# Horizons and other special surfaces on Kerr metric

The Kerr metric is

$$\begin{equation} ds^2 = - \big(1-\frac{2GMr}{\rho^2}\big)dt^2-\frac{2GMra}{\rho^2}\sin^2 \theta \big(dtd\phi+d\phi dt\big)+ \frac{\rho^2}{\Delta}dr^2+\rho^2 d\theta^2 + \frac{\sin^2 \theta}{\rho^2}\big[(r^2+a^2)^2-a^2\Delta\sin^2\theta\big]d\phi^2 \end{equation}$$

where

$$\begin{equation} \rho^2= r^2+a^2\sin^2\theta \\ \Delta = r^2 -2GMr + a^2 \end{equation}$$

The interesting positions are the points where $$g_{rr}\rightarrow \infty$$ and the ones where $$g_{tt}\rightarrow 0$$ since they are related with the surfaces where certain Killing vectors change from spacelike to timelike or viceversa (I know that making this kind of statement about the metric is actually coordinate-dependent but all books, like Carroll, or Misner, do this to find the horizons). The interesting radii are

$$\begin{equation} g_{rr}\rightarrow \infty \ \ \ \ \ \text{at} \ \ \ \ \ R^{(r)}_{\pm}=GM \pm \sqrt{(GM)^2-a^2}\\ g_{tt}\rightarrow 0 \ \ \ \ \ \text{at} \ \ \ \ \ R^{(t)}_{\pm}=GM \pm \sqrt{(GM)^2-a^2\cos^2\theta} \end{equation}$$

If I understand correctly, the biggest one is $$R^{(t)}_{+}$$, which is the beginning of the ergosphere, also called the stationary limit surface or the infinite redshift surface. Then we get $$R^{(r)}_{+}$$ which is the outer horizon, where the escape velocity becomes greater than c. Then we have the inner horizon $$R^{(r)}_{-}$$ where the metric goes "back to normal" in the sense that the radial component is spacelike again so you can exit the black hole.

This is the usual treatment in all the books I checked (Carroll, Wald, Misner, etc.). But no one seems to talk about $$R^{(t)}_{-}$$. What's the deal with that radius? Does it have some special property? Is it just an artifact of the coordinates we are using? Is it meaningless in some other way because it's too deep into the Black Hole?

• I would expect it to be the ergosphere of the next universe on the other side of the black hole. Have you looked at the Penrose diagram for the Kerr metric? It's pretty wild. – Javier Apr 17 '20 at 16:46
• Adding to my comment: it would help if you told us if you know what a Penrose diagram is, and what it looks like for the Kerr metric. Or in general, if you know the maximal extension of the Kerr black hole, with its infinite universes. – Javier Apr 18 '20 at 23:08
• Hi! I do know what a Penrose Diagram is and I also know how the maximal extension of the Kerr BH works. However, I didn't see any mention on the radial position I'm asking about in the literature. – P. C. Spaniel Apr 18 '20 at 23:39

## 2 Answers

Short answer: it's the inner border of the ergosphere (or ergoregion), $$R^{(t)}_+$$ being the outer border. Longer answer follows.

the radial component is spacelike again so you can exit the black hole.

Be careful with this: you can sort of exit the black hole, but not to the same place from which you came in. Let's tell the story carefully just in case. As you approach a Kerr black hole, there are a few different checkpoints:

• When you cross the ergosphere, you cannot help but rotate with the black hole. Your radial movement is unrestricted, so you can leave if you want, but you can't stay still (with respect to infinity): $$\partial_t$$ is spacelike, and you need to add some $$\partial_\phi$$ to it to make it timelike.
• Then you have the two horizons, one inside another. When you cross the outer one, $$r$$ becomes decreasing as time goes on, so eventually you will cross the inner one. This is sort of a transition region, which you can only cross in one direction. And this is important: since $$g_{tt}$$, $$g_{rr}$$ and $$g_{\theta\theta}$$ are all positive, no trajectory with constant $$\phi$$ can be timelike. That is, you still have to rotate with the black hole.
• Finally you reach the inside, where the ring-shaped singularity is. Now you can change your $$r$$ at will, so if you want you can go back out — but not the way you came in! After all, in the transition region you can only move inwards. If you cross the inner horizon again you'll be moving outwards through a white hole, eventually entering a different universe. You came out of the white hole, but if you want you can fall back in (since in the future it becomes a black hole), and repeat the whole thing as many times as you want.
• Now for the main point: when you enter the inner region, $$g_{tt}$$, $$g_{rr}$$ and $$g_{\theta\theta}$$ are still positive, so you're still rotating with the black hole. If you approach the singularity, you will eventually cross $$R^{(t)}_-$$, and finally you're free to move however you want. In more technical terms, you can become stationary with respect to the usual coordinates.

