Past exam of the general relativity and cosmology course of the University of Oxford / 2013 / 1 / d

For a massive particle parametrized by proper time, the geodesic equation isThe weak-field approximation alone does not imply Newtonian motion. Also take a nonrelativistic limit, $|v|/c\ll 1$, in the usual quasistatic regime so the potentials' time variation does not compensate for the small velocity factors. Reparametrizing by coordinate time and retaining the leading velocity-independent spatial term givesIn empty space the first field equation gives ${\nabla }^2\Psi =0$. For static potentials, the spatial equations imply ${\nabla }^2(\Phi -\Psi )=0$ and then ${\partial }_i{\partial }_j(\Phi -\Psi )=0$. The difference is at most affine; isolated-system boundary conditions fixing both potentials to zero at infinity remove it. Thus $\Phi =\Psi$ andthe empty-space Newton-Poisson equation. With the same quasistatic assumptions and dust present it becomes ${\nabla }^2\Phi =4\pi G\rho$. Keeping rapid time dependence instead leaves ${\nabla }^2\Phi =-3{\partial }_0^2\Psi$ in vacuum, so the static qualification matters.