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Event horizon of a Black Hole

  1. Sep 21, 2007 #1
    Hello,

    I's a spead though that black holes swallow everything in it's vicinity.
    But, something bothers me : when we close on the event horizon of a black hole from the outside, the gravitational field tends to be infinite, and at the surface of the horizon it is infinte. So to say, that spacetime warp when closing to the horizon is in such a way that the space in infinitely compressed (from the point of view of an external viewer, it might be say 1cm, but in the facbric near the horizon, il might be billion of centimeters), and the time is dialed.
    So theoriticall, an object closing to the horizon will never read it (from an external point of view).
    Even light would never reach it.

    Is that coherent ?

    -----------------------------------------------------
    Correct me if I am wrong.
    http://ghazi.bousselmi.googlepages.com/présentation2
     
  2. jcsd
  3. Sep 21, 2007 #2
  4. Sep 21, 2007 #3

    mgb_phys

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    The gravitational field at the event horizon isn't infinite.
    Since the event horizon grows with 'r' but gravitational field goes as 'r^2' as you get larger holes the field at the 'surface' gets lower.
     
  5. Sep 21, 2007 #4

    pervect

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    It depends on what you mean by "the gravitational field at the event horizon". The proper acceleration required to "hold station" at given value of the Schwarzschild r coordiante does approach infinity as r approaches the event horizon at 2M.

    See for instance this wikipedia reference for the exact formula. [add]This is also mentioned in Wald, if you want a textbook reference.

    However, tidal forces on someone free-falling through the event horizon of a black hole are finite. (This is discussed for instance in MTW's "Gravitation", I've posted the reference in the past. Since most of the people who are asking questions on these forums don't own this book, though, I won't bother looking it up again unless there is interest and the interested person can't find where this has been discussed before.)
     
    Last edited: Sep 21, 2007
  6. Sep 21, 2007 #5
    You are confusing me, but after re-reading some articles, it seems that the gravitational field at the event horizon of a black hole is indeed infinite.
    In the Schwarzschild solution, you would not have a blackhole if its radius is not less than the radius of Schwarzschild, and thus, at the event horizon (radius > Schwarzschild radius), the gravitational field is infinite (or rather say, the warp of space time is infinite).

    I reiterate my question then : it seems that no object or light can EVER reach the event horizon yet to go through ??

    -----------------------------------------------------
    Correct me if I am wrong.
    http://ghazi.bousselmi.googlepages.com/présentation2
     
    Last edited: Sep 21, 2007
  7. Sep 21, 2007 #6

    pervect

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    A free-falling object can reach and pass through the event horizon in a finite amount of proper time.

    See for instance this sci.physics.faq which I will quote in part:

    In case you're not familiar with the jargon, you can think of "proper time" as the amount of time someone with a wristwatch would measure. So if you were to jump into a black hole, you would reach it in a finite amount of time by your wristwatch.
     
    Last edited: Sep 21, 2007
  8. Sep 21, 2007 #7
    May be i'm too sleepy, but i find the information here http://www.math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html rather unsatisfying and evasive.

    * "On my worldline as I fall into the black hole, it turns out that the Schwarzschild coordinate called t goes to infinity when I go through the event horizon."

    * "So if you, watching from a safe distance, attempt to witness my fall into the hole, you'll see me fall more and more slowly as the light delay increases. You'll never see me actually get to the event horizon. My watch, to you, will tick more and more slowly, but will never reach the time that I see as I fall into the black hole. Notice that this is really an optical effect caused by the paths of the light rays."

    Explicitly and implicitly in the upper statements, the writer states that the time goes to infinity when approching the event horizon, in other words, time is dialated (compared to external obervers), the proper time of a falling object (into the hole) is slower and slower compared to the proper time of an external obs, if you excuse the language abuse.


    * "Light beams aimed directly outward from just outside the horizon don't escape to large distances until late values of t."
    If the time he is speaking of is the proper time of the falling object, this statement combined with the constancy of the speed of light, we can infer that the writer agrees that there is a spacial compression (in the spacetime of the object).
    If the time he is speaking of is the proper time of an external observer, this statement combined with the constancy of the speed of light, we can infer that the writer agrees that there is a temporal dialation (in the spacetime of the object).



