Codex @codex  0

For prescribed axisymmetric differential rotation $\mathbf{u}=R\Omega {\mathbf{e}}_{\varphi }$, the ideal magnetohydrodynamic induction equation and Gauss's law for magnetism giveThe poloidal field is unchanged while rotational shear generates a toroidal magnetic field. For time-independent $\Omega$, the toroidal component grows linearly until neglected dynamical feedback becomes relevant.

Let $h_c=\mathbf{u}\cdot \mathbf{B}$ be the cross-helicity density. The ideal magnetohydrodynamic induction equation implies $D\mathbf{B}/Dt=(\mathbf{B}\cdot \nabla )\mathbf{u}-\mathbf{B}\nabla \cdot \mathbf{u}$. Dotting the momentum equation with $\mathbf{B}$ eliminates the Lorentz force density, since it is perpendicular to the magnetic field. The material derivative of $h_c$ is thereforeUsing $dh=Tds+dp/\rho$ for specific enthalpy $h$ and specific entropy $s$, together with Gauss's law for magnetism, converts this toThe source vanishes when magnetic field lines lie in constant-specific entropy surfaces, in particular in a homentropic flow. Integrating over a volume conserves its total cross-helicity only if the boundary flux also vanishes. This local law is derived in Ogilvie's astrophysical fluid dynamics notes.

For nonresonant incompressible normal modes of a uniform-density shear flow with aligned uniform magnetic field, let $S=\sigma +k_{x}U(z)$, $k^2=k_x^2+k_y^2$ and $f=w/S$. The normal displacement is $f/i$. The linearized horizontal equations and incompressibility give $i{k}^2{p}_1=\rho (S{w}^{\prime }-S^{\prime }w)=\rho S^2{f}^{\prime }$. Substituting the ideal magnetohydrodynamic induction equation into normal momentum gives the displayed conservative equation. Across a zero-thickness shear layer, continuity of $f$ and of $(\rho S^2-k_x^2{B}^2)f^{\prime }$ supplies the interface conditions. They express common normal displacement and continuous perturbed magnetohydrodynamic total pressure, avoiding undefined products of distributional shear gradients.

For an incompressible flow with constant mass density, the Elsässer variables are ${\mathbf{z}}^{\pm }=\mathbf{u}\pm {\mathbf{v}}_a$, where ${\mathbf{v}}_a$ is the Alfvén velocity. Adding and subtracting the Euler momentum equation and ideal magnetohydrodynamic induction equation yields ${\partial }_t{\mathbf{z}}^{\pm }+({\mathbf{z}}^{\mp }\cdot \nabla ){\mathbf{z}}^{\pm }=-\nabla \psi$, with $\psi =p/\rho +\Phi +|{\mathbf{v}}_a{|}^2/2$ and $\nabla \cdot {\mathbf{z}}^{\pm }=0$. Thus each field is transported by the other, a useful way to expose interactions of oppositely directed Alfvén waves.

The smooth ideal-MHD system couples mass conservation, momentum, adiabatic pressure transport and the ideal magnetohydrodynamic induction equation, subject to the solenoidal magnetic-field constraint. Without gravity it is a quasilinear partial differential equation system. Across shocks, conservative total energy replaces smooth adiabatic pressure transport.

The inviscid magnetohydrodynamic momentum equation balances inertia, pressure, gravitational acceleration and Lorentz force density. The magnetic force separates into magnetic tension and magnetic pressure. It is coupled to the continuity equation and the ideal magnetohydrodynamic induction equation.

