Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 314 (Astrophysical Fluid Dynamics)

Use the Euler equations for an inviscid fluid with no magnetic or gravitational force. Let $w$ denote specific enthalpy. For a homentropic flow, $dw=dp/\rho$, so the pressure acceleration is $-\nabla w$. The velocity identitytherefore givesTaking the curl, using that the curl of a gradient is zero and that differentiation commutes for smooth fields, proves barotropic vorticity transport:More generally the same proof works for any barotropic fluid, replacing $w$ by a pressure potential ${\int }^{\rho }{p}^{\prime }(s)ds/s$. Here “isentropic” must supply this barotropic closure, for example through one common specific entropy throughout the fluid. Merely imposing entropy advection equation $Ds/Dt=0$ allows spatial specific entropy gradients; in that case the vorticity equation contains the additional baroclinic vorticity generation term ${\rho }^{-2}\nabla \rho \times \nabla p$.

Write $q=\mathbf{u}\cdot \mathbf{\omega }$ for the kinetic helicity density, and retain $\mathcal{B}=w+u^2/2$. Since $\mathbf{\omega }$ is a curl, $\nabla \cdot \mathbf{\omega }=0$. Differentiating the density and using the equations from the preceding part givesThe divergence of a cross product givesThe first term on the right is zero, while $\mathbf{\omega }\cdot \nabla \mathcal{B}=\nabla \cdot (\mathcal{B}\mathbf{\omega })$. Thus the kinetic helicity conservation law has fluxThe last equality for the flux uses the vector triple product identity. Integrating and applying the divergence theorem gives $d{\int }_{V}qdV/dt=-{\int }_{\partial V}{\mathbf{F}}_{H_k}\cdot \mathbf{n}dS$ for a fixed volume. Hence its total kinetic helicity is conserved when the boundary flux vanishes; for example, tangency of both velocity field and vorticity to the boundary is sufficient. The local conservation law does not by itself imply that its density follows each fluid particle unchanged.

Expanding the kinetic helicity conservation law separates advection from the other flux:Thus material conservation of kinetic helicity density holds precisely whenAn especially useful sufficient pair of conditions is incompressible flow, $\nabla \cdot \mathbf{u}=0$, and constancy of $w-u^2/2$ along integral curves of the vorticity, $\mathbf{\omega }\cdot \nabla (w-u^2/2)=0$. Together these conditions make both terms vanish without cancellation. They are not necessary individually, because the two terms in the displayed balance can cancel. An irrotational flow is a trivial conserved case with $q=0$.

For contrast, combining the balance with mass conservation givesConsequently constancy of $w-u^2/2$ along integral curves of the vorticity makes $q/\rho$ an advected scalar even in compressible flow; it does not generally make $q$ itself constant.

Work in the shock frame, assume a calorically perfect gas with fixed specific-heat ratio $\gamma >1$, and neglect magnetic stresses and body forces across the thin shock. The internal energy per unit mass is $e=p/[(\gamma -1)\rho ]$, and the specific enthalpy is $w=\gamma p/[(\gamma -1)\rho ]$. Integrating the local conservation laws across a thin stationary control volume makes the mass flux $\rho u$, momentum flux $p+\rho u^2$, and energy flux $u(\rho u^2/2+\gamma p/(\gamma -1))$ continuous. Hence the Rankine-Hugoniot conditions for a perfect gas areMultiplication of the last equality by $m$ gives continuity of the stated energy flux. These express that mass, momentum and energy cannot accumulate in an infinitesimally thin steady shock, although entropy production occurs on the physical compression branch. The same constant in the polytropic equation of state $p=K{\rho }^{\gamma }$ must therefore not be imposed on both sides. For flow from $z>0$ to $z<0$, both signed velocities are negative; the flux equations and ratios remain valid.

In the Strong-shock Rankine-Hugoniot conditions, $p_1/({\rho }_1{u}_1^2)=1/(\gamma M_1^2)$ tends to zero. Put $r={\rho }_2/{\rho }_1$. The first two flux balances give $u_2=u_1/r$ and $p_2={\rho }_1{u}_1^2(1-1/r)$ to leading order. The energy balance becomesDiscarding the unchanged-state root $r=1$ gives the compressive shock compression ratio and the required limits:

The dimensions of the injected power and ambient mass density are $[\dot{E}]=M{L}^2{T}^{-3}$ and $[{\rho }_1]=M{L}^{-3}$. Requiring ${\dot{E}}^a{\rho }_1^b{t}^c$ to have the dimensions of length givesThus dimensional analysis gives the adiabatic superbubble similarity solutionThe dimensionless coefficient $\alpha$ can depend on $\gamma$ and the specified inner structure, but dimensions alone cannot determine it. Constant power differs from a fixed-energy explosion because the energy available at time $t$ is proportional to $t$. This scaling assumes that ambient pressure, gravity, radiative losses and the finite injection radius introduce no relevant competing scale.

