Thermodynamics from the total energy of a Black hole and its temperature

Question

• Given the blackhole temperature formula, $$T =\frac{\hbar c^3}{8\pi k_BGM},$$ can I use $M=E/c^2$ (where $M$ is the mass of the Blackhole) to invert the above equation to express it as $$E=E(T)$$ and use the formula $C_V=\left(\frac{\partial E}{\partial T}\right)_{V,P}$ to calculate the specific heat of the Blackhole? This little piece of calculation leads to a negative specific heat $$C_V=-\frac{\hbar c^5}{8\pi Gk_BT^2}.$$

• Will this also be true for a rotating black hole and a charged black hole because for both of which I suspect $E\neq Mc^2$?

Answer 1

The first part is perfectly fine. Indeed, the specific heat for a black hole is negative, which is in fact usual for self-gravitating systems, such as stars. See, for example, Explanation for negative specific heat capacities in stars? and its answers.

For charged and rotating black holes the changes are a bit more subtle. You still have $E = Mc^2$, but the black hole's temperature is no longer given by the expression you wrote down. The general expression for the black hole temperature is $$T = \frac{\kappa\hbar}{2 \pi k_B c},$$ where $\kappa$ is known as the black hole's surface gravity. For a Schwarzschild black hole, it is given by $$\kappa = \frac{c^4}{4M G}$$ if I didn't miss any units. For a charged, rotating black hole (known as a Kerr—Newman black hole) it is given by (R. M. Wald's General Relativity, Eq. (12.5.4)) $$\kappa = \frac{\sqrt{M^2 - a^2 - Q^2}}{2M \left(M + \sqrt{M^2 - a^2 - Q^2}\right) - Q^2},$$ where $Q$ is the black hole's charge and $J = a \cdot M$ is its angular momentum. In the previous expression, units with $\hbar = c = G = k_b = 4\pi \epsilon_0 = 1$ are employed (see Planck units).