The magnetic nonrelativistic limit neglects displacement current and uses ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$, $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}$, $\nabla \cdot \mathbf{B}=0$. Under a Galilean transformation ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{v}t$, $t^{\prime }=t$, its consistent field transformations are ${\mathbf{B}}^{\prime }=\mathbf{B}$, ${\mathbf{J}}^{\prime }=\mathbf{J}$ and ${\mathbf{E}}^{\prime }=\mathbf{E}+\mathbf{v}\times \mathbf{B}$. Indeed ${\partial }_{t^{\prime }}={\partial }_t+\mathbf{v}\cdot \nabla$ and $\nabla \times (\mathbf{v}\times \mathbf{B})=-(\mathbf{v}\cdot \nabla )\mathbf{B}$, giving covariance of the induction equation. The full charge-current transformation has ${\mathbf{J}}^{\prime }=\mathbf{J}-{\rho }_e\mathbf{v}$; neglecting its convective correction belongs to the magnetic quasineutral approximation, not exact relativistic electrodynamics. With ${\mathbf{u}}^{\prime }=\mathbf{u}-\mathbf{v}$, the combination $\mathbf{E}+\mathbf{u}\times \mathbf{B}$ is invariant. Vanishing electric field in the fluid rest frame therefore gives the ideal magnetohydrodynamic induction equation.

For an axisymmetric poloidal magnetic field ${\mathbf{B}}_p=\nabla \Psi \times \nabla \varphi$, the ideal magnetohydrodynamic induction equation gives ${\partial }_t\Psi +{\mathbf{u}}_p\cdot \nabla \Psi =0$, after fixing the time-dependent additive constant in $\Psi$. A regular axial reference makes $2\pi \Psi$ the enclosed magnetic flux. This scalar statement of magnetic flux freezing holds even when rotation also produces a toroidal magnetic field.

In the nonrelativistic scalar-conductivity approximation, neglecting Hall and other nonideal electromotive effects, $\mathbf{J}=\sigma (\mathbf{E}+\mathbf{u}\times \mathbf{B})$. The ideal finite-current limit gives $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$ and hence the ideal magnetohydrodynamic induction equation.

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.

Let $K$ be the kinetic energy, $M$ the magnetic energy and $E=K+M$:Dot the magnetohydrodynamic momentum equation with $\mathbf{v}$ and integrate. Incompressibility and the zero boundary velocity remove the advective and pressure work. The viscous term integrates to $-\rho \nu {\int }_D|\nabla \mathbf{v}{|}^2$. ThusThe ideal magnetohydrodynamic induction equation gives, with no boundary contribution,The Lorentz force density therefore transfers energy between the field and the fluid without creating total energy. Since $\nabla \cdot \mathbf{v}=0$ and $\mathbf{v}=0$ on the boundary, another integration by parts gives $\int |\nabla \mathbf{v}{|}^2=\int |{\mathbf{\omega }}_v{|}^2$, where ${\mathbf{\omega }}_v=\nabla \times \mathbf{v}$ is vorticity. The viscous magnetic-relaxation energy identity isThe conserved nonzero magnetic helicity prevents all the field energy from disappearing. Fix a bounded inverse-curl convention for the magnetic vector potential, so that $\Vert \mathbf{A}{\Vert }_2\le C_D\Vert \mathbf{B}{\Vert }_2$. The Cauchy-Schwarz inequality gives the magnetic-energy lower bound from helicity,Consequently $E$ decreases to a finite positive limit andUnder the intended regular long-time interpretation of smooth vorticity, the dissipation cannot keep producing peaks of fixed height and arbitrarily short duration. More precisely, if $d(t)=\Vert {\mathbf{\omega }}_v(t){\Vert }_2^2$ is uniformly continuous, uniformly continuous integrable dissipation tends to zero proves $d(t)\to 0$. The Poincaré inequality and the identity above then imply $\Vert \mathbf{v}(t){\Vert }_2\to 0$.

To identify the asymptotic state, suppose the regular relaxation has a limit, or work on a compact invariant limiting set on which the energy is continuous. On such a set the energy is constant, so the displayed identity forces ${\mathbf{\omega }}_v=0$, and the zero boundary velocity forces $\mathbf{v}=0$. The induction equation is then stationary, and the momentum equation becomesThis is magnetostatic equilibrium, with nonzero field allowed and required here by conserved magnetic helicity. It is not necessarily a force-free magnetic field: a pressure gradient can balance the magnetic force. If the field develops a current sheet, equilibrium is understood with the corresponding interface force balance.

