Hawking Radiation - how could it shrink a Black Hole?

Discussion in 'Physics & Math' started by MyBrainHurts, Sep 29, 2008.

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  1. MyBrainHurts Registered Member

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    I've never understood how Hawking Radiation would work. I don't see how it could cause a Black Hole to shrink.

    I mean, don't matter and anti-matter behave in a similar way under gravity? (The answer is "yes")

    That being the case, when a particle pair spontaneously appears near a Black Hole's event horizon, it's gonna be a 50-50 chance as to which one of the two particles falls in.

    If the matter particle fell in, the black hole would gain mass. On the other hand, if the anti-matter particle fell in, it would cancel / annihilate some of the matter inside the Black Hole, so the black hole would lose mass.

    If the two processes have an equal chance of happening, there should be no overall effect upon the mass of a Black Hole, because the two would cancel each other out. i.e. over a period of time, on average, the amount of matter and anti-matter falling into the Black Hole would be the same.

    Furthermore, after one of a virtual particle pair has fallen into the black hole, what exactly makes the remaining particle wander off into the universe? Surely, being so close to the event horizon, it's likely to fall in too after a while (because it would need an escape velocity close to the speed of light to escape).

    And then there's the question of annihilation. The reduction in mass of the Black Hole appears to rely upon anti-matter particles coming into contact with the singularity, therefore annihilating the anti-matter particle and an equivalent amount of matter. However, as the particles approach the Black Hole's singularity, gravity becomes stronger, so time slows down (or would appear to slow down, if an outside observer could see in). In a massive Black Hole, the falling particles would pretty much grind to a halt. Doesn't this suggest a shell of matter and anti-matter particles pretty much frozen in time, and therefore hanging around for a very long time before colliding?

    To an outside observer, the Hawking Radiation would feed a large Black Hole faster than the matter and anti-matter inside could annihilate each other (although the outside observer could not directly see what's going on beyond the Event Horizon, any variation in gravitational strength ought to be measureable). That would suggest a Black Hole may go through a long period of increasing mass due to Hawking Radiation.

    Or am I missing something? :)
     
  2. Vkothii Banned Banned

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    First big mistake. The matter particle is 1/2 of the virtual pair - if it gets captured by the EH that created it, there is no net change in mass or energy (since they're equivalent).
    The energy it gained from the EH.

    Have a closer read of Hawking's theory. It's in the first 2 of his books, described by the man himself.
     
  3. MyBrainHurts Registered Member

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    Thanks Vkothii.

    There will be no net change in mass or energy in the universe as a whole, but there will be a change in mass within the black hole because an extra (positive) particle of matter has been added to its mass (that's assuming we take the Event Horizon as being the black hole's perimeter).

    The Event horizon does not create virtual pairs. According to the Uncertainty Principle, virtual pairs can spontaneously appear anywhere throughout the "vacuum" of space. It seems to me, the Event Horizon of a black hole is only important here because its powerful gravity pulls one particle away from the other, overcoming the matter-antimatter attractive force that would usually cause them to collide and annihilate themselves. i.e. it allows them to survive long enough to be regarded as "real".

    Does anybody here know whether a strong gravitational field increases the probability of virtual pair creation?

    I asked:
    What energy did it gain from the Event Horizon? None as far as I can see. The Event Horizon did not create the virtual pair. The Event Horizon is just a boundary, beyond which a singularity's gravity is so strong that even light does not have sufficient velocity to escape.

    Having taken another look at Hawking's A BRIEF HISTORY OF TIME as you suggested Vkothii, I see that he has vaguely touched upon the subject in the BLACK HOLES AIN'T SO BLACK section:

    I'm still left wondering: how could the forsaken partner particle have enough "positive energy" to escape from the very near vicinity of a black hole? Is it kinetic energy from when the two virtual particles initially parted? If so, what determines their speed just after they are created? And what kinds of particles might they be?

    Referring to virtual particles, Hawking states:

    Okay, basically it's saying there's no such thing as a free lunch. So far so good. According to Einstein's equation, energy and mass are interchangeable.

