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Use a planar saline Stefan problem with upward coordinate $z$, the initial contact at $z=0$, and phase boundaries $z=a(t)$ and $z=b(t)$ enclosing pure ice, where $a<b$. Fresh liquid occupies $z>b$ and brine occupies $z<a$. We assume negligible bulk flow, equal constant mass density $\rho$, specific heat capacity $c_p$ and thermal diffusivity $\kappa$ in all regions, and zero salt content and salt transport in the ice. Equal densities remove phase-change volume flow; suppressing convection isolates the molecular-transport mechanism. Different material properties would change the numerical coefficients. Set $K=\rho c_p\kappa$ for the common thermal conductivity and $\Delta T=T_m-T_{\infty }>0$. Denote the brine salt diffusion coefficient by $D$.

Use a linear ice liquidus, $T_L(C)=T_m-mC$, with $m>0$ and $C$ the salt mass fraction. Assume the initial brine is a stable liquid, $T_{\infty }>T_L(C_0)$, and all concentrations considered are below the eutectic composition. Thus $R=m{C}_0/\Delta T>1$. These assumptions exclude independent nucleation in a supercooled brine or an eutectic system event, neither of which is specified by the initial data alone. Start with a negligible seed of ice and use local phase equilibrium without interfacial kinetics or curvature. Temperature differences may be in Celsius, since only differences enter this calculation.

WriteThe paper calls $S$ a Stefan number; it is the latent-to-sensible heat ratio, reciprocal to another common Stefan number convention. At the lower phase boundary, define $T_i=T_L(C_i)$ and letThe liquidus relation is $\vartheta =Rc$.

A similarity solution gives an explicit calculation of the two positions. Putand use $u=z/(2\sqrt{\kappa t})$, $v=z/(2\sqrt{Dt})$. The temperatures and salinities areThese error function and complementary error function profiles satisfy the heat equation and the salt diffusion equation, their far-field conditions, and the interfacial temperature conditions. The fresh-water temperature is constant because the interface and the far field are both at $T_m$.

Salt conservation at the moving lower phase boundary requiresFor upward motion this is dilution by melting salt-free ice, rather than salt rejection by freezing brine. Substitution gives the dilution function for a melting saline Stefan front:For $A>0$, $0<c<1$, so the interfacial brine is fresher than the remote liquid.

The Stefan condition at the upper boundary is $\rho L\dot{b}=K{T}_{s,z}(b)$, since the fresh liquid has no temperature gradient. At the lower boundary it isThese signs follow from the jump in enthalpy: the upper front freezes liquid while a positive $\dot{a}$ melts solid. With $E=\mathrm{erf}B-\mathrm{erf}\delta$, they reduce toTogether with $\delta =ϵA$, these equations determine $A,B,\vartheta$, and hence both positions, without discarding the salt-diffusion correction.

For the stated large latent-to-sensible heat ratio, the layer is thin relative to the thermal diffusion length. Expanding the thermal equations for $B,\delta \ll 1$ givesConsequently a useful leading calculation iswhere the positive $A$ is obtained fromThe left side decreases with $A\ge 0$ while the right side increases, and their values at zero and infinity guarantee a unique positive root for large $S$. Thus this leading calculation includes the translation of the lower boundary as well as the increasing thickness.

If the scale separation also obeys $ϵSA\ll 1$, the familiar simpler result iswithThis last simplification needs the logarithmic refinement $ϵS\sqrt{\log (RS)}\ll 1$. The algebraic ordering $ϵ\ll S^{-1}$ alone should not be used to discard an arbitrarily large logarithmic correction; the preceding coupled equations remain the appropriate calculation when that refinement is unavailable. This distinction is captured by large latent heat in a freezing and melting ice layer.

The illustration uses $S=40$, $ϵ=2.5\times 10^{-4}$ and $R=2$. Solving the full equations gives $A=1.73707$, $B=0.0143053$ and $\vartheta =0.0158766$. It shows the ice layer between freshwater and cold brine, with both fronts advancing upward and a strongly diluted lower liquid boundary.

