A simple example of an expression of the second rank is afforded by the stresses in a solid or viscous fluid. The component of stress denoted by~$p_{xy}$ is the traction in the $y$-direction across an interface perpendicular to the $x$-direction. Each component is thus associated with two directions. %[** TN: Shortened running head] \Section[Inner multiplication and contraction]{24.}{Inner multiplication and contraction. The quotient law} If we multiply $A_{\mu}$ by $B^{\nu}$ we obtain sixteen quantities $A_{1}B^{1}$, $A_{1}B^{2}$, $A_{2}B^{1}$,~\dots constituting a mixed tensor. Suppose that we wish to consider the four \PageSep{53} ``diagonal'' terms $A_{1}B^{1}$, $A_{2}B^{2}$, $A_{3}B^{3}$, $A_{4}B^{4}$; we naturally try to abbreviate these by writing them~$A_{\mu}B^{\mu}$. But by the summation convention $A_{\mu}B^{\mu}$~stands for the sum of the four quantities. The convention is right. We have no use for them individually since they do not form a vector; but the sum is of great importance. $A_{\mu}B^{\mu}$~is called the \emph{inner product} of the two vectors, in contrast to the \index{Contraction of tensors}% \index{Inner multiplication}% \index{Multiplication, inner and outer}% \index{Product, inner and outer}% ordinary or \emph{outer product}~$A_{\mu}B^{\nu}$. In rectangular coordinates the inner product coincides with the \emph{scalar-product} defined in the well-known elementary theory of vectors; but the outer product is not the so-called \emph{vector-product} of the elementary theory. By a similar process we can form from any mixed tensor~$A_{\mu\nu\sigma}^{\tau}$ a ``contracted\footnotemark''\footnotetext {German, \Foreign{verjüngt}.}% tensor~$A_{\mu\nu\sigma}^{\sigma}$, which is two ranks lower since $\sigma$~has now become a dummy suffix. To prove that $A_{\mu\nu\sigma}^{\sigma}$~is a tensor, we set $\tau = \sigma$ in~\Eq{(23.3)}, \[ A_{\mu\nu\sigma}'^{\sigma} = \frac{\dd x_{\alpha}}{\dd x_{\mu}'}\, \frac{\dd x_{\beta}}{\dd x_{\nu}'}\, \frac{\dd x_{\gamma}}{\dd x_{\sigma}'}\, \frac{\dd x_{\sigma}'}{\dd x_{\delta}}\, A_{\alpha\beta\gamma}^{\delta}. \] The substitution operator $\dfrac{\dd x_{\gamma}}{\dd x_{\sigma}'}\, \dfrac{\dd x_{\sigma}'}{\dd x_{\delta}}$ changes~$\delta$ to~$\gamma$ in~$A_{\alpha\beta\gamma}^{\delta}$ by~\Eq{(22.4)}. Hence \[ A_{\mu\nu\sigma}'^{\sigma} = \frac{\dd x_{\alpha}}{\dd x_{\mu}'}\, \frac{\dd x_{\beta}}{\dd x_{\nu}'}\, A_{\alpha\beta\gamma}^{\gamma}. \] Comparing with the transformation law~\Eq{(23.22)} we see that $A_{\mu\nu\sigma}^{\sigma}$~is a covariant tensor of the second rank. Of course, the dummy suffixes $\gamma$ and~$\sigma$ are equivalent. Similarly, setting $\nu = \mu$ in~\Eq{(23.23)}, \[ A_{\mu}'^{\mu} = \frac{\dd x_{\alpha}}{\dd x_{\mu}'}\, \frac{\dd x_{\mu}'}{\dd x_{\beta}}\, A_{\alpha}^{\beta} = A_{\alpha}^{\alpha} = A_{\mu}^{\mu}, \] that is to say $A_{\mu}^{\mu}$~is unaltered by a transformation of coordinates. Hence it is an invariant.