschild radius. Does anyone have a better simple non-relativistic formula for the radius of a gravitationally formed sphere of Mass M, with a density profile of 1/r^2, and a supporting pressure of (1/3)pc^2 or (2/3)pc^2 or pc^2, where p is the radiation density?

 

[Bernie G]

Maybe the maximum fraction of energy that can exerted by a neutron to generate pressure is (1/3)Mc^2 or (2/3)Mc^2, and above that a neutron disintegrates.

 

[mjacobsca]

There are theories for quark stars and preon stars, so why not further degenerate states such as photon-degenerate stars, string-degenerate, etc...? I suppose a photon-degenerate star would technically be a radiation soup as suggested above. But my gut says that the actual center of a black hole is going to be no more interesting than the center of any other degenerate star. No infinity. No wormholes. No baby universes. Just plain old photons/other tiny particle stuck together by gravity with a big old beware-of-danger sign on their front porch. I'm no expert, but it seems a degenerate state of matter beyond from which light cannot escape is simple, logical, and plausible without blowing up physics as we know it.

 

[tom.stoer]

There are theories for quark stars and preon stars, so why not further degenerate states such as photon-degenerate stars, string-degenerate, etc...?

There are such theories, i.e. fuzzballs in string theory.

Just plain old photons/other tiny particle stuck together by gravity ... but it seems a degenerate state of matter beyond from which light cannot escape is simple, logical, and plausible without blowing up physics as we know it.

The problem is that we do not know any force in nature that is able to stabilize this state. There are indications coming from quantized gravity (strings, loops) that gravity itself could do the job, but this is still work in progress.

 

[Bernie G]

Yes, the "relativists" say that inside the Schwarzschild radius a photon must have energy greater than mc^2 if it is not to proceed inexorably towards the center. My experience is they do not accept that other forces (such as radiation pressure) can overcome or balance the force of gravity inside the Schwarzschild radius. One possible error in this way of thinking is that just inside the Schwarzschild radius of a large black hole, the gravitational force is much weaker than that just inside the Schwarzschild radius of a small black hole. This might be settled someday if and when the effects of the merger of 2 roughly equal size black holes are observed. If there is a large radiation burst the relativists will not be able to explain it. In the meantime they will give each other praise and awards.

 

[phinds]

... In the meantime they will give each other praise and awards.

This point of view seems to me to demonstrate a totally unwarrented distain for scientists and the scientific method. Is that in fact how you feel or am I misinterpreting you?

 

[tom.stoer]

My experience is they do not accept that other forces (such as radiation pressure) can overcome or balance the force of gravity inside the Schwarzschild radius. One possible error in this way of thinking is that just inside the Schwarzschild radius of a large black hole, the gravitational force is much weaker than that just inside the Schwarzschild radius of a small black hole.

Either have to explain (e.g.) the radiation pressure in terms of the energy-momentum tensor, or you have to explain weakening of the gravitational force in terms of the gravitational field. As soon as you an derive a repusive effect based on the Einstein equations this will sound convincing. But as long as no such calculation exists we have to accept the collaps.

btw.: we know about quantum corrections of Einstein equations in certain quantum gravity models (e.g. LQC) which result in a to a "short-range repulsive core of the gravitational potential".

 

[IttyBittyBit]

According to the holographic principle, every bit of information that there is to know about a black hole is encoded on it's even horizon (albeit in a very mixed-up and highly entropic way). Given that we can never probe a black hole's interior, I was under the impression that the holographic principle implies that there is no 'inside' to a black hole. Asking what's inside a black hole is like asking what was there before the big bang. Correct me if I'm wrong.

 

[Bernie G]

If two relatively small equal size black holes merge, and IF each contains a radiation star with 80% of the radius of the Schwarzschild radius, at the point of contact of each star’s surface there will be no net gravitational force and a large radiation burst will occur. Probably each star would only have to be only 70% of the radius Schwarzschild radius for a radiation burst to occur because of the bulging effect at each surface as each black hole approaches the other. Therefore its possible to get information out of a black hole, but it would require contact with another black hole. Its true I don’t have a lot of respect how awards are given out nowadays, not just scientific awards but also in many other fields. Often its a one hand washing the other relationship. The Nobel prize has even been corrupted. But I should not have made the praise and awards comment as it is a distraction from the technical discussion.