The phase diagram shows the spatial path from remote brine to the lower interface: temperature first changes over $\sqrt{\kappa t}$ at almost unchanged salinity, then salinity changes over its much smaller diffusion scale at nearly constant temperature. The liquid path ends on the liquidus at $(C_i,T_i)$; the solid has $C=0$ and its temperature rises from $T_i$ to $T_m$. The eutectic position is schematic, with $C_E=2C_0$ in the illustration, and is not used in the calculation. A point initially in the fresh layer freezes when $b$ reaches it, cools in the ice, and later melts when $a$ reaches it. The inset indicates this temporal path in the opposite direction through the solid branch and into the liquid.

The mechanism is simultaneous upper freezing and lower melting, with net growth of the ice thickness. Cold brine accepts most of the latent heat released by upper freezing. Salt reaching the lower contact lowers its equilibrium melting temperature, and melting freshens that liquid until its liquidus is close to $T_m$. The heat delivered through the ice supplies the much smaller melting demand there. The remote brine remains liquid even though it is below the pure-water melting temperature, because it lies above its own saline liquidus. Thus “cold” does not by itself decide which phase is stable. These conclusions apply while the layers remain effectively deep and the stated no-convection, pure-ice and phase-equilibrium assumptions hold.

Let $\xi =z-Vt$ increase into the liquid, with the planar solidification front at $\xi =0$. Denote the common thermal conductivity by $K=\rho c_p\kappa$. A bounded travelling temperature profile satisfies $\kappa T^{\prime \prime }+V{T}^{\prime }=0$. In the solid, boundedness as $\xi \to -\infty$ forces $T_s=T_m$; in the liquid,The Stefan condition then givesThus the travelling solidification front in a pure supercooled melt releases just enough latent heat to warm the initially supercooled liquid to $T_m$. There is no separate imposed heat flux at infinity in this bounded travelling-wave problem.

For the interface perturbation write $\xi =\zeta (y,t)=\hat{\zeta }{e}^{i\alpha y+\sigma t}$. Use the Gibbs--Thomson relation for curvature-induced melting-temperature depressionHere $\gamma$ is the solid-liquid surface energy, $T_m$ in this expression is absolute temperature in kelvin, and $\mathcal{K}$ is the sum of principal curvatures, positive for solid convex into liquid. For this graph, $\mathcal{K}=-{\zeta }_{yy}={\alpha }^2\zeta$. If mean curvature is defined as half this sum, $\mathcal{K}=2H$; this fixes the capillary factor unambiguously.

Under the small-wavelength thermal approximation for a solidification front, take harmonic perturbation temperatures that decay away from the interface:This approximates their temperature equation by Laplace's equation, neglecting perturbation time dependence and the moving-frame advection. Expanding the interfacial temperature at the displaced surface givesThe second expression includes displacement through the negative base liquid gradient. Linearizing the Stefan condition also requires evaluation of the base gradient at the displaced surface:Using $S=1$ gives the requested reduced dispersion relation:orThe destabilizing $V\alpha$ term describes a protrusion entering colder liquid. The cubic term expresses stabilization by the curvature-induced temperature decrease.

There is an important accuracy qualification. The harmonic approximation gives the leading short-wave thermal exponents, but the retained constant term is not a uniformly controlled next-order correction to the full moving-boundary problem. If perturbation diffusion is retained, its exponents areSubstitution into exactly the same linearized boundary balances gives the full thermal stability of a travelling solidification front:For example, with capillarity removed, $\sigma =V\alpha$ solves this relation exactly, whereas the harmonic reduction gives $V\alpha -V^2/\kappa$. The difference is smaller than the leading term when $\alpha \gg V/\kappa$, but demonstrates why the reduced formula should not be asserted as an exact dispersion law or a precise onset calculation. To neglect perturbation time dependence one also needs $|\sigma |\ll \kappa {\alpha }^2$; extremely large capillary damping is outside that approximation.

To analyse the displayed cubic, defineThen $\kappa \sigma /V^2=g(q)=-1+q-2q^3/R_c$. It is negative at zero and at sufficiently large $q$. Its positive stationary point is the unique maximum:Thus the capillary threshold of a reduced solidification dispersion iswith equality giving a neutral maximum and strict inequality giving a band of unstable positive wavenumbers. At equality, however, $q_{\ast }=3/2$, not $q_{\ast }\gg 1$. The number $27/2$ is therefore the threshold of the stated reduced cubic, not a controlled threshold of the full thermal problem under the stated short-wave ordering. Far above this threshold its fastest-growing scale does satisfy $q_{\ast }\gg 1$, making the short-wave mechanism meaningful.

