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Continue on the branch of steady planar ideal-MHD field-line invariants, with $k\ne 0$ and positive $\rho ,T$. Adiabatic entropy transport gives $\mathbf{u}\cdot \nabla s=0$. As all quantities are $z$-independent,The axial component of the ideal magnetohydrodynamic momentum equation has no pressure or gravitational term. Magnetic tension givesSince $k$ is constant along a magnetic field line, this becomesSubtracting the two axial invariants yields the useful compatibility relationThe squared ratio of horizontal flow speed to horizontal Alfvén speed is $M_h^2={\mu }_0{k}^2/\rho$; vector equality with the horizontal Alfvén velocity additionally requires $k>0$. If $V-U\ne 0$, the right side is a nonzero constant along the field line, so a regular finite $B_z$ cannot permit $M_h^2=1$ anywhere. This is the nondegenerate exclusion.

However, $B_z\not\equiv 0$ does not imply $V-U\ne 0$. The Alfvénic degeneracy of steady planar MHD is a genuine exception: take constant $\rho ,p,s$, $\Phi =0$, and a constant magnetic field with nonzero horizontal and axial components, and setEvery equation holds, but the horizontal velocity equals the horizontal Alfvén velocity everywhere despite $B_z\ne 0$. Therefore the printed pointwise exclusion needs a nondegeneracy hypothesis, such as $V\ne U$. The general conclusion is the boxed compatibility relation; when $V=U$, it says $(M_h^2-1)B_z=0$ and allows the Alfvénic branch.

Write horizontal components with a subscript $h$. The steady ideal magnetohydrodynamic induction equation gives $\nabla \times (\mathbf{u}\times \mathbf{B})=0$. Since ${\partial }_z=0$, its horizontal components implyThus $(\mathbf{u}\times \mathbf{B}{)}_z=C$ is a spatial constant. A vanishing axial electric field, $E_z=-C=0$, is needed to infer horizontal alignment. It is satisfied, for example, if the constant electric field vanishes at a boundary or at infinity. With this additional condition and ${\mathbf{B}}_h\ne 0$, define the magnetohydrodynamic mass loading $k$ by $\rho {\mathbf{u}}_h=k{\mathbf{B}}_h$, and put $U=u_z-k{B}_z/\rho$. ThenThe steady continuity equation and the solenoidal magnetic-field constraint giveAlso $\mathbf{u}\times \mathbf{B}=U{\mathbf{e}}_z\times \mathbf{B}$. Using the vector identity for the curl of a cross product, together with ${\partial }_z\mathbf{B}=0$ and $\nabla \cdot \mathbf{B}=0$, its curl is ${\mathbf{e}}_z\mathbf{B}\cdot \nabla U$. HenceThese are steady planar ideal-MHD field-line invariants: both quantities are constant along a magnetic field line.

The zero-electric-field condition does not follow from the printed stationarity assumptions alone. Constant $\rho ,p,s$, $\Phi =0$, $\mathbf{u}=u_0{\mathbf{e}}_x$ and $\mathbf{B}=B_0{\mathbf{e}}_y$, with $u_0{B}_0\ne 0$, satisfy every displayed ideal magnetohydrodynamics equation but have perpendicular horizontal velocity and magnetic field. Their $C=u_0{B}_0$ is nonzero, so no $k,U$ give the stated decomposition. The field-line decomposition is valid on the $E_z=0$ branch; as an unrestricted implication, the first claim has this counterexample. The subsequent calculations use that branch.

Linearizing the ideal magnetohydrodynamic induction equation about a static magnetic field and using $u^{\prime }={\partial }_t\xi$ 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 toIts divergence vanishes because it is a curl. Directly, the two products of first derivatives in ${\partial }_i{b}_i$ 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 ${\rho }^{\prime }=-\nabla \cdot (\rho \xi )$, so $\boxed{{\rho }^{\prime }=-\xi \cdot \nabla \rho }$.

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 areThere is no perturbed gravitational potential here because the field is fixed. Multiply by ${\xi }_i^{\ast }$ and integrate. The pressure term vanishes by the assumed boundary cancellation and $\nabla \cdot {\xi }^{\ast }=0$. Thuswhere $\mathcal{M}=-\int {\xi }_i^{\ast }(B_j{\partial }_j{b}_i+b_j{\partial }_j{B}_i)dV$.

