## L

vt/2

vt/2

Light clock. (a) In the rest frame of the clock, the light bounces back and forth. (b) In the laboratory frame, with the clock moving, the light beam travels a longer path. (c) Calculating the extra distance traveled.

We want to solve for t in terms of t0:

Taking the square root of both sides and solving for t, we have t =

In the frame in which the clock is moving, the light beam takes a longer path. Since the speed of light is the same in both frames, the beam must take longer to make the round trip. From the figure, we see that the distance traveled is 2[L2 + (vt/2)2]1/2, so the time is t = (2/c) [L2 + (vt/2)2]1/2

We use equation (7.1) to eliminate L, giving t2 = t02 + (v2/c2) t2

The significance of this result is that the time interval measured in the frame in which the clock is moving is greater than that measured in the frame in which the clock is at rest. Suppose we have two identical clocks. If we keep one at rest (with respect to us) and let the other one move, the moving clock appears to run slow. It is important to realize that the situation is perfectly symmetric. If there is an observer traveling with each clock, each observer sees the other clock running slow. This effect is called time dilation.

From equation (7.2), we can see that the amount of time dilation depends on the quantity 1/[1 -(v/c)2]1/2. This quantity is generally designated y and is plotted as a function of (v/c) in Fig. 7.5. Note that this quantity is close to unity for small velocities, and only becomes large when v is very close to c. This confirms our intuition that the results of

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