352
6 Special Theory of Relativity
6.78
1
2
(M
2
− m 1
2
− m 2
2 ) = E 1 E 2 − p 1 p 2 cos θ
θ = 90
0
P ± = 530 MeV/c
E ± = (P
2
± + m π
2 )
1/2
= 548
2 × 548
2
+ 2 × 140
2
= M
2
M = 800 MeV/c
It is a ρ meson.
Fig. 6.10
6.79 Using the invariance of squared four-momentum before and after the decay
E i
2
− P i
2
= E f
2
− |p f |
2
M
2
= (E 1 + E 2 )
2
− ( p 1
2
+ p 2
2
+ 2 p 1 p 2 cos θ )
= (E 1
2
− p 1
2 ) + (E 2
2
− p 2
2 ) + 2(E 1 E 2 − p 1 p 2 cos θ )
= m 1
2
+ m 2
2
+ 2(E 1 E 2 − p 1 p 2 cos θ)
Or E 1 E 2 − p 1 p 2 cos θ =
1
2
(M
2
− m 1
2
− m 2
2 ) = Invariant
6.80 s + t + u = (1/c
2 )[(P A + P B )
2
+ (P A − P C )
2
+ (P A − P D )
2 ]
= (1/c
2 )[3P A
2
+ P B
2
+ P C
2
+ P D
2
+ 2P A (P B − P C − P D )]
(1)
From four-momentum conservation, P A + P B = P C + P D
(2)
(1) becomes (s + t + u)c
2
= (3m A
2
+ m B
2
+ m C
2
+ m D
2 ) − 2P A
2
Using P A = m A c, P B = m B c, P C = m C c and P D = m D c
(s + t + u)c
2
= (m A
2
+ m B
2
+ m C
2
+ m D
2 )c
2
Or s + t + u =
i=A,B,C,D
m
2
i
6.81 t = (P A − P C )
2
/c
2
= (1/c
2 )(P A
2
+ P C
2
− 2P A P C )
= (1/c
2 )[m A
2 c
2
+ m C
2 c
2
− 2(E A E C /c
2
− P A .P C )]
For elastic scattering A ≡ C. Thus E A = E C and |P A | = |P C | = p
So that P A .P C = p
2 cos θ.
c
2 t = 2m A
2 c
2
− 2(E A
2
/c
2
− p
2 cos θ)
But E A
2
= c
2 p
2
+ m A
2 c
4
Therefore, t = −2 p
2 (1 − cos θ)/c
2
6.82 sin ϕ/2 = m π c
2
/2(E 1 E 2 )
1/2 (see Prob. 6.103 and 6.104)
Minimum angle is ϕ min = 2/γ