# Thin Shell Cavity in General Relativity

1. May 28, 2013

### Markus Hanke

I am trying to understand the implications of the Birkhoff theorem as it applies to a cavity within a spherically symmetric, stationary thin shell. Is the region of space-time within the cavity described by the actual Minkowski metric diag{1,-1,-1,-1}, or is it some other metric with constant components ? We know that whatever it is, it must be constant at all points in the cavity.

I am finding this somewhat confusing, because the vacuum metric within the cavity must smoothly connect to the interior metric of the thin shell itself, which in turn must smoothly connect to the exterior Schwarzschild metric. The idea is that, globally, space-time for this scenario is everywhere smooth and differentiable, even at the boundaries of the shell. I might be wrong here, but that seems impossible if the metric in the cavity is strictly Minkowskian, or not ? At the same time it is clear that the Riemann curvature tensor must vanish everywhere inside the cavity.

Physically this boils down to the interesting question of gravitational time dilation - is a clock at rest within the cavity gravitationally time dilated as compared to an ideal, hypothetical clock at infinity ? If the vacuum metric within the cavity is Minkowskian, then the answer is clearly no; however, this seems to be in contradiction to Newtonian physics, where the gravitational potential within the cavity is everywhere constant, but not vanishing ( I am assuming the convention that the potential vanishes at infinity ). I understand that the value of the potential is arbitrary, but the question is whether it is the same inside the cavity as it is at infinity. It would seem that this cannot be the case.

Any comments will be much appreciated !

2. May 28, 2013

### Bill_K

The spacetime inside the cavity is flat. You are not compelled to use Minkowski coordinates, you may choose some other flat metric.

Of course the interior is time dilated. Drill a tiny hole in the shell and drop a photon in, and it will be blue-shifted.

3. May 28, 2013

### Markus Hanke

Thanks Bill_K. However, if we choose to use the Minkowski metric for the interior of the cavity, then how can a clock there be gravitationally time dilated as compared to a clock at infinity, which is also in locally Minkowskian space-time ? This is what I am confused about. I am probably missing something very basic here.

4. May 28, 2013

### Staff: Mentor

Regardless of the details of your spacetime you can always use Minkowski coordinates locally. So you can use Minkowski locally inside and locally at infinity. It is only when you want a non local coordinate system that you cannot use Minkowski if there is curvature.

In your scenario, light from infinity would blueshift as it approached the shell and then stop further blue shifting once it passed through it. So a global coordinate system would not be Minkowski and would show the requisite time dilation.

5. May 28, 2013

### pervect

Staff Emeritus
The inside and outside are both flat, as the curvature tensor is zero.

There is a simple coordinate transformation that maps diag(-1,1,1,1) to a constant metric. Changing the coordinates (the labels you attach to points) doesn't change the physics.

If you want some very detailed solutions, try
https://www.physicsforums.com/showpost.php?p=4019709&postcount=40
https://www.physicsforums.com/showpost.php?p=4022914&postcount=74

To give the bare bones:

For a hollow sphere, the metric can be put in the form:
$ds^2 = - J(r) dt^2 + \frac{1}{1 - \frac{2 m(r)}{r}} dr^2 + r^2 d\Omega^2$

Let's consider a hollow sphere of constant density, the inner radius is R1, the outer is R2.
Different possible solutions exist - one of the simplest, with no radial transfer of pressure, is:

for R1<r<R2
$$m(r) = \frac{M \left( r^3 - R1^3\right)}{R2^3 - R1^3}$$
$$J(r) = K \, exp \, \left( \int_{x=R1}^{x=r}\frac{2\,m(x)\,dx}{x\left(x-2\,m(x)\right)}\right)$$

K is an arbitrary constant, it can be chosen so that J(R2) = (1-2M/R2) if one wants to make the metric appear Schwarzschild in the exterior region. Alternatively, one can chose K=1, and have J(r)=1 in the interior region.

Both values of K satisfy Einstein's field equations. Both values of K represent the same physical system.

Last edited: May 28, 2013
6. May 28, 2013

### WannabeNewton

By "outside" do you mean the vacuum region exterior to the shell? If so, I think you may have misphrased this pervect. While the space-time in the shell cavity is flat as per Birkhoff's theorem, the space-time outside the shell is not necessarily flat. It being the vacuum exterior region, we simply have that the Ricci tensor vanishes but the Riemann tensor need not vanish-recall that flatness is determined by the vanishing of the Riemann tensor (i.e. the Weyl tensor will not vanish).

