Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 3

Write $q=|r-r^{\prime }|$. Away from the source, the radial Laplacian gives ${\nabla }^2(-1/(4\pi q))=q^{-2}{\partial }_q[q^2/(4\pi q^2)]=0$. Through a small source-centred sphere, the outward flux of its gradient isThis identifies a unit Dirac delta at the source. More explicitly, for a smooth compactly supported test function $\psi$, apply Green's second identity outside the radius-$\epsilon$ ball. The inner outward normal of that punctured domain is $-e_q$. The term $H{\partial }_n\psi$ has integral $O(\epsilon )$, while $-\psi {\partial }_{n}H$ tends to $\psi (r^{\prime })$. Thus $\int H{\nabla }^2\psi dV=\psi (r^{\prime })$, proving the distributional identityWhenever the integral converges, convolution with this Green function of the Laplacian solves the Poisson equation for Newtonian gravity:The harmonic term represents the homogeneous freedom and is fixed by boundary conditions. For an isolated finite mass distribution, regularity away from the prescribed sources and the normalization $\Phi \to 0$ at infinity select $h=0$.

For a point mass at the origin, the delta integral gives the Newtonian potential of a point mass,

Choose the polar axis in the source integration along the evaluation vector $r$. The Euclidean distance then satisfies $|r-r^{\prime }{|}^2=r^2+r^{\prime 2}-2r{r}^{\prime }\cos {\theta }^{\prime }$, givingFor the uniform ball, integrate its Newtonian gravitational potential in these spherical coordinates. With $\mu =\cos {\theta }^{\prime }$,For $r,r^{\prime }>0$, the inner integral isobtained from the antiderivative $-\sqrt{r^2+r^{\prime 2}-2r{r}^{\prime }\mu }/(r{r}^{\prime })$. ThereforeFor $0<r<R$, split the radial integral at $r^{\prime }=r$. Its two pieces are ${\int }_0^{r}2r^{\prime 2}d{r}^{\prime }/r=2r^2/3$ and ${\int }_r^{R}2r^{\prime }d{r}^{\prime }=R^2-r^2$. The Newtonian gravitational potential of a uniform solid sphere is consequentlyThe centre value follows by continuity, removing the intermediate $1/r$. For $r>R$, the same integral gives $-GM/r$, so both the potential and first derivative match at the surface. Finally ${\nabla }^2\Phi =4\pi G\rho$ inside, while $-\nabla \Phi =-GM\mathbf{r}/R^3$ is directed inward and vanishes at the centre, checking the sign and normalization.