Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 69 / 2 / a / Solution

The divergence theorem states ${\int }_V\nabla \cdot \mathbf{F}dV={\oint }_{\partial V}\mathbf{F}\cdot \mathbf{n}dS$. With the Poisson equation for Newtonian gravity, this givesor gravitational-acceleration flux $\oint \mathbf{g}\cdot \mathbf{n}dS=-4\pi GM(V)$.

Use cylindrical radius $R=r\sin \theta$, so $r(1+|\cos \theta |)=r+|z|$. An arbitrary reference length in the logarithm only adds a constant potential. Off the equatorial plane, the radial part of ${\nabla }^2\log [r(1+|\cos \theta |)]$ is $1/r^2$. Writing $c=\cos \theta$, its angular part is $r^{-2}d[(1-c^2)d\log (1+|c|)/dc]/dc=-1/r^2$ on either open hemisphere. The two cancel: there is no volume density away from $z=0$.

At the plane, ${\partial }_z\varphi {|}_{0^{+}}=v_0^2/R$ and ${\partial }_z\varphi {|}_{0^{-}}=-v_0^2/R$. Applying the gravitational flux law to a thin pillbox gives $4\pi G\Sigma ={\partial }_z\varphi {|}_{0^{+}}-{\partial }_z\varphi {|}_{0^{-}}$. ConsequentlyThis is the Mestel disk potential-density pair. There is no additional point mass at the origin: the enclosed mass tends to zero linearly as the enclosing radius shrinks.

At $z=0$, $\varphi =v_0^2\log R$ up to a constant, so radial circular balance givesOn a sphere of radius $r$, ${\partial }_r\varphi =v_0^2/r$ at every nonsingular angular point. Its flux is $4\pi v_0^{2}r$, and henceDirect integration of the surface density confirms $2\pi {\int }_0^r\Sigma (R)RdR=v_0^{2}r/G$. Thus $v_{\mathrm{circ}}^2=GM(<r)/r$ even though the source is a razor-thin disk. This equality is a special scale-free property of the Mestel disk, not evidence that the source is spherical. In a general disk, matter outside the orbit and nonspherical interior forces prevent this enclosed-mass formula for the local circular speed. The model has infinite total mass, so its potential is defined by differences rather than a zero at infinity.