CTCs are found in the region where $r < 0$ . That should be just inside the ring singularity, since in Boyer Lindquist coordinate system $r = 0$ means ring singularity. Does that mean this image showing CTC outside the ringularity is wrong, or I am missing something?


  • $\begingroup$ I am hazarding a guess that CTC = Closed Timelike Curves? $\endgroup$
    – RC_23
    Commented Nov 9, 2022 at 5:32
  • $\begingroup$ yes, that's what a CTC is. $\endgroup$
    – Nayeem1
    Commented Nov 15, 2022 at 16:10

1 Answer 1


You are missing something. In a text where $r < 0$ is possible, the singular ring is a curve, not a surface: its equation in B-L coordinates is $(r = 0, \cos \theta = 0)$.

In contrast, $r = 0$ is the equation of a surface: it consists of two disks $(r = 0, \cos \theta > 0)$ and $(r = 0, \cos \theta < 0)$ together with the singular ring, their common boundary. The 2 disks are made of regular points: they are actually 2 manholes through which you can reach into negative space.

The picture is ambiguous: if the text from which you got it claims the CTC is in negative space (not sure, the pic is clipped), then it lies below the region $r = 0$ rather than just outside.

  • $\begingroup$ Just to add that the planar disk inside the ring singularity is actually a discontinuity in the metric components. This is shown in H Ohanian’s book. $\endgroup$
    – Cham
    Commented Dec 25, 2021 at 17:57
  • $\begingroup$ @Cham in such an interpretation, $r < 0$ is impossible. $\endgroup$ Commented Dec 25, 2021 at 18:14

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