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Consider this a followup question of this one

In the classical schwarszchild solution with an eternal black hole, the user falls through the event horizon in finite local time, but this event does take place for distant observers in the infinite future. As Leonard Susskind explains, measurements of objects near the horizon are subject to large uncertainties, since all the radiation used to see them is largely red-shifted and received in increasingly large intervals in the future. This is why in some still vaguely defined sense, the effective event horizon grows, even if no matter has actually yet crossed the original event horizon (the stable event horizon after the black hole collapse period)

But when the event horizon radiates, the event horizon must recede faster than any infalling observer. Any infalling matter will probably be still stir-fried by the hawking radiation over the eons the evaporation takes (which in the observer frame is intensely blue shifted). The only matter that ever gets to be inside the formal event horizon (not the effective horizon) before the final evaporation is the matter that was already in the original collapsing star, around which the event horizon formed

Is there still any room to say otherwise? can still be argued that matter will fall in the event horizon in finite time for distant observers (faster than the evaporation will shrink the event horizon)?

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I suspect this is duplicate of a set of other questions. –  Anixx Aug 9 '12 at 19:26
    
You should also note that in any collapse the horizon initially forms in one point and then grows, so there is no "initial matter" under horizon. If such matter under horizon existed, this would amount to information loss which is impossible. –  Anixx Aug 9 '12 at 19:27
    
@Anixx, the situation for the initial matter is different, because far away observers will agree that it went through the event horizon in finite time for them (during the collapse). –  lurscher Aug 9 '12 at 19:47
    
"Room to say otherwise"? Everyone say otherwise! Matter will fall in, because your idea that the horizon is impassable is confusing the horizon pctures. –  Ron Maimon Aug 10 '12 at 19:31
    
@RonMaimon, help me to understand where my confusion lies please –  lurscher Aug 10 '12 at 19:35
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2 Answers

It's key to remember that the observer outside of the black hole will never see you fall into the black hole. However, this observer can jump into the black hole himself, and see what happened to you. Outside of the black hole, since he is unable to receive light signals from you, he may say that you have stopped at the event horizon. But since he can go in and see you (assuming you haven't been crushed by the singularity yet), it is a matter of philosophy of whether the infalling observer actually fell in or if he 'froze' at the event horizon.

Analogy: If you are accelerating in a rocket, you don't have to believe that time on earth slowed to a stop corresponding to your Rindler horizon. If you wish, you can conclude no such thing. However, it is indisputable that no signals from earth past a certain time will reach you as long as you continue accelerating. But you can stop accelerating and then see what you missed. Similarly, you can 'stop accelerating away from the SC horizon' - i.e. fall into it - and then see all the history of infaller's you missed.

You should read the entry on the Usenet Physics FAQ:

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

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I think this answer makes a popular error of dealing with Swartzshield eternal black hole instead of evaporating black hole about which is the question. –  Anixx Aug 10 '12 at 14:32
    
@Anixx: The evaporating black hole is a red herring--- the answer is the same. Please stop confusing this issue--- it is perfectly possible to fall into a quantum evaporating black hole, the evaporating radiation vanishes at the boundary. –  Ron Maimon Aug 10 '12 at 19:30
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The black hole does not "evaporate before you reach it", althuogh this is a consistent (but misleading) classical picture. The problem is that any horizon that's been around for a long time has both a white hole and a black hole continuation, and the two look different in terms of horizon properties.

For a white hole horizon, you are classically smooshed on the edge until the explosion, while for a black hole horizon you fall through. In time reverse, for a black hole, you fall through, and stuff coming out is redshifted.

The back-reaction on a black hole is entirely negligible, and only identifiable globally. It's a property of the whole hole, not of one patch of the horizon. The locally Rindler form of a normal black hole shows that you fall through, and any tiny perturbation due to the evaporation is irrelevant.

This is despite the bogus intuition people here seem to have that objects freeze on the horizon and the black hole evaporates under them before they have time to fall in. This intuition is seductive, because it is partly right--- this is the white-hole picture (which is complementary to the black hole picture). But it is false in the sense that the infalling observer is not destroyed at the horizon, but feels nothing special.

