I was doing some basic algebra to estimate the range of possible masses $M$ and electric charge $Q$ for a nearly extremal Reissner-Noström black hole. I want to see if the logic is correct

the electric field near the event horizon of such a black hole would be

$$ E= \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0} \frac{q c^4}{G^2 M^2} = 10^{66} \frac{\text{Volt Kg}^2}{\text{m Coul}} \frac{q}{M^2}$$

since Schwinger-pair vacuum breakdown occurs at $10^{18} \frac{\text{Volt}}{\text{m}}$, it implies that

$$ \frac{q}{M^2} \lt 10^{-48} \frac{\text{Coul}}{\text{Kg}^{2}}$$

Would be a bound that guarantees electromagnetic vacuum stability

Question: can a charged black hole that is stable against this breakdown still be nearly extremal?

well, it seems so, because the extremality condition:

$$ r_Q = \frac{r_s}{2}$$

Implies that

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

$$ \frac{Q}{M} \approx 10^{-16} \frac{Coul}{Kg}$$

So the above two relations would imply that a black hole with a mass

$$ M \gt 10^{32} Kg \approx 100 \text{ Solar masses}$$

Can be both stable (against the specific Schwinger-pair decay) and extremal.

In particular, if the mass is $10^{32}$ kilograms, a charge around $10^{16}$ Coulomb would make the black hole arbitrarily close to extremal as one might want, and is not at all clear that it would decay from the extremality state

But probably i'm overlooking other ways than the extremality of the black hole can decay (in particular, i'm ignorant about how Hawking radiation will contribute in here, specially if the temperature is below 511 KeV). Any insights into this would be greatly appreciated

  • $\begingroup$ I'm not an expert on this, so unable to give an answer but this presentation and the paper seem to discuss the relative rates of the Hawking and Schwinger mechanisms in the cases you're interested in. $\endgroup$ – twistor59 Dec 19 '12 at 7:35
  • $\begingroup$ Note that the probability of Schwinger pair production is exponentially suppressed if the electric field is weak (meaning $E < m^2/e$), but not zero. It is always energetically possible for an extremal black hole to decay via Schwinger pair production (provided the 'electron' in your universe has $e > m$). $\endgroup$ – gj255 Apr 9 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.