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For the sextic even Landau potentialput $q={\varphi }^2$. The stationary-point equation isso the nonzero stationary points haveAt such a point, $\epsilon =-q^2-gq$, and henceA first-order transition occurs when a nonzero minimum has the same free energy as the disordered minimum $G(0)=0$. ThereforeAt $g=0,\epsilon =0$, the first-order line meets the continuous line and forms the tricritical point. For $g>0$, the transition at $\epsilon =0$ is continuous.

For $g<0$, the disordered state $\varphi =0$ is a local minimum when $\epsilon >0$ and loses stability at the disordered spinodal point $\epsilon =0$. The ordered minima exist whileand merge with the intervening maxima at the ordered spinodal $\epsilon =g^2/4$. Consequently the two spinodals enclose a region of metastability. If increasing $\epsilon$ corresponds to heating, the ordered phase can persist above the coexistence curve throughoutwhile the disordered phase can persist below it throughout

The stationary points obeyNonzero stationary points exist whenso they first appear at the ordered-phase spinodal point $a_2=9a_3^2/(32a_4)$. The disordered state $m=0$ is locally stable for $a_2>0$ and loses that stability at $a_2=0$.

The actual phase boundary is found by requiring a nonzero stationary point $m_{\ast }$ to have the same free energy as $m=0$. Solving $f^{\prime }(m_{\ast })=0$ and $f(m_{\ast })=0$ givesThe disordered state is the global minimum for $a_2>a_3^2/(4a_4)$, the ordered state is the global minimum for $a_2<a_3^2/(4a_4)$, and they coexist at equality.

Since $a_3\ne 0$, the order parameter jumps from zero to $-a_3/(2a_4)$ at coexistence. Hence this model has no continuous phase transition as the phases exchange stability, but it does have a first-order phase transition at the displayed positive value of $a_2$.

For $b<0$ at $T=T_c$, the quartic term lowers the free energy near $m=0$, while the positive sextic term stabilizes it at larger $|m|$. The nonzero stationary points haveThey are global minima. The stationary point $m=0$ has zero quadratic curvature but is a flat local maximum because its leading nonzero term is negative. Thus the system is already ordered at $T_c$, with a finite magnetization.

More explicitly, a continuous onset from $m=0$ could occur only where $f^{\prime \prime }(0)=2r$ vanishes, namely at $r=0$. But the global minima there are the finite values above. For sufficiently large positive $r$, only $m=0$ is stable. Therefore the equilibrium minimum must change at some positive $r$ by crossing between a zero and a finite nonzero order parameter. This is a first-order phase transition. Its actual temperature lies above $T_c$ and is calculated next; $T_c$ itself is the disordered spinodal point, not the coexistence temperature.

At phase coexistence, the finite ordered global minimum has the same free energy as $m=0$. With $x=m^2>0$ and $f(0)=0$, stationarity and equal values giveSubtracting eliminates $r$:Substitute back to get $r_{\ast }=3b^2/(16c)$, soThe three degenerate equilibrium values at this temperature are $0$ and $\pm m_{\mathrm{jump}}$. A useful global check, avoiding confusion with a spinodal point, is the exact factorizationAll three zeros are therefore global minima. Their curvatures are positive: $f^{\prime \prime }(0)=3b^2/(8c)$ and $f^{\prime \prime }(\pm m_{\mathrm{jump}})=3b^2/(2c)$.

The ordered local minimum first appears on cooling at $r=b^2/(4c)$, where the nonzero stationary roots merge. It becomes globally stable only at $r_{\ast }$; the disordered minimum remains metastable down to $r=0$. These different temperatures distinguish phase coexistence from loss of local stability.

For the Landau free energy $G(\varphi )={\varphi }^6/6+g{\varphi }^4/4+\epsilon {\varphi }^2/2$, positive $g$ gives a continuous phase transition at $\epsilon =0$. For negative $g$, phase coexistence between the disordered global minimum $\varphi =0$ and the ordered global minima occurs on $\epsilon =3g^2/16$, with ${\varphi }^2=-3g/4$. Indeed, on that locus the polynomial is $G(\varphi )={\varphi }^2({\varphi }^2+3g/4{)}^2/6\ge 0$, so these three zeros are global minima. For $g<0$, the disordered and ordered spinodal points are respectively $\epsilon =0$ and $\epsilon =g^2/4$; equality of stationary free energies determines phase coexistence, while loss of a local minimum determines a spinodal point.

A Van der Waals isotherm is the pressure–particle volume curve at a fixed temperature. Put $x=v/v_c$, $t=T/T_c$, and $q=p/p_c$, using the Van der Waals critical point values below. Then the existing Van der Waals law of corresponding states becomesAll curves in the figure are calculations in this reduced model.

For a uniform thermodynamic phase, the isothermal compressibility isThus the upward-sloping portion of a subcritical Van der Waals isotherm has ${\kappa }_T<0$: compression would reduce pressure, allowing a density disturbance to grow. It is mechanically unstable. The spinodal points bound that portion; setting the derivative $\partial q/\partial x$ to zero gives their curve $t=(3x-1{)}^2/(4x^3)$. Negative slope alone guarantees only local mechanical stability: portions between the spinodal and the phase coexistence endpoints have metastability. Equilibrium replaces the entire interval between those endpoints, including these metastable portions, by a Maxwell construction plateau.^[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.

The Maxwell construction presumes a pure bulk fluid at thermodynamic equilibrium. Its endpoint equality does not specify how rapidly a real compression reaches equilibrium. A homogeneous metastable sample can persist while nucleation is delayed; the spinodal marks loss of local mechanical stability in this mean-field theory, rather than the equilibrium onset of condensation. Surface tension, finite sample size and impurities can affect the observed transition. These qualifications matter when comparing a mathematical Van der Waals isotherm with a visible liquid meniscus.