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For $N$ distinguishable lattice sites occupied by $N_i$ mutually indistinguishable atoms of each species, the number of configurations isUsing Stirling's approximation,For one mole, $N=N_A$ and $k_B{N}_A=R$, soEquiatomic configurational entropy follows when the composition has $X_i=1/n$, givingThe reference is the unmixed configurational arrangement, and the model assumes random site occupation without short-range order. It counts configurational entropy, not all vibrational, electronic or magnetic entropy. A large number of elements does not by itself guarantee a stable single solid solution: competing thermodynamic phases and the enthalpy term also enter the Gibbs free energy.

At equilibrium between two thermodynamic phases of a pure substance, their molar Gibbs free energies are equal. Along a coexistence line,Subtracting the two differentials givesFor a nonzero volume change,The second equality uses $\Delta g=\Delta h-T\Delta s=0$ on coexistence. Total entropy and volume changes may be used instead if they refer to the same amount of material. The ratio must use one consistent transformation direction, so reversing the labels changes both signs and leaves the slope unchanged. This is the general Clapeyron equation; the ideal-vapour approximation often called Clausius-Clapeyron makes further assumptions not needed for two solids.

Diamond can be metastable: its Gibbs free energy is higher than graphite's under ambient conditions, but transformation is kinetically inhibited. Conversion requires substantial rearrangement of the strong carbon-bond network and the formation and growth of a new thermodynamic phase, with an activation and nucleation barrier.

At room temperature the rate can be extremely small even though the free-energy change favors graphite. Thermodynamic stability specifies the preferred equilibrium state, not the time needed to reach it. Persistence of diamond therefore does not contradict graphite's ambient equilibrium stability.

At fixed temperature, a phase's Gibbs free energy satisfies $\mathrm{d}G=V\mathrm{d}p$. For the metallic-minus-molecular difference,Increasing pressure favors the smaller-volume metallic phase. With a pressure-independent volume approximation, $\Delta G(p)\simeq \Delta G(p_0)+(p-p_0)\Delta V$, so pressure can overcome an initially positive free-energy difference. The volume comparison explains the direction of stabilization; it does not by itself determine the transition pressure or the electronic origin of metallic bonding.

For a mixture with phase compositions $x_1,x_2$, its molar Gibbs free energy is the straight-line combination $g=(1-f)g_1(x_1)+f{g}_2(x_2)$ at overall $x=(1-f)x_1+f{x}_2$. Minimizing over phase compositions gives equal slopes and equal tangent intercepts: the common-tangent construction expresses equality of both component chemical potentials. The equilibrium curve is the lower convex envelope of the candidate phase curves. Phase fractions follow the lever rule.

At $100{}^{\circ }\mathrm{C}$, the stable sequence is $(\mathrm{Bi})$, $(\mathrm{Bi})+(\beta \mathrm{Sn})$, $(\beta \mathrm{Sn})$. The stable two-solid common tangent lies below the liquid curve. There is no stable liquid interval, although a metastable extrapolated liquid Gibbs free energy can still be drawn.

At $138{}^{\circ }\mathrm{C}$, one line is tangent to all three Gibbs free energy curves at approximately $x_{\mathrm{Sn}}=0.03$ for (Bi), $0.61$ for liquid, and $0.93$ for $(\beta \mathrm{Sn})$. These three phases can coexist at the eutectic point. Between the two solid compositions, the equilibrium phase fractions depend on how much liquid has already transformed; exactly at the invariant temperature, the overall composition alone does not fix all three fractions. At the eutectic overall composition, cooling just below the temperature eliminates the liquid and leaves the two solids.

At $200{}^{\circ }\mathrm{C}$, moving from Bi-rich to Sn-rich composition gives $(\mathrm{Bi})$, $(\mathrm{Bi})+L$, $L$, $L+(\beta \mathrm{Sn})$, $(\beta \mathrm{Sn})$. Approximate liquid coexistence compositions read from the diagram are $x_{\mathrm{Sn}}\simeq 0.33$ on the left and $0.87$ on the right; the coexisting solids lie near their respective pure ends. The stable Gibbs free energy follows solid curves near the endpoints, two common-tangent constructions in the coexistence intervals, and the liquid curve between them.

The Gibbs free energy is $G=U-TS+pV$. For an extensive thermodynamic system, scaling the amount of matter at fixed pressure, temperature and composition scales $G$ by the same factor, hence $G(T,p,N)=N\mu (T,p)$. Euler's relation $U=TS-pV+\mu N$ gives the same conclusion. This uses the ordinary thermodynamic-limit extensivity assumption, rather than a universal claim about finite-size or long-range-interacting systems.

The fundamental thermodynamic identity gives $dG=-SdT+Vdp+\mu dN$. Comparing with $d(N\mu )$ at fixed $N$ yieldsThus the chemical potential changes with pressure at a rate equal to the volume per particle.

