Perhaps it might have been expected that with the advent of the electron \index{Non-Maxwellian stresses}% theory of matter it would become unnecessary to retain a separate material energy-tensor~$M^{\mu\nu}$, and that the whole energy and momentum could be included in the energy-tensor of the electromagnetic field. But we cannot dispense with~$M^{\mu\nu}$. The fact is that an electron must not be regarded as a purely electromagnetic phenomenon; that is to say, something enters into its constitution which is not comprised in Maxwell's theory of the electromagnetic field. In order to prevent the electronic charge from dispersing under its own repulsion, non-Maxwellian ``binding forces'' are necessary, and it is the energy, stress and momentum of these binding forces which constitute the material energy-tensor~$M^{\mu\nu}$. \Section{77.}{The electromagnetic energy-tensor} \index{Electromagnetic action!energy-tensor}% \index{Energy-tensor of matter!electromagnetic@of electromagnetic field}% To determine explicitly the value of~$E_{\mu}^{\nu}$ we have to rely on the relation found in the preceding section \[ E_{\mu\nu}^{\nu} = h_{\mu} = F_{\mu\nu} J^{\nu} = F_{\mu\nu} F_{\sigma}^{\nu\sigma}. \Tag{(77.1)} \] The solution of this differential equation is \[ E_{\mu}^{\nu} = -F^{\nu\alpha} F_{\mu\alpha} + \tfrac{1}{4} g_{\mu}^{\nu} F^{\alpha\beta} F_{\alpha\beta}. \Tag{(77.2)} \] To verify this we take the divergence, remembering that covariant differentiation obeys the usual distributive law and that $g_{\mu}^{\nu}$~is a constant. \begin{alignat*}{2} E_{\mu\nu}^{\nu} &= -F_{\nu}^{\nu\alpha} F_{\mu\alpha} &&- \phantom{\tfrac{1}{2}} F^{\nu\alpha} F_{\mu\alpha\nu} + \tfrac{1}{4} g_{\mu}^{\nu} (F_{\nu}^{\alpha\beta} F_{\alpha\beta} + F^{\alpha\beta} F_{\alpha\beta\nu}) \\ % &= -F_{\nu}^{\nu\alpha} F_{\mu\alpha} &&- \phantom{\tfrac{1}{2}} F^{\nu\alpha} F_{\mu\alpha\nu} + \tfrac{1}{2} g_{\mu}^{\nu} F^{\alpha\beta} F_{\alpha\beta\nu} \qquad\text{by~\Eq{(26.3)}} \\ % &= -F_{\nu}^{\nu\alpha} F_{\mu\alpha} &&- \tfrac{1}{2} F^{\beta\alpha} F_{\mu\alpha\beta} - \tfrac{1}{2} F^{\alpha\beta} F_{\mu\beta\alpha} + \tfrac{1}{2} F^{\alpha\beta} F_{\alpha\beta\mu} \\ \intertext{by changes of dummy suffixes,} &= F_{\nu}^{\alpha\nu} F_{\mu\alpha} &&+ \tfrac{1}{2} F^{\alpha\beta}(F_{\mu\alpha\beta} + F_{\beta\mu\alpha} + F_{\alpha\beta\mu}) \end{alignat*} by the antisymmetry of~$F^{\mu\nu}$. \PageSep{183} It is easily verified that \[ F_{\mu\alpha\beta} + F_{\beta\mu\alpha} + F_{\alpha\beta\mu} = \frac{\dd F_{\mu\alpha}}{\dd x_{\beta}} + \frac{\dd F_{\beta\mu}}{\dd x_{\alpha}} + \frac{\dd F_{\alpha\beta}}{\dd x_{\mu}} = 0 \] by \Eq{(30.3)} and~\Eq{(73.71)}; the terms containing the $3$-index symbols mutually cancel. Hence \[ E_{\mu\nu}^{\nu} = F_{\nu}^{\alpha\nu} F_{\mu\alpha} = J^{\alpha} F_{\mu\alpha}, \] agreeing with~\Eq{(77.1)}.