The <retarded and advanced null coordinates> invert to
$$
t=\frac{u+v}{2},\qquad z=\frac{v-u}{2}.
$$
Thus
$$
-dt^2+dz^2=-du\,dv
$$
and the <Minkowski metric> becomes
$$
\boxed{ds^2=-du\,dv+dx^2+dy^2.}
$$

The <coordinate basis> transforms by the chain rule:
$$
\boxed{
\partial_t=\partial_u+\partial_v,\qquad
\partial_z=-\partial_u+\partial_v,
}
$$
or equivalently
$$
\boxed{
\partial_u=\frac12(\partial_t-\partial_z),\qquad
\partial_v=\frac12(\partial_t+\partial_z).
}
$$
In the $(z,t)$ diagram, $\partial_t$ points vertically upward and $\partial_z$ horizontally right. The vector $\partial_u$ points along the future-left null ray and $\partial_v$ along the future-right null ray; their factors of one half affect length in the coordinate drawing but not direction.

Using these derivative relations, the <Minkowski wave operator in null coordinates> is
$$
\boxed{
\Box=-\partial_t^2+\partial_z^2+\partial_x^2+\partial_y^2
=-4\partial_u\partial_v+\partial_x^2+\partial_y^2.
}
$$

Solution

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

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>.

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

{title2=$\Box=-4\partial_u\partial_v+\partial_x^2+\partial_y^2$}

For $u=t-z$ and $v=t+z$, the four-dimensional Minkowski wave operator is
$$
\Box=-4\partial_u\partial_v+\partial_x^2+\partial_y^2.
$$
Every profile depending only on one null coordinate solves the wave equation when it is uniform in the transverse coordinates.


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