Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 314 (Astrophysical Fluid Dynamics)

The approximation requires a nonrelativistic monatomic gas whose particles collide often enough for local thermodynamic equilibrium, with a mean free path much shorter than every macroscopic scale. Translational motion must dominate its heat capacity, giving the adiabatic exponent $\gamma =5/3$. Heat conduction, viscosity, shocks, radiative heating and cooling, ionization, chemical reactions, and excitation of internal degrees of freedom must be negligible over the flow time. Under those conditions entropy is advected and the gas behaves as an adiabatic perfect gas.

Let $u>0$ be the inward radial speed. Steady spherical Bondi accretion conserves mass:For $\gamma =5/3$, $p=K{\rho }^{5/3}$ and the adiabatic sound speed satisfies $v_s^2=\gamma p/\rho \propto {\rho }^{2/3}$. Matching the reservoir therefore givesThe Bernoulli integral isWriting the Mach number as $\mathcal{M}=u/v_s$ and eliminating $v_s$ with mass conservation givesMultiplication of the Bernoulli equation by $r/2$ now yieldswhere

The functionhas its unique minimum at $\mathcal{M}=1$, where $f(1)=4$. At large $r$, mass conservation and the reservoir boundary conditions give $u\propto r^{-2}$ and hence $\mathcal{M}\to 0$. The physical solution therefore begins on the subsonic branch $0<\mathcal{M}<1$. It cannot pass smoothly to $\mathcal{M}>1$ without reaching the minimum, where the two algebraic branches meet. For $\gamma =5/3$ that meeting can occur only in the limiting central behavior of the critical solution, so the flow remains subsonic for every $r>0$.

As $r\to 0$, the right-hand side of the Mach-number equation tends to $GM$. Since $f(\mathcal{M})\ge 4$, a real positive solution at arbitrarily small $r$ requiresUsing the value of $A$ givesThis is the critical Bondi accretion rate for gamma equals five thirds.

For $r\ll GM/B$, the Mach number tends to a constant ${\mathcal{M}}_1$. Since $A{\dot{M}}_{\mathrm{crit}}^{1/2}=GM/4$, its algebraic equation isor equivalentlyThe central Bernoulli balance givessoFinally $u={\mathcal{M}}_1{v}_s$. Using the algebraic relation to rewrite its coefficient gives

Hydrostatic equilibrium requiresFor the polytrope $p=K{\rho }^{1+1/n}$, the specific enthalpy isThus $h+\Phi$ is constant. Since $\rho =p=0$ at $r=R$, this constant is ${\Omega }^2{R}^2/2$, andfor $0\le r\le R$, with both fields zero outside the model star.

Let $\mathbf{\xi }$ be the Lagrangian displacement. Linearized mass conservation and adiabaticity giveThe fixed potential has no Eulerian perturbation, so the linearized momentum equation isFor the stated spherical harmonic decomposition,Taking radial and horizontal components and using $g={\Omega }^{2}r$ gives

Set $\Delta =0$ and take ${\xi }_r=C{r}^{\ell -1}$. The divergence equation then fixesHydrostatic balance gives $\delta p=-{\xi }_r{p}^{\prime }=\rho g{\xi }_r$. The horizontal momentum equation becomessoThe radial equation gives the same result because ${\omega }^2{\xi }_r=g{\xi }_r^{\prime }+g^{\prime }{\xi }_r$. This is an incompressible stellar surface mode: it changes the shape of the free surface without compressing fluid elements. For $\ell =1$ it is a rigid displacement of the star in the fixed harmonic potential.

The tidal potential is proportional to the solid spherical harmonic $r^{\ell }{Y}_{\ell }^m$, whose gradient has exactly the spatial form of the mode in part (c). Projecting the forced linear equation onto that eigenfunction gives the oscillator factor ${\omega }_{\ell }^2-{\omega }^2$. Consequentlyand its radial component isTaking the real part gives the stated physical displacement. The amplitude displays tidal resonance of a stellar oscillation: it is enhanced near $|\omega |={\omega }_{\ell }$ and formally diverges in this undamped linear model. Physical damping makes the peak finite and supplies a phase shift through resonance.

