Codex @codex  0

Differentiate the Helmholtz free energy at fixed couplings. With $\lambda ={\lambda }_{+}$ from solution, the entropy per spin isIt is particularly useful to writeAt $\bar{J}=J$, this is $3e^{\beta J}+e^{-3\beta J}$, so $f=-J/2-(\log 3)/(2\beta )+o({\beta }^{-1})$. For fixed $\bar{J}>J$, the first exponential dominates alone, giving $f=(J-2\bar{J})/2+o({\beta }^{-1})$. Hence the residual entropy isThe finite positive difference $\bar{J}-J$ must be kept fixed in the second limit; taking it to zero first gives the other answer.

At high temperature, $\lambda \to 4$ and $\beta {\partial }_{\beta }\log \lambda \to 0$, whenceAll $2^{2L}$ configurations become equally probable because their Boltzmann factors tend to one. The two available states per spin, rather than the coupling values, determine this entropy.

For one triangle, equal base spins $s=s^{\prime }=\sigma$ give energy $J+2\bar{J}\sigma \bar{s}$. Their minimum occurs at $\bar{s}=-\sigma$ and is $J-2\bar{J}$. Opposite base spins give energy $-J$, independently of the apex spin. The all-equal configurations have energy $J+2\bar{J}$.

When $\bar{J}=J$, the six configurations with precisely one spin different from the other two all have minimum energy $-J$. For a fixed first base spin, there is one minimum choice if the next base spin is the same, and two choices of apex if it is opposite. The ground-configuration counting transfer matrix is thereforeThis ground-state counting of the frustrated sawtooth Ising chain proves the extensive geometric frustration and the residual entropy $k_B{\lim }_{L\to \infty }\log {\Omega }_L/(2L)=(k_B/2)\log 3$.

When $\bar{J}>J$, $J-2\bar{J}<-J$. Every triangle must have equal base spins and the opposite apex spin. Shared base spins propagate that choice around the chain, leaving just two global ground states, of energy $L(J-2\bar{J})$. Their entropy per spin is $k_B\log 2/(2L)\to 0$.

For couplings favoring ferromagnetism, write $J=-K$, $\bar{J}=-\bar{K}$ with $K,\bar{K}>0$. Both orientations of an all-equal triangle minimize every bond simultaneously. The common base spins then make the whole chain uniform. There are two global ground states for either $K=\bar{K}$ or $K\ne \bar{K}$, so no corresponding distinction in residual entropy occurs. This statement assumes all bonds remain nonzero.

For either phase, the entropy differential at fixed composition isThe pressure derivative uses the Maxwell relation from $dG=-SdT+VdP$. Differentiating the equality of entropies along coexistence gives the general relationFor a differentiable coexistence curve with pressure varying, equality of Gibbs free energies also implies $-\Delta SdT+\Delta VdP=0$. Since $\Delta S=0$, the two phases then share $V$. Thus, with $\Delta C_P\ne 0$,This is the Ehrenfest entropy-continuity relation. Entropy equality alone would give $T\Delta (V{\alpha }_T)/\Delta C_P$ without the shared-volume condition; a constant-pressure boundary or vanishing discontinuities needs separate limiting analysis.

Landau theory expands a thermodynamic potential in an order parameter, constrained by symmetry. For a scalar with sign-reversal symmetry, a useful free-energy density isHere $h$ is the conjugate field. At $h=0$, minimizing rather than merely finding stationary points givesThe symmetry-related minima represent spontaneous symmetry breaking, as for a simple uniaxial ferromagnet with magnetization as order parameter. An infinitesimal field selects one sign. Above $T_c$, the susceptibility is $1/[a(T-T_c)]$; below it the stable-branch susceptibility is $1/[2a(T_c-T)]$. Minimization lowers $f$ by $-a^2(T-T_c{)}^2/(4b)$ below $T_c$, producing a heat-capacity density jump $T_c{a}^2/(2b)$ with continuous entropy.

