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Chris Hillman
Nov27-07, 01:16 PM
Sci Advisor
P: 2,340
Hi, Mentz114,

Quote Quote by Mentz114 View Post
I get a different result for the Ricci scalar, viz
\frac{\partial rf}{re^{f}}
[/tex] where [itex]\partial r[/itex] is differentiation wrt r.
The Ricci scalar is [itex]R = \frac{2 f^\prime}{r} \, \exp(-2f)[/itex] in the notation I used above. (The Ricci tensor, with components evaluated wrt the coframe field I gave, is diagonal with both diagonal components equal to the Riemann curvature component I computed.)

The result of Problem 3.4-19 in Struik implies that the Gaussian curvature is
K = \frac{R_{r \phi r \phi}}{\exp(2f) \, r^2}
= \frac{r \, f^\prime}{ r^2 \, \exp(2f)}
= \frac{f^\prime}{r} \, \exp(-2f) = R_{1212}
as I claimed.

Exercise: write an orthogonal chart for the general Riemannian two-manifold in the form [itex]ds^2 = A^2 \, du^2 + B^2 \, dv^2[/itex], where A,B can be functions of u,v (although this isn't neccessary; without loss of generality we could impose further restrictions), and adopt the coframe field
[itex]\sigma^1 = A \, du, \; \sigma^2 = B \, dv[/itex]. Using the method of curvature two-forms, show that
-R_{1212} = \frac{ \left( \frac{A_u}{B} \right)_u + \left( \frac{B_v}{A} \right)_v }{AB}
This is one of the formulas offered by Gauss for what we call the Gaussian curvature in his October 1827 paper (published in 1828), with the same small change of notation that I made at the beginning of my Post #2.