The Penrose diagram for a non-spinning Schwarzschild black hole is enter image description here

Notably, there is a second universe "on the other side" of the black hole. However, actual black holes form by stellar collapse, and the collapse process leads to a different Penrose diagram:

enter image description here

A spinning Kerr black hole famously has the following crazy Penrose diagram:

enter image description here

The Kerr spacetime has two horizons, in natural units $r_\pm= M \pm \sqrt{M^2 - a^2}$ where the angular momentum is $J = aM$. $r_+$ is the regular event horizon. Inside of that is $r_-$, where you can begin navigating to leave the black hole and enter an entirely new universe. (You can also pass through the ring singularity, or the "ringularity", and enter an "antiverse.")

My question is, what is the Penrose diagram for a realistic black hole that forms via stellar collapse, and not an idealized "eternal" Kerr black hole? What happens to the parallel universes, $r_-$, and the antiverses? Do any of these features remain?

  • 2
    $\begingroup$ Great question, looking forward to seeing answers. $\endgroup$
    – user4552
    Apr 20, 2019 at 20:08
  • $\begingroup$ We have two answers that both seem plausible, but they also seem contradictory. One shows a timelike singularity, the other a spacelike singularity. $\endgroup$
    – user4552
    Apr 26, 2019 at 2:58
  • $\begingroup$ If, from our point of view, time stops at the event horizon, how is it meaningful to say that a black ole is spinning? $\endgroup$
    – R.W. Bird
    Sep 19, 2019 at 17:43
  • $\begingroup$ Let's say that we were far away from the black hole watching someone fall in. I believe what would happen is that, from our faraway perspective, we would see the infalling person's time slow down as they approach the horizon and get "pasted" right above the horizon, asymptotically approaching it. However, I think we would still see their pasted image would still be spinning with some angular velocity. $\endgroup$ Sep 19, 2019 at 18:03
  • 1
    $\begingroup$ Hey, can you tell me a good source that shows how the penrose diagram is created for a Kerr black hole? $\endgroup$
    – jaydesai10
    May 8, 2020 at 21:43

3 Answers 3


Your question basically boils down to a recognition of the following fact:

  • The Schwarzschild metric, with spacelike $r=0$, admits an "eternal" BH to form by stellar collapse, like the one you've drawn above. It forms, then maintains a permanent static state forever, but doesn't evaporate or absorb anything else.

  • BH metrics with a timelike $r=0$, like Kerr (rotating), Reissner-Nordstrom (charged), and Hayward (nonsingular) do not admit a reasonable "eternal" solution, since the region beyond the inner horizon leads to strange and unwanted parts of spacetime.

This means that for rotating BHs (as well as charged and nonsingular BHs), if we want a full Penrose diagram, we need to confront a difficult question:

  • What happens to the BH after if forms?

Most people agree that the BH evaporates by emitting Hawking radiation, but no one agrees on the correct semi-classical spacetime to model this process (google "evaporating black hole spacetime", e.g.). Moreover, not everyone even agrees that Hawking evaporation is the dominant process (popular alternatives include, e.g., remnants, quantum bounce models, mass inflation instability).

The good news about this is that we should have had to confront this anyway: if BHs evaporate the eternal diagram you drew above is wrong (at the top) anyway.

The bad news is that we can't say what the correct diagram is, we can only postulate some ideas about what a reasonable diagram might be.

Here is one toy-model possibility, assuming that the BH forms from a collapsing star, then evaporates by emitting an outgoing burst of Hawking radiation from just outside the trapping horizon, while absorbing a burst of negative mass. An evaporation model with ingoing/outgoing fluxes of negative/positive energy like this is motivated by the DFU stress tensor for Hawking radiation.

If this was for a charged nonrotating BH, which has a similar causal structure, this would be pretty satisfying. Then we could say:

  • The collapsing star surface cuts off unwanted past regions.
  • The evaporation process cuts off unwanted future regions.

Unfortunately, for the rotating BH, this only represents the $\theta=0$ axis, so there are problems:

  • Not clear what other $\theta$ values look like in diagram.
  • By going to other $\theta$ values, you can still go through the ring singularity and get to unwanted regions.

It is not obvious how these issues associated with rotation should be resolved, or even that the interior metric has to be exactly Kerr, since:

  • We don't have a simple exact metric for a rotating star collapsing to a BH. Only numerical studies.
  • The metric is only asymptotically Kerr around a rotating body, so the metric near $r=0$ might not be known.
  • See this Kerr Spacetime Introduction and the Living Reviews in Relativity for discussion of these issues.

