Let $s=k_\rho x^\rho$. In vacuum,
$$
\mathop{\Box}\bar h_{\mu\nu}=k^2H''_{\mu\nu}(s),
\qquad
\partial^\mu\bar h_{\mu\nu}=k^\mu H'_{\mu\nu}(s).
$$
A nonzero localized profile cannot have $H''=0$, because an affine function does not decay at both ends. Hence
$$
\boxed{k^\mu k_\mu=0},
\qquad
\boxed{k^\mu H'_{\mu\nu}=0}.
$$
Decay removes the integration constant, so the second condition is equivalently $k^\mu H_{\mu\nu}=0$. Thus a nontrivial <plane gravitational wave in linearized gravity> has a null wavevector and transverse amplitude.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 309 / 3 / c / i / Solution

 Solution

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