Polar molecules; orientation polarization
Next we will consider a molecule which carries a permanent dipole moment $p_0$—such as a water molecule. With no electric field, the individual dipoles point in random directions, so the net moment per unit volume is zero. But when an electric field is applied, two things happen: First, there is an extra dipole moment induced because of the forces on the electrons; this part gives just the same kind of electronic polarizability we found for a nonpolar molecule. For very accurate work, this effect should, of course, be included, but we will neglect it for the moment. (It can always be added in at the end.) Second, the electric field tends to line up the individual dipoles to produce a net moment per unit volume. If all the dipoles in a gas were to line up, there would be a very large polarization, but that does not happen. At ordinary temperatures and electric fields the collisions of the molecules in their thermal motion keep them from lining up very much. But there is some net alignment, and so some polarization (see Fig. 11–2). The polarization that does occur can be computed by the methods of statistical mechanics we described in Chapter 40 of Vol. I. To use this method we need to know the energy of a dipole in an electric field. Consider a dipole of moment $\FLPp_0$ in an electric field, as shown in Fig. 11–3. The energy of the positive charge is $q\phi(1)$, and the energy of the negative charge is $-q\phi(2)$. Thus the energy of the dipole is \begin{equation} U=q\phi(1)-q\phi(2)=q\FLPd\cdot\FLPgrad{\phi},\notag \end{equation}  or \begin{equation} \label{Eq:II:11:14} U=-\FLPp_0\cdot\FLPE=-p_0E\cos\theta, \end{equation}  where $\theta$ is the angle between $\FLPp_0$ and $\FLPE$. As we would expect, the energy is lower when the dipoles are lined up with the field. We now find out how much lining up occurs by using the methods of statistical mechanics. We found in Chapter 40 of Vol. I that in a state of thermal equilibrium, the relative number of molecules with the potential energy $U$ is proportional to \begin{equation} \label{Eq:II:11:15} e^{-U/kT}, \end{equation}  where $U(x,y,z)$ is the potential energy as a function of position. The same arguments would say that using Eq. (11.14) for the potential energy as a function of angle, the number of molecules at $\theta$ per unit solid angle is proportional to $e^{-U/kT}$. Letting $n(\theta)$ be the number of molecules per unit solid angle at $\theta$, we have \begin{equation} \label{Eq:II:11:16} n(\theta)=n_0e^{+p_0E\cos\theta/kT}. \end{equation}  For normal temperatures and fields, the exponent is small, so we can approximate by expanding the exponential: \begin{equation} \label{Eq:II:11:17} n(\theta)=n_0\biggl(1+\frac{p_0E\cos\theta}{kT}\biggr). \end{equation}  We can find $n_0$ if we integrate (11.17) over all angles; the result should be just $N$, the total number of molecules per unit volume. The average value of $\cos\theta$ over all angles is zero, so the integral is just $n_0$ times the total solid angle $4\pi$. We get \begin{equation} \label{Eq:II:11:18} n_0=\frac{N}{4\pi}. \end{equation}  We see from (11.17) that there will be more molecules oriented along the field ($\cos\theta=1$) than against the field ($\cos\theta=-1$). So in any small volume containing many molecules there will be a net dipole moment per unit volume—that is, a polarization $P$. To calculate $P$, we want the vector sum of all the molecular moments in a unit volume. Since we know that the result is going to be in the direction of $\FLPE$, we will just sum the components in that direction (the components at right angles to $\FLPE$ will sum to zero): \begin{equation*} P=\underset{\substack{\text{unit}\\\text{volume}}}{\sum} p_0\cos\theta_i. \end{equation*}  We can evaluate the sum by integrating over the angular distribution. The solid angle at $\theta$ is $2\pi\sin\theta\,d\theta$, so \begin{equation} \label{Eq:II:11:19} P=\int_0^\pi n(\theta)p_0\cos\theta\,2\pi\sin\theta\,d\theta. \end{equation}  Substituting for $n(\theta)$ from (11.17), we have \begin{equation*} P=-\frac{N}{2}\int_1^{-1} \biggl(1+\frac{p_0E}{kT}\cos\theta\biggr) p_0\cos\theta\,d(\cos\theta), \end{equation*}  which is easily integrated to give \begin{equation} \label{Eq:II:11:20} P=\frac{Np_0^2E}{3kT}. \end{equation}  The polarization is proportional to the field $E$, so there will be normal dielectric behavior. Also, as we expect, the polarization depends inversely on the temperature, because at higher temperatures there is more disalignment by collisions. This $1/T$ dependence is called Curie’s law. The permanent moment $p_0$ appears squared for the following reason: In a given electric field, the aligning force depends upon $p_0$, and the mean moment that is produced by the lining up is again proportional to $p_0$. The average induced moment is proportional to $p_0^2$.  We should now try to see how well Eq. (11.20) agrees with experiment. Let’s look at the case of steam. Since we don’t know what $p_0$ is, we cannot compute $P$ directly, but Eq. (11.20) does predict that $\kappa-1$ should vary inversely as the temperature, and this we should check. From (11.20) we get \begin{equation} \label{Eq:II:11:21} \kappa-1=\frac{P}{\epsO E}=\frac{Np_0^2}{3\epsO kT}, \end{equation}  so $\kappa-1$ should vary in direct proportion to the density $N$, and inversely as the absolute temperature. The dielectric constant has been measured at several different pressures and temperatures, chosen such that the number of molecules in a unit volume remained fixed.1 [Notice that if the measurements had all been taken at constant pressure, the number of molecules per unit volume would decrease linearly with increasing temperature and $\kappa-1$ would vary as $T^{-2}$ instead of as $T^{-1}$.] In Fig. 11–4 we plot the experimental observations for $\kappa-1$ as a function of $1/T$. The dependence predicted by (11.21) is followed quite well. There is another characteristic of the dielectric constant of polar molecules—its variation with the frequency of the applied field. Due to the moment of inertia of the molecules, it takes a certain amount of time for the heavy molecules to turn toward the direction of the field. So if we apply frequencies in the high microwave region or above, the polar contribution to the dielectric constant begins to fall away because the molecules cannot follow. In contrast to this, the electronic polarizability still remains the same up to optical frequencies, because of the smaller inertia in the electrons.