Page:WitteSagnac1.djvu/5

thus it is:

$\Delta U={\frac {v}{c}}\cdot U$

2)

and the phase change $x$ at wavelength $\lambda$, is given by:

$(\pm )x={\frac {\Delta U}{\lambda }}={\frac {v}{\lambda \cdot c}}\cdot U$

3)

This formula is also to be applied in our case, it must only be considered that the direction of the translatory velocity $v$ (tangent)

Fig.2

and that of light (polygon side) do not coincide, it is thus (as it follows from Fig. 2) substituted instead of $v$:

$v\cdot \cos {\frac {\vartheta }{2}}$

where $\vartheta$ denotes the polygon side of the corresponding central-angle.

If one substitutes this value, than the phase difference $2\cdot x$, mutually experienced by rays $R$ and $L$ due to the relative aether stream, is:

$2\cdot x=2\cdot {\frac {v}{c}}\cdot {\frac {U}{\lambda }}\cdot \cos {\frac {\vartheta }{2}}$

4)

Finally one can substitute value 1) for velocity $v$ of the relative aether stream with rotating number $N$ and circle radius $r$, then it is given:

$2\cdot x={\frac {4\pi N}{c\cdot \lambda }}\cdot U\cdot r\cdot \cos {\frac {\vartheta }{2}}$