You've eloquently explained why GR demands that the cloud be splitting apart. But your conclusion "Thus, I don't think it's any problem..." contradicts the EP, so the paradox remains. The EP tells me the laws of physics in my lab are the same as those in a horizon-less inertial frame. Then no Rindler horizon can affect my experiment and I should have no problem letting all the cloud's particles move in formation as I measure, the same as I could do in some other inertial frame. Search for "There is no horizon anywhere for them". How do you rebut this?

Even though the Rindler horizon isn't a "horizon" for an inertial observer, it's still a well-defined lightlike path through spacetime, and it's still true that slower-than-light particles on one side must inevitably get closer and closer to it even if they never cross, while STL ones on the other side must get farther from it with time. I realized there's a mistake in my answer though, because if you just copy and paste one of the Rindler observer worldlines to the other side of the horizon, the above will be true yet the distance bt. the two worldlines will be constant in the inertial frame.

If I'm summarizing your edit properly, you see no issue with all the particles moving in formation as I measure, even as they move toward multiple destinations. That contradicts the EP. Consider a lab in the ISS. If a precise velocity measurement is taken of one particle that heads toward the Moon and another particle that heads toward the Earth, their velocities will always differ; the particles won't be found to be moving in formation. How do you rebut that?

As I mentioned in my answer, just because things are moving with a constant coordinate separation between them in a local inertial frame, that doesn't imply their separation is remaining constant in a non-local, non-inertial coordinate system covering a large region of curved spacetime, like Schwarzschild coordinates. And as I said, even though any slower-than-light particle outside the event horizon must be approaching the horizon in a local inertial frame at the horizon, that same particle can still be moving outward in terms of Schwarzschild radius, and can escape to the moon or wherever.

That doesn't refute what I wrote though. The determiner is whether the things are moving toward multiple destinations in the larger system. If they are, they'll not move in formation in any inertial frame. (Throw 2 rocks in opposite directions. Can they be moving in formation as measured in any inertial frame?) Things whose coordinate separation varies in the larger system don't necessarily move toward multiple destinations in that system. If I'll measure all the things to move in formation in my inertial frame, they need be all escaping to the moon or wherever, not just some of them.

"Throw 2 rocks in opposite directions. Can they be moving in formation in any inertial frame?" - Not if they are thrown in opposite directions locally, but if you throw two rocks in the same direction as measured in local inertial frame on an infinitesimal patch of spacetime, they can certainly end up traveling in different directions over a larger region of curved spacetime, even if they both travel on geodesics through the spacetime. Do you have any argument as to why this is impossible or contradictory, or is it just a matter of it seeming intuitively wrong?

By the way, there is a close mathematical analogy between what I describe above and the behavior of locally parallel spatial geodesics in curved (non-Euclidean) space--if you're not familiar with the idea that locally parallel geodesics can diverge or converge in non-Euclidean geometries like the surface of a sphere or a hyperbolic space, then I can write up an addition to my answer discussing this analogy.

No, no argument against that; I agree with it. But if the rocks end up traveling in opposite directions over a larger region of curved spacetime, then they aren't moving in formation as measured in any inertial frame, right? My cloud's particles are traveling in opposite directions over a larger region of curved spacetime from the outset, hence I'm still at a loss as to how I could let them move in formation like the EP implies I can.

No, my point is that they can be traveling in parallel when they are infinitesimally close together (which is the only time when their paths can both be measured in the same local inertial frame), but then diverge on large scales, in just the same way that two spatial geodesics in hyperbolic geometry can be parallel when they are infinitesimally close, but diverge on large scales.

I understood your point; it was well put. My point employed yours, didn't contradict yours and still stands, as I see it.

If you agree with my point, do you agree there is no contradiction between the idea that they are moving in formation in a local inertial frame on the event horizon, and the idea that they "diverge on large scales", so one stays inside the horizon (and eventually hits the singularity), while the other escapes to some arbitrarily large distance away from the black hole? BTW I realized that "infinitesimally close together" is not quite right, it should really be more like "some arbitrarily small but finite distance apart", that's the only way they can later diverge to a large finite distance.

On your BTW, I understood your shorthand fine. I disagree they can move in formation in a local inertial frame on the event horizon. I understand "diverge on large scales". But here the divergence occurs within my lab. If I throw upward 2 rocks, 1 at exactly escape speed and the other at the same speed but thrown from a tiny bit lower point, they'll eventually diverge so that they are moving toward multiple destinations (Earth, and infinity). At that moment they aren't moving in formation in any inertial frame. And likewise is the case in my lab falling through the horizon. Do you see that?

