What happens within an extended object passing the event horizon of a black hole? [duplicate]

Question

I hope this question is not too infantile: Falling into a Black Hole (intentionally or by accident) and thereby passing the Schwarzschild radius is very often said to be "nothing special" in the view of the falling observer. Tidal forces can be neglected when the mass is only big enough. But let’s say you are in a starship which is 100m long. There is a point in time, where the head of the rocket is inside the event horizon and the tail outside. Then, it cannot be possible to switch on a light on the tail by moving a switch on the head, because this would mean, that a signal is sent out of the Schwarzschild radius. Same for communicating to the tail or even controlling the movement of your legs (outside) from the brain (inside).

I cannot imagine that this absurd situation is indeed the case - what is wrong with this funny view? Is inside and outside not defined in the view of the observer?

Answer 1

The big thing to realize here is that the "nothing special" caveat really has a "time" element to it. All of those arguments require that the spatial extent of the object is smaller than the radius of curvature of the local spacetime, AND that any experiment that is conducted is such that $ct$ is much less than the local radius of curvature.

So, for this question to make sense, you have to think a bit about what "turning on the light in the head", and "observing the light hitting the back of the spaceship" means. Look at the kruskal spacetime diagram for Schwarzschild:

Ignore the regions labeled III and IV, which don't relate to us. I is the exterior of the spacetime, II is the black hole. The upward direction is time. The key thing to see is that, after the front of the ship is in the horizon, the rest of the ship must move upward through the diagram if the ship is to stay together. Since the light ray must stay in the ship, and it must travel parallel to the dotted line, this means that the ship must fall into the black hole in such a way that, by the time the light beam has hit the back of the ship, the back of the ship is inside the black hole (since the ship would be a fixed shape with a fixed orientation, moving upward through this diagram). The only other option is for the ship to break apart, which then would make it completely reasonable that they don't observe the light otuside the black hole.

Answer 2

I find that the most intuitive approach to a wide variety of questions in this vein is to invoke the Kruskal description and realize that the dynamics at the event Horizon, locally, are identical to those in special relativity at any lightlike surface. That is, if one zooms in on the diagonal line of the event horizon at the point of crossing, then you may as well be looking at the same diagram in 2d SR. This is the sense in which crossing the event horizon is "nothing special". Light cannot escape the event horizon in exactly the same way that two light rays going in the same direction (being parallel lines) cannot cross in SR. Notice that it is the case for any lightlike surface in SR (i.e. a 45 degree line in some inertial coordinate system) that, once one crosses it, they will never be able to do so again: this property of an event horizon, then, is not nearly as unique or mysterious a thing as it is often presented to be. Taking this approach allows us to ground the problem in our everyday experience without needing to imagine that these are exotic phenomena.

So, what happens when an extended object crosses a lightlike surface in SR? Of course, we ourselves are doing this constantly and continuously (as spacetime is foliated by lightlike surfaces), and it doesn't seem to impose any constraint on the ability of our brain to transmit signals to the rest of our body or our light switches to turn on lights across the room. The resolution is as pointed out in other answers: if our brain emits a signal to wiggle our toe at an instant (according to some fixed inertial observer: remember that simultaneity isn't meaningful in relativity) when our extremities are on opposite sides of some lightlike surface, it is simply that by the time our toe receives the signal, it must have crossed to the other side of the surface as well.

Could we concoct a scenario in which our toe never crosses the surface, and so cannot receive the signal to wiggle? In principle, yes, provided we have infinite energy at our disposal. Doing so requires that we accelerate asymptotically to the speed of light sufficiently quickly. As one does this, I expect that David is correct that we would necessarily be torn apart, as eventually the force carriers could only transmit one direction (from toe to head, but not head to toe). This does not occur simply by virtue of crossing the lightlike surface / event horizon, though-- it occurs when we attempt to accelerate to the speed of light so as to straddle the surface / horizon indefinitely.

Answer 3

I'm inclined to say that that your "funny" view is correct. It is impossible to transmit information out of a black hole if the information originates inside the event horizon. You can't turn on a light on the tail by sending a signal from the head, because what will carry the signal? Light cannot escape the event horizon. Neither can electrons, which would carry electric currents. Neither can you send a signal from your brain to your legs, again because your body's molecules cannot escape the event horizon.

Of course, all known materials would be ripped apart if they straddled an event horizon. Atoms are held together by forces between the nucleus and electron, and photons of light carry the force between them. If the light cannot travel from the nucleus to the electron or vice vearsa because that atom straddles the event horizon, the atom itself would be ripped apart.