Ferroelectricity; BaTiO$_{\boldsymbol{3}}$
We want to describe now one special class of crystals which have, just by accident almost, a built-in permanent moment. The situation is so marginal that if we increase the temperature a little bit they lose the permanent moment completely. On the other hand, if they are nearly cubic crystals, so that their moments can be turned in different directions, we can detect a large change in the moment when an applied electric field is changed. All the moments flip over and we get a large effect. Substances which have this kind of permanent moment are called ferroelectric, after the corresponding ferromagnetic effects which were first discovered in iron. We would like to explain how ferroelectricity works by describing a particular example of a ferroelectric material. There are several ways in which the ferroelectric property can originate; but we will take up only one mysterious case—that of barium titanate, BaTiO$_3$. This material has a crystal lattice whose basic cell is sketched in Fig. 11–9. It turns out that above a certain temperature, specifically $118^\circ$C, barium titanate is an ordinary dielectric with an enormous dielectric constant. Below this temperature, however, it suddenly takes on a permanent moment. In working out the polarization of solid material, we must first find what are the local fields in each unit cell. We must include the fields from the polarization itself, just as we did for the case of a liquid. But a crystal is not a homogeneous liquid, so we cannot use for the local field what we would get in a spherical hole. If you work it out for a crystal, you find that the factor $1/3$ in Eq. (11.24) becomes slightly different, but not far from $1/3$. (For a simple cubic crystal, it is just $1/3$.) We will, therefore, assume for our preliminary discussion that the factor is $1/3$ for BaTiO$_3$. Now when we wrote Eq. (11.28) you may have wondered what would happen if $N\alpha$ became greater than $3$. It appears as though $\kappa$ would become negative. But that surely cannot be right. Let’s see what should happen if we were gradually to increase $\alpha$ in a particular crystal. As $\alpha$ gets larger, the polarization gets bigger, making a bigger local field. But a bigger local field will polarize each atom more, raising the local fields still more. If the “give” of the atoms is enough, the process keeps going; there is a kind of feedback that causes the polarization to increase without limit—assuming that the polarization of each atom increases in proportion to the field. The “runaway” condition occurs when $N\alpha=3$. The polarization does not become infinite, of course, because the proportionality between the induced moment and the electric field breaks down at high fields, so that our formulas are no longer correct. What happens is that the lattice gets “locked in” with a high, self-generated, internal polarization. In the case of BaTiO$_3$, there is, in addition to an electronic polarization, also a rather large ionic polarization, presumed to be due to titanium ions which can move a little within the cubic lattice. The lattice resists large motions, so after the titanium has gone a little way, it jams up and stops. But the crystal cell is then left with a permanent dipole moment. In most crystals, this is really the situation for all temperatures that can be reached. The very interesting thing about barium titanate is that there is such a delicate condition that if $N\alpha$ is decreased just a little bit it comes unstuck. Since $N$ decreases with increasing temperature—because of thermal expansion—we can vary $N\alpha$ by varying the temperature. Below the critical temperature it is just barely stuck, so it is easy—by applying an external field—to shift the polarization and have it lock in a different direction. Let’s see if we can analyze what happens in more detail. We call $T_c$ the critical temperature at which $N\alpha$ is exactly $3$. As the temperature increases, $N$ goes down a little bit because of the expansion of the lattice. Since the expansion is small, we can say that near the critical temperature \begin{equation} \label{Eq:II:11:30} N\alpha=3-\beta(T-T_c), \end{equation}  where $\beta$ is a small constant, of the same order of magnitude as the thermal expansion coefficient, or about $10^{-5}$ to $10^{-6}$ per degree C. Now if we substitute this relation into Eq. (11.28), we get that \begin{equation*} \kappa-1=\frac{3-\beta(T-T_c)}{\beta(T-T_c)/3}. \end{equation*}  Since we have assumed that $\beta(T-T_c)$ is small compared with one, we can approximate this formula by \begin{equation} \label{Eq:II:11:31} \kappa-1=\frac{9}{\beta(T-T_c)}. \end{equation}  This relation is right, of course, only for $T>T_c$. We see that just above the critical temperature $\kappa$ is enormous. Because $N\alpha$ is so close to $3$, there is a tremendous magnification effect, and the dielectric constant can easily be as high as $50{,}000$ to $100{,}000$. It is also very sensitive to temperature. For increases in temperature, the dielectric constant goes down inversely as the temperature, but, unlike the case of a dipolar gas, for which $\kappa-1$ goes inversely as the absolute temperature, for ferroelectrics it varies inversely as the difference between the absolute temperature and the critical temperature (this law is called the Curie-Weiss law).  When we lower the temperature to the critical temperature, what happens? If we imagine a lattice of unit cells like that in Fig. 11–9, we see that it is possible to pick out chains of ions along vertical lines. One of them consists of alternating oxygen and titanium ions. There are other lines made up of either barium or oxygen ions, but the spacing along these lines is greater. We make a simple model to imitate this situation by imagining, as shown in Fig. 11–10(a), a series of chains of ions. Along what we call the main chain, the separation of the ions is $a$, which is half the lattice constant; the lateral distance between identical chains is $2a$. There are less-dense chains in between which we will ignore for the moment. To make the analysis a little easier, we will also suppose that all the ions on the main chain are identical. (It is not a serious simplification because all the important effects will still appear. This is one of the tricks of theoretical physics. One does a different problem because it is easier to figure out the first time—then when one understands how the thing works, it is time to put in all the complications.) Now let’s try to find out what would happen with our model. We suppose that the dipole moment of each atom is $p$ and we wish to calculate the field at one of the atoms of the chain. We must find the sum of the fields from all the other atoms.  