Integrate the first magnetic term along $B$, using $\nabla \cdot B=0$, then insert the expression for $b$. This gives the positive term $\int |(B\cdot \nabla )\xi {|}^{2}dV$ and two mixed derivative terms. For clarity, the troublesome term transforms asThe first term on the right cancels the other mixed term. Combining the remaining derivative with the term containing ${\partial }_j{B}_k{\partial }_k{B}_i$ givesDifferentiating the equilibrium relation now combines the gradient of mass density and magnetostatic equilibrium terms:We obtain the incompressible magnetic displacement energy identityThe 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 $\boxed{{\sigma }^2\text{ is real}}$. 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 isConsequently the interchange stability of an incompressible magnetized atmosphere is governed by the numeratorFor ${\xi }_z=\sin ky$, 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 ${\rho }_z\le 0$ 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 ${\xi }_y=i{f}^{\prime }(z)e^{iky}/k$, 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 quotientunder 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 $-g{\rho }_z/\rho +k^2{v}_A^2$, with the Alfvén speed $v_A=B/\sqrt{{\mu }_0\rho }$.

For disturbances varying only along the imposed magnetic field, set $k=0$ and $x=m^2>0$. The steady horizontal-field magnetoconvection dispersion relation becomesExpand it in powers of $x$:Its derivative and second derivative areBecause $R$ diverges as $x\to 0$ and $x\to \infty$, the unique minimum is determined byFor large Chandrasekhar number, this equation first gives $x\sim {\pi }^3{Q}^{-1/2}$ and hence the requested strong-field steady magnetoconvection varying along the field selection law:

For the threshold through order unity, the leading $Qx+{\pi }^6/x$ contribution at its optimum is $2{\pi }^3\sqrt{Q}$. More precisely, the stationary equation givesThe correction to $Qx+{\pi }^6/x$ has no first-order contribution because that expression is already stationary at its leading minimizer. Also $3{\pi }^{2}x=O(Q^{-1/2})$ and $x^2=O(Q^{-1})$. ThereforeThe constant term $3{\pi }^4$ must be retained. This restricted family bends magnetic field lines, so its large stationary threshold is compatible with the unrestricted field-aligned minimum in the preceding part.

All invariants in this part refer to the selected magnetic field line. Its squared poloidal Alfvén number isAt the Alfvén surface crossing it equals unity, so ${\rho }_a={\mu }_0{k}^2$ and $M_{Ap}^2=1/y$. Subtract the two azimuthal integrals in part (a) to findFinite azimuthal fields and velocity at $y=1$ require the Alfvén-surface regularity condition for an axisymmetric wind $\ell =\omega R_a^2$. HenceHere $v_{\varphi }$ denotes the azimuthal velocity relative to field-line rotation. For gravity of a point mass at $z=0$, $\Phi =-GM/R=-GM/(R_{a}x)$.

Insert these quantities in the corotating energy integral with $w=0$, and divide by the gravitational energy scale $GM/R_a$. This gives the cold radial magnetohydrodynamic wind integral,withBoth are dimensionless. The apparent $0/0$ at $x=y=1$ must be evaluated on the smooth solution branch, not assigned an arbitrary value.

A constant-energy curve obeys $f_x+f_{y}dy/dx=0$. At a nondegenerate magnetosonic point, the vanishing coefficient of the mass density derivative requires simultaneous vanishing of its numerator:The curve must admit a real crossing slope. Expanding about such a point gives $f_{yy}{m}^2+2f_{xy}m+f_{xx}=0$, $m=dy/dx$, so a regular saddle point of a scalar function crossing has $f_{xy}^2-f_{xx}{f}_{yy}>0$. These are the form of the slow and fast regularity conditions in a warm wind, with the enthalpy contribution retained in $f$.

