 Solution

A pressureless relativistic fluid has stress-energy tensor $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$, with four-velocity normalized by $u^{\mu }{u}_{\mu }=-c^2$. In its rest frame, $u^i=0$, so $\boxed{T^0{}_0=-\rho c^2,T^i{}_j=0}$. Here $\rho$ is rest-frame mass density, or energy density divided by $c^2$.

For the linearized Ricci tensor and scalar, discard the quadratic connection terms. The contracted connection is ${\Gamma }^{\mu }{}_{\mu \nu }={\partial }_{\nu }(\Phi -3\Psi )/c^2$. Differentiation then givesThe Ricci scalar uses the zeroth-order inverse metric, $R=-R_{00}+{\delta }^{ij}{R}_{ij}$, because multiplying first-order curvature by another metric perturbation would again be second order. These are the linearized curvature of a two-potential weak-field metric in the paper's Riemann curvature tensor sign convention.