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The Clapeyron equation connects the slope of a phase coexistence curve to the latent heat and the change in volume during a first-order phase transition. A steep boundary can indicate a large latent heat, a small change in volume, or both. This lets a measured boundary constrain caloric properties and predicts how changing pressure shifts a melting temperature.

For a pure substance in bulk phase coexistence, label the initial and final thermodynamic phases $\alpha$ and $\beta$. Both have the same absolute temperature $T$, pressure $p$, and chemical potential. With $s$, $h$, and $v$ denoting entropy, enthalpy, and volume on one common basis, define $\Delta s=s_{\beta }-s_{\alpha }$, $\Delta h=h_{\beta }-h_{\alpha }$, and $\Delta v=v_{\beta }-v_{\alpha }$. ThenThe derivative follows the phase coexistence curve; it is not a partial derivative within one thermodynamic phase. Use molar volume $v_m$ with molar latent heat $L_m$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or specific volume $v$ with latent heat $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Both give $dp/dT$ in $\mathrm{Pa}{\mathrm{K}}^{-1}$. For molar mass $M$, $L_m=ML$ and $\Delta v_m=M\Delta v$, so the predicted slope is independent of the chosen basis. This formulation assumes a simple compressible system, fixed composition, and negligible interface contributions; a curved interface can instead require the Young–Laplace equation.

For positive latent heat in the stated conversion direction, the sign of the slope is the sign of $\Delta v$. Converting liquid to gas normally increases volume and gives a positive slope. The negative melting-point slope of water illustrates the opposite sign. At a thermodynamic critical point, the coexisting gas and liquid merge: both $L$ and $\Delta v$ vanish, so their ratio needs a limiting analysis rather than substitution of zero into the Clapeyron equation.^[ref]

A short derivation uses the molar Gibbs free energy $g_m$, equal to the chemical potential for a pure substance. Equality persists along the boundary, so $d{g}_{m,\alpha }=d{g}_{m,\beta }$ with $d{g}_m=-s_{m}dT+v_{m}dp$. Therefore $-\Delta s_{m}dT+\Delta v_{m}dp=0$. Equality of $g_m=h_m-T{s}_m$ also gives $\Delta h_m=T\Delta s_m$. Eliminating $\Delta s_m$ yields the Clapeyron equation.

For the stable scalar Landau free energy $f=b{\lambda }^4+c{\lambda }^3+a{\lambda }^2+d\lambda$, $b>0$, a stable zero-curvature equilibrium must have $f^{\prime }=f^{\prime \prime }=f^{\prime \prime \prime }=0$. Solving these equations givesAt this point $f=f({\lambda }_c)+b(\lambda -{\lambda }_c{)}^4$. Varying two control parameters allows a line of first-order phase transitions to terminate here. A spinodal point satisfies zero curvature but need not satisfy the third-derivative condition or remain a local minimum.

For a scalar Landau-de Gennes free energy with $\bar{b}>0$, $\bar{c}<0$, and coupling $-\chi E^2\lambda$ with $\chi >0$, the critical endpoint of a quartic Landau free energy is reached at the displayed field strength and ${\bar{a}}_c=3{\bar{c}}^2/(8\bar{b})$, ${\lambda }_c=-\bar{c}/(4\bar{b})$. It terminates the field-biased first-order phase transition when both field and quadratic coefficient can be tuned. This is a quartic small-field mean-field approximation; higher terms may alter its location.

Dependence of a system's observed state on the path by which its control parameters were reached. In a first-order phase transition, a metastable local minimum may persist after another phase becomes globally stable, so increasing and decreasing the control parameter give different switching points. Nucleation and sweep rate determine actual switching; a mean-field spinodal point is a local-stability limit, not the equilibrium phase coexistence condition.

The Landau-de Gennes free energy is a rotational symmetry preserving Landau free energy of the nematic order parameter. One three-dimensional normalization through fourth order isThe cubic invariant generically produces a first-order phase transition. Its absence in two dimensions is an algebraic identity, not a guarantee that a mean-field approximation captures all fluctuations.

