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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.

The spin of a small torque-free test gyroscope is represented by a spacelike vector orthogonal to its four-velocity, $s^{\mu }{u}_{\mu }=0$. Its temporal component vanishes in its own rest frame. Along a geodesic, $D{s}^{\mu }/d\tau =0$: parallel transport preserves the orthogonality and the invariant spin magnitude. For an accelerated observer the corresponding nonrotating transport is Fermi-Walker transport.