Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 3

Integrate each conservative ideal magnetohydrodynamics law through a thin pillbox around the stationary shock. Mass conservation givesThe MHD momentum-flux tensor isIts normal and tangential components givewhere $\Pi =p+B^2/(2{\mu }_0)$ is total pressure. The solenoidal condition gives $[B_x]=0$, while steady Faraday's law gives continuity of tangential electric field,Finally, material energy flux plus the normal Poynting vector component giveswhere $h$ is specific enthalpy per unit mass.

Gravity is a bounded volume force, so its integral across a shock whose thickness tends to zero vanishes. Viscosity and resistivity inside the layer may produce entropy; consequently entropy flux need not be equal on both sides, although the second law requires nonnegative net entropy production.

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The quantities $B_x$ and $u_x{B}_y-u_y{B}_x=-E_z$ are each constant across the shock. A single tangential Galilean boost therefore changes the common $E_z$ to zero on both sides. This is the de Hoffmann–Teller frame. Ideal MHD then gives

Let $j=\rho u_x$ and retain the common $B_x$. The independent jump conditions simplify toTerms involving the common $B_x^2$ cancel from normal momentum, and the electromagnetic term vanishes from energy because $E_z=0$.

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In the aligned frame, define the Alfvén numberwhere the second equality uses ${\mathbf{u}}_i\parallel {\mathbf{B}}_i$. Since $j={\rho }_i{u}_{xi}$ and $B_x$ are common,Therefore

Using alignment in the tangential momentum flux givesIts continuity then yieldsorwhenever the ratios are determinate.

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If $B_{y1}=B_{y2}=0$, alignment also gives $u_{y1}=u_{y2}=0$. The common normal magnetic field contributes the same constant magnetic stress on both sides and carries no Poynting flux in the aligned frame. The remaining mass, normal-momentum, and energy conditions are therefore exactlywhich are the Rankine-Hugoniot conditions for a perfect gas.

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If $B_{y2}=0$ but $B_{y1}\ne 0$, tangential momentum continuity in the formforces $A_1^2=1$. Since the normal components are positive,so the upstream velocity equals the upstream Alfvén speed. This is the limiting switch-off shock condition.

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Take $B_{y2}=-B_{y1}\ne 0$ and chooseThen both tangential momentum fluxes vanish, so that jump condition is satisfied. The density relation from part (c) gives ${\rho }_2={\rho }_1$, and mass conservation gives $u_{x2}=u_{x1}$. Alignment makes $u_{y2}=-u_{y1}$, so $u_2^2=u_1^2$.

Because $B_{y2}^2=B_{y1}^2$, normal momentum now givesThe energy condition is then automatic. Thus density and pressure are continuous while the tangential magnetic field and velocity reverse direction. This is a $180^{\circ }$ rotational discontinuity in magnetohydrodynamics, carrying no compression or entropy jump.

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