Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 309 / 2

For any vector field $Y$, apply the Leibniz rule to the scalar $\omega (Y)$:Using ${\mathcal{L}}_{V}Y=[V,Y]$ and expanding the Lie bracket in a coordinate chart leavesThe connection terms cancel because the Levi-Civita connection is torsion-free. Applying this formula to each slot of the metric tensor and using metric compatibility givesEquivalently, one may prove both identities at a point in normal coordinates; since both sides are tensors, the result then holds in every coordinate system.

The trace of the electromagnetic stress-energy tensor in four spacetime dimensions isFor its covariant divergence, the source-free Maxwell equations eliminate the derivative of the first factor. Contracting the Bianchi identity ${\nabla }_{[a}{F}_{bc]}=0$ with $F^{ac}$ giveswhich cancels the derivative of the trace term. HenceFor $J^a=T^{ab}{V}_b$, stress-energy conservation and symmetry of $T^{ab}$ now imply

At the chosen event use the orthonormal frame in spacetime from the hint. The electromagnetic energy density isand the energy flux is the Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{B}$. Depending on the index convention, the required contraction is $\rho -w{S}_1$ or $\rho +w{S}_1$. In either case,Rescaling $V$ and rotating the spatial frame covers arbitrary future timelike $V$ and future causal $W$. Thus the Maxwell field satisfies the dominant energy condition.

Since $t=u+r$, the hypersurface ${\Sigma }_t$ is an ordinary constant-Minkowski-time slice. Its future unit normal isin the $(u,r,\theta ,\varphi )$ basis. Lowering the index gives $n_a=(-1,-1,0,0)=-{\nabla }_{a}t$. Restricting the metric to $dt=du+dr=0$ gives the Euclidean spherical metricso its induced Riemannian volume form is

For $V=r{\partial }_r$, Cartan's magic formula giveswhereas contraction and exterior differentiation both vanish for $du,d\theta ,d\varphi$. ThusApplying the Lie derivative to
$g=-d{u}^2-2,du,dr+r^{2}d{\Omega }^2$ yieldsHere $n_a=-(du+dr{)}_a$ and $V_a=-r(du{)}_a$, so comparison with $2g$ givesUsing tracelessness from part (b)(i),by the dominant energy condition, because $n$ is future timelike and $V$ is future null.

The integrand defining $\epsilon (t)$ is nonnegative by the dominant energy condition, henceApply the spacetime divergence theorem to $J^a=T^{ab}{V}_b$ on the slab $S$. The flux through the cylinder at $r\to \infty$ vanishes by the stated decay. With the Lorentzian boundary orientation, the two spacelike boundary contributions givePart (ii) makes the right-hand side nonnegative, so for $t_1\ge t_0$,Therefore $\epsilon$ is a nonnegative monotone decreasing energy functional.