Kerr holes

Most of the astrophysical objects rotate, so we expect most black holes formed by gravitational collapse to be rotating. The Kerr black hole is rotating, and it is axially symmetric but not spherically symmetric. This solution of Einstein's equations, discovered in 1963 by R. Kerr (1963), was not at first recognized to be a black hole solution.

The most general stationary black hole metric, with parameters mass $M$, angular momentum $J$, and electric charge $Q$, is called the Kerr-Newman metric. A Kerr hole is the special case when $Q=0$. Black holes in nature are expected to have $Q=0$, since matter on a macroscopic scale is neutral. Written in the $(t,r,\theta ,\phi )$ coordinates of Boyer and Lindquist (1967), which is a generalization of the Schwarzschild coordinates, the Kerr metric has the form:

$$ds^2= - \left(1-\frac{r_gr}{\rho^2}\right)c^2dt^2- \frac{2acr_gr\sin^2\theta}{\rho^2}dtd\phi + \frac{\rho^2}{\Delta}dr^2 +$$

$$+ \rho ^2 d\theta ^2 + \left(r^2+a^2+\frac{r_gra^2\sin^2 \theta }{\rho^2}\right)\sin^2\theta d\phi^2.$$ (27)

$M$ is the mass of the black hole, and $a$ is its specific angular momentum divided by $c$. The black hole is rotating in the $\phi$ direction, and

$$a\equiv \frac{J}{Mc}, \quad \Delta \equiv r^2-r_gr+a^2, \quad \rho^2 \equiv r^2+a^2\cos^2\theta.$$ (28)

In textbooks one often use the so-called geometrical units, a system of units in which $c=G=1$. In these units, the Schwarzschild radius is $r_g=2M$. $M$ is thus measured in cm. Here, I will not use the geometrical unit system, but the CGS system.

There is an off-diagonal term in the metric, in contrast to the Schwarzschild metric:

$$g_{t\phi}=\frac{acr_gr\sin^2\theta}{\rho^2}.$$ (29)

When setting $a=0$ in equation (27) we get the Schwarzschild metric.

The component $g_{tt}$ in the metric (27), which determines the rate of flow of time, vanishes at:

$$\rho^2-r_gr=r^2+a^2\cos^2\theta -r_gr=0,$$

or at $r=r_1$, where $r_1$ is given by

$$r_1=\frac{r_g}{2}+\sqrt{ \left( \frac{r_g}{2} \right)^2-a^2\cos^2\theta }.$$ (30)

The boundary $r_1(\theta)$ is called the static limit (see figure 5). No material bodies can be at rest within the surface $r=r_1 (\theta )$. The reason for this is the same as in the Schwarzschild field at $r=r_g$. Namely, the world line of the observer ceases to be time-like, as indicated by the reversal of sign of $g_{tt}$ at $r<r_1$. But an essential difference in comparison with the Schwarzschild field must be emphasized, because there are some additional features caused by the rotation. As a massive object spins, it induces a rotation in the surrounding space-time, a phenomenon known as frame dragging. If $r<r_1$, all bodies are unavoidably dragged into rotation, although $r_1$ is not the event horizon because a body may escape from this region. In the metric (27), the event horizon lies at $\Delta=0$, that is, at $r=r_+$, where

$$r_+=\frac{r_g}{2}+\sqrt{ \left( \frac{r_g}{2} \right)^2-a^2 }.$$ (31)

For a rotating black hole, the Boyer-Lindquist coordinates are singular at the horizon. As in the Schwarzschild case, it requires an infinite time $t$ (measured by an observer at rest far away from the black hole), for any particle or photon to fall through the event horizon. The dragging of inertial frames forces particles and photons near the horizon to orbit the black hole with $d\phi /dt>0$. Consequently, for a particle falling through the horizon we have that $t\to \infty$ and also $\phi\to \infty$, an infinite twisting of worldlines around the horizon.

Figure: Near a rotating black hole, frame dragging is so severe that there is a nonspherical region outside the event horizon called the ergosphere where any particle must move in the same direction that the black hole rotates. The outer boundary of the ergosphere is called the static limit, so named because once inside this boundary a particle cannot possibly remain at rest there (i.e., be static) relative to the distant stars.

The region between the event horizon $r_+$ and the static limit $r_1$ is called the ergosphere. Space-time within the ergosphere is rotating so rapidly that a particle would have to travel faster than the speed of light to remain at the same angular coordinate (e.g. at the same value of $\phi$ in the coordinate system used by a distant observer). When $a=0$, the static limit and the event horizon coincide, there is no dragging of inertial frames; there is no ergosphere.

There is an upper limit of $a$ for a black hole to form, namely

$$\left( \frac{r_g}{2} \right)^2\ge Q^2+a^2.$$

If the black hole spins so fast that this constraint is violated, the centrifugal forces, will exceed the gravitational forces and halt the collapse. For a Kerr hole where $Q=0$, this implies that the spin of the black hole must be $0\le a\le r_g/2$. A black hole with $a=r_g/2$ is called an extreme Kerr hole. The event horizon for an extreme Kerr hole is situated at $r_+= r_g/2$.