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6 Special Theory of Relativity
Fig. 6.11 (a) Decay
∧ → P + π
− in flight. (b)
Momentum triangle
(a)
(b)
6.86 M 0 βγ = p 1 sin θ + p 2 cos θ (momentum conservation along x-axis)
(1)
p 1 cos θ = p 2 sin θ (momentum conservation along y-axis)
(2)
M 0 γ = E 1 + E 2 (Energy conservation)
(3)
Solving (1), (2) and (3)
M 0 = (m 1
2
+ m
2
2 + 2E 1 E 2 )
1/2
β = ( p 1
2
+ p 2
2 )
1/2
/(E 1 + E 2 )
θ = tan
−1 ( p 1 / p 2 )
Fig. 6.12 Decay
M 0 → m 1 + m 2
6.87 Under the assumption (a)
M
2
= 2(E π + + E π− − P π+ P π− cos θ) + m π+
2
+ m π −
2
(1)
E π + = ( p π
2
+ m π+
2 )
1/2
(2)
E π − = (P π −
2
+ m π−
2 )
1/2
(3)
m π + = m π − = 0.14 GeV/c
2
, θ = 15
◦
(4)
p + = 1.67, p − = 0.408 GeV/c
( 5 )
Using (2), (3), and (4) in (1) and solving for M, we find M = 0.239 GeV/c
2 ,
a value quite different from the standard value, m k0 = 0.498 GeV/c
2
Under the assumption (b)