On the gravitational collapse of a massive shell

[Mattergauge]

TL;DR Summary

The formation of a black hole as a collapsing massive shell. What happens to the inside metric? What happens as seen from the center?

Black holes form. An undeniable fact. Let's imagine a massive shell collapsing under its own weight (the exact composition of the mass is not important, so just imagine to be a continuous mass with zero thickness).

What happens if the process of collapse evolves? The time on the inside will run slower and slower, as compared to the outside. The metric inside will stay flat during collapse. Bow is this information conveyed to the inside, without resorting to hypothetical gravitons? Are the some kinds of flat gravitational waves involved? Outside of the shell the Schwarzschild metric will follow the collapsing mass. Inside, the metric is flat, while the time component of the metric follows that of the component outside. What happens if the metric reaches the artificial singularity at the Schwarzschild radius? On the inside, one can see the shell going on to collapse. It's confusing though. From the outside one sees the shell stop collapsing on the event horizon, but then the horizon has to be there in the first place, and the horizon is not yet there, as the hole hasn't formed yet. But still the shell seems to slow down in collapsing. And seen from the outside, there is nothing behind the horizon. The hole seems to have zero volume. As opposed to the inside. Any thoughts?

 

[Dale]

Summary:: The formation of a black hole as a collapsing massive shell. What happens to the inside metric? What happens as seen from the center?

Bow is this information conveyed to the inside, without resorting to hypothetical gravitons?

What information?

Inside, the metric is flat, while the time component of the metric follows that of the component outside.

What do you mean by “time component of the metric follows that of the component outside”. The metric inside is just the ordinary standard Minkowski metric. Nothing about the time component changes.

 

[Mattergauge]

What information?

What do you mean by “time component of the metric follows that of the component outside”. The metric inside is just the ordinary standard Minkowski metric. Nothing about the time component changes.

But the time component outside the shell is that of the Schwarzschild metric. Shouldn't the metric be continuous? I mean, shouldn't the inside metric "follow" the metric on the outside. "Flat-extended", so to speak?

 

[Mattergauge]

But the time component outside the shell is that of the Schwarzschild metric. Shouldn't the metric be continuous? I mean, shouldn't the inside metric "follow" the metric on the outside. "Flat-extended", so to speak?

Of course the manifold is a diffeomorpism, so the massive surface poses a problem here, being infinitely small. But you can give it a small thickness.

 

[Dale]

But the time component outside the shell is that of the Schwarzschild metric. Shouldn't the metric be continuous? I mean, shouldn't the inside metric "follow" the metric on the outside. "Flat-extended", so to speak?

Neither a vector nor a tensor has a component as an intrinsic part of the vector or tensor itself. Components describe the tensor with respect to a given basis. As you move from event to event in the manifold, the basis need not be continuous, and therefore the components need not be continuous, even if the tensor itself is. However, with a thin shell the metric will change abruptly, as dictated by the EFE.

The metric inside is simply ordinary flat spacetime. Locally nothing is changed and there is no information required.

 

[Mattergauge]

Neither a vector nor a tensor has a component as an intrinsic part of the vector or tensor itself. Components describe the tensor with respect to a given basis. As you move from event to event in the manifold, the basis need not be continuous, and therefore the components need not be continuous, even if the tensor itself is. However, with a thin shell the metric will change abruptly, as dictated by the EFE.

The metric inside is simply ordinary flat spacetime. Locally nothing is changed and there is no information required.

The manifold has to be a diffeomorphism though. A clock, as seen from the outside (thereby using coordinates and thus a preferred base, of course) on the inside of the sphere seems to tick slower. At the same rate as on the surface of the mass. So what about the coordinate independent manifold on the inside? For someone on the inside the metric is flat. With a normal +--- Minkowskian metric. But isn't that putting coordinates on it too? If a clock from the inside is compared with a clock far away from the shell, there will be a difference. Of course I compare coordinates here. The inside is just flat, but can't an equivalent flatness be achieved if you transplant the outside metric to the inside, so the metric is continuous when crossing the shell?

The manifold itself, independent of coordinates, must be a diffeomorphism, which it isn't when the metric changes abruptly. Is this because of the assumption that the shell is 2d?

 

[Dale]

The manifold has to be a diffeomorphism though.

