This must hold for all values of the arbitrary displacements~$\delta x_{\sigma}$ at all \index{Geodesic, equations of}% points, hence the coefficient in the integrand must vanish at all points on the path. Thus \[ \frac{1}{2} \frac{dx_{\mu}}{ds}\, \frac{dx_{\nu}}{ds}\, \frac{\dd g_{\mu\nu}}{\dd x_{\sigma}} - \frac{1}{2} \frac{dg_{\mu\sigma}}{ds}\, \frac{dx_{\mu}}{ds} - \frac{1}{2} \frac{dg_{\sigma\nu}}{ds}\, \frac{dx_{\nu}}{ds} - \frac{1}{2} g_{\mu\sigma}\, \frac{d^{2}x_{\mu}}{ds^{2}} - \frac{1}{2} g_{\sigma\nu}\, \frac{d^{2}x_{\nu}}{ds^{2}} = 0. \] Now\footnote {These simple formulae are noteworthy as illustrating the great value of the summation convention. The law of total differentiation for four coordinates becomes formally the same as for one coordinate.} \[ \frac{dg_{\mu\sigma}}{ds} = \frac{\dd g_{\mu\sigma}}{\dd x_{\nu}}\, \frac{dx_{\nu}}{ds} \quad\text{and}\quad \frac{dg_{\sigma\nu}}{ds} = \frac{\dd g_{\sigma\nu}}{\dd x_{\mu}}\, \frac{dx_{\mu}}{ds}. \] Also in the last two terms we replace the dummy suffixes $\mu$ and~$\nu$ by~$\epsilon$. The equation then becomes \[ \frac{1}{2}\, \frac{dx_{\mu}}{ds}\, \frac{dx_{\nu}}{ds}\, \biggl(\frac{\dd g_{\mu\nu}}{\dd x_{\sigma}} - \frac{\dd g_{\mu\sigma}}{\dd x_{\nu}} - \frac{\dd g_{\nu\sigma}}{\dd x_{\mu}}\biggr) - g_{\epsilon\sigma}\, \frac{d^{2} x_{\epsilon}}{ds^{2}} = 0. \Tag{(28.3)} \] We can get rid of the factor~$g_{\epsilon\sigma}$ by multiplying through by~$g^{\sigma\alpha}$ so as to form the substitution operator~$g_{\epsilon}^{\alpha}$. Thus \[ \frac{1}{2}\, \frac{dx_{\mu}}{ds}\, \frac{dx_{\nu}}{ds}\, g^{\sigma\alpha} \biggl(\frac{\dd g_{\mu\sigma}}{\dd x_{\nu}} + \frac{\dd g_{\nu\sigma}}{\dd x_{\mu}} - \frac{\dd g_{\mu\nu}}{\dd x_{\sigma}}\biggr) + \frac{d^{2} x_{\alpha}}{ds^{2}} = 0, \Tag{(28.4)} \] or, by~\Eq{(27.2)} \[ \frac{d^{2} x_{\alpha}}{ds^{2}} + \{\mu\nu, \alpha\}\, \frac{dx_{\mu}}{ds}\, \frac{dx_{\nu}}{ds} = 0. \Tag{(28.5)} \] For $\alpha = 1$, $2$, $3$, $4$ this gives the four equations determining a geodesic. \Section{29.}{Covariant derivative of a vector} \index{Covariant derivative of vector}% \index{Derivative!covariant}% The derivative of an invariant is a covariant vector (\SecRef{19}), but the derivative of a vector is not a tensor. We proceed to find certain tensors which are used in this calculus in place of the ordinary derivatives of vectors.