is this generalized Hawking radiation formula right?

Question

Look at equation 11.2.17 in this page. The expression is:

$$ T = 10^{-5} \text{K m} \frac{\xi}{\frac{GM}{c^2} \lbrace \frac{GM}{c^2} + \xi \rbrace - e^2 }$$

where

$$ \xi = (r_s^2 - a^2 - e^2)^{1/2}$$

and the usual parameters

$$ r_s = \frac{GM}{c^2} $$

$$ a^2 = \frac{L^2}{M^2 c^2}$$

$$ e^2 = \frac{Q^2 G}{4 \pi \epsilon_0 c^4}$$

This formula is supposed to describe the temperature of a black hole with angular momentum $L$, charge $Q$ and mass $M$

Question: is the above formula correct?

I'm trying to find the limit temperature for $a=0$ and $e=r_s$

The temperature expression could be simplified as

$$ T = 10^{-5} \text{K m} \frac{\xi}{\xi^2 + a^2 + \frac{GM}{c^2} \xi }$$

if $a=0$,

$$ T = 10^{-5} \text{K m} \frac{1}{\xi + \frac{GM}{c^2} } $$

so when the black hole has extremal charge, $\xi=0$ and the temperature looks like the normal black hole temperature for the Schwarzschild black hole, which looks very wrong

Any idea where is the mistake? I was expecting the temperature of the charged extremal black hole with $a=0$ to be infinite

Answer 1

In a previous question, Stan Liou posted an answer that had these equations:

At what rate does a rotating black hole lose mass via Hawking Radiation?

$$r_\pm = \frac{G}{c^2}\left[M\pm\sqrt{M^2-\frac{1}{4\pi\epsilon_0G}Q^2-\frac{c^2}{G^2}\frac{J^2}{M^2}}\right]$$

$$\kappa = c^2\frac{r_+-r_-}{2(r_+^2+a^2)},$$

$$T = \frac{\hbar}{c k_B}\frac{\kappa}{2\pi}.$$

Now, begin my work.

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I will use the same notation in the question here, for $r_s$, $a^2$, and $e^2$. With these, we can rewrite the above equations.

$$ r_{\pm} = r_s \pm \sqrt{ r_s^2 - e^2 - a^2 } = r_s \pm \xi$$

$$ \kappa = c^2 \frac{ r_{+} - r_- }{2 (r_+^2+a^2) } = c^2 \frac{ 2 \xi }{ 2 ( (r_s+\xi)^2 + a^2 ) } $$

Reduce the equation for kappa further. This is just algebra, substituting in xi once.

$$ \kappa = c^2 \frac{ \xi }{ 2 r_s (r_s+\xi) - e^2 } $$

Plug into temperature, group constants out in front.

$$ T = \left( \frac{ \hbar c }{ 2 \pi k_B } \right) \frac{ \xi }{ 2 r_s (r_s+\xi) - e^2 }$$

One problem is that for the group of constants I get $3.6 \times 10^{-4} m K$. Another is that this doesn't match the first equation you posted, it's off by a 2. Lastly, as we go $e=r_s$, temperature goes to zero, not infinity like you wanted.

So in short, not only did I not fix your problem, I created even more problems.