Escaping a black hole

  • I often hear that nothing can escape a black hole because its "escape velocity" is greater than c. If that is accurate, what about the following? I know that the following has a lot of most likely impossible assumptions, but I'm trying to understand whether the escape velocity explanation makes sense.


    Suppose you've got some sort of space ship that's inside the event horizon, and it hasn't been destroyed by the gravity or tidal forces. Also suppose that you've got a whole lot of reaction mass and a fantastic power supply.


    Could this space ship, through applying enough constant force escape the event horizon?


  • Stan Liou

    Stan Liou Correct answer

    9 years ago

    One of the clearest way to see what's going on is to look at a Penrose diagram of a Schwarzschild black hole. A Penrose diagram is like a map of the spacetime drawn in such a way as to preserve angles and put the light rays diagonally at $45^\circ$ angles, forming the light cones.



    Since we're mapping all of an infinite space(time) into a finite drawing, distances are necessarily distorted, but that's a small price to pay.



    Schwarzschild causal diagram, original by Andrew Hamilton



    Time goes up on the diagram, and a typical infalling trajectory is in blue.



    Because every massive object must be locally slower than light, it must stay within those light cones. Hence, no matter how you accelerate, at every point of your trajectory you must go in a direction that stays within those $45^\circ$ diagonals at that point. But once you're inside the horizon, every direction that stays within the light cones leads to the singularity.



    Accelerating this way or that just means that you get to choose where you hit the singularity--a little on the left on the diagram or a little to the right. Trying to escape using high acceleration brings you closer to the light's $45^\circ$ lines, which because of time dilation will actually shorten your lifespan. You'll actually get to the singularity sooner if you struggle in that manner.



    This particular image came from Prof. Andrew Hamilton. Note that it pictures an eternal Schwarzschild black hole, i.e. one that has always and will exist. An actual black hole is formed by stellar collapse and will eventually evaporate, so its diagram will be slightly different (in particular, there will be no "antihorizon"). However, in all respects relevant here, it's the same situation.


    Could you elaborate on how it's "the same situation" despite the black hole eventually evaporating, given that the singularity itself seems to actually reside an infinite time in the future, according to the diagram? (or at least according one understandable, though potentially naive, interpretation of the diagram)

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Content dated before 7/24/2021 11:53 AM

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