Physical interpretation of the Reissner-Nordström metric

Question

The Reissner-Nordström metric (in spherical coordinates and $c = 1$) differs from the Schwarzschild metric in a additive term $$\frac{GQ^2}{4\pi\epsilon_0 r^2},$$ so the metric is

$$ds^2 = \left(1 - \frac{2GM}{r} + \frac{GQ^2}{4\pi\epsilon_0 r^2}\right)dt^2 - \left(1 - \frac{2GM}{r} + \frac{GQ^2}{4\pi\epsilon_0 r^2}\right)^{-1} dr^2 - r^2d\Omega^2.$$

This has an effect of "repulsion".

Why adding more energy (by the addition of charge to the black hole) has this repulsive effect? What is the physical meaning of adding charge to a black hole?

Answer 1

In the Schwarzschild metric all the mass is at the centre of the black hole. So the stress-energy tensor is zero everywhere except at the singularity (where it's undefined).

In the Reissner-Nordström metric we still have the mass at the centre, but we also have the electrostatic field that exists both inside and outside the event horizon. This makes an additional contribution to the stress-energy tensor that is not present in the Schwarzschild geometry.

Suppose you consider a sphere of radius $r_-$ and ask how much of the electrostatic field is inside the sphere and how much is outside. Obviously at $r_- = 0$ all the field, and all the energy in it, is outside the sphere, and if you don't mind a rather loose metaphor you can think of the part of the field outside the sphere as pulling the surface of the sphere pulling outwards. This creates a inner horizon that grows outwards from $r = 0$ as you increase the energy in the electrostatic field. The radius of the inner horizon is at the value of $r$ where the field energy inside and outside the horizon balance out. This happens at the radius:

$$ r_- = \tfrac{1}{2}\left(r_s - \sqrt{r_s^2 - 4r_Q^2}\right) $$

where $r_s$ is the Schwarzschild radius and $r_Q^2 = Q^2G/4\pi\varepsilon_0 c^4$.

For the outer horizon we get an analogous but opposite effect. The outer radius starts at the usual Scharzschild event horizon radius of $r = 2M$. When we introduce an electrostatic field some of the field is inside the event horizon and some is outside. Because the field falls with $r$ it turns out that the net effect is that the field inside the horizon dominates and pulls the horizon inwards to a radius:

$$ r_+ = \tfrac{1}{2}\left(r_s + \sqrt{r_s^2 - 4r_Q^2}\right) $$

When $r_s^2 = 4r_Q^2$ the two horizons merge and disappear leaving a naked singularity. However this is thought to be unphysical and the charge can never get that high.

One last comment while I'm here: the Schwarzschild geometry does not literally have a mass $M$ at the centre. The point $r = 0$ is singular and the geometry is not defined there. The mass is actually zero everywhere the Schwarzschild metric applies. However we can associate a mass called the ADM mass with the geometry, and the parameter $M$ is actually this ADM mass. The same applies to the charge $Q$. This is not literally the charge at the singularity, it the charge associated with the electrostatic field.