Page:Zur Theorie der Strahlung bewegter Koerper.djvu/11

 Thus it is

$P=2\rho\frac{\mathfrak{B}_{-}^{2}\cos^{2}\phi}{\mathfrak{B}^{2}(1-\sigma^{2})}.$

If we want to introduce herein the absolute beam direction again, then we use the relation immediately given from Fig. 2

$\mathfrak{B}_{-}cos\phi=\mathfrak{B}\ cos\varphi-c=\mathfrak{B}(\cos\ \varphi-\sigma).$

And it becomes:

$P=2\rho\frac{(\cos\ \varphi-\sigma)^{2}}{1-\sigma^{2}},$|undefined

an expression already given by as well.

Now, by generalization of the thought that was first spoken out by, the same result can be derived from the standpoint of the elastic theory of light, which I would like to show soon.

We consider a light source, moving under the (absolute) angle $$i$$ against the $$X$$-axis. It is given by

$\zeta=A\cos\ m(x\ \cos\ i+y\ \sin\ i+\mathfrak{B}t).$

If this wave falls upon a mirror lying perpendicular to the $$X$$-axis, then a reflected wave is formed, which is given by

$\zeta'=A'\cos\ m'(x\ \cos\ r-y\ \sin\ r-\mathfrak{B}t).$

Herein, $$r$$ is the angle of reflection.

At the surface of the mirror it must be $$\zeta+\zeta'=0$$. If it is moving with velocity $$c$$ in the direction of the normal, i.e., in the direction of the positive $$X$$-axis, then it must be

$x=ct,\ \zeta+\zeta'=0$