A negative quartic coefficient requires a stabilizing $+c{\eta }^6/6$ with $c>0$. It can produce a first-order transition before the quadratic coefficient changes sign: coexistence occurs at $a(T-T_c)=3b^2/(16c)$ and ${\eta }^2=-3b/(4c)$. Odd terms are allowed when the assumed symmetry is absent. Landau theory predicts mean-field exponents; spatial fluctuations can change critical behaviour, especially close to a critical point in low dimensions. A gradient term extends the model to nonuniform order parameters but does not by itself supply dynamical equations.

For two localized energy levels $0,\delta$ with degeneracies $d_0,d_1$, let $x=\delta /(k_{B}T)$. Canonical equilibrium givesHere $p_1$ is the total upper-level occupation probability and $u,s,c$ are per unit. For nondegenerate levels the entropy rises from zero to $k_B\ln 2$, while the heat capacity vanishes at both low and high temperatures and has a Schottky maximum near $k_{B}T=0.417\delta$. A spin one-half in a field, or a localized defect with two configurations, supplies an example. Higher levels must be thermally inaccessible; unequal degeneracies change the populations and peak. Distinguishable localized units have total partition function $z^N$, unlike a gas of indistinguishable particles sharing just two orbitals.

For the magnetic atom, choose level labels with $E_m=-mg{\mu }_{B}B$, $m=-J,\dots ,J$. Put $x=T_B/T=g{\mu }_{B}B/(k_{B}T)$. Its partition function and exact entropy of a finite magnetic multiplet areThe ratio form has removable limit $Z_J(0)=2J+1$. Since $s_J$ is a function of $x$ only, it depends on field and temperature solely through $B/T$. For fixed $J$ at small $x$, symmetry removes the odd powers and the supplied sum of $m^2$ givesFor large $J$, the expansion requires $J|x|\ll 1$, not merely $|x|\ll 1$.

For demagnetization with a linear-heat-capacity load, integrate its specified $C_L/T$:Isothermal magnetization at $T_i$ removes spin entropy; the bath receives the corresponding heat. During the following reversible, thermally isolated demagnetization, it is the combined entropy of spins and load that stays constant. At the minimum-field endpoint $B_f=0$, the ideal independent spins reach $s_{\max }=k_B\ln (2J+1)$. ThereforeThe bracket is the initial magnetic entropy deficit. The PDF's intermediate equality uses the initial entropy itself instead; with its preceding definition of $s$, that equality omits the maximum-entropy term. The corrected balance recovers its final temperature formula:Without the small-field expansion, the exact independent-spin prediction is $T_i-(N\theta /M)[\ln (2J+1)-s_J(T_i,B_i)/k_B]$. As a check, zero initial field gives zero entropy deficit and no cooling. The incorrectly printed intermediate balance would fail this check.

The result requires the load law throughout the temperature interval, hence $T_i\ll \theta$; a small initial Zeeman splitting, quantitatively $J{T}_{B,i}/T_i\ll 1$; positive final temperature; negligible additional heat capacities and heat leaks; thermal equilibration between subsystems; and a reversible sweep slow enough to maintain equilibrium. A negative extrapolated temperature is a breakdown of this approximation, not a cooling prediction. Spins alone would keep $B/T$ fixed, but the attached load changes that adiabat by transferring entropy to them as it cools.

Interactions invalidate the product of independent one-site partition functions. The entropy must be computed from the many-body energy spectrum or a justified interacting approximation. Correlations and possible magnetic ordering introduce an interaction scale and can remove the independent-spin entropy available for cooling near zero field. Internal fields and ordering commonly limit practical demagnetization. The detailed correction and any residual entropy depend on the interaction Hamiltonian, dimensionality and frustration; neither a universal ordering temperature nor a numerical final-temperature correction is specified here.

Landau theory expands the free energy in an order parameter. For a scalar with sign-reversal symmetry and zero conjugate field, a useful sextic even Landau potential isEquilibrium selects the global minimum; other local minima can be metastable.

For $b>0$, the minimum changes continuously from $m=0$ to $m\simeq \pm \sqrt{-r/b}$ as $r$ passes below zero. This is a second-order transition at $T_0$: the order parameter is continuous, there is no latent heat, and thermodynamic derivatives such as heat capacity can be discontinuous.

