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):

${\begin{aligned}\xi _{1}&=x_{1}\mu _{1}+y_{1}\nu _{1}+z_{1}\pi _{1}-\varkappa t&&=a_{1}-\varkappa t\\\eta _{1}&=\left(x_{1}\mu _{2}+y_{1}\nu _{2}+z_{1}\pi _{2}\right)q&&=b_{1}q\\\zeta _{1}&=\left(x_{1}\mu _{3}+y_{1}\nu _{3}+z_{1}\pi _{3}\right)q&&=c_{1}q\\\tau &=t-{\frac {\varkappa }{\omega ^{2}}}(\mu _{1}x+\nu _{1}y+\pi _{1}z)&&=t-{\frac {\varkappa a_{1}}{\omega ^{2}}}{,}\end{aligned}}$

8)

where for μ_h, ν_h, π_h 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 μ_h, ν_h, π_h 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 $\zeta =\eta =0$