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I just read in Landau-Lifshitz that the Kerr metric admits closed timelike curves in the region $r \in (0, r_{hor})$ where $r_{hor}$ is the event-horizon ( I am talking about the case $|M|>|a|$ (subextremal case) here ). Now, unfortunately they don't give an example of such a curve. Could anybody of you write down explicitly such a CTC so that I could go through the computation once by myself. I would really like to see this once.

If anything is unclear, please let me know.

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    $\begingroup$ Googling "Kerr metric CTC" gives me as the first hit a PDF that contains an explicit example for a CTC. Please make an effort to find the answer to questions yourself before asking here. $\endgroup$ – ACuriousMind Mar 24 '16 at 11:29
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    $\begingroup$ @ACuriousMind they do it for $r<0$ (that's why I specified precisely the case that I am interested in) $\endgroup$ – Johnny Mar 24 '16 at 11:32
  • $\begingroup$ @Horus In the paper that ACruiousMind is implicitly quoting, they consider maximal extensions of the Kerr metric and there they also consider an extension that admits $r<0$ and construct a $CTC$ for that case. $\endgroup$ – Johnny Mar 24 '16 at 12:23
  • $\begingroup$ See also physics.stackexchange.com/questions/189239/… $\endgroup$ – Void Dec 11 '18 at 16:08
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See section 3.19 of Black Holes: An Introduction By Derek J. Raine, Edwin George Thomas

https://books.google.ca/books?id=O3puAMw5U3UC&pg=PA103&lpg=PA103&dq=kerr+schild+closed+timelike&source=bl&ots=elnzJu2ySm&sig=B4cWXIkib4fqbs0D7yA2YlZKE8A&hl=en&sa=X&redir_esc=y#v=onepage&q=kerr%20schild%20closed%20timelike&f=false

That book has an example: Boyer Lindquist coordinates: take an orbit where only phi is changing, then the proper time on that orbit is given by (a formula in the book) then we set r = just inside the ring singularity, and one gets a timelike dt > 0 path that is periodic.

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Let me write down the metric in the equatorial plane ($\vartheta = \pi/2$) of the Kerr space-time in Boyer-Lindquist coordinates: $$ d s^2 = -\left(1 - \frac{2M r}{r^2 + a^2}\right) d t^2 + \frac{r^2+a^2}{r^2 - 2Mr + a^2} d r^2 + \left(r^2 + a^2 + \frac{2M r a^2 }{r^2 + a^2}\right) d \varphi^2 - \frac{2 M r a}{r^2 + a^2} dt d \varphi $$ Now you have to trust me that going through the ring singularity at $r=0$ (remember that $r,\vartheta$ are in fact oblate ellipsoidal coordinates and $r=0$ is a disk) means going to $r$ negative. Now I would like to know if there is an $r$ such that the vector $\eta^\mu = \delta^\mu_\varphi $ is time-like. I find that $$\eta^\mu \eta_\mu = g_{\varphi \varphi} = r^2 + a^2 + \frac{2M r a^2 }{r^2 + a^2}$$ which is negative in a certain range of negative $r$ (the range has a closed and cumbersome expression corresponding to two roots of a quartic equation). Since the integral curves of $\eta^\mu$ are closed, in that range of $r$, there are closed time-like curves. When you go off the equator and plot regions where $g_{\varphi\varphi}$ is negative, you find this region is a finite "doughnut" near the ring singularity in the $r<0$ part of the Kerr space-time.

The issue is, however, that if you start with your time-like curve in this doughnut, you can also leave it, go through the singularity back to $r>0$, and all the way to the inner horizon of the Kerr black hole. Then you return back to the $g_{\varphi\varphi}<0$ doughnut, and you are allowed to circle in $\varphi$ indefinitely with a small negative drift in $t$, and eventually end up closing your time-like curve (meet yourself from your proper past). So closed time-like curves are implied by the existence of the "acausal doughnut" all the way to the inner horizon of the Kerr black hole. This is also one of the reasons one often discards the region within the inner horizon as unphysical.

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