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

  • 1
    $\begingroup$ Possible duplicate of What happens when an electric charge crosses the event horizon? $\endgroup$
    – A.V.S.
    Dec 25 '20 at 11:26
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    $\begingroup$ Because an electron has spin, the metric becomes Kerr-Newman, not Reissner-Nordstrom. The singularity must change from an infinitely long spacelike line of Schwarzschild to a finite timelike ring of Kerr-Newman. $\endgroup$
    – safesphere
    Dec 25 '20 at 14:35
  • $\begingroup$ @safesphere Thank you for reminding. You are right, I forget consider the spin. $\endgroup$ Dec 28 '20 at 1:57

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

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


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

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.

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    $\begingroup$ Also maybe worth noting that while we are bringing a charge towards the black hole, the geometry wouldn't be purely Schwarzschild either, due to the presence of the charge. So the whole process would involve a smooth, not sudden, change. $\endgroup$
    – kaylimekay
    Dec 25 '20 at 11:58
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    $\begingroup$ The OP said "electron", not just charge. The election has spin, so the metric becomes Kerr-Newman, not Reissner-Nordstrom. $\endgroup$
    – safesphere
    Dec 25 '20 at 14:39

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