But you are assuming from the start that they must be moving at different speeds (and/or directions) in your lab in order for their paths to diverge. Why? "Escape velocity" is a Newtonian concept, not a GR one, so you should throw away all references to that concept in your argument. Do you disagree that on a curved space, two geodesics can appear parallel in a local cartesian coordinate system on a very small patch of space which differs negligibly from flat space, yet diverge on large scales? If not, why do you have a problem when we replace space w/ spacetime, cartesian w/ inertial?

GR demands that their paths noticeably diverge within my lab, no matter how small in spacetime my lab is; i.e. on an arbitrarily small scale. This doesn't require that I disagree with your last two sentences. Here's how I know their paths diverge: First I devise this law of physics: "two free test objects in an inertial lab don't move in formation as measured in that lab when they aren't moving in formation as measured by one of the objects." Then I let one of the particles below the horizon have a device to measure...

...whether it's moving relative to one of the escaping particles, and if so, semaphore. A particle below the horizon will definitely measure that it's moving relative to an escaping particle, because the former is falling below the horizon at relativistic speed as measured in its own frame (GR predicts) while the latter stays above the horizon. Then the former raises its flag, which is observed by one of my lab assistants moving right past that particle at that moment. Now I know that the particles' paths are diverging within my lab. The cloud is splitting apart, in violation of the EP.

"A particle below the horizon will definitely measure that it's moving relative to an escaping particle, because the former is falling below the horizon at relativistic speed as measured in its own frame (GR predicts) while the latter stays above the horizon." Where do you get the idea that GR predicts that? Assuming the object below the horizon is using a local inertial frame, it predicts the horizon itself is moving upward at the speed of light, but the objects above and below the horizon may be at rest relative to one another.

When the objects above and below the horizon are at rest relative to one another, it follows that all the objects are moving inward in formation, toward the singularity. But that's not the case here. The escaping particles in my lab move ever outward, away from the black hole. If you were right that my cloud needn't be splitting apart, an escaping particle could receive a signal sent from a particle below the horizon. But GR predicts such signal won't be received, which makes sense because the escaping particle remains above the horizon. What am I missing?

"it follows that all the objects are moving inward in formation, toward the singularity." But "moving inward" is coordinate-dependent. In a local inertial frame straddling the event horizon, the simple fact that the EH is moving outward at the speed of light means there is no possible direction/velocity a slower-than-light object outside the EH could have that would avoid it getting closer to the EH with time on the small spacetime patch where the local inertial frame is defined. (do you agree?) But in some other system like Schwarzschild coords, its distance can be increasing on this patch.

Okay, I think I understand all that. But I still see an issue when an object is escaping. Put yourself on a particle below the horizon. All as measured in your frame: You say "small spacetime patch" but it need be only small enough, which could in principle (and for sake of example here) be a light day across, lasting for a day. Say you're initially 1m away from an escaping particle. A day later you're a light day away from it (it's still above the horizon, which is now a light day away). How do you reconcile its apparent movement with remaining at rest with respect to it?

I think the answer here is that if you have a sufficiently huge black hole that a region of spacetime 1 light-day and 1 day across differs negligibly from being flat, then if you have a particle whose distance above the horizon at the beginning of the interval is less than 1 meter in its own local inertial rest frame, then it's on a geodesic that will inevitably fall below the horizon unless it accelerates. On the other hand, if you used a local inertial frame where it was traveling at a very large fraction of c, it could start less than 1 meter and avoid falling in during the 1 day period.

Agreed. But there's a problem with that: I let the particles above the horizon be escaping. I let all the particles be moving in formation, so in their own local inertial rest frames (LIRFs) the other particles are at rest (at least until they exit my lab). Then in the LIRF of a particle below the horizon, an escaping particle is falling toward the horizon at c. When initially 1m above the horizon it'll cross the horizon within 5 nanoseconds. The particle is definitely escaping, so this frame cannot last as long as 5ns or else GR is broken. What prevents the frame from lasting that long?

