3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.