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Curvature tensor of sphere radius R

  1. Jan 29, 2010 #1
    hello! I need to find curvature tensor of sphere of R radius. How can I start? thanks!
  2. jcsd
  3. Jan 30, 2010 #2
  4. Jan 30, 2010 #3


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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.
    Last edited by a moderator: Jan 30, 2010
  5. Jan 30, 2010 #4
    thanks you!!! i finally know what to do;] i going to try to do this, i ask if get problems
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