Note that this issue really doesn't depend on the other particles, you can just consider a single particle above the horizon and ask the same question. My guess is that if the black hole is large enough that a spacetime region lasting more than 5ns can be treated as flat, then the following two statements are incompatible, although either could be true on its own: 1) In the particle's local inertial rest frame, it is only 1m above the horizon initially; 2) the particle's initial position/velocity are such that if it follows a geodesic, it will avoid falling through the horizon.

Okay, let's consider a single escaping particle. You implied that, when all my cloud's particles are moving in formation in my lab, an escaping particle would be getting closer to the EH in its own LIRF. (The same LIRF all the particles call their own as long as they remain in my lab.) But then it's not escaping, right? If it follows a geodesic it'll fall through the horizon. I declared it to be freely falling, so it's following a geodesic. How do you resolve this contradiction? Its change in distance from the EH as measured in some other system like Schwarzschild coords is ignorable here.

I re-read your posts above to see whether you'd already answered my last question. It seems I'm challenging this: "...even though any slower-than-light particle outside the event horizon must be approaching the horizon in a local inertial frame at the horizon, that same particle ... can [in principle] escape to the moon or wherever." Those are incompatible ideas as I see it. According to your first clause a free test particle above the horizon is approaching the horizon in its own LIF at the EH, in which case it's following a geodesic that crosses the horizon (i.e. it cannot be escaping).

I think you're not considering the difference between the local view and the global view. What about my earlier point about spatial geodesics in negatively curved (hyperbolic) space, and how they can be locally parallel yet diverge on large scales? Presumably depending on the curvature it would even be possible to have two geodesics that were approaching each other somewhat on a small (very nearly flat) patch, but which still diverged on large scales. Do you think local parallel/approaching paths and global divergence would be "incompatible ideas" in the case of geodesics in curved space?

The issue in your statements is all in a LIF, so why make things more complicated by considering a global view? You imply above that you could be a freely falling astronaut approaching the moon, 1m above it initially as you measure, yet in principle escape from the moon while staying in free fall. I want to know how you think that's possible. Did I miss something?

Well, the question of whether it eventually falls into the horizon (assuming it doesn't do so in the LIF) requires the global view, no? I'm not sure the moon example is comparable because the moon's surface isn't moving at the speed of light in a LIF, whereas the event horizon of a black hole must be because it's a null surface; and a particle on the moon's surface wouldn't be following a geodesic, whereas a photon that remained on the horizon would be.

I'm back. I disagree that the question requires a global view. The particle above the EH, within its own LIF straddling the EH, is approaching the horizon, as you agreed above. Then its geodesic crosses the EH within that LIF; it isn't possible for it to escape. The same is true for a particle that is 1m above & approaching the moon. Isn't it that simple? When you're in free fall and directly approaching a planet or EH, you will definitely reach it; you are definitely not escaping and it doesn't matter how fast the planet or EH is approaching you.

"The same is true for a particle that is 1m above & approaching the moon. Isn't it that simple?" - No, because there are some geodesic trajectories such that, in a local inertial frame which contained both the particle following the geodesic and the surface of the moon, the particle would be moving away from the surface. Whereas if you replace the moon's surface with an event horizon, there are no possible geodesics near the horizon such that in a local inertial frame containing the horizon, the particle is moving away from the horizon, at least not in the first-order approximation.

(continued) And that's because the event horizon is a null surface, so it should move at the speed of light in all local inertial frames--if a 2D wall is rushing towards you at the speed of light, and you can only be moving at the speed of light or slower in any inertial frame, how can you possibly be moving away from it? (in the sense of the distance between you and it growing rather than shrinking with time)

I made it a given that a particle is approaching the moon. Then it's not moving away from the surface, and then the cases are analogous. I use the moon to make the issue easier to visualize. On "if a 2D wall is rushing towards you at the speed of light", I agree, you can't be moving away from the EH as measured in your own LIF. In that LIF you're approaching the EH, in which case you'll definitely reach it and hence can't be escaping. Just like you'll definitely reach the surface of the moon when you're in free fall and directly approaching the moon. Then how are my particles escaping?

Let me summarize here. Your answer to my original question, as I see it, is that all my particles can be moving in formation as I measure in my lab, even though the particles above the EH are escaping and those below the horizon are moving toward the singularity. I need to consider the global view, you say. But you're also implying that none of my particles can possibly be escaping. To understand your answer I need resolution on how my particles above the EH can be escaping.

