Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 314 (Astrophysical Fluid Dynamics)

A magnetostatic force-free magnetic field is in equilibrium when its electric current exerts no Lorentz force density:Thus $\mathbf{J}$ is pointwise parallel to $\mathbf{B}$, so for some scalar $\alpha (\mathbf{r})$,Using the magnetostatic Ampère's law, $\nabla \times \mathbf{B}=(4\pi /c)\mathbf{J}$, givesTaking the divergence and using $\nabla \cdot \mathbf{B}=0$ yieldsThe force-free parameter is therefore constant along each magnetic field line.

Put $q(x)=4\pi \alpha (x)/c$. Because every component depends only on $x$, the solenoidal constraint gives $B_x^{\prime }=0$, while the force-free equation givesFor a general nonzero, nonconstant $q$, $B_x=0$. Eliminating $B_y=-B_z^{\prime }/q$ gives the closed equationor equivalently $(B_z^{\prime }/q{)}^{\prime }+q{B}_z=0$. Since $\nabla \alpha ={\alpha }^{\prime }(x){\mathbf{e}}_x$ and $B_x=0$, the constraint $\mathbf{B}\cdot \nabla \alpha =0$ is automatically satisfied.

Define the accumulated rotation angleThe coupled first-order equations describe a rotation of $(B_y,B_z)$ and have the general solutionDirect differentiation verifies both force-free equations, and $B_y^2+B_z^2=C_1^2+C_2^2$ is constant. Thus a one-dimensional force-free magnetic field rotates without changing its magnitude.

Hereafter choosing the integration constant so that $\theta (\pm \infty )=0$. The condition $B_z\to 0$ at both ends removes the cosine solution. Up to an overall sign and magnitude, one may writeThe angle ranges from $0$ to $I_{\max }=a^2/(2d^2)$. Neither nonzero component changes sign when the entire range lies in the first quadrant, namely $I_{\max }\le \pi /2$. Thereforefor positive length scales $a,d$. Equality allows $B_y$ to touch zero at $x=0$ without changing sign.

Project the steady Euler equations for an inviscid fluid along a streamline. With no gravity,For an isentropic fluid, $dp=c_s^{2}d\rho$, soThe scalar mass flux is $j=\rho u$, and henceIt increases with speed in subsonic flow and decreases with speed in supersonic flow.

The continuous transonic branch passes from $dj/du>0$ to $dj/du<0$, so $j(u)$ has a maximum where its derivative vanishes. At this critical speed of a polytropic flow,The maximum mass flux is therefore attained at the sonic point.

The Bernoulli function for steady unmagnetized flow without gravity isFor a polytropic equation of state, $h=c_s^2/(\gamma -1)$. Evaluation at the critical point found in part b givesConsequentlySince $C_B$ is conserved along a streamline, that streamline has a unique critical speed.

Across a stationary normal shock wave, the Rankine-Hugoniot conditions for a perfect gas conserveMomentum conservation and $1/\rho =u/j$ giveDirect elimination of the two pressures and densities from these three jump conditions givesPart c identifies the right side with the square of the critical speed, so the Prandtl shock relation isIt maps the unique upstream supersonic state on a given Bernoulli streamline to its downstream subsonic state.

Spherical symmetry reduces the Poisson equation for gravity toAfter one integration,The omitted integration constant would represent a point mass and is absent for the dark-matter configuration alone. Choosing the additive constant so that the potential tends to zero at large radius gives

Let $u>0$ denote inward radial speed. Steady spherical mass conservation and the radial momentum equation giveA regular sonic point requires both factors to vanish:so $c_{s,s}^2=4\pi G{\Psi }_0/\sqrt{r_s}$ and ${\Phi }_{\mathrm{dm}}(r_s)=-4c_{s,s}^2$.

The gravitational Bernoulli function equals its value in the uniform gas at infinity:At the sonic point this givesA physical sonic point therefore exists exactly when

Combining the two expressions for the sonic sound speed givesFor a polytropic gas, $c_s^2\propto {\rho }^{\gamma -1}$, and henceThe transonic accretion in a steep dark-matter cusp has mass accretion rateThus $\dot{M}\propto c_0^{-7}$, whereas classical Bondi accretion onto a point mass has ${\dot{M}}_{\mathrm{B}}\propto c_0^{-3}$. The steeper dependence comes from the cusp potential, whose sonic radius scales as $c_0^{-4}$ instead of $c_0^{-2}$.

The black-hole acceleration is $GM/r^2$, while the dark-matter acceleration is $8\pi G{\Psi }_0/r^{3/2}$. Their ratio decreases outwards throughout the subsonic region $r\ge r_s$, so it is enough to demand that the black hole be negligible at the sonic point:Equivalently,Under this condition the subsonic solution and its sonic transition are controlled primarily by the dark-matter cusp, justifying the neglect of the central point-mass gravity.

Use the material forms of the ideal magnetohydrodynamic induction equation and momentum equation:The magnetic force is perpendicular to $\mathbf{B}$. For $h_c=\mathbf{u}\cdot \mathbf{B}$, the two equations therefore giveSince $\nabla \cdot \mathbf{B}=0$, this is the cross-helicity conservation lawwithFor a homentropic flow, $\nabla s=0$, so the source vanishes.

If $\mathbf{u}$ is parallel to $\mathbf{B}$, write $\mathbf{u}=\kappa \mathbf{B}$. Thenand the flux from part a reduces towhere $C_B$ is the gravitational Bernoulli function. If it is constant along field lines, $\mathbf{B}\cdot \nabla C_B=0$, thenThe source also vanishes because the flow is homentropic. Hence ${\partial }_t{h}_c=0$, and every fixed volume satisfiesThus the cross-helicity is conserved even for this time-dependent aligned flow.