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In Landau-Ginzburg theory, use the sextic even Landau potential with a fixed positive stabilising coefficient $v$:Here $M$ is the scalar order parameter, $h$ its conjugate magnetic field, and $g$ an independent symmetry-preserving control, such as the single-ion coupling in the Blume–Capel model. Thus $g$ need not break $M\mapsto -M$, whereas $h$ does. The Blume–Capel mean-field tricritical parameters are one realization; generally, a tricritical point occurs whenTwo controls tune away the quadratic and quartic terms. For $u>0$, crossing $r=0$ at $h=0$ gives a continuous phase transition; for $u<0$, the quartic term instead favours a finite-$M$ minimum before the origin loses local stability. The sextic term bounds the free energy below. The tricritical point joins the continuous and discontinuous transition loci.

For the symmetric phase diagram, write $s=M^2$. A nonzero stationary point obeys $r+us+v{s}^2=0$. Its free-energy density relative to the origin is $rs/2+u{s}^2/4+v{s}^3/6$. Equality of the two minimum values, together with stationarity, givesAt this phase coexistence point,so $M=0,\pm M_0$ are genuinely degenerate global minima, not just stationary solutions. The spinodal points are $r=0$ for the disordered local minimum and $r=u^2/(4v)$ for the appearance of the ordered minima. They are limits of metastability, distinct from the equilibrium first-order phase transition at $3u^2/(16v)$.

Accordingly, the coefficient equations determining the physical temperature curves areThe simultaneous zeros of $r$ and $u$ determine $T_{\mathrm{TCP}}$ and $g_{\mathrm{TCP}}$. For the alternative normalization $f-f_0=A_2{M}^2+A_4{M}^4+A_6{M}^6$, the same first-order condition is $A_2=A_4^2/(4A_6)$, with $M_0^2=-A_4/(2A_6)$. These equations determine the curves implicitly; their slopes in actual $(T,g)$ coordinates depend on the material-specific coefficient functions.

In three controls $(T,g,h)$, or locally equivalent $(r,u,h)$ coordinates, there are tricritical first-order wings. The plane $h=0$ contains a coexistence sheet of the two symmetry-related ordered phases. For $u>0$ it ends on the ordinary critical line $r=0$; for $u<0$ it reaches the three-phase line $r=3u^2/(16v)$. Two further first-order surfaces extend from that line into $h>0$ and $h<0$. Across a wing, two minima of the same sign but different magnitudes exchange global stability; the field favours that sign. Each wing ends on an ordinary critical edge where those minima and their intervening maximum merge. The three surfaces and their critical boundaries meet at the tricritical point.

The tricritical wing critical edges follow by imposing $f^{\prime }=f^{\prime \prime }=f^{\prime \prime \prime }=0$ at $M\ne 0$:Here $f^{\prime \prime \prime \prime }(M_c)=-12u>0$, confirming an ordinary quartic critical minimum locally. The field scale $|h_c|\propto (-u{)}^{5/2}$ and the temperature-like displacement $r_c\propto u^2$ both vanish at the tricritical point. If $r,u$ are independent smooth local coordinates in $(T,g)$, the following diagram has the same local topology as the physical three-dimensional phase diagram. Its surfaces use exact equal-minimum conditions, as described by tricritical wing coexistence factorization.

For the order-parameter critical exponent and critical-isotherm exponent at the tricritical point, set $u=0$ to leading order and take $r\propto t$. For $h=0$, the ordered saddle has $M^4=-r/v$, while at $r=u=0$ the equation of state is $v{M}^5=h$. Hence the tricritical mean-field critical exponents areThese are mean-field critical exponents. Below the tricritical upper critical dimension they need not be the interacting exponents, and at that dimension logarithmic corrections can accompany the powers. Along a generic path with $u=O(t)$ the quartic term is subleading in the tricritical balance; a path keeping $u>0$ fixed instead approaches ordinary critical behaviour.

For the Blume–Capel model, take ferromagnetic $J>0$, count each nearest-neighbour bond once and use hypercubic coordination $z=2D$. The nearest-neighbour direction sum in the printed Hamiltonian is understood implicitly. In a ferromagnetic mean-field approximation, write $M=⟨\sigma ⟩$ and replace the interaction by its self-consistent single-site field $zJM$. The single-site partition function and variational free-energy density areThe first term prevents double counting the interaction energy. Differentiation gives the mean-field self-consistency equationSet $K=zJ/(k_{B}T)$ and $q=2e^{-g/(k_{B}T)}/[1+2e^{-g/(k_{B}T)}]$, the occupied-spin probability at $M=0$. Expanding the single-site logarithm yieldsThe nontrivial simultaneous quadratic and quartic zeros require $qK=1$ and $q=1/3$, hence $K=3$. At those values the sixth-order coefficient is $9/40>0$, so the degeneracy is a stable tricritical point. The mean-field tricritical parameters areThis is a mean-field result, not a dimension-independent exact lattice transition temperature. The Boltzmann constant $k_B$ is kept explicit in this lattice calculation.

