# Curvature tensor of sphere radius R

by player1_1_1
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,488 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.