The tensor of polarizability
Physicists always have a habit of taking the simplest example of any phenomenon and calling it “physics,” leaving the more complicated examples to become the concern of other fields—say of applied mathematics, electrical engineering, chemistry, or crystallography. Even solid-state physics is almost only half physics because it worries too much about special substances. So in these lectures we will be leaving out many interesting things. For instance, one of the important properties of crystals—or of most substances—is that their electric polarizability is different in different directions. If you apply a field in any direction, the atomic charges shift a little and produce a dipole moment, but the magnitude of the moment depends very much on the direction of the field. That is, of course, quite a complication. But in physics we usually start out by talking about the special case in which the polarizability is the same in all directions, to make life easier. We leave the other cases to some other field. Therefore, for our later work, we will not need at all what we are going to talk about in this chapter. The mathematics of tensors is particularly useful for describing properties of substances which vary in direction—although that’s only one example of their use. Since most of you are not going to become physicists, but are going to go into the real world, where things depend severely upon direction, sooner or later you will need to use tensors. In order not to leave anything out, we are going to describe tensors, although not in great detail. We want the feeling that our treatment of physics is complete. For example, our electrodynamics is complete—as complete as any electricity and magnetism course, even a graduate course. Our mechanics is not complete, because we studied mechanics when you didn’t have a high level of mathematical sophistication, and we were not able to discuss subjects like the principle of least action, or Lagrangians, or Hamiltonians, and so on, which are more elegant ways of describing mechanics. Except for general relativity, however, we do have the complete laws of mechanics. Our electricity and magnetism is complete, and a lot of other things are quite complete. The quantum mechanics, naturally, will not be—we have to leave something for the future. But you should at least know what a tensor is. We emphasized in Chapter 30 that the properties of crystalline substances are different in different directions—we say they are anisotropic. The variation of the induced dipole moment with the direction of the applied electric field is only one example, the one we will use for our example of a tensor. Let’s say that for a given direction of the electric field the induced dipole moment per unit volume $\FLPP$ is proportional to the strength of the applied field $\FLPE$. (This is a good approximation for many substances if $\FLPE$ is not too large.) We will call the proportionality constant $\alpha$.1 We want now to consider substances in which $\alpha$ depends on the direction of the applied field, as, for example, in crystals like calcite, which make double images when you look through them. Suppose, in a particular crystal, we find that an electric field $\FLPE_1$ in the $x$-direction produces the polarization $\FLPP_1$ in the $x$-direction. Then we find that an electric field $\FLPE_2$ in the $y$-direction, with the same strength as $\FLPE_1$, produces a different polarization $\FLPP_2$ in the $y$-direction. What would happen if we put an electric field at $45^\circ$? Well, that’s a superposition of two fields along $x$ and $y$, so the polarization $\FLPP$ will be the vector sum of $\FLPP_1$ and $\FLPP_2$, as shown in Fig. 31–1(a). The polarization is no longer in the same direction as the electric field. You can see how that might come about. There may be charges which can move easily up and down, but which are rather stiff for sidewise motions. When a force is applied at $45^\circ$, the charges move farther up than they do toward the side. The displacements are not in the direction of the external force, because there are asymmetric internal elastic forces. There is, of course, nothing special about $45^\circ$. It is generally true that the induced polarization of a crystal is not in the direction of the electric field. In our example above, we happened to make a “lucky” choice of our $x$- and $y$-axes, for which $\FLPP$ was along $\FLPE$ for both the $x$- and $y$-directions. If the crystal were rotated with respect to the coordinate axes, the electric field $\FLPE_2$ in the $y$-direction would have produced a polarization $\FLPP$ with both an $x$- and a $y$-component. Similarly, the polarization due to an electric field in the $x$-direction would have produced a polarization with an $x$-component and a $y$-component. Then the polarizations would be as shown in Fig. 31–1(b), instead of as in part (a). Things get more complicated—but for any field $\FLPE$, the magnitude of $\FLPP$ is still proportional to the magnitude of $\FLPE$. We want now to treat the general case of an arbitrary orientation of a crystal with respect to the coordinate axes. An electric field in the $x$-direction will produce a polarization $\FLPP$ with $x$-, $y$-, and $z$-components; we can write \begin{equation} \label{Eq:II:31:1} P_x=\alpha_{xx}E_x,\quad P_y=\alpha_{yx}E_x,\quad P_z=\alpha_{zx}E_x. \end{equation}  All we are saying here is that if the electric field is in the $x$-direction, the polarization does not have to be in that same direction, but rather has an $x$-, a $y$-, and a $z$-component—each proportional to $E_x$. We are calling the constants of proportionality $\alpha_{xx}$, $\alpha_{yx}$, and $\alpha_{zx}$, respectively (the first letter to tell us which component of $\FLPP$ is involved, the last to refer to the direction of the electric field). Similarly, for a field in the $y$-direction, we can write \begin{equation} \label{Eq:II:31:2} P_x=\alpha_{xy}E_y,\quad P_y=\alpha_{yy}E_y,\quad P_z=\alpha_{zy}E_y; \end{equation}  and for a field in the $z$-direction, \begin{equation} \label{Eq:II:31:3} P_x=\alpha_{xz}E_z,\quad P_y=\alpha_{yz}E_z,\quad P_z=\alpha_{zz}E_z. \end{equation}  Now we have said that polarization depends linearly on the fields, so if there is an electric field $\FLPE$ that has both an $x$- and a $y$-component, the resulting $x$-component of $\FLPP$ will be the sum of the two $P_x$’s of Eqs. (31.1) and (31.2). If $\FLPE$ has components along $x$, $y$, and $z$, the resulting components of $\FLPP$ will be the sum of the three contributions in Eqs. (31.1), (31.2), and (31.3). In other words, $\FLPP$ will be given by \begin{equation} \begin{alignedat}{4} P_x&=\alpha_{xx}&&E_x+\alpha_{xy}&&E_y+\alpha_{xz}&&E_z,\\[1ex] P_y&=\alpha_{yx}&&E_x+\alpha_{yy}&&E_y+\alpha_{yz}&&E_z,\\[1ex] P_z&=\alpha_{zx}&&E_x+\alpha_{zy}&&E_y+\alpha_{zz}&&E_z. \end{alignedat} \label{Eq:II:31:4} \end{equation}  The dielectric behavior of the crystal is then completely described by the nine quantities ($\alpha_{xx}$, $\alpha_{xy}$, $\alpha_{xz}$, $\alpha_{yx}$, …), which we can represent by the symbol $\alpha_{ij}$. (The subscripts $i$ and $j$ each stand for any one of the three possible letters $x$, $y$, and $z$.) Any arbitrary electric field $\FLPE$ can be resolved with the components $E_x$, $E_y$, and $E_z$; from these we can use the $\alpha_{ij}$ to find $P_x$, $P_y$, and $P_z$, which together give the total polarization $\FLPP$. The set of nine coefficients $\alpha_{ij}$ is called a tensor—in this instance, the tensor of polarizability. Just as we say that the three numbers $(E_x,E_y,E_z)$ “form the vector $\FLPE$,” we say that the nine numbers ($\alpha_{xx}$, $\alpha_{xy}$, …) “form the tensor $\alpha_{ij}$.”