Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 54 (Astrophysical Fluid Dynamics)

Write $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ and $\mathbf{L}=(\nabla \times \mathbf{B})\times \mathbf{B}/{\mu }_0$. The continuity equation converts a material specific-energy balance into a conservative energy density balance. Dot the ideal magnetohydrodynamic momentum equation with $\mathbf{u}$, and use the time independence of the Newtonian gravitational potential:For the energy density $U=p/(\gamma -1)$ associated with internal energy, the adiabatic pressure equation givesThe ideal magnetohydrodynamic induction equation and the cross-product divergence identity give the magnetic energy balanceIndeed $(\mathbf{u}\times \mathbf{B})\cdot (\nabla \times \mathbf{B})=-{\mu }_0\mathbf{u}\cdot \mathbf{L}$. The magnetic work cancels the kinetic magnetic work. The remaining pressure terms are $-\nabla \cdot (p\mathbf{u})$. Adding all three balances proves ideal magnetohydrodynamic energy conservation:whereThe last term is the Poynting vector with the ideal electric field $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$. A time-dependent imposed potential would instead supply the source $\rho {\partial }_t\Phi$.

Use axisymmetry and $\nabla \cdot \mathbf{B}=0$. Since the purely azimuthal velocity has zero divergence, the ideal magnetohydrodynamic induction equation becomes ${\partial }_t\mathbf{B}=(\mathbf{B}\cdot \nabla )\mathbf{u}-(\mathbf{u}\cdot \nabla )\mathbf{B}$. Differentiation of the cylindrical unit vectors contributes to its azimuthal component, givingThe azimuthal component of the magnetic tension force isThere is no azimuthal pressure or gravitational force and no azimuthal advective acceleration for this motion. Because the poloidal magnetic field is divergence-free,These coupled induction and tension equations describe a torsional Alfvén wave.

The axial angular momentum density is $\rho R{u}_{\varphi }$. The second equation in part (b) writes its conservation law with magnetic axial angular momentum fluxThe nonmagnetic part of the energy flux is azimuthal. Using $(\mathbf{u}\times \mathbf{B})\times \mathbf{B}=\mathbf{B}(\mathbf{u}\cdot \mathbf{B})-\mathbf{u}{B}^2$, its poloidal part isThe ratio is therefore $\Omega$ wherever the compared component of the angular-momentum flux is nonzero; the proportionality remains meaningful at zero flux.

A fully steady magnetic configuration requires ${\partial }_t{B}_{\varphi }=0$, hence ${\mathbf{B}}_p\cdot \nabla {\Omega }_0=0$: angular velocity is constant along poloidal field lines. This is Ferraro's law of isorotation. The steady azimuthal force additionally requires ${\mathbf{B}}_p\cdot \nabla (R{B}_{\varphi })=0$, which holds, for example, if $B_{\varphi }=0$. The remaining meridional force balance is a separate equilibrium condition.

Since $\rho$ and ${\mathbf{B}}_p$ are time independent, differentiate the angular-momentum equation once more and substitute the induction equation:This is the variable-coefficient wave equation for the torsional Alfvén wave.

In the local short-wavelength approximation, derivatives of the slowly varying coefficients are smaller than derivatives of the phase. For the perturbation $v{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, replace ${\partial }_t$ by $-i\omega$ and ${\mathbf{B}}_p\cdot \nabla$ by $i\mathbf{k}\cdot {\mathbf{B}}_p$ in the torsional Alfvén wave equation. Cancel the common nonzero amplitude and $R^2$ to obtainThus the local dispersion relation is an Alfvén wave relation with Alfvén velocity ${\mathbf{v}}_{A,p}={\mathbf{B}}_p/\sqrt{{\mu }_0\rho }$. The approximation requires wavelength small compared with the background variation scales. A wave vector exactly perpendicular to ${\mathbf{B}}_p$ gives zero leading frequency, so cannot simultaneously obey the assumed large-frequency limit.

The magnetic field satisfies $\nabla \cdot \mathbf{B}=0$. Expanding the divergence of its Maxwell stress tensor givesFor the Newtonian gravitational field $\mathbf{g}=-\nabla \Phi$, Poisson equation for Newtonian gravity gives ${\partial }_j{g}_j=-4\pi G\rho$, and the gradient representation gives ${\partial }_j{g}_i={\partial }_i{g}_j$. ConsequentlyThe negative of the Newtonian gravitational stress tensor, in this force-stress convention, therefore supplies $\rho g_i$. Adding the pressure stress supplies $-{\partial }_{i}p$, soEach summand is symmetric. Absence of external gravitational sources is needed to represent the full gravitational force by this self-gravitating stress.

Apply the continuity equation and integrate by parts, with the stated vanishing boundary terms and finite moments. The second mass moment tensor satisfiesand differentiation again givesInsert the stress-divergence equation from part (a). The two force integrals becomeThe first term is $4K_{ij}$. Hence the magnetized-fluid tensor virial theorem isHere $\mathcal{T}$ is the volume-integrated stress, whose sign differs from some gravitational potential-energy tensor conventions. The derivation also requires the advective mass-moment surface terms to vanish; this is automatic for an isolated sufficiently decaying configuration.