That doesn’t make sense as stated. A diffeomorphism is a map between two manifolds. A single manifold can’t be a diffeomorphism by itself.

A clock, as seen from the outside (thereby using coordinates and thus a preferred base, of course) on the inside of the sphere seems to tick slower.

Sure, but that is not a local property of the interior metric. The interior metric is just the ordinary flat metric.

To get from a clock in the interior to an observer in the exterior you must pass through the shell. A null geodesic will parallel transport its tangent through the shell so as to be appropriately redshifted for any hovering observer. This does not in any way imply that the interior metric is anything other than the completely standard Minkowski metric.

For someone on the inside the metric is flat. With a normal +--- Minkowskian metric. But isn't that putting coordinates on it too?

No, stating the signature and stating that the metric is flat in no way puts coordinates on it. Of course, if you wish to put coordinates on it you certainly may.

For example, on flat spacetime we can use coordinates where:

$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$

Or we can use coordinates where

$ds^2 = -dt^2 + dr^2 + r^2 d\theta^2 + r^2 \sin(\theta)^2 d\phi^2$

Or we can use coordinates where

$ds^2 = - a dT^2 + a dX^2 + a dY^2 + a dZ^2$

All these coordinates are valid coordinates on the same flat Minkowski metric.

The manifold itself, independent of coordinates, must be a diffeomorphism

Again, this doesn’t make sense as stated.

 

[PAllen]

I will give a summary of collapse of shell treated purely classically in GR (the case with even just corrections for Hawking radiation is disputed among experts; there is a majority view, but not a strong consensus). Note, that so far as I know, there are no exact solutions for collapsing shell, so all treatments use approximations or numerics. The only collapse scenario I know of that has been treated exactly is the dust collapse solution of Oppenheimer and Snyder.

Inside the shell, you have flat Minkowski spacetime. Within the shell, you have some nonzero stress energy tensor (SET), equating via the EFE, to nonzero Ricci curvature. It would typically be assumed that this SET would satisfy the dominant energy condition (or, to make it easier to draw rigorous conclusions, strong energy condition is often assumed, even though this is non quite realistic). Outside the shell, you would have Schwarzschild metric.

At the boundaries, for practical treatment, the metric is required to be continuous, but not differentiable (discontinuous first derivatives). There is nothing unusual about this in any branch of physics, e.g. in classical EM discontinuities in charge density are used at boundaries all the time. Trying to smooth out boundaries simply makes the problem mathematically intractable without changing the predictions in any meaningful way.

In GR, there are various extra conditions imposed on boundaries besides metric continuity (for physicial plausibility), but I don't think the details are relevant to this discussion.

As the collapse continues, the interior region shrinks, and the Schwarzschild boundary moves inwards, and of course, the average density of the shell increases. The absolute event horizon (as distinct from apparent horizons) starts at the center of the interior region and grows outward, until crossing the outer surface of the collapsing shell. A singularity forms shortly after the interior flat region disappears. At this point, it may be assumed that the whole spacetime is the same as the late stage of the Openheimer Snyder collapse, generally assumed to be the Schwarzschild exterior plus one half (really one sheet) of one Kruskal interior quadrant.

This is what happens, and there is no 'relativity' in what happens. Different observers will be able to observe different parts of what happens, based on the relevant causal relations. For example, making an assumption that the shell is transparent, and there is little light emitter in the center of the shell, a distant observer first sees the light emitter go dark, and a little later cannot receive signal from the shell (for practical purposes - infinite redshift and time delay can only be observed as absence of any signal). A detector in the center continues to see the outside universe evolve at the same rate throughout the collapse - while not exactly applicable here, the notion of potential difference between the center and spatial infinity gives a good approximation for time dilation. This does not change at all during the whole collapse process, so the overall blue shift of the distant universe never changes for a central observer, as long as they can exist.

I would request the OP study this description well before asking further questions, so we have a common ground for discussion.

 

[PAllen]

One key additional point is that an "inside shell" observer is not at all similar to an "outside shell stationary observer". The former would typically be taken to be inertial observers, while the latter have unbounded proper acceleration on approach to the shell, as the collapse proceeds (and also, velocity approaching c relative to free fall from infinity trajectory). Any application of features of stationary external observers to internal observers, is then just wrong on first principles.

 

[Mattergauge]

That doesn’t make sense as stated. A diffeomorphism is a map between two manifolds. A single manifold can’t be a diffeomorphism by itself.

