Page:MichelsonMorley1886.djvu/8

x.	v.
0.00 .20 .40 .60 .80 .90 .95 1.00	1.000 .993 .974 .929 .847 .761 .671 .000

The curve constructed with these numbers coincides almost perfectly with the curve

$v=(1-x^{2})^{.165}$.

The total flow is therefore $2\pi \int _{0}^{1}(1-x^{2})^{.165}=xdx={\frac {\pi }{1.165}}$. The area of the tube being $\pi$, the mean velocity $={\frac {1}{1.165}}$ of the maximum; or the maximum velocity is 1.165 times the mean. This, then, is the Number by which the velocity, found by timing the flow, must be multiplied to give the actual velocity in the axis of the tube.

Formula.

Let l be the length of the part of the liquid column which is in motion.

$\mu$ = velocity of light in the stationary liquid.

v = velocity of light in vacuo.

$\theta$ = velocity of the liquid in the axis of tube.

$\theta x$ = acceleration of the light.

The difference in the time required for the two pencils of light to pass through the liquid will be ${\frac {l}{u-\theta x}}-{\frac {l}{u+\theta x}}={\frac {2l\theta x}{u^{2}}}$ very nearly. If $\Delta$ is the double distance traveled in this time in air, in terms of $\lambda$, the wave-length, then

$\Delta ={\frac {4l\theta n^{2}x}{\lambda v}}$ whence x$={\frac {\lambda v}{4ln^{2}\theta }}\Delta$.

$\lambda$ was taken as .00057 cm.

v = 30000000000 cm

$n^{2}$ = 1.78.

The length l was obtained as follows: The stream entered each tube by two tubes a, b (figs. 1, 2) and left by two similar ones d, c. The beginning of the column was taken as the intersection, o, of the axes of a and b, and the end, as the intersection, o' , of the axes of d and c. Thus $l=oo'$. $\Delta$ is found by observing the displacement of the fringes; since a displacement of one whole fringe corresponds to a difference of path of one whole wave-length.