Using the stated <Christoffel symbols> in the definition of the <Riemann curvature tensor>, the two nontrivial contractions are
$$
\begin{aligned}
R^x{}_{uxu}
&=-\partial_u\left(\frac{f'}f\right)
-\left(\frac{f'}f\right)^2
=-\frac{f''}f,\\
R^y{}_{uyu}
&=-\partial_u\left(\frac{g'}g\right)
-\left(\frac{g'}g\right)^2
=-\frac{g''}g.
\end{aligned}
$$
Hence
$$
R_{uu}=R^\alpha{}_{u\alpha u}
=-\frac{f''}f-\frac{g''}g,
$$
and the vacuum <Einstein field equations> reduce to
$$
\boxed{\frac{f''}f+\frac{g''}g=0.}
$$
This is the diagonal <Rosen coordinates for a plane gravitational wave> equation.

For the plus-polarized wave, set
$$
H(u)=H_+e^{-i\omega u}.
$$
Comparison of the transverse metric components gives, to <linear order>,
$$
f(u)^2=1+H(u),\qquad
g(u)^2=1-H(u),
$$
or
$$
f(u)=1+\frac12H(u)+O(H_+^2),
\qquad
g(u)=1-\frac12H(u)+O(H_+^2).
$$
Therefore
$$
\frac{f''}f+\frac{g''}g
=\frac12H''-\frac12H''+O(H_+^2)
=O(H_+^2).
$$
The <Vacuum Einstein equations> are consequently satisfied at linear order. A real gravitational wave is obtained by taking the real part of the complex plane-wave notation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 357 / 3 / d / Solution

{c}

Rosen coordinates put a plane wave into the transverse-metric form
$$
ds^2=-du\,dv+g_{AB}(u)dx^A dx^B.
$$
For diagonal $g_{AB}=\operatorname{diag}(f^2,g^2)$, the only potentially nonzero <Ricci tensor> component is $R_{uu}=-f''/f-g''/g$.

Codex @codex  0

 Incoming links: Rosen coordinates for a plane gravitational wave