Sure, but that is not a local property of the interior metric. The interior metric is just the ordinary flat metric.

To get from a clock in the interior to an observer in the exterior you must pass through the shell. A null geodesic will parallel transport its tangent through the shell so as to be appropriately redshifted for any hovering observer. This does not in any way imply that the interior metric is anything other than the completely standard Minkowski metric.

No, stating the signature and stating that the metric is flat in no way puts coordinates on it. Of course, if you wish to put coordinates on it you certainly may.

For example, on flat spacetime we can use coordinates where:

$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$

Or we can use coordinates where

$ds^2 = -dt^2 + dr^2 + r^2 d\theta^2 + r^2 \sin(\theta)^2 d\phi^2$

Or we can use coordinates where

$ds^2 = - a dT^2 + a dX^2 + a dY^2 + a dZ^2$

All these coordinates are valid coordinates on the same flat Minkowski metric.

Again, this doesn’t make sense as stated.

Now that's an answer. Still... Shouldn't the manifold be differentiable everywhere? By which I mean, shouldn't it be without a sudden change in metric, as from Schwartzschild to flat Minkowskian, as happens on the shell, where the metric isn't defined?

 

[PAllen]

Now that's an answer. Still... Shouldn't the manifold be differentiable everywhere? By which I mean, shouldn't it be without a sudden change in metric, as from Schwartzschild to flat Minkowskian, as happens on the shell, where the metric isn't defined?

No. See my post, which discusses this very point. Also, please note that the smoothness of a manifold has nothing to do with the smoothness of e.g. tensor fields defined on it. A manifold can be smoothly homeomorphic to Rⁿ in any sufficiently small neighborhood, while having a metric that is not everywhere smooth.

 

[Dale]

Shouldn't the manifold be differentiable everywhere? By which I mean, shouldn't it be without a sudden change in metric, as from Schwartzschild to flat Minkowskian, as happens on the shell, where the metric isn't defined?

That only happened because you specified a thin shell. That is perfectly fine if you don’t care about what happens on the shell itself. If you want to know what happens on the shell itself then you can’t treat it as thin.

As long as you only care about length scales much larger than the thickness of the shell then you can simply allow the metric to be non-differentiable at the shell without losing any of the physics.

 

[PAllen]

Now that's an answer. Still... Shouldn't the manifold be differentiable everywhere? By which I mean, shouldn't it be without a sudden change in metric, as from Schwartzschild to flat Minkowskian, as happens on the shell, where the metric isn't defined?

I assumed a finite size thin shell. In this case, the metric is well defined everywhere. As I discussed, not only is it defined, one typically assumes it meets one of the energy conditions. There are still boundaries where the metric is not differentiable - shell to interior, and shell to exterior.

One can also treat the case of a shell that has no thickness, modeled as a type of SET delta function. Analyses of this type have also been done. I missed that you wanted zero thickness, which requires this approach. But feature of this approach is that there is effectively an SET defined on the boundary, with the Ricci curvature having a delta function description.

 

[Mattergauge]

I will give a summary of collapse of shell treated purely classically in GR (the case with even just corrections for Hawking radiation is disputed among experts; there is a majority view, but not a strong consensus). Note, that so far as I know, there are no exact solutions for collapsing shell, so all treatments use approximations or numerics. The only collapse scenario I know of that has been treated exactly is the dust collapse solution of Oppenheimer and Snyder.

Inside the shell, you have flat Minkowski spacetime. Within the shell, you have some nonzeror stress energy tensor (SET), equating via the EFE, to nonzero Ricci curvature. It would typically be assumed that this SET would satisfy the dominant energy condition (or, to make it easier to draw rigorous conclusions, strong energy condition is often assumed, even though this is non quite realistic). Outside the shell, you would have Schwarzschild metric.

At the boundaries, for practical treatment, the metric is required to be continuous, but not differentiable (discontinuous first derivatives). There is nothing unusual about this in any branch of physics, e.g. in classical EM discontinuities in charge density are used at boundaries all the time. Trying to smooth out boundaries simply makes the problem mathematically intractable without changing the predictions in any meaningful way.

In GR, there are various extra conditions imposed on boundaries besides metric continuity (for physicial plausibility), but I don't think the details are relevant to this discussion.

