Environmental Engineering Reference
In-Depth Information
Now it is time to make use of the 1st postulate which says that if Eqs. (6.25) are
true then so necessarily are
a(x
+
vt
),
x
=
(6.26a)
a
t
+
c
2
.
vx
=
t
(6.26b)
This makes manifest the equivalence of the two frames. It can be seen by consid-
ering Figure 5.1 and swapping the primed and unprimed co-ordinate labels around
whilst at the same time reversing the direction of
v
. We can determine the coefficient
a
now by substituting for
x
and
t
using Eqs. (6.25) into either of Eqs. (6.26), i.e.
a
ax
a
2
x
1
av
2
x
c
2
v
2
c
2
x
=
−
avt
+
avt
−
=
−
1
⇒
a
=
1
=
γ.
(6.27)
v
2
/c
2
−
We have succeeded in deriving the Lorentz transformations:
x
=
γ(x
−
vt),
(6.28a)
t
=
vx/c
2
),
γ(t
−
(6.28b)
y
=
y,
(6.28c)
z
=
z.
(6.28d)
Sometimes the inverse transformations will be more useful:
γ(x
+
vt
),
x
=
(6.29a)
γ(t
+
vx
/c
2
).
=
t
(6.29b)
Eqs. (6.28) are perhaps the most important equations we have derived so far in this
part of the topic.
Example 6.2.1
Use the Lorentz transformations to derive the formula for time
dilation.
Solution 6.2.1
Let us consider the situation illustrated in Figure 6.6. A clock is at
rest in S
, let's suppose it is at position x
0
. Now consider one tick of the clock. In S
,
we suppose that the tick starts at time t
1
and ends at time t
2
such that t
=
t
1
is the duration in the clock's rest frame. The question is: 'what is the duration of
the same tick as determined by an observer in S?'
There are two events to consider. Event 1 (start of the tick) has co-ordinates
(x
0
,t
1
) in S
and event 2 (end of tick) which has co-ordinates (x
0
,t
2
) in S
. We want
to know the time of each event in S. Given that we know both the location and time of
the events in S
we should use Eq. (6.29b) to give us the corresponding times in S:
t
2
−
γ(t
1
+
vx
0
/c
2
),
t
1
=
γ(t
2
+
vx
0
/c
2
).
t
2
=
