Given gravitational time dilation, under what conditions will a test particle cross an event horizon before the black hole evaporates? Assume zero background radiation.


closed as unclear what you're asking by user10851, Brandon Enright, Abhimanyu Pallavi Sudhir, akhmeteli, tpg2114 Nov 26 '13 at 4:56

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  • $\begingroup$ Possibly relevant (and possibly wrong). $\endgroup$ – Keith Thompson Aug 18 '12 at 9:33
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    $\begingroup$ This seems suspiciously like homework....is it? $\endgroup$ – Pranav Hosangadi Aug 18 '12 at 9:54
  • $\begingroup$ It's not homework. $\endgroup$ – madarab Aug 18 '12 at 11:36
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    $\begingroup$ The title isn't specific - it should tell what the question is. I'm also not sure what the question is asking. Start off by clarifying, where does the particle start out, what velocity does it start out with? Why are you then conflicted about its behavior? What scenarios are you entertaining? With more elaboration you are likely to get a good answer but for now it's too short and lacking clarity. $\endgroup$ – Alan Rominger Aug 18 '12 at 16:47
  • $\begingroup$ Does this depend on the observer? $\endgroup$ – Peter Shor Aug 18 '12 at 22:29

There is no way to answer this question, as it is mixing up a classical picture with quantum effects.

If you have a classical black hole and a particle falling into it, it will freeze on the horizon, considered from the exterior viewpoint, and when the black hole is evaporating, you might conclude that it evaporates before the particle has a chance to cross. This is a classically allowed picture, it is consistent. It is found by considering the black hole horizon to be a white hole horizon. The white hole horizon recedes from any infalling matter at the speed of light, so you can't fall in, but then all the matter that ever fell in is just compressed to the final endpoint of the white hole and it hits this endpoint at finite time. In this picture, anything coming out of the hole comes from a back region which is sensible.

But this black hole picture is not compatible with the formation point of the black hole. This formation point is just the time reverse of the annihilation point, and near this point, you must think of the black hole as a proper black hole. The two extensions are classically incoherent at different places, one near the end and one near the beginning, which is why modern people accept Hawking's identification of white hole and black hole as the same object. Susskind clarified this with black hole complementarity, which says that both the black hole and white hole extension are correct in different basis.

This shows that while it is consistent to say the particle can't cross before the black hole evaporates, and it is equally consistent to say a particle can't leave in finite time from the black hole's formation, both consistent pictures are foolish hobgoblins. The right picture allows either continuation in the domain where the black hole is static, so it's been around for a while, and will continue to be around.

  • $\begingroup$ I'm guessing he is asking about an observer falling in with the test particle just as the last of the black hole is evaporating. $\endgroup$ – jcohen79 Sep 18 '12 at 21:13

At the event horizon, a test particle will appear to be moving at the speed of light, however due to time dilation to an external observer that particle will appear to have its time slowed to the point it's time is not advancing at all. At this point it will neither appear to travel faster than the speed of light, nor continue to advance in time.

Accordingly the external observer will observe the black-hole evaporate before the particle appears to cross.


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