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At the tricritical point, tune $u=0$, hold $v>0$, and let $r=at$ cross zero. The equation of state becomes $h=rM+v{M}^5$. At $h=0,r<0$, $M^4=-r/v$, while the inverse susceptibility on the ordered branch is $r+5v{M}^4=4|r|$. On the disordered side it is $r$. At $r=0$, $M=\mathrm{sgn}(h)(|h|/v{)}^{1/5}$. Therefore $\beta =1/4$, $\gamma =1$ and $\delta =5$.

The minimized singular free-energy density isTwo temperature derivatives give a singular heat capacity proportional to $|t{|}^{-1/2}$ on the ordered side. Consequently the mean-field tricritical exponent calculation givesThe displayed scaling relations for critical exponents hold numerically:These are Landau theory values with both relevant coefficients tuned: along a generic path with fixed $u>0$, the asymptotic transition is ordinary rather than tricritical. The sextic upper critical dimension is three, so fluctuations can change the powers below three dimensions, and logarithmic corrections are possible at three. The vanishing disordered mean-field singular amplitude does not invalidate the ordered-side exponent.

Take a positive gradient coefficient ${\kappa }^{-1}$, a positive squared mass, a finite-volume regulator and an ultraviolet cutoff. Use the dimensionless statistical action convention $Z=\int \mathcal{D}\varphi e^{-H}$: inverse temperature is absorbed into the coefficients. If $H$ denotes dimensional energy instead, the exponent below gains a factor ${\beta }_{\mathrm{th}}$ and the covariance gains ${\beta }_{\mathrm{th}}^{-1}$. These conventions cannot be mixed.

Choose the Fourier transform conventionIntegration of the quadratic gradient and source terms givesComplete the square with ${\tilde{\varphi }}_{\mathrm{cl}}(p)=\tilde{\Delta }(p{)}^{-1}\tilde{J}(p)$. Translation of the regulated real Gaussian variables leaves their measure unchanged, so the Gaussian integral givesHere $Z_G=Z_0[0]$ is the source-free Gaussian partition function. For $n$ independent regulated real coordinates and a positive quadratic matrix $A$, it is $(2\pi {)}^{n/2}(\det A{)}^{-1/2}$ under Lebesgue normalization. The continuum determinant needs the stated regulator and a measure convention. Reality pairs the nonzero Fourier modes, so they must not be counted as two independent real fields.

A connected correlation function subtracts the product of the means. The logarithm $W[J]=\log Z_0[J]$ generates its cumulants: differentiating $W$ twice with respect to the source, using the measure $d^{D}p/(2\pi {)}^D$ in the functional derivative convention, yieldsThe source changes the mean but not the covariance; the stated zero-source limit therefore follows immediately. Momentum conservation uses $p+q$, appropriate for a real field without complex conjugation on either displayed factor.

For the continuum massive Gaussian field correlation tail, set $\mu =\sqrt{\kappa }m>0$ and $r=|x|$. Fourier inversion givesThe large-argument Modified Bessel function of the second kind has $K_{\rho }(z)\sim \sqrt{\pi /(2z)}{e}^{-z}$. Hence $G_0(r)$ is proportional to $r^{-(D-1)/2}{e}^{-\mu r}$ at large distance andAt zero mass the exponential length diverges. For $D>2$ the critical continuum tail is algebraic, $G_0(r)\propto r^{2-D}$; in lower dimensions infrared regulation needs additional care.

For a uniform source $h$, perform a momentum-shell renormalization group step: separate slow modes $|p|<\Lambda /b$ and fast modes $\Lambda /b<|p|<\Lambda$. In a Gaussian field theory they decouple; the uniform source couples only to the retained zero mode. Integrating fast modes contributes only a field-independent determinant to the free energy. Restore the cutoff by $x^{\prime }=x/b$, $p^{\prime }=bp$ andThe factors from the measure, gradient and two fields cancel in the gradient term. The mass term gains $b^2$, and the linear source gains $b^{(D+2)/2}$. ThereforeHere $m$ is the positive mass; for signed thermal coordinates it is the squared-mass relation that applies. The Gaussian fixed point has thermal scaling eigenvalue ${\lambda }_t=2$, source eigenvalue ${\lambda }_h=(D+2)/2$ and zero anomalous dimension.

