Geodesic deviation equation

Remarks

The proof uses that $\partial_s$ and $\partial_\tau$ commute, so $\frac{D}{d\tau} J = \frac{D}{ds} u$ because the connection is torsion-free. Then $\frac{D^2}{d\tau^2} J = \frac{D}{d\tau}\frac{D}{ds} u = \frac{D}{ds}\frac{D}{d\tau} u + R(u, J) u$, and $\frac{D}{d\tau} u = 0$ since the curves are geodesics; finally $R(u, J) = -R(J, u)$. So $J$ restricted to each geodesic is a Jacobi field, and the equation is the Jacobi equation in Lorentzian signature. Example: in Minkowski space $R = 0$, so $D^2J/d\tau^2 = 0$, and nearby free particles separate at a constant rate. In the Newtonian limit, with gravitational potential $\Phi$ and nearly static particles, $R^i{}_{0j0} \approx \partial_i \partial_j \Phi$, and the equation becomes $d^2 J^i/dt^2 = -\partial_i \partial_j \Phi \, J^j$, the Newtonian tidal acceleration.

Going deeper

For a student

The equation is how curvature becomes observable: the equivalence principle removes gravity at one point, but the relative acceleration of nearby free particles cannot be removed. A gravitational wave passing through a ring of free test masses deforms it by this equation, which is the principle of interferometric detectors. The trace of the curvature term, $R_{\mu\nu} u^\mu u^\nu$, is the term that appears in the Raychaudhuri equation for the expansion of a congruence of geodesics.