As you're freely falling through the air you don't feel any force except the wind - but you're also getting *stretched* a tiny amount because gravity is a bit stronger near your feet. This is called a 'tidal force' because it creates tides: for example, water on the side of the Earth facing the Moon is pulled toward the Moon more than water on the opposite side.
As a star falls toward a black hole it can get stretched and even destroyed by this tidal force - we've seen it happen! It can create a huge flare of radiation.
But surprisingly, the bigger the black hole, the smaller the tidal force is near the event horizon. We could be falling through the event horizon of a truly enormous black hole right now, and we'd never notice - though I consider this very unlikely.
More importantly, a star like the Sun will only get disrupted *before* it crosses the event horizon if the black hole is < 100 million solar masses. Otherwise it will get sucked in and be lost to sight without any drama!
The big black hole in the center of our galaxy is only 4 million solar masses, so this 'silent death' doesn't happen here. But it happens elsewhere. The biggest black hole known is 66 *billion* solar masses!
Black holes emit flares of light that we don't understand. Some must be from stars falling in. But many flares show very little light in hydrogen's spectral lines! This talk is pretty fun, and it's all about these mysteries.
@johncarlosbaez "We could be falling through the event horizon of a truly enormous black hole right now."
I've been always puzzled by this. Surely you'd see the horizon as a black surface coming at you, since no light can escape the black hole. And once your legs get past the horizon, you'd lose the feeling in them forever. Although the math inside a black hole gets really crazy, so I don't know how much I can trust my intuitions.
@BartoszMilewski - trust the equivalence principle: any small enough patch of spacetime is indistinguishable from Minkowski spacetime for a free-falling particle.
If you fall through the event horizon of an enormous black hole with your arm outstretched before you, your hand doesn't disappear as it crosses the horizon. But if you use rockets to hover outside the horizon and stick your arm in, it gets ripped off and disappears from view.
@johncarlosbaez
If my hand doesn't disappear, it means I can see things in front of me, including the singularity? What does the singularity look like?
@BartoszMilewski - light never comes out through the horizon, yet your outstretched hand doesn't disappear as you fall in a big black hole. Explain!
You never see the singularity, even from inside a black hole, because it's always in your future.
@BartoszMilewski - both these questions can be answered using the Penrose diagram of a black hole. Light moves at 45 degree lines. Think about what happens when you and your outstretched arm fall, at less than light speed, through the horizon! You are always looking back in the past along 45 degree lines.
@johncarlosbaez But presumably @BartoszMilewski's statement that "you'd see the horizon as a black surface coming at you" is true at some level, because what you see in the distance is not only a function of local spacetime (unless you're inside an enclosure, which is the usual conceit of equivalence principle thought experiments).
@internic @BartoszMilewski - right, as you approach the horizon it looks dark except perhaps for Einstein rings.
From a distance:
@johncarlosbaez @internic
So you see my confusion: If the approaching horizon looks like a black wall, you shouldn't be able to see your hand that has just crossed it.
@BartoszMilewski
As far as I understand GR, even long before you actually arrive at the proper horizon things will get very dark, as only the light moving in a narrower and narrower cone will be able to escape.
@johncarlosbaez @internic
@j_bertolotti @BartoszMilewski Right. The situation with your arm is different, and that's where the conformal diagram that @johncarlosbaez posted is helpful. If you imagine an extended object falling into the black hole you can see that light from the left side (closer to the center) will always reach the right side at a later time. If, however, the object stops falling inward while partially over the horizon, light from the portion inside the horizon will never reach the portion outside (correspondingly meaning no causal force law can possibly keep them from tearing apart).
I think that another way to think of it is that near the horizon light is moving outward from the center more and more slowly (in terms of Schwarzschild coordinate distance vs. coordinate time), at the horizon it's standing still, and inside the horizon it's actually falling inward. If you're stationary outside the event horizon, the light from inside never reaches you. If you're falling inward, you will catch up with with the light from stuff further toward the center, because while it is falling toward the singularity, it is doing so more slowly than you (or any massive body).
@internic @j_bertolotti @johncarlosbaez
Let's simplify the problem. Two observers are free-falling one after another. We are the second observer filming the first one. Does the first one disappear behind the horizon?
Assume this is a humongous black hole, so tidal forces are negligible.
@BartoszMilewski @j_bertolotti @johncarlosbaez No, from the perspective of the second observer the first never disappears. You can see this by taking the above conformal diagram and drawing two timelike world lines that cross the event horizon and end at the singularity. If you then draw a set of outward-pointed light rays (moving up and to the right at 45°) from events on the first world line, you will see they continue to intersect the second. However, also note that the image that the second observer receives just before crossing the event horizon is of light the first observer emitted just before crossing the event horizon.
It's also worth noting that technically even for a stationary observer an in-falling object never disappears, it just appears to gets dimmer, slower, and more redshifted as it approaches the event horizon (until it becomes imperceptibly dim as an effectively frozen image at the event horizon).
@internic @BartoszMilewski @j_bertolotti - Nick is right. You can see it clearly from here if you remember that light moves along 45 degree lines.
@johncarlosbaez @internic @j_bertolotti
I'm trying to visualize this. Right before I hit the horizon, in my free fall, I will see in front of me everything that has ever fallen into the black hole. So that's not really a black wall. It's more like a windshield of a speeding car.
@BartoszMilewski @internic @j_bertolotti - but the infalling objects are quickly redshifted to oblivion so you don't actually see them. You see the sky above you shrinking to a smaller and smaller disk. There are movies of this on Youtube.
@johncarlosbaez @j_bertolotti I would expect the movie @BartoszMilewski posted elsewhere in this discussion to be pretty accurate, as it was made by @SchnittGetsReal and co. based on simulation.
https://mathstodon.xyz/@BartoszMilewski/113525935025414414
