Throwing an electron to a Black hole

Question

if a black hole with Schwarzschild metric absorb a single electron, does the metric suddenly change to Reissner–Nordström metric?

Answer 1

The Reissner-Nordström geometry is not totally different to the Schwarzschild geometry. The Reissner-Nordström metric can be written as:

$$ ds^2=-c^2\left(1-\frac{r_s}{r}+\frac{r_q^2}{r^2}\right)dt^2 + \left(1-\frac{r_s}{r}+\frac{r_q^2}{r^2}\right)^{-1}dr^2 + r^2d\Omega^2 $$

where:

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

If we start with a charged black hole and gradually reduce the charge then $r_q \to 0$ and the Reissner-Nordström geometry becomes gradually more and more similar to the Schwarzschild geometry:

$$ ds^2=-c^2\left(1-\frac{r_s}{r}\right)dt^2 + \left(1-\frac{r_s}{r}\right)^{-1}dr^2 + r^2d\Omega^2 $$

until in the limit of zero charge they are identical.

So conversely if we start with an uncharged black hole and add an infinitesimally small charge then while the geometry is Reissner-Nordström it would be indistinguishable from Schwarzschild.

Charge is quantised of course, so we cannot add an infinitesimally small charge - the smallest charge we can add is $\pm e$. Nevertheless, if we started with an uncharged Solar mass black hole and added one electron the resulting geometry, while technically Reissner-Nordström, would in practise be indistinguishable from the Schwarzschild geometry.