= Solution
Assume $b>0$ and temperature-independent $d$ for the local model. The zero-field <stationary points> obey
$$
m(a+dm+bm^2)=0,\qquad
m_\pm=\frac{-d\pm\sqrt{d^2-4ab}}{2b}.
$$
For $d\ne0$, the favoured ordered sign is opposite to $d$. The nonzero extrema first appear at $a=d^2/(4b)$; their existence is not yet a coexistence condition. To find the equilibrium transition, require both $F'(m_*)=0$ and $F(m_*)=F(0)$. Substituting $a=-dm_*-bm_*^2$ into the <free energy> gives
$$
F(m_*)-F_0=-\frac d6m_*^3-\frac b4m_*^4.
$$
Hence
$$
\boxed{m_*=-\frac{2d}{3b},\qquad a_* =\frac{2d^2}{9b}>0.}
$$
At coexistence the entire potential factorizes:
$$
F-F_0=\frac b4m^2\left(m+\frac{2d}{3b}\right)^2.
$$
It has equal minima at zero and $m_*$ separated by a barrier. \b[Every fixed nonzero cubic coefficient produces a first-order transition in this quartic model], with a finite <magnetization> jump before the quadratic coefficient reaches zero.
For $d^2/(4b)>a>a_*$, the ordered minimum is metastable and zero is globally stable. For $a_* >a>0$, the ordered state is globally stable and zero is metastable. At $a=0$ the zero minimum loses local stability. Thus the spinodals are $a=d^2/(4b)$ and $a=0$, distinct from $a_*$. For $a<0$, the opposite-sign nonzero state can also be a <local minimum>, but the cubic term favours the state of sign $-d$.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2013/paper-1-tp1-cubic-transition.png]
{title=First-order coexistence, metastability and the magnetization jump in a cubic-quartic Landau model}
If $a=a_0(T-T_c)/T_c$, the transition is shifted to $T_*=T_c[1+2d^2/(9ba_0)]$. With $b,d$ constant and $a'(T)>0$, the <entropy> jump from the disordered to ordered phase is $-a'(T_*)m_*^2/2$, giving a <latent heat> of magnitude $T_*a'(T_*)m_*^2/2$. As $d\to0$, the jump and shift vanish and the original continuous transition is recovered.
A genuine zero-field uniaxial <ferromagnet> with exact spin-reversal <Z2 symmetry> forbids this cubic term. The modified model therefore presumes that symmetry has been changed; it must not be presented as a symmetry-allowed correction for the original <ferromagnet>.
Back to article page
