# The reality of black holes

1. Jun 22, 2015

### wolram

AFAIK black holes are factual, but i can not find physical proof of the singularity, apart from Einstein's theory of general relativity, which showed that when a massive star dies, it leaves behind a small, dense remnant core. If the core's mass is more than about three times the mass of the Sun, the equations showed, the force of gravity overwhelms all other forces and produces a black hole.
http://science.nasa.gov/astrophysics/focus-areas/black-holes/

Scientists can't directly observe black holes with telescopes that detect x-rays, light, or other forms of electromagnetic radiation.

They can only infer them by other other means
We can, however, infer the presence of black holes and study them by detecting their effect on other matter nearby.
My question is how do we know apart from the theory of relativity that the singularity exists.?

2. Jun 22, 2015

### phinds

"Singularity" is just a placeholder name meaning "the place where our math theories break down and give unphysical results", so of course it "exists" but the question is, what IS it? The consensus seems to be that if/when we develop a good theory of quantum gravity, we'll have a better idea what is at the heart of a black hole.

3. Jun 22, 2015

### PWiz

Yes, that's when the neutron degeneracy pressure is overcome. The actual mass limit is known as the Tolman-Oppenheimer-Volkoff limit, and you can read about it here: https://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_limit
I'm pretty sure that's because we don't have any definitive proof about that at all. Everything that happens beyond the EH of a black hole is unknown to us. According to classical physics (GR), the geometry of spacetime beyond the EH is well-defined. It's just that a true singularity (or so we think according to GR) that is present at the center of a black hole is the "weird" point on the manifold of spacetime which cannot be eliminated with a change of coordinates, and there is controversy regarding its existence. Most people believe that some quantum gravitational effects come into picture which prevent a singularity from forming, but we can't really confirm anything since we don't have a working quantum gravity theory (attempts to reconcile GR with quantum mechanics result in a non-renormalizable theory) or direct observational evidence.
EDIT: phinds beat me to it, arghhh...

Last edited: Jun 22, 2015
4. Jul 4, 2015

### craigi

Is this true?

For any observer, there is nothing beyond their apparent event horizon, right? No space, no time, just things approaching it, from the outside, asymptotically. Or have I misunderstood something?

The infalling observer would see this apparent event horizon recede, but still have no meaning for space and time beyond it, right?

Or are you talking about the absolute horizon?

Last edited: Jul 4, 2015
5. Jul 4, 2015

### rootone

Whatever is inside the event horizon is certainly not anything resembling matter in the usual sense.
You could think of a neutron star as attempted black hole which didn't quite have quite enough mass for the final collapse.
A neutron star is composed mainly of free neutrons with some smaller amount of ionized atomic nucleii and free electrons mixed in.
What happens when even the neutrons get crushed out of existence?
Theoretically the next stage of degeneracy has to be free quarks, yet according to the standard model (I think), 'free quarks' is an impossible state.

6. Jul 4, 2015

### phinds

There is nothing inside the EH that an outside observer can MEASURE, but that doesn't mean there's nothing there. For example, there is no reason at all to believe that something crossing the EH of a very large BH (where tidal forces are minimal) is instantly affected in ANY way other than that it can never cross back the other way. Pretty shortly, it will be affected materially, because it will get to where tidal forces ARE appreciable and it will be spaghetified and the shortly after that its constituent components will reach the center and we have no idea what happens there.

7. Jul 4, 2015

### phinds

True at the singularity but not true just inside the EH. See my previous post

8. Jul 4, 2015

### craigi

Sure, in the infalling frame, the spaghettified matter will reach a point where we have no valid predictions. But for the distant observer, nothing can cross the apparent horizon. So how can we assign meaning to spacetime beyone the apparent horizon?

The only way I can see it, is that there is no spacetime beyond the apparent horizon, but the apparent horizon recedes to a point as the infalling frame reaches the singularity.

