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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 One of my favourite black hole fun facts!
@johncarlosbaez I believe this process is called spaghettification
>water on the sunny side of the Earth is pulled toward the Sun more than water on the night side
While the Sun does cause a tidal force as you describe, the tides that we see are mainly due to the Moon (which is smaller but much closer). So water on the moony side (if you will) is pulled more, while water on the opposite side is pulled less, and the land (which moves more as a solid chunk) is pulled some amount in the middle. So we get high tide on both the moony side (where up is towards the moon) and on the opposite side (where up is away from the moon).
You probably know this, but it seems weird to focus on the relatively minor tidal effect of the Sun, so I thought that I would clarify for other readers.
@TobyBartels @johncarlosbaez Just to add some quantitative pedantry, the tidal force due to the sun is about half that due to the moon, which is why spring and neap tides are so different.
(I still remember how surprised I was when I first did the calculation to see the relative sizes of the effects, so now I take every opportunity to share the information, however marginally relevant it might be.)
@RobJLow @TobyBartels @johncarlosbaez
And the coolest tidal fact of all (which Rob once prompted me to realise) is that the near-perfect angular size match in solar eclipses means that essentially all the difference in lunar vs solar tidal strength (in the approximation that the separation distances are much larger than the radii of the bodies) is due to the ratio of densities between the sun and the moon.
@gregeganSF @TobyBartels @johncarlosbaez So cool, isn't it!
@gregeganSF @RobJLow @johncarlosbaez : Neat! Above, when I described the Moon as ‘smaller but much closer’ than the Sun, by ‘smaller’ I meant ‹less massive›. But I could have just written ‘denser’, or maybe ‘denser (and closer, but this is almost perfectly cancelled by its smaller size)’ for more detail. (Looking up these densities on Google, it looks like their ratio is about 2.4.)
@TobyBartels @gregeganSF @RobJLow @johncarlosbaez The Road to Reality makes this a cute little problem: show that our moon/sun tidal forces are in the same ratio as their densities.
@abecedarius @TobyBartels @gregeganSF @RobJLow - that's a really cool thing I'd never noticed. Now I'm looking for other examples, like this:
"This is a composite of two images to the same angular scale. The Andromeda galaxy is located over 2 million light years from Earth and was imaged by NASA’s GALEX satellite and added to an image of the crescent moon at sunset from ESO’s Paranal Observatory in Chile. From the surface of Earth, the diameter of our moon is about ½ a degree and the Andromeda galaxy is about 3 degrees across. In fact, because of the enormous difference in actual size and distance, the moon, which is only about 3,400 km in diameter and the Andromeda Galaxy, which is over 100,000 light years in diameter, seem comparable in size!"
Of course Andromeda is vastly less dense.
@johncarlosbaez @TobyBartels @gregeganSF @RobJLow At Catalina Island once I could just make out, maybe, the core of Andromeda. Horns a second off, wisp two million years away: one sky, one fall night
@abecedarius @TobyBartels @gregeganSF @johncarlosbaez I'll be honest: I'm quite chuffed that I noticed for myself something that Penrose thought interesting enough to be worth setting as an exercise :-)
@RobJLow @johncarlosbaez : So when a supervillain threatens to destroy the Moon, ending the tides and disrupting the ecology of bays, it's not that serious?
@TobyBartels @johncarlosbaez Well, it might not be top of the list of things to worry about:
@RobJLow @TobyBartels - since the Moon is held together by its gravity, I'm not sure why it would "shatter". It seems any impact intense enough to break it would actually melt it. There's a pretty believable theory that the Moon was made when a planet hit the Earth, but this involved a lot of melting. Here's a picture from a simulation, 50 minutes after the impact. Color indicates temperature. Note the Earth didn't just crack into large chunks.
https://math.ucr.edu/home/baez/earth.html
Of course SF allows for an occasional miracle.
@RobJLow @johncarlosbaez : These comic-book scenarios don't usually pay attention to debris. (See also the Endor Holocaust when the fans won't let it go.)
@TobyBartels wrote: "You probably know this, but it seems weird to focus on the relatively minor tidal effect of the Sun".
I wasn't thinking about the relative effect of the Sun and Moon on tides, and it seemed quicker, and more rapidly absorbed, to say "sunny side" and "night side" than "moon-pointing side" and "side away from the moon". "Moony side" is fun but in these quick posts I'm usually trying to keep readers focused, not distract them.
I've tried to fix my post by changing "the tides" to "tides".
@johncarlosbaez isn't the wind resistance much stronger than the gravitational difference, and so a falling object would be compressed instead?
@clusterfcku - yes, the wind resistance vastly exceeds the tidal forces for a person falling on Earth. I hoped that was clear. But it wasn't, so I've tried to improve my statement.
@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 @BartoszMilewski When I read the sentence, "We could be falling through the enormous event horizon of an enormous black hole right now", my first thought was "yes, this applies to all spacetimes whose curvature is sufficiently small". But I was still amazed by how directly we can show this with the Kruskal-Szeres Penrose diagram. Thanks for explaining!
@johncarlosbaez @BartoszMilewski also, on a side note: It just occurred to me that there is a namespace collision between Penrose (conformal) diagrams and Penrose (tensor network) diagrams...
@vacuumbubbles @BartoszMilewski - You're right, Vanessa, this business about not seeing your hand disappear is a local question, and the concept of 'horizon' is global. There's no way to tell if you've crossed a horizon at the moment you cross it - it's like "having the happiest moment of your life".
I've never heard anyone call tensor networks "Penrose diagrams". Category theorists think of them as a special case of "string diagrams" in monoidal categories, so that's what I usually call them.
