The Schwarzschild solution of Einstein's field equations clearly shows that a far away observer sees time slow down for an in-falling object. Relative to him, it takes an infinite amount of time for the object to reach the event horizon. But all black holes evaporate by Hawking radiation in a long but finite time. So wouldn't the observer see the BH completely evaporate away before any object reaches its event horizon?

What would an in-falling person experience? In a very short finite proper time, he would expect to reach the event horizon. But since the BH lost mass, the event horizon would have shrunk. It would keep shrinking away from him, until its radius is zero and the BH is gone. Would he see an explosion at this time and then be in free space in the far future?

Considering time dilation of particles collapsing to form a BH, it seems that neither an event horizon or singularity could ever form in a finite time. Astronomical BHs would really be concentric shells of matter that have almost, but not quite, shrunk within their Schwarzschild radius. Does this seem reasonable?

  • $\begingroup$ Related: Why does Stephen Hawking say black holes don't exist? $\endgroup$ Commented Jan 28, 2016 at 10:08
  • $\begingroup$ @JohnRennie: Is the paper "Hawking radiation as seen by an infalling observer", Eric Greenwood and Dejan Stojkovic Published 10 September 2009, Journal of High Energy Physics, Volume 2009 any good? The paper seems to answers the question, but it's not clear to me if the assumptions are justifiable. $\endgroup$
    – CuriousOne
    Commented Jan 28, 2016 at 10:28
  • $\begingroup$ @CuriousOne: the question isn't whether a freely falling observer sees Hawking radiation. It's whether the singularity disappears (due to the evaporation) before the freely falling observer reaches it. t'Hooft says not but I don't see from his answer how he justifies this. $\endgroup$ Commented Jan 28, 2016 at 10:56
  • $\begingroup$ One quick clarification: an observer falling into a (classical - no evaporation) Schwarzschild black hole does not cross a horizon. They see an apparent horizon retreating before them and they only reach this horizon at the moment they hit the singularity. $\endgroup$ Commented Jan 28, 2016 at 10:58
  • $\begingroup$ @JohnRennie: Yes, I understand that the "classical order" of falling observer, apparent horizon and singularity doesn't change, I am simply not convinced by information arguments (either way) whether the black hole evaporates completely or not (in which case the singularity never even forms?) without a reasonable QM model that describes the process quantitatively. I am not particularly interested in the argument between t'Hooft and Ron. I don't need a black hole to cook an astronaut, a regular star will do just fine IMHO... $\endgroup$
    – CuriousOne
    Commented Jan 28, 2016 at 11:19


Browse other questions tagged or ask your own question.