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Consider the nontrivial gravitational mode with $k\ne 0$. At ${\omega }^2=0$, horizontal momentum requires $ikW=0$, hence $W=0$ andSubstituting into the Poisson equation for Newtonian gravity and using $4\pi G{\rho }_0/c_s^2=2/H^2$ givesSet $\tau =\tanh (z/H)$ and $\nu =kH$. Since ${\partial }_z=(1-{\tau }^2){\partial }_{\tau }/H$, this becomes the associated Legendre equation of degree one and order $\nu$:

To verify the two explicit solutions, put $\eta =z/H$ and $T=\tanh \eta$. They arebecause $e^{2\eta }=(1+\tau )/(1-\tau )$. Differentiating $F_{+}$ twice giveswhich cancels identically against $[2(1-T^2)-{\nu }^2]F_{+}$. Reflection $\eta \mapsto -\eta$ proves the same for $F_{-}$. Their Wronskian is $2\nu (1-{\nu }^2)$, evaluated at $\eta =0$; thus they are generally independent but coincide or become dependent at the exceptional orders $0,\pm 1$.

Use $n=|k|H>0$ to analyze decay, which is unaffected by the sign of $k$. The solution decaying at $+\infty$ is proportional to $e^{-n\eta }(n+\tanh \eta )$. At $-\infty$ its leading term is $(n-1)e^{n|\eta |}$, so it also decays there only if $n=1$. At this valueTherefore the marginal fragmentation mode of an isothermal slab hasAt $n=1$ the second independent solution obtained by reduction of order grows at infinity, so the coincidence of the two explicit formulas does not supply an extra admissible mode. At $k=0$ a uniform vertical translation gives another neutral disturbance, and mass-preserving relabeling displacements can have $\delta \rho =\varphi =0$. Neither is the nonzero-wavenumber fragmentation threshold just calculated.

The unstable fragmentation modes have wavelengths greater than $2\pi H$. Gas-pressure support becomes less effective relative to self-gravity at long wavelength. The side of the threshold can also be checked from the energy identity: at $|k|=1/H$ the marginal mass density disturbance has zero quadratic energy. Keep that mass density disturbance and decrease $|k|$; the pressure integral stays fixed, while the gravitational term becomes more negative because the positive inverse operator $(-{\partial }_z^2+k^2{)}^{-1}$ increases as $k^2$ decreases. Thus the quadratic energy becomes negative, giving an unstable mode on the longer-wavelength side. This argument concerns the gravitational sector and its stated decay conditions.

For the Schwarzschild criterion, displace a small gas parcel upwards. Assume its composition is unchanged, it remains in pressure balance with its surroundings, and it moves quickly enough that heat exchange is negligible. The parcel obeys the adiabatic equation of state $P{\rho }^{-\gamma }=\mathrm{constant}$. Combining this with the ideal gas equation gives $T\propto P^{(\gamma -1)/\gamma }$, so its adiabatic temperature gradient is ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma =2/5$.

Let the surrounding temperature gradient be $\nabla =d\log T/d\log P$. For an upward displacement $\delta \log P<0$, the parcel's temperature relative to its new surroundings isIf $\nabla <{\nabla }_{\mathrm{ad}}$, this is negative: the parcel is cooler and, at equal pressure and composition, denser than its surroundings. It therefore sinks back; a downward displacement likewise gives a restoring buoyancy force. If the inequality is reversed, the displaced parcel continues to rise. The stability condition is thusEquality is the neutral boundary. There is no composition-gradient term because the ideal-gas parcel and the background have the same mean molecular weight.

In the thin upper stellar atmosphere, take a plane-parallel approximation with constant $g$, optical depth increasing inward, and zero overlying pressure $P(0)=0$. Hydrostatic equilibrium in optical depth gives $dP/d\tau =g/\kappa$. The ideal gas equation converts the stipulated opacity intoDefine $u=1+3\tau /2$, so the given grey atmosphere relation is $T^4=T_e^{4}u/2$. Integrating from the surface givesThis proves the required pressure law. In particular $P^2\propto g{T}_e^{-4\beta }$ times a function only of $u$ and $\beta$.

For the grey-atmosphere convection onset with density-linear opacity, differentiate the two profiles to obtain the radiative temperature gradient:It increases monotonically from zero, since $\beta >1$, and first reaches the adiabatic temperature gradient at $u_b^{\beta -1}=(4\beta +1)/5$. ThereforeBelow this point the radiative stratification is unstable and convection must transport some of the outward energy flux.

