 Solution

The hypothesis is $|{\Sigma }_M|<\Sigma$, with the physical surface density nonnegative. At height $z>0$, the vertical derivative of an isolated thin-disk gravitational potential has a strictly positive kernel:The same formula with ${\Sigma }_M$ represents ${\partial }_z{\Phi }_M$. The triangle inequality and the strict surface density bound therefore give the positive-kernel comparison of thin-disk fieldsUse ${\partial }_z{\Phi }_M=-\sqrt{4\pi G/{\mu }_0}{B}_z$. The integrand in part (c) is then $({\partial }_z{\Phi }_M{)}^2-({\partial }_z\Phi {)}^2<0$ in the upper vacuum region. Reflection symmetry gives the same result below; the infinitesimally thin disk has zero three-dimensional volume. Thus, with the finite-integral assumptions of the tensor virial theorem,Since the disk starts at rest, ${\dot{I}}_{\perp }=0$ initially and its radial second moment begins to decrease. This is magnetic subcriticality of a razor-thin disk: magnetic support cannot prevent initial contraction in the global virial sense. The conclusion concerns the mass-weighted radial size, and does not by itself prove that every fluid element accelerates inward or that contraction continues indefinitely.