Codex @codex  0

The neutral displacement equation gives, for each nonzero-frequency root,The vector displacements are parallel as complex polarization vectors, with this scalar amplitude and phase ratio.

For the rapidly damped root, $\mu -i{\omega }_d=-\mu +a^2/(4\mu )+O(a^4/{\mu }^3)$, and henceThe two fluids move almost equally and oppositely; collisions remove the counterflow at a rate close to $2\mu$.

For either weakly damped Alfvén wave, expand the ratio around unity:Equivalently, the exact slippage relation is ${\mathbf{\xi }}_i-{\mathbf{\xi }}_n=-(i\omega /\mu ){\mathbf{\xi }}_n$. Thus the relative slippage has size $|a|/(\sqrt{2}\mu )\ll 1$ and is primarily a small phase difference. The fluids almost share a common displacement, but magnetic tension acts directly on the ions and drag accelerates the neutrals. This imperfect locking causes ambipolar damping.

The sign of the damping is also checked by the perturbation energy balance. With periodic or vanishing-flux boundaries,It follows by dotting each linear momentum equation with its velocity and the induction equation with $\delta \mathbf{B}/(4\pi )$; magnetic-tension work cancels magnetic-energy change, pressure contributes only boundary flux, and the two drag terms combine into the negative square. Even though spatial diffusion was neglected, ion-neutral friction still dissipates the wave energy.

Write $u=u_x$, $w=u_y$ and $B=B_y$. The isothermal equation of state replaces the adiabatic thermal equation by $p=c_s^2\rho$. Since the fields depend only on $x,t$, the solenoidal constraint gives ${\partial }_x{B}_x=0$ and the $x$ component of the ideal magnetohydrodynamic induction equation gives ${\partial }_t{B}_x=0$. Hence $B_x$ is constant in both space and time.

The continuity, two momentum and transverse induction equations areHere $B$ is the transverse magnetic component and the constant $B_x$ is the longitudinal component. The longitudinal magnetic force is the gradient of transverse magnetic pressure; the transverse force is magnetic tension. In primitive variables the isothermal coplanar magnetohydrodynamic characteristic matrix isThe imposed $z$ components stay zero because their corresponding equations have zero right-hand sides under this reduction.

Let $c=v-u$ be a characteristic speed relative to the fluid and define the Alfvén velocity ${\mathbf{v}}_a=\mathbf{B}/\sqrt{{\mu }_0\rho }$. To derive the characteristic polynomial, an eigenvector with components $(\delta \rho ,\delta u,\delta w,\delta B)$ obeysEliminating $\delta u$ and $\delta w$ on a nondegenerate branch givesTheir consistency yields $(c^2-c_s^2)(c^2-v_{ax}^2)=v_{ay}^2{c}^2$. Equivalently, direct expansion of $\det (\mathbf{A}-vI)$ gives the same polynomial without dividing by $c$ or a possibly vanishing field component:ThusThe discriminant is $(c_s^2-v_a^2{)}^2+4c_s^2{v}_{ay}^2\ge 0$, and both squared speeds are nonnegative. These are the fast magnetosonic wave and slow magnetosonic wave families, transported with the mean flow. The equations are hyperbolic; on a generic state the four distinct real eigenvalues give four characteristic directions. Special field alignments can make speeds coincide or the slow speed vanish. There is no separate entropy characteristic because the thermal model is isothermal, and the independent out-of-plane Alfvén wave polarization is excluded by $u_z=B_z=0$.

To construct a simple wave in magnetohydrodynamics, take $\mathbf{U}=\mathbf{U}(a)$ along a state-space curve whose tangent is a right eigenvector of one selected branch:Substitution into the partial differential equations then gives the scalar transport equationSuch curves exist locally by integrating the eigenvector field away from degeneracies. For example, parameterizing by $a=\rho$ where its eigenvector density component is nonzero gives the explicit state-curve equationswith signed $c$ equal to one of the four relative wave speeds. These formulas are used only where their denominators are nonzero.

A particularly transparent nonlinear example is an isothermal perpendicular magnetosonic simple wave. Set the allowed constant $B_x=0$, choose $B=K\rho$ and keep $w$ constant. For the fast branch,These relations satisfy the eigenvector equations and therefore all four original equations wheneverThe amplitude-dependent characteristic speed has derivativeThis already proves the existence of genuinely nonlinear waves in the stated system.

