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Hawking Radiation

  1. Aug 27, 2007 #1
    Something I've been wondering for a while, but haven't been convinced by the answers I've received.

    If an astronaut falls into a black hole, from the perspective of us outside he never actually reaches the horizon. From his perspective, he falls straight through and doesn't notice any particular problems at the horizon. However, if the black hole is emitting Hawking radiation, then it only has a finite lifetime. Even if it takes 10^100 years (according to a distant observer), the black hole will evaporate, by which time the astronaut still won't have crossed the event horizon. So, my question is, does the astronaut ever, even from his own point of view, cross the horizon, or does he find himself evaporating before he gets there?

    Thinking about the mechanism of Hawking doesn't seem to help. Do the negative energy particles come out of the black hole towards the incoming astronaut to annihilate him? Do they chase him over the horizon and catch up with him while he's inside? Presumably he perceives the radiation to be being formed at a much greater rate than we do. Is it enough to destroy him before he reaches the horizon?

    Thanks to anyone who can answer these.
  2. jcsd
  3. Aug 27, 2007 #2


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    The free falling observer indeed would cross the horizon in a finite time without noticing any change. There is actually a conceptual challenge when putting both descriptions together, which is known as "black hole complementarity". You may be interested in this popular article by Leonhard Susskind:

    http://staff.science.uva.nl/~jdeboer/gr/susskind.pdf [Broken]
    Last edited by a moderator: May 3, 2017
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