CK-12-Physics - Intermediate

http://www.ck12.org Chapter 22. The Special Theory of Relativity

FIGURE 22.6

Time from the frame of reference of observer A.

tA=

d

c

The light arrives back at the clock in time 2tA.

What time interval will observerBsee? Since the mirror moves with velocityv, to the right, observerBwill see
the light move in a diagonal path as shown inFigure22.7. From their point of view, the light is travelling a longer
distance, but it is still travelling at the same speed,c, from Postulate #2 of relativity.

Einstein’s answer to the puzzle was that time works differently in this other frame of reference. For observerB, the
time for the light to go one way is a different time,tB. Using the Pythagorean Theorem, the distance the light travels
one way is:

(vtB)^2 +d^2 = (ctB)^2

d^2 = (ctB)^2 −(vtB)^2

d=tB

√

c^2 −v^2

tB=

d

√

c^2 −v^2

This is the time to go one way from the source to the mirror. The return time is the same, so the total time is 2tB. We
can relate this back to the time inA’s frame of reference.

tB=

tA

√

1 −vc^22

Since the denominator is smaller than 1, the time interval measured by observerBwill be longer than the time
measured by observerA. We can, therefore, definetime dilationas follows: