Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 309 (General Relativity)

Let ${\Phi }_s$ be the local flow generated by the vector field $X$. For a covariant tensor, its Lie derivative isFor contravariant indices one equivalently uses the differential of the inverse flow, and the construction extends to every tensor product by the Leibniz rule. It measures the infinitesimal change of $T$ under transport by the flow of $X$.

At a point where $X\ne 0$, the flow-box theorem supplies a transverse hypersurface with coordinates $x^i$. Flow each point on it for parameter $t$ and keep the $x^i$ constant along the flow. In the resulting coordinates,Because the coordinate basis is transported by this flow, the Lie derivative of any tensor is obtained by differentiating its coordinate components:

For a function, pullback by the flow givesIn flow-box coordinates $X={\partial }_t$, write $Y=Y^{\mu }{\partial }_{\mu }$. ThenBoth sides are vector fields defined invariantly, so equality in these coordinates provesin every coordinate system.

For one-forms, the defining pullback or the identitygivesApplying the Leibniz rule to both covariant slots of a type-$(0,2)$ tensor yields

Every metric coefficient is independent of $w$, sois a Killing vector field.

Forthe only potentially nonzero components of ${\mathcal{L}}_{X}g$ areandThus the necessary and sufficient conditions areThe second-order linear ODE has a two-dimensional solution space, giving a two-parameter family

Interchanging the two transverse directions changes the sign of the quadratic profile. Henceis another two-parameter family of Killing fields.

Fix $u=u_0$. Initial data $p(u_0),p^{\prime }(u_0)$ and $q(u_0),q^{\prime }(u_0)$ may be prescribed independently for the two oscillator equations. In particular, values of $p(u_0)$ and $q(u_0)$ generate arbitrary translations in $x$ and $y$, while their derivatives merely add controllable $w$ components. The independent Killing field ${\partial }_w$ supplies any remaining translation in $w$. These Killing fields span the tangent space of each constant-$u$ wavefront, so their flows act transitively. Therefore an isometry maps any point $r$ on such a surface to any other point $s$, realizing a homogeneous wavefront of a plane gravitational wave.

Assume the test body moves slowly, $d{x}^i/dt=O(ϵ)$, and that the field is quasistatic, so time derivatives are of higher order. The spatial geodesic equation, parametrized by coordinate time, then reduces at leading order toDefining the Newtonian potential bygivesthe Newtonian equation of motion. Terms involving $h_{0i}$, spatial velocity, or time derivatives enter beyond the stated order.

The nonrelativistic fluid assumptions giveAssume again a stationary field and asymptotic flatness. The harmonic-gauge Linearized Einstein equations becomewhile the other trace-reversed components are smaller. Trace reversal then gives $h_{00}={\bar{h}}_{00}/2$ to leading order, soUsing $h_{00}=-2\Phi$ yieldsin units $G=1$, which is the Poisson equation of Newtonian gravity.

In the weak-field, slow-source approximation, the leading luminosity is the quadrupole formulain units $G=c=1$, whereOnly the $\ell =2$ symmetric trace-free coefficient contributes. Indeed,and contraction with traceless $a_{kl}$ removes the first term. ThereforeThe power crossing the large sphere at time $t$ is consequentlyThe monopole is conserved total mass, the dipole is center-of-mass motion, and every $\ell \ne 2$ term is orthogonal to the trace-free quadrupole at this leading order.

The metric directly giveswith no mixed products. Henceis an orthonormal frame.

The dual orthonormal coframe isbecause $e^{\mu }(e_{\nu })={\delta }^{\mu }{}_{\nu }$.

One hasCartan's torsion-free first structure equation and metric compatibility give the only nonzero connection 1-forms:

Cartan's second structure equation givesand, for $i\ne j$,With all indices lowered and the curvature convention in the question, the independent nonzero orthonormal Riemann curvature tensor components aretogether with those obtained from the Riemann symmetries.

PutContracting the curvature components givesin the orthonormal frame, with no sum on $i$. Vacuum requires $S_1=S_2$ and $p_i(S_1-1)=0$ for every $i$. Since the $p_i$ are not all zero, some $p_i\ne 0$, forcing $S_1=1$ and then $S_2=1$. Conversely these conditions make every Ricci component vanish. Thus the Kasner metric is vacuum exactly whenso $A=B=1$.

Write $h_{ab}=\delta g_{ab}$ and $h=g^{ab}{h}_{ab}$. Standard metric variation givesand variation of metric compatibility gives

Varying the coordinate expression for the curvature and replacing partial derivatives by covariant ones givesThereforewhereThis is the metric variation of scalar curvature; its divergence contributes only a boundary term to the variation of the Einstein-Hilbert action.

Varying $\Phi$ and integrating by parts givesSince $\delta F_{ab}=2{\nabla }_{[a}\delta A_{b]}$, variation of the potential and another integration by parts giveThese are the scalar and Maxwell equations of Einstein-Maxwell-dilaton theory.

Varying the matter terms with respect to the inverse metric and writing the Einstein equation as $G_{ab}=T_{ab}$ givesThe factors reflect that the complete action, including its matter terms, carries the common prefactor $1/(16\pi )$. With the conventional definition in which a separate matter action satisfies $G_{ab}=8\pi T_{ab}^{\mathrm{conventional}}$, the corresponding tensor is rescaled accordingly.

Let $k^a$ be null and define $v_c=k^a{F}_{ac}$. The metric-proportional terms vanish after contraction with $k^a{k}^b$, leavingAntisymmetry of $F$ gives $v_c{k}^c=0$. A vector orthogonal to a null vector in Lorentzian signature has nonnegative squared norm, so $v_c{v}^c\ge 0$; also $e^{\alpha \Phi }>0$. Henceand the theory satisfies the null energy condition.