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Escaping a black hole

  1. May 3, 2009 #1
    I was listening to some GR lectures,and the description of the event horizon of a black hole was the surface at which the escape velocity becomes larger than c. (I'm paraphrasing)
    This seems like the event horizon would appear to get smaller the closer you got to it, since the velocity required to get to your location would drop below c, since there was not as much work required to arrive.
    Did I misunderstand the way it was described?
    Rockets do not start with the full escape velocity, they add velocity as they go. At some point, their V becomes larger then the escape velocity, at their altitude.
    Also, the escape velocity is larger than the orbital velocity.
    So, forgetting the whole time dilation thing for a minute, would it be possible to orbit through the event horizon at less than c?
    If so, having gone inside the event horizon and returning out, couldn't the orbiting object than emit a photon that would be able to escape, since it is now in a location for which escape V < c?
    Thanks,
    Spizmar
     
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  3. May 4, 2009 #2

    JesseM

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    "Escape velocity" is a concept drawn from Newtonian gravity, so I think your questions show the problems with explaining black holes in terms of this concept--in Newtonian physics it would presumably be true that you could escape from an object with an escape velocity of c using powered flight at a lower velocity, and that you could orbit at this height, but general relativity is a fundamentally different theory of gravity involving curved spacetime, so the same calculations don't apply here. One good way to think of why it's impossible to escape the event horizon is in terms of light cones--at the event horizon it works out that your entire future light cone lies at or inside the horizon, and in relativity the worldline of any sublight object passing through a point in spacetime is guaranteed to stay within the future light cone of that point. In this diagram drawn in Eddington-Finkelstein coordinates (from the textbook Gravitation by Misner/Thorne/Wheeler), you can see that future light cones become "tilted" as they approach the event horizon, so that at the event horizon the light cone's outer side is exactly parallel to the horizon:

    http://www.valdostamuseum.org/hamsmith/DFblackIn.gif [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. May 4, 2009 #3
    Thank you, but, ow, your hurting my head.
    OK, so the description of the event horizon I gave is wrong. It is not the surface for which the escape velocity is c, but the maximum closed surface for which all the future light cones, for all points on the surface, never cross the surface.
    I guess this is the same as saying the surface for which the orbital velocity is c.
    OK, this makes sense. The apparent horizon can change as you approach it, since you enter the FLC of points whose FLC you were not in, but the real horizon does not change.
    That is not true either. As long as the FLC diverges from the event horizon, it will end including everything away from the black hole. So you are in the FLC of everything not within the event horizon, but once inside the horizon, you are not even in your own FLC.
    How does that work? If I fall through the horizon, for a large enough BH, I am still in a valid frame of reference, and I should not be able to tell I have crossed it, but the FLC tells me I can't see my hand in front of my face, assuming I face the BH.
    Am I understanding that correctly?
    TIA,
    Spizmar
     
  5. May 4, 2009 #4

    JesseM

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    What do you mean by "apparent horizon"?
    I don't follow. Why would you be in the FLC of everything not in the event horizon when you're outside? For example, a point in spacetime right next to me as I type this does not lie in the future light cone of an event 1 year in the past of my typing this but 10 light-years distant from my current position. My worldline will eventually enter the future light cone of that event (assuming I live that long), if that's all you mean.

    And I don't understand what you mean when you say that inside the horizon "you are not even in you own FLC. For any event on your worldline, like your watch showing a time of 10 seconds, your future worldline (including events like your watch showing a time of 20 seconds) will lie in the future light cone of that event--this is just as true inside the horizon as outside. If you look at the curve in that diagram labeled "World line of infalling particle", you can see visually that the worldline always lies within all the future light cones drawn along it.
    No, you can still see your hand, why wouldn't you? If your hand is carrying a watch, then you'll see some time on that watch as soon as your eyes enter the future light cone of that watch-reading, just like it would work outside the black hole.

    It may be easier to think about in Kruskal-Szekeres coordinates, which have the nice property that the worldlines of light beams always travel at a 45 degree angle from the horizontal axis (and the vertical axis), both inside and outside the horizon. So if you want to see what's in the future light cone of any event, just draw two diagonal lines going up the page which converge on that event. The event horizon is also a pair of diagonal lines in these coordinates, which shows why any event inside the horizon will have a future light cone wholly within the horizon. But this diagram also shows more clearly that for an observer inside the horizon, the singularity is not a particular point in space, but rather a set of events lying in their future--the horizontal axis in the Eddington-Finkelstein coordinate diagram is actually a timelike dimension inside the horizon (the same would be true in Schwarzschild coordinates, where the radial coordinate is timelike inside the horizon), which may be what's confusing you, but the Kruskal-Szekeres diagram is more clear because in these coordinates the vertical axis is always timelike and the horizontal axis always spacelike. From this page, here's what the Kruskal-Szekeres diagram looks like:

    stk.gif

    The pink line represents the event horizon, with the region below the pink line being outside the horizon and the region above being inside it. The blue curve is the singularity. All those orange and yellow lines represent possible worldlines of light rays; an object moving slower than light would have a worldline that's always closer to the vertical axis than the 45-degree angles of the yellow and orange light rays. If you pick any event at the intersection of a yellow and orange light ray, the future light cone is the region lies between those two rays above the event.
     
    Last edited: May 4, 2009
  6. May 4, 2009 #5
    You will see your hand in front of your face. The light coming from your hand towards your face is actually moving towards the black hole more slowly than your face, so you catch up with it.

    Or something like that, maybe with an imaginary i here and there, but anyway you certainly will see your hand. If it hasn't already been ripped off by gravitational tidal effects, of course.
     
  7. May 4, 2009 #6

    DrGreg

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    As your hand crosses the event horizon, light starts to travel towards your eye (relative to your own frame), but by the time it reaches your eye, your eye has crossed the event horizon too. So the light never crosses the event horizon.
     
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