# What is the metric tensor on a spherical surface?

1. Aug 3, 2006

### HeilPhysicsPhysics

What is the metric tensor on a spherical surface?

2. Aug 3, 2006

### 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

$$g_{11}(\theta, \phi) = \langle \frac{\partial f}{\partial \theta}, \frac{\partial f}{\partial \theta} \rangle$$
$$g_{12}(\theta, \phi) = \langle \frac{\partial f}{\partial \theta}, \frac{\partial f}{\partial \phi} \rangle$$
$$g_{21}(\theta, \phi) = \langle \frac{\partial f}{\partial \phi}, \frac{\partial f}{\partial \theta} \rangle$$
$$g_{22}(\theta, \phi) = \langle \frac{\partial f}{\partial \phi}, \frac{\partial f}{\partial \phi} \rangle$$

The 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 gotta do is calculate the derivatives. Have fun. :p

Last edited: Aug 3, 2006
3. Aug 3, 2006