Writing $f=1-2M/R$, orthogonality from part a gives
$$
0=u^t(-fs^t+R^2\Omega s^\phi),
\qquad
s^t=\frac{R^2\Omega}{f}s^\phi.
$$
The needed <Christoffel symbols> are
$$
\Gamma^r_{tt}=\frac{fM}{R^2},
\qquad
\Gamma^r_{\phi\phi}=-fR,
\qquad
\Gamma^\phi_{\phi r}=\frac1R.
$$
The radial and azimuthal transport equations reduce to
$$
\dot s^r=\Omega(R-3M)s^\phi,
\qquad
\dot s^\phi=-\frac\Omega R s^r.
$$
Hence
$$
\ddot s^r+\Omega^2\left(1-\frac{3M}{R}\right)s^r=0.
$$
With the stated initial direction,
$$
\boxed{s^r=A\cos(\Omega' t),\qquad
s^\phi=-\frac{A\Omega}{\Omega'R}\sin(\Omega't)},
\qquad
\boxed{\Omega'=\Omega\sqrt{1-\frac{3M}{R}}}.
$$

Solution

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

 Solution

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