# Singularities in the Reissner–Nordström metric

I am doing a presentation on black holes but I'm having trouble finding information on the Reissner–Nordström metric. From the metric $$\textrm{d}s^2=\left(1-\frac{R_S}{r}+\frac{R_Q^2}{r^2}\right)c^2\textrm{d}t^2-\left(1-\frac{R_S}{r}+\frac{R_Q^2}{r^2}\right)^{-1}\textrm{d}r^2-r^2(\textrm{d}\theta^2+\sin^2\theta\textrm{d}\phi^2)$$ there seems to be three singularities, at $r=0$, $r=\frac{R_S+\sqrt{R_S^2-4R_Q^2}}{2}$ and $r=\frac{R_S-\sqrt{R_S^2-4R_Q^2}}{2}$ Is it the case as in with normal black holes that the last two singularities may be removed by a coordinate transformation or is anyone of this a trus singularity?

• You can figure out which is a true singularity by computing a curvature invariant, such as $R_{abcd} R^{abcd}$ May 3, 2015 at 22:33

The singularities are in a sense mere artefacts of your coordinate system, but do describe important surfaces.

The reason your metric was ill defined up on those surfaces is that you picked a metric that was nice when you are far far from the black hole (all those $1/r$ terms get small and it looks like the SR metric for a spherical coordinate system).

And those surfaces describe important boundaries, they are like event horizons. When you cross that outer one, you can never cross it back to get back to the outer part of ... that ... universe. And like a normal event horizon your choices become constrained. After you cross the first horizon, your r coordinate must get smaller and smaller until you cross the second horizon.

At this point things are different than for an uncharged black hole. Firstly, and this is horrific, as you cross the second horizon, you can see the singularity (and it can later see you seeing it). This is really a game ender here because the theory is crazy at this point. However, you haven't hit the singularity yet so you could say don't shoot the messenger and say that seeing the singularity isn't the same as being there, and then another option equally strange is available. If you have or acquire enough rocket fuel, you can avoid the singularity, climb back to the second horizon and which point you can see it, but it won't see you seeing it because it will die before that light gets to it (it's staying inside you are getting out). But they you can keep escaping, until you cross the first horizon again and now you finally won't see the singularity anymore.

However, there is also no guarantee that you are in the original universe. There might be many insides and outsides and you might have travelled from an earlier outside to a later outside. In a sense more than an infinite amount of time passed on your original outside while you were inside, all of the outside had it's chance to enter the first horizon by the time you crossed the second horizon the second time. That door is closed. So in the causal sense that even arbitrarily late events outside the black hole can't (might not) be affected by you means it is later. We don't know if you'd come out to a different universe. Everyone seems to assume it is a different universe. The theory is pretty crazy having a singularity you can see, so maybe you shouldn't read too much into it.

But those singularities all had a bearing on the r coordinate. The location r=0 is a true singularity. Crossing inside the outer surface means you have to have your r get smaller unless you can travel faster than light. This holds all the way up until you get to the second surface at which this is no longer true, but once you do that you can see a singularity. Maybe that's not strange if the big bang is a singularity in our past.

I'm not familiar with a name for the coordinates that work nicely. The details I recited are easy to read off the Penrose Diagram for the Reissner–Nordström metric. But you do have to learn how to read them. It is not too complicated or time consuming.