Question

In a linear dielectric, the polarization is proportional to the field:

$P={\in }_0{\chi }_{e}E$.If the material consists of atoms (or nonpolar molecules), the induced

dipole moment of each one is likewise proportional to the field$p=\alpha E$ . Question:

What is the relation between the atomic polarizabilityand the susceptibility ${\chi }_e$? Since P (the dipole moment per unit volume) is P (the dipole moment per atom)times N (the number of atoms per unit volume),$P=Np=N\alpha E$, one's first inclination is to say that

${\chi }_e=\frac{N\alpha }{{\in }_0}$

And in fact this is not far off, if the density is low. But closer inspection reveals

a subtle problem, for the field E in Eq. 4.30 is the total macroscopicfield in the

medium, whereas the field in Eq. 4.1 is due to everything except the particular atom under consideration (polarizability was defined for an isolated atom subject to a specified external field); call this field $E_{else}$· Imagine that the space allotted to each atom is a sphere of radius R ,and show that

$E=\left(1-\frac{N\alpha }{3{\in }_0}\right)E_{else}$

Use this to conclude that

${\chi }_e=\frac{N\alpha /{\in }_0}{1-N\alpha /3{\in }_0}$

Or

$\alpha =\frac{3{\in }_0}{N}\left(\frac{{\in }_r-1}{{\in }_r+2}\right)$

Equation 4.72 is known as the Clausius-Mossottiformula, or, in its application to

optics, the Lorentz-Lorenzequation.

Short answer

It is shown that $\alpha =\frac{3{\epsilon }_0}{N}\frac{\left({\epsilon }_r-1\right)}{\left({\epsilon }_r+2\right)}$.

Step-by-step solution

Define function 

Write the expression for the Polarization is proportional to the electric field.

$P={\epsilon }_0{\chi }_{e}E$ …… (1)

If the material consists of atom ( or nonpolar molecules ), the induced dipole moment of each one is likewise proportional to the field.

$p=\alpha E$ …... (2)

Here, is the dipole moment per unit volume and is the dipole moment per atom.

Write the relation between these two.

$P=Np$

$$\begin{gathered} P=N\left(\alpha E\right) \\ P=N\alpha E \end{gathered}$$

$\begin{array}{rcl}{\epsilon }_0{\chi }_{e}E& =& N\alpha E\\ {\chi }_e& =& \frac{N\alpha }{{\epsilon }_0}\\ & & \end{array}$ ……. (3)

From the above equation (3) gives the relation between atomic polazabilityand susceptibility${\chi }_e$ and his equation is not valid if the density is very low.

Determine macroscopic field

Write the expression for the density of the atoms.

$N=\frac{1}{\left(\frac{4}{3}\right)\pi R^3}$

The macroscopic field E is given by,

$E=E{}_{self}+E_{else}$ …… (4)

Here,$E_{self}$is the average field over the sphere due to atom itself.

Write the expression for $E_{self}$.

$E_{self}=-\frac{1}{4\pi {\epsilon }_0}\frac{p}{R^3}$ …… (5)

Write the expression for dipole moment per atom.

$p=\alpha E_{else}$

Write the expression for dipole moment per unit volume.

$\begin{array}{rcl}P& =& Np\\ & =& N\left(\alpha E_{else}\right)\\ & =& N\alpha E_{else}\\ & & \end{array}$ …… (6)

Thus,

Write the expression of macroscopic field.

$\begin{array}{rcl}E& =& E_{self}+E_{else}\\ & =& -\frac{1}{4\pi {\epsilon }_0}\frac{p}{R^3}+E_{else}\\ & =& -\frac{1}{4\pi {\epsilon }_0}\frac{\alpha E_{else}}{R^3}+E_{else}\\ & =& E_{else}\left[1-\frac{1}{4\pi {\epsilon }_0}\frac{\alpha }{R^3}\right]\\ & & \end{array}$

Solve as further,

$\begin{array}{rcl}E& =& E_{else}\left[1-\frac{1}{4\pi R^3}\frac{\alpha }{{\epsilon }_0}\right]\\ & =& E_{else}\left[1-\frac{N}{3}\frac{\alpha }{{\epsilon }_0}\right]\left[\because N=\frac{3}{4\pi R^3}\right]\\ & =& E_{else}\left[1-\frac{N\alpha }{3{\epsilon }_0}\right]\\ & & \end{array}$

$E_{else}=\frac{E}{\left[1-\frac{N\alpha }{3{\epsilon }_0}\right]}$

Now, substitute$\frac{E}{\left[1-\frac{N\alpha }{3{\epsilon }_0}\right]}$for$E_{else}$in equation (6)

$\begin{array}{rcl}P& =& N\alpha \left(\frac{E}{\left[1-\frac{N\alpha }{3{\epsilon }_0}\right]}\right)\\ & =& \left(\frac{N\alpha }{1-\frac{N\alpha }{3{\epsilon }_0}}\right)E\\ & & \end{array}$ …… (7)

Now comparing equation (7) with equation (1),

$$\begin{gathered} {\epsilon }_0{\chi }_e=\frac{N\alpha }{1-\frac{N\alpha }{3{\epsilon }_0}} \\ {\chi }_e=\frac{\left(\frac{N\alpha }{{\epsilon }_0}\right)}{1-\frac{N\alpha }{3{\epsilon }_0}} \\ {\chi }_e\left(1-\frac{N\alpha }{3{\epsilon }_0}\right)=\frac{N\alpha }{{\epsilon }_0} \\ {\chi }_e-{\chi }_e\frac{N\alpha }{3{\epsilon }_0}=\frac{N\alpha }{{\epsilon }_0} \end{gathered}$$

Solve as further,

$\begin{array}{rcl}{\chi }_e& =& {\chi }_e\frac{N\alpha }{3{\epsilon }_0}+\frac{N\alpha }{{\epsilon }_0}\\ {\chi }_e& =& \frac{N\alpha }{{\epsilon }_0}\left(1+\frac{{\chi }_e}{3}\right)\\ \alpha & =& \frac{{\epsilon }_0}{N}\frac{{\chi }_e}{\left(1+\frac{{\chi }_e}{3}\right)}\\ \alpha & =& \frac{3{\epsilon }_0}{N}\frac{{\chi }_e}{3+{\chi }_e}\\ & & \end{array}$.....(8)

But .${\chi }_e={\in }_r-1$

Substitute the value of ${\chi }_e$in equation (8)

$$\begin{gathered} \alpha =\frac{3{\epsilon }_0}{N}\frac{\left({\epsilon }_r-1\right)}{3+\left({\epsilon }_r-1\right)} \\ \alpha =\frac{3{\epsilon }_0}{N}\frac{\left({\epsilon }_r-1\right)}{\left({\epsilon }_r+2\right)} \end{gathered}$$