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In their paper The entropy of Hawking radiation, Juan Maldacena et al. write :

«In other words, if the black hole degrees of freedom together with the radiation are producing a pure state, then the fine-grained entropy of the black hole should be equal to that of the radiation $S_{\text{black hole}}=S_{\text{ rad}}$. But this fine-grained entropy of the black hole should be less than the Bekenstein-Hawking or thermodynamic entropy of the black hole, $S_{\text{ black hole}}≤ S_{\text{Bekenstein−Hawking}} = S_{\text{coarse−grained}}$»

My question is :

What's the difference between Bekeinstein-Hawking entropy and the black hole entropy?

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In this context, the black hole entropy is the fine-grained entropy of the black hole. Namely, if you knew all black hole microstates and hence the density matrix of the black hole, $\rho_\text{BH}$, the black hole entropy would be the von Neumann entropy associated to $\rho_\text{BH}$ $$S_\text{black hole} := S_\text{vN}(\rho_{\text{BH}}):= -\mathrm{tr}(\rho_\text{BH}\log\rho_\text{BH}).$$ The Bekenstein-Hawking entropy is the coarse-grained, or equivalently, the thermodynamic entropy of the black hole. $$ T\,\mathrm{d}S_\text{Bekenstein-Hawking} \overset{!}{=} \mathrm{d}M - \Omega\,\mathrm{d}J.$$ Knowing $\rho_\text{BH}$, this is equivalent to the following. Choose a bunch of coarse-grained observables that you want to measure $\left\{\mathcal{O}_i\right\}_i\subset\left\{\text{all observables}\right\}$ and find all density matrices $\rho_\text{BH}^\text{coarse}$ such that $$ \mathrm{tr}(\rho_\text{BH}^\text{coarse} \mathcal{O}_i)=:\left<\mathcal{O}_i\right>_\text{coarse}\overset{!}{=}\left<\mathcal{O}_i\right> := \mathrm{tr}(\rho_\text{BH}\mathcal{O}_i), \qquad \forall\mathcal{O}_i\in \left\{\mathcal{O}_i\right\}_i.$$ Then the Bekenstein-Hawking entropy is $$S_\text{Bekenstein-Hawking} := \max_{\rho_\text{BH}^\text{coarse}} S_\text{vN}(\rho_\text{BH}^\text{coarse}).$$ This turns out to be exactly what we, unknowingly, do when we compute thermodynamic entropies.

This definition also makes it clear why $$ S_\text{black hole}\leq S_\text{Bekenstein-Hawking}.$$

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  • $\begingroup$ Thanks for your intervention. I have grasped the definition of fine-grained entropy and coarse-grained entropy. My confusion is about the identification of black entropy as coarse-grained entropy (thermodynamic entropy) or as fine-grained entropy (von Neumann entropy). As you have explained, coarse-grained entropy is bigger than the fine-grained entropy because the coarse-grained entropy is the maximal one. If I have very understood, in this context the entropy outside the horizon vanishes because, in general, the coarse-grained entropy is equal to the general entropy $\endgroup$
    – Kadi Harouna Illia
    Jan 13 at 9:02

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