At the outer convection boundary, $u_b$ depends only on $\beta$, so $T_b=T_e(u_b/2{)}^{1/4}\propto T_e$ and $P_b\propto g^{1/2}{T}_e^{-2\beta }$ for fixed $\mu ,{\kappa }_0,\beta$. The matching constant in $P=K{T}^{5/2}$ is $K=P_b/T_b^{5/2}$. ConsequentlyThis temperature form of the polytropic relation follows from an ideal monatomic adiabat; its $K$ is not the density-polytrope coefficient in Question 3.

For the deep radiative boundary beneath a power-law-opacity convective envelope, negligible envelope mass and absent local energy generation give $m\simeq M$ and $L_r\simeq L$. Divide the radiative temperature equation by the pressure equation to get the stellar radiative temperature gradientThis is the gradient that would be needed if radiation alone carried $L$, and it is the quantity to compare with $2/5$ even while the actual region is convective. The deeper Kramers' opacity law gives $\kappa ={\kappa }_1\mu P{T}^{-9/2}/\mathcal{R}$. Substituting $P=K{T}^{5/2}$ yieldsAs $T$ rises inward this decreases, so the transition back to a stable radiative interior is atThus $T_{\mathrm{in}}^{7/2}\propto K^{2}L/M$. Using the outer matching and the Stefan–Boltzmann law,The mass and radius cancel. Hence, at fixed opacity coefficients and composition,This uses the same adiabat through the convection zone and distinguishes its inner transition temperature from its outer onset temperature.

For a regular, spherical star, hydrostatic equilibrium gives $dP/dr=-Gm(r)\rho (r)/r^2$. Multiply by $4\pi r^3$ and integrate to the surface. Integration by parts givesThe central boundary term vanishes for finite central pressure. Since $dm=4\pi r^2\rho dr$, the right-hand side of the integrated hydrostatic equation isThe last expression is the gravitational energy: adding a thin mass shell $dm$ to the interior mass $m$ contributes $-Gmdm/r$, counting each interacting pair once. HenceWith the specified zero surface pressure this proves the stellar virial theorem in the required form,

To derive the nonrelativistic electron-degeneracy pressure in a hydrogen-helium mixture, take fully ionized hydrogen and helium with hydrogen mass fraction $X$, the Electron number density isHere helium nuclei have mass $4m_p$ in the approximation used in the question. Filling the two spin states of each momentum mode up to the Fermi momentum $p_0$ givesThe isotropic momentum flux gives pressure $\frac{1}{3}\int pvdn$. For nonrelativistic Electrons $v=p/m_e$, so the electron degeneracy pressure isSubstitute the composition-dependent $n_e$ and simplify powers of two:Comparing $5/3=1+1/n$ gives the polytropic index $n=3/2$. The result assumes complete Electron degeneracy, $p_0\ll m_{e}c$, and negligible nuclear mass corrections.

Now adopt the stated approximate partially degenerate equation of state, keeping composition and hence $K,\mu$ fixed during contraction. Set $q=m/M$ and define the mass-averaged temperature ${\overline{T}}_M={\int }_0^{1}Tdq$. The stellar virial theorem from part (a) givesFor an index-$3/2$ stellar polytrope, the permitted gravitational-energy formula is $\Omega =-6G{M}^2/(7R)$. Also, using $\bar{\rho }=3M/(4\pi R^3)$,Thus, with $A=3K{\int }_0^1[3\rho /(4\pi \bar{\rho }){]}^{2/3}dq$,The normalized density $\rho /\bar{\rho }$ is a fixed function of $q$ for this homologous polytropic structure. Therefore its integral and $K$ fix $A$, which is independent of $M$ and $R$ for a common composition. The mean here is mass-weighted, rather than a volume average.

For the mass-averaged temperature maximum in a partially degenerate polytrope, put $\alpha =2\mu GM/(7\mathcal{R})$ and $\delta =\mu A{M}^{2/3}/(3\mathcal{R})$, both positive. As a function of $s=1/R$ the temperature is the concave quadraticIt reaches its maximum when $R=2\delta /\alpha =7A{M}^{-1/3}/(3G)$, withAt large radius, contraction increases the temperature as in a thermally supported star. Once Electron degeneracy provides a sufficiently large fraction of the support, further contraction can lower the thermal temperature. The formal zero-temperature radius is $R_{\mathrm{deg}}=\delta /\alpha =R_{\max }/2$, with $R_{\mathrm{deg}}\propto M^{-1/3}$. The continuation to negative temperatures would be unphysical; the adopted sequence only describes its positive-temperature part.

