Chaikin and Lubensky exercise: upper critical dimension of anisotropic Lifshitz potential

Exercise from Chaikin and Lubensky: Chapter 5

1. Calculate the upper critical dimension $$d_c$$ for the following critical points: b) The $$(d,m)$$ Lifshitz point described by the Hamiltonian

$$H = \int d^d x \left[ \frac{1}{2} r \phi^2 + \frac{1}{2} \sum_{i=1}^{m} (\nabla_i \phi)^2 + \frac{1}{2} \sum_{i=m+1}^{d} (\nabla_i^2 \phi)^2 + u \phi^4 \right]$$

Hint: It is useful to introduce correlation lengths $$\xi_{\parallel}\sim r^{-1/2}$$ and $$\xi_{\perp}\sim r^{-1/4}$$ to describe correlations in the two directions defined in the free energy. The correlation volume is then $$\xi_{\parallel}^m \xi_{\perp}^{d-m}$$

I confirmed the given correlation lengths by following section 4.3.1 (pg 154), but looking at two distinct Fourier-space correlation functions, $$\chi_{\parallel}$$ and $$\chi_{\perp}$$ : $$\chi_{\parallel}(\vec{q}) = \frac{1}{r + 12 u \langle \phi \rangle^2 + \frac{1}{2} \sum_{i=1}^{m}q_i^2 }$$, $$\chi_\perp(\vec{q}) = \frac{1}{r + 12 u \langle \phi \rangle^2 + \frac{1}{2} \sum_{j=m+1}^d q_j^4 }.$$ The given correlations lengths follow by defining $$\xi$$ such that $$\chi_{\parallel}(q_i)=\frac{\xi_{\parallel}}{1+(q_i\xi_{\parallel})^2}$$, $$\chi_{\perp}(q_j)=\frac{\xi_{\perp}}{1+(q_j\xi_{\perp})^4}$$

and inserting a value $$\langle \phi \rangle^2 =|r|/4u$$. I have changed notation to $$q_i$$ being those with indices between $$1$$ and $$m$$ and $$q_j$$ those with indices between $$m+1$$ and $$d$$.

Following section 5.1 (pg 214-5), next I will need to retrieve $$G(\vec{x},0) = T \chi (\vec{x},0)$$ , or at least its scaling $$\sim \xi^n$$. As I understand it we need $$G(\vec{x},0)$$ because we have an indication of the size of fluctuations and their relation to mean-field value $$\langle \delta \phi_{coh }\rangle = V^{-1} \int d^d x G(\vec{x},0) < \langle \phi \rangle^2$$ leading to a (dimension $$d$$-dependant) scaling of coherence length $$\xi$$ with temperature near the critical point.

For the isotropic Landau potential we had $$\chi(\vec{x},0) = \chi T \int \frac{d^d q}{(2\pi)^d} \frac{e^{i \vec{q}\vec{x}}}{1+ (q_i \xi_{\parallel})^2 }$$

(Eq 4.3.17 pg 153; the textbook actually has $$\frac{1}{(2\pi)^2}$$ but I think this is a typo.)

Whereas now we have $$\chi(\vec{x},0) = \chi T \int \frac{d^m q_i d^{d-m} q_j}{(2\pi)^d} \frac{e^{i \vec{q}\vec{x}}}{1+ (q_i \xi_{\parallel})^2 + (q_j \xi_{\perp})^4},$$ but I don't know how to integrate this. Conveniently we have $$\xi_\parallel^2=\xi_\perp^4$$; I don't know how to bring this into some simpler form further. I cannot sperate it into $$q_i^2$$, $$q_j^4$$ parts.

On the whole the integral looks like more work than I expect from this question and I haven't used the `correlation volume' hint, so I suspect I'm on the wrong track with going via this calculation of $$G(\vec{x},0)=T \chi (\vec{x},0)$$ .

How can I do this integral, or how else can I derive the relationship $$\xi_\perp \sim t^n, \xi_\parallel \sim t^n$$ differently?

I don't need to Fourier transform $$G(\vec{q}) \rightarrow G(\vec{x},0)$$ or $$\chi(\vec{q}) \rightarrow \chi(\vec{x},0)$$ by evaluating the integral.

In section 5.2.5 "Breakdown of-mean field theory revisited" I find an easier way to see the dimensional dependence of correlation length(s), given $$\xi \sim r^{-v}$$.

We estimate the energy of non-uniform contributions to the Hamiltonian as

$$\frac{1}{2} c \int d^d x (\nabla \phi)^2$$,

As far as I can tell estimating $$(\nabla \phi)^2 \approx \langle \phi \rangle^2 / \xi^2$$ and integrating over a coherence volume, $$\int_0^{\xi_i} \Pi dx_i$$.

In the case of the question we have $$\xi_\parallel= (\frac{c}{|r|})^{1/2}$$ in $$m$$ dimensions, $$\xi_\perp= (\frac{c}{|r|})^{1/4}$$ for the remaining $$d-m$$ directions, and $$\langle \phi \rangle^2 =\frac{|r|}{4u}$$, so that this gradient energy is $$\frac{1}{2} c \int_0^{\xi_\parallel} \int_0^{\xi_\perp} d^{d-m} x d^m x \frac{\langle \phi\rangle^2}{\xi_{\parallel}^2} \sim \frac{1}{2} c \xi_\parallel^{m-2} \xi_\perp^{(d-m)} \langle \phi \rangle^2$$ $$=\frac{c^2}{8u} \xi_\parallel^{m-4} \xi_\perp^{(d-m)}$$

I have chosen as estimate of energy of fluctuations $$q_\parallel^2 \langle \phi \rangle =\frac{1}{\xi_{\parallel}^2} \langle \phi \rangle$$ but could have written $$q_\perp^4 \langle \phi \rangle =\frac{1}{\xi_{\perp}^4} \langle \phi \rangle$$; they are the same value with $$\xi_\parallel^2 = \xi_\perp^4$$. I next insert $$\xi_\parallel=\xi_\perp^2$$ and finally obtain $$F_{nu}\sim \xi_\parallel^{1/2(d+m)-4}$$ The exponent must be positive for validity of mean field theory, $$d+m=8$$ seems to be the upper critical dimension, even though it is really two criteria.