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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.

In a smooth scalar Landau free energy, a line of coexistence between two distinct minima can end by those minima and the intervening maximum coalescing. At a generic endpoint with no competing third phase,The first condition is stationarity, the second removes the inverse susceptibility, and the third is needed because nonzero cubic curvature at a zero-quadratic point would prevent it being a stable minimum. After shifting $M=M_c+\psi$ and mixing two control parameters, the local potential is $V_c+R{\psi }^2/2+U{\psi }^4/4-H\psi +\cdots$, $U>0$. Its two-phase coexistence line $H=0,R<0$ reaches $R=H=0$, with $\Delta M=2\sqrt{-R/U}\to 0$ and diverging magnetic susceptibility. This is the familiar critical endpoint of a quartic Landau free energy.

The word “must” needs the two-phase endpoint assumptions. A first-order line can instead end at a triple point or a boundary of the available control parameters. There is an explicit scalar counterexample in the sextic potential: fix $u<0<v$ and vary $(r,h)$. The $+M/-M$ coexistence line at $h=0$, $r<3u^2/(16v)$ ends when the $M=0$ phase also becomes globally stable. At that triple point the nonzero minima remain $\pm \sqrt{-3u/(4v)}$; their jump does not vanish. Thus the literal unrestricted assertion is false. The continuous-endpoint argument above establishes the intended case and distinguishes it from this counterexample.

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.

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.