At the initially resting instant $K_{ij}=0$, and the cold-fluid assumption removes the pressure stress. Sum the $xx$ and $yy$ components of the magnetized-fluid tensor virial theorem. The magnetic and gravitational traces areSince $I_{\perp }=I_{xx}+I_{yy}$ and $g_z=-{\partial }_z\Phi$, this yields the horizontal virial balance of a cold magnetized fluid:Being at rest sets the instantaneous velocity to zero; it does not assert equilibrium or zero acceleration. The fields throughout space contribute to this stress integral, including their vacuum exterior.

Integrate Poisson equation for Newtonian gravity through a narrow slab around the disk. The horizontal derivative contributions vanish as its thickness tends to zero, leaving the normal-derivative jumpThe even function symmetry of $\Phi$ makes the derivatives opposite, soIn the current-free simply connected upper half-space, Ampère's circuital law gives $\nabla \times \mathbf{B}=0$ and permits a magnetic scalar potential. Rescale it so that $\sqrt{4\pi G/{\mu }_0}\mathbf{B}=-\nabla {\Phi }_M$. The divergence-free condition makes ${\Phi }_M$ satisfy Laplace's equation, withCompare with the gravitational jump condition. Subject to the same isolated-field boundary condition at infinity, ${\Phi }_M$ is the harmonic potential of the effective surface densityThis gravity-equivalent magnetic surface density can have either sign; it is a mathematical representation of the exterior magnetic field, not physical negative mass. An imposed nondecaying field would require additional boundary data and would not be fixed by the disk surface density alone.

The hypothesis is $|{\Sigma }_M|<\Sigma$, with the physical surface density nonnegative. At height $z>0$, the vertical derivative of an isolated thin-disk gravitational potential has a strictly positive kernel:The same formula with ${\Sigma }_M$ represents ${\partial }_z{\Phi }_M$. The triangle inequality and the strict surface density bound therefore give the positive-kernel comparison of thin-disk fieldsUse ${\partial }_z{\Phi }_M=-\sqrt{4\pi G/{\mu }_0}{B}_z$. The integrand in part (c) is then $({\partial }_z{\Phi }_M{)}^2-({\partial }_z\Phi {)}^2<0$ in the upper vacuum region. Reflection symmetry gives the same result below; the infinitesimally thin disk has zero three-dimensional volume. Thus, with the finite-integral assumptions of the tensor virial theorem,Since the disk starts at rest, ${\dot{I}}_{\perp }=0$ initially and its radial second moment begins to decrease. This is magnetic subcriticality of a razor-thin disk: magnetic support cannot prevent initial contraction in the global virial sense. The conclusion concerns the mass-weighted radial size, and does not by itself prove that every fluid element accelerates inward or that contraction continues indefinitely.

For steady spherical polytropic flow, write $u=u_r$ and use the polytropic equation of state $p=K{\rho }^{\gamma }$, with $K>0$. Mass conservation and radial Euler momentum equation areThe mass equation gives ${\rho }^{\prime }/\rho =-2/r-u^{\prime }/u$. Substitute it into momentum balance to obtainAt a sonic point the derivative coefficient vanishes. A smooth finite-slope solution must make the numerator vanish there too:Finally $dp/\rho =d[\gamma K{\rho }^{\gamma -1}/(\gamma -1)]$. Integrating momentum gives the Bernoulli equationThis is kinetic energy plus specific enthalpy plus gravitational potential per unit mass. The polytropic stellar wind selects a particular transonic branch of these equations.

The outward mass loss rate is $\dot{M}=4\pi r^2\rho u>0$. Put $\beta =4/(\gamma +1)$, $\alpha =4(\gamma -1)/(\gamma +1)$ and $y=\rho r^{\beta }$. Both terms in the Bernoulli equation then have the same radial scaling: $\beta (\gamma -1)=4-2\beta =\alpha$. Eliminate $u$ and multiply by $r^{\alpha }$ to obtainFor positive terminal speed, $C>0$. Differentiation gives $F^{\prime }=r^{\alpha -2}[\alpha Cr-(1-\alpha )GM]$. It has an interior minimum precisely when $0<\alpha <1$, or $1<\gamma <5/3$, atAt a sonic point, the Bernoulli equation and $GM/r_c=2c_{s,c}^2$ give $C=c_{s,c}^2(5-3\gamma )/[2(\gamma -1)]$, recovering this radius. The limiting $C=0$, $\gamma =5/3$ case has no isolated interior minimum selected by these formulas.

The function $G$ diverges at both ends of $y>0$ and has a unique minimum whereThis condition is $u_c^2=c_{s,c}^2$. Thus a smooth crossing requires minimum matching at a polytropic sonic point: $F(r_c)=G(y_c)$. If the minimum of $F$ is lower, an interval has no real positive solution; if it is higher, the two $y$ branches stay separate and cannot cross the sonic value. When the minima agree, their positive second derivatives give $G^{\prime \prime }(y_c)(y-y_c{)}^2\sim F^{\prime \prime }(r_c)(r-r_c{)}^2$, allowing a finite-slope branch to pass between them. The wind with finite positive terminal speed takes the lower-$y$ branch outside $r_c$, because $\rho \sim \dot{M}/(4\pi \sqrt{2C}{r}^2)$ and $\beta <2$ make $y\to 0$.

