I Is using r as a coordinate in Birkhoff's theorem a limitation?

[Pencilvester]

TL;DR

Does using r as a coordinate in Birkhoff's theorem limit the scope of applicability?

In Birkhoff’s theorem, doesn’t assuming we can use r (defined as circumference divided by $2 \pi$ for any given sphere) as a coordinate across the spacetime implicitly assume that the spheres must always be getting bigger in some specific direction? Is there a version of the proof that doesn’t have this limitation?

I’m thinking about if we made a similar move on 2-dimensional manifolds that ought to exhibit infinite order rotational symmetry. A cylinder would clearly fit, but if we limited ourselves to only looking at surfaces where we can start at a point and move out with concentric circles of increasing circumference, then this and a whole host of other manifolds would be excluded from our analysis from the get-go.

 

[Dale]

So you are thinking of manifolds where spacetime is foliated as a family of concentric spheres. But the size of those spheres, for example, gets smaller as you go away from the center.

 

[martinbn]

If $r$ increases and then decreases, you will have different events(space-time points) with the same coordinates. That is not ok for a coordinate chart.

 

[Dale]

I agree. You would have to use at least two separate charts to cover that spacetime.

 

[Pencilvester]

If r increases and then decreases, you will have different events(space-time points) with the same coordinates. That is not ok for a coordinate chart.

Right, so doesn’t making $r$ a coordinate that ought to cover the spacetime restrict the number of possible spherically symmetric spacetimes that are covered in Birkhoff’s theorem?

So you are thinking of manifolds where spacetime is foliated as a family of concentric spheres. But the size of those spheres, for example, gets smaller as you go away from the center.

Yes, that would be one potential example of a spherically symmetric spacetime that appears to me to be excluded from the application of Birkhoff’s theorem. Another one could be extending my cylinder example so that the spheres don’t even change size. Or you could have the size of the spheres oscillate.

 

[Dale]

that would be one potential example of a spherically symmetric spacetime that appears to me to be excluded from the application of Birkhoff’s theorem.

I think you are correct that such a manifold is excluded from Birkhoff’s theorem. However, I think it is not a problem with the theorem.

Basically, you can solve the EFE by specifying a stress energy tensor and calculating the metric. Or, you can specify the metric and calculate the stress energy tensor. The metric you describe would undoubtedly not be a vacuum stress energy tensor.

 

[Pencilvester]

Or, you can specify the metric and calculate the stress energy tensor. The metric you describe would undoubtedly not be a vacuum stress energy tensor.

I personally don’t doubt this either, but has this been proven anywhere? If so, I’d be interested to see the way or ways this can be proven.

 

[Dale]

has this been proven anywhere?

I think it is proven by Birkhoff’s theorem. Birkhoff’s theorem shows that the spherical vacuum metric is unique. If the metric you describe were a vacuum metric then Birkhoff’s theorem would have had two solutions, one where $r$ increased outward and one where $r$ increased inward. Then the solutions you describe would be found by stitching positive-in and positive-out sections with appropriate boundary matching.

But there aren’t multiple solutions to stitch together.

 

[Ibix]

I'd need to check details, but I think it can be shown that the most general spherically symmetric metric can be written as $e^{2A(r,t)}dt^2-e^{2B(r,t)}dr^2-r^2d\Omega^2$. If so, the proposed cylindrical metric is either not valid (although not sure what I'd mean by that) or can be rewritten in other terms so that it does fit that form.

 

[PeroK]

First, we need a definition of what it means for a manifold to be spherically symmetric. There is a formal definition here, in terms of the rotation group SO(3).

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

Birkhoff's theorem then shows that there is a unique solution to the EFE in vacuum that has the property of being spherically symmetric.

In Carroll's proof he admits to using a level of rigor that might be suitable for physicists but make a mathematician uneasy. He references Hawking and Ellis (1973) for a "more careful" treatment.

 

[PeterDonis]

Does using r as a coordinate in Birkhoff's theorem limit the scope of applicability?

No. The areal radius $r$ is an invariant. Picking it as one of the coordinates helps in making the proof of Birkhoff's Theorem look simple, but it is not necessary for the theorem to be true.

doesn’t assuming we can use r (defined as circumference divided by $2 \pi$ for any given sphere) as a coordinate across the spacetime implicitly assume that the spheres must always be getting bigger in some specific direction?

No. The range of $r$ is assumed to be $0 < r < \infty$, but that in itself says nothing about how $r$ changes in any particular direction.

 

[PeterDonis]

the size of those spheres, for example, gets smaller as you go away from the center.

This isn't possible; the "center" is defined as the locus $r = 0$. Since $r$ is the areal radius, it's impossible for it to be smaller anywhere than it is at the center.

 

[PeterDonis]

I'd need to check details, but I think it can be shown that the most general spherically symmetric metric can be written as $e^{2A(r,t)}dt^2-e^{2B(r,t)}dr^2-r^2d\Omega^2$.

This is correct. See the Insights article I referred to, or MTW for more details.

 

[Dale]

This isn't possible; the "center" is defined as the locus r=0.

In principle, you should be able to have a manifold with two distinct $r=0$ loci. Consider a sphere, latitude lines are submanifolds that form a “spherical” foliation. As you go from the South Pole to the North Pole on a longitude line (geodesic) the “areal” $r$ coordinate would first increase and then decrease, without ever turning.

