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There are about ${\eta }^{-1}\sim 10^9$ photons per baryon. Even when the typical photon energy is below the Hydrogen binding energy, the high-energy tail can provide abundant ionizing photons. The nonrelativistic electron's translational phase space also favours the ionized state. Quantitatively, the Saha equation contains both suppressing factors $\eta$ and $(T/m_e{)}^{3/2}$. Recombination needs their tiny product to be overcome by $e^{I/T}$, so $I/T$ must be several tens, not of order one.

For an illustrative ${\Omega }_B{h}^2=0.02$, the supplied baryon-to-photon relation gives $\eta =5.36\times 10^{-10}$. With $m_e=5.11\times 10^5\mathrm{eV}$, direct evaluation givesThus the equilibrium ionization fraction reaches a few per cent just below $0.3\mathrm{eV}$, while at $13.6\mathrm{eV}$ it is essentially unity. The exact temperature depends weakly on the baryon abundance and on the fraction used to define cosmological recombination. This is small baryon abundance delays hydrogen recombination. Recombination, photon decoupling and residual electron freeze-out are related but distinct: the last two involve rates and cannot be determined by chemical equilibrium alone.

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.

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 cosmological principle assumes statistical spatial homogeneity and cosmological isotropy on sufficiently large scales. Spatial homogeneity means that coarse-grained statistical properties do not depend on location; cosmological isotropy means that they do not depend on direction. The claim concerns a large-scale description, allowing individual galaxies, clusters and voids. It motivates the FRW metric and makes cosmic expansion describable by a single scale factor and a small number of matter variables.

Its adoption combines empirical large-scale regularity with the Copernican principle, the assumption that our location is typical. The resulting models provide a coherent account of cosmic expansion, the cosmic microwave background and primordial abundances. Neither the Einstein field equations alone nor cosmological isotropy around just one observer proves global spatial homogeneity: a spherically symmetric universe specially centred on that observer would also look isotropic. Observations of a finite past light cone support the principle over sampled scales without establishing it beyond the observed region.

Three supporting observations probe complementary aspects. First, after the observer-motion dipole and local foregrounds are removed, the cosmic microwave background is extremely isotropic, with intrinsic temperature fluctuations of order $10^{-5}$ and a nearly identical thermal spectrum across the sky. Second, suitably corrected all-sky counts of distant radio sources and extragalactic background intensities show no large preferred direction; Galactic obscuration and survey selection must be accounted for. Third, three-dimensional galaxy redshift surveys show that clustering and voids average toward statistically similar distributions in sufficiently large regions: large-volume number counts approach volume scaling and correlations weaken on large separation scales. The first two primarily test cosmological isotropy, while the third adds evidence for large-scale spatial homogeneity. The principle is a well-supported large-scale working hypothesis whose observational and typicality assumptions remain explicit.

Take $z$ upward from the unperturbed interface in the apparatus frame, with liquid at $z>0$ and pulling velocity $-V{\mathbf{e}}_z$. The primary ice contains no salt. We use the dilute, equal-density approximation and neglect fluid motion relative to the mould. The initial salt inventory per unit horizontal area, with a common density factor suppressed, is $N=h{C}_0$. Its conservation is essential: the far liquid has zero concentration, so an imposed nonzero far-field composition would describe a different experiment.

The steady advection-diffusion equation is $-V{C}^{\prime }=D{C}^{\prime \prime }$. Decay into the pure water and interfacial salt rejection giveSince ${\int }_0^{\infty }C(z)dz=D{C}_I/V=N$, the finite-inventory directional salt rejection solution isThe decay length is $D/V$. Local liquidus equilibrium givesThus the thermal apparatus selects the interface position at this isotherm. The stated speed inequality is precisely $C_I<C_E$, so the interface is warmer than the eutectic temperature, $T_I>T_E=-m{C}_E$. These are steady values after the initial finite saline layer has rearranged; the final exponential distribution is not confined to the original depth $h$.

For $Vh{C}_0/D>C_E$, a steady pure-ice front would require an interfacial composition beyond the eutectic composition. That branch is unavailable. During the transient, rejected brine concentrates until it reaches $C_E$; secondary salt-bearing crystals can then form and the residual brine freezes as an eutectic system. Salt is incorporated into the product rather than remaining indefinitely in a super-eutectic exponential layer. The detailed pattern can include brine pockets and a nonplanar freezing region; it depends on the thermal history and the morphological stability. The speed inequality alone does not decide that latter stability.

