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For a $y$-independent solenoidal magnetic field, the transverse components can be written $B_x=-{\psi }_z$ and $B_z={\psi }_x$. Contours of $\psi$ are transverse magnetic field lines. The independent axial component $B_y$ can twist the flux tube. Under the MHD induction equation, an additive time-dependent gauge can be chosen so that $D\psi /Dt=0$.

A compressible intensification mechanism for a thin, nearly vertical flux tube in a superadiabatic atmosphere. An unstable downflow drains material and lowers internal gas pressure; approximate lateral balance $p_i+B^2/(2{\mu }_0)=p_e$ then permits the tube to contract and its field to strengthen at nearly conserved magnetic flux. It relies on compressibility and thermal evolution, which are absent from a strict Boussinesq approximation. Dynamic photospheric structures need not satisfy thin-tube equilibrium, so this mechanism does not explain every intense field concentration.

Near the polar axis of a magnetic dipole field, $B\propto r^{-3}$. Conservation of magnetic flux in a flux tube consequently gives $A(r)\propto r^3$. A dipole magnetic-field line intersecting a sphere of radius $R_{\star }$ at small polar angle ${\theta }_{\star }$ obeysThis area is for one polar tube. The approximation requires ${\theta }_{\star }^{2}r/R_{\star }\ll 1$; the full magnetic dipole field does not remain a narrow polar tube at arbitrarily large radius.

The cross-sectional area of a narrow flux tube, measured normal to its magnetic field. Constancy of the magnetic flux gives $A|\mathbf{B}|=\mathrm{constant}$ along the tube. This area controls the mass transport when flow follows the field.

For a regular cylindrical flux tube with $\mathbf{B}=(0,B_{\varphi }(s),B_z(s))$, vanishing field beyond radius $s_0$, and exterior pressure $p_e$, define $⟨h⟩=2s_0^{-2}{\int }_0^{s_0}sh(s)ds$ and $B_p^2=2{\mu }_0{p}_e$. Integrating cylindrical magnetostatic pressure balance against $s^2$ gives the displayed identity. The toroidal magnetic pressure and hoop magnetic tension cancel in this cross-sectional axial-field average. Nonnegative gas pressure gives $⟨B_z^2⟩\le B_p^2$, with strict inequality precisely when $⟨p⟩>0$.

An open flux tube carries magnetic flux through its end caps. Its ordinary magnetic helicity can therefore change under a single-valued gauge transformation. For $\mathbf{B}=aR{\mathbf{e}}_{\varphi }+b{\mathbf{e}}_z$ inside $R<R_0$ and zero outside, a regular family of magnetic vector potentials is $A_{\varphi }=bR/2$, $A_z=C-a{R}^2/2$ inside and $A_{\varphi }=b{R}_0^2/(2R)$, $A_z=C-a{R}_0^2/2$ outside. Their tangential components agree at $R_0$. Inside, $\mathbf{A}\cdot \mathbf{B}=bC$, so the helicity per unit length is $\pi bC{R}_0^2$. Changing $C$ is the gauge transformation $\Lambda =\Delta Cz$, and changes this result whenever $b\ne 0$. The convention $A_z=0$ outside gives $C=a{R}_0^2/2$ and helicity per unit length $\pi ab{R}_0^4/2$, while $C=0$ gives zero. Without an end-cap or gauge convention the ordinary helicity is not unique.

An instability in which neighboring flux tubes exchange positions across a pressure gradient or other stratification, reducing the appropriate potential energy. Unfavorable magnetic field-line curvature can drive the exchange; parallel variation bends the magnetic field and introduces stabilizing magnetic tension. In a steep-gradient low-plasma beta local Z-pinch, ${\omega }^2=k_{\parallel }^2{v}_A^2-2p_0/({\rho }_{0}R{L}_p)$.

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.

