How gravity works beyond the Event Horizon

[phinds]

@Stephanus It is important to remember that this expression only gives the coordinate acceleration. The proper acceleration will always be zero at every point along a geodesic path in curved spacetime.

Yeah, but I think bringing that up just confuses the issue as far as the tidal force is concerned and the tidal force is the issue at hand.

EDIT: I see I should amend that statement. Your point IS a good one based on the original question, it's just that the discussion has veered of into more of a discussion of tidal forces.

 

[DrGreg]

@Stephanus It is important to remember that this expression only gives the coordinate acceleration.

Actually it's the proper acceleration of a particle hovering at constant R.

The proper acceleration will always be zero at every point along a geodesic path in curved spacetime.

True.

 

[PWiz]

Actually it's the proper acceleration of a particle hovering at constant R.

I was tempted to replace "coordinate acceleration" with "the acceleration of an object released from a point infinitely far away from the BH (where $g_{μν} \rightarrow η_{μν}$ as the spatial distance from the BH $R \rightarrow ∞$ ) as measured by a static (hovering) observer" while making that post, but I decided not to, because I wasn't 100% sure. Would this have been a legitimate statement?

 

[Stephanus]

@Stephanus It is important to remember that this expression only gives the coordinate acceleration. The proper acceleration will always be zero at every point along a geodesic path in curved spacetime.

Wow, that's new! At least for me

 

[PeterDonis]

I was tempted to replace "coordinate acceleration" with "the acceleration of an object released from a point infinitely far away from the BH (where $g_{μν} \rightarrow η_{μν}$ as the spatial distance from the BH $R \rightarrow \infty$ ) as measured by a static (hovering) observer" while making that post, but I decided not to, because I wasn't 100% sure. Would this have been a legitimate statement?

Technically, yes, but it's way more complicated than it needs to be. You're thinking of escape velocity, which is the velocity of an object that freely falls from rest at infinity, relative to a static observer. For acceleration, it doesn't matter where the object was released as long as it's freely falling; an object released from rest at infinity and an object released from rest by the static observer will both have the same coordinate acceleration relative to the static observer if they are both freely falling, even though they will have very different velocities. (Note also that, as I have just implied, this acceleration is still coordinate acceleration.)

 

[PWiz]

Technically, yes, but it's way more complicated than it needs to be. You're thinking of escape velocity, which is the velocity of an object that freely falls from rest at infinity, relative to a static observer. For acceleration, it doesn't matter where the object was released as long as it's freely falling; an object released from rest at infinity and an object released from rest by the static observer will both have the same coordinate acceleration relative to the static observer if they are both freely falling, even though they will have very different velocities. (Note also that, as I have just implied, this acceleration is still coordinate acceleration.)

I see. I guess if there's one thing I've learned hanging around these relativity threads is that you can never be too careful

 

[rrogers]

This viewpoint only works outside the horizon. At and inside the horizon, the concept of "gravitational potential energy" is not well-defined; there is no "cliff" to have a slope or a height in your analogy.
Again, this viewpoint only works outside the horizon. At and inside the horizon, there are no static observers, and "gravity" in the sense you're using it here is only well-defined relative to static observers.
Same comment here: "escape velocity" is only well-defined relative to static observers, and there aren't any at or inside the horizon.
Again, this depiction is only valid outside the horizon. At and inside the horizon, there is no way to draw a picture of "space", because that would require spacetime to be static, and spacetime is not static at or inside the horizon.

Thanks for the correction/elucidation; yes the analogy was only intended to illustrate the difference (and relationship) between local effects and global accumulation of them. There is indeed a definite "edge" beyond which it shouldn't be pushed.