 Solution

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.