In magnetically channelled accretion, a strong magnetic field guides an accretion flow along flux tubes. The component of the Lorentz force density parallel to the magnetic field vanishes, so the longitudinal flow can be treated hydrodynamically while the field controls the transverse geometry. A magnetic dipole field near a neutron star channels matter toward small polar caps.

A highly conducting solar wind carries the Sun's magnetic field outward by magnetic flux freezing. This can open magnetic field lines that would form closed loops in a static dipole model. Beyond an opening radius $r_o$, approximate the field as radial. Constancy of the magnetic flux in a fixed solid-angle flux tube gives the radial solar-wind magnetic flux lawwhich decays more slowly than the vacuum magnetic dipole field, proportional to $r^{-3}$.

For example, if the field becomes radial already near the solar radius $R_{\odot }$, normalizing both estimates to the same field there givesThus a dipole estimate $10^{-11}\mathrm{T}$ becomes of order $2\times 10^{-9}\mathrm{T}$, comparable to the stated hundredfold enhancement. More generally, matching dipolar decay inside $r_o$ to radial decay outside givesThis is a plausible scaling explanation, not a determination of the actual opening radius. The magnetic field need not be outward everywhere: opposite-polarity sectors can preserve zero net magnetic flux through a closed surface, consistent with Gauss's law for magnetism. The argument concerns the changed geometry of the field in a flowing plasma, rather than changing the dipole exponent by assumption.

The result is a fluid Eulerian perturbation generated by translating the entire equilibrium phase-space distribution function through ${\xi }_x$:Here $\Delta$ follows the displaced flux tube. Magnetic flux freezing carries the tube and its particles across the equilibrium pressure gradient. To leading order the tube changes neither its field strength nor its length by enough to alter the particle phase-space distribution function: these changes are smaller by plasma beta or $L_p/R$. Parallel streaming remains within the same displaced tube, on which the original equilibrium phase-space distribution function was uniform. It therefore cannot erase the displacement-generated contrast between adjacent tubes. This is why the term $v_{\parallel }\delta b_x{\partial }_x{f}_{0\alpha }$ in c exactly cancels the apparent parallel relaxation of $-{\xi }_x{\partial }_x{f}_{0\alpha }$. No collisional restoration of a Maxwell-Boltzmann distribution is needed.

Assume the stated outward-decreasing pressure, $L_p>0$, and setFor $|k_{\parallel }|<k_c$, the interchange instability hasAn outward-displaced flux tube carries higher pressure than its new surroundings. In the unfavorable magnetic field-line curvature of the Z-pinch, the resulting imbalance drives it farther outward. Parallel variation bends field lines, and magnetic tension supplies the stabilizing $k_{\parallel }^2{v}_A^2$ term. The fastest local interchange is at $k_{\parallel }=0$, with $\gamma ={\omega }_I$; this endpoint is understood as the limit of the stated Alfvénic ordering. There is no leading $k_y$ dependence once $k_y\gg k_{\parallel }$.

For $|k_{\parallel }|>k_c$, the two real frequencies are $\omega =\pm \sqrt{k_{\parallel }^2{v}_A^2-{\omega }_I^2}$: the displacement oscillates as a curvature-modified Alfvén wave. At $|k_{\parallel }|=k_c$, the leading equation is ${\ddot{\xi }}_x=0$, so the marginal solutions are a constant and a term linear in time; neglected corrections determine finer behavior. For a favorable reversed pressure gradient, $L_p<0$, this drive is stabilizing.

The leading dispersion relation is the same in ideal magnetohydrodynamics. Its adiabatic law gives $\delta p=-{\xi }_x{p}_0^{\prime }-\Gamma p_0\nabla \cdot \mathbf{\xi }$; the compressive part is smaller than the steep-gradient advection by $O(L_p/R)$ or $O(\beta )$, giving the same leading $\delta p=-{\xi }_x{p}_0^{\prime }$. Relaxing those orderings introduces compression, field-strength and field-length changes, and pressure anisotropy; the drift-kinetic equation can then introduce parallel kinetic phase mixing and Landau damping. At larger plasma beta, coupling to compressive branches matters; if pressure anisotropy develops, mirror instability and firehose instability may also become relevant. These are possibilities beyond the derived leading equation, not additional roots already established here.

