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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 real unrestricted scalar order parameter and $A=A_2{m}^2/2+A_3{m}^3/3+A_4{m}^4/4$ with $A_3\ne 0$ and $A_4>0$, the disordered global minimum is replaced discontinuously by a nonzero minimum at the displayed phase coexistence condition. At coexistence $A=A_4{m}^2[m+2A_3/(3A_4){]}^2/4$. Ordered stationary points first appear at the spinodal point $A_2=A_3^2/(4A_4)$, while the disordered point loses local stability at $A_2=0$; neither local limit is the equilibrium transition condition.

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.

For the Landau free energy $V(M)=r{M}^2/2+u{M}^4/4-hM$ with $r<0$ and $u>0$, stationary branches satisfy $h=rM+u{M}^3$. Their local stability requires $r+3u{M}^2>0$. At a spinodal point, $M_{\mathrm{sp}}=\pm \sqrt{|r|/(3u)}$ and $h_{\mathrm{sp}}=(2r/3)M_{\mathrm{sp}}$, giving the displayed field magnitude. Following local minima without nucleation produces a hysteresis loop switching at these limits. Equilibrium instead switches between the two global minima at $h=0$.

A metastable phase is a local but nonglobal minimum of the thermodynamic potential. A finite nucleation barrier can delay conversion to the globally stable phase until fluctuations create a sufficiently large nucleus or the metastable minimum reaches a spinodal point.

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.

Let $A(M)=V(M)+hM$ be the zero-field constrained order-parameter free energy. At coexistence in a conjugate field $h_c$, two global minima $M_1<M_2$ obeyThus the same line of slope $h_c$ is tangent to $A$ at both endpoints: the common-tangent construction for phase coexistence. Equivalently,This is the Maxwell equal-area construction on the equation-of-state curve $h=A^{\prime }(M)$: the signed areas on either side of the horizontal coexistence line cancel. Equal slope alone finds stationary phases; the integral additionally enforces equal free energy.

If the mean order parameter is constrained between the endpoints, the system can separate into two macroscopic phases. With $\lambda =(M-M_1)/(M_2-M_1)$, their bulk free-energy density is $(1-\lambda )A(M_1)+\lambda A(M_2)$. The interface contributes subextensively in the thermodynamic limit. The resulting constrained order-parameter free energy is the convex envelope of $A$, with a straight coexistence segment. This explains why a nonconvex uniform Landau free energy must be convexified when describing stable constrained equilibrium, and why a spinodal point is not the Maxwell construction boundary.

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.

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.

Use the connected correlation function $G(x)=⟨M(x)M(0)⟩-⟨M{⟩}^2$, since the disconnected constant in an ordered phase does not measure fluctuation decay. Away from criticality its large-distance exponential defines the correlation length:when that limit exists. Algebraic prefactors do not affect this definition.

Expand the Landau-Ginzburg theory about a stable homogeneous minimum $M_0$. With $A=V^{\prime \prime }(M_0)>0$, the quadratic fluctuation energy is $\frac{1}{2}\int d^{D}x[\kappa |\nabla \delta M{|}^2+A(\delta M{)}^2]$. The Gaussian functional integral givesThis Ornstein--Zernike correlation function shows why the curvature of the free-energy minimum controls correlations. At a continuous phase transition that curvature vanishes, allowing long-range fluctuations. In the ordinary quartic theory with $u>0$, $A=r$ above the transition and $A=-2r$ on either ordered branch below, so $\xi \propto |r{|}^{-1/2}$ on both sides. At an equilibrium first-order transition the competing minima generally retain nonzero curvature, and the correlation length remains finite. The limit of local stability at a spinodal point is a distinct mean-field phenomenon.

Set $h=0$ in the even sextic Landau free energy. For $u>0$, near $r=0$ the nonzero stationary branch satisfies $M^2=-r/u+O(r^2)$ for $r<0$. The minimum value is $V_{\min }=-r^2/(4u)+O(r^3)$ below the transition and zero above. The order parameter is continuous, and if $r$ crosses zero linearly with temperature, the first temperature derivative of the minimized free-energy density is continuous too. There is no latent heat; a second derivative has a finite jump at this level.