    Now all this agrees with the upper issue : "it would take an infinite time (proper time of an external observer [edited]) for an object to reach the event horizon".


    -----------------------------------------------------
    Correct me if I am wrong.
    http://ghazi.bousselmi.googlepages.com/présentation2
     
    Last edited: Sep 21, 2007
  9. Sep 21, 2007 #8
    In-falling objects will experience a finite amount of proper time (as measured by themselves) before reaching the singularity at the centre. Furthermore, they cannot tell when they fall past the event horizon, as there is nothing to distinguish it locally.

    An observer parked infinitely far away will see the in-falling object approach the event-horizon, and the radiation from it will become red-shifted until detection is difficult. From the observer's point-of-view, nothing ever crosses the event-horizon.

    Coordinates in GR almost always mean nothing, so things like the Schwarzchild coordinate t going to infinity at the event horizon just means the Schwarzchild coordinate system is a bad one. Nothing relevant about the physics can be inferred from the coordinate problems.

    Now, what's the problem?
     
  10. Sep 21, 2007 #9

    pervect

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    Huh? You'd better read it again, because it says explicitly, as I already quoted, that the proper time for the infalling observer is finite.
     
  11. Sep 21, 2007 #10
    GR Coordinates mean nothing ?
    And the event horizon is distinguishable, at least to exterval observers as the gravitational field is infinite.

    My problem is just with that singularity at the event horizon.
    Put simply, for a sperical non charged and non rotating black hole, the grvitational field is infinite at the event horizon (Schwarzchild solution to the gravitational field). From this, i was wondering if an object would ever be able to cross this horizon, since its time is infinitely dialated and its space is infinitely compressed (compared to external viewers).
    And by "never reach it", i'm not speaking of the fact that the external cannot view it becoz light would not get back to him : just the fact that it would take an infinite time (in the sense of the faraway observer).


    PS:
    yes, right, i edited the post.


    -----------------------------------------------------
    Correct me if I am wrong.
    http://ghazi.bousselmi.googlepages.com/présentation2
     
    Last edited: Sep 21, 2007
  12. Sep 21, 2007 #11

    pervect

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    The time observed by an external observer is not a proper time. If you draw a space-time diagram, any proper time can be regarded as the "length" of some particular curve on the diagram. The time computed by an external observer is not such a quantity, it is a coordinate time interval computed by subtracting a pair of time coordinates. The "proper time" is the time observed by someone present at both events, it's a time measured along his worldline. The time computed by an external observer is not such a proper time, because he is not present at one of the events, it is a distant event that happens far away.

    Note that an observer on a constantly and uniformly accelerating spaceship sees an event horizon that is mathematically and formally very similar in structure to that of a black hole. This horizon is called the Rindler horizon. On such a spaceship [add] with an acceeration of 1g, events more than 1 light year away are "below the horizon", as long as the ship keeps accelerating. But we don't argue that such events that occur "below the horizon" of the accelerating spaceship do not exist, or "never happen".

    Similarly we should not argue that the interior region of a black holes do not exist, just because an external observer can never see them.

    Note that the infinite time dilation of a static observer near a black hole is misleading, because an observer who falls into a black hole is not a static observer. In fact, the velocity of an observer who falls into a black hole approaches the speed of light relative to a static observer. If you properly take this into account, the infinity on the numerator cancels out the infinity on the denominator :-).

    If you don't like infinities, you have to use some coordinate system that is well behaved at the event horizon of a black hole, and avoid Schwarzschild coordinates. The physics of a black hole is not singular at the event horizon - but the Schwarzschild coordinates are poorly behaved there.

    Also, while the gravity as defined by Newton becomes infinite as one appraoches a black hole, the physically important Riemann tensor which can be regarded as one of the GR equivalents to the "gravitational field gradient" remains finite.
     
    Last edited: Sep 21, 2007
  13. Sep 21, 2007 #12
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