Page:Elektrische und Optische Erscheinungen (Lorentz) 112.jpg

 $$V^{2}\int[\mathfrak{d}'.\mathfrak{H}']_{n}d\sigma$$,

is the energy flux, under

$$\varkappa\ \epsilon$$

the difference calculated for the unit time, and under

$$\varkappa\ \epsilon\ dt$$

the difference calculated for the element dt, of the energy fluxes through two fixed surfaces that are mutually distant by the length ε.

Now, let $$Q\epsilon$$ be the energy, which at time t is more surrounded by our surface $$\sigma$$ in its fixed location, as when this surface would be displaced by $$\epsilon$$ in the direction of $$\mathfrak{p}$$; then we immediately see, that

By that, and furthermore by partial integration, (108) is transformed into

$$\mathfrak{p}\int_{t_{0}-T}^{t_{0}}\frac{dQ}{dt}(t_{0}-t)dt=-\mathfrak{p}TQ_{t=t_{0}-T}+\mathfrak{p}\int_{t_{0}-T}^{t_{0}}Q\ dt$$,

or

$$-\mathfrak{p}TQ_{t=t_{0}}+\mathfrak{p}\int_{t_{0}-T}^{t_{0}}Q\ dt$$,

since, except magnitudes of order $$\mathfrak{p}$$, Q has again the original value after the expiration of time T.

§ 82. Until now, we only spoke about the first member in (107). If we denote the two other members by A, then we have

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

the complete value of the energy, that travelled outwards through $$\sigma$$. If we add the increase of the energy in the interior of $$\sigma$$, and the work of the forces, by which the