(The “Alpha” in the title indicates that the Alpha rules (http://www.sciforums.com/showthread.php?t=62837) apply to this thread.)

I will prove that general relativity (GR) is self-inconsistent. First some supporting info (or skip to “Now the proof” near the end):

From pg. 1-19 of Exploring Black Holes by Taylor and Wheeler: “The spacetime arena for special relativity is the free-float (inertial) frame, one in which a free test particle at rest remains at rest and a free test particle in motion continues that motion unchanged. We call a region of spacetime flat if a free-float frame can be set up in it. ... In principle one can set up a latticework of synchronized clocks in a free-float frame. The position and time of any event is then taken to be the location of the nearest lattice clock and the time of the event recorded on that clock. The observer is the collection of all such recording clocks in a given reference frame.”

From the glossary of Exploring Black Holes:

flat spacetime: Region of spacetime in which it is possible to set up a free-float (inertial) reference frame.

horizon: One-way surface surrounding a black hole, defined by the property that anything may pass inward through the horizon, but (in the non-quantum description) nothing, not even light, may pass outward.

inertial frame (free-float frame): Generally, a reference frame in which a free test particle initially at rest remains at rest. More technically, a reference frame with respect to which relative (tidal) accelerations of test particles can be neglected for the purposes of a given experiment.

tidal acceleration [(tidal force)]: Relative acceleration of two free test particles located in different parts of a reference frame.
From the glossary (http://www.pbs.org/wgbh/nova/time/timespeak.html) of Black Holes & Time Warps by Thorne:

event: A point in spacetime; that is, a location in space at a specific moment of time. Alternatively, something that happens at a point in spacetime, for example, the explosion of a firecracker.

freely falling object: An object on which no forces act except gravity.

local inertial reference frame: A reference frame on which no forces except gravity act, that falls freely in response to gravity’s pull, and that is small enough for tidal gravitational accelerations to be negligible inside it.

tidal gravity [(tidal force)]: Gravitational accelerations that squeeze objects along some directions and stretch them along others. Tidal gravity produced by the Moon and Sun is responsible for the tides on the Earth’s oceans.

Spacetime curvature and tidal gravity [(tidal force)] are different names for the same thing. (Taylor and Wheeler concur on pg. 2-7 of Exploring Black Holes (http://www.eftaylor.com/pub/chapter2.pdf).)
It follows from the above that:

The spacetime throughout an inertial frame is flat.

The definition of an inertial frame allows them to be arbitrarily large (they need be only “small enough”).
From pg. 98 of Black Holes & Time Warps (the italicized statement is Einstein’s): “In any small, freely falling reference frame anywhere in our real, gravity-endowed Universe, the laws of physics must be the same as they are in an inertial reference frame in an idealized, gravity-free universe. Einstein called this the principle of equivalence, because it asserts that small, freely falling frames in the presence of gravity are equivalent to inertial frames in the absence of gravity. This assertion, Einstein realized, had an enormously important consequence: It implied that, if we merely give the name "inertial reference frame" to every small, freely falling reference frame in our real, gravity-endowed Universe (for example, to a little laboratory that you carry as you fall over a cliff), then everything that special relativity says about inertial frames in an idealized universe without gravity will automatically also be true in our real Universe. Most importantly, the principle of relativity must be true: All small, inertial (freely falling) reference frames in our real, gravity-endowed Universe must be "created equal"; none can be preferred over any other in the eyes of the laws of physics.”

GR predicts that an inertial frame can exist everywhere except at the center of a black hole. Here are confirmations:

From pg. 2-4 of Exploring Black Holes (http://www.eftaylor.com/pub/chapter2.pdf): “Our old, comfy, free-float (inertial) frame carries us unharmed to the center of a black hole. Well, unharmed almost to the center! ... No one can stop us from observing a black hole from an unpowered spaceship that drifts freely toward the black hole from a great distance, then plunges more and more rapidly toward the center. Over a short time the spaceship constitutes a "capsule of flat spacetime" hurtling through curved spacetime. It is a free-float frame like any other. Special relativity makes extensive use of such frames, and special relativity continues to describe Nature correctly for an astronaut in a local free-float frame, even as she falls through curved spacetime, through the horizon, and into a black hole.” From pg. 2-6: “Confronted by tidal accelerations, how can we define a free-float frame falling into a black hole? At the center of the black hole we cannot; general relativity predicts infinite tidal accelerations there. However, short of the center, [we limit] the space and the time—the region of spacetime!—in which experiments are conducted.” See also the section free-float frame on pg. 2-31.

From pg. 21 of Black Holes: A Traveler’s Guide by Pickover: “If you were approaching a 10 solar masses black hole with a radius of 30 kilometers, you would be killed long before you reached the horizon, at an altitude of 400 kilometers. However, you could reach the horizon of a 1,000 solar masses black hole, and even be able to explore the interior of a 10 million solar masses black hole. The tidal forces at the horizon of this gigantic black hole would be weaker than those produced by Earth, which are already impossible for us to feel.”

Another online reference (http://casa.colorado.edu/~ajsh/singularity.html): “In a supermassive black hole the tidal forces are weaker, and you could survive well inside the horizon of the black hole before being torn apart.”
From the Black Holes FAQ (http://cosmology.berkeley.edu/Education/BHfaq.html): “You can think of the horizon as the place where the escape velocity equals the velocity of light. Outside of the horizon, the escape velocity is less than the speed of light, ...”

A definition of escape velocity (http://en.wikipedia.org/wiki/Escape_velocity): “In physics, for a given gravitational field and a given position, the escape velocity is the minimum speed an object without propulsion needs to have to move away indefinitely from the source of the field, as opposed to falling back or staying in an orbit within a bounded distance from the source.”

Now the proof:

Let X be an inertial frame falling through the horizon of a black hole. Since the spacetime throughout an inertial frame is flat, it must be possible to set up an inertial frame Y that extends throughout X and in which a free test particle, that is above the horizon and moves away from the black hole indefinitely, stays at rest. But GR predicts that nothing may pass outward through a horizon. Then Y cannot extend below the horizon (if only because otherwise a latticework of synchronized clocks, that stays at rest with respect to Y and is spread throughout Y, would be passing outward through the horizon), and so the spacetime cannot be flat throughout X. Then X cannot be an inertial frame. That is, an inertial frame cannot fall through the horizon of a black hole. GR required this conclusion, yet the theory predicts otherwise, contradicting itself, so it is self-inconsistent.

Note that:
The definitions of flat spacetime and inertial frame in the supporting info account for the tidal force. The tidal force in X need not be nonexistent.
X can be arbitrarily small and its duration can be arbitrarily short. Then the tidal force in X can be nonexistent in the limit.

Since the spacetime throughout an inertial frame is flat, it must be possible to set up an inertial frame that extends throughout X and in which a free test particle that is above the horizon and rising indefinitely stays at rest.

You mean at rest with respect to the inertial frame X, or rest with respect to the horizon? It's not clear what you mean by "rising indefinitely."

You mean at rest with respect to the inertial frame X, or rest with respect to the horizon? It's not clear what you mean by "rising indefinitely."
The particle, which is above the horizon and rises (moves outward, away from the black hole) indefinitely, stays at rest in its own inertial frame that extends throughout X.

The particle, which is above the horizon and rises (moves outward, away from the black hole) indefinitely, stays at rest in its own inertial frame that extends throughout X.

But the rest frame of such a particle is clearly non-inertial. The only inertial frames in the vicinity of a black hole are free-fall frames, and since the particle is moving *against* the influence of gravity, it is clearly not in free-fall.

Objects in free fall can include objects that are rising ("moving *against* the influence of gravity"). An inertial frame can be rising, according the definitions in the OP (in particular see Thorne's, and see his definition for "freely falling object"). That's intuitively obvious too, since an inertial frame (and the objects in it) can at once be both falling (relative to some object) and rising (relative to some other object). For example, the Apollo module rose above the Earth as it fell toward the Moon.

Objects in free fall can include objects that are rising ("moving *against* the influence of gravity").

Okay, so the object in question has some initial velocity away from the black hole, and this velocity is furthermore greater than the relevant escape velocity. Its speed away from the black hole will constantly diminish, but never reach zero. Is this what you had in mind?

So the problem, then, is that another object with identical initial velocity, but starting below the horizon, will exhibit acceleration relative to the first object? Which would seem to contradict the idea that there's some inertial frame straddling the horizon which contains them both at the starting of the experiment?

Okay, so the object in question has some initial velocity away from the black hole, and this velocity is furthermore greater than the relevant escape velocity. Its speed away from the black hole will constantly diminish, but never reach zero. Is this what you had in mind?
Yes.

So the problem, then, is that another object with identical initial velocity, but starting below the horizon, will exhibit acceleration relative to the first object? Which would seem to contradict the idea that there's some inertial frame straddling the horizon which contains them both at the starting of the experiment?
The problem stems from the fact that the particle's inertial frame (that extends throughout X) cannot extend below the horizon, even though the spacetime throughout X is flat. If the particle's inertial frame extended below the horizon, a latticework of synchronized clocks spread throughout the frame would be passing outward through the horizon, which GR disallows.

Okay; I think that's essentially the same thing I was talking about.

I think that the resolution may lie in the fact that the horizon appears to be in different places according to the free-falling observer and the escaping object.

From http://en.wikipedia.org/wiki/Event_horizon

"Observers that fall into the hole are moving with respect to the distant observer, and so perceive the horizon as being in a different location, seeming to recede in front of them so that they never contact it."

Which is to say that it doesn't make sense to talk about an inertial frame straddling an event horizon, as from the perspective of a free-fall observer, the event horizon is never crossed.

That statement is talking about an apparent horizon, which is a relative concept (its location depends on the observer's frame), in contrast to an absolute horizon, which is the same in all frames. Almost always "horizon" or "event horizon" of a black hole refers to an absolute horizon, as it does in the OP. It does make sense to talk about an inertial frame straddling an absolute horizon.

As far as I understand it, the the absolute event horizon coincides with the apparent event horizon once the black hole has formed and had time to settle down. So I don't think it makes a difference, unless you're doing the experiment during the formation of the black hole in question.

As far as I understand it, the the absolute event horizon coincides with the apparent event horizon once the black hole has formed and had time to settle down. So I don't think it makes a difference, unless you're doing the experiment during the formation of the black hole in question.
Agreed to both. If the statement you quoted from Wikipedia is referring to an absolute horizon, then it must be referring to some optical illusion. The absolute horizon is at a specific location (unchanging for a quiescent black hole) agreed upon by all observers, so an observer can definitely cross it, and so can an inertial frame. Taylor and Wheeler and other sources confirm that, as noted in the OP.