Latent heat is the heat absorbed or released in a first-order phase transition through its entropy jump. For an equilibrium conversion from thermodynamic phase $\alpha$ to thermodynamic phase $\beta$ at fixed pressure, evaluate both thermodynamic phases at the same temperature and pressure on their phase coexistence curve:Use one normalization throughout: $L$, $S$ and $H$ may all be per mole, giving $L$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or all per unit mass, giving $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Here $S$ is entropy and $H$ is enthalpy on that basis. Equal Gibbs free energies of the coexisting thermodynamic phases give $\Delta H=T_{\mathrm{tr}}\Delta S$. Identifying this enthalpy difference with supplied heat assumes a closed system and an equilibrium conversion with only pressure-volume work. ^[ref]

For a pure substance with positive latent heat per unit mass $L$, let $x$ be the phase fraction of the higher-enthalpy thermodynamic phase, measured by mass. While both thermodynamic phases coexist at the fixed pressure,Adding heat changes $x$ rather than the temperature; this creates the horizontal portion of a heating curve. Once $x=1$, additional heat raises the temperature according to specific heat capacity at constant pressure. Reversing the conversion releases heat. The plateau represents changing enthalpy, rather than a finite single-thermodynamic phase heat capacity satisfying $Q=C_p\Delta T$. ^[ref]

For a constructed fixed-pressure model, take $m=1\mathrm{kg}$, initial temperature $280\mathrm{K}$, transition temperature $T_{\mathrm{tr}}=300\mathrm{K}$, solid specific heat capacity at constant pressure $c_s=2\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, latent heat $L=100\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$, and liquid specific heat capacity at constant pressure $c_l=3\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. These are stipulated model parameters, not empirical properties of a named material. Heating the solid to $300\mathrm{K}$ takes $40\mathrm{kJ}$ of sensible heat; converting it fully to liquid takes another $100\mathrm{kJ}$ of heat; heating the liquid to $320\mathrm{K}$ takes $60\mathrm{kJ}$. The total is $200\mathrm{kJ}$. After a total heat input of $90\mathrm{kJ}$, $x=(90-40)/100=1/2$ and the temperature is still $300\mathrm{K}$. The corresponding per-unit-mass entropy jump is $L/T_{\mathrm{tr}}=\frac{1}{3}\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$.

A continuous phase transition has no latent heat. Its heat capacity can have a jump or a singularity without a finite interval in which added heat converts coexisting thermodynamic phases at a constant temperature. A heating plateau therefore requires the stated phase coexistence conditions; it is not a feature of every phase transition. ^[ref]

In equilibrium without hysteresis, the system selects the lower free-energy minimum. At $T_2$ the order parameter jumps between $0$ and $2b/(3c)$, which identifies a first-order phase transition. Differentiating the minimized free energy does not add a $dP/dT$ contribution because $F^{\prime }=0$. Thus the polarization-dependent entropy is $S=-a{P}^2/2$, givingThe heat absorbed on heating from the ordered to the disordered phase isThe same amount is released on cooling. If $\Delta S$ is defined in the opposite direction, the signed transition heat is correspondingly negative.

Landau-Ginzburg theory describes a slowly varying order parameter by a local free-energy functional consistent with the microscopic symmetries. For a representative scalar system with spin-inversion symmetry, take the dimensionless functionalThe gradient term penalizes spatial variation, while the even local potential implements the zero-field symmetry $\varphi \mapsto -\varphi$. The positive quartic coefficient stabilizes a continuous transition. A negative quartic coefficient requires higher powers and can instead produce a first-order phase transition. A vector order parameter gives the corresponding rotationally invariant model with powers of $|\mathbf{\varphi }{|}^2$.

In the mean-field approximation, replace the field by a constant $M$ and minimize the local potential. Its equation of state and stable zero-field solutions areBelow the transition these two minima exhibit spontaneous symmetry breaking. Differentiating the equation of state gives $\chi =(r+3u{M}^2{)}^{-1}$: it is $1/r$ above the transition and $1/(2|r|)$ within either ordered phase. At $r=0$, $M=(h/u{)}^{1/3}$. The minimized zero-field free-energy density is zero above and $-r^2/(4u)$ below, apart from an analytic background; its second temperature derivative has a finite jump. Consequently

To include fluctuations, write $\varphi =M+\psi$. The quadratic part isA Gaussian functional integral therefore gives $⟨\psi (q)\psi (-q)⟩=(c{q}^2+r+3u{M}^2{)}^{-1}$, with the momentum-volume delta function understood. The Ornstein--Zernike correlation function has correlation length $\xi =\sqrt{c/(r+3u{M}^2)}$. Thus ${\nu }_{\mathrm{MF}}=1/2$, and at criticality the Gaussian covariance is proportional to $q^{-2}$, giving ${\eta }_{\mathrm{MF}}=0$. Continuous-symmetry ordered phases additionally have transverse Goldstone modes, for which the restoring mass vanishes.

