= Solution
Use the <metric signature> $\eta=\operatorname{diag}(1,-1,-1,-1)$ and $c=1$ throughout this question. The <trace-reversed metric perturbation> satisfies $\phi=\eta_{\mu\nu}\phi^{\mu\nu}=-h$, so its inverse relation is
$$
h_{\mu\nu}=\phi_{\mu\nu}-\frac12\eta_{\mu\nu}\phi.
$$
Set $C_\nu=\partial^\sigma\phi_{\nu\sigma}$. Substituting this inverse relation into the linearized field operator cancels the second derivatives of the <trace>, leaving
$$
\mathcal E_{\mu\nu}=\partial_\mu C_\nu+\partial_\nu C_\mu
-\Box\phi_{\mu\nu}-\eta_{\mu\nu}\partial^\rho C_\rho.
$$
The imposed <Lorenz gauge in linearized gravity> is $C_\nu=0$. Therefore the <Linearized Einstein equations> in vacuum reduce to
$$
\boxed{\Box\phi^{\mu\nu}=\partial_\sigma\partial^\sigma\phi^{\mu\nu}=0,\qquad
\Box=\partial_t^2-\nabla^2.}
$$
The overall sign of the field operator is immaterial for this homogeneous equation.
Back to article page
