I say "initially retrograde" because am aware that while inside the ergosphere, everything necessarily has prograde motion. So the two options I can envision are:

  1. The object approaches the black hole in retrograde orbit, but from the perspective of an outside observer, it reverses direction as it approaches the ergosphere, travels in the prograde direction within the ergosphere (but more slowly than an object which fell in with 0 angular momentum) and then reverses direction again as it exits the ergosphere, preserving its original angular momentum; or
  2. There is no such orbit. All orbits which are retrograde outside the ergosphere either do not enter the ergosphere, or enter the event horizon and thus do not emerge.

I am more interested in the case of an open orbit (i.e., a flyby) but I would also love to hear about the possibility of a closed orbit.


1 Answer 1


The prograde and retrograde circular orbit velocities relative to a local ZAMO are

$$\rm v^{+}=\frac{a^2-2 a \sqrt{r}+r^2}{\sqrt{a^2+(r-2) \ r} \left(a+r^{3/2}\right)} \ , \ \ v^{-}=\frac{a^2+2 a \sqrt{r}+r^2}{\sqrt{a^2+(r-2) \ r} \left(a-r^{3/2}\right)}$$

while the equatorial ergosphere is always at $\rm r_{E}=2$ and the horizon at $\rm r_{H}=1+\sqrt{1-a^2}$, so the condition is $\rm |v|\leq1$ in the region between the two. This is only possible for prograde orbits, see below for three examples with different spin parameter $\rm a$:

prograde and retrograde orbital velocities around a spinning black hole

The $\rm r$ coordinate is on the horizontal axis and the local velocity $\rm v$ on the vertical one, prograde is cyan and retrograde magenta. The first vertical gridline is the horizon and the second one the ergosphere. The units are natural ones with $\text{G=M=c=1}$, for a comparison with different spin parameters see here.

As you can see the last possible retrograde circular orbit for a maximally spinning black hole with $\text{a=1}$ is at $\text{r=4}$, which is twice the $\rm r$ as the ergosphere. In the prograde direction the last possible orbit with half the speed of light relative to a ZAMO is at the horizon itself. For nonspinning black holes with $\text{a=0}$ the last possible orbit in all directions is at the photon sphere at $\text{r=3}$.

If we don't restrict ourselves to circular orbits it is of course possible to move retrograde relative to a local ZAMO inside the ergosphere, but without radial propulsion you won't be able to keep a constant $\rm r$ that way, and relative to a far away observer you will still move in the prograde direction, meaning the $\rm d\phi/dt$ with $\rm t$ as the far away coordinate bookkeeper's time will still be positive, although than less than the ZAMO's.

  • $\begingroup$ The simulation video shows a test particle entering from a retrograde orbit and switching directing from an external observers perspective; however it appears to be in a trajectory which asymptotically approaches the event horizon and will never exit the ergosphere. My question was specifically whether there are trajectories which enter from a retrograde orbit, but then exit. You indicated that this is true, but the rest of your discussion is either about circular orbits (which do not exit) or the behavior of a particle inside the ergosphere, so I don't consider this a compete answer yet. $\endgroup$
    – brendan
    Dec 24, 2023 at 11:03
  • $\begingroup$ @brendan wrote: "trajectories which enter from a retrograde orbit, but then exit" - of couse that is not possible, if you require higher than the speed of light to orbit circular you'd need even more speed to escape in that direction. $\endgroup$
    – Yukterez
    Dec 24, 2023 at 15:31
  • $\begingroup$ @Yukuterez Of course, good point. In that case my option 1 is wrong, when you have said it is correct, and my option 2, which you have said is wrong, is almost correct. The "almost" is because I guessed that the trajectory would "enter" the event horizon, but it actually only approaches it asymptotically, which I should have considered. If you reword the beginning of your answer to reflect that, I'll accept it. $\endgroup$
    – brendan
    Dec 27, 2023 at 11:43
  • $\begingroup$ @brendan - true, in the option 1 I overlooked the part exiting the ergosphere and focussed on the conserved angular momentum, so I retract the paragraph stating option 1 was correct. Except for the part exiting the ergosphere it is correct though. $\endgroup$
    – Yukterez
    Dec 27, 2023 at 17:43

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