Local field and Clausius-Mossotti equation

Local field

The local field $E_{loc}$ is the actual electric field acting on an individual atom or molecule inside a dielectric. It is not equal to the applied (macroscopic) field $E$, because the induced dipoles of all the surrounding atoms also produce a field at that point. In a polarised solid, $E_{loc} > E$.

Lorentz method for the local field

Imagine a small spherical cavity around the chosen atom, inside a dielectric between capacitor plates. The field at the centre is

$$E_{loc} = E_1 + E_2 + E_3$$

• $E_1$: field due to charges on the plates and the polarization charges on the outer dielectric surfaces; together this is the applied macroscopic field $E$.
• $E_2$: field due to polarization charges on the surface of the spherical cavity.
• $E_3$: field of the dipoles inside the sphere; it is zero for a cubic (symmetric) arrangement.

  +  +  +  +  +  +  +   plate
  ------------------
      .--------.
     /  - - -  \     surface charge
    |  (atom)   |    on cavity: E2
     \  + + +  /
      '--------'
  ------------------
  -  -  -  -  -  -  -   plate

Field of cavity surface charge ($E_2$): on a ring of the sphere at angle $\theta$ to the field, the surface charge density is $P\cos\theta$. The ring of radius $r\sin\theta$, width $r\,d\theta$ has area $2\pi r^2\sin\theta\, d\theta$, and its field component along $E$ at the centre is

$$dE_2 = \frac{(P\cos\theta)(2\pi r^2 \sin\theta\, d\theta)}{4\pi\varepsilon_0 r^2}\cos\theta$$ $$E_2 = \frac{P}{2\varepsilon_0}\int_0^{\pi}\cos^2\theta\sin\theta\, d\theta = \frac{P}{2\varepsilon_0}\cdot\frac{2}{3} = \frac{P}{3\varepsilon_0}$$

So the Lorentz local field is

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

Derivation of Clausius-Mossotti equation

For electronic polarization, each atom acquires an induced dipole $p = \alpha_e E_{loc}$. With $N$ atoms per unit volume:

$$P = N\alpha_e E_{loc} = N\alpha_e\left(E + \frac{P}{3\varepsilon_0}\right)$$

Also, from the macroscopic definition, $P = \varepsilon_0(\varepsilon_r - 1)E$, so $E = \dfrac{P}{\varepsilon_0(\varepsilon_r - 1)}$. Then

$$\begin{aligned} P &= N\alpha_e\left[\frac{P}{\varepsilon_0(\varepsilon_r - 1)} + \frac{P}{3\varepsilon_0}\right] \\ 1 &= \frac{N\alpha_e}{\varepsilon_0}\left[\frac{3 + \varepsilon_r - 1}{3(\varepsilon_r - 1)}\right] = \frac{N\alpha_e}{3\varepsilon_0}\cdot\frac{\varepsilon_r + 2}{\varepsilon_r - 1} \end{aligned}$$ $$\boxed{\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha_e}{3\varepsilon_0}}$$

This is the Clausius-Mossotti equation. It links a microscopic property (polarizability $\alpha_e$ of one atom) to a macroscopic property (relative permittivity $\varepsilon_r$).

Notes:

• It is valid for non-polar solids with cubic symmetry and for gases and liquids where $E_3 = 0$.
• From it, $E_{loc} = \dfrac{\varepsilon_r + 2}{3}E$. Example: Si ($\varepsilon_r = 11.9$) has $E_{loc} = 4.63E$.
• For a dilute gas, $\varepsilon_r \approx 1$, and it reduces to $\varepsilon_r = 1 + N\alpha_e/\varepsilon_0$.