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Hawking Radiation

  1. Jul 8, 2006 #1
    I'm having trouble understanding why it is necessary for there to be a particle antiparticle pair formation in order for the black hole to evaporate.

    According to my understanding of the theory, a particle anti particle pair are formed, and rather than colliding within a given period of time (dictated by the heisenburg uncertainty principle), in order to prevent the violation of conservation of mass, one of the particles falls into the black hole. According to wikipedia-

    "In order to fill the energy 'hole' left by the pair's spontaneous creation, energy tunnels out of the black hole and across the event horizon. By this process the black hole loses mass, and to an outside observer it would appear that the black hole has just emitted a particle."

    My first question is, what was preventing the energy from tunneling out of the black hole in the first place? I mean, isn't quantum tunneling simply a matter of probability? Say you have two states, the inside of the event horizon, and the outside of the event horizon. Treat the system as an energy hill, like an activation energy in chemistry, except the hill is infinitely high, which is why nothing can escape a black hole in classical physics. However, according to the theory of quantum physics, it's merely a matter of probability that the subject exists in one state, and not the other, so there is a chance that the subject could "tunnel", or appear, on the other side of the event horizon.

    My second question is, does the energy disappear? Say an anti proton falls into the black hole, and a proton escapes the event horizon. The black hole then parcels out some ammount of energy equivalent to the combined mass of both the particle and the anti particle, and that energy disseapears from the black hole, such that, not only has the gain in energy from the anti proton falling in been negated, but also an ammount of energy has been lost equivalent to the proton being "emited". But where does that energy go?

    I suspect there's something crucial missing in my understanding of the instantaneous creation and anihalation of particle anti particle pairs. Any help?
  2. jcsd
  3. Jul 10, 2006 #2
    I agree - this is something I also never understood about Hawking radiation. The simplistic description in terms of particle-anti-particle pair creation would seem to imply that the in-falling particle has "negative mass-energy" :eek: (otherwise the Black Hole would gain mass instead of losing mass).

    In fact, Hawking himself describes it in just this way in Chapter 7 of his book A Brief History of Time :

    see the chapter here : http://www.physics.metu.edu.tr/~fizikt/html/hawking/f.html [Broken]

    I suspect the error is in the simplistic interpretation in terms of particle-anti-particle pair creation.

    Can anyone shed light on this?

    Best Regards
    Last edited by a moderator: May 2, 2017
  4. Jul 10, 2006 #3
    I think is the moot point:

    "So in empty space the field cannot be fixed at exactly zero, because then it would have both a precise value (zero) and a precise rate of change (also zero). There must be a certain minimum amount of uncertainty, or quantum fluctuations, in the value of the field. One can think of these fluctuations as pairs of particles of light..."

    You can think of them as particles. But they're fluctuations rather than "billiard ball" particles.
  5. Jul 10, 2006 #4

    George Jones

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    Last edited by a moderator: Apr 22, 2017
  6. Jul 10, 2006 #5
    Thanks for that, George. I wasn't clear on why the Black Hole can swallow a negative half of a "virtual particle pair" leaving the positive half free to radiate, but not vice versa:

    Note that this doesn't work in the other direction - you can't have the positive-energy particle cross the horizon and leaves the negative-energy particle stranded outside, since a negative-energy particle can't continue to exist outside the horizon for a time longer than h/E. So the black hole can lose energy to vacuum fluctuations, but it can't gain energy.

    Can you or anybody else rephrase this?
  7. Jul 11, 2006 #6

    That seems like a bit of a cop out explanation. It's almost just saying "the anti particle can't exist outside of the event horizon, because it's just not aloud" The fundamental process, according to my understanding, is as simple as one crossing the event horizon, and the other not. The process makes no distinction between a particle or an anti particle. The only way I could imagine there being a distinction made is if one of the particles were gravitationally repulsive. But it would be the regular particle which would have to be gravitationally repulsive in order for it to always be outside of the event horizon, and we all know that conventional matter is gravitationally attractive.
  8. Jul 11, 2006 #7
    Thanks, interesting link, but I'm still confused. The last paragraph just doesn't seem explicable :

    This simply states that the positive energy particle cannot cross the event horizon - it does not explain WHY it cannot cross the event horizon. What prevents it?