From http://en.wikipedia.org/wiki/Event_horizon

"Observers that fall into the hole are moving with respect to the distant observer, and so perceive the horizon as being in a different location, seeming to recede in front of them so that they never contact it."
I think I get what the author is talking about here. Notice “perceive” and “seeming”. The author is talking about a third type of horizon, a perceived horizon. As the observer approaches the absolute horizon, a wall of blackness looms since no light is passing outward through the horizon. When the observer crosses the absolute horizon, they do not suddenly see the singularity. Instead the wall of blackness recedes, always hiding the singularity, because all photons fall below a horizon, so no light can get to the observer from some point below. The observer could still see part of their ship that is below them, though; those photons—whose direction of motion is upward—do fall, but the observer falls faster, impacting them at the speed of light. The observer will not impact the photons below the receding perceived horizon; those photons reach the singularity before the observer does.

Yeah, I think that distinction is important. To my understanding, the absolute event horizon is defined in terms of the regions where light travelling radially outward from the black hole will eventually escape off to infinity. This would coincide with the perceived horizon for a distant observer. While all observers will agree on where this horizon is, for an observer in the vicinity of a black hole, the perceived horizon would correspond to regions from which radially directed light can reach the observer's radius (which is finite). This should be a smaller region than the absolute horizon. Which is to say that GR doesn't prevent objects from passing out of an absolute horizon, it's just that they can't do so with sufficient radial velocity to escape to infinity (perhaps someone more versed in GR could jump in to confirm/disconfirm this point?).

Another issue I think is important is to limit the free-fall inertial frame in both size and time. Generally, one must limit both of these aspects to apply the equivalence principle. Imagine a spaceship falling accross the absolute horizon, and imagine it emits two light pulses, one from the tail (which is above the horizon), and the other from the nose (which is below the horizon), both radially outward. For a short period of time, both of these pulses will seem to travel radially outward at c, exactly as they should in flat spacetime. If the spaceship waits long enough, however, he will eventually observe the nose pulse changing directions and heading back towards the black hole, while the tail pulse will fly off to infinity. Note that the spaceship's speed can be arbitrarily small as he's crossing the absolute horizon (suppose he fired retro-rockets right before reaching the horizon, resulting in a speed just barely below the relevant escape velocity). Of course, some other effect will probably tip the observer off that he's not actually in flat spacetime before he actually encounters the nose pulse, but you get the idea.

I don't think it's a problem for a hypothetical lattice of clocks to cross the event horizon, as they do not actually exist. You should rephrase the OP in terms of actual physical quantities, like the light pulses in my example. As it's written, it's dangerously close to a tautology (i.e., the assertion that because objects can't exit a horizon, an inertial frame can't straddle a horizon is what you're trying to prove, but you speak of it as given).

To my understanding, the absolute event horizon is defined in terms of the regions where light travelling radially outward from the black hole will eventually escape off to infinity. This would coincide with the perceived horizon for a distant observer.
Agreed to both.

While all observers will agree on where this horizon is, for an observer in the vicinity of a black hole, the perceived horizon would correspond to regions from which radially directed light can reach the observer's radius (which is finite). This should be a smaller region than the absolute horizon.
Agreed. But it’s a side topic, because the OP refers to only the absolute horizon. The proof in the OP does not depend on any other type of horizon.

Another issue I think is important is to limit the free-fall inertial frame in both size and time.
Yes, that limitation is given in the definition of an inertial frame. In the proof in the OP, X can be arbitrarily small (but need not be, as the supporting info points out; it need only be “small enough”), and the case is made in an arbitrary short time. If a latticework of synchronized clocks in the particle’s frame cannot pass outward through the horizon for even a moment, the case is made.

Generally, one must limit both of these aspects to apply the equivalence principle. Imagine a spaceship falling accross the absolute horizon, ...
I do get the idea that in some thought experiments no self-inconsistency is proven because of the time required. That’s why I designed the proof in the OP to require no more time than a moment.

If the spaceship waits long enough, however, he will eventually observe the nose pulse changing directions and heading back towards the black hole, ...
The nose pulse will always fall. All photons below the horizon fall, according to GR. But the nose pulse will still recede at the speed of light as measured by the crew. Consider that if the nose pulse was emitted right at the horizon, it could not rise above the horizon at all, because GR doesn’t allow that.

Note that the spaceship's speed can be arbitrarily small as he's crossing the absolute horizon (suppose he fired retro-rockets right before reaching the horizon, resulting in a speed just barely below the relevant escape velocity).
Every observer must cross the horizon at exactly c as the observer measures. (The speed can be measured in the limit relative to an object hovering at zero distance in the limit above the horizon.) That’s how the crew could measure the nose pulse to recede at c even if it was emitted right at the horizon. For more info search for “The horizon has some very strange geometrical properties” here (http://cosmology.berkeley.edu/Education/BHfaq.html).

I don't think it's a problem for a hypothetical lattice of clocks to cross the event horizon, as they do not actually exist. You should rephrase the OP in terms of actual physical quantities, like the light pulses in my example.
The lattice of clocks need only be able to exist in principle. That’s true of objects in all thought experiments in physics. The supporting info by Taylor and Wheeler in the OP shows that “In principle one can set up a latticework of synchronized clocks in a free-float frame”.

As it's written, it's dangerously close to a tautology (i.e., the assertion that because objects can't exit a horizon, an inertial frame can't straddle a horizon is what you're trying to prove, but you speak of it as given).
There’s no tautology or anything close to that in the OP. That an inertial frame cannot straddle a horizon is proven. It’s not a given. The OP starts with a given that an inertial frame is straddling a horizon, and then proves that GR disallows that. GR predicts that an inertial frame can fall through a horizon, and it predicts that nothing can pass outward through a horizon. I showed that these predictions are mutually exclusive (they cannot both be true). If you still think there’s a tautology, please elaborate on why you think so.

I didn't say it was actually a tautology, just close. "A fixed lattice of synchronized clocks" is almost the definition of an inertial frame, so simply invoking it doesn't make for a very compelling demonstration that it can't straddle an event horizon. Maybe this point is obvious to you, but it's the crucial step in your thought experiment, so it probably deserves more than a parenthetical comment. You should explicitly demonstrate why this is so. For example, you could consider an identical object with identical initial velocity that starts out just below the horizon. Then, you would demonstrate that its velocity (as seen by the escaping object) must be nonzero even on asymptotically small time scales. If you want to use an entire lattice of clocks, that's fine too, but you should introduce all of the objects in the set-up of a thought experiment, not in a parenthetical comment halfway through.

More substantially, I'm not convinced there's anything inconsistent about the rest frame of an escaping object not extending below the horizon. You started with the assumption that an infalling frame could straddle the horizon, but none of your sources say anything about the frames of escaping objects near the horizon. I don't see any reason why it can't be the case that only infalling frames can straddle the horizon. Indeed, it's seems to me that any theory of relativity sufficient for describing an event horizon should have such an asymmetry between infalling and escaping frames, as it reflects the fact that you can only cross a horizon in one direction.

"A fixed lattice of synchronized clocks" is almost the definition of an inertial frame, so simply invoking it doesn't make for a very compelling demonstration that it can't straddle an event horizon.
That is not all that is invoked in the proof. It is X, the inertial frame falling through the horizon, and not the particle’s inertial frame, that is proven to be unable to straddle the horizon, based on the conclusion that the particle’s inertial frame cannot extend throughout X. The lattice is invoked in the particle’s frame to prove that the particle’s frame cannot extend throughout X. That fact is used to show that an inertial frame cannot fall through a horizon. You seem to be treating X and the particle’s frame as the same frame. They are two different frames.

Maybe this point is obvious to you, but it's the crucial step in your thought experiment, so it probably deserves more than a parenthetical comment. You should explicitly demonstrate why this is so.
The proof already covers why the particle’s inertial frame cannot extend throughout X. If it could, then that would violate GR’s prediction that nothing can pass outward through a horizon, since in principle a latticework of synchronized clocks can be set up throughout the particle's frame, in which case the lattice would be passing outward through the horizon. And the proof already covers why X cannot fall through a horizon. If it could, then the particle’s frame could extend throughout X, because the spacetime throughout an inertial frame is flat, and X is given as an inertial frame. All the bases are covered in the proof.

For example, you could consider an identical object with identical initial velocity that starts out just below the horizon. Then, you would demonstrate that its velocity (as seen by the escaping object) must be nonzero even on asymptotically small time scales. If you want to use an entire lattice of clocks, that's fine too, but you should introduce all of the objects in the set-up of a thought experiment, not in a parenthetical comment halfway through.
An escaping object could not see any object below the horizon, because nothing can pass outward through a horizon, including light.

Einstein didn’t introduce all of the objects in his thought experiments in an initial set-up. Instead he made such introductions when needed, and in his relativity of simultaneity thought experiment he introduces the lightning strokes in an example in parentheses: “(e.g. the two strokes of lightning A and B)”.

You started with the assumption that an infalling frame could straddle the horizon, but none of your sources say anything about the frames of escaping objects near the horizon.
The sources needn’t have said anything about that explicitly. They need only provide the info from which incontrovertible conclusions are drawn. Can you show that any of the statements in the proof are incorrect or do not follow from the sources or prior statements in the proof?

I don't see any reason why it can't be the case that only infalling frames can straddle the horizon. Indeed, it's seems to me that any theory of relativity sufficient for describing an event horizon should have such an asymmetry between infalling and escaping frames, as it reflects the fact that you can only cross a horizon in one direction.
The proof in the OP shows rigorously that an infalling inertial frame (such as X) cannot straddle a horizon. If it could, then the particle’s frame could extend throughout X, since the spacetime throughout an inertial frame is flat. But the particle’s frame cannot extend throughout X, if only because otherwise a latticework of synchronized clocks spread throughout the particle’s frame would be passing outward through a horizon, which GR disallows. The asymmetry you describe can be had only via a self-inconsistency of GR. The fact that you can cross a horizon in only one direction is a symptom of the self-inconsistency.

I'm not convinced that the inertial frame of the escaping object not extending below the horizon has any consequences for X. It's not at all clear to me that the free-fall frames of objects in some inertial frame X must necessarily include all of X, particularly when said objects are moving at relativistic velocities (even without an event horizon). All your proof offers to back up this step is the statement "spacetime cannot be flat throughout X." But the universe specified by the thought experiment isn't flat *anywhere*, so this by itself doesn't contradict any assumptions. You need to show how the limited extent of the particle's frame implies measurable tidal effects in X, on an instantaneous time scale.

To give an example of why the free-fall frames of particles with different velocities would have different extents, consider two particles both falling towards a massive object with different initial speeds. The tidal forces will grow more quickly in the faster object's frame, and so the time scale on which it can be considered inertial will be shorter (even when the two particles are very close together!). Which is to say that the extent in spacetime of a freefalling object's inertial frame depends not only on the curvature of spacetime but also on the object's velocity. Thus, the escaping object in your thought experiment could well be in the inertial frame of an infalling observer, and yet the infalling observer would not be in the freefall frame of the escaping object. This seems natural to me, since the infalling observer can still receive light pulses from the escaping object, but not vice-versa.

