Equipartition and correlations This question references C.L.Henley's paper on arxiv. Page 3, section B: Effective free energy and correlations.
There is an ice polarization field $\mathbf{P}(\mathbf{r})$, that has been coarse grained such that it is smoothly varying across the crystal.
The author argues the total free energy (completely entropic in origin) as a function of coarse-grained $\mathbf{P}(\mathbf{r})$:
$$
F_{\mathrm{tot}} ( \mathbf{P}(\mathbf{r})) = \frac{T\kappa}{2v_\mathrm{cell}} \int d^3\mathbf{r} |\mathbf{P}(\mathbf{r})|^2  \tag{2.2}
$$
where $\kappa$ is some constant (stiffness), $v_\mathrm{cell}$ is the volume of unit cell, $T$ is temperature.
The Fourier transform of (2.2) gives 
$$
\tilde{F}_{\mathrm{tot}} = \sum_{\mathbf{k}} \frac{T\kappa}{2} |\mathbf{P}(\mathbf{k})|^2
$$
and it might also be relevant to note the probability distribution of the polarization field is a Gaussian distribution (ignoring the ice-rule constraints as in the paper for now):
$$
\text{Prob}( \{ \mathbf{P}(\mathbf{r}) \} ) \propto \exp(- F_{\mathrm{tot}} ( \mathbf{P}(\mathbf{r}))/T) \tag{2.4}
$$
The part that confuses me is:
In the following statement (right after Eq 2.4 in the paper), the author writes: 

..., so a naive use of equipartition would give
  $$
\left\langle P_\mu (-\mathbf{k})  P_\nu (\mathbf{k}) \right\rangle = \delta_{\mu\nu}/\kappa 
$$

but I'm not sure how equipartition comes into play to give you correlations, nor the logic jump that the author seems to make here.
 A: The equipartition of energy is familiar as a result in classical statistical mechanics. It is usually applied to the case where a coordinate or momentum appears in the system Hamiltonian as a squared term, such as a kinetic energy component $p_x^2/2m$ or a harmonic term $\frac{1}{2}\kappa x^2$. Then, in calculating the canonical ensemble average at temperature $T$, involving the Boltzmann distribution, we can use the properties of Gaussian integrals to show
$$
\left\langle \tfrac{1}{2}\kappa x^2 \right\rangle
= \frac{\int dx  \, \tfrac{1}{2}\kappa x^2\, \exp(-\tfrac{1}{2}\kappa x^2/k_BT)}{\int dx  \,  \exp(-\tfrac{1}{2}\kappa x^2/k_BT)}
= \frac{1}{2}k_B T
$$
where $k_B$ is Boltzmann's constant. Hence we get an expression for the mean-square fluctuations
$$
\left\langle x^2 \right\rangle = \frac{k_BT}{\kappa}
$$
The same trick is often applied to coarse-grained free energies, $\mathcal{F}$, when they can be expressed as sums of quadratic terms in some generalized coordinate, or as a functional of a displacement field (squared, and integrated over space) or a similar integral involving the squared gradient of a field. Often the expression is only approximate, perhaps valid at long wavelengths. Examples include the fluctuations of interfaces (capillary waves) and director fluctuations in liquid crystals: the book Principles of condensed matter physics by PM Chaikin and TC Lubensky contains several of these examples. All that is required is an argument that the variables, or coordinates, are distributed according to a Boltzmann-like expression $\exp(-\mathcal{F}(\{Q_i\})/k_BT)$ where $\{Q_i\}$ is the set of coarse-grained coordinates, and that some of the coordinates of interest appear simply as quadratic terms in $\mathcal{F}$. The algebra then proceeds exactly as in the classical equipartition case.
In this case (and in many others) this becomes clear after taking a Fourier transform. There is a set of independent quadratic terms in the free energy, each term taking the form $\frac{1}{2}\kappa T |P_\alpha(\mathbf{k})|^2$, for each vector component $\alpha$, and for each wave-vector $\mathbf{k}$. A factor of $T$ has been included in the multiplying factor $\kappa T$, and the units have been chosen to make $k_B=1$, so the factor $k_BT$ will disappear from the final result. The equipartition formula then becomes
$$
\left\langle |P_\alpha(\mathbf{k})|^2 \right\rangle = 
\left\langle P_\alpha(\mathbf{k})P_\alpha(-\mathbf{k}) \right\rangle = \frac{1}{\kappa}
$$
and the fact that each component appears separately in the free energy means that any cross terms will vanish (hence the Kronecker delta).
