Special Theory of Relativity
9
but at positions x 1 = 0 and x 2 = l respectively, the corresponding events in frame
F′ take place at
t 1 ′ = 0
2
1/ 2
2
2
2
1
vl
t
v
c
c
′ = −
−
(1.23)
i.e., the event at x 2 = l took place earlier in frame F′. In order to understand this
result a little better, let the two events correspond to emission of light signals
which travel towards the point
2
l
x =
which they will reach at
2
l
t
c
=
s.
However, as observed from frame F′, by the time these signals reach the
midpoint, the midpoint would have travelled some distance towards the point
x = 0 and away from the point x = l. Therefore, the signal from x = l travels
through a longer distance. Because the two signals reach the point
2
l
x = at the
same time, according to an observe in frame F′, the event at x = l must have
taken place at an earlier time and the events at x = 0 and x = l are not simultaneous.
He will also observe that since the local clocks at x = 0 and x = l, stationary in
frame F, show time t = 0 when the events took place, the F clock at x = l is
ahead of the F clock at x = 0.
With simultaneity being no longer a universal concept, it is necessary to set
up a system of synchronised clocks to measure the time coordinates of events
taking place at different positions. Consider now two clocks at rest in frame F,
located at a distance l apart. Let a clock at rest in the moving frame F′ record
time t 0 ′ and t 1 ′ when it passes the clocks of frame F. The corresponding times
recorded by the clocks in frame F may be designated by t 0 and t 1 . Since these
observations correspond to observations taking place at the same place in frame
F′, one obtains from Eq. (1.21)
1
1
0
1/ 2
2
2
1
0
′ − ′
− =
−
t t
t t
v
c
(1.24)
Thus, the moving clock appears to run at a slower rate. However, as
discussed before, the observer in frame F′ will find that the clocks in frame F
are not synchronised and the clock he passes later is ahead of the clock he
passed earlier by an amount δ. After taking this into account, the reciprocal
nature of the two frames then requires that
1
0
1
0
1/ 2
2
2
(
)
1
− δ −
− =
′
′
−
t
t
t t
v
c
(1.25)
9
but at positions x 1 = 0 and x 2 = l respectively, the corresponding events in frame
F′ take place at
t 1 ′ = 0
2
1/ 2
2
2
2
1
vl
t
v
c
c
′ = −
−
(1.23)
i.e., the event at x 2 = l took place earlier in frame F′. In order to understand this
result a little better, let the two events correspond to emission of light signals
which travel towards the point
2
l
x =
which they will reach at
2
l
t
c
=
s.
However, as observed from frame F′, by the time these signals reach the
midpoint, the midpoint would have travelled some distance towards the point
x = 0 and away from the point x = l. Therefore, the signal from x = l travels
through a longer distance. Because the two signals reach the point
2
l
x = at the
same time, according to an observe in frame F′, the event at x = l must have
taken place at an earlier time and the events at x = 0 and x = l are not simultaneous.
He will also observe that since the local clocks at x = 0 and x = l, stationary in
frame F, show time t = 0 when the events took place, the F clock at x = l is
ahead of the F clock at x = 0.
With simultaneity being no longer a universal concept, it is necessary to set
up a system of synchronised clocks to measure the time coordinates of events
taking place at different positions. Consider now two clocks at rest in frame F,
located at a distance l apart. Let a clock at rest in the moving frame F′ record
time t 0 ′ and t 1 ′ when it passes the clocks of frame F. The corresponding times
recorded by the clocks in frame F may be designated by t 0 and t 1 . Since these
observations correspond to observations taking place at the same place in frame
F′, one obtains from Eq. (1.21)
1
1
0
1/ 2
2
2
1
0
′ − ′
− =
−
t t
t t
v
c
(1.24)
Thus, the moving clock appears to run at a slower rate. However, as
discussed before, the observer in frame F′ will find that the clocks in frame F
are not synchronised and the clock he passes later is ahead of the clock he
passed earlier by an amount δ. After taking this into account, the reciprocal
nature of the two frames then requires that
1
0
1
0
1/ 2
2
2
(
)
1
− δ −
− =
′
′
−
t
t
t t
v
c
(1.25)