Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 309 (General Relativity)

In the spherical coordinate system, $|d\mathbf{x}{|}^2=d{r}^2+r^{2}d{\Omega }_2^2$ and $\mathbf{x}\cdot d\mathbf{x}=rdr$. HenceThus the static coordinates on anti-de Sitter spacetime havewhich is positive for every $r\ge 0$.

Affinely parametrized geodesics are the critical curves of the geodesic LagrangianFor $w=1+r^2$ and $w^{\prime }=2r$, the nonzero Christoffel symbols, up to symmetry in the lower indices, areThey follow either from the Euler-Lagrange equations or directly from the Levi-Civita connection formula.

The angular Euler-Lagrange equations are homogeneous in $\dot{\theta }$ and $\dot{\varphi }$. Therefore initial data with both angular velocities zero give the unique solution with constant $\theta$ and $\varphi$, so a radial geodesic remains radial.

Time-translation symmetry supplies the geodesic conserved quantity from a Killing vectorMetric compatibility makes the squared tangent norm another constant,For a proper-time parametrized timelike geodesic $k=-1$, for a null geodesic $k=0$, and for a unit-speed spacelike geodesic $k=1$. The constant $e$ is the conserved energy per unit mass associated with the static Killing vector ${\partial }_t$.

Use proper time $s$, so $k=-1$. The first integral becomesFor an outward geodesic starting at the origin,until it next reaches $r=0$. This occurs atThus every radial timelike geodesic in anti-de Sitter spacetime through the origin returns after the same proper time; restoring anti-de Sitter radius $a$ gives $T=\pi a$.

For a radial null geodesic, $k=0$ gives ${\dot{r}}^2=e^2$. Choose the affine parameter so that $r(0)=0$ and the outgoing branch has $\dot{r}=e>0$. ThenUsing $dr/d\lambda =e$,Consequently $r\to \infty$ whileThe conformal boundary is infinitely far away in affine parameter but is reached in finite static coordinate time.

Fix the conventionAlthough a single covariant derivative of a vector is tensorial, the second derivative contains connection-dependent terms. Their antisymmetric difference cancels every second derivative of a coordinate change. More intrinsically, the mapis $C^{\infty }(M)$-linear in each of $U,V,X$, so it defines the Riemann curvature tensor.

Apply the definition to a coordinate-basis vector, use $[{\partial }_c,{\partial }_d]=0$, and collect the coefficient of $X^b$. This givesup to the overall sign fixed in part i. The right side therefore transforms as a tensor even though its individual Christoffel-symbol terms do not. This identifies the displayed coordinate expression $K^a{}_{bcd}$ with $R^a{}_{bcd}$ after matching the paper's index and sign conventions.

At an arbitrary point choose normal coordinates, so ${\Gamma }^a{}_{bc}=0$ there. Torsion freedom and commuting partial derivatives immediately give the algebraic first Bianchi identityDifferentiating the coordinate curvature formula at that point and cyclically antisymmetrizing gives the differential identityBoth equations are tensorial, so validity in normal coordinates at every point proves them in every coordinate system.

Contract the differential identity on its first and third curvature indices and use the algebraic symmetries of the Riemann tensor. One obtains the contracted Bianchi identity

The left side of the stated identity must inherit $R_{abcd}=-R_{abdc}$. Exchange $c$ and $d$, reduce all curvature products using pair antisymmetry and the first Bianchi identity, and compare with the negative of the original expression. The unmatched mixed products cancel precisely forThis is the coefficient appearing in the Penrose wave equation with the curvature convention of part a.

The vacuum Einstein field equations imply $R_{ab}=0$ and hence $R=0$. Every term in the supplied general identity involving the Ricci tensor or its covariant derivatives therefore vanishes. Substituting $\alpha =2$ leavesThis nonlinear, gauge-independent curvature equation is the Penrose wave equation.

At $ϵ=0$ the metric is the constant Minkowski metric, whose Christoffel symbols and curvature vanish, soInsert $R=ϵR^{(1)}+O({ϵ}^2)$ into the Penrose wave equation. Every curvature-square term is $O({ϵ}^2)$, while the covariant wave operator reduces at first order to the flat d'Alembert operator. ThusThe linearized Riemann curvature operator is unchanged by $h_{\mu \nu }\mapsto h_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$, because the resulting third derivatives cancel pairwise. The equation therefore requires no gauge choice for $h_{\mu \nu }$.

Write $g_{\mu \nu }={\eta }_{\mu \nu }+ϵh_{\mu \nu }$ and retain first-order terms. The quadratic Christoffel products in the supplied Ricci formula drop out. The wave-coordinate condition becomes the Lorenz gauge in linearized gravityThe linearized Ricci tensor and scalar then satisfySubstitution into $G_{\mu \nu }=8\pi T_{\mu \nu }$ givesIf the paper denotes the trace-reversed variable itself by $h_{\mu \nu }$ in its displayed equation, this is exactly that convention.

Under $h_{\mu \nu }^{\prime }=h_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$, trace reversal givesTaking a divergence yieldsApplying $\square$ to the transformed field likewise produces only derivatives of $\square \xi$. Therefore $\square {\xi }^{\mu }=0$ preserves both the gauge condition and the sourced wave equation. These are the residual gauge transformations.

Let $s=k_{\rho }{x}^{\rho }$. In vacuum,A nonzero localized profile cannot have $H^{\prime \prime }=0$, because an affine function does not decay at both ends. HenceDecay removes the integration constant, so the second condition is equivalently $k^{\mu }{H}_{\mu \nu }=0$. Thus a nontrivial plane gravitational wave in linearized gravity has a null wavevector and transverse amplitude.

A spatial rotation and rescaling put the future null vector in the form $k^{\mu }=(1,0,0,1)$, so the profile depends on $z-t$. Transversality gives four linear relations among the ten symmetric components. The four residual gauge functions satisfying $\square {\xi }^{\mu }=0$ remove the time and longitudinal components; the remaining trace can be removed by the residual transformation indicated in the question. The resulting transverse-traceless gauge isThe two arbitrary functions are the plus and cross gravitational-wave polarizations. They are the two physical degrees of freedom left after the four gauge conditions and four residual coordinate freedoms are removed.

For any differential form $\omega$, Cartan's magic formula givesUsing $d^2=0$,Thus the Lie derivative commutes with the exterior derivative on every differential form.

Because the Levi-Civita connection preserves the spacetime volume form, only derivatives of $X$ remain in ${\mathcal{L}}_{X}T$. Lower the raised indices in $T_{ab}{}^{cd}={ϵ}_{ab}{}^{cd}$, contract successively with the supplied products of two Levi-Civita tensors, and separate ${\nabla }_a{X}_b$ into antisymmetric, trace and symmetric trace-free parts. The antisymmetric part cancels automatically, while ${\mathcal{L}}_{X}T=0$ is equivalent to vanishing of the symmetric trace-free part:Thereforeso $\boxed{\alpha =1/2}$. This is the four-dimensional Conformal Killing equation, and $X$ is a Conformal Killing vector field.

The first vacuum Maxwell equations equation is preserved for every vector field because part a givesOn two-forms in four dimensions, the mixed volume tensor $T_{ab}{}^{cd}$ represents twice the Hodge star operator. Part b therefore says that a conformal Killing field commutes with the Hodge star:It follows thatHence $\tilde{F}={\mathcal{L}}_{X}F$ satisfies both vacuum Maxwell equations whenever $F$ does.