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Take $z$ positive downwards from the roof. Write $T_0=T_m-m{C}_0$, $C_B=(T_m-T_B)/m$, $\Delta T=T_0-T_B=m(C_B-C_0)>0$ and $\Delta C=C_B-C_0$. The cooled magma follows the decreasing liquidus from $(C_0,T_0)$ towards $(C_B,T_B)$; the initially superheated liquid starts at $(C_0,T_{\infty })$ above that liquidus. The solid has zero solute concentration. At each temperature in the mushy layer, a horizontal tie line joins the pure solid to the residual liquid on the liquidus. The lever rule gives the liquid fraction from the bulk concentration. Since $T_B>T_E$, this path stops before the eutectic temperature and does not cross the eutectic composition.

The porosity $\varphi$ denotes the liquid fraction, so $1-\varphi$ is the solid fraction. Equal phase mass densities remove volume-change flow; the stable density stratification suppresses convection. In the stagnant mushy-layer model, define bulk specific enthalpy and bulk solute concentration byThe arbitrary reference of specific enthalpy has been omitted. The two conservation laws in $0<z<h(t)$ areHere $\kappa =k_T/(\rho c_p)$ is the thermal diffusivity, with $k_T$ the thermal conductivity. The positive $L{\varphi }_t$ term records increased specific enthalpy on melting; during freezing it supplies latent heat. In the liquid $z>h(t)$, the heat equation and solute diffusion equation arewhere $D$ is the solutal diffusivity. Neglecting solute diffusion inside the mushy layer does not permit dropping $D{C}_{zz}$ in the liquid at this stage.

Use $T(0,t)=T_B$, liquid $T\to T_{\infty }$ and $C\to C_0$ as $z\to \infty$, and initially $h=0$, $T=T_{\infty }$, $C=C_0$ for $z>0$. These are the half-space boundary conditions appropriate before cooling reaches the bottom of the chamber. For a closed chamber of finite depth $H$, replace the far-field conditions by the specified bottom conditions, for example $T_z=C_z=0$ at an insulated impermeable floor; the half-space similarity solution then ceases to apply when the thermal penetration depth approaches $H$.

At $z=h(t)$, let $C_i$ and $T_i=T_m-m{C}_i$ be the common liquid concentration and temperature, and let ${\varphi }_i$ be the limiting porosity on the mush side. Continuity of $T,C$, the Rankine-Hugoniot condition for solute, and the interfacial conservation of energy giveThe first condition has the sign of solute rejection into the liquid. For the sharp-edge stagnant mushy-layer model with nonzero liquid solutal diffusivity, close the free-boundary problem by marginal equilibrium at a mush-liquid boundary:This prevents constitutional supercooling immediately ahead of the mushy layer. A small solid fraction jump at its edge is allowed: imposing ${\varphi }_i=1$ as well as these finite-$D$ conditions would generally overdetermine the reduced model. At the roof, $C=C_B$ follows from the liquidus, and no additional solute-flux boundary condition is needed in the diffusion-free mush.

To obtain the large-concentration similarity solution, putThe positive superheat specifies the physically intended liquid initial state. On a similarity solution, $C_i,{\varphi }_i$ are constant, so every newly incorporated level leaves the same constant $B={\varphi }_i{C}_i$. Write $C_i=C_0+\Delta C\delta$ and $B=\Delta C(\mathcal{C}+a)$. Then throughout the mushy layerFor $a,\theta =O(1)$, $\mathcal{C}\gg 1$, and $\mathcal{S}/\mathcal{C}=O(1)$, the effective heat capacity is constant to relative error $O({\mathcal{C}}^{-1})$. Thus setIntegration of the similarity solution ordinary differential equation gives the leading analytical profilesThe temperature is $T=T_0+\Delta T\theta$. In the mushy layer, the concentration and porosity are the displayed $C,\varphi$; in the liquid,These retain the liquid solutal boundary layer. For completeness, finite-$r$ values of $\delta ,a,\lambda$ are determined analytically, to the same large-$\mathcal{C}$ order, by settingand solvingThe solute jump gives ${\varphi }_i=1-b/(\mathcal{C}+\delta )$ exactly; replacing $\mathcal{C}$ by $\mathcal{C}+\delta$ on the right of the last equation retains that jump's next-order factor, but does not make the linearized thermal profiles exact at finite $\mathcal{C}$. The branch has $0\le \delta <1$, $0<\varphi \le 1$. All these profiles are leading asymptotic expressions, not exact solutions of the nonlinear finite-$\mathcal{C}$ heat equation.

