1
$\begingroup$

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?

$$ds(r\to\infty)=ds(r<R,\,R\to r_s)$$

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.

$\endgroup$
1
$\begingroup$

Point on the infinity and the point inside the shell are the same in a sense that they do not experience the presence of the shell. Point on the infinity — because the shell is far far away. Point inside the shell — because the influence from the shell is perfectly balanced (Birkhoff theorem).

$\endgroup$
  • $\begingroup$ This assumes the existence of something (spacetime) that is "absolute" in the sense that it can remain the same as this "absolute" value in disconnected regions, if they don't "experience" (are not changed by) an external influence. However, the existence of such an "absolute" standard is not convincing, because spacetime itself is a field of the field equations and the value of "one meter" may not be the same in the regions disconnected by a coordinate singularity. $\endgroup$ – safesphere Sep 11 at 14:28
  • 1
    $\begingroup$ This becomes philosophical. The known equations describe the overseen world. By “creating” absolutely thin shell, you have created mathematical singularity, which is physically infeasible. If you want to consider a true Schwarzschild singularity, then every guess is as good as yours. Mathematically you can take any value. Physically, we don't know. $\endgroup$ – Vasily Mitch Sep 11 at 14:39
  • $\begingroup$ I have nothing more to say to you. I hope someone helps you with your question. $\endgroup$ – Vasily Mitch Sep 11 at 15:00
  • $\begingroup$ I appreciate your answer. Thank you so much! :) $\endgroup$ – safesphere Sep 11 at 15:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.