 Anisotropic Gaussian smectic scaling

For a Gaussian kernel $K{q}_{\parallel }^2+L{q}_{\perp }^4+t$, keeping both gradient coefficients fixed gives $q_{\parallel }^{\prime }=b{q}_{\parallel }$ and $q_{\perp }^{\prime }=b^{1/2}{q}_{\perp }$. The volume rescales by $b^{(d+1)/2}$. Fourier-field normalization is $m^{\prime }(q^{\prime })=m(q)/b^{(d+5)/4}$; the real-space normalization instead divides by $b^{(3-d)/4}$. The thermal coefficient scales by $b^2$ and the uniform source by $b^{(d+5)/4}$. Thus ${\xi }_{\parallel }\sim t^{-1/2}$, ${\xi }_{\perp }\sim t^{-1/4}$ and the formal Gaussian free-energy powers are $\alpha =(7-d)/4$, $\Delta =(d+5)/8$.