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This goes far beyond my limited knowledge regarding the physics of black holes, but my colleague told me this today:

I doubt that if I stand on the surface of a black hole - inside the Schwarzschild radius - and a stone falls on my head from outside the Schwarzschild radius, it cannot come out again. Then I just have to throw it back up with the energy it fell on my head so that it leaves the black hole. The stone can accept this energy because he fell on my head with it. Otherwise, the physical laws of gravity no longer apply. Just try to let a stone vibrate in a gravitational field whose generating mass has no expansion and idealizes does not collide with the vibrating body. It does not get stuck in the point x=0, but flies out again with infinite energy against an infinite acceleration. And this in clearly finite time.

If you have managed this calculation, then the question arises, where the energy should go, if the stone can fall in, but not fall out again? Energy is simply the time integral of the acceleration equation after multiplying both sides by the corresponding velocity equation. Since due to the existence of the acceleration equation also the velocity equation exists, the energy a body has in a force field is fixed for every movement. And this energy is on the way there, the same as on the way back, because it depends only on the place and not on the time. Even if time is deformed and the path is compressed, energy and place remain firmly connected. So the stone comes out of the black hole.

I wanted to share this with the community in order to get a clearer perspective to his statement. Are his assumptions correct? If so is there proof and maybe further literature?

(As far as I remember, the jets coming from super massive black holes do not contain anything from beyond the Schwarzschild radius.)

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    $\begingroup$ There is no surface on which to stand inside the event horizon. You will already be well on your way to the singularity. $\endgroup$ – Triatticus May 20 at 15:55
  • $\begingroup$ Hawking radiation can escape. $\endgroup$ – Qmechanic May 20 at 16:20
  • $\begingroup$ @Qmechanic Sort of. Hawking radiation isn't emitted from inside the event horizon. In fact, the location of Hawking radiation is rather delocalised. $\endgroup$ – PM 2Ring May 20 at 16:24
  • $\begingroup$ @Qmechanic There is no Hawking radiation. If I hover and see you killed in a free fall by a high energy Hawking photon, you cannot be alive at the same time just because this photon is not there in your frame. $\endgroup$ – safesphere May 21 at 7:16
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I don't know what surface of black hole means, but you cannot stand on the fixed radius from centre of gravitational field (center of black hole) without having superluminal speed.

The problem is that inside event horison the spacetime becomes dynamic - you can imagine space itself is collapsing and it is collapsing quickly enaugh to prevent anything that does not have superluminal speed from escaping. Yes you can throw the object away from the center, but because the space itself collapses to the center the net effect is that the object will always move to the center no matter how much energy you provide

The next problem is that there is no energy conservation in GR in the sense it is in newtonian mechanics. The energy is conserved only locally, not globally. The energy is not even defined globally - simply because due to equivalence principle you cannot define energy of gravitational field itself. In local inertial frame there is no gravitational field. The supposed energy accumulated by falling in gravitational field is not due to the fall of object, it is due to the acceleration needed to remain still in gravitational field.

The energy of an object is simply projection of objects 4-velocity onto your own 4-velocity. The objects 4-velocity remains "the same" (is parallel transported) during the fall, but your own is accelerated and thus is changing in any local inertial frame around you. The percieved kinetic energy of the object is not increasing in the free fall due to objects motion, but due to your own acceleration. Sometimes this can be mathematically reformulated so that you can associate changes of kinetic energy with freely falling object itself like in newtonian limit, but not in general.

This is all very sloppy of course, because i assume you don't know much about mathematic that is needed to formulate the ideas precisely. If you want to learn it though, i would recommend Gravitation by Misner, Wheeler and Thorne. Its huge book, but you don't need to read everything and all the ideas are explained very inuitively as well as mathematically

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    $\begingroup$ The next problem is that there is no energy conservation in GR in the sense it is in newtonian mechanics. The energy is conserved only locally, not globally. This is true but not relevant here. For a Schwarzschild black hole there is a conserved energy for test particles, and that's all that matters in the context of the OP's question. $\endgroup$ – Ben Crowell May 20 at 16:46
  • $\begingroup$ Thank you very much @Umaxo, this provided the insight I wanted to gather. Thank you also for recommendation of Literature. $\endgroup$ – Andreas Schwab May 21 at 6:56
  • $\begingroup$ @safesphere "This is incorrect" i dont think it is. One look at penrose diagram shows that for any event inside the the BH there exists spacelike path that takes you from inside the schwraschild radius to outside. "No, you cannot do this just like you cannot throw an object into the past" depends on what it means away from the center. In my example i imagined foliation of spacetime into spacelike hypersurfaces,f.e. using coordinates of penrose diagram. On this hypersurface the direction away from the black hole is spacelike and you can throw the ball with 3-velocity in this direction $\endgroup$ – Umaxo May 21 at 8:24
  • $\begingroup$ I can't quickly find the expression for what happens to radial spacelike geodesics. I need to do the math and give this more thought, so meanwhile I've deleted my comment. I think you are correct that with a superluminal speed one can escape. Although to do so, he would need to move parallel to the singularity, because this is the direction of spacelike geodesics. Then he gets out after reaching the infinite distance along the singularity. However the radial direction away from the singularity is timelike, so I still disagree with your statement that you can throw away from the center. $\endgroup$ – safesphere May 21 at 10:03
  • $\begingroup$ @safesphere do you know about penrose diagram? It is really obvious, no need to do the math (the math was done in its construction). And how do you find wheter it is parallel to singularity? Is the notion of prallelsim between two 4-vectors at different locations not dependent on the path by which you parallel transport them? Also, what direction does singularity have, since singularity is not part of the manifold? Can it be defined by some limit? $\endgroup$ – Umaxo May 21 at 10:26
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This exceprt doesn't quite get what a black hole is. Once you're inside the Schwarzschild radius, it's not that "you can't go fast enough too get out", it's that the idea of "going outside of the black hole" is a concept that doesn't make sense. Once you are inside, all timelike paths (which are the allowable physical paths of objects that move from the past into the future) end in the black hole singularity.

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    $\begingroup$ This is a nice answer, but could be improved by explaining to the OP the misconceptions involved in the idea that they would "stand on the surface of a black hole - inside the Schwarzschild radius." Misconceptions: that there is a solid surface; that the solid surface is inside the Schwarzschild radius;(and that you can stand there, but you've addressed that). $\endgroup$ – Ben Crowell May 20 at 16:49

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