I Being at the position of a singularity before it is formed

[HansH]

TL;DR Summary

being at the position at the center of a heavy cloud of iron particles with a mass sufficient to form a black hole. What do i experience in time?

Suppose we have a large sphere with thin non rotaring surface of floating iron particles (to prevent that they start a chain reaction making a star instead of a black hole) with a total mass sufficient to form a black hole. with an observer in the center of the sphere. This means the sphere starts contracting and at a certain moment the sphere gets smaller than its own event horizon. Bus as the observer is at least inside this all I assume this means there is no gravity at his location. What does the observer experience in time (what does he see and what does he feel?) and is there a moment that the sphere of mass passes his position giving a steep change in spacetime curvature?

 

[Ibix]

I haven't done the maths to prove it, but I expect the causal structure of this spacetime is the same as Oppenheimer-Snyder. I'm assuming that's correct in what follows

The key point to realize is that the singularity isn't a place - it's more like a time, although that's not entirely correct either - so you can't be at the position of the singularity.

If you were inside the shell and not at the center, you are in flat spacetime initially, as you say. You would see an increasing blue shift in star light penetrating the shell as it collapses, until it passes you. At that time, you'd start to fall towards the center. If you were far enough from the center to be outside the Schwarzschild radius of the collapsing matter you could escape given a sufficiently powerful rocket. If you were inside, you may distance yourself from the matter shell, but you will eventually strike the singularity. In either case you can pass repeatedly through the shell into flat spacetime and back out if you have sufficiently powerful rockets and enough fuel.

If you are at the exact center of the shell, I think the matter passes you at the singularity, so you basically see flat spacetime around you until you die.

 

[Dale]

I expect the causal structure of this spacetime is the same as Oppenheimer-Snyder. I'm assuming that's correct in what follows

I agree, that is how I would approach it also

 

[PeterDonis]

I expect the causal structure of this spacetime is the same as Oppenheimer-Snyder.

Yes. The initial conditions the OP describes are basically the same as Oppenheimer-Snyder except for the iron part.

If you are at the exact center of the shell, I think the matter passes you at the singularity

No, it doesn't; the matter reaches you at the exact instant that the singularity forms, and you and the matter are destroyed in the singularity at that same instant. The matter can't "pass" you at the center because there is literally nowhere it can go.

 

[Ibix]

The matter can't "pass" you at the center because there is literally nowhere it can go.

True - "passes" was sloppy wording there.

 

[Ibix]

One interesting point about this is that the OP specifies iron to avoid a stellar phase. But iron does fuse - it's just that it's an endothermic process so it doesn't generate energy to support a star against gravitational collapse. So presumably we'd start getting heavier elements forming in the shell at some point. But where does the energy to do it come from?

I think that as the shell collapses both the density and pressure rise until fusion happens, at which point the density rises (because the fused nucleus is more massive than the "fuel" nuclei) and the pressure drops (because the number density of particles drops). So the energy for the fusion comes from the pressure. And you might see a change in the rate of collapse at that time, as a result of the change in the stress-energy tensor.

...right?

 

[PeterDonis]

iron does fuse - it's just that it's an endothermic process so it doesn't generate energy to support a star against gravitational collapse. So presumably we'd start getting heavier elements forming in the shell at some point. But where does the energy to do it come from?

From the kinetic energy of infall, which, in a realistic collapse where the material can increase pressure (see below), can cause various endothermic reactions to occur.

I think that as the shell collapses both the density and pressure rise until fusion happens, at which point the density rises (because the fused nucleus is more massive than the "fuel" nuclei) and the pressure drops (because the number density of particles drops). So the energy for the fusion comes from the pressure. And you might see a change in the rate of collapse at that time, as a result of the change in the stress-energy tensor.

Yes. This part was left out of the Oppenheimer-Snyder model because they assumed zero pressure throughout. But in a more realistic collapse, as we now know from numerical simulations, yes, these things would happen and would change the details of things like the rate, though not the final result.

 

[HansH]

in addition Iam also wondering how the observer somewhere on a line experiences flow of time compared to the far away observer. I assume time flows slower the more the observer comes close to the ring of material, but what happen for the observer inside the ring. I suppose time flows same as for the far away observer as for the observer inside the ring there is no curvature?

