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Relativity and Black Hole Question

  1. Mar 3, 2014 #1
    Assume that there exist two people, person A and person B. Person B is falling into a black hole.

    I understand that due to relativity and time dilation, person A will have to spend an infinite amount of time to watch person B cross the event horizon. Person B will appear to move at a slower and slower rate towards the event horizon. (I know that person B will be gravitationally red shifted).

    So is it true, that relative to us in our universe, nothing ever falls into a black hole?

    If this is true, how is it possible that black holes ever form in the first place?
     
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  3. Mar 3, 2014 #2

    Dale

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    There is no requirement in physics that says that if something is not visually seen to happen by a particular observer then it must not have happened. Person A's visual experience has nothing to do with what happens or doesn't happen to B.
     
    Last edited: Mar 3, 2014
  4. Mar 3, 2014 #3

    phinds

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    Because relative to the black hole, stuff falls in. It doesn't care what things look like from our reference frame.
     
  5. Mar 4, 2014 #4

    BruceW

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    well, each coordinate system should give a consistent picture, right? So with the Schwarzschild metric, any test particle which starts outside the event horizon will end up at the event horizon at ##t \rightarrow \infty##. So I guess the OP's concern is how matter would end up inside the event horizon to form the black hole.

    I don't know much about GR, so I'm not really sure about this kind of stuff. But I would guess that the reason is related to the fact that it is a test particle. And if we consider the 'true' metric due to some spherical mass distribution, then it is possible for that matter to fall inwards to create the black hole. (i.e. to create the situation which is given by the Schwarzschild metric, where all the significant matter ends up inside the event horizon). I mean, when there is significant matter outside the event horizon, then that situation is not really modelled by a Schwarzschild solution, if I understand it right.
     
  6. Mar 4, 2014 #5

    Nugatory

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    For the portions of spacetime that the coordinate systems covers, yes. But there's an entire region of spacetime that isn't covered by the external Schwarzchild solution at all. I can't assign a Schwarzchild ##t## coordinate to a point on the event horizon, but that doesn't mean that a particle cannot follow a worldline to and through the event horizon, it just means that I can't use Schwarzchild coordinates to describe that worldline.
     
  7. Mar 4, 2014 #6

    BruceW

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    hmm. OK. But in the Schwarzschild coordinate system, it is true that the test particle approaches the event horizon, and gets 'stuck' just outside it, right? You can say this is not the best coordinate system to choose, but that's what happens according to this coordinate system. (unless I've not understood correctly).
     
  8. Mar 4, 2014 #7

    PeterDonis

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    No, in the Schwarzschild coordinate system, the test particle asymptotically approaches the horizon as ##t## goes to infinity; it never gets "stuck". But even the asymptotic approach is because the coordinates are singular at the horizon, not because the particle doesn't reach the horizon. The closer you get to the horizon, the more distorted Schwarzschild coordinates are, to the point of infinite distortion at the horizon.

    No, this isn't right, because you can't use coordinate-dependent quantities to tell "what happens". You have to look at invariants--things that don't depend on what coordinates you choose. The appropriate invariant here is the proper time along the infalling object's worldline from some large radius to the horizon, and that is finite.
     
  9. Mar 4, 2014 #8

    PeterDonis

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    In the idealized case of a perfectly spherically symmetric collapse, the vacuum portion of the spacetime (i.e., outside the collapsing matter) is modeled by a Schwarzschild solution (which contains both a region outside the horizon, and a region inside the horizon, once the collapsing matter falls inside the horizon radius for its mass). The region occupied by the collapsing matter is modeled by a collapsing FRW solution (the time reverse of the sort of solution that's used to model the expanding universe), which is matched to the Schwarzschild solution along the boundary at the surface of the collapsing matter. Oppenheimer and Snyder first developed this idealized solution in a 1939 paper; MTW give a good discussion of it.
     
  10. Mar 4, 2014 #9

    Dale

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    I would take the complete opposite approach. I would say that no coordinate system gives the physical "picture". The physics is in the invariants, not the coordinates.

    Each coordinate system will calculate the same invariants, but it is those invariants and not the coordinates which are important.
     
  11. Mar 4, 2014 #10

    DrGreg

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    Consider as an analogy a map of the world drawn with the Mercator projection. On such a map you have to travel an infinite distance (on the map) upwards to reach the North Pole. Yet people have travelled to the North Pole, and beyond, covering a finite distance (on the Earth's surface).
     
  12. Mar 4, 2014 #11
    I think it is necessary to include the mass of the infalling person to the mass of the black hole. If you treat the black hole as static, and the person's mass as negligible, then the person never crosses the event horizon in the Schwarzschild metric. But a real person has mass, so the event horizon increases in size when a person falls in, so the static Schwarzschild metric isn't valid.
     
  13. Mar 4, 2014 #12

    PeterDonis

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    This is incorrect. Why do you think this?

    Technically this is true, but for any black hole of practical interest, i.e., of stellar mass or larger, its mass will be so much larger than the mass of a typical infalling object that the static Schwarzschild metric works fine for analyzing the infall process.
     
  14. Mar 4, 2014 #13

    WannabeNewton

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    No that's not really what's going on. In general we go from stationary state to stationary state of the black hole i.e. a certain multi-parameter family of stationary black hole solutions and infalling mass perturbations can be taken into account using the equations of black hole dynamics but if the infalling mass is negligible in comparison to the black hole mass then the stationary state solutions are essentially the same.