The fact that you can cross a horizon in only one direction is a symptom of the self-inconsistency.

Am I to understand that a consistent theory of gravity would, in your opinion, allow crossing of event horizons in both directions? Which would be to say that there is no such thing as an event horizon.

It's not at all clear to me that the free-fall frames of objects in some inertial frame X must necessarily include all of X, particularly when said objects are moving at relativistic velocities (even without an event horizon). All your proof offers to back up this step is the statement "spacetime cannot be flat throughout X."
The step “it must be possible to set up an inertial frame that extends throughout X and in which a free test particle that is above the horizon and rising indefinitely stays at rest” is not backed up by the statement "spacetime cannot be flat throughout X"; it’s backed up by what precedes it: “Since the spacetime throughout an inertial frame is flat”, which in turn is backed up by Taylor and Wheeler’s definition of flat spacetime in the OP: “Region of spacetime in which it is possible to set up a free-float (inertial) reference frame.”

But the universe specified by the thought experiment isn't flat *anywhere*, so this by itself doesn't contradict any assumptions. You need to show how the limited extent of the particle's frame implies measurable tidal effects in X, on an instantaneous time scale.
The definitions of flat spacetime and inertial frame in the OP already take the tidal force into account. I can depend on those definitions, and need not further account for the tidal force in the proof.

To give an example of why the free-fall frames of particles with different velocities would have different extents, consider two particles both falling towards a massive object with different initial speeds. The tidal forces will grow more quickly in the faster object's frame, and so the time scale on which it can be considered inertial will be shorter (even when the two particles are very close together!). Which is to say that the extent in spacetime of a freefalling object's inertial frame depends not only on the curvature of spacetime but also on the object's velocity.
I can’t tell for sure what part of the proof in the OP you’re contesting here. Are you saying that an inertial frame cannot fall through a horizon? GR predicts that an inertial frame can fall through a horizon; multiple references in the OP support that. That’s all I need to let X be given as an inertial frame falling through a horizon. And the tidal force in X need not be nonexistent, according to the definition of an inertial frame.

And velocity relative to what? All objects cross a horizon at c in the limit relative to an object hovering at zero distance in the limit above the horizon.

Also, spacetime curvature is the tidal force; they are synonymous (see supporting info in the OP). Then “the extent in spacetime of a freefalling object's inertial frame” cannot depend on the curvature of spacetime (tidal force) and something else, where that something else is included solely because it affects the tidal force. That is double-counting the tidal force (or spacetime curvature, take your pick). Instead you measure only the tidal force to determine whether or not the spacetime in a region is flat (negligibly curved), and if it is flat then an inertial frame can extend throughout it. And the tidal force can be measured by an observer without consideration of the observer’s velocity relative to something else (makes sense, since the observer can at once have various velocities relative to various other objects).

This seems natural to me, since the infalling observer can still receive light pulses from the escaping object, but not vice-versa.
What seems natural to you is a symptom of GR’s self-inconsistency. The proof in the OP shows that GR is self-inconsistent where “not vice-versa” applies.

Am I to understand that a consistent theory of gravity would, in your opinion, allow crossing of event horizons in both directions? Which would be to say that there is no such thing as an event horizon.
The proof in the OP shows that a self-consistent theory of gravity in which SR applies locally everywhere cannot predict black holes (in particular, their horizons).

Hi zanket,
The step “it must be possible to set up an inertial frame that extends throughout X and in which a free test particle that is above the horizon and rising indefinitely stays at rest” is not backed up by the statement "spacetime cannot be flat throughout X"; it’s backed up by what precedes it: “Since the spacetime throughout an inertial frame is flat”, which in turn is backed up by Taylor and Wheeler’s definition of flat spacetime in the OP: “Region of spacetime in which it is possible to set up a free-float (inertial) reference frame.”

If Y covers the same region of spacetime as X, then there's only a short time involved...

I think that if Y covers the same space as X at a given instant, but an indefinite time, then it's not clear that the spacetime in Y is flat.

Actually, I'm not sure that it would follow even for a short time. We know that the spacetime isn't exactly flat, it's only flat enough that the difference doesn't matter... but it seems possible that while the difference doesn't matter in X, it could matter in Y.

There is another difficulty with Y - what is Y's velocity relative to X? I think that it must be c, or possibly more, which makes it very difficult to define at all. This makes sense, because with Y, you are attempting so define an inertial reference frame straddling the event horizon in which the event horizon is stationary - which I don't think can be done.

If Y covers the same region of spacetime as X, then there's only a short time involved...

I think that if Y covers the same space as X at a given instant, but an indefinite time, then it's not clear that the spacetime in Y is flat.
The duration for the thought experiment can be arbitrarily short, as noted at the bottom of the OP.

Actually, I'm not sure that it would follow even for a short time. We know that the spacetime isn't exactly flat, it's only flat enough that the difference doesn't matter... but it seems possible that while the difference doesn't matter in X, it could matter in Y.
The "not exactly flat" argument is covered at the bottom of the OP. The tidal force (synonymous with spacetime curvature) in X can be nonexistent in the limit, because X can be arbitrarily small and the duration of the thought experiment can be arbitrarily short. When the tidal force is nonexistent in the limit in X, it is also nonexistent in the limit in Y, because Y is in X. The tidal force has no effect on the conclusion of the thought experiment.

There is another difficulty with Y - what is Y's velocity relative to X? I think that it must be c, or possibly more, which makes it very difficult to define at all.
If Y's velocity relative to X must be c or more, to be at rest relative to a free test particle in X, that's GR's fault not mine. The equivalence principle says that X is equivalent to any other inertial frame J that is wholly above the horizon (the tidal force in J could even be slightly greater than in X). But an inertial frame K, that extends throughout J, can be at rest relative to any free test particle in J. You are describing a nonequivalence between inertial frames.

To be an inertial frame, Y need only meet the definition of an inertial frame, which it can, because the spacetime throughout X is flat and the definition of flat spacetime is a region of spacetime in which it is possible to set up an inertial frame. The spacetime throughout X must be flat because X is given to be an inertial frame; X can be set up only in flat spacetime.

This makes sense, because with Y, you are attempting so define an inertial reference frame straddling the event horizon in which the event horizon is stationary - which I don't think can be done.
Y is not stationary relative to the horizon. It is stationary relative to a free test particle in X that is above the horizon and moving outward (away from the black hole).

Hi zanket,
The duration for the thought experiment can be arbitrarily short, as noted at the bottom of the OP.
My apologies, I misunderstood.

The "not exactly flat" argument is covered at the bottom of the OP. The tidal force (synonymous with spacetime curvature) in X can be nonexistent in the limit, because X can be arbitrarily small and the duration of the thought experiment can be arbitrarily short. When the tidal force is nonexistent in the limit in X, it is also nonexistent in the limit in Y, because Y is in X. The tidal force has no effect on the conclusion of the thought experiment.
Hang on, you're not taking a limit... you have a reference frame of some non-zero size, with negligible (not zero) tidal forces. If you take the limit so that the tidal force is zero, then the reference frame is also of zero size.

So, I'm not sure that the tidal force in Y is necessarily negligible. I think that some maths will be necessary to be sure either way.


If Y's velocity relative to X must be c or more, to be at rest relative to a free test particle in X, that's GR's fault not mine.
Blame is irrelevant. What is relevant is that it is questionable whether Y is a valid frame of reference.

The equivalence principle says that X is equivalent to any other inertial frame J that is wholly above the horizon (the tidal force in J could even be slightly greater than in X). But an inertial frame K, that extends throughout J, can be at rest relative to any free test particle in J. You are describing a nonequivalence between inertial frames.
I don't follow what you mean. I'm happy that X, J, and K can be inertial reference frames, but I'm not sure about Y.

To be an inertial frame, Y need only meet the definition of an inertial frame, which it can, because the spacetime throughout X is flat and the definition of flat spacetime is a region of spacetime in which it is possible to set up an inertial frame. The spacetime throughout X must be flat because X is given to be an inertial frame; X can be set up only in flat spacetime.
The spacetime in question is almost flat. And regardless of local flatness of spacetime, I don't think it is meaningful to define a reference frame with a speed of c or greater relative to a given reference frame.


Y is not stationary relative to the horizon. It is stationary relative to a free test particle in X that is above the horizon and moving outward (away from the black hole).
The same point applies.
You are attempting to define an inertial reference frame straddling the event horizon in which the event horizon is moving in the wrong direction and the wrong speed - which I don't think can be done.



I think we need some quantitative analysis. I'm going to consider the velocity of the test particle in X, and see what happens.
I think you'll find that it is very tightly defined - so tightly, in fact, that tidal forces are no longer negligible

Hang on, you're not taking a limit... you have a reference frame of some non-zero size, with negligible (not zero) tidal forces. If you take the limit so that the tidal force is zero, then the reference frame is also of zero size.
When the region of spacetime is zero sized in the limit (not zero sized), the tidal force in the frame is nonexistent in the limit (not nonexistent).

So, I'm not sure that the tidal force in Y is necessarily negligible. I think that some maths will be necessary to be sure either way.
The tidal force in Y must be nonexistent in the limit when the tidal force in X is nonexistent in the limit. A region of spacetime that is zero sized in the limit in one frame is zero sized in the limit in any other frame in which that region exists, because any multiple—like a Lorentz factor for length or time, or a reciprocal thereof—of zero in the limit is zero in the limit.

Blame is irrelevant. What is relevant is that it is questionable whether Y is a valid frame of reference.

I don't follow what you mean. I'm happy that X, J, and K can be inertial reference frames, but I'm not sure about Y.
If Y is not an inertial frame, then GR violates the equivalence principle, which is GR's problem not mine. Look at the principle as stated by Einstein in the OP. X is equivalent to J, according to the principle. They would not be equivalent if K could be set up in J (where K is at rest relative to any given free test particle in J) but Y cannot be set up in X (where Y is at rest relative to any given free test particle in X).

The spacetime in question is almost flat.
The spacetime in X can be perfectly flat in the limit, which is the same as saying the tidal force in X can be nonexistent in the limit. Look at the definitions of an inertial frame in the OP. The spacetime of an inertial frame need not be perfectly flat (the tidal force can be negligible), even though the definition of flat spacetime is a region of spacetime in which it is possible to set up an inertial frame. Then a region of “flat spacetime” is negligibly curved (almost flat). And Einstein’s equivalence principle refers to “small” inertial frames, having a negligible tidal force.

