Since $dx_{\mu}$~is contravariant and $ds$~invariant, a ``velocity'' $dx_{\mu}/ds$ is a contravariant vector. Hence if $A_{\mu}$~is any covariant vector the inner product \[ A_{\mu}\, \frac{dx_{\mu}}{ds} \text{ is invariant.} \] \PageSep{61} The rate of change of this expression per unit interval along any assigned curve must also be independent of the coordinate-system, i.e.\ \[ \frac{d}{ds} \left(A_{\mu}\, \frac{dx_{\mu}}{ds}\right) \text{ is invariant.} \Tag{(29.1)} \] This assumes that we keep to the same absolute curve however the coordinate-system is varied. The result~\Eq{(29.1)} is therefore only of practical use if it is applied to a curve which is defined independently of the coordinate-system. We shall accordingly apply it to a geodesic. Performing the differentiation, \[ \frac{\dd A_{\mu}}{\dd x_{\nu}}\, \frac{dx_{\nu}}{ds}\, \frac{dx_{\mu}}{ds} + A_{\mu}\, \frac{d^{2}x_{\mu}}{ds^{2}} \text{ is invariant along a geodesic.} \Tag{(29.2)} \] From \Eq{(28.5)} we have that along a geodesic \[ A_{\mu}\, \frac{d^{2}x_{\mu}}{ds^{2}} = A_{\alpha}\, \frac{d^{2}x_{\alpha}}{ds^{2}} = -A_{\alpha}\, \{\mu\nu, \alpha\}\, \frac{dx_{\mu}}{ds}\, \frac{dx_{\nu}}{ds}. \] Hence \Eq{(29.2)}~gives \[ \frac{dx_{\mu}}{ds}\, \frac{dx_{\nu}}{ds} \left(\frac{\dd A_{\mu}}{\dd x_{\nu}} - A_{\alpha} \{\mu\nu, \alpha\}\right) \text{ is invariant.} \] The result is now general since the curvature (which distinguishes the geodesic) has been eliminated by using the equations~\Eq{(28.5)} and only the gradient of the curve ($dx_{\mu}/ds$ and $dx_{\nu}/ds$) has been left in the expression. Since $dx_{\mu}/ds$ and $dx_{\nu}/ds$ are contravariant vectors, their co-factor is a covariant tensor of the second rank. We therefore write \[ A_{\mu\nu} = \frac{\dd A_{\mu}}{\dd x_{\nu}} - \{\mu\nu, \alpha\}\, A_{\alpha}, \Tag{(29.3)} \] and the tensor~$A_{\mu\nu}$ is called the \emph{covariant derivative} of~$A_{\mu}$. By raising a suffix we obtain two associated tensors ${A^{\mu}}_{\nu}$ and ${A_{\mu}}^{\nu}$ which must be distinguished since the two suffixes are not symmetrical. The first of these is the most important, and is to be understood when the tensor is written simply as~$A_{\nu}^{\mu}$ without distinction of original position.