Page:Ueber das Doppler'sche Princip.djvu/3



If we take $$\alpha\beta\gamma$$ as given, then we have 12 constants available, so we can arbitrarily use three of them.

The solution is most comfortable when we use a temporary co-ordinate system X1, Y1, Z1, for which β and γ disappear in equations (2), α is equal to ϰ, that is, a co-ordinate system whose X1-axis falls in the direction, of which the direction cosine is proportional to X, Y, Z with α, β, γ.

Furthermore, it should be set

$$\begin{array}{clcclcclcclc} m_{h}^{2}+n_{h}^{2}+p_{h}^{2} & = & q_{h}^{2}, & m_{h}/q_{h} & = & \mu_{h}, & n_{h}/q_{h} & = & \nu_{h}, & p_{h}/q_{h} & = & \pi_{h}\\\\ a^{2}+b^{2}+c^{2} & = & d^{2}, & a/d & = & \mu, & b/d & = & \nu, & c/d & = & \pi, \end{array}$$

then μ, ν, π are the direction cosines of 4 directions, which we will denote by δ1, δ2, δ3 and δ, against the system X1, Y1, Z1.

By these introductions our equations (3), (4) and (5) will be:

$$\mu_{2}\mu_{3}+\nu_{2}\nu_{3}+\pi_{2}\pi_{3}=\mu_{3}\mu_{1}+\nu_{3}\nu_{1}+\pi_{3}\pi_{1}=\mu_{1}\mu_{2}+\nu_{1}\nu_{2}+\pi_{1}\pi_{2}=0$$

$$\mu\mu_{1}+\nu\nu_{1}+\pi\pi_{1}=\frac{\varkappa}{\omega^{2}q_{1}d},\ \mu\mu_{2}+\nu\nu_{2}+\pi\pi_{2}+\mu\mu_{3}+\nu\nu_{3}+\pi\pi_{3}=0$$

According to (4'), the three directions δ1, δ2, δ3 are perpendicular to each other, according to (5') $$\delta_1$$ falls into δ, then it must be: