1000 Solved Problems in Modern Physics

352 6 Special Theory of Relativity

6.78

1

2

(M^2 −m 12 −m 22 )=E 1 E 2 −p 1 p 2 cosθ

θ= 900

P±=530 MeV/c

E±=(P±^2 +mπ^2 )^1 /^2 = 548

2 × 5482 + 2 × 1402 =M^2

M=800 MeV/c

It is aρmeson.

Fig. 6.10

6.79 Using the invariance of squared four-momentum before and after the decay

Ei^2 −Pi^2 =Ef^2 −|pf|^2

M^2 =(E 1 +E 2 )^2 −(p 12 +p 22 + 2 p 1 p 2 cosθ)

=(E 12 −p 12 )+(E 22 −p 22 )+2(E 1 E 2 −p 1 p 2 cosθ)

=m 12 +m 22 +2(E 1 E 2 −p 1 p 2 cosθ)

OrE 1 E 2 −p 1 p 2 cosθ=

1

2

(M^2 −m 12 −m 22 )=Invariant

6.80s+t+u=(1/c

(^2) )[(P
A+PB)
(^2) +(P
A−PC)
(^2) +(P
A−PD)
(^2) ]
=(1/c^2 )[3PA^2 +PB^2 +PC^2 +PD^2 + 2 PA(PB−PC−PD)] (1)
From four-momentum conservation,PA+PB=PC+PD (2)
(1) becomes (s+t+u)c^2 =(3mA^2 +mB^2 +mC^2 +mD^2 )− 2 PA^2
UsingPA=mAc,PB=mBc,PC=mCcandPD=mDc
(s+t+u)c^2 =(mA^2 +mB^2 +mC^2 +mD^2 )c^2
Ors+t+u=

∑

i=A,B,C,Dm

2

i

6.81t=(PA−PC)^2 /c^2 =(1/c^2 )(PA^2 +PC^2 −^2 PAPC)

=(1/c^2 )[mA^2 c^2 +mC^2 c^2 −2(EAEC/c^2 −PA.PC)]

For elastic scatteringA≡C. ThusEA=ECand|PA|=|PC|=p

So thatPA.PC=p^2 cosθ.

c^2 t= 2 mA^2 c^2 −2(EA^2 /c^2 −p^2 cosθ)

ButEA^2 =c^2 p^2 +mA^2 c^4

Therefore,t=− 2 p^2 (1−cosθ)/c^2

6.82 sinφ/ 2 =mπc^2 /2(E 1 E 2 )^1 /^2 (see Prob. 6.103 and 6.104)
Minimum angle isφmin= 2 /γ