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Virtual antiparticle pairs at event horizon

  1. May 27, 2004 #1
    One of my friend had asked me one question which I did not really have a good answer for. His question was: Why does the antiparticle counterpart of the virtual antiparticle pairs (those that appear due to uncertainty between time and energy) at the event horizon fall into the black hole, instead of the 'particle' counterpart of the pair (for example: a 'particle' is an electron, an antiparticle is the positron)?

    My answer to this question was that it was to mantain the universal overall of matter and energy at the same amount, since the antiparticle counterpart actually annihilates the singularity's matter. Is there a better answer for this?
     
    Last edited: May 27, 2004
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  3. May 27, 2004 #2

    turin

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    I didn't know that the singularity was accepted to consist of matter (i.e. leptons and quarks).
     
  4. May 28, 2004 #3
    Well, according to Mister Stephen Hawking, black holes can evaporate and can losing their mass until they diasppear. This is explained either by the fact that they (the black holes) have temperature, or that the anti-particles I mentioned annihilate the matter of the Singularity gradually.

    Even if black holes have no 'hair' (the theory which states that all information of an black hole, except its mass, charge, and angular momentum, is lost), they do have matter, though the information regarding what kind of matter it is is lost. Only the mass of it is known.
     
  5. May 29, 2004 #4

    turin

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    I am still a little confused. Antiparticles have positive mass, right? If, for instance, we had a negatively charged black hole, then it would be more likely to capture a positron than an elecctron in pair production because the negative charge would repel the electrons and attract the positrons. But, once the positron hits the singularity, shouldn't it increase the mass (by at least 511 keV), even though it would cancel some of the charge? I do not see how anti-particle annihilation can explain BH evaporation. Even if the negative charge were electronic, the annihilation should produce >1 MeV gammas, which should not be able to escape the BH, and thus should still contibute to an increase in mass.
     
  6. May 30, 2004 #5
    Alright, I had made some mistake in my explanation for this problem! Sorry... :frown: The truth is that any of the antiparticle pair could enter the black hole or leave the black hole. It is the negative energy from creation of the anti-particle pairs that slowly destroy the matter inside a black hole. Just read some websites or books that explain this theory.
     
  7. May 30, 2004 #6

    turin

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    This is what I don't understand. What negative energy? I don't know very much about this pair-production stuff, but I have always been under the impression that both the particle and anti-particle carry positive energy. Is it that, since energy overall must be conserved, and this pair production is occuring out of nothing, then there must be negative energy left behind after the pair production to balance the positive energy carried by the pair to sum to zero net energy? Even in that case, though, it seems that negative energy should be repelled by the BH, not pulled into the singularity (i.e. gravity attracts positive mass and repels negative mass?).

    I'm not trying to tell you that you're wrong; I am just probably a lot more confused about this than you are and would like some clarification. Can you give some quotes that have led you to this assertion?
     
  8. May 31, 2004 #7
    This is the explanation from website http://eportfolio.babson.edu/2006/rbrennan1/BlackHoles/classifications.htm :

    "Hawking speculates that pair creation exists right around the event horizon. During the pair creation, the particle or the antiparticle has the ability to fall within the event horizon and therefore has the ability to be separated from its pair. Unfortunately, nature is very unforgiving with many of its laws and the remaining particle, whether it is the particle or antiparticle, still requires for the pair in order to release energy. Since the pair is already to far away from its match, nature gains its necessary energy from the black hole itself, which begins to radiate out tiny particles (Al-Khalili 106). This process of emitting energy allows for the black holes to be reduced in size. Remember, the e (energy) and the m (mass) in the E=mc2 equation are interchangeable and therefore, when the energy is reduced, subsequently the mass will be reduced. In addition, the radiation of these particles creates a red glow at the point of the event horizon (Al-Khalili 106). (Although this does not fit in with the topic, this red glow discredits the theory that black holes are maintaining an almost invisible state within the universe)."

    But it seems to be void of anything related to negative energy... if you want to know about negative energy, go to this fantastic website:
    http://www.physics.hku.hk/~tboyce/sf/topics/wormhole/wormhole.html
     
  9. May 31, 2004 #8

    turin

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    From that negative energy website, I found two statements that seem somewhat (though not necessarily) contradictory.
    The first quote seems to say that the negative energy from the pair production is attracted into the singularity. The second quote seems to suggest that the negative energy from pair production should be repelled.
     