What's more, regardless of whether this BH is spinning or charged, it has some weird bad properties:

  • There's a naked singularity.
  • You can fall in past all the horizons and escape without anything terrible happening.

That certainly doesn't seem right.

One way to try to resolve all of these issues at once is by assuming that instead of a singularity, BHs have an extremely tiny, extremely dense core. The assumption is that classical GR holds until densities and curvatures reach the Planck scale, at which point Quantum Gravity takes over the dynamics. It might sound like this violates the singularity theorems, but it doesn't: the energy conditions required for singularity theorems to hold are already violated by the Hawking radiation, and definitely can't be assumed to hold a priori in quantum gravity. This point of view is not widely accepted, however, although personally I think it should be.

That assumption results in the theory of nonsingular (or "regular") BHs, see e.g. nonrotating and rotating cases. The rotating variety still have some technical issues, but if some good rotating nonsingular metric does exist, then the diagram would:

  • Look basically like the diagram above, except that near $r=0$ in the Kerr metric would not be a vacuum, but would rather be an extremely dense core of matter, extending into the region between $r_{\pm}$.
  • The ring singularity is replaced by a dense matter blob (if rotating fast probably a pancake shape), no more issue of going through the ring. No more unwanted regions.
  • Anyone who falls through outer horizon falls into core and gets turned into quantum gravity soup before being emitted in the Hawking radiation.
  • No more singularity = no more naked singularity.

Like I said, this is conjecture, since the right metric for this hasn't been discovered, as far as I know.

So that's my point of view of what this diagram probably should look like, but, as I've pointed out, there may be many others. My assumptions were that Hawking evaporation dominates the late time dynamics, and that the ingoing/outgoing radiation is a reasonable semiclassical model for the evaporation process.

If anyone has an alternative reasonably self-consistent diagram for this process, I think it would be very interesting to compare. Trying to sort out these "astrophysically relevant" BH scenarios seems like a good way to weed out some of the BH nonsense...

  • $\begingroup$ Calling anything involving Hawking radiation "astrophysically relevant" is rather curious. The Hawking temperature of any astrophysical black hole is many orders of magnitude smaller than the temperature of the CMB, meaning they do not evaporate. (At least not until the universe has cooled significantly.) $\endgroup$
    – TimRias
    Apr 24, 2019 at 6:11
  • $\begingroup$ Can't argue with you there... though I did have the decency to put it in "quotes", so, have mercy. The "relevant" part is really the formation, not evaporation. Doesn't change the fact that how you sort out the issues with the diagram depends on what you assume for the end state. If you question the evaporation, I have no problem with chopping the top off Kerr and putting a question mark, but that's not much fun. And trying to model the rotating/charged evaporation highlights that some of the aspects of the usual spherical Hawking radiation diagram are questionable. $\endgroup$ Apr 24, 2019 at 6:24
  • $\begingroup$ How come it looks like all the Hawking radiation was emitted at a single event? $\endgroup$ Apr 25, 2019 at 13:40
  • $\begingroup$ It's drawn with all the Hawking radiation emitted in a single "burst", just because that's simpler. It would be more realistic for the evaporation to happen gradually, but drawing it right gets a bit tricky. It doesn't affect the overall causal structure (i.e. the overall shape of the diagram), so the simplification doesn't have any effects relevant to this question. $\endgroup$ Apr 25, 2019 at 17:34

The Penrose diagram for a spinning star will in all likelihood look remarkably similar to the one you show for a spherical (Schwarzschild) collapse:

spherical (Schwarzschild) collapse

However, many of the features are more subtle (and subject of ongoing debate). I'll discuss the main regions in order.

Inside the star - This part is easy. Assuming the matter distribution in the star is regular, the metric inside the star is regular. For the Penrose diagram this means that this region will look like a patch of the trivial Minkowski space diagram.

Outside the star and outside the outer horizon - It is very hard (if not impossible) to find a matter distribution for a collapsing star that leads to exactly the Kerr metric for the external solution. However we do know a couple of things about this solution: 1) It is asymptotically flat (fixing the Penrose diagram at future/past null infinity) and 2) after the star passes the outer event horizon the outer solution will rapidly (exponentially) approach a stationary Kerr solution. (This last bit follows from the black hole "no hair" theorems.