In the strictly cold approximation used to obtain the displayed $f$, the slow speed is zero. There is a cold-limit degeneracy of a slow magnetosonic point: a finite-speed interior slow crossing need not survive. Indeed,so this cold equation has no ordinary sub-Alfvénic slow critical point. A small retained enthalpy is needed to locate that point, which can approach the launching boundary as the sound speed tends to zero. The fast critical point still obeys the simultaneous conditions above wherever the reduced expression is nonsingular.

Let ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{v}t$, $t^{\prime }=t$ for a constant relative velocity $\mathbf{v}$. The Galilean transformation gives ${\nabla }^{\prime }=\nabla$ and ${\partial }_{t^{\prime }}={\partial }_t+\mathbf{v}\cdot \nabla$. In the magnetic Galilean limit of the pre-Maxwell equations, takeThe Ampère's law without displacement current and the solenoidal constraint are immediately unchanged. For Faraday's law, the constant-velocity vector identity is $\nabla \times (\mathbf{v}\times \mathbf{B})=\mathbf{v}\nabla \cdot \mathbf{B}-(\mathbf{v}\cdot \nabla )\mathbf{B}$. ThusThese are transformations of the magnetic nonrelativistic approximation. The exact charge-current transformation also contains the convective term $-{\rho }_e\mathbf{v}$ in ${\mathbf{J}}^{\prime }$; ignoring it requires the quasineutral magnetic limit. They are not exact transformations of the full Maxwell equations.

Since ${\mathbf{u}}^{\prime }=\mathbf{u}-\mathbf{v}$, the combination ${\mathbf{E}}^{\prime }+{\mathbf{u}}^{\prime }\times {\mathbf{B}}^{\prime }$ equals $\mathbf{E}+\mathbf{u}\times \mathbf{B}$. For an ideally conducting fluid the rest-frame electric field vanishes, so the ideal moving-conductor Ohm law is $\mathbf{E}+\mathbf{u}\times \mathbf{B}=0$. Substitution in Faraday's law givesThe ideal magnetohydrodynamic induction equation requires nonrelativistic fluid motion, negligible displacement current, and negligible nonideal terms in the electric-field relation. In particular, magnetic diffusion must be slow compared with advection: the magnetic Reynolds number $\mathrm{Rm}=UL/{\eta }_m\gg 1$, where ${\eta }_m=({\mu }_0\sigma {)}^{-1}$ is the magnetic diffusivity. The single-fluid moving-conductor Ohm law must apply on the scales considered, with corrections to that relation negligible; viscosity is not itself excluded by this induction equation.

For a smooth ideal evolution, magnetic flux freezing states that the magnetic flux through every material surface is conserved and that magnetic field lines are carried with the fluid. Their connectivity is preserved: ideal smooth advection does not reconnect them. These implications are stated here without proof, as requested.

Put $D_t={\partial }_t+\mathbf{u}\cdot \nabla$. The continuity equation and the expanded ideal magnetohydrodynamic induction equation areso $D_t(B_i/\rho )=(B_j/\rho ){\partial }_j{u}_i$. Because $\mathbf{B}=\nabla \times \mathbf{A}$, the induction equation permits the magnetic vector potential equation ${\partial }_t\mathbf{A}=\mathbf{u}\times \mathbf{B}-\nabla \phi$, with an arbitrary scalar electric potential $\phi$ carrying the gauge choice. Using $(\mathbf{u}\times \nabla \times \mathbf{A}{)}_i=u_j{\partial }_i{A}_j-u_j{\partial }_j{A}_i$ yieldsContracting with $B_i/\rho$ and differentiating that factor givesThe last two terms cancel by exchanging their dummy indices. Thus the requested material derivative isThis is material transport of magnetic helicity density.

For a material volume, mass conservation gives $d{\int }_{V(t)}\rho hdV/dt={\int }_{V(t)}\rho D_{t}hdV$. Apply this with $h=\mathbf{A}\cdot \mathbf{B}/\rho$, and use $\nabla \cdot \mathbf{B}=0$ and the divergence theorem:If the magnetic field lines do not penetrate the boundary, $\mathbf{B}\cdot \mathbf{n}=0$, so $\boxed{d{H}_m/dt=0}$.

For uniqueness under a regular, single-valued gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \Lambda$,Consequently this magnetic helicity is both conserved and gauge independent. As usual, vector potentials are understood to be globally regular potentials for the same field, rather than unrelated harmonic-circulation choices on a multiply connected restricted domain.

