Thread: Shell Theorem in General relativity

1. Originally Posted by xylophobe
Your statement that the outside observer experiences "nothing" can not be true because gravity is density dependent. For instance a gas cloud of mass "X" when viewed from an infinite distance appears in the gravitaional realm the same as a blackhole of the same mass "X". If your outside observer is anywhere closer than infinity distance then the outside observer will note a concentrating of gravitational force by the collapsing shell. The curvature of the gravitational field becomes "tighter" by the collapse to from r2 to r1.
I think the others have nicely responded to this and your later posts. The external gravitational field of a spherically symmetric mass distribution is the same no matter what the radius of the sphere is. The internal field is something else, and if sphere shrinks or expands, there will changes in the affected regions, but not externally. In Newton, that external field is the simple inverse square field, -GM/r^2 directed radially, acting as though all the mass is concentrated at the center.

The same is true in GR. A spherically symmetric mass distribution produces the Schwarzschild spacetime externally. There are differences between that and Newton, but in the weak field limit, GR reduces to Newton.

Non-spherical mass distributions are something else. But even there, while the near field is very different and more complex, far enough away, the field approaches the inverse square field of a point mass.

And finally, later on you said that inside the shell, clocks at the center tick the slowest. This is not true. Inside the hollow of a hollowed-out spherical mass distribution, all clocks tick at the same rate. The Newtonian potential is constant in that hollow, and in the weak field limit, the Newtonian potential is what determines the clock rate, dtau/dt ~ sqrt(1 + 2Phi/c^2), where Phi is the Newtonian potential, and Phi = 0 is taken for the reference clock rate (where dt is ticking).

-Richard

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Originally Posted by publius
I think the others have nicely responded to this and your later posts. The external gravitational field of a spherically symmetric mass distribution is the same no matter what the radius of the sphere is. The internal field is something else, and if sphere shrinks or expands, there will changes in the affected regions, but not externally. In Newton, that external field is the simple inverse square field, -GM/r^2 directed radially, acting as though all the mass is concentrated at the center.

The same is true in GR. A spherically symmetric mass distribution produces the Schwarzschild spacetime externally. There are differences between that and Newton, but in the weak field limit, GR reduces to Newton.

Non-spherical mass distributions are something else. But even there, while the near field is very different and more complex, far enough away, the field approaches the inverse square field of a point mass.

And finally, later on you said that inside the shell, clocks at the center tick the slowest. This is not true. Inside the hollow of a hollowed-out spherical mass distribution, all clocks tick at the same rate. The Newtonian potential is constant in that hollow, and in the weak field limit, the Newtonian potential is what determines the clock rate, dtau/dt ~ sqrt(1 + 2Phi/c^2), where Phi is the Newtonian potential, and Phi = 0 is taken for the reference clock rate (where dt is ticking).

-Richard
Then what's the difference between a super-massive blackhole and a stellar blackhole? I believe the difference is the tidal forces for a super-massive blackhole are minimized due to the near "flatness" of the large diameter which does not produce as pronounced of a tidal effect as the curvey stellar blackhole tidal forces. In this case then shape matters since size is an attribute of shape.

3. Originally Posted by xylophobe
Then what's the difference between a super-massive blackhole and a stellar blackhole? I believe the difference is the tidal forces for a super-massive blackhole are minimized due to the near "flatness" of the large diameter which does not produce as pronounced of a tidal effect as the curvey stellar blackhole tidal forces. In this case then shape matters since size is an attribute of shape.
The mass is the difference. Take a mass M in a spherical configuration. At some r outside of the mass, it doesn't matter what the size of the sphere itself it, the gravity is the same.

What you're talking about is the tidal forces experienced by a free faller near the event horizon. That decreases with mass as the radius of the horizon increases with mass, R_s = 2GM/c^2. The radial stretching tide goes as 2GM/r^3. Plugging that in gives the tide at the horizon for a free faller as c^6 / (4(GM)^2), decreasing as the square of the mass.

Note the formula for the tide (of the spherical distribution) has nothing about the shape in it all, only the total mass and the distance from it (in GR inside the horizon, this "distance" is actually a time-like thing which complexity we won't worry with here ). It's just that for larger masses, the horizon is much bigger and the point of no return occurs where the local curvature is less.

