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

For the source's definition $s^{\alpha }=\gamma (\mathbf{s}\cdot \mathbf{v}/c,\mathbf{s})$, direct contraction with $u^{\alpha }=\gamma (c,\mathbf{v})$ givesIt remains zero under the torque-free transport law. Differentiate the contraction, using the product rule and $u^2=-c^2$: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.