I was studying Killing horizons in General Relativity and I met some difficulties on the following point.

In a stationary, axisymmetric and asymptotically flat black hole, we know that there exist two Killing vectors, which are $K=\partial_t$ and $\tilde{K}=\partial_\phi$. The Killing vector field $\xi$ associated with a Killing event horizon of this black hole is a combination of the two Killing vectors: \begin{equation} \xi=\partial_t + \Omega_H \partial_\phi \,. \end{equation} The author of the lecture notes now says that $\Omega_H$ can be interpreted as the angular velocity of the black hole, in the sense that any test body dropped into it, as it approaches the horizon $r_+$, ends up circumnavigating it at this angular velocity \begin{equation} \frac{\mathrm{d}\phi}{\mathrm{d}t} \bigg |_{r \to r_+} = \Omega_H \,. \end{equation} My question is: how can I get this last statement in a rigorous way, starting from some known equations?


1 Answer 1


As a warm-up, consider the non-rotating case $\Omega_H=0$. From the perspective of a distant observer, any pointlike test object that falls into a non-rotating black hole will appear to freeze at the point where its worldline intersects the horizon. The important concept here is that this is true for any timelike worldline, whether or not it represents free-fall.

For a rotating black hole, a similar thing is true, but now the point where the object freezes (from the perspective of a distant observer) rotates around the hole with angular velocity $\Omega_H$.

To derive this, consider motion in the equatorial plane for simplicity. In Boyer-Lindquist coordinates $t,\phi,r$, the metric outside a Kerr black hole is implicit in this equation for the test object's proper time $\tau$: $$ d\tau^2 = \frac{\Delta}{r^2}(dt-a\,d\phi)^2 -\frac{1}{r^2}\big((r^2+a^2)d\phi-a\,dt\big)^2 -\frac{r^2}{\Delta}dr^2 \tag{1} $$ with $$ \Delta\equiv r^2-2Mr+a^2 \tag{2} $$ where $M$ and $a$ are constants characterizing the black hole. This is equation 12.3.1 in ref 1, specialized to motion in the equatorial plane. The event horizon corresponds to $\Delta = 0$. Multiply both sides of equation (1) by $\Delta/dt^2$ and then add the negative terms to both sides to get $$ \Delta\,\dot\tau^2 +\frac{\Delta}{r^2}\big((r^2+a^2)\dot\phi-a\big)^2 +r^2\,\dot r^2 = \frac{\Delta^2}{r^2}(1-a \dot\phi)^2 \tag{3} $$ with $$ \dot\phi\equiv\frac{d\phi}{dt} \hspace{2cm} \dot r\equiv\frac{dr}{dt}. \tag{4} $$ As the object approaches the event horizon ($\Delta=0$), the right-hand side of (3) goes to zero, so the left-hand side must also go to zero. The right-hand side goes to zero like $\Delta^2$, and since all terms on the left-hand side are positive, each one of them must go to zero as fast as $\Delta^2$ does as the object approaches the horizon. The second term on the left-hand side only has one factor of $\Delta$, so the other factor $((r^2+a^2)\dot\phi-a)^2$ must go to zero as fast as $\Delta$ does: $$ \dot\phi\to\frac{a}{r^2+a^2}, \tag{5} $$ where the $r$ on the right-hand side is evaluated on the horizon. According to equation 12.3.19 in ref 1, the right-hand side of (5) is equal to $\Omega_H$ on the horizon. This completes the proof.

For simplicity, I showed the equations for motion in the equatorial plane, but the generalization to arbitrary timelike motion is straightforward. Just replace equation (1) with the full Kerr metric and then follow the same steps.


  1. Wald (1984), General Relativity
  • 1
    $\begingroup$ A nice answer with a very good analysis. $\endgroup$
    – SG8
    Jun 29, 2021 at 19:59

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