For $u<0$, solve the two conditions $V^{\prime }(M)=0$ and $V(M)=V(0)$ with $M\ne 0$. Setting $q=M^2$ gives $r+uq+v{q}^2=0$ and $r/2+uq/4+v{q}^2/6=0$. Eliminating $r$ givesBoth the disordered and ordered minima have positive curvature there, but their values coincide. Crossing this phase coexistence boundary changes the global minimum discontinuously. The order parameter jumps; for $r=r(T)$ at fixed $u,v$, ${\partial }_T{V}_{\min }=r^{\prime }(T)M^2/2$ jumps as well, generically yielding a nonzero latent heat. Local extrema can persist past coexistence as metastable states, which explains hysteresis in a dynamically delayed transition. The spinodal points, where local curvature vanishes, do not locate the equilibrium transition.

For $u>0$, the quartic Landau free energy at $h=0$ minimizes at $m=0$ for $r>0$ and at $m=\pm \sqrt{-r/u}$ for $r<0$. The order parameter changes continuously at $r=0$, giving a continuous phase transition and a divergent response. The minimized potential is $0$ above the transition and $-r^2/(4u)$ below it, so its first temperature derivative is continuous while its second derivative jumps.

For $u<0$, retain $v>0$. A first-order phase transition occurs when the $m=0$ minimum and a nonzero minimum have equal free energy. Combining stationarity $r+u{m}^2+v{m}^4=0$ with equality of potential values givesThe order parameter jumps at a positive $r$ before the disordered minimum loses local stability. Generic temperature dependence then also produces a jump in entropy and a latent heat. Equality of global minima locates coexistence; the disappearance of a metastable minimum instead defines a spinodal point.

A first-order phase transition exchanges two distinct global minima of the Landau free energy. The equilibrium free energy remains continuous, but a first derivative such as the magnetization or entropy jumps. A temperature-driven transition generically has latent heat; crossing $h=0$ between symmetry-related ordered phases instead has a magnetization jump without requiring latent heat. A continuous phase transition has no such order-parameter jump and usually has a divergent correlation length and nonanalytic higher responses.

For $u>0$, $h=0$ and $r$ crossing zero, minimizing $V=r{M}^2/2+u{M}^4/4$ gives $M=0$ for $r>0$ and $M=\pm \sqrt{-r/u}$ for $r<0$. The minima merge continuously at zero. The curvature vanishes there, so the long-distance response becomes singular.

For $u<0<v$, put $z=M^2$. A nonzero stationary minimum satisfies $r+uz+v{z}^2=0$. Equality with the $M=0$ free energy givesThe finite separation demonstrates the first-order phase transition. Metastability occurs where a nonglobal local minimum persists; its disappearance at a spinodal point is not the equilibrium coexistence criterion.

Along an isolated two-phase coexistence line, two distinct stable minima have equal free energy. If this line ends at an ordinary interior endpoint without encountering another phase, the distinction between the two phases must disappear: their minima and the intervening barrier merge. For separated nondegenerate minima, their stationary values are smooth functions of the controls. The difference $\Delta V=V(m_1)-V(m_2)$ has ${\partial }_h\Delta V=-m_1+m_2\ne 0$. The implicit function theorem therefore continues the equation $\Delta V=0$ as a local coexistence curve. It cannot end while those distinct stable minima persist.

At the merger the potential satisfiesThe third-derivative condition is essential: a stationary zero-curvature point with nonzero third derivative is an inflection, not a stable equilibrium. Shifting to $\psi =m-m_c$ gives an ordinary quartic Landau free energy with two independent control directions, temperature-like and field-like. The curvature and the order-parameter discontinuity vanish, the magnetic susceptibility diverges, and the correlation length diverges. The endpoint is therefore associated with a continuous transition. This is the critical endpoint of a quartic Landau free energy mechanism.