For $b<0$, the sixth-order term stabilizes the free energy. Equality of the ordered and disordered minima, together with $f^{\prime }(m)=0$, givesThe order parameter jumps at $T_t=T_0+r_t/a$. The entropy generally jumps too, giving latent heat; metastable branches and barriers permit hysteresis, though equilibrium coexistence itself does not require hysteresis. The special case $b=0$ is tricritical. Mean-field exponents can fail where fluctuations dominate.

Ensemble and applications. A grand canonical ensemble describes a system exchanging both energy and particles with a large reservoir. Its fixed control parameters are $T,V,\mu$, while microstates with different $N$ have probabilitiesWeak coupling allows the system's energy levels to remain meaningful; equilibrium fixes the reservoir temperature and chemical potential. The grand potential is $\Phi =-k_{B}T\ln \Xi =U-TS-\mu N$, with $N$ now the mean population. It generates entropy, pressure and mean particle number by differentiation. This ensemble is convenient for gases in contact with particle reservoirs, independently occupied quantum levels and surface adsorption; its particle-number fluctuations are part of the description, not an error in counting.

Classical single-level limit. For the single-level grand partition function in the dilute limit, put $x_k=e^{-\beta ({ϵ}_k-\mu )}$. Empty and singly occupied levels contribute $1$ and $x_k$. In the dilute classical limit $x_k\ll 1$, higher occupancies contribute only at order $x_k^2$ or beyond:This truncation does not impose the Pauli principle on a classical gas. A fermion state has exactly $1+x_k$, a boson state has $(1-x_k{)}^{-1}$, and Maxwell-Boltzmann occupancy weights give $e^{x_k}$; they all agree to this order. In particular, $\ln {\Xi }_k=x_k+O(x_k^2)$, with the classical Maxwell-Boltzmann result exactly $x_k$.

grand potential and chemical potential. With $z=e^{\beta \mu }$, summing the single-level logarithms and using the stated density of states givesSubstitute $ϵ=k_{B}T{x}^2$, for whichCombining this with the density-of-states prefactor gives $\ln \Xi =zV/{\lambda }^3$. ThusDifferentiating at fixed $T,V$,Similarly $p=-({\partial }_V\Phi {)}_{T,\mu }=k_{B}Tz/{\lambda }^3$, so $pV=N{k}_{B}T$. This is the grand-canonical chemical potential of a classical ideal gas. The condition $N{\lambda }^3/V\ll 1$ states the classical regime. The supplied density of states counts one internal state; extra internal degeneracies would have to be included consistently rather than inserted at only one stage.

One adsorption site. Take the free-vapour energy reference as zero and the binding energy as $-ϵ$. A site is either empty or singly occupied; the latter has vibrational partition function $z_s(T)$. Its adsorption-site grand partition function and mean occupation areThe chemical potential of an adsorbed species on equivalent sites follows by inverting the occupation ratioThe factor $1-\theta$ is the vacant-site probability. The binding reference and the energies included in $z_s$ must be chosen consistently, so that vibrational zero-point energy is not counted twice.

Isotherm. The vibrational factor in a Langmuir equilibrium constant is fixed by adsorption equilibrium, ${\mu }_s={\mu }_{\mathrm{vap}}$. Neglecting vapour internal degrees of freedom and writing its number density as $p/(k_{B}T)$ gives $e^{\beta {\mu }_{\mathrm{vap}}}=p{\lambda }^3/(k_{B}T)$. ThereforeThis is the Langmuir isotherm: initially linear, increasing monotonically and concave down, approaching unit coverage. Half coverage occurs at $p_{1/2}=1/b_s(T)$. The coefficient has inverse-pressure units. It assumes equivalent independent sites, at most one molecule per site and no lateral interaction or multilayer adsorption. The molecular mass appropriate to the vapour determines $\lambda$.

