Physical significance of angular velocity of orbits around Kerr black holes

Question

For the Kerr metric $$ds^2=\left(g_{tt}-\frac{g_{t\phi}^2}{g_{\phi\phi}}\right)dt^2+g_{\phi\phi}\left(d\phi-\omega dt\right)^2+g_{rr}dr^2+g_{\theta\theta}d\theta^2$$ the angular momentum is defined as $$\Omega=\frac{u^\phi}{u^t}=-\frac{g_{t\phi}+lg_{tt}}{g_{\phi\phi}+lg_{t\phi}}$$ where $l$ is the specific angular momentum.

However, in the metric we have a term $\omega$ defined as $$\omega=\frac{\dot{\phi}}{\dot{t}}=-\frac{g_{t\phi}}{g_{\phi\phi}}$$

Observing the two expressions for $\Omega$ and $\omega$, we can see that when $l=0$, then $\Omega=\omega$. These expressions are mainly used in the study of orbits in Kerr geometry.

$\textbf{Question 1: }$ So I could not understand the difference between these two terms, $\Omega$ and $\omega$. What are their physical significance?

$\textbf{Question 2: }$ What does the specific angular momentum $l$ mean?

Answer 1

• $\Omega$ is the angular velocity of a (general) geodesic orbit, where $l$ is the ratio of the angular momentum and energy of the orbit. (Both of which are constants of motion).

• $\omega$ is the angular velocity of a Zero Angular Momentum Observer (ZAMO) orbiting the Kerr black hole. That is, it is the angular momentum of a particular type of geodesic orbit, namely those with $l=0$. Particularly, it is the limit of $\Omega$ as $l$ goes to zero. The limit of $\omega$ as $r$ goes to the outer horizon $r_{+}$ is usually interpreted as the angular velocity of the horizon.