6.3 Solutions 363
6.107 (a), (b) In the CMS,m 2 will move with the velocityβcin a direction opposite
to that ofm 1. By definition, the total momentum in the CMS before and after
the collision is zero. In natural unitsc=1.
m 1 γ∗β∗=m 2 γcβc (1)
Squaring (1) and expressing the velocities in terms of Lorentz factors
m 12 (γ∗^2 −1)=m 22 (γc^2 −1) (2)
Using the invariance
(ΣE)^2 −|ΣP|^2 =(ΣE∗)^2 −|ΣP∗|^2 =(ΣE∗)^2 (3)
(because
∑
P∗=0, in the CMS)
(m 1 γ+m 2 )^2 −m 12 (γ^2 −1)=(m 1 γ∗+m 2 γc)^2 (4)
Combining (2) and (4) and callingv=m 2 /m 1
γc=(γ+ν)/(1+ 2 γν+ν^2 )^1 /^2 (5)
γ∗=(γ+ 1 /ν)/(1+ 2 γ/ν+ 1 /ν^2 )^1 /^2 (6)
For the special case,m 1 =m 2 ,asintheP–Pcollision
γc=γ∗=[(γ+1)/2]^1 /^2 (7)
In addition ifγ 1
γc≈(γ/2)^1 /^2 (8)
(c), (d)
The Lorentz transformations are
Pcosθ=γc(p∗cosθ∗+E∗)(9)
Psinθ=p∗sinθ∗ (10)Dividing (10) by (9)
tanθ=p∗sinθ∗/γc(p∗cosθ∗+βcE∗)=sinθ∗/γc(cosθ∗+βc/β∗) (11)
(becausep∗/E∗=β∗)
From the inverse transformation
P∗cosθ∗=γc(Pcosθ−βcE) (12)
and (10) we get
tanθ∗=sinθ/γc(cosθ−βc/β) (13)6.108 At the right angle to the direction of source velocity the Doppler shift in
wavelength is calculated from
ν′=γνorλ′=λ/γ
whereγis the Lorentz factor of the carbon atoms andTis the kinetic energy
of carbon andMc^2 is the approximate rest mass energy, the quantity