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

This gives

$A\ cos\ m(t({\mathfrak {B}}+c\ \cos \ i)+y\ \sin \ i]+$ $+A'\cos \ m'[t(c\ \cos \ r-{\mathfrak {B}})-y\ \sin \ r]=0.$

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

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

(10)

and the Doppler effect

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

in full agreement with Abraham. Incidentally, equation (10) is also easily given from equation (12) of my earlier treatise.^[1]

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:

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

1. These proceedings., CXIII, p. 489.