Solution

u^{c} \nabla_{c} (s^{a} s_{a}) = 2 s_{a} u^{c} \nabla_{c} s^{a} = 0.

Using

s^{t} = R^{2} Ω s^{ϕ} / f

gives

s^{2} = \frac{( s ^{r} ) ^{2}}{f} + (R^{2} - \frac{R ^{4} Ω ^{2}}{f}) (s^{ϕ})^{2} .

Substitution of part d makes this independent of

t

precisely when

Ω^{2} = \frac{M}{R ^{3}},

the relativistic circular-orbit form of Kepler third law. One orbit takes

T = 2 π /Ω

, during which the spin phase advances by

Ω^{'} T = 2 π 1 - 3 M / R

. Relative to the radial direction, the spin therefore lags by the geodetic precession angle

α = 2 π [1 - 1 - \frac{3 M}{R}] .

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