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Is anything falling into a black hole *now*?

  1. Jun 6, 2012 #1


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    Hi. Wikipedia says, that for the external observer, time stops at the event horizon.

    Is then, for the external observer, anything falling below the event horizon? Or rather, every piece of matter stops before the horizon?
  2. jcsd
  3. Jun 6, 2012 #2
    I was just wondering the same thing myself.
  4. Jun 6, 2012 #3


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    Your emphasis on *now* suggests that you haven't wrapped your head around the relativity of simultaneity.

    In any case, if I jump into a black hole and my watch reads 12:00 the moment I cross the event horizon, then no matter how long you wait, you will never see through your telescope that reading on my watch... unless you jump into the black hole after me.
  5. Jun 6, 2012 #4
    Can you point me to the page? It's wrong and needs to be changed.
  6. Jun 6, 2012 #5


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    By "now" I mean "now for any observer around, that is reasonably far from the event horizon, and for whom the black hole's center is mostly stationary".

    So if you jump into a black hole, then you are seen a bit redder. After a year, you produce a very long wave radiation only, and your watch shows 11:59:00 to the surroundings. After the next 1000 years, your watch shows 11:59:59. Billions of years from now, and you are still very faintly visible to anyone around, your clock showing faint 11:59:59.99.

    So, in effect, for most of folks in the galaxy around, you will never fall into the black hole in the physically foreseable future, whatever it would mean? And, for the same folks, the black hole, once formed, stops to consume anything, but only produces a shell around it of anything that falls into it?

  7. Jun 6, 2012 #6


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    Oh well, so it is already answered in another thread ?t=463880, I do not post a full link as I am not allowed to do that by this forum.

    Someone claims there that it is not even decidable if the black hole forms at all for the external observer.
  8. Jun 6, 2012 #7
    Why is this so bad:

    I've seen essentially that one liner many places. Maybe "appears to stop" would be better ...but it is what a distant observer observers.

    The best physical interpretation is that, although we can never actually see someone fall through the event horizon (due to the infinite redshift), he really does. As the free-falling observer passes across the event horizon, any inward directed photons emitted by him continue inward toward the center of the black hole. Any outward directed photons emitted by him at the instant he passes across the event horizon are forever frozen there. So, the outside observer cannot detect any of these photons, whether directed inward or outward.

    There's no coordinate-independent way to define the time dilation at various distances from the horizon—a clock is ticking relative to coordinate time, so even if that rate approaches zero in Schwarzschild coordinates which are the most common ones to use for a nonrotating black hole, in a different coordinate system like Kruskal-Szekeres coordinates it wouldn't approach zero at the horizon. The Schwarzschild singularity at r = 2M (the event horizon) is a coordinate singularity, not a real, physical singularity. So a free falling observer passes right through that spherical surface without even being able to observe it.

    Kip Thorne says (Lecture in 1993 Warping Spacetime, at Stephan Hawking's 60th birthday celebration, Cambridge, England,)

  9. Jun 7, 2012 #8


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    There is no "outside". The emitter could define something which could be described as "outside", but it leads towards the center, too.

    However, this becomes more interesting together with a finite lifetime of black holes due to Hawking radiation.
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