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Do black holes vanish ?

  1. Oct 13, 2014 #1

    NTW

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    As I understand it, the Hawking radiation is produced when a pair particle-antiparticle are produced at a place close to the event horizon of a black hole, so that one of them, energy-negative for a distant observer, is swallowed by the hole, and the other appears as 'black hole radiation'. I believe that, at the moment of the 'split', the probability of a particle or antiparticle falling into the black hole is the same as the probability of escaping from it. If it is so, then why Hawking says that black holes vanish with time? They should remain stable...

     
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  3. Oct 13, 2014 #2

    phinds

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    Somehow, and I don't "get" this myself but have read it so many times it's locked in my brain as dogma, the particle that falls in is always negative in the sense that it removes mass from the BH, so over an unimaginably long time, the BH evaporates. The evaporation rate is proportional to the size, so as it gets smaller, it evaporate more quickly and really tiny BH's evaporate very quickly. "Evaporate" isn't the right word to use here, but I'm using it in the sense of "loses mass"
     
  4. Oct 13, 2014 #3

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    If I understand this correctly, from the two particles resulting from the'split', only the antiparticle falls into the black hole. And why should italways be the antiparticle? It seems reasonable to believe that theprobability of falling is the same for a particle as well as for an antiparticle... But Hawking says that it's always the antiparticle that falls... How can that 'preference', that in turn explains the 'evaporation', be understood...? It seems to me that if particles and antiparticles were to fall there with the same frequency, the net effect would be zero... IMHO...

     
  5. Oct 13, 2014 #4

    Chalnoth

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    As I understand it, this is more of a heuristic than anything, that the real derivation is somewhat different (though I don't know the details).

    But in this description, it's not that an anti-particle falls into the black hole. It's that the virtual particle pair has zero total energy. So in order for the outgoing particle to keep going and have positive energy, the particle that falls into the black hole must have negative energy. Thus the infalling particle subtracts from the mass of the black hole.

    The outgoing particles will be essentially equal numbers matter and anti-matter.
     
  6. Oct 13, 2014 #5

    NTW

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    Thus, both particles are somehow 'categorized' by the event of one of them falling into the hole. Because of its fate, the falling particle acquires a negative mass, and the other one, that flees, gets a positive mass...

    Is it so...?
     
  7. Oct 13, 2014 #6

    Chalnoth

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    Like I said, I think that this is more analogy than reality, that the true derivation is somewhat different. But suffice it to say, in this description of Hawking radiation, in order for one of a pair of virtual particles to fall in, it must have negative mass.
     
  8. Oct 13, 2014 #7

    George Jones

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    I agree with Chalnoth

     
  9. Oct 13, 2014 #8

    NTW

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    Well, if things must be that way, there's nothing to discuss... I understand that the problem is too complex to be explained in simple terms, but the trouble with those 'simple terms' is that they sometimes sound illogical...
     
  10. Oct 13, 2014 #9

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    Thanks for the explanations

    I understand that this is a complicated affair, but I can't accept that negative-mass particles may be more prone to fall into the black hole than positive-mass ones. Both particles appear 'out of the blue' because of the fluctuations of quantum vacuum, and the probability of one of the pair falling into the hole should not be higher than the probability of the other one... Mass 'sign' notwithstanding...

    But, having said that, I understand too, that the problem may be clearly seen only within a heavy mathematical context.
     
  11. Oct 13, 2014 #10

    Chalnoth

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    Right. The way the world works is very often wildly counter-intuitive.

    And it's not that the negative-energy particle is more likely to fall in. It's that the negative-energy particle can't exist outside the horizon for any significant amount of time, so the only thing that can happen is the negative energy particle falling past the horizon.
     
  12. Oct 13, 2014 #11

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    Hard to swallow, that... If the positive-energy particle is more long-lived, the likeness of it falling into the black hole must be higher than that of the negative-energy particle, which could vanish even before the fall...

    Anyway, and not wanting to be confronted with heavy math, I don't wish to pursue the matter any further...
     
  13. Oct 13, 2014 #12

    Chalnoth

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    If the negative-energy particle vanishes before the fall, so does the positive-energy particle. This is the way such virtual particle pairs work: normally, they can't exist for a time longer than about h/E (h is Planck's constant). It is only once the one member of the pair is captured by a black hole that the other member can escape. And when you look at the physics of how that capture happens, the math you get for the infalling particle makes its energy come out negative.

    The trick that was described in the link above by Steve Carlip, is that the curvature of space-time makes it so that time is switched with one dimension of space: the direction towards the center of the black hole acts like time, and the dimension we see as time acts like a dimension of space. This is relevant because in the language of General Relativity, energy is associated with time while momentum is associated with space. So while one dimension of space is swapped with time, one direction of momentum is swapped with energy. We see negative energy outside the black hole, but an observer inside would see negative momentum and positive energy. Momentum, being directional, has no problem being positive or negative.
     
  14. Oct 13, 2014 #13

    Chronos

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    It does not matter which particle 'falls' into the BH, all that matters is the ones that escape have positive energy. That energy must come from somewhere, hence, the BH loses mass.
     
  15. Oct 14, 2014 #14
    Note that the text from Carlip starts with that "[t]here are a number of ways of describing the mechanism responsible for Hawking radiation." If you insist at looking at the black hole (BH) processes locally, you will need to bother with their details.

    But the BH vanishing is thermodynamic as every other process. And you can chose to look at BHs at infinity. (This seems to be a usual method of theorists by the way, complementing the specifics with generics by switching scale.) So, given Hawking radiation, what happens? It will look like thermal radiation at infinity*. Hence the BH must loose energy and evaporate over time. (Loosing energy to evaporation will mean loosing mass, spin and charge eventually.*)

    I'm a simple (lay)man, so that has been enough explanation for me through the years.

    *Mind that "look like" is not arguable but "is" would be, because a "coded" non-thermal radiation can solve the BH information paradox of what happens in more detail with mass, spin and charge as the BH evaporates. [ http://en.wikipedia.org/wiki/Black_hole_information_paradox ] But again, unnecessary detail.
     
    Last edited: Oct 14, 2014
  16. Oct 14, 2014 #15
    So is this a likely case of an entangled quantum system or just one of the differences between how particles and antiparticles behave as LHC is beginning to see? or something else altogether?
     
  17. Oct 14, 2014 #16

    Chronos

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    'Entanglement' is an excuse. I fail to see where it has 'succeeded' in any meaningful way. You have a choice here, mathematics or observational evidence. I favor option 2.
     
  18. Oct 15, 2014 #17
    You may enjoy following Seth Lloyd, physicist at MITs recent experiments. Even skeptic Prem Kumar of Northwestern has set aside some level of skepticism since viewing Lloyd's math. Here's a teaser http://arxiv.org/abs/0803.2022
     
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