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For a small, uniform surface at temperature $T$, enclosed by much larger black walls all at temperature $T_s$, the outward net heat transfer by thermal radiation isAssume an opaque grey body that is a diffuse thermal emitter, with the same emissivity $ϵ$ and radiometric absorptance at the surface's current temperature. Its emitting area $A$ sees only the enclosure: the view factor to the walls is one, with no self-viewing surface, opening or intervening obstruction. The intervening medium must neither absorb nor emit appreciable thermal radiation; a vacuum supplies a useful idealization. The enclosure is maintained at uniform $T_s$. The small-body assumption makes that reservoir description physically reasonable; genuinely black isothermal walls supply the incoming irradiance even without an infinite area ratio. Positive ${\dot{Q}}_{\mathrm{rad}}$ means net loss by the surface; negative means net gain. This expression counts thermal radiation alone, not all heat transfer.^[ref]

The reasoning is a two-way energy balance. Black walls supply irradiance $E=\sigma T_s^4$, so the surface absorbs $P_{\mathrm{abs}}=ϵAE$ and emits $P_{\mathrm{emit}}=ϵA\sigma T^4$. Subtract these to obtain ${\dot{Q}}_{\mathrm{rad}}$. For an opaque surface the remaining fraction $1-ϵ$ of incoming radiant flux is reflected; counting emitted plus reflected outgoing radiant flux minus total incoming radiant flux gives the same result. Kirchhoff's law of thermal radiation justifies equal spectral absorption and emission factors; the grey body assumption lets one factor be taken outside both spectral integrals.

Choose $ϵ=0.8$, $A=0.01{\mathrm{m}}^2$, $T=500\mathrm{K}$ and $T_s=300\mathrm{K}$. These are stipulated model inputs, not measured material properties. Using $\sigma \approx 5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$ givesThe incoming irradiance is $459.300\mathrm{W}{\mathrm{m}}^{-2}$, corresponding to $4.5930\mathrm{W}$ incident on the surface; only $80\%$ is absorbed. At $T=T_s=300\mathrm{K}$, emitted and absorbed powers are both $3.6744\mathrm{W}$, so radiative equilibrium has zero net exchange despite continuing emission. Changing only emissivity to $0.2$ reduces both one-way powers and the net loss by a factor of four, giving $6.1694\mathrm{W}$ net. A low-emissivity coating therefore suppresses both radiative cooling to colder walls and radiative heating from hotter walls.

Near equal temperatures, write $T=T_s+\Delta T$. Taylor's theorem givesFor constant emissivity, the radiative heat transfer coefficient $h_{\mathrm{rad}}$ is approximately $4.8992\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$ at $T_s=300\mathrm{K}$ and $ϵ=0.8$. This local linear approximation is useful for small temperature differences; the $200\mathrm{K}$ difference in the worked example needs the full fourth-power expression.

A cooling experiment requires a wider first law of thermodynamics balance. In a uniform-temperature model without a heater,where $C(T)$ is the sample's heat capacity and the last two terms are signed outward convection and thermal conduction losses. A vacuum suppresses convection but does not eliminate thermal conduction through supports or sensor leads. A cooling slope therefore cannot establish emissivity unless these paths and heat capacity are independently constrained. Multiple wall temperatures, incomplete view factors, wavelength-selective materials or appreciable intervening absorption require a more complete radiative transfer calculation.

Optical-electrical heating non-equivalence means that equal absorbed optical and electrical powers do not necessarily give identical thermometer responses. Light and the heater may deposit heat in different places, producing different temperature gradients and different routes of thermal conduction to the support. Holding one thermometer reading fixed need not hold every part of the cavity at the same temperature. This differs from incomplete radiometric absorptance: one concerns where absorbed heat goes, the other how much incident light becomes absorbed heat.

A useful accounting convention iswhere the signed correction ${\delta }_{\mathrm{ne}}$ is determined by characterization of the instrument and represents the residual departure from the lumped electrical substitution heat balance. It is not a universal extra source of energy. Varying the beam location or heater location, comparing equivalent temperature excursions, and checking repeatable dark returns can test the assumptions; no single small correction proves that every other correction vanishes.