Both mechanisms can extract free energy from an equilibrium pressure gradient in unfavorable magnetic field-line curvature, but the dynamics retained here is the macroscopic electromagnetic displacement of a flux tube. It depends on the total pressure gradient, including electron and ion contributions, and it does not require an ion temperature gradient specifically. Its stable branch has $\omega \sim k_{\parallel }{v}_A$; with low plasma beta, this is much faster than ion parallel streaming at the same $k_{\parallel }$.

The familiar ion-temperature-gradient instability is a slower drift-scale instability. In its usual electrostatic ordering the Boltzmann electron response is established by rapid electron parallel motion, while ion magnetic drifts, ion temperature evolution and polarization drift provide the instability dynamics. A representative ordering isThe last two expressions specify scales rather than exact geometric factors; $L_{T_i}^{-1}=-d\ln T_{0i}/dx$. The present leading kinetic magnetohydrodynamics system drops these relative species drifts and their finite-gyroradius corrections. It also retains the perturbed field direction, which enforces the advected-distribution cancellation in c; a rapid electron streaming rate alone cannot turn that cancellation into a Boltzmann electron response on an unperturbed field.

To recover ion-temperature-gradient instability physics, take the slow, nearly electrostatic limit $|\omega |/(k_{\parallel }{v}_A)\ll 1$, retain magnetic curvature drift, gradient-B drift and polarization drift in low-flow drift kinetics or gyrokinetics, and allow a sufficiently strong destabilizing ion temperature gradient. At long perpendicular wavelengths one may still have $k_{\perp }{\rho }_i\ll 1$, provided the leading nonzero ion polarization is kept; at $k_{\perp }{\rho }_i=O(1)$ full finite Larmor radius effects are needed. There is no contradiction between a drift-scale ion-temperature-gradient instability and the macroscopic interchange instability derived here: they use different asymptotic balances.

Choose a magnetic vector potential $\mathbf{A}$ with $\nabla \times \mathbf{A}=\mathbf{B}$, and define the total magnetic helicity byThe value is unchanged by a single-valued gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$, becauseHere both the solenoidal magnetic-field constraint and the tangent boundary field are essential. In a multiply connected domain, choose the vector potential by extending the tangent field by zero to all space and fixing its decay there, or fix the equivalent circulation convention. This avoids silently varying the non-gradient harmonic ambiguity of a vector potential on the restricted domain.

The ideal magnetohydrodynamic induction equation permitsDifferentiating $H_M$ and using integration by parts for curl,The volume product vanishes algebraically, the gauge flux vanishes since $\mathbf{B}\cdot \mathbf{n}=0$, and the last boundary term vanishes since $\mathbf{v}=0$. Thus magnetic helicity is conserved even though viscosity dissipates mechanical energy.

Through magnetic flux freezing, magnetic field lines move with the fluid and retain their linkage during every smooth ideal motion. For two thin closed flux tubes carrying fluxes ${\Phi }_1,{\Phi }_2$, their mutual contribution is $2L{\Phi }_1{\Phi }_2$, where $L$ is their linking number; self-contributions measure twisting and writhing of the tubes. This explains the topological content of magnetic helicity. Conservation of this one scalar is a consequence of the frozen structure, not a complete description of all that structure.

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.

Take $u<0$ for inward radial velocity, and assume a constant specific-heat ratio $\gamma >1$. The isentropic flow has $p=K{\rho }^{\gamma }$ and adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$. Steady spherical mass conservation and the radial Euler equation for fluid motion giveEliminating the mass density derivative gives $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$, orDifferentiating $c^2\propto {\rho }^{\gamma -1}$ gives the corresponding sound speed equation:At a smooth sonic point, $u_s^2=c_s^2$. A finite derivative requires the numerator of the flow equation to vanish too. ThusThis regularity requirement concerns a smooth transonic branch, rather than an arbitrary singular solution reaching unit Mach number.

