Hi all. I saw this paradox elsewhere (no, it's not mine), but haven't seen a satisfactory answer to it yet. Do you know the solution?

A postulate of general relativity (GR), the principle of equivalence (EP), says that the laws of physics are the same in all small freely falling frames in our real, gravity-endowed universe. But the following paradox (in my own words) seems to show that GR contradicts the EP (which would invalidate the theory):

In an arbitrarily small freely falling frame "X" that is falling through the horizon of a black hole, let there be a box in free fall that straddles the horizon. Inside the box, above the horizon, let there be a particle in free fall that is escaping to infinity. The box must be falling toward the black hole's singularity, because all objects below the horizon inexorably do so, according to GR*. So the box must be moving relative to the particle, which is receding from the black hole. This seems to contradict the EP, because the laws of physics in any small freely falling frame "Y" somewhere wholly above the horizon do not preclude a box located anywhere in the frame, enclosing any given particle, from being at rest with respect to the particle.

The tidal force, which is arbitrarily weak in X, cannot be the culprit that resolves the paradox. For one thing the tidal force could be stronger in Y than in X, if only slightly. For another thing the tidal force cannot force a particle inside a box to move with respect to the box; the tidal force only stretches and squeezes objects, so if the particle was at the center of the box, say, the tidal force could not move it from the center.

Although the EP is precisely correct only at points in spacetime, the larger size of X and Y cannot be the culprit that resolves the paradox, because the EP does not precisely apply in those frames due only to the tidal force in them, according to GR, and the tidal force is already ruled out as a culprit that solves the paradox.**

So does anyone know the solution to this paradox? I've seen no direct solution to it elsewhere, only things that effectively ignore it (like "not rigorous enough") or are obviously false (things that contradict references or the laws of logic, like "thought experiments can't show a problem with a theory"). I can provide references for the basis of the paradox, if requested.

* In the frame of any material object at the horizon of a black hole, the horizon moves outward at the speed of light, to explain why light shone radially upward by an observer at a horizon moves at c in the observer's frame but does not pass outward through the horizon. Nothing, including light, can pass outward through the horizon of a black hole, according to the definition of such horizon.

** According to GR, the only detectable difference between a small freely falling frame in our real, gravity-endowed universe, and an inertial frame in an idealized gravity-free universe, is the tidal force in the former frame.

You said that the particle is escaping to infinity, and it is just outside the event horizon, so it must be moving pretty fast. If the box is at rest with respect to the particle in some local frame, then this means the box (or a part of it) is escaping to infinity too.

Yes. However, GR doesn't allow that. GR demands that the box be falling toward the black hole's singularity. Note that the paradox cannot be resolved by having only the part of the box that is above the horizon be escaping to infinity in formation with the particle, with the rest of the box falling toward the black hole's singularity, if only because in Y a whole box (located anywhere in the frame) can be at rest with respect to any given particle in the box, so the laws of physics in X and Y would still differ.

Well, then Y wouldn't be a local frame. I mean with local frames you cannot achieve this.

So the box must be moving relative to the particle, which is receding from the black hole. This seems to contradict the EP, because the laws of physics in any small freely falling frame "Y" somewhere wholly above the horizon do not preclude a box located anywhere in the frame, enclosing any given particle, from being at rest with respect to the particle.

If the particle is fixed wrt the box, it is not "escaping to infinity". The particle is moving relative to the box. Certainly that is not the contradiction. Just pick up a ball, put it in a box, and shake the box. The ball bounces around inside the box. No need for an event horizon. The first thing you need to do is explain how this scenario differs from an everyday ball bouncing around inside an everyday box situated far from a black hole.

So, back to the black hole. Suppose I am fixed with respect to the box and have a bunch of instruments with me. The particle is moving with respect to the box. I can see that motion with my instruments. While I may not be able to sense my own imminent doom, I can certainly deduce that the particle will soon leave the box. Suppose an intelligent ant decked out with miniature sensors is at rest with respect to the particle. While he doesn't sense that I am racing toward my doom, he does sees big me and the box racing away from him and the particle. After a short while, I vanish. While that certainly is surprising, where exactly is the contradiction?

Well, then Y wouldn't be a local frame. I mean with local frames you cannot achieve this.
Cannot achieve what exactly?

Although the EP is precisely correct only at points in spacetime, the larger size of X and Y cannot be the culprit that resolves the paradox
Yes, they can. The author of the paradox has to prove that the EP doesn't become vanishingly small at the event horizon. Spatial coordinates become time-like and time becomes space-like inside the event horizon. The event horizon is in a sense a singularity to the EP. No paradox, just a nasty singularity.

If the particle is fixed wrt the box, it is not "escaping to infinity". The particle is moving relative to the box.
The paradox doesn't say that the particle is fixed wrt the box. The particle is given to be escaping to infinity, so yes, it must be moving relative to the box.

So, back to the black hole. Suppose I am fixed with respect to the box and have a bunch of instruments with me. The particle is moving with respect to the box. I can see that motion with my instruments. While I may not be able to sense my own imminent doom, I can certainly deduce that the particle will soon leave the box. Suppose an intelligent ant decked out with miniature sensors is at rest with respect to the particle. While he doesn't sense that I am racing toward my doom, he does sees big me and the box racing away from him and the particle. After a short while, I vanish.
I can agree with all of this.

While that certainly is surprising, where exactly is the contradiction?
The contradiction, the paradox, is that the laws of physics in frame X seem to differ from those in frame Y, in contradiction to GR's equivalence principle (EP), which implies that the laws of physics in X are the same as those in Y. A box located anywhere in Y can be at rest wrt any given particle inside the box (including a particle that has the same velocity in Y as the escaping particle has in X). In Y, unlike in X, it is possible that no instrument fixed wrt the box or fixed wrt the particle could detect that the box and the particle are moving wrt each other.

Yes, they can. The author of the paradox has to prove that the EP doesn't become vanishingly small at the event horizon. Spatial coordinates become time-like and time becomes space-like inside the event horizon. The event horizon is in a sense a singularity to the EP. No paradox, just a nasty singularity.
The EP asserts the equivalence of all small freely falling frames, including X and Y, so a paradox arises when any difference in the laws of physics between such frames is shown. That the EP is precisely correct only at points in spacetime means that among larger frames the results of like experiments may differ at some level of precision. That does not preclude using larger frames to show that the EP is violated. (The EP applies to larger frames to some level of precision, which explains how special relativity can be experimentally confirmed in a larger frame.) I disagree that the author need do what you say.

As a hypothetical example, suppose GR predicted that a ball can be thrown at more than 50 kph in Y, but not in X. The EP would then definitely be violated (and GR would contradict itself) despite the fact that the EP is precisely correct only at points in spacetime, and the EP would be violated no matter that "Spatial coordinates become time-like and time becomes space-like inside the event horizon", because the EP implies that if a ball can be thrown at more than 50 kph in Y, then the same is true in X as well. Nothing else that GR predicts about X or Y would change that.

The EP asserts the equivalence of all small freely falling frames, including X and Y
No, it does not. It says the laws of physics are valid in each frame. It does not say the frames are equivalent. What laws of physics are different in the two frames? Moreover, you still have not addressed my resolution that the frame in which the equivalence principle applies becomes vanishingly small at the event horizon. The box X in which the equivalence principle holds becomes a point at the event horizon. The event horizon is a singularity for the EP. There is no paradox; there is ust a singularity.

To the moderator: This thread should be moved to Physics & Math (and then most likely to pseudoscience).

No, it does not. It says the laws of physics are valid in each frame. It does not say the frames are equivalent.
From Black Holes & Time Warps by Thorne, pg. 98 (emphasis author'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 the 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 important, 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.

The EP does not say that the laws of physics are valid in each frame. It clearly implies that the laws of physics are the same in any small, freely falling reference frame anywhere in our real, gravity-endowed Universe, and hence all such frames are equivalent to each other, just like Thorne points out in no uncertain terms.

What laws of physics are different in the two frames?
You asked this before, and I already answered it.

Moreover, you still have not addressed my resolution that the frame in which the equivalence principle applies becomes vanishingly small at the event horizon. The box X in which the equivalence principle holds becomes a point at the event horizon. The event horizon is a singularity for the EP. There is no paradox; there is ust a singularity.
This was already addressed above. The EP is not inapplicable inside a frame larger than a point. It holds to some precision in a larger frame; that's how special relativity can be experimentally confirmed. The box X in which the equivalence principle holds need not become a point at the event horizon. For a sufficiently large black hole, the box could be many light years across and yet the tidal force across the box could be weaker than the tidal force in any lab in which special relativity has been experimentally confirmed. (Do you think a human being would necessary be ripped apart at the horizon of any black hole? That is not the case.) The tidal force is the only detectable difference between X and an inertial frame in an idealized, gravity-free universe, according to GR.

