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For a variation of the covariant metric tensor, $\delta g^{\alpha \beta }=-g^{\alpha \mu }{g}^{\beta \nu }\delta g_{\mu \nu }$ and $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{\mu \nu }\delta g_{\mu \nu }$. Writing $\mathcal{L}=-\frac{1}{2}{g}^{\alpha \beta }{\partial }_{\alpha }\varphi {\partial }_{\beta }\varphi -V$, one obtainsThe specified metric-variation convention therefore givesFor $\varphi =\varphi (t)$, its invariant gradient square is $-{\dot{\varphi }}^2$. Hence $T^{00}={\dot{\varphi }}^2-({\dot{\varphi }}^2/2-V)={\dot{\varphi }}^2/2+V$, while $T^{ij}=a^{-2}{\delta }^{ij}({\dot{\varphi }}^2/2-V)$ and $T^{0i}=0$. Lowering one index makes $T^{\mu }{}_{\nu }=\mathrm{diag}(-\rho ,P,P,P)$, withPotential energy has negative pressure, whereas scalar kinetic energy contributes positive pressure. This sign distinction is what permits accelerated expansion.

Write $v=u_x$ and $D={\partial }_t+v{\partial }_x$. The Euler equations and conservation of mass give $D\rho =-\rho v_x$ and $\rho Du=-p_x{e}_x$. The internal-energy density of the ideal gas is $e=p/(\gamma -1)$. The adiabatic pressure equation therefore givesMultiply the momentum equation by $u$ and use continuity to obtain the kinetic-energy balanceAdding the two equations moves ${\partial }_x(pv)$ into the flux. Thus the entire system is ${\partial }_{t}Q+{\partial }_{x}F=0$, whereThis derives the conservative energy flux of a polytropic ideal gas. In particular one-dimensional spatial dependence does not remove the transverse velocities from the total energy. The energy flux can equally be written $\rho v(u^2/2+w)$ with specific enthalpy $w=\gamma p/[(\gamma -1)\rho ]$.

Set $a=v_2/v_1$. Mass and normal momentum conservation implyAfter cancelling the unchanged tangential kinetic energies, the energy condition is $v_1^2/2+\gamma p_1/[(\gamma -1){\rho }_1]=a^2{v}_1^2/2+\gamma a{p}_2/[(\gamma -1){\rho }_1]$. With $v_{s1}^2=\gamma p_1/{\rho }_1$ and normal Mach number $\mathcal{M}=v_1/v_{s1}$, rearrangement factors it asThe first root is the no-shock solution. The nontrivial perfect-gas shock jump conditions therefore giveIts density and pressure ratios arePhysical admissibility selects a compressive, entropy-increasing shock. This can be verified directly: with $X={\mathcal{M}}^2$, define $K_i=p_i/{\rho }_i^{\gamma }$. The derivative of $\ln (K_2/K_1)$ after inserting these ratios iswhere pressures are positive, and $K_2/K_1=1$ at $X=1$. Thus the entropy increases precisely on the branch $X>1$; the positive-pressure expansion branch $X<1$ decreases entropy. Since the normal speed is positive,The endpoint one is the zero-strength limit, not a finite shock. The downstream normal Mach number obeys ${\mathcal{M}}_2^2=[(\gamma -1){\mathcal{M}}^2+2]/[2\gamma {\mathcal{M}}^2-(\gamma -1)]<1$.

For fixed smooth ${\xi }_y,{\xi }_z$, the only appearances of ${\xi }_x$ are in $\delta \Pi =A-Td$ and the nonnegative tension contribution $C{k}_y^2|{\xi }_x{|}^2$. Positivity of $T$ makes the pressure square nonnegative too. Impose $d=A/T$ by choosingThen $\delta \Pi =0$ exactly and ${\xi }_x=O(k_x^{-1})$ as $|k_x|\to \infty$, removing both nonnegative contributions in the limit. Since the remaining terms do not involve ${\xi }_x$, no other choice can lower their limiting value. This is the short transverse wavelength limit for magnetic buoyancy.

Take the chosen vertical and along-field displacements compactly supported in the interior when constructing instability trials; the resulting ${\xi }_x$ also vanishes near the boundaries, satisfying the surface condition. The argument constructs an infimum, not necessarily a minimizing finite-wavelength eigenfunction. Whenever the limiting energy is strictly negative, sufficiently large finite $|k_x|$ already gives negative energy and hence the instability asserted by the supplied variational criterion.

