I've been trying to make progress on some of the smaller pieces of this question about the environment around a Kerr black hole. In order to calculate the effects of special relativistic Doppler shift on background radiation, I'm trying to find out how to calculate the effective velocity of a particle in orbit around a Kerr black hole with respect to radiation falling in from infinity, and the proper-time period of the orbit in the Zero Angular Velocity frame- i.e., the proper time it would take for an orbiting observer to see the background stars revolve once.

I have found this useful-looking web page which provides lots of formulae for things like the innermost stable orbit, horizon radius, ergosphere radius, and the Keplerian angular velocity and frame dragging angular velocity. Unfortunately, not being an expert in GR myself, I am having trouble extracting simple formulae for the actual orbital period or velocity in terms of orbital radius ($r$), black hole mass ($M$) and angular momentum ($a$).

My guess is that I should be able to use the Keplerian angular velocity to derive tangential orbital velocity with respect to radiation falling in from infinity, scaling by $r$ to convert angular velocity into true velocity. Is that correct? If so, my only remaining hurdle there is figuring out what conversions I need to insert to go from geometrized to MKS units.

In order to determine the orbital period in the ZAVO frame, I then assume that I would add the angular velocity of frame dragging to the Keplerian angular velocity, and then divide $2\pi r$ radians by that summed angular velocity to get the period.

Finally, in each case, I would then apply the time dilation formula to convert coordinate velocity and coordinate time into proper velocity and proper time.

So, am I on the right track here? If not, where am I going wrong, and what's the correct way to calculate these quantities (proper orbital period and proper orbital velocity) in terms of $r$, $M$, and $a$?

  • $\begingroup$ @Yukterez I would be interested in an answer to this Q too., $\endgroup$
    – ProfRob
    Commented May 27, 2022 at 20:22
  • $\begingroup$ @ProfRob - answer is below, I hope it helps $\endgroup$
    – Yukterez
    Commented May 28, 2022 at 0:18

1 Answer 1


Logan R. Kearsley asked: "what's the correct way to calculate these quantities (proper orbital period and proper orbital velocity) in terms of r, M, and a?"

The circular velocities relative to a ZAMO in the equatorial plane and can be found by setting the coordinate acceleration (assuming we use pseudospherical Boyer Lindquist or Doran Raindrop coordinates) and all velocities except the tangential velocity to zero and solving for the remaining velocity component, the pro- and retrograde solutions in natural units of $\rm{G=M=c=1}$ are:

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

The local velocity of a ZAMO relative to the fixed background is

$${\rm v_z} = \sqrt{g_{\rm t \phi} \ g^{\rm t \phi}} = \sqrt{1 - g_{\rm t t} \ g^{\rm t t}} = \rm \frac{2 a \ \sqrt{\frac{a^2 \ (r+2)}{r}+r^2} \ \ \sqrt{\frac{4}{a^2+(r-2) \ r}+\frac{2}{r}+1}}{a^2 \ (r+2)+r^3}$$

($\rm v_z=1$ at $\rm r=2$ where the ergosphere has its equatorial edge), while the equatorial circumference in the frame of the ZAMO is

$${\rm U}_{\phi}=2 \pi \ \sqrt{|g_{\phi \phi}|}=\rm \frac{2 \pi}{r} \ \sqrt{\left(a^2+r^2\right)^2-a^2 \left[a^2+(r-2) \ r\right]}$$

so the orbital period in terms of the proper time of the orbiter is

$$\rm \tau = \frac{U_{\phi} \ \sqrt{1-v_{\pm}^2}}{v_z + v_{\pm}}$$

Example: a prograde orbit at $\rm r=3$ with $\rm a=1$ has a local velocity of $\rm v_{+}=0.527416$, a ZAMO velocity of $\rm v_z=1/3$ and a circumference of ${\rm U}_{\phi}=20.5208$, so $\tau=20.2552$:

prograde orbit around a Kerr black hole

A retrograde orbit would not be possible at this $\rm r=3$, since for $\rm v_{-}$ to be larger than $-1$ (smaller than the speed of light) we need $\rm r>4$ in this example with $\rm a=1$ (with $\rm a=0$ we would get $\rm v_{+}=1$ and $\rm v_{-}=-1$ at $\rm r=3$, where a Schwarzschild black hole has its photon sphere).

