thermodynamics - Derivation for the temperature of Reissner-Nordström (charged) black hole

A lot of the text for this is from "How does one correctly interpret the behavior of the heat capacity of a charged black hole?" but this concerns a different question. The Reissner-Nordström black hole solution is: $$ds^2=-(1-\frac{2M}{r}+\frac{Q^2}{r^2})dt^2+(1-\frac{2M}{r}+\frac{Q^2}{r^2})^{-1}dr^2 +r^2d\Omega_{2}^2$$

Let us define $f(r)\equiv (1-\frac{2M}{r}+\frac{Q^2}{r^2})$. Clearly, the solutions to $f(r)=0$ are $r_{\pm}=M\pm \sqrt{M^2-Q^2}$, and these represent the two horizons of the charged black hole. If we are considering a point near $r_+$, we can rewrite $f(r)$ as follows: $$f(r_+)\sim \frac{(r_+ -r_-)(r-r_+)}{r_{+}^2}$$

What I don't understand is how can derive the temperature of the black hole from this relation. The temperature is given by $$T=\frac{r_+-r_-}{4\pi r_+^2}=\frac{1}{2\pi }\frac{\sqrt{M^2-Q^2}}{(M+\sqrt{M^2-Q^2})^2}$$

I couldn't find a reasonable answer as to how we can obtain the temperature from $f(r_+)$. What are the steps and reasoning that are missing when making this jump?

Answer

The temperature of a black hole is related to its surface gravity. For stationary black holes, the surface gravity is given by $$\kappa^2 = - \frac{1}{2} D^a \xi^b D_a \xi_b \big|_{r=r_+}$$ where $\xi = \partial_t$ is the time-like Killing vector of the space-time and $r=r_+$ is the horizon. For space-times that have metric of the form $$ds^2 = - f(r) dt^2 + \frac{dr^2}{f(r)} + r^2 d\Omega^2$$ This quantity works out to be $$\kappa = \frac{1}{2} f'(r_+)$$ Then the temperature of the black hole is given by $$T = \frac{\kappa}{2\pi} = \frac{1}{4\pi} f'(r_+)$$ For the RN black hole $$f(r) = \frac{1}{r^2} ( r - r_+ )( r - r_- ) \implies f'(r_+) = \frac{r_+-r_-}{r_+^2}$$ which reproduces your formula.