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Feb21-12, 06:01 AM
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Pardon the long delay in replying I'm behind grading papers.
Quote Quote by juanrga View Post
I was said in my thermo course that residual entropies are the result of ignoring some interaction that breaks the degeneracy. I.e. that those degeneracies are fictitious. In any case I cannot see how substituting the strong form by the weak form [itex]\lim_{T\to 0}S = S_0[/itex] changes anything for BH thermo.
The assertion that there exists some ignored interaction is ad hoc and not part of the theory. It is most especially invalid in the BH case. Where it makes a difference is that with a BH the entropy is increasing with decreasing temperature. Not a violation of 3rd law if you allow it to increase to a "constant" (diverge to infinity) rather than require it be zero.

The zero value clearly ignores entropy of mixing. Yes you can resolve that by asserting the system has a boundary breaking spatial symmetry and so some mixtures have higher energy than others but that's just the thing about a black hole. Its interior has no spatial boundary. The event horizon and the singularity are null surfaces.

How do you adjust the 'exceptional behaviour' of evaporating BHs with the thermodynamic properties of the supposedly emitted thermal radiation?
No adjustment is needed. You can have a BH in thermal equilibrium with its environment. [BIG box with black hole and thermal photon gas] Inject energy into the system and the system will reach a new higher entropy equilibrium with a larger black hole and a colder environment.

I spent some time this weekend seeing if I could integrate the exterior volume to replicate some quantitative figures but its a busy time in the semester and I have other pressing priorities. If I have time and come up with anything worth posting I will.

The most exceptional aspect of a BH is that it must reach infinite size to achieve zero temperature. This is not outside the practical meaning of the 3rd law which is that absolute zero is an asymptotic limit one cannot achieve via finite processes.

If the Bekenstein-Hawking formula is wrong then one should be able to generate a contradiction via some though experiment. I see nothing like this in the OP reference.