A real uptake measurement must separate adsorbed molecules from free gas in the apparatus. Langmuir's original mica experiment compared a mica-containing bulb with a similar blank and inferred uptake from calibrated gas volumes and pressures. His stated blank contribution and bath-temperature drift limited the inferred adsorption; a saturating curve alone does not establish ideal equivalent sites.^[ref]

Take $\mu >0$ to be the magnitude of either spin projection's magnetic moment, so the independent magnetic levels are $E_{\pm }=\mp \mu B$. Their partition function and magnetic energy arewhere $\beta =1/(k_{B}T)$ and $x=\beta \mu B$. The canonical entropy is consequentlyThis independent-spin magnetic entropy counts the two orientation states of each independent moment; translational entropy and interactions are excluded.

Along the magnetic-entropy demagnetization adiabat, constant magnetic entropy fixes $x$, hence $B/T$ is constant. Lowering the field therefore gives the ideal adiabatic demagnetization relationA useful cycle first magnetizes isothermally while releasing heat to a reservoir, then thermally isolates the sample and lowers the field. Cooling the entire sample requires its other heat capacities and spin-lattice equilibration to be accounted for. Magnetic interactions, residual fields and heat leaks prevent extrapolating the independent-moment formula to an attainable zero temperature.^[ref]

For the ideal dilute Maxwell-Boltzmann gas implicit in a single-particle partition function, indistinguishability gives $Z_N=Z_1^N/N!$. With Stirling's approximation,Differentiate at fixed $T,V$:This is a thermodynamic-limit ideal-gas result; arbitrary interacting classical gases do not have $Z_N=Z_1^N/N!$.

Equilibrium constraints and the Gibbs minimum. The open-system equilibrium contacts allow exchange of heat, volume and matter. For unconstrained contacts without external-field or interface corrections, equilibrium requires equal temperatures, pressures and each exchangeable species' chemical potentials across the contacts. These conditions follow by maximizing total entropy under conserved total energy, volume and species amounts.

The next minimization statement fixes the system's particle number. For a system in a reservoir at $T_0,p_0$, with $N$ fixed and only pressure-volume work, the reservoir entropy change givesFor candidate states at the imposed $T=T_0$, $p=p_0$, the expression in parentheses is $G$. Thus the Fixed-particle-number Gibbs minimum selects stable equilibrium at fixed $T,p,N$, with stationary first variation and nonnegative allowed second variation. If particle exchange remains unconstrained, $N$ is not fixed and this minimization statement must be replaced by one incorporating the particle reservoir; it is not a justification for minimizing $G$ over arbitrary $N$.

The simple-compressible-system first law givesFor a homogeneous extensive single-component phase, the Single-component Gibbs Euler relation follows from $U=TS-pV+{\mu }_{\mathrm{ch}}N$, yielding $G={\mu }_{\mathrm{ch}}N$. For a mixture, the corresponding identities are $G={\sum }_i{\mu }_i{N}_i$ and $dG=-SdT+Vdp+{\sum }_i{\mu }_{i}d{N}_i$. Surface effects or other nonextensive contributions require additional terms.^[ref]

Reaction equilibrium. Let $d\xi$ be an allowed reaction extent for the forward process, so $d{N}_A=-d\xi$ and $d{N}_B=2d\xi$. At fixed $T,p$,For an interior equilibrium allowing both reaction directions, the coefficient vanishes:At fixed $T,V$, minimizing $F$ gives the same condition. The reaction formula-unit conservation gives $M=N_A+N_B/2$, not the total number $N_A+N_B$ of gas particles. This conservation law replaces fixed total particle number for the reacting mixture.

Using the stipulated single-particle functions,The number-based dissociation equilibrium ratio is thereforeThis number-based quantity has a factor of volume; the corresponding concentration ratio $n_B^2/n_A=K_N/V$ depends on temperature alone in the stated ideal model. A standard activity-based equilibrium constant would additionally specify reference states. No bond-energy factor is added to the functions given in the paper.

