Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2022 / Section A

A fluid particle follows a Lagrangian trajectory, so its position satisfies $\dot{x}=u(x,t)$ rather than the velocity evaluated permanently at its initial position. Assume $\omega \ne 0$, and writeSet $\alpha =k{X}_0$ and $\theta =\alpha -\omega t$. A Taylor expansion of the velocity about $X_0$ givesThus $\dot{a}=\sin \theta$, andAt the next order,so the required second-order trajectory in a sinusoidal velocity field isBoth corrections vanish at $t=0$. All retained terms have units of length. The secular contribution gives the Stokes driftalthough the Eulerian time average of the given velocity is zero. The particle samples the wave at its displaced position, producing this second-order drift. The expansion is in the dimensionless parameter $|k{u}_0/\omega |\ll 1$, for fixed times; it ceases to be uniform when the accumulated drift changes the wave phase by order unity.

Interpret “reach” first as remaining an energetic, supersonic blast wave, rather than as the arrival of any acoustic wave. An atmospheric blast-wave range estimate must account for the thin atmosphere: the distance $D=8.5\times 10^6\mathrm{m}$ is much greater than its height $H=10^4\mathrm{m}$.

For an optimistic estimate, suppose the deposited energy $E$ remains in a horizontally expanding atmospheric layer. If its representative mass density is ${\rho }_a$, the swept mass is of order $\pi D^{2}H{\rho }_a$. Conservation of energy gives the self-similar blast wave scalingUsing ordinary atmospheric estimates ${\rho }_a\sim 1\mathrm{kg}{\mathrm{m}}^{-3}$ and speed of sound $c_a\sim 300\mathrm{m}{\mathrm{s}}^{-1}$, this givesThese are scaling estimates, not an exact shock wave solution. The strong shock wave approximation would already have failed atEquivalently, keeping the disturbance strongly supersonic out to $D$ would require substantially more than the sonic-transition energy scalewhich exceeds $10^{22}\mathrm{J}$ by about a factor of twenty. Reasonable atmospheric averages and order-unity geometrical coefficients do not reverse the comparison. Escape of energy through the top of the atmosphere makes the fully confined estimate optimistic; Earth's curvature changes its area estimate by an order-unity factor.

The strong atmospheric blast will not survive to Lansing in this model. A weakened acoustic wave can nevertheless travel there. The supplied energy and atmospheric height do not determine its detectable amplitude, detailed attenuation or damage: those require an atmospheric propagation model. Thus “no disturbance of any kind reaches Lansing” would be an unjustified stronger conclusion.

This is magnetic flux freezing in ideal magnetohydrodynamics. In addition to the supplied ideal magnetohydrodynamic induction equation, a physical magnetic field obeys Gauss's law for magnetism, $\nabla \cdot \mathbf{B}=0$.

Parameterize the material surface by two particle labels $a,b$, so $d\mathbf{S}=({\partial }_a\mathbf{x}\times {\partial }_b\mathbf{x})dadb$ and $D\mathbf{x}/Dt=\mathbf{u}$. Differentiating these tangents gives the material surface element evolution lawwhere $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$. Thus the material magnetic-flux transport theorem isExpand the curl in the ideal magnetohydrodynamic induction equation:Since $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ is the material derivative, the first three terms cancel the corresponding terms in the surface transport formula. The remaining integrand is $\mathbf{u}\nabla \cdot \mathbf{B}=0$. ThereforeNo incompressible flow assumption is needed: the changing surface area cancels the compression term. This proof assumes smooth velocity and magnetic field and a surface carried with the fluid; the ideal magnetohydrodynamic induction equation by itself, for an arbitrary non-solenoidal vector field, would leave the final divergence term.