= 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 gives
$$
\boxed{\begin{aligned}
R_{00}&=\frac{\nabla^2\Phi+3\partial_0^2\Psi}{c^2},\\
R_{0i}&=\frac{2\partial_0\partial_i\Psi}{c^2},\\
R_{ij}&=\frac{\partial_i\partial_j(\Psi-\Phi)+\delta_{ij}(\nabla^2\Psi-\partial_0^2\Psi)}{c^2},\\
R&=\frac{4\nabla^2\Psi-2\nabla^2\Phi-6\partial_0^2\Psi}{c^2}.
\end{aligned}}
$$
The <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.
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