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Event Horizons and Metaphors

  1. Jun 23, 2010 #1
    So I was listening to a book by Leonard Susskind and he was explaining what it would be like crossing the event horizon of a black hole from two different reference frames.

    In the first, if you were actually crossing the event horizon you would feel nothing as it is just some mathematical boundary in space. But someone from another perspective just outside of the horizon will watch you fall into it but then what does he see?

    He sees an image of you? Is the image the "information" in electromagnetic radiation?
    The image must go away because inside the event horizon things blueshift? How can an outside observer see you if the event horizon marks the point of no return? How does something escape it? Are my definitions incorrect?
     
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  3. Jun 24, 2010 #2

    tom.stoer

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    An observer located outside the event horizon (EH) will never see you crossing the EH. As you are approaching the EH, what she sees becomes affected by time dilation, so you are approaching the EH asymptotically and will eventually reach the EH in her "infinite future". In addition what she sees becomes affected by gravitational red shift, so your image gets redhsifted and evenually becomes invisible.

    An observer located exactly at the event horizon cannot exist physically. The EH is not a usual two dimensional sphere, it is a light-like sphere. One can understand what that means bylookingat light rays moving outwords. A light ray emitted from an observer during his free fall nto the BH will
    a) escape outwards with speed of light as long as the observer is located outside the EH
    b) fall in the singularity with speed of light when the observer is located inside the EH
    c) is frozen "at" the event horizont but of course keep moving with the speed of lightas it is a light ray if it is emitted exactly at the event horizon.
    So the EH is moving at the speed of light!

    An observer with non-zero rest mass therefore crosses the EH always exactly with speed of light. Therefore it is impossible for a physical observer to stay "at" EH and watch objects crossing it.
     
  4. Jun 24, 2010 #3
    A free falling observer passes the theoretical event horizon without special effect....tidal gravity increases a bit as the observer gets closer to the singularity, but to such an observer nothing special happens at the horizon.

    Tom's first paragraph describes a distant stationary observer.

    An observer just outside the event horizon, say accelerating in a spaceship to remain in place, or hanging on a long tether to a distant spaceship, observes the virtual particles as seen by a distant observer as thermal particles...and would be vaporized...burned....

    I did not know this and do not really understand....


    Is the book THE BLACK HOLE WAR????
     
    Last edited: Jun 24, 2010
  5. Jun 24, 2010 #4
    If you are into THE BLACK HOLE WAR Susskind discusses later in the book his conclusion that information is NOT lost in a black hole. His answer is YES, information is hidden (scrambled) in the radiation.

    He utilizes string wiggles as accounting for black hole information and evaporation according to work of Callahan and Maldacena...a string loop forms and can break off the event horizon...sort of like a bubble bursting (my analogy) and that's what's observed from the outside as a bit of radiation...information.

    Susskind also says of related work: "Whatever Witten and Maldacena had done, they had proved beyond any shadow of a doubt that information would never be lost behind a black hole horizon. The string theorists would understand this immediately; the relativists* would take longer..."

    * Likely the reference includes Hawking

    Also check here:
    https://www.physicsforums.com/showthread.php?p=2632279#post2632279
     
    Last edited: Jun 24, 2010
  6. Jun 24, 2010 #5

    tom.stoer

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    Your question
    regarding my statement "An observer with non-zero rest mass therefore crosses the EH always exactly with speed of light".

    The event horizon is a lightlike 2-manifold. All observers with non-zero rest mass cross light-like manifolds with the speed of light. This is nothing else but the postulate of constant speed of light for arbitrary reference frames.
     
  7. Jun 26, 2010 #6
    Tom; I still can't put your explanation into perspective...

    What does "arbitrary" reference mean?



    Are these all from the same reference frame...which one??....or any "arbitrary" one??
    Any discussions I could read oline about these ideas to better understand???



    I got this from another thread, I believe here on the forums:


    Here what Birrell and Davies, pages 268-269, has to say.

    It seemed to fit in with my general understanding that a free falling observer passed without any incident through a black hole horizon......absolute horizon, Cauchy Horizon, stretched horizon, and probably others I have never even heard about. Are you referenceing any one of these or perhaps another specific one??