And regardless of local flatness of spacetime, I don't think it is meaningful to define a reference frame with a speed of c or greater relative to a given reference frame.
Y is just an inertial frame that extends throughout X and is at rest relative to the particle. Then you’re implying that a particle can move at c or greater in X. But GR doesn’t allow a material object to move at c or greater in any inertial frame. GR cannot save itself from a self-inconsistency by contradicting itself in another way. If the particle must move at c or greater in X, then that’s GR’s fault, not mine, because GR allows X (an inertial frame falling through the horizon of a black hole) to exist and allows a free test particle that is anywhere above the horizon, including in X, to move outward (away from the black hole) indefinitely.

You are attempting to define an inertial reference frame straddling the event horizon in which the event horizon is moving in the wrong direction and the wrong speed - which I don't think can be done.
You’re implicitly assigning a special status to X, an inertial frame falling through a horizon of a black hole. But the equivalence principle says that X is equivalent to J. To make your case, you would need to show than an inertial frame K cannot be set up in J such that it extends throughout J and is at rest relative to some free test particle in J. But you can’t do that, because there is no “wrong direction” or “wrong speed” possible in J. If such is possible in X, then GR violates the equivalence principle (X would be nonequivalent to J). GR cannot save itself from a self-inconsistency by contradicting itself in another way.

I think we need some quantitative analysis. I'm going to consider the velocity of the test particle in X, and see what happens.
I think you'll find that it is very tightly defined - so tightly, in fact, that tidal forces are no longer negligible
Think about it intuitively. Call the room you’re in now inertial frame X. Let X fall through the horizon of a black hole so big that the tidal force in X is 10^7000 times weaker than the tidal force in your room now. If you did an experiment and found that a particle in X moved faster than c, or you found that a latticework of clocks at rest with respect to some particle in X could not be spread throughout X, do you think the extremely negligible tidal force in X could explain that? If it did, then we could duplicate those experiments in inertial labs near Earth, or in J, according to the equivalence principle. But while GR predicts such outcomes of experiments in X, it forbids such outcomes in J, even though the theory says that X is equivalent to J. That’s a self-inconsistency.

In any case, it would be impossible for you to show that the tidal force need be significant in either X or Y, for the reason I gave above (any multiple—like a Lorentz factor for length or time, or a reciprocal thereof—of zero in the limit is zero in the limit).

Hi zanket,
Thanks. I believe that I've reached the heart of the problem.

If X is an inertial reference frame falling across the event horizon, then all particles in X are moving toward the black hole.

I believe that it is not possible to identify a particle moving away from the black hole in X unless X is large enough that tidal forces are significant.
I.e. if the movement of the particle is known to sufficient precision that an observer in X can determine that it is moving away from the hole, then the observer has sufficient precision to detect tidal forces in X.

Maths will follow to demonstrate this.

When the region of spacetime is zero sized in the limit (not zero sized), the tidal force in the frame is nonexistent in the limit (not nonexistent).The tidal force in Y must be nonexistent in the limit when the tidal force in X is nonexistent in the limit. A region of spacetime that is zero sized in the limit in one frame is zero sized in the limit in any other frame in which that region exists, because any multiple—like a Lorentz factor for length or time, or a reciprocal thereof—of zero in the limit is zero in the limit.

I don't see where you've shown that the limit of any multiple of the frame size must also be zero. If the mulitplier goes to oo (i.e., the frames are moving at the speed of light relative to one another), you can get a nonzero limit. Recall that, from the perspective of any observer outside the event horizon, the infalling object never appears to cross the horizon; they slow down, redshift and, in the limit disappear. That is to say that a small increment of time in X (the moments right up till the infalling object crosses the horizon) gets expanded in Y by an undounded Lorentz factor, due to the relative velocity approaching c. If you define Y to include all of X (as you do in the OP), it cannot be inertial, because it is not time-limited. Only other escaping events can be time-limited in Y, and so if Y is to be inertial, it cannot straddle the horizon. The assumption that an infalling X is inertial and straddles the horizon doesn't allow you to construct an escaping frame Y that is both inertial and straddles the horizon, and so doesn't contradict the requirement that nothing can exit a horizon.

If X is an inertial reference frame falling across the event horizon, then all particles in X are moving toward the black hole.
Not according to GR, which allows a particle that is anywhere above the horizon, including in X (which is partly above the horizon), to move away from the black hole indefinitely. (The escape velocity anywhere above a horizon is less than c.) Then not all particles in X must move toward the black hole.

I.e. if the movement of the particle is known to sufficient precision that an observer in X can determine that it is moving away from the hole, then the observer has sufficient precision to detect tidal forces in X.
It would be possible in principle to detect a tidal force in X; that is possible in any nonzero-sized frame. The tidal force in an inertial frame is (by definition) negligible, not nonexistent.

I don't see where you've shown that the limit of any multiple of the frame size must also be zero. If the mulitplier goes to oo (i.e., the frames are moving at the speed of light relative to one another), you can get a nonzero limit. Recall that, from the perspective of any observer outside the event horizon, the infalling object never appears to cross the horizon; they slow down, redshift and, in the limit disappear. That is to say that a small increment of time in X (the moments right up till the infalling object crosses the horizon) gets expanded in Y by an undounded Lorentz factor, due to the relative velocity approaching c. If you define Y to include all of X (as you do in the OP), it cannot be inertial, because it is not time-limited. Only other escaping events can be time-limited in Y, and so if Y is to be inertial, it cannot straddle the horizon. The assumption that an infalling X is inertial and straddles the horizon doesn't allow you to construct an escaping frame Y that is both inertial and straddles the horizon, and so doesn't contradict the requirement that nothing can exit a horizon.
You conclude that “if Y is to be inertial, it cannot straddle the horizon”, and the proof in the OP agrees (it concludes: “Then Y cannot extend below the horizon”). But whereas you imply that this is a problem with the proof, it actually proves that GR is self-inconsistent. The proof in the OP makes the following case, rephrased: The spacetime throughout X must be flat (negligibly curved), otherwise X could not be set up in that region of spacetime, according to Taylor and Wheeler’s definition of flat spacetime. (We know that the spacetime throughout X must be flat also because the tidal force is synonymous with spacetime curvature, and by the definition of an inertial frame the tidal force in the frame is negligible.) Since the spacetime throughout X is flat, it must be possible to set up Y as described, again according to the definition of flat spacetime. If the horizon precludes that, that’s GR’s fault not mine. The definition of flat spacetime is a “region of spacetime in which it is possible to set up a free-float (inertial) reference frame”. That definition makes no exception for a horizon. If Y cannot be set up as described, then it proves that the region of spacetime in X is not flat. Then X cannot be an inertial frame, in which case no inertial frame can fall through the horizon of a black hole, contradicting GR’s prediction to the contrary. GR’s self-inconsistency itself cannot be used against the proof of it; that is implicitly what you’re doing above.

Note also that you’re implicitly assigning a special status to X that wouldn’t apply to an inertial frame J that is wholly above a horizon. The equivalence principle (stated by Einstein in the OP) says that X is equivalent to J. Can you show that an inertial frame K cannot be set up in J such that it extends throughout J and is at rest relative to some free test particle in J? You can’t do that, because your explanation depends on a horizon, and there is no horizon in J. If X is nonequivalent to J, as you imply, then GR violates the equivalence principle.

Not according to GR, which allows a particle that is anywhere above the horizon, including in X (which is partly above the horizon), to move away from the black hole indefinitely.
Not if X is small enough to be considered inertial.

It would be possible in principle to detect a tidal force in X; that is possible in any nonzero-sized frame. The tidal force in an inertial frame is (by definition) negligible, not nonexistent.
Which means that X is not inertial.


Try it this way:
Consider an immensely large black hole, so large that the event horizon is essentially a flat surface. In fact, why not take the limit... consider a black hole of infinite size, so that the event horizon is in fact flat. In this limit, no particle can depart from the event horizon.

And there's the heart of the matter - X is flat in the limit, but in the limit, no particle in X can move away from the event horizon.


To put it yet another way, the only way to model a particle in X moving away from the black hole is to model X as non-inertial.

You conclude that “if Y is to be inertial, it cannot straddle the horizon”, and the proof in the OP agrees (it concludes: “Then Y cannot extend below the horizon”). But whereas you imply that this is a problem with the proof, it actually proves that GR is self-inconsistent.
What it shows is that the event horizon is a coordinate singularity in Y. A coordinate singularity is a singularity that can be removed by changing the coordinate system. For example, The North and South Poles are coordinate singularities in Earth's regular latitude and longitude coordinates.

The Schwarzchild geometry is an example of a common coordinate system in which the event horizon is a singularity:

Schwarzschild spacetime diagram (http://casa.colorado.edu/~ajsh/schwp.html)
http://casa.colorado.edu/~ajsh/st0.gif

To put it yet another way, the only way to model a particle in X moving away from the black hole is to model X as non-inertial.
Let's cover one point at a time.

Do you agree that part of an inertial frame falling through the horizon of a black hole is above the horizon? Do you agree that the escape velocity above the horizon is less than c?

Let's cover one point at a time.

Do you agree that part of an inertial frame falling through the horizon of a black hole is above the horizon?
Yes.

Do you agree that the escape velocity above the horizon is less than c?
Not in the inertial frame. In the inertial frame, the escape velocity is c throughout. If the escape velocity varies within the frame, it means that tidal forces are involved.

Not in the inertial frame. In the inertial frame, the escape velocity is c throughout. If the escape velocity varies within the frame, it means that tidal forces are involved.
Do you agree that any frame in which the escape velocity varies is not an inertial frame? Do you agree that the escape velocity varies in any nonzero-sized frame?

Do you agree that any frame in which the escape velocity varies is not an inertial frame?
Yes, I think that's correct.

Do you agree that the escape velocity varies in any nonzero-sized frame?
Only in non-flat spacetime.
I think that the escape velocity in any reference frame in flat space will not vary.

Do you agree that the escape velocity varies in any nonzero-sized frame?
Only in non-flat spacetime.
I think that the escape velocity in any reference frame in flat space will not vary.
Do you agree that the spacetime in any nonzero-sized frame in our real, gravity-endowed universe is curved (not flat)?

Do you agree that the spacetime in any nonzero-sized frame in our real, gravity-endowed universe is curved (not flat)?

That's probably true.

Since the spacetime throughout X is flat, it must be possible to set up Y as described, again according to the definition of flat spacetime.

No, this part doesn't work. This has been demonstrated in multiple ways already, but perhaps the easiest way to see it is the following: you're trying to construct an inertial frame associated with an object moving at the speed of light. Even in a perfectly flat universe (i.e., special relativity), you can't do this. The definitions you're using are vague and heuristic, and you're misapplying them in asserting that Y can be constructed as you specify.

Do you agree that the escape velocity varies in any nonzero-sized frame?
Only in non-flat spacetime.
I think that the escape velocity in any reference frame in flat space will not vary.