Thorne goes on to say:

Formulate any law of physics in terms of measurements made in one small, inertial (freely falling) reference frame. Then, when restated in terms of measurements in any other small inertial (freely falling) frame, that law of physics must take on precisely the same mathematical and logical form as in the original frame. And this must be true whether the (freely falling) inertial frame is in gravity-free intergalactic space, or is falling off a cliff on Earth, or is at the center of our galaxy, or is falling through the horizon of a black hole.

Thorne obviously disagrees with you.

To the moderator: This thread should be moved to Physics & Math (and then most likely to pseudoscience).
The thread is already in Physics & Math. Only an unscientific person would move it to pseudoscience at this point, since no bona fide solution to the paradox has been shown, and your points against it have been soundly refuted, with references. (But unscientific people who love censorship tend to be prevalent on these forums, even mods, so this will be a test if the mod--and this forum--is scientific or not.) I think there's something wrong with the paradox, but I'm not interested in "solutions" that references obviously refute. Anyone can come up with a junk solution.

Be very specific, please. Exactly which law of physics does not hold in frame X or frame Y? Does one of the conservation laws, the standard model, or GR itself fail in one of the boxes?

The following law of physics applies in Y but not in X: (taken from where I saw the paradox; I can't put the link since I'm new here): "a free test body located anywhere in the frame can in principle be at rest relative to any possible other free test body in the frame".

Two points. Firstly, the "paradox" arises in defining frame X as crossing the event horizon and then presuming that the equivalence principle applies in this frame. You still have not addressed the fact that the locals frames in which the equivalence principle holds becomes vanishingly small at the event horizon. In other words, there is no frame Y and frame X cannot cross the event horizon.

Secondly, a free test body can be placed such that it is instantaneously at rest with respect to some another test body in frame X. Fine. Since frame X cannot cross the event horizon, there is no paradox. The paradox arises only because of an invalid interpretation of the equivalence principle. Black holes are also called singularities for a very good reason.

Two points. Firstly, the "paradox" arises in defining frame X as crossing the event horizon and then presuming that the equivalence principle applies in this frame. You still have not addressed the fact that the locals frames in which the equivalence principle holds becomes vanishingly small at the event horizon. In other words, there is no frame Y and frame X cannot cross the event horizon.
I have addressed this, by showing that it's not an issue. The "locals frames in which the equivalence principle holds" do not become "vanishingly small at the event horizon". You're contradicting Thorne, as quoted above, and many other references. I'm not going to say anything more about it. Either this is a scientific forum, or it's not.

Secondly, a free test body can be placed such that it is instantaneously at rest with respect to some another test body in frame X.
Sure, but the box straddling the horizon cannot be at rest with respect to the escaping particle.

Since frame X cannot cross the event horizon, there is no paradox.
Frame X can cross the horizon. Thorne's quote above makes that clear, and so do lots of other references.

The paradox arises only because of an invalid interpretation of the equivalence principle.
You've not identified an invalid interpretation. Thorne's statement and interpretation of the EP, above, clearly contradicts you.

Black holes are also called singularities for a very good reason.
There is no coordinate singularity in a frame falling through a horizon. Nor should such singularity be confused with the physical singularity at the center of the black hole. Nothing special need happen in a small (but not point-sized) frame falling through the horizon of a black hole; numerous references support that, including the one I gave above. Special relativity can in principle be experimentally confirmed to any precision in such a frame.

You're ignoring the references I gave and implying that Thorne, a black hole expert, is wrong. I'm not interested in that kind of discussion. I'll ignore you if you repeat your points. (Although it is common on these forums for a mod to take the side of those who disagree with reputable refererences, even when they give no references of their own--like you haven't--so we'll see whether or not this is a scientific site.)

Well you may say tidal forces don't matter, but they do. The portion of the box below the event horizon is causally disconnected from the portion above the horizon, i.e. effectively they don't even touch. So no matter how small you choose your box, the tidal forces of the black hole are stronger than any force you could use to hold the box together. If you choose a box that is at rest with respect to the escaping particle, the top portion will rip off and go with the particle to infinity, while the bottom experiences whatever happens on the other side of the event horizon.

If you want to argue that the box can be made so small that the equivalence principle applies and no tidal forces are felt, that is impossible, because the equivalence principle only really applies to infinitesimally small boxes, with the principle holding to good approximation on larger scales in weak gravitational fields. If you have an infinitesimally small box, then that box can only contain a single point of space, therefore it cannot straddle both sides of an event horizon.

Mind you, quantum physics really makes this point moot, as well as the fact that in theory it takes infinitely long to cross the event horizon as seen from the outside. But I'm assuming this topic is simply an attempt to argue against General Relativity strictly from relativistic principles, so that's what I'm trying to stick to.

If you choose a box that is at rest with respect to the escaping particle, the top portion will rip off and go with the particle to infinity, while the bottom experiences whatever happens on the other side of the event horizon.
Such need not be the case in Y, though, and therein lies the paradox. The paradox can be resolved only by showing that the laws in physics in X are the same as they are in Y. You're arguing against that here.

If you want to argue that the box can be made so small that the equivalence principle applies and no tidal forces are felt, that is impossible, because the equivalence principle only really applies to infinitesimally small boxes, with the principle holding to good approximation on larger scales in weak gravitational fields.
The box need not be infinitesimally small before the tidal force on it is insufficient to break it. So what would break the box in an arbitrarily small frame, like X is defined to be? X can be as small as you want.

The EP holds to good approximation on larger scales even in strong gravitational fields. That explains how a human could survive falling through the horizon of black hole. Many references confirm that the larger the black hole, the less tidal force the human will feel. (Google for "What would happen to me if I fell into a black hole?") It follows that, for a sufficiently large black hole, a box falling through its horizon could be of any given size (even light years across), yet the tidal force on it could still be insufficient to break it, even if it's made out of balsa wood. So it seems it would be impossible to explain how the tidal force could rip the box apart cleanly at the horizon, yet at the same moment in X be too weak to break apart the part of the box that is below the horizon.

Ok, this should resolve it once and for all.

http://en.wikipedia.org/wiki/Event_horizon#Interacting_with_an_event_horizon

From an external Earthbound observer's point of view, watching a spaceship approach a black hole's event horizon, they see the front of the spacecraft approach the horizon and gradually slow down to an asymptotic halt as it gets arbitrarily close. The rear of the spacecraft soon follows, and it too comes to an asymptotic halt. The external observer sees the spacecraft get squashed up into a nearly flat disc as it approaches the horizon, but never sees it cross. From the spaceship's point of view, the ship itself is not squashed up and remains intact, experiencing only minor tidal forces (if the event horizon is far enough from the singularity).

Suppose we put flashlights on both the front and back of the spacecraft to beam signals back to Earth, and these flashlights pulse at regular intervals. From the POV of the spacecraft, the pulses are simultaneous so that the front and back of the ship give off the same number of pulses. But from the external observer's point of view, they see more pulses (or semi-complete pulses) from the back of the ship than from the front, as it is further away from the event horizon. Suppose the distant observer in flat space observes 4 pulses from the front of the ship, and 5 from the back, and after that time slows down near the event horizon to the point that no more pulses will ever be seen unless the spaceship turns around ASAP.

From the spacecraft's POV, if they wait until they see their flashlights pulse for a 5th time, they're already screwed. At that point, they will have already crossed the event horizon without even realizing it, because they have witnessed a 5th pulse from the front of the ship when the outside world only sees 4. The only way to avoid crossing the event horizon after watching 4 pulses of their flashlights is to immediately turn their craft around and accelerate hard in the opposite direction, so that the front tip of the spacecraft is at most the only thing touching the event horizon (which would also stop the 5th pulse at the front from reaching Earth). Accelerating in this fashion, i.e. to escape to infinity, the spacecraft is no longer on a geodesic and the equivalence principle no longer applies, meaning now the spacecraft can feel some rather extreme forces, and extreme redshifts would prevent any contact with things that have fallen inside the horizon. Any portion of the spacecraft which has dipped below the event horizon cannot come back out with the rest and is therefore severed.

To quote the Wikipedia article:

Attempting to make an object approaching the horizon remain stationary with respect to an observer requires applying a force whose magnitude becomes unbounded (becoming infinite) the closer it gets.

So basically, the whole premise of the paradox is false- if you leave the box as an inertial frame and have half the box crossing the event horizon, then the other half also crosses it, and hence so does the particle inside. If you desire that the particle can escape, then it must do so before it can make causal contact with the portion of the box inside the event horizon, i.e. before a light signal from this portion can reach the particle as seen by an observer travelling inside the box.

This is the very problem that brought our esteemed overlord BenTheMan to Sciforums.

Previous discussion here:
General relativity dissatisfies the equivalence principle
General relativity is self-inconsistent
General relativity is self-inconsistent v2

I scanned through the other topics to check the history of the debate, but I didn't see anyone make the same argument as what I had in my last post.

