Page:LorentzGravitation1915.djvu/11

 a consequence of which is

$\psi_{ba}^{*}=-\psi_{ab}^{*}$

and we shall complete our definition by

The term we are considering then becomes

$\begin{array}{l} \Sigma(\overline{ab})\psi_{a'b'}^{*}\left(\dfrac{\partial q_{b'}}{\partial x_{a'}}-\dfrac{\partial q_{a'}}{\partial x_{b'}}\right)=\Sigma(\overline{ab})\psi_{ab}^{*}\left(\dfrac{\partial q_{b}}{\partial x_{a}}-\dfrac{\partial q_{a}}{\partial x_{b}}\right)=\\ \\ =\tfrac{1}{2}\Sigma(ab)\psi_{ab}^{*}\left(\dfrac{\partial q_{b}}{\partial x_{a}}-\dfrac{\partial q_{a}}{\partial x_{b}}\right)=-\Sigma(ab)\psi_{ab}^{*}\dfrac{\partial q_{a}}{\partial x_{b}}=\\ \\ =-\Sigma(ab)\dfrac{\partial\left(\psi_{ab}^{*}q_{a}\right)}{\partial x_{b}}+\Sigma(ab)\dfrac{\partial\psi_{ab}^{*}}{\partial x_{b}}q_{a}, \end{array}$|undefined

so that, using (14), we obtain for (38)

where we have taken into consideration that

$\Sigma(\overline{ab})\overline{\psi_{ab}}\left(w_{b}\delta x_{a}-w_{a}\delta x_{b}\right)=\Sigma(ab)\overline{\psi_{ab}}w_{b}\delta x_{a}.$|undefined

If we multiply (40) by $$dS$$ and integrate over the space $$S$$ the first term on the right hand side vanishes. Therefore (12) requires that in the subsequent terms the coefficient of each $$q_a$$ and of each $$\delta x_{a}$$ be 0. Therefore

and

by which (40) becomes

In (41) we have the second set of four electromagnetic equations, while (42) determines the forces exerted by the field on the charges. We see that (42) agrees with (19) (namely in virtue of (31)).

§ 12. To deduce also the equations for the momenta and the energy we proceed as in § 6. Leaving the gravitation field unchanged we shift the electromagnetic field, i. e. the values of $$w_a$$ and $$\psi_{ab}$$ in the direction of one of the coordinates, say of $$x_c$$, over a distance defined by the constant variation $$\delta x_{c}$$ so that we have