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

${\begin{array}{c}\Xi =4\pi V^{2}\int {\mathfrak {d}}_{x}\left({\frac {\partial {\mathfrak {d}}_{x}}{\partial x}}+{\frac {\partial {\mathfrak {d}}_{y}}{\partial y}}+{\frac {\partial {\mathfrak {d}}_{z}}{\partial z}}\right)\ d\ \tau +\\+{\frac {1}{4\pi }}\int \left\{{\mathfrak {H}}_{z}\left({\frac {\partial {\mathfrak {H}}_{x}}{\partial z}}-{\frac {\partial {\mathfrak {H}}_{z}}{\partial x}}\right)-{\mathfrak {H}}_{y}\left({\frac {\partial {\mathfrak {H}}_{y}}{\partial x}}-{\frac {\partial {\mathfrak {H}}_{x}}{\partial y}}\right)\right\}\ d\ \tau +\\+\int \left({\mathfrak {H}}_{y}{\frac {\partial {\mathfrak {d}}_{z}}{\partial t}}-{\mathfrak {H}}_{z}{\frac {\partial {\mathfrak {d}}_{y}}{\partial t}}\right)\ d\ \tau .\end{array}}$

(14)

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