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# 3.5: Appendix(A) - The Onsager Problem

An interesting variant of the problem of a sphere in a uniform field has been discussed by Onsager in connection with the calculation of the dielectric constant of a material from its atomic polarizability; L.Onsager, J.Amer.Chem.Soc.58, 1486-1493 (1936). When an isolated atom is placed in a uniform external electric field it develops a dipole moment, pa, that is proportional to the applied field E0;

$\mathrm{p}_{\mathrm{a}}=\alpha \epsilon_{0} \mathrm{E}_{0}, \nonumber$

where the polarizability $$\alpha$$ has the dimensions of a volume, and can in principle be calculated using quantum mechanics. In a solid or a liquid the atom is not isolated, but its electric moment is influenced by the electric fields due to its neighbours. As a crude approximation one may imagine that the atom plus its associated electric moment is located at the center of a spherical cavity of radius R cut out of an otherwise homogeneous dielectric material characterized by a dielectric constant $$\epsilon$$, see Figure (3.4.18). Far from the cavity the electric field is E0 and directed along the z-axis corresponding to the potential function

$\mathrm{V}=-\mathrm{E}_{0} \mathrm{z}=-\mathrm{E}_{0} \mathrm{r} \cos \theta, \nonumber$

where r and θ are spherical polar co-ordinates. The problem is to determine the field inside the cavity that acts to polarize the atom. The externally applied electric field is derived from a potential function whose angular dependence is proportional to cos (θ); one is therefore motivated to seek a solution of this problem that corresponds to the use of the terms proportional to cos (θ) in the expansion for the potential, Equation (3.2.19). Inside the cavity the potential near r=0 must be dominated by the dipole potential

$\frac{\mathrm{p}_{\mathrm{a}}}{4 \pi \epsilon_{0}} \frac{\cos \theta}{\mathrm{r}^{2}}. \nonumber$

One is therefore led to try

Inside: r < R

$\mathrm{V}_{\mathrm{i}}(\mathrm{r}, \theta)=\left(\frac{\mathrm{p}_{\mathrm{a}} \cos \theta}{4 \pi \epsilon_{0}}\right) \frac{1}{\mathrm{r}^{2}}-A \mathrm{r} \cos \theta. \label{3.73}$

and

Outside: r > R

$\mathrm{V}_{\mathrm{o}}(\mathrm{r}, \theta)=-\mathrm{E}_{0} \mathrm{r} \cos \theta+\frac{b \cos \theta}{\mathrm{r}^{2}}. \label{3.74}$

The requirements that the potential function and the normal components of $$\vec D$$ be continuous across the surface of the sphere, r=R, lead to the two equations

\begin{aligned} A+\frac{b}{\mathrm{R}^{3}} &=\left(\frac{\mathrm{p}_{\mathrm{a}}}{4 \pi \epsilon_{0}}\right) \frac{1}{\mathrm{R}^{3}}+\mathrm{E}_{0} \\ -A+\frac{2 \epsilon_{r} b}{\mathrm{R}^{3}} &=\left(\frac{2 \mathrm{p}_{\mathrm{a}}}{4 \pi \epsilon_{0}}\right) \frac{1}{\mathrm{R}^{3}}-\epsilon_{r} \mathrm{E}_{0}, \end{aligned} \nonumber

where $$\epsilon_{r}=\epsilon / \epsilon_{0}$$. From these two equations one finds

$A=\left(\frac{3 \epsilon_{r}}{2 \epsilon_{r}+1}\right) \mathrm{E}_{0}+\left(\frac{\epsilon_{r}-1}{2 \epsilon_{r}+1}\right) \frac{2 \mathrm{p}_{\mathrm{a}}}{4 \pi \epsilon_{0} \mathrm{R}^{3}}, \label{3.75}$

and

$\frac{b}{\mathrm{R}^{3}}=\left(\frac{1-\epsilon_{r}}{2 \epsilon_{r}+1}\right) \mathrm{E}_{0}+\left(\frac{3}{2 \epsilon_{r}+1}\right) \frac{\mathrm{p}_{\mathrm{a}}}{4 \pi \epsilon_{0} \mathrm{R}^{3}}. \label{3.76}$

But A is just the value of the uniform field inside the cavity that is responsible for the induced dipole moment on the atom, therefore from the definition of the polarizability one has

$\mathrm{p}_{\mathrm{a}}=\alpha \epsilon_{0} A. \label{3.77}$

This value can be substituted into Equation (\ref{3.75}) for the constant A to obtain

$A=\left(\frac{3 \epsilon_{r}}{2 \epsilon_{r}+1}\right) \mathrm{E}_{0}+\left(\frac{\epsilon_{r}-1}{2 \epsilon_{r}+1}\right) \frac{2 \alpha A}{4 \pi \mathrm{R}^{3}}. \label{3.78}$

Eqn.(\ref{3.78}) can be solved for A in terms of the applied electric field E0, and this result can be used in Equation (\ref{3.77}) to calculate the atomic dipole moment pa:

$\mathrm{p}_{\mathrm{a}}=\left(\frac{3 \epsilon_{r} \epsilon_{0}}{2 \epsilon_{r}+1-\left(\frac{2 \alpha}{4 \pi \mathrm{R}^{3}}\right)\left(\epsilon_{r}-1\right)}\right) \alpha \mathrm{E}_{0}. \label{3.79}$

But the dipole moment per atom can be used to calculate the dipole moment per unit volume, $$\vec P$$:

$|\overrightarrow{\mathrm{P}}|=\mathrm{P}=\mathrm{N} \mathrm{p}_{\mathrm{a}}, \label{3.80}$

where N is the number of atoms per unit volume. From the definition

$\mathrm{D}=\epsilon_{0} \mathrm{E}_{0}+\mathrm{P} \nonumber$

one has

$\mathrm{P}=\left(\epsilon_{r}-1\right) \epsilon_{0} \mathrm{E}_{0}.\label{3.81}$

(Notice that one can drop the vector signs on D, E0, and P because all of these vectors are parallel with the z-axis). Using Equations (\ref{3.81},\ref{3.80}, and \ref{3.79}) one can obtain a relation between the relative dielectric constant, $$\epsilon$$r and the polarizability $$\alpha$$:

$\epsilon_{r}-1=\left(\frac{3 \epsilon_{r}}{2 \epsilon_{r}+1-\left(\epsilon_{r}-1\right)\left(\frac{\alpha}{2 \pi R^{3}}\right)}\right) \mathrm{N} \alpha. \nonumber$

The latter expression can be solved to obtain the polarizability in terms of the relative dielectric constant, $$\epsilon$$r:

$\alpha=\frac{\left(2 \epsilon_{r}+1\right)}{\left(\epsilon_{r}-1+\left(\frac{3 \epsilon_{r}}{\epsilon_{r}-1}\right) 2 \pi \mathrm{NR}^{3}\right)} 2 \pi \mathrm{R}^{3}. \label{3.82}$

Eqn.(\ref{3.82}) can be used to calculate the atomic polarizability from measured values of the relative dielectric constant, $$\epsilon$$r. These values of $$\alpha$$ can then be compared with values calculated from atomic theory.