Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 64 (Astrophysical Fluid Dynamics)

Take $u<0$ for inward radial velocity, and assume a constant specific-heat ratio $\gamma >1$. The isentropic flow has $p=K{\rho }^{\gamma }$ and adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$. Steady spherical mass conservation and the radial Euler equation for fluid motion giveEliminating the mass density derivative gives $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$, orDifferentiating $c^2\propto {\rho }^{\gamma -1}$ gives the corresponding sound speed equation:At a smooth sonic point, $u_s^2=c_s^2$. A finite derivative requires the numerator of the flow equation to vanish too. ThusThis regularity requirement concerns a smooth transonic branch, rather than an arbitrary singular solution reaching unit Mach number.

Integrating the momentum equation using the specific enthalpy $c^2/(\gamma -1)$ yields the Bernoulli equationFor a finite positive mass accretion rate, its large-radius solution hasThese expressions satisfy both the flow equations asymptotically and show $u\to 0$ and $\rho \to {\rho }_{\infty }$. In particular $u^{\prime }/u\sim -2/r$ agrees with the first boxed equation. Combining the Bernoulli equation with sonic regularity additionally gives $c_s^2=2c_{\infty }^2/(5-3\gamma )$, so a finite positive sonic point occurs in the range identified below.

For the small-radius inward free fall branch, neglecting enthalpy against $GM/r$ in the Bernoulli equation predicts $u^2\sim 2GM/r$. The continuity equation then fixes $\rho \propto r^{-3/2}$, and consequently $c^2\propto r^{-3(\gamma -1)/2}$. The neglected-to-leading ratio isIt tends to zero precisely whenThis makes the assumed supersonic flow and leading free fall self-consistent. At equality, the ratio is of constant order rather than vanishing; above it, the proposed asymptotic balance fails. The isothermal limit $\gamma =1$ also permits free fall, with logarithmic rather than power-law enthalpy.

For the stellar funnel, write the magnetic scalar potential as $\psi =\mu \cos \theta /r^2$, where $R=r\sin \theta$ and $z=r\cos \theta$. Its magnetic dipole field has $B_r=-2\mu \cos \theta /r^3$ and $B_{\theta }=-\mu \sin \theta /r^3$, so a magnetic-field-line equation isThe edge of the polar flux tube meets the stellar surface at ${\theta }_{\ast }\simeq a/R_{\ast }$. Thus, near the pole, ${\theta }^2\simeq {\theta }_{\ast }^{2}r/R_{\ast }$, and its transverse radius is $R\simeq r\theta$. This proves the dipolar flux-tube area lawEquivalently magnetic flux conservation gives $A|\mathbf{B}|=\mathrm{constant}$ and $|\mathbf{B}|\propto r^{-3}$. The stipulated small cross section keeps the tube close to the polar axis, where its direction is approximately radial; the approximation is not valid all the way to arbitrarily large $r$.

Outside the star the prescribed field is a gradient, hence $\nabla \times \mathbf{B}=0$ and $\mathbf{j}=0$. Its Lorentz force density is therefore $\mathbf{j}\times \mathbf{B}=0$. The magnetic-pressure and magnetic-tension parts cancel in this exact current-free dipole. Field-aligned velocity also has $\mathbf{u}\times \mathbf{B}=0$, consistent with a steady prescribed field. The following calculation uses the narrow polar funnel approximation.

Now steady mass conservation reads $\rho uA=\mathrm{constant}$. With $A\propto r^3$, replace the spherical factor 2 by 3:The Bernoulli equation is unchanged because the magnetic field supplies no work or force. On a gravity-dominated free fall branch, $u^2\sim 2GM/r$, but $\rho \propto r^{-5/2}$ and $c^2\propto r^{-5(\gamma -1)/2}$. HenceThe same bound follows from regular transonic accretion in a power-law tube: $u_s^2=c_s^2=GM/(3r_s)$ and the reservoir-matched Bernoulli equation gives $c_s^2=2c_{\infty }^2/(7-5\gamma )$. The stellar surface cuts off the mathematical small-$r$ limit; the criterion refers to a funnel that becomes gravity dominated and crosses its sonic point before reaching that surface.

The equilibrium isothermal equation of state is $p=c_s^2\rho$. Hydrostatic equilibrium gives $p_z=-\rho {\Omega }^{2}z$, henceHere $\Omega \ne 0$ is assumed. This is the same local harmonic-gravity balance as isothermal vertical structure in stellar gravity.

For $\gamma >1$, this equilibrium is stably stratified under adiabatic fluid perturbations. Its squared buoyancy frequency isA general perturbation can therefore support acoustic waves restored by compression and internal gravity waves restored by buoyancy, with nonoscillatory entropy or vortical disturbances also possible. The prescribed velocity is purely vertical and horizontally uniform. It describes vertical acoustic waves; it does not contain the horizontally varying circulation required for an oscillatory internal-gravity branch. For zero horizontal wavenumber, that branch has zero frequency. An isothermal equilibrium must not be confused with an isothermal perturbation: the latter would remove the buoyancy restoring force.