Integrating the momentum equation using the specific enthalpy $c^2/(\gamma -1)$ yields the Bernoulli equationFor a finite positive mass accretion rate, its large-radius solution hasThese expressions satisfy both the flow equations asymptotically and show $u\to 0$ and $\rho \to {\rho }_{\infty }$. In particular $u^{\prime }/u\sim -2/r$ agrees with the first boxed equation. Combining the Bernoulli equation with sonic regularity additionally gives $c_s^2=2c_{\infty }^2/(5-3\gamma )$, so a finite positive sonic point occurs in the range identified below.

For the small-radius inward free fall branch, neglecting enthalpy against $GM/r$ in the Bernoulli equation predicts $u^2\sim 2GM/r$. The continuity equation then fixes $\rho \propto r^{-3/2}$, and consequently $c^2\propto r^{-3(\gamma -1)/2}$. The neglected-to-leading ratio isIt tends to zero precisely whenThis makes the assumed supersonic flow and leading free fall self-consistent. At equality, the ratio is of constant order rather than vanishing; above it, the proposed asymptotic balance fails. The isothermal limit $\gamma =1$ also permits free fall, with logarithmic rather than power-law enthalpy.

For the stellar funnel, write the magnetic scalar potential as $\psi =\mu \cos \theta /r^2$, where $R=r\sin \theta$ and $z=r\cos \theta$. Its magnetic dipole field has $B_r=-2\mu \cos \theta /r^3$ and $B_{\theta }=-\mu \sin \theta /r^3$, so a magnetic-field-line equation isThe edge of the polar flux tube meets the stellar surface at ${\theta }_{\ast }\simeq a/R_{\ast }$. Thus, near the pole, ${\theta }^2\simeq {\theta }_{\ast }^{2}r/R_{\ast }$, and its transverse radius is $R\simeq r\theta$. This proves the dipolar flux-tube area lawEquivalently magnetic flux conservation gives $A|\mathbf{B}|=\mathrm{constant}$ and $|\mathbf{B}|\propto r^{-3}$. The stipulated small cross section keeps the tube close to the polar axis, where its direction is approximately radial; the approximation is not valid all the way to arbitrarily large $r$.

Outside the star the prescribed field is a gradient, hence $\nabla \times \mathbf{B}=0$ and $\mathbf{j}=0$. Its Lorentz force density is therefore $\mathbf{j}\times \mathbf{B}=0$. The magnetic-pressure and magnetic-tension parts cancel in this exact current-free dipole. Field-aligned velocity also has $\mathbf{u}\times \mathbf{B}=0$, consistent with a steady prescribed field. The following calculation uses the narrow polar funnel approximation.

Now steady mass conservation reads $\rho uA=\mathrm{constant}$. With $A\propto r^3$, replace the spherical factor 2 by 3:The Bernoulli equation is unchanged because the magnetic field supplies no work or force. On a gravity-dominated free fall branch, $u^2\sim 2GM/r$, but $\rho \propto r^{-5/2}$ and $c^2\propto r^{-5(\gamma -1)/2}$. HenceThe same bound follows from regular transonic accretion in a power-law tube: $u_s^2=c_s^2=GM/(3r_s)$ and the reservoir-matched Bernoulli equation gives $c_s^2=2c_{\infty }^2/(7-5\gamma )$. The stellar surface cuts off the mathematical small-$r$ limit; the criterion refers to a funnel that becomes gravity dominated and crosses its sonic point before reaching that surface.