@johncarlosbaez @BartoszMilewski I just looked it up, and I seem to have misremembered: the GR-specific tensor diagrams are called "Penrose graphical notation" (https://en.wikipedia.org/wiki/Penrose_graphical_notation)
@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 some lensing.
From a distance:
@johncarlosbaez @internic
This is NASA simulation of a fall into a black hole. It shows a black wall.
@BartoszMilewski @johncarlosbaez Incidentally, that video is the work of @SchnittGetsReal
@BartoszMilewski - I hope you read this... another branch of this sprawling thread:
@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 @internic @j_bertolotti
So back to the original question: Can you tell whether you're approaching the event horizon, even if the black hole is enormous?
@BartoszMilewski @internic @j_bertolotti - the larger the black hole the harder it is to tell if you're falling through the horizon, especially if everything you know is falling in with you.
@johncarlosbaez @internic @j_bertolotti
So what do you see when you look towards the singularity? You're looking into the past of the black hole, seeing all the stuff that's fallen into it--progressively red shifted. Sort of like looking at the Big Bang. So all these animations are wrong.
@BartoszMilewski Your personal clock is only a tiny bit redshifted from the things closer to the boundary. In your personal coordinate system, and theirs, things are happily passing through the boundary.
Which is to say, I thought this is merely a coordinate singularity, no?
@4raylee @BartoszMilewski @internic @j_bertolotti - one crucial thing, when doing these mental visualizations, is to decide whether you're freely falling into the black hole or powering yourself with rockets to hover it at a fixed height. The results are different. I can't tell which one you folks are talking about here.
(Let's ignore that either way, you die a miserable death before you get too close to the horizon, unless you're freely falling into a very large black hole, in which case the pain only comes *after* you cross the horizon.)
I recommend these animations:
@BartoszMilewski @johncarlosbaez @internic @j_bertolotti a little bit off topic: I've always found that textbooks on GR arrive to black holes after such a long way behind that they're like exhausted, and deal with the topic lightly. Perhaps this is not the case with an specific book, like the one by Chandrasekhar.
@davidsuculum - have you read Gravitation by Misner, Thorne and Wheeler, or General Relativity by Wald? I think both of these handle black holes in quite a lot of detail, though in such different ways that you need to read both to get the full picture.
These are my two favorite books on general relativity.
@BartoszMilewski @johncarlosbaez @internic @j_bertolotti the event horizon is a not a local concept, it only makes sense in the context of the full spacetime. It's the surface from which nothing can escape to infinity. This is not the same as nothing can make it to you.
Think of the following situation: You're falling to a small black hole outside its event horizon. So clearly, you can see your outstretched arm. Later, I.e in your future, a gigantic object falls in. This now extends the event horizon, and you're suddenly inside. This is because the horizon is a spacetime property, not a local one.
Only in a static spacetime do local and global horizons agree. But the local one is kind of fictitious
@sergedroz @BartoszMilewski @internic @j_bertolotti - right! Your example shows why crossing the horizon is like 'having the happiest day in your life' - it's not something you can know is happening until later: if you get hit by a car next week we may suddenly realize that yesterday was the happiest day in your life.
@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
@BartoszMilewski @johncarlosbaez @internic : You never see your hand (or anything) the way it looks right now; you see it as it appeared in the past, when the light left it. As you fall in, hand first, your hand passes the event horizon (entering the black hole) while you're receiving the light from before it entered. Then when your head enters, you receive the light from when your hand entered. Later, you'll see the light that left your hand at the time when your head entered.
@TobyBartels @BartoszMilewski @internic - I told Bartosz to contemplate the Penrose diagram while keeping in mind that light moves along 45 degree lines, but I guess it takes practice to read Penrose diagrams. So yes: as you fall in the black hole, Toby's scenario takes place, and you never lose sight of your hand.
Alternatively you can use rocket thrusters to permanently keep your head and body out of the black hole while you lower your hand through the horizon. Then it will inevitably get ripped off, and you can pull back the bloody stump of your arm if you're strong enough.
@johncarlosbaez @TobyBartels @BartoszMilewski @internic
I did a long set of calculations exploring various scenarios for lowering things through a horizon. These calculations use a Rindler Horizon, but that works as a good approximation to the horizon of a black hole large enough that tidal effects are irrelevant.
It’s not exhaustive, but it’s nice to have some explicit calculations that are near enough to a few of the “Yes, but what if I did X?” questions that people sometimes propose as ways to “cheat” an event horizon.
https://www.gregegan.net/SCIENCE/Rindler/RindlerHorizon.html
@gregeganSF - by the way, your page makes me a bit sad that I went along with Chris Hillman when he demanded that I remove his page "Relativity on the World-Wide Web" from my website. He was being harassed by people whose work he had criticized, and he wanted to simply disappear. Now I'm trying to see if I kept a backup.
Yes.
@johncarlosbaez @gregeganSF : It's also on the Wayback Machine (just barely, with the first archive only 3 months before it was removed).
@TobyBartels @gregeganSF - thanks, Toby. I have no idea what became of Chris. The last emails I received from him, years ago, were quite sad.
@johncarlosbaez @TobyBartels @gregeganSF Oh no! I'd no idea that had happened, that's terrible.
@johncarlosbaez @TobyBartels @internic
Yes, it takes practice to read Penrose diagrams. The difficulty for me is to figure out what the world looks like from the point of view on a given observer. I think the "dust trail" in @gregeganSF visualization comes closest to the situation I'm interested in. I think, for a very large black hole, the dust lines would be practically vertical. Is that right?
@BartoszMilewski @johncarlosbaez @TobyBartels @internic
If you mean the world lines of infalling dust particles on that Penrose diagram, I’m not sure; I don’t know exactly what coordinate system is used there.
BTW, this other page I wrote on things falling into black holes might also be of interest: