Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 314 (Astrophysical Fluid Dynamics)

In a static spherical star, hydrostatic equilibrium and the Poisson equation giveor equivalently $r^{-2}d(r^{2}g)/dr=4\pi G\rho$.

Let $\mathbf{\xi }$ be the fluid displacement field, so the velocity perturbation is $\delta \mathbf{u}={\partial }_t\mathbf{\xi }$. Linearizing the ideal-fluid momentum equation about the static state and cancelling the background hydrostatic terms givesConservation of mass says that the Lagrangian density perturbation is ${\Delta }_L\rho =-\rho \nabla \cdot \mathbf{\xi }$. The relation between Eulerian and Lagrangian fluid perturbations and adiabatic compression gives, with $\Delta =\nabla \cdot \mathbf{\xi }$,while linearized self-gravity gives $\boxed{{\nabla }^2\delta \Phi =4\pi G\delta \rho }$.

For the stated spherical harmonic displacement, the radial divergence is $r^{-2}d(r^2{\tilde{\xi }}_r)/dr$, while $\nabla {\tilde{\xi }}_h$ is radial and orthogonal to the angular gradient of $Y_l^m$. HenceEquating the radial and horizontal coefficients of $Y_l^m$ and $\nabla Y_l^m$, and applying the separated Laplacian in spherical coordinates to $\widetilde{\delta \Phi }(r)Y_l^m$, gives

Define the stellar buoyancy frequency byEliminating $\tilde{\Delta }$ between the density and pressure perturbations givesSubstitution in the radial equation, followed by use of $dp/dr=-\rho g$, yieldsRegular spherical profiles near the center have $\rho ={\rho }_c+O(r^2)$ and $p=p_c+O(r^2)$, while $m(r)=4\pi {\rho }_c{r}^3/3+O(r^5)$ and hence $g=O(r)$. Both logarithmic gradients in the definition of $N^2$ are $O(r)$, so $N^2=A{r}^2+O(r^4)$. A Sun-like radiative central stratification is stable, making $A>0$.

For one-dimensional compressible flow, define the total energy densityThe conservative mass, momentum, and energy equations areA monatomic nonrelativistic perfect gas has $\gamma =5/3$, an ultrarelativistic gas or radiation-dominated fluid has $\gamma =4/3$, and a rotationally active diatomic gas has $\gamma =7/5$.

Integrating the three conservation laws across a stationary normal shock wave gives the Rankine-Hugoniot conditions for a perfect gasEliminate $u_{x2}$ with the conserved mass flux and write $v_s^2=\gamma p/\rho$ and $M_x=u_x/v_s$. Solving the remaining two algebraic equations givesFor a compressive shock, $M_{x1}>1$, the density and pressure increase, and the downstream normal flow is subsonic in the shock frame.

For an oblique shock, boost parallel to the front by the upstream tangential speed. The transformed upstream flow is normal, so part b applies. Inviscid momentum balance has no tangential stress and therefore makes the tangential velocity continuous. Transforming back giveswhile all normal velocity, density, and pressure relations from part b remain unchanged with $M_x$ the normal Mach number.

The pressure jump relation with $p_2/p_1=1+ϵ$ givesIf ${\beta }_1$ is the angle between the upstream velocity and the shock front, then $M_{x1}=M_1\sin {\beta }_1$. To leading order,the Mach angle. Expansion of the compression ratio givesBecause $u_y$ is continuous and $\tan \beta =u_x/u_y$,Writing $\theta ={\beta }_1-{\beta }_2$ and linearizing the tangent about $\sin {\beta }_1=1/M_1$ gives the weak-oblique-shock deflectionThe normal component decreases while the tangential component is unchanged, so ${\beta }_2<{\beta }_1$: the flow turns toward the shock front.

Since $\nabla \varphi ={\mathbf{e}}_{\varphi }/r$, the vector field has cylindrical componentsThe divergence in cylindrical coordinates is thereforeDirect use of the curl in cylindrical coordinates givesandThus, definingwe obtain

For the stated magnetic vector potential, part a givesApplying the same identity once more and using the Ampère-Maxwell equation in the magnetostatic limit givesExpanding the Lorentz force density ${\mathbf{F}}_m=\mathbf{J}\times \mathbf{B}$, using $\nabla \varphi ={\mathbf{e}}_{\varphi }/r$ and the vector triple-product identity, separates its poloidal and azimuthal parts:where

In an axisymmetric magnetostatic state, pressure and gravitational forces are poloidal, so the azimuthal component of the Lorentz force density must vanish. Hence $f=0$ andThe two gradients are locally parallel, so $L\alpha$ is constant on each regular level surface of $\beta$. Thereforefor an arbitrary flux function $F$.

For a barotropic fluid, $\nabla p=\rho \nabla h$. Magnetostatic force balance and $L\alpha =F(\beta )$ giveThe left side is a gradient multiplied by $\rho$, so taking the curl shows that $(L\beta -F{F}^{\prime })/(r^2\rho )$ is constant on each $\beta$ surface. Absorbing the fixed factor ${\mu }_0$ into an arbitrary function $G$ gives the Axisymmetric magnetostatic Grad-Shafranov system

The divergence-free poloidal field can be represented by the poloidal magnetic flux functionIn a steady axisymmetric ideal magnetohydrodynamics flow, the azimuthal component of $\nabla \times (\mathbf{u}\times \mathbf{B})=0$ makes ${\mathbf{u}}_p$ parallel to ${\mathbf{B}}_p$. WriteMass conservation and $\nabla \cdot \mathbf{B}=0$ then imply $\mathbf{B}\cdot \nabla k=0$, so the magnetohydrodynamic mass loading $k=k(\psi )$ is constant along each magnetic line.

The poloidal part of $\mathbf{u}\times \mathbf{B}$ isIts curl vanishes only if its coefficient is a flux function, giving the field-line angular velocityThe azimuthal component of momentum conservation is a divergence of matter and magnetic angular-momentum flux. Dividing its field-line constant by the mass loading yields the magnetohydrodynamic angular-momentum invariantThe conservative total-energy equation similarly gives the magnetohydrodynamic Bernoulli invariantFinally, the entropy advection equation and ${\mathbf{u}}_p\parallel {\mathbf{B}}_p$ imply $s=s(\psi )$. Thus $k,\omega ,\ell ,ϵ$, and $s$ are constant along each magnetic field line.

Introduce the squared poloidal Alfvén numberSolving the two linear azimuthal invariants givesAt an Alfvén surface, $M_A^2=1$. Smooth passage through the apparent singularity requires both numerators to vanish at the same cylindrical radius $r_A$, giving the Alfvén-surface regularity condition for an axisymmetric windThe finite limiting values of $u_{\varphi }$ and $B_{\varphi }$ then follow by l'Hopital's rule from the local variation of $M_A$ and $r$ along the field line; the algebraic invariants alone fix their combination rather than each value separately at the critical point.

Along the open line, $B_p\sim F/r^2$ and the mass-loading relation with $u_p\to u_{\infty }$ givesThe solutions of part b consequently haveThe azimuthal Alfvén speed therefore approaches the nonzero constantFor an unconfined outflow it is natural to take $\Phi \to 0$ and $h\to 0$ at infinity; also $u_{\varphi }\to 0$. The magnetic term in the magnetohydrodynamic Bernoulli invariant tends toHence the asymptotic energy of a radial magnetohydrodynamic wind is