Page:Zur Theorie der Strahlung in bewegten Körpern.djvu/15

 where $$\epsilon_{0}$$ has the meaning given by (20), while we obtain for $$\tau$$ after execution of the integration:

When we neglect terms starting with order $$\beta^4$$, then it becomes

As already mentioned several times, the energy quantity $$h\epsilon'$$ present in space $$R$$, is gained from mechanical work. Yet, since no work is altogether performed at uniform translation of our system, then this energy quantity must be the equivalent of a work which is performed at the acceleration of it, and which is thus added to the work against the ordinary resistance of inertia. One can easily illustrate this to oneself in the following way: Imagine a system being at rest in the beginning, and the surfaces are somehow hindered of radiating energy, so that space $$R$$ contains no energy. Now, if we bring the system suddenly to velocity $$w$$, and simultaneously set free the radiation of $$A$$ and $$B$$, then the work $$w\cdot 2\pi\ p_{1}\ \cos\psi\ \sin\psi\ d\psi$$ (14) in unit time must immediately be performed against the radiation emanating from $$A$$. As soon as the radiation arrives in $$B$$, the same amount of work is gained there; though the time $$h/c_{-} \cos\psi$$ passes until this happens; during this time, the external work (15) is thus uncompensated. Accordingly, the radiation emanating from $$B$$ immediately performs the work $$w\cdot 2\pi\ p_{2}\ \cos\psi\ \sin\psi\ d\psi$$, and now the time $$h/c_{+} \cos\psi$$ passes before this work is compensated by an equal amount of work of the external forces in $$A$$. Thus altogether, the work

$2\pi\cos\psi\sin\psi\ d\psi\ w\left(p_{1}\frac{h}{c_{-}\cos\psi}-p_{2}\frac{h}{c_{+}\cos\psi}\right)$

must be provided from the outside. If we integrate this expression from 0 to $$\pi/2$$ and if we consider (19), then we recognize that the performed work has indeed the amount $$h\epsilon'$$.