Page:Zur Theorie der Strahlung bewegter Koerper.djvu/12

 This gives

This equation can only then be satisfied for all values of $$y$$ and $$t$$, when

$\begin{array}{rl} A= & -A',\\ m(\mathfrak{B}+c\ \cos\ i)= & m'(\mathfrak{B}-c\ \cos\ r),\\ m\ \sin i= & m'\ \sin r.),\end{array}.$

These equations give us the law of reflection

and the Doppler effect

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

in full agreement with. Incidentally, equation (10) is also easily given from equation (12) of my earlier treatise.

Now, to derive the value of the pressure from it, we have to assume that the energy density of a light wave is proportional (at equal amplitude) to the -2. power of the wave length, thus to the quantity $$m^2$$. Thus when we denote by $$\rho$$ the density of the incident wave, by $$\rho'$$ that of the reflecting one, then it is

$\frac{\rho'}{\rho}=\left(\frac{m'}{m}\right)^{2}=\left(\frac{1+\sigma\ \cos\ i}{1-\sigma\ \cos\ r}\right)^{2}.$

Now, the energy falling upon the mirror in unit time, is contained in space $$\mathfrak{B}\ cos\ i+c$$; the reflected energy in space $$\mathfrak{B}\ cos\ r-c$$. Thus the amount the incident energy per second is:

$\rho(\mathfrak{B}\ \cos\ i+c)=\rho\mathfrak{B}(\cos\ i+\sigma)$