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.