For rectangular coordinates the raising or lowering of a suffix leaves the components unaltered in three-dimensional space\footnotemark;\footnotetext {If $ds^{2} = dx_{1}^{2} + dx_{2}^{2} + dx_{3}^{2}$, $g_{\mu\nu} = g^{\mu\nu} = g_{\mu}^{\nu}$ so that all three tensors are merely substitution-operators.} and it merely reverses \index{Components, covariant and contravariant}% the signs of some of the components for Galilean coordinates in four-dimensional space-time. Since the elementary definitions of physical quantities refer to rectangular axes and time, we can generally use any one of the associated tensors to represent a physical entity without infringing pre-relativity definitions. This leads to a somewhat enlarged view of a tensor as having in itself no particular covariant or contravariant character, but having \emph{components} of various degrees of covariance or contravariance represented by the whole system of associated tensors. That is to say, the raising or lowering of suffixes will not be regarded as altering the individuality of the tensor; and reference to a tensor~$A_{\mu\nu}$ may (if the context permits) be taken to include the associated tensors~$A_{\mu}^{\nu}$ and~$A^{\mu\nu}$. It is useful to notice that dummy suffixes have a certain freedom of movement between the tensor-factors of an expression. Thus \[ A_{\alpha\beta} B^{\alpha\beta} = A^{\alpha\beta} B_{\alpha\beta},\quad A_{\mu\alpha} B^{\nu\alpha} = {A_{\mu}}^{\alpha} {B^{\nu}}_{\alpha}. \Tag{(26.3)} \] The suffix may be raised in one term provided it is lowered in the other. The proof follows easily from \Eq{(26.1)} and~\Eq{(26.2)}. In the elementary vector theory two vectors are said to be \emph{perpendicular} \index{Length of a vector}% \index{Perpendicularity of vectors}% \index{Self-perpendicular vector}% if their scalar-product vanishes; and the square of the \emph{length} of the vector is its scalar-product into itself. Corresponding definitions are adopted in the tensor calculus. The vectors $A_{\mu}$ and~$B_{\mu}$ are said to be \emph{perpendicular} if \[ A_{\mu} B^{\mu} = 0. \Tag{(26.4)} \] If $l$~is the \emph{length} of~$A_{\mu}$ (or~$A^{\mu}$) \[ l^{2} = A_{\mu} A^{\mu}. \Tag{(26.5)} \] A vector is self-perpendicular if its length vanishes. \PageSep{58} The interval is the length of the corresponding displacement~$dx_{\mu}$ because \begin{align*} ds^{2} &= g_{\mu\nu}\, (dx)^{\mu} · (dx)^{\nu} \\ &= (dx)_{\nu} (dx)^{\nu} \end{align*} by~\Eq{(26.2)}. A displacement is thus self-perpendicular when it is along a light-track, $ds = 0$.