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 $$ \tilde{F}_{\mathrm{tot}} = \sum_{\mathbf{k}} \frac{\kappa}{2} |\mathbf{P}(\mathbf{k})|^2 $$

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 $$ \langle P_\mu (-\mathbf{k}) P_\nu (\mathbf{k}) \rangle = \delta_{\mu\nu}/\kappa $$

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.

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.

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 $$ \tilde{F}_{\mathrm{tot}} = \sum_{\mathbf{k}} \frac{\kappa}{2} |\mathbf{P}(\mathbf{k})|^2 $$

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, the author writes:

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

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.

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 $$ \tilde{F}_{\mathrm{tot}} = \sum_{\mathbf{k}} \frac{\kappa}{2} |\mathbf{P}(\mathbf{k})|^2 $$

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 $$ \langle P_\mu (-\mathbf{k}) P_\nu (\mathbf{k}) \rangle = \delta_{\mu\nu}/\kappa $$

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.