For $k_y=0$, the minimized energy density from part (c) isIf the bracket is nonnegative throughout, this density and the omitted pressure square are nonnegative for every displacement, so there is no negative-energy trial. Conversely, a strict negative value at a point of a smooth equilibrium persists in a small interval. Choose a nonzero smooth ${\xi }_z$ supported there, set ${\xi }_y=0$, and choose ${\xi }_x$ as in part (c). This makes the pressure square zero and gives a strictly negative integral. Therefore the localized interchange criterion for a magnetized atmosphere isThese disturbances interchange flux tubes without along-field variation; both gas and magnetic pressure contribute to their effective compressional stiffness. Equality alone is marginal and does not give a negative energy integral.

Set $P=\gamma p$, so $T=P+C$, and for $k_y\ne 0$ put $q=i{k}_y{\xi }_y$. This variable can be chosen freely. The minimized transverse-displacement energy from part (c) has densityComplete the square in $q$ to obtainThis identity also applies at a zero of $B$, where $C=0$ and the first square contributes nothing. If $-(\ln \rho {)}^{\prime }\ge \rho g/P$ everywhere, every term is nonnegative, including the pressure and transverse-tension terms removed earlier. Thus no nonzero along-field wavenumber gives an unstable trial.

For the converse, suppose the strict reverse inequality holds at an interior point. On a sufficiently small closed interval around it, $-g{\rho }^{\prime }-{\rho }^2{g}^2/P$ is uniformly negative and $C$ is bounded. Choose a sufficiently small but nonzero $|k_y|$ so that addition of $C{k}_y^2$ keeps it negative there. Choose a smooth supported ${\xi }_z$ in that interval and set $q=\rho g{\xi }_z/P$, making the first square vanish. This determines a finite ${\xi }_y=q/(i{k}_y)$. Finally choose ${\xi }_x$ by part (c) with sufficiently large finite $|k_x|$; its remaining tension energy can be made smaller than the strict negative margin. All displacements vanish near the boundaries. The full energy is therefore negative.

We have proved the long-wavelength undular magnetic buoyancy criterion:Under the assumed absence of interchange instability, the new unstable range is $\rho g/(\gamma p+C)\le -(\ln \rho {)}^{\prime }<\rho g/(\gamma p)$. Along-field motion allows material to drain and removes magnetic pressure from the limiting buoyancy stiffness, while tension at nonzero $k_y$ remains stabilizing. The statement is about existence of an allowed nonzero wavenumber; a fixed prescribed $k_y$ retains the term $C{k}_y^2$, and a finite imposed periodic length could exclude the required long wavelengths.

Use $\mathcal{R}$ for the stellar gas constant, reserving $R$ for the stellar radius. The specific internal energy of a monatomic gas obeying the ideal gas law is $e=P/[(\gamma -1)\rho ]=3P/(2\rho )$. At fixed enclosed mass, the first law of thermodynamics gives the heat available to the outgoing luminosity as $ϵ=-{\partial }_{t}e-P{\partial }_t(1/\rho )$ when there is no stellar nuclear fusion heating. ThereforeThe second coefficient includes both the mass density dependence of $e$ and pressure work. This is gravitational energy generation in a homologously contracting ideal-gas star.

Write the characteristic scales asThen $\rho ={\rho }_{0}b$, $P=P_{0}p$, $T=T_{0}p/b$, $m=Mq$ and $L_r=Ll$. Since $d/dr=R^{-1}d/dx$, direct substitution in the stellar hydrostatic equation and mass conservation givesFor radiative diffusion in a star, dividing the temperature equation by $T_0/R$ givesIn particular the fourth power here is of $\mathcal{R}$, not of the radius.

Because $q(x)$ is fixed and $M$ is constant, a fixed mass label has fixed $x$. Thus stellar homology gives ${\partial }_t\rho =-3\rho \dot{R}/R$ and ${\partial }_{t}P=-4P\dot{R}/R$. The heating rate reduces to $ϵ=-3P\dot{R}/(2\rho R)$, positive during Kelvin-Helmholtz contraction. Substitution in $d{L}_r/dr=4\pi r^2\rho ϵ$ now yieldsThe dimensionless profiles fix $D$ and $E$. In a common stellar homology family with fixed stellar composition and opacity normalization, the expression for $D$ impliesIt is independent of radius and constant in time for a particular fixed-mass star. The formula for $E$ gives $\dot{R}=-2EL{R}^2/(3G{M}^2)$, which integrates toWhen the initial-radius term is negligible,This is the Kelvin-Helmholtz contraction time law in the prescribed stellar homology model. Infinite initial radius is a limiting idealization; a finite initial radius retains the first term.

