# Black hole entropy

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Problem P16.7.

Bizarre as it seems, an entropy can be defined for a black hole. From thermodynamics (which we haven’t covered yet, so you’ll need to look elsewhere for a derivation for now), the entropy ${S}$ of a system is defined in terms of its temperature ${T}$ and internal energy ${U}$ by

$\displaystyle \boxed{\frac{1}{T}\equiv\frac{\partial S}{\partial U}} \ \ \ \ \ (1)$

A black hole’s internal energy is just its mass, so ${U=M}$, and we found an expression for its temperature earlier:

$\displaystyle T=\frac{\hbar}{8\pi k_{B}GM} \ \ \ \ \ (2)$

We can thus work out the entropy as a function of mass:

 $\displaystyle \frac{\partial S}{\partial M}$ $\displaystyle =$ $\displaystyle \frac{8\pi k_{B}G}{\hbar}M\ \ \ \ \ (3)$ $\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{4\pi k_{B}G}{\hbar}M^{2} \ \ \ \ \ (4)$

where the last line assumes that ${S=0}$ at ${M=0}$. From this, we can see that if we combine two black holes with masses ${M_{1}}$ and ${M_{2}}$, the total entropy of the system increases.

 $\displaystyle S_{tot}$ $\displaystyle =$ $\displaystyle \frac{4\pi k_{B}G}{\hbar}\left(M_{1}+M_{2}\right)^{2}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi k_{B}G}{\hbar}\left(M_{1}^{2}+M_{2}^{2}+2M_{1}M_{2}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle S_{1}+S_{2}+\frac{8\pi k_{B}G}{\hbar}M_{1}M_{2} \ \ \ \ \ (7)$

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