# is this generalized Hawking radiation formula right?

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

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

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A disturbance in the force tells me that your original equation should have a factor of two on the brackets in the denominator. Your intuition problem, however, might be a problem with handling the fraction as $\xi$ limits to zero. If I look at your original equation, the temperature would go to zero. To get to your final state you would have to divide both sides of the fraction by zero... angering the math gods. quickmeme.com/meme/36gxmd –  AlanSE Dec 19 '12 at 16:52
Schwarzschild radius is actually $2GM/c^2$ but i relabeled $r_s$ without the factor 2, but that is ok since extremality is supposed to happen at $2 r_Q= r_s$, so the relabeled condition looks like $r_Q = r_s$, at which $\xi$ should tend to zero. –  lurscher Dec 19 '12 at 16:56
sorry, maybe i misunderstood where exactly the factor of two is missing? In any case, i don't see any factor of 2 in the denominator in the linked page. I've been looking a separate source of the temperature expression for checking but i can't find any. –  lurscher Dec 19 '12 at 16:59
@AlanSE it tends to zero for me too, which contradicts what is said later on that page (that the temperature should tend to infinity at extremality) which is precisely the reason that leds me to think that, maybe the whole expression is wrong? –  lurscher Dec 19 '12 at 17:01
There's a derivation of the temp for Kerr-Newman in this thesis on page 36. –  twistor59 Dec 19 '12 at 21:14

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.

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.

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