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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    $\begingroup$ "Is the spacetime flat" seems like a strange and possibly misleading way to put it... if something analogous to Newton's Shell Theorem is applicable here, that tells you only that the net curvature caused by the shell is zero... even if that's the case, that still wouldn't imply spacetime is flat inside, since there may be other objects in the universe, outside the shell, that are exerting forces and warping that region of space, right? $\endgroup$
    – Don Hatch
    Jul 28, 2018 at 6:03
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    $\begingroup$ The GR case isn't quite as analogous to the Newtonian case as you might think, or as important. GR is nonlinear, so you can't treat a spherically symmetric mass distribution as the sum of concentric shells. And there is an ambiguity in how we pose the problem. For example, the spacetime of a freely collapsing shell of dust is given by an Oppenheimer-Snyder solution, whereas the spacetime for a shell held in static equilibrium by internal forces is different. Things like redshifts at infinity aren't well defined if it's not static. $\endgroup$
    – user4552
    Apr 20, 2019 at 15:04

1 Answer 1


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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    $\begingroup$ 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? $\endgroup$ Feb 23, 2013 at 22:23
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    $\begingroup$ A simple and beautiful answer indeed! $\endgroup$
    – Friedrich
    Feb 24, 2013 at 0:13
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    $\begingroup$ @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? $\endgroup$
    – safesphere
    Nov 29, 2018 at 3:40
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    $\begingroup$ @Qmechanic Here is the correct solution: arxiv.org/abs/1203.4428 $\endgroup$
    – safesphere
    Nov 29, 2018 at 3:57
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    $\begingroup$ @safesphere: Thanks for the feedback. Indeed the relations between coordinate systems inside, outside and on the infalling thin shell is non-trivial. $\endgroup$
    – Qmechanic
    Nov 29, 2018 at 13:14

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