= Solution
Linearizing the <ideal magnetohydrodynamic induction equation> about a static <magnetic field> and using $u'=\partial_t\xi$ gives the <fluid Lagrangian displacement>-generated <magnetic field> perturbation $b=\nabla\times(\xi\times B)$. The two <solenoidal vector field> conditions reduce the <curl> identity to
$$
\boxed{b_i=B_j\partial_j\xi_i-\xi_j\partial_jB_i.}
$$
Its <divergence> vanishes because it is a <curl>. Directly, the two products of first derivatives in $\partial_ib_i$ cancel after exchanging their indices, while the remaining terms differentiate $\nabla\cdot\xi$ or $\nabla\cdot B$, both zero. Similarly the integrated linear <continuity equation> gives $\rho'=-\nabla\cdot(\rho\xi)$, so \b[$\boxed{\rho'=-\xi\cdot\nabla\rho}$].
Let $\Pi=p+B^2/(2\mu_0)$ be the <magnetohydrodynamic total pressure>. <Magnetostatic equilibrium> and the momentum equation in the <linearized ideal magnetohydrodynamic equations> are
$$
\partial_i\Pi=-\rho\partial_i\Phi+\mu_0^{-1}B_j\partial_jB_i,
$$
$$
-\sigma^2\rho\xi_i=-\partial_i\pi'-\rho'\partial_i\Phi
+\mu_0^{-1}(B_j\partial_jb_i+b_j\partial_jB_i),\qquad
\pi'=p'+B\cdot b/\mu_0.
$$
There is no perturbed <gravitational potential> here because the field is fixed. Multiply by $\xi_i^*$ and integrate. The <pressure> term vanishes by the assumed boundary cancellation and $\nabla\cdot\xi^*=0$. Thus
$$
\sigma^2I=-\int\xi_i^*\xi_j\rho_{,j}\Phi_{,i}\,dV+\mu_0^{-1}\mathcal M,
\qquad I=\int\rho|\xi|^2dV,
$$
where $\mathcal M=-\int\xi_i^*(B_j\partial_jb_i+b_j\partial_jB_i)dV$.
Integrate the first magnetic term along $B$, using $\nabla\cdot B=0$, then insert the expression for $b$. This gives the positive term $\int|(B\cdot\nabla)\xi|^2dV$ and two mixed derivative terms. For clarity, the troublesome term transforms as
$$
-\int(B_k\partial_k\xi_i^*)\xi_j\partial_jB_i
=\int\xi_i^*B_k(\partial_k\xi_j)\partial_jB_i
+\int\xi_i^*\xi_jB_k\partial_k\partial_jB_i.
$$
The first term on the right cancels the other mixed term. Combining the remaining derivative with the term containing $\partial_jB_k\partial_kB_i$ gives
$$
\mathcal M=\int|(B\cdot\nabla)\xi|^2dV
+\int\xi_i^*\xi_j\partial_j(B_k\partial_kB_i)dV.
$$
Differentiating the equilibrium relation now combines the gradient of <mass density> and <magnetostatic equilibrium> terms:
$$
-\rho_{,j}\Phi_{,i}+\mu_0^{-1}\partial_j(B_k\partial_kB_i)
=\Pi_{,ij}+\rho\Phi_{,ij}.
$$
We obtain the <incompressible magnetic displacement energy identity>
$$
\boxed{\sigma^2I=\int\xi_i^*\xi_j\Pi_{,ij}dV
+\int\rho\xi_i^*\xi_j\Phi_{,ij}dV
+\frac1{\mu_0}\int|(B\cdot\nabla)\xi|^2dV.}
$$
The two <Hessian matrices> are real symmetric, making their <Hermitian form> evaluations real. The last term is a squared <norm>, and $I>0$ for a nonzero displacement in fluid with positive <mass density>. Thus \b[$\boxed{\sigma^2\text{ is real}}$]. It may be negative, in which case one temporal branch has exponential growth; real squared <frequency> is not itself a proof of stability.
For the atmosphere with a horizontal <magnetic field>, $\Phi=gz$ has zero <Hessian matrix> and the equilibrium equation is
$$
\boxed{\frac{d}{dz}\left(p+\frac{B^2}{2\mu_0}\right)=-\rho g.}
$$
Consequently the <interchange stability of an incompressible magnetized atmosphere> is governed by the numerator
$$
Q[\xi]=-g\int\rho_z|\xi_z|^2dV+\frac1{\mu_0}\int B^2|\partial_x\xi|^2dV.
$$
For $\xi_z=\sin ky$, the <magnetic tension> term is zero: these <fluid Lagrangian displacements> interchange neighboring <magnetic field lines> without varying along them. A <mass density> increasing with height gives negative energy and instability; decreasing <mass density> gives positive energy for these vertical interchanges. If $\rho_z\leq0$ everywhere, the displayed form is nonnegative for every admitted incompressible displacement. If the gradient changes sign, suitably localized <divergence>-free interchange trials detect an unstable inverted layer; For example, take $\xi_z=f(z)e^{iky}$ and $\xi_y=if'(z)e^{iky}/k$, with $f$ supported in that layer. Their <divergence> is zero, and neither component varies along the <magnetic field>.
For $\xi_z=\sin kx$, <magnetic field lines> are bent. Averaging over a full horizontal <wavelength> gives the trial <Rayleigh quotient>
$$
\frac{Q}{I}=\frac{-g\int\rho_z\,dz+(k^2/\mu_0)\int B^2\,dz}{\int\rho\,dz},
$$
under the stipulated admissible boundary/integrability conditions. It is a trial quotient, not a claim that this displacement is an exact <eigenfunction> for every profile. <mass density> decreasing upward is stable here too. For inverted <mass density>, <magnetic tension> can stabilize sufficiently short waves along the <magnetic field>, but does not remove the transverse interchange instability. Locally the two contributions have the familiar scale $-g\rho_z/\rho+k^2v_A^2$, with the <Alfvén speed> $v_A=B/\sqrt{\mu_0\rho}$.
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