Relativity Lite

To c or Not to c  | 5 It is clear from figure 2 that the hypotenuse d of the triangle is longer than the vertical leg H . Then since the speed of light c is the same for both the hypotenuse and the verti- cal leg, t must be larger than τ . But the hypotenuse of any right triangle you can draw is always longer than each leg or is equal to the length of one leg if the length of the other is zero. (Draw several examples to prove this to yourself.) This means that t is always larger than τ for nonzero v . We call this minimum time value τ proper time . It is simply the time measured in any frame in which the two events are measured at the same place , such as the emission and the return points of the bouncing beam of light, or two clicks of a clock. The coordinate time interval t is measured in a frame of reference moving at velocity v relative to the frame in which the clock is stationary. (If we want to be precise, τ in figure 1 is actually half of the proper-time interval for a round trip 2 τ . Likewise, in figure 2, t is half of the corresponding coordinate time 2 t .) HOW DOES THIS RELATE TO ME? We say that the coordinate time is dilated . What exactly does that mean? It means that if your twin sister drives a fast car while you walk everywhere, she will live longer than you do—her lifetime will be dilated relative to yours! By how much? We could find how much larger t is than τ by using the Pythagorean theorem and some algebra * but there is a simple way to get the time-dilation factor by drawing the spaceship picture carefully, with v properly proportional to c as demonstrated in figure 3 : 1. Start with a square that is 10 cm on each side (the bottom side of which represents the distance light travels in the coordinate frame of reference). 2. Now express the velocity of the spaceship as a fraction of the speed of light and draw a rectangle that is that same fraction of 10 cm wide and is the full 10 cm high. † Suppose v is 3 5 6 10 = of the speed of light (111,600 miles/second); then the width of the rectangle is 6 cm, shown as red as in figure 3. * Suppose v c = 3 5 ; then the ratio of the flight path (in the Earth’s frame of reference) to the light path is also 3 5 . Let us use 3 cm and 5 cm, respectively. We can use the Pythagorean theorem to relate the two times, τ and t : (flight path) 2 + (vertical light path) 2 = (diagonal light path), 2 or (vertical light path) 2 = (diagonal light path) 2 − (flight path) 2 = 25 cm 2 − 9 cm 2 = 16 cm 2 . This means that the vertical light path is 4 cm. Then t/τ = 5 4 or 1.25. † Suppose the speed is v = 111,600 miles/second. Then the fraction v c = 111600 186000 miles second miles second / / = 0.6. The width of the rectan- gle would then be 0.6 × 10 cm = 6 cm.