Bel–Robinson tensor

Remarks

The tensor $T_{abcd}$ is totally symmetric and trace-free in every pair of indices. If the Ricci tensor of $g$ vanishes, then the Bianchi identities give $\nabla^a T_{abcd} = 0$. For future-pointing causal vectors $u, v, w, x$ one has $T_{abcd} u^a v^b w^c x^d \ge 0$. Example: for the Schwarzschild metric the Weyl tensor is nonzero outside the singularity, so $T_{abcd} u^a u^b u^c u^d > 0$ for a static observer with four-velocity $u$. By contrast, in Minkowski space $C_{abcd} = 0$, so $T_{abcd} = 0$. In a vacuum the Weyl tensor equals the Riemann tensor, and then $T_{abcd}$ is often written with $R_{abcd}$ in place of $C_{abcd}$.