In a steady flow the ideal magnetohydrodynamic induction equation gives $\nabla \times (\mathbf{u}\times \mathbf{B})=0$. Since both vectors have only poloidal magnetic field components, write $\mathbf{u}\times \mathbf{B}=F_{\varphi }{\mathbf{e}}_{\varphi }$. Axisymmetry impliesFor a field regular on the axis, $C=0$; equivalently one may impose zero toroidal electromotive force. Under this physical condition, $\mathbf{u}$ is parallel to $\mathbf{B}$. Write $\rho \mathbf{u}=k\mathbf{B}$. Continuity and $\nabla \cdot \mathbf{B}=0$ then give $\mathbf{B}\cdot \nabla k=0$. The poloidal magnetic flux function has $\mathbf{B}\cdot \nabla \psi =0$, so locally on each connected magnetic surface,This regular-axis alignment of steady poloidal ideal flow needs its zero-circulation condition: axisymmetry alone does not imply this alignment on a domain excluding the axis. An explicit counterexample is available in a cylindrical annulus. Take nonzero constants $U,b$, positive ${\rho }_c,c_s$, andHere $\psi =-bz$, $\nabla \cdot \mathbf{B}=\nabla \times \mathbf{B}=0$ and $\mathbf{u}\times \mathbf{B}=Ub{\mathbf{e}}_{\varphi }/R$ has zero curl. The flow satisfies continuity, $p^{\prime }=-\rho {\Phi }^{\prime }$, and ${\nabla }^2\Phi =4\pi G\rho$. Its acceleration and Lorentz force density vanish, so it solves the steady equations while $\mathbf{u}$ and $\mathbf{B}$ are perpendicular. The nonzero toroidal circulation and the axial singularity explain why this counterexample is excluded from a regular polar accretion column. The remaining derivation uses the regular, aligned branch.

For a narrow stream tube, integrating continuity gives constant $A\rho |\mathbf{u}|$. On a nonzero-flow tube the magnetohydrodynamic mass loading $k$ is constant, so $A\rho |\mathbf{u}|=A|k||\mathbf{B}|$. Therefore $A|\mathbf{B}|$ is constant and $A\propto |\mathbf{B}{|}^{-1}$. This also follows directly from conserved magnetic flux through a flux tube.

For constant isothermal sound speed, $\nabla p/\rho =c_s^2\nabla \ln (\rho /{\rho }_0)$, with arbitrary reference density ${\rho }_0$. Project momentum along $\mathbf{B}$. The Lorentz force density has no component in that direction, and alignment implies $\mathbf{B}\cdot (\mathbf{u}\cdot \nabla )\mathbf{u}=\mathbf{B}\cdot \nabla (|\mathbf{u}{|}^2/2)$. Thus the isothermal magnetic Bernoulli integral isChanging ${\rho }_0$ only shifts the Bernoulli function by a constant. Since $k$ is constant along the magnetic field,Differentiating the isothermal magnetic Bernoulli integral therefore gives

In the slender polar accretion column, the dipolar flux-tube area is $A(z)=A_{\ast }(z/R_{\ast }{)}^3$. The decreasing axial field is a leading approximation, not an exactly solenoidal field $(0,0,B_z(z))$ throughout a cylinder. Indeed $\nabla \cdot \mathbf{B}={\partial }_z{B}_z\ne 0$ for that literal field. Near the axis a small radial component $B_R\simeq -(R/2){\partial }_z{B}_z=3R{B}_z/(2z)$ supplies the required radial divergence. Its magnitude is smaller by $R/z$, even though its divergence is leading order. Keeping this expanding flux tube while neglecting transverse forces yields the intended one-dimensional model.

