Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 42 / 1 / iv / Solution

Critical exponents describe leading power-law singularities as the reduced temperature $t=(T-T_c)/T_c$ and the conjugate field tend to zero. Their amplitudes depend on microscopic details, while the exponents are characteristic of a universality class. Define $m_0\sim (-t{)}^{\beta }$ on a selected ordered branch, $\chi \sim |t{|}^{-\gamma }$ at zero field, $C_{V,s}\sim |t{|}^{-\alpha }$ for the singular heat capacity, and $m(t=0,h)\sim \mathrm{sgn}(h)|h{|}^{1/\delta }$. Also define $\xi \sim |t{|}^{-\nu }$ and $G(r,t=0)\sim r^{-(D-2+\eta )}$. In these statements $\beta$ is the order-parameter critical exponent, not inverse temperature.

Assume $r=r_{t}t$ with $r_t>0$ and a nonzero positive quartic coefficient $u$. Minimizing the quartic potential gives $m_0=\sqrt{-r/u}$, hence $\beta =1/2$. Differentiating its equation of state givesso $\gamma =1$ with different amplitudes. At $r=0$, $h=u{m}^3$, giving $\delta =3$. The minimized potential is zero above the transition and $-r^2/(4u)$ below, so its second temperature derivative has a finite jump: $\alpha =0$.

The quadratic fluctuation kernel around the stable uniform minimum is $\kappa p^2+V^{\prime \prime }(m_0)$, so ${\xi }^2=\kappa /V^{\prime \prime }(m_0)$. Since $V^{\prime \prime }$ is $r$ above and $-2r$ below, $\nu =1/2$. Its critical momentum dependence is $p^{-2}$, giving $\eta =0$. Thus the ordinary mean-field critical exponents areThese are predictions of the Landau approximation, whose validity is constrained by the Ginzburg criterion; they need not be the exponents of the fluctuating theory below its upper critical dimension.