(Alpha) The Existence of Black Holes

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

  1. RJBeery Natural Philosopher Valued Senior Member

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    I preface this post by saying that I am not a Physicist and am very likely to mangle terminology, so I ask for your patience. I've made this an Alpha thread because I want to avoid ad hominems and other derailments. If you ask me questions as a point of clarification of something I've written or asked I shall do my best to get my meaning across. In return, I ask that you make a sincere attempt to understand my layman's terminology. Any posts, even PART of a post, not directly related to the OP (i.e. the existence of black holes and the various attempts to explain that existence) shall be reported to the moderator for deletion.

    In the past I've asked (what I feel are) some pretty basic, logical questions about black holes to many people that I would assume would be able to supply straight-forward, well established, community-accepted answers. Here's a version of my logical conundrum:
    =========================================
    #1 Do all frames outside of the BH calculate that no mass crosses the EH using Schwarzschild coordinates (or, more specifically, mass crosses the EH at t=infinity)?
    #2 Does the EH expand only after mass has crossed it?

    If you concur with #1 and #2, then run the clock backwards in your mind and describe to me how this theoretical black hole formed in the first place...
    =========================================

    To this point I've received many responses including the following:

    1) Backreaction occurs before EH crossing
    2) Unruh radiation
    3) Vaidya metric
    4) Kruskal coordinates
    5) The finite proper time of the in-falling body proves the BH's existence
    6) "A very famous man told me so" (this is my favorite)
    7) Speculative "mystery mechanisms"
    8) "Here, check out my class notes" (implying, to me, that they feel the answer was explained to them in class but they didn't understand it either)

    ...and in most cases, when I ask more questions, I generally get the feeling that the respondent is speaking with less authority than I would expect from someone that was 100% sure of what they were talking about. [It goes without saying that my authority is exactly ZERO, save for rational thought and the ability to ask questions; if black holes cannot be explained rationally then I have a problem with them] If nothing else, I find it odd that there isn't more consistency in the explanations.

    Anyway, my most recent discussion involved #4 (Kruskal coordinates), although posters are welcome to discuss any explanation they wish. If you believe that Kruskal coordinates allow for the existence of black holes, and you feel up to the challenge of explaining this to a layman, please explain to me how they do so. I understand that Kruskal coordinates avoid the mathematical singularity associated with the Schwarzschild event horizon by selecting new space and time components, U and V, to replace R and T. This, I am OK with and believe I understand. Next, the claim is made that, because these new components are finite at the crossing of the EH, infalling bodies cross that EH in finite time from the outsider's perspective. This, I am not getting...a change in coordinates should not change conclusions about whether an infalling body makes it beyond a certain threshold...please help. :shrug:
     
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  3. Guest254 Valued Senior Member

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    Who said this?
     
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  5. przyk squishy Valued Senior Member

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    I addressed this in my last post in the other thread. Here's the relevant part reproduced:

    No, I'm saying that the question of when an in-falling body crosses the event horizon according to an outside observer isn't a meaningful one. You need to explain what convention you're using to attribute the horizon crossing event a time on the outside observer's clock. And I don't know any such convention that wouldn't be arbitrary. That's why, when you raise the question, I keep substituting it with different questions formulated in terms of concepts such as trajectories (eg. of light signals) and light cones, which are better defined (ie. independent of the choice of coordinate system).

    For example, take the fact that an in-falling object eventually becomes irretrievable to an outside observer, which I illustrated on a Kruskal chart. This result isn't meaningful just because I happen to like the Kruskal chart. It's meaningful because I was careful to pick a question whose answer was independent of any particular choice of coordinates. I then used the Kruskal chart to illustrate the result only because light cones have a particularly simple appearance in that chart. Proper times and light cones are invariant concepts, so a complete calculation in some other chart is guaranteed to produce the same result.

    Depends on the context. In general, it's only meaningful to consider the "clock time" of an event occurring at the same place as the clock (ie. the event has to occur on the clock's world line). Attributing a time on a clock to an event that takes place somewhere else isn't so straightforward.
     
