Page:Zur Theorie der Strahlung bewegter Koerper.djvu/13

 and the amount of the energy reflected in the same time

$\rho'(\mathfrak{B}\ \cos\ r-c)=\rho'\mathfrak{B}(\cos\ r-\sigma).$

The difference of these two expressions must be equal to the work of the radiation pressure per unit time. Thus it must be

$Pc=\rho'\mathfrak{B}(\cos\ r-\sigma)-\rho\mathfrak{B}(\cos\ i+\sigma)$

Or

$cP=\mathfrak{B}\rho(\cos\ i+\sigma)\left[\left(\frac{1+\sigma\cos\ i}{1-\sigma\cos\ r}\right)^{2}\frac{\cos\ r-\sigma}{\cos\ i+\sigma}-1\right].$

With the aid of the easily derivable relation, already given by :

$\frac{1+\sigma\cos\ i}{1-\sigma\cos\ r}=\frac{\sigma+\cos\ i}{\sigma-\cos\ r}=\frac{\sin\ i}{\sin\ r}$

and the equation:

$\sin\ r=\frac{\sin\ i(1-\sigma^{2})}{1+\sigma^{2}+2\sigma\ \cos\ i}$

it becomes

$cP=\mathfrak{B}\rho(\cos\ i+\sigma)\left[\frac{1+\sigma^{2}+2\sigma\ \cos\ i}{\ i-\sigma^{2}}-1\right].$|undefined

and

$P=2\rho\frac{(\cos\ i+\sigma)^{2}}{1-\sigma^{2}}.$|undefined

The agreement with the value given by is thus a complete one (here, $$\sigma$$ has the opposite sign as earlier.)

Of course, upon this foundation one can calculate the values $$p_{1}$$ and $$p_{2}$$, when one assumes that a moving body is emanating waves, whose amplitude is the same as in the stationary state, while the density of the radiation energy is inversely proportional to the square of the wavelength.