-Richard

4. Yes, the normal use of the word "shape" has removed from it the concept of "size", expressly so as to be able to treat these two rather separate concepts in a separated way. So we would say the Earth has the shape of a basketball but not its size, and a basketball has the size of a football but not its shape.

5. Originally Posted by Ken G
So we would say the Earth has the shape of a basketball but not its size, and a basketball has the size of a football but not its shape.
Took me a minute to figure out what you're saying here: where I'm from, football = soccer.

I think Publius has answered all Xylophobe's outstanding questions for me, but to summarize, yes, non-spherical mass distributions will have an effect. We were however talking about differences in density, which will not for (approximately) spherical things like stars and black holes.

Note also that the supermassive black hole v. stellar-mass black hole comparison violates both conditions I stated for the force to be the same - the objects have different mass, and we're considering observers at different distances (the SMBH having its event horizon further way, and we're looking at observers a bit outside the event horizon). And it's anyway not the gravitational force you're interested in here, but in the tide.

6. Originally Posted by AndreasJ
Took me a minute to figure out what you're saying here: where I'm from, football = soccer.
Oops, ironically, I first said "soccer ball", but decided its shape wasn't all that different!

7. Xylophobe-

You're confused. The external observer feels attraction to a point at center of the inside of the shell. If the shell collapses and he remains at the same radius, he has the same distance from the center before and after the collapse, and the shell has the same mass before and after the collapse; he feels nothing different.

The escape velocity at the surface of the shell is indeed higher after the collapse, but our external observer did not move. If he had collapsed with the shell, then indeed, it would require more force after the collapse to counteract the acceleration of gravity. Since he maintained his radius, there is no change.

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Originally Posted by Roobydo
Xylophobe-

You're confused. The external observer feels attraction to a point at center of the inside of the shell. If the shell collapses and he remains at the same radius, he has the same distance from the center before and after the collapse, and the shell has the same mass before and after the collapse; he feels nothing different.

The escape velocity at the surface of the shell is indeed higher after the collapse, but our external observer did not move. If he had collapsed with the shell, then indeed, it would require more force after the collapse to counteract the acceleration of gravity. Since he maintained his radius, there is no change.
I understand that the centerpoint of gravity remains the same for all occasions, the mass being identical: spherical gas cloud, star, blackhole but the fact that the mass is at different densities in each instance affects the way that gravity influences the outside observer (in a minute way but for thought-experimental purposes it is real). I am working on a geometric graphic to show my meaning but it is taking a while and I'm out of time today.

9. Originally Posted by xylophobe
I understand that the centerpoint of gravity remains the same for all occasions, the mass being identical: spherical gas cloud, star, blackhole but the fact that the mass is at different densities in each instance affects the way that gravity influences the outside observer (in a minute way but for thought-experimental purposes it is real). I am working on a geometric graphic to show my meaning but it is taking a while and I'm out of time today.
Actually, the (somewhat surprising) result is that it doesn't. In an earlier post, you were imagining the object divided into left and right hemispheres, and suggesting that since some of the gravitational force from each hemisphere would not be directed radially (and would cancel with some of the force from the other hemisphere), that the total force might be a little bit less. One could imagine an extreme case of this, measuring the gravitational force from a rigid rod, perpendicular to our line of sight. And indeed, if we were talking about a rigid rod, the gravitational force is not the same as a point object of the same mass as the rod. There's a deviation from the simple inverse square law that becomes more pronounced as you get closer.

But notice that a rigid rod is not a spherically symmetric object. In the case of a sphere, there's not just some of the matter over on the left and some on the right. There's also some matter closer to you, which will exert a stronger pull, and some further away, which will exert a weaker pull. Since the force goes like the inverse square of the distance, there's not just a simple linear relationship. Intuitively, it seems likely that you'd have to know the exact distribution of matter to know the force it exerts, and that there could be a big difference. However, if you work out the math, you discover that as long as the distribution is spherically symmetric, it doesn't matter whether it's a tiny object with all of the mass concentrated at the center, or a big diffuse cloud with the mass all spread out, or even a hollow sphere, with all the mass concentrated on the surface, the resultant gravitational force is the same. It really is a pretty surprising result, but it's also really true, and pops out at you if you work out the math. You can see the derivation here.