At main sequence arrival, CNO cycle heating with fixed stellar composition has luminosityThe integral is dimensionless and fixed within the stellar homology family. The constant-opacity radiative scaling still gives $L\propto M^3$. Equating the stellar nuclear fusion and transported luminosities givesFinally the Kelvin-Helmholtz contraction law evaluated at $R_{\mathrm{MS}}$ gives $t_{\mathrm{MS}}\propto M^2/(L{R}_{\mathrm{MS}})\propto M^{-34/19}$. In a coeval star cluster, the mass just arriving at the main sequence consequently satisfiesThese results comprise constant-opacity homologous contraction to CNO ignition; their mass exponents compare fixed-composition models with common dimensionless profiles.

Consider a small outward displacement $\delta r$ of a fluid element. It adjusts to the ambient pressure while retaining its specific entropy and stellar composition, so an adiabatic process gives $d\log {\rho }_{\mathrm{parcel}}=d\log P/\gamma$. Its excess mass density over the new surroundings is, to first order,Restoring buoyancy requires this bracket to be positive. With $\gamma =5/3$, the Schwarzschild criterion is thereforeEquality is neutral stability. For uniform mean molecular weight, the ideal gas law gives $d\log \rho =d\log P-d\log T$. Since hydrostatic equilibrium pressure decreases outward, the inequality is equivalent to $\nabla =d\log T/d\log P<{\nabla }_{\mathrm{ad}}=2/5$. Dividing the radiative diffusion in a star and hydrostatic equilibrium stellar structure equations givesThis is the stellar radiative temperature gradient test for stellar convective stability.

In the thin upper stellar atmosphere take $m\simeq M$, $L_r\simeq L$ and use gas pressure dominance. The prescribed opacity becomes $\kappa ={\kappa }_0\mu P{T}^{12}/\mathcal{R}$. Dividing the two structure equations then yieldsAt zero optical depth, the atmospheric boundary has $T_s^4=T_e^4/2$ and negligible gas pressure. Integrating from that boundary, with $T_s^{-8}=4T_e^{-8}$, givesTo locate the onset, put $y=T_e^8/T^8$. Logarithmic derivative gives $d\log P/d\log T=4y/(4-y)$, henceThis starts at zero at the outer boundary and increases inward. It first reaches $2/5$ atThe corresponding optical depth is ${\tau }_b=\frac{4}{3}(\sqrt{13/20}-\frac{1}{2})>0$, so the onset lies inside the stellar atmosphere. This is grey-atmosphere convection onset with thirteenth-power opacity.

Let the interior of the fully convective star have $P=K_T{T}^{5/2}$, with $K_T$ spatially constant. At its boundary $T_b/T_e$ is fixed, so the atmospheric pressure formula givesThe Stefan–Boltzmann law, $L=4\pi R^2\sigma T_e^4$ with $\sigma =ac/4$, consequently impliesThe ideal gas equation transforms the interior relation intoFor completeness, the mass-radius scaling follows by taking $\rho ={\rho }_c{\theta }^{3/2}$ and $r=\alpha \xi$ with ${\alpha }^2=5K_{\rho }{\rho }_c^{-1/3}/(8\pi G)$. Hydrostatic equilibrium and mass conservation reduce to the Lane-Emden equation $({\xi }^2{\theta }^{\prime }{)}^{\prime }=-{\xi }^2{\theta }^{3/2}$, with $\theta (0)=1$, ${\theta }^{\prime }(0)=0$. Its fixed dimensionless profile gives $R\propto \alpha$ and $M\propto {\alpha }^3{\rho }_c$. Eliminating ${\rho }_c$ gives $R\propto K_{\rho }{G}^{-1}{M}^{-1/3}$, or $K_{\rho }\propto G{M}^{1/3}R$. Thus $K_T^2\propto M^{-1}{R}^{-3}$ at fixed stellar composition. Equating this with the atmospheric expression gives $L^{17/4}\propto M^2{R}^{19/2}$, and henceThis Hayashi relation with thirteenth-power opacity concerns the interior of a fully convective star beneath a thin stellar atmosphere. The coefficient $K_T$ may vary between stars; spatially constant specific entropy does not mean equal specific entropy across the whole sequence.