For the stability calculation use an interface displacement $\eta =\hat{\eta }{e}^{ikx+st}$ and a liquid perturbation $c=\hat{c}{e}^{ikx+st-qz}$. Here $k>0$ is the wavenumber, $s$ is the temporal rate and $\Re q>0$ gives spatial decay. The frozen thermal field is $T_I+Gz$. Define $\Gamma ={\gamma }_{sl}{T}_m/(\rho L)>0$, where ${\gamma }_{sl}$ is solid-liquid surface energy, $T_m$ is absolute pure-solvent melting temperature, $\rho$ is mass density and $L$ is specific latent heat. For a solid protrusion into liquid the curvature is $\mathcal{K}=-{\eta }_{xx}=k^2\eta$, and the Gibbs-Thomson effect gives local equilibrium $T=-mC-\Gamma \mathcal{K}$. We assume isotropic surface energy and instantaneous interface kinetics, so no kinetic undercooling term is present.

Linearizing the equilibrium condition at the displaced interface givesThe perturbation advection-diffusion equation givesThe local solute balance at a moving graph, to first order, isIts perturbation is $(V-Dq)\hat{c}+s{C}_I\hat{\eta }=0$, because the terms $D\eta C^{\prime \prime }(0)$ and $V\eta C^{\prime }(0)$ cancel. Consequently the full frozen-temperature solutal interface dispersion isNo quasistatic approximation to solute diffusion was made here. For a growing mode the unique admissible liquid root isWriting $b=(G+\Gamma k^2)/(m{C}_I)$ and eliminating $s$ gives $(q-V/D)(q-V/D+b)=k^2$. The root continuously connected to the growing interface branch is thereforeIn particular, unstable wavelengths exist exactly when $mV{C}_I/D>G$, and their band isThis is the constitutional supercooling criterion: immediately above the plane, the local liquidus rises at rate $-m{C}^{\prime }(0)=mV{C}_I/D$, so it overtakes the imposed thermal gradient when that rate exceeds $G$. The Gibbs-Thomson effect cuts off short waves. At long wavelength $s\sim D(V/D-b)k^2/b$ for positive $G$. The exactly uniform inventory-changing displacement is not an instability of a fixed-inventory experiment. On stable branches the full diffusion problem also admits relaxation transients; the root selection above is sufficient to identify every exponentially growing interface mode.

If one additionally neglects the perturbation time derivative in the liquid diffusion equation, the alternative reduced rate isIt has the same onset and cutoff, but need not have the same rate away from the slow-growth regime. The frozen-temperature approximation by itself does not justify this extra reduction.

After onset, protrusions reach more weakly saline liquid and advance faster, while grooves accumulate rejected brine. The planar interface develops cells and, as growth becomes sufficiently strong, crystal dendrites with concentrated liquid between them. Nonlinear competition changes the selected spacing; linear rates alone cannot fix the final geometry. The interdendritic liquid forms a mushy layer, concentrates toward the eutectic composition, and ultimately freezes into a salt-bearing eutectic product when it reaches $T_E$. This replaces the pure planar salt-rejection picture with trapping and freezing of brine between crystals.

Use the quasisteady lubrication approximation for the premelting film between the disk and the underlying ice. The ice is anchored to the lower plane; new ice formed from water supplied through this upper film raises the disk. Let $U$ be its upward separation rate, $d(r)$ the premelted thickness, $\mu$ the liquid dynamic viscosity, $\rho$ a common ice/water mass density, $L$ the specific latent heat of fusion, and $T_m$ the absolute bulk melting temperature. As usual in this leading model, neglect the ice–water density difference, pressure-dependent changes in $T_m$ due to a common phase pressure, inertia and edge corrections. The imposed thermal field is taken as maintained by an external heat supply/removal system, so no additional heat-conduction bottleneck is introduced.

Write $\Delta =T_m-T_0>0$, $K_T=\rho L/T_m$ and $\Delta (r)=\Delta (1-r^2/a^2)$. The Clapeyron pressure relation for equal-density phases makes the pressure difference across the solid–liquid interfaceThis thermomolecular pressure is the repulsive stress transmitted through the film. To calculate a definite rate one also needs a film-thickness constitutive law. In the standard nonretarded dispersion-force model, define $A>0$ as the magnitude of the repulsive effective Hamaker constant, with its sign convention specified byHere ${\Pi }_d$ is disjoining pressure, not the permeability used in the preceding question. The sign of an effective Hamaker coefficient depends on the three materials; the positive parameter here denotes an interaction that stabilizes a premelted film. The problem does not specify $A$ numerically, so it must remain a defined material parameter.

