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Hawking radiation and Black hole evaportion

  1. Feb 3, 2013 #1
    When the 'evaporation' of a black hole supposedly occurs, it is always attributed to 'Hawking radiation.' As I understand it, Hawking radiation occurs when a virtual particle pair is 'split' at the exact edge of an event horizon. When this occurs, one of the particle pair escapes to 'normal' spacetime, while the other is pulled down into the hole. When this occurs, it is logical to assume that the black hole actually gains a very small amount of matter (the doomed companion of the free particle). As I understand 'evaporation,' it occurs when a portion of a 'thing' or condition actually yields part of its being or totality. With Hawking radiation, this is not actually happening, since the Black hole/singularity is not giving anything of its 'self' but is actually gaining (the sad particle of the pair). How can we speak of the black hole evaporating when in fact it is 'gaining' matter during Hawking radiation?
     
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  3. Feb 3, 2013 #2

    Bill_K

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    Quoting from Birrell and Davies, "The average wavelength of the emitted quanta is ~M, i.e comparable with the size of the hole. As it is not possible to localize a particle to within one wavelength, it is therefore meaningless to trace the origin of the particles to any particular region near the horizon." ...
    "Virtual particle pairs with wavelength λ separate temporarily to a distance ~ λ. For λ ~ M, strong tidal forces operate to prevent reannihilation. One particle escapes to infinity with positive energy to contribute to the Hawking flux, while its corresponding antiparticle enters the black hole trapped by the deep gravitational potential well on a timelike path of negative energy relative to infinity."
     
  4. Feb 3, 2013 #3
    Thanks. So would it be accurate to assume that the antiparticle is the pair partner that is always caught in the gravity well of the black hole, thereby diminishing the mass of the singularity?
     
  5. Feb 3, 2013 #4

    Bill_K

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    That's correct. Of course the term "anti" is used here in a relative sense. It can just as well be the antiparticle that escapes and the normal particle that falls in.
     
  6. Feb 3, 2013 #5

    tom.stoer

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    It's rather difficult: afaics Hawking's interpretation (particle-antiparticle pairs, virtual particles) cannot be derived from his (brilliant) calculation. There is no particle-antiparticle pair, and there is no one-loop integral, therefore no virtual particles (in the sense of the formalism).

    Nevertheless Hawking's interpretation became standard.
     
  7. Feb 3, 2013 #6

    phinds

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    To say what Bill said in a slightly different way, the only thing that matters is that WHICHEVER of the virtual pair falls into the EH, it has negative energy.

    Also, as I understand it from posts on this forum, there is some confusion around the actual operation of Hawking radiation and at least one post here said that Hawking himself said that the "virtual particle pair" is just a conceptual device to describe something that is a bit more complex than that.

    EDIT: apparently it's another one of those cases where the math doesn't really translate well into English.
     
  8. Feb 3, 2013 #7

    PeterDonis

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    The Usenet Physics FAQ entry takes a similar position:

    http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html

    It says that nobody has ever worked out a local description of how Hawking radiation is generated that corresponds to the "virtual particle pair" description, and describes in a bit more detail how the actual computations are done.
     
  9. Feb 3, 2013 #8

    tom.stoer

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    I have to find a paper which does not use virtual particles but at least tunneling.
     
  10. Feb 4, 2013 #9

    atyy

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    So you wouldn't object if one just said "real" particle creation instead of "virtual" particle?
     
  11. Feb 4, 2013 #10

    tom.stoer

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    Yes, there are real particles, but their properties depend on the spacetime background; in some sense Hawking radiation is like distorted wave approximation; the interpretation of a mode depends where you look at it (Hawking radiation is studied at null-infinity and cannot be 'localized' at the horizon); the interpretation uses nothing else but free particles described by distorted waves, no interaction, no higher order, ... is used.

    I recommend to study Hawking's original paper:

    http://www.itp.uni-hannover.de/~giulini/papers/BlackHoleSeminar/Hawking_CMP_1975.pdf

    The idea is very simpe, the math is not very complicated. The main problem is that math and its interpretation start to deviate already in this paper! That's why I don't like Hawking's popular books; he doesn't care much for exactness when he starts to spin a yarn.
     
  12. Feb 4, 2013 #11

    Bill_K

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    Tom, I hope you weren't expecting a Feynman diagram. :smile: The point is that particles are present in the vacuum state, and some of them head outward to future null infinity, while others head inward and cross the black hole's horizon. There's no suggestion that they were created pairwise at a definite location.
     
  13. Feb 4, 2013 #12
    Descriptions I have seen:

    posted above:
    it is not possible to localize a particle to within one wavelength,

    virtual particle separation,

    tunneling


    and another


    from Leonard Susskind: pieces of long strings break off from the stretched horizon due to quantum fluctuations...and appear as real particles...... from his book THE BLACK HOLE WAR.
     
    Last edited: Feb 4, 2013
  14. Feb 4, 2013 #13

    tom.stoer

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    Of course I don't expect Feynman diagrams; all what I am saying is that Hawking's interpretation of his own calculation is missleading.
     
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