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, p_a, 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 p_a:

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

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.