The analogy for a snowflake is dendritic growth of a snowflake. A protrusion has better access to the driving supersaturation or undercooling and grows faster than a sheltered region; this amplifies tips and promotes branching into crystal dendrites. Interfacial capillarity suppresses arbitrarily fine tips. In an actual snowflake, the main growth is deposition from water vapour, so the diffusion field and its driving variable differ from this pure-melt thermal example. The hexagonal ice crystal anisotropy supplies six equivalent preferred arm directions; an isotropic planar stability calculation alone cannot predict six arms. Nonlinear tip selection, vapour transport and side branching are needed for the later detailed morphology. The analysis explains the competition between diffusion-driven protrusion growth and capillary smoothing, with this physically necessary qualification.

Take $A>0$ for the repulsive molecular interaction, $p_0$ as the common reference pressure, and let $T_i=T_m-\delta T$ be the ice-film temperature. All ratios involving $T_m$ use its absolute value, approximately $273.15\mathrm{K}$, rather than the numerical Celsius value zero. The premelting film permits water to reach and freeze at the bottom of the ice while the upper surface is displaced upward without freezing there.

For equal phase mass densities, equality of the solid and liquid chemical potentials gives the Clapeyron pressure relation for equal-density phases. Indeed, $d\mu =-sdT+dp/\rho$ and $s_l-s_s=L/T_m$, so expansion around coexistence at $(T_m,p_0)$ yieldsThis is a pressure difference between phases, not a same-pressure Clausius-Clapeyron slope. The molecular disjoining pressure supports this difference:Treat the normal solid load as its effective pressure in this planar model. The gravitational load corresponding to hydrostatic pressure is $p_s=p_0+\rho gh$, so $p_l=p_0+\rho gh-p_T$. Refer the bath pressure to the sheet's upper gravitational datum, or neglect the sheet-scale hydrostatic head. The Darcy flow law then gives the upward supply and the ice growth rateHere $\Pi$ is permeability of a porous medium, $\mu$ is dynamic viscosity, and equal densities identify supplied liquid volume with added ice volume to leading order in $a/h$.

To recover the printed law, use the usual hydraulic control of frost heave idealization: the sheet's top is at $T_m$, the liquid film and ice have a common thermal conductivity $K$, and the flow is slow enough that latent-heat production is small compared with the conductive heat passing through the layer. In addition to a quasi-steady ice temperature field, this needsThe leading heat flux is consequently continuous through the film and ice. Their thermal resistances giveLet $B_T=\rho L\Delta T/T_m$, a pressure scale. The thermomolecular pressure and the molecular law then giveSubstitute this pressure into the Darcy flow law:The first term draws water toward the undercooled ice; the increasing gravitational load eventually cancels it.

Quasi-steady temperature alone is not sufficient to fix this exact prefactor. With film conductivity $K_l$, ice conductivity $K_s$, and nonnegligible latent heat, the appropriate additional balance isTogether with the Darcy equation, this is the implicit growth model. Even in the slow-flow limit, a conductivity ratio $K_s/K_l=r$ changes the leading driving pressure by $r^{3/4}$. Taking $r=2$ is a counterexample to obtaining the printed prefactor from quasi-steadiness alone. Thus the boxed equation is the intended equal-conductivity, thin-film, hydraulically limited model, rather than a consequence of only the stated quasi-steady assumption. This distinction is the heat-balance correction to a premelted-film growth model.

For that reduced equation defineThe scale $h_{\ast }$ balances $F{h}_{\ast }^{-3/4}$ against $\rho g{h}_{\ast }$; $F$ has units of pressure times length to the power $3/4$, so $h_{\ast }$ is a length and $t_{\ast }$ is a time. They yield

Set $u={\eta }^{7/4}$. The equation becomes linear:For initial thickness $\eta (0)={\eta }_0\ge 0$, the explicit solution of gravity-limited frost heave isThe usual zero-initial-thickness sketch usesIts early and late behaviour areFor $0<\eta <1$, the derivative is positive and ${\eta }^{\prime \prime }=(-\frac{3}{4}{\eta }^{-7/4}-1){\eta }^{\prime }<0$, so the curve is increasing and concave downward. It approaches the stable thickness $h_{\ast }$ exponentially. Initial thickness above $h_{\ast }$ instead relaxes downward; at $h_{\ast }$ it is stationary.