Under the formal homogeneous Gaussian scaling hypothesis, substitution of those eigenvalues givesThe last two values are independently visible in $\xi \propto t^{-1/2}$ and $\chi ={\tilde{G}}_0(0)=1/m^2\propto t^{-1}$, agreeing with Landau-Ginzburg theory. The specific-heat power follows directly from the determinant: after two thermal derivatives its singular contribution is proportional toThe change of variables $p=\sqrt{\kappa t}q$ proves this power because the rescaled integral converges at large $q$ exactly when $D<4$. The two-dimensional free energy itself has a $t\log t$ resonance, while its second derivative still has the indicated $t^{-1}$ power.

The printed magnetization exponent is a formal Gaussian scaling index, not a proved ordered-phase law of the purely quadratic model. The ordered-phase obstruction in a Gaussian scalar model is explicit: the uniform field has energy density $m^2{M}^2/2-hM$. At $h=0,m^2<0$ it tends to $-\infty$ as $|M|\to \infty$, and the partition function diverges. For $m^2>0$, the mean is $M=h/m^2$ and tends to zero as $h\to 0$, so there is no spontaneously ordered branch. Thus no literal $M\sim |t{|}^{\beta }$ for $t<0$ can be established from this Statistical Hamiltonian alone. The value $(D-2)/4$ follows from the scaling dimension of the field if one assumes such a branch; for $D\le 2$ its nonpositive value further warns against that interpretation. Adding a stabilizing quartic interaction defines an ordered phase, but below four dimensions that interaction is relevant and generally leads to the Wilson-Fisher fixed point, not these Gaussian thermodynamic powers.

For the Gaussian specific-heat infrared threshold, at $D=4$ the same determinant integral grows like $\log ({\Lambda }^2/t)$, so $\alpha =0$ as a power exponent, with a logarithmic divergence. For $D>4$, it has a finite cutoff-dependent limit; the leading heat capacity is a regular constant and has no divergent power, again described by $\alpha =0$ in the requested convention. After subtraction of this background the Gaussian singular term can still have the negative power index $2-D/2$; it is not correct to erase that distinction. In a stable quartic Landau-Ginzburg theory above four dimensions, the quartic coefficient is a dangerously irrelevant coupling: it is needed for the ordered phase and contributes $f_s\propto -t^2/u$, giving the mean-field specific-heat jump and $\alpha =0$ while invalidating naive hyperscaling relation. At four dimensions, marginal interactions can add further logarithmic corrections. The agreement with Landau theory is in the leading heat-capacity power; it does not assert the absence of logarithms or supply a missing stable Gaussian ordered phase.

Use the planar geometry indicated by the supplied $x,z$ equation and streamfunction; the source heat input is then measured per unit out-of-plane span. Let $\Theta =T-T_0$ and defineThis is the buoyancy part of Darcy law. Let $C_h$ denote the constant heat capacity per volume consistent with the given transport equation; for the fluid heat-flux convention $C_h={\rho }_0{c}_p$. Porosity or matrix heat-storage factors, if retained, must be used consistently in this coefficient and the effective transport parameters.

For plume width $b\ll z$, continuity gives $u\sim wb/z$. Both $u{\Theta }_x$ and $w{\Theta }_z$ are consequently of order $w\Theta /z$: lateral advection must not be discarded. Lateral thermal diffusion is of order $\kappa \Theta /b^2$, while vertical diffusion is smaller by $(b/z{)}^2$. The slender steady porous thermal plume model is thereforeWriting it in conservative form,and integrating across the plume, where $\Theta$ and lateral diffusive/advective heat transport vanish at large $|x|$, provesIn steady state with no lateral heat loss and negligible vertical conductive flux, this equals the heat supplied at the source. Define $Q=\beta F_h/C_h=\int w^{2}dx>0$.

