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

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

where $\mathrm {V}$ is the velocity of light, $v$ the velocity of the earth's surface at the element of path $ds$, and $\theta$ the angle between $v$ and $ds$.

If the circuit be horizontal, and $x$ and $y$ denote distances east and west and north and south respectively, and $\phi$ the latitude of the origin, and $\mathrm {R}$ the radius of the earth, then for small values of $y/\mathrm {R}$ we have approximately

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

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

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

The corresponding difference of path for equal times expressed in light-waves of length $\lambda$ is

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

Thus, for latitude 45° $\sin \phi ={\sqrt {1/2}},\ {\frac {v_{0}}{\mathrm {R} }}={\frac {2\pi }{\mathrm {T} }}$; the velocity of light is $3\times 10^{8}$ in the same units, and the length of a light-wave is $5\times 10^{-7}$: which approximate values substituted in the preceding formula give

$\Delta =7\times 10^{-7}\mathrm {A} .$

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