The singular slope at zero belongs to the formal reduced solution. Since $a/h\propto h^{-3/4}$, the thin-film assumption eventually fails as $h\to 0$. The early power law therefore describes an intermediate continuum regime after any microscopic startup, not arbitrarily early physical times. A finite positive ${\eta }_0$ supplies a regular initial condition when that regime begins. Gravity limits the ultimate thickness; molecular suction and the Darcy resistance set the growth toward it.

Use seawater specific heat capacity $c_w\simeq 4.0\times 10^3\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, which is an additional standard material approximation: the paper supplies the ice heat capacity, not the seawater heat capacity. Relative to freezing, the assumed uniform column has $\Delta T=7-(-1.8)=8.8\mathrm{K}$. Its ocean heat content per horizontal area isThis is heat above the reference freezing state, not the absolute internal energy of seawater.

If all this heat reaches ice already at its fusion temperature, divide by the supplied latent heat:For a chosen ice mass density ${\rho }_i=917\mathrm{kg}{\mathrm{m}}^{-3}$, this corresponds to about $5.86\mathrm{m}$ of ice. Density was not specified for this conversion, so the mass per area is the result that needs no further ice-density assumption.

Cold ice must first be warmed. If its initial temperature is $T_i<T_m$, the corresponding ideal melt mass is $Q/[L_f+c_i(T_m-T_i)]$, using the supplied ice specific heat capacity $c_i=2100\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. No initial ice temperature or salinity-dependent fusion temperature is supplied. The latent-only result is therefore an upper bound; actual melting is smaller when sensible warming, ocean or atmospheric losses and incomplete heat transfer matter.

In stellar homology, the dimensionless radial profiles are the same after scaling radius, enclosed mass, pressure, temperature and luminosity. At corresponding radii let $r^{\prime }=\xi r$, $m_r^{\prime }=m{m}_r$, $T^{\prime }=tT$, $P^{\prime }=pP$ and $L_r^{\prime }=l{L}_r$. The printed $R^{\prime }=\xi r$ is understood as this local radial scaling, with surface radius $R^{\prime }=\xi R$.

Let $q={\rho }^{\prime }/\rho$ and keep composition, opacity coefficient and nuclear coefficient fixed first. Mass conservation gives $q=m/{\xi }^3$, and hydrostatic equilibrium gives $p=mq/\xi =m^2/{\xi }^4$. The ideal gas law then gives $t=p/q=m/\xi$. Scaling the stellar energy-generation rate equation and the stellar radiative temperature gradient gives respectivelyThe second follows from $dT/dr\propto -\kappa \rho L_r/(r^2{T}^3)$, not from energy production. Equating the two luminosity scalings givesThis assumes the denominator is nonzero and a consistent homologous family exists. If the mean molecular weight differs by $u={\mu }^{\prime }/\mu$, then $t=um/\xi$ andRatios of the opacity and energy-generation coefficients multiply the right-hand side. Thus fixed composition is a real restriction, not an automatic property of every stellar sequence.

Use the Stefan–Boltzmann law $L=4\pi R^2\sigma T_{\mathrm{eff}}^4$ to locate the family on a Hertzsprung-Russell diagram. It implies $T_{\mathrm{eff}}\propto M^{(Y-2X)/4}$ andFor proton–proton chain burning take the usual local approximation $\eta =4$, with Kramers' opacity law $\lambda =1$, $\nu =-7/2$. ThenFor the CNO cycle with electron-scattering opacity, $\lambda =\nu =0$, soA conventional local choice $\eta =16$ gives $R\propto M^{15/19}$ and slope $76/9\simeq 8.44$. Choosing $\eta =17$ instead gives slope $60/7\simeq 8.57$. The source specifies no numerical nuclear exponents, and the effective exponent changes with temperature; the general expression is the unambiguous answer.

The Hertzsprung-Russell diagram places hotter stars to the left. Its branches rise toward higher luminosity and mass, and the CNO cycle/electron-scattering opacity branch has the steeper logarithmic slope. Their illustrative joining point and normalization are arbitrary because the proportional opacity and reaction laws do not specify absolute stellar scales. These are the fully radiative ideal-gas homology predictions, not exact observed main sequence relations: convection and increasing radiation support limit those assumptions in real stars.

