Your "rock-solid logic" makes at least one incorrect assumption about local inertial frames in curved spacetime.
However, if the diagram is just supposed to be a diagram of "space" at the horizon (not spacetime), then it is not correct because the direction to the left of the diagram, towards the singularity, is not a spacelike direction; that direction is supposed to be the inward radial direction, but that direction is a timelike direction, not a spacelike direction.
In short, the diagram of Frame X can't be correct, so no valid logical argument can be based on it.
[Edit: The argument given in words has the same problems, but it's easier to illustrate them in the diagram.]
> that direction is supposed to be the inward radial direction, but that direction is a timelike direction, not a spacelike direction.
It's an inward radial direction according to the diagram. The "but" part of your sentence doesn't offer evidence to invalidate that or otherwise show a problem. There's a coffee mug in front of me now. Regardless whether it's a spacelike or timelike direction, the mug is in that direction. Likewise to someone in that frame X the singularity is definitely in a particular direction. I'm not seeing a problem with the diagram.
No, it wouldn't. That's the point. The horizon is an outgoing lightlike surface; with respect to Frame X (or to any inertial frame that straddles the horizon), it's moving radially outward at the speed of light. It's not a plane stationary in space.
The singularity would be in one direction perpendicular to the horizon
No, it wouldn't. The singularity is to the future of the horizon; the direction from the horizon to the singularity is timelike, not spacelike.
There's a coffee mug in front of me now. Regardless whether it's a spacelike or timelike direction, the mug is in that direction.
This is not correct; timelike directions are fundamentally different from spacelike directions. What direction is next Tuesday from you now? That's a timelike direction, and it's not the same direction as the direction of a coffee mug, or any object that is in the space around you now. It's not even the same kind of direction. But that kind of direction--the kind that next Tuesday is in from you now--is the direction the singularity is in from the horizon.
I'm not seeing a problem with the diagram.
That's because you evidently don't understand what the curved spacetime around a black hole is like. It's a perfectly self-consistent model; but it doesn't work the way the web page author, and evidently you, think it works.
"Lightlike surface" and plane are not incompatible ideas. Surface = plane in this case. The diagram or its caption doesn't say the horizon is stationary, it says the frame is falling through the horizon. Presumably the horizon is indeed moving in that frame at the speed of light (an outgoing lightlike surface).
> What direction is next Tuesday from you now?
The singularity is not a date; it's an object, like my coffee mug is. It's not invalid to show a direction to an object just because you'll reach the object next Tuesday.
> That's because you evidently don't understand what the curved spacetime around a black hole is like.
I've take a couple years of classes on this stuff. You're using the terminology inappropriately. The terms spacelike and timelike are not limited to black holes. By saying the diagram incorrectly shows the direction to the singularity you're implying that neither can I point to my coffee mug and say it's in that direction. But I can point to my coffee mug as validly as someone in frame X could point to the singularity.
I can even point to a real purported singularity by walking outside and pointing to the direction of the center of the Milky Way. It's no more complex than that. Can I validly point to the center of the Milky Way? If yes, then there's no problem with the diagram in that regard.
> It's a perfectly self-consistent model
An assumption that should be set aside when considering an objection to it.
If that's true, the the diagram that's shown cannot portray a steady state of Frame X, as it claims to do. At most it can show how things look in Frame X at a single instant of time. The horizon plane is moving to the right in the diagram at the speed of light, and none of the cloud particles can move at the speed of light, so the horizon plane will quickly overtake them and be to the right of the entire cloud. The argument in the text is not consistent with this.
The singularity is not a date; it's an object, like my coffee mug is.
No, the singularity is not an object like your coffee mug. That's the point. The singularity is a moment in time, not a place in space. If you don't know this then you need to do some review; it's a basic fact about the black hole solution in General Relativity.
I've take a couple years of classes on this stuff.
What material have you covered? Have you specifically covered the Schwarzschild solution to the Einstein Field Equation, and the description of it in the Kruskal coordinate chart? (Or the Painleve chart, or the Eddington-Finkelstein chart, or any other chart which is not singular at the horizon?) If not, I suggest you wait until those things are covered, or read ahead in your course material.
You're using the terminology incorrectly. The terms spacelike and timelike are not limited to black holes.
I didn't say they were. You're correct, they are general terms describing different kinds of directions in spacetime. ("Lightlike" or "null" is the third kind of direction.) A black hole is a spacetime, therefore the terms apply in it, and I was using them correctly.
Can I validly point to the center of the Milky Way?
