Page:Ueber das Doppler'sche Princip.djvu/8

 If we substitute according to (10), it is, if $$\textstyle{\sqrt{1-\frac{\varkappa^{2}}{\omega^{2}}}=q}$$ is set:

$(W)=Ae^{\frac{2\pi(\mu y+\nu z)q}{T\omega}}\sin\frac{2\pi}{T}\left[t\left(1+\frac{\varkappa\sigma}{\omega}\right)-x\left(\frac{\sigma}{\omega}+\frac{\varkappa}{\omega^{2}}\right)\right].$|undefined

This gives for x = ϰt, if we write $$\frac{\mu}{q}=\mu'$$, $$\frac{\nu}{q}=\nu'$$:

$(\overline{W})=Ae^{\frac{2\pi(\mu'y+\nu'z)}{\omega T'}}\sin\frac{2\pi t}{T}\text{, where }T'=\frac{T}{1-\frac{\varkappa^{2}}{\omega^{2}}},$|undefined

thus we have an oscillating and at the same time propagating plane; however, the propagated displacement reads:

where we now have $$\sigma=\sqrt{1+(\mu^{'2}+\nu^{'2})q^{2}}$$.

We notice that different laws as the Doppler principle are given, even if we limit ourselves to the first approximation, and ϰ²ω² is neglected compared to 1.

3) If the illuminating surface is a very small sphere of radius R, which oscillates according to the law for the rotation angle

$\overline{\psi}=A\sin\frac{2\pi t}{T}$

around the X-axis, then, at the distance $$r=\sqrt{x^{2}+y^{2}+z^{2}}$$ from the center of the sphere, the propagated rotations ψ are given by

where

$\frac{2\pi(r-R)}{T\omega}=\operatorname{ctg}\frac{2\pi\eta}{T}$ 