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

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\frac{\beta\gamma}{\omega^{2}}=m_{2}m_{3}+n_{2}n_{3}+p_{2}p_{3}

\frac{\gamma\alpha}{\omega^{2}}=m_{3}m_{1}+n_{3}n_{1}+p_{3}p_{1}

\frac{\alpha\beta}{\omega^{2}}=m_{1}m_{2}+n_{1}n_{2}+p_{1}p_{2}

4)
\frac{\alpha}{\omega^{2}}=am_{1}+bn_{1}+cp_{1}

\frac{\beta}{\omega^{2}}=am_{2}+bn_{2}+cp_{2}

\frac{\gamma}{\omega^{2}}=am_{3}+bn_{3}+cp_{3}

5)

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:

1-\omega^{2}d^{2}=q_{1}^{2}-\frac{\varkappa^{2}}{\omega^{2}}=q_{2}^{2}=q_{3}^{2} 3')


\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


\text{that is }\cos(\delta_{2},\delta_{3})=\cos(\delta_{3},\delta_{1})=\cos(\delta_{1},\delta_{2})=0 4')


\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


\text{that is }\cos(\delta,\delta_{1})=\frac{\varkappa}{\omega^{2}q_{1}d},\ \cos(\delta,\delta_{2})=\cos(\delta,\delta_{3})=0. 5')

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:

\mu=\mu_{1},\ \nu=\nu_{1},\ \pi=\pi_{1}\ \text{ and }\ \frac{\varkappa}{\omega^{2}q_{1}d}=1. 6)