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So how DOES a black hole form?

Roboramma said:
Regarding the neutron star, if I understand correctly, even the part of the neutron star that is initially at a distance less than 2MG from the center of the star at the time the collapse begins will also be spread out over the surface of the event horizon from the perspective of a distant observer.

[...]

By the ways, Hans, this lecture by Leonard Susskind goes into pretty much exactly the issues that you're asking about, you'll probably find it very interesting:
https://www.youtube.com/watch?v=EJrgKI8aXgQ
Click to expand...

Thank you, it was very interesting indeed and I learned some new stuff. Including what you say in the first paragraph. I hadn't even thought of that before.

And I really like the guy. He's good at explaining stuff. I'll look for more of his lectures.
 
Matter gets inside the abstract mathematical sphere defined by the Schwarzschild radius not only by inward flow of matter, but also by outward expansion of the sphere. So the last bit of matter you're thinking of as needing to fall in so the black hole can become a black hole doesn't need to fall in; it just needs to be close enough while the Schwarzschild sphere comes up to meet it because of the increase in the Schwarzschild radius that was caused by the last bit of matter to truly fall in before. Then the surface of that sphere becomes the event horizon after enveloping enough matter.
 
Hmm, really? I had just understood that all the matter seems to be on the surface (from the point of view of a distant observer, which is why I'm asking) even if it was inside the schwarzschild radius before the black hole was formed.

Besides, I think that wouldn't work even in newtonian mechanics, because of Newton's XXXI theorem, a.k.a., shell theorem. The only mass that matters for gravity is the one closer to the centre. As long as the new mass doesn't get to a lower radius than the existing mass in the accretion disk (and it can't because the old one has a massive head start there) the old mass won't notice the increase in gravity. I.e., it would result in the layer cake mentioned earlier, with the new mass smeared on the apparent horizon of the mass that came before it, and so on.

'Course, in GR it seems like I was wrong and everything ends up smeared on the same event horizon, but same idea. To actually observe the old mass getting behind the event horizon, you'd still need some way to move information faster than light.

Well, the way I understand it, anyway.
 
To paraphrase Hawkins we are seeing the breakdown of English in describing reality when discussing black holes, English is simply not up to the task!
 
HansMustermann said:
Not entirely sure what you mean by "Your frame doesn't apply to him." Please explain.

I mean, I can dig that x, y, z and t are different in his frame. But I would ASSUME that in my own x, y, z and t, I can put a dot on the graph for where the guy is from my perspective. At least as long as in THAT perspective the guy isn't already past the door.

Also not sure what you mean about those other coordinates, since well, I never went to a physics college, so I'm clueless there. You may need to explain those too. But, unless I got the whole idea of physics wrong, there must be SOME coordinate transform back to x, y, z and time. So it may be easier to calculate stuff using different coordinates, but eventually you must make a prediction like "it's gonna happen over there at that time".
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I think you know this, but I'm repeating it because it's relevant: Not only is there no preferred coordinate system, there may not be a single coordinate system that covers all of spacetime.

In particular: the coordinate system in which time slows down as you approach the event horizon and appears to stop at the event horizon does not cover all of spacetime. So no, you can't always "put a dot on the graph for where the guy is from my perspective". That's the fault of the coordinate system you prefer to use.

This isn't all that mysterious. The longitude/latitude coordinate system has a similar problem at the poles and at zero longitude. At the north pole, the "go south" direction isn't well-defined, which breaks the mathematical rules for what counts as a legitimate coordinate system. At zero longitude, the longitude coordinate changes discontinuously, which also breaks the mathematical rules for what counts as a legitimate coordinate system.

Mathematically, this is solved by using an atlas of coordinate systems, with each coordinate system covering only part of the earth's surface, and with well-defined continuous transformations between every pair of coordinate systems (which are called maps) on the part of the surface where they intersect. The idea of an atlas, and the continuous transformations between pairs of coordinate systems, are key elements of the mathematical definition of a manifold, which is the mathematical generalization of Euclidean space needed to discuss general relativity (and black holes in particular).
 
