Missing proof of the Shell theorem in General Relativity

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
34 replies · 3K views
PeroK said:
The Schwarzschild solution is valid outside an object, as long as the radius of the object is greater than its Schwarzschild radius.
Actually it has to be greater than 9/8 of the Schwarzschild radius, per Buchdahl's Theorem.
 
Physics news on Phys.org
Bosko said:
Do you think that for ##r_1 < r_s## the mathematical model does not correctly represent physical reality?
That's not the issue. The issue is that you don't understand what physical reality the mathematical model represents for values of ##r## less than ##r_s##. For such values of ##r## the model represents the interior of a black hole. It does not represent the vacuum region outside an ordinary gravitating object with that radius.
 
PeroK said:
For ##r < r_s##, the coordinate ##r## is timelike.
This is true for Schwarzschild coordinates, but there are other charts for which it is not true. You give a much better description of the actual problem (which can be stated in a form that's invariant, so it doesn't actually depend on your choice of coordinates) here:

PeroK said:
The region of spacetime represented by ##0 < r < r_1 < r_s## for any ##\phi, \theta## and fixed ##t## is not a spatial volume. It cannot be the volume of a massive object.
 
Bosko said:
There is no parameter of a spherically symmetric object in the solution. For example, the radius .
The solution does not depend on the size of the object.
Let's find some detailed explanation of the formula and see when the radius of the object vanishes .

Is this from Wikipedia good for analysis?
https://en.wikipedia.org/wiki/Derivation_of_the_Schwarzschild_solution

I'm looking for a detailed derivation available on the internet.

Can we agree that the solution does not depend on the radius of a spherically symmetric object?
I think that this is some form of the shell theorem.
You assume nothing but spherical symmetry, and vacuum outside some r. You assume nothing about inside r (except spherical symmetry throughout). You derive that the solution is unique up to one parameter - a mass parameter associated with everything inside r. Thus you have derived, not assumed, that nothing about the nature or size of what is inside r can matter. There is simply no place for it in the solution. The shell theorem in GR is thus derived, not assumed. It is also different from the Newtonian statement ( you can’t talk about a mass point). The GR statement is: if all vacuum outside of r, nothing about the inside matters except one parameter describing total mass.