In the propagated frame $T=e_0$, and the <constant sectional curvature> formula reduces <geodesic deviation> to
$$
\ddot y^0=0,
\qquad
\ddot y^i=K y^i.
$$
The stated temporal initial data give $y^0(t)=0$. Writing $\omega=\sqrt{-K}$ for $K<0$ and $\kappa=\sqrt K$ for $K>0$, the spatial displacement is
$$
\boxed{
y^i(t)=
\begin{cases}
Y^i\cos(\omega t)+(V^i/\omega)\sin(\omega t),&K<0,\\
Y^i+V^it,&K=0,\\
Y^i\cosh(\kappa t)+(V^i/\kappa)\sinh(\kappa t),&K>0.
\end{cases}}
$$
These formulas are valid to first order in the initial separation and relative velocity.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 / 1 / d / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-309/1/d/ii/solution