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Possible method to escape a black hole?

  1. Dec 20, 2005 #1

    ACG

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    What's wrong with the following method of escaping from inside an event horizon (and possibly exploring physics inside it?)

    1. A rocket with a very strong engine slips below the event horizon -- just barely.

    2. Fire the rocket engine so it would travel at c-epsilon were it not for the hole. Since you are inside the horizon, you will slowly start to fall in. But you can make it arbitrarily slow.

    3. The black hole's radius will decrease over time due to Hawking radiation. If the rocket falls inward more slowly than the event horizon, suddenly it will wind up outside the hole as the horizon moves back over it.

    4. The rocket escapes with its observations.

    Ignoring the fact that such an escape would be impractical, where does it go wrong when it comes to special relativity? My guess is that the event horizon can never retreat inward faster than the falling object, but who knows...

    Thanks in advance,

    ACG
     
  2. jcsd
  3. Dec 21, 2005 #2

    Garth

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    Inside the event horizon the space dimensions become time-like and the time dimension becomes space-like.

    Inside the event horizon you can no more prevent yourself from falling into the singularity - with exceeding haste - any more than outside the event horizon can you prevent yourself going forward in time.

    Garth
     
  4. Dec 21, 2005 #3

    pervect

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    Here's the problem - you can't make it arbitrarily slow. Your velocity when you cross the event horizon will always be exactly 'c', as judged by a local observer who is "holding station" at an infinitesimal distance above the event horizon.

    This was discussed in a thread a long while back, the supporting math is at

    https://www.physicsforums.com/showthread.php?p=621802#post621802
     
  5. Dec 21, 2005 #4
    I dont see a reason why one cant escape from a black hole as described by ACG. The escape velocity is greater then c but that is not a problem. Escape velocity refers to projectiles which have been launched and then no extra force is applied after the initial burst. This is not whats happening in our case. The scenario described has the rocket constantly acceleating and thus it will escape. Infact, a from the point of view of someone who's in a rocket which crosses the event horizon, nothing 'special' happens when they cross it. What's intersting is what an external observer observes of all this.
     
  6. Dec 21, 2005 #5

    ACG

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    I don't think the rocket's acceleration is going to help -- it's still going to be going downwards because it would have to be moving at c to stay put at the event horizon. It can burn the engine as long as it wants but it will still fall downwards.

    The scenario I have is based on the fact that black holes' radii can decrease -- something which has nothing to do the with the rocket or its engine. If the radius decreases at a rate (m/s) larger than the rocket's "fall in" speed, the event horizon will pass back over the rocket and the rocket would be able to escape. In effect, you'd have a case where the force of gravity DECREASES as you get closer to the hole (r decreases for the GmM/r^2, but if M decreases faster than r (Hawking radiation) then conceivably gravity can decrease despite the lessening r)

    There's also the arguments made by the first two posters.

    ACG
     
  7. Dec 21, 2005 #6

    JesseM

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    As Garth said, once you pass the event horizon, the radial spatial dimension of the black hole as seen by outsiders becomes your time dimension, with the singularity lying in your future and the event horizon in your past. Based on this, I'd guess that escaping back out the event horizon would probably be equivalent to travelling backwards in time.
     
  8. Dec 21, 2005 #7

    pervect

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    A rocket ship, no matter how much it accelerates, is never going to be able to catch up to a photon starting at the same point.

    If the rocketship ever drops below the horizon (the point where even an outgoing photon strikes the singularity rather than escaping) then the rocketship is doomed (it will hit the singularity before the photon).

    Because the black hole is shrinking, the actual horizon will be slightly inside the ideal position of R=2M (for an ideal non-rotating Scwarzschild black hole in geometric units).

    Thus the rocketship might be able to reach R=2M depending on how/when M was measured. But if the rocket is able to escape that location, a photon must also be able to escape, and that location must be outside the actual postion of the event horizon horizon for that evaporating black hole.

    Basically, the rocketship has gotten very, very, very close to the event horizon, but not quite crossed beyond it.

    The horizon is defined as the region at which a photon can't escape. If a photon can't escape, the rocket can't either.
     
  9. Dec 22, 2005 #8

    robphy

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    In this discussion, it might be helpful to refer to this spacetime diagram
    https://www.physicsforums.com/attachment.php?attachmentid=3320&d=1114623912
    from my post in this earlier thread Black Hole XVI.

    I'm not that well-versed in black hole physics (and its dynamics)... but it seems to me what needs to happen to escape is that some of the light cones of events in the chronological future of the escapee tip back over enough to let him out. A shrinking horizon might not be sufficient for escape if the light cones continue to tip inward.
     
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