Electrostatics of Microscopic Media

${\overrightarrow{p}}_{mol}={\epsilon }_0{\gamma }_{mol}\overrightarrow{E}$. C.f. $\overrightarrow{P}={\epsilon }_0{\chi }_e\overrightarrow{E}$ for macroscopic media.

Consider a polar molecule in an electric field. The polar molecule tries to align along the field and experiences a torque. $⟨\overrightarrow{p}⟩$ becomes more along the $z$-axis, assuming $\overrightarrow{E}$ is along the $z$-axis, over time due to the torque felt by the polar molecules.

$p_i\propto \exp (-E_i/kT)$. Thus, lower energies are most likely. The leading constant is, $1=\sum p_i=A\sum _i\exp (-E_i/kT)=AZ$. Thus, $p_i=\frac{\exp (-E_i/kT)}{Z}$.

$H=H_0-{\overrightarrow{p}}_0\cdot \overrightarrow{E}$. $\overrightarrow{p}=q\overrightarrow{d},\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=q\frac{\overrightarrow{d}}{2}\times \overrightarrow{E}={\overrightarrow{p}}_0\times \overrightarrow{E}=-\hat{y}{P}_{0}E\sin \theta =-\frac{\partial U}{\partial \theta }$. So, $U=-p_{0}E\cos \theta$.

In the classical case, we have a continuous energy spectra, so $Z=\iiint \iiint \exp (-E(p,q)/kT)d^{3}p{d}^{3}q$.

Using $p_z=p_0\cos \theta$, $⟨p_{mol}⟩=\frac{\int p_0\cos \theta \exp \left(\frac{-H}{kT}\right)d\mathrm{\Omega }}{Z}=p_0(\coth \alpha -1/\alpha ),\alpha =p_{0}E/kT=p_{0}E\beta$. Note that $\alpha$ is the ratio between the thermal and interaction energies.

For $T\to 0$, $⟨p_{mol}⟩\approx p_0$. Thus, the particles are very well aligned.

For $kT\gg p_{0}E$, $\alpha \ll 1$. So, $\coth \alpha \approx \frac{1}{\alpha }\frac{(1+\frac{1}{2}{\alpha }^2)}{(1+\frac{1}{6}{\alpha }^2)}\approx \frac{1}{\alpha }(1+\frac{1}{2}{\alpha }^2)(1-\frac{1}{6}{\alpha }^2)\approx \frac{1}{\alpha }(1+\frac{1}{3}{\alpha }^2)$. Then, $⟨p_{mol}\approx p_0\frac{1}{3}\alpha =\frac{1}{3}\frac{p_0^{2}E}{kT}={\epsilon }_0{\gamma }_{mol}E$. So, ${\gamma }_{mol}\approx \frac{p_0^{2}E}{3{\epsilon }_{0}kT}$.