The Gaussian integration contributes $\frac{1}{2}\int d^{d}q(2\pi {)}^{-d}\log (c{q}^2+r)$ to the disordered-phase free-energy density. Differentiating twice with respect to $r$ gives a critical contribution proportional to $\int d^{d}q(c{q}^2+r{)}^{-2}$; for $d<4$ its singular part grows as $r^{(d-4)/2}$ and at four dimensions it is logarithmic. Interacting fluctuations thus become important close to the critical point below four dimensions. The Ginzburg criterion tests this failure of the mean-field approximation; the renormalization group then explains modified exponents and universality. Mean-field theory finds the competing minima, while fluctuations determine whether its predicted critical behavior survives at long distance.

A first-order phase transition has a discontinuity in a first derivative of the equilibrium free energy, such as entropy or the order parameter; a thermal first-order transition normally has latent heat. At a continuous phase transition the order parameter goes continuously to zero and there is no latent heat, although response functions and higher derivatives of the free energy can be singular.

For the symmetric scalar Landau free energy at zero conjugate field, minimizeIf $u>0$, the sixth-order term is unnecessary near $r=0$. For $r>0$ the minimum is $M=0$, whereas for $r<0$ there are two symmetry-related minima with $M^2=-r/u$ to leading order. The order parameter vanishes continuously as $r\uparrow 0$. The equilibrium free-energy density changes from zero to $-r^2/(4u)$, so its first temperature derivative is continuous and its second derivative has a jump. This is the ordinary mean-field continuous phase transition.

If $u<0$, keep $v>0$ for stability. A nonzero stationary order parameter, with $s=M^2$, obeys $r+us+v{s}^2=0$. For phase coexistence with $M=0$ its free energy must also vanish. Substituting $r=-us-v{s}^2$ into $V$ givesThe nonzero solution of $V=0$ is thereforeThe competing minima exchange stability at positive $r_t$, with a finite jump of the order parameter. This is a first-order phase transition. The metastability region is also visible: nonzero stationary points first appear at the spinodal $r=u^2/(4v)$, and the disordered local minimum loses stability at $r=0$.

For example, holding $u,v$ fixed and taking $r=a(T-T_0)$, the ordered phase has entropy lower than the disordered phase by $a{M}_t^2/2$ at coexistence. Heating through the transition absorbs latent heat per volume $T_{t}a{M}_t^2/2$. More generally the temperature dependence of all coefficients contributes to this entropy difference. With no inversion symmetry, a cubic term can also produce a first-order phase transition, as comparison of the competing minima of the Landau free energy shows.

A tricritical point is the meeting of a line of continuous phase transitions and a line of first-order phase transitions. In the symmetric scalar Landau free energy, it occurs at $r=u=0$ with $v>0$. Two independent controls are needed to tune both coefficients. The continuous transition lies at $r=0,u>0$, and the tricritical three-phase line found above has $r=3u^2/(16v),u<0$; both approach the tricritical point. At zero conjugate field the three coexisting states on this first-order line are the disordered state and the two opposite ordered states.

Setting $u=H=0$ makes the leading stabilizing interaction sextic. The nonzero minimum obeys $M^4=-r/v$, so the order-parameter critical exponent along this tuned direction is ${\beta }_{\mathrm{tri}}=1/4$. Substitution givesand two temperature derivatives give the heat-capacity critical exponent ${\alpha }_{\mathrm{tri}}=1/2$. At $r=u=0$, the equation of state is $H=v{M}^5$, giving the critical-isotherm exponent ${\delta }_{\mathrm{tri}}=5$.

For magnetic susceptibility, differentiating $H=rM+v{M}^5$ gives ${\chi }^{-1}=r+5v{M}^4$. This is $r$ above the transition and $-4r$ below, hence ${\gamma }_{\mathrm{tri}}=1$. The quadratic fluctuation kernel still has a mass proportional to $|r|$, giving ${\nu }_{\mathrm{tri}}=1/2$ and ${\eta }_{\mathrm{tri}}=0$. Thus the mean-field tricritical exponent calculation yieldsDetuning $u$ gives tricritical crossover scaling; one must specify the approach to the tricritical point when quoting these critical exponents. The sextic interaction has upper critical dimension three, so sufficiently strong long-distance fluctuations change these mean-field predictions below three dimensions.