Do you agree that the spacetime in any nonzero-sized frame in our real, gravity-endowed universe is curved (not flat)?
That's probably true.
Do you agree that the escape velocity varies in any nonzero-sized frame in our real, gravity-endowed universe?

No, this part doesn't work. This has been demonstrated in multiple ways already, but perhaps the easiest way to see it is the following: you're trying to construct an inertial frame associated with an object moving at the speed of light. Even in a perfectly flat universe (i.e., special relativity), you can't do this. The definitions you're using are vague and heuristic, and you're misapplying them in asserting that Y can be constructed as you specify.
No object in the OP is moving at the speed of light, unless GR is self-inconsistent. What do you think is moving at c, and relative to what?

The definitions I'm using are vague only if Einstein, Taylor, Thorne, and Wheeler are vague; all the definitions in the OP come from them. I don't see how the definitions are vague. They've been around for decades and no doubt used by thousands of people.

To make your case, you need to be more specific about what I'm misapplying, and you need to resolve any contradiction with GR. For example, if you think the particle is moving at c, then you need to explain how a particle above the horizon cannot move away from the black hole indefinitely, which contradicts GR's prediction of an escape velocity less than c above the horizon. As far as I can tell, you think I'm misapplying something only because I show a self-inconsistency of GR. But that self-inconsistency is not my problem; it's GR's problem.

No object in the OP is moving at the speed of light, unless GR is self-inconsistent. What do you think is moving at c, and relative to what?

As has been mentioned repeatedly, the escaping object is moving at (or extremely close to) c relative to X (actually, relative to *any* inertial frame, since c is frame-invariant). We know that the escape velocity approaches c at the horizon, so if X straddles the horizon and contains the escaping object, the escaping object must be moving at c. If you look at what happens to the Lorentz transforms as the velocity approaches c, you'll find that the entire universe gets length-contracted down to a single point. It becomes impossible, then, to limit the spacetime extent of such a frame.


The definitions I'm using are vague only if Einstein, Taylor, Thorne, and Wheeler are vague; all the definitions in the OP come from them.

Yes, they are still vague. Einstein wrote lots of material for non-physicists, and every book on GR that I've ever seen contains introductory sections with hand-waving descriptions of the type you use here. When you introduce a definition that is stated in terms of differential geometry, then I will be satisfied.

As has been mentioned repeatedly, the escaping object is moving at (or extremely close to) c relative to X (actually, relative to *any* inertial frame, since c is frame-invariant). We know that the escape velocity approaches c at the horizon, so if X straddles the horizon and contains the escaping object, the escaping object must be moving at c.
No, the escaping object (the particle) need not move at c to escape. The escape velocity is less than c above the horizon, where the particle is. There's no problem with the particle moving at extremely close to c relative to X.

Yes, they are still vague. Einstein wrote lots of material for non-physicists, and every book on GR that I've ever seen contains introductory sections with hand-waving descriptions of the type you use here. When you introduce a definition that is stated in terms of differential geometry, then I will be satisfied.
You need to prove that the definitions are vague. Prove that something about them is ambiguous. Where they are contained is not such proof.

No, the escaping object (the particle) need not move at c to escape. The escape velocity is less than c above the horizon, where the particle is. There's no problem with the particle moving at extremely close to c relative to X.

The fact that the Lorentz transforms blow up in the limit is, by definition, a problem for constructing inertial frames for objects with speeds approaching c. The Lorentz multipliers are, after all, continuous functions of velocity. X is defined as an asymptotically small interval of spacetime centered on the event horizon. The event horizon is moving at c in such a frame, so any escaping object must also be going at least that fast relative to X. The assumption that X is effectively flat implies that the escape velocity is effectively constant throughout X (as Pete has pointed out).

Also, the definition of "escape velocity" is in terms of objects than can escape off to infinity, so it's an inherently non-local property. Any experiment to determine escape velocity necessarily requires unlimited time, and so can't be carried out in X (which was defined to have asymptotically small extent in both space and time). So, as far as X is concerned, Y may as well be moving at c.

Do you agree that the escape velocity varies in any nonzero-sized frame in our real, gravity-endowed universe?

Sure. Do you agree that X is not inertial?

I don't think you can have it both ways, zanket... as soon as you acknowledge that the escape velocity varies in X, then you are acknowledging that X is not inertial.

If you take the limit of X being inertial then you are also implicitly taking the limit of zero variation in escape velocity.

The fact that the Lorentz transforms blow up in the limit is, by definition, a problem for constructing inertial frames for objects with speeds approaching c.
There is no such problem for speeds less than c. The Lorentz transforms work fine for a velocity less than c, no matter how close to c the velocity is. The particle’s measured velocity is less than c. It must be less than c, unless GR is self-inconsistent, because GR allows the particle to escape and does not allow the particle’s measured velocity to be >= c.

X is defined as an asymptotically small interval of spacetime centered on the event horizon.
No, the definition of an inertial frame allows them to be arbitrarily large, as noted in the OP. Taylor, Thorne, and Wheeler all agree that an inertial frame need only be "small enough" that the tidal force in the frame is negligible. For a sufficiently large black hole, X could be a cube whose proper length per side is a million light years, where the tidal force in the frame is 10^7000 times weaker than the tidal force in your room now. SR, whose arena of applicability is an inertial frame, is experimentally confirmed to high precision in inertial frames whose size is larger than asymptotically small and whose duration is longer than asymptotically short. To make your case you need to show that SR is not experimentally confirmed, due to our measuring equipment being larger than asymptotically small. The size of an inertial frame can be larger than asymptotically small and the duration of an inertial frame can be longer than asymptotically short, or else SR is not experimentally confirmed to any precision. You can’t have it both ways. Still, the OP works fine when the size of X is asymptotically small and its duration is asymptotically short.

The event horizon is moving at c in such a frame, so any escaping object must also be going at least that fast relative to X.
Not unless GR is self-inconsistent. GR allows an inertial frame to fall through the horizon of a black hole, in which case part of the frame is above the horizon. And GR allows a particle above a horizon to escape. And GR does not allow the particle’s measured velocity to be >= c. If you were right, then GR would contradict itself, because the particle’s measured velocity would be >= c.

The assumption that X is effectively flat implies that the escape velocity is effectively constant throughout X (as Pete has pointed out).
The escape velocity is not constant in X. The definitions by Taylor, Thorne, and Wheeler in the OP are unambiguous: The spacetime in an inertial frame need not be perfectly flat; it need only be negligibly curved. An inertial frame need not be asymptotically small; it need only be “small enough” that the tidal force (spacetime curvature) in the frame is negligible. The escape velocity in an inertial frame need not be constant; it need only negligibly vary. Even when X is asymptotically small and its duration is asymptotically short, the escape velocity in it still varies and the spacetime in it is still curved; the variance of the escape velocity is zero in the limit (not zero) and the spacetime in it is flat in the limit (not flat).

Also, the definition of "escape velocity" is in terms of objects than can escape off to infinity, so it's an inherently non-local property. Any experiment to determine escape velocity necessarily requires unlimited time, and so can't be carried out in X (which was defined to have asymptotically small extent in both space and time). So, as far as X is concerned, Y may as well be moving at c.
No, escape velocity is predicted by GR. The OP can depend on that prediction. The OP need not show that the prediction can be experimentally confirmed in X. Your conclusion is a non-sequitur.

Do you agree that any frame in which the escape velocity varies is not an inertial frame?
Yes, I think that's correct.

Do you agree that the escape velocity varies in any nonzero-sized frame in our real, gravity-endowed universe?
Sure.
Do you agree that any nonzero-sized frame in our real, gravity-endowed universe is not an inertial frame?

Do you agree that X is not inertial?
No. There’s no reason for me to believe that. The supporting info in the OP is clear that an inertial frame can fall through the horizon of a black hole, and that is all that X is given to be.

I don't think you can have it both ways, zanket... as soon as you acknowledge that the escape velocity varies in X, then you are acknowledging that X is not inertial.
No. You’ve given nothing to support that. And it contradicts what you’ve agreed to, as I’ll show with further questioning. You are the one who is trying to have it both ways.

If you take the limit of X being inertial then you are also implicitly taking the limit of zero variation in escape velocity.
A limit of zero variation is not no variation. Nor need an inertial frame have zero variation of escape velocity even in the limit. SR is experimentally confirmed to high precision in far less perfect inertial frames.

I'm satisfied that the problems with Zanket's proof have been identified and amply explained. Zanket's responses indicate a willful lack of understanding both of the refutations and the relevant physics and mathematics, and so I see no need to pursue this any further. You can lead a horse to water, but you can't make him drink. I'm confident that any unbiased reader following this thread will see the relative merits of each argument and reach an appropriate conclusion.

It may be worth reconsidering certain parts of the Alpha rules to reflect the fact that most posters who originate these threads engage in argumentative reasoning, as opposed to scientific inquiry. No matter how correct a refutation is, or how cogently it is explained, the originator can always simply refuse to accept it and plunge ahead as if nothing was wrong (in the process burying the correct refutation under repetitive, confusing restatements of the OP). This approach is fine for a court of law, wherein an overseer ensures that both sides are adequately represented, calls an end to arguments, and then relies on an impartial body to decide which side was right. It is problematic for a forum, where there is no limit on arguments or final decision by an impartial body. Lacking such a process for resolution, these threads become nothing more than a contest of wills: as long as the originator never admits he's wrong, he can continue to assert the backing of an Alpha thread for his position. All that matters is having the last word, in order to create the impression that the other side has given up. The effect is that, instead of encouraging legitimate scientific outcomes, the Alpha rules become a free pass for any crackpot to use in bolstering his claims.

A perfect solution to this situation would probably require more invasive and impartial moderation (like the judge and jury in a court of law). Threads should probably be locked if they reach a point where the originator is not making a good faith effort to keep an open mind. Lacking this, some improvement could be gained by adding a stipulation that, for claims which clearly go against accepted physics, the burden of proof is on the originator. This would at least prevent the cheap rhetorical device of shifting the burden of proof to people who support standard physics and math. But unless something is done to address the abuse of Alpha threads by people who do not have an open mind about their claims, the Alpha rules will be little more than a prohibition on directly ridiculing the crackpots in question.

quadraphonics, I haven't simply refused to accept your refutations. I've refuted your refutations with ample evidence. A lot of what you've said directly contradicts the supporting info in the OP (which comes from Einstein, Taylor, Thorne, and Wheeler), as I've pointed out.

Do you agree that any nonzero-sized frame in our real, gravity-endowed universe is not an inertial frame?
I'll accept that for the purpose of this discussion. What about you?

I'll accept that for the purpose of this discussion.
Do you agree that your definition of an inertial frame is more restrictive than the definitions in the OP by Taylor, Thorne, and Wheeler? Their definitions allow an inertial frame to be nonzero-sized. In Einstein’s statement of the equivalence principle in the OP, he also talks about “small”—not zero-sized—inertial frames.

