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

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


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

and the schwarzchild radius is:

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

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?

  • 3
    $\begingroup$ do you recognize the formula that came out of your calculations? $\endgroup$
    – paulina
    Commented May 16 at 10:05
  • $\begingroup$ Yes it is similar than equation for time dilation for object that has velocity v_esc in special relativity. $\endgroup$
    – Sami M
    Commented May 17 at 13:45
  • 1
    $\begingroup$ 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. $\endgroup$
    – safesphere
    Commented May 17 at 14:40

1 Answer 1


If the question is about black holes: Yes, for a neutral test body.

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.

This shows that the relation is also satisfied for the Kerr and Reissner-Nordström metrics.

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).

But as I pointed out at the beginning this is only true for electrically uncharged bodies in this form.


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