If you have the usual Reissner-Nordström metric of a charged black hole, is the electromagnetic potential of the black hole still:

$$A_0(r,\theta,\phi) = \frac{Q}{r}$$

in these units?


1 Answer 1


A Schwarzschild black hole has no charge and no electrostatic potential. This is the potential of a Reissner-Nordström black hole, in Gaussian units.

If you are talking about the potential of a point charge at rest outside a Schwarzschild black hole, with $r$ being some measure of distance from the charge rather than the Schwarzschild radial coordinate, then this is not its potential. The potential and electrostatic field of a charge at rest outside the hole are not spherically symmetric around the charge as in flat spacetime, because the curvature of the black hole warps the field.

The electrostatic potential of a point charge at rest outside a Schwarzschild black hole was investigated by Copson in 1928 and his solution was corrected by Linet in 1976. The solution is

$$V=\frac{q}{b r}\left[ \frac{(b-M)(r-M)-M^2\cos\theta}{\sqrt{(r-M)^2+(b-M)^2-2(b-M)(r-M)\cos\theta-M^2\sin^2\theta}}+M \right]$$

where $M$ is the mass of the black hole, $r$ and $\theta$ are Schwarzschild coordinates, $q$ is the charge of the point particle outside the horizon, and the point particle is at rest at $r=b$, $\theta=0$. This is written in units where $G=c=1$.

Here is a contour plot of the potential. The larger blue circle is the black hole. The smaller blue circle is the point particle. The white area is an artifact of the rendering, where the contour lines get too close together. The horizon is an equipotential surface.

enter image description here

The asymmetry of the field causes a gravitationally-induced electrostatic self-force on the charged particle. It is directed away from the hole and, in the frame of a freely-falling observer instantaneously at rest at the position of the charge, has magnitude


This was first calculated exactly by Smith and Will in 1980.


Interestingly, as $b\rightarrow2M$ so that the charge is just outside the horizon, the potential does become symmetrically centered on the hole, rather than on the charge, and is just $q/r$ outside and 0 inside.

  • $\begingroup$ Interesting answer. Which tells us that when a charge falls onto the event horizon then it's as if the charge is spread out over the event horizon. Thanks. $\endgroup$
    – user84158
    Feb 9, 2019 at 18:14
  • $\begingroup$ It suggests that, but idoesn’t prove it. This potential is for a charge at rest outside the hole, which would need a little rocket or a “sky hook” to keep it from falling in, except at one particular distance where the repulsive electrostatic self-force could balance the gravitational attraction. It is not the potential of a freely-falling charge. $\endgroup$
    – G. Smith
    Feb 9, 2019 at 18:25
  • $\begingroup$ Very interesting answer indeed! How is it possible that the horizon is a equipotential surface? Doesn't this imply that there is a charge inside horizon? Unless the electric field is exactly 0 on the horizon $\endgroup$
    – magma
    Feb 10, 2019 at 7:13
  • $\begingroup$ Is it possibile to see the derivation of this formula somewhere? $\endgroup$
    – magma
    Feb 10, 2019 at 7:15
  • 1
    $\begingroup$ Copson’s paper is here: royalsocietypublishing.org/doi/abs/10.1098/rspa.1928.0044 Linet’s paper is here: iopscience.iop.org/article/10.1088/0305-4470/9/7/010/pdf $\endgroup$
    – G. Smith
    Feb 10, 2019 at 17:33

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