The contraction-limited central temperature has the same mass scaling. Indeed, if the total polytropic pressure is $P=K_{\mathrm{tot}}{\rho }^{5/3}$, the approximate equation of state gives $T=(\mu /\mathcal{R})(K_{\mathrm{tot}}-K){\rho }^{2/3}$. The fixed density profile therefore fixes $T_c/{\overline{T}}_M$. A sufficiently small mass never reaches the central temperature required for sustained hydrogen burning, producing a hydrogen-burning minimum mass. Such an object becomes a cooling, degeneracy-supported brown dwarf instead of settling onto the main sequence. More massive contracting objects can ignite hydrogen before reaching this degeneracy-limited maximum. This conclusion uses the nonrelativistic, fixed-composition model; it does not require extrapolation into the relativistic regime.

Here $\alpha$ is the thermal expansion coefficient and $\rho$ is the mass density. Let $K=|\mathbf{k}|$, $h^2=K^2-k_z^2$, $q=\mathbf{B}\cdot \mathbf{k}$, and $a^2=q^2/({\mu }_0\rho )$, the squared Alfvén frequency. We initially consider a nonzero frequency. The temperature and magnetic equations for a Fourier mode giveIn particular, the minus sign in the magnetic relation follows from $-i\omega \hat{\mathbf{b}}=iq\hat{\mathbf{u}}$. Substitution into the magnetohydrodynamic momentum equation, with $P=p/\rho$, givesFor incompressible flow, $\nabla \times (\hat{\mathbf{z}}\times \mathbf{u})=-{\partial }_z\mathbf{u}$. Taking the vertical component of the curl eliminates pressure and gives, with $\hat{\zeta }=(i\mathbf{k}\times \hat{\mathbf{u}}{)}_z$,Taking another curl and its vertical component uses $\nabla \times \nabla \times \mathbf{u}=-{\nabla }^2\mathbf{u}$ and $(\nabla \times \nabla \times (\theta \hat{\mathbf{z}}){)}_z=h^2\hat{\theta }$. Thusor, after eliminating temperature,The two equations for vertical velocity and vertical vorticity have a nontrivial solution only when their determinant vanishes. This proves the stratified magneto-Coriolis wave dispersion relationNo division by $\Delta$ was necessary, so a possible $\Delta =0$ is retained. Division by $\omega$, however, was necessary. Stationary disturbances must be checked directly in the original equations: for example, with $q\ne 0$, $\mathbf{u}=0$ and an appropriate solenoidal magnetic perturbation can balance temperature-induced buoyancy and pressure. Such static balances are separate from the nonzero-frequency wave branches.

Magnetoconvection is the competition between thermally driven convection and the force and transport of a conducting fluid's magnetic field. An imposed magnetic field resists motions that bend its field lines through magnetic tension, while magnetic diffusion permits slippage. The competition changes the onset threshold, preferred spatial scale, temporal behaviour and finite-amplitude organization. The geometry matters: a disturbance nearly invariant along a horizontal imposed magnetic field can avoid much of its tension, whereas an impermeable layer with a vertical imposed magnetic field cannot generally avoid bending it.

A useful quantitative example is a nonrotating Boussinesq layer of depth $d$, heated from below with temperature drop $\Delta T>0$ and threaded by a vertical magnetic field $B_0\hat{\mathbf{z}}$. Let $\nu$, $\kappa$, $\eta$ denote kinematic viscosity, thermal diffusivity and magnetic diffusivity. In thermal-diffusion units define the Rayleigh number, Prandtl number, magnetic-to-thermal diffusivity ratio and Chandrasekhar number byUse stress-free impermeable boundaries, fixed temperature perturbation zero, and magnetic boundary conditions keeping the field vertical. More explicitly, the normal modes below satisfy $w={\partial }_z^{2}w=\theta =b_x=b_y={\partial }_z{b}_z=0$ at $z=0,1$. For a horizontal wavenumber $a>0$, take $w,\theta \propto \sin \pi z$, $b_z\propto \cos \pi z$, and put $s=a^2+{\pi }^2$. The linear temperature and resistive induction equation give $(\lambda +s)\theta =w$ and $(\lambda +\zeta s)b_z={\partial }_{z}w$. Applying the solenoidal projection to the magnetohydrodynamic momentum equation removes pressure; the magnetic coupling is $Q\zeta {\partial }_z\mathbf{b}$, and the vertical projected buoyancy has coefficient $R{a}^2/s$. Substituting the temperature and magnetic responses gives the vertical-field magnetoconvection dispersion relationThese stated boundary conditions make the single vertical mode exact in this model; other boundaries change numerical thresholds.

