Calculating the elastic constants
As our last topic on elasticity we would like to show how one could try to calculate the elastic constants of a material, starting with some knowledge of the properties of the atoms which make up the material. We will take only the simple case of an ionic cubic crystal like sodium chloride. When a crystal is strained, its volume or its shape is changed. Such changes result in an increase in the potential energy of the crystal. To calculate the change in strain energy, we have to know where each atom goes. In complicated crystals, the atoms will rearrange themselves in the lattice in very complicated ways to make the total energy as small as possible. This makes the computation of the strain energy rather difficult. In the case of a simple cubic crystal, however, it is easy to see what will happen. The distortions inside the crystal will be geometrically similar to the distortions of the outside boundaries of the crystal. We can calculate the elastic constants for a cubic crystal in the following way. First, we assume some force law between each pair of atoms in the crystal. Then, we calculate the change in the internal energy of the crystal when it is distorted from its equilibrium shape. This gives us a relation between the energy and the strains which is quadratic in all the strains. Comparing the energy obtained this way with Eq. (39.13), we can identify the coefficient of each term with the elastic constants $C_{ijkl}$. For our example we will assume a simple force law: that the force between neighboring atoms is a central force, by which we mean that it acts along the line between the two atoms. We would expect the forces in ionic crystals to be like this, since they are just primarily Coulomb forces. (The forces of covalent bonds are usually more complicated, since they can exert a sideways push on a nearby atom; we will leave out this complication.) We are also going to include only the forces between each atom and its nearest and next-nearest neighbors. In other words, we will make an approximation which neglects all forces beyond the next-nearest neighbor. The forces we will include are shown for the $xy$-plane in Fig. 39–10(a). The corresponding forces in the $yz$- and $zx$-planes also have to be included. Since we are only interested in the elastic coefficients which apply to small strains, and therefore only want the terms in the energy which vary quadratically with the strains, we can imagine that the force between each atom pair varies linearly with the displacements. We can then imagine that each pair of atoms is joined by a linear spring, as drawn in Fig. 39–10(b). All of the springs between a sodium atom and a chlorine atom should have the same spring constant, say $k_1$. The springs between two sodiums and between two chlorines could have different constants, but we will make our discussion simpler by taking them equal; we call them $k_2$. (We could come back later and make them different after we have seen how the calculations go.) Now we assume that the crystal is distorted by a homogeneous strain described by the strain tensor $e_{ij}$. In general, it will have components involving $x$, $y$, and $z$; but we will consider now only a strain with the three components $e_{xx}$, $e_{xy}$, and $e_{yy}$ so that it will be easy to visualize. If we pick one atom as our origin, the displacement of every other atom is given by equations like Eq. (39.9): \begin{equation} \begin{aligned} u_x&=e_{xx}x+e_{xy}y,\\ u_y&=e_{xy}x+e_{yy}y. \end{aligned} \label{Eq:II:39:42} \end{equation}  Suppose we call the atom at $x=y=0$ “atom $1$” and number its neighbors in the $xy$-plane as shown in Fig. 39–11. Calling the lattice constant $a$, we get the $x$ and $y$ displacements $u_x$ and $u_y$ listed in Table 39–1. Now we can calculate the energy stored in the springs, which is $k/2$ times the square of the extension for each spring. For example, the energy in the horizontal spring between atom $1$ and atom $2$ is \begin{equation} \label{Eq:II:39:43} \frac{k_1(e_{xx}a)^2}{2}. \end{equation}  Note that to first order, the $y$-displacement of atom $2$ does not change the length of the spring between atom $1$ and atom $2$. To get the strain energy in a diagonal spring, such as that to atom $3$, however, we need to calculate the change in length due to both the horizontal and vertical displacements. For small displacements from the original cube, we can write the change in the distance to atom $3$ as the sum of the components of $u_x$ and $u_y$ in the diagonal direction, namely as \begin{equation*} \frac{1}{\sqrt{2}}\,(u_x+u_y). \end{equation*}  Using the values of $u_x$ and $u_y$ from the table, we get the energy \begin{equation} \label{Eq:II:39:44} \frac{k_2}{2}\!