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Consider a spherical symmetric thin cell of photons converging to a point. At some moment, there is a formation of an horizon and a black hole. But each black hole is evaporating,and so, after some time, all the black hole has evaporated.

So, we could consider an "initial state", well before the creation of the horizon, and a "final state", well after the end of the evaporation of the black hole.

If we call $E$ the total energy of the converging photon cell, what is the difference of entropy between the "initial state" and the "final state"?

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(Note: I throughout use Planck units).

The initial state is very well defined, I would say it has zero entropy. Let us assume that a black hole of mass $M$ (where $M=E$ is the initial radiation energy), is created from this initial state. It has zero momentum and zero charge, it's a Schwarzschild black hole: its only parameter is $M$. The black hole has radius $R=2M$ and entropy $$S_H=A/4=\pi R^2=4\pi M^2 ,$$ where $A=4\pi R^2$ is the horizon area of the black hole (Hawking's formula). Then the black hole evaporates, which means that it is radiating as a black body at temperature $$T=\frac{1}{8\pi}M^{-1}.$$ During a time $\def\d{\mathrm d}\d t$, the black hole radiates an energy $\d W=P\d t$, so its mass is reduced by $\d M=-P\d t$. The entropy of the black hole decreases by $$\d S_{\rm BH}=\d A/4=8\pi M\,\d M$$ while the entropy of the rest of the universe increases by $$\d S_0=\d W/T=-(8\pi M)\d M.$$ The total universe entropy variation is $$\d S=\d S_{\rm BH}+\d S_0=8\pi M\,\d M-8\pi M\d M=0.$$ So, when the black hole is fully evaporated, the entropy of the universe has increased by $$\Delta S=4\pi M^2,$$ the entropy created at the formation of the black hole.

During this story, the initial photons in a perfect order have been transformed into different particles, randomly radiated by the black hole at its horizon.

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  • $\begingroup$ It is not correct, IMHO, to assert that the total universe entropy variation (from the beginning to the end of the evaporation) is zero . It is certainly stricly positive, the evolution is irreversible. In fact, I would expect that the variation of entropy between the very beginning and the formation of the black hole $\sim S_H$ is also the variation of the entropy during the evaporation, so that the total entropy variation is $2 S_H$ $\endgroup$
    – Trimok
    Jan 16 '14 at 17:23
  • $\begingroup$ @Trimok. It actually may looks surprising. However the evaporation is very slow, it is possible that it is adiabatic. $\endgroup$
    – Tom-Tom
    Jan 16 '14 at 17:59
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    $\begingroup$ @Trimok. My answer does not say anything about the dynamics and the "death" of a black hole is singular. In his lectures notes on black holes, t'Hooft remarks on page 40 that a better understanding for this event requires a more advanced understanding of quantum gravity, since the size of the dying black hole makes it a quantum object, but the energy released is "formidable", as he says. $\endgroup$
    – Tom-Tom
    Jan 16 '14 at 20:19
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This is merely a suggestion for an answer (I can't post comments yet), but you could use the entropic uncertainty inequality to figure out a lower bound for (Shannon) entropy.

Take a coherent spherically symmetric EM beam. If I'm not mistaken, the smallest volume you can focus the beam to has a diameter of at least the wavelength (or in that order of magnitude), i.e. for the Schwarzschild radius $$\frac{2GM}{c^2}\ge\lambda$$

where $M$ denotes the mass of the corresponding black hole. On the other hand, for the energy of a photon we have $$\frac{2\pi\hbar c}{\lambda}=\frac{Mc^2}{N}\ge\frac{c^4\lambda}{4GN}$$ where $N$ is the number of photons in the beam, so $$N\ge\frac{c^3\lambda^2}{8\pi\hbar G}=\frac{1}{8\pi}\left(\frac{\lambda}{l_P}\right)^2$$ Even for gamma rays ($\lambda\approx10^{-12}m$), $$N\ge \sim 1.5\cdot 10^{44}$$ and for common wavelengths this number will be much higher. So the chance that the black hole thus formed evaporates into few photons (or into photons alone) is slim. All this to say you can assume nearly perfect spherical symmetry of evaporation, but the 'final state' won't look much like the 'initial state'.

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    $\begingroup$ IIRC, the radiation from an event horizon is a thermal spectrum of all particle species. Also, there is no such thing as a spherically symmetric EM beam. $\endgroup$ Nov 17 '13 at 10:51
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    $\begingroup$ What Lionel means is that spherical symmetric transverse vector fields are impossible in 3 dimensions by the hairy ball theorem, i.e. the far fields are perfectly transverse. Also symmetric near fields are impossible: they would be monopole fields which would violate charge conservation see en.wikipedia.org/wiki/… $\endgroup$ Dec 17 '13 at 13:35
  • $\begingroup$ I'm quite familiar with the hairy ball theorem. Mine was just a working assumption to clarify the OP's assumption. Why didn't you write your comments at the question? How can you have a 'spherical symmetric thin cell of photons converging to a point' save for a superposition of coherent states? Your thoughts on the remainder of my post? $\endgroup$
    – suissidle
    Feb 3 '14 at 1:27

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