## Timing the ferry

The principle of the Michelson-Morley experiment can be understood by taking the analogy of two identical ferries which bring tourists on river trips. One ferry makes return trips across a wide river and the other makes return trips along the same river (Figure 15.3). Both ferries start from and return to the same jetty. There is a current flowing in the river with velocity v from left to right, as illustrated below. Each ferry is capable of the same speed, V, relative to still water and travels a total distance of 2D, but will the two trips take the same time?

Intuitively, the answer is not obvious. Is it more difficult to fight the cross-current both going and coming back, or to have the advantage of a following current on the outward journey, but then pay the penalty of having to struggle head-on against it on the way back?

In order to travel directly across the river, ferry 1 will have to aim at an angle upstream to allow for the current, so that the combination of the velocity v of the current and the velocity V of the ferry gives a resultant velocity directed straight across the river.

Ferry 1

The vector triangle gives the magnitude R of the resultant velocity of v and V.

r Total distance = 2D.

Total time taken T1 =

r Total distance = 2D.

Total time taken T1 =

Ferry 2

Velocity downstream = V + v

î ■ V v outgoing-»-► Velocity upstream = V - v

Total distance = 2D

return

Total time taken T2 =

D D 2DV

Since the ship must travel faster than the current, V is always greater than v and therefore v 2/V2 < 1 and T2/T1 > 1.

T2 > Tv The round trip on ferry 2 takes longer.

Which ferry trip takes more time, and which takes less, is not important. What is important is that the times are not equal.

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