The region beyond the ring singularity in the maximal Kerr spacetime is described as having closed timeline curves. Why and/or how is the question.enter image description here

Now if you look a Kruskal-Szkeres Diagram (or a Penrose Diagram as above) you can see that the Kerr singularity (right) is timelike but the Schwarzschild singularity is spacelike.

Inside the Schwarzschild event horizon curves with constant longitude, latitude, and areal radius are actually spacelike so areal radius is actually a time direction. So you could claim that since r is the timelike direction there are curves that start and stop at the same t (since t is a spacelike direction) but I've never seen anyone claim there are CTC inside the event horizon of a Schwarzschild solution.

And even if we interpreted it like that, the region in Kerr where r (not areal r in Kerr, but the usual r for Kerr) is timelike is the region between the two horizons. And the lines making an X to the right of our universe are the outer horizon (see Penrose diagram) whereas the lines making an X to the left of the right most singularity is the inner horizon so the region between where r is timelike isn't connected to the singularity except in its infinite past (where we won't go).

So over where the singularity is, the singularity is a vertical line and is r=0 so r looks pretty spacelike there. We can avoid the singularity since that vertical line is r=0 which includes the whole disk that has the ring as its edge.

So we can get to the region the diagram labels as the weird space. And people usually just cite Hawking and Ellis for the existance of closed timelike curves instead of working it out but on page 164 the existence of closed timelike curves is asserted, but then it seems like it is just a discussion of the ergoregion and the two horizons, but I don't see any more mention of closed timelike curves until the section on Gödel's solution which is a different solution, not the Kerr solution.

So I'd like to know why and/or how there are closed timelike curves in the negative r region of the Kerr solution. And if someone knows why people cite Hawking and Ellis for that fact that would be interesting too.

  • $\begingroup$ I am sure you are aware that none of the objects you are categorizing here is actually physically realized? They are all just different border cases of how general relativity breaks and we don't have the slightest idea of how to repair it correctly. Just in case you weren't aware, I think you may want to think about that a little before you go off to create a new theory that mathematically models the breakdown of another... rather than the actual behavior of nature. $\endgroup$
    – CuriousOne
    Jun 13, 2015 at 5:40
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    $\begingroup$ I don't see anything wrong with this question. The Kerr metric is a solution to Einstein's equations and it seems to me perfectly proper to ask about the properties of the solution. To downvote or VTC this question on the grounds that the Kerr metric doesn't reflect reality is making the (rather arrogant) assumption that you know what reality is. $\endgroup$ Jun 13, 2015 at 6:19
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    $\begingroup$ @Timaeus: the Penrose diagrams show only the $u$ and $v$ coordinates and ignore the angular coordinates. The CTCs (one class of them at least) require moving in a ring in the equatorial plane, i.e. constant $u$ and $v$ at $\theta = \pi/s$, so you can't plot them on the diagram you show. On your diagram they would be represented by a single point. $\endgroup$ Jun 13, 2015 at 6:53
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    $\begingroup$ @Timaeus: It is at the bottom of page 162. $\endgroup$
    – MBN
    Jun 13, 2015 at 7:31
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    $\begingroup$ @CuriousOne: The question is very clear, where are the CTC in the Kerr solution and why people cite Hawking and Ellis. The theory doesn't break down, it gives a logically consistent answer. I don't see what you rant is all about! $\endgroup$
    – MBN
    Jun 13, 2015 at 8:53

3 Answers 3


I"m pretty sure that this discussion does appear in Hawking and Ellis, though I admit that it's been a while since I looked. It's not done through a Penrose diagram, though.

The argument really comes down to the fact that for sufficiently small $r$, $d\phi$ is timelike. But, by construction, the orbits of $\phi$ are closed curves. When $d\phi$ is spacelike outside the horizon, this just generates the axisymmetry of the Kerr solution. But, for these small values of $r$, it becomes timelike, and therefore, these orbits represent closed timelike curves. You avoid intersection with the horizon so long as your value of $\theta$ doesn't put you on the same plane as the ring singularity, so these curves are not incomplete geodesics or anything like that.

You can't see this on any of those penrose diagrams, because they all suppress $\theta$ and $\phi$

  • $\begingroup$ Oh yes, the $\phi$ orbit is really the most elegant way to do this. It should be just noted that $d \phi$ can be timelike only for $r<0$ and some $\theta$ reasonably around $\pi/2$. $\endgroup$
    – Void
    Jun 13, 2015 at 17:24
  • $\begingroup$ It is at the bottom of page 162. $\endgroup$
    – MBN
    Jun 13, 2015 at 19:32
  • $\begingroup$ The claim that it is only in the case of negative r that dϕ can be timelike and CTCs are present is true only for the simple Kerr spacetime. In the Kerr-Newman spacetime, they are already realized in certain positive r ranges. @Void $\endgroup$ Nov 18, 2022 at 23:57

This is really one google search away, see e.g. page 26 (marked 64) here.