The original PDF supplies the intended profiles $\rho ={\rho }_{1}f(\eta )$, $p={\rho }_1{\dot{R}}^{2}g(\eta )$ and $u=\dot{R}h(\eta )$; the TeX has corrupted the first two. At fixed $r$, ${\partial }_t\eta =-\eta \dot{R}/R$, and at fixed $t$, ${\partial }_r\eta =1/R$. The spherical continuity equation is thereforeSince $\dot{R}/R=3/(5t)$ is nonzero, the continuity equation for a superbubble similarity solution isThe ambient gas is at rest. In the locally stationary shock frame, its velocity is $-\dot{R}$; the downstream velocity is $u-\dot{R}$. Using the shock compression ratio from part (a) converts the downstream speed back to the laboratory frame and gives the inner shock boundary conditionsImmediately outside the shock, the corresponding unperturbed profiles are $f=1$, $g=0$ and $h=0$, with $g=0$ understood in the negligible-ambient-pressure limit.

The inner boundary must supply the injected power. If the source is idealized as a point and the profiles extend all the way inward, central energy input in a similarity solution requiresor equivalently ${\lim }_{\eta \downarrow 0}{\eta }^{2}h(f{h}^2/2+\gamma g/(\gamma -1))=125/(108\pi {\alpha }^5)$. A finite source region or an inner wind/contact region supplies a different matching description. One cannot impose a regular zero-flux centre on a source-free interior and still inject nonzero power.

The conservation of energy to use here is the total kinetic energy plus internal energy inside the shock, not just the internal energy of a uniform-pressure core. With no radiative loss and negligible initial ambient energy,Substitute the adiabatic superbubble similarity solution and define the finite positive dimensionless integralThen $\dot{E}t=4\pi {\rho }_1{\dot{R}}^2{R}^{3}I$. Since $\dot{R}=3R/(5t)$ and $R^5={\alpha }^5(\dot{E}/{\rho }_1)t^3$,This is the energy normalization of a continuously driven spherical shock. It fixes $\alpha$ once the similarity profiles are determined. If the profiles describe only the shocked ambient layer, the energy of the interior hot bubble must also be included before using this normalization. A radiatively cooled shell has a different energy budget; its familiar numerical coefficient should not be substituted into this loss-free model.

Set $A={\Phi }_0{R}_0^{\beta }<0$, $q={\lambda }^2$ and $s=R^2+q{z}^2$. The flattened power-law gravitational potential is $\Phi =A{s}^{-\beta /2}$. In axisymmetric cylindrical coordinates, the Poisson equation for Newtonian gravity givesThe first derivatives are ${\partial }_R\Phi =-\beta AR{s}^{-(\beta +2)/2}$ and ${\partial }_z\Phi =-\beta Aqz{s}^{-(\beta +2)/2}$. Differentiating again and collecting powers givesTherefore the required mass density isThis is a formal density for arbitrary $\lambda$, but a physical dark matter distribution must be nonnegative. The density positivity for a flattened power-law potential condition isNecessity follows by evaluating on the midplane and symmetry axis; sufficiency follows because both numerator coefficients are then nonnegative. In this range the origin is a locally integrable density cusp: the mass enclosed near radius $r$ scales as $r^{1-\beta }$ and tends to zero, so no point mass needs adding. The scale-free distribution has infinite total mass at large radius and represents an idealized background, not a finite isolated halo.

For the cold, non-self-gravitating disk, radial balance of a circular orbit is $R{\Omega }^2={\partial }_R\Phi (R,0)$. ThusThe displayed positive root chooses the rotation orientation; the opposite orientation has the negative of this angular frequency.

Use the poloidal magnetic flux function convention ${\mathbf{B}}_p=R^{-1}\nabla \Psi \times {\mathbf{e}}_{\varphi }$, so $2\pi \Psi$ is magnetic flux up to a reference constant. Its poloidal magnetic field components areThe prescribed surface flux therefore givesThe constant offset ${\Psi }_1$ does not affect either field component. If instead $\Psi$ denotes the full physical flux rather than flux divided by $2\pi$, both component formulas acquire a common $1/(2\pi )$ factor; the subsequent logarithmic derivative relation is unchanged.