For $ab\ne 0$, a magnetic field line stays at fixed $R$ and satisfies $d\varphi /dz=B_{\varphi }/(R{B}_z)=a/b$. It is therefore a helix with axial pitch $2\pi b/a$; the sign of $a/b$ fixes its handedness. The axial magnetic flux is $\pi b{R}_0^2$, and the toroidal magnetic field supplies the twist. If $a=0$ the field is untwisted; if $b=0$ it consists of azimuthal circles rather than helical lines carrying axial flux.

A convenient globally regular magnetic vector potential has $A_R=0$ andIndeed, the cylindrical curl gives $B_z=R^{-1}d(R{A}_{\varphi })/dR$ and $B_{\varphi }=-d{A}_z/dR$, with $B_R=0$. The tangential potential components are continuous at $R_0$, so their derivatives introduce no unwanted delta-function magnetic field at the tube boundary. The $C$ term is a constant axial potential.

The most general regular potential is this one plus $\nabla \Lambda (R,\varphi ,z)$ for an arbitrary single-valued gauge function. A putative extra $C_{\varphi }/R$ term in the inner $A_{\varphi }$ is excluded by regularity at the axis; the exterior $1/R$ circulation is already fixed by the enclosed magnetic flux. Since the potential is globally defined across the axis and exterior, any remaining difference is a curl-free vector field, hence a gradient. The constant $C$ can itself be absorbed into a gauge function proportional to $z$.

For the displayed gauge family, inside the tubeand henceFor example, requiring $A_z=0$ outside sets $C=a{R}_0^2/2$ and gives $\boxed{H_m/L=\pi ab{R}_0^4/2}$. The equally regular choice $C=0$ gives zero. More generally, for a length-$L$ segment the gauge transformation contributesIn particular, $\Lambda =\Delta Cz$ shifts $H_m/L$ by $\pi b{R}_0^2\Delta C$. This gauge dependence of helicity in an open magnetic flux tube is possible because field lines penetrate the end caps when $b\ne 0$. There is no contradiction with part (b), whose no-penetration condition fails at those caps. A gauge or end-cap convention is needed to give a unique value per unit length. For $b=0$ the ordinary helicity is zero and this end-cap ambiguity disappears.

Matter corotating with the footpoint has fixed angular velocity ${\Omega }_0=(GM/R_0^3{)}^{1/2}$. Its effective potential in that rotating frame is the sum of Newtonian gravitational potential and centrifugal potential,Along an outward-inclined straight magnetic field line, write $R=R_0+s\sin i$ and $z=s\cos i$, where $s$ is distance from the footpoint. The exact potential along it isIts first derivative at $s=0$ is zero by the Keplerian rotation balance. At the footpoint the meridional Hessian is $\mathrm{diag}(-3{\Omega }_0^2,{\Omega }_0^2)$ with zero mixed derivative. HenceWhen $0\le i<30^{\circ }$, the potential initially rises and a cold parcel must overcome a barrier. When $i>30^{\circ }$, its quadratic curvature is negative, so the footpoint is a local maximum along the field and a small outward displacement is accelerated further outward:This is the thirty-degree magnetocentrifugal launching criterion. It describes the lack of a local cold-launch barrier, not a nonzero force on a parcel exactly at rest at $s=0$. At exactly $30^{\circ }$ the quadratic term vanishes; for the specified straight line the next term is $-7{\Omega }_0^2{s}^3/(16R_0)$, as in marginal straight-line launch at thirty degrees. Global wind escape additionally depends on conditions farther along the line.

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.

In magnetostatics, $-\nabla p-\rho \nabla \Phi +{\mathbf{f}}_L=0$. The plane-parallel Newtonian gravitational potential and all thermodynamic fields are independent of $y$, so the $y$ component of the Lorentz force must vanish:Away from a null of $\nabla \psi$, this says that $B_y$ is constant along each connected contour of the Cartesian magnetic flux function. Thus, on a regular flux region,This statement is local; disconnected contours with the same numerical flux label need not share one global function without the usual flux-tube connectivity assumption.

