It will be seen by reference to \Eq{(73.41)} and~\Eq{(73.42)} that Maxwell's equations are equivalent to \begin{gather*} \frac{\dd F_{\mu\nu}}{\dd x_{\sigma}} + \frac{\dd F_{\nu\sigma}}{\dd x_{\mu}} + \frac{\dd F_{\sigma\mu}}{\dd x_{\nu}} = 0, \Tag{(73.71)}\displaybreak[0] \\ \frac{\dd F^{\mu\nu}}{\dd x_{\nu}} = J^{\mu}. \Tag{(73.72)} \end{gather*} The first comprises the four equations \Eq{(73.61)} and~\Eq{(73.64)}; and the second comprises \Eq{(73.62)} and~\Eq{(73.63)}. \PageSep{173} On substituting $F_{\mu\nu} = \dd\kappa_{\mu}/\dd x_{\nu} - \dd\kappa_{\nu}\dd x_{\mu}$ in~\Eq{(73.71)} it will be seen that the equation is satisfied identically. Also \Eq{(73.72)}~is the simplified form for Galilean coordinates of $(F^{\mu\nu})_{\nu} = J^{\mu}$. Hence Maxwell's laws reduce to the simple form \begin{align*} F_{\mu\nu} &= \frac{\dd\kappa_{\mu}}{\dd x_{\nu}} - \frac{\dd\kappa_{\nu}}{\dd x_{\mu}}, \Tag{(73.73)} \\ F_{\nu}^{\mu\nu} &= J^{\mu}, \Tag{(73.74)} \end{align*} which are \emph{tensor equations}. By~\Eq{(51.52)} the second equation becomes \[ \frac{\dd\mf{F}^{\mu\nu}}{\dd x_{\nu}} = \mf{J}^{\mu}. \Tag{(73.75)} \] Owing to the antisymmetry of~$\mf{F}^{\mu\nu}$, $\dd^{2}\mf{F}^{\mu\nu}/\dd x_{\mu}\, \dd x_{\nu}$ vanishes, the terms in the summation cancelling in pairs. Hence \[ \frac{\dd^{2}\mf{F}^{\mu\nu}}{\dd x_{\mu}\, \dd x_{\nu}} = \frac{\dd\mf{J}^{\mu}}{\dd x_{\mu}} = 0, \Tag{(73.76)} \] whence, by~\Eq{(51.12)}, \[ (J^{\mu})_{\mu} = 0. \Tag{(73.77)} \] The divergence of the charge-and-current vector vanishes. \index{Charge, electric, conservation of}% \index{Electric charge!conservation of}% For our original coordinates \Eq{(73.77)}~becomes \[ \frac{\dd\sigma_{x}}{\dd x} + \frac{\dd\sigma_{y}}{\dd y} + \frac{\dd\sigma_{z}}{\dd z} + \frac{\dd\rho}{\dd t} = 0. \Tag{(73.78)} \] If the current is produced by the motion of the charge with velocity $(u, v, w)$, we have $\sigma_{x}, \sigma_{y}, \sigma_{z} = \rho u, \rho v, \rho w$, so that \[ \frac{\dd(\rho u)}{\dd x} + \frac{\dd(\rho v)}{\dd y} + \frac{\dd(\rho w)}{\dd z} + \frac{\dd\rho}{\dd t} = 0, \] which is the usual equation of continuity (cf.~\Eq{(53.71)}), showing that electric \index{Continuity, equation of!in electric flow}% charge is conserved.