 Newtonian tidal curvature

Using $[{\nabla }_c,{\nabla }_d]V^a=R^a{}_{bcd}{V}^b$, the Newton–Cartan gravitational connection has only the displayed Hessian components and their antisymmetric partners in the last two indices. The geodesic deviation equation gives relative acceleration $-{\partial }_i{\partial }_j\Phi \delta x^j$. The Poisson equation for Newtonian gravity is therefore $R_{ab}=4\pi G\rho (dt{)}_a(dt{)}_b$. A spatially uniform gravitational acceleration has zero tidal curvature even if the connection coefficients do not vanish in the original coordinates.