Environmental Engineering Reference
In-Depth Information
not all massless and moving in the same direction. In that case
E
i
=
c
|
p
i
|
0 so the result still holds.
12.4 It is not possible simultaneously to conserve both energy and momentum. A
quick way to see it is to note that the momentum four-vector of the photon
has zero length whilst that of the electron-positron pair cannot be less than
2
mc
(see previous question) where
mc
2
and since all the momenta are parallel
M
≥
is the mass of the electron/positron.
γ(y
+
Vt
)
,
x
x
,
t
γ(t
+
Vy
/c
2
)
and re-arrange to
12.5 Substitute
y
=
=
=
provided
k
x
=
k
,
k
y
=−
ωγ V /c
2
uV γ k/c
2
obtain
the
answer
=−
and
ω
=
γω
.For
K
to be a four-vector it must transform accordingly, i.e. we
require that
k
x
=
k
x
,
k
y
=
Vω/c
2
)
and
ω
=
Vk
y
/c
2
)
.These
γ(k
y
−
γ(ω
−
0. Speed of propagation in
S
is
equations are satisfied since
k
x
=
k
and
k
y
=
ω
k
=
γu
u
=
γ
2
u
2
V
2
/c
4
)
1
/
2
.
(
1
+
c
we obtain
u
=
0 we obtain
u
→
=
→
For
u
c
and for
u
u
.
12.6 Conservation of four-momentum states that
P
π
=
P
µ
+
P
ν
. This can be writ-
P
π
+
P
µ
−
ten as
P
ν
=
P
π
−
P
µ
and squaring both sides gives 0
=
2
P
π
·
P
µ
. This leads to the result since
P
π
m
π
c
2
and
P
µ
=
m
µ
c
2
. Since the
=
E
π
E
µ
/c
2
.
muon travels at right angles to the pion we can write
P
π
·
P
µ
=
(m
π
+
m
µ
)c
4
/(
2
E
π
)
. Substituting for
E
π
=
=
Hence
E
µ
143
.
6 MeV gives
a muon kinetic energy equal to 1.4 MeV. The neutrino emerges at an angle
θ
29
.
6
◦
.
=
arctan
(p
µ
/p
π
)
. Substituting for
p
µ
=
17
.
0MeV
/c
gives
θ
=
12.7 Use
p
x
=
vE/c
2
)
and
p
y
=
γ(p
x
−
p
y
with
cp
x
=
E
cos
θ
and
cp
y
=
E
sin
θ
. Thus
p
y
p
x
=
sin
θ
γ(
cos
θ
tan
θ
=
−
v/c)
68
◦
. The frequency is obtained from
E
=
which gives
θ
=
γ(E
−
vp
x
)
and
10
14
Hz.
12.8 (a) In the phi rest frame, momentum conservation dictates that the kaons
must be emitted in opposite directions with equal energies. Energy conser-
vation fixes the value, i.e.
E
K
=
hf
,
f
=
(v/c)
cos
θ)
,i.e.
f
=
since
E
=
γf (
1
−
7
.
4
×
m
φ
c
2
/
2
=
510 MeV. The momenta are
obtained using
(cp
K
)
2
E
K
−
(m
K
c
2
)
2
127 MeV
/c
.(b)
v
must
point in the opposite direction to
n
. Its magnitude can be determined using
E
φ
=
=
and are
±
c
2
p
φ
+
m
φ
c
4
(γ m
φ
c
2
)
2
=
which can be solved to give
|
v
|=
0
.
947
c
.
(c) Use
p
la
K
=
γ(p
K
+
E
K
/c)
for each kaon in turn and with
γ
=
3
.
107 to
1895 MeV
/c
.
12.9 At threshold, the particles are produced at rest in the zero momentum
frame, i.e. the invariant mass is
(
P
π
+
obtain momenta of
+
1106 MeV
/c
and
+
m
)
2
c
2
. Since this
is invariant we can also compute it in the neutron's rest frame where
P
π
+
P
n
)
2
=
(m
K
+
m
n
c)
2
p
π
m
)
2
c
2
.
P
n
=
(E
π
/c
+
m
n
c,
p
π
)
,i.e.
(E
π
/c
+
−
=
(m
K
+
Substituting for
p
π
E
π
/c
2
m
π
c
2
m
)
2
c
2
(m
π
+
=
−
gives
E
π
=
((m
K
+
−
m
n
)c
2
)/(
2
m
n
)
897 MeV.
12.10 In the kaon rest frame the pions are produced back-to-back. Suppose the
pion that moves in the lab frame has a speed
u
in the kaon rest frame. Then
=