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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.

Let the fixed direction of motion be the unit vector $\mathbf{n}$ and decompose $\mathbf{s}=s_{\parallel }\mathbf{n}+{\mathbf{s}}_{\perp }$. In ii the two parallel contributions cancel, whereas the transverse equation isFor the stated speed law, $1-v^2/c^2=e^{-2\Gamma \tau }$, hence $\gamma =e^{\Gamma \tau }$. Since the particle is initially at rest, $\gamma (0)=1$, andThe corresponding spin four-vector has components $s^0=\gamma \beta s_{\parallel }(0)$, $s_{\parallel }^{\mathrm{spatial}}=\gamma s_{\parallel }(0)$, and ${\mathbf{s}}_{\perp }^{\mathrm{spatial}}={\mathbf{s}}_{\perp }(0)$. Its squared norm is therefore $s_{\parallel }(0{)}^2+|{\mathbf{s}}_{\perp }(0){|}^2$, independent of time. The transverse decay of the source's laboratory three-vector is a boost-normalization effect, not a loss of proper angular momentum or a torque.

For a real increasing speed take $\Gamma >0$ and $\tau \ge 0$. The prescribed profile has a singular initial acceleration: $v\sim c\sqrt{2\Gamma \tau }$ near zero. The differential transport calculation applies for $\tau >0$ and the solution has the continuous initial limit displayed above. This trajectory should not be silently described as motion at constant proper acceleration.

Take the dot here to mean $d/d\tau$, as in the source. Differentiate $u^{\alpha }=\gamma (c,\mathbf{v})$ and the given spin components. Their contraction simplifies because the terms involving $\dot{\gamma }$ cancel:The spatial part of the transport equation consequently readsUse $\dot{\gamma }/\gamma ={\gamma }^2\mathbf{v}\cdot \dot{\mathbf{v}}/c^2$ and divide by $\gamma$. The result isNo assumption of acceleration perpendicular to the velocity was made. In particular the second term cannot be dropped for a particle speeding up along a straight line. This $\mathbf{s}$ is the three-vector in the source's particular spin four-vector parametrization; its Euclidean magnitude need not be the invariant proper-spin magnitude.