Horizons and other special surfaces on Kerr metric

Question

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?

Answer 1

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 $\theta=\pi/2$.

Answer 2

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).