 

[Bernie G]

I think the following figures are roughly correct: One of the sources of gamma ray bursts may be the merger of orbiting black hole - neutron star pairs in other galaxies, and perhaps about a hundred of these occur annually. If about 1% of this number is the merger of orbiting black hole - black pairs, then about one of these observable BH-BH mergers should occur annually.

 

[Bernie G]

The last sentence should have read: If about 1% of this number is the merger of orbiting black hole - black pairs, then about one of these BH-BH mergers should occur annually, and possibly are observable by a gamma ray burst.

 

[tom.stoer]

According to the holographic principle, every bit of information that there is to know about a black hole is encoded on it's even horizon (albeit in a very mixed-up and highly entropic way). Given that we can never probe a black hole's interior, I was under the impression that the holographic principle implies that there is no 'inside' to a black hole. Asking what's inside a black hole is like asking what was there before the big bang. Correct me if I'm wrong.

You must distinguish between "representation" of the information inside the BH on the EH and the interior itself. According to GR an astronaut could fall into a BH and would while crossing the horizon of a sufficiently large BH) not feel or see anything special.

 

[genneth]

You must distinguish between "representation" of the information inside the BH on the EH and the interior itself. According to GR an astronaut could fall into a BH and would while crossing the horizon of a sufficiently large BH) not feel or see anything special.

I've always thought that this might just be a pathology of classical GR.

As Ashtekar is keen to point out, the correct concept to think about wrt. horizons is that of isolated or dynamic horizon, if for no other reason than that it is possible to do Hamilotian mechanics this way (conserved Louiville form, etc.). From the point of view of an asymptotic observer, nothing ever falls through the horizon. The quantum description of the spacetime for that observer should simply not include the space inside the horizon.

For an in-falling observer, the horizon should shrink due to radiative loss. This should mean that there are no isolated horizons, but *only* dynamic ones. Is it known what an in-falling observer would see of that horizon, as it evaporates? Does the observer then still ever cross it? My gut feeling is that actually, no --- the quantum effects will always hide the inside of the horizon from view, so all observers will only need a description of the outside, which is known (i.e. conjectured from non-quantum GR) to have a Hamiltonian description and thus be described by a quantum theory.

If you know of literature to answer this, I'd be fascinated. My Google-fu is weak, and I have yet to find anything...

 

[tom.stoer]

Afaik LQG does not say anything else but classical GR. It provides a quantum description of isolated horizons, but I see no reason why the large scale dynamics should change.

Regarding radiative loss: afaik there is no theory which is able to predict this radiation in the quantum gravity regime; Hawking result is restricted to classical GR. Regarding time scales: you can calculate the time for an infalling observer to cross the horizon and compare it with the time for complete evaporation. You will find that the time to cross the horizon is much smaller than the evaporation time.

The asmptotic observer at infinity is of no relevance for the pure observer crossing the horizon in finite proper time.

I don't think that any theory of quantum gravity will change this picture

 

[genneth]

Regarding radiative loss: afaik there is no theory which is able to predict this radiation in the quantum gravity regime; Hawking result is restricted to classical GR. Regarding time scales: you can calculate the time for an infalling observer to cross the horizon and compare it with the time for complete evaporation. You will find that the time to cross the horizon is much smaller than the evaporation time.

That evaporation time you refer to is measured by an asymptotic observer, but the proper time of an infalling observer is clearly not --- these two are not comparable. My point is that I think (and would like to be educated) that the calculations do not exist, but my grasp of the subject is not good enough to simply go and calculate it myself, or understand why such a calculation might be hard/ill-posed.

 

[IttyBittyBit]

You must distinguish between "representation" of the information inside the BH on the EH and the interior itself. According to GR an astronaut could fall into a BH and would while crossing the horizon of a sufficiently large BH) not feel or see anything special.

At the event horizon gravity becomes as strong as all the other forces. It is by definition in the realm of quantum gravity. I don't think GR, by itself at least, is really applicable to studying it, even though the concept of black holes originally arose from GR.

For an in-falling observer, the horizon should shrink due to radiative loss. This should mean that there are no isolated horizons, but *only* dynamic ones. Is it known what an in-falling observer would see of that horizon, as it evaporates? Does the observer then still ever cross it? My gut feeling is that actually, no --- the quantum effects will always hide the inside of the horizon from view, so all observers will only need a description of the outside, which is known (i.e. conjectured from non-quantum GR) to have a Hamiltonian description and thus be described by a quantum theory.

That's actually a very interesting viewpoint, and if true it strengthens my argument.

