Talk:Birkhoff's theorem (relativity)

message: Does asymtotically flatness follow from staticity?
I think Birkhoff has proven that spherical symmetric vacuum solutions of Einsteins equations are static.
So they of course correspond to the SS-solutions in that point, but SS solutions are asymtotically flat too. Where is the asymtotic flatness in Birkhoff's theorem or does it really follow from Birkhoffs assumptions as stated in the article?
 (student access at University of Vienna)
:Is this your question?: "is every spherically symmetric static vacuum solution a Schwarzschild vacuum solution, or must one add the additional hypothesis of asymptotic flatness?"  ---CH 21:38, 6 June 2006 (UTC)
thread_title:  Aymptotically Flatness 

message: I had been monitoring this article, but I am leaving the WP and am now abandoning this article to its fate.
Just wanted to provide notice that I am only responsible (in part) for the last version I edited; see User:Hillman/Archive.
I emphatically do not vouch for anything you might see in more recent versions.
Good luck in your seach for information, regardless!---CH 22:43, 30 June 2006 (UTC)
thread_title: Students beware

message: The first paragraph says: Birkhoff's theorem states that any spherically symmetric solution of the vacuum field equations must be stationary and asymptotically flat. This is not correct as the solution is obviously not static inside the horizon. The statement should say Birkhoff's theorem states that any spherically symmetric solution of the vacuum field equations must be the Schwarzschild solution. Basically, it says that the solution when written in Schwarzschild's coordinates has coefficients independent of the t-coordinate. Outside the horizon this requirement together with the diagonal form of the metric do translate into "static" of course, but not inside. The t-independence inside means the metric is spatially homogeneous there. JanBielawski 19:07, 25 April 2007 (UTC)
replies: text: No response to this over 5 years later? All that's happened in that time is that "stationary" has been replaced by "static", but the current statement "any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat" is clearly incorrect as the Schwarzschild solution is a non-static counterexample inside the event horizon. Can someone who has access to a reliable source say what Birkhoff's theorem really says? The only (non-rigorous) version I have seen is "Schwarzschild's solution is the unique vacuum solution with spherical symmetry". -- Dr Greg  talk  17:01, 12 September 2012 (UTC)
replies: text:  I'll fix it tonight if I remember it. JanBielawski (talk) 21:43, 6 October 2016 (UTC)

text: Your statement above doesn't seem to make sense. In what way is the Minkowski metric a counterexample? It simply represents the Schwarzschild solution with zero mass, so it's not a counterexample, it confirms Birkhoff's theorm. JanBielawski (talk) 21:43, 6 October 2016 (UTC)
thread_title: Something is wrong with the statement

message: In the Implications section, it is asserted that the interior solution to a spherical shell corresponds to Minkowski space. 
But should not the metric be at least continuous when one passes through the shell? 
Consider for a moment the gravitational forces one will calculate with a discontinuous metric!  A mathematically discontinuous metric has infinite derivatives.
Infinitely far away, we have Minkowski space. But near the outside of the shell, we have a bit of curved space leading (for example) to a gravitational redshift of light leaving the shell's surface and going on to infinity, i.e., $g_{00}\ne 1$. 
Inside the shell, there is indeed no curvature, i.e., spacetime is flat, and thus there are no gravitational forces; but that doesn't mean that g_{00} must be equal to 1, i.e., that it is Minkowskian spacetime.
Should it, and the rest of the metric, instead, match the value right on the outside of the shell?
The implication of that is, of course, that time inside the shell should be dilated relative to the time infinitely far from the shell. Specifically, time should flow more slowly inside the shell than far from the shell where the space is flat and Minkowskian.
That is, the time dilation should be the same everywhere inside as just outside the shell. There is a gravitational force outside, but none inside. But it is not Minkowskian inside.
Since charges are also mentioned, the same kind of thing should happen with a charged shell (with or without mass). There should be a time 
dilation within the shell equal to that just outside. But, inside the shell, there is no gravitational force, nor any electromagnetic field. 
Next consider two static, massless, concentric shells, with equal but opposite charges on them. There would be an electric force (acting on a test charge) between the shells, but none inside the inner one, nor outside the outer one. There is also a gravitational field (Schwarzchild metric) outside the outer shell (even if the shells are massless). Solving the Einstein equations inside the inner shell should give flat space, but time dilated relative to the space far from this charged capacitor. 
(Incidentally, that capacitor is the basis of a purely electric (no massive blackholes, etc) time machine that will hurl travelers into the future [slightly faster than the rest of us] without experiencing the destructive forces of blackholes [giant masses] or traveling far from Earth at high speeds [as one of the twins].)  —Preceding unsigned comment added by 222.130.176.89 (talk) 04:13, 16 July 2008 (UTC)
replies: text: Hi would you like to talk, during my lost time I thought about the same thing, one years ago. I just hae one remark, the metric can be flat or curved anywhere, depending on the frame is "accelerating" or not. If we can always find a frame where metric is minkowsky, then we always can find a frame where metric is curved.Klinfran (talk) 12:25, 4 April 2011 (UTC)
thread_title:  Inside Spherical Shell