Page:Elektrische und Optische Erscheinungen (Lorentz) 113.jpg

 aether is acting on the ponderable body, then we must, if the energy theorem shall be satisfied, obviously obtain zero.

The increase energy in a full period T would be zero, if the surface $$\sigma$$ together with the body K would be displaced over the distance $$\mathfrak{p}T$$, and at this place would have taken the location $$\sigma$$; it factually consists of the energy amount, which, at time $$t_0$$, is more contained in $$\sigma$$ than in $$\sigma$$. This is now, as it follows form the definition given for Qε, exactly

$$\mathfrak{p}TQ_{t=t_{0}}$$.

The work mentioned above can be expressed, as we will see soon, by an expression of the form

$$\int_{t_{0}-T}^{t_{0}}dt\int S\ d\sigma$$;

thus the energy law requires that

$$\mathfrak{p}\int_{t_{0}-T}^{t_{0}}Q\ dt+\int_{t_{0}-T}^{t_{0}}dt\int A\ d\sigma+\int_{t_{0}-T}^{t_{0}}dt\int Sd\sigma=0$$

If it is additionally achieved, to represent Q as an integral over $$\sigma$$, e.g. in the form

$$Q=\int\ q\ d\sigma$$

and to show, that

then we have achieved our goal.

§ 83. From the definition given for $$Q\epsilon$$ we derive, that by $$q\epsilon d\sigma$$ we have to understand the energy content of the space, which is traversed by the element during the displacement ε, and namely we have, depending on whether the displacement takes place with respect to the inner- or the outer-side of $$\sigma$$, to apply the positive or the negative sign. Thus we have

$$q\epsilon d\sigma=-\epsilon\cos(\mathfrak{p},n)\left(2\pi V^{2}\mathfrak{d}^{2}+\frac{1}{8\pi}\mathfrak{H}^{2}\right)d\sigma$$,

and

$$\mathfrak{p}q=-\mathfrak{p}_{n}\left(2\pi V^{2}\mathfrak{d}^{2}+\frac{1}{8\pi}\mathfrak{H}^{2}\right)$$.