The inner horizon - The inner horizon in the Kerr solution is a so-called Cauchy horizon, which means that as you pass through it you are exposed to the entire history of the universe outside the black hole. This also means that all signals reaching you will be infinitly blue-shifted. This has led to the conjecture that the inner-horizon in the Kerr solution is actually unstable with an outside perturbation leading to its collapse. This conjecture is closely related to the strong cosmic censorship conjecture, which says that timelike singularities (like in the interior region of Kerr) should not be able to form in nature.

The exact status of this conjecture and the exact nature of the resulting singularity is subject of ongoing debate. The current status seems to be that for generic perturbations the curvature will diverge at the inner horizon. However, if you allow metrics which are not twice differentiable (not $C^2$) you can still extend the spacetime beyond this singularity. (see Dafermos's website for a nice discussion of the status). In physical terms, I believe this can be summarized roughly as: general relativity must breakdown at the inner horizon, but a theory of quantum gravity may not.

Put together, these three statements lead to a Penrose diagram that is qualitatively similar to that of a collapsing spherical (Schwarzschild) spacetime.

  • $\begingroup$ Good answer. Not every mathematical solution is real +1 $\endgroup$
    – safesphere
    Jun 16, 2021 at 14:49
  • $\begingroup$ I think what is interesting is that in quantum gravity, when you can have a holographic duality between a bulk and a CFT on the boundary (say AdS/CFT), if you have two independent CFTs, the corresponding bulk duals must also be independent from the No Transmission Principle. I find it curious as to how this applies when one says that there is the possibility that quantum gravity resolves the singularity. Perhaps there is an algebraic aspect to consider? But even then I am wary of this. $\endgroup$
    – VaibhavK
    Sep 21, 2023 at 4:25

Since there are some good responses, let me put my two cents on a particular point you mentioned, with the inner Cauchy horizon $\mathcal{CH}_{I}$ and strong cosmic censorship.

In an "ideal" case, one would expect this to hold for some reasons, one of them being that in the Christodoulou formulation a generic initial data problem would essentially result in an unstable decay of the horizon. SCC holds. However, this is usually not easy, since in a "realistic" model, you can throw in all sorts of conditions that could keep $\mathcal{CH}_{I}$ stable. In semiclassical gravity, this is an issue. However, Marija Tomasevic and Roberto Emparan had a nice paper that showed that this could be preserved in nonlinear orders. This is in the perturbative picture, but what about quantum gravity? If one goes a step further and tries to make sense of this in a holographic theory of quantum gravity. Suppose we're in AdS/CFT, and wanted to work out the stability problem of $\mathcal{CH}_{I}$. Maybe one can do a brute-force analysis (which is usually more deterministic; see Oscar Dias' CMSA talk), but on a fundamental level, if the CFTs are independent, the bulk duals must also be independent to preserve the No Transmission principle; and therefore, in principle, SCC should hold. This doesn't mean SCC is proved, but one can only conjecture. Maybe there are some algebraic discrepancies to consider (Netta Engelhardt and Hong Liu will be having a paper on algebraic ER=EPR soon, as she talked about at the Strings 2023 meeting), but nonetheless by looking at the simplest Penrose diagram of the BTZ black hole, immediately this is the answer that resonates. Add in other factors, but the basis of the conjecture is still the same. Of course, one could conjecture that in classical limit SCC could be violated, but it could be preserved in semiclassical or quantum gravity (I had written an essay for GRF on this arXiv:2304.01292), but I don't see how this is an exact result.

What I am trying to say is, if one does find the precise Penrose diagram accounting for all sorts of things (add in mass, charge, maybe different geometry, etc.), some of the fundamental questions will still remain somewhat unresolved. The Penrose diagram in itself (i.e. the infinities and overall outlook) will remain the same, so the infinities and causal structure would remain the same. However, there is usually some discontinuity between how Penrose diagrams look in semiclassical gravity and in classical gravity. Not to say that the Penrose diagrams used in semiclassical calculations are "classical", since there is no dependence on the exact regime you are working in. However, as discussed in arXiv:2309.08116, there are regions where semiclassical analysis fails. So, the bottom line is, we simply start from the ideal cases where the Penrose diagrams correspond to things that are much simpler to take into account, and then find out more numerical extensions to other sorts of cases.


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