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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

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    $\begingroup$ 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? $\endgroup$ Aug 29, 2017 at 14:18
  • $\begingroup$ 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)!! $\endgroup$ Aug 29, 2017 at 14:21
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    $\begingroup$ Which makes sense only if the metric is assimptotically flat, for one. $\endgroup$ Aug 29, 2017 at 14:25
  • $\begingroup$ @Uldreth Can you please enlighten me on this point? I would like to know more why it has to be assymptotically flat! $\endgroup$ Aug 29, 2017 at 14:26
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    $\begingroup$ @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. $\endgroup$ Aug 29, 2017 at 15:11

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This is certainly not possible in general. The topology of the spacetime is not determined by the Einstein equations, so you could have solutions in situations for which there is no "infinity" to which you could escape. Think something like $R \times T^3$, where $R$ is an unbounded, time-like direction and $T^3$ is a 3-torus. You could find solutions to the Einstein equations for that, but what would it mean to "escape to infinity"?

Someone else might try to make that precise, but the basic idea should be right.

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