Page:Elektrische und Optische Erscheinungen (Lorentz) 023.jpg

We replace in (9), by (4) and (III)

$4\pi \rho {\mathfrak {v}}_{x}{\text{. u. s. w.}}$

by

${\frac {\partial {\mathfrak {H}}_{z}}{\partial y}}-{\frac {\partial {\mathfrak {H}}_{y}}{\partial z}}-4\pi {\frac {\partial {\mathfrak {d}}_{x}}{\partial t}}{\text{, u. s. w.,}}$

(10)

and transform the parts of the integral, that contain derivatives of ${\mathfrak {H}}_{x},{\mathfrak {H}}_{y},{\mathfrak {H}}_{z}$, by partial integration.

By consideration of equation (IV) we will find

$d\ A=d(L+U)+V^{2}d\ t\int \ [{\mathfrak {d.H}}]_{n}d\ \sigma {,}$

(11)

where

$U=2\pi V^{2}\int \ {\mathfrak {d}}^{2}d\ \tau +{\frac {1}{8\pi }}\int {\mathfrak {H}}^{2}d\ \tau$

(12)

At first is should be assumed, that the electric motions are restricted to a certain finite space, and that surface $\sigma$ is entirely outside of that space. Then at the surfaces it will be ${\mathfrak {d}}=0,{\mathfrak {H}}=0$, and

$d\ A=d(L+U).$

Therefore the magnitude L + U really applies, whose increase is equal to the work of the external forces, and which therefore is denoted by the expression "energy". It is composed of the ordinary mechanical energy L and the "electrical" energy U, and as regards the latter we find again the value given by Maxwell.

The theorem of Pointing.

§ 14. Even if we abandon the previously made assumption about $\sigma$, formula (11) allows of a simple interpretation. With Maxwell we not only assume that the electric force would have the value (12), but also, that it is really distributed over the space as it is expressed by the formula, i.e. that it amounts for unit volume

$2\pi V^{2}{\mathfrak {d}}^{2}+{\frac {1}{8\pi }}{\mathfrak {H}}^{2}$