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What is the metric tensor on a spherical surface?

  1. Aug 3, 2006 #1
    What is the metric tensor on a spherical surface?
     
  2. jcsd
  3. Aug 3, 2006 #2

    quasar987

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    I learned this like 2 minutes ago but I believe the following is correct:

    A parametrisation of the sphere of radius [itex]\rho[/itex] centered on the origin is

    [tex]f(\theta, \phi)=(\rho \sin(\theta) \cos(\phi) , \rho \sin(\theta)\sin(\phi),\rho \cos(\theta))[/tex]

    where I am using this convention for the spherical angles : http://en.wikipedia.org/wiki/Spherical_coordinates#Spherical_coordinates

    The components of the metric tensor are then

    [tex]g_{11}(\theta, \phi) = \langle \frac{\partial f}{\partial \theta}, \frac{\partial f}{\partial \theta} \rangle[/tex]
    [tex]g_{12}(\theta, \phi) = \langle \frac{\partial f}{\partial \theta}, \frac{\partial f}{\partial \phi} \rangle[/tex]
    [tex]g_{21}(\theta, \phi) = \langle \frac{\partial f}{\partial \phi}, \frac{\partial f}{\partial \theta} \rangle[/tex]
    [tex]g_{22}(\theta, \phi) = \langle \frac{\partial f}{\partial \phi}, \frac{\partial f}{\partial \phi} \rangle[/tex]


    The matrix form is then

    [tex]G(\theta, \phi)=\left( \begin {array} {cc} g_{11}(\theta, \phi) & g_{12}(\theta, \phi) \\ g_{21}(\theta, \phi) & g_{22}(\theta, \phi) \end {array} \right)[/tex]

    All you gotta do is calculate the derivatives. Have fun. :p
     
    Last edited: Aug 3, 2006
  4. Aug 3, 2006 #3
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