Michelson and Morley used an interferometer to show that the speed of light is constant, regardless of Earth's motion through any perceived luminiferous aether. An analogy can be understood from the different times it takes for a rowboat to travel two different round-trip paths in a river that flows at a constant velocity \((u)\) downstream. Let one path be for a distance \(D\) directly across the river, then back again; and let the other path be the same distance \(D\) directly upstream, then back again. Assume that the rowboat travels at constant speed, \(v\) (with respect to the water), for both trips. Neglect the time it takes for the rowboat to turn around. Find the ratio of the cross-stream time divided by the upstream-downstream time, as a function of the given constants.

Step by step solution

01

Calculate cross-stream time

First, let's find the time it takes for the rowboat to travel directly across the river and back (cross-stream path). The total distance traveled is \(2D\). The rowboat's velocity in relation to the water is \(v\), but it must also factor in the river velocity \(u\). To do this, we can use the Pythagorean theorem to find the boat's effective velocity: \(v_{eff} = \sqrt{v^2 - u^2}\) Then, we can find the time it takes: \(t_{cross} = \frac{2D}{v_{eff}}\)

02

Calculate upstream-downstream time

Now, let's find the time it takes for the rowboat to travel directly upstream and then back downstream (upstream-downstream path). The total distance traveled is also \(2D\). However, the boat's velocity is now affected by the river's velocity differently: When going upstream, the effective velocity is: \(v_{up} = v - u\) The time it takes to travel upstream is: \(t_{up} = \frac{D}{v_{up}}\) When going downstream, the effective velocity is: \(v_{down} = v + u\) The time it takes to travel downstream is: \(t_{down} = \frac{D}{v_{down}}\) The total time of the upstream-downstream path is: \(t_{up-down} = t_{up} + t_{down}\)

03

Calculate the ratio of cross-stream time and upstream-downstream time

Finally, we can find the ratio of the cross-stream time divided by the upstream-downstream time: \(ratio = \frac{t_{cross}}{t_{up-down}}\) Plug in our expressions for \(t_{cross}\), \(t_{up}\), and \(t_{down}\) into the ratio: \(ratio = \frac{\frac{2D}{\sqrt{v^2 - u^2}}}{\frac{D}{v - u} + \frac{D}{v + u}}\) Simplify the expression: \(ratio = \frac{2(v^2 - u^2)}{2D(v^2 - u^2)}\) The final ratio as a function of the given constants is: \(ratio = \frac{2(v^2 - u^2)}{(v - u)(v + u)}\)

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