For steady marginal convection, set $\lambda =0$. The threshold curve isThe imposed magnetic field adds a positive penalty. Differentiating with respect to $x=a^2$ shows that the minimizing wavenumber obeysAt $Q=0$ this gives $a_c=\pi /\sqrt{2}$ and $R_c=27{\pi }^4/4$. With increasing $Q$, the root moves to larger $x$ because the left-hand side is strictly increasing for $x\ge {\pi }^2/2$. The strong-field wavenumber selection in magnetoconvection limit isThus a strong vertical imposed field raises the steady threshold and selects narrower convection cells. At small horizontal wavenumber the incompressibility constraint makes a given vertical motion require large horizontal displacement and magnetic bending. Increasing horizontal wavenumber reduces this magnetic penalty, but increases viscous and thermal diffusion. Their balance selects the narrow cells. This statement is specific to the vertical-field geometry: wavenumber selection in oblique-field magnetoconvection can instead favour nearly field-invariant motions.

Oscillatory marginality in vertical-field magnetoconvection occurs because magnetic and temperature perturbations need not relax at the same rate. To exhibit the phase balance, substitute $\lambda =i\omega$ with $\omega \ne 0$ and write the neutral condition asThe imaginary part must vanish because $R$ is real. Dividing that imaginary part by $\omega s$ givessoA nonzero oscillation consequently requires $0<\zeta <1$ and a sufficiently strong imposed magnetic field. In this regime thermal diffusion can erase the temperature anomaly faster than magnetic diffusion relaxes bent field lines. The magnetic restoring force can outlast the buoyant anomaly, reverse the motion, and maintain the phase lag needed for oscillatory instability. The real part gives the oscillatory neutral curveOnly its portions with ${\omega }^2>0$ are physical. The first onset is the lower of the minimized steady and admissible oscillatory thresholds; the existence of an oscillatory neutral solution at a given wavenumber alone does not establish that it is first. For example, at fixed $P>0$ and $0<\zeta <1$, both minimizing wavenumbers grow like $Q^{1/6}$ for large $Q$, while $R_s\sim {\pi }^{2}Q$ and $R_o\sim D{\pi }^{2}Q$ with $D=\zeta (P+\zeta )/(1+P)<1$. At those oscillatory minima, the positive order-$Q$ contribution to ${\omega }^2$ dominates its order-$Q^{2/3}$ negative contribution, so the oscillatory branch is genuinely admissible and becomes the lower threshold. An overstability therefore precedes steady onset in this limit. More generally nonlinear convection can oscillate even when linear onset is steady; the linear calculation is not a classification of every time-dependent state.

The nonlinear regime introduces a different feedback. The magnetic Reynolds number $\mathrm{Rm}=Ud/\eta$ measures induction relative to diffusion for a finite convective velocity $U$. At large but finite $\mathrm{Rm}$, persistent convection rolls can redistribute magnetic flux. In a two-dimensional incompressible roll, a flux function $\mathcal{A}$ obeysCirculation repeatedly stretches the flux pattern, building gradients on which even small magnetic diffusivity acts efficiently. Over the flux-rearrangement time this permits flux expulsion: the interior of a vigorous roll becomes weakly magnetized, while magnetic flux is concentrated near its edges. The weak-field interior then suffers less magnetic inhibition and can support broader, stronger convection than the linear narrow-cell prediction for the original uniform magnetic field. Exact zero resistivity would preserve material flux and cannot by itself justify the relaxed flux-expelled state.

