\question
{Yuki Tatsumi}
{01/16/12}
{tatsumi@flex.phys.tohoku.ac.jp}
{4-2}
%question
{Obtain Eq. (4.7) from Eq. (4.6).}
%answer
{
Eq. (4.6) is given by

\begin{equation}
\chi \equiv \frac{P}{E} = \frac{{\displaystyle \sum_{j}N_{j}\alpha_{j} }}{1 - {\displaystyle \frac{4\pi}{3}\sum_{j}N_{j}\alpha_{j} }} ,
\label{eq:Q4-2-1}
\end{equation}

where $\chi$, $P$, $E$, $N_{j}$ and $\alpha_{j}$  are the susceptibility, the polarization of crystal, the macroscopic electric field, the atomic concentration and the polarizability of each species, respectively. Using the relation of the dielectric constant and the susceptibility, $\varepsilon = 1 + 4\pi\chi$ (CGS),  we calculate Eq. (4.6) as follows:

\begin{eqnarray}
\frac{\varepsilon - 1}{4\pi}
& = &
\frac{{\displaystyle \sum_{j}N_{j}\alpha_{j} }}{1 - {\displaystyle \frac{4\pi}{3}\sum_{j}N_{j}\alpha_{j} }}
\nonumber \\
\left(\varepsilon - 1\right)\left(1 - \frac{4\pi}{3}\sum_{j}N_{j}\alpha_{j}\right)
& = &
4\pi\sum_{j}N_{j}\alpha_{j}
\nonumber \\
\varepsilon - 1
& = &
\left(\frac{4\pi}{3}\varepsilon - \frac{4\pi}{3} + 4\pi\right)\sum_{j}N_{j}\alpha_{j}
\nonumber \\
\varepsilon - 1
& = &
\frac{4\pi}{3}\left(\varepsilon + 2\right)\sum_{j}N_{j}\alpha_{j}
\nonumber \\
\frac{\varepsilon - 1}{\varepsilon + 2}
& = &
\frac{4\pi}{3}\sum_{j}N_{j}\alpha_{j}.
\label{eq:Q4-2-2}
\end{eqnarray}

Then we obtain `` the Clausius-Mossotti relation'' Eq. (4.7) which relates the dielectric constant and the electric polarizability, but only for crystal structures for which the Lorentz local field relation applies.
}