Categories: Electromagnetism, Physics.

Clausius-Mossotti relation

The polarizability $\alpha$ of a small dielectric body (e.g. an atom or molecule) is defined to relate the electric field $\vb{E}$ applied to that body to the resulting dipole moment $\vb{p}$ :

\begin{aligned} \vb{p} = \alpha \varepsilon_0 \vb{E} \end{aligned}

If there are $N$ such bodies per unit volume, the polarization density $\vb{P} = \varepsilon_0 \chi_e \vb{E}$ with $\vb{P} = N \vb{p}$ suggests that $\chi_e = N \alpha$ . However, this is an underestimation: each body’s induced dipole creates its own electric field, weakening the field felt by its neighbors. We need to include this somehow, but $\alpha$ is defined for a single dipole in a vacuum.

Let $\vb{E}_\mathrm{int}$ be the uniform internal field excluding the dipoles’ contributions, and $\vb{E}(\vb{r})$ the net field including them. Assume that the dipoles $\vb{p}_i$ are arranged in a regular crystal lattice at sites $\vb{R}_i$ . Then $\vb{E}(\vb{r})$ is the sum of $\vb{E}_\mathrm{int}$ and all the dipoles’ fields:

\begin{aligned} \vb{E}(\vb{r}) = \vb{E}_\mathrm{int} + \sum_{i} \vb{E}_i(\vb{r} - \vb{R}_i) \end{aligned}

Where the individual contribution $\vb{E}_i(\vb{r})$ of each dipole $\vb{p}_i$ is as follows:

\begin{aligned} \vb{E}_i(\vb{r}) = - \frac{1}{4 \pi \varepsilon_0} \nabla \bigg( \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \bigg) \end{aligned}

Proof

Proof. The atoms or molecules $\vb{p}_i$ need not be perfect dipoles, as long as they approximate one when viewed from a distance much smaller than the crystal’s lattice constant. Clearly, in a multipole expansion of the true charge distribution $\rho_i(\vb{r})$ ’s electric potential $V_i(\vb{r})$ , the dipole term will be dominant in that case, given by:

\begin{aligned} V_i(\vb{r}) \approx \frac{1}{4 \pi \varepsilon_0} \frac{1}{|\vb{r}|^2} \int_{-\infty}^\infty \rho_i(\vb{r}') \: |\vb{r}'| \cos{\theta} \dd{\vb{r}'} \end{aligned}

Where $\theta$ is the angle between $\vb{r}$ and $\vb{r}'$ , so this can be rewritten as a dot product with the unit vector $\vu{r}$ , normalized from $\vb{r}$ :

\begin{aligned} V_i(\vb{r}) = \frac{1}{4 \pi \varepsilon_0} \frac{1}{|\vb{r}|^2} \: \vu{r} \cdot \!\!\int_{-\infty}^\infty \vb{r}' \rho_i(\vb{r}') \dd{\vb{r}'} \end{aligned}

The integral is a more general definition of the dipole moment $\vb{p}_i$ . You can convince yourself of this by defining $\rho_i(\vb{r})$ as two opposite charges $+q$ and $-q$ respectively located at $+\vb{d}_i/2$ and $-\vb{d}_i/2$ ; evaluating the integral then yields $q \vb{d}_i = \vb{p}_i$ . Therefore:

\begin{aligned} V_i(\vb{r}) = \frac{1}{4 \pi \varepsilon_0} \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \end{aligned}

Then the corresponding electric field $\vb{E}_i$ is given by $- \nabla V_i$ as is well known.

The dipole $\vb{p}_0$ at $\vb{r} = 0$ feels a net local field $\vb{E}_\mathrm{loc}$ , given below. The crystal’s symmetry ensures that all its neighbors’ fields cancel out: since $\vb{E}_i(-\vb{r}) = -\vb{E}_i(\vb{r})$ , each dipole has a counterpart with the exact opposite field. We exclude the singular term $\vb{E}_0(0)$ :

\begin{aligned} \vb{E}_\mathrm{loc} \equiv \vb{E}(0) = \vb{E}_\mathrm{int} + \sum_{i \neq 0} \vb{E}_i(-\vb{R}_i) = \vb{E}_\mathrm{int} \end{aligned}

Even if there is no regular lattice, this result still holds well enough, as long as the dipoles are uniformly distributed over a large volume.

So what was the point of including $\vb{E}_i(\vb{r})$ in the first place? Well, keep in mind that the sum over neighbors is nonzero for $\vb{r} \neq \vb{R}_i$ , which does affect the macroscopic field $\vb{E}$ , defined as:

\begin{aligned} \vb{E} = \vb{E}_\mathrm{loc} + \frac{1}{\Omega} \int_\Omega \sum_i \vb{E}_i(\vb{r} - \vb{R}_i) \dd{\vb{r}} \end{aligned}

Where $\Omega$ is an arbitrary averaging volume, large enough to contain many dipoles, and also small enough to assume that the polarization is uniform within. Thanks to linearity and symmetry, we only need to evaluate this volume integral for a single dipole:

\begin{aligned} \int_\Omega \vb{E}_i(\vb{r}) \dd{\vb{r}} &= - \frac{1}{4 \pi \varepsilon_0} \int_\Omega \nabla \bigg( \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \bigg) \dd{\vb{r}} \\ &= - \frac{1}{4 \pi \varepsilon_0} \oint_{\partial \Omega} \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \dd{\vu{n}} \end{aligned}