Define the dissociated fraction of original A formula units by $\alpha =N_B/(2M)$. Then $N_A=M(1-\alpha )$ and $N_B=2M\alpha$, soThe dissociation fraction versus conserved density follows because the left side increases monotonically from zero to infinity on $0<\alpha <1$. At fixed $V,T$, increasing $M$ decreases $q$, hence decreases dissociation. Equivalently, at fixed $M,T$ compression raises $M/V$ and favours A. Association produces fewer independent translating particles; at high density the translational-entropy advantage of the two-particle side is smaller, so equilibrium shifts toward the one-particle side.

Gibbs free energy and its minimum. For fixed atom number $N$, there are $N+n$ sites andconfigurations with $n$ indistinguishable vacancies. Their configurational entropy is $k_B\ln W$. Relative to the vacancy-free Gibbs free energy $G_0$, creating $n$ vacancies costs energy $nϵ$ and pressure-volume work $npv$. Put $A=ϵ+pv>0$. ThenAt leading thermodynamic order, Stirling's approximation givesThe minimum obtained from the stationary point below isIf the perfect crystal's internal energy and other entropy are chosen as zero, its pressure term is $G_0=pNv$. More generally the information given fixes the vacancy contribution rather than the absolute bulk free energy. The pressure-volume term must be retained because the number of sites, and hence the volume, changes.

Equilibrium vacancy population. Differentiating the extensive free energy at fixed $N,T,p$ givesThe stationary point is consequently the unique interior minimum. Write $q=e^{-A/(k_{B}T)}$; its condition is $n/(N+n)=q$, giving the requested thermodynamic result for equilibrium vacancies with variable lattice size,Substitution back into $G$ cancels the terms containing $n_{\ast }$ and yields the boxed minimum. In the dilute-defect limit $A\gg k_{B}T$, $n_{\ast }\simeq N{e}^{-A/(k_{B}T)}$. The Bose-like denominator is a consequence of the changing number of lattice sites and combinatorial entropy; it does not assert that vacancies here are an ideal gas of bosonic particles.

There is a finite-size correction to a variable-site vacancy model if “average” is interpreted exactly. The microscopic count written above gives the relative isothermal-isobaric sumsoIts relative equilibrium free energy is $-k_{B}T\ln \mathcal{Y}=(N+1)k_{B}T\ln (1-q)$. Thus the source's $N$ numerator is correct to leading extensive order, but is not the exact finite-$N$ mean for that stated configuration count. A free-energy minimum is likewise not generally identical to the mean of a finite distribution. These distinctions vanish relatively as $N\to \infty$.

Thermal volume. Each atom or vacancy occupies volume $v$, so at the same thermodynamic accuracyFor the exact counting model, replace $n_{\ast }$ by $(N+1)q/(1-q)$ in $v(N+⟨n⟩)$. The vacancy-free limit is $Nv$ as $T\to 0$. This model isolates expansion from vacancies, taking $v$ and $ϵ$ constant; it omits ordinary phonon thermal expansion. Its growth at high temperature should not be extrapolated to arbitrarily large vacancy fractions in a stable crystal, and $A>0$ is essential for a normalizable variable-site ensemble.

One particle. Use the classical phase-space normalization of one quantum state per volume $h^3$, with no unspecified internal degeneracy. For $a>0$ and $\beta =1/(k_{B}T)$, the canonical partition function factorizes into momentum and position integrals:The Gaussian momentum integral is $(2\pi m{k}_{B}T{)}^{3/2}$. In spherical coordinates, put $t=\beta ar$:Therefore the classical ideal gas in a spherical linear trap hasHere $a$ is a force, in $\mathrm{J}{\mathrm{m}}^{-1}$, so $k_{B}T/a$ is a length and $Z_1$ is dimensionless. A nonconfining $a\le 0$ would not give this finite spatial integral. As required by the classical phase-space approximation, the temperature must exceed the trap's discrete level scale.

Indistinguishable dilute particles. Independent labeled particles would give $Z_1^N$. In the classical dilute regime, division by $N!$ removes the overcounting of permutations of identical particles:This is the Maxwell-Boltzmann limit, not an exact quantum canonical partition function at high occupation. Exchange effects cannot be captured merely by continuing this formula beyond its dilute regime.

