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> What material have you covered? The material sufficient for this discussion. I know the material well enough to point out contradictions in your thinking. >

by genwin 14y ago

> What material have you covered? 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.

pdonis 14y ago
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. 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?
- genwin 14y ago
  > Because spacetime is curved, and the curvature limits the size of a local inertial frame straddling the horizon. 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.
  - pdonis 14y ago
    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. 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?

by genwin 14y ago

pdonis 14y ago

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. 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?

genwin 14y ago
> Because spacetime is curved, and the curvature limits the size of a local inertial frame straddling the horizon. 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.
- pdonis 14y ago
  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. 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?