As the collapse continues, the interior region shrinks, and the Schwarzschild boundary moves inwards, and of course, the average density of the shell increases. The absolute event horizon (as distinct from apparent horizons) starts at the center of the interior region and grows outward, until crossing the outer surface of the collapsing shell. A singularity forms shortly after the interior flat region disappears. At this point, it may be assumed that the whole spacetime is the same as the late stage of the Openheimer Snyder collapse, generally assumed to be the Schwarzschild exterior plus one half (really one sheet) of one Kruskal interior quadrant.

This is what happens, and there is no 'relativity' in what happens. Different observers will be able to observe different parts of what happens, based on the relevant causal relations. For example, making an assumption that the shell is transparent, and there is little light emitter in the center of the shell, a distant observer first sees the light emitter go dark, and a little later cannot receive signal from the shell (for practical purposes - infinite redshift and time delay can only be observed as absence of any signal). A detector in the center continues to see the outside universe evolve at the same rate throughout the collapse - while not exactly applicable here, the notion of potential difference between the center and spatial infinity gives a good approximation for time dilation. This does not change at all during the whole collapse process, so the overall blue shift of the distant universe never changes for a central observer, as long as they can exist.

I would request the OP study this description well before asking further questions, so we have a common ground for discussion.

As if one is there to see it, very illuminating! So seen from the center, nothing happens until the shell arrives? This will happen in a comparatively small time. Compared to a distant observer looking at the collapsing shell, to whom it looks, in his own time, that it takes a whole lot more time. How much time would it take for an observer falling along with the shell? Is that undefined? For an observer stationary in the space near the mass, the collapse takes less and less time, if the distance to the shell decreases (but the observer is stationary). Is the objective answer that there just are a lot of ways to observe, and no "bird perspective" is possible? If the horizon is not yet present, can we say that a distant observer sees the shell to get frozen on the horizon? That the shell "becomes" the horizon" and that it's frozen in time? Or sees the distant observer only empty space on which later infalling matter gets stuck? But then again, this would only increase the Schwartzschild radius. There is no difference between the inside flat spacetime and the spacetime far away from the shell (assuming no further mass). But isn't the metric on the inside, as compared to the flat metric far away (although both flat and Minkowskian), somehow different from it (on their own they are the same indeed)?

 

[Mattergauge]

I assumed a finite size thin shell. In this case, the metric is well defined everywhere. As I discussed, not only is it defined, one typically assumes it meets one of the energy conditions. There are still boundaries where the metric is not differentiable - shell to interior, and shell to exterior.

One can also treat the case of a shell that has no thickness, modeled as a type of SET delta function. Analyses of this type have also been done. I missed that you wanted zero thickness, which requires this approach. But feature of this approach is that there is effectively an SET defined on the boundary, with the Ricci curvature having a delta function description.

Thanks for the great replies! I'll delete this comment later, as it doesn't contribute to the discussion. Just wanted to let you know.

I'm not sure why a Dirac delta is used though. The mass on the shell is finite everywhere. There is just a sudden jump to a flat metric or vice-versa.

 

[ergospherical]

I'm not sure why a Dirac delta is used though. The mass on the shell is finite everywhere. There is just a sudden jump to a flat metric or vice-versa.

For the thin shell? If you put $\rho(r) = a \delta(r-R)$, then constrain\begin{align*}
M \overset{!}{=} \int_{\mathbf{R}^3} \rho(r) dV = a \int_{\mathbf{R}^3} \delta(r-R) r^2 dr d\Omega = 4\pi a R^2
\end{align*}you find $\rho(r) = \dfrac{M}{4\pi R^2} \delta(r-R)$ as the density function

 

[Dale]

But isn't the metric on the inside, as compared to the flat metric far away (although both flat and Minkowskian), somehow different from it (on their own they are the same indeed)?

The metric is locally the same. All local invariants are the same. All local admissible charts on one are admissible in the other. To get any difference requires a non-local operation.

 

[PAllen]

As if one is there to see it, very illuminating! So seen from the center, nothing happens until the shell arrives?

Correct.

This will happen in a comparatively small time.

Yes.

Compared to a distant observer looking at the collapsing shell, to whom it looks, in his own time, that it takes a whole lot more time.

Depends what you mean. The shell becomes a thousand time blacker than CMB in quite a short time for a distant observer.

