Сарычева Л.И. Физика фундаментальных взаимодействий
Подождите немного. Документ загружается.
α = e
2
/¯hc =
1/137
σ
γp
σ
γγ
Xs
ε
+ Y s
η
X = 0.0677, ε = 0.0808
Y = 0.129, η = 0.4525.
E ,
γp
γγ
β
β
e
ν
β n
p
G
F
= 1.436 · 10
−49
·
3
n → pe
−
¯ν
e
ν
e
n → pe
−
˜ν
e
p → ne
+
K
+
τ ∼ θ
β
K
CP
W
±
Z
0
W
±
Z
0
r ∼
1
M
2
M
2
∼ Q
2
∼
1
r
W
±
Z
0
π
+
→ µ
+
ν
µ
π
−
→ µ
−
˜ν
µ
ν
µ
ν
e
L
µ
L
e
Z
0
j
0
= ¯ee+ ¯µµ+ ¯ττ + ˜ν
e
ν
e
+ ˜ν
µ
ν
µ
+ ˜ν
τ
ν
τ
+ ˜uu+
˜
dd+ ˜ss + ˜cc+
˜
bb+
˜
tt
j = ¯eν
e
+ ¯µν
µ
+ ¯τν
τ
+V
¯
du
·
¯
du+V
¯su
·¯su+V
¯
bu
·
¯
bu+V
¯
dc
·
¯
dc+V
¯sc
·¯sc +
+ V
¯
bc
·
¯
bc + V
¯
dt
·
¯
dt + V
¯st
· ¯st + V
¯
bt
·
¯
bt
V
¯
du
V
¯su
. . .
j
¯pp
W
±
Z
0
W
±
Z
0
m
W
= 80.22±0.26
m
Z
= 91.1884±0.0022 W
±
Z
0
m
Z
= m
W
/ cos θ
W
sin
2
θ
W
=
0.23 θ
W
sin
2
θ
W
Z
0
ν
e
ν
µ
ν
τ
e
+
e
−
Z
0
m
H
> 114 c
2
ν
τ
ν
τ
τ
2 · 10
6
ν
µ
+ e
−
→ µ
−
+ ν
e
σ = ( 14.2 ± 2.9) · 10
−42
· E
ν
2
sin
2
θ
W
= 0.2553 ± 0.0019.
K
0
CP
K
L
K
S
R =
Γ(K
L
→ π
0
π
0
)
Γ(K
S
→ π
0
π
0
)
,
Γ(K
L
→ π
−
π
+
)
Γ(K
S
→ π
−
π
+
)
= 1 − 6
ε
′
ε
.
(ε
′
/ε) 6= 0
CP
(ε
′
/ε) = (17 ±6) ·10
−4
(ε
′
/ε) = (19.2 ±4.6) · 10
−4
CP
B
0
B
0
σ =
G
2
F
m
e
· Eν
2π
(g
νe
v
± g
νe
A
)
2
+
1
3
(g
νe
v
∓ g
νe
A
)
2
;
g
νe
v
= ρ
νe
−
1
2
+ 2K
νe
· sin
2
θ
w
, g
νe
A
= ρ
νe
−
1
2
.
ν
e
↔ ν
µ
↔ ν
τ
ν
1
↔ ν
2
L =
4πE
ν
(m
ν
1
−m
ν
2
)
2
.
m
ν
6= 0
m
ν
= 0
ν
µ
→ ν
τ
hE
ν
i ∼ 10 ∆m
2
ν
= 3 · 10
−3
c
2
)
2
ν
µ
→ ν
e
L ∼ 250
”
“
W Z
W
±
Z
0
W Z