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

The expansion is a local phenomenological description in a small scalar order parameter near a transition. Its coefficients summarize microscopic physics and generally depend smoothly on temperature. At zero field, spin-reversal Z2 symmetry requires even powers of magnetization; the conjugate field supplies the term $-hm$. The additive $F_0$ does not affect minimization with respect to $m$.

For the ordinary continuous-transition quartic model, $b>0$ ensures stability, and the quadratic coefficient changes sign, $a=a_{0}t+O(t^2)$ with $a_0>0$. If $b$ vanishes or is negative, stabilizing higher powers are required and the phase structure can change. Treating one uniform magnetization neglects spatial gradients, fluctuations and domains; the resulting exponents are predictions of mean-field theory.

Equilibrium requires $h=am+b{m}^3$. On the critical isotherm $a=0$, soHere $\delta$ is defined by $h\propto \mathrm{sgn}(m)|m{|}^{\delta }$ at the critical temperature, or equivalently $|m|\propto |h{|}^{1/\delta }$. The positive quartic coefficient selects a stable equilibrium, including the flat but stable zero-field thermodynamic critical point.

Differentiate the equation of state at fixed temperature:Above the transition, zero-field equilibrium has $m=0$. Below it the stable pure phases have $m^2=-a/b$. ThusThis is the mean-field magnetic-susceptibility amplitude ratio. Below the transition, take a one-sided response in a selected ordered phase: the globally minimizing magnetization switches discontinuously between opposite phases when $h$ crosses zero. Its two-sided derivative at precisely zero field is therefore not the finite pure-phase magnetic susceptibility used in the ratio.

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.