Curvature tensor of sphere radius R

[player1_1_1]

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

 

[player1_1_1]

hello

 

[HallsofIvy]

Are you talking about the sphere of radius R in three dimensions?

Start by writing out x, y, and z in spherical coordinates with $$\rho$$ taken as the constant R:

$$x= Rcos(\theta)sin(\phi)$$
$$y= Rsin(\theta)sin(\phi)$$
$$z= R cos(\phi)$$

Calculate the differentials:
$$dx= - R sin(\theta)sin(\phi)d\theta+ Rcos(\theta)cos(\phi)d\phi$$
$$dy= R cos(\theta)sin(\phi)d\theta+ Rsin(\theta)cos(\phi)d\phi$$
$$dz= -R sin(\phi)d\phi$$

Find $$ds^2= dx^2+ dy^2+ dz^2$$ in terms of spherical coordinates:
$$dx^2= R^2 sin^2(\theta)sin^2(\phi)d\theta^2$$$$- 2R^2sin(\theta)cos(\theta)sin(\phi)cos(\phi)d\theta d\phi$$$$+ R^2cos^2(\theta)cos^2(\phi)d\phi^2$$
$$dy^2= R^2cos^2(\theta)sin^2(\phi)d\theta^2$$$$+ 2R^2sin(\theta)cos(\theta)sin(\phi)cos(\phi)d\thetad\phi$$$$+ R^2sin^2(\theta)cos^2(\phi)d\phi^2$$
$$dz^2= R^2 sin^2(\phi)d\phi^2$$

Adding those
$$ds^2= R^2sin^2(\phi)d\theta^2+ R^2 d\phi^2$$
which gives us the metric tensor:


$$g_{ij}= \begin{pmatrix}R^2sin^2(\phi) & 0 \\ 0 & R^2 \end{pmatrix}$$

You can calculate $$g^{ij}$$, the Clebsh-Gordon coefficients, and the curvature tensor from that.

 

[player1_1_1]

thanks you! i finally know what to do;] i going to try to do this, i ask if get problems