Initially I was tempted to dismiss muinrat's paradox on the grounds that near an event horizon, you have to restrict the region you can consider to be locally flat in the equivalence principle, because the gravitational fields are so powerful. But muinrat had a good comeback which is that in theory, you can set up an event horizon sufficiently far from the singularity so that this is no longer a concern.

This last argument like muinrat made seemed to be an impediment to finding a resolution to the paradox in previous topics, and it looks to me like there were a lot of people not satisfied with the conclusions drawn in the previous debates. I'm proposing that the equivalence principle can be applied, and yet still give reasonable results for all external observers. A box straddling an event horizon from the perspective of a local reference frame is such that if the half of the box inside the event horizon is able to communicate with the other half, the whole box is doomed. The only way for the half of the box and its contents on the outside of the event horizon to escape to infinity is if, in the local reference frame, it accelerates quickly enough to avoid light signals from the doomed half, in which case no contact whatsoever is made with objects inside the black hole.

So basically, the whole premise of the paradox is false- if you leave the box as an inertial frame and have half the box crossing the event horizon, then the other half also crosses it, and hence so does the particle inside. If you desire that the particle can escape, then it must do so before it can make causal contact with the portion of the box inside the event horizon, i.e. before a light signal from this portion can reach the particle as seen by an observer travelling inside the box.
I don't see how this shows that the premise of the paradox is false. The particle need not escape to infinity for the paradox to be established. The paradox depends on the particle escaping to infinity, not having escaped to infinity. The paradox is that in X the box must move relative to the particle, whereas that is not the case in Y (in Y, a freely falling box located anywhere in the frame can be at rest relative to any possible freely falling particle inside the box). If the laws of physics in X were the same as they are in Y, then the box in X could be at rest relative to the escaping particle, escaping to infinity in formation with the particle, in which case the particle need not hit the inside of the box and fall through the horizon, right? It seems that your "hence so does the particle inside" depends on the paradox being true, in which case that conclusion of yours cannot logically be used to show that "the whole premise of the paradox is false".

Although not required to save the paradox, it might help you to imagine that there's a hole in the box that the escaping particle passes through as the box falls toward the black hole's singularity. Then the particle need not fall through the horizon along with the box, so that your attempted solution to the paradox fails that way.

I agree that "the other half also crosses it". There really was no question about that. The only way the whole box could not be falling is if there was a rocket or something attached to the part of the box that is above the horizon, to break apart the box and keep that part from falling too. But then the box would not be in free fall as given, so that's not the case. Even if the box did somehow break apart so that the part of it that is above the horizon can be at rest relative to the escaping particle, the laws of physics in X would still differ from those in Y, because in Y every part of a freely falling box located anywhere in the frame can be at rest relative to any possible freely falling particle inside the box.

Let me know if you think I missed something. I appreciate you making your arguments as well as you are, because I do want to know what's wrong with this paradox. I'm supporting the paradox only to find out what's really wrong with it (what can withstand all scrutiny).

I don't see how this shows that the premise of the paradox is false. The particle need not escape to infinity for the paradox to be established. The paradox depends on the particle escaping to infinity, not having escaped to infinity. The paradox is that in X the box must move relative to the particle, whereas that is not the case in Y (in Y, a freely falling box located anywhere in the frame can be at rest relative to any possible freely falling particle inside the box). If the laws of physics in X were the same as they are in Y, then the box in X could be at rest relative to the escaping particle, escaping to infinity in formation with the particle, in which case the particle need not hit the inside of the box and fall through the horizon, right?

Well, there's a false premise here. If you presume that X has already crossed the event horizon, and that the particle is at rest with respect to X, then the particle could not really have escape velocity. The box could be firing planet-sized nuclear engines thinking it's going to get away from the singularity, but the singularity will just keep getting closer no matter what, and the particle inside the box will endure the same fate even if it suffers from the delusion that it's achieved escape velocity.

The problem seems to be with how we presume we can simply set X up to have an escape velocity like the particle. Plop a box in the middle of an event horizon, and set its initial conditions so that Y has escape velocity. It may well be that from the perspective of X, were it to attempt to keep up with Y while Y was achieving escape velocity, X might see Y as going faster than light. Hence you couldn't physically set up such a box with both Y having escape velocity and X lying beyond the event horizon but in the same inertial frame. Pick one or the other. Now I haven't actually checked to see if X would indeed see Y as going faster than light if it was attempting to keep up, but I'm pretty sure that's what would happen. I could probably do the calculations myself, but I'm lazy and want to see what everyone else says first.

Although not required to save the paradox, it might help you to imagine that there's a hole in the box that the escaping particle passes through as the box falls toward the black hole's singularity. Then the particle need not fall through the horizon along with the box, so that your attempted solution to the paradox fails that way.

Then you're removing the restriction that the particle must be at rest relative to Y and hence to X, thus there's still no paradox.

Let me know if you think I missed something. I appreciate you making your arguments as well as you are, because I do want to know what's wrong with this paradox. I'm supporting the paradox only to find out what's really wrong with it (what can withstand all scrutiny).

It's really good practice for developing physical reasoning. I tried searching for an answer to this problem in my GR textbook, but the one I normally used only really talks about the motions of point particles, not extended objects. So rather than going digging, sometimes you can work it out on your own.

I'm not at all saying my argument is flawless and correct (after all, you have convinced me that my first argument is irrelevant), I'll need to see what arguments others make or if you still have any objections. If more knowledgeable people agree with my logic or can put forth a better argument, I'm sure we'll all come away from this with a better intuitive understanding of GR.

Well, there's a false premise here. If you presume that X has already crossed the event horizon, and that the particle is at rest with respect to X, then the particle could not really have escape velocity.
There's no presumption in the paradox that "the particle is at rest with respect to X". Where do you see that? It is a given that the particle in X is escaping to infinity, so the particle is definitely escaping to infinity.

The box could be firing planet-sized nuclear engines thinking it's going to get away from the singularity, but the singularity will just keep getting closer no matter what, and the particle inside the box will endure the same fate even if it suffers from the delusion that it's achieved escape velocity.
I agree about the box's fate, but the particle is given to be escaping to infinity, so it must be doing that, and nothing in the paradox prevents that, like states or suggests that the particle is at rest relative to the box or X.

Here's an elaboration of the paradox:

Imagine that frame X is a lab. You're a scientist sitting at your desk, which is fixed to the floor of the lab. The lab, X, is in free fall, falling through the horizon of the black hole. But you don't notice any painful tidal force because the black hole is so large that the tidal force on the entire lab is many orders of magnitude less than it would be if the lab was on the Earth. With instruments you are monitoring a small box that is in free fall in the lab. The box straddles the horizon (that's a given). Inside the box, above the horizon, is a particle that is escaping to infinity (that's a given, and nothing is preventing the particle from doing that; it hasn't yet reached the interior surface of the box).

Since both you and the box must be falling toward the black hole's singularity, yet the particle is moving away from the black hole, the particle must be moving relative to both you and the box. Thus you can deduce, using pure thought, that the laws of physics in X do not allow a freely falling box, located anywhere in the frame, to be at rest with respect to a some given freely falling particle inside the box. Your instruments should show that the box is moving relative to the particle.

Now you ask yourself: is that also true in Y, a lab in free fall that is wholly above the horizon of a black hole, like in intergalactic space? Is there anything about the laws of physics in Y that would prevent a freely falling box, located anywhere in Y, from being at rest relative to a freely falling particle in the box that has the same velocity with respect to Y as the escaping particle has with respect to X, or any other freely falling particle in the box? It seems clear that the answer is no. If so, that is a violation of the EP, GR's own postulate. Since GR itself demanded that the EP be violated, GR contradicts itself. Or so it seems.

Then you're removing the restriction that the particle must be at rest relative to Y and hence to X, thus there's still no paradox.
There's no restriction in the paradox that "the particle must be at rest relative to Y".

..., I'm sure we'll all come away from this with a better intuitive understanding of GR.
That's what I'm hoping for too. If the paradox is false, then I think my understanding of GR must be wrong.

Previous discussion here:
General relativity dissatisfies the equivalence principle
General relativity is self-inconsistent
General relativity is self-inconsistent v2
Very interesting, thank you.

I took a look at each one but haven't read through all of them. Hopefully the solution is not buried somewhere in the long (second) thread. Do you know if so? The first one employs acceleration (thrust). The last one seems to be the most similar to the paradox, but still not the same.

The paradox I posted here comes from an anonymous blog. It's as good as any relativity paradox I've seen, but more maddening since I know the solution to many others but not this one.

Pardon me if this is obviously wrong, but how could the part of the box on the inside of the event horizon even be logically compared to the particle and the part of the box on the outside? The whole idea of a black hole is that nothing can escape from it (pass the event horizon from within). So in essence everything on the inside of the event horizon is isolated from the outside in totality. So the part of the box that has crossed the EH is no longer part of the box. If the particle where replaced with rockets it would surely take the part of the box still outside of the EH with it. From this, there is no paradox.