The Rosseland mean opacity is the transport opacity appropriate to an optically thick region close to local thermodynamic equilibrium. If ${\kappa }_{\nu }$ is monochromatic transport opacity and $B_{\nu }(T)$ the Planck function, thenIt is a weighted harmonic mean: low-opacity frequency windows carry much of the diffusive flux. One must combine the monochromatic processes before averaging, rather than adding their separately averaged opacities.

In hot fully ionized material, electron-scattering opacity supplies an almost density- and temperature-independent contribution at fixed composition. In the nonrelativistic Thomson scattering limit, ${\kappa }_{\mathrm{es}}={\sigma }_T{n}_e/\rho \simeq 0.2(1+X)\mathrm{c}{\mathrm{m}}^2{\mathrm{g}}^{-1}$. Free-free opacity, the absorption counterpart of thermal bremsstrahlung, comes from photon absorption during encounters between free electrons and ions. Its approximate mean obeys Kramers' opacity law, ${\kappa }_{\mathrm{ff}}\propto \rho T^{-7/2}$, with composition and quantum correction factors.

When ions retain bound electrons, bound-free absorption is photoionization; edges, occupations and ionization fractions strongly affect its frequency dependence. Bound-bound absorption produces absorption lines. Numerous metal lines can close low-opacity windows and produce important opacity features in partially ionized layers. Line populations depend on excitation as well as ionization, while Doppler broadening, pressure broadening and overlapping lines change the transport windows. Stimulated inverse processes must be included consistently with detailed balance and the local thermal radiation field.

In cooler hydrogen-rich atmospheres, negative hydrogen ion opacity from both bound-free and free-free processes is often important. Its strong temperature dependence helps determine convective surface relations. In still cooler layers, molecular bands, and at sufficiently low temperatures dust absorption/scattering, may be important. These processes cannot be represented by one Kramers power law over the whole density-temperature plane.

An opacity calculation therefore needs the chemical abundances, the equation of state, electron density and bound-state populations. Saha ionization equation and thermal excitation populations give useful dilute LTE estimates; high-density pressure ionization, nonideal effects and altered level populations require corrections. Composition matters even when one writes only $\kappa (\rho ,T)$: hydrogen, helium and the metal abundance pattern are additional fixed inputs.

In dense degenerate cores, thermal conduction by electrons can dominate energy transport. An equivalent conductive opacity may be introduced so that $1/{\kappa }_{\mathrm{eff}}=1/{\kappa }_R+1/{\kappa }_{\mathrm{cond}}$, because the two diffusive fluxes add; conduction is not itself a photon absorption process. Very hot regimes also require departures from the simple nonrelativistic scattering approximation. Near the optically thin photosphere, a Rosseland diffusion description alone is inadequate, and frequency-dependent radiative transfer with appropriate boundary conditions is required. These regime changes are why realistic stellar opacity tables, rather than one global power law, enter stellar structure.

For a stationary film of uniform thickness, thermal conduction gives $T_{zz}=0$, so $T=T_0+cz$. The cooling boundary condition determines $c$:Consequently the surface temperature isThis is the quasistatic temperature of a conductively cooled film, with a weak-cooling expansion when $\alpha h/\kappa \ll 1$.

Let the thickness and longitudinal scales be $h_{\ast },L$, with $ϵ=h_{\ast }/L\ll 1$. Incompressibility gives transverse speed of order $ϵU$, and film evolution has the kinematic time scale $L/U$. For the temperature variation scale $\Theta$, longitudinal diffusion divided by transverse diffusion is $h_{\ast }^2/L^2={ϵ}^2$. Each advective term, and the time derivative on that evolution scale, divided by transverse diffusion isThus ${ϵ}^2\ll 1$ and $Pe\ll 1$ make the leading thermal profile locally linear and quasistatic. Retaining the weak-cooling condition gives the same surface-temperature approximation with the local $h(x,t)$. For any independently imposed faster time variation one would also require its thermal relaxation ratio $h_{\ast }^2/(\kappa t_{\mathrm{evol}})$ to be small.

