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Do event horizons appear to be surrounded by all the stuff that has ever fallen in?

  1. Jan 6, 2012 #1
    Hi everyone:

    Do event horizons appear to be surrounded by all the stuff that has ever fallen in? As I understand it, if you watch something fall into a black hole's event horizon, it will appear to take forever.

    To A/B it, if Alice starts watching the horizon at t=0, and something falls in at t=1, Alice will observe it taking forever to fall in.

    But what if Bob starts watching the horizon at t=2 (after the object fell in)? Will he see the object taking forever to fall in, or will he not see the object at all (because it already fell in)? Thanks for your time!

    Rob
     
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  3. Jan 7, 2012 #2

    Simon Bridge

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    Welcome to PF.
    It's true that you get lots of stuff around a BH that has never fallen in and still has a long time yet before it does - these are called accretion disks, and are the main way we detect BHs.

    However, this is not what you mean.

    Recall that the radius of the event horizon is proportional to the mass enclosed.
    When an object gets close enough to the even horizon to be part of a black-hole, what happens is the event horizon expands to cover the object. The BH sort-of "gulps down" things that get too close.

    Be aware - when we try to describe things like black holes using simple analogies - we are going to get it wrong in important ways. The math for black holes is not as bad as you think so if you are really really interested, they make a decent study.)

    10 things you don't know about black holes (Bad Astronomy blog.)
     
  4. Jan 7, 2012 #3

    tom.stoer

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    As seen by an outside, stationary observer it will take forever. But note that in reality he will see nearly nothing (absolutely nothing) i.e. nearly no (no) light from the objects closed to (at) the horizon b/c there is a large (infinite) red shift for light emitted near (at) the horizon.

    For the objects themselves it takes finite proper time to cross the horizon and to fall into the singularity, so they will not stay at the horizon (they will cross the horizon at the speed of light!)

    The horizon marks the boundary between a region where light rays can escape to infinity and a region where all light rays will converge towards the singularity. That means that light emitted radially outwards exactly at the horizon will neither fall into the singularity nor escape to infinity - it will stay at the horizon! But b/c light always moves at c the entire horizon will move at c. That means that a free falling observer will cross the horizon at c simply b/c the entire horizon is a so-called light-like surface moving with the speed of light.

    No back to the accretion disks: b/c of the tidal forces becoming large at the horizon of solar black holes, infalling matter becomes extremly hot and radiates X- and γ-rays. Rays emitted radially outwards directly at the horizon will stay at the horizon. Rays emitted tangentially at a larger radius (1.5 * Schwarzschild radius for Schwarzschild black holes) can orbit the black hole at constant radius (like satellite) and form the so-called photon-orbit.
     
  5. Jan 7, 2012 #4

    PeterDonis

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    The part in parentheses is not correct. The infalling object (assuming it is not itself an ingoing light beam) moves on a timelike worldline, so it never moves "at the speed of light".

    The horizon will move outward *past* the object at the speed of light, but that's because the *horizon* is a lightlike surface, as you note:

    It's the *horizon* that's moving at c, not the object. We don't say we are "moving at c" when a light ray passes us going in the exact opposite direction; that's what the horizon is doing relative to the infalling object.
     
  6. Jan 7, 2012 #5

    tom.stoer

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    The infalling object will cross the horizon at the speed of light i.e. it moves at the speed of light w.r.t. the horizon - and the horizon moves at the speed of light w.r.t. the object.

    So the part in parentheses is correct; the next sentences provide the explanation.
     
  7. Jan 7, 2012 #6
    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    Nothing light-like can act as an observer, so "at the speed of light w.r.t. the horizon" is not a valid expression.
     
  8. Jan 8, 2012 #7

    tom.stoer

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    It cannot act as a physical observer, that's correct; the definition of reference frames becomes 'singular'.