Kalster, suppose you’re in a freely falling lab that is falling through the horizon of a black hole, with the end of the lab where you’re at still above the horizon. At that moment, given what you wrote, you might think that the part of the lab that is below the horizon will be dark as you observe, because, after all, no light from that part of the lab can reach you. But, according to many GR texts, you need not notice anything strange. The other end of the lab will not be dark, because the light reaching your eyes at that moment left that part of the lab before that part of the lab crossed the horizon. You’ll be at the horizon, or have crossed it, before you see any light that left the other end of the lab after it reached or crossed the horizon. (GR predicts that the horizon crosses the lab at precisely the speed of light.) Your lab is analogous to frame X in the paradox. As long as the black hole is large enough that the tidal force on the lab is negligible, you can do good experiments throughout the lab even as you fall through the horizon, so say GR texts, including the quote I gave from Thorne on page 1.

GR predicts that the horizon crosses the lab at precisely the speed of light.Sure about this? As I have it, you can cross the EH at any arbitrary speed with a source of thrust providing an opposite force to the gravity of the singularity. At the EH though, space-time is warped in such a manner that no trajectory (at any speed) leads to the outside. So a spaceship can in theory (as I have it) descend at any arbitrary speed, but any part of it that crosses the EH would stay there, including any light emitted from a point beyond. That would give plenty of time for the last photons from the far end of the ship to reach you and then stop being able to do so. Then that part of the ship/box is irrevocably isolated and cannot be considered in terms of the outside part anymore.

As far as i can tell, the Thorn quote does not deal with event horizons, which are a whole other bowl of wax.

Well I do believe Muinrat is actually right about entering a black hole. You can go to locally inertial coordinates and follow a spaceship beyond the event horizon, without the ship being ripped up or any regions of darkness appearing inside the ship.

However, Muinrat, what I think you are still missing is that you can't have Y moving at escape velocity and still place X in the same inertial frame. I thought about it some more and looked into it some more, and you can see that anything with escape velocity must lie outside the light cone of any event inside the event horizon. It's pretty obvious intuitively, but you can look into the extended Schwarzschild black hole solutions if you want to check it rigorously.

This means that if you set Y to have escape velocity, it travels outside any light cone from X, which is equivalent to saying that from the perspective of X, it travels faster than light. So when Y has escape velocity, you cannot boost X to a physical reference frame which is stationary with respect to Y. The reason you can have a spaceship falling into a black hole and crossing the event horizon without any trouble is because you can set X and Y to be in the same frame, but only if Y is moving slower than the speed of light with respect to X and is falling into the black hole.

Sure about this? As I have it, you can cross the EH at any arbitrary speed with a source of thrust providing an opposite force to the gravity of the singularity.
Suppose the horizon crossed your lab at only 100 kph. You turn on a flashlight directed radially upward (directly away from the black hole) at the exact moment you cross the horizon. Special relativity demands that the speed of the light coming off the flashlight, as measured by you, is exactly c, the speed of light (the light moves radially outward at c). But then the light must be passing outward through the horizon, in defiance of the definition of such horizon, because the horizon is moving through the lab at less than c (it moves radially outward at less than c).

From Ted Bunn's Black Hole FAQ (google for it):

[Once] you get close to the horizon, you realize that it has a very large velocity. In fact, it is moving outward at the speed of light! That explains why it is easy to cross the horizon in the inward direction, but impossible to get back out. Since the horizon is moving out at the speed of light, in order to escape back across it, you would have to travel faster than light. You can't go faster than light, and so you can't escape from the black hole.

However, Muinrat, what I think you are still missing is that you can't have Y moving at escape velocity and still place X in the same inertial frame.
I'm not sure where you're getting this from. It's not in the paradox. Y is a frame, which could be orbiting a planet, or be in intergalactic space. It's just some frame that isn't straddling the horizon of a black hole, or below one. The frame Y need not be moving at escape velocity. It's only the particle in X that is given to be escaping to infinity. The particle in Y can be moving at the same velocity (with respect to Y) as the escaping particle in X is moving (with respect to X), if you want, but the particle in Y need not be thought of as escaping from anything.

X is a frame, so if you placed X in the "same inertial frame", you'd have a frame within a frame, but there's nothing like that in the paradox either.

The frames X and Y could be in different galaxies, keep in mind. X and Y are like labs in which you can do experiments. The only connection you should make between X and Y is to compare the laws of physics in them. It seems that you are making more of a connection between them than that.

I thought about it some more and looked into it some more, and you can see that anything with escape velocity must lie outside the light cone of any event inside the event horizon.
That's true.

This means that if you set Y to have escape velocity, it travels outside any light cone from X, which is equivalent to saying that from the perspective of X, it travels faster than light. So when Y has escape velocity, you cannot boost X to a physical reference frame which is stationary with respect to Y.
I can't follow this, because X and Y are frames that could be in different galaxies. They aren't related as you are relating them. Treat X and Y as labs, and don't consider anything beyond the confines of each lab except to compare the laws of physics between the labs. Consider only whether the box in X must move relative to the particle in that box. Then consider whether a box located anywhere in Y (where Y is perhaps orbiting a planet in another galaxy) must move relative to some particle in that box.

Makes sense?

I'm sorry, I got confused by your initial statement, so I thought X was the half of the box below the event horizon, and Y was the top half on the other side. So let Y be located far away from the black hole; the simplest way would be to place it at infinity. You can see that no matter how small the box at X is, light cones on one half of the horizon cannot contain escaping trajectories on the other half. From this vantage point, the portion of X above the event horizon cannot keep up with the particle and stay attached to the bottom half at the same time. The tidal forces become infinite, because one requires an infinite amount of energy to escape the event horizon. You can't apply the equivalence principle from the vantage point of Y and say that all parts of X should be able to move in unison regardless of what velocity you choose for the top half, because the equivalence principle only says that space looks flat in a small neighbourhood of Y, as seen by Y. Y would still see an infinite amount of spacetime curvature near X, including the event horizon slicing it in half.

If Y is chosen to be inside the event horizon near X, then my previous argument still holds- you still can't choose the two halves of the box to be in the same inertial frame while the top half escapes, because the bottom half can't be boosted faster than light. If Y is chosen to be just above the event horizon but again near X, Y can only apply the equivalence principle to X insofar as Y is falling into the black hole with X, in which case it would see the particle boosted beyond light speed. Otherwise, if Y is on an escape trajectory, then it can see the event horizon and therefore the equivalence principle doesn't apply to a frame extending beyond it. You might object that Y can have escape velocity but still be in freefall. However, space will only appear flat in a region containing Y, which does not include the infinitely curved event horizon. That infinite curvature only disappears when Y is falling into the black hole.

You can't apply the equivalence principle from the vantage point of Y and say that all parts of X should be able to move in unison regardless of what velocity you choose for the top half, because the equivalence principle only says that space looks flat in a small neighbourhood of Y, as seen by Y. Y would still see an infinite amount of spacetime curvature near X, including the event horizon slicing it in half.
You're still mixing experiments in X and Y. The paradox gives no indication that you should do that. Maybe you should forget about Y. Strike it from the paradox. Instead compare the laws of physics in X to the laws of physics in an inertial frame in an idealized, gravity-free universe.

Well, the space around X is not locally flat unless X is falling into the black hole. Otherwise, if some portion of X has escape velocity, that portion will see an infinitely curved event horizon and no matter what, you cannot include it in a local frame of reference with stuff on the other side of the horizon, no matter how small you choose X to be. The infinite curvature only smoothes out when you're travelling into the black hole on a geodesic.

Basically, all the equivalence principle says is that in any reference frame, the spacetime metric expressed in that frame's coordinates is always Minkowskian at the origin and has vanishing first derivatives. In no way does that guarantee that you can always draw a box of some minimal size, i.e. 3cm wide, that will always enclose a region of flat space in that frame. To apply the equivalence principle to an object straddling an event horizon, with the top half escaping off to infinity, you have to apply it either to the top half or bottom half separately, because of the infinitely curved horizon. There is no reference frame, period, that would see the top and bottom halves of the box moving uniformly together unless both of those halves are tumbling into the black hole. This is the only, I repeat the only situation in which the infinities cancel out and you can apply the EP to the whole box. If you don't agree with this basic fact, there's nothing more I can do to convince you, you'll have to look up the actual calculations.

Well, the space around X is not locally flat unless X is falling into the black hole.
X is falling through the horizon of a black hole. That's a given in the paradox.

To apply the equivalence principle to an object straddling an event horizon, with the top half escaping off to infinity, you have to apply it either to the top half or bottom half separately, because of the infinitely curved horizon.
It is not given in the paradox that the top half of the box is escaping to infinity. Only the particle in the box is given to be escaping to infinity.

