The Central Region of Black Holes: Composition, Energetics, and Equilibrium

[stewart brands]

What is the central region of the so called black hole composed of? Are there intense energetic waves caused by thermonuclear fissile processes? Does the high intensity ray conflagration coupled with immense crushing cause electron repulsion?Does this process also cause nuclear disintegration into a Bose Einstein condensate? For a nuclear condensate to be created,is not close to an Absolute zero state necessary? So is the opposite also not true that the existence at the center of a black hole of a Bose Einstein condensate implies that the central volume is near absolute zero?
Would the released energy have a motion vector toward this absolute cool center(which is colder than space)by the Thermodynamic Law? Would not this inward force of the heat and other energies coupled with the gravitational energies not increase the condensate? Would this not then reach equilibrium into an ice cold core surrounded by an in-escaping fire shell?

 

[PAllen]

First, you have to decide are you interested in what classical GR says ( or classical GR as a background for established theories matter), or what is likely given that both the standard model of particle physics and GR lose their accuracy somewhere inside a sufficiently collapsed object? In the latter case, the obvious answer is nobody knows and I won't speculate.

Even in the GR as semi-classical background, the more you involve QM the more disputed the results become (e.g. firewalls; whether evaporation beats collapse). So sticking with 'almost classical GR':

1) In a supermassive BH horizon, the interior involves no unusual physics, at first. You can have completely normal stars, well separated, inside the horizon.

2) Independent of whether a BH ultimately forms, the mass/energy density and pressure determine the character of local physics. Thus, the regime of neutron star physics applies whether you have an isolated, largely stable, neutron star or the same conditions somewhere inside of a collapsing body that is fated to form a BH, or that already has and these conditions exist somewhere inside the horizon.

3) Somewhere inside any BH, eventually, you get energy densities and distance scales such that known theories are expected to break down, so nothing further can be said (except speculation).

 

[stewart brands]

condensate

Does a black hole have a condensate core and is this at absolute zero temp? Could the Sun also have an absolute zero temp. condensate core?(a Bose Einstein condensate )

 

[Nugatory]

Does a black hole have a condensate core and is this at absolute zero temp?

As PAllen said above, we have no theory that describes the point at the center of a black hole, although general relativity does tell us that much of the space inside the event horizon can be a perfectly reasonable and unsurprising vacuum.

Could the Sun also have an absolute zero temp. condensate core? (a Bose Einstein condensate)

<understatement>
Seems unlikely, as the core of the Sun is a region of high pressure and high temperatures with a fair number of internal degrees of freedom.
</understatement>

 

[stewart brands]

Are nucleii destroyed(reduced to sub elementary particles) near the sun's center?

 

[Nugatory]

Are nucleii destroyed(reduced to sub elementary particles) near the sun's center?

No. They fuse into larger nuclei. Google for "stellar evolution", "carbon cycle fusion", "fusion stars" and similar phrases and you'll find lots of information.

 

[stewart brands]

The definition of a star would imply thermonuclear processes in which there is a net loss of energy.How then could a star produce heavier atoms,when its raison detre is to destroy them and dump their energy into space?

 

[PeterDonis]

The definition of a star would imply thermonuclear processes in which there is a net loss of energy.

A net loss of energy is not necessarily the same as going from heavier atoms to lighter atoms. If it were, stars could not exist at all, since they run on fusion reactions, which make heavier atoms out of lighter atoms.

The key is that, for atoms lighter than iron, heavier nuclei are more tightly bound than lighter ones (it's actually more complicated than that, for details see here, but the complications aren't necessary for now), so taking, for example, four hydrogen atoms and fusing them into one helium atom is an exothermic reaction--there is a net release of energy. Similarly for fusing helium to carbon, and so on up the periodic table until you get to iron. (For nuclei heavier than iron, heavier nuclei are less tightly bound, so fission reactions have a net release of energy.)

 

[audioloop]

What is the central region of the so called black hole composed of? Are there intense energetic waves caused by thermonuclear fissile processes? Does the high intensity ray conflagration coupled with immense crushing cause electron repulsion?Does this process also cause nuclear disintegration into a Bose Einstein condensate? For a nuclear condensate to be created,is not close to an Absolute zero state necessary? So is the opposite also not true that the existence at the center of a black hole of a Bose Einstein condensate implies that the central volume is near absolute zero?
Would the released energy have a motion vector toward this absolute cool center(which is colder than space)by the Thermodynamic Law? Would not this inward force of the heat and other energies coupled with the gravitational energies not increase the condensate? Would this not then reach equilibrium into an ice cold core surrounded by an in-escaping fire shell?

https://www.physicsforums.com/showthread.php?t=622022

 

[Nugatory]

The definition of a star would imply thermonuclear processes in which there is a net loss of energy.How then could a star produce heavier atoms,when its raison detre is to destroy them and dump their energy into space?