In the symmetric quartic example, the coexistence line is $h=0,r<0$. Its jump $2\sqrt{-r/u}$ vanishes at $(r,h)=(0,0)$, where $V=u{m}^4/4$. This supplies a concrete realization. The word “must” requires the stated ordinary-endpoint assumptions: a first-order line can instead meet a third phase, a multicritical point, a boundary of parameter space, or continue without a finite endpoint. A spinodal point alone is not an equilibrium critical endpoint.

A first-order phase transition has a discontinuity in a first derivative of the equilibrium free energy, such as the entropy or order parameter. It can have latent heat when the entropy jumps. A continuous phase transition has a continuously vanishing order parameter and no latent heat, with singular higher derivatives and a diverging correlation length. The LG theory compares global minima, not merely the points where a local minimum loses stability.

For the uniform quartic Landau free energythe equation of state is $rm+u{m}^3=h$. At zero field the stable minimum is $m=0$ for $r>0$, and $m=\pm \sqrt{-r/u}$ for $r<0$. Thus tuning $r$ through zero gives a continuous transition. For a fixed $r<0$, varying $h$ through zero instead switches between the two ordered minima and makes $m$ jump, giving a first-order phase transition in the conjugate field.

A temperature-like first-order transition can occur at zero field when $u<0$ and a positive sextic term $v{m}^6/6$ stabilizes the potential. Write $q=m^2$. Stationarity of a nonzero phase gives $r+uq+v{q}^2=0$, while equality with $V(0)=0$ gives $rq/2+u{q}^2/4+v{q}^3/6=0$. Solving these two conditions yieldsThe order parameter jumps from zero to $\pm \sqrt{q_{\mathrm{coex}}}$. At this pointso the competing stationary points are genuinely global minima. This phase coexistence condition differs from the spinodal points $r=0$ and $r=u^2/(4v)$, which mark loss or creation of local stability, not equilibrium coexistence.

If the symmetry permits a cubic term $w{m}^3/3$, a positive quartic coefficient does not preclude a first-order transition. For $V=r{m}^2/2+w{m}^3/3+u{m}^4/4$ with $w\ne 0,u>0$, stationarity and coexistence give $m=-2w/(3u)$ and $r=2w^2/(9u)$. This is the first-order transition in a cubic-quartic Landau potential; the symmetry restriction on the expansion is therefore part of the prediction.

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.

Assume the scalar order parameter can take either sign. The cubic term breaks $m\mapsto -m$ symmetry, and a positive quartic coefficient bounds the Landau free energy below. Stationary points satisfyso, besides the disordered point $m=0$, the possible ordered points areThe disordered point has curvature $A^{\prime \prime }(0)=A_2$ and is locally stable when $A_2>0$. Nonzero stationary points first appear whenThat condition alone does not make them the equilibrium phase: one must compare their free energies.

At a nonzero stationary point, substitute $A_2=-A_{3}m-A_4{m}^2$ to obtainFor an ordered state to coexist with $A(0)=0$, this requiresBoth curvatures at phase coexistence are positive and equal to $2A_3^2/(9A_4)$. The barrier is at $m=-A_3/(3A_4)$. In fact the potential at phase coexistence factorizes:The two distinct minima are therefore explicit, with a finite jump in the order parameter whenever $A_3\ne 0$.

To state the equilibrium inequalities for either sign of $A_3$, put $m=-\mathrm{sgn}(A_3)\rho$ with $\rho \ge 0$. This sign always has lower energy than the opposite sign of equal magnitude. ThenThe bracket's minimum is $A_2/2-A_3^2/(9A_4)$. It follows thatThe favored ordered minimum, when it exists, isThus a first-order phase transition is possible and occurs when the coefficients cross the coexistence relation. There is no continuous transition between $m=0$ and the equilibrium ordered state while $A_3$ remains nonzero and $A_4$ remains finite and positive. The ordered state becomes globally favorable while $A_2$ is still positive, before the disordered curvature can vanish. Although a stationary solution tends to zero at $A_2=0$, it is not the global transition branch. A continuous transition would require eliminating the cubic term, outside the stated $A_3\ne 0$ condition.

