Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 54 / 2

The magnetic field satisfies $\nabla \cdot \mathbf{B}=0$. Expanding the divergence of its Maxwell stress tensor givesFor the Newtonian gravitational field $\mathbf{g}=-\nabla \Phi$, Poisson equation for Newtonian gravity gives ${\partial }_j{g}_j=-4\pi G\rho$, and the gradient representation gives ${\partial }_j{g}_i={\partial }_i{g}_j$. ConsequentlyThe negative of the Newtonian gravitational stress tensor, in this force-stress convention, therefore supplies $\rho g_i$. Adding the pressure stress supplies $-{\partial }_{i}p$, soEach summand is symmetric. Absence of external gravitational sources is needed to represent the full gravitational force by this self-gravitating stress.

Apply the continuity equation and integrate by parts, with the stated vanishing boundary terms and finite moments. The second mass moment tensor satisfiesand differentiation again givesInsert the stress-divergence equation from part (a). The two force integrals becomeThe first term is $4K_{ij}$. Hence the magnetized-fluid tensor virial theorem isHere $\mathcal{T}$ is the volume-integrated stress, whose sign differs from some gravitational potential-energy tensor conventions. The derivation also requires the advective mass-moment surface terms to vanish; this is automatic for an isolated sufficiently decaying configuration.

At the initially resting instant $K_{ij}=0$, and the cold-fluid assumption removes the pressure stress. Sum the $xx$ and $yy$ components of the magnetized-fluid tensor virial theorem. The magnetic and gravitational traces areSince $I_{\perp }=I_{xx}+I_{yy}$ and $g_z=-{\partial }_z\Phi$, this yields the horizontal virial balance of a cold magnetized fluid:Being at rest sets the instantaneous velocity to zero; it does not assert equilibrium or zero acceleration. The fields throughout space contribute to this stress integral, including their vacuum exterior.

Integrate Poisson equation for Newtonian gravity through a narrow slab around the disk. The horizontal derivative contributions vanish as its thickness tends to zero, leaving the normal-derivative jumpThe even function symmetry of $\Phi$ makes the derivatives opposite, soIn the current-free simply connected upper half-space, Ampère's circuital law gives $\nabla \times \mathbf{B}=0$ and permits a magnetic scalar potential. Rescale it so that $\sqrt{4\pi G/{\mu }_0}\mathbf{B}=-\nabla {\Phi }_M$. The divergence-free condition makes ${\Phi }_M$ satisfy Laplace's equation, withCompare with the gravitational jump condition. Subject to the same isolated-field boundary condition at infinity, ${\Phi }_M$ is the harmonic potential of the effective surface densityThis gravity-equivalent magnetic surface density can have either sign; it is a mathematical representation of the exterior magnetic field, not physical negative mass. An imposed nondecaying field would require additional boundary data and would not be fixed by the disk surface density alone.

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.