[fargolime]

I do appreciate your effort. Unfortunately I only have like half an hour a day to devote to this kind of stuff. So I have to keep this short.

> the size of the LIF is much, much smaller than the predicted distance, extrapolated from within the LIF, that it will take for the horizon to catch the probe.

I'm discussing the simple picture in the blog and its caption. Not a more complex puzzle, or Laws J & K. The cloud of particles is demanded by GR to be splitting in two along the horizon, when all the particles above the horizon are let to be escaping. This "splitting in two" contradicts the equivalence principle and occurs in every arbitrarily short period of time in the life of frame X. There is always an arbitrarily short duration of time available in any LIF, so frame X's size in spacetime is not an issue.

When you talk about "global coordinate chart" and a "lightlike curve" you're being unnecessarily complex, I say. We're talking a simple inertial frame of special relativity here, plus one basic prediction of GR as it relates to the horizon, namely that everything below the horizon moves inexorably inward toward the singularity. This discussion can be much simpler than you're making it. After reading your explanation I shouldn't be left wondering which sentence in the blog is incorrect or doesn't follow from its premises.

The cloud splits in two and that's a "bug". To prove the blog wrong you need to show that the cloud doesn't split in two, that all the cloud's particles can in fact move outward in formation. Can you do that in a way that clearly shows which sentence in the picture's caption is incorrect?

[pdonis]

Since you're keeping it short, I will as well. I don't think it's worth discussing the blog post because it's too vague and it makes too many misstatements about what relativity says. That's why I tried to take out what I saw as the essential points and put them into puzzles that were properly stated. If you can't or won't discuss the puzzles as I presented them, I won't be able to give much of a response. All I can do is point out some particular items that seem to me to capture the mistakes you and the blog post author are making.

The cloud of particles is demanded by GR to be splitting in two along the horizon

This is wrong; GR "demands" no such thing. The blog post's author has misunderstood GR in this respect. See below.

When you talk about "global coordinate chart" and a "lightlike curve" you're being unnecessarily complex, I say

And I say that you are making a big mistake here, because a proper understanding of those concepts and how they relate to the LIF centered on the horizon is crucial to properly understanding and stating what GR says about this scenario. You'll see several examples of this below.

one basic prediction of GR as it relates to the horizon, namely that everything below the horizon moves inexorably inward toward the singularity

That means everything below the horizon must decrease its r coordinate in the global coordinate chart. It does not mean that everything below the horizon must decrease its x coordinate in the LIF. Remember: the horizon itself is the line t = x in the LIF; i.e., the horizon is moving in the positive x direction in the LIF at the speed of light, and it passes through the origin of the LIF at t = 0, x = 0. So a "cloud" particle that is at some negative x value at t = 0 in the LIF will be inside the horizon, and a "cloud" particle that is at some positive x value at t = 0 in the LIF will be outside the horizon; and it is perfectly possible for the first particle (inside the horizon) to be moving in the positive x direction in the LIF faster than the second particle. And this can be true even if the second particle (the one outside the horizon) is moving at "escape velocity". And it can also be true even though the first particle's r coordinate is decreasing and the second particle's r coordinate is increasing. My previous post gave more details on why those statements are true.

Oh, and one other thing: what does "toward the singularity" mean in the LIF? Which direction is the singularity in? The answer is, it's in the positive t direction--i.e., it's in the future. So any object in the LIF is moving "toward the singularity" as far as the LIF is concerned, since all objects move towards the future. There is no way to tell, from within the LIF, which objects are going to ultimately hit the singularity and which ones are not. That's a global concept.

To prove the blog wrong you need to show that the cloud doesn't split in two, that all the cloud's particles can in fact move outward in formation.

My previous posts (plus all the ones in the earlier HN thread I linked to, plus what I posted on PF) have already done that, multiple times. However, I've given a quick recap above.

This discussion can be much simpler than you're making it.

If only that were true.

[fargolime]

> That means everything below the horizon must decrease its r coordinate in the global coordinate chart.

Agreed.

> It does not mean that everything below the horizon must decrease its x coordinate in the LIF.

Agreed.

> ...it is perfectly possible for the first particle (inside the horizon) to be moving in the positive x direction in the LIF faster than the second particle.

Agreed.

> And this can be true even if the second particle (the one outside the horizon) is moving at "escape velocity".

Agreed.

> And it can also be true even though the first particle's r coordinate is decreasing and the second particle's r coordinate is increasing.

Disagree. There is no way you could show this for an inertial frame falling in the Earth's atmosphere, like for a skydiver (ignoring air friction). The EP demands that the laws of physics in the skydiver's frame and frame X are the same, so what you say here should be the same for the skydiver. Of course in the skydiver's frame you wouldn't use terms like "global coordinate chart" and a "lightlike curve", as that would be unnecessarily complex.