Define the positive surface-tension gradient coefficientThe weak-cooling approximation gives ${\gamma }_s\simeq {\gamma }_0+Bh$, so the Marangoni stress is ${\gamma }_{s,x}=B{h}_x$. The Young–Laplace equation gives the leading liquid pressure $p=p_a-{\gamma }_0{h}_{xx}$, with constant ambient pressure. In lubrication theory, $p_z=0$ and $\mu u_{zz}=p_x$. The no-slip boundary condition at the wall and the tangential surface condition $\mu u_z(h)={\gamma }_{s,x}$ giveHence the volume flux isConservation of mass gives the thermocapillary thin-film equationThe replacement of surface tension by ${\gamma }_0$ in the capillary term needs $B{h}_{\ast }/{\gamma }_0\ll 1$. For comparable capillary and thermocapillary fluxes,so the fractional tension change is only $O({ϵ}^2)$. Both the tension correction to $\gamma h_{xxx}$ and the differentiated-tension contribution ${\gamma }_x{h}_{xx}$ are then higher-order. Weak cooling alone would not justify that replacement if $|{\gamma }^{\prime }|\Delta T/{\gamma }_0$ were allowed to grow without bound.

The capillary term smooths the film: a crest has negative curvature and higher liquid pressure, causing flow away from it. The thermal term has the opposite effect. A thicker region has a colder surface and therefore higher surface tension; Marangoni stress pulls surface fluid toward that region, reinforcing the thickness disturbance. These mechanisms give fourth-order damping and second-order growth respectively.

Use $H=h/\hat{h}$, $X=ϵx/\hat{h}$, and $\tau =t/\hat{t}$. The two dimensionless flux-divergence coefficients are $B{ϵ}^2\hat{t}/(2\mu )$ and ${\gamma }_0{ϵ}^4\hat{t}/(3\mu \hat{h})$. Setting both equal to one gives the balanced scales for a thermocapillary film:These choices yield $H_{\tau }+(H^2{H}_X{)}_X+(H^3{H}_{XXX}{)}_X=0$. The predicted $ϵ$ must be small for the assumed long-wave approximation.

Linearize about $H=1$ using a normal mode of amplitude $\delta$. The resulting equation is $s-k^2+k^4=0$, so the dispersion relation isModes with $0<|k|<1$ grow, while $|k|>1$ modes decay. Differentiating $s$ gives the most unstable wavenumbersThe physical fastest wavenumber is $ϵ/(\sqrt{2}\hat{h})$.

For a positive steady film with zero volume flux, $H^2{H}_X+H^3{H}_{XXX}=0$. Divide by $H^3$ and integrate once:Choosing the thickness scale so that $H=1$ when $H_{XX}=0$ sets $C=0$. Multiply $H_{XX}=-\ln H$ by $H_X$ and integrate to obtain the zero-flux thermocapillary film profiles relationThus the stated potential is $V(H)=H(\ln H-1)$ and the integration constant $E$ is the corresponding first integral.

Cooling from above makes dense cold fluid overlie lighter warm fluid, so the upper boundary layer is unstable and can mix the interior. Assume high Rayleigh number, rapid interior mixing, constant properties, negligible wall heat capacity and a quasi-steady cooling flux obeying the four-thirds convective heat-transfer law. Write $\theta (t)=T_{\mathrm{bulk}}(t)-T_0$ and use a coefficient $C$ appropriate to this single cooling boundary. The volumetric heat capacity is $\rho c_p=k/\kappa$, so the area-wise heat balance isWith $K=C\kappa [g\beta /(\nu \kappa ){]}^{1/3}/h$, integration gives convective cooling under the four-thirds heat-transfer law:The power-law decay is a bulk approximation between the initial boundary transient and the later stage when the decreasing temperature contrast no longer maintains strong convection. It is not an exact spatially uniform solution satisfying the cold boundary itself.

