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> 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 t

• pdonis 14y ago
  Presumably the horizon is indeed moving in that frame at the speed of light 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.
  • 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?