The Gauss's law for magnetism constraint is $R^{-1}{\partial }_R(R{B}_R)+{\partial }_z{B}_z=0$. Since $\alpha$ is constant with radius on the surface, it gives the power-law poloidal field near a disk surface relationFor a smooth one-sided field above the surface, $B_z(R,z)=B_z(R,0)[1-(\delta -1)\tan \alpha z/R]+O(z^2)$. Thus the field initially decreases with height if $\delta >1$, increases if $0<\delta <1$, and has zero first height derivative if $\delta =1$.

For local cold magnetocentrifugal acceleration, the field must guide the gas and enforce approximate corotation with its disk footpoint in the sub-Alfvénic launching region. Let that footpoint be $R=R_f$, and keep its field-line angular velocity ${\Omega }_f=\Omega (R_f)$ constant along the field. In this rotating frame the relevant effective potential isThe footpoint is a critical point because ${\partial }_R\Phi (R_f,0)=R_f{\Omega }_f^2$ and ${\partial }_z\Phi (R_f,0)=0$. The meridional Hessian matrix there isTake distance $\ell$ along a smooth outward magnetic field line whose initial tangent is $(\sin \alpha ,\cos \alpha )$ in the $(R,z)$ plane. Since the gradient of $V$ vanishes at the footpoint, field-line curvature does not contribute to its second derivative. ConsequentlyA negative quadratic coefficient gives a downhill displacement and acceleration without thermal assistance. The magnetocentrifugal launching criterion in a flattened power-law potential is thereforeAngles are measured from the vertical, with $0\le \alpha <\pi /2$. Below the critical angle the footpoint is a local minimum along the field, so a cold particle cannot cross the initial potential barrier. At equality the quadratic test is marginal; higher derivatives and the unspecified field curvature can decide the outcome. The threshold is a local condition for launch, not proof that the wind escapes along an arbitrary global field geometry. The spherical point-mass limit $\lambda =1$, $\beta \to 1$ recovers ${\alpha }_{\mathrm{crit}}=30^{\circ }$, consistent with Ogilvie's discussion of cold disk winds, section 9.9.

For an axisymmetric vector field, write $\mathbf{B}=B_R{\mathbf{e}}_R+B_{\varphi }{\mathbf{e}}_{\varphi }+B_z{\mathbf{e}}_z$. Pure differential rotation givesThis vector has no azimuthal component and is independent of $\varphi$, so its curl has zero radial and vertical components. Its azimuthal component isThe final braces are $R\nabla \cdot \mathbf{B}=0$ by Gauss's law for magnetism. The ideal magnetohydrodynamic induction equation therefore yields axisymmetric magnetic winding:Differential rotation creates a toroidal magnetic field from the fixed poloidal magnetic field. A stationary field under the same purely rotational assumptions requires Ferraro's law of isorotation, ${\mathbf{B}}_p\cdot \nabla \Omega =0$.

Let $M$ be the central stellar mass and take positive Keplerian rotation, $\Omega (R)=\sqrt{GM/R^3}$. The initial poloidal magnetic field stays fixed by part (a), while $R{\Omega }^{\prime }(R)=-(3/2)\Omega (R)$. Integrating the toroidal induction equation with $B_{\varphi }(R,0)=0$ givesThe initial field is a solenoidal vector field on the disk region $R>0$, since ${\partial }_R(R{B}_R)=0$. Assuming $B_0\ne 0$, the amplitude ratio isIt reaches one atThus the equality time is the same fraction $1/(3\pi )\approx 0.106$ of the orbital period at every radius. If “orbital time” instead means ${\Omega }^{-1}$, the corresponding fraction is $2/3$. These are kinematic solutions with prescribed rotation and the neglected magnetic back-reaction.

In the Gaussian units of the paper, the total magnetic pressure is $p_B=(B_R^2+B_{\varphi }^2)/(8\pi )$. Define the initial plasma beta ${\beta }_p=8\pi p_0/B_0^2\gg 1$, assuming a nonzero initial field. The solution from part (b) givesFor any $t>0$ this ratio decreases strictly with $R$, tends to infinity as $R\downarrow 0$ and tends to $1/{\beta }_p<1$ as $R\to \infty$. The magnetic-pressure equality radius under Keplerian winding is therefore unique in the formal profile on $R>0$:The same expression proves $p_B>p$ for $R<R_{\mathrm{eq}}$ and $p_B<p$ for $R>R_{\mathrm{eq}}$. At equality, $|B_{\varphi }|/|B_R|=\sqrt{{\beta }_p-1}\gg 1$, justifying the toroidal-dominated approximationThe source's existence claim refers to the indefinitely extended radial profile and the stated neglect of back-reaction. For a disk with a finite inner and outer edge, an equality radius lies in the disk only while the displayed value lies between those edges. If the initial field were zero, no magnetic winding or equality radius would occur.