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  7. RJBeery Natural Philosopher Valued Senior Member

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    Here you used the word eventually (emphasis mine), yet you also said
    Maybe this is my problem. You are saying that there exists a time, call it T1, that a distant observer would read on his clock after which time he cannot rescue the infalling observer. We don't need to calculate it because you can see it on the graph (is that what you meant?). Whether or not we "need" to calculate it, are we able to calculate it? I'm asking because frankly this contradicts my understanding of black holes.

    Another question: were U and V assigned the labels of time and space components arbitrarily, or could the labels be reversed?
     
  8. RJBeery Natural Philosopher Valued Senior Member

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    This seems to me to contradict the notion that there exists a time T1 on the the distant observer's clock after which he can no longer rescue the infalling body. In other words, changing the question as you did to that of one involving light cones and causal relationships should be able to provide us with a definitive time, if it exists.
     
  9. przyk squishy Valued Senior Member

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    Yes. At least, the Kruskal chart allows us to see that T1 is finite without calculating it.

    Yes.

    They're not arbitrary. I also gave an answer to this in my last post in the other thread:
    No. General relativity provides a clear distinction between space-like, time-like, and light-like vectors (and therefore coordinates) at every point. It inherits these notions from SR. From the perspective of any locally inertial frame at a point, a particle following the Kruskal V coordinate (ie. U = cst) would be travelling at less than the speed of light as it passed that point. A point following the U coordinate (V = cst) would conversely be moving faster than light. Basically, if you imagine drawing the "coordinate grid", a material point could travel along the time-like coordinate lines but not the space-like ones.

    There's no contradiction. To clarify, consider the case in flat space-time. Suppose someone tells you there's going to be a Big Event one light-minute from where you live, at exactly 12 o'clock (all in your rest frame). Then:
    • At 11:59, it's too late to decide you want to participate, because you can't travel faster than light.
    • At 12:00, the Event is supposedly occurring "right now".
    • At 12:01, light emitted or reflected by the Big Event reaches you and you see it on your telescope. You can confirm that the Event took place.
    Getting back to the in-falling observer crossing the event horizon, I'm saying that for the outside observer there's an analogue of (a) which occurs after a finite time on his clock, no well defined analogue of (b), and he never experiences the analogue to (c) unless he himself crosses the event horizon.
     
    Last edited: Oct 8, 2010
  10. prometheus viva voce! Registered Senior Member

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    2,045
    If you went to Japan and spoke to them in Afrikaans the Japanese people (probably) wouldn't understand you. The same applies here. If you want to order a drink from a bar in "physicsland," you need to learn to speak the language, at least to a passable level.

    The answer to this question is no. An observer in the "Painleve coordinate frame," that is one either falling towards or moving away from the black hole sees matter crossing the horizon at a finite time.

    That depends on what you mean by "after." In special relativity, simultaneity is tricky, but in general relativity as you might imagine it's even worse. I will admit now that I don't know too much about this, but I know of a good example where the black hole grows as matter crosses the horizon.

    Since this is an alpha I shall give you a reference: here.

    I'm almost certain this is a waste of the time I just spent to look that up again because I've provided that link to you before and I'm certain you haven't got it. If you're super lazy go straight to figure 2. That shows the collapse of a continuous set of light shells, and the line labelled R shows the horizon.

    I think a lot of these have been lost in translation, for example what does
    Have to do with black hole formation? Unruh radiation is the radiation observed by an accelerating observer. It is related to Hawking radiation but only because the mathematical technique you use to calculate it is very similar.

    Also, the
    is the metric of a radiating star.

    Who's digging out the ad hominems now? Consider this post reported.
     