For $r\to 0$ at fixed positive $\lambda$, the complementary error function asymptotic gives $K(\lambda /\sqrt{r})\sim \sqrt{\pi }\lambda /\sqrt{r}$, hence $\delta =O(r)$, $b=O(r)$, $a\to 0$, and ${\varphi }_i\to 1$. The result simplifies toIn particular, $\varphi (0,t)=C_0/C_B=\mathcal{C}/(\mathcal{C}+1)$ and $\varphi (h^{-},t)=1$. There is no finite interfacial latent heat jump in this limit: the latent heat is released continuously throughout the mushy layer. Matching the two temperature gradients givesThe printed final equation is missing the square on $\Omega$ in the exponential. The corrected exponent follows directly from differentiating $\mathrm{erf}(\Omega \eta )$ and is consistent with ${\Omega }^2=1+\mathcal{S}/\mathcal{C}$. The positive root is unique: the mush-side gradient decreases from infinity to zero with $\lambda$, whereas ${\theta }_{\infty }K(\lambda )$ increases. The zero-superheat limit is singular and is not described by a finite positive $\lambda$ in this two-region construction.

Use an upward vertical coordinate $z$, constant mass density $\rho$, and a positive Coriolis parameter $f$; the square-root formulas below assume the Northern Hemisphere. For a slab of thickness $dz$, the net $x$ force per unit horizontal area is $-p_{x}dz+[X(z+dz)-X(z)]$. Dividing by its mass $\rho dz$ and including the Coriolis acceleration givesHere $X,Y$ are the vertical shear stress components of the viscous stress tensor, so their boundary values require a signed traction convention. Linearization removes advective acceleration. Decompose the horizontal velocity into an exterior pressure response and an Ekman layer correction. On an f-plane these satisfyandThe exterior pressure response is in geostrophic balance when steady. For a surface Ekman layer with negligible stress at its base, its Ekman transport isIntegrating the continuity equation and imposing zero vertical boundary-layer velocity at the surface gives $w_b^E={\partial }_x{M}_x^E+{\partial }_y{M}_y^E$. Thus the base velocity, positive upward, isThe PDF prints the opposite sign. Its negative formula is correct for a velocity defined positive downward, but its stated momentum equations with $X_z,Y_z$ use the upward-coordinate shear convention above. This sign must be accounted for when interpreting the subsequent pumping equation.

For constant kinematic viscosity, write $W_E=u^E+i{v}^E$ and $\delta =\sqrt{2\nu /f}$. The steady Ekman layer equation is $\nu W_E^{\prime \prime }=if{W}_E$. Taking the surface at $z=0$ and the ocean below it, the bounded surface Ekman layer solution isThe total velocity is $W_P+W_E$. Prescribed wind stress determines the coefficient of this Ekman layer correction; it does not determine a relation between the wind and the independent pressure-driven velocity. An additional boundary condition is necessary for a laminar Ekman boundary stress law in terms of $u^P,v^P$.