    I've also come across this weird paper :

    http://arxiv.org/PS_cache/hep-th/pdf/0402/0402145.pdf [Broken]

    which suggests that black holes do NOT in fact evaporate, but they simply accumulate "negative energy" as they radiate :

    ???? :surprised
    Last edited by a moderator: May 2, 2017
  9. Jul 11, 2006 #8
    Further problems trouble me with Hawking Radiation, as follows :

    (1) The HUP says that the particle-anti-particle pair will exist for a maximum time t given by

    t = h/(4.Pi.E)

    where E is the total energy of the pair. If one particle has negative energy and the other has positive energy, then surely the total energy of the pair could be zero, hence the pair could continue to exist indefinitely..... ?

    (2) The energies involved in Hawking radiation are so small (for massive black holes) that the radiation is most likely to be in the form of photons. The energy of a photon is simply related to it's frequency (f) as follows :

    E = h.f

    Thus the HUP-lifetime of a pair of virtual photons (if we consider them both to have positive energy) is given by :

    t = 1/(8.Pi.f)

    In other words, the HUP-lifetime will be approximately 4% of the time required for the photons to "exist for one wavelength". These virtual photons therefore pop into, and out of, existence in (at most) 4% of their wavelengths.......

  10. Jul 12, 2006 #9
    I doubt the radiation is in the form of photons

    I don't think there's such thing as a photon/anti-photon pair
  11. Jul 12, 2006 #10


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    The photon is its own antiparticle.
  12. Jul 13, 2006 #11

    For a black hole with mass the same as earth, the event horizon would have a radius of just 8.9mm, and would emit Hawking radiation with frequency around 1.35Ghz (similar in frequency to the radio waves used in mobile phones). This corresponds to an energy of approx 5.6 x 10^-6 eV.

    The rest mass of an electron is 5.1 x 10^5 eV - almost one trillion times more than the above Hawking radiation energy. Only some neutrinos, gravitons (if they exist) and photons, are less massive than the electron (in fact photons and gravitons are massless).

    Thus the only known way that a black hole with the mass of the earth could emit Hawking radiation would be in the form of either photons, some neutrinos, or gravitons (and gravitons I'm not sure about).

    More massive black holes would emit even lower energy radiation.

    Best Regards

    see : http://library.thinkquest.org/C007571/english/advance/english.htm
    Last edited by a moderator: Apr 22, 2017
  13. Jul 16, 2006 #12
    Actually, it's even worse than I suggested.

    For any particle to be emitted as Hawking radiation, the particle would need to have a velocity close to the speed of light in order to ensure it did not get sucked back into the black hole. For any particle with non-zero rest mass this implies a total energy far in excess of its rest mass

    Best Regards
  14. Jul 18, 2006 #13
    I think the reason why it is always given that the black hole absorbs the negative energy is because the particle/antiparticle pairs are entangled. Hence, one half of the entangled pair crosses the horizon while the other propagates out into space. If the "free" half of the pair is detected then it will give a positive energy in the detector meaning that the other particle has negative energy.
  15. Jul 19, 2006 #14
    I don't think it's that simple. If one half of an entangled pair has negative energy then the total energy of the pair will be zero and the permitted lifetime (t) of that virtual pair will be infinite, by the Heisenberg Uncertainty Principle :

    t = h/(4.Pi.E)

    where E is the total energy of the pair.

    The only reason that virtual particle pairs have very short lifetimes (which we can calculate using the above equation) is because the total energy of the pair is positive.

    Best Regards
  16. Mar 29, 2010 #15
    Last edited by a moderator: Apr 24, 2017
  17. Mar 29, 2010 #16
    There was a previous thread about HR, and one of the staff (Maybe Doc Al? Or Mentz... I forget) pointed out that this really takes specialized math and lectures to grasp. The concept as it is usually disseminated (particle/anti-particle fail to annihilate) isn't wrong per se, but it's not right either. This is just one of the very VERY technical theories out there, and one which is still being developed.
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