Under Newton, the escape velocity is

$$v_{esc} = \rm c \ \sqrt{r_s/r}$$

where $\rm r_s=2 \ GM/c^2$. In the nonrotating relativistic case (the Schwarzschild case) the radial escape velocity is the same:

$$v^{\parallel}_{esc} = \rm c \ \sqrt{r_s/r}$$

which is the speed of light, $v={\rm c}$, at the event horizon at $\rm r=r_s$. Since the photon sphere, where an object with $v={\rm c}$ would not escape, but be captured into orbit, is at $r=3 \ {\rm GM/c^2}$, the transverse escape velocity is

$$v^{\perp}_{esc} = \rm c \ \sqrt{r_s/r}\div \sqrt{1-r_s/r}$$

and for the observed (shapiro delayed) velocity ${\rm v}$ the transformation is

$${\rm v}^{\perp}=v^{\perp} \ \sqrt{\rm 1-r_s/r} \ , \ \ {\rm v}^{\parallel}=v^{\parallel} \ (\rm 1-r_s/r)$$

so the observed angular motion is the same as with Newton, yet locally higher, while the observed radial motion is slowed down because of the radial length contraction and time dilation.

But what about the Kerr case where the rotation is significant and the event horizon is at $\rm r=(1+\sqrt{1-a^2}) \ {\rm GM/c^2}$ instead of $\rm r=r_s$?

A testparticle which is locally at rest and therefore corotating with the angular velocity of the frame dragging

$${\Omega = \rm d \phi/ {\rm d t} = {\rm 2 \ a \ r/((a^2+r^2)^2-a^2 (a^2+(r-2) \ r) \sin ^2(\theta ))}}$$

should experience the time dilation factor

$$\rm \dot t = {\rm d t}/ {\rm d}\tau = \sqrt{{\rm ((a^2+(r-2) \ r) (a^2 \cos ^2(\theta )+r^2))/((a^2+r^2)^2-a^2 (a^2+(r-2) \ r) \sin ^2(\theta ))}}$$

relatively to the far away observer, so multiplying $\Omega$ with $\rm \dot t$ should give the local frame dragging angular velocity, which, multiplied with the $\rm \sqrt{x^2+y^2}$ value of the Boyer-Lindquist to cartesian transformation should give the transverse velocity - this is slightly below $\rm c$ for a maximaly rotating black hole (with $\rm a = 1$ - at the edge of the outer equatorial ergosphere where $\rm r=1$ and $\rm R=\sqrt{r^2+a^2}$).

But how do I find the radial and total escape velocity, and what are the rules to convert the shapiro-delayed coordinate velocities $\rm d r/d t \ , \ d \phi/d t \ , d \theta/d t$ to the local 3-velocity components $v^{\parallel}$ and $v^{\perp}$ of a test particle (local meaning relatively to a frame dragged probe with contant $\rm r$)? The radial length contraction seems a bit different than in the Schwarzschild-case, where it is the same as the gravitational time dilation factor...


2 Answers 2


2 hours left to award the bounty, but no answear so far. At least I found out that the local 3-velocity $$\rm v=\sqrt{v_r^2+v_{\theta}^2-v_{\phi}^2}=\sqrt{v_x^2+v_y^2-v_z^2}$$ is related to the coordinate derivatives $$\rm \dot r, \ \dot \theta, \ \dot \phi$$ by $$\rm \frac{v_{r}}{\sqrt{1-v^2}} = \dot r \ \sqrt{\frac{\Sigma}{\Delta}}$$ for the radial component, $$\rm \frac{v_{\theta} \ \sqrt{\Sigma}}{\sqrt{1-v^2}} = \dot \theta \ \Sigma$$ with the radius of gyration $$\bar R = \sqrt{\frac{\left(a^2+r^2\right)^2-a^2 \Delta \sin ^2 \theta }{a^2 \cos ^2 \theta +r^2}}$$ for the motion along the latitude and $$\rm \frac{v_{\phi} \ \bar{R}}{\sqrt{1-v^2}} = \dot \phi \ \Sigma$$ for the motion along the axis of symmetry, with the terms $$\rm \Sigma = r^2 + a^2 \cos^{2} \theta, \quad \Delta = r^2 - 2 \ r + a^2$$ and $\rm x, \ y, \ z$ as the cartesian transformation of the Boyer-Lindquist coordinates.

With the gravitational time dilation

$$\rm \varsigma =\sqrt{\frac{\left(a^2+r^2\right)^2-a^2 \left(a^2+(r-2) r\right) \sin ^2(\theta )}{\left(a^2+(r-2) r\right) \left(a^2 \cos ^2(\theta )+r^2\right)}}$$


$$\rm \varsigma = \frac{1}{\sqrt{1-v_{esc}^2 }}$$

the radial escape velocity should be

$$\rm v_{esc}=\frac{\sqrt{\varsigma ^2-1}}{\varsigma }$$

which, in the limit of $a\to 0$ gives the same result as Schwarzschild.

Example of a testparticle being thrown radially upwards with the local escape velocity by a corotating ZAMO sitting close above the horizon at θ=45°:

enter image description here

(The Spin parameter in the animation is a=Jc/G/M²=0.998)

  • $\begingroup$ Can you explain the units of $r$. Is it expressed as multiples of $2GM/c^2$? $\endgroup$
    – ProfRob
    Commented Mar 16, 2018 at 8:27
  • $\begingroup$ r is in units of 1GM/c², where M is the gravitational mass (not to be confused with the irreducible mass). $\endgroup$
    – Yukterez
    Commented Mar 18, 2018 at 5:51

This question has a complex answer, as the Kerr orbits cannot be integrated as easily as the schwarzschild orbits. Your answer will depend on where you sit on the horizon, and in what direction you move. I don't really know if the notion of "escape velocity" really works in relativity like it does classically, especially in the case of a Kerr hole.

  • $\begingroup$ @СимонТыран: OK, if you are OK with an "escape velocity" that has a six-parameter (radius from star, angular position, and three velocity components) dependence, I guess you can define it. But that's hardly an "escape velocity" in my mind, becaue the key part of that concept, to me, is the direction-independence. $\endgroup$ Commented Aug 7, 2017 at 14:39
  • $\begingroup$ By the way, one definition of the event horizon is that the escape velocity converges to c (that goes for Schwarzschild as well as for Kerr), so one can surely call it escape velocity, although it's always direction-dependend even in the full symmetric Schwarzschild case. $\endgroup$
    – Yukterez
    Commented Aug 8, 2017 at 5:03
  • $\begingroup$ @СимонТыран : I woudln't call it "escape velocity" in the schwarzschild case, either. $\endgroup$ Commented Aug 8, 2017 at 14:51
  • $\begingroup$ But if you google for "Schwarzschild escape velocity" you will find a bunch of reputable sources who call it the same way, and with the definition given on Wikipedia: "In physics, escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body" there is no need for angle-independence. $\endgroup$
    – Yukterez
    Commented Aug 13, 2017 at 18:24
  • $\begingroup$ @СимонТыран: minimum speed implies minimizing over all possible angles, so this is an angle independent definition. $\endgroup$
    – A.V.S.
    Commented Mar 16, 2018 at 5:21

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