Write $r_i={\mu }_i^2$ and assume $g>0$, as required for quartic stability. In the mean-field approximation, the two-component radial quartic Landau potential isSet $x={\varphi }_1^2\ge 0$, $y={\varphi }_2^2\ge 0$. At fixed $\rho =x+y$, the quartic term is fixed and the quadratic term is minimized by assigning all of $\rho$ to the component with smaller $r_i$. Minimizing over $\rho$ then givesThe resulting phase diagram has these regions:

On $r_1=r_2=r<0$, the symmetry is enhanced to the orthogonal group $O(2)$, and the minima form the circleMixed-component minima occur on this line, rather than in an open coexistence region. Crossing the negative diagonal swaps the selected component discontinuously: this is a first-order phase transition, also called a spin-flop transition. For example, holding $r_2=r<0$ fixed and moving $r_1$ across $r$, the derivative $\partial f_{\min }/\partial r_1$ jumps from $-r/(8g)$ to zero. The angular degeneracy exactly on the line does not remove this discontinuity.

The positive half-axis $r_1=0$, $r_2>0$ is a continuous phase transition, as is $r_2=0$, $r_1>0$. Their endpoint at $(0,0)$ is a bicritical point, where the two continuous boundaries meet the first-order negative diagonal. Approaching the origin radially makes the ordered amplitude vanish continuously. The negative half-axes lie inside ordered phases and are not additional transition lines. There are two continuous half-axis boundaries and one first-order negative diagonal, meeting at the origin.

Before the ice reaches the salty layer, the state moves vertically in a temperature--concentration phase diagram along $C=0$: cooling crosses the pure-water melting point and produces pure ice. After the encounter, the liquid state starts at $(T_{\infty },C_0)$ and approaches the interface stateon the liquidus $T_L=-mC$. The solid remains on the $C=0$ axis.

When $ϵ\ll 1$, the solutal diffusion length $\sqrt{Dt}$ is much smaller than the thermal diffusion length $\sqrt{\kappa t}$. Immediately ahead of the interface, rejected salt makes $C$ fall steeply from $C_i$ to $C_0$, so the local liquidus $T_L=-mC$ rises steeply with depth while the actual temperature changes comparatively slowly. Constitutional supercooling occurs whereverIts local onset criterion at the interface isThe supercooled zone lies in the thin salty boundary layer immediately ahead of the planar front. A small forward protrusion then enters liquid that is already below its local freezing point, so the planar interface is susceptible to a morphological instability.

A phase diagram partitions a space of thermodynamic variables into regions occupied by different equilibrium thermodynamic phases and records the phase coexistence curves between them. For a pure, one-component substance, a pressure–temperature phase diagram gives one point for each pair $(T,p)$. In a single-thermodynamic phase region both controls can vary independently. On a two-thermodynamic phase phase coexistence curve, they obey one relation; at a triple point, both are fixed. These statements assume bulk thermodynamic equilibrium, fixed composition and no additional applied fields.^[ref]

At fixed temperature and pressure, the stable bulk thermodynamic phase minimizes Gibbs free energy for a fixed amount of substance. For a pure substance this is a comparison of chemical potentials on a common normalization. Two candidate thermodynamic phases coexist where ${\mu }_{\alpha }(T,p)={\mu }_{\beta }(T,p)$ and neither is exceeded in stability by a third thermodynamic phase. Differentiating their equality gives the Clapeyron equation. Thus a phase coexistence curve is a locus of equal stability, rather than a trajectory that every heated sample must follow.^[ref]

The following original cartoon uses dimensionless positive coordinates $\theta =T/T_{\ast }$ and $\pi =p/p_{\ast }$, with arbitrary reference scales $T_{\ast },p_{\ast }$. It represents no particular substance and contains no measured data. Stipulate a triple point at $(1,1)$, a liquid-gas critical point at $(3,5)$, and these drawing rules:They specify sublimation, liquid–gas phase coexistence and melting boundaries for the picture. They are not an equation of state, do not predict latent heat or mass density, and do not supply a quantitative critical theory. In particular, the melting slope is merely an illustrative choice.

For example, at fixed $\pi =3$ the melting boundary is at $\theta =1.2$ and the liquid–gas boundary is at $\theta =2$. Heating from $(\theta ,\pi )=(1.6,3)$ to $(3.6,3)$ therefore starts in liquid, crosses phase coexistence at $(2,3)$, and ends in gas. In an equilibrium conversion at that crossing, added heat can change the phase fraction at fixed temperature through latent heat. The pressure–temperature point alone does not specify how much of each thermodynamic phase is present: for that one needs an additional amount or volume constraint.

