Does the physical singularity of the Reissner-Nordstrom metric have a ring structure?

Question

The physical singularity of the Kerr metric has a ring structure due to the axi-symmetric nature of the metric.

The Reissner-Nordstrom metric is the solution for a non-spinning, electrically charged black hole, and has two horizons: an event horizon and a Cauchy surface, the locations of both depend on the black hole's mass and charge.

Question: mathematically what is the structure of the physical singularity in the Reissner-Nordstrom metric? Is it a point like the Schwarzschild case? Is it a ring like the Kerr case? Or is it different from both? And why?

I've done some digging but cannot find a concrete explanation. The answer at this SE question is helpful, since it shows that even theoretically, the existence of the Kerr singularity could be a mathematical artifact. But I'm curious about if such a mathematical feature exists for the Reissner-Nordstrom metric, and why?

Answer 1

There is no physical singularity. $r{=}0$ isn't part of the manifold. This isn't just nitpicking, because the behavior you get in the $r{\to}0$ limit doesn't really fit with the idea of the singularity as a point at the center.

In the Schwarzschild case, two geodesics that approach $r{=}0$ at the same Schwarzschild $t$ but different $θ,\phi$ end up out of causal contact at the end: they don't see each other's entire history, even though the distance between them goes to zero. (This is like the time reversal of the horizon problem in positively curved FLRW cosmologies.) Because of this, it makes more sense to think of the singularity at fixed $t$ as a sphere – a sphere of no defined size, but topologically a sphere, not a point. When you add the $t$ direction, which is spacelike inside the event horizon, the singularity as a whole is topologically a cylinder with spherical cross section.

As far as I know, the same thing happens in the Reissner-Nordström case, except that the $t$ direction is timelike, so the singularity can again be viewed as a cylinder, but with a well defined time ordering on the spherical cross sections. (And the Kerr singularity is similar, but with toroidal cross sections.)

It's likely that a more realistic model of a black hole interior would have a singularity just outside the infinite-blueshift inner horizon, due to what's called the mass inflation instability. If the singularity is indeed there, then all varieties of hairless black hole have a singularity with essentially the same spacelike cylindrical "shape" as the Schwarzschild singularity, except that you can assign a specific nonzero size to the spherical cross sections.

Answer 2

The singularity in a Reissner-Nordström geometry is at $r = 0$. With a little assistance from Mathematica I find the Kretschmann scalar to be:

$$ K = \frac{56r_q^4}{r^8} - \frac{48 r_q^2 r_s}{r^7} + \frac{12 r_s^2}{r^6} $$

and this is finite everywhere except at $r = 0$.

theoretically, the existence of the Kerr singularity could be a mathematical artifact.

Neither the RN nor the Kerr metric can represent a real black hole since they are both time independent and have existed for an infinite time in the past, which is obviously not possible in a universe $13.8$ billion years old. So we can be confident they are both a mathematical ideal not a real object. In addition it is widely believed (though I think not proved) that the geometry inside the inner horizon is unstable to perturbations. So even if a real RN or Kerr geometry could be constructed it would immediately evolve into some other geometry in some way that is not fully understood.

Answer 3

It's worth noting that the Reissner-Nordstrom singularity is timelike, like how the Kerr singularity is. (and unlike the Schwarzschild singularity, which is spacelike)

but the metric is spherically symmetric, so the singularity cannot be a ring, because that would choose a special plane in the spacetime.