504
9 Particle Physics – I
9.4 In natural units τ = 2/mα
5 . Introduce the factors and c
Let τ = 2
x c
y
/m α
5
Find the dimensions: [] = [Energy × Time], [c] = [velocity]
[τ ] = [1/m][ML
2 T
−1 ]
x [LT
−1 ]
y
[T ] = [M]
x−1 [L]
2x+y [T ]
−x−y
Equating the coefficients on both sides of the equation,
−x − y = 1, 2x + y = 0, x − 1 = 0
we get x = 1, y = −2
Therefore τ = 2c
−2
/mα
5
= 2 /mc
2
α
5
τ = 2 × (0.659 × 10
−21 MeV-s) × (137)
5
/0.511 MeV
= 1.245 × 10
−10 s
9.5 Γ μ = G F
2 m μ
5
/192π
3
(1)
Now, the dimensional formula for G F is
[Energy]
−2 [ c]
3 and [Γ μ ] = [Energy]
Introduce and c in (1) and take dimensions on both sides.
Γ μ = G F
2 mμ
5
x c
y
/192π
3
(2)
[M L
2 T
−2 ] = [M L
2 T
−2 ]
−4 [M]
5 [M L
2 T
−1 ]
x+6 [L T
−1 ]
y+6
or [M L
2 T
−2 ] = [M]
7+x [L]
10+2x+y [T ]
−4−x−y
Equating powers of M, L and T , we find
x = −6 and y = 4
and (2) becomes
Γ μ = G F
2 m μ
5
−6 c
4
/192π
3
Γ μ = /τ = G F
2 (m μ c
2 )
5
/192 (c)
6
π
3
= (1.116 × 10
−5 GeV
−2 )
2 (105.659 × 10
−3 )
5
/192π
3
τ = 2.39 × 10
−6 s
9.3.2 Production
9.6 (a) The relation for lab angle θ and the C.M.S. angle θ
∗ is given by
tan θ = sin θ
∗
/γ c (cos θ
∗
+ β c /β
∗ )
( 1 )
where β c = v c /c is the CMS velocity and γ c is the corresponding Lorentz
factor (see summary of Chap. 6). For photon β
∗
= 1 and β c = 1 as the
electron is ultra relativistic. Dropping off the subscript c, (1) becomes
tan θ = sin θ
∗
/γ (cos θ
∗
+ 1) = tan (θ
∗
/2)/γ
(2)
(b) Assuming that the photons are emitted isotropically, half of the photons will
be contained in the forward hemisphere in the CMS, that is within θ
∗
= 90
◦ .
Substituting θ
∗
= 90
◦ in (2)
tan θ = 1/γ
As the incident electron is ultrarelativistic the photons in the lab would come
off at small angles so that tan θ ≈ θ = 1/γ . Thus half of the photons will be
emitted within a cone of half angle θ ≈ 1/γ .
9 Particle Physics – I
9.4 In natural units τ = 2/mα
5 . Introduce the factors and c
Let τ = 2
x c
y
/m α
5
Find the dimensions: [] = [Energy × Time], [c] = [velocity]
[τ ] = [1/m][ML
2 T
−1 ]
x [LT
−1 ]
y
[T ] = [M]
x−1 [L]
2x+y [T ]
−x−y
Equating the coefficients on both sides of the equation,
−x − y = 1, 2x + y = 0, x − 1 = 0
we get x = 1, y = −2
Therefore τ = 2c
−2
/mα
5
= 2 /mc
2
α
5
τ = 2 × (0.659 × 10
−21 MeV-s) × (137)
5
/0.511 MeV
= 1.245 × 10
−10 s
9.5 Γ μ = G F
2 m μ
5
/192π
3
(1)
Now, the dimensional formula for G F is
[Energy]
−2 [ c]
3 and [Γ μ ] = [Energy]
Introduce and c in (1) and take dimensions on both sides.
Γ μ = G F
2 mμ
5
x c
y
/192π
3
(2)
[M L
2 T
−2 ] = [M L
2 T
−2 ]
−4 [M]
5 [M L
2 T
−1 ]
x+6 [L T
−1 ]
y+6
or [M L
2 T
−2 ] = [M]
7+x [L]
10+2x+y [T ]
−4−x−y
Equating powers of M, L and T , we find
x = −6 and y = 4
and (2) becomes
Γ μ = G F
2 m μ
5
−6 c
4
/192π
3
Γ μ = /τ = G F
2 (m μ c
2 )
5
/192 (c)
6
π
3
= (1.116 × 10
−5 GeV
−2 )
2 (105.659 × 10
−3 )
5
/192π
3
τ = 2.39 × 10
−6 s
9.3.2 Production
9.6 (a) The relation for lab angle θ and the C.M.S. angle θ
∗ is given by
tan θ = sin θ
∗
/γ c (cos θ
∗
+ β c /β
∗ )
( 1 )
where β c = v c /c is the CMS velocity and γ c is the corresponding Lorentz
factor (see summary of Chap. 6). For photon β
∗
= 1 and β c = 1 as the
electron is ultra relativistic. Dropping off the subscript c, (1) becomes
tan θ = sin θ
∗
/γ (cos θ
∗
+ 1) = tan (θ
∗
/2)/γ
(2)
(b) Assuming that the photons are emitted isotropically, half of the photons will
be contained in the forward hemisphere in the CMS, that is within θ
∗
= 90
◦ .
Substituting θ
∗
= 90
◦ in (2)
tan θ = 1/γ
As the incident electron is ultrarelativistic the photons in the lab would come
off at small angles so that tan θ ≈ θ = 1/γ . Thus half of the photons will be
emitted within a cone of half angle θ ≈ 1/γ .