General relativity is self-inconsistent v3

This seems okay.

The rod mentioned in the OP is not qualified as infinitely long.

Let's not go any further though; at least I won't. The argument in the OP of this old thread was too contentious. I've sinced evolved the argument to one that raises much less objection, which I put here in the other thread.

 

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There's no such thing as "escaping to infinity" in a local Lorentz approximation which straddles a black hole horizon because the approximation is only good when local and is only local to distance ε in space and ε/c in time.

A tiny mathematical concept to add a tiny bit more rigor.

 

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Even distance ε is sufficient for part of the frame to be above the horizon, where the escape velocity is less than c. ε is not zero.

 

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But the frame itself is falling into the horizon -- an imaginary line which sweeps over your local approximation at about the speed of light -- and this is precisely true when the inward particle is exactly at 2R. If the frame is not freely-falling then it isn't approximately Lorentzian.

 

The frame is freely falling. The horizon sweeps over the frame; even at the speed of light, this takes some time (however short) as measured in the frame. Then there will be a moment in the frame when part of it is above the horizon (where a particle can be escaping to infinity) and part of it is below the horizon (where all particles must be falling relative to the horizon).

I'm not sure what you mean by "inward particle". This is an old thread someone else resurrected, that contains an argument I no longer desire to argue. A current post & argument that covers your objection (search for "A common objection claims") is here.

 

In all 3 of the "GR is self-inconsistent" threads you have posted, you are assuming a radial differential for two test particles above a gravity source. These two test particles are not in the same inertial frame. One is being accelerated more than the other - this was my point with the rocket ships. Showing that physics differs when one frame is being accelerated more than the other could be demonstrated quite easily in your car.

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The issue isn't that the spatial extent of the frame doesn't include the horizon, but that its temporal extent is not long enough to talk about "escape" events. I.e., the only thing that happens while the approximate frame is valid is that both of the test particles and the horizon all move a short distance in the same direction. Their final positions end up as if they all moved with constant velocities, and this matches the non-approximate GR solution over that same time interval to within some arbitrarily small approximation error. The smaller you make the approximation error, the shorter the time interval where the approximation is valid becomes.

In order to talk about "escape," which is a global property of the system, you must use a reference frame that is defined for an infinitely long time (otherwise, how can you tell whether a particle has truly escaped, or just hasn't yet fallen into the horizon?). If you try to use the equivalence principle to create an approximately flat frame valid for an infinite amount of time, it follows that the approximation error will be infinite (or the frame will have zero spatial extent, which is equally useless).

 

By definition, an inertial frame need only be "small enough", not a point. It's perfectly okay to have a tidal force in an inertial frame, as long as it's negligible for the purposes of the experiment in question (limiting the number of significant digits in results). How do you think SR was experimentally confirmed, if not in frames with a tidal force present?

 

Note that I'm arguing only this post, not the OP. (Nothing about a rod or rope.)

To make my case there, the escaping particle doesn't need to actually escape to infinity. It need only be escaping to infinity, or even just rising away from the black hole, even for an arbitrarily short duration of X. The two free test particles in X cannot have the same velocity with respect to X (to any number of significant digits) for even an arbitrarily short duration of X, because one is falling toward the black hole's central singularity and the other is rising away from the black hole. As the post elaborates, this means that GR requires that the laws of physics among inertial frames radically differ, in contradiction to GR's own equivalence principle. I could have substituted a particle "rising away from the black hole" for a particle "escaping to infinity".

 

Given the unfounded censorship here on Physics & Maths (mod thinks if his profs disagree with me, that's sufficient), discussion here is a waste of my time, but anyone can PM me if you want.

 

The black hole itself (i.e, the singularity) is not in any of the approximately inertial frames you can construct, so you can't even really define motion relative to it in that context. It's only the event horizon that is in the frames.

So presumably what you want is to see the distance between from the escaping particle to the event horizon increasing, and the distance from the trapped particle and the event horizon also increasing. Or, at least, neither of them should be decreasing. And that is exactly what you will see in the approximation, up to some specified error: both particles and the horizon move with constant velocities in the same direction. If the test particles are photons, the relative positions of them and the horizon will not change at all in the approximately flat frame.

Not in the reference frame in question. According to an infalling observer, both test particles are moving away from the singularity. The reason the one doesn't escape is because the event horizon is also moving in this frame, at velocity c, and so the trapped particle can't ever catch up with it.

If you transform this result back to the frame of a distant observer, then it will show up as the escaping particle moving away from the singularity, and the other falling towards it (likewise the event horizon appears stationary in the distant observer's frame). The reversal in particle direction may seem strange (it's not the kind of thing that happens when one changes frames in SR, for example), but that is exactly the sort of weird thing that occurs around black holes.

Your problem seems to be an assumption that the event horizon is sitting at some fixed position in the infalling frame. It emphatically is not, and so there is no contradiction between a particle moving "away from" the singularity and the particle never exiting the event horizon.