How much time would it take for an observer falling along with the shell? Is that undefined?

It is perfectly well defined, and would be quite short.

For an observer stationary in the space near the mass, the collapse takes less and less time, if the distance to the shell decreases (but the observer is stationary).

No. For any stationary external observer, no matter how close to the horizon, the shell is never 'visually seen' to cross the horizon. It get arbitrarily black (e.g. 1000 times blacker than CMB) in a short time per this observer, but nothing is ever seen from the moment of shell and horizon crossing.

Is the objective answer that there just are a lot of ways to observe, and no "bird perspective" is possible?

Correct.

If the horizon is not yet present, can we say that a distant observer sees the shell to get frozen on the horizon?

Depends what you mean. Mathematically, you can say that the past lightcone of the distant observer always includes events from the shell being just outside the horizon. Visually, using any plausible model of photon emission, the last photon of any frequency ever received by a distant observer occurs quite soon after the beginning of catastrophic collapse. The only common source I know of which discusses this anyalysis is MTW.

That the shell "becomes" the horizon" and that it's frozen in time? Or sees the distant observer only empty space on which later infalling matter gets stuck? But then again, this would only increase the Schwartzschild radius. There is no difference between the inside flat spacetime and the spacetime far away from the shell (assuming no further mass). But isn't the metric on the inside, as compared to the flat metric far away (although both flat and Minkowskian), somehow different from it (on their own they are the same indeed)?

Now, you lose me. Whatever any different observers see, the horizon and shell cross, the shell continues collapsing, and the horizon eventually has fixed area.

An observer infalling with the shell would see the distant universe proceeding normally (moderate blueshift) until singularity, and then nothing more.

As best I can tell from confused comments in this paragraph, you are mixing up local geometry (local metric) properties versus global properties. The global past light cone of some observer is affected in large part by global properties of a manifold, including its topology. You can take any small region of the past light cone and replace the geometry with flat Minkowski, with appropriate junction conditions, and this will have no affect on the global properties of the past light cone.

 

[Dale]

I'm not sure why a Dirac delta is used though. The mass on the shell is finite everywhere. There is just a sudden jump to a flat metric or vice-versa.

The mass is finite, but the mass density is not. (For a thin shell)

 

[PAllen]

The mass is finite, but the mass density is not. (For a thin shell)

And, of course, it is densities that appear in the SET, thus a delta function is required for it.

 

[Mattergauge]

The mass is finite, but the mass density is not. (For a thin shell)

You mean that the mass density is infinite on the 2d shell? The surface mass density is finite, so isn't the volume density also? Or do you mean the density is zero (which is finite though)? Or is it the density in the radial direction only?

 

[PAllen]

You mean that the mass density is infinite on the 2d shell? The surface mass density is finite, so isn't the volume density also? Or do you mean the density is zero (which is finite though)? Or is it the density in the radial direction only?

Say what? What is the volume of a zero thickness shell? Density has only one definition (for most purposes in physics) - quantity divided by volume.

 

[Mattergauge]

You mean that the mass density is infinite on the 2d shell? The surface mass density is finite, so isn't the volume density also? Or do you mean the density is zero (which is finite though)? Or is it the density in the radial direction only?

Yes indeed, that's clear. Like the mass density of a point mass is infinite, while the mass is finite. I guess I mixed up the mass with the whole volume it's in. The surface density is finite (like the mass of the point particle) but the density per volume is infinite (like the mass of the point particle per meter is). All clear!

 

[martinbn]

That doesn’t make sense as stated. A diffeomorphism is a map between two manifolds. A single manifold can’t be a diffeomorphism by itself.

Off topic but his phrasing reminds me of the joke where on an exam problem two manifolds X and Y are given. The question is: Are they diffeomorphic? A student answers: X is diffeomorphic, Y is not.

 

[Mattergauge]

Off topic but his phrasing reminds me of the joke where on an exam problem two manifolds X and Y are given. The question is: Are they diffeomorphic? A student answers: X is diffeomorphic, Y is not.

Haha! That's exactly what I thought. Nevertheless. If they are not diffeomorphic that means one of them has non-differentiable properties. Its form (morphic) is non-differentiable (non-diffeo) at places, like on the surface of the collapsing 2d shell. Of course you can't differentiate manifolds, but still it makes sense (at least to me) to speak of diffeomorphic manifolds... Looking at their form, I can tell imediately. If they are not diffeomorphic then it's a non-physical manifold (as the one discussed here).

