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

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?

• Commented Aug 12, 2021 at 6:58
• There is a point in time, where the head of the rocket is inside the event horizon and the tail outside.” - This is incorrect. In the frame of the tail, the head does not cross until the tail does. So they cross at the same proper time and at the same coordinate time $r=r_s$ (starting from the horizon, $r$ is a time coordinate). See Kevin Brown: “Everything that falls into a black hole falls in at the same local time, although spatially separated, just as everything in our city is going to enter tomorrow at the same time” - mathpages.com/rr/s7-03/7-03.htm Commented Aug 13, 2021 at 5:56

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.

• Suppose the guy touches the switch when the horizon is at the middle of the ship. The signal has to reach the back of the ship after or at the same time than the back of the ship reaches the horizon. Would not this imply faster than light travel by the back of the ship? assume the black hole is huge and space looks flat at the scale of the ship
– user65081
Commented Aug 11, 2021 at 20:09
• @Wolphramjonny: the key to the riddle is basically "inside the horizon, moving closer to the singularity IS moving forward in time. If the ship is to stay together, any "outbound" signal from inside the horizon cannot reach the end of the ship until the end of the ship is inside the horizon. And the why is as simple as drawing three line segments (representing the ship at different times) on the diagram above, each of them intersecting the horizon, and seeing that there is no way to draw a 45 degree line that goes from the inside to the outside. Commented Aug 11, 2021 at 21:26
• Thanks for your answer, I dont doubt you are right, but I am still new to these diagrams. Is it the answer to my question a yes or a no?
– user65081
Commented Aug 12, 2021 at 4:50
• @Wolphramjonny The answer to your question is "no", and I'd say that the fundamental issue is in imagining the event horizon as being some point in space, rather than a surface in spacetime. What one even means by the "the guy touches the switch when the horizon is at the middle of the ship" is ambiguous: the simultaneity of these events is frame-dependent. If one locally fixes the notion of simultaneity by working in the rest frame of the ship, the back of the ship does not move in space at all, but it has still crossed the event horizon by the time the light has reached it. Commented Aug 12, 2021 at 8:18
• @Wolphramjonny “the horizon is at the middle of the ship” - This is impossible. See my comment above for details. Commented Aug 13, 2021 at 6:05

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.

• Hello old friend :) Your answer is misleading on too many levels to list in one comment. “Locally at the horizon”? The horizon is global. “Any lightlike surface”? It is clear on the Penrise diagram that the horizon is a null infinity, which no $r=const$ line crosses, unlike any other lightlike surface. Crossing the horizon is equivalent to crossing your own future light cone, not just any lightlike surface. You cannot place the horizon in the middle of the ship. And so on. Hope you are doing well :) Commented Aug 13, 2021 at 6:31
• @safesphere Haha, hello there, safesphere. Suffice it to say we still vehemently disagree on many points. My answer is entirely in line with standard GR formalism. Notice I've taken the validity of the Kruskal coordinate system for granted, which I know you take issue with. If one accepts this, my claims are quite straightforward. The null infinities have a definition that the horizon doesn't satisfy-- it's not a matter of looking similar on a Penrose diagram. Commented Aug 13, 2021 at 13:10
• Good to hear from you Sir! Ah, yes, the "standard" unphysical formalism :) Diagrams aside, the horizon is a null infinity by its physical meaning. Can you please clarify what definition does it not satisfy? (I suspect this definition is artificially exclusive). Thanks! Commented Aug 13, 2021 at 15:03
• @safesphere Nice to hear from you as well. See page 16 here: arxiv.org/pdf/0811.0354v1.pdf . The definition isn't artificially exclusive-- it's simply designed to capture what the concept is meant to describe, points infinitely far away along null geodesics. You're welcome to maintain that standard GR is unphysical, but you should be honest in your comments that you do not take the mainstream view. Not being up front only sows confusion amongst readers who are expecting to learn about what mainstream GR says. Commented Aug 13, 2021 at 15:30
• @safesphere Again, we simply disagree. GR is the mathematical structure together with physical interpretation. Without standard physical interpretation, it's just semi-Riemannian geometry, not GR. "Mainstream" is a term you've used to describe it repeatedly. The $r$ coordinate is not arbitrary-- it can be characterized entirely geometrically, and this characterization is what makes the definition natural. Did you read that "$I^+$ is an idealization of far away observers who can receive radiation from the system"? Commented Aug 13, 2021 at 21:58

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.

• I thing with regard to ripping you are not right. Tidal forces are inevitably vey high as you move towards the singularity, but not necessarily at the event horizon. Commented Aug 11, 2021 at 19:42
• Of course, all known materials would be ripped apart if they straddled an event horizon. This directly contradicts descriptions of the happenings at event horizons like this books.google.de/… or this en.wikipedia.org/wiki/… or any other description I have ever seen.
– AnoE
Commented Aug 13, 2021 at 6:47