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
=
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