Sure, but you won't be pointing to the singularity in the center of the black hole there. Once again, that would be like pointing to next Tuesday--or, since the center of the galaxy is some thirty thousand light years from us, it would be like pointing to a moment thirty thousand years or more into your future.
An assumption that should be set aside when considering an objection to it.
Only if the purported criticism represents the model correctly. It doesn't.
The material sufficient for this discussion. I know the material well enough to point out contradictions in your thinking.
> If that's true, the the diagram that's shown cannot portray a steady state of Frame X, as it claims to do.
I don't see such claim made there. Presumably it's a snapshot of X at a single instant of time, the initial state of the thought experiment. The author should be clearer however.
> The horizon plane is moving to the right in the diagram at the speed of light, and none of the cloud particles can move at the speed of light, so the horizon plane will quickly overtake them and be to the right of the entire cloud.
GR allows particles above the horizon to be escaping to infinity (since the escape velocity there is less than c), in which case the horizon plane will never overtake them. GR allows the escaping particles to never be overtaken by the horizon even though GR also demands that those particles move slower than the horizon does. If all the particles above the horizon are escaping then the cloud is splitting along the horizon.
> Sure, but you won't be pointing to the singularity in the center of the black hole there.
That's a contradiction. You can't say it has no location in space and then refer to its location in space even in relative terms like "in the center of". The singularity has a location in both space and time (a location in spacetime, if we ignore the technicality that spacetime breaks down there), and I could validly point to it now, no matter its location in my future. I suspect you're getting your idea from a cosmic singularity, like for the big bang. In that case there would be no single location in space you could point to now; it has a specific location in time only.
I agree, if this is the author's intent it should be clearer. However, I'm willing to interpret the diagram that way for this discussion.
GR allows the escaping particles to never be overtaken by the horizon even though GR also demands that those particles move slower than the horizon does.
Because spacetime is curved, and the curvature limits the size of a local inertial frame straddling the horizon.
Consider a "cloud" particle in the right half of cloud in the Frame X diagram. In order to be moving at escape velocity at its location, it must be moving to the right at just a smidgen less than c. So it will move off the right edge of the diagram, and out of the region covered by Frame X, before the horizon catches it. Once it's out of Frame X, its worldline doesn't have to look like a straight line in Frame X, so it can continue to avoid the horizon; in fact it will gradually get farther and farther away from it.
Another way to see what's going on is to ask what a line of constant radius r looks like in Frame X. An observer at constant r close to the horizon has to accelerate very hard to stay at the same radius; that means a line of constant r in Frame X is going to look like a hyperbola in a spacetime diagram. If we imagine a bunch of observers at constant r passing through the cloud in the Frame X diagram as shown, they will all be accelerating very hard to the right, hard enough to keep the horizon from catching up with them; but a cloud particle moving at escape velocity just a little to the right of the horizon will be moving to the right faster than any of the constant-r observers, so he will pass them one by one as they accelerate; they won't be able to catch him. That means the escape velocity particle is increasing its radius with time, i.e., escaping.
If all the particles above the horizon are escaping then the cloud is splitting along the horizon.
Because the particles that are escaping have to be moving to the right at very, very close to the speed of light, whereas the particles that are not escaping are not. This is not a contradiction.
[Edit: I suppose the intent could be that all of the cloud particles are moving to the right at just a smidgen less than c; but if that's the case, the ones to the right of the horizon line will make it out of Frame X before the horizon catches them, as above, but the ones at or to the left of the horizon line obviously won't, because the horizon will move ahead of them.]
You can't say it has no location in space and then refer to its location in space even in relative terms like "in the center of".
I haven't done any such thing. You appear to be reading things into my posts that I am not saying. I have said, consistently, that the singularity is a moment in time. More precisely, the singularity is represented by a spacelike line. A "place in space" is represented by a timelike line, not a spacelike one. A point in space at an instant of time is just a point, not a line.
The singularity has a location in both space and time
No, it has a location in time only; it is all of space (or at least all of it inside the horizon) at that moment in time.
I suspect you're getting your idea from a cosmic singularity, like for the big bang.
The black hole singularity is indeed very similar to the big bang singularity; both of them are spacelike lines. The only real differences are that the black hole singularity is hidden behind a horizon, and that the black hole singularity is a future singularity while the big bang singularity is a past one.
You say you have taken classes on this material; have you raised the issues you are raising here in class? If so, what response did you get?
True but irrelevant. The thought experiment needs only an arbitrarily small time interval as measured in X to make its case, so the size of X doesn't matter. That said, X can be arbitrarily large in principle, even light years across, just make the black hole as massive as needed. When X is light years across the escaping particles needn't exit X for years, yet the thought experiment is concluded in as little as a nanosecond.