There is no reference frame, period, that would see the top and bottom halves of the box moving uniformly together unless both of those halves are tumbling into the black hole.
This is a complex way of saying that the whole box must be falling toward the black hole's singularity. Which is what the paradox already concludes. GR demands that the part of the box that is below the horizon must fall toward the black hole's singularity. There's no reason to think that the box is breaking apart, since there's no indication of that in the paradox.

This is the only, I repeat the only situation in which the infinities cancel out and you can apply the EP to the whole box. If you don't agree with this basic fact, there's nothing more I can do to convince you, you'll have to look up the actual calculations.
I don't know what you're trying to convince me of. All you've done, that I can see, is come to conclusions that are already givens or conclusions of the paradox. You haven't dealt with the paradox itself yet, which is this:

Since in X at least part of the box is falling toward the black hole's singularity, whereas the particle is receding from the black hole, those two objects must be moving relative to each other. But in an inertial frame in an idealized, gravity-free universe, a whole freely falling box located anywhere in the frame can be at rest relative to any possible freely falling particle in the box. This difference between X and an inertial frame in an idealized, gravity-free universe contradicts the EP, or so it seems.

If you think you've resolved the paradox, if only by finding some problem with its premises, I really don't see it. If you have found a problem, I'd like to understand it.

Since in X at least part of the box is falling toward the black hole's singularity, whereas the particle is receding from the black hole, those two objects must be moving relative to each other. But in an inertial frame in an idealized, gravity-free universe, a whole freely falling box located anywhere in the frame can be at rest relative to any possible freely falling particle in the box. This difference between X and an inertial frame in an idealized, gravity-free universe contradicts the EP, or so it seems.
You have 2 seperate referance frames if you specify the particle as escaping and the box as infalling, and the scenario is irrelevant.
The particle and the box, if located in the same referance frame, CAN still be at rest however; in this case they are both falling into the black hole.

The problem is because of the assumption (by the author of the 'paradox') that any 2 different objects, under specific conditions, can be said to be in the same referance frame. This is simply not true, even when thier spatial coordinates occupy the same place. A bullet and it's firearm can be said to have the same referance frame, up untill the point when the bullet is shot, after which, even when it's still in the barrel, it has a different frame of referance.
It is quite obviouse that our box and particle will have very different views of the universe too.

That is, assuming my understanding is correct; no guarentees :p
-Andrew

You have 2 seperate referance frames if you specify the particle as escaping and the box as infalling, and the scenario is irrelevant.
There can be more than one object in a frame. A frame is like a lab. An experiment in a lab can involve multiple objects. The objects need not be at rest relative to each other, nor need they be at rest with respect to the lab.

The particle and the box, if located in the same referance frame, CAN still be at rest however; in this case they are both falling into the black hole.
Sure, but that doesn’t resolve the paradox, because in the paradox it is a given that the particle is escaping to infinity. It is definitely not falling into the black hole.

This is simply not true, even when thier spatial coordinates occupy the same place. A bullet and it's firearm can be said to have the same referance frame, up untill the point when the bullet is shot, after which, even when it's still in the barrel, it has a different frame of referance.
You’re talking about an object’s own frame, a frame in which that object is at rest. There can be another frame in which the object is moving. The paradox doesn’t say that X is either the box’s or the particle’s own frame. It specifies only that the box and the particle are in X.

I don't know what you're trying to convince me of. All you've done, that I can see, is come to conclusions that are already givens or conclusions of the paradox. You haven't dealt with the paradox itself yet, which is this:

Since in X at least part of the box is falling toward the black hole's singularity, whereas the particle is receding from the black hole, those two objects must be moving relative to each other. But in an inertial frame in an idealized, gravity-free universe, a whole freely falling box located anywhere in the frame can be at rest relative to any possible freely falling particle in the box. This difference between X and an inertial frame in an idealized, gravity-free universe contradicts the EP, or so it seems.

If you think you've resolved the paradox, if only by finding some problem with its premises, I really don't see it. If you have found a problem, I'd like to understand it.

Well basically I say I'm ready to give up because I'm running out of ways to explain this to you without doing some tedious math, which shouldn't be necessary. I'll try to focus my arguments now that I think I understand what your basic problem is, without throwing all these other stipulations in like observer Y, etc.

Here is one problematic statement:
But in an inertial frame in an idealized, gravity-free universe, a whole freely falling box located anywhere in the frame can be at rest relative to any possible freely falling particle in the box.

So, what if I stipulate that the particle is a tachyon, travelling faster than light? Then indeed your statement is false, there are restrictions which strictly forbid your inertial frame from being at rest with respect to this particle. In fact, forget tachyons. Imagine the particle is a simple photon. Then there is no physical frame anywhere in the universe in which the photon is locally at rest- it always has velocity c, no matter which frame you choose (as long as that frame has the particle at the origin and space is flat at the origin in these coordinates). So even if space were 100% flat and gravity-free, if your particle is a photon then there is no frame whatsoever in which that particle is (locally) at rest.

So now imagine you have your nearly infinitesimal box X straddling the event horizon. Since X is a physical observer who can be considered to be at rest in an inertial frame, X must be travelling slower than lightspeed, and therefore falling into the black hole (pretty obvious so far). If you place any particle in the box and give it any speed less than c relative to the box (as measured in this inertial frame), there is no problem in boosting the box so that it matches the velocity of this particle. However, when you demand that the particle have escape velocity as seen from the outside, that is equivalent to telling observer X that the particle is moving faster than light.

After all, from the perspective of X as it falls into the black hole (instantaneously straddling the event horizon), anything moving slower than light relative to X will fall into the black hole as well. An observer at X could see a particle leaving the box at sublight speed and assume that it's escaping the black hole, but it will only appear that way to the box because the box is descending more quickly to the singularity. So again, for you to demand that the particle in the box is escaping to infinity, it's equivalent to telling X that it should see the particle as going faster than light. That's why X can't be at rest with respect to an escaping particle, because you can't take X and boost it to another frame moving faster than light with respect to its original frame.

Your problem seems to be the misconception that you can take an inertial frame and boost it to any velocity you want. You can boost it to any velocity slower than light, but no faster. There's nothing in the equivalence principle saying that escape velocity has to be slower than light. The EP doesn't say anything about escape velocity at all.

Ultimately, if you still believe there is a paradox, then since black holes and the EP can be expressed mathematically, you should be able to demonstrate the paradox mathematically as well. Switch to the reference frame of observer X and go to locally inertial coordinates, express the spacetime metric in those coordinates and show how, from the perspective of X, escape velocity should be a perfectly reasonable velocity to obtain. I would, however, imagine this to be an impossible task to complete, because general relativity is specifically constructed so as to be manifestly covariant, meaning these paradoxes are eliminated in the maths by default.

There can be more than one object in a frame. A frame is like a lab. An experiment in a lab can involve multiple objects. The objects need not be at rest relative to each other, nor need they be at rest with respect to the lab.
...
You’re talking about an object’s own frame, a frame in which that object is at rest. There can be another frame in which the object is moving. The paradox doesn’t say that X is either the box’s or the particle’s own frame. It specifies only that the box and the particle are in X.
Alright, I will agree with that.

Sure, but that doesn’t resolve the paradox, because in the paradox it is a given that the particle is escaping to infinity. It is definitely not falling into the black hole.
This is where a problem lies.
It is begging the question (logical fallacy); you assert that the box is infalling (by saying it has crossed the horizon)
Then you assert that the particle is moving away from the box.
Then you say that this contradicts EP because they are not at rest with each other.

Essentialy, what the 'paradox' has set up is analogous to:
for all a+b, there exists some value c.
then it decides that a=1, b=2, and c=5
then finding that, according to standard mathematical axioms, a+b=3, and therefore a+b does not equal c, it then says that a+b=c must be false, and that therefore, mathematics is self-contradictory.

But, there is some value of c which does work: 3. Similarily, there is some system which works for the 'paradox,' namely: that the particle is infalling with the box.

-Andrew

So, what if I stipulate that the particle is a tachyon, travelling faster than light? Then indeed your statement is false, there are restrictions which strictly forbid your inertial frame from being at rest with respect to this particle. In fact, forget tachyons. Imagine the particle is a simple photon. Then there is no physical frame anywhere in the universe in which the photon is at rest- it always has velocity c, no matter which frame you choose. So even if space were 100% flat and gravity-free, if your particle is a photon then there is no frame whatsoever in which that particle is at rest.
But the particle is presumably neither a tachyon nor a photon.

So now imagine you have your nearly infinitesimal box X straddling the event horizon. Since X is a physical observer who can be considered to be at rest in an inertial frame, X must be travelling slower than lightspeed, and therefore falling into the blackhole (pretty obvious so far). If you place any particle in the box and give it any speed less than c relative to the box (as measured in this inertial frame), there is no problem in boosting the box so that it matches the velocity of this particle. However, when you demand that the particle have escape velocity as seen from the outside, that is equivalent to telling observer X that the particle is moving faster than light.
The sentence I bolded is false, or cannot be true unless GR contradicts itself. If the particle has a speed of c or greater relative to the box, that would be a problem of GR, not the paradox, because GR allowed the box and the particle to exist as given in the paradox, and it was deduced that they must be moving relative to each other.