More generally, the method of characteristics applied to initial data $a(x,0)=a_0(x)$ givesIf the initial speed decreases towards increasing $x$ somewhere, the denominator reaches zero in finite time. When the negative minimum is attained, the first steepening time isFaster trailing states catch slower leading states, giving shock formation by characteristic intersection. For the right-moving perpendicular fast example, a negative density gradient provides such a compression. Beyond $t_{\ast }$ a smooth simple wave is no longer sufficient and an appropriate shock solution is required. Rarefactive profiles instead spread. A linearly degenerate characteristic field has $v^{\prime }=0$ along its state curve and avoids this mechanism; the conclusion is that compressive nonlinear magnetosonic simple waves steepen, not that every possible MHD waveform must do so.

The magnetorotational instability extracts energy from differential rotation through magnetic tension. It can destabilize a disk whose specific angular momentum increases outward, even though such a disk satisfies Rayleigh's circulation criterion. The distinction is that an unmagnetized displaced parcel approximately conserves its own angular momentum, whereas a field line transfers angular momentum between parcels. If angular velocity decreases outward, an outward-displaced parcel lags its inner partner. Magnetic tension transfers angular momentum to the outer parcel and removes it from the inner one, making their radial separation increase. A sufficiently stiff magnetic connection suppresses the separation instead. The instability therefore requires a finite range of field-line-bending strengths.

Make this mechanism quantitative in the local shearing sheet of a cylindrical rotation profile $\Omega (r)$. Put $q=-d\log \Omega /d\log r$, and take $0<q<2$ so the unmagnetized flow is stable. Its radial epicyclic frequency is ${\kappa }^2=2(2-q){\Omega }^2$. Let the uniform field be vertical, $B_z{\mathbf{e}}_z$, with constant mass density, and use an axisymmetric horizontal disturbance proportional to $e^{st+ikz}$. Write $\mathbf{h}=\mathbf{b}/\sqrt{{\mu }_0\rho }$ and ${\omega }_A=k{B}_z/\sqrt{{\mu }_0\rho }$. The linearized ideal magnetohydrodynamics equations areThe first pair contains Coriolis acceleration and the epicyclic response; the last equation is the shear-winding of a radial field perturbation into an azimuthal one. Omitting that winding term would remove the driving mechanism.

Eliminating $h_r,h_{\varphi }$ gives $(s^2+{\omega }_A^2)u_r=2\Omega s{u}_{\varphi }$ and $s(s^2+{\omega }_A^2)u_{\varphi }+\Omega [(2-q)s^2-q{\omega }_A^2]u_r=0$. Their determinant is the ideal magnetorotational dispersion relationIts discriminant as a quadratic in $s^2$ is ${\kappa }^4+16{\Omega }^2{\omega }_A^2>0$. Since $2{\omega }_A^2+{\kappa }^2>0$, a positive $s^2$ exists precisely when its constant term is negative:Thus angular velocity, rather than specific angular momentum, must decrease outward. A Keplerian disk has $q=3/2$, ${\kappa }^2={\Omega }^2$ and is hydrodynamically stable, yet magnetic modes with $0<k^2{v}_A^2<3{\Omega }^2$ grow.

The growing root is $s^2=-{\omega }_A^2-{\kappa }^2/2+\frac{1}{2}\sqrt{{\kappa }^4+16{\Omega }^2{\omega }_A^2}$. Differentiation with respect to ${\omega }_A^2$ gives the maximum growth rate of ideal magnetorotational instability:For Keplerian shear these are $3\Omega /4$ and $15{\Omega }^2/16$. The ideal maximum is set by shear; reducing field strength reduces the fastest wavelength instead of reducing this maximum growth rate. This is why the exactly zero-field problem differs from a sequence of weak-field problems with progressively larger wave number. At fixed wave number, growth does tend to zero as the field vanishes. The wavelength must fit the actual disk, and diffusion eventually matters at sufficiently short scales.

The instability has the required angular-momentum-transfer sign. In a growing mode, $u_{\varphi }/u_r=(s^2+{\omega }_A^2)/(2\Omega s)>0$. At the fastest mode this ratio is one, while induction gives $h_{\varphi }=-h_r$ after allowing for their common Fourier phase. Thus the averaged Reynolds stress minus Maxwell stress tensor, $\rho ⟨u_r{u}_{\varphi }-h_r{h}_{\varphi }⟩$, is positive: angular momentum moves outward. Its product with $q\Omega$ is the energy supplied by the background shear. These local mechanisms and coefficient conventions are also checked in the primary MRI stability calculation.