According to this page: http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html, the event horizon keeps receding until it's a point, and that is the precise moment you 'fall' in. By the time you're inside it, it is a 0-dimensional point without any inside. You are then promptly converted to Hawking radiation and ejected. So, if the inside of the event horizon remains forever beyond the grasp of any observer, there is no reason to think that the inside of it exists at all.

 

[tom.stoer]

That evaporation time you refer to is measured by an asymptotic observer, but the proper time of an infalling observer is clearly not

You have to transform the result accordingly

 

[tom.stoer]

At the event horizon gravity becomes as strong as all the other forces. It is by definition in the realm of quantum gravity. I don't think GR, by itself at least, is really applicable to studying it, even though the concept of black holes originally arose from GR.

Sorry to say that, but this is totally wrong.

It is not even true that gravity becomes strong at the horizon. The larger the black hole the smaller the surface gravity is. Closed to a sufficiently large black hole the surface gravity is very small, the observer feels nothing special, not even when he crosses the horizon.

There are numerous papers regarding black hole geometries (Schwarzschild, Kerr, ...), exact calculations (free falling observer, stable and unstable orbits, ...), numerical calculations (infalling matter, accretion discs, black hole merger, ...) all based on GR.

There is no indication that GR does break down and requires correction near the event horizon.

 

[Bernie G]

"It is not even true that gravity becomes strong at the horizon. The larger the black hole the smaller the surface gravity is. Close to a sufficiently large black hole the surface gravity is very small, the observer feels nothing special, not even when he crosses the horizon."

Yes. This is probably why the largest black holes "turn off" in terms of visibility. Weird.

 

[IttyBittyBit]

It is not even true that gravity becomes strong at the horizon. The larger the black hole the smaller the surface gravity is. Closed to a sufficiently large black hole the surface gravity is very small, the observer feels nothing special, not even when he crosses the horizon.

Maybe I phrased myself incorrectly. I should have said that at the event horizon, the curvature of spacetime becomes so huge that it can no longer be ignored at small scales. This is regardless of the size of the black hole. To study the event horizon, GR itself is insufficient - it doesn't predict Hawking radiation, for example, which is emitted `from' the event horizon.

 

[Bernie G]

"It is not even true that gravity becomes strong at the horizon. The larger the black hole the smaller the surface gravity is. Close to a sufficiently large black hole the surface gravity is very small, the observer feels nothing special, not even when he crosses the horizon."

I think you said it right. At the event horizon of the largest black holes the gravity and curvature is probably small enough that in-falling material doesn't radiate.

 

[genneth]

You have to transform the result accordingly

I know... but that transform is non-trivial! E.g. the part of the world line of the infalling observer which is inside the horizon is not even in the spacetime of the asymptotic observer, so the transform must have some singularities; my question is whether they are physical. Searching for literature on this gives remarkably thin results (i.e. none). I really would like to know the answer, but I don't think anyone has it --- I would be happy to be shown otherwise.

 

[Bernie G]

BTW, maybe its not appropriate here, but I went to a talk a few days ago where the Nasa speaker said there may be far distant black holes with up to 10^12 solar masses. If true, that's about the equivalent of 1000 Milky Ways. Wow.

 

[tom.stoer]

E.g. the part of the world line of the infalling observer which is inside the horizon is not even in the spacetime of the asymptotic observer ...

I was only talking about the evaporation time compared to the time it takes for the infalling observer to cross the horizon; not to hit the singularity.

You only need a rough estimate.

The black hole evaporation time for a black hole of one solar mass (calculated based on Hawking radiation - which is definately incorrect as soon as the hole becomes smaller and has Planck size) is approx. 10⁶⁷ years. Now think about an observer not located at infinity but e.g. at the Earth orbit. The result is approx. the same (the gravity of the sun at the Earth orbit is small, therefore time dilation due to the gravitational field is very small). And now think about this observer falling into the black hole. It will definity take less than 10⁶⁷ years ...

http://en.wikipedia.org/wiki/Hawking_radiation

 

[IttyBittyBit]

tom.stoer, I think what genneth is getting at is that near the event horizon, gravitational time dilation increases without bound.

http://en.wikipedia.org/wiki/Gravitational_time_dilation#Outside_a_non-rotating_sphere

No matter how long it took for the black hole to evaporate, there is some finite distance from the event horizon where you would experience this time to be very short.

Think about it this way. As you fall into the event horizon, the Hawking radiation from the black hole is blue-shifted to such a high energy that it appears that the black hole is evaporating very quickly.

