(Alpha) The Existence of Black Holes

Discussion in 'Physics & Math' started by RJBeery, Oct 8, 2010.

  1. przyk squishy Valued Senior Member

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    What's the relevance of this? I don't even know whether I should agree with "there is a finite time after which there is no source of those photons" or not. As presented, that statement is too vague and ill-defined to even discuss in the context of GR.

    I've always been specific about what I was claiming: the point where the in-falling observer crosses the event horizon drops out of the outside observer's future light cone after a finite time. Even with that settled, your statement doesn't make sense: let
    X = "light emitted by the in-falling observer will continue to be received by the outside observer indefinitely"​
    and
    Y = "the in-falling observer becomes irretrievable to the outside observer in a finite time on the outside observer's clock".​
    If you say "X, and it doesn't follow that Y", then read literally you're saying "X doesn't imply Y". Now that's true, but irrelevant: nobody is trying to argue that Y is true *because* X is. I'm claiming that Y is true, independently of X, because I've depicted the situation on a Kruskal chart.

    That's obvious, so I think it's more likely you meant "X implies Y is false". And that simply isn't true. X and Y together mean that the horizon crossing event falls out of the outside observer's future light cone after a finite time, but never falls into his past light cone. There's really no reason X should invalidate Y. Both are visible on the Kruskal diagram for instance.

    Why not? What's happening is simply that light emitted by the in-falling observer has a progressively harder time climbing out of the black hole's gravitational well and reaching the outside observer. Another point of view is that the outside observer is accelerating away from the light, which is how the situation appears on a Kruskal diagram. He'd get all the light in a finite time if he simply allowed himself to free-fall past the event horizon.
     
    Last edited: Oct 10, 2010
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  3. RJBeery Natural Philosopher Valued Senior Member

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    With respect, this is not what others in this thread are saying, but I did leave this possibility open when I said
    What I was saying is that there is a big difference between no longer seeing photons because the light source has passed the EH vs no longer seeing them because the light source's emitted energy does not suffice to produce a photon.
    This is quite likely, but in fairness I don't feel I've been given "the" explanation, but rather many disparate ones, which is one of the confounding reasons I keep asking questions. You say the photons will stop coming; Pete and przyk say they won't; Guest appears to question whether przyk's argument is valid (by implication, here, when he asked who made that statement in post #2); prometheus says the answer lies in the "Painleve coordinate frame". I am only really able to concentrate on a single line of reasoning at a time which is why I want my discussion with przyk and Pete to run its course before moving on to something else (such as "illusions" and "Painleve coordinate frame", which I guess I shall need to add as #9 and #10 in my OP

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    )
     
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  5. AlexG Like nailing Jello to a tree Valued Senior Member

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    Which is why I suggest you research the issue yourself. Try websites ending in .edu.
     
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  7. przyk squishy Valued Senior Member

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    No they're not. If a clock is moving with a speed \(\beta = \frac{v}{c}\) its apparent time dilation factor could be anything between \(\frac{\sqrt{1 - \beta}}{\sqrt{1 + \beta}}\) and \(\frac{\sqrt{1 + \beta}}{\sqrt{1 - \beta}}\) depending on whether it's approaching or receding from you. In the former case, an approaching clock will appear to be ticking faster than yours. It's only when you correct for the Doppler effect that you get the familiar factor of \(\frac{1}{\sqrt{1 - \beta^{2}}}\).
     
  8. M00se1989 Banned Banned

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    what's all this nonsense about a big blob of energy?

    Don't blame the clock for getting stretched out a bit. It runs off light which slows down the time when it gets a little "bent".

    and so does any theory.

    Larger means more heat, and therefore more gravity, and more Light.

    Just because you don't see a photon does not mean it isn't still there.

    but it would swallow you up and spit you right back out in "space"

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    its only when your in its "sphere of influence" that you are part of it. And that is determined entirely by your relative speed inside the sphere. So if you have stopped you probably went too far too fast and crashed.
     
  9. przyk squishy Valued Senior Member

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    They aren't all so disparate actually:

    He's right in at least one sense: if the in-falling observer emits N photons before crossing the horizon, the outside observer will receive N photons. One of these will be the last photon and the outside observer will detect it after a finite time. How long that time is depends on how close the in-falling observer was to the horizon before emitting it. The time approaches infinity as the point of emission approaches the event horizon.

    So far I've just been discussing in terms of continuous, classical light rays.

    Guest asked who claimed "infalling bodies cross that EH in finite time from the outsider's perspective". I've never claimed this. I've repeatedly stated claims of this sort aren't well defined.

    I'm not familiar with Painleve coordinates, but (emphasis added):
    This sounds like it probably is consistent with what I'm saying: a free-falling outside observer does see matter cross the horizon after a finite time (in the sense of receiving light emitted or reflected by the in-falling mass). He sees the mass cross the horizon when he himself crosses the horizon.

    prometheus may also have meant the mass crosses the horizon in finite coordinate time in Painleve coordinates, in the same way the mass crosses the horizon in finite Kruskal time. I don't know.
     
