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The question is this: will the thermodynamic properties of a black hole (Hawking radiation spectra and temperature, entropy, area, etc.) depend if the black hole sits in a DeSitter or an Anti-DeSitter space-time?

why and how?

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1 Answer 1

The answer is trivially yes. When large AdS blackholes radiate, the Hawking radiation is 'reflected' at infinity, and you eventually end up with a sort of thermodynamic equilibrium state where the blackhole becomes stable and eternal as it reabsorbs its radiation.

The details are a little bit more involved, but suffice it to say nothing like that happens for DeSitter blackholes, at least on timescales that are similar.

Now, there are many things that are superficially similar (the area dependance, the semi-classical leading order corrections and so forth) but the dynamics will change based on these asymptotic values and thus the statistical mechanics will also differ.

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Beyond this, the surface gravity is $\Lambda$ dependent, so this makes the temperature an explicit function of the cosmological constant. –  Jerry Schirmer Dec 20 '12 at 1:46
    
what does it mean 'reflected at infinity'? any radiating black hole will be finite in time (they will form and then they'll evaporate), are you referring to an eternal black hole solution? –  lurscher Dec 20 '12 at 16:10
    
In AdS space, every large black hole is eternal. See hep-th/0106112. –  Columbia Jan 4 '13 at 8:23

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