Since $\nabla B_y=b^{\prime }(\psi )\nabla \psi$, the transverse magnetostatic equilibrium equation becomesThis is a Cartesian magnetostatic flux-function equilibrium. Projecting it along a transverse magnetic field line gives hydrostatic balance along that line; projecting across the line balances the gas pressure and weight against magnetic pressure and magnetic tension.

Use Faraday's law ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$ and the solenoidal magnetic-field constraint $\nabla \cdot \mathbf{B}=0$. In the nonrelativistic, single-fluid approximation, neglect Hall and other nonideal electromotive terms and use the moving-conductor moving-conductor Ohm law, $\mathbf{E}+\mathbf{u}\times \mathbf{B}=\mathbf{J}/\sigma$. Infinite electrical conductivity with finite current gives $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, henceThis is the ideal magnetohydrodynamic induction equation. For comparison, neglecting displacement current in Ampère-Maxwell equation gives $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$; with uniform finite electrical conductivity it produces magnetic diffusion ${\eta }_m{\nabla }^2\mathbf{B}$, where the magnetic diffusivity is ${\eta }_m=1/({\mu }_0\sigma )$. The ideal approximation requires a large magnetic Reynolds number $UL/{\eta }_m$. Dropping displacement current is useful for this finite-conductivity comparison, but Faraday's law and the ideal Ohm relation already suffice for the ideal induction equation.

Expanding the curl and using the solenoidal magnetic-field constraint gives the material derivative formCombine this with mass conservation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$, to obtainNow parametrize a material curve by a fixed label $a$: $\mathbf{X}(a,t)$ obeys ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$. Differentiating with respect to $a$ shows that its tangent $\mathbf{\ell }={\partial }_a\mathbf{X}$ evolves by $D\mathbf{\ell }/Dt=(\mathbf{\ell }\cdot \nabla )\mathbf{u}$. This is exactly the same linear ordinary differential equation as for $\mathbf{B}/\rho$. Initially parallel tangents remain parallel by uniqueness, with a label-dependent proportionality factor constant along each particle trajectory. Thus magnetic field lines are transported as material curves, wherever the field and fluid flow are smooth and the field is nonzero. This is the field-line form of magnetic flux freezing.

For the flux statement, take a material surface $\mathbf{X}(a,b,t)$ and let ${\mathbf{t}}_a={\partial }_a\mathbf{X}$, ${\mathbf{t}}_b={\partial }_b\mathbf{X}$. Both tangents obey $D{t}_i/Dt=({\partial }_j{u}_i)t_j$. Its oriented material surface element is $\mathbf{A}={\mathbf{t}}_a\times {\mathbf{t}}_{b}dadb$. Differentiating the cross product, rather than assuming its transport rule, givesEquivalently, $D\mathbf{A}/Dt=(\nabla \cdot \mathbf{u})\mathbf{A}-(\nabla \mathbf{u}{)}^T\mathbf{A}$, with $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$. Contracting this derived rule with the induction equation gives a pointwise cancellation:Integrating over the fixed material labels therefore proves conservation of flux through an open material surface:The surface need not be closed; its boundary is carried with the fluid. The result follows from material transport, rather than from the zero flux through a closed surface.

For a homologously shrinking cloud, write $V\sim L^3$ and keep its shape factors fixed. Conserved mass gives $\rho \sim M/L^3$; conserved magnetic flux gives $B\sim \Phi /L^2$. Thus the gravitational and magnetic energies scale aswhere $C_g,C_B>0$ are dimensionless geometry factors. Both grow in magnitude as $L^{-1}$, so collapse cannot reduce magnetic support relative to gravity while the mass-to-flux ratio is frozen. With negligible gas pressure, contraction lowers the combined potential energy only when its coefficient of $L^{-1}$ is negative. Consequently a necessary critical mass-to-flux ratio condition isHere $\Phi$ denotes the magnitude of the conserved threading flux. The numerical coefficient depends on geometry and boundary conditions; the scaling argument does not determine it or make the condition sufficient in the presence of other support.