    Last edited: Oct 10, 2010
  10. RJBeery Natural Philosopher Valued Senior Member

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    ...and my argument is even clearer in that context. A discrete event horizon crossing would result in an abrupt end to the classical continuous light ray from the outsider's perspective. This is not what is predicted from Schwarzschild coordinates. This directly exposes my (unsophisticated) view that there is a problem here, because unless someone is claiming that Schwarzschild coordinates are invalid or wrong (meaning specifically that they would hold inaccurate predictive power), then I am left with a cognitive disconnect.
    You're right, this is my mistake. Despite you trying to clarify this point, I still have trouble seeing the difference between saying that something happens in finite time on my watch vs saying that something is eventually outside of my future light cone. There should be a specific point on my watch after which that event is outside my light cone, correct?
     
  11. przyk squishy Valued Senior Member

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    It's not predicted period. It doesn't matter which coordinate system you use. You get the same result using the Kruskal chart for instance.

    There really isn't anything remarkable about this result. It's even possible to describe a very similar situation in SR: suppose you start from rest at (t = 0, x = 1) and travel along the trajectory
    \(x(t) \,=\, \sqrt{1 \,+\, c^{2} t^{2}}\)​
    (in the coordinates of some given inertial frame), and suppose I emit a signal from the spatial origin x = 0 starting at ct = -1 up until ct = 0. Then you'll receive the beginning of my signal just as you start to leave, but your curve accelerates away from me in such a way that you'll never, by your watch, receive the end of it.

    Roughly, here's what the situation looks like on a Minkowski diagram:

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    Here the red dotted lines represent the propagation of the beginning and the end of my signal. The blue curve gives an idea of your hyperbolic trajectory. It asymptotically approaches, but never touches, the world line depicting the end of the signal. The upper diagonal line is effectively an event horizon to you: no signal emitted from above and to the left of it will ever reach you.

    In Kruskal coordinates, a diagram of the signal sent by the in-falling observer approaching the event horizon and received by the outside observer would look very similar to this.

    Yes. The reason I'm picking on this is that, referring to the analogy I gave in post #6, saying that a distant event happens in finite time on your watch would normally mean the analogue of (b).
     
    Last edited: Oct 11, 2010
  12. RJBeery Natural Philosopher Valued Senior Member

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    OMG przyk, I'm so grateful you said this. This talk of alternate coordinate choices to solve problems seriously had me feeling like higher mathematics demanded a forfeiture of reasoning. I've been interpreting some posters' comments as implying that a change in coordinates allows for a change in prediction of what "actually happens", and this was deeply troublesome for me to understand.

    Your illustration is very, well, illustrative by the way. I see your point and after seeing your graph did I realize something that I should've thought of earlier - the proper time of the infalling object is indisputably finite before crossing the EH. This in itself necessitates (in my mind) a finite number of photons to be emitted, period.

    It's a busy week for me but I'm going to try posting a picture of my own later. I'd really like to see an application of the Kruskal coordinates with, say, a 1 solar mass BH to calculate T1 after which the outside observer could not rescue the infalling body. I won't ask you to do it (unless it's trivial); I'll try to make it to campus and read what I can on the subject.
     
  13. AlphaNumeric Fully ionized Registered Senior Member

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    This demonstrates the point made by myself, Guest and others, that because you haven't had any experience with this area of mathematics nor anything which is considered a requirement for said area of mathematics you lack the ability to make an informed opinion or view on the matter.

    One of the founding principles of theoretical physics, in fact pretty much any area of physics striving for quantitative models, is that the outcomes should not depend on our method of description. Its from this you get the notion of covariant formalisms, both in terms of space-time coordinates and gauge potentials. High level general relativity is often done in a way which doesn't even need you to define a set of coordinates and the best derivations or proofs are those which don't require them. Most advanced courses on GR go down that route because it presents more powerful and rigorous approaches the problems.

    If you'd done any reading on the role of coordinates in physics you'd have found this out. Instead it's several threads and dozens of posts for you to find out something which you should have known before forming an opinion.

    I don't think its against the Alpha rules to say that perhaps it would be wise if you spent a little less time arguing about something you haven't done any reading on and a little more time actually doing the reading. That isn't an insult, its a piece of honest advice.
     
  14. RJBeery Natural Philosopher Valued Senior Member

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    So, what you mean to say is that I was right to trust my gut rather than my (possible) misinterpretation of what others were telling me? Thanks for the encouragement!

    Also, what I'm doing with przyk is not arguing, it's questioning and answering. I'm very impressed with his style, and others on this forum would do well to take note.
     

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