 Solution

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,The orthogonality condition says the temporal spin component vanishes in the body's rest frame, not necessarily in Schwarzschild coordinates. Here it givesIn addition to the radial Christoffel symbols in part (a), use ${\Gamma }_{tr}^t=\mu /(r^{2}f)$ and ${\Gamma }_{r\varphi }^{\varphi }=1/r$. On the equator the parallel transport equations becomeThe radial equation follows by substituting $s^t$ into $d{s}^r/dt+{\Gamma }_{tt}^r{s}^t+\Omega {\Gamma }_{\varphi \varphi }^r{s}^{\varphi }=0$. Eliminating $s^{\varphi }$ gives $d^2{s}^r/d{t}^2+{\Omega }^2{q}^2{s}^r=0$.

Let $s_0$ be the invariant spin magnitude. Initially radial spin means $s^r(0)=s_0\sqrt{f}$ and $s^t(0)=s^{\varphi }(0)=s^{\theta }(0)=0$. The complete solution isTo 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 areTheir squared sum is $s_0^2$, and their angle relative to the outward radial direction decreases uniformly at rateThe differing coordinate amplitudes in $s^r,s^{\varphi }$ 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.