Fusing lighter elements into heavier (hydrogen into helium, helium into lithium, ...) generates energy until you get up to iron. Take two nuclei lighter than iron, fuse them into one heavier nucleus that's still no heavier than iron, and energy will be released to be radiated out into space.

Google for "nuclear binding energy" to see how this works, and if you haven't already tried "stellar evolution" as I suggested above, I'll repeat my advice that you google for that too.

 

[BruceW]

Does the high intensity ray conflagration coupled with immense crushing cause electron repulsion?Does this process also cause nuclear disintegration into a Bose Einstein condensate? For a nuclear condensate to be created,is not close to an Absolute zero state necessary? So is the opposite also not true that the existence at the center of a black hole of a Bose Einstein condensate implies that the central volume is near absolute zero?
Would the released energy have a motion vector toward this absolute cool center(which is colder than space)by the Thermodynamic Law? Would not this inward force of the heat and other energies coupled with the gravitational energies not increase the condensate? Would this not then reach equilibrium into an ice cold core surrounded by an in-escaping fire shell?

what? as others said, inside the event horizon, things happen 'normally', except there is gravitational pull towards the singularity. And what goes on inside the singularity is speculation. So there is no equilibrium, just stuff falls into the singularity. Also, the entire point of a Bose-Einstein condensate is that it happens not near absolute zero, even though all the Bosons are in the ground state. Well, it is typically a lot closer to absolute zero than room temperature, I suppose. Also, I don't know why you suggest that a high-pressure system would cause a Bose-Einstein condensate anyway.

edit: I really don't know much astronomy, so don't take me too seriously. I guess when matter is 'sucked' in by the black hole, there might be a lot of heating as the matter is pushed into a smaller volume, so this probably produces a lot of radiation, but I doubt that any kind of equilibrium could be formed which would prevent matter from falling into the singularity.

 

[Ookke]

How it would look like inside black hole, any educated guess? The singularity is not visible from above event horizon, but how about below it? Might look interesting... (or not)

 

[Nugatory]

The singularity is not visible from above event horizon, but how about below it?

Not visible. It's in the future of any event inside the horizon.

(That's the answer for the Schwarzschild solution, which assumes zero angular momentum. For any realistic black hole, there will be some non-zero angular momentum so we have to use the Kerr solution - and it's not at all certain that the Kerr solution works inside the horizon)

 

[The17YearOld]

Well this is my first post on the website. I read articles on it quite a bit, but I finally made an account to share my thoughts.
If the question is what is inside a black hole, then obviously the answer is going to speculative as there would be no way to prove it, but just by using what is known about black holes one can make many hypothesizes. One of things that is known about a black hole is that gravity begins to get particularly strong at one time like hitting a wall of extreme gravity rather than slowly easing into it like most objects falling into a gravitational field. It is generally said that when this wall known as the event horizon is reached, time slows to nothing, when judged from a reference point outside of the black hole. It would seem that the event horizon is the real "beginning" or "shell" of the black hole. To understand what is inside the black hole, one would need an explanation for what causes this extreme change in gravity so suddenly and warps space enough to cause time to stop at one moment and causing acceleration of an object to increase impossibly rapidly. Having an explanation for that would allow the inside of a black hole to be better imagined and understood.

 

[BruceW]

I think the event horizon is not so special as you make it seem to be. gravity is 'normal' even when you pass over the event horizon. There is no sharp difference in gravity either side as you suggest.

Edit: ah, also, welcome to physicsforums :)

 

[The17YearOld]

Is it not true that once light crosses the event horizon it can never escape due to the gravity that the black hole has. The event horizon is the only place in the universe that once crossed, light the fastest known thing in the universe, cannot escape. The gravity beyond the event horizon pulls light back with more force than than the light has to escape. It seems to me that the event horizon is a very dramatic change in gravity if it has the power to hold light back from its path. Even the largest objects in the universe only curve space enough to make light bend, not pull it backwards.

 

[WannabeNewton]

The metric tensor is continuous in going from $r > 2M$ to $r < 2M$ (as long as you write it down in coordinates non-singular at $r = 2M$). There is no extreme change in it.

