Radial temperature gradient of a black hole

  1. Assuming that the accretion disk has been totally consumed by the black hole, does the temperature of the black hole due to Hawking radiation vary with respect with proximity with the black hole? For example, if I were next to the black hole, would this radiation would have a higher temperature than I was far away?

    Should the temperature gradient, in effect, be corrected for gravitational redshift, such that the temperature declines as distance from the blackhole's center decreases? Or should the temperature gradient not get corrected for the gravitational redshift, such that the temperature of the blackhole at some location is determined by the local observer at its local coordinate frame, rather than from a global coordinate frame of reference?

    If light cannot escape beyond the black hole's event horizon. Doesn't that make it a heat sink? If it is a heat sink, musn't that mean that, as far as thermodynamics are concerned, that it must be treated as a colder body, and not a hotter one, so a black hole does not have infinite temperature and entropy? Also, wouldn't an evaporating black hole be evaporating due to the universe heating it up?
  2. jcsd
  3. DaleSpam

    Staff: Mentor

    That is an interesting question. Is a redshifted black body spectrum even a black body spectrum?
  4. Bill_K

    Bill_K 4,160
    Science Advisor

    Yes indeed.
  5. DaleSpam

    Staff: Mentor

    Then it would definitely seem that the temperature of a black hole, as determined by the black body spectrum of Hawking radiation, would depend on the distance from the black hole. I don't know if the equation for Hawking radiation is at the event horizon or at infinity.
  6. Pengwuino

    Pengwuino 7,118
    Gold Member

    If I understand Hawking's derivation, it's the temperature seen at infinity.
  7. phinds

    phinds 8,339
    Gold Member

    Hawking radiation is not dependent on the surrounding temperature, so no I don't think so.
  8. Bill_K

    Bill_K 4,160
    Science Advisor

    I think this is the answer. It increases! This is taken from Birrell & Davies, p.282.

    "For the Unruh vacuum, FU(E)/unit proper time = 1/E(eE/kT - 1) where kT = [64π2M2(1 - 2M/R)]. As the detector approaches the horizon (R → 2M) the temperature of the flux determined by the detector diverges. This is due to the fact that the detector must be noninertial to maintain a fixed distance from the black hole The magnitude of the acceleration relative to the local freely-falling frame is M/[R2(1 - 2M/R)½]. Such acceleration gives rise to the detection of additional particles. As the horizon is approached, the acceleration diverges, as does the temperature."
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