[SOLVED] Is spacetime flat inside a spherical shell?

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?

@Qmechanic 2012-11-07 16:11:07

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.

@Carl Brannen 2013-02-23 22:23:33

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?

@Friedrich 2013-02-24 00:13:47

A simple and beautiful answer indeed!

@asmaier 2017-01-10 23:08:50

Does this conclusion still hold if we have a non-vanishing positive cosmological constant?

@Qmechanic 2017-01-11 10:38:16

Yes, but the role of Minkowski space is then replaced by de Sitter space.

But Leos also asked about the redshift of the light from the cavity.

@Yukterez 2018-10-02 12:02:59

It's the same as the redshift from the surface

@safesphere 2018-11-29 03:40:31

@Qmechanic Substituting $R=0$ In $(1)$ does not produce the correct time dilation inside the shell. In other words, this metric does not satisfy the junction conditions across the shell. What is the correct formula for the metric inside?

@safesphere 2018-11-29 03:57:04

@Qmechanic Here is the correct solution: arxiv.org/abs/1203.4428

@Qmechanic 2018-11-29 13:14:06

@safesphere: Thanks for the feedback. Indeed the relations between coordinate systems inside, outside and on the infalling thin shell is non-trivial.

@Ben Crowell 2019-04-20 12:20:07

The OP asked "How is signal from the cavity redshifted to an observer at infinity?," but the answer doesn't address this. The Zhang and Yi paper does discuss this.

@Ben Crowell 2019-04-20 12:45:15

@safesphere: Substituting R=0 In (1) does not produce the correct time dilation inside the shell. In other words, this metric does not satisfy the junction conditions across the shell. What is the correct formula for the metric inside? There is nothing incorrect about writing the metric in this form. The Zhang paper just points out that the coordinates used in writing the metric this way cannot be connected in a natural and continuous way to the Schwarzschild coordinates in the exterior region.

@safesphere 2019-04-20 16:08:16

@BenCrowell The paper clarifies that "discontinuous time term of the metric at all interior boundaries, i.e., the clocks are defined differently at both sides of a boundary [...] is clearly nonphysical, and also mathematically incorrect".