Cooling from below is different: dense cold fluid now lies below warm fluid and stabilizes the layer. With no other imposed motion, cooling is by thermal conduction, and temperature becomes depth dependent. Let $z=0$ be the cooled bottom and $z=h$ the insulating top. Solve ${\theta }_t=\kappa {\theta }_{zz}$ with $\theta (0,t)=0$, ${\theta }_z(h,t)=0$ and $\theta (z,0)=\Delta T$. The mixed-boundary eigenfunctions have ${\lambda }_n=(n+1/2)\pi /h$, givingIts volume-averaged temperature isThe late conductive decay is exponential on the timescale $4h^2/({\pi }^2\kappa )$, rather than the convective $t^{-3}$ bulk law. This assumes the usual positive thermal expansion coefficient and no destabilizing compositional stratification.

A spatially uniform pressure is a useful leading approximation when the sound speed makes the acoustic crossing time much shorter than the radiative cooling and thermal conduction times. The motions must remain at low Mach number, and gravity and other stresses must be negligible. Sound then redistributes pressure before appreciable thermal evolution occurs. Small pressure corrections can still drive the slow motion; imposing exactly zero pressure gradient in the full momentum equation at every order would be a stronger assumption.

The equation of state gives the specific internal energy of an ideal gas as $e=p/[(\gamma -1)\rho ]$. With $D/Dt={\partial }_t+v{\partial }_x$, the continuity equation is $D\rho /Dt=-\rho v_x$. Differentiating $e$ therefore yieldsInsert this into the internal energy equation, whose compressional term is $-p{v}_x$, and use spatially uniform pressure, so $Dp/Dt=p_t$. The result isFor temporally constant pressure put $A=\mu p/\mathcal{R}$, so the mass density is $\rho =A/T$. The continuity equation now gives $v_x=T^{-1}DT/Dt$. Multiplication of the last equation by $(\gamma -1)T/(\gamma p)$, with $\lambda ={\lambda }_0{T}^{\alpha }$, givesThis is the isobaric conductive cooling in mass coordinates transformation before changing coordinates.

There is a boundary assumption implicit in the proposed Lagrangian coordinate. For a mass integral from a fixed origin, integration of the continuity equation givesThus $m$ is material when the base mass flux vanishes, as appropriate for the stationary cold reservoir below. If the origin permits mass flux, replace $m$ by $m-{\int }^t\rho (0,s)v(0,s)ds$; this corrected label has zero material derivative. In a material mass coordinate, $D/Dt={\partial }_t{|}_m$ and ${\partial }_x=(A/T){\partial }_m$. In particularThe conduction coefficient therefore becomes $C{\lambda }_0$, with the same $C$ as the cooling coefficient. With $\tau =Ct$, the equation is

In the hot gas there is no volumetric radiative cooling. The initial temperature is uniform and the cold boundary introduces no finite mass length. The remaining nonlinear diffusion equation is invariant under $m\mapsto bm$, $\tau \mapsto b^2\tau$ with the same temperature scale. Its natural similarity solution is thereforeThe far-field condition also recovers the initial temperature at every fixed $m>0$ as $\tau \downarrow 0$. Substitution gives the ordinary differential equationFor $\alpha =1$, this reduces to $f^{\prime \prime }+\xi f^{\prime }/2=0$. Integrating first for $f^{\prime }$ gives $f^{\prime }=K\exp (-{\xi }^2/4)$. The two boundary conditions fix $K=1/\sqrt{\pi }$, hence the error-function isobaric cooling front isHere the error function has normalization $2/\sqrt{\pi }$, as printed in the original PDF; the converted TeX's $2/\pi$ is incorrect. The following original sketch shows the temperature profiles. The cooling layer broadens as $\sqrt{\tau }$, and at fixed positive mass label the temperature falls as time increases, while the remote hot gas remains at $T_0$.

For $\alpha =1$, the magnitude of the heat flux into the cold reservoir isAlthough the thermal conductivity tends to zero there, the physical temperature gradient diverges: their product has this finite limit. The stationary reservoir has no incoming advective energy flux and radiates the conducted energy immediately. Since $T_m(0,\tau )=T_0/\sqrt{\pi {\lambda }_0\tau }$, the radiated power per unit area isThus the requested exponent is $\boxed{k=1/2}$. The initial divergence is the usual idealization of an instantaneous temperature discontinuity in a diffusion equation.

