Page:Eddington A. Space Time and Gravitation. 1920.djvu/33

CHAPTER I

THE FITZGERALD CONTRACTION

In order to reach the Truth, it is necessary, once in one's life, to put every thing in doubt—so far as possible.

Descartes.

Will it take longer to swim to a point 100 yards up-stream and back, or to a point 100 yards across-stream and back?

In the first case there is a long toil up against the current, and then a quick return helped by the current, which is all too short to compensate. In the second case the current also hinders, because part of the effort is devoted to overcoming the drift down-stream. But no swimmer will hesitate to say that the hindrance is the greater in the first case.

Let us take a numerical example. Suppose the swimmer's speed is 50 yards a minute in still water, and the current is 30 yards a minute. Thus the speed against the current is 20, and with the current 80 yards a minute. The up journey then takes 5 minutes and the down journey 1 1/4 minutes. Total time, 6 1/4 minutes.

Going across-stream the swimmer must aim at a point $E$ above the point $B$ where he wishes to arrive, so that $OE$ represents his distance travelled in still water, and $EB$ the amount he has drifted down. These must be in the ratio 50 to 30, and we then know from the right-angled triangle $QBE$ that $OB$ will correspond to 40. Since $OB$ is 100 yards, $OE$ is 125 yards, and the time taken is 2 1/2 minutes. Another 2 1/2 minutes will be needed for the return journey. Total time, 5 minutes.

In still water the time would have been 4 minutes.

The up-and-down swim is thus longer than the transverse swim in the ratio 6 1/4: 5 minutes. Or we may write the ratio ${\frac {1}{\sqrt {\left(1-\left({\frac {30}{50}}\right)^{2}\right)}}}$