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Curvature tensor of sphere radius R

by player1_1_1
Tags: curvature, radius, sphere, tensor
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Jan29-10, 08:06 PM
P: 118
hello! I need to find curvature tensor of sphere of R radius. How can I start? thanks!
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Jan30-10, 05:32 AM
P: 118
Jan30-10, 06:53 AM
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Are you talking about the sphere of radius R in three dimensions?

Start by writing out x, y, and z in spherical coordinates with [tex]\rho[/tex] taken as the constant R:

[tex]x= Rcos(\theta)sin(\phi)[/tex]
[tex]y= Rsin(\theta)sin(\phi)[/tex]
[tex]z= R cos(\phi)[/tex]

Calculate the differentials:
[tex]dx= - R sin(\theta)sin(\phi)d\theta+ Rcos(\theta)cos(\phi)d\phi[/tex]
[tex]dy= R cos(\theta)sin(\phi)d\theta+ Rsin(\theta)cos(\phi)d\phi[/tex]
[tex]dz= -R sin(\phi)d\phi[/tex]

Find [tex]ds^2= dx^2+ dy^2+ dz^2[/tex] in terms of spherical coordinates:
[tex]dx^2= R^2 sin^2(\theta)sin^2(\phi)d\theta^2[/tex][tex]- 2R^2sin(\theta)cos(\theta)sin(\phi)cos(\phi)d\theta d\phi[/tex][tex]+ R^2cos^2(\theta)cos^2(\phi)d\phi^2[/tex]
[tex]dy^2= R^2cos^2(\theta)sin^2(\phi)d\theta^2[/tex][tex]+ 2R^2sin(\theta)cos(\theta)sin(\phi)cos(\phi)d\thetad\phi[/tex][tex]+ R^2sin^2(\theta)cos^2(\phi)d\phi^2[/tex]
[tex]dz^2= R^2 sin^2(\phi)d\phi^2[/tex]

Adding those
[tex]ds^2= R^2sin^2(\phi)d\theta^2+ R^2 d\phi^2[/tex]
which gives us the metric tensor:

[tex]g_{ij}= \begin{pmatrix}R^2sin^2(\phi) & 0 \\ 0 & R^2 \end{pmatrix}[/tex]

You can calculate [tex]g^{ij}[/tex], the Clebsh-Gordon coefficients, and the curvature tensor from that.

Jan30-10, 01:00 PM
P: 118
Curvature tensor of sphere radius R

thanks you!!! i finally know what to do;] i going to try to do this, i ask if get problems

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