More general fields require separating field-line bending from field gradients and curvature. At leading local order, for a stationary compatible field in an incompressible flow with axisymmetric wavevector $(k_r,0,k_z)$, the bending frequency isPressure elimination projects the radial force by $f=k_z^2/k^2$. The radial momentum equation becomes $s{u}_r-2f\Omega u_{\varphi }=i{\omega }_A{h}_r$; the azimuthal and induction equations retain the form above. Eliminating the amplitudes gives the oblique axisymmetric magnetorotational dispersion relationHere $k_z\ne 0$. The unstable range is $0<{\omega }_A^2<-fd{\Omega }^2/d\log r$. The poloidal magnetic field therefore determines whether an axisymmetric perturbation bends a field line. A locally uniform toroidal magnetic field does not enter $\mathbf{k}\cdot \mathbf{B}$ for axisymmetry, and can contribute magnetic pressure that is projected out of this incompressible leading-order problem. Thus a toroidal component does not suppress the weak-field MRI merely by being present. Conversely a purely toroidal field has no ordinary axisymmetric local bending MRI in this approximation. This does not imply that every globally curved toroidal equilibrium is stable.

A steady general poloidal magnetic field must also be compatible with the equilibrium rotation. The azimuthal ideal magnetohydrodynamic induction equation contains ${\partial }_t{B}_{\varphi }=r{\mathbf{B}}_p\cdot \nabla \Omega$. A stationary axisymmetric equilibrium with no meridional flow consequently requires ${\mathbf{B}}_p\cdot \nabla \Omega =0$, Ferraro's law of isorotation. A radial field crossing cylindrical differential-rotation surfaces would otherwise wind secularly; it cannot simply be substituted into a stationary normal-mode calculation. Boundary conditions and the along-field eigenvalue problem replace a freely chosen local $k$ in nonuniform or bounded equilibria. Strong enough field-line tension can then stabilize all admissible wavelengths.

An explicit axisymmetric effect of nonuniform field strength and curvature is obtained with a purely toroidal field $B_{\varphi }(r){\mathbf{e}}_{\varphi }$, constant mass density and cylindrical rotation. A radial fluid displacement changes the field byThis follows directly from $\mathbf{b}=\nabla \times (\mathbf{\xi }\times \mathbf{B})$ and incompressibility. Its radial curvature force, after absorbing the magnetic-pressure perturbation, is $-2B_{\varphi }{b}_{\varphi }/({\mu }_0\rho r)$. The axisymmetric azimuthal momentum equation is $s{u}_{\varphi }+(2\Omega +r{\Omega }^{\prime })u_r=0$. Combining it with the radial force and then projecting out pressure givesThis is the local Michael criterion for axisymmetric toroidal-field interchange. The new term can either stabilize or destabilize: for $B_{\varphi }\propto r^p$, $\mathcal{D}={\kappa }^2+2(1-p)v_{A\varphi }^2/r^2$. A current-free $B_{\varphi }\propto r^{-1}$ is stabilizing; $B_{\varphi }\propto r$ leaves this axisymmetric interchange coefficient unchanged; a sufficiently steep increase with $p>1$ can destabilize even a centrifugally stable rotation law. That last instability can persist without differential rotation, so its energy source is the magnetic-current equilibrium rather than the ordinary MRI. Nonaxisymmetric current-driven modes are outside the requested essay scope.

For mixed poloidal and toroidal fields, the full induction perturbation contains $(\mathbf{B}\cdot \nabla )\mathbf{\xi }-(\mathbf{\xi }\cdot \nabla )\mathbf{B}-\mathbf{B}\nabla \cdot \mathbf{\xi }$, and the linear Lorentz force contains both $(\nabla \times \mathbf{b})\times \mathbf{B}$ and $(\nabla \times \mathbf{B})\times \mathbf{b}$. These terms explain why gradients and curvature cannot be replaced everywhere by a scalar $k^2{v}_A^2$: they can draw on equilibrium-current energy and couple the radial and vertical motions. Compressibility introduces additional magnetic pressure and magnetosonic responses; a strong toroidal field can alter the dispersion even though it disappeared from the leading incompressible weak-field result.