The two printed laminar stress formulas and the positive pressure-Laplacian pumping formula are instead the standard Bottom Ekman layer relations. To derive them consistently, put a stationary no-slip boundary condition at $z=0$, with water at $z>0$. ThenConsequentlyThese are shear stress values; the actual bottom traction on the fluid has the opposite sign. Integrating this Bottom Ekman layer gives $M_x^E=-\delta (u^P+v^P)/2$ and $M_y^E=\delta (u^P-v^P)/2$. Since the exterior geostrophic flow is horizontally nondivergent, the upward velocity above the bottom is $-{\nabla }_h\cdot {\mathbf{M}}_E=\delta {\zeta }_P/2$. Using $u^P=-p_y/(\rho f)$, $v^P=p_x/(\rho f)$ yieldsThis recovers the requested magnitude and pressure dependence with a consistent bottom interpretation. The surface version would require its own specified boundary velocity and corresponding signs.

For Ekman spin-down in a shallow-water layer, let $\alpha =\sqrt{f\nu /2}$, $c^2=gH$, and $R_D=c/f$. Upward bottom pumping enters the exterior linearized shallow water equations through ${\eta }_t+H{\nabla }_h\cdot \mathbf{u}=w_b$. Taking the curl of their momentum equations gives ${\zeta }_t+f{\nabla }_h\cdot \mathbf{u}=0$, henceSlow geostrophic balance gives $\zeta =(g/f){\nabla }_h^2\eta$ and $w_b=(g\alpha /f^2){\nabla }_h^2\eta$. Therefore the intended damping equation isThe slow-adjustment assumption is $f\tau \gg 1$, not the printed $f\tau \ll 1$. The printed negative pumping term in continuity can alternatively describe downward extraction at the upper boundary, but it cannot be combined unchanged with the upward bottom pumping just derived. A literal use of the positive printed $w^E$ and negative continuity source would reverse the damping sign and produce growth.

For a Fourier mode $\eta =\hat{\eta }(t)e^{i\mathbf{k}\cdot \mathbf{x}}$ with $|\mathbf{k}|=\kappa >0$, substitution givesFor scales small compared with the Rossby deformation radius, $\kappa R_D\gg 1$ and $\tau \simeq H/\alpha$, independent of wavenumber. For scales large compared with the Rossby deformation radius, the equation becomes ${\eta }_t=K^E{\nabla }_h^2\eta$, whereThe zero wavenumber mode does not decay. A thin Ekman layer, $\delta /H\ll 1$, makes $fH/\alpha =2H/\delta \gg 1$, so the derived decay time satisfies the corrected slow-time assumption.

Let the constant mass density be $\rho >0$ and put ${\mathbf{v}}_a=\mathbf{B}/\sqrt{{\mu }_0\rho }$, the Alfvén velocity. Since both the velocity and the magnetic field have zero divergence, the magnetic tension and magnetic pressure decomposition of the Lorentz force givesThe Newtonian gravitational potential $\Phi$ remains in $\psi$; uniform mass density does not justify dropping a prescribed gravitational acceleration. Adding and subtracting the two equations proves the Elsässer variable equationsHere ${\mathbf{z}}^{\pm }=\mathbf{u}\pm {\mathbf{v}}_a$; each Elsässer variable is transported by the other.

Define $I_{\pm }={\int }_V|{\mathbf{z}}^{\pm }{|}^{2}dV$. Taking the dot product with $2{\mathbf{z}}^{\pm }$ gives the Elsässer energy invariant balanceThe divergence theorem proves $d{I}_{\pm }/dt=0$ whenever the net boundary flux vanishes. In particular, $\mathbf{u}\cdot \mathbf{n}=\mathbf{B}\cdot \mathbf{n}=0$ makes ${\mathbf{z}}^{\pm }\cdot \mathbf{n}=0$, so both fluxes vanish pointwise. Periodic boundary conditions or decay at infinity are also sufficient. These boundary conditions matter: fixed volume alone gives no conservation.

The kinetic energy plus magnetic energy isExpanding $|\mathbf{u}\pm {\mathbf{v}}_a{|}^2$ and using the cross-helicity definition yieldsBoth energy and cross-helicity therefore follow from the two Elsässer energy invariants. The quantity $E$ here is exactly the requested kinetic energy plus magnetic energy, without an additional gravitational term.

