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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.

Use the usual homogeneous, local, analytic Landau-Ginzburg theory; rotational symmetry and Z2 symmetry alone do not specify locality, analyticity, or the signs of the coefficients. The local derivative expansion contains every rotational scalar with an even total number of fields:The omitted terms include all allowed higher field powers and contracted spatial derivatives. Terms differing by a total derivative are equivalent for bulk behavior with suitable boundary conditions. The displayed gradient energy assumes spatial homogeneity. If only rotations about an origin are imposed, rotationally invariant position-dependent coefficients are also possible.

In the mean-field approximation, take a spatially uniform order parameter and minimize its Landau free energy. For $\gamma >0$, $u>0$ and $r=a(T-T_c)$ with $a>0$, keeping the quadratic and quartic terms givesThe single minimum splits into two minima as the temperature passes below the critical temperature. Choosing one minimum gives spontaneous symmetry breaking of the Z2 symmetry and a continuous phase transition. The order parameter is the expectation in a selected ordered state; averaging equally over both states gives zero. A Z2 symmetry does not guarantee a continuous phase transition: a negative quartic coefficient stabilized by a positive sextic coefficient can instead give a first-order phase transition.

Relaxing the Z2 symmetry permits odd terms. A nonzero linear source generally rounds this ordinary continuous phase transition. A cubic term can instead give a first-order phase transition: for $V(\varphi )=r{\varphi }^2/2+w{\varphi }^3/3+u{\varphi }^4/4$, $u>0$ and $w\ne 0$, equal minima occur atThe order parameter jumps from zero to ${\varphi }_{\ast }$. A continuous phase transition without an exact Z2 symmetry is still possible with additional tuning; odd terms do not logically exclude it.

For the multicritical even Landau potential, set $r={\mu }^2=a(T-T_c)$ and assume an integer $n\ge 2$, ${\lambda }_n>0$ and $\gamma >0$. For $n>2$, all lower even interactions must be absent or tuned to zero to obtain these mean-field critical exponents. The stationary equation isIts nonzero solutions are minima because $V^{\prime \prime }({\varphi }_0)=-(2n-2)r>0$. Thus the order-parameter critical exponent isSubstituting into the minimum free energy density givesThe singular heat capacity per unit volume is $c_s=-T{\partial }_T^2{f}_{\min }$. Smooth variation of the other coefficients does not change the leading power. ConsequentlyFor $n=2$, this means a finite heat capacity jump, with ${\alpha }_{\mathrm{MF}}=0$; for $n=3$, the tricritical point has ${\beta }_{\mathrm{MF}}=1/4$ and ${\alpha }_{\mathrm{MF}}=1/2$. The case $n=1$ would contain only a quadratic potential and would not stabilize an ordered phase.

The Gaussian fluctuation correction near a critical point is more singular than the mean-field approximation contribution whenThis is the upper critical dimension of an even scalar interaction. Equivalently, its coupling has engineering dimension $2n-(n-1)d$, positive below $d_c$. Dominant fluctuations signal breakdown of the mean-field approximation; the Gaussian expression is not then a calculation of the exact interacting heat-capacity critical exponent. At $d=d_c$, equality of the powers allows logarithmic corrections.

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.

The nematic order parameter is a symmetric second-rank tensor and a traceless second-rank tensor describing orientational anisotropy. For molecular unit axes $\mathbf{s}$, $Q_{ij}$ is proportional to $⟨s_i{s}_j-{\delta }_{ij}/3⟩$. It vanishes in an isotropic phase and is unchanged by head-tail reversal $\mathbf{s}\mapsto -\mathbf{s}$, unlike a polar order parameter. For uniaxial nematic order, the nematic director is thus an unoriented axis.

One convention for the Landau-de Gennes free energy through fourth order, respecting rotational symmetry, isFor a traceless second-rank tensor in three dimensions, the Cayley-Hamilton theorem gives $\mathrm{Tr}{Q}^4=(\mathrm{Tr}{Q}^2{)}^2/2$, so there is only one independent quartic rotational invariant of a symmetric traceless tensor. The only possible linear scalar is $\mathrm{Tr}Q=0$; without an external anisotropy no linear term survives.

The cubic rotational invariant of a symmetric traceless tensor distinguishes prolate and oblate forms of uniaxial nematic order. Head-tail reversal leaves $Q$ unchanged and does not impose $Q\mapsto -Q$. Thus a cubic term is allowed in three dimensions and generically produces a first-order phase transition, rather than a symmetry-enforced continuous onset. In two dimensions a symmetric traceless second-rank tensor has eigenvalues $s,-s$, hence $\mathrm{Tr}{Q}^3=0$: the cubic invariant vanishes identically in two dimensions. This is a statement about the Landau free energy; fluctuations in two dimensions require a separate treatment.