At the porous thermal front, the cold polymer solution behind has dynamic viscosity $\mu$, while the hot solution ahead has dynamic viscosity $b\mu >\mu$. Thus a more mobile region pushes a less mobile one. A forward protrusion locally reduces the length of high-resistance hot fluid and concentrates Darcy flux into itself; it advances faster, amplifying the protrusion. This is the mechanism of the Saffman–Taylor instability.

The leading polymer–oil interface can nevertheless be stable when the hot polymer dynamic viscosity is at least the oil dynamic viscosity. Stability of that leading interface does not guarantee stability of the trailing porous thermal front. Slow thermal conduction makes the latter sharp and offers weak short-wave smoothing; in practice it is the adverse dynamic viscosity contrast across this front that can invalidate the assumed planar stable-flow calculation.

Take constant Darcy velocity $U$ and work locally near a planar porous thermal front moving at $V_T=\Gamma U$. Let $\xi =x-V_{T}t$ and perturb its position by $\eta ={\eta }_0{e}^{\sigma t+iay}$, with wavelength $\lambda =2\pi /|a|$. Neglect thermal conduction and regard the adjoining layers as semi-infinite on this disturbance scale. In each region, Darcy's law and incompressibility imply that the pressure perturbation is harmonic, so the bounded normal modes areHere the minus side is cold with dynamic viscosity $\mu$, and the plus side is hot with dynamic viscosity $b\mu$. The base pressure gradients are $p_{-,x}^0=-\mu U/k$ and $p_{+,x}^0=-b\mu U/k$. Linearizing pressure continuity at the displaced front yieldsContinuity of the perturbed normal Darcy velocity givesso $A_{+}=-b{A}_{-}$. Therefore $\delta U=U|a|(b-1){\eta }_0/(b+1)$. The thermal-front kinematic law is $\sigma {\eta }_0=\Gamma \delta U$, proving the thermal-front viscous-fingering dispersion relationIt is positive for $b>1$, zero for $b=1$, and negative for $b<1$. Without thermal conduction the model has no finite fastest wavelength: the positive growth rate increases without bound as $\lambda \to 0$. A resolved thermal transition and diffusion regularize that short-wave limit.

For a material interface the kinematic coefficient would be $1/\varphi$ rather than $\Gamma$. Thus, if the interface of interest were instead the hot polymer–oil interface, the same calculation would give $\sigma =2\pi (U/\varphi )({\mu }_h-b\mu )/[\lambda ({\mu }_h+b\mu )]$, the usual planar viscous-fingering dispersion relation. This distinguishes the two interfaces and their speeds. The boxed thermal result uses the local planar approximation; distant wells or the other front modify the pressure eigenfunctions for wavelengths comparable to those separations.

Write $ϵ=1-{\rho }_s/\rho >0$ and $\Delta v=1/{\rho }_s-1/\rho =ϵ/{\rho }_s$. When the ice sphere grows by volume $4\pi a^2\dot{a}dt$, it consumes liquid volume $({\rho }_s/\rho )4\pi a^2\dot{a}dt$. The difference is expelled from the fixed cavity. Thus mass conservation fixes the outward Darcy flux in the surrounding porous medium:The Darcy law $u_r=-(\Pi /\mu ){\partial }_{r}p$ and $p(\infty )=p_m$ implyMatching at the cavity wall, whose internal pressure is uniform, givesThe solid and liquid pressures are equal when the Gibbs--Thomson relation is neglected; nevertheless the coexistence temperature changes with their common pressure because their specific volumes differ. Along coexistence, equality of chemical potentials givesHence the common-pressure melting-point slope is $d{T}_m/dp=-T_m\Delta v/L$. To first order in the pressure change,The $T_m$ multiplying this coefficient is absolute temperature, approximately $273.15\mathrm{K}$ for water; a temperature difference may equally be measured in kelvin or degrees Celsius. Substituting the numerical value $0$ from the Celsius scale into a thermodynamic denominator would be incorrect.

