Past exam of the symmetry and relativity course of the University of Oxford / 2015 / 2 / d

First note that the supplied spin equation preserves $s\cdot u=0$: its electromagnetic bracket has contraction $u_a{F}^{a\lambda }{s}_{\lambda }+(u^2/c^2)F^{\mu \nu }{u}_{\mu }{s}_{\nu }=0$, and the remaining transport term cancels $s\cdot \dot{u}$ as in 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 writtenContract with $2s_a$: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 $s^a{s}_a$ 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 $gq/(2m)$.