For characteristic velocity $W$ and width $b$, flux conservation gives $W^{2}b\sim Q$ and the advection-diffusion balance gives $W{b}^2\sim \kappa z$. ConsequentlyThe dimensional scales are $b\sim ({\kappa }^2{z}^2/Q{)}^{1/3}$ and $W\sim (Q^2/(\kappa z){)}^{1/3}$. The plume becomes relatively more slender, $b/z\propto z^{-1/3}$, away from the source.

To find the profile, put $\eta =x/b(z)$ and $\psi =-W(z)b(z)f(\eta )$, so $w=W{f}^{\prime }$ andInserting into $u{w}_x+w{w}_z=\kappa w_{xx}$ givesIts coefficient is constant under the derived scaling. Choose the width normalization $W{b}^2=6\kappa z$. Integrating once with decaying velocity derivatives gives $f^{\prime \prime }+2f{f}^{\prime }=0$. A second integration gives $f^{\prime }+f^2=1$ after normalizing the centreline velocity to $W$; symmetry has $f(0)=0$. The decaying, positive solution is $f=\tanh \eta$. Hence the hyperbolic-secant porous plume profile isSince ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^4\eta d\eta =4/3$, the precise amplitudes areThey satisfy both $W{b}^2=6\kappa z$ and $(4/3)W^{2}b=Q$. The derived $u$ supplies lateral entrainment; setting $u=0$ would not reproduce this profile or satisfy continuity.

A heat-weighted head speed of a porous plume estimate follows by filling the column below a head height $Z_h(t)$ with this steady profile. Its excess heat per unit height isConservation of the total injected heat gives $F_{h}t\simeq {\int }_0^{Z_h}\mathcal{H}(z)dz$. Differentiating,Thus a heat-weighted head advances at a speed of order the local centreline velocity and decelerates as $Z_h^{-1/3}$. With $K=(3Q^2/(32\kappa ){)}^{1/3}$,The factor $2/3$ is the energy-conserving estimate for a truncated steady column; the foremost centreline parcels have characteristic speed $W(Z_h)$ instead. The steady calculation does not resolve the transient nose or define an exact sharp temperature front. A truly axisymmetric point-source plume would require different geometry and cannot use this planar profile unchanged.

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.

For a stationary, uniformly permeable skeleton, Darcy law gives $\mathbf{u}=-(\Pi /\mu )\nabla p$. Constant-density conservation of mass gives $\nabla \cdot \mathbf{u}=0$, hence the pressure is harmonic:With uniform gravitational acceleration, the same argument applies to pressure after subtracting its hydrostatic part. The spherical calculation below neglects body-force variation and uses the excess pressure relative to the far field.

Set $ϵ=1-{\rho }_s/\rho >0$ and $\beta =\varphi ϵ$. When a shell freezes, its pore mass decreases at rate $4\pi a^2\varphi (\rho -{\rho }_s)\dot{a}$; the difference must leave as liquid. Since Darcy velocity is volume discharge per total area, not pore-fluid speed, the interfacial and radial balances giveThe positive sign means outward expulsion by the expansion on freezing. Integrating Darcy law with $p(\infty )=0$ givesThere is no liquid Darcy velocity in the ice-filled region.

To state a closed Stefan problem, let $k_{-},k_{+}$ be the effective thermal conductivities of the frozen and unfrozen composites and $C_{-},C_{+}$ their volumetric heat capacities. If the matrix mass density is ${\rho }_m$, the common specific heat gives $C_{-}=c_p[(1-\varphi ){\rho }_m+\varphi {\rho }_s]$ and $C_{+}=c_p[(1-\varphi ){\rho }_m+\varphi \rho ]$. These material coefficients are not numerically specified. With $\mathcal{L}=\varphi {\rho }_{s}L$, the thermal equations areThe exterior equation retains advected sensible heat. A more detailed mechanical-energy model adds the Darcy dissipation source $\mu u^2/\Pi$; its smallness at leading order is checked below. At the origin require ${\partial }_r{T}_{-}(0)=0$, and at infinity require $T_{+}\to T_{\infty }$, $p\to 0$. At the phase boundary impose temperature continuity, phase equilibrium and the Stefan condition:These equations, the radial liquid-mass balance and Darcy law, together with an initial radius and compatible initial temperature fields, specify the moving-boundary model. No curvature correction is imposed. A vanishing initial seed corresponds to the leading growth law below with $a(0)=0$.