After all, from the perspective of X as it falls into the black hole (instantaneously straddling the event horizon), anything moving slower than light relative to X will fall into the black hole as well.
If this is true, then a particle in X that is escaping to infinity would be moving at c or greater relative to X. But then GR would contradict itself, because it allows the particle and X to exist as given, and special relativity (SR), which applies in X according to GR, demands that the particle’s velocity must be less than c in X. If GR allows us to set up a thought experiment that violates the theory, its goose is cooked, rather than the thought experiment having a problem.

An observer at X could see a particle leaving the box at sublight speed and assume that it's escaping the black hole, but it will only appear that way to the box because the box is descending more quickly to the singularity. So again, for you to demand that the particle in the box is escaping to infinity, it's equivalent to telling X that it should see the particle as going faster than light. That's why X can't be at rest with respect to an escaping particle, because you can't take X and boost it to another frame moving faster than light with respect to its original frame.
How do you resolve the issue that GR allows the particle to be escaping to infinity? GR cannot both allow and disallow the particle to be escaping to infinity, right?

The EP doesn't say anything about escape velocity at all.
No, but GR as a whole does, and GR must not contradict its own EP or else the theory is invalid. GR allows the particle to be escaping to infinity, so it can't be a problem of the paradox.

Ultimately, if you still believe there is a paradox, then since black holes and the EP can be expressed mathematically, you should be able to demonstrate the paradox mathematically as well. Switch to the reference frame of observer X and go to locally inertial coordinates, express the spacetime metric in those coordinates and show how, from the perspective of X, escape velocity should be a perfectly reasonable velocity to obtain.
Yes the paradox could be expressed mathematically, but that would be overkill. Nothing can escape a black hole, where the black hole is delimited by a horizon by definition. Above the horizon there is no black hole, so objects can escape to infinity from there. Do you really doubt this?

I would, however, imagine this to be an impossible task to complete, because general relativity is specifically constructed so as to be manifestly covariant, meaning these paradoxes are eliminated in the maths by default.
You’re assuming that GR doesn’t contradict itself. But the paradox shows otherwise, or so it seems. You can’t show a problem with the paradox by showing that GR contradicts itself, which is what you have done by showing that a particle above the horizon of a black hole (namely, the particle in X) cannot be escaping to infinity.

If you think I made a mistake somewhere, let me know. I'm not trying to be pig-headed here.

It is begging the question (logical fallacy); you assert that the box is infalling (by saying it has crossed the horizon)
It’s a deduction, not a given, based on the box’s location in X.

Then you assert that the particle is moving away from the box.
It’s a deduction, not a given, that the particle is moving relative to the box, because the box is infalling (it was deduced), whereas the particle is moving away from the black hole (that’s a given).

Then you say that this contradicts EP because they are not at rest with each other.
Note that the paradox did not specify the box’s velocity relative to X or the particle. The only thing the paradox specifies about the box is its location in X. In an inertial frame in an idealized, gravity-free universe, you cannot tell that any part of a freely falling box and a freely falling particle inside the box are moving relative to each other, given no more information about the box than its location in the frame. That difference between those frames violates the EP, or so it seems.

Essentialy, what the 'paradox' has set up is analogous to:
for all a+b, there exists some value c.
then it decides that a=1, b=2, and c=5
then finding that, according to standard mathematical axioms, a+b=3, and therefore a+b does not equal c, it then says that a+b=c must be false, and that therefore, mathematics is self-contradictory.
The problem with your analogy is that the paradox deduced from GR’s predictions—it was not a given, as you allude—that the box and the particle are moving relative to each other.

But, there is some value of c which does work: 3. Similarily, there is some system which works for the 'paradox,' namely: that the particle is infalling with the box.
The particle is given to be escaping to infinity, so the only system that would work to resolve the paradox is that the whole box is also escaping to infinity, moving in formation with the particle in X. But GR doesn’t allow that, which we know given no more information about the box than its location in X. Review bolded statement above to see the paradox.

If this is true, then a particle in X that is escaping to infinity would be moving at c or greater relative to X. But then GR would contradict itself, because it allows the particle and X to exist as given, and special relativity (SR), which applies in X according to GR, demands that the particle’s velocity must be less than c in X. If GR allows us to set up a thought experiment that violates the theory, its goose is cooked, rather than the thought experiment having a problem.

Maybe this is where you're having the most conceptual difficulty. The speed of light isn't an absolute constant in the sense that if you were to track a beam of light through space, you would observe that it travelled at 3.00x10^8 m/s the whole way. For instance, if you shine a beam of light at a black hole and track it, you will notice the beam slowing down more and more, and coming to an asymptotic halt at the horizon. You could then argue that you go faster than that beam of light even when you're just taking your morning jog. However, if you were able to somehow catch up with this beam and then come to rest in some chosen inertial frame located near the beam, you'd still see it travelling at 3.00x10^8 m/s.

From the perspective of someone approaching the event horizon, infinite amounts of time pass by in the outside world before they even cross the horizon. So strange things like objects moving faster than light can be perceived from inside the event horizon. If X shines a beam of light out to the escaping particle, that beam of light will never be able to catch up, so this is certainly one sense in which X sees the particle as going faster than light.

In general, you can have objects going faster than light if the spacetime surrounding them is stretching out faster than light. In that local spacetime, c is still measured as 3.00x10^8 m/s, but beams of light in that spacetime would appear to be going much faster as seen by distant observers. This idea is encapsulated in many theories of inflationary cosmology, where the assumption is that the expansion of the universe will continue to accelerate until we could even (in principle) start to see galaxies move away from us faster than light. However, one caveat of these theories is that once a galaxy is moving away from us at lightspeed, spacetime will be stretching so fast that even light from this galaxy won't be able to reach us, effectively forming an event horizon.

So one key point I think you're missing is that c is only 3.00x10^8 m/s as measured in a local frame, and c can be varied on a global scale. So if X sees the escaping particle going faster than 3.00x10^8 m/s, it would still see escaping light going faster than the particle, but both speeds can be greater than what X measures as c in its own local frame. So if you try to apply SR to this picture, you have to treat the escaping particle as some kind of tachyon, because that's how you've set it up from the very beginning by insisting that the particle escapes.

So strange things like objects moving faster than light can be perceived from inside the event horizon. If X shines a beam of light out to the escaping particle, that beam of light will never be able to catch up, so this is certainly one sense in which X sees the particle as going faster than light.
This is not the case in an inertial frame in an idealized, gravity-free universe. The EP tells us that the laws of physics in X are the same as they are in an inertial frame in an idealized, gravity-free universe. So here you're arguing for the paradox.

In that local spacetime, c is still measured as 3.00x10^8 m/s, but beams of light in that spacetime would appear to be going much faster as seen by distant observers.
I would completely ignore what a distant observer would measure or see in X. The paradox is about the laws of physics in X, as measured by an observer who is in X and at rest with respect to X. The paradox is not at all about the laws of physics for a distant observer who measures what happens in X.

This idea is encapsulated in many theories of inflationary cosmology, where the assumption is that the expansion of the universe will continue to accelerate until we could even (in principle) start to see galaxies move away from us faster than light. However, one caveat of these theories is that once a galaxy is moving away from us at lightspeed, spacetime will be stretching so fast that even light from this galaxy won't be able to reach us, effectively forming an event horizon.
You can't resolve the paradox by referencing what GR predicts about non-inertial frames, like you're doing here. According to the EP, X is equivalent to an inertial frame in an idealized, gravity-free universe.

So one key point I think you're missing is that c is only 3.00x10^8 m/s as measured in a local frame, and c can be varied on a global scale.
But that point is irrelevant, because X is a local frame, according to the EP. The paradox doesn't reference measurements made on a global scale. It references only measurements that could be made in X, by an observer at rest with respect to X, about objects wholly within X.

So if X sees the escaping particle going faster than 3.00x10^8 m/s, it would still see escaping light going faster than the particle, but both speeds can be greater than what X measures as c in its own local frame.
I don't know what you mean by "X sees", because X is a frame. If you mean "an observer in X measures", then your statement is false. An observer in X, at rest with respect to X, would measure the speed of light in X to be exactly c, according to SR. The observer would measure the speed of the particle to be less than c, according to SR. The observer would measure the box to be moving relative to the particle, according to GR. Then it is impossible for the observer to measure the box moving at the same velocity as the particle, according to GR. That contradicts GR's own EP, or so it seems.

