If we calculate from~\Eq{(57.6)} the value of \[ \frac{\dd}{\dd x_{\alpha}} (h_{\mu}^{\alpha} - \tfrac{1}{2} \delta_{\mu}^{\alpha} h), \] the operator~$\dd/\dd x_{\alpha}$ indicates a displacement in space and time of the point considered, involving a change of~$r'$. We may, however, keep $r'$ constant on the right-hand side and displace to the same extent the element~$dV'$ where $(T_{\mu}^{\alpha})'$~is calculated. Thus \[ \frac{\dd}{\dd x_{\alpha}} (h_{\mu}^{\alpha} - \tfrac{1}{2} \delta_{\mu}^{\alpha} h) = -4\int \left\{\frac{\dd}{\dd x_{\alpha}}(T_{\mu}^{\alpha})\right\}' \frac{dV'}{r'}. \] But by~\Eq{(55.2)} $\dd T_{\mu}^{\alpha}/\dd x_{\alpha}$ is of the second order of small quantities, so that to our approximation \Eq{(57.4)}~is satisfied. \PageSep{130} The result is that \[ \Wave h_{\mu\nu} = 2G_{\mu\nu} \Tag{(57.7)} \] satisfies the gravitational equations correctly to the first order, because both the equations into which we have divided~\Eq{(57.2)} then become satisfied. Of course there may be other solutions of~\Eq{(57.2)}, which do not satisfy \Eq{(57.31)}~and \Eq{(57.32)} separately. For a static field \Eq{(57.7)}~reduces to \begin{align*} -\nabla^{2} h_{\mu\nu} &= 2G_{\mu\nu} \\ &= -16\pi (T_{\mu\nu} - \tfrac{1}{2} \delta_{\mu\nu}T) \quad\text{by~\Eq{(54.5)}.} \end{align*} Also for matter at rest $T = T_{44} = \rho$ (the \emph{inertial} density) and the other components of~$T_{\mu\nu}$ vanish; thus \[ \nabla^{2} (h_{11}, h_{22}, h_{33}, h_{44}) = 8\pi\rho(1, 1, 1, 1). \] For a single particle the solution of this equation is well known to be \[ h_{11}, h_{22}, h_{33}, h_{44} = -\frac{2m}{r}. \] Hence by~\Eq{(57.1)} the complete expression for the interval is \[ ds^{2} = -\left(1 + \frac{2m}{r}\right)(dx^{2} + dy^{2} + dz^{2}) + \left(1 - \frac{2m}{r}\right) dt^{2}, \Tag{(57.8)} \] agreeing with~\Eq{(46.15)}. But $m$~as here introduced is the inertial mass and not \index{Gravitational mass of sun!equality with inertial mass}% \index{Inertial mass!equal to gravitational mass}% \index{Mass!gravitational and inertial}% merely a constant of integration. We have shown in \SecRefs{38}, \SecNum{39} that the~$m$ in~\Eq{(46.15)} is the gravitational mass reckoned with constant of gravitation unity. Hence we see that inertial mass and gravitational mass are equal and expressed in the same units, when the constant of proportionality between the world-tensor and the physical-tensor is chosen to be~$8\pi$ as in~\Eq{(54.3)}.