Environmental Engineering Reference
In-Depth Information
(x
−
ct)
where
x
=
x
B
−
x
A
etc. Putting the numbers in gives
γ
=
2 and hence
v/c
0
.
866. Note that you could get this directly from the
length contraction formula since the events occur at the same time in one
of the frames. We can make use of “
t
=
=
vx/c
2
)
” to determine the
γ(t
−
corresponding time interval, i.e.
2
0
10
8
1000
t
=
10
−
6
s
.
−
0
.
866
×
=−
5
.
77
×
3
×
The minus sign means that the event at larger
x
is observed first in the primed
frame.
6.14 The emission of the two pulses corresponds to two events with
x
=
4km
5
µ
s. Given
t
=
0 we need to use
t
=
vx/c
2
)
to
and
t
=
γ(t
−
0
.
375.
6.15 Key here is to identify the relevant events. Let event A be the impact of
the comet and event B be the event in the party that is simultaneous with
event A according to an observer whizzing past the Earth. Then
t
=
determine the speed, i.e.
v/c
=
ct/x
=
0
by definition (primes indicate the arbitrary inertial frame) and we want
to know the corresponding time interval in the Earth frame (
t
). Since
we know
t
and
x
the relevant equation is
t
=
vx/c
2
)
.
γ(t
−
The extreme values of
t
occur when
v
=±
c
whence
t
=±
x/c
. Thus
10
11
/(
3
10
8
)
s
89 minutes. The
impact of the comet occurs (in the Earth frame) at the midpoint of the
party whilst the party ends with the students observing the impact using
their telescopes.
6.16 A diagram will help to clarify the way in which you must use the velocity
addition formula. The speed is
(
0
.
70
the party must last a time 2
×
8
×
×
=
0
.
97
c
.
6.17 As in the last question, a diagram will help. Required speed is
(
0
.
5
+
0
.
85
)c/(
1
+
0
.
7
×
0
.
85
)
=
−
0
.
8
)c/(
1
−
0
.
5
×
0
.
8
)
=−
0
.
5
c
. The minus sign implies the rocket moves
towards the Earth.
6.18 Velocity addition gives
V
=
(v
+
c/n)/(
1
+
v/(nc))
.Noteif
n
=
1 then
V
=
c
. Fizeau would have worked with
v
nc
hence the denominator can
v/(nc))
−
1
be simplified using
(
1
+
≈
1
−
v/(nc)
. This yields the result after
neglecting terms suppressed in
v/c
.
6.19 Work in A's rest frame and determine the velocity of B in that frame, then
use tan 30
◦
=
v
By
/v
Bx
to determine
u
. In more detail: use velocity addition
to obtain
v
Bx
=
=
√
3 which implies
u
and
v
By
=
u/γ (u)
and hence
γ(u)
=
√
2
/
3.
u/c
PROBLEMS 7
10
−
27
10
8
)
2
/(
1
.
6022
10
−
13
)
MeV
7.1 Use
E
=
1
.
673
×
×
(
2
.
9979
×
×
=
938
mc
2
. Hence
m
938 MeV
/c
2
.
MeV
=
=
7.2 (a)
γ
−
1
=
1 hence
v
=
0
.
866
c
.(b)
γ
−
1
=
5 hence
v
=
0
.
986
c
.
c
2
d
m/
d
t
. Hence rate of mass loss is 4.2 million tonnes per
7.3 Power output
=
second.