10. In the OP scenario, would the tidal forces experienced by the outside observer change as the shell collapses?

11. Originally Posted by loglo
In the OP scenario, would the tidal forces experienced by the outside observer change as the shell collapses?
No.

12. Originally Posted by xylophobe
...spherical gas cloud, star, blackhole but the fact that the mass is at different densities in each instance affects the way that gravity influences the outside observer.
No, I'm afraid that is not the fact. The fact is that an observer outside of a gas cloud will experience an acceleration proportional to the product of his mass and the mass of the gas cloud divided by the square of his distance from the center of the cloud.

As the cloud shrinks into a star, he will be attracted to the same point. The product of the masses will not change, so assuming he maintains his position, the force pulling him toward the center will also remain unchanged. Note that at this point he is much further away from the surface of the star than he was from the surface of the gas could.

As the star shrinks into a black hole, he will be attracted to the same point, which is now the singularity of the black hole. The product of the masses does not change, nor the radius, so again, his acceleration will also remain unchanged. Note that at this point he is much further away from the surface of the black hole than he was from the surface of the star.
Last edited by Andrew D; 2010-Oct-01 at 03:29 PM. Reason: fixed horrendous typo

13. Originally Posted by Roobydo
No, I'm afraid that is not the fact. The fact is that an observer outside of a gas cloud will experience an acceleration proportional to the product of his mass and the mass of the gas cloud divided by his distance from the center of the cloud.
Obviously just a typo: divided by square of distance ?

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This is the best link I've seen on the situation inside and outside of the spherical shell: Gravitational field due to rigid bodies

Check out Figure 9, amongst the others.

CC

15. I think you mean Figure 6, which is for the hollow sphere. Figure 9 is for a solid sphere, somewhat like the Earth.

16. You know, I don't know if anyone reading this would be interested, but while the external Schwarzschild metric is well known, the so-called "interior" for a spherically symmetric fluid/solid body is not as well known. For example, as can be seen in Coldcreation's links, we know the Newtonian solution for a sphere of constant density. The field increases linearly from the center up to the surface, then falls off in inverse square fashion above. Thus the potential increases as the square of radius from a minimum value at the center, then joins to a -1/r function on the outside which increases asymptotically to zero at infinity.

But what about the GR version? We'd expect g_00 to have something in it that goes at r^2 up to the surface. And that is the "interior" solution. Things depend on the fluid and its equation of state, but a simple version based on a constant "proper" density (which isn't really realistic) can be found. Things get complicated because of pressure, which contributes to the gravity, and the "binding energy", meaning the coordinate rest energy of something down in a gravity well is different than at infinity, which means the gravitational mass is less. It's sort of a mess, but it can be worked out. The "proper density" is the density of an observer on the scene, comoving with the local mass element, and that is what one declares to be constant, not the coordinate density as seen by the observer at infinity.

So what one does is declare the proper density to be a constant, and lets the pressure at radius be whatever is has to be to support the weight of the fluid above that radius. And, in the GR version, that pressure contributes to the gravity, so we get a far more complex set of equations to solve that in Newton.

Anyway, the metric takes the form:

ds^2 = A(r) (c dt)^2 - [B(r) dr^2 + dSpherical^2], where dSpherical is the spherical angular part. There are no functions of r to worry about on that part in spherically symmetric forms.

Now, let R be the radius of the sphere, and let S be the familiar Schwarzschild radius, 2GM/c^2 (I usually use R for that, but alas, I need another R and don't want to mess with subscripts), where M is the total Schwarzschild mass, which is not the same as the rest mass of all the pieces at infinity . It's funny how the Schwarzschild radius always pops up easily in these things:

Now, B(r), the spatial part is given by:

1/[1 - Sr^2/R^3]

Now, knowing S = 2GM/c^2, we can see that dividing by R^3 looks like a density, so this looks pretty close to the Newtonian potential term. Note that for regular Schwarzschild, B = 1/A, but alas that turns out NOT to be the case here for the interior.

A(r) is a bit messy and is given by:

1/4 * [ 3*sqrt(1 - S/R) - sqrt(1 - Sr^2/R^3) ]^2

Note that we have a constant term inside the brackets. If that were zero, the A would be 1/B, save for the factor of 4 in front of the whole thing.

There is something interesting about A(r) and what it says about R vs M. Consider the clock rate at r = 0. What does this imply?

