Dipole moment and Clausius-Mossotti equation

Electric dipole moment

Two equal and opposite charges $+Q$ and $-Q$ separated by a small distance $a$ form an electric dipole. Its dipole moment is a vector from $-Q$ to $+Q$:

$$\mathbf{p} = Q\mathbf{a} \qquad (\text{unit: C m})$$

When an atom is placed in a field, its electron cloud shifts relative to the nucleus, creating an induced dipole moment proportional to the field acting on it:

$$p = \alpha_e E_{loc}$$

where $\alpha_e$ is the electronic polarizability. The polarization $P$ is the dipole moment per unit volume, $P = Np$.

 no field:   ( +)  cloud centred on nucleus, p = 0
 field E --> 
             (-- +)  cloud shifted left
              <-a->   p = Q a  (points along E)

Local electric field

The local field $E_{loc}$ is the actual field experienced by one atom inside a dielectric. It is the applied field plus the fields of all the other induced dipoles around it. In a solid it is larger than the applied macroscopic field $E$.

Lorentz expression for the local field

Cut an imaginary sphere around the reference atom, large compared with atomic size. The field at its centre is

$$E_{loc} = E + E_{sph} + E_{in}$$

• $E$: macroscopic field (from plate charges and outer surface polarization charges).
• $E_{sph}$: field from bound charges on the inner surface of the spherical cavity.
• $E_{in}$: field from dipoles inside the sphere; zero for cubic symmetry or random arrangement.

            E -->
         .------.
       +'        '-       cavity surface charge
      +    atom    -      magnitude P cos(theta),
      +     .  ->  -      theta measured from E
       +.        .-
         '-....-'

The bound charge on the cavity wall at angle $\theta$ (measured from $E$) has density of magnitude $P\cos\theta$. It is positive on the wall facing the incoming field and negative on the far wall, so its field at the centre points along $E$. A ring between $\theta$ and $\theta + d\theta$ has area $dA = 2\pi r^2\sin\theta\,d\theta$. The field component along $E$ at the centre is

$$dE_{sph} = \frac{P\cos\theta \cdot 2\pi r^2\sin\theta\, d\theta}{4\pi\varepsilon_0 r^2}\cos\theta = \frac{P}{2\varepsilon_0}\cos^2\theta\sin\theta\, d\theta$$ $$E_{sph} = \frac{P}{2\varepsilon_0}\int_0^{\pi}\cos^2\theta\sin\theta\, d\theta = \frac{P}{2\varepsilon_0}\left[-\frac{\cos^3\theta}{3}\right]_0^{\pi} = \frac{P}{3\varepsilon_0}$$

Hence

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

Clausius-Mossotti equation

With $N$ atoms per unit volume:

$$P = N\alpha_e E_{loc} = N\alpha_e\left(E + \frac{P}{3\varepsilon_0}\right) \quad\cdots (1)$$

From electrostatics, $D = \varepsilon_0E + P = \varepsilon_0\varepsilon_rE$, so

$$P = \varepsilon_0(\varepsilon_r - 1)E \quad\Rightarrow\quad E = \frac{P}{\varepsilon_0(\varepsilon_r - 1)} \quad\cdots (2)$$

Put (2) in (1) and divide by $P$:

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

Meaning and use

• It relates the microscopic polarizability $\alpha_e$ to the macroscopic, measurable $\varepsilon_r$.
• It gives $E_{loc} = \frac{\varepsilon_r + 2}{3}E$; for Si ($\varepsilon_r = 11.9$, $N = 5 \times 10^{28}$ m⁻³) it gives $\alpha_e = 4.17 \times 10^{-40}$ F m² and $E_{loc} = 4.63E$.
• At optical frequency $\varepsilon_r = n^2$, giving the Lorentz-Lorenz equation for refractive index.
• Valid for non-polar dielectrics with cubic symmetry, gases and non-polar liquids; not for polar materials where permanent dipoles interact strongly.