Where we have used the divergence theorem. Let us define $\Omega$ as a sphere with radius $R$ centered at $\vb{r} = 0$ . In spherical coordinates, the surface integral then becomes:

\begin{aligned} \int_\Omega \vb{E}_i(\vb{r}) \dd{\vb{r}} &= - \frac{1}{4 \pi \varepsilon_0} \int_0^\pi \int_0^{2 \pi} \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \: \vu{r} \: |\vb{r}|^2 \sin{\theta} \dd{\varphi} \dd{\theta} \end{aligned}

The radial coordinate disappears, so in fact the radius $R$ is irrelevant. We choose our coordinate system such that $\vb{p}_i$ points in the positive $z$ -direction, i.e. $\vb{p}_i = (0, 0, |\vb{p}_i|)$ , leaving:

\begin{aligned} \int_\Omega \vb{E}_i(\vb{r}) \dd{\vb{r}} &= - \frac{|\vb{p}_i|}{4 \pi \varepsilon_0} \int_0^\pi \int_0^{2 \pi} \begin{pmatrix} \sin{\theta} \cos{\varphi} \\ \sin{\theta} \sin{\varphi} \\ \cos{\theta} \end{pmatrix} \sin{\theta} \cos{\theta} \dd{\varphi} \dd{\theta} \end{aligned}

Now, consider the following two straightforward indefinite integrals:

\begin{aligned} \int \cos{x} \sin^2{x} \dd{x} &= \:\:\,\, \frac{1}{3} \sin^3{x} \\ \int \cos^2{x} \sin{x} \dd{x} &= -\frac{1}{3} \cos^3{x} \end{aligned}

Applying these to our integral over $\theta$ , the expression is reduced to just:

\begin{aligned} \int_\Omega \vb{E}_i(\vb{r}) \dd{\vb{r}} &= \frac{|\vb{p}_i|}{12 \pi \varepsilon_0} \int_0^{2 \pi} \begin{bmatrix} \sin^3{\theta} \cos{\varphi} \\ \sin^3{\theta} \sin{\varphi} \\ \cos^3{\theta} \end{bmatrix}_0^\pi \dd{\varphi} \\ &= - \frac{|\vb{p}_i|}{6 \pi \varepsilon_0} \int_0^{2 \pi} \begin{pmatrix} 0 \\ 0 \\ \:1\: \end{pmatrix} \dd{\varphi} \\ &= - \frac{\vb{p}_i}{3 \varepsilon_0} \end{aligned}

Inserting this result into the macroscopic field $\vb{E}$ , and recognizing the second term as the linear polarization density $\vb{P}$ , we arrive at the key result:

\begin{aligned} \vb{E} = \vb{E}_\mathrm{loc} - \frac{1}{3 \varepsilon_0} \sum_i \frac{\vb{p}_i}{\Omega} \qquad \implies \qquad \boxed{ \vb{E}_\mathrm{loc} = \vb{E} + \frac{1}{3 \varepsilon_0} \vb{P} } \end{aligned}

Whereas the individual dipoles are polarized by $\vb{E}_\mathrm{loc}$ , the macroscopic polarization density is defined from $\vb{E}$ , so we now know that:

\begin{aligned} \vb{P} = \varepsilon_0 \chi_e \vb{E} = \varepsilon_0 N \alpha \vb{E}_\mathrm{loc} = \varepsilon_0 N \alpha \bigg( \vb{E} + \frac{1}{3 \varepsilon_0} \vb{P} \bigg) \end{aligned}

Isolating this equation for $\vb{P}$ , and using that $\vb{P} = \varepsilon_0 \chi_e \vb{E}$ , we get a more accurate expression for the electric susceptibility $\chi_e$ :

\begin{aligned} \boxed{ \chi_e = \frac{N \alpha}{1 - N \alpha / 3} } \end{aligned}

Notice that for $N \alpha \ll 1$ this reduces to our earlier naive estimate: if each dipole’s neighbors are far away, they do not provide much shielding.

The corresponding macroscopic dielectric function $\varepsilon_r$ is given by:

\begin{aligned} \boxed{ \varepsilon_r = 1 + \chi_e = \frac{3 + 2 N \alpha}{3 - N \alpha} } \end{aligned}

Experimentally, $\varepsilon_r$ can be measured directly, but $\alpha$ cannot. By isolating $\varepsilon_r$ for $N \alpha$ , we finally arrive at the Clausius-Mossotti relation for calculating $\alpha$ :

\begin{aligned} \boxed{ \frac{1}{3} N \alpha = \frac{\varepsilon_r - 1}{\varepsilon_r + 2} } \end{aligned}

Since many theoretical models only calculate $\alpha$ (e.g. the Lorentz oscillator model), this result is useful for relating theory to experimental results.

References

M. Fox, Optical properties of solids, 2nd edition, Oxford.
D.E. Aspnes, Local-field effects and effective-medium theory: a microscopic perspective, 1982, American Journal of Physics 50.
A. Zangwill, Modern Electrodynamics, Cambridge.
D.J. Griffiths, Introduction to Electrodynamics, 5th edition, Cambridge.