Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 52 (Astrophysical Fluid Dynamics)

Write $\mathbf{u}={\partial }_t\mathbf{\xi }$. In the uniform equilibrium, the linearized ideal magnetohydrodynamic equations reduce toIntegrating from the undisplaced reference state, and using the uniform background magnetic field, givesThese are the perturbations induced by the fluid displacement; independent time-independent changes to the reference state are excluded. The linear Lorentz force density is $(\nabla \times \delta \mathbf{B})\times \mathbf{B}/{\mu }_0$. Its magnetic pressure and magnetic tension parts giveFor a Fourier mode, put $s=\mathbf{k}\cdot \mathbf{\xi }$, $q=\mathbf{B}\cdot \mathbf{\xi }$ and $K=\mathbf{k}\cdot \mathbf{B}$. Then $\delta p=-i\gamma ps$ and $\delta \mathbf{B}=i(K\mathbf{\xi }-\mathbf{B}s)$. Substitution givesIf $s=q=0$, the fluid displacement is perpendicular to both the wave vector and the magnetic field. Only magnetic tension restores it, andThis is the Alfvén wave. The Alfvén velocity is the field-directed propagation vector: $\omega =\pm \mathbf{k}\cdot {\mathbf{v}}_a$. The signed phase velocity normal to the wavefront is $\pm v_a\cos \theta \hat{\mathbf{k}}$; the group velocity is $\pm {\mathbf{v}}_a$. The distinction matters for oblique propagation. For generic directions the Alfvén wave has one transverse polarization, with $\delta \rho =\delta p=0$.

To obtain the other magnetohydrodynamic waves, take the scalar products with $\mathbf{k}$ and $\mathbf{B}$:The determinant condition for a nonzero pair $(s,q)$ isThus the fast magnetosonic wave and slow magnetosonic wave haveTheir fluid displacements lie in the plane of $\mathbf{k}$ and $\mathbf{B}$ and are generally compressive. In the fast magnetosonic wave, gas and magnetic pressure provide the stronger restoring combination; in the slow magnetosonic wave, their perturbations oppose one another. The phase speeds depend on direction. They satisfy $v_{\mathrm{sl}}\le \min (c_s,v_a)$ and $v_{\mathrm{f}}\ge \max (c_s,v_a)$. For parallel propagation the two speeds are $c_s$ and $v_a$, with a degeneracy between a transverse branch and the Alfvén wave. For perpendicular propagation $v_{\mathrm{f}}^2=c_s^2+v_a^2$ and $v_{\mathrm{sl}}=0$; the latter is a nonpropagating limiting disturbance. These special directions require interpreting the polarizations by continuity rather than assuming three distinct nonzero frequencies.

For this geometry $\nabla \cdot \mathbf{u}=0$, $\nabla \cdot \mathbf{B}=0$ and $\mathbf{u}\cdot \nabla =0$. Consequently constant mass density and pressure satisfy the full nonlinear continuity and adiabatic pressure equations. Write ${\mathbf{B}}_{\perp }=(B_x,B_y)$ and ${\mathbf{u}}_{\perp }=(u_x,u_y)$. The ideal magnetohydrodynamic induction equation and the transverse ideal magnetohydrodynamic momentum equation becomeThe longitudinal momentum equation isThus constant transverse magnetic magnitude eliminates the otherwise unavoidable magnetic pressure acceleration. Differentiating the ideal magnetohydrodynamic induction equation in time gives, for either transverse component,A constant-magnitude nonlinear Alfvén wave is obtained by taking $B_z\ne 0$ andwhere $A$ and ${\mathbf{U}}_0$ are constants and $F$ is any sufficiently differentiable real function. Both first-order equations hold, and $B_x^2+B_y^2=A^2$. Taking $F(\zeta )=k\zeta$ gives an exact finite-amplitude circularly polarized wave. More general traveling rotations also work; an arbitrary superposition of oppositely traveling solutions of the wave equation need not preserve transverse magnetic magnitude and hence need not solve the full nonlinear system. If $B_z=0$, the coupled first-order equations instead require time-independent transverse fields and velocities, with spatially constant transverse magnetic magnitude; there is no propagating Alfvén wave in the $z$ direction.

