# Page:Michelson1904.djvu/2

717
Relative Motion of Earth and Æther.

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

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

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

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

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

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

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

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

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

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

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

Thus if the circuit be one kilometre square

${\displaystyle \Delta =0.7.}$

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

Phil. Mag. S. 6. Vol. 8. No. 48. Dec. 1904.
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