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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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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

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

in some radial interval $r \in I:=[r_1, r_2]$. Here the constant $R$ is the Schwarzschild radius.

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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Thanks. What is OP? –  Leos Ondra Nov 7 '12 at 19:42
    
@Leos Ondra: OP=Original Poster. In this case: you. –  Qmechanic Nov 7 '12 at 19:55
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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

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