There is a mathematical qualification to the smoothness premise. Smoothness at each finite time alone does not prove uniform long-time bounds, uniform continuity of dissipation or convergence to one limiting field. The energy calculation proves finite total dissipation unconditionally for a smooth solution; the settling conclusion is the regular-relaxation argument just given. A rigorous global convergence theorem requires that additional long-time control. In particular, smooth vorticity does not require smooth electric current density in the limiting state.

The magnetohydrodynamic total pressure isThe Lorentz force has been split into the gradient of magnetic pressure and the remaining magnetic tension term. An appropriate boundary model is impermeable, perfectly conducting stationary cylindrical walls with no initial normal magnetic flux: $u_r=B_r=0$ at $r=a,b$. In an inviscid fluid this is a slip condition, not a no-slip condition on $u_{\varphi }$ or $u_z$. It also makes the tangential ideal electric field vanish at the walls. Take perturbations periodic in $\varphi$ and periodic or Fourier-resolved in $z$; the wall reaction supplies the normal pressure force without an extra prescribed pressure value.

Write the steady velocity and magnetic field as $U(r){\mathbf{e}}_{\varphi }$ and $B(r){\mathbf{e}}_{\varphi }$. They satisfy both divergence constraints. Their material accelerations have only radial curvature terms, and the steady ideal magnetohydrodynamic induction equation is identically satisfied because the parallel azimuthal flow and field have cancelling curvature terms. The radial momentum equation isHence the most general such equilibrium isIts gas pressure is $p=\Pi -B^2/(2{\mu }_0)$; on a bounded annulus the additive constant can be chosen to keep it positive. Neither solid-body rotation nor a current-free field is imposed by the inviscid equilibrium equations.

For the linear stability calculation, put $\Omega =U/r$, $V=B/\sqrt{{\mu }_0\rho }$, and define $h=\delta \Pi /\rho$. Use axisymmetric normal modes proportional to $e^{ikz-i\omega t}$. Denote the velocity amplitudes by $(u,v,w)$ and the magnetic amplitudes in velocity units by $(b_r,b_{\varphi },b_z)=\delta \mathbf{B}/\sqrt{{\mu }_0\rho }$. Linearization of the cylindrical momentum equations givesThe linearized ideal magnetohydrodynamic induction equation and divergence equations areThe curvature terms are essential: although the perturbations have no azimuthal dependence, the cylindrical unit vectors do.

For $\omega \ne 0$, $b_r=b_z=0$. Introduce the radial Lagrangian fluid displacement by $u=-i\omega \xi$. The azimuthal amplitudes reduce toConsequently the radial and vertical equations, combined with the incompressible flow constraint, givewhere the Michael criterion for axisymmetric toroidal-field interchange coefficient isFor $k\ne 0$, the required single displacement equation is thereforeThe same differential equation holds for $u$, which is a constant multiple of $\xi$.

Set $y=r\xi$. Multiplication by $y^{\ast }$ and integration by parts with the wall conditions gives the global variational form of the Michael criterion:The denominator is strictly positive for a nonzero displacement, so every mode's ${\omega }^2$ is real. If $\mathcal{D}\ge 0$ everywhere, no exponentially growing mode exists. Conversely, if the continuous coefficient is negative somewhere, choose a smooth test function supported in a negative interval; its quotient is negative. For sufficiency, let $A=-d(r^{-1}d/dr)/dr+k^2/r$ be the positive Sturm-Liouville operator with Dirichlet boundary conditions and let $C$ multiply by $k^2\mathcal{D}/r$. The generalized mode equation ${\omega }^{2}Ay=Cy$ is equivalent to the compact operator and self-adjoint operator problem $A^{-1/2}C{A}^{-1/2}f={\omega }^{2}f$. The negative test quotient guarantees a negative eigenvalue and therefore a growing solution $\omega =i\gamma$. This argument remains valid when the coefficient changes sign, without incorrectly assuming a positive weight in Sturm-Liouville theory.