  11. RJBeery Natural Philosopher Valued Senior Member

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    I'm sorry but what does cst stand for?
    Excellent, good analogy. OK, let's discuss #3, because confirming the Big Event in your telescope here is analogous to no longer seeing the in-falling body due to the fact that it has crossed the event horizon. The problem is that I contend this "never" happens. What is actually predicted to happen is that the in-falling body appears dimmer and dimmer, and more and more red-shifted. If we had perfect-precision instruments that could detect "arbitrarily red-shifted" photons then I'm not sure there is ever a time that they would ever stop coming, even if we were required to wait longer and longer and longer for them to arrive. (I could be wrong on this point, but even so it could be explained by the quantum nature of photons rather than the fact that the light is no longer there)
     
  12. przyk squishy Valued Senior Member

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    Constant.

    This is correct. Why do you consider this a problem? It wouldn't be much of an event horizon otherwise.
     
  13. rpenner Fully Wired Valued Senior Member

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    A Schwarzschild chart is a set of spherically static coordinates -- an artifice which may or may not be useful in describing the physics of space time, depending on the question you are asking.

    Assuming we are talking about static, uncharged and finite black holes, the coordinates you speak of allow hypersurfaces of constant time coordinate values which far from the black hole approximate the hypersurfaces of constant time of a Lorentzian flat-space frame at rest with respect to the black hole. But since even in flat-space the stationary observer has no privilege to refer to his chosen time coordinates as "time" itself. In Special Relativity, different states of motion have different time coordinates. In General Relativity, even observers with constant separation may experience different time coordinates. And in both, two observers may synchronize clocks, take different paths through space-time, and rendezvous at some future point in space and time with different values on their clocks.

    In the usual Schwarzschild coordinates, the exterior of the black hole is labeled with \((r_A, \theta, \phi, t)\) with \(r_A \in \left( + \frac{2 G M}{c^2}, +\infty \right)\), \(\theta \in \left[0, 2 \pi \right)\), \(\phi \in \left[0, \pi \right] \) and \(t \in \left(-\infty,+\infty \right)\). But, even though the angles seem to make sense, and the circumference for every radius is \(2 \pi r\), the speed of light is not the same in every direction with the line element of \( c^2 \left(\Delta \tau \right)^2 = 0 = \left(c^2 - \frac{2 G M}{r_A} \right) \left(\Delta t \right)^2 - \frac{1}{1 - \frac{2 G M}{c^2 r_A}} \left(\Delta r_A \right)^2 - r_A^2 \left(\Delta\phi\right)^2 - r_A^2 \sin^2\phi \left(\Delta\theta\right)^2 \).

    Even if you use Eddington's isotropic Schwarzschild coordinates, the coordinate speed of light is now the same in every direction for every point, but now the speed is different at different radii, when physically the speed of light is an assumed physical constant. The physics are distinct from the coordinates, and sticking to one global chart is going to be crippling to your understanding of General Relativity.

    It's even crippling to use one chart on Earth, since Australians don't enforce their traffic regulations based on what a chartist in England calculates their coordinate speed to be.

    So perhaps your first question needs clarification.
     
  14. RJBeery Natural Philosopher Valued Senior Member

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    OK...now, if photons continue to reach the outside observer indefinitely, regardless of their degree of red-shifting or the amount of time it takes between photon arrival events, then we cannot say that a time T1 exists on the outside observer's clock after which he cannot rescue the infalling body. The reason is because if photons can travel from the infalling body to the outside observer then that body can escape with a sufficiently powerful jetpack. In a trivial example we could define "rescuing" the infalling observer as simply shining a light toward him that would ignite it. (I presume we are not going to disagree that the photons shining toward the infalling body shall never have a problem reaching him.) We've already agreed that photons emitted near the EH, eventually reaching the outside observer, will do so indefinitely from the observer's perspective (albeit requiring more and more patience and precise detection instrumentation), and this forces us to admit that the return path away from the EH back to the outside observer is also available indefinitely from the outsider's perspective.
     
  15. AlexG Like nailing Jello to a tree Valued Senior Member

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    No, this isn't quite correct. There is a point at which we no longer receive photons from the infalling body. It's not that the photons are no longer perceptible due to red-shifting. There are no more photons, and what we would detect is a frozen image which fades to nothing.

