I agree. But that’s the opposite of what you said to which I replied. You said “The essential problem with your argument is that your thought experiment is defined in such a way that no matter how small the tidal acceleration is, it is never negligible for the purpose of your particular experiment”. That’s saying that that no matter how small the tidal acceleration is, it is always significant for the purpose of my experiment. But all my experiment does is measure the tidal force. And the value of a measurement does not affect the measurement of that value.In the real world, all instruments have limited precision, so a small enough tidal force is not detectable, and is therefore negligible for the purpose of the measurement.

In the world of "in principle", perfect instruments will register zero tidal force in perfectly flat space, which thus has negligible tidal force.

My thought experiment requires a measurable tidal force only because that shows that the spacetime in X cannot be flat, which is not an experiment but rather only a conclusion. If a thought experiment makes its case by showing that the tidal force is above a certain value, that does not prove that the thought experiment does more than just measure the tidal force.True enough, for the purpose of this discussion... but your thought experiment is notonlya measurement of the tidal force.

Your thought experiment is specifically set up torequirea measurable tidal force.

What is your evidence? It is given that the particle is in X and escaping from the black hole; that’s all the information about its position and motion that is needed for the thought experiment to make its case. GR allows the particle to be in X and be escaping from the black hole; the supporting info in the OP shows that.The particle's position and motion can not be specified in X unless the tidal force is known.

The self-inconsistency you describe is GR’s fault, not mine. According to Taylor and Wheeler, GR predicts that Y can cover the same region of spacetime as X. A self-inconsistency of GR cannot be used to refute a proof of that self-inconsistency. More on this below.In that case, Y can not cover the same region of space-time as X.

You have tried to define Y using three self-inconsistent conditions: there is no inertial frame in which the particle is at rest that covers the same region of space-time as X. To see why, you only need to consider the distance from the particle to the event horizon in the particle's inertial rest frame... in that frame, the event horizon isinfinitelydistant from the particle.

No math is necessary. Do you agree that a strain gauge can show the tidal force on an object? Do you think the value shown on the gauge is affected by your velocity relative to the gauge? (That is analogous to asking if the ocean tides are altered by your velocity relative to the Earth.) The particles’ relative acceleration is measured in the frame of one of the particles, just like their relative velocity would be.Since relative acceleration depends on both distance and time, which are certainly dependent on the chosen reference frame, it is not at all clear that tidal acceleration is not frame dependent.

Applying some maths will be necessary to know for sure, I think.

They do imply that. It would be nice if they were more specific. I know I’m right because flat spacetime is the arena of SR, and in SR any inertial frame can be set up in any other inertial frame. In SR, the frame of any free test particle in an inertial frame J is likewise inertial when extended throughout J.You misread Taylor and Wheeler. They do not imply that it is possible to set upanyinertial frame in a region of spacetime that can be considered flat.

You’re taking Taylor and Wheeler too literally. If we took them literally, then an inertial frame could contain a black hole, because they don’t specify that the tidal force must be negligibleAccording to the literal meaning of the quoted extract, it is sufficient that one inertial frame exists across a given region of space-time for that region to be considered flat. (Don't read too much into that, either... that introductory paragraph needs to be understood in the context of the rest of the text.)

So, if X is inertial, then (according to the extract) the space-time in X can be considered flat, regardless of whether and other frame in that space-time is inertial or not.

*throughout*the frame, but rather only

*in*the frame, which could be taken to be just in a small part of it. (Thorne is likewise not specific enough.) To make your case, you need to show that the frame of a free test particle in an inertial frame J can be noninertial when extended throughout J. But to do that you would need to refute SR.

Yes, they all seem to be moving at c in X, or close to it. Nevertheless, the particle is escaping, and the rod is at rest w.r.t. the particle and straddles the horizon. And the tidal force in X is negligible since X is an inertial frame. Then the rod must be passing outward through the horizon; it must be escaping too. But GR doesn’t allow that, and so the rod cannot exist. If there is any contradiction there then that’s GR’s fault, not mine.Without doing the maths, I can't be sure, but here's what I suspect:

During the time covered by X, the rod is (close enough to) at rest w.r.t. the escaping particle. It is also close enough to at rest w.r.t. the event horizon - ie in X, the particle, the rod, and the EH all seem to be moving at c.