Do you agree that part of an inertial frame falling through the horizon of a black hole is above the horizon?
Yes.
Do you agree that a zero-sized inertial frame falling through the horizon of a black hole cannot be partly above the horizon?

What about you?
I don’t accept your definition of an inertial frame, which doesn’t agree with the definition from any of my sources. Nor does it make sense to me to accept your definition “for the purpose of this discussion”, which to me is trying to have it both ways. (Like, inertial frames can be nonzero-sized for experimentally confirming SR, but must be zero-sized to prevent a self-inconsistency of GR from being shown.) The definition of an inertial frame should be consistent for all applications, whether or not a horizon is involved.

It is also is trying to have it both ways to say that inertial frames are zero-sized but can still somehow be partly above a horizon as they fall through a horizon. X is given to be an inertial frame falling through the horizon of a black hole. If X is partly above the horizon, then the frame must be nonzero-sized and the escape velocity in the frame must vary, if only slightly. And the escape velocity above the horizon is less than c, according to GR.

You’re basically arguing that any tidal force in X is significant due to the horizon. But there is no reason that justifies using a different definition of an inertial frame for X than for an inertial frame elsewhere. And in the OP Taylor and Wheeler confirm that a nonzero-sized inertial frame can fall through the horizon of a black hole.

Hi zanket,
This is a most engaging discussion. Thanks for the exercise!
Do you agree that your definition of an inertial frame is more restrictive than the definitions in the OP by Taylor, Thorne, and Wheeler? Their definitions allow an inertial frame to be nonzero-sized.
No, actually... not that it matters. I don't think we need to quibble over semantics. I don't care much whether we call X inertial or not, but I do care whether any curvature in X (however small) makes a difference to your thought experiment.

Do you agree that a zero-sized inertial frame falling through the horizon of a black hole cannot be partly above the horizon?
Yes. But again, don't get too hung up on the semantics. The relavent question is whether X is flat enough to be considered inertial for the purpose of your experiment - do you agree?

Here's what I think we've established:
X is not perfectly flat unless it's zero sized, but we can make it as flat as we like.
If X is not zero sized, there will be tidal forces in X, but they can be as small as we like.
Since X is not perfectly flat, it is possible for a particle in [X] to be moving away from the black hole, even though X crosses the event horizon
If X was perfectly flat (either because X was zero sized, or because the black hole was infinitely large), then it would not be possible for a particle in [X] to be moving away from the black hole
Do you agree with those points?

Now, the big question... can the relative (tidal) accelerations of test particles in X be neglected for the purposes of your thought experiment?

In context, that question is equivalent to this one:
Will the escaping particle above the horizon in X accelerate relative to a particle with similar initial velocity below the horizon in X?

It's clear to me that those two particles must accelerate relative to each other, since they must depart from the event horizon in opposite directions.

So I think it's clear that the tidal accelerations in X are precisely what allows the particle in question to be escaping the black hole. ... therefore the tidal accelerations can clearly not be neglected for the purpose of your thought experiment.

You’re basically arguing that any tidal force in X is significant due to the horizon.
Actually, I'm arguing that any tidal force is X is significant due to the way you've set up the experiment. A similar experiment could be set up without an event horizon.

I think that what you are doing is constructing a frame that is almost flat and squeezing some non-inertial behaviour into that almost.
Do you agree that the only thing that allows the escaping particle to exist in X is the (arbitrarily small) curvature of X?

This is a most engaging discussion. Thanks for the exercise!
Likewise!

The relavent question is whether X is flat enough to be considered inertial for the purpose of your experiment - do you agree?
I agree that’s a pertinent question. The definition of an inertial frame requires that the tidal force in the frame be negligible for the purposes of a given experiment.

Here's what I think we've established:
X is not perfectly flat unless it's zero sized, but we can make it as flat as we like.
If X is not zero sized, there will be tidal forces in X, but they can be as small as we like.
Since X is not perfectly flat, it is possible for a particle in [X] to be moving away from the black hole, even though X crosses the event horizon
If X was perfectly flat (either because X was zero sized, or because the black hole was infinitely large), then it would not be possible for a particle in [X] to be moving away from the black hole
Do you agree with those points?
An infinitely large black hole would be problematic (e.g. no room for the particle), but other than that I agree.

Will the escaping particle above the horizon in X accelerate relative to a particle with similar initial velocity below the horizon in X?

It's clear to me that those two particles must accelerate relative to each other, since they must depart from the event horizon in opposite directions.

So I think it's clear that the tidal accelerations in X are precisely what allows the particle in question to be escaping the black hole. ... therefore the tidal accelerations can clearly not be neglected for the purpose of your thought experiment.
What is clear to you is a symptom of GR’s self-inconsistency. Think about what you’re saying here: you’re saying that two free test particles within X must move away from each other in opposite directions, therefore the tidal force cannot be neglected, and so X is not an inertial frame. All you’ve done with those two particles is measure the tidal force in X (if you disagree then see the definition of tidal force by Taylor and Wheeler in the OP), so your logic applies to any experiment in X, in which case X cannot be an inertial frame for the purposes of any experiment in X. That is, no inertial frame can fall through the horizon of a black hole. But GR predicts that an inertial frame can do that (multiple sources in the OP support that). Do you agree that your two particles simply measured the tidal force? Do you agree that the tidal force in X cannot be neglected for the purposes of any experiment in X? If yes, do you agree that X cannot be an inertial frame for the purposes of any experiment in X? If yes, do you agree that this conclusion contradicts those multiple sources in the OP?

You came to the same conclusion as the proof in the OP does: X cannot be an inertial frame for the purposes of any experiment in X. But whereas you see that being a problem with the proof, it actually proves GR’s self-inconsistency, because our joint conclusion was required by GR even though the theory contradictorily predicts that X can be an inertial frame.

Actually, I'm arguing that any tidal force is X is significant due to the way you've set up the experiment.
You are arguing that any tidal force is X is significant for the purposes of any experiment in X, and the proof in the OP agrees. Your two particles serve the same function as the latticework of synchronized clocks mentioned in the OP: they both measure the tidal force in X and find that the spacetime in X cannot be flat (negligibly curved). But GR disagrees.

A similar experiment could be set up without an event horizon.
Yes. Above a horizon no self-inconsistency of GR is noticed. In an inertial frame J that is wholly above a horizon, a free test particle can stay essentially at rest with another free test particle elsewhere in the frame. Only in X, which is nothing more than an inertial frame falling through the horizon of a black hole—which GR predicts an inertial frame can do—is it required by GR that some free test particles necessarily move apart from each other (indeed, in the lower particle’s frame the upper particle necessarily recedes at c, because GR predicts that all material objects cross the horizon at c as they can measure in the limit). What does that say about GR's adherence to the equivalence principle?

I think that what you are doing is constructing a frame that is almost flat and squeezing some non-inertial behaviour into that almost.
The OP notes that the tidal force in X can be nonexistent in the limit. Even then the proof still shows that GR is self-inconsistent. The tidal force in X cannot be made negligible for the purposes of any experiment in X.

Do you agree that the only thing that allows the escaping particle to exist in X is the (arbitrarily small) curvature of X?
Yes. Yet you and I both concluded that GR requires that a frame falling through the horizon of a black hole cannot be an inertial frame for the purposes of any experiment in X. Then X cannot exist as defined. But GR disagrees, and so the theory is self-inconsistent.

What is clear to you is a symptom of GR’s self-inconsistency. Think about what you’re saying here: you’re saying that two free test particles within X must move away from each other in opposite directions, therefore the tidal force cannot be neglected, and so X is not an inertial frame. All you’ve done with those two particles is measure the tidal force in X (if you disagree then see the definition of tidal force by Taylor and Wheeler in the OP), so your logic applies to any experiment in X, in which case X cannot be an inertial frame for the purposes of any experiment in X.
Not so!
The acceleration of the two particles relative to each other is arbitrarily small.
What this means is that in X, it is virtually impossible to distinguish the two. We only know that there is a difference because of the artifice of the event horizon.

Think of it like this:
If a large number of particles are passing through X at (apparently) the same very high velocity, such that:
- at least one of those particles is in X at the same time as X crosses a large event horizon, and
- that particle will not fall into the black hole,

then:
it is not possible to determine in X which is the last particle to escape the black hole and which is the first to be trapped,

unless:
The equipment available in X is sufficiently precise to measure the tidal acceleration across X at that time.

To put it another way, an observer in X can't tell if the experiment was set up correctly or not.

Sorry to intrude again, but:

There's no crossing the event horizon. The time dilation goes total at the event horizon. That means it takes all of time to cross it. I know it sounds like there's an infinity here, and we're always suspicious of infinities in nature. But actually it's not an infinity, it's a zero - a zero c. That's why Black Holes used to be known as Frozen Stars. You might talk about crossing it within a reference frame, but it's more like a brick wall. Crossing it is hypothetical only, because you're talking about what will happen after the end of time. By the way, all this is why I'm suspicious of central singularities and yet more infinities. You might be tempted to brush this off, but it's serious stuff, worthy of your consideration. Here's a Scientific American article:

http://www.sciam.com/article.cfm?articleID=00012DEF-46AA-1F04-BA6A80A84189EEDF

"Demolishing stars, powering blasts of high-energy radiation, rending the fabric of spacetime: it is not hard to see the allure of black holes. They light up the same parts of the brain as monster trucks and battlebots do. They explain violent celestial phenomena that no other body can. They are so extreme, in fact, that no one really knows what they are.

Most researchers think of them as microscopic pinpricks, the remnants of stars that have collapsed under their own weight. But over the past couple of years, a number of mavericks have proposed that black holes are actually extended bodies, made up of an exotic state of matter that congeals, like a liquid turning to ice, during the collapse. The idea offers a provocative way of thinking about quantum gravity, which would unify Einstein's general theory of relativity with quantum mechanics...."

... therefore the tidal accelerations can clearly not be neglected for the purpose of your thought experiment.

The acceleration of the two particles relative to each other is arbitrarily small.
How do these statements of yours not contradict each other? In the first, you say the tidal force is significant. In the second, you say the tidal force is negligible. Which is it? The tidal force is the relative acceleration of two free test particles located in different parts of a frame.

What this means is that in X, it is virtually impossible to distinguish the two.
Two what?

unless:
The equipment available in X is sufficiently precise to measure the tidal acceleration across X at that time.
Do you think that an experiment that measures the “acceleration of the two particles relative to each other” is different than an experiment that measures the tidal force (or tidal acceleration, as you put it)?

To put it another way, an observer in X can't tell if the experiment was set up correctly or not.
What experiment?

Are you arguing that an observer in X cannot tell whether or not the tidal force throughout X is negligible for the purposes of any experiment spread throughout X?

