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I'm a third year maths undergrad and I'm trying to find (and follow) a curvature based derivation of the Schwarzschild metric, if there exists such a proof?

user12262
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Sarah Jayne
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  • it is conventonially derived as the only spherically symmetric metric solution for the vacuum. This seems to me like perfectly simple approach. Does that not satisfy you? – Danu Feb 09 '14 at 10:36
  • If you like this question you may also enjoy reading this Phys.SE post. – Qmechanic Feb 09 '14 at 10:44
  • Unfortunately not! I'm writing a project on minimal surfaces - so the curvature plays a big role :-) – Sarah Jayne Feb 09 '14 at 10:46
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    So what would you mean by a 'curvature based derivation'? – Danu Feb 09 '14 at 12:15
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    Since the idea of a geometric manifold with associated metric tensor, as well as other derived quantities such as the Riemann curvature tensor, Ricci tensor, etc., it would seem hard to find a derivation that IS NOT a curvature based derivation. So what exactly are referring to when you say curvature-based? – Bruce Dean Feb 09 '14 at 19:02

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