# Page:Michelson1904.djvu/2

measurement of the difference of time required for the two pencils to traverse the circuit would furnish a quantitative test of the entrainement.

But it is not necessary that the path should encircle the globe, for there would still be a difference in time for any position of the circuit.

This difference is given by the formula

$T=\frac{2}{V^{2}}\int v\ \cos\ \theta\ ds$

where V is the velocity of light, v the velocity of the earth's surface at the element of path ds, and θ the angle between v and ds.

If the circuit he horizontal, and x and y denote distances east and west and north and south respectively, and φ the latitude of the origin, and R the radius of the earth, then for small values of y/R we have approximately

$T=\frac{2v_{0}}{V^{2}}\int\left(\cos\phi-\frac{y}{R}\sin\phi\right)dx.$

The integral being taken round the circuit the first, term vanishes, and if $A=\int ydx=area$ of the circuit,

$T=\frac{2v_{0}A}{V^{2}R}\sin\phi.$

The corresponding difference of path for equal times expressed in light-waves of length λ is

$\Delta=\frac{2v_{0}A}{VR\lambda}\sin\phi.$

Thus, for latitude 45° $\sin\phi=\sqrt{1/2},\ \frac{v_{0}}{R}=\frac{2\pi}{T}$; the velocity of light is $3\times10^{8}$ in the same units, and the length of a light-wave is $5\times10^{-7}$: which approximate values substituted in the preceding formula give

$\Delta=7\times10^{-7}A.$

Thus if the circuit be one kilometre square

$\Delta=0.7.$

The system of interference-fringes produced by the superposition of the two pencils — one of which has traversed the circuit clockwise, and the other counterclockwise — would be shifted through seven-tenths of the distance between the fringes, in the direction corresponding to a retardation of