In a linear dielectric, the polarization is proportional to the field: P = 0χeE. If the material consists of atoms (or nonpolar molecules), the induced dipole moment of each one is likewise proportional to the field p = αE. Question: What is the relation between the atomic polarizability α an

Question

In a linear dielectric, the polarization is proportional to the field: [latex] P =\epsilon_{0} \chi_{e} E[/latex] . If the material consists of atoms (or nonpolar molecules), the induced dipole moment of each one is likewise proportional to the field [latex]p =\alpha E[/latex] . Question: What is the relation between the atomic polarizability α and the susceptibility [latex]\chi_{e}[/latex] ?

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 = NαE, one’s first inclination is to say that

[latex]\chi_{e}=\frac{N \alpha}{\epsilon_{0}}[/latex] (4.70)

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 macroscopic field 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 [latex]E _{ ext {else }}[/latex] . Imagine that the space allotted to each atom is a sphere of radius R, and show that

[latex]P =\epsilon_{0} \chi_{e} E[/latex] (4.30)

p = αE. (4.1)

[latex]E =\left(1-\frac{N \alpha}{3 \epsilon_{0}} ight) E _{ ext {else }} [/latex] (4.71)

Use this to conclude that

[latex]\chi_{e}=\frac{N \alpha / \epsilon_{0}}{1-N \alpha / 3 \epsilon_{0}}[/latex] ,

or

[latex]\alpha=\frac{3 \epsilon_{0}}{N}\left(\frac{\epsilon_{r}-1}{\epsilon_{r}+2} ight)[/latex] . (4.72)

Equation 4.72 is known as the Clausius-Mossotti formula, or, in its application to optics, the Lorentz-Lorenz equation.

Accepted Answer

The density of atoms is [latex]N=\frac{1}{(4 / 3) \pi R^{3}}[/latex] . The macroscopic field [latex]E ext { is } E _{ ext {self }}+ E _{ ext {else }}, ext { where } E _{ ext {self }}[/latex] is the average field over the sphere due to the atom itself.

[latex]p =\alpha E _{ ext {else }} \Rightarrow P =N \alpha E _{ ext {else }}[/latex] .

[Actually, it is the field at the center, not the average over the sphere, that belongs here, but the two are in fact equal, as we found in Prob. 3.47d.] Now

[latex]E _{ self }=-\frac{1}{4 \pi \epsilon_{0}} \frac{ p }{R^{3}}[/latex]

(Eq. 3.105), so

[latex]E _{ ave }=-\frac{1}{4 \pi \epsilon_{0}} \frac{ p }{R^{3}}[/latex] (3.105)

[latex]E =-\frac{1}{4 \pi \epsilon_{0}} \frac{\alpha}{R^{3}} E _{ ext {else }}+ E _{ ext {else }}=\left(1-\frac{\alpha}{4 \pi \epsilon_{0} R^{3}} ight) E _{ ext {else }}=\left(1-\frac{N \alpha}{3 \epsilon_{0}} ight) E _{ ext {else }}[/latex] .

So

[latex]P =\frac{N \alpha}{\left(1-N \alpha / 3 \epsilon_{0} ight)} E =\epsilon_{0} \chi_{e} E[/latex] ,

and hence

[latex]\chi_{e}=\frac{N \alpha / \epsilon_{0}}{\left(1-N \alpha / 3 \epsilon_{0} ight)}[/latex] .

Solving for α:

[latex]\chi_{e}-\frac{N \alpha}{3 \epsilon_{0}} \chi_{e}=\frac{N \alpha}{\epsilon_{0}} \Rightarrow \frac{N \alpha}{\epsilon_{0}}\left(1+\frac{\chi_{e}}{3} ight)=\chi_{e}[/latex] ,

or

[latex]\alpha=\frac{\epsilon_{0}}{N} \frac{\chi_{e}}{\left(1+\chi_{e} / 3 ight)}=\frac{3 \epsilon_{0}}{N} \frac{\chi_{e}}{\left(3+\chi_{e} ight.} . \quad ext { But } \chi_{e}=\epsilon_{r}-1, ext { so } \alpha=\frac{3 \epsilon_{0}}{N}\left(\frac{\epsilon_{r}-1}{\epsilon_{r}+2} ight)[/latex] . qed

Question 4.41: In a linear dielectric, the polarization is proportional to ...

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