 Solution

Use the metric signature $\eta =\mathrm{diag}(1,-1,-1,-1)$ and $c=1$ throughout this question. The trace-reversed metric perturbation satisfies $\varphi ={\eta }_{\mu \nu }{\varphi }^{\mu \nu }=-h$, so its inverse relation isSet $C_{\nu }={\partial }^{\sigma }{\varphi }_{\nu \sigma }$. Substituting this inverse relation into the linearized field operator cancels the second derivatives of the trace, leavingThe imposed Lorenz gauge in linearized gravity is $C_{\nu }=0$. Therefore the Linearized Einstein equations in vacuum reduce toThe overall sign of the field operator is immaterial for this homogeneous equation.