[fargolime]

What sentence in that picture's caption is wrong, or what doesn't follow from its premises? Asking for extraordinary evidence isn't a scientific thing. Valid logic is all the evidence needed.

Yes it's my argument in that way. I'll challenge a refutation or accept it, but that's difficult when the counter argument is a vague wall of text. If I were to try to summarize what you think is wrong I couldn't do it. I can't decipher your points to see how they attempt to refute any particular sentence in the blog. Please be way more clear and short and to the point, and I'll agree, or disagree with my reasoning. (For example check out Millstone's reasoning above. He/she disagrees that an inertial frame can cross the horizon. One short sentence is enough to summarize!)

> the blog post's claims about what the theory of relativity actually says are not correct

Anyone can say that; be more specific. What the blog says in that way sources from experts in relativity. Its statement of the equivalence principle, for example, is Kip Thorne's.

> Agree with what?

They agree that frame X in the blog post is validly defined and validly used in the thought experiment. Their quote in the blog makes that clear.

[pdonis]

I can't decipher your points to see how they refutes any particular sentence in the blog.

That's because the blog post itself is not very well written, and doesn't state correctly what relativity actually says.

Please be way more clear and short and to the point

I'll give it another shot below.

They agree that frame X in the blog post is validly defined and validly used in the thought experiment. Their quote in the blog makes that clear.

It does no such thing. Taylor and Wheeler do not say that the blog post's "Law J" and "Law K" are correct statements of what relativity actually says. They're not; they're not even stated precisely enough to have a definite meaning that can be compared with what relativity says.

Rather than try to go through the blog post in detail, let me try stating two simpler puzzles that bring out what seem to me to be the essential points. Here's the setting for both puzzles: an astronaut is falling through the event horizon of a black hole. Just before reaching the horizon, he launches a probe outward at escape velocity, which will be just a smidgen less than the speed of light.

Now consider the astronaut's local inertial frame (LIF) as he crosses the horizon: i.e., the origin (t = 0, x = 0) of this frame is the event at which the astronaut crosses the horizon, and the time axis of the frame is the astronaut's worldline as he falls in. According to relativity, the following are all true statements:

(1) The horizon is an outward-moving lightlike curve that passes through the frame's origin; i.e., it is the line t = x in that frame.

(2) The probe's worldline starts at t = some value just a little bit less than zero, x = 0, and moves in the positive x direction at just a bit less than the speed of light.

(3) Therefore, in this LIF, the horizon is moving outward faster than the probe.

The first puzzle is simple: if all of the above are true, how can the probe ever escape? Won't the horizon catch it? (Meaning, won't it end up below the horizon, not escaping to infinity?)

The resolution of this puzzle is equally simple: 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. Therefore, within the LIF, the probe remains outside the horizon. And once we're outside the LIF, tidal gravity is not negligible, and it will "pull" the horizon down with respect to the probe, so the horizon will never catch the probe.

The calculations underlying what I just said about the size of the LIF vs. the distance required for the horizon to catch the probe are on PhysicsForums here:

http://www.physicsforums.com/showpost.php?p=4285946&postcoun...

The second puzzle is a bit more subtle. Consider the following statements, which are also true according to relativity:

(4) With respect to a global coordinate chart describing the black hole spacetime, the horizon is at a constant radial coordinate r.

(5) With respect to the same global coordinate chart, the probe's radial coordinate r is increasing.

(6) However, the distance from the probe to the horizon, in the astronaut's LIF, is decreasing, at least while the probe and horizon remain within the LIF. (This is obvious from statement #3 given earlier.)

The second puzzle then is, how can #5 and #6 be reconciled? How can the probe have increasing r when it's getting closer to the horizon, which is at constant r?

The best way I know of to see the resolution this second puzzle is to actually draw a spacetime diagram of the LIF, with curves of constant r correctly drawn. If you do that, you will see that the curves of constant r are drawn in such a way that the probe's worldline has increasing r even though its distance from the horizon, in terms of the LIF's x coordinate, decreases. I can't draw such a diagram here, but the key fact that makes this true can be seen by considering statement #4 above, combined with statement #1: the horizon is a curve of constant r, but in the LIF, it is a 45 degree line moving up and to the right! (It's the line t = x in the LIF, per #1.) A curve of constant r lying just outside the horizon will slope up and to the right, a bit more vertical than 45 degrees (which makes it timelike); whereas a curve of constant r just inside the horizon will slope up and to the right a bit more horizontal than 45 degrees (which makes it spacelike). The slope of the lines of constant r that pass through the probe's worldline, which also slopes up and to the right at almost 45 degrees (since the probe is moving outward at almost the speed of light) turn out to be sloped a little closer to 45 degrees than the probe's worldline is; so the probe ends up crossing curves with gradually increasing values of r.

[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.