 

[Dale]

it makes sense (at least to me) to speak of diffeomorphic manifolds

No, it really doesn’t. The word you are looking for is “differentiable”. A single manifold can be differentiable, but diffeomorphic refers to a mapping between two.

 

[Mattergauge]

No, it really doesn’t. The word you are looking for is “differentiable”. A single manifold can be differentiable, but diffeomorphic refers to a mapping between two.

I wrote "to me". If there is a diffeomorphism between two manifolds, this has implications for the manifolds themselves. Like being differentiable or not. Can you construct diffeomorphisms between two manifolds that contain non-differentiable regions? Yes, of course. But how look the maps involved? Are these maps differentiable? Not if one manifold is not. If one manifold is the one we consider here, and the other is the the quasi Riemannian Minkowkski manifold, the map from the non-differentiable part of the manifold (the 2d shell) will the map be differentiable?

 

[Mattergauge]

I wrote "to me". If there is a diffeomorphism between two manifolds, this has implications for the manifolds themselves. Like being differentiable or not. Can you construct diffeomorphisms between two manifolds that contain non-differentiable regions? Yes, of course. But how look the maps involved? Are these maps differentiable? Not if one manifold is not. If one manifold is the one we consider here, and the other is the the quasi Riemannian Minkowkski manifold, the map from the non-differentiable part of the manifold (the 2d shell) will the map be differentiable?

Of course, this does injustice to the term "diffeomorphism".

 

[PAllen]

Two smooth (infinitely differentiable) manifolds need not be diffeomorphic (among other obstacles, there may topological obstructions to the existence of a diffeomorphism).

Further, the smoothness of a manifold is independent of whether a metric you choose to impose on it is everywhere differentiable. Thus you can have two smooth manifolds connected by a diffeomorphism which carries the nondifferentiable metric from one to the other.

There really is just no sensible way to talk about one manifold being diffeomorphic, by itself. It is like asking if a set, by itself, is a differentiable function ??!

 

[PeterDonis]

I wrote "to me".

You can't expect to communicate effectively with others about physics if you make up your own terminology. You are being told the standard terminology. Please use it.

 

[Mattergauge]

Two smooth (infinitely differentiable) manifolds need not be diffeomorphic (among other obstacles, there may topological obstructions to the existence of a diffeomorphism).

Further, the smoothness of a manifold is independent of whether a metric you choose to impose on it is everywhere differentiable. Thus you can have two smooth manifolds connected by a diffeomorphism which carries the nondifferentiable metric from one to the other.

There really is just no sensible way to talk about one manifold being diffeomorphic, by itself. It is like asking if a set, by itself, is a differentiable function ??!

I was exactly wondering about that. Two smooth manifolds can indeed be non-diffeomorphic. Thanks!

 

[Mattergauge]

Two smooth (infinitely differentiable) manifolds need not be diffeomorphic (among other obstacles, there may topological obstructions to the existence of a diffeomorphism).

Further, the smoothness of a manifold is independent of whether a metric you choose to impose on it is everywhere differentiable. Thus you can have two smooth manifolds connected by a diffeomorphism which carries the nondifferentiable metric from one to the other.

There really is just no sensible way to talk about one manifold being diffeomorphic, by itself. It is like asking if a set, by itself, is a differentiable function ??!

Of course the term diffeomorphism has a strict meaning. I was wondering: can the Schwarzscild metric, with artificial singularities at the horizon, be replaced with one that has a singularity at the center of the hole only?

 

[ergospherical]

Yes. Viz, ingoing Eddington-Finkelstein coordinates.

 

[PeterDonis]

the Schwarzscild metric, with artificial singularities at the horizon

The "singularities" at the horizon of Schwarzschild spacetime are coordinate singularities only, not "artificial". They are not singularities of the spacetime geometry.

be replaced one that has a singularity at the center of the hole only?

No "replacement" is necessary; the Schwarzschild spacetime already only has a singularity of the spacetime geometry at $r = 0$. See above. (Note that this singularity is not "at the center", however; it's a moment of time, not a place in space.)

 

[Mattergauge]

Yes. Viz, ingoing Eddington-Finkelstein coordinates.

What does "ingoing" stand for? Freely falling?