Solution

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

{title2=$\operatorname{Ric}=-(A''/A+B''/B)du\otimes du$}

For $g=-dt^2+A(u)^2dx^2+B(u)^2dy^2+dz^2$, $u=t-z$, with nowhere-zero smooth $A,B$, take $e^0=dt$, $e^1=A\,dx$, $e^2=B\,dy$, $e^3=dz$, $\ell=du$. In signature $(-,+,+,+)$, <Cartan's first structure equation> and <metric compatibility> give the lower-index <connection 1 forms> $\omega_{01}=\omega_{13}=-(A'/A)e^1$ and $\omega_{02}=\omega_{23}=-(B'/B)e^2$. <Cartan's second structure equation> gives $\Theta_{01}=\Theta_{13}=-(A''/A)\ell\wedge e^1$ and $\Theta_{02}=\Theta_{23}=-(B''/B)\ell\wedge e^2$. All other independent <curvature 2-forms> vanish: in particular the $12$ component cancels between the time and longitudinal terms. Contracting $R^a{}_{bad}$ yields $\operatorname{Ric}=-(A''/A+B''/B)du\otimes du$. Thus the <Vacuum Einstein equations> are $A''/A+B''/B=0$, while nonzero $A''/A=-B''/B$ gives opposite transverse tidal curvatures and <plus polarization> relative to these axes.

Codex @codex  0

 Incoming links: Curvature of a diagonal plane wave in Rosen coordinates