In the skydiver's LIF under the conditions above, the first particle (the lower particle) would always be moving in the positive x direction in the LIF slower than the second (upper) particle, regardless of the skydiver's speed relative to the Earth, and regardless of the second particle's speed relative to the Earth (i.e. it doesn't need to be escaping).

[pdonis]

> Disagree.

Then you are disagreeing with the theory of relativity, because what I've said is what the theory of relativity says. See below.

> The EP demands that the laws of physics in the skydiver's frame and frame X are the same

True.

> so what you say here should be the same for the skydiver

False, because there is no law of physics that says the r coordinate has to behave the same in every LIF. The r coordinate is a global coordinate, not a coordinate in the LIF; so as soon as you talk about the r coordinate, you are not just talking about the LIF, you are talking about the relationship between the LIF and a global coordinate chart. And there is no law of physics that says that relationship must be the same for every LIF. In fact that relationship is very different for the skydiver LIF as compared to the LIF that is falling through the horizon of a black hole. So any reasoning you do based on the assumption that that relationship is the same for both is simply wrong.

> Of course in the skydiver's frame you wouldn't use terms like "global coordinate chart"

As soon as you talk about the r coordinate, you are using a global coordinate chart, whether you realize it or not. So by not using such terms, you are failing to understand a key aspect of the scenario.

> as that would be unnecessarily complex.

It's (somewhat) complex, yes, but it's not "unnecessarily" complex. As I've said several times, understanding the proper relationship between the LIF and the global r coordinate is crucial if you want to correctly state what relativity says about this scenario. You and the blog post author have given excellent demonstrations of the mistakes you make if you don't have that understanding.

[pdonis]

> In the skydiver's LIF under the conditions above, the first particle (the lower particle) would always be moving in the positive x direction in the LIF slower than the second (upper) particle, regardless of the skydiver's speed relative to the Earth, and regardless of the second particle's speed relative to the Earth (i.e. it doesn't need to be escaping).

I'm responding to this separately because it was easier than trying to cram this plus my other responses into one post. If you are going to talk about what is or is not the same in the skydiver LIF and the LIF falling through the black hole's horizon, you have to first make sure the initial conditions are set up the same. Here's how you would do that:

(1) The LIF is in free fall, i.e., the astronaut/skydiver who is at rest in the LIF is freely falling in the gravitational field of some central body.

(2) At time t = 0 in the LIF, the astronaut/skydiver meets an outgoing light ray. (In the LIF falling through the black hole's horizon, this outgoing light ray is the horizon; in the skydiver LIF, it's just whatever outgoing light ray happens to be passing him at t = 0. Within the LIFs, there is no way to distinguish the two.)

(3) At some time t = minus epsilon in the LIF, the astronaut/skydiver releases a probe that flies outward at nearly the speed of light. (This is a key point that I don't think you understand: the initial condition in the LIF is that the relative velocity of the probe and the astronaut/skydiver must be the same. It is not that the probe's initial velocity is escape velocity. "Escape velocity" is a global concept, not a local concept; it has no meaning within the LIF. It so happens that, in the LIF falling through the black hole horizon, the first probe gets launched at a velocity that, globally, is just sufficient for it to escape to infinity, whereas in the skydiver LIF, the probe's initial velocity is way, way more than needed for it to escape; but there's no way to tell that from within the LIF.)

(4) At some time t = plus epsilon in the LIF, the astronaut/skydiver releases a second probe that flies outward at a speed even closer to the speed of light than the first probe.

These conditions are perfectly possible to set up in both LIF's (the skydiver LIF and the LIF falling through the black hole's horizon), and within the LIF's, there is no way to tell which LIF you are in; every observation within the LIF will be the same for both. The second probe will move closer to the first probe (while they are both within the LIF); but the second probe will be falling behind the light ray that passes the astronaut/skydiver at t = 0.

It's true that, once all these objects exit the LIF, things will be very different in the two cases. In the skydiver case, the outgoing light ray will catch up with and pass the first probe. In the black hole horizon case, it won't. But there's no way to tell that from within the LIF.

It's also true that the r coordinates of these objects behave very differently, even within the LIF. In the skydiver case, all three of the objects that are moving outward (the first probe, the light ray, and the second probe) are increasing their r coordinates (and rather rapidly at that). In the black hole case, the first probe has (very slowly) increasing r, the light ray (the horizon) has constant r, and the second probe has (very slowly) decreasing r. But as I said in the other post I made in response to your latest, the r coordinate is a global coordinate, not a coordinate in the LIF; and there is no law of physics that says the relationship between local coordinates within an LIF and global coordinates must be the same for every LIF. In fact, it obviously can't be, because the whole point of the equivalence principle is that LIFs that look the same locally can occur in parts of spacetime that look very different on a global scale.

[fargolime]

Thanks for that explanation. Most of it I agree with.

> There is no way to tell which LIF you are in

There is a way.

