Past exam of the theoretical physics I course of the University of Cambridge / 2013 / Paper 1 / 6 / d / Solution

Assume $b>0$ and temperature-independent $d$ for the local model. The zero-field stationary points obeyFor $d\ne 0$, 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 $F^{\prime }(m_{\ast })=0$ and $F(m_{\ast })=F(0)$. Substituting $a=-d{m}_{\ast }-b{m}_{\ast }^2$ into the free energy givesHenceAt coexistence the entire potential factorizes:It has equal minima at zero and $m_{\ast }$ separated by a barrier. 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 $d^2/(4b)>a>a_{\ast }$, the ordered minimum is metastable and zero is globally stable. For $a_{\ast }>a>0$, 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 $a_{\ast }$. For $a<0$, the opposite-sign nonzero state can also be a local minimum, but the cubic term favours the state of sign $-d$.

If $a=a_0(T-T_c)/T_c$, the transition is shifted to $T_{\ast }=T_c[1+2d^2/(9b{a}_0)]$. With $b,d$ constant and $a^{\prime }(T)>0$, the entropy jump from the disordered to ordered phase is $-a^{\prime }(T_{\ast })m_{\ast }^2/2$, giving a latent heat of magnitude $T_{\ast }{a}^{\prime }(T_{\ast })m_{\ast }^2/2$. As $d\to 0$, 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.