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?
 A: The equation is derived from the relationship between the temperature of the black hole and the surface gravity:
$$ T = \frac{\kappa}{2\pi} $$
where $\kappa$ is the surface gravity. See the Wikipedia article on Hawking radiation for more on this, and in particular the section titled Emission process.
For the RN black hole this expression gives:
$$ T = \left. \frac{1}{2\pi} \frac{1}{\sqrt{g_{rr}}} \frac{d\sqrt{-g_{tt}}}{dt} \right|_{r=r_+} $$
Converting this into the expression you cite is just algebra.
A: 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. 