> Because the particles that are escaping have to be moving to the right at very, very close to the speed of light, whereas the particles that are not escaping are not. This is not a contradiction.
The particles below the horizon can have a higher velocity in the outward direction than the escaping particles do. The velocities of the particles below the horizon is not specified in the diagram, and there's no limit to how close any of the particles' velocities can be to c, as measured in X. When all the particles above the horizon are escaping and all the particles below the horizon are let to have a velocity closer to c in the outward direction than the escaping particles have, GR still demands that the cloud splits, unlike what would be predicted for an inertial frame with the same velocities for particles specified. In an inertial frame under those conditions the particles with higher velocities would be moving closer to the other particles, not getting further away from them. There is a contradiction.
> Another way to see what's going on is to ask what a line of constant radius r looks like in Frame X.
I understand your example here. Nevertheless the cloud splits under conditions where it wouldn't split in an inertial frame in special relativity, in contradiction with the principle of equivalence. Unless that can be disproved it seems the blog raises a valid issue with GR.
> I haven't done any such thing.
You said "you won't be pointing to the singularity in the center of the black hole there". When you say that singularities have a location in time only you can't then refer to their location in space without contradicting yourself. The "in the center of the black hole there" references a place in space. You'd have to instead be careful to say something like "you won't be pointing to the singularity that you say is in the center of the black hole there".
> You say you have taken classes on this material; have you raised the issues you are raising here in class?
I first saw this blog yesterday.
This is actually a better statement of the argument than the one given on the web page (better because it's more concise and doesn't get bogged down in irrelevancies); in fact it makes it clearer to me just what the argument is actually supposed to be. However, it's still not a valid argument against GR. Here's why: GR agrees that, with respect to Frame X, the particle above the horizon and the particle below the horizon get closer together! But only within the confines of Frame X, and only with respect to distance relative to Frame X.
Distance is frame-dependent; distance relative to Frame X is not the same thing as distance with respect to the radial coordinate r that is used in a global coordinate chart. The fact that the particles are getting closer with respect to distance relative to Frame X does not mean that they are getting closer together with respect to the coordinate r; in fact they are not. The particle inside the horizon has an r coordinate that decreases with time, while the one outside the horizon has an r coordinate that increases with time (here "time" is time with respect to Frame X).
Also, the fact that the particles are getting closer together with respect to Frame X does not mean that they will be getting closer together with respect to some other local inertial frame centered on an event further along the horizon. Local inertial frames centered on different events don't "line up" in curved spacetime the way they do in flat spacetime; that's one way of expressing what it means for a spacetime to be curved. So you can't assume that, just because the particles are getting closer together with respect to Frame X, they will continue to get closer together once they've left Frame X, much less that they will continue to get closer together until they meet. That deduction would be valid in flat spacetime, but it isn't in curved spacetime.
When you say that singularities have a location in time only you can't then refer to their location in space without contradicting yourself. The "in the center of the black hole there" references a place in space. You'd have to instead be careful to say something like "you won't be pointing to the singularity that you say is in the center of the black hole there".
Fine, consider my statement suitably modified (it should have been obvious from context that that's what I meant anyway, but if you insist on quibbling over language, so be it). As modified, the statement is true.
Instead of quibbling over language, you could have asked a relevant question: what are you pointing at when you point at the center of the galaxy? The problem with this question is that it assumes that there is a unique answer to it; that there is some unique "object" sitting "at the center of the galaxy". In an ordinary spacetime without a black hole present, that would be a valid assumption; but it isn't when there is a black hole present. In one sense, when you point at the center of the galaxy, you could be said to be pointing at the object or objects that originally collapsed to form the black hole there; in another sense, you could be said to be pointing at the hole's horizon. Once again, a spacetime with a black hole in it doesn't work like the ordinary flat spacetime you are used to, or even like an ordinary curved spacetime with a planet or star in the center instead of a black hole (so that there is no event horizon and no singularity).
I first saw this blog yesterday.
But haven't scenarios at least something like this come up in your classes?
> The fact that the particles are getting closer with respect to distance relative to Frame X does not mean that they are getting closer together with respect to the coordinate r; in fact they are not.
