bei48482_FM

Because the equations of physics must have the same form in both Sand S , we need

only change the sign of (in order to take into account the difference in the direction

of relative motion) to write the corresponding equation for xin terms of x and t :

xk(x   t ) (1.34)

The factor kmust be the same in both frames of reference since there is no difference

between Sand S other than in the sign of .

As in the case of the Galilean transformation, there is nothing to indicate that there

might be differences between the corresponding coordinates y,y and z,z which are

perpendicular to the direction of . Hence we again take

y   y (1.35)

z   z (1.36)

The time coordinates tand t , however, are notequal. We can see this by substi-

tuting the value of x given by Eq. (1.33) into Eq. (1.34). This gives

xk^2 (xt)kt

from which we find that

t ktx (1.37)

Equations (1.33) and (1.35) to (1.37) constitute a coordinate transformation that

satisfies the first postulate of special relativity.

The second postulate of relativity gives us a way to evaluate k. At the instant t0,

the origins of the two frames of reference Sand S are in the same place, according to

our initial conditions, and t   0 then also. Suppose that a flare is set off at the com-

mon origin of Sand S at tt 0, and the observers in each system measure the

speed with which the flare’s light spreads out. Both observers must find the same speedc

(Fig. 1.23), which means that in the Sframe

xct (1.38)

and in the S frame

x   ct (1.39)

Substituting for x and t in Eq. (1.39) with the help of Eqs. (1.33) and (1.37) gives

k(xt)cktcx

and solving for x,

xct

ct^

1 



c





1 

k

1

 2  (^1) 
c

k

c
k

k
1 
k
k^2
c
cktkt

k
1 
k
k^2
c
1 k^2

k
1 k^2

k
The Lorentz Transformation 39
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