    E = m times c(squared)

    It seems to follow that the particle possessing "negative energy" (E-) will have negative mass (m-). That way an m+ particle and an m- particle can annihilate/nullify each other's mass when they meet.

    Surely, if a particle of positive energy or mass falls into the Event Horizon, the black hole will increase in mass. If a particle of negative energy or mass falls into the Event Horizon, the black hole will decrease in mass. In the latter case, the decrease in mass of the black hole, and therefore its decrease in gravitational strength (to an outside observer), would only occur after this negative particle has annihilated some positive mass/matter within the black hole - presumably this could be via collision with matter falling towards the singularity at the centre of the Black Hole, or via contact with the singularity itself. This reduction of black hole mass by contact with negative energy (or mass) appears to be generally supported by Hawking's statement:

    I said:
    I think I may have cracked this one. Perhaps I'm mistaken in assuming gravity is the only strong force near a black hole. Perhaps matter-antimatter attraction tips the balance, increasing the chance of antimatter falling across the Event Horizon, in comparison to matter which is attracted by gravity alone. Anyone care to comment?

    However, just when I thought I had that bit sussed, I read the following Hawking statement in his book:

    Doesn't it then follow that any positive matter falling into a black hole may be converted into negative energy, and therefore reduce the mass of the black hole rather than increasing it?

    Frustratingly, Hawking does not appear to explain why or how "the gravitational field inside a black hole is so strong that even a real particle can end up having negative energy".

    Can somebody please explain it in plain English, or refer me do an article that does?

    MyBrainHurts
    (it really does)
     
  4. Vkothii Banned Banned

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    An event horizon isn't just a boundary, it's the surface between ordinary 3-dimensional space, and whatever the black hole is. This surface is stretched very tight - a lot tighter than ordinary space.

    That's why when virtual pairs "pop up" like they do all over the universe (i.e. throughout space), one of the pair can gain mass, or enough energy to escape the BH's gravity well, but it has to be at the horizon.
    Any pair at the horizon has a chance of becoming a single real particle that enters the universe, plus a single 'negative energy' particle that enters the BH and reduces its mass.
     
  5. MyBrainHurts Registered Member

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    Nope, I'm still confused!

    Hawking said:

    How can an antiparticle have positive energy? Isn't an antiparticle supposed to be the incarnation of negative energy, as half of a virtual pair? :shrug:

    MyBrainHurts
    (it really does)
     
  6. MyBrainHurts Registered Member

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    Hi Vkothii,

    My last post wasn't a reply to your comment (which I've just seen). It was just a continuation of my thoughts.

    I'll have a think about what you've said.

    Thanks.
     
  7. Vkothii Banned Banned

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    No, an antiparticle is the same particle with a different sign for a certain quantum number. These numbers determine the state of any particle.

    It's called a CP number, for particles and corresponding antiparticles. But either state for a particle has real mass, and positive energy, or it wouldn't be a particle, right?
     
  8. MyBrainHurts Registered Member

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    As I've already mentioned, with regard to virtual particles Hawking states:

    How does that fit? If one of the particles is not innately negative in terms of energy (or mass, since they appear to be interchangeable), what sort of negative energy (and positive energy) are we talking about?
     
  9. AlphaNumeric Fully ionized Moderator

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    Antimatter are the negative energy mode solutions of the Dirac Operator. This is not the same as having negative energy in terms of the black hole's gravitational field.
     
  10. Vkothii Banned Banned

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    You're taking that sentence out of context; Hawking means what happens to a VP which is at, or on the horizon, doesn't he?

    In normal 3d space, a virtual pair can't gain energy to become real, it only happens at event horizons. Loosely speaking, a negative energy particle is never seen in real space, it's assumed to exist inside event horizons.
     
  11. MyBrainHurts Registered Member

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    Okay. I'm clearly in over my head, but I think I'm getting there.