The uniform stationarity equation and local stability condition areFor $u>0$, $h=0$, and small negative $r$, the stable nonzero solution satisfies $M^2=-r/u+O(r^2)$. Thus the order parameter vanishes continuously as $r\uparrow 0$. The equilibrium Landau free energy has leading singular contribution $f_{\min }=-r^2/(4u)$ below the transition and zero above it. Its first thermal derivative is continuous, while its second derivative has a finite jump: the mean-field approximation predicts a continuous phase transition with no latent heat. The magnetic susceptibility follows by differentiating the stationarity equation:It diverges on approaching the ordinary thermodynamic critical point, and the quadratic Landau scalar correlation length behaves as $\xi \sim \sqrt{\kappa /|r|}$, with a different amplitude on the two sides.

For $u<0$, a nonzero stationary point at $h=0$ has $r+u{M}^2+v{M}^4=0$. Substitution into $V(M)-V(0)$ gives $-u{M}^4/4-v{M}^6/3$. Setting this to zero yieldsThe finite jump of $M$ makes this a first-order phase transition. For a thermal path $r=r(T)$ with $r^{\prime }(T)\ne 0$ and the other coefficients fixed, $\partial f_{\min }/\partial T=r^{\prime }(T)M^2/2$ jumps, giving an entropy discontinuity and nonzero latent heat. The Landau free energy itself remains continuous. The disordered minimum loses stability at $r=0$; the two nonzero stationary solutions merge at $r=u^2/(4v)$. These spinodal points bound metastability, but neither replaces the equilibrium phase coexistence condition.

Writing $r=at$ with $a>0$ near an ordinary thermodynamic critical point, the zero-field equation gives $M\sim (-t{)}^{1/2}$. At $r=0$, small $h$ obeys $h=u{M}^3+O(M^5)$, hence $M\sim \mathrm{sgn}(h)|h{|}^{1/3}$. The ordinary mean-field critical exponents are therefore $\beta =1/2$ and $\delta =3$. These are Landau theory predictions, not dimension-independent values for an interacting theory below its upper critical dimension.

At the tricritical point, both the quadratic and quartic even coefficients vanish, while the sextic coefficient remains positive. Two even control parameters must therefore be tuned, in addition to setting the symmetry-breaking conjugate field to zero. The continuous phase transition line and first-order phase transition line meet, the coexistence order parameter jump shrinks to zero, and the tricritical wings end there. The balance governing the equation of state changes from quartic to sextic stabilization.

Along the tricritical thermal path $u=0$, $r=at$, $h=0$, minimizing $V$ gives $M^4=-r/v$ on the ordered side. At $r=u=0$, the equation of state is $h=v{M}^5$. ConsequentlyThese mean-field critical exponents require the tuned tricritical path; a path with fixed $u>0$ instead has the ordinary exponents derived above. The corresponding equilibrium Landau free energy is $f_{\min }=-(-r{)}^{3/2}/(3\sqrt{v})$ below the transition. Its first thermal derivative is continuous but its second derivative diverges, giving the tricritical mean-field approximation a heat-capacity exponent $\alpha =1/2$. Fluctuations are negligible asymptotically only above the tricritical upper critical dimension three; at three dimensions logarithmic corrections can accompany these powers.

A phase diagram divides a space of control parameters into equilibrium thermodynamic phases, marking phase coexistence boundaries and thermodynamic critical points. Its dimension counts independently variable controls, not spatial dimensions. An explicit three-dimensional example uses the controls $(r,u,h)$ of a sextic even Landau potential:Here $M$ is a scalar order parameter, $h$ its conjugate field, and $r$ a thermal control. The equilibrium magnetization minimizes $V$ globally. At $h=0$, $u>0$, the boundary $r=0$ is a continuous phase transition. At $h=0$, $u<0$, the boundary $r=3u^2/(16v)$ is a first-order phase transition between $M=0$ and $M=\pm \sqrt{-3u/(4v)}$. The two boundaries meet at the tricritical point $r=u=h=0$. A fixed strictly positive quartic coupling alone cannot produce this tricritical point; the stabilizing sextic term is essential when the quartic coefficient is tuned through zero.