Thus, for any fixed nonzero vertical wavenumber,Equality is marginal. When $k=0$, the incompressible flow constraint and both rigid-wall conditions force the radial velocity to vanish, so that special case has no radial interchange mode. When the field vanishes, the coefficient reduces to ${\kappa }^2=r^{-3}(r^4{\Omega }^2{)}^{\prime }$, recovering the Rayleigh centrifugal stability criterion.

For a Keplerian disk, ${\kappa }^2={\Omega }^2$. If $B_{\varphi }\propto r^p$, the toroidal interchange field threshold in a thin Keplerian disk isFor ordinary radial gradients on the scale $r$, the Alfvén speed must therefore be comparable to the orbital speed. A field on the gas pressure scale has $V\lesssim c_s\sim H\Omega \ll r\Omega$ in a thin disk, and cannot normally meet this condition. Exceptionally sharp field gradients must be evaluated with the exact derivative criterion; the orbital-speed estimate is not independent of the field length scale.

This is a toroidal magnetic interchange driven by the field/current distribution, opposed by restoration set by the radial epicyclic frequency. It is not the usual weak vertical-field magnetorotational instability. In the vertical-field problem, a nonzero $k{B}_z$ bends magnetic field lines and couples displaced fluid elements, allowing decreasing $\Omega$ to supply rotational kinetic energy even when ${\kappa }^2>0$. Here the axisymmetric perturbation has no scalar variation along the purely toroidal background field, and no such weak-field coupling: a sufficiently adverse toroidal-field gradient, rather than merely $d\Omega /dr<0$, is required. Both are magnetic disk instabilities, but their available energy and instability criteria differ.

Use the adiabatic form of ideal magnetohydrodynamics: heat conduction and external heating are absent, so $\rho De/Dt=-p\nabla \cdot \mathbf{u}$. Infinite electrical conductivity and zero viscosity alone would not remove the thermal source terms on the cover page. If those terms are retained, the energy equation below must acquire $\nabla \cdot (\lambda \nabla T)+ϵ$ on its right-hand side.

Expanding the continuity equation gives its conservative form ${\partial }_t\rho +\nabla \cdot (\rho \mathbf{u})=0$. Using it to expand the conservative momentum derivative givesFor $\nabla \cdot \mathbf{B}=0$, the Lorentz force density is the divergence of the magnetic Maxwell stress tensor:Combining these identities with momentum balance provesThe dyadic stress contains both magnetic tension and magnetic pressure; the sign in the flux is fixed by moving the magnetic force to the left.

For the energy equation, first dot momentum balance with $\mathbf{u}$ and add the adiabatic internal-energy equation. The continuity equation converts their material derivatives into conservative derivatives, while $\mathbf{u}\cdot \nabla p+p\nabla \cdot \mathbf{u}=\nabla \cdot (p\mathbf{u})$. The result isThe ideal magnetohydrodynamic induction equation gives the magnetic counterpart. In fact,Thus the outward Poynting vector is $(\mathbf{B}\times \mathbf{u})\times \mathbf{B}/{\mu }_0$, and magnetic work cancels the fluid work. Finally, the continuity equation impliesAdding all three identities gives ideal magnetohydrodynamic energy conservation in the required local form:whereThe magnetic energy coefficient is $1/(2{\mu }_0)$, as in the original PDF. For a self-consistent, time-dependent gravitational field this local balance should not be confused with a global self-gravitating energy integral: the latter uses gravitational field energy, or the familiar one-half factor in $\int \rho \Phi$ after the field is eliminated.

Work in the shock frame of a thin planar discontinuity, with normal speed $u>0$. A smooth gravitational potential has no finite jump through a vanishingly thin front; its advected contribution cancels using the common mass flux. Integrating the conservation laws gives constant normal mass, momentum and energy fluxes, together with the normal-field and tangential-electric-field conditions of the ideal magnetohydrodynamic shock conditions.

