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There's something which I don't quite get about black holes and event horizons (feel free to tell me if I got something completely backwards at any step along the way):

Say we have a singularity and we launch a particle from the singularity's Schwarzchild radius R at the speed of light directed outwards from the black hole - wouldn't that particle be able to escape beyond R, slowly decelerating in a parabolic path to a halt somewhere outside it (at a radius of, say, R1)?

If so - wouldn't an observer A between R and R1 be able to see that particle as it decelerates? And if so, we couldn't define the event horizon to be at R, because light from R can affect A who's outside it. We couldn't define the event horizon to be at R1 either, because light from the observer could escape R1 entirely.

So where would we define the event horizon in the case presented above? How can there be an event horizon if light can be launched slightly below it and take an parabolic path outside it, even if only for a moment?

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marked as duplicate by John Rennie, Brandon Enright, Dan, Qmechanic Feb 3 at 0:54

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The thing is, once you're on the horizon, there does not exist an "outward" direction anymore. –  Raskolnikov Feb 1 at 22:14
    
Isn't escape velocity defined in the outward direction? Care to expand on that a little? –  saarraz1 Feb 1 at 22:21
    
Escape velocity is really a Newtonian concept. The event horizon is not defined in terms of escape velocity, but rather as the boundary of a region beyond which the outside world can not be affected anymore. Another way to express this is by looking at what happens with light cones in the geometry of a black hole. As you get closer to the whole, the future light cone tips towards the black hole. On the event horizon, it lies completely inside the boundary. –  Raskolnikov Feb 1 at 22:23
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Read up on lightcones a bit first, the last picture shows what happens in the black hole case. If that doesn't answer your question yet, I'll try to write a full reply. –  Raskolnikov Feb 1 at 22:33
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possible duplicate of Why is a black hole black? –  John Rennie Feb 2 at 9:49

1 Answer 1

It is an interesting point you are making. I think if you can launch light or a particle moving at the speed of light from 'just' below the event horizon (in the order of a Planck length), then you are correct, it will have a chance to escape. However, the only question is how do you launch light from just under the event horizon? The laws of physics inside the black hole are completely different than the laws of physics in the universe, and it seems that it would be impossible to have a man made launching mechanism inside the black hole - as any matter or electromagnetic radiation inside would collapse instantly into the uncertain lump that composes the interior of the black hole. It is similar to saying that if we can accurately measure the position and momentum of a particle simultaneously to infinite precision, then we can make a device that would violate the second law of thermodynamics. Or if we can travel faster than light, then we can go back in time.

In a classical sense the escape velocity from a black hole would exceed the speed of light, so nothing, including light can escape the black hole.

However, in light of your question, there is an exception in which radiation does escape the black hole. It involves quantum effects near the event horizon. It is called Hawking radiation. Please check the following link. I hope you find it interesting.

http://en.m.wikipedia.org/wiki/Hawking_radiation

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