Write $\Delta =T_m-T_{\infty }$ and $S=L/(c_p\Delta )$, the latent-to-sensible heat ratio. A conductive interfacial balance gives $a\dot{a}\lesssim k_{+}\Delta /\mathcal{L}$. On length scale $a$, the ratio of thermal diffusion time to growth time is consequently bounded byIt is small for $S\gg 1$ with fixed porosity and comparable constituent densities and thermal properties. Very small porosity requires the sharper condition $C_{\pm }\Delta /(\varphi {\rho }_{s}L)\ll 1$; the bare heat-capacity ratio alone should not be used uniformly as $\varphi \to 0$. The advective-to-conductive ratio isEven if mechanical heating is retained, its size relative to latent-heat release is $p_i\beta /\mathcal{L}=(T_m-T_i)/T_m\le \Delta /T_m$. For fixed water properties this is also small as $S\to \infty$, consistently with the linearized Clausius-Clapeyron relation. Thus unsteadiness, thermal advection and mechanical heating are higher-order effects. The local leading temperature fields satisfy Laplace's equation. The approximation describes radii of order $a$; the distant transient diffusion tail supplies the far-field matching and need not be globally steady.

Regularity at zero and the interface/far-field values giveThe Stefan condition becomes $\mathcal{L}a\dot{a}=k_{+}(T_i-T_{\infty })$. Introduce the dimensionless pressure-feedback factorCombining this heat balance with pressure-dependent phase equilibrium yields the growth and interface values:This is pressure-limited spherical freezing in a porous medium. The interface values are constant during this leading solution, although $p(r,t)$ changes at a fixed exterior radius.

As permeability of a porous medium decreases, $\Lambda$ increases: the interface pressure rises toward the finite value $\Delta /\gamma$, while the interface temperature falls toward $T_{\infty }$. Hydraulic resistance suppresses freezing through pressure-induced freezing-point depression. In the limit $\Pi \to 0$, $a\dot{a}\sim \Pi \Delta /(\gamma \mu \beta )$; a zero-seed radius grows as $\sqrt{\Pi t}$ and tends to zero at fixed time. A finite seed instead hardly changes. At exactly zero permeability an incompressible rigid model cannot expel the expansion volume, so sustained growth stops; deformation or compressibility would require a different model.

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

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

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

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

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

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

At an ordinary zero-field continuous phase transition, the positive quartic coefficient remains nonzero while ${\mathcal{A}}_2\to 0$. Approaching from the ordered side therefore sends the variable $X$ in the tricritical crossover scaling form to $+\infty$, rather than to zero. With $Y=0$ the dimensionless minimizer obeys ${\psi }^4+X{\psi }^2-1=0$, soAt that stationary point, substitution gives ${\Phi }_{-}(X,0)=-{\psi }^2/4-{\psi }^6/12$. Its large-quartic asymptotic of tricritical scaling is consequently ${\Phi }_{-}(X,0)\sim -1/(4X)$, andIf ${\mathcal{A}}_2$ crosses zero linearly with temperature, this is quadratic in $T-T_c$, giving a finite heat capacity jump rather than a power divergence. ThereforeFor locally constant ${\mathcal{A}}_4>0$ and slope $a={\partial }_T{\mathcal{A}}_2{|}_{T_c}$, the leading singular heat capacity jump from this term is $\Delta C=T_c{a}^2/(2{\mathcal{A}}_4)$. Treating $\Phi$ as a nonzero constant while $X\to \infty$ would incorrectly assign the tricritical exponent to this ordinary transition.

