The energy ellipsoid
We want now to get some experience with tensors. Suppose we ask the interesting question: What energy is required to polarize the crystal (in addition to the energy in the electric field which we know is $\epsO E^2/2$ per unit volume)? Consider for a moment the atomic charges that are being displaced. The work done in displacing the charge the distance $dx$ is $qE_x\,dx$, and if there are $N$ charges per unit volume, the work done is $qE_xN\,dx$. But $qN\,dx$ is the change $dP_x$ in the dipole moment per unit volume. So the energy required per unit volume is \begin{equation*} E_x\,dP_x. \end{equation*}  Combining the work for the three components of the field, the work per unit volume is found to be \begin{equation*} \FLPE\cdot d\FLPP. \end{equation*}  Since the magnitude of $\FLPP$ is proportional to $\FLPE$, the work done per unit volume in bringing the polarization from $\FLPzero$ to $\FLPP$ is the integral of $\FLPE\cdot d\FLPP$. Calling this work $u_P$,2 we write \begin{equation} \label{Eq:II:31:6} u_P=\tfrac{1}{2}\FLPE\cdot\FLPP=\tfrac{1}{2} \sum_iE_iP_i. \end{equation}  Now we can express $\FLPP$ in terms of $\FLPE$ by Eq. (31.5), and we have that \begin{equation} \label{Eq:II:31:7} u_P=\tfrac{1}{2}\sum_i\sum_j\alpha_{ij}E_iE_j. \end{equation}  The energy density $u_P$ is a number independent of the choice of axes, so it is a scalar. A tensor has then the property that when it is summed over one index (with a vector), it gives a new vector; and when it is summed over both indexes (with two vectors), it gives a scalar.  The tensor $\alpha_{ij}$ should really be called a “tensor of second rank,” because it has two indexes. A vector—with one index—is a tensor of the first rank, and a scalar—with no index—is a tensor of zero rank. So we say that the electric field $\FLPE$ is a tensor of the first rank and that the energy density $u_P$ is a tensor of zero rank. It is possible to extend the ideas of a tensor to three or more indexes, and so to make tensors of ranks higher than two. The subscripts of the polarization tensor range over three possible values—they are tensors in three dimensions. The mathematicians consider tensors in four, five, or more dimensions. We have already used a four-dimensional tensor $F_{\mu\nu}$ in our relativistic description of the electromagnetic field (Chapter 26). The polarization tensor $\alpha_{ij}$ has the interesting property that it is symmetric, that is, that $\alpha_{xy}=\alpha_{yx}$, and so on for any pair of indexes. (This is a physical property of a real crystal and not necessary for all tensors.) You can prove for yourself that this must be true by computing the change in energy of a crystal through the following cycle: (1) Turn on a field in the $x$-direction; (2) turn on a field in the $y$-direction; (3) turn off the $x$-field; (4) turn off the $y$-field. The crystal is now back where it started, and the net work done on the polarization must be back to zero. You can show, however, that for this to be true, $\alpha_{xy}$ must be equal to $\alpha_{yx}$. The same kind of argument can, of course, be given for $\alpha_{xz}$, etc. So the polarization tensor is symmetric. This also means that the polarization tensor can be measured by just measuring the energy required to polarize the crystal in various directions. Suppose we apply an $\FLPE$-field with only an $x$- and a $y$-component; then according to Eq. (31.7), \begin{equation} \label{Eq:II:31:8} u_P=\tfrac{1}{2}[\alpha_{xx}E_x^2+(\alpha_{xy}+\alpha_{yx})E_xE_y+ \alpha_{yy}E_y^2]. \end{equation}  With an $E_x$ alone, we can determine $\alpha_{xx}$; with an $E_y$ alone, we can determine $\alpha_{yy}$; with both $E_x$ and $E_y$, we get an extra energy due to the term with $(\alpha_{xy}+\alpha_{yx})$. Since the $\alpha_{xy}$ and $\alpha_{yx}$ are equal, this term is $2\alpha_{xy}$ and can be related to the energy. The energy expression, Eq. (31.8), has a nice geometric interpretation. Suppose we ask what fields $E_x$ and $E_y$ correspond to some given energy density—say $u_0$. That is just the mathematical problem of solving the equation \begin{equation*} \alpha_{xx}E_x^2+2\alpha_{xy}E_xE_y+\alpha_{yy}E_y^2=2u_0. \end{equation*}  This is a quadratic equation, so if we plot $E_x$ and $E_y$ the solutions of this equation are all the points on an ellipse (Fig. 31–2). (It must be an ellipse, rather than a parabola or a hyperbola, because the energy for any field is always positive and finite.) The vector $\FLPE$ with components $E_x$ and $E_y$ can be drawn from the origin to the ellipse. So such an “energy ellipse” is a nice way of “visualizing” the polarization tensor. If we now generalize to include all three components, the electric vector $\FLPE$ in any direction required to give a unit energy density gives a point which will be on the surface of an ellipsoid, as shown in Fig. 31–3. The shape of this ellipsoid of constant energy uniquely characterizes the tensor polarizability. Now an ellipsoid has the nice property that it can always be described simply by