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Let's assume an observer looking at a distant black hole that is created by collapsing star.

In observer frame of reference time near black hole horizon asymptotically slows down and he never see matter crossing event horizon. So black hole is visible as some kind of frozen object - falling matter almost stops near horizon and light emitted or reflected by it becomes more and more redshifted as it approaches horizon.

On the other hand black hole emits thermal radiation that causes it's evaporation. After some time it will finally disappear.

As these two points seems extremely incompatible with each other, my question is how transition from such frozen object to state of no black hole (complete evaporation) looks for distant observer?

What I'm curious about is where this Hawking radiation comes from. I suppose it will be surface of the horizon, but in fact from observer frame of reference there is nothing at this surface yet - matter didn't fall on it yet, it's just approaching it asymptotically.

What is strange from classical point of wiew is how Hawking radiation can be visible being emitted from place where there is nothing? Or what's more interesting how the matter observer see falling on BH will dissappear as evaporation is in progress?

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Related: physics.stackexchange.com/q/11576/2451 –  Qmechanic Jan 11 '13 at 17:15
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2 Answers

The fact that a black hole horizon appears frozen to a distant observer is a consequence of the fact that the black hole horizon is a null surface--it represents the path that light follows as it tries to escape the black hole's gravity and just exactly fails.

If a black hole is evaporating, the horizon is no longer a null surface, but rather becomes a timelike surface--analogous to that of a sphere of massive observers travelling forward in time. There is no ban on signals leaving this timelike surface. In particular, that null ray that tried ot leave the black hole and just exactly failed would find itself outside of the horizon and able to escape to infinity after the shrinking event. So, the distant observer will see the black hole evaporate.

I should also say that, since there is no singularity at late times, a black hole that completely evaporates has no event horizon at all by the strict mathematical definition of an event horizon, which is dependent on the existence of a late time black hole singularity.

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Thank you for the answers. I think I didn't formulated my question precisely enough, so I expanded it a little bit (added last 2 paragraphs) to better show my point. –  krzysztoft Jan 12 '13 at 16:33
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Considering that the field of quantum gravity is a huge questionmark, there might be some effects we don't know about. However, let us assume that there are there are no surprises, and a black hole gradually evaporates until there is nothing left of it. If it does so in finite proper time, then the cause of the "freezing"-phenomenon close to the horizon, namely the strong curvature of spacetime, also diminishes in finite time. Therefore a distant observer will notice the evaporation of the black hole.

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to : " the cause of the "freezing"-phenomenon close to the horizon, namely the strong curvature of spacetime" : For very large black holes the curvature near the horizon is not strong. A free falling observer sees nothing special at the horizon. –  jjcale Jan 12 '13 at 18:37
I am aware of that, maybe I should have phrased it differently. –  Frederic Brünner Jan 12 '13 at 20:22
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