If the “outer end of the rod will be slowed down enough for the EH to catch up and pass it”, then the rod does not stay at rest w.r.t. to the escaping particle, and does not exist as given. That would make my case. Because Y is inertial, the rod, or a latticework of synchronized clocks, canAt a later time, however, the outer end of the rod will be slowed down enough for the EH to catch up and pass it - unless the rod breaks. I don't know how strong the force acting on the rod will be. I don't know if it's negligible or huge, so I don't know if breakage is likely or not.

I'm fuzzy on the precise details here... like I said, a mathematical analysis would be good.

*stay*at rest w.r.t. to the escaping particle. If you were right then Y could not exist in X, and then the spacetime in X could not be flat, and then GR would be self-inconsistent.

All that matters is that the particle is escaping, and GR allows that. The rod, or a latticework of synchronized clocks, isIn X, the escaping particle is moving at effectively the same speed as the horizon. I'm not sure if it will be negligibly less than c, or if the negligible curvature allows it to be moving negligibly faster than c. Maths...

*given*to stay at rest w.r.t. to the escaping particle, and GR allows that in Y, an inertial frame. If the curvature of spacetime is negligible, which it is because X is an inertial frame, then the tidal force on the rod is negligible, because spacetime curvature = tidal force. Then the rod would be passing outward through the horizon, which GR does not allow, and so the rod cannot exist. Hence GR is self-inconsistent. If you find some inconsistency regarding the velocity of the particle, that’s GR’s fault, not mine.

The rod cannot be at once both negligibly stretched and tremendously stretched by a tidal force. A strain gauge on the rod cannot show two values at once.The rod is negligibly stretched (I think) in X.

In the hovering observer's frame it is clearly either stretched considerably, or undergoes tremendous tidal acceleration, or both.

The situationThe situation isn't symmetrical. I don't see that the same logic should apply in each direction. I think that the box frame is inertial, while the rod's rest frame is not.

*is*symmetrical. Please provide evidence to the contrary. The lack of evidence to the contrary proves my point that the situation is symmetrical. To prove your point you need to show something asymmetrical about the situation.

I gave the equation for the tidal force, direct from Taylor and Wheeler, and velocity was not an input. (In the link I gave you, Taylor and Wheeler are clear that the equation I gave you is the one and only equation for spacetime curvature = tidal force in Schwarzschild geometry.) Then my case is made that the tidal force is not dependent on velocity. It is impossible to crunch numbers with velocity on an equation that does not take velocity as an input.Perhaps if you crunched the numbers you'd discover otherwise. Or not.

Feel free to crunch them and present your results.

It’s aNotice that the units of the curvature factor are neither force nor acceleration.

I don't know how to extract tidal acceleration from the equations for the Schwarzachild metric, but it's clear that it's not simply the curvature factor.

*factor*; it’s dimensionless, unitless, like the Lorentz factor for time dilation and length contraction. It tells you the degree of the tidal force relative to the tidal force given by the equation using other inputs. Do you think the Lorentz factor does not indicate time dilation just because it’s unitless?

Yes. Also the object is uncharged and nonrotating.I don't know how to use the Schwarzschild metric, but I have the impression that it's based around a single spherically symmetrical massive body in an otherwise empty universe...

I don’t know what you’re getting at here. The curvature factor that I gave is all you need to know the degree of tidal force in Schwarzschild geometry. The incorporation of the curvature factor in the Schwarzschild metric is the sole difference between the Schwarzschild metric and the metric for flat spacetime, the metric of SR, in which the tidal force is nonexistent. This link says: “The tidal force goes as M / r^3 at distance r from a black hole of mass M.” You won’t find any reputable reference that supports a claim that tidal force is frame or velocity dependent. Nor would that make sense, since a strain gauge can show the degree of tidal force, and shows the same value to observers in all frames.... so I'd guess that the dr term in the equation is the distance from the center of that body.

In the language of the metric, you're not looking for a velocity term, you're looking for dr, dt, and d(phi), I think.

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