Tensors of higher rank
The stress tensor $S_{ij}$ describes the internal forces of matter. If the material is elastic, it is convenient to describe the internal distortion in terms of another tensor $T_{ij}$—called the strain tensor. For a simple object like a bar of metal, you know that the change in length, $\Delta L$, is approximately proportional to the force, so we say it obeys Hooke’s law: \begin{equation*} \Delta L=\gamma F. \end{equation*}  For a solid elastic body with arbitrary distortions, the strain $T_{ij}$ is related to the stress $S_{ij}$ by a set of linear equations: \begin{equation} \label{Eq:II:31:26} T_{ij}=\sum_{k,l}\gamma_{ijkl}S_{kl}. \end{equation}  Also, you know that the potential energy of a spring (or bar) is \begin{equation*} \tfrac{1}{2}F\,\Delta L=\tfrac{1}{2}\gamma F^2. \end{equation*}  The generalization for the elastic energy density in a solid body is \begin{equation} \label{Eq:II:31:27} U_{\text{elastic}}=\sum_{ijkl}\tfrac{1}{2}\gamma_{ijkl}S_{ij}S_{kl}. \end{equation}  The complete description of the elastic properties of a crystal must be given in terms of the coefficients $\gamma_{ijkl}$. This introduces us to a new beast. It is a tensor of the fourth rank. Since each index can take on any one of three values, $x$, $y$, or $z$, there are $3^4=81$ coefficients. But there are really only $21$ different numbers. First, since $S_{ij}$ is symmetric, it has only six different values, and only $36$ different coefficients are needed in Eq. (31.27). But also, $S_{ij}$ can be interchanged with $S_{kl}$ without changing the energy, so $\gamma_{ijkl}$ must be symmetric if we interchange $ij$ and $kl$. This reduces the number of different coefficients to $21$. So to describe the elastic properties of a crystal of the lowest possible symmetry requires $21$ elastic constants! This number is, of course, reduced for crystals of higher symmetry. For example, a cubic crystal has only three elastic constants, and an isotropic substance has only two. That the latter is true can be seen as follows. How can the components of $\gamma_{ijkl}$ be independent of the direction of the axes, as they must be if the material is isotropic? Answer: They can be independent only if they are expressible in terms of the tensor $\delta_{ij}$. There are two possible expressions, $\delta_{ij}\delta_{kl}$ and $\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}$, which have the required symmetry, so $\gamma_{ijkl}$ must be a linear combination of them. Therefore, for isotropic materials, \begin{equation*} \gamma_{ijkl}=a(\delta_{ij}\delta_{kl})+ b(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}), \end{equation*}  and the material requires two constants, $a$ and $b$, to describe its elastic properties. We will leave it for you to show that a cubic crystal needs only three. As a final example, this time of a third-rank tensor, we have the piezoelectric effect. Under stress, a crystal generates an electric field proportional to the stress; hence, in general, the law is \begin{equation*} E_i=\sum_{j,k}P_{ijk}S_{jk}, \end{equation*}  where $E_i$ is the electric field, and the $P_{ijk}$ are the piezoelectric coefficients—or the piezoelectric tensor. Can you show that if the crystal has a center of inversion (invariant under $x,y,z\to-x,-y,-z$) the piezoelectric coefficients are all zero?