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My understanding is that time slows and approaches stopping when approaching the event horizon of a black hole. I have seen this explained several places, including a brief explanation in the last paragraph under:, quoted below:

Oppenheimer and his co-authors interpreted the singularity at the boundary of the Schwarzschild radius as indicating that this was the boundary of a bubble in which time stopped. This is a valid point of view for external observers, but not for infalling observers. Because of this property, the collapsed stars were called "frozen stars",[17] because an outside observer would see the surface of the star frozen in time at the instant where its collapse takes it inside the Schwarzschild radius.

Does this mean then that no matter actually falls into a black hole (except possibly what was there at its formation)? Would this also mean matter is accumulating just outside its event horizon? As I understand it, this would be the perspective from outside the black hole. If this is the case, I wonder if we would observe a tremendous amount of matter surrounding the event horizon, but it would be extremely red shifted?

Edit: I noticed an answer to a different question, especially the end portion, provides some insight here as well:

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You should quote where you read it. However, I guess you are talking about relativistic effects (delay) observed from a distant observer. Is it correct? – Py-ser May 15 '14 at 3:09
My personal opinion: That's the reason (together with Hawking radiation making the BH vanishing over finite time, as seen from outside), why an event horizon never can form. But that's not (yet?) the main-stream opinion. – Gerald May 15 '14 at 9:31
@Py-ser - Yes, this is correct, I am talking about the relativistic effects. – Jonathan May 15 '14 at 14:57
up vote 7 down vote accepted

What you're describing is basically the "frozen star" interpretation of black holes that was common prior to the late 1950s. It was a mistake.

Suppose you are distant and stationary relative to the black hole. You will observe infalling matter asymptotically approaching the horizon, growing ever fainter as it redshifts. Does it mean that matter "clumps" around the horizon? To find out, suppose you throw yourself towards the black hole to try to catch the matter that you see. What you will find is that it fell into the black hole long ago

In other words, the most sensible way to answer whether or not infalling matter clumps on the horizon is to look at the situation from the frame of that infalling matter. And there, it is clear: no, it does not clump, as it crosses the horizon in finite proper time. (As an aside, for a Schwarzschild black hole, falling from rest is exactly Newtonian in Schwarzschild radial coordinate and proper time.)

Ok, but since it still takes an infinite amount time in the frame adapted to a stationary distant observer, does that mean that the horizon never forms in that frame? It does form: the underlying assumption in the argument that it does not would be either that the infalling matter needs to reach the center for the horizon to form or cross a pre-existing horizon to make it expand. But that assumption is simply not true.

An event horizon is defined in terms of future lightlike infinity, roughly speaking in terms of whether or not light rays escape if one waits an infinite amount of time. That means the location of the horizon at any time depends on not just what has happened, but also what will happen in the future. In the frame of the distant stationary observer, as matter falls towards the event horizon, it does slow down to asymptotically approach... but the horizon also expands to meet it. Similarly, the initial collapsing matter does not need to collapse all the way to the center for the event horizon to form.


How can the finite life-time of the Black hole due to Hawking radiation be made consistant with the infinite amount of time (future) needed for the expansion of the event horizon (in the outer time-frame)?

There's no need to: [edit]that a particular time coordinate doesn't cover the full manifold is a fault of the coordinate chart, not of spacetime[/edit]. From every event, send out an omnidirectional locus of idealized light rays. The event horizon is the boundary of the spacetime region from which none of these light rays escape to infinity. This question has an objective answer--for any given light ray, either it will escape or it won't.

An external observer would need to wait infinitely long to know for sure where the event horizon is exactly, but that's a completely different issue. With Hawking radiation, the black hole shrinks, but it doesn't change the fact that light rays from some events will fail to escape, and thus that an event horizon will exist.

Here's a Penrose diagram of a spherically collapsing star forming a black hole that subsequently evaporates:

Penrose diagram of an evaporating black hole

Light rays run diagonally at ±45° on the diagram. Note that there is a region from which outgoing light rays (running diagonally lower-left to upper-right) don't escape and instead meet the $r = 0$ singularity (the bolded, undashed horizontal line). The horizon itself is the the $r = 2m$ line marked on the diagram and its extension into the star: it should actually go from the (dashed, vertical) $r = 0$ line on the left, rather than extending from the star's collapsing surface. That's because some of the (idealized, noninteracting) light rays from inside the star will also fail to escape to infinity.

Now suppose that on this diagram you draw timelike curves that stubbornly stay away from the horizon, and you insist on using a parameter along them as a time coordinate. Does the fact that you've chosen coordinates that exclude the horizon needs to be made consistent with whether or not the event horizon actually exist? The resolution is simple: if you want to talk about the horizon, stop using coordinates that exclude it.