giving the directions of three “principal axes” and the diameters of the ellipse along these axes. The “principal axes” are the directions of the longest and shortest diameters and the direction at right angles to both. They are indicated by the axes $a$, $b$, and $c$ in Fig. 31–3. With respect to these axes, the ellipsoid has the particularly simple equation \begin{equation*} \alpha_{aa}E_a^2+\alpha_{bb}E_b^2+\alpha_{cc}E_c^2=2u_0. \end{equation*}  So with respect to these axes, the dielectric tensor has only three components that are not zero: $\alpha_{aa}$, $\alpha_{bb}$, and $\alpha_{cc}$. That is to say, no matter how complicated a crystal is, it is always possible to choose a set of axes (not necessarily the crystal axes) for which the polarization tensor has only three components. With such a set of axes, Eq. (31.4) becomes simply \begin{equation} \label{Eq:II:31:9} P_a=\alpha_{aa}E_a,\quad P_b=\alpha_{bb}E_b,\quad P_c=\alpha_{cc}E_c. \end{equation}  An electric field along any one of the principal axes produces a polarization along the same axis, but the coefficients for the three axes may, of course, be different. Often, a tensor is described by listing the nine coefficients in a table inside of a pair of brackets: \begin{equation} \label{Eq:II:31:10} \begin{bmatrix} \alpha_{xx} & \alpha_{xy} & \alpha_{xz}\\ \alpha_{yx} & \alpha_{yy} & \alpha_{yz}\\ \alpha_{zx} & \alpha_{zy} & \alpha_{zz} \end{bmatrix}. \end{equation}  For the principal axes $a$, $b$, and $c$, only the diagonal terms are not zero; we say then that “the tensor is diagonal.” The complete tensor is \begin{equation} \label{Eq:II:31:11} \begin{bmatrix} \alpha_{aa} & 0 & 0\\ 0 & \alpha_{bb} & 0\\ 0 & 0 & \alpha_{cc} \end{bmatrix}. \end{equation}  The important point is that any polarization tensor (in fact, any symmetric tensor of rank two in any number of dimensions) can be put in this form by choosing a suitable set of coordinate axes. If the three elements of the polarization tensor in diagonal form are all equal, that is, if \begin{equation} \label{Eq:II:31:12} \alpha_{aa}=\alpha_{bb}=\alpha_{cc}=\alpha, \end{equation}  the energy ellipsoid becomes a sphere, and the polarizability is the same in all directions. The material is isotropic. In the tensor notation, \begin{equation} \label{Eq:II:31:13} \alpha_{ij}=\alpha\delta_{ij} \end{equation}  where $\delta_{ij}$ is the unit tensor \begin{equation} \label{Eq:II:31:14} \delta_{ij}= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}. \end{equation}  That means, of course, \begin{equation} \begin{aligned} &\delta_{ij}=1,\quad\text{if}\quad i=j;\\[1mm] &\delta_{ij}=0,\quad\text{if}\quad i\neq j. \end{aligned} \label{Eq:II:31:15} \end{equation}  The tensor $\delta_{ij}$ is often called the “Kronecker delta.” You may amuse yourself by proving that the tensor (31.14) has exactly the same form if you change the coordinate system to any other rectangular one. The polarization tensor of Eq. (31.13) gives \begin{equation*} P_i=\alpha\sum_j\delta_{ij}E_j=\alpha E_i, \end{equation*}  which means the same as our old result for isotropic dielectrics: \begin{equation*} \FLPP=\alpha\FLPE. \end{equation*}  The shape and orientation of the polarization ellipsoid can sometimes be related to the symmetry properties of the crystal. We have said in Chapter 30 that there are $230$ different possible internal symmetries of a three-dimensional lattice and that they can, for many purposes, be conveniently grouped into seven classes, according to the shape of the unit cell. Now the ellipsoid of polarizability must share the internal geometric symmetries of the crystal. For example, a triclinic crystal has low symmetry—the ellipsoid of polarizability will have unequal axes, and its orientation will not, in general, be aligned with the crystal axes. On the other hand, a monoclinic crystal has the property that its properties are unchanged if the crystal is rotated $180^\circ$ about one axis. So the polarization tensor must be the same after such a rotation. It follows that the ellipsoid of the polarizability must return to itself after a $180^\circ$ rotation. That can happen only if one of the axes of the ellipsoid is in the same direction as the symmetry axis of the crystal. Otherwise, the orientation and dimensions of the ellipsoid are unrestricted. For an orthorhombic crystal, however, the axes of the ellipsoid must correspond to the crystal axes, because a $180^\circ$ rotation about any one of the three axes repeats the same lattice. If we go to a tetragonal crystal, the ellipse must have the same symmetry, so it must have two equal diameters. Finally, for a cubic crystal, all three diameters of the ellipsoid must be equal; it becomes a sphere, and the polarizability of the crystal is the same in all directions. There is a big game of figuring out the possible kinds of tensors for all the possible symmetries of a crystal. It is called a “group-theoretical” analysis. But for the simple case of the polarizability tensor, it is relatively easy to see what the relations must be.