    I'd give you a link but I haven't looked it up anywhere on the web. The best I can do is refer you to pages 244-249 in Kip Thorne's Black Holes & Time Warps

    Given that, then
    is not valid.
     
  16. Pete It's not rocket surgery Registered Senior Member

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    This argument seems incomplete.
    If the infalling body could escape with a sufficiently powerful jetpack at the time it emitted a photon, then how does it follow that the outside observer who receives that photon at a later time can reach the infalling body before it crosses the horizon?

    I had a vague idea that photons emitted from arbitrarily close to the event horizon (in schwarzschild coordinates) take an arbitrarily long time to escape... so the outside observer will keep receiving photons from the infalling object for an arbitrarily long time after it has crossed the event horizon.
    Is there a negative too many in that sentence?

    I do question whether photons shining toward the infalling body can reach it before it crosses the horizon.
     
  17. M00se1989 Banned Banned

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    On black holes... what physical elements do we believe it is inherently made of?

    is it a larger element than is observable to us? It takes in mass when mass is to delicate to pierce through its entire volume. It also emits light at a constant, relative to its mass and volume. Light is the only "constant" that we can travel both toward and away from. Any atom has electrons that create space inherently. Is it not possible to say the elements within its composition are greater than those of our known values?

    If the elements were making themselves they would have done it in secret. That means there is a potential to do it in front of you, but also without you knowing.
     
  18. AlexG Like nailing Jello to a tree Valued Senior Member

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    It's not made of any physical elements. What we commonly call the black hole is the event horizon, that point where gravitational forces are strong enough so that the escape velocity is the speed of light. Everything and anything can be swallowed by the BH. NO light is emitted.

    Friends don't let friends drink and post.
     
  19. przyk squishy Valued Senior Member

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    True but irrelevant. I specifically attempted to clarify this in an edit to [POST=2629283]this[/POST] post:
    I specifically showed the opposite: the point where the in-falling observer crosses the event horizon drops out of the outside observer's future light cone. After that point, the outside observer can't send signals that will intercept the in-falling observer before he crosses the horizon.

    If you program a probe to fire its engines and return to you only when it receives a signal from you, and allow that probe to fall toward a black hole, you only have a finite time to send that signal before the probe becomes irretrievable.
     
  20. RJBeery Natural Philosopher Valued Senior Member

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    If you both agree that photons will continue to be received by the outside observer indefinitely then it does not follow that there is a finite time after which there is no source of those photons. There cannot be some sort of "infinite cache" of light waiting to make its journey to the outside observer (unless that "cache" is the infalling body itself either reflecting or emitting those photons, which is what the Schwarzschild coordinate calculates).
     
  21. AlexG Like nailing Jello to a tree Valued Senior Member

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    The apparent freezing of the infalling body is an illusion generated by the difference in relativistic frames.


    http://casa.colorado.edu/~ajsh/schwp.html

     
  22. RJBeery Natural Philosopher Valued Senior Member

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    This technically doesn't eliminate my objection. All photons must have an emission source, and I doubt the author referred to above is suggesting a "ghost source" that hangs around indefinitely providing any supposed illusions.

    Also, in all other aspects of Relativity, light itself is the ultimate arbiter of measurement (e.g. in SR, "seeing" a passing clock move slowly and deducing that it is moving slowly are equivalent; in GR, "seeing" the light being bent by the sun in the eclipse of 1919 is what confirmed Einstein's theory over Newton). Given this, it is difficult for me to accept any such explanation as being a mere illusion.
     
  23. AlexG Like nailing Jello to a tree Valued Senior Member

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    Yes, but the photons are not received indefinitely. The last photon emitted before the body passes the event horizon is frozen in time, as seen from the external frame, but it is the last photon received.

    Research it more. The site I linked to is a place to start. If you have difficulty accepting an explanation which has universal acceptance among those who have studied the subject, you might consider that the problem is a deficiency in your understanding, rather than the explanation is wrong.
     

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