# How is the gravitational time dilation related to escape velocity in general relativity?

In schwarzschild metric the gravitational time dilation at distance r in the viewpoint of an observer at infinity is:

$$\frac{\tau }{t} = \sqrt{1-\frac{r_s}{r}}$$

I here define that escape velocity is a velocity of a test object in some direction (it is a vector) that is needed for small test mass to reach infinity with velocity v=0. In Schwarzschild metric the escape velocity at the radius r is

$$v_{esc} = \sqrt{\frac{2GM}{r}}$$

therefore

$$r = \frac{2GM}{v_{esc}^2}$$

$$r_s = \frac{2GM}{c^2}$$

using these two equations i get:

$$\frac{\tau }{t} = \sqrt{1-(\frac{v_{esc}}{c})^2}$$

Is this relation between time dilation and escape velocity valid more generally? What is the generalization of this relation?

• do you recognize the formula that came out of your calculations? Commented May 16 at 10:05
• Yes it is similar than equation for time dilation for object that has velocity v_esc in special relativity. Commented May 17 at 13:45
• It is not just similar, it is the generalization you are looking for. To escape, the velocity time dilation must match the gravitational time dilation. Commented May 17 at 14:40

I was planning a rather long derivation, but it turns out that it doesn't take much effort. The most common black hole metric is the Kerr-Newman black hole (see https://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric), where you can write it up to be a gravitational time dilation parameter (important: this is the reciprocal of what you wrote!): $$\zeta = \sqrt{g^{tt}}$$ and so the escape velocity: $$v_{esc} = \frac{\sqrt{\zeta^2 - 1}}{\zeta}$$ So the exact same relationship.
Of course, in the case of black holes, this will obviously be true concretely at the outer event horizon, where the escape velocity is (by definition) $$c$$, and the gravitational time dilation is necessarily $$\infty$$ (or zero if you look at reciprocal).