Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 56 / 2

Describe the geodesic variation by a smooth map $F(s,\tau )$, with each curve $\tau \mapsto F(s,\tau )$ a timelike geodesic parametrized by proper time. Its tangent vector is $T=F_{\ast }{\partial }_{\tau }$ and its deviation vector is $S=F_{\ast }{\partial }_s$. Thus $S$ measures the first-order separation of neighboring members at the same value of proper time.

Mixed coordinate derivatives on the parameter surface commute. Consequently, along the variation,For an embedded two-dimensional variation this is the usual Lie bracket of vector fields on its image; more generally the same identity is read along the map $F$ using its commuting parameter derivatives. Since the Levi-Civita connection is torsion-free, it follows that ${\nabla }_{T}S={\nabla }_{S}T$.

Use precisely the stated convention for the Riemann curvature tensor, and let $D={\nabla }_T$. The geodesic equation says ${\nabla }_{T}T=0$ because proper time is an affine parameter. The commuting variation and the torsion-free connection giveThus the geodesic deviation equation isHere $(R(X,Y)Z{)}^a=R^a{}_{bcd}{Z}^b{X}^c{Y}^d$. Declaring this index convention fixes the sign: writing $R(S,T)T$ instead would require a minus sign. A deviation vector satisfying this equation is a Jacobi field.

For every member of the geodesic variation, proper time normalization gives $g(T,T)=-1$. By metric compatibility, the geodesic equation, and ${\nabla }_{T}S={\nabla }_{S}T$,The last equality uses normalization across the whole variation, not merely constancy along one curve. Hence $g(S,T)$ is constant along each timelike geodesic, provingThis is orthogonality of proper-time deviation vectors.

Contracting the assumed Riemann curvature tensor gives the Ricci tensorThe contracted Bianchi identity and metric compatibility now implyThus ${\nabla }_{b}R=0$, so the scalar curvature is constant on each connected component of spacetime. The sectional curvature is consequently the constant $\kappa =R/12$. This is the four-dimensional case of the Schur theorem in pseudo-Riemannian geometry.

Let $\kappa =R/12$, which is constant by the preceding argument. Substitution into geodesic deviation givesusing $g(T,T)=-1$ and the established orthogonality. With $V=DS$, metric compatibility yields $\dot{f}=2g(S,V)$ andThereforeBoth $S$ and $DS$ are orthogonal to $T$, so their squared norms are nonnegative. The minus sign in this conserved deviation energy in constant curvature reflects $D^{2}S=+\kappa S$, rather than an ordinary oscillator when $R>0$.

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.