Let $e=\rho u^2/2+p/(\gamma -1)$ be the total gas energy density, with $\gamma >1$. The one-dimensional continuity and momentum equations areUsing continuity to put the momentum equation into conservative form givesThe internal energy density $e_{\mathrm{int}}=p/(\gamma -1)$ satisfiesby the adiabatic pressure equation. Multiplying the momentum equation by $u$ and using continuity givesAdding these equations combines the pressure work into $-{\partial }_z(pu)$. The conservation law variables and conservation law fluxes are thereforeThe three rows express conservation of mass, momentum and total energy. Their integral conservation laws remain meaningful across a shock wave, where the differential pressure and velocity equations cannot be applied pointwise.

Integrating each conservation law across a vanishingly thin interval around the stationary shock wave leaves equal conservation law fluxes on its two sides. HenceThese are the Rankine-Hugoniot conditions for a perfect gas. Signed velocities may both be negative when the material travels from positive to negative $z$; no sign change is needed in the conservation law fluxes.

Put $r={\rho }_2/{\rho }_1=u_1/u_2$, $P=p_2/p_1$ and $m=M_1^2={\rho }_1{u}_1^2/(\gamma p_1)$. Momentum conservation givesDivide the energy condition by the nonzero mass flux. Equality of kinetic energy plus specific enthalpy givesSubstituting $P$ and multiplying by $2(\gamma -1)r^2/\gamma$ yieldsThe factor $r-1$ is the continuous, no-shock solution. On the nontrivial normal shock wave branch,An admissible compressive gas shock wave has $M_1>1$, so $r>1$ and $P>1$. The algebraic jump equations alone also allow a reversed expansive discontinuity; the entropy production in a perfect-gas shock excludes that branch. At $M_1=1$ the nontrivial branch joins the continuous solution.

Let $D=u_1>0$ denote the front speed in the stationary upstream frame. In the shock frame, the upstream and downstream velocities are $-D$ and $-D/r$. Taking $M_1\to \infty$ in the Rankine-Hugoniot conditions for a perfect gas givesThe downstream laboratory velocity is $D-D/r=2D/(\gamma +1)$, which distinguishes the gas speed from the front speed.

For the planar blast-wave energy scaling of a self-similar blast wave, integration of the total energy density over the shocked interval $0<z<Z$ gives the energy per unit area on this side:The similarity solution makes $I$ time-independent; a finite positive explosion energy requires $0<I<\infty$. Conservation of energy therefore gives $\dot{Z}=(E_{+}/({\rho }_{1}I){)}^{1/2}{Z}^{-1/2}$ for the expanding front. Integrating from $Z(0)=0$,If $E$ denotes the one-sided energy, $C=(9/(4I){)}^{1/3}$. If the released energy $E$ feeds two symmetric fronts, $E_{+}=E/2$ and $C=(9/(8I){)}^{1/3}$. In either convention the requested scaling is $Z=C(E/{\rho }_1{)}^{1/3}{t}^{2/3}$. The constant depends on the similarity profiles and the energy convention; energy conservation determines the exponent without solving those profiles. The Strong-shock Rankine-Hugoniot conditions additionally fix $f(1)=(\gamma +1)/(\gamma -1)$ and $g(1)=h(1)=2/(\gamma +1)$.

