1000 Solved Problems in Modern Physics

336 6 Special Theory of Relativity

or

tanθμ∗= 1 /βcγc (3)

tanθμ=sinθμ∗/γc(cosθμ∗+βc/βμ∗)= 1 /γc^2 βc(1+ 1 /β∗μ)(4)

(Sinceθμ∗=π−θν∗) and we have used (2)

But β∗μ=(mπ^2 −mμ^2 )/(mπ^2 +mμ^2 )(5)

Substituting (5) in (4) and simplifying

tanθμ=(mπ^2 −m^2 μ)/ 2 γ^2 cβcm^2 π

6.9 Thez-component of velocity is zero; hence the particle must be moving in the

xy=plane. Further, they-component of velocity is unchanged. This implies

that the Lorentz transformation is to be made alongx-axis

cPx=γ(cPx′+βE′)(1)

c^2 mβxγ 0 =γ(c^2 mβ′xγ′+mβγ′c^2 )(2)

βx= 1 / 21 /^2 ,γ 0 = 21 /^2 ,β′x=− 1 / 21 /^2 ,γ′= 21 /^2 ,γ= 1 /(1−β^2 )^1 /^2 (3)

Using (4) in (1) and simplifying we getβ= 2 × 21 /^2 / 3

6.10 Energy conservation gives

T 1 +T 2 =Q (1)

Momentum conservation gives

P 1 +P 2 = 0

orp 12 =p 22

T 12 + 2 T 1 m 1 =T 22 + 2 T 2 m 2 (2)

Solving (1) and (2)

T 1 =Q(Q+ 2 m 2 )/2(m 1 +m 2 +Q);T 2 =Q(Q+ 2 m 1 )/2(m 1 +m 2 +Q)

6.11γπ= 1 +Tπ/mπ= 1 + 140 / 140 = 2
βπ=(γπ^2 −1)^1 /^2 /γπ=(2^2 −1)^1 /^2 / 2 = 0. 866
By Problem 6.54 ,Tμ∗= 4 .0 MeV therefore
γμ∗= 1 +Tμ∗/mμ= 1 +(4/106)= 1. 038
βμ∗=(1. 037772 −1)^1 /^2 / 1. 0377 = 0. 267
γμ=γγμ∗(1+βπβμ∗cosθ∗)
γμ(max)=γπγμ∗(1+βπβμ∗)= 2 × 1 .038(1+ 0. 866 × 0 .267)= 2. 556
(Becauseθ∗=0)
Tμ(max)=(γμ(max)−1)mμ=165 MeV
Using the formula for optical Doppler effect
Tν(max)=γπTν∗(1+βπ)= 2 × 29 .5(1+ 0 .866)=110 MeV

6.12βc=|p++p−|/(E++E−)
Using the invariance principle
(total energy)^2 −(total momentum)^2 =invariant
(E++E−)^2 −|p++p−|^2 =(E 1 ∗+E 2 ∗)^2 −|p 1 ∗+p 2 ∗|^2
ButE 1 ∗=E 2 ∗since the particles have equal masses. Also by definition of
center of mass,|p 1 ∗+p 2 ∗|= 0
Therefore,E 1 ∗^2 =E 2 ∗^2 =^14 [E++E−)^2 −(p+^2 +p−^2 + 2 p+p−cosθ)]
whereθis the angle betweene+−e−pair