In the quasi-stationary thermal conduction approximation, neglect advective heat transport and use the stated effective conductivity $k$ outside the sphere. The radial Laplace equation gives the bounded solid temperature and the liquid/porous-region temperature:This treats the exterior as a uniform conducting medium, as required by the printed reduction; distinct cavity and porous-medium conductivities would give an additional thermal resistance. There is no solid-side heat flux in the bounded quasi-stationary solution. The Stefan condition therefore becomesEliminate $T_i$ and $p$ to obtainFor $x=a/R$, $t=t_0\tau$ and $t_0=R^2{\rho }_{s}L/(k\Delta T)$, this isThe pressure follows either from the Darcy flux or the Stefan condition:In particular $T_i-T_{\infty }=\Delta TK/(x+K)$: rising pressure progressively reduces the available undercooling. This is spherical freezing inside a porous cavity, whose fixed hydraulic outlet radius $R$ distinguishes it from freezing directly through a porous matrix.

For the idealized zero-radius start, integration givesA finite seed replaces the right side by its difference from the initial value. The reduction ends when the sphere fills the cavity.

For $K\gg 1$, hydraulic drainage is easy and the pressure feedback is small throughout the cavity. The leading expressions areThis is ordinary conduction-controlled spherical solidification, with a completion time asymptotic to $t_0/2$.

For $K\ll 1$, distinguish the initial crossover from the pressure-dominated regime. At $x\ll K$, equivalently $\tau \ll K^2$, the same square-root growth applies and $p-p_m\simeq p^{\ast }x/K$. At $K\ll x\le 1$, equivalently $K^2\ll \tau \lesssim 1/K$,The interface approaches $T_{\infty }$, the pressure approaches the finite coexistence value $p_m+p^{\ast }$, and growth becomes hydraulically limited. The leading completion time is $t_0/(3K)$. The cube-root approximation is not valid at the instant of nucleation, where the initial square-root regime must be retained.

Use the one-sided thermal conduction model encoded by the requested dispersion relation: neglect solid-side perturbation heat flux, take equal phase densities, and let the normal point from solid into liquid. Write the isotropic linear kinetic undercooling law using the solidification attachment mobility $\mathcal{M}>0$:Here $\gamma$ is solid-liquid surface energy, $\mathcal{K}$ is the sum of principal curvatures, and $T_m$ is absolute temperature. Positive curvature convex into liquid lowers the equilibrium freezing temperature through the Gibbs--Thomson relation. The planar base state has $\mathcal{K}=0$.

Set $\Delta T=T_m-T_{\infty }>0$ and defineThe paper's $S$ is the latent-to-sensible heat ratio, the reciprocal of the common Stefan number convention. Below, velocities, lengths and growth rates are dimensionless in these scales. In the travelling coordinate $\xi =X-Vt$, the liquid heat equation gives ${\theta }_0^{\prime \prime }+V{\theta }_0^{\prime }=0$, while the bounded solid steady field is constant. Matching a decaying liquid profile to the Stefan condition $SV=-{\theta }_0^{\prime }(0)$ givesThe kinetic undercooling condition is ${\theta }_i=1-V$. ThusThe speed is constant for $S<1$, determined by $S$ and the attachment coefficient in the specified scales. Thermal diffusivity sets the exponential decay length $\kappa /U$, not the selected speed. Including solid-side perturbation conduction would change the stability equation below.

For linear stability analysis, corrugate the interface to $\xi =\zeta e^{\sigma t+i\alpha y}$ and write the liquid perturbation as $A_1{e}^{-p\xi }{e}^{\sigma t+i\alpha y}$. The liquid heat equation requiresFor this graph the linearized curvature is ${\alpha }^2\zeta e^{\sigma t+i\alpha y}$. Evaluate the temperature on the displaced interface: its perturbation is $A_1+{\theta }_0^{\prime }(0)\zeta =A_1-SV\zeta$. The kinetic undercooling law therefore givesLikewise the derivative evaluated at the displaced surface is ${\theta }_0^{\prime }(0)+{\theta }_0^{\prime \prime }(0)\zeta -p{A}_1$, with ${\theta }_0^{\prime \prime }(0)=S{V}^2$. The Stefan condition at perturbation order readsElimination proves the one-sided kinetic solidification dispersion:The destabilizing term comes from a protrusion encountering colder liquid; capillarity and kinetic relaxation oppose it.

