Does a charged particle respond to a charged black hole both gravitationally and electromagnetically?

Question

The Reissner–Nordström metric is given by:

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

Which is intuitive even though the derivation is ridiculously hard. The metric is similar to the Schwarzschild metric except with the addition of an extra term containing charge.

Now obviously, a charged black hole contains more energy, so since it's the energy that curves spacetime, it curves spacetime more than a typical black hole. This means that any particle would experience a stronger gravitational pull, whether charged or neutral. However, there is still Maxwell's equations into play, which affects charged particles. Does that mean a charged black hole not only has a gravitational effect on charged particles, but also an electromagnetic effect?

Answer 1

The charge itself is gravitationally repulsive with 1/r³, while the attraction from the mass goes with 1/r², so the closer you get the smaller the net gravitational attraction compared to an uncharged black hole with the same mass (see the Reissner Nordström repulsion).

Additionally a charged test particle will also be subject to a force which goes with 1/r², so depending if his charge has the same or opposite sign to the black hole the force will be attractive or repulsive.

For the equations of motion in standard coordinates see here, and if you want to go below the horizon here.

Answer 2

Yes. The geodesic equation for a particle of mass $m$ and charge $q$ is given by: $$\ddot{x}^\rho +\Gamma_{\mu\nu}^\rho\ddot{x}^\mu\ddot{x}^\nu =\frac{q}{m}F_{\;\;\sigma}^\rho\ddot{x}^\sigma.$$ The charge $Q$ of the Reissner-Nordstrom black hole appears in the Christoffel symbol $\Gamma_{\mu\nu}^\rho$ as well as the electromagnetic tensor $F_{\;\;\sigma}^\rho$, but the latter only affects the geodesic of the particle, when $q\neq 0$.

An electromagnetic field has energy, which is equivalent to a mass, which causes gravity. The solution is therefore not a vaccum, but an electrovacuum solution. This does indeed affect non-charged particles.

What is also interesting about this contribution to gravity, is, that it is repulsive as seen by the different signs in front of the terms $-r_\mathrm{S}/r$ and $+r_\mathrm{Q}^2/r^2$ in the Reissner-Nordstrom metric. Unfortunately, we will never experience this repulsive gravity caused by charge. For an electron, the radius, where attractive gravity changes to repulsive gravity is exactly the classical electron radius. (You can derive it with $\frac{\mathrm{d}}{\mathrm{d}r}g_{tt}\stackrel{!}{=}0$.) For a charged black hole of mass $M$ and charge $Q$, we have $|Q|\leq M$ as there would be a naked singularity otherwise. The radius, where attractive gravity changes to repulsive gravity is therefore inside the black hole.

Answer 3

The Reissner-Nordström metric is a solution to both Einstein equation and Maxwell equation in vacuum (that metric is called "electro-vac solution"). And yes, a charged particle around that black hole would feel an electric force (attractive or repulsive), in addition to the gravitational pull.

By the way, you could also add the effect of the vacuum cosmological constant $\Lambda$, which gives you the Reissner-Nordström-deSitter metric: $$\tag{1} ds^2 = \mathcal{G}(r) \, dt^2 - \frac{1}{\mathcal{G} \smash{(r)}} \: dr^2 - r^2 \, d\Omega^2, $$ where $$\tag{2} \mathcal{G}(r) = 1 - \frac{2 G M}{r} + \frac{G \alpha Q^2}{r^2} - \frac{\Lambda}{3} \: r^2. $$ On a side note, I suggest that you use units (or scale all quantities) such that $c \equiv 1$. Relativity is a bit simpler when we get rid of the pesky $c$ that litters every equation of physics!