 

[HansH]

here a picture to previous post. and of course how does this behavior evolve in time when the rings contracts to a singularity. up to which moment can you speak of a ring of falling material with nothing in it as the singularity would mean a single point so could where does the situation change over from nothin in it to the 1 point of singularity? or is the situation with nothing in it always remaining but at a continuously reducing scale?

 

[PeterDonis]

Iam also wondering how the observer somewhere on a line experiences flow of time compared to the far away observer.

There is no way to compare the two. More precisely, any comparison between the two will depend on choosing a simultaneity convention, and there are an infinite number of different possible conventions you could choose.

 

[HansH]

There is no way to compare the two. More precisely, any comparison between the two will depend on choosing a simultaneity convention, and there are an infinite number of different possible conventions you could choose.

not sure if I can follow you. What I mean is that an observer at a certain point ages slower than an observer at infinite distance. so if you bring back the observer from the ring back to infinity you could compare both watches and make a conclusion about the relative flow of time?

 

[Ibix]

There is no way to compare the two. More precisely, any comparison between the two will depend on choosing a simultaneity convention, and there are an infinite number of different possible conventions you could choose.

Not even if the shell is zero thickness? That should define a unique relationship between the inside and outside planes of simultaneity, no?

 

[PeterDonis]

f you bring back the observer from the ring back to infinity

You can't. The observer at the center is inside the event horizon, so he can't escape to infinity.

 

[PeterDonis]

Not even if the shell is zero thickness? That should define a unique relationship between the inside and outside planes of simultaneity, no?

What "planes of simultaneity"? The spacetime is not stationary.

 

[Ibix]

What I mean is that an observer at a certain point ages slower than an observer at infinite distance. so if you bring back the observer from the ring back to infinity you could compare both watches and make a conclusion about the relative flow of time?

The problem is that when passing through the shell there's no global definition of "time" because it's not a stationary spacetime. You can certainly push someone inside and pull them back out and there's a unique answer to how old they are, but how much of it is attributable to gravitational time dilation and how much to kinematic time dilation is not well defined. You can make cases for different splits (or argue that splitting things up like that is illegitimate on this case), and no one split is picked out by the physics.

I think you can get away with it if the shell is infinitesimally thin, but maybe not - delta function mass distributions are not really physical.

 

[PeterDonis]

Not even if the shell is zero thickness? That should define a unique relationship between the inside and outside planes of simultaneity, no?

For a stationary shell, it would. But this shell is not stationary. It's collapsing.

 

[Ibix]

What "planes of simultaneity"? The spacetime is not stationary.

The Minkowski bit is and so is the Schwarzschild bit. In the surface of the shell it's not, sure. But if the shell is infinitely thin, is that a problem?

 

[PeterDonis]

In the surface of the shell it's not, sure. But if the shell is infinitely thin, is that a problem?

Yes. See post #16.

 

[PeterDonis]

You can certainly push someone inside and pull them back out

The question was about comparing aging at infinity. An observer inside the event horizon can't get back out to infinity.

 

[Ibix]

The question was about comparing aging at infinity. An observer inside the event horizon can't get back out to infinity.

I was assuming the shell had not yet collapsed inside its own Schwarzschild radius. In that case you can, if you don't dawdle too much, drop in from far away then turn around and come back. Once the shell radius is less than the Schwarzschild radius you can't, sure.

 

[PeterDonis]

I was assuming the shell had not yet collapsed inside its own Schwarzschild radius. In that case you can, if you don't dawdle too much, drop in from far away then turn around and come back.

Not necessarily. It depends on how far inside the shell you go. The event horizon initially forms at the center and moves outward until it encounters the shell just as the shell's surface area equals $16 \pi M^2$, where $M$ is the externally measured mass of the shell.

 

[Ibix]

Well, yes, you have to be able to exit the shell before it crosses its own horizon.

 

[HansH]

so you could do this comparison at least for certain not too extreme situations? The reason why I ask this is because I would expect time flows the same inside the ring at at infinite distance as there is no curvature inside the ring. so we could at least conclude if this is true for a ring much larger than its event hozizon. Even if the ring crosses its own event horizon there is still no curvature inside the ring, so I would expect from inside the ring it does not make any difference how big the event horizon is. only problen is that the observer cannot get out of the ring then anymore to compare his flow of time with that of infinite distance. but not clear if this makes any difference as curvature inside the ring remains zero?

