Clausius-Mossotti equation for ionic polarization

Local field and polarization

Polarization $P$ is the net induced dipole moment per unit volume of a dielectric: $P = N p$, where $N$ is the number of dipoles per unit volume and $p$ the average dipole moment.

The local field $E_{loc}$ is the field that actually acts on an ion or atom. It equals the applied field $E$ plus the field produced by all the surrounding polarized ions. Since the dipoles of the neighbours point the same way, $E_{loc}$ is greater than $E$, and the dipole moment of each unit is $p = \alpha E_{loc}$, not $\alpha E$.

Ionic polarization and ionic polarizability

In an ionic crystal such as NaCl, an applied field pushes the cations along the field and the anions against it. Each ion pair is displaced from its equilibrium spacing by a small amount $x$, which induces a dipole.

  No field:   Na+  ---a---  Cl-
  With field E  -->
              Na+ -->   <-- Cl-
              (spacing changes by x)

The restoring force behaves like a spring of constant $\beta$: $F = \beta x = eE_{loc}$. The induced dipole is

$$p_i = e x = \frac{e^2}{\beta}E_{loc} = \alpha_i E_{loc}, \qquad \alpha_i = \frac{e^2}{\beta}$$

Each ion is also electronically polarized, so the total polarizability per ion pair is

$$\alpha = \alpha_i + \alpha_{e+} + \alpha_{e-}$$

Lorentz local field

Take a spherical cavity around the reference ion. The field at its centre is

$$E_{loc} = E + E_{sphere} + E_{inside}$$

The polarization charge density on the cavity wall at angle $\theta$ is $P\cos\theta$. Summing the axial field of all rings:

$$E_{sphere} = \int_0^{\pi}\frac{P\cos\theta \cdot 2\pi r^2\sin\theta\, d\theta}{4\pi\varepsilon_0 r^2}\cos\theta = \frac{P}{3\varepsilon_0}$$

For a cubic crystal the dipoles inside the sphere give $E_{inside} = 0$. Hence

$$E_{loc} = E + \frac{P}{3\varepsilon_0}$$

Clausius-Mossotti equation

With $N_i$ ion pairs per unit volume:

$$P = N_i\alpha E_{loc} = N_i\alpha\left(E + \frac{P}{3\varepsilon_0}\right) \;\Rightarrow\; P = \frac{N_i\alpha E}{1 - \dfrac{N_i\alpha}{3\varepsilon_0}}$$

Macroscopically $P = \varepsilon_0(\varepsilon_r - 1)E$. Equating:

$$\begin{aligned} \varepsilon_0(\varepsilon_r - 1) &= \frac{N_i\alpha}{1 - N_i\alpha/3\varepsilon_0} \\ (\varepsilon_r - 1)\left(1 - \frac{N_i\alpha}{3\varepsilon_0}\right) &= \frac{N_i\alpha}{\varepsilon_0} \\ \varepsilon_r - 1 &= \frac{N_i\alpha}{3\varepsilon_0}\left(3 + \varepsilon_r - 1\right) = \frac{N_i\alpha}{3\varepsilon_0}(\varepsilon_r + 2) \end{aligned}$$ $$\boxed{\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N_i(\alpha_i + \alpha_{e+} + \alpha_{e-})}{3\varepsilon_0}}$$

Low and high frequency forms

• At low frequency (DC up to infrared), both ionic and electronic polarizations follow the field: $\varepsilon_r = \varepsilon_r(0)$ and the full $\alpha$ is used.
• At optical frequencies the heavy ions cannot follow, so only the electronic terms remain, and $\varepsilon_r = \varepsilon_{op} = n^2$ (refractive index squared):

$$\frac{\varepsilon_{op} - 1}{\varepsilon_{op} + 2} = \frac{N_i(\alpha_{e+} + \alpha_{e-})}{3\varepsilon_0}$$

Subtracting the two equations gives $\alpha_i$ alone. Example: NaCl has $\varepsilon_r(0) \approx 5.9$ and $\varepsilon_{op} \approx 2.3$ ($n \approx 1.54$), showing that ionic polarization contributes more than half of its static permittivity.

Significance

The equation relates the microscopic polarizability of each ion pair to the measurable permittivity, explains why $E_{loc} = \frac{\varepsilon_r + 2}{3}E$, and shows why ionic crystals have a higher static permittivity than optical permittivity. It holds for cubic ionic crystals; it fails for polar liquids with permanent dipoles, where dipole interactions are strong.