    In the freely falling frame of the radially infalling test particle, there are no "issues": the Schwarzschild metric can perfectly well describe the physics as observed in this freely falling frame and we simply find that the freely falling particle falls past the event horizon in finite time as read by a comoving clock. For static observers there is an infinite redshift of light signals being sent from the infalling particle, simple as that-it's only ostensibly an "issue".
     
    Last edited: Mar 4, 2014
  15. Mar 4, 2014 #14

    BruceW

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    Awesome, thanks. I believe this is the kind of answer the OP (and I) was looking for. I don't have that book, but I found this webpage http://www.aei.mpg.de/~rezzolla/lnotes/mondragone/collapse.pdf Which gives more detail on this kind of stuff (although probably not as much detail as that book gives, but oh well).

    So anyway, the Oppenheimer and Snyder solution seems to be a specific case for when the matter distribution is uniform within some region, and zero outside. But that's pretty good. I think that gives a good idea of the sort of behaviour to expect. And as you say, in this case, the region occupied by the collapsing matter is modelled by an FRW metric, not a Schwarzschild metric.

    I think this is the key point, because in the FRW metric, the geodesics will fall through the event horizon, to the centre of the mass distribution. So this gives an answer as to how matter can form the black hole in the first place. (Because the metric is not a Schwarzschild metric in the first place). When the matter distribution is spread out, the event horizon is well within the region of matter. Therefore, this matter can pass inwards through the event horizon (which is contained in the region described by FRW metric). And as matter passes inside the event horizon, the event horizon grows, until eventually, the event horizon is greater than the matter region. And it is only now, that the event horizon is in the region described by the Schwarzschild metric.

    Also, the region which is described by the Schwarzschild metric is only outside of any non-negligible amount of matter. So, if we have formed a black hole, yet there is more material which is outside the event horizon, then the metric describing the larger region of matter is not a Schwarzschild metric. And this is the reason that the black hole is able to pull the extra material inside the event horizon (even according to some far-away stationary observer).
     
  16. Mar 4, 2014 #15

    BruceW

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    Just to reiterate the point I'm trying to say here: I agree that we can just use Invariants, or some other coordinate system, and everything will work out fine. What I was trying to figure out is how the Schwarzschild coordinates are reconcilable to what is actually going on. In other words, how it can be reconciled with the fact that matter must be inside the black hole, yet in the Schwarzschild coordinates, geodesics do not pass through the event horizon. And what I think is the reconciling answer, is that the FRW metric is the correct metric, when there is still infalling, non-negligible matter. (And I think this is the kind of answer the OP was looking for). And thanks to PeterDonis, who actually explained this first.

    edit: well, the FRW metric is only for a specific case when the infalling matter is homogeneous within some region. For a more general matter distribution, there would be a more general metric. But anyway, the FRW metric is a good example of the kind of metric tensor which would apply to the infalling matter.
     
  17. Mar 4, 2014 #16
    I think the best way to try to understand that is to forget about the black hole and study Rindler coordinates instead. Rindler coordinates is a set of coordinates for flat Minkosky space (special relativity) which exhibits a horizon (with hawking radiation and all). It comes to show how nothing special really happens at a horizon. Rindler coordinates are useful for uniformly accelerated motion (constant proper acceleration). It turns out that if you accelerate, part of minkowsky space becomes invisible to you (because it is behind the horizon).
     
  18. Mar 4, 2014 #17

    PeterDonis

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    This is not correct. The correct statement is that geodesics reach the "edge" of Schwarzschild coordinates (i.e, the limit as ##t \rightarrow \infty##) in finite proper time, and at that point, Schwarzschild coordinates can't describe what happens. That's not the same as what you said in the quote above; what you said in the quote above would only be valid if Schwarzschild coordinates could actually cover the horizon, and they can't.

    Obviously the Schwarzschild metric is not valid when there is non-negligible matter present, since the Schwarzschild solution is a vacuum solution. However, you are wrong in thinking that this means there is anything to "reconcile". See above.
     
  19. Mar 4, 2014 #18

    BruceW

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    according to a far-away observer, if we used Schwarzschild solution, infalling matter would stop at the event horizon. This is a problem. But Schwarzschild solution is not the correct solution when there is still infalling matter. Therefore, no problem. This is what I would say to the OP. Am I right in saying this? Or do I still have it wrong?
     
  20. Mar 4, 2014 #19

    WannabeNewton

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    This is wrong. It doesn't stop at the event horizon. The distant static observer simply asymptotically loses communication with it because incoming light signals from it to the distant static observer get asymptotically infinitely redshifted.
     
  21. Mar 4, 2014 #20

    PeterDonis

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    No, this is *not* correct. You are mistaking an artifact of coordinates for an actual physical phenomenon.

    If the infalling matter is negligible in mass compared to the black hole, the Schwarzschild solution is not *exactly* correct, but it's a very, very good approximation, certainly good enough that if it did in fact predict that matter could not fall through the horizon, that would be a problem. There is no problem because it does *not* predict that.

    If the infalling matter is *not* negligible in mass compared to the black hole, then the solution is certainly more complicated, yes. But if such a solution were the only way to describe matter falling into a black hole's horizon, that would be a problem, since there are many cases of interest in which the mass of the infalling matter *is* negligible. There is no problem because, as above, the Schwarzschild solution works fine for such cases.
     
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