Radiative transport supplies the temperature gradient used below. A purely radiative equilibrium model also needs stellar convective stability; the transport assumption alone is not a general proof of stability.

A stellar equation of state supplies pressure and internal energy as functions of density, temperature and composition, together with thermodynamic derivatives needed for stability and transport. Hydrostatic equilibrium fixes the pressure gradient, but does not determine which microscopic components provide the pressure. In ordinary dense interiors local thermodynamic equilibrium is a useful starting point. A consistent mixture iswhere ions, electrons, radiation and interaction corrections are distinguished. The classical electron pressure and electron degeneracy pressure are two limits of the same electron contribution, and must not be added as if they belonged to different particles. The finite-temperature electron equation of state interpolates between them.

In a fully ionized, nondegenerate, nonrelativistic gas, $P_g=\rho k_{B}T/(\mu m_u)$ and specific thermal energy is $u_g=3k_{B}T/(2\mu m_u)$. With nuclear mass fractions $X_A$, charges $Z_A$ and mass numbers $A_A$, the mean molecular weight satisfies ${\mu }^{-1}={\sum }_A{X}_A(1+Z_A)/A_A$, and ${\mu }_e^{-1}={\sum }_A{X}_A{Z}_A/A_A$ is the mean molecular weight per electron. This regime describes much of an ordinary main sequence interior. Toward cooler layers, ionization and molecular dissociation change particle numbers and consume heat. The Saha equation relates ionization to both temperature and electron density: there is no universal horizontal ionization boundary. These regions have larger heat capacity and can have a reduced stellar adiabatic exponent. The simple fully ionized formula is then insufficient.

Equilibrium photons give radiation pressure $P_{\gamma }=a_r{T}^4/3$ and energy per volume $a_r{T}^4$, or specific energy $u_{\gamma }=a_r{T}^4/\rho$. In the nondegenerate gas regime, equality with gas pressure gives the radiation-to-gas pressure boundarya line of slope $1/3$ on a $(\log \rho ,\log T)$ plot. Higher temperatures at fixed mass density favor photon support. A monatomic gas has stellar adiabatic exponent $5/3$, while radiation alone has $4/3$; their adiabatic exponents of a monatomic gas-radiation mixture are not obtained by assuming a fixed pressure fraction during compression. Radiation support is particularly important in massive stars.

For electrons, Pauli exclusion principle and the Fermi-Dirac distribution determine occupation numbers. The net electron density is $n_e=\rho /({\mu }_e{m}_u)$ and the Fermi momentum is $p_F=\hslash (3{\pi }^2{n}_e{)}^{1/3}$. Define the kinetic electron Fermi temperatureFor $T\gg T_F$ the electrons are nearly classical; for $T\ll T_F$ they are strongly degenerate and their pressure depends primarily on density. The intermediate region requires finite-temperature electron equation of state integrals, not a discontinuous switch of formulas.

The equation of state of a cold electron gas gives, in its two limits,These are respectively the $n=3/2$ and $n=3$ pressure-density powers, explaining the approximate white dwarf polytropic mass-radius relation sequence and the Chandrasekhar limit. The kinetic energy per volume is $3P_e/2$ in the nonrelativistic limit and $3P_e$ in the ultrarelativistic limit. Ions can still supply much of the heat capacity even when the electron pressure supplies the mechanical support.

The electron degeneracy crossover $T\sim T_F$ has slope $2/3$ at low mass density and $1/3$ at high mass density. The electron relativistic density threshold isa vertical marker where $p_F=m_{e}c$. It is different from the thermal electron relativistic threshold $k_{B}T\sim m_e{c}^2$, near $5.93\times 10^9\mathrm{K}$, a horizontal temperature scale. Hot dilute matter can have relativistic thermal electrons without degeneracy; cold dense matter can have relativistic degenerate electrons without reaching that temperature.

Electron degeneracy also does not automatically imply that electrons dominate the total pressure. Comparing the cold electron limit with radiation gives the radiation-to-degeneracy pressure boundaryIts logarithmic slopes are $5/12$ for nonrelativistic electrons and $1/3$ for ultrarelativistic electrons. One must compare the pressures separately from the degeneracy criterion; extrapolating the classical gas-radiation line into a degenerate region is incorrect.

