6.3 Solutions 335
∂φ/∂x=γ(∂φ/∂x′−βc∂φ/∂t′)(9)
∂φ/∂t=γ(∂φ/∂t′−βc∂φ/∂x′) (10)
∂^2 φ
∂x^2=γ^2∂^2 φ
∂x′^2+(γ^2 β^2 c^2 )∂^2 cφ
∂t′^2−(2γβc)∂^2 φ′
∂x′∂t′(11)
∂^2 φ
∂t^2=
γ^2 ∂^2 φ
∂t′^2+(γ^2 β^2 c^2 )∂^2 φ
∂x′^2−(2γβc)∂^2 φ′
∂x′∂t′(12)
Dividing (12) throughc^2 and subtracting the resulting equation from (11)
∂^2 φ
∂x^2 −( 1
c^2)∂ (^2) φ
∂t^2
=(γ^2 −γ^2 β^2 )
∂^2 φ
∂x′^2
−
(
1
c^2)
(γ^2 −γ^2 β^2 )∂^2 φ
∂t′^2=∂^2 φ
∂x′^2−
(
1
c^2)
∂^2 φ
∂t′^2
sinceγ^2 −γ^2 β^2 =γ^2 (1−β^2 )= 1
Similarly, the Klein–Gordon equation
(∇^2 −(1/c^2 )∂^2 /∂t^2 +m^2 c^2 /^2 )=0 is Lorentz invariant.6.7 The only wayπ−is emitted at rest in the lab system is when it is emitted at
θ 1 ∗= 180 ◦in the CMS (rest frame ofK◦) with with the same speed asK◦in
the lab system. In that caseπ−will be emitted atθ∗ 2 = 0 ◦in the CMS.
The energy releasedQ= 498 − 2 × 140 =218 MeV
As the product particles are identical, each pion carries half of the enrgy,
109 MeV
γ∗= 1 +T∗/mπ= 1 + 109 / 140 = 1. 778
From the above discussion
γc=γ∗,βc=β∗
γ=γ∗γc(1+β∗βc)=γ∗^2 (1+β∗^2 )
=γ∗^2 (1+(γ∗^2 −1)/γ∗^2 )= 2 γ∗^2 − 1
= 2 × 1. 7782 −^1 =^5.^3266
T=(γ−1)mπ=(5. 3266 −1)× 140
= 605 .7MeV6.8tanθν
∗=sinθ
ν/γc(cosθν−βc/βν
∗)=− 1 /γ
cβc (1)
(Becauseθν= 90 ◦andβν∗=1). Hereβcis the velocity of the pion.
It follows that
sinθν∗= 1 /γcand cosθν∗=−βc (2)
In the CMS (the system in which the pion is at rest)
θμ∗=π−θν∗, because the muon and neutrino must fly in the opposite direc-
tion to conserve momentum.
tanθμ∗=tan(π−θν∗)=−tanθν∗=−^1 /γcβc