 

[HansH]

NThe event horizon initially forms at the center and moves outward until it encounters the shell

but how is that possible when inside the shell there is no matter so no curvature? so how can an event horizon than be inside the shell?

 

[Ibix]

The reason why I ask this is because I would expect time flows the same inside the ring at at infinite distance as there is no curvature inside the ring.

No, this is wrong. With a static shell where time dilation is well defined, the time dilation factor between the interior of the shell and infinity is the same as that at the surface of the shell. This is because time dilation depends on gravitational potential differences, not curvature.

 

[Ibix]

but how is that possible when inside the shell there is no matter so no curvature? so how can an event horizon than be inside the shell?

The event horizon is the surface dividing events that can send light signals to infinity (future null infinity, to be pedantic) and those that cannot. An any point within the shell there is a last chance to send signals to infinity, which is the last time you could send a signal that reaches the shell just before it crosses its own Schwarzschild radius. After that, you cannot send signals to infinity; ergo you are within the event horizon.

 

[HansH]

No, this is wrong. With a static shell where time dilation is well defined, the time dilation factor between the interior of the shell and infinity is the same as that at the surface of the shell. This is because time dilation depends on gravitational potential differences, not curvature.

Thanks. This was exactly why I asked the question. so time dilatation can only count up going inwards and not becoming lower even if curvature inside the shell is zero. so that also explains that if time dilatations gets infinite at the event hozizon this also holds for further going inwards.

 

[Ibix]

Thanks. This was exactly why I asked the question. so time dilatation can only count up going inwards and not becoming lower even if curvature inside the shell is zero. so that also explains that if time dilatations gets infinite at the event hozizon this also holds for further going inwards.

Time dilation isn't defined at the horizon, rather than being infinite. And note that what I said only applies to a static shell, not a collapsing one. You can't define gravitational time dilation for a collapsing shell of finite thickness (and Peter says not for one of zero thickness either - I'm not completely convinced, but Peter's right far more often than not).

 

[HansH]

The event horizon is the surface dividing events that can send light signals to infinity (future null infinity, to be pedantic) and those that cannot. An any point within the shell there is a last chance to send signals to infinity, which is the last time you could send a signal that reaches the shell just before it crosses its own Schwarzschild radius. After that, you cannot send signals to infinity; ergo you are within the event horizon.

it is a bit difficult for me to understand what you are saying but do you mean that within the shell one can send light signals outwards but not further than where the event horizon is when the lightbeam reaches it?

 

[Ibix]

it is a bit difficult for me to understand what you are saying but do you mean that within the shell one can send light signals outwards but not further than where the event horizon is when the lightbeam reaches it?

When the shell collapses so that it is completely contained within its own event horizon, an observer at the shell can no longer send signals to infinity: they are now inside the event horizon. That's the deadline for sending signals to infinity.

If an observer inside the shell wishes to send a signal to infinity they must send it early enough that it passes through the shell before that deadline. If they wait too long they can no longer send signals to infinity - which is to say that they are inside the horizon, even though they are in flat spacetime.

 

[Ibix]

For a stationary shell, it would. But this shell is not stationary. It's collapsing.

OK, here's my argument.

Inside the shell we have Minkowski spacetime, with a particular simultaneity convention picked out by the symmetry of the shell. Outside we have Schwarzschild spacetime. So I have two regions with well defined notions of simultaneity.

Consider a shell of finite thickness. I can establish a notion of simultaneity between a clock just above the shell and infinity. I cannot establish unique simultaneity between a clock just inside the shell and infinity. However, any definition of simultaneity I do define between the clock just inside and just outside the shell requires simultaneous events to be outside each others' light cones.

If I consider the same scenario with a thinner shell, I can have my just-inside and just-outside clocks closer together, so I have a narrower range of events I can consider simultaneous.

So if I consider a shell of delta-function thickness I can have my clocks arbitrarily close together. So do I have any choice about which event just inside and just outside the shell are simultaneous? You seem to think so, if I understand you correctly, but I don't see why.

