= Solution
Use the <Fourier transform> convention
$$
\phi(\mathbf r)=V^{-1/2}\sum_{\mathbf q}\phi_{\mathbf q}e^{i\mathbf q\cdot\mathbf r},
\qquad
\phi_{-\mathbf q}=\phi_{\mathbf q}^*.
$$
The 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 into
$$
H_{\rm G}
=\frac12\sum_{\mathbf q}
\left(a+\kappa q^2+\gamma q^4\right)|\phi_{\mathbf q}|^2
=\sum_{\mathbf q}^{+}G(q)|\phi_{\mathbf q}|^2,
\qquad
G(q)=a+\kappa q^2+\gamma q^4.
$$
Here $\sum_{\mathbf q}^{+}$ is the <positive-wavevector sum for a real field>. Each independent complex amplitude has density proportional to $e^{-G(q)|\phi_{\mathbf q}|^2}$, so its elementary <Gaussian integral> gives the <static structure factor>
$$
S(q)=\langle|\phi_{\mathbf q}|^2\rangle=\frac1{G(q)}
$$
whenever $G(q)>0$.
The stationary points of the <Brazovskii model> kernel obey
$$
G'(q)=2q(\kappa+2\gamma q^2)=0.
$$
Because $\kappa<0<\gamma$, the nonzero minimum is the <nonzero-wavevector soft-mode sphere>
$$
q_0^2=-\frac{\kappa}{2\gamma}.
$$
At this <wavevector>,
$$
G(q_0)=a-\frac{\kappa^2}{4\gamma}.
$$
The first <Gaussian field theory> divergence therefore occurs at
$$
a_c=\frac{\kappa^2}{4\gamma}.
$$
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