W.D.Clinger said:
I think you know this, but I'm repeating it because it's relevant: Not only is there no preferred coordinate system, there may not be a single coordinate system that covers all of spacetime.

In particular: the coordinate system in which time slows down as you approach the event horizon and appears to stop at the event horizon does not cover all of spacetime. So no, you can't always "put a dot on the graph for where the guy is from my perspective". That's the fault of the coordinate system you prefer to use.

This isn't all that mysterious. The longitude/latitude coordinate system has a similar problem at the poles and at zero longitude. At the north pole, the "go south" direction isn't well-defined, which breaks the mathematical rules for what counts as a legitimate coordinate system. At zero longitude, the longitude coordinate changes discontinuously, which also breaks the mathematical rules for what counts as a legitimate coordinate system.

Mathematically, this is solved by using an atlas of coordinate systems, with each coordinate system covering only part of the earth's surface, and with well-defined continuous transformations between every pair of coordinate systems (which are called maps) on the part of the surface where they intersect. The idea of an atlas, and the continuous transformations between pairs of coordinate systems, are key elements of the mathematical definition of a manifold, which is the mathematical generalization of Euclidean space needed to discuss general relativity (and black holes in particular).
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I can understant that my coordinate system doesn't cover the inside of the black hole, because there's essentially a division by zero in the coordinate transform at the event horizon. Which I guess is the whole point of an event horizon: there's no way to observe anything behind it. So, yes, it doesn't cover everything.

BUT, the space outside such areas I would assume I can still convert back to my own X, Y, Z and time, right? For example some chunk of matter in the accretion disk, that's still going to be observed somewhere that I can give a coordinate and a time in my own frame, right?

I'm only guessing though, so please correct me if I'm wrong.
 
HansMustermann said:
I can understant that my coordinate system doesn't cover the inside of the black hole, because there's essentially a division by zero in the coordinate transform at the event horizon. Which I guess is the whole point of an event horizon: there's no way to observe anything behind it. So, yes, it doesn't cover everything.

BUT, the space outside such areas I would assume I can still convert back to my own X, Y, Z and time, right?
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If you use Shwarzchild coordinates, you can convert everything except the event horizon into x, y, z, and t. But you're talking about this as if coordinates have some absolute meaning. They do not. What is your t? Ar your location, it may mean your time. But if you go someone else, or move in a different way, it doesn't. Moreover, there isn't even a unique way to create a coordinate system where t is your time, but t for other events changes. So t isn't their time, and you can't even assign a unique value of t to these other events. So which value should you assign, and what does it mean? There is no tidy answer.
 
Delvo said:
Matter gets inside the abstract mathematical sphere defined by the Schwarzschild radius not only by inward flow of matter, but also by outward expansion of the sphere. So the last bit of matter you're thinking of as needing to fall in so the black hole can become a black hole doesn't need to fall in; it just needs to be close enough while the Schwarzschild sphere comes up to meet it because of the increase in the Schwarzschild radius that was caused by the last bit of matter to truly fall in before. Then the surface of that sphere becomes the event horizon after enveloping enough matter.
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Essentially what's called the Hoop Conjecture.

https://en.wikipedia.org/wiki/Hoop_Conjecture

Though by that, not just meet it but fit inside it.

Key point being that the Schwarzschild radius isn't an event horizon until that condition is met.
 
Let's play around a little with coordinates and metrics. We'll start off with an easy one. I've got two coordinates, x and y. In order to describe the geometry of space described by my coordinates, I'll give you what's called the metric, which describes how to measure distances in this space. Here's my first metric:

ds2 = dx2 + dy2
OK, now to explain this in more detail. If I take a tiny little step in the x direction (I'll call it dx) and a tiny little step in the y direction (dy), I will have moved a tiny little distance ds. I can measure the length of a path by adding up all my little ds bits (or integrate using calculus to get it exact). Now, you might recognize this formula as the Pythagorean theorem:

a2 + b2 = c2
That's not a coincidence: the Pythagorean theorem tells you how to measure distances in Euclidean space. And that's the space I've created: it's just flat Euclidean space.

