= Solution
Write $g_{\mu\nu}=\eta_{\mu\nu}+\epsilon h_{\mu\nu}$ and retain first-order terms. The quadratic Christoffel products in the supplied Ricci formula drop out. The wave-coordinate condition becomes the <Lorenz gauge in linearized gravity>
$$
\partial^\mu\bar h_{\mu\nu}=0,
\qquad
\bar h_{\mu\nu}=h_{\mu\nu}-\frac12\eta_{\mu\nu}h.
$$
The <linearized Ricci tensor and scalar> then satisfy
$$
G^{(1)}_{\mu\nu}=-\frac12\mathop{\Box}\bar h_{\mu\nu}.
$$
Substitution into $G_{\mu\nu}=8\pi T_{\mu\nu}$ gives
$$
\boxed{\mathop{\Box}\bar h_{\mu\nu}=-16\pi T_{\mu\nu}},
\qquad
\boxed{\partial^\mu\bar h_{\mu\nu}=0}.
$$
If the paper denotes the trace-reversed variable itself by $h_{\mu\nu}$ in its displayed equation, this is exactly that convention.
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