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So is this correct? From a reference point outside the black hole, matter does indeed accumulate (or clump together) approaching the event horizon, but eventually the event horizon expands to engulf it when more matter is accumulating? – Jonathan May 15 '14 at 15:43
If you insist on defining 'clumping' that way, yes, though I wouldn't. As for the latter question, actually, no: as the horizon expands, it carries the frozen, redshifted images of the stuff that has fallen in the past outward with it. That's one reason I wouldn't call the former case 'clumping'; rather, the Schwarzschild time coordinate (or appropriate generalization for distant stationary observers) is badly behaved at the horizon at so simply shouldn't be used there. – Stan Liou May 15 '14 at 15:52
I don't agree that the external time reference should not be used, as that is what we would "see" if we look at a black hole. It is an interesting point you made that the "image" of all the matter that has fallen in before moves outward when the event horizon expands. Thank you for taking the time to provide a detailed answer too, very thought provoking! – Jonathan May 15 '14 at 15:59
@StanLiou How can the finite life-time of the Black hole due to Hawking radiation be made consistant with the infinite amount of time (future) needed for the expansion of the event horizon (in the outer time-frame)? – Gerald May 15 '14 at 16:25
@Gerald edited to add response in post. – Stan Liou May 15 '14 at 21:39

The logical consequence is, that an event horizon cannot form, since the first particle slows down asymptotically to zero, just before the event horizon forms (Fermat's infinite descent).

The emergence of the event horizon therefore takes infinite time seen from outside. But due to Hawking radiation a black hole exists only a finite time. Hence an event horizon doesn't form.

The frustrating thing about this is, that you need to be at least Stephen Hawking, to not be called a geek.

The current mainstream way to circumvent this paradoxon is to switch to a purely general relativistic geometry of infalling space-time, which doesn't experience the event horizon. That way you avoid the event horizon as a pole, but you get the singularity at the center of the black hole, governed by yet to investigate physical laws of quantum gravity.

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That is an interesting point, and very thought provoking. It will be interesting to see what further discoveries are made about black holes. I wonder still about the matter that was "inside" the black hole when it formed (e.g. I would think this matter is indeed inside the black hole / event horizon). Although, if it is correct that the "image" of the matter expands with the event horizon, even that matter could be on the edge of the event horizon from an external view point. – Jonathan May 15 '14 at 16:03
@Jonathan If you assume e.g. the Schwarzschild solution, the simplest form of a black hole, from an outside observer you need to distinguish three zones: the space-like, the light-like, and the time-like zone. The light-like zone corresponds to the event horizon. If you transform properties of matter between these zones they change their physical properties so much, that the term "the matter is" doesn't make much sense, neither "matter" nor "is". One space dimension changes roles with time. – Gerald May 15 '14 at 16:18
@Jonathan One way of thinking may be, that the information of the matter is stored at the event horizon, some fluid-simulations indicate a fractal structure of the event horizon due to infalling matter; this might be a way to overcome the information paradox. That's neither the Schwarzschild nor the Kerr solution. – Gerald May 15 '14 at 16:48

We need to think about just where the time dilation occurs and by then thinking about the observations from each point of view, the falling object and the external observer, we can come to terms with just what is happening as opposed to what appears to be happening.

Experience of time

We must remember that an object moving at a certain speed will also be travelling through time at a certain slower rate. This does not mean that it moves slower, otherwise it would obviously not be going "at a certain speed".

Where time slows is in the ticking of the physical processes of the object itself. In other words, my clock would tick twice as slow according to you as I flew past you at 87% the speed of light.

The falling object's point of view

If you were the object falling into the black hole, you would accelerate as you approached the event horizon, but you would take longer and longer to react to the approach to the point where you would fall into the black hole seemingly in no time at all. From your perspective, it's approach would take a very short amount of time.

In other words, you would fall incredibly fast into the black hole, but you would barely have registered the final plunge in your brain because there just wasn't enough time.

The stationary observer's point of view

Now, the stationary observer outside the black hole's influence would observe something very different. The light (or rather, information) about your descent would become more and more redshifted, but also take longer and longer to actually reach you.

This means that according to you, I would slow down to a halt at the event horizon and have disappeared.

So what really happened?

  • The falling object fell in very quickly, but hardly realised it happening
  • The stationary observer would think that the object disappeared and never reached the event horizon.
  • Cooper taps on some gravity books and saves the human race.
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How then can the observer see a black hole at all, if from his position, never enough mass falls into it for it to take shape and exist to begin with? – LocalFluff Nov 27 '15 at 8:07
@LocalFluff What does "see a black hole" mean? If by that you mean observe its gravitational effects, I don't see the problem. – Rob Jeffries Nov 27 '15 at 19:30
Your first section is mistaken. It takes a finite amount proper time to fall in, i.e. time experiences by the object, such as you. As a cute coincidence illustrating this, for radial freefall from rest into a Schwarzschild black hole, the time it takes to reach the horizon (or any particular Schwarzschild radial coordinate) happens to exactly match the prediction of Newtonian gravity. – Stan Liou Nov 28 '15 at 19:52
@RobJeffries But then it would remain a neutron star for all outside observers. Black holes would never form for outside observers regardless of how they are observed. One can potentially see a black hole transiting background objects. A non-accreting SMBH doesn't shine at all, while a neutron star with millions of Solar masses very near its surface would be pretty wild. – LocalFluff Dec 2 '15 at 8:06
@LocalFluff A neutron star and a black hole are completely different. No neutron star can exist with a radius anywhere near the Schwarzschild radius. That's why you can see a neutron star. – Rob Jeffries Dec 2 '15 at 8:18

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