Under $h'_{\mu\nu}=h_{\mu\nu}+\partial_\mu\xi_\nu+\partial_\nu\xi_\mu$, trace reversal gives
$$
\bar h'_{\mu\nu}=\bar h_{\mu\nu}
+\partial_\mu\xi_\nu+\partial_\nu\xi_\mu
-\eta_{\mu\nu}\partial_\rho\xi^\rho.
$$
Taking a divergence yields
$$
\partial^\mu\bar h'_{\mu\nu}
=\partial^\mu\bar h_{\mu\nu}+\mathop{\Box}\xi_\nu.
$$
Applying $\Box$ to the transformed field likewise produces only derivatives of $\Box\xi$. Therefore $\Box\xi^\mu=0$ preserves both the gauge condition and the sourced wave equation. These are the <residual gauge symmetry of linearized gravity>[residual gauge transformations].

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-309/3/b/solution