Substituting in~\Eq{(77.2)} we find \begin{align*} E_{1}^{1} = -E_{2}^{2} = -E_{3}^{3} = E_{4}^{4} &= \tfrac{1}{2} e^{-\lambda-\nu} \kappa_{4}'^{2} \\ &= \frac{1}{2}\, \frac{\epsilon^{2}}{r^{4}}. \Tag{(78.5)} \end{align*} By \Eq{(77.7)} we have to substitute $-8\pi E_{\mu\nu}$ for zero on the right-hand side of (\EqNo{38.61}--\EqNo{38.64}). The first and fourth equations give as before $\lambda' = -\nu'$; and the second equation now becomes \begin{align*} e^{\nu} (1 + r\nu') - 1 &= -8\pi g_{22} E_{2}^{2} \\ &= -4\pi \epsilon^{2}/r^{2}. \end{align*} Hence writing $e^{\nu} = \gamma$, \[ \gamma + r\gamma' = 1 - 4\pi \epsilon^{2}/r^{2}, \] so that \[ r\gamma = r + 4\pi \epsilon^{2}/r - 2m, \] where $2m$~is a constant of integration. Hence the gravitational field due to an electron is given by \[ ds^{2} = -\gamma^{-1}\, dr^{2} - r^{2}\, d\theta^{2} - r^{2}\sin^{2}\theta\, d\phi^{2} + \gamma\, dt^{2}, \] with \[ \gamma = 1 - \frac{2m}{r} + \frac{4\pi\epsilon^{2}}{r^{2}}. \Tag{(78.6)} \] This result appears to have been first given by Nordström. I have here followed the solution as given by G.~B. Jeffery\footnotemark.\footnotetext {\Title{Proc.\ Roy.\ Soc.}\ vol.~99\Vol{A}, p.~123.} \PageSep{186} The effect of the term $4\pi\epsilon^{2}/r^{2}$ is that the effective mass decreases as $r$~decreases. This is what we should naturally expect because the mass or energy is spread throughout space. We cannot put the constant~$m$ equal to zero, because that would leave a \emph{repulsive} force on an uncharged particle varying as the inverse cube of the distance; by \Eq{(55.8)} the approximate Newtonian potential is $m/r - 2\pi\epsilon^{2}/r^{2}$. The constant~$m$ can be identified with the mass and $4\pi\epsilon$ with the electric charge of the particle. The known experimental values for the negative electron are \begin{align*} m &= 7 · 10^{-56} \text{ cm.}, \\ a &= \frac{2\pi\epsilon^{2}}{m} = 1.5 · 10^{-13} \text{ cm.} \end{align*} The quantity~$a$ is usually considered to be of the order of magnitude of the radius of the electron, so that at all points outside the electron $m/r$~is of order $10^{-40}$ or smaller. Since $\lambda + \nu = 0$, \Eq{(78.4)}~becomes \[ F_{41} = \kappa_{4}' = \frac{\epsilon}{r^{2}}, \] which justifies our identification of~$4\pi\epsilon$ with the electric charge. This example shows how very slight is the gravitational effect of the electronic energy. We can discuss most electromagnetic problems without taking account of the non-Euclidean character which an electromagnetic field necessarily imparts to space-time, the deviations from Euclidean geometry being usually so small as to be negligible in the cases we have to consider.