The weak interior field also follows from a steady leading-order balance for a coherent two-dimensional roll. Let $\psi$ be its streamfunction and suppose its interior streamlines are nested closed curves with $\nabla \psi \ne 0$. In the small-diffusivity interior, $\mathbf{u}\cdot \nabla {\mathcal{A}}_0=0$ implies ${\mathcal{A}}_0=F(\psi )$. Integrate the full steady flux equation over the region enclosed by one streamline. Because $\mathbf{u}\cdot \mathbf{n}=0$ there, the advection integral vanishes; the divergence theorem then gives $\oint {\partial }_n\mathcal{A}ds=0$. In a regular interior expansion this requires $F^{\prime }(\psi )\oint {\partial }_n\psi ds=0$. The normal derivative of $\psi$ has one nonzero sign around a regular nested streamline, so $F^{\prime }(\psi )=0$. Hence the leading interior flux function is constant, and its spatial gradient, which gives the two-dimensional magnetic field, vanishes. The mismatch with external flux is accommodated by magnetic boundary layers. This argument assumes the coherent steady roll and a regular interior; it does not prove that every turbulent flow expels flux.

The mean imposed vertical magnetic flux is conserved for horizontally periodic boundaries. In a schematic two-region picture, let a fraction $f$ have weak field $B_c$ in vigorous convection and the remainder have field $B_q$. ThenThe intensified field outside the active cells inhibits surrounding convection. This gives magnetic flux separation: vigorous weak-field patches coexist with strong-field regions whose motion is small or occurs on a finer scale. The redistribution is a consequence of nonlinear induction and back reaction, rather than disappearance of the net imposed flux.

This feedback can permit subcritical magnetoconvection. Infinitesimal perturbations decay below the conductive state's linear threshold because they encounter the full imposed field. A finite-amplitude roll can nevertheless persist by first expelling enough flux to reduce its own magnetic restraint. A sufficient finite disturbance can then switch between the stable conductive state and a convective state, producing hysteresis. A schematic amplitude equation makes the distinction between local linear stability and finite-amplitude persistence explicit:Here $\mu$ measures distance from linear onset; this is an illustrative normal form, not a claim that these coefficients have been derived for every magnetic layer. Writing $r=|A{|}^2$, the nonzero equilibria satisfyThey appear at a saddle-node bifurcation at $\mu =-c^2/(4d)$. For $-c^2/(4d)<\mu <0$, the zero state is stable, the lower radial-amplitude branch is unstable, and the upper branch is stable because its radial derivative has sign $c-2d{r}_{+}<0$. Thus finite-amplitude convection can coexist with a linearly stable conductive state below the onset threshold. Large $\mathrm{Rm}$ facilitates the flux-expulsion mechanism but is not a universal sufficient condition: geometry, the imposed magnetic field, the capacity to form coherent recirculating cells, and magnetic back reaction all matter. The broad nonlinear cells in weak-field regions can coexist with much narrower cells in magnetically dominated regions, linking the nonlinear flux distribution back to scale selection.

Reflection symmetry gives ${\Phi }_z(R,0)=0$. In a vertically stable thin disk, expand the vertical force as ${\Phi }_z(R,z)={\Omega }_z^{2}z+O(z^3)$, where ${\Omega }_z^2={\Phi }_{zz}(R,0)$ is the vertical epicyclic frequency squared. For local hydrostatic equilibrium and height-independent isothermal sound speed, $p=c_s^2\rho$ andThus $\rho ={\rho }_0\exp [-z^2/(2H^2)]$ with $H=c_s/{\Omega }_z$. Integrating the pressure and mass density through the disk gives the vertically integrated pressure of an isothermal disk:The force expansion is approximate in $H/R$, while $P=c_s^2\Sigma$ is exact for a height-independent sound speed.

In a nearly Keplerian disk, the vertical oscillation and radial epicyclic motion frequencies are close to the orbital frequency. The exact leading orbital differences are $\Omega -{\Omega }_z$ and $\Omega -\kappa$, soThus ${\omega }_1$ is the nodal precession rate of a slightly tilted orbit, and ${\omega }_2$ is the apsidal precession rate of a slightly eccentric orbit. Their signs distinguish prograde from retrograde precession. This also agrees with the sign convention in the complex tilt: a freely precessing ring obeys $W_t=i{\omega }_{1}W$.

