Page:Zur Theorie der Strahlung bewegter Koerper.djvu/14

 Incidentally, the values of $$p_{1}$$ and $$p_{2}$$ have been derived by by a peculiar consideration. However, confines himself to the case of perpendicular emission. If one accordingly put $$\cos\varphi = \cos\phi = 1$$ in the expressions given here, then the agreement is complete.

4.

Now, we have to provide the proof of the assertion made in § 1.

Our system given by Fig. 1 shall be at rest at the beginning. Then the energy quantity

$E_{0}=\frac{4\pi i_{0}}{\mathfrak{B}}\cdot D.$|undefined

is in space $$R$$. Now it is the question, what is happening with the energy, when the system is suddenly brought to velocity $$c$$. We can assume from the outset, that only a fraction of it becomes absorbed, while another fraction is transformed into mechanical work.

Now we have to notice in particular, that this energy is uniformly distributed into all absolute directions. Thus when we maintain our earlier mode of expression, then the density of energy whose absolute direction of propagation encloses an angle between $$\phi\,$$ and $$\phi+d\phi\,$$ with the normal, is:

$\frac{2\pi i_{0}}{\mathfrak{B}}\sin\varphi\ d\varphi.$|undefined

As soon as the system is in motion, it is about the distribution in terms of the relative propagation direction. Then the density of energy, whose relative propagation direction encloses an angle between $$\phi\,$$ and $$\phi+d\phi\,$$ with the normal, is evidently:

$\frac{2\pi i_{0}}{\mathfrak{B}}\sin\varphi\ \frac{d\varphi}{d\phi}d\phi=-\frac{2\pi i_{0}}{\mathfrak{B}}\ \frac{d\ \cos\varphi}{d\phi}d\phi.$|undefined