It is usually assumed that the non-Maxwellian stresses are confined to the interior, or the close proximity, of the electrons, and do not wander about in the detached way that the Maxwellian stresses do, e.g.\ in light-waves. I shall adopt this view in order not to deviate too widely from other writers, although I do not see any particular reason for believing it to be true\footnotemark.\footnotetext {We may evade the difficulty by extending the definition of electrons or matter to include all regions where Maxwell's equations are inadequate (e.g.\ regions containing quanta).} If then all non-Maxwellian stresses are closely bound to the electrons, it follows that in regions containing no matter $E_{\mu}^{\nu}$~is the entire energy-tensor. Then \Eq{(54.3)}~becomes \[ G_{\mu}^{\nu} - \tfrac{1}{2} g_{\mu}^{\nu} G = -8\pi E_{\mu}^{\nu}. \Tag{(77.6)} \] Contracting, \[ G = 8\pi E = 0, \] and the equation simplifies to \[ G_{\mu\nu} = -8\pi E_{\mu\nu} \Tag{(77.7)} \] for regions containing electromagnetic fields but no matter. We may notice that the Gaussian curvature of space-time is zero even when electromagnetic energy is present provided there are no electrons in the region. Since for electromagnetic energy the invariant mass,~$m$, is zero, and the relative mass,~$M$, is finite, the equation~\Eq{(12.3)} \[ M = m\, \Chg{dt/ds}{\frac{dt}{ds}} \] shows that $ds/dt$~is zero. Accordingly free electromagnetic energy must always have the velocity of light. \PageSep{185} \Section{78.}{The gravitational field of an electron} \index{Electron!gravitational field of}% This problem differs from that of the gravitational field of a particle (\SecRef{38}) \index{Gravitational field of a particle!of an electron}% in that the electric field spreads through all space, and consequently the energy-tensor is not confined to a point or small sphere at the origin. For the most general symmetrical field we take as before \[ g_{11} = -e^{\lambda},\quad g_{22} = -r^{2},\quad g_{33} = -r^{2} \sin^{2}\theta,\quad g_{44} = e^{\nu}. \Tag{(78.1)} \] Since the electric field is static, we shall have \[ F, G, H = \kappa_{1}, \kappa_{2}, \kappa_{3} = 0, \] and $\kappa_{4}$~will be a function of $r$~only. Hence the only surviving components of~$F_{\mu\nu}$ are \[ F_{41} = -F_{14} = \kappa_{4}', \Tag{(78.2)} \] the accent denoting differentiation with respect to~$r$. Then \[ F^{41} = g^{44} g^{11} F_{41} = -e^{-(\lambda+\nu)} \kappa_{4}', \] and \[ \mf{F}^{41} = F^{41} \sqrt{-g} = -e^{-\frac{1}{2}(\lambda+\nu)} r^{2} \sin\theta · \kappa_{4}'. \] Hence by~\Eq{(73.75)} the condition for no electric charge and current (except at the singularity at the origin) is \[ \frac{\dd\mf{F}^{41}}{\dd x_{1}} = -\sin\theta\, \frac{\dd}{\dd r}(e^{-\frac{1}{2}(\lambda+\nu)} r^{2}\kappa_{4}') = 0, \Tag{(78.3)} \] so that \[ \kappa_{4}' = \frac{\epsilon}{r^{2}}\, e^{\frac{1}{2}(\lambda+\nu)}, \Tag{(78.4)} \] where $\epsilon$~is a constant of integration.