There's no crossing the event horizon.
GR predicts that someone above the horizon cannot observe anything to cross a horizon, and predicts that an observer can observe itself to cross a horizon.

How do these statements of yours not contradict each other? In the first, you say the tidal force is significant. In the second, you say the tidal force is negligible. Which is it? The tidal force is the relative acceleration of two free test particles located in different parts of a frame.
The tidal acceleration can be arbitrarily small, but it can still be measurable with arbitrarily precise instruments.

Two what?
...All you’ve done with those two particles is measure the tidal force in X...The acceleration of the two particles relative to each other is arbitrarily small.
What this means is that in X, it is virtually impossible to distinguish the two...

Do you think that an experiment that measures the “acceleration of the two particles relative to each other” is different than an experiment that measures the tidal force (or tidal acceleration, as you put it)?
No, it's precisely the same thing.
BTW, I'm using the term "tidal acceleration" because that's the term used in your opening post. Tidal force is derived from tidal acceleration.

What experiment?
Your thought experiment.

Are you arguing that an observer in X cannot tell whether or not the tidal force throughout X is negligible for the purposes of any experiment spread throughout X?
No, and observer in X can certainly do that. All they need do is measure the relative acceleration of freely-falling particles.
If they can't detect any relative acceleration, then the tidal force is negligible.

The tidal acceleration can be arbitrarily small, but it can still be measurable with arbitrarily precise instruments.
OK, if you think the tidal acceleration can be arbitrarily small (i.e. negligible), then why did you say “... therefore the tidal accelerations can clearly not be neglected for the purpose of your thought experiment”? If it can’t be neglected, then it cannot be arbitrarily small, right?

No, it's precisely the same thing.
Agreed.

BTW, I'm using the term "tidal acceleration" because that's the term used in your opening post. Tidal force is derived from tidal acceleration.
OK.

No, and observer in X can certainly do that. All they need do is measure the relative acceleration of freely-falling particles.
If they can't detect any relative acceleration, then the tidal force is negligible.
Agreed, unless there’s a self-inconsistency of the theory, which can prevent the tidal force from being measured. For example, the tidal force in X cannot be measured by an observer in X who looks at a tidal strain gauge on a free-floating rod that straddles the horizon and is at rest with respect to the escaping particle, because GR does not allow such rod to exist.

Your thought experiment.
I don’t see how you’ve shown that “an observer in X can't tell if [your thought] experiment was set up correctly or not.” Can you elaborate? You said:

Think of it like this:
If a large number of particles are passing through X at (apparently) the same very high velocity, such that:
- at least one of those particles is in X at the same time as X crosses a large event horizon, and
- that particle will not fall into the black hole,

then:
it is not possible to determine in X which is the last particle to escape the black hole and which is the first to be trapped,

unless:
The equipment available in X is sufficiently precise to measure the tidal acceleration across X at that time.
You agree that an “observer in X can certainly” measure the tidal force. Then your “unless” condition can always be satisfied in principle. So what’s the problem?

In any case, an observer in X need not measure the tidal force. The OP proves that X is not an inertial frame for the purposes of any experiment. If it was, then the spacetime in X would be flat (i.e. negligibly curved, synonymous with a negligible tidal force), in which case Y could be set up in X’s flat spacetime, all according to Taylor and Wheeler. But Y cannot be set up as described, for the reason given in the OP. The attempt to set up Y is not an experiment that depends on whether the tidal force in X is negligible. Rather, it is an experiment that determines whether the spacetime in X is flat, which is synonymous with an experiment that determines whether the tidal force in X is negligible. The thought experiment tests the flatness of X’s spacetime, and finds that its spacetime is not flat for the purposes of any experiment, in which case X cannot be an inertial frame for the purposes of any experiment.

OK, if you think the tidal acceleration can be arbitrarily small (i.e. negligible), then why did you say “... therefore the tidal accelerations can clearly not be neglected for the purpose of your thought experiment”? If it can’t be neglected, then it cannot be arbitrarily small, right?
The operative phrase is "for the purpose of your thought experiment". Your thought experiment is defined with arbitrary precision. The closer the tidal acceleration is to zero, the more precisely the position and velocity of the test particle is defined.

Agreed, unless there’s a self-inconsistency of the theory, which can prevent the tidal force from being measured. For example, the tidal force in X cannot be measured by an observer in X who looks at a tidal strain gauge on a free-floating rod that straddles the horizon and is at rest with respect to the escaping particle, because GR does not allow such rod to exist.
Such a rod can certainly exist. It will necessarily be stretched, of course, and if the rod doesn't break the tension in the rod will pull the "escaping" end across the horizon.

I don’t see how you’ve shown that “an observer in X can't tell if [your thought] experiment was set up correctly or not.” Can you elaborate?

You agree that an “observer in X can certainly” measure the tidal force. Then your “unless” condition can always be satisfied in principle. So what’s the problem?
Well, part of the conditions for your thought experiment is that X is inertial... so you've essentially denied the "unless" condition.

In any case, an observer in X need not measure the tidal force. The OP proves that X is not an inertial frame for the purposes of any experiment.
I really don't see how you reach that conclusion. For example, if the strain gauge rod wasn't moving at such fantastic speeds in X, it would show zero tidal force throughout X. A ball bouncing in a box at rest in X wouldn't show any unusual behaviour.

If it was, then the spacetime in X would be flat (i.e. negligibly curved, synonymous with a negligible tidal force), in which case Y could be set up in X’s flat spacetime, all according to Taylor and Wheeler.
It does not follow that Y is an inertial frame, or even a meaningful one. Even using Taylor and Wheeler's simplified explanations, they do not imply that all reference frames in flat space-time are inertial.

The thought experiment tests the flatness of X’s spacetime, and finds that its spacetime is not flat for the purposes of any experiment
The thought experiment finds that the spacetime is not flat for the purposes of that experiment, and that experiment alone. How do you conclude otherwise?

The operative phrase is "for the purpose of your thought experiment". Your thought experiment is defined with arbitrary precision. The closer the tidal acceleration is to zero, the more precisely the position and velocity of the test particle is defined.
It sounds like you’re talking about the uncertainty principle, not GR. In any case, this doesn’t tell me why “the tidal accelerations can clearly not be neglected for the purpose of your thought experiment”. But it doesn’t matter. The thought experiment simply measures the tidal force; if that measurement shows that the tidal force cannot be neglected, that doesn’t refute the thought experiment. A thought experiment that measures something is not refutable by pointing to the value of the measurement. Do you disagree?

Such a rod can certainly exist. It will necessarily be stretched, of course, and if the rod doesn't break the tension in the rod will pull the "escaping" end across the horizon.
X is given to be an inertial frame, with a negligible tidal force by the definition of an inertial frame. So why would the rod “necessarily be stretched”? If the tidal force isn’t stretching it, what is? Why does the tension in the rod pull the rod down (toward the black hole) only? Why not up instead? If the tension in the rod pulls “the "escaping" end across the horizon” then the rod is not at rest with respect to the escaping particle, and does not exist as given (I said the rod “is at rest with respect to the escaping particle”). If the rod exists as given, then the tidal force on the rod is negligible (it’s negligible on any object in X, since X is an inertial frame), and the rod is at rest with respect to the escaping particle, in which case the rod would be passing outward through the horizon. GR does not allow anything to pass outward through the horizon, and so the rod cannot exist according to GR.

Well, part of the conditions for your thought experiment is that X is inertial... so you've essentially denied the "unless" condition.
The tidal force in X is negligible, not necessarily undetectable, so the “unless” condition is not denied. (The tidal force doesn’t necessarily become significant just because finer instruments for detecting the tidal force become available.) So how have you shown that “an observer in X can't tell if [your thought] experiment was set up correctly or not”?

I really don't see how you reach that conclusion. For example, if the strain gauge rod wasn't moving at such fantastic speeds in X, it would show zero tidal force throughout X. A ball bouncing in a box at rest in X wouldn't show any unusual behaviour.
Why do you think the gauge on the rod is affected by the rod’s speed in X? The definition of the tidal force in the OP says nothing about the particles’ speeds in the frame in which the tidal force is measured (the tidal force is their relative acceleration, not their speeds). In the rod’s frame the box is moving at a fantastic speed. Then why wouldn’t a strain gauge on the box read a nonzero value, using your logic? If the rod “will necessarily be stretched”, then why wouldn’t the box necessarily be stretched too? Keep in mind that the proper length of the rod can be less than the proper length/width/height of the box. The only requirements for the rod are that it stays at rest with respect to the escaping particle and straddles the horizon.

It does not follow that Y is an inertial frame, or even a meaningful one. Even using Taylor and Wheeler's simplified explanations, they do not imply that all reference frames in flat space-time are inertial.
Y is given to be an inertial frame. Then there’s no question whether Y is an inertial frame. Taylor and Wheeler define flat spacetime as a “region of spacetime in which it is possible to set up a free-float (inertial) reference frame”. Then it must be possible to set up Y in X, if X’s spacetime is flat. But GR does not allow Y to be set up in X. More on that below.

The thought experiment finds that the spacetime is not flat for the purposes of that experiment, and that experiment alone. How do you conclude otherwise?
The thought experiment tests the flatness of X for the purposes of an unspecified experiment, and not for the purposes of the thought experiment itself, which would make no sense. An experiment that measures spacetime curvature is not an experiment that requires flat spacetime for the purposes of that experiment. An experiment that measures the tidal force is not an experiment that requires a negligible tidal force for the purposes of that experiment. Do you disagree? If so, then you must agree that the tidal force can be measured only if it’s negligible, right?

There are two ways to test the flatness of a region of spacetime. One, you can measure the tidal force; if it’s undetectable or negligible then the spacetime is flat, because a negligible tidal force = negligible spacetime curvature = flat spacetime. Two, you can try to set up an inertial frame in that region; if it cannot be set up then the spacetime in the region is not flat, according to Taylor and Wheeler. The two methods are equivalent because, by the definition of an inertial frame, the tidal force in an inertial frame is negligible.

One way to measure the tidal force is to look at the gauge on the rod. But the rod cannot exist as given, or else it would be passing outward through the horizon, which GR disallows. (And if the tension in the rod pulls “the "escaping" end across the horizon” then the rod is not at rest with respect to the escaping particle, and does not exist as given.) If the rod cannot exist as given then Y (an inertial frame with respect to which the escaping particle stays at rest) cannot extend throughout X. If Y extended throughout X then a rod in Y could stay at rest with respect to Y, have a negligible tidal force strain on it, and straddle the horizon (in which case the rod would be passing outward through the horizon). If Y cannot extend throughout X, then the spacetime in X cannot be flat throughout X, according to Taylor and Wheeler. GR demands that Y cannot extend throughout X, and so the spacetime in X cannot be flat, and so X cannot be an inertial frame, which contradicts GR’s prediction to the contrary. Then GR is self-inconsistent.

