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In a perfectly symmetrical spherical hollow shell, there is a null net gravitational force according to Newton, since in his theory the force is exactly inversely proportional to the square of the distance.

What is the result of general theory of relativity? Is the spacetime flat inside (given the fact that orbit of Mercury rotates I don't think so)? How is signal from the cavity redshifted to an observer at infinity?

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up vote 27 down vote accepted

Here we will only answer OP's two first question(v1). Yes, Newton's Shell Theorem generalizes to General Relativity as follows. The Birkhoff's Theorem states that a spherically symmetric solution is static, and a (not necessarily thin) vacuum shell (i.e. a region with no mass/matter) corresponds to a radial branch of the Schwarzschild solution

$$\tag{1} ds^2~=~-\left(1-\frac{R}{r}\right)c^2dt^2 + \left(1-\frac{R}{r}\right)^{-1}dr^2 +r^2 d\Omega^2$$

in some radial interval $r \in I:=[r_1, r_2]$. Here the constant $R$ is the Schwarzschild radius, and $d\Omega^2$ denotes the metric of the angular $2$-sphere.

Since there is no mass $M$ at the center of OP's internal hollow region $r \in I:=[0, r_2]$, the Schwarzschild radius $R=\frac{2GM}{c^2}=0$ is zero. Hence the metric (1) in the hollow region is just flat Minkowski space in spherical coordinates.

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Nice job. Sigh. It really annoys me that someone can write a nice reply to a nice question and end up with a "0" score even after it's selected as the answer to the question. Do people think they have to pay for +s out of their bank account? – Carl Brannen Feb 23 '13 at 22:23
A simple and beautiful answer indeed! – Friedrich Feb 24 '13 at 0:13
@Qmechanic GR is not my field, yet something caught my eye here. I think there should be a $c^2$ in front of the first term on the RHS to match the dimensions, or is it omitted on purpose? Apart from that, is the last term with the differential solid angle element squared equal to $d\theta^2 + \sin^2\theta d\phi^2$, I looked it up on Thanks for the feedback. – Vesnog Jun 15 at 15:23
@Vesnog: I updated the answer. – Qmechanic Jun 16 at 11:55

I've been looking for an answer to this question since a week ago.

I found a paper which I'm not sure if it's correct or not. Mei Xiaochun proved in his paper that the metric inside a hollow sphere is not flat but almost flat. He proved that if you assume $R$ is zero inside hollow sphere you face problems satisfying boundary conditions.

From what I understood from his proof, if you assume that $V_{sphere}=\dfrac{4\pi r^3}{3}$ in curved space then $R=0$ but since this argument is not true in general you can't assume $R=0$. He also showed that though the difference between $V$ and $V_0=\dfrac{4\pi r^3}{3}$ is so small ($\dfrac{\Delta V}{V_0}=1.8\times 10^{-4}$ for neutron star and $\dfrac{\Delta V}{V_0}=1.8\times 10^{-2}$ for our universe as a spherical shell); $R$ it is not exactly zero.

So I'm confused now...

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$R$ is just a parameter in the theory. It comes out when you assume spherical symmetry. YOu don't have to assume anything special about the coordinate system at all. See Weinberg's Cosmology book for explicit detail about this. – Jerry Schirmer May 12 '15 at 19:53
@JerrySchirmer yes I already know about that, but my point was that $R$ is not equal to the Schwarzschild radius. So the argument that he used for showing $R=0$ is wrong. – Mohammadreza Zakeri May 13 '15 at 1:21
I don't even get what you're talking about. Birchoff's theorem is exactly true in general relativity. Who cares if you call the parameter the "Schwarzschild radius" or whatever. it's a constant in the solution, and it is zero inside a spherical cavity. And you don't assume anything about the volume of a sphere when deriving these results. – Jerry Schirmer May 13 '15 at 1:25
@JerrySchirmer Okay, The answer is that because "we shouldn't have any singularities inside the cavity", $R=0.$ Which is just a physical argument. I agree with it. So maybe that guy who wrote the paper I mentioned earlier made a mistake somewhere. – Mohammadreza Zakeri May 13 '15 at 2:05
I"m reading the paper you linked now, but I have to say that it contradicts relatively simple arguments derived from just using Einstein's equations directly, as in Weinberg's book. And your answer seems to confuse some core concepts of how you paste together solutions. – Jerry Schirmer May 13 '15 at 2:10

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