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Insights A Short Proof of Birkhoff's Theorem - Comments

  1. Aug 18, 2015 #1

    PeterDonis

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    Last edited: Jun 27, 2016
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  3. Aug 18, 2015 #2

    bcrowell

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    Nice job. For comparison, I have a proof in my GR book http://www.lightandmatter.com/genrel/ , section 7.4, but it's really very similar. Some other statements and proofs of the theorem that I've seen:

    Birkhoff's original proof, in Birkhoff, Relativity and Modern Physics, 1923. A horrible, long monstrosity with an out of date attitude toward the significance of coordinates.

    Hawking and Ellis: "Any C^2 solution of Einstein's empty space equations which is spherically symmetric in an open set V, is locally equivalent to part of the maximally extended Schwarzschild solution in V." The part about "maximally extended" is a good point -- I always tend to think about just part of the Schwarzschild spacetime (2 of the 4 regions) and forget that it can be extended.

    http://arxiv.org/abs/gr-qc/0408067 -- "Schwarzschild and Birkhoff a la Weyl," Deser and Franklin. Birkhoff's thm is equivalent to proving that the m in the Schwarzschild metric is constant.

    As you point out, the existence of the ##\partial_t## Killing vector doesn't mean that the spacetime is static. However, it *is* asymptotically static, which is kind of the only nontrivial thing being proved. If we knew in advance that it was asymptotically static, then Birkhoff's theorem would amount to no more than the usual derivation of the Schwarzschild metric. Essentially we're seeing that there's no such thing as gravitational monopole radiation.

    For a really rigorous proof, I think one needs to deal with the possibility that the metric coefficients blow up or go to zero, and show that these would be only coordinate singularities-- but I don't do that either, just mention it in a footnote.
     
  4. Aug 18, 2015 #3

    PeterDonis

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    Thanks!

    Yes, this is true, and IIRC MTW do deal with this in their proof. I think what they do still goes the same way once the wart in their proof is removed as I (and you) remove it. :wink:
     
  5. Aug 18, 2015 #4
    Nice work Peter!
     
  6. Aug 21, 2015 #5

    vanhees71

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    Very nice indeed, but you can save MTW's ansatz (which I never understood, why it is made in so many textbooks and not your way, which is more general) by allowing the exponents to become complex. Since the exponentials must be real in the pseudometric, this amounts to adding ##2 n \pi \mathrm{i}## (which changes nothing) or ##(2n+1) \pi \mathrm{i}## with ##n \in \mathbb{Z}##. Then the Minkowskian signature of the metric constrains these possibilities to the solutions you gave. Of course, it's much more simple to just use your ansatz and staying with real quantities all the time during the derivation.
     
  7. Aug 24, 2015 #6

    stevendaryl

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    Here's a question: How does this compare with Schwarzschild's original derivation? Is the only difference that Schwarzschild started out assuming a time-independent solution, while this derivation proves that time-independence follows from the assumption of spherical symmetry?
     
  8. Aug 24, 2015 #7

    PeterDonis

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    No, although that's one difference. The other, more important difference is that Schwarzschild's original derivation used different coordinates; his radial coordinate, which I'll call ##\rho##, was defined in such a way that ##\rho = 0## corresponded to the horizon, not the singularity at what we now call ##r = 0##. This led to several decades of confusion because it was not fully appreciated that (a) a given coordinate chart might not cover all of a given spacetime, and (b) coordinates in themselves have no physical meaning; the physics of any solution is contained in the invariants.
     
  9. Aug 24, 2015 #8

    PeterDonis

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    Here's a previous PF discussion on Schwarzschild's original solution:

    https://www.physicsforums.com/threads/schwarzschilds-metric-1916.708045/page-5

    As you'll see from some of the links and posts in this thread, there are still people who make the same error that Schwarzschild originally made, thinking that if they defined a "radial coordinate" that had value zero at the horizon, that somehow meant that, physically, there couldn't be any other region of spacetime beneath the horizon.
     
  10. Aug 25, 2015 #9

    stevendaryl

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    Thanks. With the hindsight of 100 years of GR, it's hard to get back into the frame of mind that glosses over the distinction between coordinate-dependent features of a solution and physically meaningful features.
     
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