1000 Solved Problems in Modern Physics

350 6 Special Theory of Relativity

6.73 Expressingθin terms ofθ∗
tanθ=sinθ∗/γc(cosθ∗+βc/β∗)(1)

Differentiating with respect toθ∗and setting∂tanθ/∂θ∗=0, gives

cosθ∗=−β∗/βc (2)

And sinθ∗=(β^2 c−β∗^2 )^1 /^2 /βc (3)

Using (2) and (3) in (1), and the equations

β∗^2 γ∗^2 =γ∗^2 −1(4)

βc^2 γc^2 =γc^2 −1(5)

as well as

m 1 β∗γ∗=m 2 βcγc(momentum conservation in the CMS) (6)

and simplifying, we get

tanθmax=[m 22 /(m 12 −m 22 )]^1 /^2

6.74 LetP 1 be the electron four-momentum before the collision andP 2 the final
four-momentum. IfQis the four-momrntum transfer then the conservation of
energy and momentum requires that
Q=P 2 −P 1 (1)
andQ^2 =Q.Q=P 22 +P 12 − 2 P 1 P 2 (2)
ButP=(p,E/c)(3)
P^2 c^2 =E^2 −p.pc^2 =me^2 c^4 (4)
For stationary electron, the initial four-momrntum is
P 1 =(0,mec)(5)
and final four-momentum
P 2 =(p 2 ,mec+ν/c)(6)
Therefore
P 1 .P 2 =mec(mec+ν/c)(7)
Substituting (7) in (2) and using (4)
Q^2 = 2 me^2 c^4 − 2 mec(mec+ν/c)
=− 2 meν

6.75 (a) As the collision is elastic the total initial energy=total final energy.

E+Ee=E′+Ee′ (1)

Butp′=p (2)

∴E′=E (3)

It follows thatE′e=Ee (4)

Consequentlype′=pe (5)

Momentum conservation gives

p−pe=−p′+pe′ (6)