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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: http://en.wikipedia.org/wiki/Black_hole#General_relativity, 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: http://astronomy.stackexchange.com/a/1009/1386

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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
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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 4 down vote accepted
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Yes, you are absolutely right, from OUR VIEWPOINT it does.

From Kip Thorne's book "Black Holes and Time Warps: Einstein's Outrageous Legacy."

“Like a rock dropped from a rooftop, the star’s surface falls downward (shrinks inward) slowly at first, then more and more rapidly. Had Newton’s laws of gravity been correct, this acceleration of the implosion would continue inexorably until the star, lacking any internal pressure, is crushed to a point at high speed. Not so according to Oppenheimer and Snyder’s relativistic formulas. Instead, as the star nears its critical circumference, its shrinkage slows to a crawl. The smaller the star gets, the more slowly it implodes, until it becomes frozen precisely at the critical circumference. No matter how long a time one waits, if one is at rest outside the star (that is, at rest in the static external reference frame) one will never be able to see the star implode through the critical circumference. That is the unequivocal message of Oppenheimer and Snyder’s formulas.”

“Is this freezing of the implosion caused by some unexpected, general relativistic force inside the star? No, not at all, Oppenheimer and Snyder realized. Rather, it is caused by gravitational time dilation (the slowing of the flow of time) near the critical circumference. Time on the imploding star’s surface, as seen by static external observers, must flow more and more slowly, when the star approaches the critical circumference, and correspondingly everything occurring on or inside the star including its implosion must appear to go into slow motion and then gradually freeze.”

“As peculiar as this might seem, even more peculiar was another prediction made by Oppenheimer and Snyder’s formulas: Although, as seen by static external observers, the implosion freezes at the critical circumference, it does not freeze at all as viewed by observers riding inward on the star’s surface. If the star weighs a few solar masses and begins about the size of the sun, then as observed from its own surface, it implodes to the critical circumference in about an hour’s time, and then keeps right on imploding past criticality and on in to smaller circumferences.”

“By looking at Oppenheimer and Snyder’s formulas from the viewpoint of an observer on the star’s surface, one can deduce the details of the implosion, even after the star sinks within its critical circumference; that is one can discover that the star gets crunched to infinite density and zero volume, and one can deduce the details of the spacetime curvature at the crunch.” P217-218

OK, so from our perspective all the matter will be clustered around the critical circumference and no further. That's fine, this shell in theory can exert all the forces required on the external universe such as gravitational attraction, magnetic field etc. The point like singularity which is in the indefinite future of the black hole, (from our point of view) indeed in the indefinite future of the universe itself could not exert such forces on this universe. This singularity is only "reached" as an observer rides in past the critical circumference and, through the process of time dilation, reaches the end of the universe.

This is obviously an area of active research and thinking. Some of the greatest minds on the planet are approaching this issue in different ways but so far have not reached a consensus but intriguingly a consensus appears to be beginning to emerge.

http://www.sciencealert.com/stephen-hawking-explains-how-our-existence-can-escape-a-black-hole

Stephen Hawking said at a conference in August 2015 that he believes that "information is stored not in the interior of the black hole as one might expect, but on its boundary, the event horizon." His comment refers to the resolution of the "information paradox," a long-running physics debate in which Hawking eventually concedes that the material that falls into a black hole isn't destroyed, but rather becomes part of the black hole.

Read more at: http://phys.org/news/2015-06-surface-black-hole-firewalland-nature.html#jCp

In the mid-90s, American and Dutch physicists Leonard Susskind and Gerard 't Hooft also addressed the information paradox by proposing that when something gets sucked into a black hole, its information leaves behind a kind of two-dimensional holographic imprint on the event horizon, which is a sort of ‘bubble’ that contains a black hole through which everything must pass.

What occurs at the event horizon of a black hole is very hard to understand. What is clear, and what proceeds from General Relativity, is that from the viewpoint of an external observer in this universe, any infalling matter cannot proceed past the critical circumference. Most scientists then change the viewpoint to explain how, from the viewpoint of an infalling observer, they will proceed in a very short period of time to meet the singularity at the centre of the black hole. This has given rise to the notion that there is a singularity at the centre of every black hole.