9. Jul 4, 2015

### PWiz

I don't get you. I'm talking about the event horizon of a black hole; it has a physical meaning and all observers are going to agree upon it, so I don't know what you mean by 'apparent event horizon'. Moreover, spacetime is continuously described beyond the EH as well, so we can't say 'no space, no time'. The only point where this description would be apt will be the (true) singularity at the center of the black hole (whose existence is being questioned in the OP, a question which does not have any definitive answer as of now). The interior of a static, spherically symmetrical uncharged black hole can be described by the Schwarzschild metric (in standard spherical coordinates). While the usual form of the metric is $ds^2 = - (1- \frac{2GM}{rc^2}) c^2 dt^2 + (1- \frac{2GM}{rc^2})^{-1} dr^2 + r^2 (d {\theta}^2 + sin^2 \theta d{\phi}^2 )$ where $\frac{2GM}{c^2}$ represents the event horizon (note that the expression for the Schwarzschild radius / event horizon only consists of invariant quantities, and therefore all observers will agree upon its value), if you impose the condition that $0 < r < \frac{2GM}{c^2}$ , the metric tensor modifies to give $ds^2 = (\frac{2GM}{rc^2} -1) c^2 dt^2 - ( \frac{2GM}{rc^2} -1)^{-1} dr^2 + r^2 (d {\theta}^2 + sin^2 \theta d{\phi}^2 )$ . As you can clearly see, the only main difference is that $r$ has become a timelike coordinate now, which implies that any object which has crossed the EH is bound to inevitably move towards the singularity at the BH center as time passes. I repeat: spacetime is well-defined beyond the EH of a black hole. The only region where the description of spacetime breaks down is when r=0 and r=2GM/c^2. The latter is a mere coordinate singularity which does not allow the interior and exterior of the black hole to be described continuously in standard Schwarzschild coordinates, and it can be resolved by using alternate coordinate systems like Kruskal coordinates.
The only thing that I'm saying is that the EH is a one-way boundary (an outgoing null surface); the escape velocity at the EH is undefined (it tends to $c$ as we move toward the EH), and nothing escapes the BH once it has crossed the EH.
You're talking about quark stars. Though they haven't been observed yet, the concept isn't entirely ruled out. Moreover, it is theorized that free unbound quarks (and other fermions including leptons and their antiparticles) existed during the quark epoch as temperatures were too high for hadron formation at that time.
This is just redshifting in play; it does not mean that the topological description of spacetime is going to be any different.

No; I've clarified this above.
P.S. This discussion is quickly moving away from the domain of cosmology. I recommend that anyone who has relativity doubts should post them in its dedicated forum.

Last edited: Jul 4, 2015
10. Jul 4, 2015

### craigi

The Apparent Horizon is a well established concept in GR.

My first post in the thread really was motivated by ensuring that we're not providing incorrect information, but as you suggest, I'm really not confident enough in my GR knowledge to debate this, nor do I wish to confuse the original poster further.

Last edited: Jul 4, 2015
11. Jul 4, 2015

### PWiz

It is, but an 'apparent event horizon' is not. An event horizon is an absolute (coordinate independent) horizon.

12. Jul 4, 2015

### Staff: Mentor

You've misunderstood several things. First, as PWiz has pointed out, "apparent horizon" and "event horizon" are two different things. For an idealized, perfectly spherically symmetric, non-rotating black hole, they happen to coincide, but that doesn't make them the same. The one that is relevant for this discussion is the event horizon, since that's the boundary of the black hole.

Second, the event horizon is not observer-dependent; it's the same surface for all observers. (Apparent horizons are observer-dependent, but, as above, they don't define the boundary of the black hole.)

Third, there is indeed spacetime inside the event horizon. See below.

This isn't true for either the apparent horizon or the event horizon. (If we try to visualize what an infalling observer would actually see, as in the actual light rays he would receive, it is true that, even after he falls through the event horizon, light coming from the horizon can appear to be coming from in front of him, because of the way light paths are bent inside the event horizon. But that doesn't mean the event horizon is actually receding from him. It's just an optical illusion.)

