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

The scaling hypothesis for critical phenomena says that the singular part of the equilibrium free energy is a generalized homogeneous function of its thermal and field-like controls. Smooth backgrounds must first be removed; the hypothesis does not claim that the entire free energy, including arbitrary regular terms, has a pure scaling form.

For the ordinary scalar quartic LG theory, set $r=r_{t}t$ with $r_t>0$ and $u>0$, and rescale the uniform order parameter asThe order-parameter-dependent free-energy density becomesDefine $f_{\pm }(H)$ as the minimum of the braces, with the sign of the quadratic term respectively positive or negative. ConsequentlyFor an extensive $A_s$, $a$ additionally contains the system volume. The printed double-inequality subscript labels the two temperature branches: the upper-temperature function is $f_{+}$ and the lower-temperature function is $f_{-}$. It does not classify positive and negative magnetic fields. In particular $f_{+}(0)=0$ and $f_{-}(0)=-1/4$; below the transition the field dependence has a cusp at zero, so derivatives are taken on a selected branch.

For the following derivatives, take $A_s$ to be a density, so $m$ is the magnetization density; for total free energy the derivative gives total magnetization instead. Differentiate this mean-field scalar free-energy scaling form. The magnetization is $m=-\partial A_s/\partial h=-ab|t{|}^{1/2}{f}_{\pm }^{\prime }(H)$, giving $m_0\propto (-t{)}^{1/2}$ and $\beta =1/2$. The magnetic susceptibility is $\chi =-{\partial }_h^2{A}_s=-a{b}^2|t{|}^{-1}{f}_{\pm }^{\prime \prime }(H)$, giving $\gamma =1$. At zero field the nonzero curvature amplitudes are $-f_{+}^{\prime \prime }(0)=1$ and $-f_{-}^{\prime \prime }(0^{+})=1/2$.

To allow nonclassical critical exponents, replace the fixed powers byThe heat-capacity critical exponent is defined by $C_{V,s}\sim |t{|}^{-\alpha }$; temperature differentiation gives the thermal exponent $2-\alpha$ in $A_s$. The order-parameter critical exponent has $m_0\sim (-t{)}^{\beta }$, and the magnetic-susceptibility critical exponent has $\chi \sim |t{|}^{-\gamma }$. Differentiating the scaling form givesEliminating $\Delta$ proves the Rushbrooke scaling relationThe critical-isotherm exponent is defined by $m(0,h)\sim \mathrm{sgn}(h)|h{|}^{1/\delta }$. At fixed small $h$, the $t\to 0$ limit requires $f_{\pm }(H)\sim c|H{|}^{(2-\alpha )/\Delta }$, so that the temperature factors cancel. Therefore $1/\delta =(2-\alpha -\Delta )/\Delta =\beta /\Delta$, yielding $\Delta =\beta \delta$. But the two differentiated identities also give $\Delta =\beta +\gamma$, and hence the Widom scaling relationThese are relations among the leading power indices. At marginal dimensions, multiplicative logarithms can accompany them, and an additive analytic background must not be mistaken for the singular scaling contribution.