Why domain of Kerr black hole includes negative values for $r$ coordinate?

Question

I understand the domain of $t$ is all real numbers but mathematically, how to prove the domain of $r$ coordinate is also all real numbers except $r=0$ when $\theta = \pi/2$. I know that we get two disks when $\theta < \pi/2$ and $\theta > \pi/2$ for $r=0$ but could we use this result and say $r<0$ as well? Secondly these two regions are timelike or spacelike? And if they are timelike can we go from one disk to other disk easily because there is a ring at the center which is curvature singularity?

Answer 1

The range and topology of coordinates can often be simply chosen when interpreting the metric in question. This is a crucial point when talking about the analytical extensions of Kerr.

For example, you can take the Kerr metric, transform to Cartesian-like coordinates by setting $$x=r\cos\phi \sin \theta,\;y=r\sin\phi \sin \theta,\;z=r\cos \theta$$ Then you can identify $z=r_+ +l$ with $z=-(r_+ +l)$ with some $l>0$ and $r_+$ the horizon radius (effectively leaving out the $z \in (-r_+ -l,r_+ +l)$ region out of the space-time). As a result, you get a space-time without a black hole and a metric jump at the identification. This jump can be interpreted as generated by a razor-thin disk of finite total mass and angular momentum equal to $M,a$ of the Kerr metric. (Source: Bičák, J., & Ledvinka, T. (1993). Relativistic disks as sources of the Kerr metric. Physical review letters, 71(11), 1669.)

So you see, you can have the absolutely identical metric locally but, if you modify the topology of the coordinates, you get no black hole, no curvature singularity, no extra universes etc., just an extended thin rotating disk. This kind of freedom is ubiquitous in relativity.

Now for the central disk of Kerr. In Schwarzschild we say that $r\in (0,\infty)$ necessarily because there is a full-blown curvature singularity at $r=0$, no matter from which direction you approach. In Kerr this is not the case. In particular, by transforming to Kerr-Schild coordinates, one sees that Boyer-Lindquist $r=const.$ surfaces are oblate ellipsoids, and the $r=0$ surface corresponds to the disk at the origin in oblate spheroidal coordinates: Another way to show that $r=0$ is not a point is to observe the fact that physical scalars such as the Kretschmann scalar are dependent on $\theta$ at $r=0$.

Now you can say that the space-time ends at $r=0$ and the two sides of the disk are identified, this is allowed. Geodesics will then be possible to extend smoothly through the disk, and there will be a jump in the metric when passing through the disk. The jump in the metric can be viewed as sourced by a razor-thin negative-mass density on the surface of the disk, but I guess black holes are weird, so ok.

But you can also try and avoid this negative mass density by saying that the two sides of the disk are not identified. Then you see that you can extend through $r=0$ to negative values and keep the metric smooth. This leads you to the "antigravity universe" at $r<0$, closed time-like curves and all that. This is nicely discussed by Misner, Thorne & Wheeler's Gravitation if you want to see details.

Nevertheless, the modern point of view is that none of this is particularly relevant, since the inner horizon of Kerr is a Cauchy horizon. In essence this means that real black holes will be approximated well by the Kerr metric up to the neighborhood of the inner horizon. At the inner horizon real black holes will be crumpled and weird, and nothing like the Kerr metric. (See Dafermos & Luk, 2017 for more.)