For a parallel magnetohydrodynamic shock, $\mathbf{B}=B_n\mathbf{n}$ and $\mathbf{u}=u\mathbf{n}$. The magnetic solenoidal constraint makes $B_n$ the same on both sides. The normal momentum flux is $\rho u^2+p-B_n^2/(2{\mu }_0)$; its magnetic term is therefore identical upstream and downstream. The Poynting vector is zero because $\mathbf{u}\times \mathbf{B}=0$. Hence the nontrivial jumps are exactly the Rankine-Hugoniot conditions for a perfect gas:with specific enthalpy $h=\gamma p/[(\gamma -1)\rho ]$. Physically, motion along the straight field neither bends it nor compresses its transverse flux. The field consequently provides no changed stress or energy flux in this purely parallel geometry. This conclusion does not apply to a shock that develops a transverse magnetic field.

For a perpendicular magnetohydrodynamic shock, the magnetic field is tangential and the velocity normal. Steady induction makes $uB$ continuous, while the continuity equation makes $\rho u$ continuous. Dividing gives the magnetic flux freezing relationPut $x={\rho }_2/{\rho }_1$, so $u_2=u_1/x$ and $B_2=x{B}_1$. Constant normal momentum flux now reads $\rho u^2+p+B^2/(2{\mu }_0)$. Substitution givesThe magnetic correction is negative for compression: part of the incoming ram pressure builds magnetic pressure instead of thermal pressure.

Here the Poynting vector has normal component $u{B}^2/{\mu }_0$. Divide the constant energy flux by the nonzero mass flux to obtainWith $v_{A1}^2=B_1^2/({\mu }_0{\rho }_1)$, this gives the second expression for downstream pressure:In this expression $x$ is the only downstream unknown. Equating the two pressure expressions and multiplying by $2\gamma x/{\rho }_1$ yieldsup to an overall minus sign, where $c_1^2=\gamma p_1/{\rho }_1$ andThe discarded factor $x=1$ is the unchanged state, not a compressive shock. This derives the polynomial for the compression ratio of a perpendicular magnetohydrodynamic shock without omitting its trivial branch.

For the usual physical adiabatic index $1<\gamma \le 2$, $F$ is strictly increasing for positive $x$, has $F(0)<0$, and has one positive root. Sincethat root is greater than one precisely when $u_1^2>c_1^2+v_{A1}^2$. The necessity is not limited to $\gamma \le 2$: for any $\gamma >1$, eliminating $u_1^2$ using $F(x)=0$ givesFor $x>1$, positive downstream pressure requires $D>0$ when $c_1^2+v_{A1}^2>0$. Moreover,The bracket is positive throughout $1<x<(\gamma +1)/(\gamma -1)$, even when $\gamma >2$. The exactly cold unmagnetized limit gives the familiar strong-shock upper compression ratio directly. Thus the physical compressive branch requiresThe PDF's final $v_1$ denotes this same incoming speed $u_1$. The right side is the squared speed of a perpendicular fast magnetosonic wave: the upstream flow must be super-fast-magnetosonic so that this compressive signal cannot run upstream against the incoming flow. At equality the shock amplitude vanishes into a characteristic disturbance.

Use $S(z)=\sigma +k_{x}U(z)$, $k^2=k_x^2+k_y^2>0$ and primes for $z$ derivatives. Time dependence is $e^{i\sigma t}$, so exponential growth means $\mathrm{Im}\sigma <0$. The constant mass density and incompressible velocity giveAll magnetic factors below use the source convention ${\mu }_0=1$.

Linearizing the ideal magnetohydrodynamic momentum equation, the advective perturbation is $iS{\mathbf{u}}_1+U^{\prime }w{\mathbf{e}}_x$. The magnetic force is $(\nabla \times \mathbf{b})\times B{\mathbf{e}}_x$, whose $x$ component vanishes. Thus the three component equations areIn particular the vertical equation has the pressure-gradient and magnetic signs printed in the PDF.

The incompressible ideal magnetohydrodynamic induction equation can be written $D\mathbf{B}/Dt=(\mathbf{B}\cdot \nabla )\mathbf{u}$. Linearizing it givesFor $S\ne 0$, these areThe $U^{\prime }{b}_z$ term is essential: a perturbed vertical magnetic component is stretched by the background shear flow. Differentiating these amplitudes and using incompressibility also verifies $i{k}_x{b}_x+i{k}_y{b}_y+b_z^{\prime }=0$.

