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I am confused about the Boyer Lindquist System. $r = 0$ at the ringularity, and outside that $r$ should be positive (intuitively). But this paper on closed CTCs in Kerr Spacetime claims that $r < 0$ just outside the ringularity.

https://dx.doi.org/10.13140/RG.2.2.19468.31365

Can someone please explain to me where r is negative and where it’s positive in Boyer-Lingquist System? A diagram will help a lot.

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The Kerr metric has a singularity at $r=0, \theta=\pi/2$, in standard Boyer-Lindquist coordinates. Because the singularity manifests only at $\theta=\pi/2$, it divides the $r=0$ hypersurface in two disjoint parts (which I will refer to as the upper and lower hemisphere). One can therefore ask what would to an observer following a geodesic that approaches the upper hemisphere. Since the metric is not singular on the upper hemisphere, it should be possible to extend this geodesic beyond the upper hemisphere. So where does it go?

A naive guess might be that the geodesic will exit form the antipodal point on the lower hemisphere. However, if one tries to make this identification between the hemispheres, one ends up with a metric that is not smooth (and not a vacuum solution) on the interface between the two hemispheres. This can therefore not be the correct solution.

Correctly analytically extending the metric across the upper-hemisphere instead leads us to a new region of spacetime, which is described by the Kerr metric in Boyer-Lindquist coordinates with $r<0$. This region is curious in various ways, e.g. it has an asymptotically flat limit at $r\to-\infty$, but there are no event horizons shielding the singularity from it. Analytically extending the metric across the lower hemisphere leads us into a similar region. One can arrange it so that this is the same region as reach from the upper hemisphere, but there is no mathematically compelling reason to make this particular choice, and it could simply be a different region with the same geometry.

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