# How does the iron shell know about infinity?

While this question involves some physics terms, its nature is purely mathematical of differential geometry.

Consider a spacetime defined by a static heavy thin shell that is somewhat larger than its Schwarzschild radius ($$R>r_s$$). According to the Birkhoff theorem, spacetime is flat inside while curved outside the shell.

The following paper On a common misunderstanding of the Birkhoff theorem clarifies that:

the time term of the metric is always maintained continuous, but the space term is not

The space term is discontinuous at the shell. Specifically, inside, there is no length contraction or expansion and the radial interval is the same as at infinity:

$$ds^2=dr^2$$

In contrast, outside the shell, the radial interval is:

$$ds^2 = \left(1-\frac{r_s}{r}\right)^{-1} \,dr^2$$

that diverges at the shell $$r=R\,$$ when $$R\to r_s\,$$ (where $$r_s\,$$ is the Schwarzschild radius).

Here the connection between the space term at infinity and inside the shell is unclear. Intuitively, why exactly is the following true?

$$ds(r\to\infty)=ds(r

The space term at infinity is defined by the chosen coordinate system. This term expands at a smaller radius $$r$$ and diverges outside at $$r=R\,$$ when $$R\to r_s$$. Then suddenly and abruptly it again becomes the same as at infinity. What makes it become exactly the same? Why does it not have an arbitrary value inside? There seems to be no intuitive connection between infinity and the inside of the shell through the coordinate singularity at the Schwarzschild radius.

I realize that the rigor of this question is given by taking the Birkhoff theorem to the limit. What I am looking for is the intuition behind it to see what the connection is between the infinity and the inside of the shell for a better understanding.