 

[BruceW]

oh, yeah. It's a removable singularity, right? So I guess in some coordinate systems there is a sharp change. I tend to think of such things as not 'physically important'. Is this about right, Newton?

Another way to picture what's going on: if you are standing at some fixed distance from the black hole, and a light beam is emitted from a place just outside the event horizon, then when the light travels up to you, it will be very red-shifted. In other words, it will only just be able to escape.

So in a sense, the energy of the light which escapes makes a smooth transition at the event horizon. i.e. if the light is emitted from the inside, zero of its energy will reach you. And if the light was emitted from only just outside, only a very small fraction of its energy will reach you. (And this limit can be taken to zero, as the place of emission tends to the event horizon).

 

[The17YearOld]

How concrete is the theory that at the center of a black hole lies a singularity. Is this just the most reasonable theory right now or is a singularity the only way a black hole could exist?

 

[Nugatory]

How concrete is the theory that at the center of a black hole lies a singularity. Is this just the most reasonable theory right now or is a singularity the only way a black hole could exist?

Neither.

A Schwarzschild black hole will be produced by any (spherically symmetric, non-rotating, non-charged) distribution of matter concentrated inside its Shwarzchild radius, whether it's collapsed into a singularity or not; so the singularity certainly is not required.

On the other hand, although the equations of general relativity say that there will be a singularity, that's often regarded as an argument that GR is incomplete, not that the singularity exists.

 

[WannabeNewton]

oh, yeah. It's a removable singularity, right? So I guess in some coordinate systems there is a sharp change. I tend to think of such things as not 'physically important'. Is this about right, Newton?

Yeah. They are usually called coordinate singularities, to contrast with geometric singularities which are intrinsic to the space-time and cannot be removed by a change of coordinates (such as the one at $r = 0$).

 

[BruceW]

yes. I don't fully understand the r=0 singularity in the Schwarzschild metric. (and this overlaps with 17's question). As I understand it, the Schwarzschild metric only applies to the region outside of the spherical matter distribution. So if the matter distribution has a finite size (lets say R), then there is no problem at r=0, because the Schwarzschild metric is only valid at r>R, and so there is no singularity at r=0.

Instead, if we say that the matter distribution is literally a point particle, then the Schwarzschild metric would be correct for r>0. Again, no problem, since at r=0, the Schwarzschild metric is not applicable. Or maybe the Schwarzschild metric is also supposed to be correct at the boundary of the matter distribution? So then we expect the Schwarzschild solution to give us the correct answer at r=0. And then we find that according to the Schwarzschild solution, the Kretschmann scalar diverges at r=0, so the singularity is a physical singularity.

So there is a physical singularity at the origin simply because we have assumed a point mass at the origin. This seems kind of obvious when it is said in this way. The last thing I was wondering about, is about the dynamic motion of the matter when it is all inside its event horizon. It will all reach r=0 within a finite time, right? So I guess if we do not feed the black hole any more matter, then eventually all of its matter will eventually end up at the r=0 point. So I suppose this is kind of a justification for saying that we will have a point mass at r=0.

 

[WannabeNewton]

yes. I don't fully understand the r=0 singularity in the Schwarzschild metric. (and this overlaps with 17's question). As I understand it, the Schwarzschild metric only applies to the region outside of the spherical matter distribution.

For Schwarzschild black hole space-times we have a maximal extension that covers everything: http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

 

[BruceW]

These coordinates solve the problem of eliminating the 'removable singularity' at the event horizon. But I'm pretty sure they assume a point mass (not a finite distribution of matter). Otherwise, there would have to be some function for how the matter density varies.

 

[WannabeNewton]

Well the coordinates don't assume anything about mass. They just extend the solution so as to cover everything other than $r = 0$; they doesn't assume anything about the distribution of matter content inside the black hole. All we can say is that the black hole has a total mass $M$. The misunderstanding seems to come from thinking that the singularity $r = 0$ is a single point in space. It isn't.

 

[BruceW]

Still, the solution only applies to the region where there is no matter, right? In other words, all the matter has to be at r=0, for the solution to apply to r>0. Also, I'm not sure what you mean that r=0 is not a single point in space. I can kind of imagine how r=0 could define some extended region which is not part of the coordinate system we are using. But is this really a meaningful concept?

 

[WannabeNewton]

$r = 0$ is a space-like hypersurface. The solution is still a vacuum solution if that's what you are asking.