Now let's try another metric:

ds2 = da2 + a2db2
If I only move in a, then the distance is proportional to my change in a. But if I only move in b, distance is proportional to my change in b, but it's ALSO proportional to my current position in x. So this is a very different metric.

But here's where it starts to get interesting. Is this a different kind of space than my first metric? No, actually, it's not. It's the exact same space, just with different coordinates. What are these coordinates? Polar coordinates. a is the radius, b is the angle. Choose a different set of coordinates and you get a different metric, but the space is the same. Furthermore, no matter which coordinate system I use, the distances I calculate using the appropriate metric will be the same as well.

OK, now for a third metric:

ds2 = da2 + sin2(a)db2
At very small values of a, sin(a) ~ a, so this looks like our second metric. But as a gets larger, then sin(a) < a. Things change. So this isn't a flat space anymore. What is it? It's the surface of a sphere. a is now your latitude (distance from the "north pole"), b is now your longitude (angle),

Now let's get a weird metric:

ds2 = dx2 - dt2
The first thing that should jump out at you here is that ds2 can now be negative. Weird, right? Well, this is the metric for special relativity (using just one spatial dimension for simplicity). In this space, we've got two categories of distances: positive and negative. Positive distances are space-like (|dx| > |dt|). Turns out, if the distance between two events is space-like, they can be simultaneous in some reference frame, and they are never located at the same place in any frame. Negative distances are time-like. If the distance between two events is time-like, they happen in the same place in some reference frame, but they can never be simultaneous in any reference frame. If the distance is null, then it is null in all reference frames.

With me so far?
 
HansMustermann said:
Hmm, really? I had just understood that all the matter seems to be on the surface (from the point of view of a distant observer, which is why I'm asking) even if it was inside the schwarzschild radius before the black hole was formed.

Besides, I think that wouldn't work even in newtonian mechanics, because of Newton's XXXI theorem, a.k.a., shell theorem. The only mass that matters for gravity is the one closer to the centre. As long as the new mass doesn't get to a lower radius than the existing mass in the accretion disk (and it can't because the old one has a massive head start there) the old mass won't notice the increase in gravity. I.e., it would result in the layer cake mentioned earlier, with the new mass smeared on the apparent horizon of the mass that came before it, and so on.

'Course, in GR it seems like I was wrong and everything ends up smeared on the same event horizon, but same idea. To actually observe the old mass getting behind the event horizon, you'd still need some way to move information faster than light.

Well, the way I understand it, anyway.
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Well, technically (in a purely Newtonian sense) what the shell theorem says is that a spherical symmetrical body, like a matter shell, effects external bodies as if all the matter were concentrated at the center. Not that matter closer to the center matters more. For a spherical symmetrical shell there is no matter at the center. Also it is the effect on external bodies that appears concentrated at the center. A body at the center of the shell would feel no net gravitational force and matter on the inside of the shell, closer to the center, would experience less net force to the center due to the opposing attraction of the matter above it.


For GR it gets different as the more localized the concentration of matter the greater the resulting curvature of space-time. So just a matter shell generates a different space-time curvature (locally) than everything concentrated as a solid homogeneous sphere. However, at a distance the shell theorem is effectively valid as those local differences between shell and solid become more diminutive.
 
Ziggurat said:
If you use Shwarzchild coordinates, you can convert everything except the event horizon into x, y, z, and t. But you're talking about this as if coordinates have some absolute meaning. They do not. What is your t? Ar your location, it may mean your time. But if you go someone else, or move in a different way, it doesn't. Moreover, there isn't even a unique way to create a coordinate system where t is your time, but t for other events changes. So t isn't their time, and you can't even assign a unique value of t to these other events. So which value should you assign, and what does it mean? There is no tidy answer.
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I THINK I understand what you're going with it. And I won't disagree one bit.

BUT, and here's the big BUT (I like big BUTs and I cannot lie;)) where you're going with it is basicaly what my coordinates mean for that object. Well, yeah, my {x, y, z, t} can mean lots of different things for it (depending on its relative speed, acceleration, etc) or nothing at all (if I'm behind some kind of horizon).