-Richard

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Originally Posted by publius
Consider the clock rate at r = 0. What does this imply?
If I'm not mistaken it would imply that the clock runs faster at the center of the sphere, i.e, according to an observer located at r = 0, her clock runs at a "standard" rate, but as she looks out at other clocks located at some distance inside the sphere (if that were possible) she notices that they run at a slower rate and do so right up to the surface of the sphere.

Likewise, if those other clock-locations emitted a light signal, the observer at r = 0 (the origin) would receive the signals with spectral lines shifted towards the red end of the spectrum. Thought the redshift (and the associated time dilation factor) would be a linear function that increases with distance, she could conclude that either spacetime inside the sphere was curved (and that she was sitting on 'top' of the field slope or gradient), or that the sphere itself was expanding.

Interesting.

CC

18. Originally Posted by xylophobe
I understand that the centerpoint of gravity remains the same for all occasions, the mass being identical: spherical gas cloud, star, blackhole but the fact that the mass is at different densities in each instance affects the way that gravity influences the outside observer (in a minute way but for thought-experimental purposes it is real). I am working on a geometric graphic to show my meaning but it is taking a while and I'm out of time today.

The others are referring to the fact that under Newtonian gravity, the gravitational effect of a spherically symmetric mass is the same as if it were completely concentrated at its center. That is often shown in early calculus classes.

19. Originally Posted by Coldcreation
If I'm not mistaken it would imply that the clock runs faster at the center of the sphere, i.e, according to an observer located at r = 0, her clock runs at a "standard" rate, but as she looks out at other clocks located at some distance inside the sphere (if that were possible) she notices that they run at a slower rate and do so right up to the surface of the sphere.
No, the clock at the center runs the slowest. The expression for A(r) is complicated but we can reduce the term in brackets to:

D - sqrt(1 - kr^2)

As r increases, that term increases. The term under the square root decreases with r, but that is subtracted from the constant D. And this is a completely static solution, mind you. There is no expansion or contraction of anything.

What I'm getting at is look at the expression for g_00. Set r = 0 and ask what that says about the clock rate at the center. That clock rate can't be zero or negative and that implies a constraint on R vs M. This is a surprising result and has a name.

-Richard

20. A(0) has a zero for R=1.125·S (and goes negative for smaller R).

I would have expected the funny stuff to start at R=S.

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Originally Posted by publius
No, the clock at the center runs the slowest.
...and I thought it was a trick question . There seems to be nothing unusual happening. Gravitational redshift and time dilation would work similarly to those effects out side the sphere.

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Originally Posted by Ken G
No-- a field takes on a vector value at every point, so it has to point in some direction. When there's no direction to point, the field must be zero. You are thinking of a potential, but a region of constant potential has no field, and no forces. It is only gradients in potential that have dynamical consequences-- like riding a bike in a mountain range, you don't care if the elevation stays constant, you care when it steeply changes.

A homogeneous field is something different, it is a field that points all in the same direction (or is zero field). You get a homogeneous field on one side of an infinite plane of mass/charge.

It's because Newtonian gravity has an infinite range. That's a problem that has to be handled in somewhat ersatz ways (same for the electric field, as shows up in the "Coulomb logarithm").

This is what isn't true-- a "classical homogeneous field" is no field at all, or it is the field next to a plane of sources. GR handles both of those cases just fine, and the zero-field case is handled by flat spacetime.
Both of those are perfectly correct.
Ok, now we're getting right down to the meat of the subject.

Let's take your mountain bike example. Say the plateau (at a certain elevation above sea level) on which the cyclist is riding continues for a while. The cyclist continues riding along, without having to go uphill or downhill, since the surface is flat. After a few days, she notices that she can continue on this trail, seemingly endlessly. And after about of month, she realizes her journey took her all the way around the sphere without having to ride up or downhill. (The trail on which she rides continues on a plateau over the entire spherical surface). For the cyclist, the trail was, in every practical sense, flat. Yet globally the entire surface is curved (this would be a Riemannian manifold of constant positive Gaussian curvature (K = 1). Locally, every point was flat, and yet, and yet.

In other words (now tuning from the analogy back to the homogeneous field concept) the gravitational field is locally flat, to any observer and at any location. But globally, the manifold may be intrinsically curved. The manifold, then, needs not be uniquely Minkowskian.

