Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 309 / 1

Under parallel transport, the gyroscope spin obeys $u^b{\nabla }_b{s}^a=0$, and its freely falling circular orbit is a geodesic, $u^b{\nabla }_b{u}^a=0$. Metric compatibility therefore givesThus $s^a{u}_a$ is constant along the orbit.

Initially $s^a$ is radial, whereas the circular-orbit four-velocity has only $t$ and $\varphi$ components. The diagonal Schwarzschild metric therefore gives $s^a{u}_a=0$ at $t=0$. By part i,throughout the orbit.

In a coordinate basis, parallel transport isAlong the circular orbit, $dr=d\theta =0$ and $d\varphi /dt=\Omega$. Dividing by $dt/d\tau$ gives

On the equatorial plane, the only potentially relevant angular connection coefficient is ${\Gamma }_{\varphi \varphi }^{\theta }=-\sin \theta \cos \theta$, which vanishes at $\theta =\pi /2$. The $\theta$ transport equation is consequently $d{s}^{\theta }/dt=0$. Since the initially radial spin has $s^{\theta }(0)=0$,

Writing $f=1-2M/R$, orthogonality from part a givesThe needed Christoffel symbols areThe radial and azimuthal transport equations reduce toHenceWith the stated initial direction,

Metric compatibility and parallel transport implyUsing $s^t=R^2\Omega s^{\varphi }/f$ givesSubstitution of part d makes this independent of $t$ precisely whenthe relativistic circular-orbit form of Kepler third law. One orbit takes $T=2\pi /\Omega$, during which the spin phase advances by ${\Omega }^{\prime }T=2\pi \sqrt{1-3M/R}$. Relative to the radial direction, the spin therefore lags by the geodetic precession angle