Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 37D (General Relativity)

By spherical symmetry, place the geodesic in the equatorial plane $\theta =\pi /2$. Withthe geodesic Lagrangian iswhere $Q$ is the constant squared norm of the tangent: $Q=-1$ for a unit-speed timelike geodesic and $Q=0$ for a null geodesic. The cyclic coordinates $t$ and $\varphi$ give the conserved quantitiesSubstitution into the normalization givesMultiplying by $f$ and rearranging producesThis is the timelike geodesic effective potential form, with the same derivation applying to any affinely parametrized causal geodesic.

The radial equation for a circular geodesic, or equivalently $V^{\prime }(r)=0$, gives the Angular velocity of a circular Schwarzschild geodesicOne orbit therefore takes Schwarzschild coordinate time

Observer $O_3$ is static at infinity, where coordinate time equals proper time, so

For the static observer $O_2$ at radius $r$, $dr=d\theta =d\varphi =0$, and the Schwarzschild spacetime metric givesThus

Along $O_1$,where $d\varphi /dt=\Omega$ was used. Hence the Proper period of a circular Schwarzschild geodesic is

Since $r>3M$,Relative to the distant observer, the static clock $O_2$ suffers gravitational time dilation. The orbiting clock $O_1$ suffers the same gravitational effect together with the kinematic time dilation caused by its orbital motion, so it accumulates the least proper time.