= Solution
For the source's definition $s^\alpha=\gamma(\mathbf s\cdot\mathbf v/c,\mathbf s)$, direct contraction with $u^\alpha=\gamma(c,\mathbf v)$ gives
$$
s\cdot u=-\gamma^2\mathbf s\cdot\mathbf v+\gamma^2\mathbf s\cdot\mathbf v=0.
$$
It remains zero under the torque-free transport law. Differentiate the contraction, using the product rule and $u^2=-c^2$:
$$
\boxed{\frac d{d\tau}(s\cdot u)
=\frac{s\cdot\dot u}{c^2}u^2+s\cdot\dot u=0.}
$$
Thus an initially spatial <spin> in the <instantaneous rest frame> remains orthogonal to the velocity. This is the orthogonal-spin specialization of <Fermi-Walker transport>. The sign is tied to the negative timelike norm; using the opposite metric sign without changing the transport formula would spoil the cancellation.
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