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Schwarzschild metric as induced metric

 
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Jun25-10, 11:28 AM   #1
 

Schwarzschild metric as induced metric


According to Nash theorem Nash theorem 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?
 
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Jun25-10, 11:38 AM   #2
 
Mentor
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
 
Jun25-10, 06:52 PM   #3
 
You need 6 dimensions to embed a Schwarzschild solution. I think that all GR solutions can be (locally) embedded in 10 dimensions.
 
Jun26-10, 05:34 AM   #4
 

Schwarzschild metric as induced metric


Quote by Passionflower View Post
You need 6 dimensions to embed a Schwarzschild solution.
How do you know it?
 
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