1
$\begingroup$

I've been presented the Clausius-Mossotti equation in my Electromagnetic Optics class, but I was not given any derivation of the expression:

$$\vec{P}=N\alpha \left(\vec{E}+\frac{\vec{P}}{3\varepsilon_0}\right)$$

where:

  • $N$: number of molecules per unit volume in the dielectric
  • $\alpha$: molecular polarizability

I've been doing a bit of research in this stack and I've found this post, which, although gives a taste of why polarization plays a role in the macroscopic electric field, I feel like it fails to explain why it is precisely $\frac{\vec{P}}{3\varepsilon_0}$ the term we choose to add, which corresponds to the field inside a uniformly polarized sphere (or rather, its opposite). We were told this is because, when averaging the microscopic field inside the dielectric, we choose a spherical volume to calculate said average. All this sounds a bit confusing to me and I can't quite understand how I can derivate the Clausius-Mossotti equation using just classical Electrostatics.

$\endgroup$

1 Answer 1

1
$\begingroup$

In order to investigate the polarizability of a molecule, we have to determine the local microscopic electric field that acts on the molecule. This is not the same as the macroscopic field, $\bf E_M$, which is usually just called $\bf E$. The field that acts on an individual molecule must be a the microscopic electric field $\bf E_\mu$, that includes the separate contribution of each individual charge.

It would be too hard to include all of the huge number of charges individually, so we break the problem of finding the microscopic field into two parts. We can write \begin{equation} {\bf E}_\mu= {\bf E}_{\rm near} +{\bf E}_{\rm far}, \end{equation} where ${\bf E}_{\rm near}$ comes from those molecules within a radius R, and ${\bf E}_{\rm far}$ comes from everything beyond the radius R. The radius $R$ is chosen to include a large number of molecules, but to be small enough that the macroscopic field $\bf E_M$ can be assumed constant within the sphere of radius $R$.

For most materials the symmetry of the charges in the near region is such that the field ${\bf E}_{\rm near}$ vanishes.
This is also true for most isotropic materials, and we will assume that this near field is zero for the following derivation.

The remaining electric field at the molecule is due to everything beyond the radius R. The only change in this field is the fact that there is now a surface charge $\sigma_b$ on the surface at radius R. This surface charge density is given by \begin{equation} \sigma_b = -{\bf{\hat r}\cdot P}. \label{sb} \end{equation} The $-\bf\hat r$ is the normal vector out of the dielectric at the surface of the sphere. The electric field due to this bound surface charge should be added to the original macroscopic field $\bf E_M$, so the microscopic electric field at the molecule at the center of the sphere is given by \begin{eqnarray} {\bf E_\mu}(0)&=&{\bf E_M}-\oint\frac{\sigma({\bf r}){\bf\hat r}\, dA}{R^2}\nonumber\\ &=& {\bf E_M}+\oint d\Omega\, {\bf{\hat r}{\hat r}\cdot P}\nonumber\\ &=& {\bf E_M}+\frac{4\pi}{3}\bf P. \label{emic} \end{eqnarray} In doing the angular integral, we used the fact that the radius R is small enough that the polarization $\bf P$ can be assumed constant on the surface.

We see that the actual microscopic field acting on a single molecule is stronger by $\frac{4\pi}{3}\bf P$ than the macroscopic field $\bf E_M$. Note that the microscopic field $\bf E_{\mu}$ is not the same as the electric field that would exist inside a cavity of radius $R$ in the material.

To put into SI units, use $4\pi\rightarrow 1/\epsilon_0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.