In a steady flow the ideal magnetohydrodynamic induction equation gives $\nabla \times (\mathbf{u}\times \mathbf{B})=0$. Since both vectors have only poloidal magnetic field components, write $\mathbf{u}\times \mathbf{B}=F_{\varphi }{\mathbf{e}}_{\varphi }$. Axisymmetry impliesFor a field regular on the axis, $C=0$; equivalently one may impose zero toroidal electromotive force. Under this physical condition, $\mathbf{u}$ is parallel to $\mathbf{B}$. Write $\rho \mathbf{u}=k\mathbf{B}$. Continuity and $\nabla \cdot \mathbf{B}=0$ then give $\mathbf{B}\cdot \nabla k=0$. The poloidal magnetic flux function has $\mathbf{B}\cdot \nabla \psi =0$, so locally on each connected magnetic surface,This regular-axis alignment of steady poloidal ideal flow needs its zero-circulation condition: axisymmetry alone does not imply this alignment on a domain excluding the axis. An explicit counterexample is available in a cylindrical annulus. Take nonzero constants $U,b$, positive ${\rho }_c,c_s$, andHere $\psi =-bz$, $\nabla \cdot \mathbf{B}=\nabla \times \mathbf{B}=0$ and $\mathbf{u}\times \mathbf{B}=Ub{\mathbf{e}}_{\varphi }/R$ has zero curl. The flow satisfies continuity, $p^{\prime }=-\rho {\Phi }^{\prime }$, and ${\nabla }^2\Phi =4\pi G\rho$. Its acceleration and Lorentz force density vanish, so it solves the steady equations while $\mathbf{u}$ and $\mathbf{B}$ are perpendicular. The nonzero toroidal circulation and the axial singularity explain why this counterexample is excluded from a regular polar accretion column. The remaining derivation uses the regular, aligned branch.

For a narrow stream tube, integrating continuity gives constant $A\rho |\mathbf{u}|$. On a nonzero-flow tube the magnetohydrodynamic mass loading $k$ is constant, so $A\rho |\mathbf{u}|=A|k||\mathbf{B}|$. Therefore $A|\mathbf{B}|$ is constant and $A\propto |\mathbf{B}{|}^{-1}$. This also follows directly from conserved magnetic flux through a flux tube.

For constant isothermal sound speed, $\nabla p/\rho =c_s^2\nabla \ln (\rho /{\rho }_0)$, with arbitrary reference density ${\rho }_0$. Project momentum along $\mathbf{B}$. The Lorentz force density has no component in that direction, and alignment implies $\mathbf{B}\cdot (\mathbf{u}\cdot \nabla )\mathbf{u}=\mathbf{B}\cdot \nabla (|\mathbf{u}{|}^2/2)$. Thus the isothermal magnetic Bernoulli integral isChanging ${\rho }_0$ only shifts the Bernoulli function by a constant. Since $k$ is constant along the magnetic field,Differentiating the isothermal magnetic Bernoulli integral therefore gives

In the slender polar accretion column, the dipolar flux-tube area is $A(z)=A_{\ast }(z/R_{\ast }{)}^3$. The decreasing axial field is a leading approximation, not an exactly solenoidal field $(0,0,B_z(z))$ throughout a cylinder. Indeed $\nabla \cdot \mathbf{B}={\partial }_z{B}_z\ne 0$ for that literal field. Near the axis a small radial component $B_R\simeq -(R/2){\partial }_z{B}_z=3R{B}_z/(2z)$ supplies the required radial divergence. Its magnitude is smaller by $R/z$, even though its divergence is leading order. Keeping this expanding flux tube while neglecting transverse forces yields the intended one-dimensional model.

Use the positive inward speed $v=-u_z$; the signed sonic velocity is $u_z=-c_s$. Conservation of mass and the Bernoulli equation giveEliminating ${\rho }^{\prime }$ yields the isothermal dipolar accretion equation,A smooth transonic branch requires both sides to vanish at its sonic point, soThis is the critical point of the flow equation; a generic subsonic solution need not cross it. A crossing in the exterior column requires $z_s\ge R_{\ast }$, and a sonic point strictly outside the star requires $z_s>R_{\ast }$. Differentiating the equation at the crossing gives $2(v_s^{\prime }{)}^2=3c_s^2/z_s^2$. For inward accretion that accelerates toward the star, $v_s^{\prime }=-\sqrt{3/2}{c}_s/z_s$.