\biggl(\!\frac{u_x\!+u_y}{\sqrt{2}}\!\biggr)^2\!\!\!=\! \frac{k_2a^2}{4}(e_{xx}\!+e_{yx}\!+e_{xy}\!+e_{yy})^2. \end{equation}  For the total energy for all the springs in the $xy$-plane, we need the sum of eight terms like (39.43) and (39.44). Calling this energy $U_0$, we get \begin{align} U_0=\,&\frac{a^2}{2}\bigg\{\!k_1e_{xx}^2\!+\! \frac{k_2}{2}(e_{xx}\!+e_{yx}\!+e_{xy}\!+e_{yy})^2\notag\\[-2pt] &+k_1e_{yy}^2\!+\! \frac{k_2}{2}(e_{xx}\!-e_{yx}\!-e_{xy}\!+e_{yy})^2\notag\\ &+k_1e_{xx}^2\!+\! \frac{k_2}{2}(e_{xx}\!+e_{yx}\!+e_{xy}\!+e_{yy})^2\notag\\ \label{Eq:II:39:45} &+k_1e_{yy}^2\!+\! \frac{k_2}{2}(e_{xx}\!-e_{yx}\!-e_{xy}\!+e_{yy})^2\!\biggr\}. \end{align}  To get the total energy of all the springs connected to atom $1$, we must make one addition to the energy in Eq. (39.45). Even though we have only $x$- and $y$-components of the strain, there are still some energies associated with the next-nearest neighbors off the $xy$-plane. This additional energy is \begin{equation} \label{Eq:II:39:46} k_2(e_{xx}^2a^2+e_{yy}^2a^2). \end{equation}  The elastic constants are related to the energy density $w$ by Eq. (39.13). The energy we have calculated is the energy associated with one atom, or rather, it is twice the energy per atom, since one-half of the energy of each spring should be assigned to each of the two atoms it joins. Since there are $1/a^3$ atoms per unit volume, $w$ and $U_0$ are related by \begin{equation*} w=\frac{U_0}{2a^3}. \end{equation*}  To find the elastic constants $C_{ijkl}$, we need only to expand out the squares in Eq. (39.45)—adding the terms of (39.46)—and compare the coefficients of $e_{ij}e_{kl}$ with the corresponding coefficient in Eq. (39.13). For example, collecting the terms in $e_{xx}^2$ and in $e_{yy}^2$, we get the factor \begin{equation*} (k_1+2k_2)a^2, \end{equation*}  so \begin{equation*} C_{xxxx}=C_{yyyy}=\frac{k_1+2k_2}{a}. \end{equation*}  For the remaining terms, there is a slight complication. Since we cannot distinguish the product of two terms like $e_{xx}e_{yy}$, from $e_{yy}e_{xx}$, the coefficient of such terms in our energy is equal to the sum of two terms in Eq. (39.13). The coefficient of $e_{xx}e_{yy}$ in Eq. (39.45) is $2k_2$, so we have that \begin{equation*} (C_{xxyy}+C_{yyxx})=\frac{2k_2}{a}. \end{equation*}  But because of the symmetry in our crystal, $C_{xxyy}=C_{yyxx}$, so we have that \begin{equation*} C_{xxyy}=C_{yyxx}=\frac{k_2}{a}. \end{equation*}  By a similar process, we can also get \begin{equation*} C_{xyxy}=C_{yxyx}=\frac{k_2}{a}. \end{equation*}  Finally, you will notice that any term which involves either $x$ or $y$ only once is zero—as we concluded earlier from symmetry arguments. Summarizing our results: \begin{equation} \begin{aligned} C_{xxxx}&=C_{yyyy}=\frac{k_1+2k_2}{a},\\[-.5ex] C_{xyxy}&=C_{yxyx}=\frac{k_2}{a},\\[-.5ex] C_{xxyy}&=C_{yyxx}=C_{xyyx}=C_{yxxy}=\frac{k_2}{a},\\[.5ex] C_{xxxy}&=C_{xyyy}=\text{etc.}=0. \end{aligned} \label{Eq:II:39:47} \end{equation}  We have been able to relate the bulk elastic constants to the atomic properties which appear in the constants $k_1$ and $k_2$. In our particular case, $C_{xyxy}=C_{xxyy}$. It turns out—as you can perhaps see from the way the calculations went—that these terms are always equal for a cubic crystal, no matter how many force terms are taken into account, provided only that the forces act along the line joining each pair of atoms—that is, so long as the forces between atoms are like springs and don’t have a sideways part such as you might get from a cantilevered beam (and you do get in covalent bonds). We can check this conclusion with the experimental measurements of the elastic constants. In Table 39–2 we give the observed values of the three elastic coefficients for several cubic crystals.2 You will notice that $C_{xxyy}$ and $C_{xyxy}$ are, in general, not equal. The reason is that in metals like sodium and potassium the interatomic forces are not along the line joining the atoms, as we assumed in our model. Diamond does not obey the law either, because the forces in diamond are covalent forces and have some directional properties—the bonds would prefer to be at the tetrahedral angle. The ionic crystals like lithium fluoride, sodium chloride, and so on, do have nearly all the physical properties assumed in our model, and the table shows that the constants $C_{xxyy}$ and $C_{xyxy}$ are almost equal. It is not clear why silver chloride should not satisfy the condition that $C_{xxyy}=C_{xyxy}$.