= 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 $\Delta=\delta+\xi\cdot\nabla$ gives
$$
\delta\rho=-\xi\cdot\nabla\rho-\rho\nabla\cdot\xi.
$$
Adiabatic 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 yields
$$
\delta B=-\xi\cdot\nabla B+B\cdot\nabla\xi-B(\nabla\cdot\xi).
$$
This 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 $\delta\Pi=\delta p+B\cdot\delta B/\mu_0$ gives
$$
\boxed{\rho\partial_t^2\xi=-\nabla\delta\Pi-g\delta\rho e_z
+\frac1{\mu_0}(\delta B\cdot\nabla B+B\cdot\nabla\delta B).}
$$
Together 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.
Back to article page