We will first calculate the field from the dipoles in only one vertical chain; we will talk about the other chains later. The field at the distance $r$ from a dipole in a direction along its axis is given by \begin{equation} \label{Eq:II:11:32} E=\frac{1}{4\pi\epsO}\,\frac{2p}{r^3}. \end{equation}  At any given atom, the dipoles at equal distances above and below it give fields in the same direction, so for the whole chain we get  \begin{equation} \label{Eq:II:11:33} E_{\text{chain}}=\frac{p}{4\pi\epsO}\,\frac{2}{a^3}\cdot \biggl(2+\frac{2}{8}+\frac{2}{27}+\frac{2}{64}+\dotsb\biggr)= \frac{p}{\epsO}\,\frac{0.383}{a^3}. \end{equation} 
 \begin{align} E_{\text{chain}}&=\frac{p}{4\pi\epsO}\,\frac{2}{a^3}\cdot \biggl(2+\frac{2}{8}+\frac{2}{27}+\frac{2}{64}+\dotsb\biggr)\notag\\[1.5ex] \label{Eq:II:11:33} &=\;\,\frac{p}{\epsO}\,\frac{0.383}{a^3}. \end{align}   It is not too hard to show that if our model were like a completely cubic crystal—that is, if the next identical lines were only the distance $a$ away—the number $0.383$ would be changed to $1/3$. In other words, if the next lines were at the distance $a$ they would contribute only $-0.050$ unit to our sum. However, the next main chain we are considering is at the distance $2a$ and, as you remember from Chapter 7, the field from a periodic structure dies off exponentially with distance. Therefore these lines contribute much less than $-0.050$ and we can just ignore all the other chains. It is necessary now to find out what polarizability $\alpha$ is needed to make the runaway process work. Suppose that the induced moment $p$ of each atom of the chain is proportional to the field on it, as in Eq. (11.6). We get the polarizing field on the atom from $E_{\text{chain}}$ using Eq. (11.32). So we have the two equations \begin{equation*} p=\alpha\epsO E_{\text{chain}} \end{equation*}  and \begin{equation*} E_{\text{chain}}=\frac{0.383}{a^3}\,\frac{p}{\epsO}. \end{equation*}  There are two solutions: $E_{\text{chain}}$ and $p$ both zero, or \begin{equation*} \alpha=\frac{a^3}{0.383}, \end{equation*}  with $E_{\text{chain}}$ and $p$ both finite. Thus if $\alpha$ is as large as $a^3/0.383$, a permanent polarization sustained by its own field will set in. This critical equality must be reached for barium titanate at just the temperature $T_c$. (Notice that if $\alpha$ were larger than the critical value for small fields, it would decrease at larger fields and at equilibrium the same equality we have found would hold.) For BaTiO$_3$, the spacing $a$ is $2\times10^{-8}$ cm, so we must expect that $\alpha=21.8\times10^{-24}$ cm$^3$. We can compare this with the known polarizabilities of the individual atoms. For oxygen, $\alpha=30.2\times10^{-24}$ cm$^3$; we’re on the right track! But for titanium, $\alpha=2.4\times10^{-24}$ cm$^3$; rather small. To use our model we should probably take the average. (We could work out the chain again for alternating atoms, but the result would be about the same.) So $\alpha(\text{average})=16.3\times10^{-24}$ cm$^3$, which is not high enough to give a permanent polarization. But wait a moment! We have so far only added up the electronic polarizabilities. There is also some ionic polarization due to the motion of the titanium ion. All we need is an ionic polarizability of $9.2\times10^{-24}$ cm$^3$. (A more precise computation using alternating atoms shows that actually $11.9\times10^{-24}$ cm$^3$ is needed.) To understand the properties of BaTiO$_3$, we have to assume that such an ionic polarizability exists.  Why the titanium ion in barium titanate should have that much ionic polarizability is not known. Furthermore, why, at a lower temperature, it polarizes along the cube diagonal and the face diagonal equally well is not clear. If we figure out the actual size of the spheres in Fig. 11–9, and ask whether the titanium is a little bit loose in the box formed by its neighboring oxygen atoms—which is what you would hope, so that it could be easily shifted—you find quite the contrary. It fits very tightly. The barium atoms are slightly loose, but if you let them be the ones that move, it doesn’t work out. So you see that the subject is really not one-hundred percent clear; there are still mysteries we would like to understand. Returning to our simple model of Fig. 11–10(a), we see that the field from one chain would tend to polarize the neighboring chain in the opposite direction, which means that although each chain would be locked, there would be no net permanent moment per unit volume!  (Although there would be no external electric effects, there are still certain thermodynamic effects one could observe.) Such systems exist, and are called antiferroelectric. So what we have explained is really an antiferroelectric. Barium titanate, however, is really like the arrangement in Fig. 11–10(b). The oxygen-titanium chains are all polarized in the same direction because there are intermediate chains of atoms in between. Although the atoms in these chains are not very polarizable, or very dense, they will be somewhat polarized, in the direction antiparallel to the oxygen-titanium chains. The small fields produced at the next oxygen-titanium chain will get it started parallel to the first. So BaTiO$_3$ is really ferroelectric, and it is because of the atoms in between.  You may be wondering: “But what about the direct effect between the two O-Ti chains?” Remember, though, the direct effect dies off exponentially with the separation; the effect of the chain of strong dipoles at $2a$ can be less than the effect of a chain of weak ones at the distance $a$. This completes our rather detailed report on our present understanding of the dielectric constants of gases, of liquids, and of solids.