1
$\begingroup$

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

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

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?

$\endgroup$
3
  • 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

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.