But it's not the question I was asking. In fact it's the converse of what I was asking. I was just asking where it is on myspace... err... in my space.

That said, yeah, I'm still with you on polar coordinates. I did a bit of computer graphics, so, yeah, I can grasp some basic coordinate transformation. And on SR. THAT part I can still understand. It's only GR (well, and most of QM) that, shall we say, I'm not even competent enough in to be a crackpot theorist :p
 
The Man said:
Well, technically (in a purely Newtonian sense) what the shell theorem says is that a spherical symmetrical body, like a matter shell, effects external bodies as if all the matter were concentrated at the center. Not that matter closer to the center matters more. For a spherical symmetrical shell there is no matter at the center. Also it is the effect on external bodies that appears concentrated at the center. A body at the center of the shell would feel no net gravitational force and matter on the inside of the shell, closer to the center, would experience less net force to the center due to the opposing attraction of the matter above it.


For GR it gets different as the more localized the concentration of matter the greater the resulting curvature of space-time. So just a matter shell generates a different space-time curvature (locally) than everything concentrated as a solid homogeneous sphere. However, at a distance the shell theorem is effectively valid as those local differences between shell and solid become more diminutive.
Click to expand...

Hmm, really? I thought that Birkhoff's theorem said the exterior of a spherical shell is described by the Schwarzschild metric. So basically it's really indistinguishable from all mass being concentrated in its centre, or for that matter inside the event horizon corresponding to its mass.

Of course, I'm just learning this stuff, so chances are high that I might be wrong. I'd appreciate it if you took the time to point out what I misunderstood.
 
HansMustermann said:
Hmm, really? I thought that Birkhoff's theorem said the exterior of a spherical shell is described by the Schwarzschild metric. So basically it's really indistinguishable from all mass being concentrated in its centre, or for that matter inside the event horizon corresponding to its mass.
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That's true. But the point about the event horizon inside the body is it's not a Schwarzchild event horizon.

Let's take a really artificial example: all the mass of a body confined to an infinitesimally thin spherical shell. Outside the shell, we use the Schwarzchild metric. Inside the shell, it's a completely flat spacetime. Further suppose that this mass is transparent, so we can see what's going on inside it.

Now let's start to shrink the shell. Shrink it enough to get it just below the Schwarzchild radius, and it's a black hole. At this point, we know we have an event horizon at the Schwarzchild radius.

Now suppose that we chose our mass such that the Schwarzchild radius was one light-second. Suppose that there's a bunch of massless fireflies buzzing around inside the shell, and we can watch them as they move about blinking on and off (since the shell is transparent). Suppose that one of these fireflies is right at the center.

Let's say the black hole forms at t = 0 seconds. After t = 0 second, we can't see any light from the firefly in the center. But when was this last light emitted? Was it emitted at t=0? No, it was emitted at t = -1 seconds. Any light emitted after t = -1 seconds can't get out before the object is a black hole. What this means is that, at t = -1 seconds, the firefly at the center is actually already inside an event horizon. Spacetime is still flat inside the shell, so this event horizon is different than the one that will exist after collapse, but it's still a boundary between what light can escape and what light can't escape.

Now let's look at a firefly that's halfway between the edge and the center, or 0.5 light seconds from the edge. The last light to escape from that firefly will have been emitted at t= -0.5 seconds. After that, this firefly is also within the event horizon, though spacetime is still flat for him as well.

In fact, the event horizon forms at t=-1 seconds at the center of the sphere, expands outwards at the speed of light, and then stops when it reaches the Schwarzchild radius at t=0, when the shell has collapsed to this point. This horizon is formed by future events. Its formation does not reflect any change in local conditions (spacetime remains flat inside the sphere until after the black hole is formed). It's really just an abstract surface, marking the separation between points from which light will be able to escape and the points from which light cannot escape.
 
HansMustermann said:
Hmm, really? I thought that Birkhoff's theorem said the exterior of a spherical shell is described by the Schwarzschild metric. So basically it's really indistinguishable from all mass being concentrated in its centre, or for that matter inside the event horizon corresponding to its mass.

