 Solution

Differentiate $f=g(S,S)$ twice, use geodesic deviation, and eliminate $g(DS,DS)$ with the conserved $K$:The most informative way to interpret this linear differential equation is to express $S$ in a parallel-propagated orthonormal frame along the timelike geodesic. Its three spatial components satisfy ${\ddot{S}}^i=\kappa S^i$.

If $R>0$, put $H=\sqrt{R/12}$. With constant spatial vectors $A,B$ in that frame,Thus generic deviations grow in length like $e^{H\tau }$, and their squared separation grows like $e^{2H\tau }$. Initially comoving neighbors, $DS(0)=0$, have $S(\tau )=S(0)\cosh (H\tau )$ and exhibit this growth. The unqualified printed assertion is too strong: the nonzero Jacobi field $S=B{e}^{-H\tau }$ has $K=0$ and shrinks exponentially. Such a Jacobi field is realizable by varying initial position and initial unit timelike velocity, with initial relative velocity $DS(0)=-HB$.

If $R<0$, put $\omega =\sqrt{-R/12}$. ThenThe deviations are bounded and oscillatory, rather than exponentially growing. The squared separation is periodic with period $\pi /\omega$ or is constant. Initially comoving neighbors have $S(\tau )=S(0)\cos (\omega \tau )$ and focus at $\tau =\pi /(2\omega )$; arbitrary vector initial data need not focus together. For completeness, $R=0$ gives $S=A+B\tau$ and $f=|A+B\tau {|}^2$. These conclusions concern infinitesimal geodesic deviation on the domain of the variation, not arbitrarily large finite separations.