
#1
Jan2910, 08:06 PM

P: 118

hello! I need to find curvature tensor of sphere of R radius. How can I start? thanks!




#2
Jan3010, 05:32 AM

P: 118

hello




#3
Jan3010, 06:53 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,902

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 ClebshGordon coefficients, and the curvature tensor from that. 



#4
Jan3010, 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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