The full phase diagram also contains the $h=0$ phase coexistence sheet between positive and negative ordered phases. Crossing it changes the sign of $M$ discontinuously. For $u<0$, the three-phase line $r=3u^2/(16v)$ is the junction of this sheet and two tricritical wings, on which two unequal same-sign equilibrium magnetizations coexist at nonzero $h$. The tricritical wing critical edges are lines of ordinary thermodynamic critical points; they terminate the first-order sheets and meet at the tricritical point. Away from these boundaries, varying $h$ produces a smooth crossover.

For an explicit construction of the tricritical wings, write their two coexisting order parameters as $M_1=s-d$ and $M_2=s+d$, with $s>0$ and $0<d\le s$. Equality of their Landau free energies and first derivatives givesIndeed, for these controls the tricritical wing coexistence factorization isThe last factor is $(M+2s{)}^2+s^2-d^2\ge 0$, proving that both specified minima are global. The reflected wing has negative $h$ and negative $M$. Letting $d\to 0$ gives the tricritical wing critical edge $u=-10v{s}^2/3$, $r=5v{s}^4$, $h=8v{s}^5/3$; there $V^{\prime \prime }(s)=V^{\prime \prime \prime }(s)=0$ and $V^{\prime \prime \prime \prime }(s)=40v{s}^2>0$. Letting $d=s$ recovers the three-phase line. This supplies the first-order sheets and their continuous boundary curves, rather than only a zero-field section.

A continuous phase transition has a continuously vanishing order parameter and no latent heat. A first-order phase transition exchanges globally stable phases with discontinuous first derivatives of a thermodynamic potential; the order parameter generally jumps, and an entropy jump gives latent heat. In Landau theory the distinction is determined by global minima, not just by the appearance of additional stationary points.

For the uniform scalar potential $U(M)=r{M}^2/2+u{M}^4/4$ with $u>0$ and $h=0$, its stationary equation is $M(r+u{M}^2)=0$. For $r>0$, $M=0$ is the minimum; for $r<0$, the minima are $M=\pm \sqrt{-r/u}$. They approach zero continuously as $r\uparrow 0$, giving an ordinary continuous transition.

For $u<0$ and $v>0$, use the sextic potential. Write $y=M^2$. Stationarity of a nonzero phase requires $r+uy+v{y}^2=0$, while equality of its potential with $U(0)=0$ requires $ry/2+u{y}^2/4+v{y}^3/6=0$. Eliminating $r$ yieldsAt this value the potential factors as $U(M)=\frac{v}{6}{M}^2(M^2+3u/(4v){)}^2\ge 0$, proving that the three stationary zeros really are degenerate global minima. Crossing this line makes the order parameter jump: the transition is first order. The disordered and ordered spinodal points instead occur at $r=0$ and $r=u^2/(4v)$ respectively, delimiting metastability; these are not the equilibrium transition. If the symmetry allows a cubic term, that term can also produce a first-order transition.

In a first-order phase transition, two distinct global minima of the Landau free energy have equal values and exchange stability. The equilibrium order parameter jumps, and a first derivative of the equilibrium free energy is discontinuous. Depending on the direction through the phase diagram, that derivative can be magnetization, entropy or another conjugate quantity; a first-order transition need not have latent heat along every chosen path.

In a continuous phase transition, the equilibrium order parameter changes continuously while the curvature at the critical minimum vanishes. For $u>0$ and $h=0$, $M=0$ is stable for $r>0$, but for $r<0$ the stable minima have $M\simeq \pm \sqrt{-r/u}$. The order parameter therefore goes continuously to zero at $r=0$; the inverse magnetic susceptibility vanishes there.

For $u<0$, the sextic term stabilizes the Landau free energy. The nonzero minima can become favorable while $M=0$ still has positive curvature. At $r=3u^2/(16v)>0$ they coexist with $M=0$, so the order parameter jumps to magnitude $\sqrt{-3u/(4v)}$. Finding a stationary point alone does not locate the transition: equilibrium requires comparing the values of all competing minima. Spinodal boundaries describe loss of local stability, not coexistence.

A phase diagram partitions a space of thermodynamic controls into regions with distinct equilibrium phases. Its boundaries indicate phase coexistence or continuous phase transitions; a critical point is a termination of coexistence where the distinction between phases disappears. A concrete three-dimensional control space is $(T,\Delta ,h)$ for the Blume–Capel model: a spin can be $0$ or $\pm 1$, the exchange favors aligned nonzero spins, and $\Delta$ controls the energetic cost of nonzero spins. Varying $\Delta$ changes the balance between a continuous ordering transition and a first-order phase transition. Their meeting is a tricritical point.

