= Solution
{c}
Let $\mu=GM/c^2$, $f=1-2\mu/r$, $q=\sqrt{1-3\mu/r}=1/u^0$, and choose $\Omega>0$. A torque-free test <gyroscope> moving on a <geodesic> has a <spin four-vector> obeying <parallel transport>,
$$
\frac{ds^\mu}{dt}+\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{dt}s^\beta=0,
\qquad g_{\mu\nu}u^\mu s^\nu=0.
$$
The orthogonality condition says the temporal spin component vanishes in the body's <rest frame>, not necessarily in Schwarzschild coordinates. Here it gives
$$
s^t=\frac{r^2\Omega}{fc^2}s^\phi.
$$
In addition to the radial <Christoffel symbols> in part (a), use $\Gamma^t_{tr}=\mu/(r^2f)$ and $\Gamma^\phi_{r\phi}=1/r$. On the equator the <parallel transport> equations become
$$
\frac{ds^t}{dt}=-\frac{\mu}{r^2f}s^r,\qquad
\frac{ds^r}{dt}=r\Omega q^2s^\phi,\qquad
\frac{ds^\phi}{dt}=-\frac{\Omega}{r}s^r,\qquad
\frac{ds^\theta}{dt}=0.
$$
The radial equation follows by substituting $s^t$ into $ds^r/dt+\Gamma^r_{tt}s^t+\Omega\Gamma^r_{\phi\phi}s^\phi=0$. Eliminating $s^\phi$ gives $d^2s^r/dt^2+\Omega^2q^2s^r=0$.
Let $s_0$ be the invariant spin magnitude. Initially radial spin means $s^r(0)=s_0\sqrt f$ and $s^t(0)=s^\phi(0)=s^\theta(0)=0$. The complete solution is
$$
\boxed{\begin{aligned}
s^r(t)&=s_0\sqrt f\cos(\Omega qt),\\
s^\phi(t)&=-\frac{s_0\sqrt f}{rq}\sin(\Omega qt),\\
s^\theta(t)&=0,\\
s^t(t)&=-\frac{s_0r\Omega}{c^2\sqrt f\,q}\sin(\Omega qt).
\end{aligned}}
$$
To interpret the orientation, convert coordinate components into the orbiting <orthonormal frame>. The <Lorentz factor> relative to a <Schwarzschild static observer> is $\gamma=\sqrt f/q$, so the body's measured radial and tangential spin components are
$$
S^{\hat r}=\frac{s^r}{\sqrt f}=s_0\cos(\Omega qt),\qquad
S^{\hat\phi}=\frac{rs^\phi}{\gamma}=-s_0\sin(\Omega qt).
$$
Their squared sum is $s_0^2$, and their angle relative to the outward radial direction decreases uniformly at rate
$$
\boxed{\Omega_r=\Omega q=\frac{\Omega}{u^0}.}
$$
The differing coordinate amplitudes in $s^r,s^\phi$ must not be interpreted as a nonuniform physical spin rotation. After one orbital period the radial basis has turned by $2\pi$ while the spin has turned by $-2\pi q$ relative to it. The residual <geodetic precession> is $\Delta\psi=2\pi(1-q)$ in the orbital direction.
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