See the derivation which begins at the start of page 5 in these notes: https://ocw.mit.edu/courses/mechanical-engineering/2-58j-radiative-transfer-spring-2006/readings/chap6_solid_prop.pdf
I went through and understood this a while ago, but have recently found a problem.
We say that each atom experiences a local field (dropping vector notation) $$E_{local}=E_{ex}+E_{ind}$$
which is the sum of the external field and the field of surrounding atomic dipoles. It is possible to calculate that $E_{ind}=\frac{P}{3\epsilon_{0}}$. Then we can use $$P=(\epsilon-\epsilon_{0})E_{ex}$$ However, why can we use $E_{ex}$ in this relation? I thought that the relation between polarisation and electric field was such that the electric field had to be the total electric field, which here would be $E_{local}$. See for example the bottom of page 179 of Griffiths http://kestrel.nmt.edu/~mce/Electromagnetics_by_Griffiths-1.pdf
Best Answer
The macroscopic electric field $\textbf{E}$ which should appear in the relation $\textbf{P} = (\epsilon - \epsilon_0)\textbf{E}$ is the spatial average of the microscopic (and rapidly fluctuating in space) electric field $\textbf{E}_{micro}(\textbf{r})$ over some sufficiently large region. It is important to note that $\textbf{E}_{micro}(\textbf{r})$ is not the same as the $\textbf{E}_{local}$ which appears in the Clausius-Mosotti arguments. The reason is that $\mathbf{E}_{local}$ is, by definition, the field that would be "felt" by a dipole placed at some position. The key point is that this does not include the field generated by that dipole itself.
So the actual microscopic field near a particular dipole at $\textbf{r} = 0$ would be $$\textbf{E}_{micro}(\textbf{r}) = \textbf{E}_{local} + \textbf{E}_{dipole}(\textbf{r})$$
where $\textbf{E}_{dipole}(\textbf{r})$ is the field generated by the dipole at $\textbf{r}=0$, and $\textbf{E}_{local}$ is supposed to represent field generated by all the other dipoles, as well as the external field. In fact, $\textbf{E}_{dipole}$ is very strong near the origin, and this gives a contribution to the spatial average of $\textbf{E}_{micro}$.
In fact, if one analyses this contribution quantitatively one finds that it is precisely $-\frac{\textbf{P}}{3\epsilon_0}$. (Why the minus sign? Remember that if the dipole moment points in the $+z$ direction, for example, then there is a positive charge at $z=+\epsilon$ and a negative charge at $z=-\epsilon$. So very close to the origin -- which is the main contribution to the spatial average -- the electric field is actually in the $-z$ direction, since it must point from positive charge to negative charge).
So by spatial averaging one finds that $$\textbf{E} = \textbf{E}_{local} - \frac{\textbf{P}}{3\epsilon_0},$$ which is precisely the Clausius-Mosotti relation! I personally find this "inverse" derivation of Clausius-Mosotti rather more transparent than the conventional one. I learned about it from this paper: http://iopscience.iop.org/article/10.1088/0143-0807/4/3/003/pdf