For exactly Keplerian motion, ${\Omega }_z=\kappa =\Omega$, so ${\omega }_1={\omega }_2=0$ and $P/\Sigma =c_s^2$. In the local approximation, freeze the slowly varying coefficients and take $W,G\propto e^{i(kR-\omega t)}$. The inviscid warp equations of a nearly Keplerian disk becomeEliminating $G$ gives ${\omega }^2=k^{2}P/(4\Sigma )=k^2{c}_s^2/4$. HenceThese are bending waves of a Keplerian disk, carrying changes in orbital tilt. The local approximation needs wavelength small compared with $R$; the secular thin-disk warp equations also describe the long-vertical-wavelength regime $H\ll \lambda \ll R$.

For a nonspinning hole, $a=0$ gives ${\omega }_1=0$. A steady disk warp then satisfies $G_R=0$. The zero inner torque forces $G=0$ throughout, and the second warp equation, with positive $P$, forces $W_R=0$. The outer orientation fixes $W(R)=W_{\infty }$ everywhere. An apsidal precession term by itself therefore does not require a tilt warp.

For nonzero spin, set the time derivatives to zero. The two equations giveEliminating the internal torque yieldsUse ${\omega }_2=3\Omega /r$, ${\omega }_1=2a\Omega r^{-3/2}$, ${\Omega }^2{R}^3=GM$, and $P\simeq {\Omega }^2\Sigma H^2$. The first coefficient becomes $GM\Sigma H^{2}r/12$ and the second becomes $2aGM\Sigma r^{-3/2}$. ThereforeThis is the stationary disk warp equation under the stated weak-precession approximations. The inner condition is $W_R(R_{\mathrm{in}})=0$ when its coefficient is finite and nonzero, because $G=iGM\Sigma H^{2}r{W}_R/12$.

Let $R_g=GM/c^2$, so $R=R_{g}r$, and impose $H=ϵR$, $\Sigma \propto R^{-3/4}$. Converting both radial derivatives carefully to $r$ and dividing out the common positive factor givesor $W_{rr}+9W_r/(4r)+(24a/{ϵ}^2)r^{-9/2}W=0$. For $a>0$, introduceSince $x_r$ is proportional to $r^{-9/4}$, the first-derivative term cancels exactly, leaving $W_{xx}+W=0$. Thus $W=A\cos x+B\sin x$.

The outer boundary $r\to \infty$ is $x\to 0$, giving $A=W_{\infty }$. The specified inner-radius relation is exactly $x_{\mathrm{in}}=\pi$. The inner zero-torque condition $W_x(\pi )=0$ then gives $B=0$. Hence the stationary bending warp of a Kerr disk isThe tilt magnitude is $\beta (r)=|W_{\infty }||\cos x|$. Far out it tends to the imposed tilt. Moving inward, it decreases to zero atthen grows again with opposite signed tilt, reaching $W(r_{\mathrm{in}})=-W_{\infty }$. All tilt vectors lie on the same line in the complex $W$ plane. Their azimuth is $\arg W_{\infty }$ outside the node and differs by $\pi$ inside; azimuth is undefined at the aligned node itself. The tilt vector remains smooth there. Thus this is an oscillatory standing bending warp, with equal outer and inner tilt magnitudes and reversed inner orientation, rather than monotone inner alignment.

The prescribed real inner radius assumes $a>0$: for $a<0$, its square root is not real, so that particular boundary prescription does not apply. With a separately chosen positive inner radius and negative spin, put $x=4\sqrt{24|a|}{r}^{-5/4}/(5ϵ)$. The equation becomes $W_{xx}-W=0$, and the same outer tilt and zero inner torque giveThis negative-spin solution decreases smoothly toward the inner disk without a sign reversal. The specified positive-spin solution is the cosine profile above; the $a=0$ case is the flat solution already derived.

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.

At leading anisotropic order, perpendicular Alfvénic turbulence evolves independently of the compressive fluctuations. The slow modes and the entropy mode are transported by that turbulence. The fast magnetosonic frequency is of order $k_{\perp }(v_A^2+c_s^2{)}^{1/2}$, much larger than the reduced Alfvén frequency; the low-frequency reduced magnetohydrodynamics ordering filters out that fast-wave sector.

