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Schwarzschild metric as induced metric

  1. Jun 25, 2010 #1
    According to Nash theorem http://en.wikipedia.org/wiki/Nash_embedding_theorem" [Broken] every Riemannian manifold can be isometrically embedded
    into some Euclidean space. I wonder if it's true also
    in case of pseudoremanninan manifolds. In particular is it possible to find
    a submanifold in pseudoeuclidean space that, the metric induced on it will be
    Schwarzschild metric? How many dimensions we need?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 25, 2010 #2

    George Jones

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    Chris Clarke* showed that every 4-dimensional spacetime can be embedded isometically in higher dimensional flat space, and that 90 dimensions suffices - 87 spacelike and 3 timelike. A particular spacetime may be embeddable in a flat space that has dimension less than 90, but 90 guarantees the result for all possible spacetimes.

    * Clarke, C. J. S., "On the global isometric embedding of pseudo-Riemannian
    manifolds," Proc. Roy. Soc. A314 (1970) 417-428
  4. Jun 25, 2010 #3
    You need 6 dimensions to embed a Schwarzschild solution. I think that all GR solutions can be (locally) embedded in 10 dimensions.
  5. Jun 26, 2010 #4
    How do you know it?
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