1
$\begingroup$

From the point of view of a far-away observer of a black hole, is the gravity on an infalling object near the event horizon very strong or very weak?

There are two competing effects: On the one hand, the gravity felt by the infalling object is increasing with decreasing distance, but on the other hand, gravitational time dilation approaches infinity near the event horizon, that means that the clock of the infalling object is slowed down close to zero with respect to the clock of the far-away observer. Which effect is dominant at the event horizon?

$\endgroup$
4
  • $\begingroup$ How are you defining gravity? (Surface gravity, scalar curvature, etc?). Moreover, what mass black hole (for very large black holes the results will be considerably different)? $\endgroup$
    – Eletie
    Nov 14, 2022 at 16:16
  • $\begingroup$ How would you quantify the relative strength of gravity from an arbitrary observer's point of view? Forget black holes and event horizons - imagine an observer watching a particle move under the influence of gravity. What would be your metric for deciding whether gravity was strong or weak? $\endgroup$
    – J. Murray
    Nov 14, 2022 at 16:16
  • $\begingroup$ The Schwarzschild metric describes the spacetime curvature and the strength of gravity. I do not want any exact number, I only want to know how the strength of gravity is behaving near the event horizon. $\endgroup$
    – Moonraker
    Nov 14, 2022 at 16:27
  • $\begingroup$ @Moonraker I know that, but you aren't specifying what you mean by the strength of gravity. One intuitive measure would be how much force you'd need to exert on an object to keep it from falling towards the singularity - which would be the object's mass multiplied by the $a$ in Dale's answer. But you could also be talking about the tidal forces which act to distort a body, which is a different effect and would depend on the size of the black hole. $\endgroup$
    – J. Murray
    Nov 14, 2022 at 18:13

1 Answer 1

1
$\begingroup$

There are two common ways of specifying the "gravity". One would be the proper acceleration of a hovering object. This is given by $$a=\frac{R}{2 r^{3/2}\sqrt{r-R}}$$ which goes to infinity as $r$ approaches the event horizon at $R$. The other would be the $tt$ Christoffel symbols. For the Schwarzschild metric there is only one $$\Gamma^r_{tt}=\frac{c^2 R(r-R)}{2r^3}$$ which goes to zero as you approach the event horizon.

$\endgroup$
5
  • $\begingroup$ Very interesting, two contradicting results - Thank you! $\endgroup$
    – Moonraker
    Nov 14, 2022 at 17:11
  • 1
    $\begingroup$ Well, they are not actually contradictory. You just didn’t specify which meaning you intended in the question so I just answered with both common meanings $\endgroup$
    – Dale
    Nov 14, 2022 at 17:25
  • $\begingroup$ As I did not mention any hovering object in my question, I think I will buy your second solution. - Does the proper acceleration take into account the effect of gravitational time dilation? $\endgroup$
    – Moonraker
    Nov 14, 2022 at 17:29
  • $\begingroup$ I am not sure what it would mean for a Christoffel symbol to take into account the effect of time dilation or not $\endgroup$
    – Dale
    Nov 14, 2022 at 17:44
  • $\begingroup$ The answer doesn't address what an observer would see about the speed at which the infalling object is pulled into the event horizon. Does the immense time dilation counteract the object's gravitational plummet or not? $\endgroup$
    – user610620
    Nov 27, 2022 at 8:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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