From~\Eq{(56.6)} \begin{align*} mg_{\mu\nu}\, \frac{dx_{\nu}}{ds} · \frac{d}{ds} \left(m\, \frac{dx_{\mu}}{ds}\right) &= -m^{2} [\alpha\beta, \nu]\, \frac{dx_{\nu}}{ds}\, \frac{dx_{\alpha}}{ds}\, \frac{dx_{\beta}}{ds} \\ &= -\tfrac{1}{2} m^{2} \, \frac{\dd g_{\alpha\nu}}{\dd x_{\beta}}\, \frac{dx_{\beta}}{ds}\, \frac{dx_{\alpha}}{ds}\, \frac{dx_{\nu}}{ds} \\ &= -\tfrac{1}{2} m^{2} \, \frac{dg_{\alpha\nu}}{ds}\, \frac{dx_{\alpha}}{ds}\, \frac{dx_{\nu}}{ds} \\ &= -\tfrac{1}{2} m^{2} \, \frac{dg_{\mu\nu}}{ds}\, \frac{dx_{\mu}}{ds}\, \frac{dx_{\nu}}{ds}. \end{align*} Adding the same equation with $\mu$ and $\nu$ interchanged %[** TN: Not broken in the original] \begin{multline*} g_{\mu\nu} · m\, \frac{dx_{\nu}}{ds} · \frac{d}{ds} \left(m\, \frac{dx_{\mu}}{ds}\right) + g_{\mu\nu} · m\, \frac{dx_{\mu}}{ds} · \frac{d}{ds} \left(m\, \frac{dx_{\nu}}{ds}\right) \\ + m\, \frac{dx_{\mu}}{ds} · m\, \frac{dx_{\nu}}{ds} · \frac{dg_{\mu\nu}}{ds} = 0 \end{multline*} or \[ \frac{d}{ds} \left(g_{\mu\nu} · m\, \frac{dx_{\mu}}{ds} · m\, \frac{dx_{\nu}}{ds}\right) = 0. \] By~\Eq{(22.1)} this gives $dm^{2}/ds = 0$. Accordingly the invariant mass of an isolated particle remains constant. The present proof does not add very much to the argument in \SecRef{17} that the particle follows a geodesic because that is the only track which is absolutely defined. Here we postulate symmetrical properties for the particle (referred to proper-coordinates); this has the effect that there is no means of fixing a direction in which it could deviate from a geodesic. For further enlightenment we must wait until Chapter~\ChapNum{V}\@. \PageSep{128} \Section[Equality of gravitational and inertial mass]{57.}{Equality of gravitational and inertial mass. Gravitational \index{Gravitation, Newtonian constant of}% \index{Inertial mass}% \index{Mass!gravitational and inertial}% waves}