I What Happens When an Object Accelerating Away from a Black Hole Stops?

[Ivo Draschkow]

"You need to put in a lot of effort and learn som geometry and general relativity"

You're right, that's why. When I asked the question, I was hoping that maybe there was an easy way to think of it. Now I know that's not the case.

 

[martinbn]

"You need to put in a lot of effort and learn som geometry and general relativity"

You're right, that's why. When I asked the question, I was hoping that maybe there was an easy way to think of it. Now I know that's not the case.

This is the easy way.

 

[Ibix]

almost everything is curved in such a way that it appears to be slowing down,

That's the bit that's wrong.

Things deep in a gravity well will appear to move slower than the same processes next to you. But they only get slower if the thing you're watching goes deeper into the well. Both the objects in your last scenario are going deeper into the gravity well, one much faster than the other. So the visual slowing effect is more pronounced in that one - but it is only more pronounced because it is deeper in the gravity well. The very reason it appears to be slowed more is because it is ahead of the other object. And the fact that it's only depth in the well that matters is why you must see the object start to fall in order for the increased sliwing to happen.

Remember that the slowing effect can be directly attributed to the way light travels in curved spacetime from the objects to your observer. Seen locally, your scenario is (qualitatively) the same as if you crouch holding a ball while I stand over you and drop a ball and you release yours when mine passes your hand. My ball is always ahead of yours. It's just that it takes longer and longer for the light to reach us. But we could never see your ball catch mine because it's the position that defines the amount of extra delay. If we saw the slower catch the faster then they'd have the same degree of slowing and it'd genuinely be paradoxical for the slower one to have caught up.

 

[Ivo Draschkow]

Thanks, I think based on your last explanation I'm starting to see the error in my thinking. I was too caught up in the infinitesimal. It also occurred to me that the object could, for example, not fully release the gas and would still ultimately accelerate towards the BH.

 

[PAllen]

A simple way to picture this is to imagine the scene filmed from just above the action, noting that (if near a supermassive BH horizon), this situation is just like being above the surface of a large planet (except proper acceleration to hover is enormously larger). Then, what you see far away is nothing more than what that film would show slowed down e.g. a million times, and red shifted a million times as well (so you visible light would be radio waves). Everything in the same vicinity is slowed and red shifted by the same proportion.

 

[PAllen]

Since this thread is labeled intermediate, we should be able to use a little math. And the math can be made really simple for the cases under discussion.

A general formula for r' (derivative of Schwarzschild r coordinate by time per very distant observer - Schwarzschild time) for ingoing timelike geodesics is for the exterior coordinate patch (R is Schwarzschild radius):

$$r'=(R/r-1)\sqrt{(R/r-1)/e^2+1}$$

For free fall from infinity, take $e^2=1$, giving
$$r'=(R/r-1)\sqrt{R/r}$$
For free drop from some $r_0>R$, take $e^2=1-R/r_0$. For a "thrown" free fall, where the particle, extrapolated back in time, would have some velocity at infinity, take $e^2=\gamma(v_\infty)^2$.

Without even integrating, it is clear that r' for free fall from infinity becomes zero at infinity and also on approach to the horizont (r=R). The maximum magnitude r' for this case is at r=3R.

For a free faller (from infinity) passing a platform drop, you see r' is nonzero for the free faller at $r_0$, and that r'=0 at $r_0$ for the drop from this altitude. At lower altitudes, you see that for the platform drop, r' increases in magnitude, then approaches zero at the horizon, but any any altitude from $r_0$ down to R, r' for free fall from infinity is larger in magnitude. Thus, the free faller (from infinity) has always increasing difference in r coordinate from the platform drop, following the coincidence event where it is zero.

While there is a closed form for t=f(r), the integration is quite messy, and doesn't really tell us more than can be deduced from the simple expressions for r'.