However this is is an illusion, as the time it will take to reach the singularity is essentially infinite to us in the external universe.

The fact that the matter cannot proceed past the critical circumference is perhaps not an “illusion” but very real. The matter must from OUR VIEWPOINT become a “shell” surrounding the critical circumference. It will never fall through the circumference while we remain in this universe. So to talk of a singularity inside a black hole is incorrect. It has not happened yet.

The path through the event horizon does lead to a singularity in each case, but it is indefinitely far in the future in all cases. If we are in this universe, no singularity has yet been formed. If it has not been formed yet, where is the mass?  The mass is exerting pull on this universe, correct?  Then it must be IN this universe.  From our point of view it must be just this side of the event horizon.

ASTONISHINGLY IT MAY BE POSSIBLE TO PROVE THIS. The recent announcement of gravitational waves detected on the merger of 2 black holes was accompanied by an unverified but potentially matching gamma ray burst from the same area of the sky. This is inexplicable from the conventional viewpoint which holds that all the matter would be compressed into a singularity and would be incapable of coming out again.

If 2 black holes merge and emit gamma rays… the above is certainly an explanation which is also consistent with General Relativity. The mass never quite made it through the event horizon (from our viewpoint) and was perturbed by the huge violence of the merger, some escaping. It may be a deep gravitational well, but a very powerful gamma ray should just be able to escape given the right kick (attraction by an even larger black hole approaching).

Further more refined observations of similar events, which are likely to be reasonably frequent, may provide more evidence. There is not likely to be any other credible explanation.

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Thank you for your answer, I would like to see if this generates further discussion! – Jonathan Mar 19 at 16:07
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One more comment on your original question. The black hole would start like a tiny vapour bubble in the middle of the imploding star which had reached a sufficiently strong gravitational "pressure". It would then expand as surrounding matter and energy fell in and reached its critical circumference, therefore I do not think any matter from the viewpoint of an external observer would be "inside" the critical circumference. – ctrebor Mar 19 at 16:40
    
FYI, I am looking for proof / references to award the bounty. – Jonathan Mar 23 at 15:32
    
Proof/references for what? There was a considerable reference to Oppenheimer & Snyder. Do you want more? – ctrebor Mar 24 at 16:02
    
Actually, it appears your answer is now good as far as references go, and I may well end up awarding you the bounty. I am also looking for further proof (e.g. observations or other types of proof) that matter does or does not collect near the event horizon of a black hole. I realize such proof may not exist yet, so I will in that case reward whoever backs up their statement the best with references, etc..., and so far you have done a good job of including references. – Jonathan Mar 24 at 19:00

What you're describing is basically the "collapsed star" (Eng) or "frozen star" (Rus) interpretation of black holes that was common prior to the late mid-1960s. 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.)

The "comoving viewpoint" was recognized by Oppenheimer and Snyder in 1939, but it was not until the 1960s, with the work of Zel'dovich, Novikov, et al., that it was generally recognized as truly significant in the community. In 1965, Penrose introduced conformal diagrams based on the Eddington-Finkelstein coordinates (1924/1958) that showed quite clearly that the stellar collapse is not slowed, but instead continues to a singularity. For an overview of the history of this change of viewpoint, cf. Kip Thorne, et al., The Memberane Paradigm (1986). These topics are commonly covered in many relativity textbooks.

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
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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
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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
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@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

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

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
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@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
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I wonder if we could actually "peek under the skirts" of a black hole if there'd be any 'there' at all. We can't of course, and anyone that asserts a singularity exists inside a black hole is simply saying that the mathematical model they're using says there is one. If all of the mass/energy that makes up a black hole was compressed into a two dimensional surface at the event horizon, is there any way to observationally tell the difference? Swiss cheese has holes in it, but no one asserts that the holes are Swiss cheese. – Howard Miller Mar 22 at 3:19
    
@Gerald FYI, I am looking for proof / references to award the bounty. – Jonathan Mar 23 at 15:33

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