Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 70 / 4

Let $\xi$ be the fluid Lagrangian displacement, so $\delta u={\partial }_t\xi$ about the static background. A displaced material volume changes by fractional amount $\nabla \cdot \xi$. Mass conservation therefore gives the fluid Lagrangian perturbation of density $\Delta \rho =-\rho \nabla \cdot \xi$. Converting to the fluid Eulerian perturbation by $\Delta =\delta +\xi \cdot \nabla$ givesAdiabatic conservation of $p/{\rho }^{\gamma }$ gives $\Delta p/p=\gamma \Delta \rho /\rho$, and consequently $\delta p=-\xi \cdot \nabla p-\gamma p\nabla \cdot \xi$.

Integrating the linearized ideal induction equation for a perturbation generated by the displacement gives $\delta B=\nabla \times (\xi \times B)$. Since the equilibrium field is solenoidal, expansion yieldsThis is the displacement form of magnetic flux freezing. It also preserves $\nabla \cdot \delta B=0$.

Rewrite the magnetic force as $-\nabla (B^2/(2{\mu }_0))+{\mu }_0^{-1}B\cdot \nabla B$. Linearizing the full momentum equation, subtracting static force balance, and setting $\delta \Pi =\delta p+B\cdot \delta B/{\mu }_0$ givesTogether these are the linear displacement equations for a magnetized atmosphere. The density multiplying acceleration is the equilibrium value, because a perturbation of that factor would multiply a first-order acceleration and contribute only at second order.

Use complex Fourier amplitudes and let primes denote $z$ derivatives. Put $C=B^2/{\mu }_0$, $T=\gamma p+C$ and $d=i{k}_x{\xi }_x+i{k}_y{\xi }_y+{\xi }_z^{\prime }$. Static balance gives $(p+B^2/(2{\mu }_0){)}^{\prime }=-\rho g$. The perturbations from part (a) then implyFor the last identity, combine $\delta p$ with $B\delta B_y/{\mu }_0$ and use static balance; this is where the background magnetic-pressure gradient cancels.

Multiply the Fourier momentum equation by $-{\xi }^{\ast }$ and integrate from $a$ to $b$. Integrating the pressure-gradient term by parts gives $[{\xi }_z^{\ast }\delta \Pi {]}_a^b-\int d^{\ast }\delta \Pi dz$. The gravity term is $\int [-g{\rho }^{\prime }|{\xi }_z{|}^2-\rho g{\xi }_z^{\ast }d]dz$. The magnetic terms reduce toIndeed the term from $\delta B_z{B}^{\prime }{e}_y$ cancels exactly the $B^{\prime }$ term arising from $i{k}_{y}B\delta B$. This establishes, before completing the square,Now $\delta \Pi =A-Td$, soThus the magnetohydrodynamic energy principle has the requested quadratic formAll squares are squared complex moduli. With the prescribed vanishing surface term the quadratic form is real, as required for the self-adjoint stability problem.

For fixed smooth ${\xi }_y,{\xi }_z$, the only appearances of ${\xi }_x$ are in $\delta \Pi =A-Td$ and the nonnegative tension contribution $C{k}_y^2|{\xi }_x{|}^2$. Positivity of $T$ makes the pressure square nonnegative too. Impose $d=A/T$ by choosingThen $\delta \Pi =0$ exactly and ${\xi }_x=O(k_x^{-1})$ as $|k_x|\to \infty$, removing both nonnegative contributions in the limit. Since the remaining terms do not involve ${\xi }_x$, no other choice can lower their limiting value. This is the short transverse wavelength limit for magnetic buoyancy.

Take the chosen vertical and along-field displacements compactly supported in the interior when constructing instability trials; the resulting ${\xi }_x$ also vanishes near the boundaries, satisfying the surface condition. The argument constructs an infimum, not necessarily a minimizing finite-wavelength eigenfunction. Whenever the limiting energy is strictly negative, sufficiently large finite $|k_x|$ already gives negative energy and hence the instability asserted by the supplied variational criterion.

For $k_y=0$, the minimized energy density from part (c) isIf the bracket is nonnegative throughout, this density and the omitted pressure square are nonnegative for every displacement, so there is no negative-energy trial. Conversely, a strict negative value at a point of a smooth equilibrium persists in a small interval. Choose a nonzero smooth ${\xi }_z$ supported there, set ${\xi }_y=0$, and choose ${\xi }_x$ as in part (c). This makes the pressure square zero and gives a strictly negative integral. Therefore the localized interchange criterion for a magnetized atmosphere isThese disturbances interchange flux tubes without along-field variation; both gas and magnetic pressure contribute to their effective compressional stiffness. Equality alone is marginal and does not give a negative energy integral.

Set $P=\gamma p$, so $T=P+C$, and for $k_y\ne 0$ put $q=i{k}_y{\xi }_y$. This variable can be chosen freely. The minimized transverse-displacement energy from part (c) has densityComplete the square in $q$ to obtainThis identity also applies at a zero of $B$, where $C=0$ and the first square contributes nothing. If $-(\ln \rho {)}^{\prime }\ge \rho g/P$ everywhere, every term is nonnegative, including the pressure and transverse-tension terms removed earlier. Thus no nonzero along-field wavenumber gives an unstable trial.

For the converse, suppose the strict reverse inequality holds at an interior point. On a sufficiently small closed interval around it, $-g{\rho }^{\prime }-{\rho }^2{g}^2/P$ is uniformly negative and $C$ is bounded. Choose a sufficiently small but nonzero $|k_y|$ so that addition of $C{k}_y^2$ keeps it negative there. Choose a smooth supported ${\xi }_z$ in that interval and set $q=\rho g{\xi }_z/P$, making the first square vanish. This determines a finite ${\xi }_y=q/(i{k}_y)$. Finally choose ${\xi }_x$ by part (c) with sufficiently large finite $|k_x|$; its remaining tension energy can be made smaller than the strict negative margin. All displacements vanish near the boundaries. The full energy is therefore negative.

We have proved the long-wavelength undular magnetic buoyancy criterion:Under the assumed absence of interchange instability, the new unstable range is $\rho g/(\gamma p+C)\le -(\ln \rho {)}^{\prime }<\rho g/(\gamma p)$. Along-field motion allows material to drain and removes magnetic pressure from the limiting buoyancy stiffness, while tension at nonzero $k_y$ remains stabilizing. The statement is about existence of an allowed nonzero wavenumber; a fixed prescribed $k_y$ retains the term $C{k}_y^2$, and a finite imposed periodic length could exclude the required long wavelengths.