Page:Elektrische und Optische Erscheinungen (Lorentz) 025.jpg

 where the integrals only have to be extended over the ponderable body, but like in § 13, it should taken for the entire space enclosed by $$\sigma$$.

At first, we replace $$4\pi\rho\mathfrak{v}_{x}$$, etc. by the expressions (10), and, because of (I), $$\rho$$ by

$$\frac{\partial\mathfrak{d}_{x}}{\partial x}+\frac{\partial\mathfrak{d}_{y}}{\partial y}+\frac{\partial\mathfrak{d}_{z}}{\partial z}$$

thus

Furthermore, a partial integration and application of (IV) and (II) gives (when we denote the direction constants of the perpendicular to $$\sigma$$ by $$\chi,\beta,\gamma$$)

$$\begin{array}{cl} \int\mathfrak{d}_{x}\frac{\partial\mathfrak{d}_{y}}{\partial y}\ d\ \tau & =\int\beta\mathfrak{d}_{x}\mathfrak{d}_{y}\ d\ \sigma-\int\mathfrak{d}_{y}\frac{\partial\mathfrak{d}_{x}}{\partial y}\ d\ \tau=\\ & =\int\beta\mathfrak{d}_{x}\mathfrak{d}_{y}\ d\ \sigma-\int\mathfrak{d}_{y}\frac{\partial\mathfrak{d}_{y}}{\partial x}\ d\ \tau-\frac{1}{4\pi V^{2}}\int\mathfrak{d}_{y}\frac{\partial\mathfrak{H}_{z}}{\partial t}\ d\ \tau{,}\\ \int\mathfrak{d}_{x}\frac{\partial\mathfrak{d}_{z}}{\partial z}\ d\ \tau & =\int\gamma\mathfrak{d}_{x}\mathfrak{d}_{z}\ d\ \sigma-\int\mathfrak{d}_{z}\frac{\partial\mathfrak{d}_{x}}{\partial z}\ d\ \tau=\\ & =\int\gamma\mathfrak{d}_{x}\mathfrak{d}_{z}\ d\ \sigma-\int\mathfrak{d}_{z}\frac{\partial\mathfrak{d}_{z}}{\partial x}\ d\ \tau+\frac{1}{4\pi V^{2}}\int\mathfrak{d}_{z}\frac{\partial\mathfrak{H}_{y}}{\partial t}\ d\ \tau{,} \end{array}$$

$$\int\left(\mathfrak{H}_{y}\frac{\partial\mathfrak{H}_{x}}{\partial y}+\mathfrak{H}_{z}\frac{\partial\mathfrak{H}_{x}}{\partial z}\right)\ d\ \tau=\int\left(\beta\mathfrak{H}_{x}\mathfrak{H}_{y}+\gamma\mathfrak{H}_{x}\mathfrak{H}_{z}\right)\ d\ \sigma-$$

$$-\int\mathfrak{H}_{x}\left(\frac{\partial\mathfrak{H}_{y}}{\partial y}+\frac{\partial\mathfrak{H}_{z}}{\partial z}\right)\ d\ \tau=\int\left(\beta\mathfrak{H}_{x}\mathfrak{H}_{y}+\gamma\mathfrak{H}_{x}\mathfrak{H}_{z}\right)\ d\ \sigma+$$

$$+\int\mathfrak{H}_{x}\frac{\partial\mathfrak{H}_{x}}{\partial x}\ d\ \tau.$$

If we substitute this value into (14), then several terms occur, that can be completely integrated, and eventually by a simple transformation