Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 / 2

Vary $g^{ac}{g}_{cb}={\delta }_b^a$. The product rule givesand multiplication by $g^{bd}$ yields the metric variation

The determinant identity $\delta \log |\det g|=g^{ab}\delta g_{ab}$ impliesHence the metric volume form varies as

Varying the scalar and integrating by parts givesbecause the boundary term vanishes by the support assumption. The fundamental lemma of the calculus of variations therefore gives the Klein-Gordon equation

Using both metric-variation formulas from part a gives the Klein-Gordon scalar stress-energy tensorIts divergence isbecause the two Hessian terms cancel by symmetry. It vanishes on every solution of the Klein-Gordon equation.

Killing equation is ${\nabla }_{(a}{K}_{b)}=0$. For the stress-energy current from a Killing vector $J^a=T^a{}_b{K}^b$,The first term vanishes on shell, while the second contracts the symmetric tensor $T^{ab}$ with the antisymmetric derivative selected by Killing's equation. Thus $\boxed{{\nabla }_a{J}^a=0}$.

The coordinate formula for the Lie derivative isFor $X={\partial }_t$, its components are constant and every metric coefficient is independent of $t$, so ${\mathcal{L}}_{{\partial }_t}g=0$. Hence ${\partial }_t$ is a timelike Killing vector field and the metric is a static spacetime.

Let $K={\partial }_t$ and let the future unit normal to ${\Sigma }_{\tau }$ be $n=f^{-1}{\partial }_t$. The conserved Killing energy is the fluxSubstitution of the Klein-Gordon scalar stress-energy tensor gives exactlyApply the divergence theorem to the slab $[{\tau }_1,{\tau }_2]\times {\mathbb{R}}^3$. The spatial-boundary flux vanishes because $\psi$ decays, and ${\nabla }_a{J}^a=0$ makes the two time-slice fluxes equal. Thus this conserved scalar-field energy in a static spacetime is independent of $\tau$.