Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 (Astrophysical Fluid Dynamics)

The boundary is material precisely when its material derivative vanishes on $f=0$. Direct differentiation givesFor this to vanish at every point of the ellipsoidal free surface, each coefficient must vanish:With these relations the expression vanishes for every interior point as well, sothroughout the affine stellar model. Thus $f$ is a Lagrangian label carried by each fluid element.

Since $Df/Dt=0$, the density ansatz hasThe velocity divergence isThe mass conservation equation therefore givesSimilarly,and hence

The $x$ component of the material acceleration isBecause $d\hat{p}/df=\hat{\rho }$,Combining this with $-{\partial }_x\Phi =-{\Omega }^{2}x$ givesThe $y$ and $z$ components give the analogous equations for $b$ and $c$. Finally, $\hat{p}(0)=0$ makes $p=0$ on the material free surface, so both dynamic and kinematic boundary condition requirements are satisfied.

Let $V=abc$ and write $T=K{V}^{-(\gamma -1)}$ for a constant $K$. The axis equations are Newton equations in the effective potentialbecauseand similarly for $b,c$. Thereforeis conserved. Multiplying by the fixed profile-dependent mass moment converts $\mathcal{E}$ into the physical kinetic, trapping-potential, and internal energy of the star, so it is proportional to total energy.

Let the equilibrium radius be $a_0=b_0=c_0$ and $2T_0={\Omega }^2{a}_0^2$. Linearizing the $a$ equation giveswith cyclic analogues. The temperature scaling gives

For the affine breathing mode of a star, all three fractional axis changes equal $q$. ThenThis is the homologous compressional mode, which changes volume, density, and temperature.

For either independent affine quadrupole mode of a star, the three fractional changes sum to zero. Then $\delta T=0$ andThese two degenerate modes deform the sphere into an ellipsoid while preserving its volume to first order.

The additional acceleration isIts three fractional-axis forcing terms are therefore proportional to $(1,1,-2)$ and have zero sum. Consequently it has no projection on the affine breathing mode of a star, which is not forced at linear order.

It lies entirely in the affine quadrupole mode of a star subspace. PutThenso away from resonanceup to free oscillations. The tidal forcing resonates with the quadrupole mode when ${\Omega }_o=\sqrt{2}\Omega$; in the ideal undamped model the resonant amplitude grows secularly.

Let $p=K{\rho }^{1+1/m}$. Hydrostatic equilibrium in the uniform gravitational field givesSubstitution and integration with $\rho =0$ at $z=0$ giveswhere $H$ fixes the normalization. Integrating the hydrostatic equation from $z$ to the free surface gives the pressure directly:This indeed has $p\propto {\rho }^{1+1/m}$ and $p(0)=0$, as required for a polytropic atmosphere.

Linearize the inviscid momentum equation in the uniformly rotating frame. The Coriolis acceleration is $2\mathbf{\Omega }\times \delta \mathbf{u}$, and the equilibrium pressure gradient cancels gravity. For perturbations proportional to $e^{i{k}_{x}x-i\omega t}$, the horizontal components areThe vertical component is

Linearizing mass conservation,givesFinally, linearizing the adiabatic pressure equation givesThese are the stated five equations. Self-gravity contributes no perturbation because it is neglected, and the equilibrium centrifugal term has already been absorbed or omitted.

Set $\delta \mathbf{u}=-i\omega \mathbf{\xi }$. The continuity and adiabatic equations give the Lagrangian perturbationsBecause the equilibrium is a neutrally stratified polytropic atmosphere, $p\propto {\rho }^{\gamma }$, so the displacement terms cancel andUsing $dp/dz=-\rho g$ then gives

Eliminating ${\xi }_y$ from the two horizontal momentum equations and combining gravity with the vertical pressure force yieldsThusand the requested coupled first-order system is

For an incompressible perturbation, $\Delta =0$, so $\Psi =g{\xi }_z$. The vertical equation givesIncompressibility together with the horizontal equation also givesEquating the two expressions yieldsThe positive-frequency branch is the surface gravito-inertial waveIts vertical displacement decays as $e^{{\omega }^{2}z/g}$ into the atmosphere $z<0$.

