# Polarizability and the Clausius-Mossotti Relation

There seems to be a fairly large inconsistency in various textbooks (and some assorted papers that I went through) about how to define the Clausius-Mossotti relationship (also called the Lorentz-Lorenz relationship, when applied to optics). It basically relates the polarizability of a material to it's relative permittivity (and hence it's refractive index).

Now the confusion arises because in Griffith's Introduction to Electrodynamics he defines the induced dipole moment of an atom/molecule in the presence of an external electric field to be given by

$$\vec{p} = \alpha \vec{E}$$ where $\alpha$ is the polarizability. From this definition, he derives the Clausius-Mossotti relation to be $$\alpha = \frac{3\epsilon_0}{N}\left(\frac{\epsilon_r -1}{\epsilon_r + 2}\right)$$

but in Panofsky and Phillips' Classical Electricity and Magnetism they've defined the induced dipole moment to be $$\vec{p} = \alpha \epsilon_0 \vec{E_{eff}}$$ where $$\vec{E_{eff}} = \vec{E} + \frac{\vec{P}}{3\epsilon_0}$$ the total electric field acting on a molecule.

Using this definition, they've arrived at the relationship $$\alpha = \frac{3}{N} \left(\frac{\epsilon_r -1}{\epsilon_r + 2}\right)$$ which is missing by a factor of $\epsilon_0$.

I've seen various sources deriving relations that both equations are right, but I can't quite figure it out. Everyone seems to be working in SI units as far as I can tell. The wikipedia article on the Lorentz-Lorenz equation (which is the same thing) has an extra factor of $4\pi$.

I tried working it out, but got lost because I don't really understand how the two differing definitions of $\vec{E}$ and $\vec{E_{eff}}$ are related. How can all these different versions of the equation be consistent with each other?

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The clearest explanation of the Clausius-Mossotti (CM) relation I have ever come across is this paper by Aspnes. The correct definition of the dipole moment must always relate to the microscopic field acting on the individual lattice sites. It is this microscopic field which induces the dipole moments. The microscopic field is different from the apparent macroscopic externally applied field. The latter is the sum of the microscopic applied field and the volume averaged dipole field (related to the macroscopic polarisation field). This is exactly what your second source is saying with $$\mathbf{E}_{eff} = \mathbf{E} + \frac{\mathbf{P}}{3\epsilon_0}.$$ $\mathbf{E}_{eff}$ is the microscopic field acting on each dipole, written in terms of the macroscopically averaged electric field $\mathbf{E}$ and polarisation $\mathbf{P}$. The factor $\frac{1}{3}$ accompanying $\mathbf{P}$ arises due to the volume averaging. I don't know the details of Griffith's derivation, but his symbol $\mathbf{E}$ must denote this microscopic field also, or he has done something dodgy.
The rest of your confusion appears to stem from definitions and units. You are free to define the polarisability $\alpha$ so that $\mathbf{p} = \alpha \epsilon_0 \mathbf{E}$ or so that $\mathbf{p} = \alpha^{\prime} \mathbf{E}$. You convert from one to the other by $\alpha^{\prime} = \alpha \epsilon_0$, exactly as you convert between your corresponding CM expressions. The appearance of a $\frac{1}{4\pi}$ in place of $\epsilon_0$ is common when converting from SI units to other unit systems common in electromagnetism, where often $\epsilon_0 = 1$ by definition.
I understand that one can absorb $\epsilon_0$ into $\alpha$, but then the units of $\alpha$ will have to change... which means that we won't be talking about the same "quantity" even in the same units system! I still have to work through the paper you linked (which looks promising!) but I suspect that the problem with $\epsilon_0$ arises due to different definitions of $E_{eff}$, rather than $\alpha$. –  Kitchi Apr 1 '13 at 15:19
@Kitchi No, the "problem with $\epsilon_0$" arises due to the two differing definitions for $\alpha$. Yes, the two definitions give $\alpha$ different units within the SI system, making them technically different quantities, but clearly describing the same physics. It is very easy to see that the two CM relations that you posted define $\alpha$ to have different units, just work out the dimensions of the right-hand sides and you will see that they are different! –  Mark Mitchison Apr 1 '13 at 16:55