Take the reservoir pressure at the rim to be $p_{\infty }$. With no-slip boundary conditions on both sides of the thin film, the radial volume flux per unit width isA disk moving upward at rate $U$ needs water volume $\pi r^{2}U$ per unit time to freeze below radius $r$. Regularity on the axis and mass conservation therefore give $2\pi r{q}_r=-\pi r^{2}U$, or $q_r=-Ur/2$. ThusDefine $B=A/(6\pi K_T)$, so $d^3=B/[\Delta (1-r^2/a^2)]$. Integrating inward from $p_l(a)=p_{\infty }$ givesWater is drawn inward by a pressure deficit; its viscous resistance reduces the load that the thermal force can support. The net normal stress on the disk is $p_l-p_{\infty }+P_T$. Let $W$ be the downward apparent load after subtracting ordinary buoyancy. For an ideal thin disk with negligible buoyant volume, $W=Mg$, where $g$ is gravitational acceleration. If a finite displaced volume is known, use its submerged weight instead; the stated mass and radius alone do not determine that volume.

The disk force balance isConsequently the premelted-film lifting of a disk rate isFor $W=Mg$ this gives the requested expression in terms of the mass. Every additional symbol in this formula has been defined above. The growing-ice lifting branch exists only when the bracket is positive.

At the threshold the required water flux and viscous pressure deficit vanish. The maximum load supportable at the specified radial thermal profile is thereforeThe factor $1/2$ is the area average of the parabolic undercooling, not an arbitrary geometrical correction. At equality the lift rate is zero; strictly positive lift requires $W<W_{\max }$. There is no radius-only, temperature-independent maximum load in this ideal model: one must specify the imposed undercooling or a physically admissible range of it.

For a fixed positive load define ${\Delta }_c=2W{T}_m/(\pi a^2\rho L)$ and $U_{\infty }=A/(6\pi \mu a^2)$. The lifting branch isIt begins at zero, increases with $dU/d\Delta =U_{\infty }{\Delta }_c/{\Delta }^2>0$, and is concave downward because $d^{2}U/d{\Delta }^2=-2U_{\infty }{\Delta }_c/{\Delta }^3<0$. Below threshold no positive steady heave is possible: the thermomolecular force is too weak to support the load. The mathematical continuation to negative $U$ is not a prediction of upward ice-driven separation there.

Increasing undercooling increases the available thermal pressure, but it also thins the premelting film. In the inverse-cube disjoining pressure model, $d^3\propto {\Delta }^{-1}$, so the hydraulic conductance decreases in exact inverse proportion to the driving scale. The remaining load correction tends to zero and the rate approaches a plateau:For positive load this is approached from below, not attained at finite undercooling. With zero load the reduced model gives the same constant rate wherever its thin-film assumptions hold.

Two limits of the calculation are worth making explicit. First, the formal $d(r)$ diverges at the warm rim, so a narrow outer connection to bulk water replaces the thin-film solution there; this leading model assumes that its contribution to the integrals is negligible. Second, very large undercooling produces a molecularly thin film and eventually invalidates continuum lubrication theory and the simple dispersion-force law. Thus the plateau is the maximum within the stated ideal constitutive model, not an unrestricted extrapolation for real water.

If a different molecular interaction is intended, the result is not universal. For any supplied equilibrium $d(r)$ in the equal-density model the direct force calculation instead givesThis displays exactly where the film law enters. The closed formula and plateau above use the conventional repulsive inverse-cube law; premelting alone, without that material law, would not determine a unique rate curve.

Take any fixed material subvolume $B_0$ of the reference configuration, with outward normal vector $N$, current particle velocity $v_i={\dot{x}}_i$, and positive temperature $\theta$. The first law of thermodynamics, including kinetic energy, isThe heat-flux sign is negative because $N_I{q}_I^0$ is outward. The Second law of thermodynamics supplies the corresponding integral Lagrangian entropy inequality:The temperature in each supply or flux term is the local temperature, not a common temperature pulled outside the integrals.