 

[HansH]

I have another point bothering me since I saw a video (sorry I don't know the link anymore) probably not completely on topic but could help to get the picture more clear.

in the video it was told that in our universe you can calculate the amount of matter inside a certain volume and that the conclusion is that for a certain sphere of volume of the universe the corrsponding event horzon is within that shpere, drawing the conclusion that we could live inside a black hole. But if that would make sense, why is all the matter inside that event horizon than not forming a singularity?

or is what they say therefore simply not true?

 

[Ibix]

in the video it was told that in our universe you can calculate the amount of matter inside a certain volume and that the conclusion is that for a certain sphere of volume of the universe the corrsponding event horzon is within that shpere,

They (or you) are mixing up the Schwarzschild radius and the event horizon. You can calculate the Schwarzschild radius for any lump of matter, but it doesn't mean there's an event horizon there. For example, the Schwarzschild radius for the Sun is about 3km, but there is not an event horizon in the solar core.

Yes, if you pick a large enough region of the universe it's wholly contained within its Schwarzschild radius, but that doesn't automatically make it a black hole. On cosmological scales there is matter everywhere with more or less the same density. That is very different from a black hole spacetime, which has the mass concentrated in one region. So no, we don't live in a black hole (or at least, the line of argument you cite does not lead to that conclusion).

 

[PeterDonis]

you have to be able to exit the shell before it crosses its own horizon.

No, it's more stringent than that. You have to avoid crossing the horizon while you are inside the shell; that can happen even if the shell itself has not yet crossed the horizon. And there is no way to know, locally, where the horizon is inside the shell; you can't even use the local spacetime geometry to estimate it because the local spacetime geometry is flat everywhere inside the shell.

 

[PeterDonis]

Inside the shell we have Minkowski spacetime, with a particular simultaneity convention picked out by the symmetry of the shell.

Yes, this part is fine.

Outside we have Schwarzschild spacetime.

Yes, but bounded on the "inside" by the shell, whose areal radius $r$ is decreasing.

So I have two regions with well defined notions of simultaneity.

Not quite. For the region occupied by simultaneity surfaces that cross the shell before the shell crosses the horizon, the Schwarzschild region is entirely exterior to the horizon, so yes, it will have a well-defined set of simultaneity surfaces picked out by orthogonality to the timelike Killing vector field in the exterior region. But you still have to match the surfaces, and there is an issue there that you do not appear to have recognized; see below.

But once the shell crosses the horizon, the Schwarzschild region includes the horizon and a portion inside the horizon, and Schwarzschild spacetime is not stationary there (the KVF is null on the horizon and spacelike inside it), so there are no well-defined surfaces of simultaneity defined by the KVF.

if I consider a shell of delta-function thickness I can have my clocks arbitrarily close together. So do I have any choice about which event just inside and just outside the shell are simultaneous? You seem to think so, if I understand you correctly

Sort of. The issue I am thinking of is that, heuristically, the Schwarzschild simultaneity surfaces outside the shell and the Minkowski simultaneity surfaces inside the shell are not parallel--except possibly at one particular instant of time. Heuristically, you can see this by imagining trying to match Rindler simultaneity surfaces with Minkowski (inertial frame) simultaneity surfaces, across a boundary at some finite value of Minkowski $x$ to the right (positive $x$ direction) of the origin. Except for the surface at Minkowski $t = 0$, the surfaces are not parallel. Schwarszschild simultaneity surfaces, heuristically, behave like Rindler simultaneity surfaces, so the same issue will arise with them.

You could argue that the "kink" in each simultaneity surface is due to the idealization of a shell of zero thickness, and that for a shell of finite thickness the kink would be smoothed out into a curve. But then there is ambiguity about how to do the smoothing.

 

[HansH]

if you pick a large enough region of the universe it's wholly contained within its Schwarzschild radius, but that doesn't automatically make it a black hole.

Here I cannot follow you anymore. As far as I understand is that for any mass a Schwarzschild radius can be calculated and when alle this mass is compressed withing this Schwarzschild radius then the Schwarzschild radius becomes an event horizon. For the sun this is of course not the case because within this 3km radius only a small fraction of the solar mass is in. sou you first must compress the sun to 3km radius to get that done.
For me however it is not clear if it is possible to have a region of normal space (which has on average a density of almost nothing) with all its mass within its own Schwarzschild radius. I understand you indicate that this is not possible as I also would expect.
But why does someone then tell such thing in a video Is there some equation that shows that the larger the volume, the lower the average mass density required to form a black hole that could trigger such person to tell something like that?

