The cross product
We should point out that we have been using tensors of the second rank since Chapter 20 of Volume I. There, we defined a “torque in a plane,” such as $\tau_{xy}$ by \begin{equation*} \tau_{xy}=xF_y-yF_x. \end{equation*}  Generalized to three dimensions, we could write \begin{equation} \label{Eq:II:31:22} \tau_{ij}=r_iF_j-r_jF_i. \end{equation}  The quantity $\tau_{ij}$ is a tensor of the second rank. One way to see that this is so is by combining $\tau_{ij}$ with some vector, say the unit vector $\FLPe$, according to \begin{equation*} \sum_j\tau_{ij}e_j. \end{equation*}  If this quantity is a vector, then $\tau_{ij}$ must transform as a tensor—this is our definition of a tensor. Substituting for $\tau_{ij}$, we have \begin{align*} \sum_j\tau_{ij}e_j&=\sum_jr_iF_je_j-\sum_jr_je_jF_i\\[1ex] &=r_i(\FLPF\cdot\FLPe)-(\FLPr\cdot\FLPe)F_i. \end{align*}  Since the dot products are scalars, the two terms on the right-hand side are vectors, and likewise their difference. So $\tau_{ij}$ is a tensor. But $\tau_{ij}$ is a special kind of tensor; it is antisymmetric, that is, \begin{equation*} \tau_{ij}=-\tau_{ji}, \end{equation*}  so it has only three nonzero terms—$\tau_{xy}$, $\tau_{yz}$, and $\tau_{zx}$. We were able to show in Chapter 20 of Volume I that these three terms, almost “by accident,” transform like the three components of a vector, so that we could define \begin{equation*} \FLPtau=(\tau_x,\tau_y,\tau_z)=(\tau_{yz},\tau_{zx},\tau_{xy}) \end{equation*}  We say “by accident,” because it happens only in three dimensions. In four dimensions, for instance, an antisymmetric tensor of the second rank has up to six nonzero terms and certainly cannot be replaced by a vector with four components. Just as the axial vector $\FLPtau=\FLPr\times\FLPF$ is a tensor, so also is every cross product of two polar vectors—all the same arguments apply. By luck, however, they are also representable by vectors (really pseudo vectors), so our mathematics has been made easier for us.  Mathematically, if $\FLPa$ and $\FLPb$ are any two vectors, the nine quantities $a_ib_j$ form a tensor (although it may have no useful physical purpose). Thus, for the position vector $\FLPr$, $r_ir_j$ is a tensor, and since $\delta_{ij}$ is also, we see that the right side of Eq. (31.20) is indeed a tensor. Likewise Eq. (31.22) is a tensor, since the two terms on the right-hand side are tensors.