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A horizontal flux tube displaced without bending preserves $B/\rho$ by magnetic flux freezing. If thermal exchange keeps its temperature equal to its new surroundings, lateral magnetic pressure balance gives $(c_T^2+v_A^2)({\rho }_i^{\prime }-{\rho }^{\prime })=\rho B(B/\rho {)}^{\prime }/{\mu }_0$. Here $c_T^2=p/\rho$, $v_A^2=B^2/({\mu }_0\rho )$, and primes denote the response to a small vertical displacement from an equilibrium. Its linear buoyancy acceleration is $-gB(B/\rho {)}^{\prime }\xi /[{\mu }_0(c_T^2+v_A^2)]$. Thus local interchange is unstable when $B(B/\rho {)}^{\prime }<0$, equivalently $(B^2/{\rho }^2{)}^{\prime }<0$. This fast-thermal-exchange criterion differs from the adiabatic localized interchange criterion for a magnetized atmosphere.

A density deficit in a flux tube is not by itself an instability calculation. For example, a straight magnetized tube in otherwise field-free gas at the same temperature has lateral balance $p_i+B_i^2/(2{\mu }_0)=p_e$. The perfect gas law then gives ${\rho }_i<{\rho }_e$, so the tube rises unless some other force balances its weight deficit. This configuration simply lacks magnetostatic equilibrium. A magnetic buoyancy instability, in contrast, starts from a force-balanced atmosphere and asks whether a small allowed displacement grows. A balanced horizontal magnetic layer can be stable or unstable; the presence of magnetic pressure alone does not decide this.

Assume a smooth equilibrium, with downward gravitational acceleration $g>0$, and write $\mathcal{R}$ for the gas constant per unit mass. The ambient total pressure and force balance areWe use ideal magnetic flux freezing on the displacement time scale and rapid lateral pressure equilibration. A bodily displacement perpendicular to a straight horizontal tube does not change its length or bend its field. If its area is $\mathcal{A}$, conservation of magnetic flux and mass per unit length gives $B\mathcal{A}=\mathrm{constant}$ and $\rho \mathcal{A}=\mathrm{constant}$. Thus the tube preservesEfficient heat exchange fixes its temperature to the ambient temperature at its new height, not to its original temperature. This is the thermal assumption that distinguishes the requested criterion from the adiabatic one.

Take a tube initially at $z_0$ and let $f_0=B_0(z_0)/\rho (z_0)$. After a small upward displacement $\xi$, let ${\rho }_i(z)$ be the density it would have in lateral balance at $z=z_0+\xi$. Its preserved $f_0$ and imposed local temperature giveAt the initial position ${\rho }_i=\rho$. Differentiate this relation at $z_0$:For the ambient atmosphere the same differentiation must also differentiate $f(z)=B_0(z)/\rho (z)$. It givesSubtracting isolates the actual displaced density contrast:The pressure-gradient force on the tube balances the weight of the displaced ambient gas, so its leading vertical buoyancy acceleration isThe coefficient multiplying $\xi$ is positive precisely when the tube becomes lighter on upward displacement. Thus the isothermal interchange criterion for magnetic buoyancy isWhere $B_0>0$, this is simply $d(B_0/\rho )/dz<0$, or $d\log B_0/dz<d\log \rho /dz$. The squared form is independent of the field's choice of direction. Equality is neutral at this order, and the reversed inequality is restoring for these displacements. A layer with an interval satisfying the strict inequality admits local growing interchange disturbances under the assumed pressure/thermal equilibration. The calculation does not address field-bending undular modes, finite heat-transfer times or magnetic diffusion; those can change the criterion. In particular, the localized interchange criterion for a magnetized atmosphere with an adiabatic gas response is not interchangeable with this fast-thermal-exchange result.

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.