You would have to use two charts with that $r$ coordinate, joined with appropriate matching conditions at the equator. But as you say, the $r$ coordinate is an invariant of the submanifold. So in an invariant sense, after you cross the equator you are continuing to smaller $r$ as you go away from the South Pole $r=0$. The coordinates have trouble with that, but not the manifold. And you should even be able to remove the poles so that there is not even a $r=0$ locus

I don’t think that a manifold like @Pencilvester describes is impossible. It is just undoubtedly not a vacuum metric.

 

[PeterDonis]

In principle, you should be able to have a manifold with two distinct $r=0$ loci.

Yes, that's true, you can--for example, maximally extended Schwarzschild spacetime, with black hole and white hole singularities.

I don’t think that a manifold like @Pencilvester describes is impossible.

Nor is it ruled out a priori by the proof of Birkhoff's Theorem. That proof places no restrictions whatever on the behavior of the areal radius $r$ at the outset. And, as my example above shows, the theorem itself does not require that the manifold have only one $r = 0$ locus, or that $r$ must be always increasing.

In other words, @Pencilvester has not thought through what the result of Birkhoff's Theorem actually says and does not say.

 

[Dale]

Nor is it ruled out a priori by the proof of Birkhoff's Theorem. That proof places no restrictions whatever on the behavior of the areal radius r at the outset.

Agreed. Such manifolds are not excluded a priori. And since the result of the theorem does not contain such solutions, the conclusion is that such metrics can not be solutions of the EFE in vacuum.

 

[Pencilvester]

The areal radius r is an invariant. Picking it as one of the coordinates helps in making the proof of Birkhoff's Theorem look simple, but it is not necessary for the theorem to be true.

So, to make sure I'm understanding: any and all statements that can be made about the behavior of the invariant $r := C / 2 \pi$ across the manifold in Birkhoff's theorem are derived and not assumed (either explicitly or implicitly). I think my problem was that Schutz's section 10.1 "Coordinates for spherically symmetric spacetimes" seems to have seeped its way into my brain as a layman's version of the proof, so I'll need to fix that.

No. The range of r is assumed to be 0<r<∞, but that in itself says nothing about how r changes in any particular direction.

You may have misunderstood my point due to my misunderstanding of Birkhoff's theorem. I was thinking that if we want to first start with only the assumption of spherical symmetry, then apply the restriction of vacuum, then there would be spherically symmetric spacetimes in this first step for which we couldn't use $r$ as a coordinate across the entire spacetime (essentially any spacetime where, after foliating it in some way that preserves the spherical symmetry in each leaf, there are at least 2 distinct spheres in any given leaf with the same circumference; then $r$ would not be a good coordinate, and at most, it could be used in patches). And on the other side of that, if we do insist on using $r$ as a coordinate that ought to cover the whole spacetime, then that excludes those very spacetimes. I understand now, however, that this is not how Birkhoff's theorem proceeds.

the theorem itself does not require that the manifold have only one r=0 locus, or that r must be always increasing.

I assume that it also doesn't require the $r=0$ locus to even exist at all from the outset?

 

[PeterDonis]

any and all statements that can be made about the behavior of the invariant $r := C / 2 \pi$ across the manifold in Birkhoff's theorem are derived and not assumed (either explicitly or implicitly)

That's correct.

I was thinking that if we want to first start with only the assumption of spherical symmetry, then apply the restriction of vacuum, then there would be spherically symmetric spacetimes in this first step for which we couldn't use $r$ as a coordinate across the entire spacetime

There are. One of them is...Schwarzschild spacetime! As I already pointed out in post #15.

You are misunderstanding how Birkhoff's theorem works--and more generally how the Einstein Field Equation works. The EFE is local. It applies at an event (or more precisely an open neighborhood of an event, since it's a differential equation so derivatives have to be well-defined). The theorem says that in any such open neighborhood of a spherically symmetric spacetime which is vacuum, the spacetime geometry is a piece of Schwarzschild spacetime. The proof involves applying the EFE in such an open neighborhood.

The theorem, by itself, tells you nothing about the global structure of the spacetime. The piece of Schwarzschild spacetime that the theorem says is there could be (mathematically speaking) a piece of the maximally extended Schwarzschild spacetime, with vacuum everywhere and both black hole and white hole regions. Or it could be a piece of the vacuum region surrounding a spherically symmetric star. Or a piece of a vacuum region of any other spacetime that meets the requirements of the theorem.

Thus, the theorem, by itself, does not require that $r$ is usable as a coordinate in a single chart that covers the entire spacetime. It only requires, if you do the proof using Schwarzschild coordinates, that $r$ is usable as a coordinate in a single chart that covers the open region in which you are doing the proof. That will be true in any spherically symmetric spacetime--it is always possible to find an open region in which you can construct a Schwarzschild-type chart that uses $r$ as a coordinate. So there is no restriction whatever imposed by doing that.

I understand now, however, that this is not how Birkhoff's theorem proceeds.

More precisely, the conclusion of Birkhoff's theorem does not require that the entire spacetime must be covered by a single chart that uses $r$ as a coordinate. That has to be the case because, as above, Schwarzschild spacetime itself can't be entirely covered by such a chart!

I assume that it also doesn't require the $r=0$ locus to even exist at all from the outset?

That's correct. And of course in Schwarzschild spacetime, it doesn't; the $r = 0$ locus is not part of the manifold.

 

[PeterDonis]

the result of the theorem does not contain such solutions

Yes, it does! That was my point in post #15. Maximally extended Schwarzschild spacetime, which is a solution that meets the requirements of the theorem, is such a solution!