bei48482_FM

In order to find Lx 2 x 1 , the length of the rod as measured in the stationary frame Sat the

time t, we make use of Eq. (1.41) to give

x   1  x   2 

Hence Lx 2 x 1 (x    2 x    1 )  1 ^2 c^2 L 0  1 ^2 c^2

This is the same as Eq. (1.9)

Inverse Lorentz Transformation

In Example 1.9 the coordinates of the ends of the moving rod were measured in the

stationary frame Sat the same time t, and it was easy to use Eq. (1.41) to find Lin

terms of L 0 and . If we want to examine time dilation, though, Eq. (1.44) is not con-

venient, because t 1 and t 2 , the start and finish of the chosen time interval, must be

measured when the moving clock is at the respective differentpositions x 1 and x 2. In

situations of this kind it is easier to use the inverse Lorentz transformation,which

converts measurements made in the moving frame S to their equivalents in S.

To obtain the inverse transformation, primed and unprimed quantities in Eqs. (1.41)

to (1.44) are exchanged, and is replaced by :

x (1.45)

yy (1.46)

z   z (1.47)

t (1.48)

Example 1.10

Derive the formula for time dilation using the inverse Lorentz transformation.

Solution

Let us consider a clock at the point x in the moving frame S. When an observer in S finds

that the time is t  1 , an observer in Swill find it to be t 1 , where, from Eq. (1.48),

t 1 

After a time interval of t 0 (to him), the observer in the moving system finds that the time is now

t   2 according to his clock. That is,

t 0 t 2 t (^1)
t 1 
c
x
2
^

 1 ^2 c^2
t 

c
x
2


 1 ^2 c^2
x t

 1 ^2 c^2
Inverse Lorentz
transformation
x 2 t

 1 ^2 c^2
x 1 t

 1 ^2 c^2
42 Appendix to Chapter 1
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