Summing the fields of force of a number of particles, we obtain \[ ds^{2} = -(1 + 2\Omega) (dx^{2} + dy^{2} + dz^{2}) + (1 - 2\Omega) dt^{2}, \Tag{(46.2)} \] \PageSep{102} where \[ \Omega = \sum \frac{m}{r} = \text{Newtonian potential at the point considered.} \] The inaccuracy in neglecting the interference of the fields of the particles is of the same order as that due to the neglect of~$m^{2}/r^{2}$, if the number of particles is not unduly large. Now calculate the~$G_{\mu\nu}$ for the expression~\Eq{(46.2)}. We have %[** 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_{\rho}\, \dd x_{\sigma}} + \frac{\dd^{2} g_{\rho\sigma}}{\dd x_{\mu}\, \dd x_{\nu}} - \frac{\dd^{2} g_{\mu\sigma}}{\dd x_{\rho}\, \dd x_{\nu}} - \frac{\dd^{2} g_{\rho\nu}}{\dd x_{\mu}\, \dd x_{\sigma}}\right) \Tag{(46.3)} \end{align*} by~\Eq{(34.5)}. The non-linear terms are left out because they would involve~$\Omega^{2}$ which is of the order~$(m/r)^{2}$ already neglected. The only terms which survive are those in which the $g$'s have like suffixes. Consider the last three terms in the bracket; for $G_{11}$ they become \[ \frac{1}{2}\biggl(g^{11}\, \frac{\dd^{2} g_{11}}{\dd x_{1}^{2}} + g^{22}\, \frac{\dd^{2} g_{22}}{\dd x_{1}^{2}} + g^{33}\, \frac{\dd^{2} g_{33}}{\dd x_{1}^{2}} + g^{44}\, \frac{\dd^{2} g_{44}}{\dd x_{1}^{2}} - g^{11}\, \frac{\dd^{2} g_{11}}{\dd x_{1}^{2}} - g^{11}\, \frac{\dd^{2} g_{11}}{\dd x_{1}^{2}}\biggr). \] Substituting for the $g$'s from~\Eq{(46.2)} we find that the result vanishes (neglecting~$\Omega^{2}$). For $G_{44}$ the result vanishes for a different reason, viz.\ because $\Omega$~does not contain $x_{4}$ ($= t$). Hence \[ G_{\mu\nu} = \tfrac{1}{2} g^{\sigma\rho}\, \frac{\dd^{2} g_{\mu\nu}}{\dd x_{\sigma}\, \dd x_{\rho}} = \tfrac{1}{2} \Wave g_{\mu\nu}\quad\text{as in \Eq{(30.65)}.} \Tag{(46.4)} \] Since time is not involved \begin{gather*} \Wave = - \nabla^{2}, \\ \begin{aligned} G_{11},\ G_{22},\ G_{33},\ G_{44} &= -\tfrac{1}{2} \nabla^{2}(g_{11}, g_{22}, g_{33}, g_{44}) \\ &= \nabla^{2}\Omega \qquad\text{by \Eq{(46.2)}.} \end{aligned} \end{gather*} Hence, making at this point the transition to continuous matter, \[ G_{11},\ G_{22},\ G_{33},\ G_{44} = -4\pi\rho \qquad\text{by \Eq{(46.1)}.} \Tag{(46.5)} \] Also \begin{align*} G = g^{\mu\nu} G_{\mu\nu} &= -G_{11} - G_{22} - G_{33} + G_{44} \\ &= 8\pi\rho \end{align*} to the same approximation. Consider the tensor defined by \index{T@$T_{\mu\nu}$ (energy-tensor)}% \[ -8\pi T_{\mu\nu} = G_{\mu\nu} - \tfrac{1}{2} g_{\mu\nu} G. \Tag{(46.6)} \] We readily find \[ T_{\mu\nu} = 0,\quad\text{except $T_{44} = \rho$,} \] and raising the suffixes \[ T^{\mu\nu} = 0,\quad\text{except $T^{44} = \rho$,} \Tag{(46.7)} \] since the $g^{\mu\nu}$~are Galilean to the order of approximation required.