What is inside the event horizon is vacuum. A black hole is not a smaller version of a static object like a neutron star, with some even more exotic state of matter inside it (see below). It's pure spacetime geometry.

As far as we know, there is no stable state of matter with higher density than neutron star matter. So once we reach densities at which neutron star matter cannot support itself against gravity, there is no other stable state possible; the only possibility is complete collapse, forming a black hole.

No, this is not correct. The correct statement is that the distant observer cannot see anything cross the horizon. (This is true of both kinds of horizon, apparent horizon and event horizon.) But the fact that the distant observer cannot see it does not mean it doesn't happen.

13. Jul 4, 2015

### craigi

Thanks for the responses.

Are you saying that in a distant observer's frame an object can pass through the event horizon?

As I understand it, in the distant observer's frame, the object approaches asymptotically, but never actually crosses the event horizon. If this is incorrect then I know what I need to investigate.

If not, then I cannot understand how we can meaningfully say that there exists spacetime within the event horizon, in the distant observer's frame.

Last edited: Jul 4, 2015
14. Jul 4, 2015

### Staff: Mentor

It depends on how you define the distant observer's frame. In the usual Schwarzschild coordinates, what you say is true; but that's an artifact of those coordinates. It doesn't tell you whether the object actually crosses the horizon or not. And there are other coordinates (such as Painleve coordinates) in which things look the same in the vicinity of the distant observer, but the coordinates are able to cover the region at and below the horizon, so the entire trajectory of the infalling object, all the way down to $r = 0$, can be described.

A better way to describe what's going on, that doesn't depend on coordinates, is what I said before: the distant observer can never see the infalling object cross the horizon, because light from the infalling object at and below the horizon can never reach the distant observer. But that, in itself, obviously does not mean the infalling object does not cross the horizon. It just means the distant observer can't see it do so. Other observers (those who fall into the hole themselves) can.

If you are tempted to argue about this, I strongly suggest that you don't. There have been hundreds of threads on PF on this topic; we probably need to just write a FAQ entry and be done with it. What I've said above is the mainstream answer given by GR, and it has remained the answer after those hundreds of threads. There's no point in another thread rehashing the same thing.

15. Jul 4, 2015

### craigi

Regardless of whether the distant observer could see an object pass the event horizon, they could in principle, see an object approach it asymptotically. The two are mutually exclusive.

I don't intend to argue about. It just doesn't make sense to me. I'll investigate elsewhere, why this is the mainstream answer in GR in order to try make sense of it. Thanks for your time.

Last edited: Jul 4, 2015
16. Jul 4, 2015

### phinds

I think the assumption is that the distant observer is not an idiot and knows that what he sees and what an infilling observer would see are two different things. That is, as for what he can SEE, you are right that he can't see anything falling in, but he knows that it is.

17. Jul 4, 2015

### Staff: Mentor

As I mentioned, there have been plenty of threads on PF discussing this. Also, the Usenet Physics FAQ has an article on this:

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html

The article also has references to the chapters in Misner, Thorne, & Wheeler, the classic GR textbook, where this issue is discussed.

18. Jul 4, 2015

### craigi

I would say that nothing is passing through the event horizon in the distant observer's frame, but perhaps we're using different terminology.

19. Jul 4, 2015

### Staff: Mentor

That's because "the distant observer's frame" (at least in the coordinate chart you are implicitly using--note that there are others, as I mentioned before, that don't have the same properties) does not cover the horizon or the region inside it. But once again, that's an artifact of that particular coordinate chart. It doesn't tell you whether the horizon exists or not, or whether objects can fall through it. You can never draw conclusions like that in GR just from looking at coordinates. You have to look at invariants.

20. Jul 4, 2015

### craigi

The article in that link agrees precisely with what I've been saying.

This is why I'm saying that there is no spacetime beyond the event horizon in the distant observer's frame.