7. May 28, 2013

### Markus Hanke

This makes a lot of sense, thank you everyone. So one of the simpler ways to look at it would be the metric in post 5; externally this can be chose to be just the usual Schwarzschild metric, and in the interior of the cavity it then becomes constant, whereby it is the same everywhere as it is on the interior surface of the shell. Globally then, we have a space-time which asymptotically approaches Minkowski at infinity, but a clock in the interior cavity is still time dilated compared to an ideal counterpart at infinity.

This is actually completely intuitive, so long as one does not get "local" and "global" wrong.

8. May 28, 2013

### Markus Hanke

Depends on what you mean by "flat". The exterior vacuum will be "Ricci flat", but not "Riemann flat", as you rightly say.

9. May 28, 2013

### pervect

Staff Emeritus
I'm not sure what I was thinking - yes, only the inside of the shell is flat. The outside is asymptotically flat, but that's not the same thing at all, certainly not what I said.

10. May 29, 2013

### pervect

Staff Emeritus
Anyway, I think my first post, in spite of its other problems, did a good job of describing what J(r) had to look like in the metric $ds^2 = - J(r) dt^2 + \frac{1}{1 - \frac{2 m(r)}{r}} dr^2 + r^2 d\Omega^2$ for a spherical shell.

What's missing is a more detailed explanation of "why" you can multiply J(r) by a constant. Other than observing that the metric still satisfies Einstein's equations after the multiplication.

The same "why" explains why there isn't any fundamental difference between diag(-1,1,1,1) and diag(-K,1,1,1) where K is any (positive) constant.

This seems to be the issue that's puzzling the OP - unfortunately, I'm not sure how to explain in more detail.

Last edited: May 29, 2013
11. May 29, 2013

### WannabeNewton

Perhaps if you know of a paper/text that talks about it, that may shed further light on things as far as details are concerned. I'm not sure how easy it would be to find such a reference however :p

12. May 29, 2013

### pervect

Staff Emeritus
I'm sure I read something, somewhere, remarking that manifolds that are diffeomorphic should be in the same "equivalence class". Unfortunately, even if I could recall exactly where I read this and and provide a reference , I don't think the approach would be too helpful.

13. May 29, 2013

### WannabeNewton

Regardless, I think your explanation is more than clear enough and takes care of the main confusion. When it comes to topics like this, MTW seems the best place to check for explicit details on the matter but that book is waaay too scary to navigate through lol.

14. May 30, 2013

### Markus Hanke

Yes, I understand this now. My confusion was initially that I equated "flat" with only diag(-1,1,1,1); but clearly that is not the case, because, as you rightly state, one can multiply the metric by a constant, and still get a valid, flat solution of the EFEs. Once I realized that, the question was answered.

Again, thank you everyone for your time.

15. May 30, 2013

### George Jones

Staff Emeritus

No.

First, consider something similar in electromagnetism, a thin shell of charge, i.e., a surface charge. Surface charge is often given as a surface charge density $\sigma$, but, in three spatial dimensions, a the charge density $\rho$ of a surface charge involves a delta function. Guass's law, $\nabla \cdot \mathbf{E} = \rho /\epsilon_0$, expresses charge density in terms of derivatives of the electric field, so, in order to give a delta function, the electric field must involve a step discontinuity st the surface. This discontinuity is normal to surface, while the part of the electric field parallel to the surface is continuous. See Griffiths or Jackson.

"thin shell" in general relativity means a surface layer of matter/energy. Since this layer persists in time, this is a 3-dimensional hypersurface in 4-dimensional spacetime. This means that the stress-energy tensor $T_{\mu \nu}$ (which includes matter/energy density) involves a delta function. Einstein's equation expresses the stress-energy tensor in terms of second derivatives of the metric, so (some) first derivatives of the spacetime metric have step discontinuities, and the metric itself has. The discontinuities (as for the elEctric field) in the derivatives of the metric must be normal to the hypersurface.

For the general theory of the thin shell formalism, see

1) section 3.7 in A Relativists Toolkit by Poisson;

2) draft version of 1): http://www.physics.uoguelph.ca/poisson/research/agr.pdf;

3) section 12.5 in Gravitation by Padmanabhan;

4) chapter 16 in Einsteins General Theory of Relativity by Gron and Hervik.

These references all treat as an example a thin shell of pressureless dust for which spacetime is Schwarzschild outside and Minkowski inside. The shell either expands or contracts, depending on initial conditions.

Last edited: May 30, 2013
16. May 30, 2013

### WannabeNewton

Aaah I didn't even know these two texts had devoted sections on the thin shell setup. This is brilliant, thank you very much George!

17. May 30, 2013

### pervect

Staff Emeritus
MTW has a section on thin shells and "juctions conditions" as well - I found it rather opaque, though.

18. May 31, 2015

### JulianStuff

For clarification and education, could you explain or provide a link to or cite the derivation of J(r)?

Thanks!