At the matching point,Insert these into $\dot{M}=4\pi r_c^2{\rho }_c{c}_{s,c}$:The positive-$C$, $1<\gamma <5/3$ range is part of this finite-radius transonic wind result, rather than an unrestricted formula for every $\gamma >1$.

For cylindrical rotation $u_{\varphi }=R\Omega (R)$, the equilibrium Euler momentum equation is $-R{\Omega }^2{\mathbf{e}}_R=-\nabla p/\rho -\nabla \Phi$. HenceThe integral is the centrifugal potential of cylindrical rotation. On a regular barotropic fluid branch define the specific enthalpy $h(\rho )$ by $dh=dp/\rho$. ThenFor a regular equation of state with $dp/d\rho >0$, $h$ is invertible, so $\rho$ and $p=p(\rho )$ depend only on $\Psi$ within a connected equilibrium region. In particular $dp/d\Psi =-\rho$. This local conclusion uses an invertible barotropic branch; distinct disconnected fluid regions can have different integration constants.

The linearized azimuthal Euler momentum equation isFor the time dependence $e^{-i\omega t}$ and $u_R^{\prime }=-i\omega {\xi }_R$, it givesThis expresses conservation of the displaced element's specific angular momentum. It applies directly to nonzero-frequency modes, with the zero-frequency limit taken in the displacement formulation.

The radial advective acceleration supplies $-2\Omega v_{\varphi }$, while the pressure force perturbation is $\delta \rho \nabla p/{\rho }^2-\nabla \delta p/\rho$. Eliminate $v_{\varphi }$ and retain the vertical equation. Under the Cowling approximation, $\delta \Phi =0$, soThe continuity equation gives $\delta \rho =-\nabla \cdot (\rho \mathbf{\xi })$. The Lagrangian adiabatic relation $\Delta p/p=\gamma \Delta \rho /\rho$, with $\Delta \rho =-\rho \nabla \cdot \mathbf{\xi }$, givesHere $\kappa$ is the radial epicyclic frequency. These formulas define the axisymmetric adiabatic displacement operator.

Let $a_{\xi }=\mathbf{\xi }\cdot \nabla \Psi$, $d_{\xi }=\nabla \cdot \mathbf{\xi }$ and ${\rho }_{\Psi }=d\rho /d\Psi$. Since $\nabla p=-\rho \nabla \Psi$, the pressure and mass density perturbations are $\delta p_{\xi }=\rho a_{\xi }-\gamma p{d}_{\xi }$ and $\delta {\rho }_{\xi }=-\rho d_{\xi }-{\rho }_{\Psi }{a}_{\xi }$. Integrating the pressure-gradient term by parts in the mass density-weighted inner product givesThe boundary term vanishes for regular admissible displacements because $p,\rho$ vanish there and $\delta p=\rho a-\gamma pd$. Completing the pressure square giveswhereThis effective-potential stratification coefficient is real. All coefficients of the bilinear form are real, so $⟨\eta ,\mathcal{F}\xi ⟩=⟨\mathcal{F}\eta ,\xi ⟩$: the operator is symmetric on the stated boundary domain, giving the usual self-adjoint realization of the stellar normal-mode problem.

Set $\eta =\xi$ and use $\mathcal{F}\xi =-{\omega }^2\xi$. The Cowling energy principle for a rotating barotropic star isThe Rayleigh quotient $Q[\xi ]/\int \rho |\xi {|}^2$ tests stability. If $Q\ge 0$ for every admissible displacement, no mode has ${\omega }^2<0$, so there is no exponentially growing mode. If an admissible trial displacement has $Q<0$, the Rayleigh-Ritz variational principle puts negative spectrum below zero; in the usual discrete stellar mode problem this gives a mode with $\omega =i\sigma$, $\sigma >0$. Equality allows neutral modes, rather than establishing strictly positive frequencies. A locally negative coefficient alone is not a complete instability proof: a trial function must also control the pressure and other positive terms.

The pressure-square term in the Cowling energy principle for a rotating barotropic star is nonnegative, and ${\mathcal{N}}^2>0$ makes its stratification term nonnegative. Write $j=R^2\Omega$ for specific angular momentum. ThenFor a regular star reaching the rotation axis, $j(0)=0$. The given $dj/dR>0$ therefore makes $j>0$ for $R>0$, and consequently ${\kappa }^2>0$. Equivalently, with the usual nonnegative angular-velocity convention the sign follows directly. All three energy terms are nonnegative, soThis is stability within the Cowling approximation used throughout. The orientation-independent rotational condition is $d(j^2)/dR\ge 0$, the Rayleigh discriminant criterion. If a fluid region excludes the axis and negative angular velocity is allowed, $dj/dR>0$ alone needs the additional sign of $j$; the squared-angular-momentum criterion avoids that ambiguity.