= Longitudinal displacement in linear dust perturbations
{title2=$\ddot\psi_L+2H\dot\psi_L-4\pi G\bar\rho\psi_L=0$}
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 $\nabla\Phi_0=+4\pi G\bar\rho r/3$.
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