Now connect exactly the same endpoints by raising $\pi$ from $3$ to $6$ at $\theta =1.6$, moving to $\theta =3.6$ at $\pi =6$, then lowering $\pi$ to $3$. The first leg stays on the liquid side because ${\theta }_{\mathrm{melt}}$ ranges only from $1.2$ to $1.5$ and ${\pi }_{\mathrm{vap}}(1.6)=2.2<3$. The second leg lies above the entire liquid–gas boundary, whose largest drawn pressure coordinate is $5$; the third lies beyond its largest temperature coordinate, $3$. No leg crosses a first-order phase transition boundary. The middle leg passes through a supercritical fluid region, with $\theta >3$ and $\pi >5$, and the final leg returns to a gaslike state continuously.

Avoiding a first-order phase transition does not mean that no heat or work is exchanged. Along the detour, mass density, enthalpy and heat capacity may change substantially but continuously. The liquid-gas critical point itself has singular behavior, so the detour goes around it rather than through it. This connected fluid region is the striking lesson of a phase diagram: a familiar change from liquid to gas need not involve boiling across a sharp boundary.

An equilibrium phase diagram also does not predict how rapidly a conversion occurs. Metastability and a barrier described by classical nucleation theory can delay the observed phase transition. Nor should a binary-composition phase diagram be read as the same picture: its coexistence regions and lever rule encode composition and phase fractions, and can occupy an area even though pure-substance two-thermodynamic phase coexistence occupies a curve in the pressure–temperature plane.

The solidus bounds the entirely solid equilibrium region of a phase diagram. For a linear liquidus $T_L(C_l)=-m{C}_l$ and a constant segregation coefficient $C_s=k_D{C}_l$, the corresponding solidus is $T_S(C_s)=-m{C}_s/k_D$. The concentration argument is the solid concentration, not the neighboring liquid concentration.

A state of matter is a broad category of material behavior. A solid resists a sustained change of shape, a liquid flows while remaining difficult to compress, and a gas expands to fill its container. These distinctions are useful first descriptions, but they do not enumerate all thermodynamic phases. Plasma adds collective behavior of charged particles, while liquid crystals show that flow and structural order need not disappear together.

Distinguish three questions: what broad state of matter does the material resemble, which thermodynamic phase is present, and what are its temperature, pressure and other state variables? Two samples at different temperatures can be in the same liquid thermodynamic phase. Conversely, crystal polymorphism gives different solid thermodynamic phases of the same substance, distinguished by their crystal structures. A phase transition between them need not change the broad state-of-matter category.

For an imagined substance with crystalline forms A and B, melting A produces a solid–liquid phase transition, while converting A to B produces a solid–solid phase transition. A sample containing A and B has two thermodynamic phases even though all its material is solid. Cutting one uniform crystal into two pieces does not create a second thermodynamic phase: count distinct equilibrium forms, rather than pieces or grains. This distinction matters when applying the phase rule.

A phase diagram locates stable thermodynamic phases using temperature and pressure. Its liquid–gas phase coexistence curve ends at a liquid-gas critical point, around which a path connects liquid and gas without a first-order phase transition. Broad categories can therefore remain useful without a sharp boundary.^[ref]

Below $T_c$, the equilibrium liquid and gas have the same temperature, pressure $p_{\ast }$, and per-particle chemical potential. For macroscopic bulk thermodynamic phases, neglect interface contributions. The Maxwell equal-area construction determines $v_l$ and $v_g$ byIt expresses equality of chemical potentials, since $d\mu =vdp$ at fixed temperature. Equivalently, the equilibrium Helmholtz free energy per particle follows a common tangent line between the two endpoint states, lying below the homogeneous continuation between them. The construction uses that mathematical continuation to compare endpoints; it is not a physical compression path through negative isothermal compressibility states.^[ref]

Here is a reproducible calculated example at $t=0.9$. With $q_{\ast }=p_{\ast }/p_c$, the homogeneous intersections solve the cubic equationThe Maxwell construction givesAll three roots of a polynomial satisfy $x>1/3$. Only the smallest and largest are the phase coexistence endpoints; the middle one has negative isothermal compressibility. Equal pressures at three admissible values of volume therefore do not imply three stable thermodynamic phases. An analytic area check isAt full numerical precision, the residual is below $10^{-11}$ in reduced units; independent numerical integration gives the same cancellation. At $t=0.9$, the spinodal points are instead $x\simeq 0.718597189$ and $1.528504964$.

The equilibrium plateau changes the phase fraction rather than the endpoint densities. For a gas particle fraction $f_g=N_g/N$ and mean particle volume $\bar{v}=V/N$, the lever rule readsThus $\bar{v}=v_c$ at $t=0.9$ gives $f_g\simeq 0.227219$: this is a particle fraction, not the volume fraction occupied by gas. Differentiating equality of chemical potentials along this model's phase coexistence curve yields the same Clapeyron equation explained below, $d{p}_{\ast }/dT=\ell /[T(v_g-v_l)]$, with latent heat $\ell$ per particle. Both endpoint differences vanish as the Van der Waals critical point is approached; the equation's critical limit must be taken as a ratio.