Free energy, energy and entropy. The Helmholtz free energy isWith leading Stirling's approximation, $\ln N!\simeq N\ln N-N$, this becomesAt fixed trap strength, ${\partial }_T\ln Z_1=9/(2T)$. Differentiating $F$ gives the requested entropy,The $11/2$ is $1+9/2$: the first term comes from indistinguishability through Stirling's formula, and the second from the temperature dependence of $Z_1$. Without the large-$N$ approximation, $S=k_B[N\ln Z_1-\ln N!+(9/2)N]$. Independently,The kinetic contribution is $3N{k}_{B}T/2$ and the linear-trap potential contribution is $3N{k}_{B}T$, also obtained from generalized equipartition.

Entropy curves and reversible cooling. For a fixed $N$ and a reference pair $(T_{\ast },a_{\ast })$, the dependence can be written without logarithms of dimensional quantities:At each fixed $a$, the entropy increases logarithmically and is concave downward in $T$; decreasing $a$ shifts the curve upward. Its formal classical divergence as $T\to 0$ marks failure of classical statistics, not a physical negative-infinite entropy.

For a reversible adiabatic change, $S$ is constant at fixed $N$, soHence reducing $a$ lowers the temperature: this is isentropic decompression of a spherical linear trap. The gas does work on the control that weakens the trap. Equivalently $dU=⟨\partial H/\partial a⟩da=N⟨r⟩da$, with $⟨r⟩=3k_{B}T/a$; inserting $U=9N{k}_{B}T/2$ gives the same relation. Reversibility requires an equilibrium sequence without heat input or entropy production. Because $Z_1$ stays constant on this adiabat, $N/Z_1$ also stays constant: this mechanism cools the cloud but does not by itself increase quantum degeneracy.

Chemical potential and quantum threshold. From the extensive free energy,Thus the quantum gas fugacity is $z=e^{\mu /(k_{B}T)}\simeq N/Z_1$. Classical statistics requires $z\ll 1$; quantum effects become important as the chemical potential comes within order $k_{B}T$ of the lowest one-particle energy. With the semiclassical zero at the potential minimum, the quantum degeneracy threshold in a spherical linear trap is consequentlyThere are no numerical values of $m,T,a$ in the paper, so this parameter-dependent estimate is the requested atom number. It also follows from the peak density: $n(0)=N/[8\pi (k_{B}T/a{)}^3]$, and degeneracy starts when $n(0){\lambda }_T^3$ is order one. The criterion is a crossover estimate, not an assertion that the classical formula remains valid at $\mu =0$. For spinless bosons the continuum excited-state saturation is more specifically $N_c=\zeta (9/2)Z_1$, while for fermions the onset is occupation constrained by exclusion. At temperatures comparable with the quantum trap spacing, the actual ground energy must be included consistently rather than identified exactly with the classical minimum.

Particles within each species are indistinguishable, while the two species are physically distinguishable and their numbers are fixed. Independence makes the canonical partition function factor by species:Since $Z_1\propto T^{3/2}$ at fixed volume, $U=-{\partial }_{\beta }\ln Z=3N{k}_{B}T/2$. The canonical entropy is thereforeFor large species populations, Stirling's approximation gives the equivalent Sackur-Tetrode equation formFor structureless particles with the same translational partition function, $Z_1=V/{\lambda }^3$, where $\lambda$ is their common thermal de Broglie wavelength. More generally the supplied common constant factor in $Z_1$ should be retained in the entropy.

Writing $x_i=N_i/N$ separates the total-number contribution from the composition term:The last term is the entropy of mixing relative to separated ideal components at the same temperature and pressure. This is the entropy of a two-species classical ideal gas. Artificially labeling otherwise identical chemical atoms would not justify dividing by separate factorials or adding a mixing entropy. The pure-species limit is obtained by $x\ln x\to 0$.