 

[Ibix]

As far as I understand is that for any mass a Schwarzschild radius can be calculated and when alle this mass is compressed withing this Schwarzschild radius then the Schwarzschild radius becomes an event horizon.

If it's in an otherwise empty spacetime, sure. Cosmological spacetime is filled with matter everywhere, though. So by your argument you can pick any sufficiently large spherical volume and argue that it "ought to be a black hole". Even overlapping volumes "ought to be black holes", and you are simultaneously in infinitely many such volumes. Which direction do you imagine you will fall if you are simultaneously trapped inside infinitely many black holes?

For me however it is not clear if it is possible to have a region of normal space (which has on average a density of almost nothing) with all its mass within its own Schwarzschild radius.

The Schwarzschild radius is $R_S=2GM/c^2$. The mass contained in a spherical volume is $\frac 43\rho r^3$. You want to know if it's possible for the radius containing a mass to be less than the Schwarzschild radius associated with that mass:$$\begin{eqnarray*}
r&<&R_S\\
&<&\frac{2G}{c^2}\frac 43\rho r^3\\
\frac{3c^2}{8G\rho}&<&r^2
\end{eqnarray*}$$so, as I said, any sufficiently large radius is larger than the Schwarzschild radius of the mass contained within it if $\rho$ is constant throughout space, no matter how small that value is. But the region still won't turn into a black hole because "all the mass inside the Schwarzschild radius" may be a necessary condition for a black hole to form, but it is not a sufficient condition on its own.

 

[PeroK]

Here I cannot follow you anymore. As far as I understand is that for any mass a Schwarzschild radius can be calculated and when alle this mass is compressed withing this Schwarzschild radius then the Schwarzschild radius becomes an event horizon. For the sun this is of course not the case because within this 3km radius only a small fraction of the solar mass is in. sou you first must compress the sun to 3km radius to get that done.
For me however it is not clear if it is possible to have a region of normal space (which has on average a density of almost nothing) with all its mass within its own Schwarzschild radius. I understand you indicate that this is not possible as I also would expect.
But why does someone then tell such thing in a video Is there some equation that shows that the larger the volume, the lower the average mass density required to form a black hole that could trigger such person to tell something like that?

Understanding the nature of the Schwarzschild spacetime is not trivial. It helps to understand the mathematics to some extent, at least. Understanding the formation of a black hole from a collapsing star is yet more complex. And it's almost impossible to convey this adequately using videos or non-mathematically in general.

From your posts generally, I suspect you haven't fully digested the nature of spacetime as a four-dimensional manifold; but, rather, are thinking of space and then a separate, dissociated time dimension. For example, the singularity is not a point in space where you can sit and wait while universal time ticks by until that point in space becomes a singularity. This is not the right way to think about things at all! In fact, singularity really means a failure of the mathematical model. One could argue there is no such thing as a singularity, other than a deficiency in the mathematical model (i.e. GR as we know it today).

We all hope that eventually a theory of quantum gravity will resolve what really happens in the case of a collapsing star.

 

[Nugatory]

But why does someone then tell such thing in a video Is there some equation that shows that the larger the volume, the lower the average mass density required to form a black hole that could trigger such person to tell something like that?

That part is indeed right - as the mass increases the "density" required to fit all that mass inside the event horizon decreases. However, there are a bunch of complexities here, starting with the scare-quotes around the word "density" - that quantity does not correspond to the classical notion of density as mass per unit volume. However, this is somewhat of a distraction when considering the question about whether the universe could be the interior of a black hole.

More important is that everything you see about black hole even horizons is based on the the Schwarzschild and other solutions of the Einstein field equations for a mass sitting in an otherwise empty and asymptotically flat universe. This is not an exactly accurate description of anything, but it turns out to be a really good approximation for stellar black holes - there's nothing else nearby and the gravitational effects of more distant objects can be ignored (this is somewhat analogous to the way that we don't worry about distant starts when we're calculating planetary orbits inside the solar system).

However, it is a completely inaccurate description of an entire universe filled with a large-scale uniform, small scale lumpy distribution of mass. Thus none of the Schwarzschild spacetime stuff applies to the universe as a whole. It's a different solution of the field equations; we can plug the mass into the formula for the Schwarzschild radius but that calculation isn't telling us anything useful.