The $$r = R^{(t)}_-$$ surface is sort of an ellipsoid just like $$R^{(t)}_+$$, but tall instead of wide: it touches the inner horizon at the poles and then becomes thinner, eventually touching the singularity at $$\theta=\pi/2$$.

• "You came out of the white hole, but if you want you can fall back in (since in the future it becomes a black hole)" - you can't go in again, since after you left the white hole your τ increases while the new universes t decreases, so you would travel backwards in bookkeeper time while your proper time moves forward (assuming you survive the infinite blueshift when you leave the Cauchy horizon, since while you cross it you move all the way to infinite future and back in a small amount of proper time and all the infalling light hits you at once, but that's another story) – Gendergaga Apr 20 '20 at 1:46
• @Yukterez I don't think that's true - I also don't think it's too useful to keep using $t$ as a reasonable coordinate. If you look at a Penrose diagram, you can see that after crossing the inside of the black hole and coming out from the white hole, moving into the future can eventually lead you to another black hole. Space-wise it's in the same place as the white hole, but it leads further into the future. – Javier Apr 20 '20 at 1:57
• When you use Finkelstein-like Kerr-Schild coordinates you can see that the t blows up to infinity when you exit the inner horizon, and the more t elapses the more photons from the outside fall in and hit you. The t is a reasonable time coordinate, far away from the black hole it is the proper time of an observer, he can send photons into the black hole in t-intervals and they all hit you at once when you exit the inner horizon, that can be seen when solving the geodesics. Raindrop-like Doran coordinates also show that you get hit by infinitely many raindrops when you exit the inner horizon – Gendergaga Apr 20 '20 at 3:06
• Because of the cosmic redshift there won't be infinitely many photons and raindrops, but all the photons and raindrops that fall in until future infinity will hit you in an infinitesimal period of proper time τ nevertheless, so the blueshift is still infinite although the total energy should be finite - see the Penrose diagram at youtube.com/watch?v=ETn6aOzwmr4&t=19m46s – Gendergaga Apr 20 '20 at 3:13
• @Yukterez that's a good point, though couldn't the same be said for the horizon of a Schwarzschild black hole? But anyway, I don't need a realistic observer to survive -- I was just describing the different regions inside a Kerr black hole. – Javier Apr 20 '20 at 13:12

When $$g_{rr} \to \infty$$ or $$g^{tt} \to \infty$$ you get the horizon where there is the point of no return where even a test particle travelling with the speed of light and radially outwards can not get out.

With $$g_{tt} \to 0$$ you get the ergosphere, which is the radius where an observer which is stationary with respect to the fixed stars or the asymtotically flat background would have to locally travel with the speed of light in the retrograde direction relative to a local and frame dragged ZAMO in order to keep a constant radial coordinate.

So that is the radius below which you can no longer stay at rest relative to a far away observer because therefore you'd need a relative velocity of $$v \geq c$$ with respect to a local and corotating observer.

Below $$g_{rr} \to \infty$$ or $$g^{tt} \to \infty$$ you can no longer keep a fixed radial coordinate, and below $$g_{tt} \to 0$$ you can no longer keep a fixed angular coordinate since $$g_{t \phi} \neq 0$$ (where $$t$$ and $$\phi$$ are the time and angle observed by the far away bookkeeper, whos frame of reference is used for the Boyer Lindquist coordinates).

Another way to look at it is the time dilation, which becomes infinite for a corotating ZAMO when $$g^{tt} \to \infty$$, and infinite for a stationary observer when $$g_{tt} \to 0$$ (stationary with respect to the fixed stars, which requires a local retrograde velocity higher than $$c$$, therefore the infinite time dilation).

• "particle travelling with the speed of light and radially outwards can not get out" , I would say that: the light cone is so tilted that, unless you can catch and pass a light wavefront, there is no way to increase r and access the outside. – rrogers Jul 30 '20 at 15:04