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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?

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2 Answers 2

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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.

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  • $\begingroup$ Would $T = \frac{1}{4\pi} f'(r_+)$ hold true for a Schwarzschild black hole in $AdS_{5}$ where $f(r)=-g_{00}=1-\frac{2m}{r^2}+\frac{r^2}{b^2}$? $\endgroup$
    – user280325
    Feb 22, 2016 at 12:30
  • $\begingroup$ @user280325 Yes. I think it is still true that $T \propto f'(r_+)$ though the overall constant might depend on the dimension of the spacetime. $\endgroup$
    – Prahar
    Aug 26, 2016 at 22:20
  • $\begingroup$ @user280325 - Actually, I just checked this and we still have $T = \frac{1}{4\pi} f'(r_+)$. $\endgroup$
    – Prahar
    Aug 26, 2016 at 22:23
  • $\begingroup$ I’m trying to calculate $\kappa$ but what is $\xi$? Is it just $(1, 0,0,0)$? $\endgroup$
    – Y2H
    Nov 17, 2018 at 12:16
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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.

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    $\begingroup$ The second equation should be corrected to being a derivative along the radial direction as the metric is stationary. $\endgroup$
    – Dhuality
    Mar 24, 2023 at 20:30

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