At $S=0$, $V=1$ and the base liquid has no thermal contrast generated by latent heat. The interface obeys $\boxed{\sigma =-\Gamma {\alpha }^2}$: positive surface energy damps all nonzero modes, while $\Gamma =0$ is neutral at this order. Here the thermal perturbation amplitude vanishes, so the auxiliary decay exponent is immaterial to that pure kinetic relaxation. At $S=1$, $V=0$ and the formal limit givesFor $\Gamma >0$ it has a negative real branch $-{\alpha }^2<\sigma <0$ and no growing branch; at $\Gamma =0$ the interface branch is neutral. Strictly, $V=0$ is a singular endpoint of the travelling profile: no finite-width stationary exponential connects ${\theta }_i=1$ to the far-field value zero. These stability statements describe the limit of the travelling-front model.

For $0<S<1$, assume $\sigma =O({\alpha }^2)$ at small wavenumber. ExpandSubstituting into the dispersion relation cancels $S\sigma /V$ on both sides and givesAt the capillary stability threshold for a kinetic freezing front $\Gamma =S/V$, continuing the same expansion gives $\sigma =-2S{\alpha }^4/V^3+O({\alpha }^6)$: the nonzero long waves are damped even at threshold.

For large wavenumber, multiply the dispersion relation by $p$ and eliminate $\sigma$ using the thermal equation. The exact algebraic form isIt prevents the uncontrolled replacement $p\simeq \alpha$ when $\sigma$ itself is of order ${\alpha }^2$. In the usual small-capillarity range the leading results areFor the second result, set $p=\sqrt{1-\Gamma }\alpha +d+O({\alpha }^{-1})$ in the cubic; its quadratic-order balance gives $2d-V=S\Gamma /(1-\Gamma )$. Thus positive capillarity stabilizes sufficiently short waves. If also $\Gamma \ll 1$ and $|\sigma |\ll {\alpha }^2$, the useful reduced approximation isits finite terms should not be treated as a uniform expansion for arbitrary fixed $\Gamma$.

For completeness, large dimensionless capillarity has different balances. At $\Gamma =1$, $p\sim S^{1/3}{\alpha }^{2/3}$ and $\sigma =-{\alpha }^2+S^{2/3}{\alpha }^{4/3}+o({\alpha }^{4/3})$. For $1<\Gamma <1+2S/V$, $p\to p_{\ast }\equiv S/(\Gamma -1)>V/2$, so $\sigma =-{\alpha }^2+p_{\ast }^2-V{p}_{\ast }+O({\alpha }^{-2})$. At $\Gamma =1+2S/V$, the real branch reaches $p=V/2$: for $S<1/3$ its growth rate is $-{\alpha }^2-V^2/4+O({\alpha }^{-4})$, and for $S=1/3$ that endpoint is exact. For $S>1/3$ at this boundary, or for $\Gamma >1+2S/V$, the selected real-$p$ normal mode ceases to exist at sufficiently large wavenumber; the full diffusion problem includes a thermal continuous spectrum and cannot be assigned a real isolated interface growth rate by this ansatz there. Indeed, writing the bulk thermal perturbation as ${\theta }_1=e^{-V\xi /2}u$ reduces its heat equation to $u_t=u_{\xi \xi }-({\alpha }^2+V^2/4)u$, whose half-line diffusion modes have decay rates $-{\alpha }^2-V^2/4-k^2$ for $k\ge 0$. The thermal branch endpoint is $p=V/2$. Loss of a damped isolated interface mode is not a morphological instability.