A fully ionized helium nucleus supplies two Electrons for approximately $4m_p$ of mass, so $n_e=\rho /(2m_p)$. Filling all Electron momentum states up to the Fermi momentum givesFor a nonrelativistic Electron, the speed is $p/m_e$. The three mean squared Cartesian components of momentum are equal by rotational symmetry and sum to $p^2$, so each is $p^2/3$. This is the isotropic tensor integral. Thus the momentum flux, or electron degeneracy pressure, isSubstituting the expression for $p_0$ and simplifying the powers of two givesHere $h$ is the Planck constant. This is the pure-helium specialization of nonrelativistic electron-degeneracy pressure in a hydrogen-helium mixture.

For a cold helium white dwarf, neglect ion thermal pressure and use this electron degeneracy pressure as the support against Newtonian gravity. Combining hydrostatic equilibrium with mass conservation givesSet $\rho ={\rho }_c{\theta }^{3/2}$ and $r=\alpha \xi$, where ${\alpha }^2=5K{\rho }_c^{-1/3}/(8\pi G)$. Direct substitution gives the Lane-Emden equationIts dimensionless solution is independent of ${\rho }_c$. Let ${\xi }_1$ be its first zero and ${\omega }_1=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)={\int }_0^{{\xi }_1}{\xi }^2{\theta }^{3/2}d\xi$. Integrating mass conservation givesEliminating ${\rho }_c$ yields the polytropic mass-radius relationThus a more massive nonrelativistic white dwarf is smaller at fixed stellar composition. The assumptions require $p_0\ll m_{e}c$ and $k_{B}T\ll p_0^2/(2m_e)$ in the region providing most support; the law is not extended into the relativistic or thermally supported regimes.

Outside the emitting layer, take $m\simeq M_c$, $L_r\simeq L$ and $P=\mathcal{R}\rho T/\mu$. The Kramers' opacity law then reads $\kappa =({\kappa }_0\mu /\mathcal{R})P{T}^{-9/2}$. Division of the radiative diffusion in a star equation by the stellar hydrostatic equation givesWith the outer pressure constant of integration neglected, integration yieldsSubstitute this into hydrostatic equilibrium:Define $B=4\mu G{M}_c/(17\mathcal{R})$. The general antiderivative is $T=B/r+T_0$. The Kramers radiative-zero envelope around a stellar core sets $T_0=0$, givingThis is exact for the idealized boundary $T\to 0$ as $r\to \infty$. With a finite outer radius $R_s$ and temperature $T_s$, instead $T=T_s+B(1/r-1/R_s)$; the displayed profile is the deep-envelope approximation when $R_s\gg R_c$ and $|T_s-B/R_s|\ll B/R_c$. Small outer pressure alone does not eliminate this temperature constant of integration.

The specific hydrogen burning rate becomesConsequentlyIts local radial e-folding length is $R_c/20$, so burning is concentrated within a small fraction of the stellar core radius. The volume heating is even steeper: $\rho ϵ\propto C^2{T}^{93/4}$. Using the exterior envelope profile to estimate the thin shell's luminosity givesA finite but very extended upper limit changes the factor by $1-(R_c/R_s{)}^{81/4}$. The large exponent makes this correction small and further verifies the hydrogen burning thin-shell approximation. Since $C^2\propto M_c/L$ and $B\propto M_c$, the integral implies $L^2\propto M_c^{97/4}{R}_c^{-81/4}$. HenceThis is the fixed-radius-core shell-burning luminosity relation. The proportionality coefficient is a thin-shell estimate: luminosity rises from its value beneath the shell to $L$ through the emitting layer, so treating it as constant there uses the exterior profile rather than resolving the burning region. The scaling holds while its dimensionless shell structure and stellar composition remain fixed.