    If there is a flash of light then the speed of the flash w.r.t. to me is c and my speed w.r.t. the flash is c, too. This is a perfectly valid statement. In GR I always have to say "speed w.r.t. something" and if this "something" is light, it will be clear that my speed is c.

    But I think this discussion is quibbling and totally irrelevant for the original question.

    My idea was to explain why it is difficult to say that an object 'stays at the horizon' b/c a) the horizon is moving at c and therefore no massive body can stay there and b/c b) this statement would depend on the reference frame and a stationary observer outside the horizon will soemthing totally dfferent than the free-falling observer who crosses the horizon

    Btw.: the observer will cross the surface of the horizon, so there is a relative speed of the observer w.r.t. to the horizon; what is this speed? c? something else? no speed at all? not defined?
     
  9. Jan 8, 2012 #8
    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    How is horizon moving at the speed of light? It stays (in Schwarchild coordinates) at [tex]r=\frac{2GM}{c^2}[/tex]. And it is not an object with a null worldline.
     
  10. Jan 8, 2012 #9

    tom.stoer

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    It is!

    We start with the Schwarzschild metric

    [tex]ds^2 = \left(1-\frac{r_s}{r}\right)\,dt^2 - \left(1-\frac{r_s}{r}\right)^{-1}\,dr^2 - r^2\,d\Omega^2[/tex]

    Now we look at radial, light-like motion, i.e. d\Omega = 0 and ds² = 0

    [tex]0 = \left(1-\frac{r_s}{r}\right)\,dt^2 - \left(1-\frac{r_s}{r}\right)^{-1}\,dr^2 [/tex]

    This can be solved for dr/dt

    [tex]\frac{dr}{dt} = 1-\frac{r_s}{r}[/tex]

    From this differential equation all radial null-geodesics can be derived.

    A special solution is

    [tex]r(t) = r_s = \text{const.}[/tex]

    for which both l.h.s. and r.h.s. of the diff eq. are zero
     
  11. Jan 8, 2012 #10
    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    It is a surface, not a point however. I don't know how you can define the speed of an extended non-rigid entity.

    BTW: isn't there a difference between null geodesics and null curves?
     
  12. Jan 8, 2012 #11

    tom.stoer

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    I derived a family of null-geodesics for all points r(t) = rS with arbitrary \Omega defining the horizon. Therefore the horizon is defined by

    [tex]x^\mu = (t,r_s,\Omega)[/tex]

    So for each t you have a 2-surface of fixed r(t) = rS with coordinates \Omega on the surface
     
  13. Jan 8, 2012 #12
    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    Whatever.
    We were simply arguing about semantics, which is neither fun nor meaningful.
     
  14. Jan 8, 2012 #13

    tom.stoer

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    Sorry, but here I totally disagree!

    What we are doing here is to start investigating what 'horizons' are. There are several different definitions (apparent horizons in terms of trapped null-surfaces = closed surfaced on which outward-pointing light rays are converging), absolute horizons, isolated horizons, ...

    This is highly non-trivial and still subject to current research.

    You may want to have a look at

    • Hawking & Ellis: The large scale structure of space-time
    • Wald: General Relativity
    • Thorne, Misner & Wheeler: Gravitation
    • Ashtekar & Krishnan: Isolated and Dynamical Horizons and Their Applications; http://relativity.livingreviews.org/Articles/lrr-2004-10/ [Broken]

    Of course all these guys neither had fun nor did they work on something meaningful (sorry for the irony ;-)
     
    Last edited by a moderator: May 5, 2017
  15. Jan 8, 2012 #14
    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    My point was that null surface as it is, the "speed" of event horizon is not a well-defined and physically useful concept. And you were trying to prove that the horizon is moving at the speed of light without defining the speed of a surface. It's not about the properties of event horizon; it's about the definition of speed.
     
    Last edited by a moderator: May 5, 2017
  16. Jan 8, 2012 #15

    tom.stoer

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    Sorry, but I think the math in post #9 is rigorous. The horizon moves at the speed of light in just the manner as a spherical wave front in Minkowski space does. Where's your problem? Which equation or derivation is wrong?
     