This is false. Frame X is equivalent to an inertial frame says the principle of equivalence. Nothing about an inertial frame prevents one of the higher-velocity particles from passing by one of the lower-velocity (escaping) particles, like cars passing each other on the highway, which could happen in as little as a nanosecond as measured in X. In that case the r-coordinate of the higher velocity particle would not only have gotten closer to the r-coordinate of the escaping particle, it would have now the higher r-coordinate of the two particles, have passed outward through the horizon and be escaping itself.
> The particle inside the horizon has an r coordinate that decreases with time, while the one outside the horizon has an r coordinate that increases with time (here "time" is time with respect to Frame X).
R-coordinates aren't limited to discussion of black holes of course. Your head has an r-coordinate, as do your feet. Assuming your feet are lower initially, if the r-coordinate of your feet is decreasing while the r-coordinate of your head is increasing, you are stretching and your feet are not getting closer to your head in any inertial frame. Conversely if the r-coordinate of your feet is increasing faster than the r-coordinate of your head increases, your feet are getting closer to your head in every inertial frame that contains you. Eventually your feet will pass by and be higher than your head; you will have flipped upside down.
I think we've exceeded the limits of HN discussion. If you want I'll email you; let me know.
Probably, but I do see a couple of things in your post worth a brief comment:
You seem to be under the impression that the r coordinate is the space coordinate of the local inertial frame. It's not; it's a coordinate in a global coordinate chart. Two objects can be approaching each other with respect to the space coordinate of the LIF but be separating with respect to the r coordinate.
You also seem to be under the impression that the predicted trajectories in the LIF, if it could be extended indefinitely (which it can't), would have the particle inside the horizon moving outside it. That's not correct; the predicted trajectories in the LIF would have the particle outside the horizon falling inside it, then falling past the one that started out inside the horizon. (This has to be true because the horizon is moving outward at the speed of light, so neither particle can outrun it from the standpoint of the LIF.) The resolution of the puzzle does not depend on explaining why the particle that starts inside the horizon doesn't escape out; it depends on explaining why the particle that starts outside the horizon doesn't fall in even though the LIF predicts that it should.
No need. What you're saying is still incorrect, but the correct answer is already in what I've posted here and what's posted in the PF thread. If that doesn't answer it for you, I don't think there's any point in further discussion.
Since it's Saturday and I have some time I'll respond to this as well, just for the record. This is wrong in two ways:
(1) I did say what was wrong with the thought experiment. You just don't want to accept that what I said is right. Of course that's your decision to make, but the fact that you don't agree with my criticism doesn't mean I didn't make one.
(2) Your view of how things work in these kinds of arguments is backasswards. Extraordinary claims demand extraordinary evidence. You are making an extraordinary claim: that the theory of General Relativity, which has been developed and refined and tested against evidence for almost a century now, and which has passed every test with flying colors, somehow has a basic error of logic in one of its fundamental principles. For a comprehensive study of how well tested GR is experimentally, see here:
http://relativity.livingreviews.org/Articles/lrr-2001-4/
The burden of proof is on you to back up this extraordinary claim. You don't do that by making or referring to a quick blog post that makes elementary errors and misstatements about what GR actually says. You do that the way Galileo and Einstein did it: by having extremely detailed knowledge of the theory you are attacking--knowing it better than its proponents--and by making sure that every single thing you say about that theory is correct according to its proponents. It's up to you to dot every i and cross every t and dig into the textbooks and the literature to make sure you are representing the theory correctly.
You have not even come close to doing all that, and neither has the author of the blog post. When I see elementary misstatements like the ones you and the blog post author have made, that tells me it's not even worth my time to read through the blog post in detail to try to figure out what's wrong with it, because it's not even wrong. That happens quite a bit on PF as well; the things you think are "rarely really answered" there are examples of the same kind of not even wrongness.
For the specific scenario under discussion here, this is what extraordinary evidence would look like: you would be able to construct, in detail, the local inertial frame in question, centered around the event of an infalling observer just crossing the horizon. By "construct, in detail", I mean you would be able to draw a spacetime diagram of it showing all the curves of interest: the horizon itself, the worldlines of all objects, curves of constant radial coordinate r, etc. You would also be able to exhibit the mathematical transformation from the coordinates in the local inertial frame to and from a global coordinate chart, such as the Painleve chart. And you would be able to exhibit the mathematical equations that give the size of the local inertial frame, and demonstrate, mathematically, that the event of two cloud particles crossing (one that starts inside the horizon and one that starts outside) must happen within that size. (When I do this computation, I find that the crossing event cannot happen within the LIF--the distance that would be required for the two particles to meet is much, much larger than the distance the LIF can cover. If you want to prove me wrong, first you have to get to that point, and you have given me no evidence that you have done so.)