So if you try to apply SR to this picture, you have to treat the escaping particle as some kind of tachyon, because that's how you've set it up from the very beginning by insisting that the particle escapes.
You're implying that the paradox does something that GR does not allow, by having the particle be escaping to infinity. But GR does allow the particle to be escaping to infinity. The escape velocity anywhere above the horizon is less than c, according to GR. Only at or below the horizon can objects not be escaping to infinity, according to GR. The particle is above the horizon. If GR requires the particle to be treated as "some kind of tachyon" in X, then GR contradicts itself.

The paradox cannot be resolved by contradicting GR's predictions, namely the predictions of SR in X, a frame that the EP says is equivalent to an inertial frame in an idealized, gravity-free universe, or the prediction of GR that objects above a horizon can be escaping to infinity. Nor can the paradox be resolved by saying that it's doing something that it doesn't, namely measuring things globally rather than locally. That is what I see that you have done above (but if you can show that I'm wrong, my mind could be changed). The paradox does not "insist" that the particle is escaping to infinity, in defiance of GR's predictions. Instead it lets the particle be escaping to infinity, which GR allows. And all of the measurements in the paradox can be made locally.

I don't know what you mean by "X sees", because X is a frame. If you mean "an observer in X measures", then your statement is false. An observer in X, at rest with respect to X, would measure the speed of light in X to be exactly c, according to SR. The observer would measure the speed of the particle to be less than c, according to SR. The observer would measure the box to be moving relative to the particle, according to GR. Then it is impossible for the observer to measure the box moving at the same velocity as the particle, according to GR. That contradicts GR's own EP, or so it seems.

Well it seems that your difficulty accepting this fact is the root of your belief that there's a paradox here. Nothing forbids an observer at X from seeing the escaping particle move faster than light, because that particle isn't located at the origin of X's coordinate system. In other words, the particle is allowed to move at arbitrary speeds as seen by X, so long as everything at the origin of X is moving at light speed or slower. The EP only applies to an infinitesimal volume centred at X. The EP can be extended to a larger object such as a spacecraft, provided the whole craft is falling into the black hole. Otherwise, insofar as you want to extend the EP to cover a frame with objects both falling into the black hole and escaping, you get strange events such as X seeing escaping particles going faster than 3.00x10^8 m/s. Black holes are strange no doubt and give unusual results, but unusual does not a paradox make.

Think of it this way: if X has already crossed the event horizon, then it's perfectly reasonable to place a particle in the centre of frame X and demand that X be at rest relative to this particle. But since in this case, both the box and particle have already crossed the event horizon, it's completely unreasonable to ask that the particle be able to have escape velocity.

Now let's go back to your example, where X only straddles the event horizon and the particle sits just outside the horizon. Give the particle some escape velocity slower than light, as seen by a distant observer. Then you can always postulate a particle with this same velocity (i.e. if it was shot out of the same cannon from X), but closer to the event horizon so that it does not escape. Then it's reasonable to ask that X be stationary with respect to this second particle, but not the first, which we can now see is located far enough away from the event horizon that we can't approximate it as being inside X's locally inertial frame. So depending on whether the particle in the box at X is escaping or not, you have to limit the volume you can consider to be locally inertial. This volume must always be small enough that it cannot contain any escaping particles, otherwise you have to treat those particles as tachyons like I said.

The event horizon is a region of infinite curvature. There is no such thing as a minimal volume in which it the EP can be applied to a given event horizon for all observers. The EP can only be considered valid in a finite region near a black hole insofar as everything in that finite region is staying on the same geodesic.

The EP only applies to an infinitesimal volume centred at X.
Who says the box can't be infinitesimally small? X is arbitrarily small in the paradox. X can be the same size as the box.

The EP can be extended to a larger object such as a spacecraft, provided the whole craft is falling into the black hole. Otherwise, insofar as you want to extend the EP to cover a frame with objects both falling into the black hole and escaping, you get strange events such as X seeing escaping particles going faster than 3.00x10^8 m/s.
The EP makes no mention that all the objects in the frame need be moving in the same direction or whatever. Can you provide a reference?

Think of it this way: if X has already crossed the event horizon, then it's perfectly reasonable to place a particle in the centre of frame X and demand that X be at rest relative to this particle. But since in this case, both the box and particle have already crossed the event horizon, it's completely unreasonable to ask that the particle be able to have escape velocity.
You're arguing against the EP, and for the paradox here. The EP makes no exception for a frame falling through the horizon of a black hole. Thorne is explicit about that in the quote of his that I gave on page 1. You can't resolve the paradox by contradicting the EP.

Now let's go back to your example, where X only straddles the event horizon and the particle sits just outside the horizon. Give the particle some escape velocity slower than light, as seen by a distant observer. Then you can always postulate a particle with this same velocity (i.e. if it was shot out of the same cannon from X), but closer to the event horizon so that it does not escape. Then it's reasonable to ask that X be stationary with respect to this second particle, but not the first, which we can now see is located far enough away from the event horizon that we can't approximate it as being inside X's locally inertial frame. So depending on whether the particle in the box at X is escaping or not, you have to limit the volume you can consider to be locally inertial. This volume must always be small enough that it cannot contain any escaping particles, otherwise you have to treat those particles as tachyons like I said.
The problem with this is that the same thing does not apply in a freely falling frame that does not straddle the horizon of a black hole, as the EP requires. So again you're arguing against the EP, and for the paradox. In a freely falling frame "Y" that does not straddle the horizon of a black hole, no matter how small Y is, a particle in Y can have the same velocity relative to Y as the first particle in your example has (relative to X), and yet a box wholly in the frame, located anywhere in the frame, and that surrounds that particle, can be at rest relative to the particle. The EP tells us that the laws of physics in X are the same as those in Y, but you're telling me that the laws of physics in X are different than those in Y, and saying that that's a problem with the paradox.

Also note that the box in X cannot be at rest with any particle inside the box that is moving away from the black hole, not just an escaping particle. A particle inside the box that is above the horizon but not escaping to infinity can be moving away from the black hole, no matter how close to the horizon that particle is. Then it is not necessarily "reasonable to ask that X be stationary with respect to this second particle".

The event horizon is a region of infinite curvature.
Spacetime curvature is infinite only at the black hole's singularity, not at its horizon.

There is no such thing as a minimal volume in which it the EP can be applied to a given event horizon for all observers.
Spacetime curvature is not infinite at a horizon, or else a human could not survive falling through one, as many GR texts tell us a human can do that. (Spacetime curvature is the same thing as the tidal force. An infinite spacetime curvature means that even atoms are ripped apart.) The EP can be applied to a frame falling through a horizon. For a given sized frame falling through the horizon, the larger the black hole, the flatter the spacetime is in the frame. Then the spacetime in X could be any number of orders of magnitude flatter than the spacetime in the room in which you're reading this. So how could SR be experimentally confirmed in labs on Earth, but cannot be experimentally confirmed in X?

The EP can only be considered valid in a finite region near a black hole insofar as everything in that finite region is staying on the same geodesic.
The EP applies to any small, freely falling frame, like X is, just like Einstein and Thorne say in the quote I gave on page 1. The smaller the frame, the better the EP applies; e.g. the more precisely SR can be experimentally confirmed in the frame. X is defined as arbitrarily small in the paradox; that is, the paradox holds, or seems to, no matter how small X is.

Who says the box can't be infinitesimally small? X is arbitrarily small in the paradox. X can be the same size as the box.
No, it can't. The equivalence principle does not hold in a region that encompasses the inside and outside of the event horizon because spacetime has infinite curvature at the event horizon; i.e., a singularity. We already know that black holes represent a singularity. That is exactly why another name for black holes are called singularities for short. Does this singularity represent a paradox? No. Does it represent a problem? Of course it does. Physicists don't like singularities.

The EP applies to any small, freely falling frame
Stop that! This is not true. There does exist a neighborhood of a point in which the EP holds. This does not mean the EP applies to any small freely falling frame. The neighborhood of a point in which the EP does hold becomes vanishingly small as the point approaches the event horizon.

Who says the box can't be infinitesimally small? X is arbitrarily small in the paradox. X can be the same size as the box.

If the box is infinitesimally small, then escape velocity is lightspeed, so you still wouldn't be able to trap the particle in a stationary frame.

The EP makes no mention that all the objects in the frame need be moving in the same direction or whatever. Can you provide a reference?

Look it up in a standard GR textbook- Sean Carroll's is the one I spent most of my time working with. You can approximate a system of objects as being located in flat space provided all of those objects are moving on similar geodesics that remain similar for long periods of time. That is to say, unless you're in perfectly flat space, objects do eventually drift apart over time due to gravity, so treating a region of space as being perfectly flat is always an approximation.

You're arguing against the EP, and for the paradox here. The EP makes no exception for a frame falling through the horizon of a black hole. Thorne is explicit about that in the quote of his that I gave on page 1. You can't resolve the paradox by contradicting the EP.