Use the positive inward speed $v=-u_z$; the signed sonic velocity is $u_z=-c_s$. Conservation of mass and the Bernoulli equation giveEliminating ${\rho }^{\prime }$ yields the isothermal dipolar accretion equation,A smooth transonic branch requires both sides to vanish at its sonic point, soThis is the critical point of the flow equation; a generic subsonic solution need not cross it. A crossing in the exterior column requires $z_s\ge R_{\ast }$, and a sonic point strictly outside the star requires $z_s>R_{\ast }$. Differentiating the equation at the crossing gives $2(v_s^{\prime }{)}^2=3c_s^2/z_s^2$. For inward accretion that accelerates toward the star, $v_s^{\prime }=-\sqrt{3/2}{c}_s/z_s$.

The reservoir boundary condition fixes $ϵ=c_s^2\ln ({\rho }_{\infty }/{\rho }_0)$. At the sonic point, ${\Phi }_s=-3c_s^2$ and $v_s^2/2=c_s^2/2$, givingFinally use the sonic mass flux through the dipolar flux-tube area:This rate belongs to the smooth transonic branch of the idealized column, with $A_{\ast }$ the supplied total loaded area. The reservoir condition alone does not force every steady solution onto this branch. A real narrow dipolar column must match an outer flow where the slender approximation ceases to apply.

The solenoidal constraint on the magnetic field reduces to ${\partial }_x{B}_x+{\partial }_z{B}_z=0$. Locally, or globally in a simply connected cross-section, it permits a Cartesian magnetic flux function satisfyingThe transverse magnetic field lines are contours of $\psi$, because $\mathbf{B}\cdot \nabla \psi =0$. The independent component $B_y$ supplies the twist of the flux tube; it is not restricted by the solenoidal constraint.

For the Lorentz force, take the curl explicitly:Its cross product with $\mathbf{B}$ has transverse components $-({\nabla }^2\psi )\nabla \psi -B_y\nabla B_y$ and $y$ component ${\psi }_x{B}_{y,z}-{\psi }_z{B}_{y,x}$. ConsequentlyThe term $-B_y\nabla B_y/{\mu }_0$ is the transverse gradient of the magnetic pressure associated with the axial field, while the other terms include magnetic tension.

A toroidal magnetic field introduces an inward hoop force. In cylindrical coordinates, the radial magnetic tension is $-B_{\varphi }^2/({\mu }_{0}s)$, so cylindrical magnetostatic pressure balance in SI units becomesAt the field-free boundary $P(s_0)=p_e$. HenceRegularity on the axis gives $B_{\varphi }=O(s)$, which makes the integral finite and gives $B_{\varphi }(0)=0$. Evaluating at the axis proves the central-field identity:The positive hoop-tension contribution allows the central axial field to exceed $B_p$ without negative pressure. For an explicit regular example, put $x=s^2/s_0^2$, choose a constant $a>0$ and $0<\epsilon <1$, and take inside the flux tubeSubstitution verifies the radial balance, with continuous zero field at $s=s_0$ and positive gas pressure. For $a=3$, $\epsilon =1/4$, the central value is $B_z(0{)}^2=3B_p^2/2$.

To obtain the cross-sectional average, multiply the radial balance by $s^2$ and integrate by parts:The toroidal contributions cancel. The resulting flux-tube axial field virial identity isThis is at most $B_p^2$, and is strictly smaller whenever the weighted pressure integral is positive. In particular, continuous matching with $p_e>0$ guarantees strictness. If only $p\ge 0$ is assumed without a nondegenerate gas or continuous positive boundary pressure, the justified conclusion is the non-strict bound; the identity specifies exactly when equality occurs.

In SI units, magnetostatic equilibrium without gravity is $\nabla p={\mu }_0^{-1}(\nabla \times \mathbf{B})\times \mathbf{B}$. For a purely axial magnetic field, the magnetic tension term $(\mathbf{B}\cdot \nabla )\mathbf{B}$ vanishes. Thus the radial balance isMatching the magnetohydrodynamic total pressure to the field-free exterior gives the gas pressure and axial field bound:The inequality follows directly from nonnegative gas pressure. It is the magnetic pressure limit for an untwisted flux tube, without inward hoop tension.