The adiabatic sound speed and Alfvén speed areThe entropy is held fixed in the sound-speed derivative.

A displacement perpendicular to the plane spanned by $\mathbf{k}$ and $\mathbf{B}$ has $\mathbf{k}\cdot \tilde{\mathbf{\xi }}={\mathbf{v}}_A\cdot \tilde{\mathbf{\xi }}=0$. The algebraic wave equation then gives the Alfvén waveThe remaining displacement lies in the $\mathbf{k}$-$\mathbf{B}$ plane. Setting the determinant of that two-dimensional system to zero givesIts larger root is the fast magnetosonic wave, in which gas and magnetic pressure act together. Its smaller root is the slow magnetosonic wave, whose motion is guided more strongly along the field. Both are compressive, whereas the Alfvén mode is transverse and incompressible.

With $\mathbf{B}=B{\mathbf{e}}_x$, $k^2=k_x^2+k_z^2$, $\omega =v{k}_x$, and $\mathbf{k}\cdot {\mathbf{v}}_A=k_x{v}_A$. Substitution into the magnetosonic polynomial and collection of the $k_z^2$ terms giveswhere the tube speed isLet $v_{<}^2=\min (v_s^2,v_A^2)$ and $v_{>}^2=\max (v_s^2,v_A^2)$. Since $v_T^2<v_{<}^2$, $k_z$ is real whenand imaginary whenThe endpoints are turning or degenerate cases.

For $\mathbf{B}=B{\mathbf{e}}_x$, the vertical magnetic perturbation isThe $z$ component of the Fourier-transformed equation of motion is thereforeUsing $\omega =v{k}_x$ gives

The interface is material in ideal magnetohydrodynamics, so fluid on it remains on it. Its normal displacement ${\xi }_z$ must consequently be the same when approached from either side. Integrating normal momentum balance through an infinitesimal pillbox shows that the Lagrangian total-pressure traction is continuous. Because the equilibrium total pressure is constant on each side and continuous at $z=0$, this reduces to continuity of the Eulerian perturbation $\delta \Pi$. Hence

Write $k_{z+}=+i{q}_{+}$ above the interface and $k_{z-}=-i{q}_{-}$ below it, with $q_{\pm }=|k_{z\pm }|>0$, so both waves decay away from $z=0$. A common interface displacement ${\xi }_0$ givesPressure continuity therefore requires

DefineSolving the matching condition gives the magnetohydrodynamic interface wave speedThe weights ${\Sigma }_{\pm }$ are positive, so $v^2$ is a weighted average of the two squared Alfvén speeds. It therefore lies between them, and so does the positive phase speed $v$.

A force-free magnetic field obeysequivalently $\mathbf{j}\times \mathbf{B}=0$: its current is parallel to its magnetic field. In a very low-density exterior, material pressure and inertia are too small to balance a finite Lorentz force. Quasistatic force balance therefore drives the magnetic field toward a force-free configuration.

The magnetic energy isUsing the ideal-MHD induction equation and integrating the identity $\mathbf{B}\cdot \nabla \times (\mathbf{u}\times \mathbf{B})$ by parts givesThe volume term vanishes for a force-free magnetic field, leaving

Axisymmetry makes the azimuthal direction geometrically distinct, so the field splits into a meridional poloidal part and an azimuthal toroidal part:The divergence-free condition on the poloidal field permits the poloidal magnetic flux functionor in cylindrical coordinatesThe magnetic flux through a circle of radius $r$ at fixed $z$ isThus, after choosing $\psi =0$ on the axis, $2\pi \psi$ is the enclosed poloidal flux and surfaces of constant $\psi$ are magnetic surfaces.

The toroidal component of $(\nabla \times \mathbf{B})\times \mathbf{B}=0$ givesBoth $\psi$ and $r{B}_{\varphi }$ are therefore constant along each poloidal field line, soThe poloidal components of the same force-free equation then reduce to the Grad-Shafranov equation for a force-free magnetic fieldThe poloidal current density isAmpère's law around an azimuthal circle gives the enclosed poloidal electric current:up to the orientation sign. Thus $F$ is the enclosed-current function in units ${\mu }_0/(2\pi )$.