Let $A^{\mu}$~be the velocity-vector of the electron ($A^{\mu} = dx_{\mu}/ds$), and $\rho_{0}$~the proper-density of the charge, then by~\Eq{(73.82)} \[ J^{\mu} = \rho_{0} A^{\mu}, \Tag{(80.21)} \] and \[ \frac{d^{2}x_{\mu}}{ds^{2}} + \{\alpha\beta, \mu\}\, \frac{dx_{\alpha}}{ds}\, \frac{dx_{\beta}}{ds} = A^{\nu} (A^{\mu})_{\nu}, \Tag{(80.22)} \] as in~\Eq{(33.4)}. Considering the verification of~\Eq{(80.1)} by experiment we remark that $X$ or $F_{\mu\nu}$ refers to the applied external field, no attention being paid to the possible disturbance of this field caused by the accelerated electron itself. To distinguish this we denote the external field by~${F'}_{\mu\nu}$. The equation to be explained accordingly becomes \[ mA^{\nu}(A^{\mu})_{\nu} = -{F^{\mu}}_{\nu} (\rho_{0} A^{\nu}), \] or, lowering the suffix~$\mu$, \[ mA^{\nu} A_{\mu\nu} = -{F'^{\mu}}_{\nu} eA^{\nu}. \Tag{(80.3)} \] We have replaced the \emph{density}~$\rho_{0}$ by the \emph{quantity}~$e$ for the reason explained in the footnote. Consider now the field due to the electron itself in its own neighbourhood. This is determined by~\Eq{(74.41)} \[ \Wave F_{\mu\nu} = J_{\mu\nu} - J_{\nu\mu} - G_{\mu}^{\epsilon} F_{\epsilon\nu} + G_{\nu}^{\epsilon} F_{\epsilon\mu} + 2B_{\mu\nu\alpha\epsilon} F^{\epsilon\alpha}. \] The discussion of \SecRef{78} shows that we may safely neglect the gravitational field caused by the energy of the electron or of the external field. Hence approximately \[ \Wave F_{\mu\nu} = J_{\mu\nu} - J_{\nu\mu}. \] The solution is as in~\Eq{(74.72)} \begin{align*} F_{\mu\nu} &= \int \frac{de\, (A_{\mu\nu} - A_{\nu\mu})}{4\pi\beta r} \\ &= \frac{1}{4\pi\beta}(A_{\mu\nu} - A_{\nu\mu}) \int \frac{de}{r}, \Tag{(80.4)} \end{align*} if all parts of the electron have the same velocity~$A^{\mu}$. This result is obtained primarily for Galilean coordinates; but it is a tensor equation applying to all coordinate-systems provided that $\int de/r$~is treated as an invariant and calculated in natural measure. We shall reckon it in proper-measure and accordingly drop the factor~$\beta$. \PageSep{191} Now suppose that the electron moves in such a way that its own field on the average just neutralises the applied external field~${F'}_{\mu\nu}$ in the region occupied by the electron. The value of~$F_{\mu\nu}$ averaged for all the elements of charge constituting the electron is given by \begin{align*} eF_{\mu\nu} &= \frac{1}{4\pi}(A_{\mu\nu} - A_{\nu\mu}) \iint \frac{de_{1}\, de_{2}}{r_{12}} \\ &= \frac{1}{4\pi}(A_{\mu\nu} - A_{\nu\mu}) \frac{e^{2}}{a}, \end{align*} where $1/a$~is an average value of~$1/r_{12}$ for every pair of points in the electron. We may leave indeterminate the exact weighting of the pairs of points in taking the average, merely noting that $a$~will be a length comparable with the radius of the sphere throughout which the charge (or the greater part of it) is spread.