# How is the logarithmic correction to the entropy of a non extremal black hole derived?

Ive just read, that for non extremal black holes, there exists a logarithmic (and other) correction(s) to the well known term proportional to the area of the horizon such that

$S = \frac{A}{4G} + K \ln \left(\frac{A}{4G}\right)$

where K is a constant.

How is this logarithmic (and other) correction term(s) derived generally? Or how can I see that there has to be such a logarithmic correction?

Im wondering if there is some kind of a general makroscopic thermodynamic or semiclassical argument (in analogy to some derivations of the first term) that motivates the appearance of the second logarithmic term and does not depend on how the microstates are quantum gravitationally implemented.

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Link to Ashoke Sen's recent paper: arxiv.org/abs/1205.0971 –  Qmechanic May 7 '12 at 21:15
Huh, interesting... I have not heard of this correction term before. –  David Z May 8 '12 at 4:32
Yep @Qmechanic a discussion of this paper is what made me asking this. Basically, I`m wondering if there is some kind of a general makroscopic thermodynamic or semiclassical argument (in analogy to some derivations of the first term) that motivates the appearance of the second logarithmic term and does not depend on how the microstates are quantum gravitationally implemented. –  Dilaton May 8 '12 at 8:32
I wonder what this looks like in non-Planck units. There must be a bunch of constants inside that $\ln$ to make its argument dimensionless, and knowing what they are might give some insight into your question. (I know this is old, but just saying.) –  Nathaniel Jul 6 '14 at 6:33
Ashoke Sen uses the entropy function formalism that he has developed along with his collaborators. This agrees with Wald's formula in cases where they can be compared. For supersymmetric, extremal black holes his methods have yielded results that agree with exact microscopic counting to sub-leading order. In this paper, he is extending those methods to compute entropy for non-extremal black holes. Of course, the microscopic counting in these cases have not been done. I don't think there is any thermodynamic argument that I know of which explains the log. –  suresh Sep 8 '14 at 11:46