Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 / 2

The collapse is perpendicular to the initial field, so magnetic flux freezing preserves the mass-to-flux ratio of each material flux tube:With no toroidal field, radial magnetostatic equilibrium isDefine the effective polytropic constantThen gas and magnetic pressure combine as $P+B_z^2/(8\pi )=K_{\mathrm{eff}}{\rho }^2$. Dividing equilibrium by $\rho$, differentiating, and using the cylindrical Poisson equationgivesPut $a^2=K_{\mathrm{eff}}/(2\pi G)$. This is the order-zero Bessel differential equation, and regularity on the axis together with $\rho (0)={\rho }_1$ yields

The physical filament ends when the mass density first reaches zero. If $x_1\simeq 2.4$ is the first positive zero of the Bessel function $J_0$, thenThe first zero must be used because continuing to the next lobe would make the density negative.

The line mass isUsing $d[x{J}_1(x)]/dx=x{J}_0(x)$ and $J_1(x)=-J_0^{\prime }(x)$ givesSince $x_1\simeq 2.4$ and $J_0^{\prime }(x_1)\simeq -0.52$,

Write $Y=B_{\varphi }^2$, $x=R^2/L^2$, and retain the frozen axial field $B_z=(B_0/{\rho }_0)\rho$. The radial magnetostatic equilibrium equation now includes both the gradient of the toroidal magnetic pressure and the inward magnetic tension:For $\rho ={\rho }_1{e}^{-x}$, cylindrical gravity givesMultiplication by the integrating factor $R^2$ therefore producesThe integration constant must vanish for regularity on the axis. Direct integration givesIts Taylor expansion at the axis isso the regular field behaves as $B_{\varphi }=O(R)$. At large radius,Nonnegativity at infinity requiresThis condition is also sufficient: at the limiting value, the braces divided by $K_{\mathrm{eff}}{L}^2$ reduce to $2e^{-x}[1-(1+x)e^{-x}]\ge 0$, and decreasing $L^2$ only increases them. Thus the toroidal magnetic field is real and regular at every radius exactly in the stated range.