Finally, a vertically or radially stratified general field also supports part of the equilibrium weight through magnetic pressure. A displaced flux tube carries its frozen-in mass-to-flux relation, so its mass density at the surrounding total pressure can differ from its new environment. This introduces magnetic buoyancy instability and competition with ordinary buoyancy. Axisymmetric interchange displacements can access this energy without an azimuthal wave number, while poloidal tension resists variation along the field. Growth that survives when $d\Omega /dr=0$ must not be attributed to shear-driven MRI. The general-field stability problem therefore combines shear, field-line tension, current/curvature forces, compressibility, buoyancy and boundary constraints; the uniform-field result isolates the basic angular-momentum feedback mechanism.

Write $\mathbf{u}={\partial }_t\mathbf{\xi }$. In the uniform equilibrium, the linearized ideal magnetohydrodynamic equations reduce toIntegrating from the undisplaced reference state, and using the uniform background magnetic field, givesThese are the perturbations induced by the fluid displacement; independent time-independent changes to the reference state are excluded. The linear Lorentz force density is $(\nabla \times \delta \mathbf{B})\times \mathbf{B}/{\mu }_0$. Its magnetic pressure and magnetic tension parts giveFor a Fourier mode, put $s=\mathbf{k}\cdot \mathbf{\xi }$, $q=\mathbf{B}\cdot \mathbf{\xi }$ and $K=\mathbf{k}\cdot \mathbf{B}$. Then $\delta p=-i\gamma ps$ and $\delta \mathbf{B}=i(K\mathbf{\xi }-\mathbf{B}s)$. Substitution givesIf $s=q=0$, the fluid displacement is perpendicular to both the wave vector and the magnetic field. Only magnetic tension restores it, andThis is the Alfvén wave. The Alfvén velocity is the field-directed propagation vector: $\omega =\pm \mathbf{k}\cdot {\mathbf{v}}_a$. The signed phase velocity normal to the wavefront is $\pm v_a\cos \theta \hat{\mathbf{k}}$; the group velocity is $\pm {\mathbf{v}}_a$. The distinction matters for oblique propagation. For generic directions the Alfvén wave has one transverse polarization, with $\delta \rho =\delta p=0$.

To obtain the other magnetohydrodynamic waves, take the scalar products with $\mathbf{k}$ and $\mathbf{B}$:The determinant condition for a nonzero pair $(s,q)$ isThus the fast magnetosonic wave and slow magnetosonic wave haveTheir fluid displacements lie in the plane of $\mathbf{k}$ and $\mathbf{B}$ and are generally compressive. In the fast magnetosonic wave, gas and magnetic pressure provide the stronger restoring combination; in the slow magnetosonic wave, their perturbations oppose one another. The phase speeds depend on direction. They satisfy $v_{\mathrm{sl}}\le \min (c_s,v_a)$ and $v_{\mathrm{f}}\ge \max (c_s,v_a)$. For parallel propagation the two speeds are $c_s$ and $v_a$, with a degeneracy between a transverse branch and the Alfvén wave. For perpendicular propagation $v_{\mathrm{f}}^2=c_s^2+v_a^2$ and $v_{\mathrm{sl}}=0$; the latter is a nonpropagating limiting disturbance. These special directions require interpreting the polarizations by continuity rather than assuming three distinct nonzero frequencies.

Use axisymmetry and $\nabla \cdot \mathbf{B}=0$. Since the purely azimuthal velocity has zero divergence, the ideal magnetohydrodynamic induction equation becomes ${\partial }_t\mathbf{B}=(\mathbf{B}\cdot \nabla )\mathbf{u}-(\mathbf{u}\cdot \nabla )\mathbf{B}$. Differentiation of the cylindrical unit vectors contributes to its azimuthal component, givingThe azimuthal component of the magnetic tension force isThere is no azimuthal pressure or gravitational force and no azimuthal advective acceleration for this motion. Because the poloidal magnetic field is divergence-free,These coupled induction and tension equations describe a torsional Alfvén wave.

Set $\xi =\tanh (z/H)$, so $h=1-{\xi }^2$ and $d/dz=(1-{\xi }^2)H^{-1}d/d\xi$. The vertical equation becomesFor $|\xi |<1$, division by the common factor gives the Legendre differential equation with eigenvalue $k^2{H}^2$. Requiring boundedness at both surfaces $\xi =\pm 1$ selectsThe second independent solution is unbounded at an endpoint. These are the bounded vertical modes of a sech-squared magnetized disk. The division by $H$ is outside the square root, as in the original PDF; the converted TeX places it incorrectly. The magnetic perturbation involves $F_n^{\prime }$, which tends to zero at either surface.