  17. Jan 8, 2012 #16

    Simon Bridge

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    To drag this back to OP's question ... there is a lot of talk about objects "falling in" to a black hole and of black-holes growing over time ... of "feeding".
    [eg. Bad Astronomy - 10 things ... about Black Holes #9
    ... Even the Milky Way has a black hole at its core ... it can’t really harm us. Unless it starts actively feeding. Which it isn’t. But it might start sometime, if something falls into it.]

    Since it takes an infinite amount of time, from the POV of an observer a long way from the BH, surely you'd never see one grow and the stuff "fallen in" never actually does this but gets stuck just "outside"... so how can it be said to "grow"?

    Considering the level the question is being asked for it is probably safe to take "event horizon" as the locus in space about the BH where the escape velocity is equal to the speed of light in a vacuum.

    The use of the word suggests a Schwartzchild Hole rather than a Kerr Hole.

    The question appears to be trying for is, given it takes an infinite amount of time to fall into a black hole, how come black holes can be said to grow? Note: I'm being deliberately sloppy with language here - the question is hard to define in sensible terms for all the reasons you guys have been discussing and more. I'm setting up the following, more accessible, description, to be found here.
    ... which should be a reasonably complete answer for OP.

    Note: All observers are moving at c wrt light. Ergo: meaningless observation.
    FWIW: I'm gonna have to agree with that one.
     
  18. Jan 8, 2012 #17

    George Jones

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    So, you're saying that in special relativity, when an observer and a photon are coincident, the observer moves with speed of light with respect to the frame of the photon? Also in special relativity, that when an observer crosses a Rindler Killing horizon, the observer moves with the speed of light with respect to the Rindler horizon?

    Aren't these examples of the type of language that we try to discourage on this forum?
     
  19. Jan 8, 2012 #18

    timmdeeg

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    PeterDonis, sorry if I come back to this, but perhaps I can improve my understanding.
    The Newtonian gravitational force is proportional to (1 - Rs/R)^-1/2, implying that an infalling mass is accelerated to c in the moment Rs = R and consequently it's future light cone is tangent to the event horizon.

    Why can we not say something moves at c, in the very special case it is accelerated to c?
     
  20. Jan 8, 2012 #19

    Simon Bridge

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    With respect to what exactly?
    (Is Newtonian Gravity a useful model for behavior close to the event horizon of a black-hole?)
     
  21. Jan 8, 2012 #20

    tom.stoer

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    Re: Do event horizons appear to be surrounded by all the stuff that has ever fallen i

    Again, I tried to rephrase and explain mathematically – see post #9 which is mathematically rigorous – please tell me if I am wrong – what it means that “the horizon moves at c – a radially emitted photon at the horizon does exactly this! I tried to explain that this can be compared to an ordinary wave front in Minkowski space – the main difference is that in Minkowski space we have r=ct whereas for the EH we have r = const. I tried to explain that a free-falling observer will cross the horizon at the speed of light – horizon and observer move at c w.r.t. each other. I explained that the horizon (a null-surface) cannot be realized as a physical reference frame i.e. cannot be realized as a massive body. I tried to explain what the “speed of a surface” means – every point on the surface follows a null-line, i.e. the whole surface is a null-surface. I explained that the perception of a free-falling observer crossing the horizon in finite proper time and an asymptotic observer seeing objects to never cross the horizon are different.

    George, Peter, I am perfectly aware of the fact that words are dangerous in this context. Nevertheless we have to try to explain what the formulas mean. In a certain sense the observer crosses the horizon at c – you can do or find calculations showing exactly this in terms of free fall w.r.t. a ‘singular unphysical reference frame hovering at the horizon’. In a sense it’s the horizon that moves at c – this is the translation of null-surface – and it can be realized physically by photons emitted radially outwards exactly at the horizon.
     
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