Sorry if this question gets a little long; I want to explain why I'm asking it.

The usual Schwarzchild metric

$$ds^2 = -\left(1-\frac{2M}{r}\right) dt^2 + \left(1-\frac{2M}{r}\right)^{-1} dr^2 + r^2(d\theta^2+\sin^2\theta\ d\phi^2)$$

makes sense when $t,r$ lie in the ranges $-\infty<t<\infty$, $0<r<\infty$. We can change to Kruskal–Szekeres coordinates $u,v$ (which Wikipedia calls $X$ and $T$), which are nonsingular at the horizon. The corresponding range for them is $u+v>0$. But if we draw spacetime on the $u,v$ plane, we can see that there is no problem extending it to all possible values of the coordinates. This adds a symmetric half to spacetime, with a past singularity that behaves like a white hole: any object or light ray will eventually exit the horizon.

When I learned this in General Relativity class, the professor said that the usual physical reasoning for doing this is that the good old Schwarzschild metric is geodesically incomplete: if we imagine a particle falling into the black hole and try to trace back its path into the past, it looks like (as long as it doesn't have too much energy) it should have come up from the black hole, stopped, and then proceeded to fall in. I wasn't too convinced of this for two reasons: One, the black hole hasn't existed for all eternity so whatever is falling into it can have its origin somewhere else. Two, we haven't actually observed any white holes.

This remained a mathematical curiosity until we analyzed a Penrose diagram of the Reissner-Nordstrom metric for a charged black hole. This spacetime has two horizons and a singularity inside. But now there are timelike curves that end at some point in the future without hitting a singularity. To me this seems like a much bigger deal, since I can perfectly well imagine something falling into a charged black hole as a physically realistic situation.

Kerr metric Penrose diagram taken from Wikimedia

The extension in this case is something much weirder: an infinte chain of universes. You can enter a black hole here and come out at some other universe, and proceed to do the same until you get tired and settle down on some planet on whatever universe you happen to be on. This is as far from physically realistic as it gets, and yet it seems unavoidable if we want to have a charged black hole (I think the same thing happens for a rotating black hole too).

Let me state my question, then: is the incompleteness of timelike curves in a charged black hole a real thing? Is it a problem? Is the infinite tower of universes the only way to make the problem go away, and if so, wouldn't that imply that it exists in the universe, since charged and rotating black holes do exist?

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    $\begingroup$ Nothing that predicts unmeasurable things in physics is convincing. As pretty as Penrose diagrams are, they don't trump the definition of science. If you can't measure it, it's not science. $\endgroup$
    – CuriousOne
    Jun 23, 2015 at 0:25
  • $\begingroup$ @CuriousOne: I don't disagree, but I'm not sure how that applies here. What I'm asking about is a definite prediction of GR. Saying "you can't know because you can't measure it" seems to me to be a bit of a copout. After all, why do we spend so much time studying the interior of a Schwarzschild black hole, for example? $\endgroup$
    – Javier
    Jun 23, 2015 at 1:01
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    $\begingroup$ A note - he Schwarzschild black hole has existed forever. It is an eternal black hole. $\endgroup$
    – Prahar
    Jun 23, 2015 at 1:09
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    $\begingroup$ "Knowing" has a certain connotation to it which requires that evidence for a statement exist. This is not the case here. We don't know what happens below the event horizon of black holes (technically we don't even know, as of today, what happens close to the event horizon) and past the singularity that's just a rabbit's hole. Knowledge it certainly ain't. To set you straight: nobody has ever studied the interior of black holes, Penrose et al. have been studying the ways Einstein's equations break down for decades, though, without any progress into actual physics. $\endgroup$
    – CuriousOne
    Jun 23, 2015 at 1:11
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    $\begingroup$ @CuriousOne It's certainly important to keep in mind that, until experimental verification, all theoretical physics is just physically motivated mathematics. I disagree, however, with the position that these motivated calculations "aren't physics" in some meaningful sense; the vast majority of modern, concrete and experimentally verified quantum physics was originally motivated by these mathematical considerations, particularly quantum field theories. The esoteric nature of the thoughts involved creates a potential for new experiments, specifically designed to verify the predictions they make. $\endgroup$
    – Alec Rhea
    Nov 8, 2017 at 0:32

3 Answers 3


Even if such a solution existed, it would be unstable because of infinite blue shift at the inner horizon of any perturbation that you introduce to the RN Black hole solution. Since the space between inner horizon and outer horizon is finite compared to the past null infinity where the space in infinite. An observer in between these horizons can collect radiation in a finite time of a radiation that has infinite space time volume at past null infinity. This will cause an infinite blue shift at the inner horizon causing a singulaity there. This was shown by Poisson and Israel in 1990. So, even if such connections to other patches existed, they would close very quickly. It is quite a graphical explaination, which is why I am being very concise here. You can find Poisson and Israel's paper at http://journals.aps.org/prd/abstract/10.1103/PhysRevD.41.1796

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    $\begingroup$ You have inner and outer horizons even in standard Schwarzschild coordinates in the RN spacetime, though. This answers a different question than the OP. $\endgroup$ Feb 8, 2018 at 14:38
  • $\begingroup$ True. I was responding to OP's observation about the Penrose diagram of an RN blackholes being 'realistic' i.e. something that exists in the real world. In retrospect, perhaps this should have been only a comment. $\endgroup$ Feb 9, 2018 at 16:26

This is the unique way of completing the coordinate system of the reissner-nordstrom spacetime. There are obvious physical issues with it, but if you go far enough into the past, the density of big bang matter becomes large enough that it is important relative to the background geometry, and the solution is no longer a good approximation of what is going on physically.

For an intuition about what happens when the matter distribution in the past "cuts off" the past horizon, see the Vaidya metric.


There seem to be a couple of misunderstandings involved in the question.


the usual physical reasoning for doing this is that the good old Schwarzschild metric is geodesically incomplete

A spacetime can be geodesically incomplete and inextensible, or it can be geodesically incomplete but the incompleteness can be removed by an analytic extension.


The spacetime of an astrophysical black hole does not look the same as, say, the maximal extension of the Schwarzschild metric. In the former case, there is a strong real-world justification for extending the metric smoothly through the coordinate singularity at the horizon. In the latter case, we're just talking about a spacetime that happens to have nice mathematical characteristics, but that can't form by gravitational collapse.


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