 

[HansH]

This is not an exactly accurate description of anything, but it turns out to be a really good approximation for stellar black holes - there's nothing else nearby and the gravitational effects of more distant objects can be ignored

Do you mean that the formation of a black hole in this way is applicable to the situation where the mass density in the region where the black hole forms is much larger than the mass density outside that region? such as valid for a collapsing star, but not the case in a more or less equal distribution of mass in the universe? That sounds logical to me because then you could choose any arbitrary sphere to define your event horizon but that is of course not valid.
so just for my understanding: if this is true and I would remove all the mass outside a sphere in the universe where the mass inside causes the Schwarzschild radius to be that sphere, do I then get an event horizon at that radius and a black hole?

If this is true and I then think one step further then the whole universe must be more or less having an equal density of matter because otherwise we would get spontanuous black holes in certain regions with a large mass bubble and sufficient empty space around it, because the equetion you refer makes it more likely to form black holes in a very large structure of space.

 

[PeroK]

If this is true and I then think one step further then the whole universe must be more or less having an equal density of matter because otherwise we would get spontanuous black holes in certain regions with a large mass bubble and sufficient empty space around it, because the equetion you refer makes it more likely to form black holes in a very large structure of space.

The density of matter in any structure other than a star or planet is extremely low. The only object that could collapse to a black hole is a large star.

 

[Ibix]

Do you mean that the formation of a black hole in this way is applicable to the situation where the mass density in the region where the black hole forms is much larger than the mass density outside that region?

Not quite. The Schwarzschild solution is an eternal black hole - it just exists forever, it does not form from anything. It's not a realistic solution, but it's probably the simplest known solution to the Einstein field equations. And once a black hole has formed it won't be very different from a Schwarzschild black hole for most purposes.

if this is true and I would remove all the mass outside a sphere in the universe where the mass inside causes the Schwarzschild radius to be that sphere, do I then get an event horizon at that radius and a black hole?

What you are describing is exactly the time-reversal of the Oppenheimer-Snyder black hole model. Oppenheimer-Snyder is a simple model of a stellar collapse into a black hole that is literally a spherical region of collapsing FLRW spacetime with vacuum outside. You've just replaced the collapsing region with an expanding region, so you are describing a white hole, which has an anti-horizon instead of an event horizon.

If this is true and I then think one step further then the whole universe must be more or less having an equal density of matter because otherwise we would get spontanuous black holes in certain regions with a large mass bubble and sufficient empty space around it, because the equetion you refer makes it more likely to form black holes in a very large structure of space.

No, we model the universe as having more or less uniform mass density everywhere because that's what we see. You can add perturbations that tend to evolve into areas of high density that could form structures like galaxies and stars and eventually black holes. Sufficiently large perturbations might form "primordial" black holes directly, and people have looked for signatures of such things. No success, as far as I am aware.

 

[PeterDonis]

As far as I understand is that for any mass a Schwarzschild radius can be calculated and when alle this mass is compressed withing this Schwarzschild radius then the Schwarzschild radius becomes an event horizon.

That is not correct. What is correct is that if the matter is collapsing, and it collapses to within the Schwarzschild radius for its mass, it becomes a black hole and an event horizon forms around it. But our universe is not collapsing; it is expanding.

Also, as @Ibix has already pointed out, the "collapse" process described above assumes an isolated system of finite extent surrounded by empty space. Our universe is no such thing.

So there are two reasons why the heuristic rule of thumb you describe does not apply to our universe as a whole.

 

[PeterDonis]

For me however it is not clear if it is possible to have a region of normal space (which has on average a density of almost nothing) with all its mass within its own Schwarzschild radius.

Yes, it is possible. A black hole has vacuum inside; all of the matter that originally collapsed to form the hole disappears into the singularity. (At least, that's what classical GR says; we don't have a theory of quantum gravity so we don't know how quantum effects might change this.)

One common misconception is that a black hole has to be "made of" matter in order to have mass. It doesn't. The "mass" of a black hole is a property of its spacetime curvature; spacetime curvature is what the hole is "made" of. As noted above, the hole is vacuum inside, i.e., "empty space".

