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Criteria for thermodynamical stability is the convex of entropy. But for black hole entropy is non-additive.

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Welcome to Physics.SE! This question is unclear, can you expand on what you are asking and give relevant context? –  zhermes May 11 '13 at 23:21
    
In particular, thermo stability requires a maximum of entropy, which in general, I think, is non-additive. What do those facts have to do with eachother? And how do they apply to black-holes? What does this have to do with dynamical stability? –  zhermes May 11 '13 at 23:23
    
Yes, the maximum of entropy -- it's reformulation of criteria convexity of entropy's surface -- in usual thermodynamics. For this reformulation you can look chapter 8, page 206 at "Thermodynamics and an Introduction to Thermostatistics" by Callen (it's avalible in net). If this criteria is performed, then system is thermodynamically stable. For calculations it's clear how to calculate convexity of entropy's surface -- and this way is usually choosen in usual thermodynamics. But for this reformulation we use hypothesis of entropy additivity(if i right). So this way is incorrect for black holes? –  drobnbobn May 12 '13 at 0:32
    
Why can't you use the standard Hawking Entropy? –  zhermes May 12 '13 at 0:38
    
I start with (Bekenstein)Hawking formula for entropy. Also i use Smarr formula for black hole's mass. Both this formulaes gives the expression for S (entropy) through black hole's parameters M, Q, J. The criteria of stability (maximum of entropy, i.e. the convexity of entropy's surface) is positive Hessian for S. I do it all because it's usual way in usual thermodynamics (look book above). And my first problem: is this way possible for black holes? Because i heard that in such way (in usual thermodynamics) we use assumption of entropy's additivity; and that it's false for black hole's entropy. –  drobnbobn May 12 '13 at 0:55
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