Past exam of the mathematical physics course of the University of Oxford / 2007 / B / 7 / Solution

The coordinate derivatives of the smooth two-parameter map commute. With a torsion-free Levi-Civita connection,because ${\Gamma }^a{}_{bc}$ is symmetric in its lower indices. Each member of the family is a geodesic, so $DV/Ds=0$. ConsequentlyThe commutator of covariant derivatives is the action of the Riemann curvature tensor. With the source's index and sign convention this givesThis is geodesic deviation: free fall relative acceleration measures curvature, even where an individual observer can eliminate the connection at one point.

For Earth's exterior Newtonian field, $\varphi =-GM/r$ and acceleration is $a_i=-{\partial }_i\varphi$. Nearby freely falling particles have relative acceleration $\delta a_i=-{\varphi }_{,ij}{Y}^j$. At the observation point align $z$ radially upward and $x,y$ horizontally. Since $GM/R_E^2=g$,The weak-field geodesic-deviation equation therefore identifiesIn these units time is measured in seconds as in the stated Newtonian comparison; if the zeroth coordinate is $ct$ instead, the curvature components acquire the corresponding $c^{-2}$ factor. A small freely falling cloud is squeezed in the two horizontal directions and stretched radially. These are tidal effects, not the uniform acceleration removed by a freely falling frame.

Their characteristic inverse-square time is $g/R_E$, so $\boxed{t_{\mathrm{tide}}=\sqrt{R_E/g}\simeq 8.1\times 10^2\mathrm{s}\simeq 13.4\min }$. One concrete thought experiment is an ideal satellite moving in a circular orbit just above a spherical Earth, ignoring atmosphere and terrain. Its orbital angular frequency is $\sqrt{GM/R_E^3}=\sqrt{g/R_E}$ and its period is $2\pi t_{\mathrm{tide}}\simeq 84$ minutes. Equivalently, initially nearby free fall trajectories develop order-one tidal changes on this characteristic scale. This is the surface tidal time scale of a spherical gravitating body.