Assume the vertical dynamical timescale of a disk is shorter than the thermal timescale of an accretion disk, which is itself shorter than the viscous timescale. This ordering is controlled in a thin disk with $\alpha \ll 1$: $t_{\mathrm{dyn}}\sim {\Omega }^{-1}$, $t_{\mathrm{th}}\sim (\alpha \Omega {)}^{-1}$ and $t_{\nu }\sim (r/H{)}^2{t}_{\mathrm{th}}$. Therefore compare heating and cooling at fixed surface density of a disk, radius, $\alpha$ and composition. Vertical hydrostatic equilibrium adjusts on the vertical dynamical timescale of a disk, giving $H\propto T^{1/2}$ and $\rho \propto T^{-1/2}$. Consequently the stipulated negative hydrogen ion opacity varies as $\kappa \propto T^{10-1/6}=T^{59/6}$. The rates per disk face then scale asThus heating increases while cooling decreases after a positive temperature perturbation. At equilibrium $Q^{+}=Q^{-}=Q_0$, a small change gives $\delta (Q^{+}-Q^{-})=(41/6)(Q_0/T)\delta T$, which drives the perturbation further from equilibrium for positive effective heat capacity. A negative temperature perturbation likewise increases the net cooling. This proves thermal instability from negative hydrogen ion opacity within the local closure; it is the temperature-slope test of thermal stability of an accretion disk. Holding volume density rather than surface density of a disk fixed would give a different cooling exponent, but still the same qualitative instability; the fixed-column comparison is the relevant thin-disk thermal test.

Use the usual homogeneous, local, analytic Landau-Ginzburg theory; rotational symmetry and Z2 symmetry alone do not specify locality, analyticity, or the signs of the coefficients. The local derivative expansion contains every rotational scalar with an even total number of fields:The omitted terms include all allowed higher field powers and contracted spatial derivatives. Terms differing by a total derivative are equivalent for bulk behavior with suitable boundary conditions. The displayed gradient energy assumes spatial homogeneity. If only rotations about an origin are imposed, rotationally invariant position-dependent coefficients are also possible.

In the mean-field approximation, take a spatially uniform order parameter and minimize its Landau free energy. For $\gamma >0$, $u>0$ and $r=a(T-T_c)$ with $a>0$, keeping the quadratic and quartic terms givesThe single minimum splits into two minima as the temperature passes below the critical temperature. Choosing one minimum gives spontaneous symmetry breaking of the Z2 symmetry and a continuous phase transition. The order parameter is the expectation in a selected ordered state; averaging equally over both states gives zero. A Z2 symmetry does not guarantee a continuous phase transition: a negative quartic coefficient stabilized by a positive sextic coefficient can instead give a first-order phase transition.

Relaxing the Z2 symmetry permits odd terms. A nonzero linear source generally rounds this ordinary continuous phase transition. A cubic term can instead give a first-order phase transition: for $V(\varphi )=r{\varphi }^2/2+w{\varphi }^3/3+u{\varphi }^4/4$, $u>0$ and $w\ne 0$, equal minima occur atThe order parameter jumps from zero to ${\varphi }_{\ast }$. A continuous phase transition without an exact Z2 symmetry is still possible with additional tuning; odd terms do not logically exclude it.

For the multicritical even Landau potential, set $r={\mu }^2=a(T-T_c)$ and assume an integer $n\ge 2$, ${\lambda }_n>0$ and $\gamma >0$. For $n>2$, all lower even interactions must be absent or tuned to zero to obtain these mean-field critical exponents. The stationary equation isIts nonzero solutions are minima because $V^{\prime \prime }({\varphi }_0)=-(2n-2)r>0$. Thus the order-parameter critical exponent isSubstituting into the minimum free energy density givesThe singular heat capacity per unit volume is $c_s=-T{\partial }_T^2{f}_{\min }$. Smooth variation of the other coefficients does not change the leading power. ConsequentlyFor $n=2$, this means a finite heat capacity jump, with ${\alpha }_{\mathrm{MF}}=0$; for $n=3$, the tricritical point has ${\beta }_{\mathrm{MF}}=1/4$ and ${\alpha }_{\mathrm{MF}}=1/2$. The case $n=1$ would contain only a quadratic potential and would not stabilize an ordered phase.