I'm saying none of the particles outside the horizon can be increasing their distance from their horizon in a local inertial frame that covers the section of their worldline very close to the horizon (do you disagree that this follows from the fact that the event horizon moves at the speed of light in local inertial frames?), but they can nevertheless be escaping in global terms (which is why I stressed the need to distinguish the local and global views, and the fact that something approaching the horizon in inertial coords can be moving away from it in another system like Schwarzschild's).

I agree that "this follows from the fact..." I disagree that they can nevertheless be escaping in global terms. I say if you are freely falling and approaching a body (if only as you measure), you aren't escaping, whether locally or globally. If you were right, a skydiver wouldn't need to pull their ripcord to avoid dying. If you were right, there would be some way a skydiver could stay in free fall and survive. What is that way?

On "the fact that something approaching the horizon in inertial coords can be moving away from it in another system like Schwarzschild's", while this is possible in principle, it's impossible when that something measures itself approaching the horizon or surface. In that case the something is always moving toward the horizon or surface in Schwarzschild coords, e.g. what the skydiver observes.

There is no privileged coordinate system in general relativity that corresponds to an object "measuring itself", you can use any coordinate system you like to measure yourself, including Schwarzschild coordinates. I guess you just mean "measured in your own local inertial rest frame"? Even so, I don't see why you think approaching a horizon in your local inertial frame is incompatible with moving away from it or maintaining a constant distance in some other frame.

To pick a simpler flat spacetime example, an accelerating Rindler observer always sees the Rindler horizon approaching them at c in their instantaneous comoving inertial frame, and thus at any given instant the horizon must be approaching them in their comoving inertial frame, but they never actually cross the horizon, and in Rindler coordinates their coordinate distance from the horizon is constant (which also implies that an object accelerating a bit more than the Rindler observer at the same position could be moving away from the horizon in Rindler coordinates, but not in inertial coords).

Yes, measured in your own LIF is what I mean, and also should've noted the observer is in free fall, like my particles, in which case your example with an accelerating Rindler observer doesn't apply. Can you get the free-falling skydiver to be moving away from the Earth in a Schwarzschild coord system? My mind would be blown if you could do that. The floors of a building may serve as Schwarzschild coords, if you wish. You could have a base jumper falling beside the building, measuring herself approaching the Earth in her own LIF. Can she move to a higher-numbered floor? I say no.

Obviously you can be moving away from the surface in Schwarzschild coordinates while in freefall--just think of someone shot out of a cannon, for example--so the question is just whether you can be doing this while at the same time the surface of the planet is actually approaching you in your local inertial frame. I don't know whether it would be possible on Earth, but I would guess it should be at least be possible for an object whose surface was sufficiently close to the Schwarzschild radius, like a neutron star.

(continued) Do you have any logical argument as to why this should be impossible, or is it just an intuition of yours? For exmaple, I wonder if you can provide any detailed justification for the rule that while it's possible for an accelerating object to move away from a horizon in one coordinate system while approaching it in their inertial rest frame, this should be impossible for a freefalling object.

A logical argument can appeal to physical experiments, so I do have one, which is simply that the base jumper's case is perfectly analogous. There's no good reason for making relevant a difference in the 2 cases (e.g. horizon vs. ground). Therefore I can confidently make a law of physics: "A free-falling observer who approaches a body as measured in her own LIF also approaches the body as measured in its Schwarzschild coords."

The reason it's possible in principle for you as an accelerating observer is because the Schwarzschild system may possibly be length-contracting in your own frame as you reach higher Schwarzchild coords (e.g. higher numbered floors of a building). When you are in free fall and reaching higher Schwarzchild coords, the Schwarzschild system is always length-expanding in your own frame.

But you haven't given a logical argument about any physical experiment. Why are you sure it's not possible for a freefalling guy to be moving towards the surface of a massive body at an instant in their local inertial frame while moving away in Schwarzschild coordinates at the same instant? I don't understand your argument about length expansion vs. length contraction, since two observers who share the same instantaneous velocity at an instant will both measure the same thing in their local inertial frame at that instant, even if one is moving on a geodesic and one has proper acceleration.

By the way, it may be that both of us are mistaken in imagining it's even meaningful to talk about local inertial frames that cover small but finite patches of spacetime--looking at discussions of the definitions like the ones here, it sounds as though it may only be possible to define coordinate systems that are "normal" at a single point in spacetime, and then presumably light (or an event horizon) would have an instantaneous velocity of c at that point in such a coordinate system. I had been assuming

(cont.) that one could define some kind of "approximately inertial" coordinate system on a very small region and some kind of approximate version of the equivalence principle would hold in these coordinates, but I'm not sure any such idea is defined in the literature (although the comment about a "small coordinate patch" here seems to suggest something like this).