The statement 'you would not notice anything while falling into a large black hole' is not technically true. I would hardly call being blasted by intense gamma radiation, increasing in energy to infinity, 'not noticing anything'.

Of course, this is just a re-stating of the trans-Planckian problem. Which indicates the difficulty current physics has with event horizons. There are proposed solutions of course, fuzzball being one of them. At the end of the day you need some form of quantum gravity to explain event horizons adequately.

 

[tom.stoer]

tom.stoer, I think what genneth is getting at is that near the event horizon, gravitational time dilation increases without bound.

...

No matter how long it took for the black hole to evaporate, there is some finite distance from the event horizon where you would experience this time to be very short.

That's true for a stationary observer but not for the infalling one. For him it takes finite proper time to cross the event horizon.

As you fall into the event horizon, the Hawking radiation from the black hole is blue-shifted to such a high energy that it appears that the black hole is evaporating very quickly.

This is wrong! (the blue shift is correct but the effect is tiny)

Of course, this is just a re-stating of the trans-Planckian problem. Which indicates the difficulty current physics has with event horizons.

There is no problem with event horizons in general relativity. They are well-understood and well-behaved.

There are proposed solutions of course, fuzzball being one of them. At the end of the day you need some form of quantum gravity to explain event horizons adequately.

All these proposals are attempts to resolve the singularity-issue. But there is no horizon-issue. They all agree that near the horizon of large black holes GR is still the correct low-energy limit.

Have you ever made a single calculation in general relativity by yourself?

 

[genneth]

That's true for a stationary observer but not for the infalling one. For him it takes finite proper time to cross the event horizon.

Finite proper time if the horizon is eternal --- but the point is that it isn't.

Consider the following statements, and tell me where the logic goes off the rails:

1. An asymptotic observer never sees an infalling observer cross the event/dynamical horizon.
2. The horizon evaporates in a finite time.
3. The asymptotic observer will see the infalling observer still there after the horizon evaporates.
4. Therefore from the asymptotic observer's point of view, she doesn't cross the horizon either, and will live to see it completely evaporate.

This calculation can indeed be pushed all the way until the semi-classical approximation breaks down, and I think it's correct. I think this paper by Krauss (http://arxiv.org/abs/gr-qc/0609024 or Phys.Rev.D76:024005,2007) says the same thing, though I'm not sure I entirely agree with the details (event horizon vs. dynamical horizon, and therefore the interpretation).

(Btw, I am in no way invested in the original genesis of this problem --- I just think this scenario is worth thinking about as a thought experiment and might be informative on matters in general, not necessarily including the issue of what replaces a singularity...)

 

[phinds]

Only thing that occurs to me is that is would appear this argument requires the the two observers see the same event as though it were happening at the same time for both of them. I'm not sure I've said that right, but my point is that it seems to merge the two reference frames in a way that is not correct.

 

[tom.stoer]

Finite proper time if the horizon is eternal --- but the point is that it isn't.

Consider the following statements, and tell me where the logic goes off the rails:

1. An asymptotic observer never sees an infalling observer cross the event/dynamical horizon.
2. The horizon evaporates in a finite time.
3. The asymptotic observer will see the infalling observer still there after the horizon evaporates.
4. Therefore from the asymptotic observer's point of view, she doesn't cross the horizon either, and will live to see it completely evaporate.

The first flaw is that the asymptotic observer sees the infalling one approaching the horizon and standing still only if the horizon does not change. But as soon as you let the black hole evaporate the horizon will shrink and the infalling observer will no longer be frozen at the horizon.

The general flaw is that you mix two scenarios, namely arguments for a static spacetime with arguments for a dynamic spacetime with an evaporating BH.

The third flaw is that you don't calculate (or believe) what the infalling observer will actually see. The free-fall time is much smaller than the evaporation time.

 

[suprised]

All these proposals are attempts to resolve the singularity-issue. But there is no horizon-issue. They all agree that near the horizon of large black holes GR is still the correct low-energy limit.

This is not correct. Practically all important discussions and confusions turn around the horizon, and almost not at all around the singularity. The point seems to be that despite the horizon is weakly curved, quantum effects are strong and emphatically quantum gravity effects must play a crucial role there. The fuzzballs were invoked to implement the required macroscopic non-locality within string theory and this is definitely a horizon issue. Indeed in certain circumstances, quantum gravity effects are very relevant in the IR, while many approaches too naively just concentrate on the UV. The whole last week of our quantum gravity workshop was, in fact, devoted to precisely this issue.