Metric compatibility and parallel transport imply
$$
u^c\nabla_c(s^as_a)=2s_a u^c\nabla_cs^a=0.
$$
Using $s^t=R^2\Omega s^\phi/f$ gives
$$
s^2=\frac{(s^r)^2}{f}
+\left(R^2-\frac{R^4\Omega^2}{f}\right)(s^\phi)^2.
$$
Substitution of part d makes this independent of $t$ precisely when
$$
\boxed{\Omega^2=\frac{M}{R^3}},
$$
the relativistic circular-orbit form of <Kepler third law>. One orbit takes $T=2\pi/\Omega$, during which the spin phase advances by $\Omega'T=2\pi\sqrt{1-3M/R}$. Relative to the radial direction, the spin therefore lags by the <geodetic precession> angle
$$
\boxed{\alpha=2\pi\left[1-\sqrt{1-\frac{3M}{R}}\right]}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-309/1/e/solution