The spinodal points clarify the metastable region: for $2A_3^2/(9A_4)<A_2<A_3^2/(4A_4)$ the ordered minimum is metastable; for $0<A_2<2A_3^2/(9A_4)$ the disordered minimum is metastable. At $A_2=0$ the disordered state loses local stability. These local-stability limits should not be mistaken for the first-order transition in a cubic-quartic Landau potential.

Discuss the zero-field phase boundaries, taking $h=0$ and writing $r={\mathcal{A}}_2$, $q={\mathcal{A}}_4$, $s={\mathcal{A}}_6>0$. This qualification matters because a conjugate field generally rounds the zero-field continuous transition. Nonzero stationary points satisfy $r+q{m}^2+s{m}^4=0$.

For $q>0$, the continuous phase transition occurs atOn the $r>0$ side the only stable phase is $m=0$; on the $r<0$ side the equilibrium global minima are $\pm m_0$ with $m_0^2\sim -r/q$. The magnetization therefore grows continuously from zero as $r$ changes sign.

For $q<0$, the first-order phase transition instead occurs while $r$ is positive. At phase coexistence, $m=0$ and $m=\pm m_0$ are equally deep global minimum points. With $a=m_0^2>0$, stationarity and equality of free energies giveEliminating $r$ gives $-q{a}^2/4-s{a}^3/3=0$, henceAt this coexistence value, the sextic even Landau potential factors as $f(m)=s{m}^2(m^2+3q/(4s){)}^2/6\ge 0$. This verifies that the three equal stationary values are global minima. Above this coexistence value of $r$ the disordered phase wins; below it the ordered phase wins, with a finite jump in the phase-selected magnetization. The values $r=0$ and $r=q^2/(4s)$ are the disordered and ordered spinodal points, respectively, not the coexistence boundary: the latter follows from the repeated root of $r+qa+s{a}^2=0$.

The two transition loci meet at the tricritical point,Two independent control parameters such as $T$ and $g$ can tune these two conditions. In a generic local phase diagram their coefficient map has a Jacobian matrix of rank two at the intersection; unspecified functions ${\mathcal{A}}_2(T,g)$ and ${\mathcal{A}}_4(T,g)$ do not by themselves determine the geometric orientation of the boundaries. The sextic term stabilizes this intersection and the negative-quartic first-order side.

Within this analytic quartic Landau theory, equilibrium of a nonconserved scalar order parameter requires $f^{\prime }=0$. At a continuous phase transition, its restoring curvature vanishes, so $f^{\prime \prime }=0$. A nonzero $f^{\prime \prime \prime }$ would leave a cubic leading term in the Taylor expansion about that state and prevent it from being a local minimum. Thus a stable critical endpoint of a quartic Landau free energy requires all three derivatives to vanish, with $f^{\prime \prime \prime \prime }>0$. The condition $f^{\prime \prime }=0$ alone can instead identify a spinodal point.

For $f=\bar{b}{\lambda }^4+\bar{c}{\lambda }^3+\bar{a}{\lambda }^2+\bar{d}\lambda$, $\bar{b}>0$, the successive equations areThey giveAt these values, the free energy is exactly $f({\lambda }_c)+\bar{b}(\lambda -{\lambda }_c{)}^4$, proving stability. With no electric field, $\bar{d}=0$, so necessarily $\bar{c}=0$, followed by ${\lambda }_c={\bar{a}}_c=0$.

For $\bar{c}<0$ and $\chi >0$, the electric-field coupling to nematic order supplies the required negative ${\bar{d}}_c$ atTuning the quadratic coefficient to ${\bar{a}}_c$ simultaneously reaches the field-induced critical endpoint of the isotropic-nematic transition. The PDF's $\tilde{c}$ in the last clause is undefined; it is interpreted here as the previously defined $\bar{c}$. The endpoint is a prediction of the stable quartic, small-field mean-field approximation; higher-order terms or a field beyond the expansion's validity can shift it.