The term gravitational mass can be used in two senses; it may refer to \Item{(a)}~the response of a particle to a gravitational field of force, or \Item{(b)}~to its power of producing a gravitational field of force. In the sense~\Item{(a)} its identity with inertial mass is axiomatic in our theory, the separation of the field of force from the inertial field being dependent on our arbitrary choice of an abstract geometry. We accordingly use the term exclusively in the sense~\Item{(b)}, and we have shown in \SecRefs{38}, \SecNum{39} that the constant of integration~$m$ represents the gravitational mass. But in the present discussion the $\rho_{0}$ which occurs in the tensor~$T_{\mu\nu}$ refers to inertial mass defined by the conservation of energy and momentum. The connection is made \Foreign{via} equation~\Eq{(54.3)}, where on the left the mass appears in terms of~$g_{\mu\nu}$, i.e.\ in terms of its power of exerting (or being accompanied by) a gravitational field; and on the right it appears in the energy-tensor which comprises~$\rho_{0}$ according to~\Eq{(53.1)}. But it will be remembered that the factor~$8\pi$ in~\Eq{(54.3)} was chosen arbitrarily, and this must now be justified\footnotemark.\footnotetext {It has been justified in \SecRef{46}, which has a close connection with the present paragraph; but the argument is now proceeding in the reverse direction.} This coefficient of proportionality corresponds to the Newtonian constant of gravitation. The proportionality of gravitational and inertial mass, and the ``constant of gravitation'' which connects them, are conceptions belonging to the approximate Newtonian scheme, and therefore presuppose that the gravitational fields are so weak that the equations can be treated as linear. For more intense fields the Newtonian terminology becomes ambiguous, and it is idle to inquire whether the constant of gravitation really remains constant when the mass is enormously great. Accordingly we here discuss only the limiting case of very weak fields, and set \[ g_{\mu\nu} = \delta_{\mu\nu} + h_{\mu\nu}, \Tag{(57.1)} \] where $\delta_{\mu\nu}$~represents Galilean values, and $h_{\mu\nu}$~will be a small quantity of the first order whose square is neglected. The derivatives of the~$g_{\mu\nu}$ will be small quantities of the first order. We have, correct to the first order, %[** TN: Not broken in the original] \begin{align*} G_{\mu\nu} &= g^{\sigma\rho} B_{\mu\nu\sigma\rho} \\ &= \tfrac{1}{2}g^{\sigma\rho} \left( \frac{\dd^{2} g_{\mu\nu}}{\dd x_{\sigma}\, \dd x_{\rho}} + \frac{\dd^{2} g_{\sigma\rho}}{\dd x_{\mu}\, \dd x_{\nu}} - \frac{\dd^{2} g_{\mu\sigma}}{\dd x_{\nu}\, \dd x_{\rho}} - \frac{\dd^{2} g_{\nu\rho}}{\dd x_{\mu}\, \dd x_{\sigma}}\right) \Tag{(57.2)} \end{align*} by~\Eq{(34.5)}.