Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 309 / 4

Choose signature $(-,+,+,+)$, with $e^0$ timelike. The orthonormal coframe in spacetime reconstructs the Lorentzian metric asBecause $A$ and $B$ are nowhere zero on the domain, the coframe is invertible and the metric nondegenerate; either sign of each function is allowed. In retarded and advanced null coordinates $u=t-z$, $v=t+z$ the metric is $-dudv+A(u{)}^{2}d{x}^2+B(u{)}^{2}d{y}^2$, a diagonal example of Rosen coordinates for a plane gravitational wave. Multiplication by $-1$ would give the opposite overall signature convention; all subsequent signs here use the displayed one.

Put $u=t-z$, $\ell =du=e^0-e^3$, $p=A^{\prime }/A$, $q=B^{\prime }/B$, with primes denoting $d/du$. The exterior derivatives are $d{e}^0=d{e}^3=0$, $d{e}^1=p\ell \wedge e^1$, $d{e}^2=q\ell \wedge e^2$. For a Levi-Civita connection, the connection 1-forms obey both Cartan's first structure equation and metric compatibility. With lower indices ${\omega }_{ab}={\eta }_{ac}{\omega }^c{}_b$, the result isAll other lower-index forms vanish or follow from ${\omega }_{ab}=-{\omega }_{ba}$. Raising the first time index changes its sign: for example ${\omega }^0{}_1={\omega }^1{}_0=p{e}^1$, whereas ${\omega }^1{}_3=-p{e}^1$ and ${\omega }^3{}_1=p{e}^1$.

For direct verification, ${\omega }^1{}_0\wedge e^0+{\omega }^1{}_3\wedge e^3=-p\ell \wedge e^1=-d{e}^1$, and the same holds for index two. For indices zero and three each term wedges $e^1$ or $e^2$ with itself, so it is zero. These forms are metric-compatible, and uniqueness follows from the existence and uniqueness of the Levi-Civita connection. The printed uniqueness statement implicitly includes metric compatibility: the torsion equation alone permits adding nonzero forms $C^a{}_{bc}{e}^c$ with $C^a{}_{bc}=C^a{}_{cb}$, since their wedge with $e^b$ vanishes.

Keep $\ell =e^0-e^3$ and set $P=p^{\prime }+p^2=A^{\prime \prime }/A$, $Q=q^{\prime }+q^2=B^{\prime \prime }/B$. Differentiation gives $d(p{e}^1)=P\ell \wedge e^1$ and $d(q{e}^2)=Q\ell \wedge e^2$. In Cartan's second structure equation, the quadratic terms for these components vanish. The apparently possible transverse component cancels:Thus the potentially nonzero mixed-index curvature 2-forms areAll other components vanish. Equivalently, the independent lower-index forms are ${\Theta }_{01}={\Theta }_{13}=-P\ell \wedge e^1$ and ${\Theta }_{02}={\Theta }_{23}=-Q\ell \wedge e^2$, with the rest supplied by lower-index antisymmetry. These signs use ${\Theta }^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{e}^c\wedge e^d$ and the curvature convention in question 2(c). This is the curvature of a diagonal plane wave in Rosen coordinates.

Let $E_a$ denote the frame dual to the coframe. For the convention just stated, the Ricci tensor is obtained from the curvature 2-forms through the Ricci one-forms ${\mathcal{R}}_b={\sum }_a{\iota }_{E_a}{\Theta }^a{}_b=R_{bd}{e}^d$. In particular ${\iota }_{E_1}(\ell \wedge e^1)=-\ell$ and likewise for index two. It follows thatThusThe scalar curvature vanishes even before imposing vacuum because $du$ is null, but the full tensor vanishes exactly whenThis is the Vacuum Einstein equations in this family. The nonzero-coframe hypothesis makes division by $A$ and $B$ legitimate. A nonzero $P=-Q$ can leave the Riemann curvature tensor nonzero even in vacuum.

Use linearization around Minkowski spacetime after constant rescaling of $x,y$: set $A=1+\epsilon a(u)$, $B=1+\epsilon b(u)$, $u=t-z$. To first order,Hence $s=a+b$ is affine in $u$, while $h=a-b$ is the freely propagating profile. The affine transverse trace in a diagonal linearized plane wave is pure gauge: it has zero linearized curvature and is locally a coordinate artifact. To see this without imposing extra boundary conditions, use a first-order linearized coordinate gauge transformation $h_{\mu \nu }\mapsto h_{\mu \nu }-{\partial }_{\mu }{\xi }_{\nu }-{\partial }_{\nu }{\xi }_{\mu }$, withThe $tx,ty,zx,zy$ changes cancel and the remaining time/longitudinal changes are proportional to $s^{\prime \prime }=0$. The transverse perturbation becomeswith all time and longitudinal components zero. It is a transverse-traceless gauge plane gravitational wave travelling in the positive $z$-direction, with plus polarization and no cross polarization in these transverse axes.

For freely falling nearby particles, geodesic deviation gives opposite transverse tidal accelerations, since $R_{txtx}=-\epsilon a^{\prime \prime }$ and $R_{tyty}=-\epsilon b^{\prime \prime }=+\epsilon a^{\prime \prime }$ in vacuum. Thus the transverse geodesic deviation eigenvalues have opposite signs; the stretching and compressing roles interchange when the curvature profile changes sign. If $h^{\prime \prime }=0$ as well, there is no linearized tidal wave: that special profile is also locally flat to this order. The exact diagonal metric describes a fixed transverse polarization through its opposite vacuum curvature eigenvalues; the linearized calculation identifies it as plus relative to the chosen axes.