Director field fluctuations in a liquid crystal in a magnetic field
If we consider the <PERSON> elastic free energy in the equal constants limit (for more details, see here Chapter 4.1.) with an external magnetic field, we have
$$ F = \frac{1}{2}\int \text{d}^3 r\left ( K|\nabla \cdot n|^2 + K|\nabla\times n|^2 -\Delta\chi |n\cdot H|^2 \right), $$ where $K$ is the elastic constant, $n$ is the unit director field ($n\cdot n =1$), $\Delta \chi$ is the magnetic susceptibility anisotropy and $H$ denotes the magnetic field.
Suppose we take the ground state director field in de z-direction: $n_0 = (0,0,1)$. A fluctuation on his field will be of the form $\delta n \approx (\delta n_x, \delta n_y,0)$ for small $\delta n_x(r)$ and $\delta n_y(r)$. In the case of zero magnetic field, I understand the system. We introduce the Fourier transform $$ \delta n_{\alpha}(r) = \frac{1}{V}\sum_q \delta\tilde{n}{\alpha}(q)e^{iqr} $$ and its inverse $$ \delta \tilde{n}(q)=\int \text{d} r \delta n{\alpha}(r)e^{-iqr}, $$ to obtain
$$ F \approx \frac{K}{2V}\sum_q \left( \delta \tilde{n}^2_x(q) + \delta \tilde{n}^2_y(q) \right)q^2. $$ Now one can impose equipartition of free energy: every mode in the Fourier decomposition carries $k_B T/2$ of energy. This yields the correlators $$ \langle \delta \tilde{n}{\alpha}^2(q) \rangle = \frac{k_B T V}{Kq^2} \Rightarrow \langle \delta n{\alpha}(0)\delta n_{\alpha}(r) \rangle = \frac{k_B T}{4\pi K |r|}. 649
$$ However, I do not understand how this works in the case of a magnetic field. Suppose we take the initial magnetic field also in the z-direction. The free energy then becomes $$ F \approx \frac{K}{2}\sum_q \left( \delta \tilde{n}^2_x(q) + \delta \tilde{n}^2_y(q) \right)q^2 - \frac{1}{2}\Delta \chi V H_z^2. $$
I do not understand how I can invoke the equipartition theorem to find the correlation functions in this new situation. Can anyone help?
EDIT: the form of the fluctuations is derived as follows. Say we have $\delta n = (\delta n_x, \delta n_y, \delta n_z)$, then we must have $|n+\delta n|^2 = 1$, yielding a conditions $1+\delta n_z = \sqrt{1-\delta n_x^2 - \delta n_y^2} \approx 1-\frac{1}{2}(\delta n_x^2+\delta n_y^2) \approx 1$ for small $\delta n_x$ and $\delta n_y$. 649