As already noted by John Rennie, Penrose diagrams are not suited for the analysis of Kerr CTCs because they show a $\phi = const., \theta=\pi/2$ slice of the global structure. The $r<0$ region is however accessible only through $\theta \neq \pi/2$. The Boyer-Lindquist coordinates actually misrepresent the central singularity but you can see the singularity "unwrapped" locally by understanding Boyer-Lindquist as oblate spheroidal coordinates.

The $r<0$ region can be essentially covered by the $r>0$ Kerr metric with $M \to -M$. Here you find cases where the $g_{tt}>0$ and you can thus choose a time-like four-velocity to point in the negative time direction with respect to $r \to \infty$. It is obvious that curves that spend some time in this region and then go "outside" towards $r \to \infty$ can be CTCs.

The Gödel solution is so often cited in this context because it is the historically first solution in which this rather unsatisfactory possibility of relativity was shown and discussed.

  • $\begingroup$ I think your answer's the best and upvoted it, but, re your 3rd paragraph, wouldn't it be more correct to say that CTCs, being timelike, occupy some space in the Kerr BH (where they're "closed") and extend somewhere else as well? $\endgroup$
    – Edouard
    Oct 9, 2018 at 16:48
  • $\begingroup$ @Edouard CTCs means "closed time-like curves". In the usual sense of the word, you cannot close a curve and then extend it somewhere else. $\endgroup$
    – Void
    Oct 10, 2018 at 7:39
  • $\begingroup$ I'd meant "closed" as an adjective, not verbally. I'm picturing the universe as past- and future-eternal, with that balancing (required, for past eternality, by the BGV Theorem) between expansion and contraction being maintained by changes in the scale of space and time, and resulting in an impossibility of distinguishing between them from within any of its iterations, that would be temporal but might be represented as concentric within each other. Like Kerr, I'm considering each rotating BH's inner event horizon to be a singularity. $\endgroup$
    – Edouard
    Oct 12, 2018 at 16:32

Timaeus : I can't answer your question, but would like to comment. However what I want to say is to big for a comment, so I'm using the answer facility. Apologies in advance, feel free to downvote.

The region beyond the ring singularity in the maximal Kerr spacetime is described as having closed timelike curves.

Let's imagine we're out in space, at a safe distance from a non-spinning black hole. What's the speed of light at the event horizon? Zero. From where we're standing the the "coordinate" speed of light is zero. (See John Rennie saying that here). Now let's say our black hole spins at half the speed of light. But half of zero is zero. So it isn't spinning. Or it's spinning infinitely faster than light. Something's not right. Something's not right with the closed timelike curve too. See this page of Palle Yourgrau's book A World Without Time: The Forgotten Legacy of Gödel and Einstein:

"Wheeler, unfortunately has conflated a temporal circle with a cycle, precisely missing the force of Gödel's conclusion that the possibility of closed, future-directed, timelike curves, ie time travel, proves that space-time is a space, not a time in the intuitive sense. Whereas a circle is a figure in space, a cycle is a journey undertaken along a circular path, one that can repeated, in Wheeler's words, "over and over again". Exactly how many times, one wants to ask Wheeler, is the journey supposed to be repeated? The question clearly cannot be answered, since the time traveller's journey is not over time, along the closed timelike curve: it is the curve itself".

The moot point is that you don't travel along a worldline. You travel through space over time. The worldline is a static representation of this. So a CTC doesn't represent time "travel". So what does it represent? Groundhog Day? No. If your CTC worldline is 24 hours long it's more like Mayfly Day. Your life is 24 hours long and it's causeless, you hatch from your own egg or something. Much as I love all those science fiction movies, I'm afraid time travel is a fantasy.

Let's move on to the Penrose diagrams, wherein "space is uni-directional within the horizon, just as time is uni-directional outside the horizon)". Whoa, hang on a minute. Einstein said a gravitational field is a place where "the speed of light is spatially variable". And if the speed of light is zero at the event horizon, how can it go slower than that? And since nothing can go faster than light, how can an object ever pass the horizon? And how can we say it will inevitably hit the singularity if this only occurs at future infinity? And where has this unidirectional space come from? A gravitational field is not a place where space is moving inwards, we do not live in some Chicken-Little world. A gravitational field is a place where space is "neither homogeneous nor isotropic". So what's all this about connecting two separate universes? Where does all this Schwarzschild wormhole stuff come from, wherein "particles from the interior white hole region can escape into either universe"? From Einstein's 1935 paper with Rosen? No way. Einstein refers to a singularity at r=2M, at the event horizon. That's where the gravitational field ends, because light can't go slower than stopped. That's the boundary of the space. And this paper is The Particle Problem in General Relativity. It's about particles like the electron, and how they can't be point-particle singularites. It isn't about wormholes to another universe. How did we get to an anti-gravity universe, and to three other universes? Methinks what we have here contradicts Einstein whilst appealing to his authority, and that it isn't just time travel that's a fantasy.


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