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Hawking Radiation and the Decay of Black Holes

  1. May 8, 2014 #1
    Hawking Radiation and the "Decay" of Black Holes

    I have been doing a lot of reading and thinking about certain quantum mechanics so that I can try and wrap my head around how it all works. However I have come upon something that I cannot find a good explanation for.

    I was reading about Hawking Radiation and the associated idea that Black Holes "decay or evaporate"
    The explanations I have read all seem to agree on the mechanics of Hawking radiation. However I have not been able to find a single source that answers a fundamental question for me.

    My understanding of quantum mechanics is that pairs of particles(one particle matter and the other anti-matter) are constantly popping up all over our universe. Now a I understand that virtually all of these particle pairs touch each other and are utterly converted to energy in the destruction of the particles.

    Now according to my understanding of the theory which predicts Hawking Radiation, when the pair of particles come into being, the matter particle is always outside the event horizon and the anti-matter twin particle is created inside the event horizon. Because the anti-matter particle is inside the event horizon it cannot touch the matter particle and destroy it. Therefore there is an extra particle which appears to be given off by the black hole and the black hole swallows a particle of anti-matter which destroys a particle of matter in the black hole. With this happening the black hole is one particle smaller and is also seeming to emit a particle and when you combine these two it appears that the black hole is evaporating and eventually will cease to exist.
    I have a pretty straightforward question that I hope either someone does know the answer to or that there is no correct answer to this question.
    Here is my question: Why does the anti-matter particle always have to be created inside the event horizon? Why can't the matter particle be inside the black hole and the anti-matter be outside? If that were the case then the black hole would be one particle bigger and the outside universe would be one particle smaller when the anti-particle collides with matter outside the black hole. If I am right and either one can be inside the black hole then it stands to reason that over the very long term of cosmic time that the numbers of each kind of particles inside the black hole would be approximately equal and the black hole would be in equilibrium over the long haul and would not "evaporate".

    Thank you all your time, effort and knowledge,
  2. jcsd
  3. May 8, 2014 #2


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    The creation of a pair could be either way as you suggest. However, the one outside is still escaping from the black hole. If it is anti-matter it will encounter a matter particle outside and end up as a pair of photons. It is still a net decrease in the black hole mass.
  4. May 8, 2014 #3
    You're confusing Antimatter with negative energy. The particles that fall in the black hole don't have to be antimatter. They are just as likely to be antimatter as to be matter. In fact the distinction between matter and antimatter is quite arbitrary. We chose to call electrons, protons, and neutrons matter and positrons, anti-protons, and anti-neutrons antimatter. but there is nothing fundamental about that choice. What's happening in the Hawking radiation is that the particles that fall in the black hole are virtual particles. The energy of real particles must satisfy two conditions.
    positive energy E > 0.
    and the dispersion relation E2 = (mc2)2 + (cp)2,
    where E is energy, m is mass, p is momentum, and c is the speed of light.
    In physicist's lingo these conditions are called shell conditions and real particles are said to be "on shell".
    The particles that fall in the black hole are virtual particles. They are not required to be on shell. In fact the ones that fall in the black hole must have negative energy because the ones that leave the hole have positive energy (they are real particles) and by energy conservation the total energy of the pair is zero since they appeared out of nothing and nothing has zero energy. Those negative energy particles fall in the whole therefore reducing its overall mass.

    Another misconception in your OP is the idea that there are particles inside of the black hole waiting to be annihilated. In fact there isn't anything inside of the hole except for the singularity at the very center. As particles fall in the hole they reach the singularity after a finite amount of time and than cease to exist. In fact there is no difference between a black hole created out of regular matter and a black hole created out of anti-matter. That's one reason it's known for sure that baryon number is not a conserved quantity - at least not in a black hole.
  5. May 8, 2014 #4


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    My understanding is that Hawking has said that the concept of "virtual particles" as an explanation for Hawking Radiation is an inexact analogy because the math does not lend itself to English. There are not really any particle-pairs involved, at least not in the way that you normally see described.

    I regret that I cannot give a citation for this but I believe it has been pointed out on this forum before.
  6. May 8, 2014 #5
    Yes, that is true. The virtual particles are a middle level explanation. Better than a hand wave but not as good as the real math.
  7. May 10, 2014 #6
    Is this creation in any way connected to quantum entanglement which Einstein dealt as spooking action at a distance in EPR paper.
  8. May 10, 2014 #7


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    There is a connection in that the two photons are correlated. (My knowledge about "action at a distance" is weak).
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