I'm getting lost in this discussion. I can't remember which points related to what. The semantics of "negligible" and "inertial" are being used in odd and inconsistent ways, and it isn't worth sorting it out.

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.

Y is given to be an inertial frame.
No it is not. You have defined Y as a frame in which the escaping particle is at rest, and covering the same region of spacetime as X.
You have stated without proof that Y is an inertial frame, and then proved that it is not.
You haven't accounted for relativistic differences between the reference frames, and assumed without proof that "flat enough" in one frame equates to "flat enough" in the other.


X is given to be an inertial frame, with a negligible tidal force by the definition of an inertial frame. So why would the rod “necessarily be stretched”? If the tidal force isn’t stretching it, what is?
The rod is necessarily stretched because the tidal force is not zero. The amount of stretch may be arbitrarily small, but it is not zero.

Why does the tension in the rod pull the rod down (toward the black hole) only? Why not up instead?
The upper end is pulled downward, the lower end is pulled upward.

If the tension in the rod pulls “the "escaping" end across the horizon” then the rod is not at rest with respect to the escaping particle, and does not exist as given (I said the rod “is at rest with respect to the escaping particle”).
The rod is (close enough to) at rest with respect to the escaping particle for the duration of the experiment. In X, both the rod and the escaping particle are moving at a speed indistinguishable from c (the particle and the upper end of the rod are slightly faster, the lower end of the rod is slightly slower).

If the rod exists as given, then the tidal force on the rod is negligible (it’s negligible on any object in X, since X is an inertial frame), and the rod is at rest with respect to the escaping particle, in which case the rod would be passing outward through the horizon. GR does not allow anything to pass outward through the horizon, and so the rod cannot exist according to GR.
The upper end of the rod is moving away from the horizon for the duration of the experiment (it's speed is very slightly higher than the horizon's speed in X), but no part of the rod is crossing the horizon the wrong way.

Why do you think the gauge on the rod is affected by the rod’s speed in X?
Well, I'm not sure that the rod actually does show much strain after all. It might... we'd have to do the maths to see.
But, the speed of the rod is relevant. In the rod's rest frame, it is much longer than it is in X, and there will be much longer between events happening at each end. Again, you'd need to crunch the numbers to figure exactly what happens.


I'm not sure how to approach it mathematically. I'll have a go if I find time.

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 a non sequitur. The tidal force cannot be “never negligible for the purpose of” an experiment that only measures the tidal force. The value of a measurement does not affect the measurement of that value. If you disagree, then you must agree that the tidal force can be measured only if it’s negligible, right? That’s not a rhetorical question.

No it is not. You have defined Y as a frame in which the escaping particle is at rest, and covering the same region of spacetime as X.
Y is given as an inertial frame. The OP says “an inertial frame Y”. Y is explicitly defined as an inertial frame.

You have stated without proof that Y is an inertial frame, and then proved that it is not.
No proof is needed for a given. Y is not proven to not be an inertial frame; it is always an inertial frame in the proof. X is the frame that is proven to not be an inertial frame. It is okay for X to be given as an inertial frame, and then proven that it cannot be an inertial frame, because that is how GR is shown to be self-inconsistent. GR predicts that X (an inertial frame falling through the horizon of a black hole) can exist, but also does not allow it to exist.

Notice that you didn’t prove that your ball in your box is in the box. Why didn’t you need to prove it? You didn’t need to prove it because it was a given. If you still think I need to prove that Y is an inertial frame, then prove to me that your ball is in the box.

You haven't accounted for relativistic differences between the reference frames, and assumed without proof that "flat enough" in one frame equates to "flat enough" in the other.
I assume by “relativistic differences” you mean relative speed, since that’s all you mentioned in that regard. Relative speed does not affect the tidal force in a region of spacetime. The definition of the tidal force makes no mention of relative speed. More on that below.

If X is an inertial frame, then the tidal force on all objects in X is negligible, by the definition of an inertial frame. The tidal force is synonymous with spacetime curvature. Then the spacetime in X is flat (negligibly curved). Then it must be possible to set up the inertial frame Y in X, according to Taylor and Wheeler’s definition of flat spacetime. Then I have proven that “"flat enough" in one frame equates to "flat enough" in the other”. To make your case, you need to show that “flat enough” in X does not necessarily equate to “flat enough” in Y. But to do that you need to refute Taylor and Wheeler. Your “relative speed” argument doesn’t suffice.

The rod is necessarily stretched because the tidal force is not zero. The amount of stretch may be arbitrarily small, but it is not zero.
You said:

Such a rod can certainly exist. It will necessarily be stretched, of course, and if the rod doesn't break the tension in the rod will pull the "escaping" end across the horizon.
If you meant that the rod is only negligibly stretched, by a negligible tidal force, then I agree. But then why say “if the rod doesn’t break”? That implies to me that you’re talking about a significant stretch, by a significant tidal force. If the tidal force on the rod is negligible, then there is no question that the rod will not break, any more than there is a question that you will break due to the tidal force on your body now. If the rod can exist, and if the rod is only negligibly stretched, then there is no reason why the rod cannot stay at rest with respect to the escaping particle, in which case the rod is passing outward through the horizon. But GR disallows that, and so the rod cannot exist.

The upper end is pulled downward, the lower end is pulled upward.
OK. So the tension in the rod does not “pull the "escaping" end across the horizon”, right? Instead the rod is negligibly stretched, both upward and downward, and the escaping end stays essentially at rest with respect to the escaping particle. The escaping particle moves at nearly c relative to the horizon. Then the rod must be passing outward through the horizon. But GR disallows that, and so the rod cannot exist.

The rod is (close enough to) at rest with respect to the escaping particle for the duration of the experiment. In X, both the rod and the escaping particle are moving at a speed indistinguishable from c (the particle and the upper end of the rod are slightly faster, the lower end of the rod is slightly slower).
Neither the rod nor the particle can be measured to move at c or greater in X, or else SR is violated. (If that presents a problem, that’s GR’s fault not mine.) If the “rod is (close enough to) at rest with respect to the escaping particle” for any duration, and the particles that make up the rod are not stretching apart at nearly c relative to each other (the velocity of the particle relative to the horizon), which they are not doing because the tidal force in X is negligible, then the rod is passing outward through the horizon. But GR doesn’t allow that, and so the rod cannot exist.

The upper end of the rod is moving away from the horizon for the duration of the experiment (it's speed is very slightly higher than the horizon's speed in X), but no part of the rod is crossing the horizon the wrong way.
The horizon’s speed in X is exactly c. That can be measured in the limit relative to an object hovering just above the horizon, at the horizon in the limit. GR predicts that all observers cross a horizon at exactly c as they observe. The directly-measured speed of the particle or any part of the rod is less than c in X, or else SR is violated. Both the rod and the particle move outward at nearly c relative to the horizon, which an object hovering just above the horizon could measure. If the rod straddles the horizon, and is negligibly stretched by the tidal force, and is moving outward at nearly c relative to an object hovering just above the horizon, then it is passing outward through the horizon just as surely as water erupting from a geyser passes outward through the Earth’s surface.

Well, I'm not sure that the rod actually does show much strain after all. It might... we'd have to do the maths to see.
No math is necessary to determine the tidal strain. That’s the beauty of an inertial frame: the tidal force is negligible on everything in it. X is given to be an inertial frame falling through the horizon of a black hole, which GR allows. You keep mentioning “for the purposes of an experiment”, as if the tidal force could be negligible in X for a particular experiment, but cannot be guaranteed to be negligible for anything else. That’s taking the phrase too literally. The tidal force need only be measured once to determine whether it’s negligible for the purposes of any number of experiments. If the tidal force in an inertial frame is measured to be less than x, then that frame is suitable for all y number of experiments that require an inertial frame with a tidal force less than x.

But, the speed of the rod is relevant. In the rod's rest frame, it is much longer than it is in X, and there will be much longer between events happening at each end. Again, you'd need to crunch the numbers to figure exactly what happens.
In the box’s rest frame, the box is much longer than it is in the rod’s frame, “and there will be much longer between events happening at each end”. So again I ask:

Then why wouldn’t a strain gauge on the box read a nonzero value, using your logic?
It should, according to your logic. But you said the ball “wouldn't show any unusual behaviour.”

The speed of the rod is irrelevant; no number crunching is required. The tidal force on an object is unaffected by its speed relative to other objects. The tidal force is a property of a region of spacetime that equally affects all objects in that region. The supporting info in the OP notes that the tidal force is synonymous with spacetime curvature. Then you’re telling me that the curvature of a region of spacetime is affected by an object’s speed in that region. Like, you’re suggesting that a region of spacetime can change from flat to not flat if a test particle moves fast enough through the region. But that isn’t so. An object’s mass affects the curvature of the spacetime around the object (that’s why the proof in the OP uses a test particle, whose mass is negligible), but its speed does not. For a confirmation of that, look at the curvature factor in the Schwarzschild metric; search for “curvature factor” in this excerpt (http://www.eftaylor.com/pub/chapter2.pdf) by Taylor and Wheeler. That’s a spacetime curvature factor, synonymous with a tidal force factor. The curvature factor = (1 – (2M / r)). Notice that there’s no velocity variable in that equation. Nor would that make sense, since an object has various velocities relative to various other objects—which velocity would you choose to input into the equation?

In your GR studies you've probably heard the phrase by Taylor and Wheeler, "Spacetime tells mass how to move; mass tells spacetime how to curve". Note there that mass tells spacetime how to curve, not velocity. Then velocity does not affect the tidal force = spacetime curvature.

And think about this: If velocity affected the tidal force = spacetime curvature, then SR could not be experimentally tested at relativistic velocities, because such velocity would change an inertial frame into a noninertial frame, in which SR does not apply.

That’s a non sequitur. The tidal force cannot be “never negligible for the purpose of” an experiment that only measures the tidal force.
Certainly it can.
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.

The value of a measurement does not affect the measurement of that value.
True enough, for the purpose of this discussion... but your thought experiment is not only a measurement of the tidal force.
Your thought experiment is specifically set up to require a measurable tidal force. The particle's position and motion can not be specified in X unless the tidal force is known.

Y is given as an inertial frame. The OP says “an inertial frame Y”. Y is explicitly defined as an inertial frame.
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 is infinitely distant from the particle.

I assume by “relativistic differences” you mean relative speed, since that’s all you mentioned in that regard. Relative speed does not affect the tidal force in a region of spacetime. The definition of the tidal force makes no mention of relative speed.
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.

If X is an inertial frame, then the tidal force on all objects in X is negligible, by the definition of an inertial frame. The tidal force is synonymous with spacetime curvature. Then the spacetime in X is flat (negligibly curved). Then it must be possible to set up the inertial frame Y in X, according to Taylor and Wheeler’s definition of flat spacetime. Then I have proven that “"fla