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Use geometric units $G=c=1$, metric signature $(-+++)$, and a four-velocity normalized by the physical metric. Write $\varphi =1+\Phi$, where $|\Phi |\ll 1$. For a pressureless perfect fluid, the dust stress-energy tensor is $T_{ij}=\rho U_i{U}_j$ and $g^{ij}{U}_i{U}_j=-1$. Since ${\eta }^{ij}={\varphi }^2{g}^{ij}$, its background trace isConsequently the field equation is $(-{\partial }_t^2+\Delta )\varphi =4\pi \rho {\varphi }^3$. Neglecting time derivatives and higher weak-field orders gives the Newtonian limit of Nordstrom gravity:The constant shift between $\varphi$ and $\Phi$ does not affect the Poisson equation or the force. Here $\rho$ is the physical rest density; if fluid indices are instead defined using the background metric, the intermediate powers of $\varphi$ change but the same leading-order equation results.

For the motion, the Levi-Civita connection of the conformally flat metric isIn particular ${\Gamma }^{\alpha }{}_{00}={\partial }_{\alpha }\log \varphi$. Replacing the affine parameter in the geodesic equation by coordinate time gives, with $w^i=(1,\mathbf{v})$ and $\mathbf{v}=d\mathbf{x}/dt$,At small speeds and for a nearly static field, this becomesThus the Newtonian gravitational potential and freely falling trajectories agree to the stated weak-field, slow-motion order. The comparison is an approximation, not an exact equivalence of relativistic trajectories.