    Thanks,,,,,
     
  8. Jun 26, 2010 #7

    tom.stoer

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    I just tried to explain what the EH really IS.

    First of all it is a purely mathematical entity where some coordinates are ill-defined. This can be cured by chosing appropriate coordinates, so one may think that everything is standard.

    Then one may think that there is nothing special physically at the horizon. Locally there is nothing special, an observer would not feel some special force, everything is fine for her.

    But there is something special: crossing the horizon means that there is no way to return; one can cross the horizon only in one direction. So there is something special at the horizon ...

    ... but this is not visible locally, but globally. One may think about the horizon as the 2-sphere consisting of all light rays unable to escape from the BH, but at the same time not doomed to get sucked into the BH. The EH consists of all light rays staying at "constant distance" from the BH. This is a geometrical idea; it is true even if there is no physical light ray.

    So crossing the horizon means crossing a sphere consisting of light rays. But crossing a sphere of light rays always means that the light rays moving at the speed of light. So the EH is moving at the speed of light.
     
  9. Jun 26, 2010 #8
    Tom, do I understand correctly that you are saying "the velocity of the EH with respect to the observer when it crosses the EH is the speed of light" ? (this velocity would obviously be the same as that of the observer w.r.t. the EH at crossing, so I rephrased the statement). I think the only misleading part is the velocity w.r.t. what/who. It is not a velocity w.r.t. an outside observer (who could never measure the crossing anyway)
     
  10. Jun 27, 2010 #9

    tom.stoer

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    You got it. The velocity of the EH w.r.t. any observer crossing the EH is just the speed of light.

    I tried to explain why this is physically reasonable. The EH consists of light rays not escaping to infinitiy and not falling into the singularity. Nevertheless it consists of light rays which means that it is a light-like 2-surface. But then an observer will always cross the EH with the speed of light (an observer will measure always the same speed of light, regardless where, and when and under which specific circumstances the experiment is set up. As the EH is just a special set of light rays it will move with sthe speed of light w.r.t. to the observer)

    Of course it is difficult (if not impossible) to talk about the speed of an object which is far away from the observer. I think one can no longer unambiguously define a velocity in the global sense in GR. It is for example possible to have objects moving faster than the speed of light which we can nevertheless see. They appear to move faster than the speed of light simply because our local definition of velocity is longer applicable. Locally everything is fine.

    Perhaps the following paper is interesting

    http://arxiv.org/abs/astro-ph/0310808
    Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe
    Tamara M. Davis, Charles H. Lineweaver
    (Submitted on 28 Oct 2003 (v1), last revised 13 Nov 2003 (this version, v2))
    Abstract: We use standard general relativity to illustrate and clarify several common misconceptions about the expansion of the Universe. To show the abundance of these misconceptions we cite numerous misleading, or easily misinterpreted, statements in the literature. In the context of the new standard Lambda-CDM cosmology we point out confusions regarding the particle horizon, the event horizon, the ``observable universe'' and the Hubble sphere (distance at which recession velocity = c). We show that we can observe galaxies that have, and always have had, recession velocities greater than the speed of light. We explain why this does not violate special relativity and we link these concepts to observational tests. Attempts to restrict recession velocities to less than the speed of light require a special relativistic interpretation of cosmological redshifts. We analyze apparent magnitudes of supernovae and observationally rule out the special relativistic Doppler interpretation of cosmological redshifts at a confidence level of 23 sigma.
     
  11. Jun 28, 2010 #10
    Humano posted:

    clear explanation, Thanks!!!! sure, an observer always measures light at "c" locally. horizons do not violate this principle.

    In fact it even seems analogous to the cosmological horizon which recedes from us a "c"....I guess assuming overall flat space....

    Leonard Susskind points out (Black Hole War, page 38):


    pg 257-259:
    so right now I am thinking this thread discussion is observer dependent, just like two observers in relative motion trying to measure a distant object...in general, they'll not agree....

    and Tom seems to confirm that here
    thanks guys.
     
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