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 which have thus far been obtained. In the same letter it was also stated that the reason why such measurements could not be made at the earth's surface was that we have thus far no method for measuring the velocity of light which does not involve the necessity of returning the light over its path, whereby it would lose nearly as much as was gained in going.

The difference depending on the square of the ratio of the two velocities, according to Maxwell, is far too small to measure.

The following is intended to show that, with a wave-length of yellow light as a standard, the quantity — if it exists — is easily measurable.

Using the same notation as before we have $$T=\tfrac{D}{V-v}$$ and $$T_{1}=\tfrac{D}{V+v}$$. The whole time occupied therefore in going and returning $$T+T_{1}=2D\tfrac{V}{V^{2}-v^{2}}$$. If, however, the light had traveled in a direction at right angles to the earth's motion it would be entirely unaffected and the time of going and returning would be, therefore, $$2\tfrac{D}{V}=2T_{0}$$. The difference between the times T+T1 and 2T0 is

or nearly $$2T_{0}\tfrac{v^{2}}{V^{2}}$$. In the time $$\tau$$ the light would travel a distance $$V\tau=2VT_{0}\tfrac{v^{2}}{V^{2}}=2D\tfrac{v^{2}}{V^{2}}$$.

That is, the actual distance the light travels in the first case is greater than in the second, by the quantity $$2D\tfrac{v^{2}}{V^{2}}$$.

Considering only the velocity of the earth in its orbit, the ratio $$\tfrac{v}{V}=\tfrac{1}{10\ 000}$$ approximately, and $$\tfrac{v^{2}}{V^{2}}=\tfrac{1}{100\ 000\ 000}$$. If D = 1200 millimeters, or in wave-lengths of yellow light, 2 000 000, then in terms of the same unit, $$2D\tfrac{v^{2}}{V^{2}}=\tfrac{4}{100}$$.

If, therefore, an apparatus is so constructed as to permit two pencils of light, which have traveled over paths at right angles to each other, to interfere, the pencil which has traveled in the direction of the earth's motion, will in reality travel $$\tfrac{4}{100}$$ of a wave-length farther than it would have done, were the earth at rest. The other pencil being at right angles to the motion would not be affected.