Use $v=|u_r|>0$ for the radial speed and distinguish the central mass $M$ from the Mach number $\mathcal{M}=v/c$. In each smooth nonzero-flow region, mass conservation, the adiabatic equation of state, and the Bernoulli equation giveHere $K>0$ is constant on a smooth isentropic flow branch, $c$ is the adiabatic sound speed, and $\mathcal{E}$ is the Bernoulli function. Either sign of $u_r$ is possible. The positive-radius nondimensionalization in the question requires $\mathcal{E}>0$; flows with $\mathcal{E}\le 0$ do not have these positive $r_s,x,\lambda$ and require another scaling.

Indeed, the supplied adiabatic process equations give $D(p{\rho }^{-3/2})/Dt=0$, so $K$ is spatially constant along a steady nonzero radial fluid flow. The adiabatic sound speed is $c^2=(\partial p/\partial \rho {)}_s$, and the specific enthalpy is $h=c^2/(\gamma -1)=2c^2$. The radial Euler momentum equation, $vdv/dr=-{\rho }^{-1}dp/dr-GM/r^2$, integrates to $\mathcal{E}$ because $dh=dp/\rho$ on this isentropic flow branch.

Put $D=[j(3K/2{)}^2{]}^{2/5}$. Eliminating the mass density from the continuity equation givesWith $y={\mathcal{M}}^{2/5}$, the Bernoulli equation becomesTherefore the physical constants of the spherical polytropic flow with adiabatic exponent three halves are

For $t>0$, $F^{\prime }(t)=4(t^5-1)/t^2$, so $F$ decreases from infinity to its unique minimum $F(1)=5$, then increases to infinity. This completely determines the algebraic curves. For $0<\lambda <1$, there are two disconnected branches at every $x>0$, one subsonic flow with $y<1$ and one supersonic flow with $y>1$. For $\lambda =1$, the line $y=x$ and a decreasing branch meet at $(1,1)$. Their slopes are $+1$ and $-1$, since the leading expansion is $F(1+\eta )=5+10{\eta }^2+O({\eta }^3)$. For $\lambda >1$, solutions exist only where $F(x)\ge 5\lambda$: if $x_{-}<1<x_{+}$ solve $F(x_{\pm })=5\lambda$, there is a forbidden interval $x_{-}<x<x_{+}$. At either endpoint the two branches turn at $y=1$ and have an infinite slope.

A smooth sonic point needs $F^{\prime }(x)=F^{\prime }(y)=0$; hence a regular sonic transition requires $\lambda =1$ and $r=r_s$. The critical speed of a polytropic flow is then $c_s^2=GM/(2r_s)=2\mathcal{E}$, and the allowed mass flux is $j=(2\mathcal{E}{)}^{5/2}{r}_s^2/(3K/2{)}^2$. Here the subscript on $c_s$ denotes evaluation at the sonic point, rather than the isothermal sound speed convention used in the next solution.

The simple transonic branch is $y=x$. Direct substitution givesThus the radial velocity has constant magnitude, while the mass density and pressure scale as $r^{-2}$ and $r^{-3}$. For an outward fluid flow, this is a subsonic flow becoming a supersonic flow as its adiabatic sound speed decreases.

On the other transonic branch, $y$ decreases as $x$ increases. At large $r$, the terms $4/y$ and $x^4$ dominate, giving $y\sim 4x^{-4}$. At small $r$, the terms $y^4$ and $4/x$ dominate, giving $y\sim 4^{1/4}{x}^{-1/4}$. The resulting asymptotic expansions areFor inward fluid flow, these are the nearly static reservoir at infinity and the inner free fall region of Bondi accretion. The specific enthalpy is smaller than $GM/r$ in the inner limit, which explains free fall. Reversing the radial velocity gives a decelerating outflow with the same profiles. The formal small-$r$ limit describes the point-mass model: if the spherical gravitating body has a finite surface, the exterior solution stops there.