For consistency, $M_c/R_c$ must provide a hot hydrogen burning shell while the stellar core remains inert to core helium burning; the radiative envelope mass must be much less than $M_c$, and $R_c$ must be well inside an extended radiative envelope. At the shell base the gas must remain a nondegenerate gas and dominated by gas pressure, in particular $a{T}_b^4/3\ll C{T}_b^{17/4}$ with $T_b=B/R_c$. The assumed stellar core sequence must genuinely have nearly fixed $R_c$ over the mass interval considered. A cold helium stellar core supported by a nonrelativistic degenerate electron gas instead has the mass-radius law from part (a), so substituting that law would describe a different model and change the luminosity exponent.

The numerical fractions describe physically very different components. Use $c=1$ so that pressure and energy density have the same units. The stress-energy conservation equation for a homogeneous component, together with its equation of state $p_i=w_i{\rho }_i$, givesNonrelativistic baryons and cold dark matter have $w\simeq 0$ and dilute as $a^{-3}$; radiation has $w=1/3$ and dilutes as $a^{-4}$; vacuum energy has $w=-1$ and remains constant. The curvature term scales as $a^{-2}$ in the Friedmann equation but is a geometric contribution, not an additional material fluid.

There is consequently a coincidence in the present mixture. Normalize $a_0=1$. ThenAt early times vacuum is negligible, while at late times it dominates overwhelmingly. Observing order-unity matter and vacuum fractions during their comparatively recent crossover motivates the cosmic coincidence problem: why should this crossover occur near our observing epoch?

The absolute scale gives a second puzzle. Restoring SI units, $H_0\simeq 2.27\times 10^{-18}{\mathrm{s}}^{-1}$ yieldsIts energy density is ${\rho }_{\Lambda ,0}{c}^2\simeq 5.8\times 10^{-10}\mathrm{J}{\mathrm{m}}^{-3}$. Quantum vacuum contributions associated with ordinary particle-physics scales can be far larger than this observed value. Explaining the tiny net value, including cancellations among contributions, is the cosmological constant problem. The required negative pressure also differs qualitatively from ordinary dilute matter.

The matter budget supplies another explanatory gap: only $0.045$ is baryonic, while $0.255$ must be nonbaryonic matter in this benchmark. Together with $0.7$ in vacuum energy, most of the cosmic density is in components whose microscopic explanation is not supplied by fitting their fractions. Finally, near-zero curvature raises the flatness problem. For nonzero $k$,During radiation domination $H\propto a^{-2}$, so $|{\Omega }_k|\propto a^2$; during matter domination $H\propto a^{-3/2}$, so $|{\Omega }_k|\propto a$. Near-flatness today therefore requires extremely small early curvature in an unmodified decelerating history. Cosmic inflation offers a mechanism that reduces this curvature fraction through accelerated expansion.

The apparent contrivances are the recent matter-vacuum coincidence, the tiny vacuum-energy scale, the near-flat initial geometry and the unexplained microscopic dark components. Their observationally inferred values impose concrete requirements on a deeper physical account.

Work in the frame of the steadily advancing mush, so the rigid primary-crystal skeleton moves with $-V{\mathbf{e}}_z$. Let $\varphi$ be solid fraction, $C$ the pore-liquid concentration, $\mathbf{q}=(u,w)$ the Darcy flux relative to that skeleton, and $K$ the common thermal conductivity. Set $\kappa =K/(\rho c_p)$, the thermal diffusivity. The actual pore velocity is $\mathbf{q}/(1-\varphi )-V{\mathbf{e}}_z$. Primary crystals are taken to exclude solute; the residual pore liquid freezes into a two-solid eutectic product at the bottom. These are the standard binary mushy layer assumptions implicit in using a single liquidus and an eutectic floor.

It is convenient to use hydraulic pressure $P=p+\rho gz$, absorbing the constant-density gravitational term. With equal phase properties, no solute diffusion, negligible fluid inertia and local liquidus equilibrium, a complete mush model isThe last equation follows from conservation of energy with mixture specific enthalpy $c_{p}T-L\varphi$. Sensible heat is transported by the total volume flux $\mathbf{q}-V{\mathbf{e}}_z$; the primary solid carries the latent-enthalpy flux $VL\varphi {\mathbf{e}}_z$. In particular the sign of the latent-heat term is fixed by the downward-moving skeleton.