For a fixed amount of gas, the first law of thermodynamics gives $dU=TdS-pdV$. Define the Gibbs free energy $G=U+pV-TS$, so $dG=-SdT+Vdp$. Equality of its mixed partial derivatives gives the Maxwell relation

The statistical Hamiltonian in this question is already in thermal units, as indicated by $e^{-H}$ without an additional ${\beta }_T$. Define $W[h]=\log Z[h]$ and $F[h]=-W[h]$. The mean field and its linear response areThese are functional derivatives of the connected generating functional; the subtraction defines the connected correlation function. In the Landau approximation, neglect loop corrections and evaluate the field integral at a stable saddle $m_h$. It satisfies the Euler-Lagrange equationThis follows by varying the gradient term and integrating by parts, with periodic, decaying, or otherwise appropriate boundary conditions.

The two requested free energy functionals, following the source and Legendre conventions of the question, areIn the scalar-field source Legendre transform on the chosen stable branch, $h$ is chosen to produce $m$, and $\delta {\Gamma }_{\mathrm{L}}/\delta m=h$. Thus the imposed-source Helmholtz free energy and the fixed-order-parameter Gibbs free energy have the appropriate opposite source derivatives. These names are used in the question's magnetic-ensemble convention; the defining sign relation is what fixes the calculation. At leading Landau approximation there is no fluctuation-determinant term in $F_{\mathrm{L}}$.

Differentiate the saddle equation with respect to $h(y)$. The response obeysThereforeEquivalently, the inverse Hessian relation for a connected two-point function states that $G$ is the inverse kernel of ${\delta }^2{\Gamma }_{\mathrm{L}}/\delta m(x)\delta m(y)$. The factor $u_0/2$ follows from twice differentiating the quartic term $u_0{m}^4/4!$; it is not $u_0/6$. This tree-level connected response is obtained by varying the saddle. It does not require replacing the exact connected correlator by a product of the saddle values, which would incorrectly give zero.

For the requested single-momentum formula, assume a homogeneous source and a translationally invariant equilibrium phase, so $m_h(x)=m_0$ and $G(x,y)=G(x-y)$. SetThe Fourier transform with the printed positive sign sends $-{\nabla }^2$ to $q^2$, while the Dirac delta function transforms to one. Hence the Ornstein--Zernike correlation function hasThe inverse convention is $G(x)=\int d^{D}q(2\pi {)}^{-D}{e}^{-iq\cdot x}\tilde{G}(q)$. For general inhomogeneous $h(x)$, $K$ has nonconstant coefficients and $G$ depends separately on its two positions; the displayed momentum-diagonal formula then does not follow. The preceding differential equation still holds in that case.

At zero source and for $u_0>0$, the homogeneous saddle is $m_0=0$ for $r_0>0$, and $m_0^2=-6r_0/u_0$ on either selected stable ordered branch for $r_0<0$. Thus the Landau scalar correlation length isIn the ordered phase the negative bare quadratic coefficient is compensated by the positive curvature at the nonzero saddle. Retaining $m_0=0$ below the transition would instead give an unstable kernel and is not a physical correlation length. With the usual analytic thermal tuning $r_0=a_0(T-T_c)$, $a_0>0$, both branches diverge as $|T-T_c{|}^{-1/2}$, so the correlation-length critical exponent isThe high-temperature amplitude is $\sqrt{2}$ times the low-temperature amplitude for the same $a_0$. This statement is within Landau theory; fluctuations can change critical behavior outside the mean-field regime.

Assume a single-phase ideal gas mixture in a closed parcel, with common temperature $T$, total pressure $P$, and molar species amounts $N_i$. Define $N={\sum }_i{N}_i$ and volume mixing ratios $x_i=N_i/N$, so ${\sum }_i{x}_i=1$. Ideal mixing makes partial pressure $P_i=x_{i}P$.

For a pure ideal species, $d{\mu }_i=-{\bar{S}}_{i}dT+{\bar{V}}_{i}d{P}_i$, with molar volume ${\bar{V}}_i=\mathcal{R}T/P_i$. At fixed temperature, integration from standard pressure $P^{\circ }$ givesThe same logarithmic composition term follows from the ideal mixing entropy $\Delta S_{\mathrm{mix}}=-\mathcal{R}{\sum }_i{N}_i\log x_i$. Extensivity gives $G={\sum }_i{N}_i{\mu }_i$. Hence the Gibbs free energy of an ideal-gas mixture isEquivalently, write each standard chemical term as ${\mu }_i^{\circ }=h_i^{\circ }-T{s}_i^{\circ }$, using molar enthalpy and entropy. For zero abundance the limit $x\log x\to 0$ defines the extensive expression continuously.