A stationary, nonradiative normal shock wave conserves the mass flux and the sum $v^2/2+h$, while the Newtonian gravitational potential is continuous across its thin layer. Hence $\mathcal{E}$ and $r_s$ are unchanged. The entropy production in a perfect-gas shock increases $K=p/{\rho }^{3/2}$, soMore explicitly, the Rankine-Hugoniot condition for $\gamma =3/2$ gives, with upstream $m={\mathcal{M}}_1^2>1$,Thus the shock wave jumps vertically, at the same $x$, from the supersonic flow to a subsonic flow on a larger-$\lambda$ curve. The figure illustrates an outward flow on $y=x$ with a shock wave at $x=1.4$; the downstream subsonic flow has $\lambda \simeq 1.224$ and continues to larger $x$. The shock wave radius is an additional boundary-data choice, rather than being determined by the smooth equations alone.

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

For a static ideal magnetohydrodynamics equilibrium with conservative gravitational acceleration, the magnetohydrodynamic energy principle gives a Hermitian operator for the restoring force in the mass density weighted inner product, provided the surface terms vanish. The factor $e^{z/(2H)}$ removes the exponential mass density weight. This can be checked directly, without assuming stability: set $s=v_s^2$, $b=v_a^2$, and $q=s+b$. Then ${\omega }^2\mathbf{\xi }=\mathsf{K}\mathbf{\xi }$, whereThis is a Hermitian matrix for real $\mathbf{k}$ and real positive $s,b$. Therefore every squared frequency ${\omega }^2$ is real; negative values give exponential growth, rather than an oscillation with a complex squared frequency. In an infinite atmosphere these are generalized plane waves, not finite-total-energy global modes.

At ${\omega }^2=0$ and $k_y\ne 0$, the $y$ equation, after cancelling its magnetic terms, gives $s\Delta =g{\xi }_z$. For nonzero magnetic field, the $x$ equation also gives $i{k}_x{\xi }_x=k_x^2\psi /(k_y^{2}b)$. Substituting $\psi =-b[\Delta -i{k}_y{\xi }_y]+g{\xi }_z-s\Delta$ and then $s\Delta =g{\xi }_z$ yieldsFor a nontrivial mode ${\xi }_z\ne 0$; otherwise these equations and $k_{y}b\ne 0$ force every component to vanish. Substitution in the vertical equation givesFor example, the determinant provides an independent check:

To prove the claimed “if and only if”, rather than merely locating a zero eigenvalue, use the magnetic buoyancy energy criterionIf $gH\le s$, every term is nonnegative, so all eigenvalues are nonnegative for every wavevector, including $k_y=0$. If $gH>s$, choose $k_z=0$, any fixed $k_x\ne 0$, and ${\xi }_z=1$, ${\xi }_x=-a/(i{k}_x)$, ${\xi }_y=g/(si{k}_y)$. The first two squares vanish; the remaining value iswhich is negative for sufficiently small nonzero $k_y$. The Rayleigh quotient of this displacement is therefore negative, proving a negative eigenvalue and an unstable normal mode. This uses precisely the assumption that boundary conditions do not exclude the chosen wavelengths. Equality is a neutral threshold, not exponential growth.

Finally the plasma beta is $\beta =p/[B^2/(2{\mu }_0)]=2c_s^2/v_a^2$. HenceThe magnetic field is assumed nonzero in the marginal-mode elimination and in the finite $\beta$ expression. The positive-sound-speed, nonmagnetic limit remains covered by the energy identity; for the usual $\gamma >1$ it is stable. Parker instability occurs because magnetic pressure supports part of the atmosphere's weight while the gas can drain along gently bent magnetic field lines. The resulting buoyant displacement can release Newtonian gravitational potential energy; choosing long wavelengths along the magnetic field makes the stabilizing magnetic tension small. The completed-square calculation quantifies this competition, including the adiabatic process restoring force.

