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Take the tensor contraction of the Einstein field equations with the inverse metric $g^{\mu \nu }$. In four-dimensional spacetime, $g^{\mu \nu }{g}_{\mu \nu }=4$, henceHere $T$ is the tensor contraction of the stress-energy tensor. Substituting this result back into the Einstein field equations, while keeping both free indices lowered, gives the trace-reversed stress-energy tensor:The factor $1/2$ uses the dimension four: in dimension $n>2$, the Trace-reversed Einstein field equations in D dimensions have $1/(n-2)$ in place of $1/2$.

Use $x^0=ct$ and metric signature $(-,+,+,+)$. With the source's curvature sign convention, the Ricci tensor isIts time-time component is thereforeFor a slowly moving, pressure-negligible source, the stress-energy tensor is approximated by relativistic dust: $T^{\mu \nu }=\rho U^{\mu }{U}^{\nu }$, where $\rho$ is mass density and $U^{\mu }$ is the four-velocity. In its approximate rest local inertial frame, $U^{\mu }\simeq (c,0,0,0)$ and $U_0\simeq -c$. Thusand the trace-reversed stress-energy tensor satisfies

Independently, introduce the metric perturbation $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$, with $|h_{\mu \nu }|\ll 1$ and constant Minkowski metric ${\eta }_{\mu \nu }$. The Christoffel symbols are first order in $h$, so their products are discarded in the weak-field approximation. With a comma denoting a partial derivative, the linearized Riemann curvature tensor isTaking the tensor contraction gives, to this order,In the static Newtonian limit, all time partial derivatives vanish. The constant ${\eta }_{00}$ also has zero partial derivatives, leavingHere ${\nabla }^2={\sum }_i{\partial }_i^2$ is the spatial Laplace operator in the approximately Cartesian background coordinates. This sign follows the paper's curvature sign convention.