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Use $A_{\beta }={\partial }_{\mu }{h}^{\mu }{}_{\beta }$ and $\square ={\eta }^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }$. The linearized Ricci tensor and scalar obeyThe background Ricci tensor and Ricci scalar vanish, so varying the metric in the scalar-curvature term contributes nothing at first order beyond this trace. The Einstein tensor perturbation is consequentlyAll contractions use the Minkowski metric. Direct differentiation gives ${\partial }^{\alpha }\delta G_{\alpha \beta }=0$, the linearized contracted Bianchi identity, which checks the relative signs. No time derivatives or gauge-dependent terms have been discarded.

The Levi-Civita connection isFor $g_{\alpha \beta }={\eta }_{\alpha \beta }+h_{\alpha \beta }$, retain only first-order terms in $h$. Substitution into the curvature formula and contraction gives the linearized Ricci tensor and scalarHence

Write $g_{\mu \nu }={\eta }_{\mu \nu }+ϵh_{\mu \nu }$ 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 gravityThe linearized Ricci tensor and scalar then satisfySubstitution into $G_{\mu \nu }=8\pi T_{\mu \nu }$ givesIf the paper denotes the trace-reversed variable itself by $h_{\mu \nu }$ in its displayed equation, this is exactly that convention.

For a point mass, $\Phi =-GM/r$, and part (b) givesThe linearized Ricci tensor and scalar obeyFor a static perturbation with $h_{0i}=0$, this becomesAway from the source, ${\nabla }^2{h}_{00}=0$. For $h_{ij}=f(r)x_i{x}_j$, one has $h_{ii}=r^{2}f$, so the two supplied identities giveThe vacuum trace equation $R=0$ therefore reduces towhose general solution for $r>0$ isThe scalar equation alone leaves $C$ arbitrary. Imposing the remaining vacuum components $R_{ij}=0$ and matching the mass fixes $C=2GM$, as in the linearized point-mass metric in radial Cartesian form.