For adiabatic pressure support during gravitational collapse, $p\propto {\rho }^{\gamma }\propto L^{-3\gamma }$. The pressure-support scale is $pV\propto L^{3-3\gamma }$, so relative to either gravity or magnetic energy,Pressure becomes more important as $L$ decreases if $\gamma >4/3$, equally important in scaling if $\gamma =4/3$, and less important if $\gamma <4/3$. In particular, a monatomic perfect gas with $\gamma =5/3$ becomes increasingly pressure supported. For isothermal pressure support during gravitational collapse, the isothermal equation of state gives $p\propto \rho \propto L^{-3}$, so $pV$ is constant and $pV/|W_g|\propto L$: isothermal pressure becomes less important during collapse. The same comparisons hold against magnetic support because its energy has the same $L^{-1}$ scaling as gravity.

Write $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative and $q=\mathbf{u}\cdot \mathbf{B}$ for the cross-helicity density, so $H_c={\int }_{V}qdV$. The ideal magnetohydrodynamic induction equation of ideal magnetohydrodynamics, with $\nabla \cdot \mathbf{B}=0$, givesThe dot product of the Euler momentum equation with the magnetic field has no Lorentz force contribution, because $\mathbf{B}\cdot [(\nabla \times \mathbf{B})\times \mathbf{B}]=0$. Consequently,Here $h=e+p/\rho$ is the specific enthalpy, $T$ the temperature, and $s$ the specific entropy; $e$ is the internal energy per unit mass. The first law of thermodynamics in this convention is $de=Tds-pd(1/\rho )$, so $dh=Tds+dp/\rho$. Substitution, followed by $\nabla \cdot \mathbf{B}=0$, yields the cross-helicity conservation lawThe vector triple product identity $\mathbf{u}\times (\mathbf{u}\times \mathbf{B})=\mathbf{u}q-u^2\mathbf{B}$ puts this in the equivalent form

For the fixed volume, the divergence theorem gives the exact cross-helicity balanceThus cross-helicity is constant when the net source equals the outward flux. Simple sufficient hypotheses are $\mathbf{B}\cdot \nabla s=0$ throughout the volume and $\mathbf{u}\cdot \mathbf{n}=\mathbf{B}\cdot \mathbf{n}=0$ on its boundary. The first condition includes a homentropic flow; the second prevents both relevant boundary fluxes. Periodic boundary conditions, or sufficiently rapid decay at infinity, provide alternatives. These statements concern smooth ideal magnetohydrodynamics: material specific entropy conservation alone, $Ds/Dt=0$, does not eliminate spatial specific entropy gradients along the magnetic field lines.

Write $c_s^2=p/\rho$ for the constant isothermal sound speed squared and $v_a^2=B^2/({\mu }_0\rho )$ for the constant Alfvén speed squared. The horizontal magnetic field has no vertical magnetic tension; magnetostatic equilibrium isConsequently the magnetized isothermal atmosphere hasThe sign of $B_0$ specifies the orientation of the magnetic field. The adiabatic sound speed is $v_s^2=\gamma p/\rho =\gamma c_s^2$, also constant.

For the proposed normal mode, let $A=\exp (i\mathbf{k}\cdot \mathbf{x}+z/(2H)-i\omega t)$, $a=1/(2H)$, and temporarily write $\mathbf{\xi }$ for its constant complex amplitude. The divergence amplitude is $\Delta =i{k}_x{\xi }_x+i{k}_y{\xi }_y+(i{k}_z+a){\xi }_z$. Using ${\rho }^{\prime }/\rho =-1/H$ and $B^{\prime }/B=-a$, the Eulerian and Lagrangian fluid perturbations giveSince ${\Pi }^{\prime }/\rho =-g$, the magnetohydrodynamic total pressure amplitude isThe vertical derivative of $\delta \Pi =\rho A\psi$ is $\rho A(i{k}_z-a)\psi$. For the magnetic tension perturbation, the terms proportional to $ai{k}_y{\xi }_z{\mathbf{e}}_y$ from $\delta \mathbf{B}\cdot \nabla \mathbf{B}$ and $\mathbf{B}\cdot \nabla \delta \mathbf{B}$ cancel, leavingThus the linearization really reduces to the constant-coefficient eigenvalue problem