Free-falling from rest into a Kerr black hole Is it impossible for a particle (with zero angular momentum) to free-fall from rest at infinity into the ergosphere of a Kerr black hole? It seems like it is very easy to show this is the case, but most of what I've read seems to skirt over this.
 A: From this paper$^1$ we have the equations for a particle of zero total energy on an infalling trajectory in the equatorial plane:
$$\begin{align}
 \Sigma\frac{d\theta}{d\tau} &= 0 \\
 \Sigma\frac{dr}{d\tau} &= -\sqrt{2Mr(r^2 + a^2)} \\
 \Sigma\frac{dt}{d\tau} &= -a^2\sin^2\theta + \frac{(r^2 + a^2)^2}{r^2-2Mr+a^2} \\
 \Sigma\frac{d\phi}{d\tau} &= -a \left( 1 - \frac{r^2 + a^2}{r^2-2Mr+a^2} \right) \\
 \Sigma &= r^2 + a^2\cos^2\theta \\
 a &= \frac{J}{M}
\end{align}$$
If you're happy to assume that the proper time remains finite, then to find out if the coordinate time becomes infinite anywhere we simply ask if $dt/d\tau$ becomes infinite anywhere. If we use the simplification that the trajectory is equatorial ($\theta = \pi/2$) then $dt/d\tau$ simplifies to:
$$  r^2\frac{dt}{d\tau} = -a^2 + \frac{(r^2 + a^2)^2}{r^2-2Mr+a^2} $$
and this becomes infinite when:
$$ r^2-2Mr+a^2 = 0 $$
which has the two solutions:
$$ r = M \pm \sqrt{M^2 - a^2} $$
or in a more familiar form:
$$ r = \frac{r_s \pm \sqrt{r_s^2 - 4a^2}}{2} $$
But the ergosphere is at $r = r_s$ in the equatorial plane, so $dt/d\tau$ is not infinite at the ergosphere and this implies the infalling particle will reach the ergosphere in finite coordinate time.

$^1$ there's nothing special about this paper, it was just the first paper I found when doing a Google search.
