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3.5: Appendix(A) - The Onsager Problem

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    22808
  • 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.