  10. Jun 3, 2004 #9
    Well, I don't know but I suggest the gravitational attraction of the black hole is strong enough to overcome that "repulsive" force. :approve:
     
  11. Jun 3, 2004 #10

    chroot

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    Either the particle or the antiparticle can fall into the whole. From the outside, in either case, it appears as though the black hole has emitted a particle (or antiparticle) and thus has lost mass. Antiparticles have the same mass as particles -- there is no such thing as negative mass.

    - Warren
     
  12. Jun 4, 2004 #11

    turin

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    I think I'm starting to get it. I should have taken the "virtual" in the title more seriously. Physically, we're really just talking about the observation of positive massive particle emission by a BH. The discussion of particle-antiparticle pairs is to help us understand why we see particles flying out of a BH, not to help us directly understand why the mass decreases. Then, simple conservation of energy tells us that the mass should decrease.

    I'm still a little shady on why these pairs should be created just outside the event horizon.
     
  13. Jun 4, 2004 #12

    chroot

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    A virtual particle is only virtual until it interacts with something -- then it's pretty darned real. Virtual pairs are created everywhere, all the time, including around the edge of a black hole's event horizon.

    - Warren
     
  14. Jun 4, 2004 #13

    turin

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    OK, well, I'm shady on this, then. Of course, I'm expecting the basic answer to be "Quantum," but I still don't get it.
     
  15. Jun 4, 2004 #14

    hellfire

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    I read often about this explanation of the Hawking radiation in terms of virtual particles becoming "real". To let such a mechanism explain the radiation as well as the loss of mass of the black hole, it seams to me that the incoming particle must have indeed a negative energy (otherwise I see no possibility of mass decrease). However, in the usual creation of virtual pairs as described by quantum field theory on flat spacetime, both particle and antiparticle have positive energy (as already pointed out).

    Anyway, I went once through the derivation of the Hawking radiation and there is nothing there about virtual pair creation. The radiation is due to a difference in the definition of the creation and annihilation operators on both sides of the horizon for a field beeing quantized in a Schwarzchild spacetime. The resulting ground state of the Hamiltonian (the energy of the vacuum) is not the same on both sides of the horizon, leading thus to a thermal emission.

    How to reconcile this with the explanation of virtual pair creation puzzles me also, but I would guess it is not that easy. My first impression is that this explanation is rather heuristic and does not really describe a physical process. But I may be wrong.

    Regards.
     
    Last edited: Jun 4, 2004
  16. Jun 4, 2004 #15

    turin

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    This doesn't sound like it would be physically significant, though. Just because we define the operators a certain way ... Also, I didn' think we could play with GR and QFT in the same sandbox, anyway, for the exact reason that things become ill-defined.

    Care to elaborate?




    I don't know much about this stuff, but can you brief me on what the ground state of the Hamiltonian has to do with thermal emission?
     
  17. Jun 4, 2004 #16
    Can I ask a related question here?

    Unruh's Law says that an observer will detect particles only at an accelerating state n temperature of surroundings proportional to acceleration. I thought particles and anti-particles annihilate and recreate in space constantly? Does that mean they are undetectable unless there's a gravitational field or something equivalent (an accelarating state) to it around that magnifies this constant activity?
     
  18. Jun 5, 2004 #17

    hellfire

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    turin: why should this be ill defined? May be you mean that we cannot quantize gravity yet, but here we are dealing with quantum field theory on a classical (and static) background. Usually the Minkowski background is used to define QFT, but one can, in principle, make also use of curved spacetimes.

    To solve the relativistic equations for fields one makes a spatial Fourier decomposition. In the canonical quantization, the Fourier coefficients are identified with creation and annihilation operators, but they depend on the Fourier modes choosen. This leads not to an ambiguity in the definition of the vacuum, because one takes the true vacuum as the one with no particles. But in some situations "other" possible vacuum states may indeed correspond to zero-particle states for special kind of observers in the background spacetime. If one computes the expectation value of the number operator (according the "other" definition of creation and annihilation) for the vacuum (the first one), one gets a thermal distribution.

    I know this description may be short and poor (I do not have a deep knowledge of QFT), but I can give you a reference:

    http://www.theorie.physik.uni-muenchen.de/~serge/T6/

    Take a look to chapter 9, where Hawking radiation is derived. In chapter 8 the Unruh effect is described. This may be also interesting because the same problem with the ground state of the Hamiltonian arises there and the problem of the definition of creation and annihilation operators is treated in a more simple and generic way.

    Regards.
     
    Last edited: Jun 5, 2004
  19. Jun 5, 2004 #18

    turin

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    Thanks, hellfire. I'll try to read those notes soon.
     
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