Electrostatics of Microscopic Media


\(\vec{p}_{mol} = \varepsilon_0\gamma_{mol}\vec{E}\). C.f. \(\vec{P} = \varepsilon_0\chi_e\vec{E}\) for macroscopic media.

Consider a polar molecule in an electric field. The polar molecule tries to align along the field and experiences a torque. \(\langle \vec{p}\rangle\) becomes more along the $z$-axis, assuming \(\vec{E}\) 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, \(p_i = \frac{\exp(-E_i/kT)}{Z}\).

\(H = H_0 - \vec{p}_0\cdot\vec{E}\). \(\vec{p} = q\vec{d},\vec{\tau} = \vec{r}\times\vec{F} = q\frac{\vec{d}}{2}\times\vec{E} = \vec{p}_0\times\vec{E} = -\hat{y}P_0E\sin\theta = -\frac{\partial U}{\partial\theta}\). So, \(U = -p_0E\cos\theta\).

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

Using \(p_z = p_0\cos\theta\), \(\langle p_{mol}\rangle = \frac{\int p_0\cos\theta \exp\left(\frac{-H}{kT}\right)d\Omega}{Z} = p_0(\coth\alpha - 1/\alpha),\alpha=p_0E/kT=p_0E\beta\). Note that \(\alpha\) is the ratio between the thermal and interaction energies.

For \(T\to 0\), \(\langle p_{mol}\rangle \approx p_0\). Thus, the particles are very well aligned.

For \(kT\gg p_0 E\), \(\alpha\ll 1\). So, \(\coth\alpha\approx \frac{1}{\alpha}\frac{(1+\frac{1}{2}\alpha^2)}{(1+\frac{1}{6}\alpha^2)}\approx \frac{1}{\alpha}(1+\frac{1}{2}\alpha^2)(1-\frac{1}{6}\alpha^2) \approx \frac{1}{\alpha}\left(1+\frac{1}{3}\alpha^2\right)\). Then, \(\langle p_{mol}\approx p_0\frac{1}{3}\alpha = \frac{1}{3}\frac{p_0^2E}{kT} = \varepsilon_0\gamma_{mol}E\). So, \(\gamma_{mol} \approx \frac{p_0^2E}{3\varepsilon_0 kT}\).

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:33