I'm not contradicting the EP, you are- or so I claim at any rate. You can't create a paradox by contradicting the EP.


The problem with this is that the same thing does not apply in a freely falling frame that does not straddle the horizon of a black hole, as the EP requires. So again you're arguing against the EP, and for the paradox. In a freely falling frame "Y" that does not straddle the horizon of a black hole, no matter how small Y is, a particle in Y can have the same velocity relative to Y as the first particle in your example has, and yet a box wholly in the frame, located anywhere in the frame, and that surrounds that particle, can be at rest relative to the particle. The EP tells us that the laws of physics in X are the same as those in Y, but you're telling me that the laws of physics in X are different than those in Y, and saying that that's a problem with the paradox.

No, I'm telling you that you have to restrict the size of X on which you can apply the EP, because you are postulating a particle with escape velocity, meaning you are assigning it a very different geodesic than the rest of the box. Give me any size box X with your so-called paradox, and I'll give you a smaller reference frame in which the EP is still valid. Make X infinitely small, and the escaping particle is a photon.

Spacetime curvature is infinite only at the black hole's singularity, not at its horizon.

It's true that the Ricci scalar doesn't blow up to infinity at the horizon, so in that sense the curvature is finite. But in terms of external, locally Minkowskian coordinate systems, the event horizon is a legitimate singularity, where time comes to a complete halt.

Spacetime curvature is not infinite at a horizon, or else a human could not survive falling through one, as many GR texts tell us a human can do that. (Spacetime curvature is the same thing as the tidal force. An infinite spacetime curvature means that even atoms are ripped apart.)

Like I told you before, that infinite curvature only disappears when you're on a geodesic heading into the black hole. If you're on a spaceship and every part of the spaceship is falling into the black hole on nearly identical geodesics, that's when you don't feel a tidal force. Demanding that a particle in your box/ship is escaping to infinity is equivalent to placing it on a geodesic which diverges radically from the rest (after all, you set it up to escape!). In this situation, you cannot apply the EP to the box and particle in the same frame, for the reasons I explained above.


The EP applies to any small, freely falling frame, like X is, just like Einstein and Thorne say in the quote I gave on page 1. The smaller the frame, the better the EP applies; e.g. the more precisely SR can be experimentally confirmed in the frame. X is defined as arbitrarily small in the paradox; that is, the paradox holds, or seems to, no matter how small X is.

And this flaw in your reasoning is going to continue to produce "paradoxes" until you understand why your assertion is not true. If you're arbitrarily close to an event horizon, the frame in which the EP applies universally to all objects in your neighbourhood is also arbitrarily small. Just because ships can fall into a black hole without feeling anything, that doesn't mean they won't feel some very powerful tidal forces if they change their minds at the last minute and decide to escape.

No, it can't. The equivalence principle does not hold in a region that encompasses the inside and outside of the event horizon because spacetime has infinite curvature at the event horizon; i.e., a singularity.
Then how do you explain numerous references that contradict this? Google for "At the center of a black hole lies the singularity, where matter is crushed to infinite density, the pull of gravity is infinitely strong, and spacetime has infinite curvature". The horizon of a black hole is not at the center of it.

We already know that black holes represent a singularity. That is exactly why another name for black holes are called singularities for short.
The part of the black hole that is the horizon is not the same part of the black hole that is the singularity.

Stop that! This is not true. There does exist a neighborhood of a point in which the EP holds. This does not mean the EP applies to any small freely falling frame.
Einstein specifically says that it applies to any small freely falling frame. Why should I believe you over the inventor of the EP?

The neighborhood of a point in which the EP does hold becomes vanishingly small as the point approaches the event horizon.
That's false, as shown above. It's contradicted by Thorne and numerous other references. Spacetime curvature is not infinite at the horizon.

If the box is infinitesimally small, then escape velocity is lightspeed, so you still wouldn't be able to trap the particle in a stationary frame.
No, no matter how small the box is, it can still straddle the horizon, with part of it above the horizon and part of it below the horizon. The horizon is at a point along a radius. Then the horizon in X is a plane of points. An infinitesimally small thing is not as small as a point. Anywhere above a horizon, no matter how close to it, a particle can escape to infinity, according to GR.

Look it up in a standard GR textbook- Sean Carroll's is the one I spent most of my time working with. You can approximate a system of objects as being located in flat space provided all of those objects are moving on similar geodesics that remain similar for long periods of time. That is to say, unless you're in perfectly flat space, objects do eventually drift apart over time due to gravity, so treating a region of space as being perfectly flat is always an approximation.
Saying that "objects do eventually drift apart over time due to gravity" is different than saying that objects in a frame must be moving in the same direction (or whatever) before the EP applies to that frame, which is what you suggested before. The original post points out that the tidal force, which is the same thing as spacetime curvature, cannot force a particle at the center of a box to move away from the center. Then even in imperfectly flat spacetime, a particle at the center of a box can remain at the center indefinitely. The edges of the box move toward or away from the particle as the box is stretched & squeezed by the tidal force (same thing as spacetime curvature), but that movement is symmetrical such that the particle can stay at the center. For a reference google for "If we subtract the center of mass force, we see the differential force acting on it".

No, I'm telling you that you have to restrict the size of X on which you can apply the EP, because you are postulating a particle with escape velocity, meaning you are assigning it a very different geodesic than the rest of the box. Give me any size box X with your so-called paradox, and I'll give you a smaller reference frame in which the EP is still valid. Make X infinitely small, and the escaping particle is a photon.
An infinitesimally small object is not as small as a point. The horizon in X is a plane of points. Then even when X is infinitesimally small, it can still straddle the horizon such that part of it is below the horizon and part of it is above the horizon. How can you refute this?

It's true that the Ricci scalar doesn't blow up to infinity at the horizon, so in that sense the curvature is finite. But in terms of external, locally Minkowskian coordinate systems, the event horizon is a legitimate singularity, where time comes to a complete halt.
You're talking about a coordinate singularity, but there is no coordinate singularity for an observer in X. Time comes to a halt at the horizon for distant observer, but not for an observer in X who is at rest with respect to X. The laws of physics in X that the paradox references are those that apply to an observer in X who is at rest with respect to X, not those that apply to a distant observer who measures what happens in X.

Like I told you before, that infinite curvature only disappears when you're on a geodesic heading into the black hole. If you're on a spaceship and every part of the spaceship is falling into the black hole on nearly identical geodesics, that's when you don't feel a tidal force. Demanding that a particle in your box/ship is escaping to infinity is equivalent to placing it on a geodesic which diverges radically from the rest (after all, you set it up to escape!). In this situation, you cannot apply the EP to the box and particle in the same frame, for the reasons I explained above.
Can you give any reference that says that the EP does not apply to a small, freely falling frame that is falling through the horizon of a black hole? That's what you're implying here (how does it not?). Thorne directly contradicts you, in the quote I gave on page 1.

If you're arbitrarily close to an event horizon, the frame in which the EP applies universally to all objects in your neighbourhood is also arbitrarily small.
You're suggesting that the tidal force becomes infinite at the horizon. But numerous references tell us that a human can survive falling through a horizon. Then the tidal force does not become infinite at the horizon, and the EP can apply to a human-sized freely frame that falls through the horizon. But even so, X is defined as arbitrarily small, so what's the problem with the paradox?

Just because ships can fall into a black hole without feeling anything, that doesn't mean they won't feel some very powerful tidal forces if they change their minds at the last minute and decide to escape.
It doesn't matter what tidal force the ship would feel if it tried to be at rest relative to the particle. No degree of tidal force can force the particle to move away from the center of the box, so the tidal force is not the reason why the particle must move relative to the box.

I have to move on now, as long as we're just regurgitating arguments. Thank you for the input given here.

I don't see that the paradox has been resolved here. I do think there's a solution. But the solution must show that the laws of physics in X are the same as those in an inertial frame in an idealized, gravity-free universe. The only difference between those frames is the tidal force, according to GR. But the tidal force cannot force a particle at the center of a box to move away from the center. The tidal force is the same thing as spacetime curvature, and spacetime curvature is finite at a horizon (to explain why a human can survive falling through one). I don't see any attempted solution above that is not ruled out by these facts, or that does not contradict GR, including its EP.

I have to move on now, as long as we're just regurgitating arguments.
The only one regurgitating arguments is you. Cpt Bork fully refuted the "paradox". You ignored his arguments (as well as mine).

Well at the end of the day, I'm mortal and can be wrong. The easiest and most inarguable way to demonstrate a paradox of this nature is to do the actual calculations. However, it doesn't seem like you truly believe in such a paradox, but merely want an intuitive understanding of why this supposed paradox doesn't really exist. If you find an explanation that makes more sense to you of any sort, whether it bears any resemblance to my arguments or not, I'd be very interested in hearing what that explanation is, as it would help my understanding as much as it would help yours. So if you ask around and find someone gives you a much more satisfactory answer, please come back here to share it with us.