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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.

Under the convention $\delta h_{\mu \nu }={\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$, choose ${\xi }^{\mu }=b^{\mu }\sin (k\cdot x)$ with constant $b^{\mu }$ and null $k^{\mu }$. The trace-reversed metric perturbation has amplitudeIt satisfies $k_{\mu }{ϵ}_{\mathrm{gauge}}^{\mu \nu }=0$ and $\square {\xi }^{\mu }=0$, so it is an allowed residual gauge symmetry of linearized gravity. Its linearized Riemann curvature operator is zero: the four wavevector factors in each second-derivative term cancel in the antisymmetric curvature combinations. Nonzero transverse-traceless curvature distinguishes the two physical gravitational wave polarizations from these coordinate modes.^[ref]

For a compact slowly moving source in the weak-field approximation, with no incoming homogeneous wave and $G=c=1$, the retarded Linearized Einstein equations give this leading far-zone trace-reversed metric perturbation. The key consequence of stress-energy conservation is ${\ddot{I}}_{ij}=2\int T_{ij}{d}^{3}x$. The physical gravitational wave is obtained by the transverse-traceless projector.