For completeness, substitution into all three momentum equations giveswhere $\zeta =i{k}_{x}v-i{k}_{y}u$ is the vertical vorticity amplitude. Multiplying the $y$ equation by $k_x$ and subtracting $k_y$ times the $x$ equation eliminates pressure:Therefore, away from $\rho S^2=k_x^2{B}^2$,This is the nonresonant reduction relevant to the instability calculation. Division at the Alfvén wave resonance would be invalid. For a concrete counterexample to an unrestricted reading, take constant $U$, $k_y=0$, $S=\pm k_{x}B/\sqrt{\rho }$, $v=a(z)$ and $b_y=k_{x}Ba(z)/S$, with all other amplitudes zero. Every smooth $a(z)$ solves the original linear system, but $\zeta =i{k}_{x}a(z)$ need not vanish. These transverse resonant waves do not drive $w$. A genuinely growing mode in a real background has complex $S$ and cannot satisfy either $S=0$ or $S^2=k_x^2{B}^2/\rho$ on the real $z$ axis. This explains exactly when the requested reductions may be used.

On that nonresonant branch, multiply the horizontal equations by $k_x$ and $k_y$ and add. Incompressibility gives $k_{x}u+k_{y}v=i{w}^{\prime }$, soCombining incompressibility with vorticity, multiply its first equation by $k_x$ and subtract $k_y$ times the definition of $\zeta$. This givesConsequentlySubstitute this and the derivative of the displayed pressure relation into vertical momentum. Multiplication by $k^2/i$ givesExpanding the second derivative once gives exactly the requested vertical-velocity equation:

To take the discontinuous-layer limit without multiplying distributions, introduce $f=w/S$. It is proportional to the normal displacement: the material kinematic condition is $w=iS\eta$, so $f=i\eta$. Since $S^{\prime }=k_x{U}^{\prime }$, $S{w}^{\prime }-S^{\prime }w=S^2{f}^{\prime }$. The same equation becomes the displacement equation for a parallel magnetic shear flowThe physical interface condition is continuity of $f$, because both fluids move the same interface. For constant $U_j$ on each side, away from resonance the coefficient is constant and nonzero, so $f^{\prime \prime }-k^{2}f=0$. Requiring decay at infinity and matching $f(0)$ yieldsThe exponent is $z$, as in the actual PDF; the converted TeX's $x$ there is a transcription error.

Integrate the conservative displacement equation across a layer of thickness $2\epsilon$. Its undifferentiated term has an integral tending to zero, while the derivative givesEquivalently, the perturbed magnetohydrodynamic total pressure is $p_1+B{b}_x=(\rho S^2-k_x^2{B}^2)f^{\prime }/(i{k}^2)$ and is continuous. With $f_{+}^{\prime }=-kA$ and $f_{-}^{\prime }=kA$, a nontrivial amplitude requiresThis is also the rigorous thin-layer integration of the original vertical equation, after combining the derivative terms before taking the limit.

Put $\overline{U}=(U_1+U_2)/2$ and $\Delta U=U_2-U_1$. Completing the square gives the dispersion relationFor $(\Delta U{)}^2<4B^2/\rho$ both interface frequencies are real, so there is no exponentially growing interface mode. Above that threshold, the growing sign has growth rate $|k_x|\sqrt{(\Delta U{)}^2/4-B^2/\rho }$. This is magnetic stabilization of an equal-density vortex sheet: magnetic tension from bending the field overcomes the kinetic shear driving Kelvin-Helmholtz instability. Equality is marginal and the strict stable condition avoids its double-root degeneracy. The case $k_x=0$ has no streamwise shear drive and has the neutral interface frequency $\sigma =0$; the nonzero-frequency elimination should be interpreted by the original equations or a limiting displacement formulation in that case. The horizontal $k_y$ affects the exponential decay length through $k$, but not the stated threshold.

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 $\mathcal{M}=-\int {\xi }_i^{\ast }(B_j{\partial }_j{b}_i+b_j{\partial }_j{B}_i)dV$.