Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 56 / 2 / c / Solution

For every member of the geodesic variation, proper time normalization gives $g(T,T)=-1$. By metric compatibility, the geodesic equation, and ${\nabla }_{T}S={\nabla }_{S}T$,The last equality uses normalization across the whole variation, not merely constancy along one curve. Hence $g(S,T)$ is constant along each timelike geodesic, provingThis is orthogonality of proper-time deviation vectors.