 Solution

Keeping both gradient coefficients fixed requires $z^2=B{b}^2=B{c}^4$. HenceEquivalently the real-space field normalization is $z_x=z/B=b^{(3-d)/4}$. The transformed thermal and source couplings areThis is the Gaussian fixed point at $t=h=0$, with anisotropic effective volume exponent $D_{\mathrm{eff}}=(d+1)/2$. The stiffnesses remain positive parameters labeling the normalization of coordinates and fields. A strictly positive $t$ is a relevant perturbation away from the renormalization-group fixed point, not itself a critical renormalization-group fixed point.

The two correlation lengths also follow directly by balancing the kernel: $K{q}_{\parallel }^2\sim t$ gives ${\xi }_{\parallel }\sim (K/t{)}^{1/2}$, while $L{q}_{\perp }^4\sim t$ gives ${\xi }_{\perp }\sim (L/t{)}^{1/4}$. Their different exponents are consistent with the required anisotropic scaling. An omitted scalar quartic interaction has scaling eigenvalue $4-D_{\mathrm{eff}}=(7-d)/2$, so the quadratic renormalization-group fixed point alone does not establish the interacting smectic's critical behavior below physical dimension seven.