Problem No 9
Assume that after the electron hits the positron, photon 1 deflects β angle and photon 2 deflects θ
angle from the electron’s original path. This way, β + θ = α .
The electron gains the energy ∆E = e ⋅ U = km0 c 2 . Therefore, its total energy is E = ( k + 1)m0 c 2 .
And, it has the momentum (using energy
energy-momentum relation):
E 2 = m0 2c 4 + p 2c 2
(k + 1)2 m0 2c 4 = m0 2c 4 + p 2c 2
p 2c 2 = m0 2c 4 (k 2 + 2k )
p = m0 c k 2 + 2k
When the electron hits the positron, the momentum and energy is conserved. The energy and the
momentum of photon 1 and photon 2 are E1 = hf1 and E2 = hf 2 , and p1 =
hf1
hf
and p2 = 2 ,
c
c
respectively.
The equation of the conservation of momentum of the axis perpendicular
perpendicular to the original path of the
electron:
p= p
hf1
hf
sin β − 2 sin θ
c
c
f1 sin β = f 2 sin θ
(1)
0=
The equation of the conservation of momentum of the axis parallel to the original path of the
electron:
p= p
hf
hf
m0c k 2 + 2k = 1 cos β + 2 cos θ
c
c
m0c 2 k 2 + 2k = hf1 cos β + hf 2 cos θ (2)
The equation of the conservation of the energy:
E=E
(k + 1)m0c + m0c 2 = hf1 + hf 2
2
m0 c 2 (k + 2) = h( f1 + f 2 )
(3)
Substituting (3) into (2), and substitute (1) afterwards:
( f1 + f 2 ) k 2 + 2k = ( f1 cos β + f 2 cos θ )(k + 2)
f2 (
sin θ
sin θ
+ 1) k 2 + 2k = ( f 2
cos β + f 2 cos θ )(k + 2)
sin β
sin β
(sin θ + sin β ) k 2 + 2k = (sin θ cos β + cos θ sin β )(k + 2)
(sin θ + sin β ) k 2 + 2k = sin(θ + β )(k + 2)
Using the relation β + θ = α , we can get:
(sin θ + sin β ) k 2 + 2k = sin(θ + β )(k + 2)
(sin(α − β ) + sin β ) k 2 + 2k = (k + 2) sin α
(4)
If we differentiate implicitly with respect to β:
(sin(α − β ) + sin β ) k 2 + 2k = (k + 2) sin α


 dα

dα
− 1 + cos β  k 2 + 2k = (k + 2) cos α
 cos(α − β ) 
dβ
 dβ



The implicit graph of the equation (4) >> using WolframAlpha. In this graph, α=B and β=A, and k=1.
The smallest possible angle of α will be the angle that
dα
= 0 (look at the red circle on the graph).
dβ


 dα

dα
− 1 + cos β  k 2 + 2k = ( k + 2) cos α
cos(α − β ) 
dβ
 dβ



[ − cos(α − β ) + cos β ]
k 2 + 2k = 0
cos(α − β ) = cos β
The solutions for the above equations are α = 0 , α = 2β , etc. But if we examine the graph, the
only correct solution is α = 2 β .
For α = 2β , :
(sin(α − β ) + sin β ) k 2 + 2k = (k + 2) sin α
(sin
α
2
+ sin
α
2sin
2sin
α
2
) k 2 + 2k = (k + 2) sin α
α
k 2 + 2k = 2(k + 2) sin
2
α
2
α

k 2 + 2k − cos (k + 2)  = 0
2 
2

Thus:
k 2 + 2k − cos
α
2
(k + 2) = 0
cos
α
2
=
k 2 + 2k
(k + 2)
 k 2 + 2k 

 (k + 2) 
α = 2 cos −1 
For k=1:
 k 2 + 2k 
 3
−1
 = 2 cos   = 1.91 rad
 3 
 (k + 2) 
α = 2 cos −1 
ADRIAN NUGRAHA UTAMA
INDONESIA
cos
α
2