# Riemann tensor, Ricci tensor of a 3 sphere

by dingo_d
Tags: ricci, riemann, sphere, tensor
 Share this thread:
 P: 211 1. The problem statement, all variables and given/known data I have the metric of a three sphere: $g_{\mu \nu} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2\sin^2\theta \end{pmatrix}$ Find Riemann tensor, Ricci tensor and Ricci scalar for the given metric. 2. Relevant equations I have all the formulas I need, and I calculated the necessary Christoffel symbols, by hand and by mathematica and they match. There are 9 non vanishing Christoffel symbols. Some I calculated and for others I used the symmetry properties and the fact that the metric is diagonal (which simplifies things). But when I go and try to calculate Riemman tensor via: $R^{a}_{bcd}=\partial_d \Gamma^a_{bc}-\partial_c\Gamma^a_{bd}+\Gamma^m_{bc}\Gamma^a_{dm}-\Gamma^m_{bd}\Gamma^a_{cm}$ I get all zeroes for components :\ And I kinda doubt that every single component is zero. The Christoffel symbols are: $\begin{array}{ccc} \Gamma _{\theta r}^{\theta } & = & \frac{1}{r} \\ \Gamma _{\phi r}^{\phi } & = & \frac{1}{r} \\ \Gamma _{r\theta }^{\theta } & = & \frac{1}{r} \\ \Gamma _{\theta \theta }^r & = & -r \\ \Gamma _{\phi \theta }^{\phi } & = & \cot (\theta ) \\ \Gamma _{r\phi }^{\phi } & = & \frac{1}{r} \\ \Gamma _{\theta \phi }^{\phi } & = & \cot (\theta ) \\ \Gamma _{\phi \phi }^r & = & -r \sin ^2(\theta ) \\ \Gamma _{\phi \phi }^{\theta } & = & -\cos (\theta ) \sin (\theta ) \end{array}$
 P: 161 that is the metric of R^3 in sphereical coordinates with one of the angles constant. hense....a plane.
 P: 211 Oh, so it's normal for Riemann tensor to be zero since there is no curvature. Phew, I thought I might be doing something wrong xD Is there some easier way of showing that all of the components of Riemann tensor are zero, rather than manually calculating them all? And how come when I look at 2-sphere $g_{\mu \nu} = \begin{pmatrix} r^2 & 0 \\ 0 & r^2\sin^2\theta \end{pmatrix}$ I get several non vanishing components of Riemann tensor? Is it because I'm now looking it from my 3d perspective so I can see that the plane has to be curved to form a sphere?
 P: 161 Riemann tensor, Ricci tensor of a 3 sphere but now you can pull out r^2 which is clearly nonflat conformal. you can't just throw away terms in the metric and think you're looking at an orthogonal projection onto the remaining coordinates. if you are in a flat geometry in curvilinear coordinate systems then transforming the coordinate system back into flat coordinates might help ease the pain.
 Sci Advisor P: 2,953 What you have at the top is not the metric of a 3-sphere, but simply the metric of R^3 in spherical polar coordinates. A 3-sphere metric will have r fixed and 3 angles to describe where you are (a sphere is hollow remember), just as a 2-sphere has r fixed and 2 angles to describe where you are.

 Related Discussions Differential Geometry 4 Calculus & Beyond Homework 2 Differential Geometry 4 Advanced Physics Homework 0 Differential Geometry 0