Locally, the same structure is displayed by the Landau free energywith three controls $(r,u,h)$. For $u>0$, the line $r=h=0$ consists of ordinary critical points. For $u<0$, equal free energies of $M=0$ and a nonzero stationary point requireEliminating $r$ gives the tricritical three-phase lineHere $M=0$ and both signs of $M$ coexist, with a discontinuous order parameter. Below this line, or below $r=0$ when $u>0$, the plane $h=0$ is an ordered phase coexistence sheet: crossing it reverses the sign of $M$. Two tricritical wings also extend to nonzero $h$, separating weakly and strongly magnetized states of the same sign. Their ordinary critical points satisfy $f^{\prime }=f^{\prime \prime }=f^{\prime \prime \prime }=0$. Since $f^{\prime \prime \prime }=6uM+20v{M}^3$, their tricritical wing critical edges areAt these edges $f^{\prime \prime \prime \prime }=-12u>0$, giving an ordinary quartic critical expansion around $M_c$. All these structures end at the tricritical point $r=u=h=0$. The diagrams show both the zero-field section and the genuinely three-dimensional coexistence structure.

For the final critical exponent calculation, take $r=at$ with $a>0$, $t=(T-T_c)/T_c$, and hold the other controls fixed. At an ordinary critical point, the quartic term dominates the sextic term. For $h=0$ and $r<0$,whereas on the disordered side the singular contribution is zero. Two temperature derivatives give a finite jump in heat capacity, so $\alpha =0$. At $r=0$, the equation of state is $h=u{M}^3+O(M^5)$, so $\delta =3$.

Along the tricritical trajectory $u=0$, the nonzero minimum obeys $M^4=-r/v$ andThus two temperature derivatives produce $|t{|}^{-1/2}$. At $r=u=0$, $h=v{M}^5$. Therefore the mean-field critical exponents areWith $A$ the thermodynamic free energy, the physical sign is $C=-T{\partial }_T^{2}A$, rather than the positive sign printed in the paper. This correction affects the sign of the heat capacity, not its critical exponent. The tricritical result assumes that the independent quartic control is tuned to zero; a generic trajectory through nearby ordinary critical points need not have these exponents.

At a continuous phase transition the equilibrium order parameter tends continuously to its critical value, there is no latent heat, and the correlation length diverges. For $u>0$ at zero field, minimizing $V(M)=r{M}^2/2+u{M}^4/4$ gives $M=0$ when $r\ge 0$ and $M=\pm \sqrt{-r/u}$ when $r<0$. These minima join continuously at $r=0$. The minimized singular free-energy density is zero above and $-r^2/(4u)$ below, so its first thermal derivative is continuous, while its second derivative has a finite jump in mean-field approximation.

A first-order phase transition instead has competing minima whose free energies cross while their order parameters remain distinct. The first derivatives of the equilibrium free energy generally jump; along a temperature-driven crossing this gives latent heat. For the symmetric sextic potential with $u<0,v>0$, let $y=M^2>0$. At phase coexistence, the stationary points and equality of the two free energies requireEliminating $r$ gives $-u{y}^2/4-v{y}^3/3=0$, henceAt coexistence $V(M)=v{M}^2(M^2+3u/(4v){)}^2/6$, proving that all three degenerate minima are global. The equilibrium order parameter jumps from zero to one of the two nonzero values even though $r_{\mathrm{coex}}>0$. The disordered minimum loses local stability at $r=0$, and the nonzero stationary points first appear at $r=u^2/(4v)$. These spinodal points delimit metastability and are not the equilibrium transition line. Cubic terms, when symmetry permits them, provide another common route to a first-order transition.

A first-order phase transition exchanges distinct global minima of the Landau free energy; the order parameter jumps and a thermal crossing generically has latent heat. A continuous phase transition has minima merging continuously and a diverging correlation length. A local loss of stability at a spinodal point is not the equilibrium coexistence condition.