For the stated uniaxial nematic order normalization, the nematic director is an eigenvector of $Q$ with eigenvalue $\lambda$, and the two perpendicular eigenvalues are $-\lambda /2$. ThereforeUsing precisely the Landau-de Gennes free energy convention in the preceding solution givesIf the quartic invariant were instead normalized as $b\mathrm{Tr}{Q}^4/4$, its coefficient would be $\bar{b}=9b/32$; the physical predictions are unchanged after redefining $b$.

For a nonzero global minimizer of the free energy, compare opposite values of the scalar nematic order parameter:The lower one has $\bar{c}\lambda <0$, soThe claim concerns the stable ordered phase, not every metastable stationary point. For $\bar{c}=0$ the two signs are degenerate. For $\bar{c}\ne 0$, equality of the ordered and isotropic free energies, together with stationarity, gives ${\lambda }_t=-\bar{c}/(2\bar{b})$ and ${\bar{a}}_t={\bar{c}}^2/(4\bar{b})>0$. The finite jump ${\lambda }_t$ exhibits the first-order phase transition driven by the cubic invariant.

In the usual three-dimensional isotropic setting, the transition is expected to be discontinuous, by the Brazovskii fluctuation-induced first-order transition mechanism.

If $\bar{\kappa }<0$, write $q_{\ast }^2=-\bar{\kappa }/(2\gamma )$ and $r=\bar{a}-{\bar{\kappa }}^2/(4\gamma )$. Then $J(q)=r+\gamma (q^2-q_{\ast }^2{)}^2$. A continuous finite-wavevector transition would require $r\to 0$ with $q_{\ast }>0$. But the soft-mode shell integral behaves asThe positive fluctuation corrections in the self-consistency equations diverge in that limit. In particular $\bar{\kappa }=\kappa +2B{I}_0$ cannot stay at a finite negative value while $I_0$ diverges. Thus the isotropic disordered state cannot lose stability through a zero-mass mode on a fixed nonzero shell in this approximation.

Ordering is instead expected through a free-energy crossing with a finite-amplitude smectic phase, giving a first-order phase transition. This is a reasoned fluctuation prediction for the finite-wavevector transition; the disordered Gaussian variational approximation alone does not locate the coexistence point or calculate the amplitude jump. A zero-wavevector instability or a lower-dimensional setting would require a separate analysis.

Landau theory represents the appropriate thermodynamic potential as an analytic function of a small order parameter, retaining every low-order term allowed by the system's symmetry. At fixed control parameters, the equilibrium phase is the global minimum of the Landau free energy; local minima describe metastable phases. A continuous phase transition occurs when a minimum changes continuously and loses stability, whereas a first-order phase transition occurs when distinct minima exchange global stability.

Forthe stationary points satisfyIf $\epsilon >0$, $\varphi =0$ is the unique minimum. If $\epsilon <0$, $\varphi =0$ is a maximum and the two degenerate minima areThus the reflection symmetry $\varphi \mapsto -\varphi$ is spontaneously broken below $\epsilon =0$, and

At the first-order phase transition, the central minimum $m=0$ and two nonzero minima $m=\pm m_0$ are degenerate, with barriers between them. Writing $x=m_0^2>0$, stationarity and equal free energies giveTheir difference gives $a_{4}x+2a_6{x}^2=0$, soThe magnitude of the magnetization therefore jumps from zero toThe two possible signs are related by the model's spin-reversal symmetry.

A first-order phase transition occurs where two phases have equal Gibbs free energy but a first derivative of that free energy is discontinuous. Typical jumps occur inA nonzero entropy jump gives a nonzero latent heat $T\Delta S$.

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 every $T<T_c$, the two zero-field minima $m_{+}$ and $m_{-}$ coexist. A positive field selects $m_{+}$ and a negative field selects $m_{-}$, so crossing $B=0$ makes the equilibrium magnetization jump between nonzero values. Hence $B=0$, $T<T_c$, is a line of first-order phase transitions, ending at the continuous critical point $(T_c,0)$.

Write $r_a={\mu }_{0,a}^2$. The stationary equations forareFor $r_1,r_2>0$, the disordered minimum is $(0,0)$. If $r_1<0$ and $r_1<r_2$, the first component orders with ${\varphi }_1^2=-r_1/(4g_0)$ and ${\varphi }_2=0$. If $r_2<0$ and $r_2<r_1$, the second component orders analogously.

The positive $r_2$ half of $r_1=0$ and the positive $r_1$ half of $r_2=0$ are continuous-transition lines. Along the negative diagonal $r_1=r_2<0$, the potential has an enhanced $O(2)$ symmetry and a circle of minima. Crossing that diagonal exchanges the two ordered axes and makes derivatives of the minimum free energy jump, so it is a first-order phase transition line. The origin, where the two continuous lines meet the first-order line, is a bicritical point.