The approximate instability condition is already clear from the long-wave coefficient: $\Gamma <S/V$, equivalently $S>\Gamma /(1+\Gamma )$. In fact it is exact for growing modes of the present one-sided model. At $\sigma =0$, let $p_0=[V+\sqrt{V^2+4{\alpha }^2}]/2$, so ${\alpha }^2=p_0(p_0-V)$. The right side of the dispersion relation becomesFor real $\sigma \ge 0$, the residual $F=\sigma (1+S/p)-SV+\Gamma {\alpha }^2+S{V}^2/p$ is strictly increasing, sinceTherefore a unique positive root exists precisely when $\Gamma <SV/p_0^2$. For $0<\Gamma <S/V$, the unstable band isThere are no unstable oscillatory roots hidden by this real-root test. Indeed, if $p=a+ib$ and $\mathrm{Re}\sigma \ge 0$, the thermal equation implies $a\ge V$. For $b\ne 0$, the imaginary parts of the two equations would require $2a-V=S(V-\Gamma {\alpha }^2)/|p+S{|}^2\le SV<V$, contradicting $2a-V\ge V$. Thus growing modes have real $p$ and real $\sigma$.

The sketch shows zero capillarity approaching the positive plateau $SV$, weak capillarity giving a positive band followed by short-wave damping, and threshold/stronger capillarity giving damping at every nonzero wavenumber. The zero-wavenumber mode is a translation. As $S\to 1$ with fixed positive $\Gamma$, both the unstable bandwidth and the positive growth rates shrink to zero; the small-wavenumber expansion is not uniform in that limit.

The three basic modes are radiation, convection and thermal conduction. Radiative transfer dominates transparent atmospheric layers; convection carries heat through sufficiently unstable fluid regions; thermal conduction matters especially in solids or highly conducting dense material.

Consider a chemically homogeneous ideal gas atmosphere and a small, dry parcel displacement, with rapid pressure equilibration and negligible heat exchange during the displacement. The parcel's first law of thermodynamics per unit mass isFor an adiabatic process, $ds=0$. Using hydrostatic equilibrium, $dP/dz=-\rho g$, the parcel followsAfter an upward displacement $\delta z>0$, its temperature relative to its new surroundings isHere $C_p$ is the specific heat capacity at constant pressure, and $N$ is the buoyancy frequency. At the same pressure, a warmer parcel has lower density and accelerates upward. Thus strict instability requires $dT/dz<-g/C_p$. Equivalently, the parcel equation is $\ddot{\delta z}=-N^2\delta z$, withNegative $N^2$ gives growing displacements. The printed non-strict inequality includes the onset boundary: equality is neutrally stable, not a strictly growing instability. This dry parcel buoyancy in a homogeneous atmosphere is the dry Schwarzschild criterion; condensation and composition gradients require modified criteria.

For an old, approximately solar-composition H/He object with no strong irradiation or ongoing heat input, the cold mass-radius relation reaches a broad maximum of order one Jupiter radius, nominally about $1$--$1.2R_J$, at a few Jupiter masses. This is a property of that composition and low-entropy sequence, not a composition-independent theorem from hydrostatic equilibrium alone. Young isolated substellar objects can be larger while cooling; strict thermal steady state of a non-burning isolated body would require its internal luminosity to vanish.

At low mass, an incompressible fixed-composition approximation gives $M\simeq 4\pi \bar{\rho }{R}^3/3$, so $R\propto M^{1/3}$. Ice giants contain dense heavy material and modest H/He envelopes, so real radii depend strongly on composition and are not obtained by extending a pure-hydrogen curve through them.

Gas giants approach a regime in which Coulomb interactions, compression and partial electron degeneracy pressure compete. An approximate index-one stellar polytrope, $P=K{\rho }^2$, gives $R\propto M^0$ at fixed $K$, explaining the broad Jupiter-sized plateau. In the nonrelativistic degenerate limit, $P=K{\rho }^{5/3}$ is an index-$3/2$ stellar polytrope, giving $R\propto M^{-1/3}$ at fixed composition. This is a limiting slope; finite-temperature brown dwarfs do not follow that exact power law over their entire observed mass range.

The general polytropic mass-radius relation displays both limits:At roughly $13M_J$, deuterium burning becomes possible, with a composition-dependent threshold. This is a conventional planet/brown-dwarf division rather than a unique formation definition. At roughly $0.075$--$0.08M_{\odot }$, about $75$--$85M_J$, sustained hydrogen burning can support a low-mass star. Its nuclear energy supply maintains a specific entropy that prevents indefinite cooling onto the cold substellar sequence; its approximate main-sequence relation is $R\propto M$ over a limited low-mass interval.