# Bekenstein entropy black hole v.s Hawking entropy black hole

Historically, Bekenstein estimated the entropy associated with a black hole in 1973, obtaining: $$S_B = \frac{\ln(2)k_Bc^3}{8\pi\hbar G}A.$$ He already acknowledges in his article that his estimates are based on classical principles and that a quantum mechanical treatment will yield a different constant, though within a factor of order one the same. A year later Hawking derived: $$S_H = \frac{k_Bc^3}{4G\hbar}A,$$ i.e. $S_B = (\ln(2)/2\pi) S_H$, such that $S_B<S_H$.

I am wondering if there could have been examples, showing that $S_B$ was not correct. So, without knowing Hawking's results, can we see that $S_B$ cannot be correct, maybe by giving a certain counter example, or using the fact that $S_B<S_H$?

• Why are you sure about correctness of Hawking's calculation? In principle, it is no more correct than Bekestein's one, and the hypothesis of both of them are pretty arguable. These results are indicative and their agreement up to a factor of order one (moreover a constant factor!) may reveal a notable behaviour of the Nature, but one can't conclude that one of them is absolutely correct, since we doesn't have experimental confirmations. – Federico Nov 3 '13 at 19:14
• I do not agree. Of course, both expressions have not been verified experimentally, but I do think that one can consider a quantum mechanical treatment better, as the nature of the problem is really quantum. The Bekenstein entropy is only an educated guess based on classical considerations. – Funzies Nov 4 '13 at 8:15
• Well, Hawking tried to do quantum mechanics in the vicinity of a black hole. In order to replace Minkowski metric, he imposed $\eta_{\mu \nu} \to g_{\mu \nu} + \hat{h}_{\mu \nu}$, where $g_{\mu \nu}$ is the Schwarzschild metric ed $\hat{h}_{\mu \nu}$ an operator by means of which he tried to "extract" quantum mechanics. All that is arbitrary as well as classical calculations. Also technically is arguable. Of course, his result was of great importance and surely makes sense, but in the lack of a complete theory we can't say it is more correct, but only that it appears more reasonable. – Federico Nov 4 '13 at 10:01
• I think your question is really hard. We know that Bekestein provided a few different classical arguments leading to the same result. In particular, one of them relies on a well established theorem of general relativity, which states the horizon of a BH ($A$ in your formulas) is non-decreasing. If we think at this, and at the fact that the relation with $S$ is linear, it seems clear that $S$ is good entropy. (i.e. satisfies the Second Law.) I think it should be enough significant, provided the limitations due to the present state of the theory. – Federico Nov 4 '13 at 15:00
• P.S. If you thought our debate was interesting, I will made up an answer, adding some other detail/consideration. – Federico Nov 4 '13 at 15:09

I recently went to a colloquium with the theme "98 years of black hole physics" by string theorist Jan de Boer from the university of Amsterdam. I asked him this question and he replied that there have been lattice computations for black hole thermodynamics, yielding precisely Hawking's factor of $1/4$. Furthermore the result has been obtained using different methods, strengthening its reliability.
In short, Bekenstein's entropy is integer amount of bits, which cannot be true for variable measured at Planck scale, where the fundamental unit of information is nat, that is $1/\ln 2$ bit