Let the free surface be at $r=R_1$ and take the Newtonian gravitational potential to vanish at infinity. Mass conservation fixes $R_1=(3M_1/(4\pi {\rho }_1){)}^{1/3}$. The enclosed mass is $m(r)=4\pi {\rho }_1{r}^3/3$, and hydrostatic equilibrium requires $d{p}_0/dr=-{\rho }_{1}Gm(r)/r^2$. With vacuum outside, $p_0(R_1)=0$, givingThe pressure is zero outside. A prescribed constant external pressure would simply add that constant to $p_0$.

The bulk Eulerian and Lagrangian fluid perturbations obey $\delta \rho =0$ because the mass density is uniform and the Lagrangian displacement has zero divergence. The linearized Euler momentum equation in the interior isTaking the curl gives ${\omega }^2\nabla \times \mathbf{\xi }=0$. Thus every nonzero-frequency normal mode has an irrotational flow displacement; the simply connected space formed by the interior admits $\mathbf{\xi }=\nabla U$ and incompressibility gives the Laplace equation ${\nabla }^{2}U=0$. For a regular center, one spherical harmonic component is $U=A_l{r}^l{Y}_l^m$, with $|m|\le l$.

It is essential to retain the surface gravity perturbation of a uniform-density sphere. Although the bulk mass density perturbation is zero, the displaced density discontinuity givesThe Poisson equation implies that $\delta \Phi$ is continuous and its outward radial derivative has the jumpUsing regularity at the center and decay at infinity, writeThe derivative jump is $-(2l+1)C_l{R}_1^{l-1}{Y}_l^m$, so $C_l=-4\pi G{\rho }_{1}l{A}_l/(2l+1)$.

Integrating the bulk Euler momentum equation gives $\delta p/{\rho }_1={\omega }^{2}U-\delta \Phi$ for $l\ge 1$. At the free surface the Lagrangian pressure perturbation vanishes:Combining these relations yields the Kelvin stellar mode frequencyThe $2l+1$ choices of $m$ are degenerate because of spherical symmetry. For $l=1$, $r{Y}_1^m$ is a linear combination of coordinates in a Cartesian coordinate system, so $\nabla U$ is a constant displacement: the zero frequency is rigid translation of the whole isolated body. Such a translation has no restoring force. The constant $l=0$ potential generates no displacement; a radial breathing motion is excluded by incompressibility and a regular center.

The printed irrotational assertion needs the nonzero-frequency qualification, or restriction to the potential incompressible stellar surface modes. It is false for all neutral displacements: $\mathbf{\xi }=\mathbf{\Omega }\times \mathbf{r}$ has zero divergence, zero normal displacement at the surface, and $\delta p=\delta \Phi =0$, so it is a zero-frequency displacement, but $\nabla \times \mathbf{\xi }=2\mathbf{\Omega }\ne 0$. This does not change the Kelvin stellar mode spectrum just derived.

Assume a Newtonian, spherically symmetric, nonrotating body in hydrostatic equilibrium, supported by a polytropic equation of state $P=K{\rho }^{1+1/n}$ with constant $K>0$ and $n>0$. The enclosed mass and hydrostatic pressure support equation areEliminate $m$ by first writing $r^2{\rho }^{-1}dP/dr=-Gm$ and then differentiating:Let ${\rho }_c$ be central mass density, and introduce the Lane-Emden variables for a stellar polytropewhereThen $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Substitution cancels the dimensional factors and yields the Lane-Emden equationThe central conditions enforce the chosen central mass density and regular spherical symmetry. The local regular expansion is $\theta =1-{\xi }^2/6+n{\xi }^4/120+\cdots$. If a finite first zero ${\xi }_1$ exists and surface pressure is negligible, the physical radius is $R=a{\xi }_1$, and the Lane-Emden mass formula is $M=4\pi a^3{\rho }_c[-{\xi }_1^2{\theta }^{\prime }({\xi }_1)]$. For $0<n<5$ the standard isolated solution has a finite surface; $n=5$ has infinite extent, so a finite surface must not be assumed for all indices. Irradiation, composition stratification and non-polytropic equations of state require more general structure equations.