To keep it simple, let's assume the probes are test particles. In the skydiver's frame the second probe will overtake the first probe, given a sufficiently small epsilon. (We can always make that epsilon small enough that it's within the duration of the LIF.) In the astronaut's frame, for the same epsilon, the second probe won't overtake the first probe. The same experiment, different results, violating the equivalence principle.

[pdonis]

> (We can always make that epsilon small enough that it's within the duration of the LIF.)

NO, YOU CANNOT.

Sorry to shout, but not only have I already said this is false (several times if you include the previous thread I linked to), I have linked to a computation that proves it's false. So once again, you are basing your reasoning on a false assumption, and therefore you are naturally reaching false conclusions. (Note that my computation proves something stronger: that the distance required, extrapolated from the LIF, for the outgoing light ray the astronaut/skydiver passes at t = 0 in the LIF to catch the first probe, is much larger than the size of the LIF. If this is true, it must also be true that the second probe can't catch the first probe within the LIF.)

To briefly expand on what the computation I linked to shows: the smaller you make epsilon, the smaller the difference in velocities between the two probes can be (because the first probe has to be launched at escape velocity, and the smaller you make epsilon, the closer escape velocity gets to the velocity of light). And the smaller the velocity difference, the larger the catch-up distance, in the same proportion. So decreasing epsilon increases the catch-up distance extrapolated from the LIF such that the extrapolated catch-up distance remains much larger than the size of the LIF. (Again, my computation proves something stronger: that the catch-up distance required for the outgoing light ray, extrapolated from the LIF, increases as epsilon decreases, such that the catch-up distance remains much larger than the size of the LIF.)

To expand on the expansion just a bit more: remember that, in order for there to be any potential issue to discuss at all, two things must be true: (1) the initial conditions must match in both LIFs; (2) the global prediction of whether or not the second probe catches the first must be different for the two scenarios. Requirement #2 is what forces us to change the initial velocity of the first probe when we change epsilon; requirement #1 is what forces us to change the initial velocity of the first probe in both LIFs when we change epsilon.

(Actually, to expand one more bit, there is a third condition: the LIF size over which we can do the comparison at all must be the smaller of the two LIF sizes. Otherwise there would be no point to the comparison, since we could always just call globally flat spacetime an "LIF" and find some difference between it and an LIF in any curved spacetime by looking at effects that happen outside the range of the curved spacetime LIF.)

[pdonis]

Quick clarification:

the smaller you make epsilon, the smaller the difference in velocities between the two probes can be

s/can be/has to be/

[fargolime]

Again I agree with most of this.

> So decreasing epsilon increases the catch-up distance extrapolated from the LIF such that the extrapolated catch-up distance remains much larger than the size of the LIF.

A fatal problem though: you arbitrarily chose the size of your LIF to make it too small.

The size is determined by the accuracy the experimenter wishes to achieve, e.g. 6 significant digits. When I compare the laws of physics between two LIFs I have even greater freedom to choose the LIF size. I choose both LIFs to be the same size and as large as needed for overtake to happen in the skydiver's frame (e.g. the width of a particle and extending a megaparsec outward into empty space). I choose a black hole massive enough that the tidal force in the astronaut's frame is less than in the skydiver's frame.

Now, when overtake happens in the skydiver's frame but not the astronaut's, I know the difference isn't due to the tidal force, because tidal force has no ability to cause one particle to overtake another (it only stretches and squeezes objects or systems), and the tidal force in the astronaut's frame can't be the reason overtake didn't happen there, because the tidal force is less there (the system of particles was stretched less there, than in the skydiver's frame). Having ruled out the tidal force completely, my result definitely shows a violation of the EP.

> There is no way to tell which LIF you are in

There are ways that allow the LIF to be arbitrarily small. For example, let the probes be tiny rods. Fire them vertically-oriented, the first probe completely above the horizon, the second initially stradding the horizon and partially alongside the first probe. In the skydiver's frame the second probe will be overtaking the first probe, moving outward faster. In the astronaut's frame the second probe, unlike the first probe, cannot reach a higher r-coordinate (lest it be moving outward through the horizon, which GR disallows), hence it won't be overtaking the first probe.

Also I still agree with the picture in the blog. The cloud splits apart at the horizon, in violation of the EP. Go back to your original probes experiment, as test particles. You say the probes can approach each other in the astronaut's frame. I say they can't. When the first probe's r-coordinate is increasing and the second probe's is decreasing, they are receding from each other as measured in any LIF that wholly contains them. You obviously wouldn't be able to devise an experiment to show otherwise in the skydiver's frame, and there's no relevant difference for the astronaut's frame.

We needn't refer to r-coordinates at all. Here is a simple law of physics (I just devised) for an LIF that you seemingly disagree with: Two free test particles each moving toward destinations beyond opposite sides of the LIF recede from each other.