For the sextic even Landau potential, write $y=M^2$. At a nonzero zero-field stationary point, $r+uy+v{y}^2=0$. To coexist with $M=0$, also require $ry/2+u{y}^2/4+v{y}^3/6=0$. Eliminate $r$ to obtain $y=-3u/(4v)$ and $r=3u^2/(16v)$, which requires $u<0$. On this locusproving that all three equal minima are globally stable. Their curvatures are $3u^2/(16v)$ at zero and $3u^2/(4v)$ at the nonzero minima, so the correlation length is finite when $u<0$ is fixed. For $u>0$, by contrast, the nonzero minimum approaches zero as $r\uparrow 0$, giving the ordinary continuous transition. At fixed $u>0$ and $r<0$, crossing $h=0$ exchanges the two ordered minima discontinuously even though the zero-field thermal transition at $r=0$ is continuous.

A phase diagram locates stable pure thermodynamic phases and their phase coexistence boundaries in a space of control parameters. For a concrete three-dimensional example, take the temperature, crystal-field parameter, and magnetic field of the Blume–Capel model. Its spins take values $0,\pm 1$; changing the cost of the zero-spin state can turn a continuous ordering transition into a discontinuous one. A local scalar description near the resulting tricritical point is the Landau-Ginzburg theoryThe three local control coordinates are $(r,u,h)$, with $r$ varying linearly with the thermal scaling field. For $u>0$, $h=0$, the transition at $r=0$ is continuous. For $u<0$, zero-field coexistence is at $r=3u^2/(16v)$, and the magnetization jumps between zero and $\pm \sqrt{-3u/(4v)}$. The two loci meet at $r=u=h=0$, the tricritical point.

The full three-dimensional phase diagram has an ordered coexistence sheet at $h=0$, where the two opposite magnetizations exchange stability when $h$ changes sign. On its three-phase line it meets two nonzero-field tricritical wings. These wings are first-order phase transition surfaces between minima of the same sign but different magnitude, terminating at ordinary thermodynamic critical points. Their tricritical wing critical edges are found by $V^{\prime }=V^{\prime \prime }=V^{\prime \prime \prime }=0$:Indeed $V^{\prime \prime \prime }=6uM+20v{M}^3$ fixes $M_c^2$, and substituting into $V^{\prime \prime }=0$ and $V^{\prime }=0$ gives $r_c,h_c$; $V^{\prime \prime \prime \prime }=-12u>0$ ensures a stable critical quartic minimum. The two signs give the two wings. All transition surfaces and critical edges end together at the tricritical point. The following plot uses $v=1$; its wing surfaces come from the exact tricritical wing coexistence factorization, rather than schematic unconstrained surfaces.

The unheaded requests for critical exponents follow from the uniform equation of stateFor an ordinary thermodynamic critical point, $u>0$ is fixed and $r=At$, $A>0$. At $h=0$, $t<0$, the small ordered minimum obeys $M^2\sim -r/u$, so the order-parameter critical exponent is $\beta =1/2$. Above the transition $\chi =(\partial M/\partial h{)}_{h=0}=1/r$, giving the magnetic-susceptibility critical exponent $\gamma =1$. At $r=0$, $h\sim u{M}^3$, so the critical-isotherm exponent is $\delta =3$. At the tuned tricritical point, $u=0$, instead $M^4=-r/v$ below the transition and $h=v{M}^5$ on its critical isotherm, while the disordered susceptibility remains $1/r$. ThereforeBelow the transition the inverse susceptibility is $r+3u{M}^2+5v{M}^4$, giving $2|r|$ in the ordinary limit and $4|r|$ in the tricritical limit. Thus both sides give the same susceptibility power, with different amplitudes.

To derive the general Ginzburg criterion, tune away all lower even nonlinearities and let the leading stabilizer be $v_n{M}^{2n}/(2n)$, $n\ge 2$. The equation of state gives $M_0^2\asymp |r{|}^{1/(n-1)}$, and its curvature at the ordered saddle is $2(n-1)|r|$, so $\xi \asymp |r{|}^{-1/2}$. The fluctuations of the order parameter averaged over a correlation volume are estimated by its long-wavelength quadratic covariance:The integral has a fixed dimensionless upper limit after $q=p/\xi$; this is a coarse-volume fluctuation estimate, not the ultraviolet-dominated point variance. Consequently the multicritical Ginzburg ratio isThe Landau approximation is asymptotically controlled by this test only when the ratio tends to zero. This gives the upper critical dimensionThe same threshold follows by giving $M$ its Gaussian engineering dimension $(D-2)/2$: the coupling to $M^{2n}$ has scaling exponent $D-n(D-2)=2n-(n-1)D$, which becomes relevant below $D_c$.