The reservoir boundary condition fixes $ϵ=c_s^2\ln ({\rho }_{\infty }/{\rho }_0)$. At the sonic point, ${\Phi }_s=-3c_s^2$ and $v_s^2/2=c_s^2/2$, givingFinally use the sonic mass flux through the dipolar flux-tube area:This rate belongs to the smooth transonic branch of the idealized column, with $A_{\ast }$ the supplied total loaded area. The reservoir condition alone does not force every steady solution onto this branch. A real narrow dipolar column must match an outer flow where the slender approximation ceases to apply.

Let primes denote radial derivatives of the equilibrium and put $\Delta u=u^{\prime }+2u/r$. The linearized continuity, adiabatic pressure and Poisson equation areTake the time derivative of the Poisson equation and substitute continuity. Integration in radius givesCentre regularity excludes a changing point mass at the origin, so $C(t)=0$ andThe perturbed self-gravity must be retained; this calculation does not make the Cowling approximation.

The radial linear momentum equation is $\rho {\partial }_{t}u=-{\partial }_r\delta p-\delta \rho {\Phi }^{\prime }-\rho {\partial }_r\delta \Phi$. Differentiate it in time and use the preceding equations:The equilibrium has $p^{\prime }=-\rho {\Phi }^{\prime }$ and ${\Phi }^{\prime \prime }+2{\Phi }^{\prime }/r=4\pi G\rho$. The remaining terms simplify asConsequently the radial stellar oscillation equation isThe derivative acts on the full variable coefficient $\gamma (r)p(r)$; discarding ${\gamma }^{\prime }$ would change the result.

For $u=r\xi (r)e^{-i\omega t}$, $\Delta u=(3\xi +r{\xi }^{\prime })e^{-i\omega t}$. Substitute and multiply by $-r^3$. Combining the two terms proportional to ${\xi }^{\prime }$ into a total derivative gives the radial stellar pulsation equation,

Set $P=\gamma p{r}^4$, $W=\rho r^4$ and $Q=r^3[(4-3\gamma )p{]}^{\prime }$. The Sturm-Liouville operator isFor regular physical perturbations, $\xi$ is finite at the centre and $P{\xi }^{\prime }\to 0$ at the free surface. For a nonzero-frequency mode the fluid displacement is $u/(-i\omega )$, so vanishing Lagrangian pressure perturbation is equivalent to $\gamma p(3\xi +r{\xi }^{\prime })\to 0$. With $p(R_s)=0$ and finite $\xi$, this gives $P{\xi }^{\prime }\to 0$. Zero-frequency modes use the same displacement boundary condition directly. Assume bounded $\gamma$, positive $\rho ,p,\gamma$ in the interior and the usual finite-energy endpoint domain. Integration by parts givesThus this physical self-adjoint differential operator has real eigenvalues ${\omega }^2$. The endpoint domain is essential: the fact that $p$ vanishes does not by itself allow arbitrary singular trial functions. The regular free-surface realization of the Sturm-Liouville problem is the one used here.

The Rayleigh-Ritz variational principle gives the fundamental squared frequency as the infimum of the weighted stellar pulsation Rayleigh quotient,The infimum is over admissible finite-energy functions satisfying the physical endpoint conditions. If this quotient is nonnegative for every such function, all radial frequencies are real and there is no exponentially growing radial mode. A negative value for even one trial function proves a negative eigenvalue and an exponentially growing solution, because ${\omega }^2<0$. A zero lowest value is marginal and needs separate treatment of neutral motion.

Choose the homologous trial function $\xi =1$, which corresponds to radial velocity proportional to $r$. Its gradient contribution is zero, andThe denominator $\int \rho r^{4}dr$ is positive. HenceThis pressure-weighted radial instability criterion is sufficient, not necessary: another trial function can detect instability even if this one does not. For a constant stellar adiabatic exponent it recovers instability below $4/3$. At constant $\gamma =4/3$, the homologous mode is neutral. For constant $\gamma >4/3$, $Q=(4-3\gamma )r^3{p}^{\prime }>0$ in a normally stratified hydrostatic equilibrium, so the quotient is positive and the star is radially stable.