Curvature: Riemann, Ricci & Einstein Tensors
Curvature: Riemann, Ricci & Einstein Tensors
The Riemann curvature tensor encodes how parallel transport is path-dependent, measuring the failure of covariant derivatives to commute. Contracting the Riemann tensor yields the Ricci tensor and Ricci scalar, which combine into the Einstein tensor — the geometric side of the field equations.
Riemann Curvature Tensor
\(R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}\). Commutator form: \([\nabla_\mu, \nabla_\nu]V^\rho = R^\rho{}_{\sigma\mu\nu}V^\sigma\). Symmetries: \(R_{\rho\sigma\mu\nu} = -R_{\sigma\rho\mu\nu} = -R_{\rho\sigma\nu\mu} = R_{\mu\nu\rho\sigma}\). In 4D: 20 independent components.
Bianchi Identity & Einstein Tensor
Second Bianchi identity: \(\nabla_\lambda R_{\rho\sigma\mu\nu} + \nabla_\rho R_{\sigma\lambda\mu\nu} + \nabla_\sigma R_{\lambda\rho\mu\nu} = 0\). Contracting twice yields \(\nabla_\mu G^{\mu\nu} = 0\), where \(G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R\) is the divergence-free Einstein tensor.
Example 1: Riemann Tensor on \(S^2\)
Compute \(R^\theta{}_{\phi\theta\phi}\) for the unit 2-sphere.
Solution: Using \(\Gamma^\theta_{\phi\phi}=-\sin\theta\cos\theta\), \(\Gamma^\phi_{\theta\phi}=\cot\theta\): \(R^\theta{}_{\phi\theta\phi} = \sin^2\theta\). The Ricci scalar \(R = 2\), confirming constant positive curvature \(K=1\).
Example 2: Ricci Scalar of Flat Space
Show that \(R = 0\) in flat Minkowski space.
Solution: All \(\Gamma^\lambda_{\mu\nu} = 0\) in Cartesian coordinates (constant metric). Thus \(R^\rho{}_{\sigma\mu\nu} = 0\), so \(R_{\mu\nu} = 0\) and \(R = 0\). Flat spacetime has zero curvature as expected.
Practice
- Prove \(R_{\rho\sigma\mu\nu} = R_{\mu\nu\rho\sigma}\) using the full symmetry properties.
- Show the Ricci tensor is symmetric: \(R_{\mu\nu} = R_{\nu\mu}\).
- How many independent components does \(G_{\mu\nu}\) have in 4D?
- Compute \(G_{\mu\nu}\) for a maximally symmetric spacetime with cosmological constant \(\Lambda\).
Show Answer Key
1. Riemann tensor symmetries: (1) $R_{\rho\sigma\mu\nu}=-R_{\sigma\rho\mu\nu}$ (antisymmetric in first pair). (2) $R_{\rho\sigma\mu\nu}=-R_{\rho\sigma\nu\mu}$ (antisymmetric in second pair). (3) $R_{\rho\sigma\mu\nu}=R_{\mu\nu\rho\sigma}$ (pair symmetry). Proof of (3): use (1), (2), and the first Bianchi identity $R_{\rho[\sigma\mu\nu]}=0$. Expanding: $R_{\rho\sigma\mu\nu}+R_{\rho\mu\nu\sigma}+R_{\rho\nu\sigma\mu}=0$. Write three such identities and combine to prove pair symmetry.
2. $R_{\mu\nu} = R^\sigma_{\;\mu\sigma\nu} = g^{\sigma\lambda}R_{\lambda\mu\sigma\nu}$. By pair symmetry of Riemann: $R_{\lambda\mu\sigma\nu} = R_{\sigma\nu\lambda\mu}$. So $R_{\mu\nu} = g^{\sigma\lambda}R_{\sigma\nu\lambda\mu} = R^\sigma_{\;\nu\sigma\mu}... $ Hmm, more directly: $R_{\mu\nu} = R^\alpha_{\;\mu\alpha\nu}$. Swap first pair: $R^\alpha_{\;\mu\alpha\nu} = g^{\alpha\beta}R_{\beta\mu\alpha\nu} = g^{\alpha\beta}R_{\alpha\nu\beta\mu} = R^\alpha_{\;\nu\alpha\mu} = R_{\nu\mu}$. ✓
3. $G_{\mu\nu} = R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R$ is symmetric (both $R_{\mu\nu}$ and $g_{\mu\nu}$ are). In 4D: a symmetric 2-tensor has $\frac{4\cdot5}{2}=10$ independent components. But $\nabla_\mu G^{\mu\nu}=0$ (Bianchi identity) gives 4 constraints, leaving 6 independent equations — matching the 6 physical degrees of freedom of the gravitational field (metric has 10 components minus 4 coordinate freedoms).
4. Maximally symmetric: $R_{\mu\nu\rho\sigma} = \frac{R}{12}(g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho})$ with $R = 4\Lambda$ (constant). $R_{\mu\nu} = \frac{R}{4}g_{\mu\nu} = \Lambda g_{\mu\nu}$. $G_{\mu\nu} = R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R = \Lambda g_{\mu\nu}-2\Lambda g_{\mu\nu} = -\Lambda g_{\mu\nu}$.