The fractions alone do not specify the extensive Gibbs free energy: the total amount $N$ or an equivalent elemental inventory must also be fixed. Moreover, in a chemically closed reacting system $N$ need not remain constant; for example, hydrogenation of carbon monoxide changes the total molecular count while conserving atoms. Minimization is therefore over species amounts subject to elemental constraints, rather than over fractions with an artificially fixed molecular count. Nonideal gases require activities or fugacities; condensates contribute separate phase terms.

Let the closed atmospheric parcel exchange heat and volume work with reservoirs at fixed temperature $T$ and pressure $P$, while reactions rearrange its conserved elemental inventory. For a reversible differential between equilibrium states, the first law of thermodynamics, with chemical work included, isThe Gibbs free energy $G=U+PV-TS$ therefore obeysThe first law of thermodynamics supplies this differential but does not alone choose the direction of a reaction. The Second law of thermodynamics does: for the parcel plus its reservoirs, at fixed $T,P$ and with no non-volume work,Thus spontaneous reactions decrease Gibbs free energy, and stable thermochemical equilibrium is its constrained minimum. For an allowed reaction of extent $d\xi$, let $d{N}_i={\nu }_{i}d\xi$, with positive stoichiometric vector entries for products and negative entries for reactants. At an interior minimum,The condition must hold for every independent allowed reaction, with fixed total atoms of each element and any relevant charge constraint. At a boundary with absent species, only feasible one-sided variations are permitted. These are the hypotheses of constrained Gibbs minimization for chemical equilibrium.

For an ideal gas mixture, ${\mu }_i={\mu }_i^{\circ }(T)+\mathcal{R}T\log (x_{i}P/P^{\circ })$, where $x_i$ is the volume mixing ratio and $\mathcal{R}$ the molar gas constant. ThereforeThis is a dimensionless activity-based equilibrium constant, not a bare product of dimensional pressures.

The three parameter groups are local temperature, local pressure, and elemental composition. The elemental composition includes both atmospheric metallicity of a giant planet and atmospheric carbon-to-oxygen ratio; these are independent compositional controls, so a specification by $T$ and $P$ alone is incomplete. Given these inputs, thermochemical equilibrium minimizes Gibbs free energy with element conservation.

Three useful chemical observables in a hydrogen-rich hot Jupiter are:

CO/CH4, H2O, and CO2 bands provide complementary temperature and elemental-abundance diagnostics, but their observed strengths also depend on the atmospheric pressure-temperature profile, exoplanet clouds and observing geometry. A spectral feature is not by itself a direct equilibrium abundance.

Three routes away from local thermochemical equilibrium are:

Vertical mixing, horizontal advection, and photochemistry supply three mechanisms and examples in both exoplanets and Solar-System atmospheres. A photochemical steady state balances production and loss; it is different from a Gibbs free energy minimum. These examples describe mechanisms and model expectations rather than asserting unique observational attribution for every planet.

At fixed temperature and pressure, a closed reacting system is in thermochemical equilibrium when its Gibbs free energy is minimal subject to elemental-abundance constraints. For every independent reaction,with a positive second variation in every allowed direction.

An atmosphere can be driven into disequilibrium chemistry in an exoplanet atmosphere by vertical mixing faster than chemical conversion, which causes chemical quenching; ultraviolet atmospheric photochemistry; and atmospheric escape. Lightning, energetic particles, horizontal transport, condensation, and rainout provide further mechanisms.

At fixed temperature and pressure, thermochemical equilibrium minimizes the Gibbs free energy subject to elemental conservation. For every independent reaction with stoichiometric coefficients ${\nu }_i$,Equivalently, forward and reverse rates satisfy detailed balance. For ideal gases,

The fundamental thermodynamic relation isBecause $E(S,V,N)$ is extensive, the Euler theorem for homogeneous functions givesSubstitution into the Gibbs free energy $G=E+pV-TS$ therefore yieldsFor a one-component equilibrium phase, the intensive chemical potential depends on the intensive state variables $T$ and $p$.

For $A+B\rightleftharpoons C+D$, the forward and reverse rates areAt thermochemical equilibrium, $r_f=r_r$ by detailed balance, whileafter the appropriate standard-concentration factors are included. The standard reaction Gibbs free energy isConsequentlyFor other stoichiometries, the same argument uses the corresponding activity product and its standard-state factors.

If a physical tension $f$ acts on a state of extension $l_i$, its mechanical energy is $-f{l}_i$. The canonical ensemble therefore assigns that state the Boltzmann factorComparison with the statistical weight $e^{\tau l_i}$ givesEquivalently, $f$ is the thermodynamic conjugate variable to $L$ in the Gibbs free energy, just as pressure is conjugate to volume.