The reason one can be sure is because the future (and past) continuations are available when the black hole has been around, and qualitatively insensitive to small perturbations. What you are doing is making a small perturbation and using this as an excuse to switch to a white hole picture, which is not an excuse at all.

The idea that black holes cannot form is the analog of the argument that white holes cannot form by collapse. This doesn't matter, as when the black hole has been sitting there for eons, you can't tell whether it's a black hole or white hole. These things are only made clear once you accept Susskind complementarity.

Mooshing on horizon

The mooshing on the horizon picture is only valid for a proper time that terminates for an infalling observer. The observer is redshifted to oblivion, and merges with the BH horizon (in an exterior picture) after a finite proper time.

But in the observer's local frame, there is nothing singular for an extremely long period of affine parameter as the path becomes null. The argument that a black hole has an interior requires the assumption that when the final explosion is far away, the Hawking radiation behaves semiclassically, it becomes invisible for the infalling observer, so this observer falls through. This is a little bit of a religious point of view in the classical world, because there is no evidence for the interior beyond what you can see in the exterior, but it is justified by the consistency of the quantum picture it gives.

Without knowing that the local equivalence principle holds at the horizon, the argument for Hawing evaporation becomes suspicious. You can use the t-independence of the BH to make what is called a "Boulware vacuum" which is nonradiating, because it conserves the t-notion of energy. This Boulware vacuum was believed to describe QFT around black holes for a long time. It corresponds to the spacetime around a Schwarzschild black hole which is surrounded by a perfect ideal mirror for everything at (R=2M). This thing is thermodynamically ridiculous in the usual picture, the mirror absorbs thermal energy and doesn't heat up to equilibrium. But this Boulware idea is resurrected every once in a while, in t'Hooft's idea that the black hole has double the correct temperature, for example, because the interior and exterior are identified by a gluing map.

The evidence for the falling-through picture, which is Susskind's, comes most persuasively from the quantum theory. It is this picture that produces AdS/CFT. Without it, it is impossible to understand how black holes become so regular and ordinarily quantum in the extremal limit, where the horizon is still present, but the Hawking radiation goes away.

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Why do you call Hawking radiation "small perturbation"? Any observer outside the BH horizon experiences Hawking radiation. As the observer approaches the BH the BH radius decreases (he can verify that by communicating with a distant observer). And the smaller radius the stronger Hawking radiation and the shorter its wavelength. –  Anixx Aug 10 '12 at 19:58
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Note also that Hawking showed that time reversal of a black hole gives the same solution. In his own words, "Black holes behave in a completely random and time-symmetric way and are indistinguishable, for an external observer, from white holes." But as all observers are external initially this means that for ANY observer BH is indistinguishable from a WH. –  Anixx Aug 10 '12 at 20:05
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Ron, i don't understand your argument, if the black hole evaporates, there will be no black hole at a finite time $t_E$. What part of your argument is about how the observer will traverse the EH in a finite time $t < t_E$? The argument of Susskind that you are both frozen at the horizon and falling inside is ad hoc, and sounds a bit like a religious statement –  lurscher Aug 10 '12 at 21:10
    
@lurscher: The problem is the notion of "finite time". "finite time" for a horizon is not "finite time" for the black hole solution. The observer falls into a black hole in finite proper time, and gets mooshed on a white hole, meeting the endpoint at the same finite proper time. The argument is not religious--- it's based on the principle made clear in the 1960s that if a solution is locally Minkowski, it always obeys the equivalence principle. It also predicted and suggested AdS/CFT, so it has firm evidence. It allowed a consistent reconstruction of blackhole physics and strings. –  Ron Maimon Aug 11 '12 at 3:41
    
"The observer falls into a black hole in finite proper time" - but the BH will evaporate completely even in shorter proper time which he can verify by contacting the external observer. I wonder why are you so stubbornly reject to account for the BH evaporation. –  Anixx Aug 11 '12 at 3:57
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