The point is that the statement, "a classical homogeneous field is no field at all" cannot be interpreted as obligatorily correct. The zero-field case is not necessarily handled by flat spacetime, in terms of general relativity (when gravity is considered a curved spacetime phenomenon). In the example above, the geometry is locally flat (curvature vanishes in the vicinity of the observer), yet the manifold is globally curved. In Newtonian terms, the potential is everywhere the same, and in relativistic terms the field gradient or magnitude is locally flat with a Minkowski signature, and globally the manifold is Riemannian surface of constant positive Gaussian curvature.

Had our cyclist been riding on a surface of constant negative Gaussian curvature (K = -1) the result would be more complicated, but essentially the same (though she would not come back to her starting point).

So how does this relate to the OP?

The OP is an attempt (making use of the shell theorem) to explore the possible solutions for describing a classical homogeneous gravitational field in general relativistic terms. My hunch is that a Minkowskian metric is only one of two (or several) possibilities that would describe a globally homogeneous field.

I linked above a work entitled Can the notion of a homogeneous gravitational field be transferred from classical mechanics to the Relativistic Theory of Gravity where some of the difficulties in finding a general solution are discussed.

See too what Cromwell writes here, about a uniform field (not quite the same as a homogeneous field, but nevertheless, the question applies globally. So it is of interest). There is no global solution to the Einstein field equations that uniquely and satisfactorily embodies Newtonian ideas about a uniform field, i.e, the desired properties for a uniform gravitational field in GR cannot all be satisfied at once (without a metric that becomes degenerate). In general relativistic terms, a necessary condition for any spacetime that would represent a globally homogeneous uniform field is that the scalar curvature should be constant, and four translation symmetries would be expected. The problem of finding a homogeneous gravitational field in general relativity remains open. That might be interpreted as saying that the global field that emerges from GR can only be considered, at best a special case. And at worst, an erroneous extrapolation of local physics to global physics.

CC

23. Originally Posted by Coldcreation
Let's take your mountain bike example. Say the plateau (at a certain elevation above sea level) on which the cyclist is riding continues for a while. The cyclist continues riding along, without having to go uphill or downhill, since the surface is flat. After a few days, she notices that she can continue on this trail, seemingly endlessly. And after about of month, she realizes her journey took her all the way around the sphere without having to ride up or downhill. (The trail on which she rides continues on a plateau over the entire spherical surface). For the cyclist, the trail was, in every practical sense, flat. Yet globally the entire surface is curved (this would be a Riemannian manifold of constant positive Gaussian curvature (K = 1). Locally, every point was flat, and yet, and yet.
I think you are confusing "flat" with "curved but with a large enough radius of curvature that it isn't obvious."

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Originally Posted by Ken G
I think you are confusing "flat" with "curved but with a large enough radius of curvature that it isn't obvious."
On the contrary. What I explained (using your analogy) was that you may be confusing flat (or Minkowski space) with what potentially amounts to a Riemannian (or pseudo-Riemannian) manifold of constant Gaussian sectional curvature with a Lorentzian signature. If this is indeed what could be considered a homogeneous spacetime in general relativistic terms (comparable to some extent with the classical notion of homogeneous field), then the spacetime manifold under consideration could be intrinsically curved (without excluding a priori the Minkowski metric as a special case).

CC

25. Originally Posted by Strange
Obviously just a typo: divided by square of distance ?
obviously.

26. Originally Posted by Coldcreation
On the contrary. What I explained (using your analogy) was that you may be confusing flat (or Minkowski space) with what potentially amounts to a Riemannian (or pseudo-Riemannian) manifold of constant Gaussian sectional curvature with a Lorentzian signature.
But that is not what is happening inside the sphere-- the spacetime is strictly Minkowskian in there, as publius explained. It is not like the bicycle circumnavigating a perfectly spherical Earth, as that situation has actual curvature that could be measured, for example, by noticing that triangles add up to more than 180 degrees.
If this is indeed what could be considered a homogeneous spacetime in general relativistic terms (comparable to some extent with the classical notion of homogeneous field), then the spacetime manifold under consideration could be intrinsically curved (without excluding a priori the Minkowski metric as a special case).
The only reason a classical "homogeneous field" applies here is because that field is identically zero inside the sphere, so is homogeneous because it isn't there. The key point is, the spacetime inside the sphere is completely indistinguishable from having no sphere there at all, regardless of how much you shrink that sphere-- and that holds in both classical gravity and GR. The spacetime outside the sphere does reflect the presence of that sphere, but it does not change in any way when the sphere shrinks.

