1000 Solved Problems in Modern Physics

354 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 sinθ+p 2 cosθ (momentum conservation alongx-axis) (1)

p 1 cosθ=p 2 sinθ (momentum conservation alongy-axis) (2)

M 0 γ=E 1 +E 2 (Energy conservation) (3)

Solving (1), (2) and (3)

M 0 =(m 12 +m^22 + 2 E 1 E 2 )^1 /^2

β=(p 12 +p 22 )^1 /^2 /(E 1 +E 2 )

θ=tan−^1 (p 1 /p 2 )

Fig. 6.12Decay
M 0 →m 1 +m 2

6.87 Under the assumption (a)

M^2 =2(Eπ++Eπ−−Pπ+Pπ−cosθ)+mπ+^2 +mπ−^2 (1)

Eπ+=(pπ^2 +mπ+^2 )^1 /^2 (2)

Eπ−=(Pπ−^2 +mπ−^2 )^1 /^2 (3)

mπ+=mπ−= 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 forM, we findM= 0 .239 GeV/c^2 ,

a value quite different from the standard value,mk 0 = 0 .498 GeV/c^2

Under the assumption (b)