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 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 Mar 16 '18 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). $\endgroup$ – Gendergaga Mar 18 '18 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.

  • $\begingroup$ The relation between gravitational time dilation and escape velocity always works, so of course it also works in Kerr spacetime, see the penultimate equation at en.wikipedia.org/wiki/… and for the animation of an object leaving the ergosphere with the escape velocity at different angles see notizblock.yukterez.net/viewtopic.php?p=361#p361 $\endgroup$ – Gendergaga Aug 5 '17 at 1:41
  • $\begingroup$ By the way, it already depends on the angle when you use Schwarzschild, since a photon can radially escape above r=2GM/c² (the event horizon), while a transverse photon will already be captured below r=3GM/c² (the photon sphere) even in the nonrotating case. Escape velocity is always in terms of radial motion (in the Kerr-case relative to a local ZAMO), only under Newton the direction does not matter. $\endgroup$ – Gendergaga Aug 5 '17 at 1:51
  • $\begingroup$ But you're right, the answear depends on where you sit, therefore the dependence on θ in the equation for ς and v_esc. $\endgroup$ – Gendergaga Aug 5 '17 at 2:02
  • $\begingroup$ @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. $\endgroup$ – Jerry Schirmer Aug 7 '17 at 14:39
  • $\begingroup$ But you won' get direction indepence, even with Schwarzschild there is a dependence. As you can see in the animation above, if the testparticle starts with v local = v escape, it keeps v local = v escape all the way. The fact that in the ZAMO frame of reference the direction needs to be purely radial simplifies it a little bit, if you define the initial velocity relative to him you don't need to worry about other components than radial. You can also use the rule if E kinetic + E potential = 0 and the angular momentum = 0, this gives the radial escape velocity. $\endgroup$ – Gendergaga Aug 8 '17 at 4:55

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