Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 (General Relativity)

Fix the conventionEquivalently,This defines the Riemann curvature tensor because it is $C^{\infty }(M)$-linear in $U,V,X$: derivatives of a multiplying function cancel between the three terms. Thus its value at a point depends only on the three tangent vectors there, rather than their extensions.

Apply the definition to the coordinate basis, for which $[{\partial }_{\mu },{\partial }_{\nu }]=0$ and ${\nabla }_{\mu }{\partial }_{\tau }={\Gamma }^{\rho }{}_{\tau \mu }{\partial }_{\rho }$. Comparing coefficients givesequivalent to the paper's ordering after commuting scalar factors. Although the individual Christoffel symbols are not tensors, the preceding intrinsic definition proves that this complete combination is tensorial.

At an arbitrary point choose normal coordinates, so the Christoffel symbols vanish there. Torsion freedom and commutation of partial derivatives then give the algebraic first Bianchi identityDifferentiate the coordinate curvature expression and cyclically antisymmetrize. Third derivatives cancel, giving the second Bianchi identityBoth statements are tensorial and hence hold in every coordinate system. Contracting the differential identity, using the curvature symmetries and metric compatibility, yieldsthe contracted Bianchi identity.

Contract the isotropic-curvature formula on $a,c$. In dimension $n$ it givesSubstitution into the contracted Bianchi identity givesSince $n>2$, ${\nabla }_{b}K=0$, so $K$ is constant on each connected component. Thus this is constant sectional curvature, withThis argument is Schur theorem in pseudo-Riemannian geometry.

The vector $T^a$ is the tangent to a reference member of a one-parameter family of affinely parametrized geodesics. The Jacobi field $Y^a$ is the infinitesimal connecting vector from that geodesic to a neighboring one at equal parameter. The geodesic deviation equation states that curvature determines their relative acceleration.

Parallel propagation means ${\nabla }_T{e}_{\alpha }=0$, where $T=\dot{\lambda }$. By metric compatibility,The inner products therefore retain their initial values ${\eta }_{\alpha \beta }$, so the parallel-propagated orthonormal frame remains orthonormal.

In the propagated frame $T=e_0$, and the constant sectional curvature formula reduces geodesic deviation toThe 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 isThese formulas are valid to first order in the initial separation and relative velocity.

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 $\tau$.

Write $g_{\mu \nu }={\eta }_{\mu \nu }+ϵh_{\mu \nu }$. The linearized inverse metric is $g^{\mu \nu }={\eta }^{\mu \nu }-ϵh^{\mu \nu }$, and all quadratic Christoffel products are $O({ϵ}^2)$. The wave-coordinate condition linearizes toUsing the supplied Ricci formula then gives $G_{\mu \nu }^{(1)}=-\frac{1}{2}\square {\bar{h}}_{\mu \nu }$. Consequently the Linearized Einstein equations in Lorenz gauge in linearized gravity areThis Lorenz condition is the first-order form of the harmonic coordinate equations.

To first order in the angular velocity, $T^{00}=\rho$ and $T^{i0}=\rho v^i$ with $v=(-\Omega y,\Omega x,0)$. The conservation equation ${\partial }_{\mu }{T}^{\mu 0}=0$ and time independence giveThus $\boxed{x{\partial }_y\rho -y{\partial }_x\rho =0}$: the density is invariant under rotations about the z-axis.

Stationarity changes $\square$ to the spatial Laplace operator. The Green function of the Poisson equation therefore givesHere the bar denotes the trace-reversed metric perturbation; this is the variable denoted by $h$ in the displayed field equation of the question.

For $r\gg R$, the multipole expansion isThe center of mass condition removes the mass dipole. Axisymmetry makes the mixed second moments vanish and gives $\int \rho x^{\prime 2}=\int \rho y^{\prime 2}$. Undoing trace reversal therefore yieldswhereThus $a=J_z/m$ is the specific angular momentum, and the cross term is the dipole part of the slowly rotating weak-field metric.

Contract ${\nabla }_a{T}^{ab}=0$ for the perfect fluid in general relativity with $u_b$. Using $u_b{u}^b=-1$ and $u_b{\nabla }_a{u}^b=0$ gives the relativistic perfect-fluid energy equationthe local first law of thermodynamics. Project instead with $g^a{}_b+u^a{u}_b$ to obtain the relativistic Euler equation

Direct differentiation gives, for example,Differentiating ${\sigma }^1,{\sigma }^2$ similarly and substituting the definitions yields $d{\sigma }^1={\sigma }^2\wedge {\sigma }^3$ and $d{\sigma }^2={\sigma }^3\wedge {\sigma }^1$. Hence these Maurer-Cartan forms on the three-sphere obey

Since $d{e}^0=0$ and $d{e}^i=\frac{1}{2}{ϵ}^i{}_{jk}{e}^j\wedge e^k$, Cartan's first structure equation givesCartan's second structure equation then givesUsing ${\Theta }^i{}_j=\frac{1}{2}{R}^i{}_{jkl}{e}^k\wedge e^l$ yieldswith no time-index curvature components. Contraction givesand therefore

In the orthonormal frame, the comoving perfect fluid in general relativity has $T_{00}=\rho$ and $T_{ij}=p{\delta }_{ij}$. The Einstein field equations with cosmological constant therefore giveThusChoosinggives $p=0$ and $\rho =1/(16\pi )>0$. The metric is then the Einstein static universe supported by pressureless matter and positive vacuum energy.