Chaikin and Lubensky exercise: upper critical dimension of anisotropic Lifshitz potential Exercise from Chaikin and Lubensky: Chapter 5


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*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?
 A: 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.
