(Alpha) General relativity is self-inconsistent

zanket

Human
Valued Senior Member
(The “Alpha” in the title indicates that the Alpha rules 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 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.)
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: “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: “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: “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: “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.
 
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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.
 
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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.

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.
 
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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).
 
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Hi zanket,
zanket said:
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).
 
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