If you jump out of a plane a mile over the earth, you’ll freefall until you hit the surface. If you don’t want to fall, you have to accelerate upwards (say with a jetpack).
If you jump out of a rocket a mile over the event horizon of a black hole, you’ll freefall toward it. Because there’s no surface at the event horizon, you’ll freefall through it. If you don’t want to fall, you have to accelerate upwards. The closer you get to the event horizon, the more thrust your jetpack has to provide to oppose your gravitational acceleration. At the event horizon, the required thrust becomes infinite.
So if you turn on your jetpack when you’re waist deep in the event horizon, it has to deliver a huge thrust to stop your torso from dropping in. But your legs, already having crossed the horizon, need infinite thrust. As the jetpack pushes on your torso, your body is structurally unable to transmit an infinite force – your legs get ripped off. This happens even if you descend very slowly with your jetpack firing continuously.
If you freefall across the event horizon, the nerve signals from your toes can reach your head because your head is moving at the same speed as your feet. But if you accelerate your head so that it doesn’t fall in, the parts of you inside the event horizon fall away from your head so fast the nerve signals can’t ever reach their destination.
Okay; now if we consider the unobtanium rod, you can drop it into the black hole and it will stay in one piece because it is in freefall; but if you tug on the outermost end to stop it from falling it will be under acceleration, so it would break. (despite being unbreakable).
Hence the proposition in the original post is an impossibilty; that is what Stranger, Chronos and the rest were saying all along.
If you’re in orbit, you should be able to approach the event horizon as slowly as you like.
I think it might be true that an object can’t get back from the event horizon once its centre of mass has crossed - but that would mean that if you still have one arm sticking up above the event horizon, and someone threw you a rope from a ship in higher orbit, you would on grasping the rope become part of a larger object with a centre of mass outside of the event horizon, which would be able to pull away (including you)
It seems to me that from the perspective of the outside observer, whatever you lower down will never cross the horizon, due to gravitational time dilation. No?
Can you be in orbit that close to an event horizon? I seem to recall that photons orbit a black hole some significant distance from the event horizon, and no object below that form a stable orbit.
I’ve never understood the obsession with the event horizon of a black hole - that’s the radius at which light, a massless particle can no longer escape…if something actually has mass, it’s doomed a long way further out, unless FTL is possible.
I’ve got a question about the terminology. Why is the spherical shell surrounding surrounding a black hole, at the radius where the escape velocity = c, called the “event horizon”?
I’ve seen the phrase “event horizon” also used, meaning the conical shell forming the boundary of a light-time cone (Wiki pages: Light cone and Event horizon.)
Applying the term to the boundary of a light cone seems to make a lot of intuitive sense to me. Using it for the boundary of a black hole seems a more specialize usage. (I’ve also seen the term “black hole” used inconsistently, sometimes meaning the entire volume inside the event horizon, which makes sense since that’s the “black” region as seen (or unseen) from the outside; but I’ve also seen “black hole” sometimes used to refer only to the massive object or singularity in the middle.)
I have to say that many of the answers here are circular - basically no signal can escape from a black hole because nothing can escape from a black hole. It seems very few posters have actually explained why.
So to simplify the OP. I have a wire with one end inside the event horizon and one end outside the event horizon. And there are black holes big enough that we can do this without tidal forces destroying the cable. Why can’t electricity travel from the inside to the outside? It seems there are a couple of ideas:
The other end will never enter the event horizon from the perspective of the outside end.
The electrical signal will be red-shifted into nothingness (but does gravity really effect electricity this way?)
The wire will be stretched until it breaks (but why can’t it send a signal until it breaks?)
Or how about this: we have a chain of people from outside to inside the event horizon. Again, tidal forces will not tear us apart for a while. I am inside the EH and squeeze your hand. When I do, you squeeze he hand of the next person in line and so forth. What stops that signal from getting out of the event horizon?
Gravity affects light less than it affects anything else. I’m not sure if red-shift is the best way to describe what happens to the electricity, but it’s not going anywhere.
If it helps, just focus on the issue of time rather than the issue of gravity. At the event horizon, to an outside observer, time stops. Once time is stopped, I’m sure you agree that it would be akin to time travel for anything to happen afterward. If the observer sees time for the probe frozen at 12:34:56, the observer simply cannot receive a signal from 12:34:57, ever. The event horizon, by definition, has cut off all of the light cones that connect :57 to the rest of the universe.
The wire might provide the illusion that things can travel across it through space, but if they can’t travel across it through time, they’re just as trapped.
So… the next question, I suppose, is what happens if you send a time machine into a black hole? :eek:
If light can escape, then anything else can, too, because anything else can move at a speed arbitrarily close to the speed of light. It might be really, really difficult for a massive particle to escape from the very close proximity of the horizon, but it’s possible.
And the event horizon is not just the distance at which the escape speed is c. If you take the standard (non-relativistic) formulas for escape speed, and naively set it equal to c, the distance you’ll find will be the same as the radius of the event horizon, but that’s actually only a coincidence. There are two different places in the calculation where, if you take relativity into account, you’re off by a factor of 2, and it just so happens that the factors of 2 cancel out.
The more accurate description of the event horizon is that there simply isn’t a path from anywhere inside the horizon to anywhere outside of the horizon. Going from inside to outside, or from a point deeper inside to a point closer to the horizon, is exactly equivalent to traveling backwards in time.
What if that other something was another black hole? Would the curvature that black hole A combined with the curvature of black hole B make what once were event horizons that had nasty curvature a lot flatter and allow energy to leave the system of 2 black holes?
When two black holes collide, they merge into a single black hole. And yes, this process does allow a heck of a lot of energy to leave the system: The relevant rule is that the total area of the event horizons cannot decrease, but since the area of the event horizon is proportional to the mass squared, the total mass can decrease considerably (with the remainder all being released in gravitational waves). For instance, if you have two nonspinning black holes of equal mass that collide, you can release up to about 60% of one of the initial masses (2-sqrt(2)) as gravitational waves.
This still does not let any information out, however.