    I suspect my big mistake has been to regard virtual particles associated with light or gravity as being the same as virtual pairs of matter particles. Now I understand Hawking's comment:

    They really are identical! And they only become real if one of them falls into the Event Horizon.

    I guess that raises the question: how does the remaining particle know that it is no longer virtual? Is it some kind of entanglement?

    And what happens when both particles from a virtual pair enter the Event Horizon simultaneously? Would they both become negative energy (a free lunch)? Or is it not possible to predict what happens?
     
  12. AlphaNumeric Fully ionized Moderator

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    A virtual pair is made, particle and antiparticle, just above the EH. The effect the gravitational field has is to make the outgoing particle, be it matter or antimatter, have positive energy. The infalling particle will experience the opposite effect, it will have negative energy, be it matter or antimatter. If the pair both fall into the event horizon, nothing changes.
     
  13. MyBrainHurts Registered Member

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    So the positive energy or negative energy nature possessd by a virtual particle is determined purely by its direction of travel relative to the black hole?
     
  14. MyBrainHurts Registered Member

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    Sorry, that was a dumb question. Let me try something else:

    Does the negative energy of the infalling particle come about because all energy inside a black hole is negative?

    I mean that in the sense that light traveling away from the vicinity of a black hole will have positive energy (though red shifted), light trapped on the Event horizon will have zero energy (because it has been red shifted so much as it tried to escape the black hole), therefore light inside the black hole ought to have a value of less than zero. Or am I digressing?
     
  15. Vkothii Banned Banned

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    Perhaps you require a better understanding of Hawking's ideas, to answer your questions.
    wikipedia's entry has a lot of references to external stuff, some available online.

    One detail that might be distracting you is what 'radiation' is.
    Lots of things radiate - nuclear reactors radiate neutrons, along with EM radiation across the spectrum. So radiation can be a particle with mass.

    These ideas are about processes that 'must' occur, if the SM is correct.
    They're speculative in a strict sense, however, they are based on some fairly firm theories.

    Hawking radiation has never been observed, it's assumed to be a real process.
    Or it's a contingent belief (he he), philosophically speaking, with a good expectation that it will be observed (or another explanation will be developed).
     
  16. Walter L. Wagner Cosmic Truth Seeker Valued Senior Member

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    As real particles, both the particle and the anti-particle [matter and antimatter] have positive mass. When they come in contact, they annihilate and release their mass/energy as photons.

    As virtual particles, one of the particle pair [particle or anti-particle] has positive mass, and the other has negative mass; for a net mass of zero [i.e. they don't exist and are virtual]

    Hawking postulates that the negative mass particle [whether matter or anti-matter] falls into the black hole, while the positive mass particle moves away. Ergo, someone far away observes positive mass coming from the vicinity of the black hole, while the black hole in turn loses mass [because the negative mass particle cancels some of the mass of the black hole].

    It is, however, not a proven theory, and others [most notably Higgs] have suggested it is wrong.
     
  17. prometheus viva voce! Moderator

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    No, you are wrong.

    One virtual particle has negative energy, not mass. It has been shown by Frank Wilczek and Maulik Parikh that the brief history of time description that you have just given is incomplete - the negative energy particle must tunnel into the black hole. They did the calculation in a completely different way to Hawking and got the same result.

    Reference
     
  18. MyBrainHurts Registered Member

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    Thanks very much for all your help folks. I think it's clear that much of my problem has been a confusion of terminology, which has led to misinterpretations. I'll have all look round elsewhere to hopefully get a firmer grasp of some of the issues - if my brain hasn't annihilated itself beforehand. :)

    MyBrainHurts
    (it really does)
     
  19. Walter L. Wagner Cosmic Truth Seeker Valued Senior Member

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    I use mass and energy interchangeably; as the mass of a proton is 0.93 GeV [when technically it is 0.93 GeV/c^2].
     
  20. rpenner Fully Wired Valued Senior Member

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    Prometheus' point is that in QFT, virtual particles are off the mass-shell and you cannot use mass and energy interchangeably until the particle is far from (in this case) the black hole.
     
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