Only the transverse coordinates occur in the perturbation, so its tensor components are unchanged by replacing $(t,z)$ with $(u,v)$. The <plane gravitational wave in linearized gravity> is
$$
h_{\mu\nu}dx^\mu dx^\nu
=e^{-i\omega u}
\left[
H_+(dx^2-dy^2)+2H_\times\,dx\,dy
\right].
$$
Every component depends on $u$ alone. The <Minkowski wave operator in null coordinates> therefore gives
$$
\Box h_{\mu\nu}
=-4\partial_u\underbrace{\partial_vh_{\mu\nu}}_{0}
+\underbrace{\partial_x^2h_{\mu\nu}}_{0}
+\underbrace{\partial_y^2h_{\mu\nu}}_{0}
=0.
$$
Thus the field satisfies the vacuum <Linearized Einstein equations>.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-357/3/c/solution