Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 37B (General Relativity)

Metric compatibility and the ordinary chain rule givewhich in components is the required identity.

The vectors $T={\partial }_{\tau }x$ and $S={\partial }_{\sigma }x$ are coordinate vector fields on the parameter surface, so their ordinary commutator vanishes. Torsion-freeness then givesSince each $\tau$-curve is geodesic, ${\nabla }_{T}T=0$. Apply the Ricci identity to $V=T$ and use ${\nabla }_{S}T={\nabla }_{T}S$:Thus the desired form holds with

First,Hence orthogonality at one point persists. Substituting the constant-curvature tensor into geodesic deviation givesbecause $T\cdot S=0$ and $T\cdot T=-1$. Thereforealong the whole geodesic.