Environmental Engineering Reference
In-Depth Information
z
S
S
′
z
′
u
y
y
′
u
x
x
′
O
x
1
O
′
x
2
Figure 6.8
A ball bouncing back and forth between two points.
Subtracting these two equations gives
x
=
x
2
−
x
1
=
γx
1
γ
x
,
⇒
x
=
which is the required result.
Example 6.2.3
A ball is rolled at speed u from the point x
1
on the x-axis to the
point x
2
=
L at which point it is reflected back again elastically, as illustrated
in Figure 6.8. In a frame moving with speed v along the positive x-axis compute:
x
1
+
(a) The spatial separation between the point where the ball starts its journey and
the point where it is reflected;
(b) The time taken for the outward part of the ball's journey;
(c) The time taken for the return part of the ball's journey.
Solution 6.2.3
(a) Event 1 is when the ball starts on its journey and has co-
ordinates (x
1
,t
1
) in S. Event 2 is when the ball arrives at the point of reflection. It
has co-ordinates (x
2
,t
1
+
x
1
. Note that it is
not going to be given by the length contraction formula since the two events are not
simultaneous in either S or S
. We know both x
L/u). We are asked to find x
=
x
2
−
=
x
2
−
x
1
=
L and t
=
L/u
and need x
. We therefore need to use Eq. (6.28a) which informs us that
x
=
γ(x
−
vt),
=
γL(
1
−
v/u)
(6.30)
and γ is of course evaluated using the relative speed of the two frames v.
(b) To get the time taken for the outward part of the journey we should use
Eq. (6.28b) (we hope that by now the reader is getting the hang of selecting the
correct equation to use), i.e.
t
out
=
vx/c
2
),
γ(t
−
γL
u
uv/c
2
).
=
(
1
−
(6.31)