The stellar equation-of-state regime diagram uses an illustrative fully ionized carbon composition, ${\mu }_e=2$, $\mu =12/7$. It displays the electron-degeneracy crossover, both radiation-pressure comparisons, and distinct thermal and density-driven relativity scales. The curves are limiting-model comparisons, not sharp phase boundaries or a calibrated complete equation of state. Partial ionization, molecular physics and interactions modify the low-temperature regions indicated on the plot.

At sufficiently high temperature, electron-positron thermal pair abundance can become important. The pair abundance depends on density and chemical potential as well as temperature; $k_{B}T=m_e{c}^2$ is not a universal onset line. In the dilute ultrarelativistic limit, both pair species together add energy density $7a_r{T}^4/4$ to the photons' $a_r{T}^4$, and have pressure one third of their energy density. While pairs are being created, thermal energy is spent on rest mass, which can reduce the stellar adiabatic exponent below $4/3$ and contribute to pair-instability supernova physics.

At high mass density and low temperature, Interactions governed by Coulomb's law invalidate the noninteracting-ion approximation. The ionic Coulomb coupling parameter $\Gamma =Z^2{e}^2/(4\pi {ϵ}_0{a}_i{k}_{B}T)$, where $a_i=(3/(4\pi n_i){)}^{1/3}$, grows as ${\rho }^{1/3}/T$. Corrections become significant when $\Gamma$ is of order one, and a sufficiently strongly coupled plasma can crystallize. At still greater mass density, electron capture alters ${\mu }_e$ and nuclear matter replaces the ideal electron-ion model; neutron star interiors require strong-interaction and relativistic equations of state. These further regimes lie beyond the simple pressure curves plotted here. A useful stellar EOS is thermodynamically consistent across the crossovers, rather than just the maximum of unrelated pressure laws.

Take a long periodic section of length $L$, or neglect end effects and choose an integer number of wavelengths. The amplitude $u_q$ here is a single real sine amplitude, fixing the normalization of the thermal membrane undulation spectrum. For a lipid vesicle with an axisymmetric membrane deformation, conservation of the enclosed volume givesThe mean radius therefore decreases at second order. This adjustment is essential: fixing the mean radius instead of the volume would miss the unstable term in the fixed-volume capillary spectrum of a cylindrical membrane.

The surface area of an axisymmetric graph is $\mathcal{S}=2\pi \int R\sqrt{1+R_z^2}dz$. Expanding for $|u_q|\ll R_0$ and $q|u_q|\ll 1$ givesMultiplication by the membrane tension yields a quadratic potential energy $\Delta E=K_q{u}_q^2/2$, whereFor $q{R}_0>1$, the equipartition theorem gives the stable-mode fixed-volume capillary spectrum of a cylindrical membraneHere $k_B$ is the Boltzmann constant and $T$ is the temperature. The cosine amplitude has the same variance and is an independent real coordinate at quadratic order. If instead $R-{\rho }_0={\sum }_{q\ne 0}{a}_q{e}^{iqz}$ with $a_{-q}=a_q^{\ast }$, the corresponding positive-$q$ complex coefficient has $⟨|a_q{|}^2⟩=k_{B}T{R}_0/[2\pi \sigma L((q{R}_0{)}^2-1)]$. This is the same thermal membrane undulation spectrum in a different Fourier mode normalization.

For $q{R}_0<1$, the cylinder is unstable rather than having a negative fluctuation variance. The quadratic surface area change is negative: a sufficiently long-wavelength modulation reduces area while retaining volume. This is the Rayleigh–Plateau instability, expressed here as pearling of a tension-dominated lipid vesicle. A cylinder supporting such modes has no unconstrained harmonic thermal equilibrium about the uniform state. The $q=0$ uniform radius change is forbidden by fixed volume, and $q{R}_0=1$ is marginal in the tension-only approximation. The finite length and endpoint constraints determine which nonzero Fourier modes are allowed.

