Clausius-Mossotti relation

In physics, the Clausius-Mossotti relation connects the relative permittivity ε_r of a dielectric to the polarizability α of the atoms or molecules constituting the dielectric. The relative permittivity is a bulk (macroscopic) property and polarizability is a microscopic property of matter; hence the relation bridges the gap between a directly-observable macroscopic property with a microscopic molecular property.

The Clausius-Mossotti equation for dielectric matter consisting of atoms or (non-polar) molecules is

${\frac {\epsilon _{r}-1}{\epsilon _{r}+2}}={\frac {1}{3\epsilon _{0}}}N\alpha ,$

where ε₀ is the electric constant (permittivity of the vacuum) and N is the number density (number of atoms or molecules per volume).

The index of refraction n of a dielectric is given by

$n={\sqrt {\epsilon _{r}\mu _{r}}}\approx {\sqrt {\epsilon _{r}}},$

where we used that for most dielectric matter the relative magnetic permeability μ_r is very close to unity. Substitution of this value of n gives the Lorentz-Lorenz relation

${\frac {n^{2}-1}{n^{2}+2}}={\frac {1}{3\epsilon _{0}}}N\alpha .$

Derivation

Consider N atoms or molecules, constituting the dielectric, in an external electric field E. The sum of this field and an internal field E_int induces a dipole p_ind on each molecule. The polarization vector P is the sum of the induced dipoles,

$\mathbf {P} =N\mathbf {p} _{\mathrm {ind} }.$

From the relation between P, the electric displacement, and the electric field follows

$\mathbf {P} =\epsilon _{0}(\epsilon _{r}-1)\mathbf {E} .$

To get an expression for p_ind, we consider a single molecule in a little spherical cavity inside the dielectric. It is shown below that the total field inside this cavity is

$E_{\mathrm {tot} }=E+E_{\mathrm {int} }=E+{\frac {1}{3\epsilon _{0}}}P,$

where we introduced absolute magnitudes, which is allowed because all vectors lie along the z-axis, the direction of E. Now,

$p_{\mathbf {ind} }=\alpha E_{\mathrm {tot} }=\alpha (E+{\frac {1}{3\epsilon _{0}}}P).$

Further,

$P=\epsilon _{0}(\epsilon _{r}-1)E=N\alpha (E+{\frac {1}{3}}(\epsilon _{r}-1)E)\quad \Longrightarrow \quad \epsilon _{0}(\epsilon _{r}-1)={\frac {N\alpha }{3}}(3+\epsilon _{r}-1),$

from which the Clausius-Mossotti relation follows directly.

We neglected here the contributions to the total field in the cavity from the other molecules in the cavity. It can be shown that these contributions average out in most dielectrics.

Internal electric field in dielectric

Slab of dielectric (green) in external electric field E. Microscopic spherical cavity (size greatly exaggerated in drawing) of radius r in dielectric.

A macroscopic slab of dielectric is placed in an outer electric field E (the z-direction) that polarizes the dielectric, so that a surface charge density σ_p is created. Since E "pushes" positive charge and "pulls" negative charge, the sign of the charge density on the outer surface is as is indicated in the figure. The polarization vector P points by definition from negative to positive charge. The surface charge density σ_p is in absolute value equal to |P|.

An infinitesimally small spherical cavity of radius r is made in the dielectric and inside this cavity there is vacuum with permittivity (electric constant) ε₀. Because of electric neutrality the surface of the cavity has the charge density (in absolute value),

$\sigma (\theta )=\mathbf {n} \cdot \mathbf {P} =P\cos \theta ,$

where n is a unit length vector perpendicular to the surface of the cavity at the point θ. By definition, the positive direction of the normal n is outward.

An infinitesimal element $d\Omega =r^{2}\;\sin \theta \;d\theta d\phi$ on the surface of the cavity gives a contribution to the internal electric field E_int parallel to n,

$d\mathbf {E} _{\mathrm {int} }=\mathbf {n} {\frac {\sigma (\theta )d\Omega }{4\pi \epsilon _{0}r^{2}}}=\mathbf {n} {\frac {P\cos \theta d\Omega }{4\pi \epsilon _{0}r^{2}}}=\mathbf {n} {\frac {P\cos \theta \sin \theta d\theta d\phi }{4\pi \epsilon _{0}}}.$

The z-component of n is cosθ and hence the contribution to the field in the z direction is

$dE_{\mathrm {int} ,z}={\frac {P}{4\pi \epsilon _{0}}}(\cos \theta )^{2}\sin \theta d\theta d\phi .$

Contributions in other than the z direction cancel mutually. Integration over the whole surface gives

$E_{\mathrm {int} ,z}=\iint dE_{\mathrm {int} ,z}=-{\frac {P}{4\pi \epsilon _{0}}}\int _{0}^{\pi }(\cos \theta )^{2}d\cos \theta \int _{0}^{2\pi }d\phi ={\frac {1}{3\epsilon _{0}}}P.$

The total electric field in the cavity is in the z direction and has magnitude

$E_{\mathrm {tot} }=E+E_{\mathrm {int} }=E+{\frac {1}{3\epsilon _{0}}}P$