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Black Hole Evaporation

  1. Feb 11, 2016 #1
    HawkingRadiation.jpg
    This picture confuses me. It showed here that the red particle at top right get suck in to the black hole. But to a distant observer, everything near the black hole get time dilation and finally freeze at the horizon, right? So, "when" the black hole evaporates?
    Moreover, why the black hole likes the particle with negative energy (in order to evaporate) rather than the one with positive energy?
     
  2. jcsd
  3. Feb 11, 2016 #2

    PeterDonis

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    It's a heuristic picture at best, and is not really a good one to use to draw conclusions from.

    No. Things near the hole appear to a distant observer to have time dilation, but that's an optical illusion caused by the increased time it takes light to get out from the region close to the horizon. The objects themselves, near the hole, don't feel any "time dilation"; they can fall into the hole perfectly normally.

    The short answer is, because only a positive energy particle can fly away and escape; a negative energy particle can't. But, as I said above, this whole picture of positive and negative energy particles is heuristic at best, and if it confuses you, the best thing to do is to throw it away and just think of black hole evaporation as a quantum field process in curved spacetime, without trying to impose a (strained) particle interpretation.
     
  4. Feb 11, 2016 #3
    I don't think that the difference point of view between those observers are being considered as optical illusion. In fact, time dilation happens to inertial observers (in SR) too. Some people said, this is a black hole paradox...

    Ah, I'm not finish with elementary quantum mechanics yet, let alone QFT. So, when I reach a certain level, I'll come back to this...
     
  5. Feb 11, 2016 #4

    phinds

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    @Narasoma, it is based on a misconception fostered unfortunately by Hawking himself who has said that his whole heuristic of virtual particles explaining Hawking Radiation was just the only way he could think of to translate into English something that really can only be understood in math. He did this for popular science explanations and it is not refereed to that way in serious physics discussions.
     
  6. Feb 11, 2016 #5

    phinds

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    No, in fact it does not. It does not happen to anyone, anytime, anywhere. It is only something perceived by a person in a different frame of reference, never to an object itself.
     
  7. Feb 12, 2016 #6
    Could you give me the math behind black hole evaporation?
     
  8. Feb 12, 2016 #7
    I mean, our measurement results of length and elapsed time could be different, and it's depend on our frame. Right?
     
  9. Feb 12, 2016 #8

    Nugatory

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  10. Feb 12, 2016 #9

    Nugatory

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  11. Feb 12, 2016 #10

    A/4

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    I've always felt black hole evaporation is better thought of as a quantum tunneling process than pair creation. In this case, there is no ambiguity as to why it's only the positive energy particle that escapes. The thermodynamic characteristics also follow naturally from this approach.

    Also, it should be noted that a more modern view is that Hawking radiation does not come from the horizon itself, but rather from a region on the order of the Schwarzschild radius from the horizon (the black hole "atmosphere"). See e.g. Steve Giddings' work on the subject: http://arxiv.org/abs/1511.08221
     
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