The large membrane tension assumption controls modes with $q{R}_0$ of order one. For large $q$ the omitted membrane bending modulus contribution grows approximately as $\kappa q^4$ and overtakes the tension term when $\kappa q^2\gtrsim \sigma$. Thus the displayed tension-only spectrum requires $q\ll \sqrt{\sigma /\kappa }$ as well as small amplitudes and a positive stiffness. Close enough to the marginal wavelength, bending corrections must also be retained; their relative scale at $q{R}_0$ of order one is $\kappa /(\sigma R_0^2)$.

For the circular phase boundary in a lipid bilayer, the energy is its perimeter times the line tension $\gamma$. Let $\phi$ be the polar angle and write $r(\phi )={\rho }_0+u_m\sin (m\phi )$, where the periodicity requires an integer $m=q{R}_0\ge 1$. Fixed enclosed area impliesThe perimeter, including the mean-radius change, isThus the fixed-area capillary spectrum of a circular boundary, again for a single real sine or cosine amplitude, isThe complex Fourier mode coefficient convention again divides this result by two. The mode $m=0$ is excluded by the fixed-area constraint, so the negative-stiffness interval $0<m<1$ is not an allowed mode of a closed circular boundary.

At $q{R}_0=1$, the zero stiffness is the translation mode of a circular boundary, not a shape instability. Displacing the centre by a small distance $h$ changes the radius to first order by $h\cos \phi$ or $h\sin \phi$ while leaving area and perimeter unchanged. For example, an exactly translated circle hasThe higher harmonics complete a true translation mode of a circular boundary. The $m=1$ denominator therefore represents free centre motion: an unconfined domain's centre has no restoring force or finite equilibrium variance in an infinite membrane. Radius fluctuations about a recentered domain omit this mode; external confinement would give it a separate restoring stiffness.

Write $\mathcal{R}=k/(\mu H)$, where $H$ is the atomic mass unit used in the mean molecular weight convention, and let $\kappa$ be the Rosseland mean opacity per unit mass. For a static spherical star with negligible thermal storage, the stellar structure equations areHere $ϵ$ is the net stellar energy-generation rate per unit mass; $a$ is the radiation energy-density constant and $c$ the speed of light. The last differential equation follows from radiative diffusion, $F=-(c/3\kappa \rho )d(a{T}^4)/dr$, with $L=4\pi r^{2}F$. There is no gravothermal storage term in the specified stellar energy balance equation. Neglecting radiation pressure in force balance does not remove radiation as an energy carrier. Regular central conditions are $m(0)=L(0)=0$; take negligible external pressure, $P(R)=0$, for the analytic hydrostatic equilibrium model below.

Set $x=r/R$. Integrating the enclosed mass directly givesThusThe stellar hydrostatic equation then givesFactorizing makes the surface behaviour transparent:The ideal gas equation of state gives the temperature profileNo small-$x$ approximation was needed. In particular, close to the centre,These exact hydrostatic formulas reveal a limitation of the prescribed linear-density stellar model. Its mass density has a central cusp, and $T^{\prime }(0^{+})=T_c/R>0$. A positive opacity with positive outward luminosity would instead require $T^{\prime }<0$ in radiative diffusion. Moreover, regular positive nuclear heating gives $L=O(r^3)$ and therefore $T^{\prime }=O(r)$ at the centre. Thus the imposed profile is a hydrostatic toy model, not a complete positive-heating radiative-equilibrium solution of all the stellar structure equations. The requested algebraic profiles and nuclear-luminosity integral can still be computed consistently as properties of that toy model.

Since $ϵ$ is per unit mass, each shell contributes $dL=ϵdm=4\pi r^2\rho ϵdr$. Both factors of mass density must be retained. Substituting the linear-density stellar model and its temperature profile givesWriting $I_{\nu }(z)$ for this incomplete integral, with $I_{\nu }=I_{\nu }(1)$, the luminosity isThis supplies both the local $L_r$ and the total luminosity associated with the printed integral extending to one. For fixed composition and fixed $\nu$, the total scales as $M^{2+\nu }{R}^{-3-\nu }$. The physical nuclear-heating exponents are positive; mathematically the surface integral converges for $\nu >-3$. As a simple normalization check, $I_0=1/30$, hence $L(R)=6{ϵ}_0{M}^2/(5\pi R^3)$ for temperature-independent heating. This is the generated luminosity; the radiative-transport incompatibility noted above is not removed by integrating the heating.