An alternative way is to set $\rm \ddot{r}=\dot{r}=\dot{\theta}=0$ and solve for $\dot{\phi}$ (overdot is differentiation by proper time $\tau$), then you get $\dot{\phi}_{+}=0.310202$ for the prograde solution, if you divide $2 \pi$ by that you get the same result for the time:

$$\dot{\phi}_{\pm} = \rm \frac{\pm \sqrt{a^2 \ (r+2) \ r^5+r^8} \pm a^2 \left[\sqrt{a^2 \ (r+2) \ r+r^4}+2 \sqrt{\frac{a^2 \ (r+2)}{r}+r^2}\right]}{\sqrt[4]{r} \ \surd \left[ \pm 2 a+(r-3) \surd r \right] \ \left[ a^2 \ (r+2)+r^3\right]^{3/2}}$$

The proper time in the ZAMO's frame is longer by a gammafactor of $\rm 1/\sqrt{1-v_{\pm}^2}$, but note that this is not the time for the orbiter to meet the ZAMO twice, but the time for the orbit to make a revolution of $360°$ where he sees the same stars as at the beginning of his orbit (the ZAMO himself has moved on in the meantime as you can see on the smaller dot in the animation).

If you are not looking for the time it takes for one full revolution, but the time it takes the orbiter to catch up with the ZAMO he overtook at the start position the proper time in the orbiter's frame is

$$\rm \tau = \frac{U_{\phi} \ \sqrt{1-v_{\pm}^2}}{v_{\pm}}$$

(which is $\tau=33.0567$ in this case), and without the gamma factor in the ZAMO's frame (that is $\rm {}_T=38.9082$). At this position they don't see the same stars they saw at their initial position.

In the frame of the coordinate bookkeeper at infinity the ZAMO's time is multiplied with

$$\sqrt{g^{\rm tt}}=\rm \sqrt{\frac{4}{a^2+(r-2) \ r}+\frac{2}{r}+1}$$

so the time for one $360°$ revolution in this frame and with said parameters is $\rm t=38.9316$ and $\rm t=63.5368$ until the orbiter meets the ZAMO for a second time.

The orbit velocity $\hat{\rm v}_{\pm}$ relative to the fixed stars or a locally stationary observer not corotating with the black hole (who can only exist outside the ergosphere) is calculated with the relativistic velocity addition formula of the ZAMO's velocity and the orbit velocity relative to the ZAMO:

$$\hat{\rm v}_{\pm} = \rm \frac{v_z + v_{\pm}}{1 + v_z \ v_{\pm}}$$

which in this case is $\hat{\rm v}_{+} = 0.732051$ for the prograde solution.

All $\rm v$ are in units of $\rm c$, the angular velocity $\dot{\phi}$ in $\rm c^3/G/M$, the $\rm r$ in $\rm G M/c^2$ and times in $\rm G M/c^3$. The equations here assume $\theta=\pi/2=90°$ (the equatorial plane), if we also allow polar velocity components the equations become much longer.

For more information see Bardeen at al (1972), page 14+. In case I made a typo in the Latex or if you get confused by the $\pm$ and $\mp$ in the solutions click here for a screenshot of the working worksheet.

  • $\begingroup$ I just found out that I already answered a similar question, the notation is a little different (vφ instead of v±) and it's not about the orbital period, but it shows the ISCO and the relation to the constants of motion like E and Lz that I didn't show here: physics.stackexchange.com/a/484455/24093 and for the first link in your question about the planet in the movie Interstellar we also have something that might help: physics.stackexchange.com/a/557210/24093 which gives you the gravitational blueshift which you doppler with your local velocity relative to the ZAMO. $\endgroup$
    – Yukterez
    Commented May 28, 2022 at 6:57
  • $\begingroup$ Still struggling. Applying your equation for $v_z$ to the Miller's planet black hole ($r=1.0000379$, $a=1-1.3\times 10^{-14}$) gives me $v_z = 52769$ ? $v_+=0.500014$, which doesn't give a blueshift sufficient to boost 61,000 to 275,000. $\endgroup$
    – ProfRob
    Commented May 28, 2022 at 19:09
  • 1
    $\begingroup$ @ProfRob - that sounds good, since this radius is well below the ergosphere where the tangential velocity of the the ZAMO becomes larger than c, at the horizon it goes to infinity $\endgroup$
    – Yukterez
    Commented May 28, 2022 at 19:11
  • $\begingroup$ OK, I see that now. So you mean boost the time dilation factor using $v_+$. But that only seems to give a gain of 1.73. The claim in the discussed Miller's planet paper is that it boosts the 61,000 time dilation up to 275,000 ?? $\endgroup$
    – ProfRob
    Commented May 28, 2022 at 19:14
  • 1
    $\begingroup$ At the end of the paper arxiv.org/pdf/1601.02897v1.pdf#page=8 that you cited they say "To avoid significant analytical difficulties associated with such an approach, we used our relativistic ray-tracing code LSDplus, which perform time-reverse direct numerical integration of the equations (A2). [...] The code also calculates relativistic frequency-ratio factor g by (A7) for ray bundles coming from distant universe." - as far as I know Boyer Lindquist coordinates can run into numerical problems when you raytrace too close above the horizon, I can't reproduce their result of z=275000 $\endgroup$
    – Yukterez
    Commented May 29, 2022 at 3:56

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