Introduce the fluid Lagrangian displacement $\xi (z)e^{i\sigma t}$, so $w=i\sigma \xi$. Use $\delta \rho ,\delta p$ for fluid Eulerian perturbations and $\Delta \rho ,\Delta p$ for fluid Lagrangian perturbations. The continuity equation and the adiabatic relation areThe fixed external gravitational potential is not perturbed. The vertical linearized Euler equations consequently giveSubstituting the preceding relations, and using $p_z=-\rho g$ and $p_{zz}=-{\rho }_{z}g-\rho g_z$, reduces its right-hand side toSince $g={\Omega }^{2}z$ and $g_z={\Omega }^2$, the same equation holds for $w$:The rigid fixed base supplies the essential condition $w(0)=0$. At infinity the condition is finite kinetic energy, not bounded velocity amplitude; a polynomial can grow while the Gaussian mass density makes its energy finite.

Put $x=z/H$ and $\lambda =({\sigma }^2/{\Omega }^2-1)/\gamma$. The equation becomes the Hermite differential equation in its probabilists' convention,For $w={\sum }_{j\ge 0}{a}_j{x}^j$, its recurrence isThe base condition makes $a_0=0$ and therefore every even coefficient zero. An odd series terminates at degree $2n+1$ when $\lambda =2n+1$. It remains to prove that a nonterminating odd solution cannot have finite energy; termination alone would not establish the requested exhaustion of modes.

For that purpose, set $y=e^{-x^2/4}w$. The finite-energy condition is $y\in L^2(0,\infty )$, and its equation isBecause the coefficients are even and the solution is odd at the base, extend $y$ oddly to the whole real line. Define lowering and raising operatorsIntegration by parts gives $\Vert ay{\Vert }^2=\lambda \Vert y{\Vert }^2$, so $\lambda \ge 0$. These identities apply to the square-integrable solution: multiplying its equation by a compactly cut off conjugate and taking larger cutoffs bounds $y^{\prime }$ and $xy$ in $L^2$, and removes the endpoint terms. The same argument applies to each lowered eigenfunction.

If $\lambda$ is not a nonnegative integer, repeatedly apply $a$. Each application lowers $\lambda$ by one and remains nonzero whenever its current value is positive, by the norm identity. After finitely many applications it would yield a nonzero square-integrable eigenfunction with negative $\lambda$, contradicting that identity. Thus $\lambda =N$ is a nonnegative integer. At $N=0$, $ay=0$ gives $y\propto e^{-x^2/4}$, which is even. Lowering reverses parity, so an initially odd eigenfunction can reach this even ground state only if $N$ is odd. Consequently $N=2n+1$. The recurrence then terminates and uniquely determines the odd solution from $a_1$.

The complete adiabatic vertical modes of a Gaussian atmosphere are thereforeHere ${\mathrm{He}}_j$ is the Probabilists' Hermite polynomial. For example the first shapes are $x$, $x^3-3x$, and $x^5-10x^3+15x$, with frequencies squared ${\Omega }^2(1+\gamma )$, ${\Omega }^2(1+3\gamma )$, and ${\Omega }^2(1+5\gamma )$. Either sign of each real frequency gives a time-harmonic solution. Gaussian moments make every displayed polynomial have finite kinetic energy. Without the fixed-base condition, finite energy alone would also admit even polynomials, which is why that condition is necessary.

Let $p_1(z)$ denote the Eulerian perturbation of a fluid variable that is the pressure amplitude, and keep $\rho$ constant. The linearized Euler equations and incompressible flow constraint areThe last equation gives $u=i{w}^{\prime }/k$. Substituting it into the first yieldsDifferentiating this formula and using the vertical momentum equation produces the Rayleigh equation for inviscid shear flow:For a growing normal mode, $c$ is nonreal, so $U-c$ never vanishes on the real $z$ axis. In each open region the base velocity is constant or linear and $U^{\prime \prime }=0$; therefore $w^{\prime \prime }-k^{2}w=0$. Decay away from the layer selects $e^{-kz}$ above and $e^{kz}$ below, while both exponentials are retained inside. Conveniently normalize these asThese are the spatially localized discrete modes relevant to instability, rather than all possible generalized neutral disturbances at a critical layer in a shear flow.

At each corner $U$ is continuous and $U^{\prime }$ jumps. The two sides have the same normal velocity and normal stress: there is no membrane, density jump or singular external force supplying a pressure jump. Thus $w$ and $p_1$ must be continuous. The pressure formula proves the required continuity of $(c-U)w^{\prime }+U^{\prime }w$. This is pressure matching at a piecewise-linear shear interface; it does not require $w^{\prime }$ to be continuous. Equivalently integrating the Rayleigh equation for inviscid shear flow through a corner gives $(U-c)[w^{\prime }]=[U^{\prime }]w$.

