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Radial temperature gradient of a black hole

by kmarinas86
Tags: black, gradient, hole, radial, temperature
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kmarinas86
#1
Oct4-11, 03:04 PM
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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?
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DaleSpam
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Oct4-11, 03:18 PM
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Quote Quote by kmarinas86 View Post
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?
That is an interesting question. Is a redshifted black body spectrum even a black body spectrum?
Bill_K
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Oct4-11, 04:09 PM
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Is a redshifted black body spectrum even a black body spectrum?
Yes indeed.

DaleSpam
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Oct4-11, 05:53 PM
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Radial temperature gradient of a black hole

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.
Pengwuino
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Oct4-11, 06:09 PM
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Quote Quote by DaleSpam View Post
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.
If I understand Hawking's derivation, it's the temperature seen at infinity.
phinds
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Oct4-11, 08:31 PM
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Quote Quote by kmarinas86 View Post
... Also, wouldn't an evaporating black hole be evaporating due to the universe heating it up?
Hawking radiation is not dependent on the surrounding temperature, so no I don't think so.
Bill_K
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Oct5-11, 02:53 PM
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Should the temperature gradient, in effect, be corrected for gravitational redshift, such that the temperature declines as distance from the black hole's center decreases?
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)]. 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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