For $D<D_c$, the fluctuation ratio diverges and the saddle calculation cannot predict the generic interacting critical exponents. At $D=D_c$ the test is marginal, not a proof of changed power exponents: the ratio has zero thermal power and marginal interactions can produce logarithmic corrections. Thus the PDF's inclusion of equality in a blanket claim about wrong power exponents needs this qualification. The Landau-Ginzburg theory as a fluctuating field theory remains useful; what loses validity is its fluctuation-free Landau approximation.

A phase diagram partitions a space of control parameters into regions with distinct equilibrium phases. Its boundaries mark nonanalyticities of the thermodynamic-limit free energy; a first-order phase transition has a jump in a first derivative, whereas a continuous phase transition has a continuously vanishing order parameter and singular fluctuations.

An explicit three-parameter example is the scalar Landau free energywith parameter axes $(r,u,h)$. These can be realized locally by the temperature, single-ion anisotropy and magnetic field of the Blume–Capel model. The three dimensions here refer to control-parameter space, not necessarily to three spatial dimensions. At $h=0$, minimizing this potential gives a continuous phase transition at $r=0$ for $u>0$. For $u<0$, coexistence occurs at $r=3u^2/(16v)$, whereThere are three global minima, $M=0$ and $M=\pm \sqrt{-3u/(4v)}$. Their order parameter jump decreases to zero as $u\uparrow 0$. Thus $(r,u,h)=(0,0,0)$ is a tricritical point. The two transition curves in the $h=0$ plane meet there.

In the three-dimensional phase diagram, crossing $h=0$ inside the ordered region switches between opposite signs of $M$, so that region is itself a first-order phase transition sheet. For $u<0$, two tricritical wings at opposite nonzero fields separate weakly and strongly magnetized phases. Their wing critical edges can be obtained rather than merely sketched: require $V^{\prime }=V^{\prime \prime }=V^{\prime \prime \prime }=0$. Since $V^{\prime \prime \prime }=6uM+20v{M}^3$, the nonzero solutions satisfyThe fourth derivative there is $-12u>0$, giving ordinary continuous critical endpoints. The wings meet the zero-field coexistence sheet along the tricritical three-phase line and terminate at these edges. Their exact tricritical wing coexistence factorization supplies the surfaces in the figure: for two nonnegative minima $a\le b$, set $s=a+b$, $p=ab$ andDirect expansion gives $V(M)-V(a)=(v/6)(M-a{)}^2(M-b{)}^2[(M+s{)}^2+p]\ge 0$, proving that these surfaces describe coexistence of global, not merely local, minima.

For the unheaded Ising model calculation, assume ferromagnetic exchange $J>0$ and let $M$ denote the mean Ising spin. On the hypercubic lattice the coordination number of a lattice is $z=2D$, even though the original positive-direction bond sum counts each bond only once. The mean-field approximation neglects products of deviations from $M$:There are $ND$ bonds, so the approximate energy is $H_{\mathrm{MF}}=-2DJM{\sum }_i{\sigma }_i+NDJ{M}^2$. Each independent Ising spin therefore sees the effective field $2DJM$. With ${\beta }_{\mathrm{th}}=1/(k_{B}T)$, averaging its two Boltzmann weights gives the self-consistency equationThe slope of its right side at zero is $T_c/T$, whereAbove $T_c$, $\tanh x<x$ for $x>0$ excludes a nonzero solution. Below $T_c$ the initial slope exceeds one, and concavity of the hyperbolic tangent on the positive axis gives one positive solution and its negative partner. The Ising auxiliary mean-field free energy $DJ{M}^2-k_{B}T\log [2\cosh (2{\beta }_{\mathrm{th}}DJM)]$ has positive quartic coefficient near $T_c$, while its quadratic coefficient changes sign there. Its stable minima therefore move continuously away from zero; this is a continuous phase transition in the mean-field theory of the Ising model.

Put $\tau =(T_c-T)/T_c>0$ and $c=T_c/T=(1-\tau {)}^{-1}$. The Taylor expansion $\tanh x=x-x^3/3+O(x^5)$ gives, on the nonzero branch,Consequently $M\sim \sqrt{3}{\tau }^{1/2}$ and the mean-field order-parameter exponent is $\beta =1/2$. The inverse-temperature symbol ${\beta }_{\mathrm{th}}$ used above is distinct from this order-parameter critical exponent.