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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.