Eliminate ${\xi }_z$ from the coupled system to obtainWriting $\Psi =\psi e^{kz}$ givesFor the neutral polytrope found in part (a),If $\psi$ is a polynomial of degree $n$ with leading term $z^n$, the coefficient of the highest power $z^n$ in the ODE isA polynomial solution therefore requiresThe factor $e^{kz}$ traps every mode below the free surface. The $n=0$ solution is the incompressible surface gravito-inertial or $f$ mode of part (d). The $n\ge 1$ solutions are vertically structured acoustic p modes; neutral stratification leaves no buoyancy-driven $g$-mode family.

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.

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 $E_z=0$.

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.

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.

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.

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.

The representationgivesHence $\mathbf{B}\cdot \nabla \psi =0$: the level sets of the poloidal magnetic flux function are magnetic surfaces. Moreover, the magnetic flux through a circular disc isThus $2\pi \psi$ is the enclosed poloidal magnetic flux up to an additive axial reference.

The flux-function form solves $\nabla \cdot \mathbf{B}=0$ identically. With $k=k(\psi )$, the poloidal relation $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$ and axisymmetry solve steady mass conservation. The ideal induction equation is solved by the velocity representation with the field-line angular velocity $\omega =\omega (\psi )$. The azimuthal momentum equation integrates once to the magnetohydrodynamic angular-momentum invariant $\ell =\ell (\psi )$.

What remains is the pressure or entropy equation and the two poloidal components of momentum. Projecting poloidal momentum along a field line produces the Bernoulli invariant in part (c); projecting across magnetic surfaces produces the transfield or Grad-Shafranov equation that determines their shape. An equation of state and boundary conditions complete the problem. The stellar gravitational potential is prescribed, so no self-gravity Poisson equation remains to solve.

Although the Lorentz force can exchange energy between poloidal and toroidal motion, it does no net work in the frame rotating with a magnetic surface. Dotting steady momentum with the poloidal field and using the mass-loading, isorotation, and angular-momentum invariants combines the magnetic work with the toroidal kinetic term to giveThereforeis constant on each magnetic surface. Hereis the gravitational potential plus the centrifugal potential in the frame corotating at the field-line angular velocity. This is the modified Bernoulli function of the wind.

Define the poloidal Alfvén number bywhere $u_p=k{B}_p/\rho$. Thus, along a given field line,density decreases as the Alfvén number grows.

The azimuthal velocity representation givesSubstitution into $\ell =r{u}_{\varphi }-r{B}_{\varphi }/({\mu }_{0}k)$ yieldsAt the Alfvén surface, $A^2=1$. Smoothness requires the numerator of the solution for $B_{\varphi }$ to vanish there, soSolving away from the surface then gives

Ideal flux freezing anchors a field line in the highly conducting disc, so its field-line angular velocity naturally equals the angular velocity of its footpoint. A circular orbit in the central potential is Keplerian:

Put $r=r_0+\delta r$ and expandThe linear radial term vanishes because $GM/r_0^2={\Omega }_0^2{r}_0$. The radial and vertical second derivatives are respectively $-3{\Omega }_0^2$ and ${\Omega }_0^2$, so

If the field line makes angle $\theta$ from the vertical, then locally $\delta r=z\tan \theta$. Its quadratic potential change iswhich is negative exactly whenThis is the local magnetocentrifugal acceleration launching criterion.

The velocity representation givesThereforeIn the launching region, $r\simeq r_0\ll r_a$ and $A\ll 1$, so part (d) givesConsequently

The modified Bernoulli kinetic energy isThus a decrease in ${\Phi }_{cg}$ is shared between poloidal acceleration and relative toroidal motion. If $R$ is approximately constant, the same potential drop produces a poloidal kinetic increase smaller by the factor $1/(1+R^2)$ than the estimate that neglects the toroidal term. If $R$ varies strongly, that term can absorb or return energy, so monotonic decrease of ${\Phi }_{cg}$ alone no longer proves an equally direct increase of $u_p^2/2$.