At the liquid roof, imposeHere $P_0=p_0+\rho gh$. The gradient condition matches heat flux into the imposed liquid field, since there is no phase-fraction jump at this roof and the phase conductivities are equal. The pore velocity there equals the relative Darcy velocity. At the floor,There is no liquid penetration into fully solid material. The unspecified solid fraction at the floor is determined by the solute balance, not set to one: remaining eutectic liquid freezes as it crosses the floor. If $C_s$ denotes the bulk concentration of the final solid product, the floor solute balance and Stefan condition areMaintaining the floor at $T_E$ permits the solid-side bath to remove this final latent heat. Equivalently its required gradient is $T_z(0^{-})=T_z(0^{+})+\rho LV[1-\varphi (0)]/K$. No additional imposed solid-side gradient is specified. Laterally we take the quadratic-pressure strain solution, without sidewalls; finite sidewalls would require additional flow conditions and generally break this horizontal invariance.

For uniform permeability of a porous medium, incompressible flow gives ${\nabla }^{2}P=0$. The roof pressure and zero normal flow at the floor are satisfied byHorizontal outflow is balanced by downward inflow. The parameter $\alpha$ is an inverse time; $E^2$ in the printed pressure condition already has dimensions of pressure divided by length squared, so no extra density factor belongs in this definition.

With $T=T(z)$, $C=C(z)$ and $\varphi =\varphi (z)$, the $x$-dependence disappears from both scalar balances. Although $u$ is nonzero, its horizontal divergence cancels $w_z$ in $\nabla \cdot (\mathbf{q}C)$. The exact strain-driven steady mush equations becomeTogether with $T(0)=T_E$, $T(h)=T_0$, $T^{\prime }(h)=G$ and $\varphi (h)=0$, they determine the scalar fields and the unknown thickness. This reduction proves that the equations admit horizontally invariant scalar solutions despite the imposed straining flow. An equivalent useful form, obtained by differentiating the solute equation, isThis exact form also shows why one must not silently discard latent heat in an arbitrary strong-flow calculation.

Define $\Delta T=T_0-T_E>0$, $R=\alpha h/V$ and $S=L/(c_p\Delta T)$. For the stated regular ordering $R\gg S\gg 1$, strain transport dominates skeleton transport and the bounded phase-change enthalpy contribution. More precisely, at fixed finite thermal strain number $\alpha h^2/\kappa$, the dimensionless energy equation isso its skeleton and integrated latent-heat corrections are small in $1/R$ and $S/R$ respectively. The leading strain-dominated mush temperature profile obeys $\kappa T^{\prime \prime }+\alpha z{T}^{\prime }=0$. Put $\delta =\sqrt{2\kappa /\alpha }$ and $H=h/\delta$. Integrating first for $T^{\prime }$ and then for $T$ givesThe error function profile follows from integrating a Gaussian, with $\mathrm{erf}x=(2/\sqrt{\pi }){\int }_0^x{e}^{-s^2}ds$. Matching the roof gradient gives the requested transcendental thickness relationFor prescribed $G>0$ and $\Delta T>0$ the right side decreases continuously from infinity to zero as $H$ increases, so it fixes one positive leading-order thickness. The resulting $h$ must still satisfy the assumed ordering and the phase-fraction constraints.

To find the product composition, integrate the exact solute balance from the roof to an arbitrary depth. With $f=1-\varphi$,At $z=0$, the exact floor composition is thereforeThe minus sign has a direct physical meaning: the horizontal outflow removes more salt than arrives at the same depth from a uniform-$C_0$ liquid, leaving a depleted solid product. Using the leading error function profile andyieldsThe full solid fraction follows by substituting this same profile into the preceding integral formula for $f(z)$. It has $\varphi (h)=0$ and $1-\varphi (0)=C_s/C_E$. A physical branch requires $0\le C_s\le C_E$ and $0\le \varphi \le 1$. In this monotone profile, $C_s\ge 0$ ensures a nonnegative liquid fraction everywhere; it is not permissible to interpret a negative formal result as a material composition.

The error-function reduction is a regular leading approximation, not a uniform statement for every possible scaling of the thermal strain number. If $\alpha h^2/\kappa$ becomes singularly large along the ordering limit, a thin basal layer can amplify the discarded terms. The printed two inequalities alone do not specify that extra scaling. In such a regime the displayed exact scalar equations and boundary conditions, rather than an unqualified zero-latent-heat profile, provide the appropriate thickness and composition calculation.