# What is really going on in the ergosphere of a Kerr black hole?

Considering the Kerr metric with $GM>a$, we can compute 2 event horizons:

$r_\pm=GM\pm \sqrt{G^2M^2-a^2}$

These event horizons are null surfaces, and trajectories are timelike between $r_+$ and $r_-$. My understanding so far is that if an observer is approaching the BH and crosses the $r_+$ surface, it must keep going until it crosses $r_-$.

However, because Kerr is not stationary, these surfaces are not Killing horizons for $K=\partial_t$ and so new surfaces arise, namely the ergosurfaces.

I don't really understand what happens in the ergosphere. From the Penrose diagram I would say that nothing special actually happens, but I read that an observer cannot hover there. Also the phenomena of frame dragging was mentioned.

Can you please explain what are the consequences of having a Killing horizon (that is not an event horizon)? And what really happens to the trajectory of a particle as it crosses the Killing horizon? Namely in the ergosphere

## 1 Answer

Let's consider Kerr spacetime in Boyer-Lindquist coordinates and units where $G=M=c=1$.

The 4-velocity of any observer must be timelike, $u^\mu u^\nu g_{\mu\nu} = -1$. This immediately leads to the fact no observer can hover on or inside the ergosphere.

For a static observer (ie. one who's hovering at fixed $r$,$\theta$,$\phi$), the above normalization requires $g_{tt}(u^t)^2 = -1$. At the so-called "static limit" (the edge of the ergosphere), $g_{tt}=0$, and inside this surface $g_{tt} > 0$. Therefore, no static observers can exist here. You can think of the ergosphere as a region where you would have to move faster than the speed of light to stand still.

Inside the ergosphere, observers are forced to co-rotate with the black hole.

A useful concept for thinking about trajectories in Kerr spacetime are the orthonormal "zero angular momentum observer" (ZAMO) frames. See http://cdsads.u-strasbg.fr/abs/1972ApJ...178..347B where they are referred to as "locally nonrotating observers".

The existence of a Killing vector associated with axisymmetry allows us to define a conserved angular momentum (per unit mass) $l=u_\phi$. A ZAMO has $l=0$ and rotates around the black hole with angular velocity $u^\phi/u^t=-g_{t\phi}/g_{\phi\phi}$. This is referred to as frame dragging.

In the local orthonormal ZAMO basis, the metric takes the form $\text{diag}(-1,1,1,1)$. Let's define $\beta^\phi=u^\phi/u^t$ and $\Omega=-g_{t\phi}/g_{\phi\phi}$. A transformation from the Boyer-Lindquist coordinate basis to the ZAMO basis gives

$$\beta^{\phi^\prime}=\frac{\left(\beta^\phi-\Omega\right)\sqrt{g_{\phi\phi}}}{\alpha}$$

where $\alpha=\sqrt{-g_{tt}+\Omega^2 g_{\phi\phi}}$, and a prime on the index denotes a quantity in the ZAMO frame. From the requirement that $-1\leq\beta^{\phi^\prime}\leq 1$, we find that

$$\Omega-\alpha/\sqrt{g_{\phi\phi}}\leq \beta^\phi \leq \Omega+\alpha/\sqrt{g_{\phi\phi}}$$

In the ergosphere, $\beta^\phi\geq 0$ for all $\beta^{\phi^\prime}$. Also note that at the horizon $\alpha=0$, $\Omega\equiv\Omega_H$, and so $\beta^\phi=\Omega_H$. That is, observers (and particles) are forced to rotate with the same angular velocity as the horizon.