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.