Write $\alpha =kd$, $y=c/U_0$, and $E=e^{2\alpha }$. Continuity of normal velocity givesAt $z=d$, the outside derivatives are $w_{+}^{\prime }=-kA$, $U_{+}^{\prime }=0$, while the inside derivatives are $w_{-}^{\prime }=-kB+kEC$, $U_{-}^{\prime }=U_0/d$. Hence pressure matching reduces toSimilarly at $z=-d$, the outside derivative is $kD$ and the inside derivative is $-kEB+kC$. Matching givesThe two equations can be written asA nonzero mode therefore requiresThis is the dispersion relation for the unbounded piecewise-linear shear layer.

With the phase convention $e^{ik(x-ct)}$, positive $\mathrm{Im}c$ gives exponential temporal growth. Since the displayed $c^2$ is real, a growing/decaying pair occurs precisely whenFor $0<\alpha \le 1/2$, $e^{-2\alpha }>1-2\alpha$ by the strict exponential tangent inequality, so every such mode is unstable. For $\alpha \ge 1/2$, define $f(\alpha )=2\alpha -1-e^{-2\alpha }$. Its derivative $2+2e^{-2\alpha }$ is positive, while $f(1/2)<0$ and $f(1)>0$. There is consequently exactly one root in that interval, andAt the cutoff the discrete roots merge at $c=0$; above it the roots are real and have no exponential growth. The physical wavelengths in the unstable band satisfy ${\lambda }_x=2\pi /k>2\pi d/{\alpha }_s\simeq 9.83d$.

This is a finite-thickness Kelvin-Helmholtz instability: long disturbances roll up the layer and draw kinetic energy from its shear. The off-diagonal $e^{-2kd}$ terms describe coupling of the two vorticity discontinuities. Short disturbances decay across the layer before strongly coupling its edges and are neutral. Indeed at $kd\to \infty$ the two phase speeds approach the velocities $\pm U_0$ of the separated edges. At $kd\to 0$, expansion of the numerator gives $c^2/U_0^2\to -1$, recovering the vortex sheet result $c=\pm i{U}_0$; the growth rate then behaves as $k|U_0|$.

For the basic axisymmetric flow, ${\mathbf{u}}_0=\Omega R{\mathbf{e}}_{\varphi }$. The steady radial Euler equation for fluid motion gives $-{\Omega }^{2}R=-p_{0,R}/\rho$; the other components give $p_{0,z}=p_{0,\varphi }=0$. Thus, in the absence of an additional body force,The additive constant is undetermined without an imposed reference pressure; the rigid boundary does not prescribe a uniform boundary pressure. Its shape is fixed by the container rather than by a free-surface pressure condition.

Let $v_R,v_{\varphi },v_z$ be the small velocity amplitudes and $W=p_1/\rho$ the pressure amplitude. Linearizing the cylindrical Euler equations for an inviscid fluid about uniform rotation, the material derivative is ${\partial }_t+\Omega {\partial }_{\varphi }$. The intrinsic frequency for the stated phase convention is thereforeThe radial and azimuthal curvature terms supply the Coriolis acceleration terms, givingFor $s\ne 0,\pm 2\Omega$, solving these equations givesThe incompressible flow constraint isOn substitution, the terms proportional to $m\Omega W_R$ cancel. The remaining equation iswhich provesThis is the pressure equation for inertial modes in a rigid rotating sphere. The exceptional frequencies used in the divisions must instead be treated in the original momentum system; they are not needed for the mode below.

A fixed rigid sphere is impermeable, so its normal velocity vanishes:Using the solved amplitudes and multiplying by $s(s^2-4{\Omega }^2)/i$ givesEquivalently divide by ${\bar{\sigma }}^2$ for $\bar{\sigma }\ne 0$. Smoothness means tangential slip is allowed: there is no no-slip condition on these inviscid normal modes. Nor is there a condition $W=0$, since the wall supplies the required varying normal pressure.

For an axisymmetric oscillation, $m=0$ and $s=\sigma$. Write $\beta =1-4{\Omega }^2/{\sigma }^2$. For $W=z(A{R}^2+B{z}^2+C)$ the radial and vertical operators areThe interior pressure equation therefore requiresThe boundary equation becomesSubstituting $R^2=r_0^2-z^2$, this polynomial must vanish for every boundary height. Its constant and $z^2$ coefficients consequently giveTogether with the interior relation the latter condition is $(4+\beta )A=0$. A nontrivial nonsingular oscillatory solution needs $A\ne 0$: if $A=0$ and $\beta \ne 0$, the other equations force $B=C=0$. Hence $\beta =-4$, givingThe cubic axisymmetric inertial mode of a rigid sphere is explicitlyIts cubic pressure is regular at the axis and centre, and its reconstructed velocity is divergence free and tangent to the sphere by the preceding identities. The frequency lies within the inertial wave band $|\sigma |<2|\Omega |$: Coriolis acceleration, rather than compression or a moving surface, supplies the restoring dynamics. The amplitude $A$ is arbitrary in this linear problem.