In physics it isn't necessary to show why some behavior isn't possible when that behavior hasn't been observed in nature. I'm free to make a law of physics that describes observed behavior, and the law stands until another observation contradicts it or it's shown to be illogical. All of our laws of physics are like that. The physical experiment I referenced in my logical argument is the base jumper's jump.

Yes, two observers who share the same instantaneous velocity at an instant will both measure the same thing in their LIF at that instant. But I'm talking about a change over many instants. Using a rocket you can accelerate relative to a building beside you as you move to a higher floor. The building then length-contracts in your frame, as SR predicts. You can possibly accelerate such that the ground approaches you via that length-contraction. A free-falling observer who moves to a higher floor is decelerating relative to the building. In that observer's LIF the building length-expands.

It must be meaningful to talk about LIFs that cover small but finite patches of spacetime. In science, theories are required to be falsifiable. GR's EP postulates that the laws of physics are the same in all such LIFs. We must be able to test that, which we can do by letting the tidal force in the LIF be negligible, which I did. It doesn't matter whether or not the EH moves at c in the LIF of one of my particles above the EH. Given my law of physics above, if the particle is approaching the EH as measured in its own LIF, at any speed, it can't be escaping.

"In physics it isn't necessary to show why some behavior isn't possible when that behavior hasn't been observed in nature." But coordinate systems are not part of "nature", they are human creations, part of mathematical physics. "I'm free to make a law of physics that describes observed behavior, and the law stands until another observation contradicts it or it's shown to be illogical." Not if it contradicts another well-established theory and fails to reproduce all the successful predictions of that theory. Besides, are you actually trying to propose a "new law"?

(continued) I thought you were just asking about what the theoretical answer would be in general relativity, which is where the equivalence principle comes from. "The physical experiment I referenced in my logical argument is the base jumper's jump." But a base jumper is on a geodesic that objectively hits the surface, I thought your question about the black hole was about what would happen with an object escaping from a region just outside the event horizon, which as I said would be more analogous to a person shot out of a cannon.

(continued) Do you have any logical argument to show it's impossible that a person shot from a cannon from just above the surface of a massive body (whether the Earth or a neutron star), on a geodesic trajectory that escapes to infinity, can't temporarily be approaching the surface in their local inertial rest frame at the moment after they have been fired?

"But I'm talking about a change over many instants." Well, the equivalence principle doesn't deal with that, it only deals with first-order effects like velocity, not higher-order ones like acceleration. See my answer here on the subtleties in the actual technical definition of the equivalence principle--somehow I thought I had linked to it earlier in our discussion, but looking back it looks like I didn't.

"It must be meaningful to talk about LIFs that cover small but finite patches of spacetime. In science, theories are required to be falsifiable." I don't think the equivalence principle would really be understood as a "theory" of its own by physicists, more like a mathematical feature of general relativity, for example this book I linked to earlier defines it in terms of having a coordinate system where a Taylor expansion of the GR metric about a chosen origin takes the form of eq. 5.1.16.

On "Not if it contradicts...", agreed, as long as that theory isn't illogical. I brought up the base jumper analogy only because you took the position that a particle above the EH approaches it in its own LIF. If we switch to the incompatible viewpoint of "shot out of a cannon" then the EH is receding as the particle measures in its own LIF. The cannon can be the surface for sake of a logical argument. Your initial distance to the cannon / surface is 0. The distance you'd measure to the cannon once you left its barrel would be > 0. Hence you're receding from the surface as you measure.

SR (hence the EP) handles constant acceleration; e.g. see the relativistic rocket equations. But that's probably a moot point now that we've switched to the cannon / receding viewpoint. Any idea in science, both theories and their postulates like the EP, are required to be falsifiable. Can you show that all of my cloud's particles can be moving in formation in my lab, even as the escaping particles are receding from the EH as measured in their own LIFs?

On your book link, search exactly for "But the inertial frame of general relativity is only valid locally". It's perfectly valid locally, in a frame larger than a point. The key point is that the tidal force in your lab must be negligible, such that the results of the experiments are unaltered by the tidal force after rounding for significant digits. In that case you'll get (or should get) the exact same results predicted by SR.