= Solution
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 $V(M)=rM^2/2+uM^4/4$ gives $M=0$ when $r\geq0$ 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> require
$$
r+uy+vy^2=0,\qquad \frac r2y+\frac u4y^2+\frac v6y^3=0.
$$
Eliminating $r$ gives $-uy^2/4-vy^3/3=0$, hence
$$
\boxed{M^2=-\frac{3u}{4v},\qquad r_{\rm coex}=\frac{3u^2}{16v}.}
$$
At coexistence $V(M)=vM^2(M^2+3u/(4v))^2/6$, proving that all three degenerate minima are global. The equilibrium <order parameter> jumps from zero to one of the two nonzero values even though $r_{\rm coex}>0$. 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-51-landau-potentials.png]
{title=Quartic Landau minima through a continuous transition and three degenerate sextic minima at first-order coexistence}
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