 Solution

ID: past-exam-of-the-symmetry-and-relativity-course-of-the-university-of-oxford/2015/2/c/i/solution

s^{α} = γ (s \cdot v / c, s)

, direct contraction with

u^{α} = γ (c, v)

gives

s \cdot u = - γ^{2} s \cdot v + γ^{2} s \cdot v = 0.

It remains zero under the torque-free transport law. Differentiate the contraction, using the product rule and

u^{2} = - c^{2}

\frac{d}{d τ} (s \cdot u) = \frac{s \cdot u ˙}{c ^{2}} u^{2} + s \cdot \overset{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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