Entropy of a black hole after evaporation

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  • #1
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Black holes have an entropy, but they evaporate. At the end of the evaporation, the entropy is greater than the entropy at the beginning of the evaporation. I am looking for an example of a quantitative result for the entropy of the black hole after evaporation (or the entropy difference between the beginning and the end of the evaporation). You can use your favorite theory (General relativity, f(R), String theory, Loop Quantum Gravity, etc..), you can use your favorite kind of black hole (Schwarzschild, rotating, charged, extremal, BTZ, etc.... ), and you can use your favorite dimension (from 4 to 11...), but I am looking for a quantitative result.
 

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  • #2
tom.stoer
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I would try a semiclassical description (Hawking radiation + classical spacetime; after evaporation this is Hawking radiation + flat spacetime). For Hawking radiation the entropy calculation is done using Bose-Einstein statistics for non-interacting photons of given temperature. There are two problems: i) one has to adapt the standard plane wave formalism to (distorted) spherical waves (have a look at Hawking's paper); ii) geometry and therefore temperature are not constant.
 
  • #3
Don't all our current theories break down for a black hole smaller than a Planck mass (or a ratio thereof, such as √∏), thus we can't arrive at a quantitative (integrative) value for the entropy of complete evaporation?
 
  • #4
tom.stoer
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Don't all our current theories break down for a black hole smaller than a Planck mass (or a ratio thereof, such as √∏), thus we can't arrive at a quantitative (integrative) value for the entropy of complete evaporation?
The semiclassical calculation (Hawking) is certainly invalid; there are proposals for microscopic state counting in LQG or string theory (but I think that evaporation with thermal radiations + corrections is not understood); anyway - why not try to calculate the BH entropy using Bekensteins formula, and then calculate the entropy for the thermal radiation?
 
  • #5
Does the Planck scale represent the realm where the horizon of a black hole is basically the same as the singularity itself, at least in regards to entropy?

With a sense of humor I hope others can appreciate, the theory I choose to answer this question in is that the universe has a ground state energy/entorpy (or related frequency) that's closely proportional to the Planck mass. Taking that same mass/energy/frequency for my Schwarzchild black hole, then the entropy change for such a black hole evaporating is quantitatively zero. Funny, that same ground state also gives (in a space domain) a value for dark energy that's pretty close to what we observe.

I obviously think I might need some help with my thinking.
 
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  • #6
Chronos
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Typically, black hole entropy calculations assume it all resides at the event horizon. I'm unconvinced that is correct.
 

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