# Page:Ueber das Doppler'sche Princip.djvu/4

Substituting this in (3'), q1, q2, q3 are determined.

At first we obtain, since only positive signs are meaningful:

$q_{1}=1\text{ or }\frac{\varkappa}{\omega}$ $d=\frac{\varkappa}{\omega^{2}}\text{ or }\frac{1}{\omega}.$

I will only use the first solution, the second is of no interest;[1] it follows from it:

 $d=\frac{\varkappa}{\omega^{2}},\ q_{1}=1,\ q_{2}=q_{3}=\sqrt{1-\frac{\varkappa^{2}}{\omega^{2}}}=q.$ 7)

Consequently, we can write equations (2):

 \begin{align} \xi_{1} & =x_{1}\mu_{1}+y_{1}\nu_{1}+z_{1}\pi_{1}-\varkappa t & & =a_{1}-\varkappa t\\ \eta_{1} & =\left(x_{1}\mu_{2}+y_{1}\nu_{2}+z_{1}\pi_{2}\right)q & & =b_{1}q\\ \zeta_{1} & =\left(x_{1}\mu_{3}+y_{1}\nu_{3}+z_{1}\pi_{3}\right)q & & =c_{1}q\\ \tau & =t-\frac{\varkappa}{\omega^{2}}(\mu_{1}x+\nu_{1}y+\pi_{1}z) & & =t-\frac{\varkappa a_{1}}{\omega^{2}}{,} \end{align} 8)

where for μh, νh, πh no more other conditions apply than those which result from their meaning as direction cosines of three successive perpendicular but otherwise quite arbitrary directions.

Therefore, the aggregates designated by $a_{1}\ b_{1}\ c_{1}$ can be considered as the coordinates of the point $x_{1}\ y_{1}\ z_{1}$ in relation to a coordinate system, which falls into the direction $\delta_{1}\ \delta_{2}\ \delta_{3}$.

Any such system μh, νh, πh gives a solution (U), (V), (W) from given U, V, W. If U, V, W adopt on a surface f(x, y, z) = 0 the given values $\overline{U}$, $\overline{V}$, $\overline{W}$, so (U), (V), (W) from those derivable $(\overline{U}), (\overline{V}), (\overline{W})$ to the surface $(f)=f(\overline{\xi_{1}},\overline{\eta_{1}},\overline{\zeta_{1}})=0$, which because of the values of ξ1, η1, ζ1 has the property to move with uniform velocity ϰ parallel to a direction δ1 or A given by direction cosines ϰ. The solutions (U), (V), (W) give thus the laws by which certain surfaces in progressive motion are shining, if they only comply with the condition

$\frac{(\partial U)}{\partial x}+\frac{\partial (V)}{\partial y}+\frac{\partial (W)}{\partial z}=0$
1. It follows from it $q_{2}=q_{3}=0$, as well as $m_{2}\ n_{2}\ p_{2}$, $m_{3}\ n_{3}\ p_{3}$ and therefore $\zeta=\eta=0$