Page:Ueber das Doppler'sche Princip.djvu/9

 is set. It is therefore $$\textstyle{\eta=\frac{T}{4}}$$ for r = R and η = 0m if r is very great compared with the wavelength Tω.

The propagated displacements follow from ψ by:

$U=0,\ V=-\psi z,\ W=\psi y;$

we briefly set

$U=0,\ V=MC,\ W=NC.$

Substituting herein for x, y, z, the values ξ, η, ζ according to (10), then the periodic part C is given by:

if $\operatorname{cotg}\frac{2\pi(\eta)}{T}=\frac{2\pi}{T\omega}\left(\sqrt{(x-\varkappa t)^{2}+y^{2}+z^{2}}-R\right)$|undefined

For $$(x-\varkappa t)^{2}+y^{2}+z^{2}=R^{2}$$, i.e., on the surface of a sphere which is displaced parallel to the X-axis, this becomes

$(\overline{C})=\sin\frac{2\pi}{T}\left(t\left(1-\frac{\varkappa^{2}}{\omega^{2}}\right)-\frac{\varkappa}{\omega^{2}}\sqrt{R^{2}-y^{2}-z^{2}}\right),$|undefined

thus, since under the assumption $$\textstyle{\frac{\varkappa^{2}}{\omega^{2}}}$$ and $$\textstyle{\frac{\varkappa R}{\omega^{2}}}$$ of second-order it follows:

$(\overline{C})=\sin\frac{2\pi t}{T}.$

(M) and (N) have the same value, as if the little sphere would oscillate as a state of equilibrium around the attained position $$x_{0}=\varkappa t$$ at time t. Therefore, by (U), (V), (W) we get the motion that was submitted by a rotating "illuminating point" with translational speed $$\varkappa$$ parallel to the direction of the rotation axis.

The propagated wave surfaces are assessed according to the value (17) for (C), which can be written by introduction of relative coordinates against the moving luminous point $$\xi=x-\varkappa t$$, $$y=\eta$$, $$z=\zeta$$ (neglecting $$\textstyle{\frac{\varkappa^{2}}{\omega^{2}}}$$ against 1) and for r which is great against $$T\omega$$:

$(C)=\cos\frac{2\pi}{T}\left(t-\frac{\varkappa\xi}{\omega^{2}}-\frac{1}{\omega}\left(\sqrt{\xi^{2}+\eta^{2}+\zeta^{2}}-R\right)\right).$|undefined