 Solution

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.