Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 46 / 2 / Solution

Use ${\beta }_{\mathrm{th}}=1/(k_{B}T)$ for inverse temperature to distinguish it from the order-parameter exponent. The decaying two-point correlation function needed for the response identity is connected:Away from the critical point in a short-range Ising model pure phase, the correlation length controls the exponential long-distance falloff, $G(r)\sim A(r)e^{-|r|/\xi }$ with a possible algebraic prefactor. The full two-spin expectation below the transition instead tends to $M^2$. The spin-mixture contribution to zero-field susceptibility explains why a symmetric mixture is unsuitable for defining a finite connected correlation length in the ordered phase.

Let $F=-{\beta }_{\mathrm{th}}^{-1}\log Z$ be the total Helmholtz free energy. Take magnetization and susceptibility per site, soDifferentiating the Boltzmann weights at fixed temperature givesTranslation invariance then yields the correlation-function susceptibility sum ruleTotal magnetization and total susceptibility multiply these expressions by $N$; per physical volume they have an additional factor $a^{-D}$ relative to the per-site convention. These normalizations do not change the critical powers.

For the subsequent scaling requests, let $F_s$ denote singular free energy per site; the fixed extensive normalization is suppressed. The labels $+$ and $-$ denote $t>0$ and $t<0$, with distinct scaling functions and amplitudes on the two sides. Below the transition a one-sided derivative at $h=0$ selects a spontaneous-magnetization branch. Writing $x=h/|t{|}^{\Delta }$, differentiation of the scaling hypothesis for critical phenomena givesThe field-independent $I_{\pm }$ makes no contribution to these field derivatives, although it can change heat-capacity amplitudes. Assuming nonzero leading amplitudes and the indicated pure-phase limits, comparison with the exponent definitions givesThis is the Rushbrooke scaling relation. A temperature-independent critical isotherm requires the large-$|x|$ magnetization scaling function to grow as $\mathrm{sgn}(x)|x{|}^{\beta /\Delta }$: otherwise $M_s=|t{|}^{\beta }{\mathcal{M}}_{\pm }(x)$ would retain a temperature power as $t\to 0$ at fixed small $h$. Therefore $1/\delta =\beta /\Delta$. Since the preceding two equalities also give $\Delta =\beta +\gamma$, the Widom scaling relation follows:For the hyperscaling relation, additionally assume that one correlation volume supplies an order-one singular free-energy contribution and that there is a single isotropic correlation length, with no dangerously irrelevant coupling spoiling this argument. Then $F_s\sim {\xi }^{-D}\sim |t{|}^{D\nu }$. Two temperature derivatives give the singular heat capacity, soHyperscaling is an additional physical assumption, not a consequence of the two-variable equation alone, and generally fails above the upper critical dimension.

Finally the correlation scaling function $f_G(x)$ falls exponentially, up to algebraic factors, for $x\to \infty$ in a massive short-range pure phase. At $x\to 0$ it approaches a finite nonzero value in the scaling regime. Approximating the lattice sum at distances much larger than $a$, and denoting the area of the unit sphere by $S_{D-1}$, givesIf $\eta <2$ and the integral has a finite nonzero limiting value, microscopic distances contribute only a regular background and the leading divergence is ${\xi }^{2-\eta }$. Thus the Fisher scaling relation isThese statements concern leading singular powers; logarithmic corrections and cancellations of amplitudes require separate treatment.