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The covariance of a centered multivariate Gaussian distribution is the inverse of its quadratic kernel. Writing $A(q)=a+\kappa q^2+\gamma q^4$, one findsConsequently the composition static structure factor is

For a damped ferromagnetic spin wave, shifting both poles into the lower half-plane preserves causality:Its spectral function isThe classical Fluctuation-dissipation theorem then givesHence the zero-frequency static structure factor iswhich has the stated proportionality after absorbing the normalization $4/s$.

Write $D={\rho }_s/s$, so ${\omega }_k=D{k}^2$, and take ${\Gamma }_k=\gamma |k|$. The small-wavenumber static structure factor behaves asThe spatial Fourier transform of $|k{|}^{-1}$ in $d$ dimensions scales as $|x{|}^{1-d}$. Thusand in three dimensions $C(x)\sim |x{|}^{-2}$. Since this two-point correlation function tends to zero rather than to a nonzero constant at large separation, it is incompatible with true long-range ferromagnetic order and hence with spontaneous symmetry breaking at this higher temperature.

Use the Fourier transform conventionThe zero mode vanishes because the compositional order parameter has zero spatial average. Parseval identity and the Fourier transform of a derivative turn the quadratic part of the dimensionless free energy intoHere ${\sum }_{\mathbf{q}}^{+}$ is the positive-wavevector sum for a real field. Each independent complex amplitude has density proportional to $e^{-G(q)|{\varphi }_{\mathbf{q}}{|}^2}$, so its elementary Gaussian integral gives the static structure factorwhenever $G(q)>0$.

The stationary points of the Brazovskii model kernel obeyBecause $\kappa <0<\gamma$, the nonzero minimum is the nonzero-wavevector soft-mode sphereAt this wavevector,The first Gaussian field theory divergence therefore occurs at

Write the optimized kernel asNear the soft-mode sphere, the supplied asymptotic result saysThus the self-consistency equationcannot reach $\bar{a}=a_c$ for any $g>0$: the fluctuation correction diverges first. The isotropic static structure factor therefore remains finite, and the isotropic state never undergoes the continuous Gaussian instability predicted in part (a).

The modulated minimum from part (b), however, has negative free energy for sufficiently low $a$ and amplitude $A_{\ast }=\pi (a_c-a)/(8g)$. For small positive $g$, its free energy must cross that of the still locally stable isotropic variational state. At the crossing the isotropic inverse susceptibility is positive and the smectic amplitude is nonzero. The order parameter consequently jumps, which is the Brazovskii fluctuation-induced first-order transition.