= Solution
First note that the supplied <spin> equation preserves $s\cdot u=0$: its electromagnetic bracket has contraction $u_aF^{a\lambda}s_\lambda+(u^2/c^2)F^{\mu\nu}u_\mu s_\nu=0$, and the remaining transport term cancels $s\cdot\dot u$ as in <2/c/i>.
For an explicitly electromagnetic acceleration, set $\dot u^a=(q/m)F^{a\lambda}u_\lambda$ and define $C=F^{\mu\nu}u_\mu s_\nu$. Antisymmetry gives $s\cdot\dot u=-(q/m)C$, so the <spin> equation can be written
$$
\dot s^a=\frac qm\left[\frac g2 F^{a\lambda}s_\lambda
+\left(\frac g2-1\right)\frac C{c^2}u^a\right].
$$
Contract with $2s_a$:
$$
\boxed{\frac d{d\tau}(s^as_a)
=\frac{gq}{m}s_aF^{a\lambda}s_\lambda
+\frac{2q}{mc^2}\left(\frac g2-1\right)C(s\cdot u)=0.}
$$
The first term vanishes because a symmetric product of the two <spin> covectors contracts an antisymmetric tensor; the second vanishes by <spin>–velocity orthogonality. In the <instantaneous rest frame> $s^0=0$, so $s^as_a$ is precisely the squared proper spatial-spin magnitude. It remains constant for arbitrary $g$ under this transport, though its direction can precess. In this equation $g$ is the dimensionless <g-factor>; the dimensional <gyromagnetic ratio> includes $gq/(2m)$.
Back to article page