Of course, I'm just learning this stuff, so chances are high that I might be wrong. I'd appreciate it if you took the time to point out what I misunderstood.
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I'll try.

Yes, the exterior of a spherical shell is described by the Schwarzschild metric. The interior however, is described by the he Minkowski metric. At least if we're talking about something that isn't a black hole yet. In that case the Schwarzschild radius is interior to the outer limit of the shell. Remember you were talking about matter closer to the center than other matter. So still within outer limit of the shell even if inside the Schwarzschild radius. That was the interior/exterior point being made.


For the other point (local/distant) probably a poor choice of wording and lack of detail on my part. I should have been clearer about what I meant by local and distant versus interior and exterior. As well as indicated that spherically symmetrical is more of an idealized condition. The more distant you are from a just a general gravitating body the more spherically symmetrical it can seem. However, locally all those variations in density, deviations from symmetry and such are apparent. Particularly if you are talking about material internal to the body (surrounded by other material). That was the distance point, that the further from the material the more spherically symmetrical it can seem.

If we are discussing how does a black hole form then we are talking about something that isn't a black hole yet and the Schwarzschild radius is not an event horizon and is for the most part interior to the outer limit of the body. If you are talking about some material of the body vs other material closer to the center it still isn't a black hole yet, I recall some explanation of the black hole have no hair conjecture with the hoop conjecture that those local variations, like from spherical symmetry, get radiated away as gravitational waves when it does collapse to form a black hole and an event horizon which is, well, spherically symmetrical.

Now some material outside the body as it collapses may also get pushed away by the gravitational waves.

If I recall correctly one of the things of the hoop conjecture was that while gravity might tend to make a body more spherically symmetrical being spherically symmetrical isn't a requirement for black hole formation. Just that it fit within the hoop.
 
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I will concede as rather obvious that, yes, there is no actual event horizon inside the shell, since it's all flat space in there. We can calculate where that horizon would be, but yep, it ain't there yet.

BUT that brings me back to the original question. From the frame of a distant observer, that would kinda be what a "black hole" really looks like, innit? Mind you, the shell would probably be within planck lengths of the Schwarzschild radius, if it's got billions of years (again, in the frame of an external and distant observer) to fall in, and that shell's been red-shifted into being blacker than black, so essentially it would look indistinguishable from an actual black hole by any practical measurement. You couldn't really tell if the Schwarzschild metric begins at the actual event horizon or picometres from it by any measurement from light years away.

But my question is really a theoretical one, really. And to make it clearer about which frame and whatnot, it's really this that got me thinking: let's say we have a star orbiting a black hole. Like, say S2, a.k.a. S0-2, one of the stars orbiting real close to Sagittarius A, the supermassive black hole at the centre of our galaxy. Well, about 17 light hours way, so time and space will get WEIRD there, but that's not relevant.

And my question was really just this: from the frame of S0-2, is it observing that it orbits around an actual black hole, with the mass inside its Schwarzschild radius, or around an empty shell just outside Sagittarius A's Schwarzschild radius? I'm thinking that from the perspective of S2, it's orbitting around that shell, innit?

Of course, from the perspective of something that already fell into Sagittarius A, it's inside the event horizon of a proper black hole and it probably hit the singularity at the centre billions of years ago. But that's not what I'm asking.
 
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Also, let me try to clarify it a bit, because I think my poor choice of wording for the thread title is misleading people. I think some might expect me to go with it towards something silly like flat-out "black holes don't exist" or "black holes never finish forming." But as I was saying, I'm not even competent enough to be that kinda crackpot. I'll need to learn some more before I get there. Baby steps :p

What struck me as interesting (for me, as a newbie) is a much more constrained / less general proposition. Basically. "black holes don't exist" or "back holes never finish forming" strictly from the perspective of a distant observer. No more, no less. And the answers so far (filtered through my very limited knowledge) lead me to think that that's actually correct.

Of course from other perspectives, such as of a chunk of matter falling it, sure, the black hole is real and it formed a long time ago. In fact, from inside, it has always existed. But I'm not arguing any of that.
 
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