Apply the divergence theorem and use the reference momentum conservation equation ${\rho }_0{\dot{v}}_i=P_{Ii,I}+{\rho }_0{g}_i$. In the total-energy equation, the kinetic derivative cancels the body-force work and the $v_i{P}_{Ii,I}$ term. Since $v_{i,I}={\dot{F}}_{iI}$, the remaining local Lagrangian internal energy balance and Lagrangian entropy inequality areThey hold pointwise because $B_0$ was arbitrary. In particular,

The specific Helmholtz free energy is $\psi =u-\theta \eta$. Substitution yields its balanceand elimination of the heat supply yields the Clausius-Duhem inequalityFor $\psi =\psi (F,\theta ,{\xi }_r)$, expand the derivative to obtainThe usual Coleman-Noll procedure admits independent reversible choices of $\dot{F}$ and $\dot{\theta }$ at a fixed state. Their coefficients must vanish, since either sign and arbitrary magnitude would otherwise violate the inequality. ThusInternal-variable evolution and heat conduction must satisfy this remaining inequality. Substituting the constitutive relations back into the energy balance givesThe last printed equation is missing a factor ${\rho }_0$ on its internal-variable term. The PDF defines $f_r=-{\psi }_{,{\xi }_r}$ with $\psi$ per unit mass, so the expression above is the dimensionally consistent consequence. Equivalently define the volumetric thermodynamic force conjugate to an internal variable ${\hat{f}}_r=-{\rho }_0{\psi }_{,{\xi }_r}$; the term is then ${\hat{f}}_r{\dot{\xi }}_r$ with no extra density. The two conventions must not be mixed.

For thermoelasticity with a temperature-dependent constraint, admissible rates obeyThe reversible coefficient functional need now vanish only on this tangent hyperplane. Hence it can be a scalar multiple $q$ of the constraint normal:ThereforeThe reaction terms cancel on admissible rates, leaving the same internal-variable/conduction inequality and entropy heating equation. The multiplier $q$ is determined by equilibrium and boundary conditions rather than by the free energy alone.

For linear thermoelasticity with prescribed thermal volume, write $F=I+H$, $H_{ij}=u_{i,j}$, and $\vartheta =\theta -{\theta }_0$. Hold the internal variables fixed in this reversible linearization, and evaluate all derivatives below at $(I,{\theta }_0)$ and zero reaction. The stress-free reference state has ${\psi }_{,F}=0$. Since $\partial \det F/\partial F=I$ there, first-order nominal and Cauchy stress tensors agree. Their constitutive laws becomewhere equality of mixed free-energy derivatives givesThe linearized constraint is $u_{i,i}=h^{\prime }({\theta }_0)\vartheta$. Thus volume is thermally prescribed; strict temperature-independent incompressibility corresponds to $h^{\prime }=0$. The coefficient $C_e$ is an entropy slope; the corresponding volumetric heat capacity at fixed deformation is ${\theta }_0{C}_e$.

Together with ${\rho }_0{\ddot{u}}_i={\sigma }_{ji,j}+{\rho }_0{g}_i$, the first-order thermal balance is ${\theta }_0(C_e\dot{\vartheta }-{\beta }_{ji}{\dot{u}}_{i,j}+h^{\prime }({\theta }_0)\dot{q})={\rho }_{0}r-q_{I,I}^0$. A heat-flux constitutive law completes the evolution problem. The paired signs of the coupling tensor follow from the same free energy, rather than from separately postulated mechanical and thermal equations.

Take $a_1,a_2>0$ and $p_1+p_2=1$. Write $g^{(k)}=\nabla u$ in phase $k$ and $h^{(k)}=a_k{g}^{(k)}$. Here $h$ is the positive conductivity flux; physical heat flux is $-h$. Continuity of temperature on a planar interface implies continuity of its tangential derivatives, while conservation gives continuity of normal heat flux. With interface normal $e_1$, these conditions areThe mean gradient is $E=p_1{g}^{(1)}+p_2{g}^{(2)}$. ThereforeBy the definition of effective conductivity, $⟨h⟩=a^{\ast }E$. ThusThe laminate conductivity is a weighted harmonic mean across the layers and a weighted arithmetic mean along them. This also explains why isotropic constituents can yield an anisotropic thermal conductivity tensor: flux must cross each resistance in sequence in the normal direction, whereas tangential transport occurs in parallel.

Within each constant-conductivity region the temperature is harmonic. For $u=v(r)\cos \theta$, the spherical Laplacian givesTrying $v=r^s$ yields $s(s-1)+2s-2=(s-1)(s+2)=0$. Thus $v=Cr+D{r}^{-2}$. Boundedness at the origin removes the interior $r^{-2}$ term, while the prescribed far gradient fixes the exterior growing term:Continuity of temperature and normal flux at $r=r_1$ giveSolving these two equations yields $B=-r_1^3(a_1-a_0)/(a_1+2a_0)$ and $C_{\mathrm{in}}=3a_0/(a_1+2a_0)$. ConsequentlyPositive conductivities make the denominator nonzero. The matched solution is unique in the usual finite-energy decaying-perturbation class: the difference of two solutions has zero far data and interface jumps, and an energy integration gives zero gradient.

