Longitudinal displacement in linear dust perturbations (${\ddot{\psi }}_L+2H{\dot{\psi }}_L-4\pi G\bar{\rho }{\psi }_L=0$)

Mass conservation in a Lagrangian dust displacement gives $1+\delta =\det (I+{\nabla }_q\psi {)}^{-1}$, hence $\delta =-{\nabla }_q\cdot \psi$ to first order. The perturbing gravitational potential determines the longitudinal part of the force, so the displayed equation holds for the curl-free displacement under the usual homogeneous boundary convention. A divergence-free transverse displacement instead satisfies ${\ddot{\psi }}_T+2H{\dot{\psi }}_T=0$. Taking the divergence gives $\ddot{\delta }+2H\dot{\delta }-4\pi G\bar{\rho }\delta =0$ without assuming the full displacement is longitudinal. The background Poisson equation with positive source requires $\nabla {\Phi }_0=+4\pi G\bar{\rho }r/3$.