Escape velocity for an arbitrary metric in General Relativity

It's a known calculation how to derive the escape velocity of a system following the Schwarschild Metric. It comes out to be, if I am not wrong $$v = c\sqrt{2GM/Rc^2 - (2GM/Rc^2)^2}$$

So, is there a general method of finding out the escape velocity for a given general metric, without assuming any symmetry considerations?

A mathematical help would be highly appreciated

• I might be totally wrong, but I highly doubt "escape velocity" makes sense in a general context. I mean, if you have a vacuum spacetime containing an impulsive gravitational wave or something like that, what do you want to escape from? – Bence Racskó Aug 29 '17 at 14:18
• The curvature of spacetime created at the place of the gravitational wave, thats what I would like to escape! Because the curvature would cause a change in my trajectory, and I would like to resist the change and go to infinity (mathematically)!! – Yuzuriha Inori Aug 29 '17 at 14:21
• Which makes sense only if the metric is assimptotically flat, for one. – Bence Racskó Aug 29 '17 at 14:25
• @Uldreth Can you please enlighten me on this point? I would like to know more why it has to be assymptotically flat! – Yuzuriha Inori Aug 29 '17 at 14:26
• @YuzurihaInori: because there has to be a notion of a "place to escape to". Also, even in the case of a Kerr hole, "escape" velocity will depend on all three spatial coordinates and on the initial velocity direction. – Jerry Schirmer Aug 29 '17 at 15:11

The equation you cite gives the coordinate velocity for an object falling freely from rest at infinity in the Schwarzschild metric. Most of us will be more familiar with it in the form:

$$v = -\, c \sqrt{ \left(1 - \frac{r_s}{r}\right) \frac{r_s}{r}}$$

The velocity is also the escape velocity because we can simply reverse time, in which case the object moves outwards and eventually comes to rest at infinity, which is exactly how the escape velocity is defined.

However the coordinate velocity is the velocity as observed by an observer far from the black hole, and it has some strange properties. For example the escape velocity calculated in this way has a maximum at $r = 2r_s$ and falls to zero as we approach the event horizon. This is an odd way to define an escape velocity.

A more sensible approach would be to ask what is the escape velocity at a distance $r$ as measured by an observer stationary at the distance $r$ (this type of observer is usually called a shell observer). This turns out to be:

$$v_\text{shell} = c \sqrt{\frac{r_s}{r}}$$

which behaves much more as you would expect an escape velocity to behave as it goes smoothly to $c$ at the event horizon.

So the answer to your question is that to calculate the escape velocity we start with the observer at rest at infinity then calculate its geodesic as it falls towards the mass (or whatever). And I'm afraid there isn't a simple way to do this. For an observer falling radially into a static spherically symmetric metric there are shortcuts because the ratio of total energy to mass $E/m$ is constant and we can extract the $r$ dependence of the velocity from this. For more general cases the calculation will be harder and might have to be done numerically.