The passive evolution can be made explicit for a uniform ideal adiabatic plasma. Let $c_s^2=\gamma p_0/{\rho }_0$, $\delta =\delta \rho /{\rho }_0$, $b=\delta B_{\parallel }/B_0$, and define the normalized entropy perturbationUse $D={\partial }_t+{\mathbf{u}}_{\perp }\cdot {\nabla }_{\perp }$ and ${\nabla }_{\parallel }={\partial }_z+(\delta {\mathbf{B}}_{\perp }/B_0)\cdot {\nabla }_{\perp }$. Perpendicular force balance makes the leading total-pressure perturbation vanish:The ideal entropy equation gives $Ds=0$. The leading continuity and parallel-induction equations have the same compressive divergence term; subtracting them gives $D(\delta -b)=-{\nabla }_{\parallel }{u}_{\parallel }$. Parallel momentum gives $D{u}_{\parallel }=-{\nabla }_{\parallel }\delta p/{\rho }_0$. PutThe two compressive equations therefore becomeLet $c_T=v_A{c}_s/(v_A^2+c_s^2{)}^{1/2}$ be the tube speed and $h=c_s^{2}r/c_T$. ThenThus the slow and entropy fluctuations in reduced magnetohydrodynamics have characteristic amplitudes $f^{\pm }=u_{\parallel }\pm h$ obeyingThe two slow-wave populations propagate in opposite directions along the bent field lines while being mixed across them by ${\mathbf{u}}_{\perp }$. The entropy perturbation is carried by the fluid without a restoring-wave frequency. A pure entropy perturbation has $r=b=u_{\parallel }=0$ and $\delta =-s/\gamma$, so its density variation is accompanied by a temperature change with no leading pressure variation.

The coefficients and advecting fields in these three scalar equations come from the Alfvénic cascade; the scalars do not enter the leading perpendicular Alfvén equations. They therefore develop small perpendicular scales through passive mixing. If a scalar $f$ has constant variance flux ${\mathcal{F}}_f$ and is mixed on the Goldreich–Sridhar time ${\tau }_{\ell }\sim {\mathcal{E}}^{-1/3}{\ell }_{\perp }^{2/3}$, thenThe flux and amplitude need not equal the Alfvénic energy flux. This passive-spectrum statement uses the same locality and stationary inertial-range assumptions as part (a), and ignores additional damping. If a passive mode is absent initially and is not forced, its homogeneous leading equation keeps it absent; the Alfvénic cascade does not have to generate every compressive mode.

In the ideal, leading anisotropic system, there are five separately conserved quadratic cascade quantities: the two oppositely propagating Alfvén populations, the two slow-wave populations, and the entropy fluctuation. With the amplitudes of part (b), one convenient normalization iswhere ${\mathbf{z}}_{\perp }^{\pm }={\mathbf{u}}_{\perp }\pm \delta {\mathbf{B}}_{\perp }/\sqrt{4\pi {\rho }_0}$. Constant background factors convert the velocity variances into modal energies and the entropy variance into an entropy-related free-energy contribution. These are the five quadratic cascade invariants of anisotropic magnetohydrodynamic turbulence.

To verify conservation rather than just count polarizations, the perpendicular Elsässer variables obeyDot each equation with $2{\mathbf{z}}_{\perp }^{\pm }$ and integrate. The propagation and nonlinear terms become surface fluxes of $|{\mathbf{z}}_{\perp }^{\pm }{|}^2$, and the pressure term becomes a surface flux because the corresponding field is solenoidal. For periodic boundaries, or boundaries with vanishing invariant flux, all these terms vanish and $d{I}_A^{\pm }/dt=0$. Their sum is proportional to perpendicular kinetic plus magnetic energy, while their difference is proportional to perpendicular cross-helicity; the two Elsässer energy invariants contain both pieces of information.

Likewise the slow-mode equations giveHere ${\mathbf{u}}_{\perp }$ is divergence-free and ${\nabla }_{\parallel }$ differentiates along the divergence-free vector $\hat{\mathbf{z}}+\delta {\mathbf{B}}_{\perp }/B_0$, so both integrals are boundary fluxes. Finally $Ds=0$ gives $d{I}_e/dt=-\int {\mathbf{u}}_{\perp }\cdot {\nabla }_{\perp }{s}^{2}dV=0$. This proves all five quadratic conservation laws under the same boundary assumptions.

The two Alfvén cascades interact dynamically: each population strains the other and is needed for its nonlinear transfer between scales. Their integrated invariants are nevertheless conserved separately, so this interaction does not transfer one population's invariant into the other. The slow and entropy cascades are driven by the Alfvénic mixing and field-line motion, with no feedback on the perpendicular cascade at the retained order. Their injection rates set independent fluxes; those fluxes need not be equal, even when the resulting perpendicular spectra share the $-5/3$ scaling.