For the polarization field of a conductivity inclusion, use the sign convention consistent with the perturbation equation:ThusFor a general incident constant gradient $E$, rotational symmetry and linearity replace $e_3$ by $E$. The polarization vanishes outside the sphere because its local conductivity equals the reference value. The temperature perturbation outside is dipolar; this is the spherical conductivity inclusion response used in the remaining parts.

Let $K$ denote isotropic permeability of a porous medium, $\mu$ the fluid dynamic viscosity, ${\rho }_0$ the reference fluid mass density, $\alpha >0$ its volumetric thermal-expansion coefficient and $g$ gravitational acceleration. The Boussinesq approximation uses $\rho ={\rho }_0[1-\alpha (T-T_{\ast })]$ only in the gravitational force. With upward coordinate $z$, hydraulic pressure $P$ and Darcy flux $\mathbf{q}$, the equations arewhere $\theta$ is the perturbation to the conductive temperature. To keep porous heat-capacity conventions explicit, let $C_e$ be effective heat capacity per bulk volume, $C_f={\rho }_0{c}_f$ the liquid volumetric heat capacity, $k_e$ effective thermal conductivity, $\sigma =C_e/C_f$, and $\kappa =k_e/C_f$. The thermal equation is $\sigma T_t+\mathbf{q}\cdot \nabla T=\kappa {\nabla }^{2}T$. If the usual unit-capacity convention is adopted, $\sigma =1$ and $\kappa$ is the thermal diffusivity. Thresholds below do not depend on $\sigma$; it only changes temporal rates.

For a conductive state with upward temperature decrease $\Gamma =-{\overline{T}}_z>0$, linearization giveswhere $w=q_z$. The second equation follows by taking the divergence of Darcy law to eliminate $P$ and then differentiating its vertical component. It is important that the Laplacian on its right is horizontal.

In the horizontal layer, both boundary planes prescribe the temperature and are impermeable, so $\theta =w=0$ at $z=0,h$. Use a normal mode proportional to $e^{st+i\mathbf{k}\cdot \mathbf{x}}\sin (n\pi z/h)$, with horizontal wavenumber $k=|\mathbf{k}|$, $n\ge 1$. Writing $b=n\pi /h$, the momentum relation gives $w=B{k}^2\theta /(k^2+b^2)$, and the thermal relation givesA neutral mode therefore requires $B\Gamma /\kappa =(k^2+b^2{)}^2/k^2$. Since $k^2+2b^2+b^4/k^2\ge 4b^2$, the smallest value is $4{\pi }^2/h^2$, attained at $n=1$ and $k=\pi /h$. This derives the onset of convection in a horizontal Darcy layer rather than simply importing its critical number. ThusEvery small disturbance decays for $T_m<T_{m,\mathrm{crit}}$; at equality the critical roll is neutral. Strict decay has this threshold as a supremum, not an attained maximum. The first roll wavelength is $2h$, compatible with the very large horizontal extent. Cooling from above gives the opposite, stable buoyancy feedback.

For the tall square column, define $a$ as its side length and $H\gg a$ as its height. The insulated vertical walls impose ${\partial }_n\theta =0$, while impermeability imposes zero normal Darcy flux. Away from the remote ends, a nearly height-independent mode has hydrostatic pressure adjustment and $w=B\theta$, with zero cross-sectional mean. The transverse temperature modes arewhere $m,n\ge 0$ but $(m,n)=(0,0)$ is excluded: a closed column cannot support a net vertical discharge. Its horizontally uniform density perturbation is balanced by pressure, rather than driving overturning. The rate of a nonconstant mode isThe smallest allowed eigenvalue is ${\pi }^2/a^2$, so the insulated square-column Darcy convection onset isAgain strict decay requires $\Gamma <{\Gamma }_{\mathrm{crit}}$. This is the leading long-column result. End conditions that maintain the basic gradient require the vertical velocity to turn around near the ends and give small finite-height corrections; their exact coefficient cannot be inferred from insulation of the vertical walls alone. For a separated vertical mode of small wavenumber $k_z$, the neutral expression is $B\Gamma /\kappa =(k_{\perp }^2+k_z^2{)}^2/k_{\perp }^2$, which tends to the displayed value as $H/a\to \infty$.