All time components of the perturbation vanish. For $\mu=t$ this immediately gives
$$
\partial^\nu h_{\mu\nu}=0.
$$
For a spatial index $\mu=i$,
$$
\partial^\nu h_{i\nu}
=-\partial_t h_{it}+\partial_mh_{im}
=\partial_mh_{im}=0
$$
by the stated property. Thus the perturbation is transverse.

Its Minkowski trace is purely spatial:
$$
h=\eta^{\mu\nu}h_{\mu\nu}=h_{xx}+h_{yy}+h_{zz}.
$$
Adding the displayed components and collecting the coefficients of the independent functions $A$, $B$, and $C$ makes each coefficient vanish separately, so
$$
\boxed{\eta^{\mu\nu}h_{\mu\nu}=0}.
$$
Together with $h_{t\mu}=0$ and $\partial^\nu h_{\mu\nu}=0$, this proves that the wave is in <transverse-traceless gauge>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 357 / 1 / c / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-357/1/c/i/solution