What is the metric tensor on a spherical surface?

[HeilPhysicsPhysics]

What is the metric tensor on a spherical surface?

 

[quasar987]

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

A parametrisation of the sphere of radius \rho centered on the origin is

f(\theta, \phi)=(\rho \sin(\theta) \cos(\phi) , \rho \sin(\theta)\sin(\phi),\rho \cos(\theta))

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

g_{11}(\theta, \phi) = \langle \frac{\partial f}{\partial \theta}, \frac{\partial f}{\partial \theta} \rangle
g_{12}(\theta, \phi) = \langle \frac{\partial f}{\partial \theta}, \frac{\partial f}{\partial \phi} \rangle
g_{21}(\theta, \phi) = \langle \frac{\partial f}{\partial \phi}, \frac{\partial f}{\partial \theta} \rangle
g_{22}(\theta, \phi) = \langle \frac{\partial f}{\partial \phi}, \frac{\partial f}{\partial \phi} \rangleThe matrix form is then

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)

All you got to do is calculate the derivatives. Have fun. :p

 

[yenchin]

We discussed this only a few threads down... https://www.physicsforums.com/showthread.php?t=126648