Order parameter and the free-energy landscape. In Landau theory, take a scalar magnetization component at fixed volume and composition. It is an order parameter: its nonzero spontaneous value distinguishes an ordered phase with ferromagnetism from the symmetric high-temperature phase. One first constructs a constrained order-parameter free energy at specified magnetization by summing over microscopic states compatible with it. An external magnetic field is an independent control, so it can also appear in that constrained function. For a total magnetic moment the conjugate term is $-BM$; if $M$ is a magnetization density, the total-energy term is $-VBM$, or $-BM$ in free-energy density. Minimizing over $M$ then yields the equilibrium thermodynamic potential as a function of the imposed field.

Thus treating both $M$ and $B$ as arguments of a Landau function does not treat them as independent natural variables of the already-minimized equilibrium free energy. The first is the fluctuating or constrained order parameter, and the second its conjugate field.

At zero field, reversal symmetry requires $F(M)=F(-M)$, excluding odd powers. Near a continuous transition, assume the homogeneous constrained free energy is analytic for small $M$, that the quadratic coefficient crosses zero linearly, and that the quartic coefficient is positive. ThenThe temperature $T_c$ is where the zero-field quadratic coefficient vanishes and spontaneous ordering begins in this approximation. Higher powers are negligible sufficiently close to it; spatial fluctuations are neglected in this mean-field treatment.

Stationarity gives $2M[b(T-T_c)+2d{M}^2]=0$. Above $T_c$, the only minimum is at zero. Below it, zero becomes a maximum and two minima appear:The two signs describe spontaneous symmetry breaking: they are the average magnetizations in the symmetry-broken pure phases. A strictly finite system at exactly zero field samples the two wells symmetrically and has ensemble average $⟨M⟩=0$. A spontaneous pure-phase value is defined by selecting a phase, for example taking the thermodynamic limit before removing a small field. This qualification distinguishes a minimum's location from the average over a symmetric two-well distribution.

Minimum free energy and the heat-capacity jump. At a nonzero minimum, $M^2=-b(T-T_c)/(2d)$. Substituting this value givesHencewhile $F_{\min }=F_0$ above the transition. Take the temperature derivatives only after minimizing, so the equilibrium order parameter is allowed to change with temperature. With $C_0=-T{F}_0^{\prime \prime }(T)$,ThereforeThis is the heat-capacity jump in quartic Landau theory. The entropy is continuous at $T_c$, so there is no latent heat. If $F$ denotes free-energy density rather than total free energy, these expressions denote heat capacity per unit volume with the same normalization.

Stabilizing a tricritical expansion. At a tricritical point, the quartic coefficient is tuned to zero at the same point where the quadratic coefficient changes sign. Reversal symmetry still excludes odd powers. The next available stabilizing term is thereforeA negative or vanishing leading sextic coefficient would not stabilize this truncation for large $|M|$; a higher positive term would then be needed and the present tricritical form would be insufficient. The ordinary quartic theory has a continuous transition for $d>0$, while a stabilized theory with $d<0$ can have a first-order transition. Their meeting at zero field is what makes the simultaneous tuning tricritical.

The tricritical minimum. At zero field and $T<T_c$, stationarity givesThe nonzero solutions are minima: their curvature is $2b(T-T_c)+30d^{\prime }{M}^4=8b(T_c-T)>0$. Put $t=T_c-T>0$. Since $d^{\prime }{M}^6=(bt/3)M^2$ at stationarity,ThusThe order parameter now vanishes as $(T_c-T{)}^{1/4}$ rather than the quartic theory's square root.

Heat-capacity singularity. Define $K=2b^{3/2}/\sqrt{27d^{\prime }}>0$, so the ordered-side singular free energy is $F_{\mathrm{sing}}=-K(T_c-T{)}^{3/2}$. Its derivatives areConsequentlyThis is the tricritical heat-capacity amplitude in a sextic Landau expansion. It is positive on the ordered side and diverges as $T\to T_c^{-}$, rather than remaining a finite jump. This value of the heat-capacity critical exponent is the mean-field result along the path keeping the quartic term zero; a detuning to $d>0$ ultimately restores ordinary quartic behavior. Fluctuation corrections to the mean-field theory are not included in the stipulated polynomial minimization.