 

[HansH]

What you are describing is exactly the time-reversal of the Oppenheimer-Snyder black hole model. Oppenheimer-Snyder is a simple model of a stellar collapse into a black hole that is literally a spherical region of collapsing FLRW spacetime with vacuum outside. You've just replaced the collapsing region with an expanding region, so you are describing a white hole, which has an anti-horizon instead of an event horizon.

Not sure If I made myself sufficiently clear. I mean all mass is still withing the sphere and no mass outside the sphere. So then I do not understand why you talk about an expanding region instead of a collapring region as there are only attracting forces between mass particles making the particles move towards each other.

I am basically looking for a proper criterion to be able to determine when in such theoretical situation a black hole forms.(for me the goal is not to understand all the details, but to get a general idea of how black holes could form)
In the first post I used a spherical shell of iron particles with 'sufficient' mass to form a black hole. Now we are talking about what is 'sufficient' and is there the relation between radius of the contracting shell, the moment that an event horizon forms and the minimum amount of mass of the shell to be able to form a black hole at all. and does it make a difference if this mass is all concentrated in a thin shell or evenly dense within the whole sphere. And then some posts back I was wandering if we could extend this situation to very large structures because of the equations posted in #37. but the answers on that still confuse me.

 

[PeroK]

I am basically looking for a proper criterion to be able to determine when in such theoretical situation a black hole forms.

See here for details of the minimum mass of star for a stellar black hole to form:

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

Note that if we take the Newtonian gravity for a star of radius $R$ and fixed density $\rho$, then the gravitational acceleration on the surface is:
$$g = \frac{GM}{R^2} = \frac{4\pi G\rho R^3}{3R^2} = \frac{4\pi G\rho R}{3}$$Where I've used the formula for the volume of a sphere to get the mass in terms of the radius and density. From this you can see that the larger a sphere becomes, the greater its surface gravity becomes (approximately a linear function of $R$). Eventually, if it is large enough, the surface gravity will cause the sphere to start to collapse. The final state for such a sphere depends on how big it is: e.g. it could collapse to a neutron star:

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

Or, if it is even larger and the collapse can overcome neutron pressure, then it may collapse to a black hole.

 

[Ibix]

Not sure If I made myself sufficiently clear. I mean all mass is still withing the sphere and no mass outside the sphere.

It depends on your initial conditions, then. If the matter is already expanding rapidly enough (which, on the large scale, matter in the universe is doing) then it will not collapse into a black hole - instead, that's the dying moments of a white hole. If the matter is static then yes, it'll collapse into a black hole.

I am basically looking for a proper criterion to be able to determine when in such theoretical situation a black hole forms.

Gravitational collapse happens as long as the matter isn't moving outwards fast enough. The collapse can be halted by various processes (e.g. forming a planet or a star) but may start up again if that process is time limited (a planet can support itself against its own gravity and does not collapse further, but a star is supported by emissions from its fusion reactions, which eventually burn out). The last process we're aware of that can stop a star collapsing is neutron degeneracy pressure. If the mass is sufficiently great to overcome this, there's nothing we know of that can stop the star collapsing into a black hole.

So generally, you need about 1.4 solar masses of matter and you need it not to be expanding initially. Or you can use less matter and provide some mechanism to slam it together really hard (which is why people got worried about the LHC possibly producing a black hole, although in reality it's far too low energy to do so).

 

[Hornbein]

The last process we're aware of that can stop a star collapsing is neutron degeneracy pressure. If the mass is sufficiently great to overcome this, there's nothing we know of that can stop the star collapsing into a black hole.

So generally, you need about 1.4 solar masses of matter

Actually about 2 solar masses to overcome neutron degeneracy pressure. The minimal mass of a neutron star is 1.4.

 

[PeterDonis]

If the matter is already expanding rapidly enough (which, on the large scale, matter in the universe is doing) then it will not collapse into a black hole - instead, that's the dying moments of a white hole.

This is not quite true. Our universe did not start out as a white hole. A white hole is an isolated region surrounded by empty space, just like a black hole. Our universe is not like that.

 

[PeterDonis]

I mean all mass is still withing the sphere and no mass outside the sphere.

This is not true of our universe; there is no way to draw a sphere around any region occupied by mass in our universe and have all the mass inside and no mass outside. So the reasoning you are trying to use simply cannot be applied to our universe.