What I find most interesting about all this is that the observer inside the sphere will still perceive the outer observer's clock as speeding up as the sphere shrinks, despite no change in either of their local spacetimes. This is a clear indication that the apparent "rate of flow of time" has nothing to do with anything that is actually happening to any of the clocks, but rather is a function of the connections between the clocks, and all the causality that is interpreted as "affecting" the comparison between clocks is a causality that refers strictly to these connections, and not to the clocks themselves. To me, that's a clear manifestation of the Machian character of GR, in the absence of spin. I remain unclear on whether or not GR can retain its Machian character when there is spin, but I'll bet that if it can't, it isn't being done right!

27. Originally Posted by AndreasJ
A(0) has a zero for R=1.125·S (and goes negative for smaller R).

I would have expected the funny stuff to start at R=S.
Bingo! The solution breaks down at R = 9/8 S. Yes, we would expect something funny to happen when R = S, but lo and behold, it happens slightly above that, at 9/8th the Schwarzschild radius. It turns out that what happens there is the pressure at the center goes to infinity. This is a consequence of the pressure increasing the gravity and hence the weight of all the mass above. The more mass, the more pressure, but the more pressure, the more gravity and yet more pressure meaning yet more gravity. That process runs away and blows up at R = 9/8 S, and not R = S as we might expect. What it means is there is no longer a static solution, and the sphere must collapse to a black hole.

This was for an ideal incompressible fluid, but amazingly it turns out this 9/8 S result is *independent of the equation of state* (well, equations of state that sastify the various energy conditions -- that is non-"exotic" matter). This is known as Buchdahl's Theorem, and it falls right out of the EFE, althought the proof is beyond my scope as they say.

Again, this is a general result of General Relativity independent of the type of matter. You can't pack a (non-exotic) mass into 9/8 of its Schwarzschild radius without it collapsing to a black hole.

-Richard

28. Originally Posted by publius
Again, this is a general result of General Relativity independent of the type of matter. You can't pack a (non-exotic) mass into 9/8 of its Schwarzschild radius without it collapsing to a black hole.
That's interesting, I hadn't heard that. I wonder if neutron stars count or not-- their equation of state from the degenerate neutrons is not particularly exotic, but there's a lot more going on in there, so the real equation of state is not known and whether or not it counts as "exotic" I couldn't say. But if it doesn't count as exotic, then we'd have to say that a neutron star probably cannot be smaller than 9/8 S.

29. Ken,

IIRC, a naive application of this to neutron stars yields a mass limit of around 35 solar masses.

The thing is that the Buchdahl limit is for spherical symmetry and, of course, non-rotation. That is going to modify things in a complex way and I have no idea what the results are, or if they're even known -- surely, someone has looked into it and via ballpark numerical approximations if nothing else. But I just don't know. We know that for Kerr/rotating black holes, the horizon is below the spherical Schwarzschild radius, so I would naively expect that the rotating Buchdahl limit would be reduced as well, and thus the mass limit higher.

-Richard

30. Ken,

I looked around for a possible "rotating Buchdahl limit" and it turns out there is none, actually. The language this question is framed in is it possible to have a smooth, continuous transition from the fluid body state to the black hole state.

For non-rotating spheres, the answer to this no. We can't compress such a body down as close as we want to the Schwarzschild radius as we hit the Buchdahl 9/8 limit. Below that, we are unstable and can't have something in between 9/8S and S, the black hole state.

But in the case of rotation, and axis symmetry, we can apparently get as close as we like by adjusting the spin/L parameter. This does not mean there's some abrupt difference between the rotating and non-rotating state (ala a recent thread we had with Russ T, IIRC), just that we can get around the Buchdahl limit by spinning it fast enough.

When I think about, this should not be surprising. We have the "extremal" limit on the L/M ratio, where enough L prevents it from collapsing to a black hole at all. So if enough spin can prevent collapse completely, then surely enough spin can override the higher Buchdahl limit.

-Richard