350
6 Special Theory of Relativity
6.73 Expressing θ in terms of θ
∗
tan θ = sin θ
∗
/γ c (cos θ
∗
+ β c /β
∗ )
( 1 )
Differentiating with respect to θ
∗ and setting ∂ tan θ/∂θ
∗
= 0, gives
cos θ
∗
= −β
∗
/β c
(2)
And sin θ
∗
= (β
2
c − β
∗2 )
1/2
/β c
(3)
Using (2) and (3) in (1), and the equations
β
∗2
γ
∗2
= γ
∗2
− 1
( 4 )
β c
2
γ c
2
= γ c
2
− 1
( 5 )
as well as
m 1 β
∗
γ
∗
= m 2 β c γ c (momentum conservation in the CMS)
(6)
and simplifying, we get
tan θ max = [m 2
2
/(m 1
2
− m 2
2 )]
1/2
6.74 Let P 1 be the electron four-momentum before the collision and P 2 the final
four-momentum. If Q is the four-momrntum transfer then the conservation of
energy and momentum requires that
Q = P 2 − P 1
(1)
and Q
2
= Q.Q = P 2
2
+ P 1
2
− 2P 1 P 2
(2)
But P = (p, E/c)
( 3 )
P
2 c
2
= E
2
− p.pc
2
= m e
2 c
4
(4)
For stationary electron, the initial four-momrntum is
P 1 = (0, m e c)
( 5 )
and final four-momentum
P 2 = ( p 2 , m e c + ν/c)
( 6 )
Therefore
P 1 .P 2 = m e c(m e c + ν/c)
( 7 )
Substituting (7) in (2) and using (4)
Q
2
= 2m e
2 c
4
− 2m e c(m e c + ν/c)
= −2 m e ν
6.75 (a) As the collision is elastic the total initial energy = total final energy.
E + E e = E
+ E e
(1)
But p
= p
(2)
∴ E
= E
(3)
It follows that E
e = E e
(4)
Consequently p e
= p e
(5)
Momentum conservation gives
p − p e = −p
+ p e
(6)