The constant $n=0$ function has $k=0$. The specified magnetic ansatz contains $1/(ik)$ and must not be applied to it. A uniform horizontal velocity with no magnetic perturbation is an epicyclic motion, not a growing magnetic mode; the zero-wavenumber formal degeneracy does not provide growing MRI at arbitrary field strength. Nontrivial magnetically coupled vertical modes have $n\ge 1$.

For $A>0$, the magnetorotational instability grows when $A<3{\Omega }^2$. Indeed, the quadratic in $s^2$ then has a negative constant term, and one positive root. For $A\ge 3{\Omega }^2$, the two roots are nonpositive, since their sum is negative, their product nonnegative, and their discriminant is ${\Omega }^4+16{\Omega }^{2}A>0$. The smallest nonzero vertical eigenvalue is $k_1^2=2/H^2$, so all vertical modes are stable ifThe strict inequality requested gives stability, and equality is marginal for $n=1$. A sufficiently strong field increases magnetic tension. The unstable MRI needs a sufficiently long vertical wavelength to exchange angular momentum without excessive restoring tension; bounded vertical structure imposes a smallest nonzero effective wavenumber. Beyond the finite-thickness magnetorotational instability criterion, no allowed magnetic mode is long enough to remain unstable.

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 $D={\partial }_t+w{\partial }_z$, $w=v_z$, and a prime for ${\partial }_z$. For the plane-parallel magnetohydrodynamic flow with imposed shear, $\nabla \cdot \mathbf{u}=w^{\prime }$, whileThe solenoidal magnetic field constraint gives $B_z^{\prime }=0$. The $z$ component of the MHD induction equation then gives ${\partial }_t{B}_z=0$. Thus $B_z$ is constant in both space and time.

Use the magnetic tension and magnetic pressure decompositionThe continuity equation and momentum equation now reduce toThe transverse components of the MHD induction equation areThe terms involving $a$ respectively accelerate the flow across the imposed shear flow and wind the transverse magnetic field. The closure is the isothermal equation of state $p=c_s^2\rho$, as printed in the PDF.

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.

Use perturbations $\mathbf{v},\mathbf{b},p,\vartheta$ for velocity, magnetic field, magnetohydrodynamic total pressure and buoyancy displacement variable. SetThis is advection by the background Keplerian shearing sheet. Subtracting the equilibrium before retaining first-order terms gives the nine linearized equations:The azimuthal coefficient $\Omega /2$ combines the Coriolis acceleration with perturbation advection of the background shear. The $-3\Omega b_x/2$ term is the winding of radial magnetic field by that shear. The tidal force cancels when the equations at a fixed position are subtracted; it does not add a separate Eulerian perturbation force. Magnetic pressure is already included in $p$, leaving only linear magnetic tension. In this normalization $\vartheta$ has dimensions of length because $D_0\vartheta =v_x$.

Let $A=k^2{v}_A^2>0$ and $C=N^2+{\Omega }^2$. Solving the radially stratified magnetorotational dispersion relation as a quadratic in $s={\sigma }^2$ givesBoth roots are real. If $A+N^2-3{\Omega }^2<0$, their product is negative, so exactly one root is positive and gives exponential growth. Conversely, when $A+N^2-3{\Omega }^2\ge 0$, the coefficient $2A+N^2+{\Omega }^2\ge A+4{\Omega }^2>0$, so both roots are nonpositive. Thus the instability criterion for a specified nonzero wavenumber isIn a local continuum of allowed wavenumbers, some unstable mode exists exactly when $N^2<3{\Omega }^2$. A finite disk only permits wavelengths fitting its vertical boundaries, so this existence statement also requires an allowed mode in the interval.

The growing root is $s_{+}$, whose derivative and curvature areSetting the derivative to zero gives the fastest-growing interior mode:For a real, nonzero wavenumber, this expression requires $|N^2+{\Omega }^2|<4{\Omega }^2$, equivalentlyNonnegative right-hand side permits the endpoints, but there the stationary point is at $k=0$, outside the nonzero-wavenumber mode used above. Substituting the interior maximizing value givesThis is the maximum growth rate of radially stratified magnetorotational instability, taking $\Omega >0$. For $N^2=0$ it reduces to the usual $3\Omega /4$, at $k^2{v}_A^2=15{\Omega }^2/16$. Negative $N^2$ enhances growth and extends the unstable band; positive $N^2$ suppresses growth and eventually eliminates it at $N^2\ge 3{\Omega }^2$. Magnetic tension enables angular-momentum exchange between displaced parcels, weakening the rotational stabilization that protected the adverse hydrodynamic stratification.