Pair production

This process is related to a high-energy photon, that is entirely transformed into an electron and a positron as it is represented in this reaction: γ → e ^- + e⁺
This is a creation of two particled from a wave.

This latter wave must have an energy E_g, at least equal to the two resting masses of electron and positron, that is 511 keV each. Thus, the minimum photon energy required for pair production to occur is 1.022 MeV.

The mechanism is as follows:


Let:
E_g the photon incident energy,
P_g its linear momentum

E_e- the total energy of the produced electron
P_e- its linear momentum

E_e+ the total energy of the produced positron
P_e+ its linear momentum

The concervation of the momentum is:
P_g = P₁ + P₂ (1)
The horizontal projection gives:
p_g= hν/c = p₁cos a + p₂ cos b (2)
The vertical projection gives:
p₁ sin a = p₂ sin b (3)

The concervation of the energy is:
E_g= E_e- + E_e+or E₁ + E₂ (4)
With :
E₁² = p₁²c ² + m₀² c⁴ (5)
E₂ ² = p₂²c ² + m₀² c⁴ (6)

We have then:
hν = (p₂² c² + m₀ c⁴ )^1/2 + (p₂² c² + m₀ c⁴ )^1/2 (7)

From (3) , The relationship (2) becomes`
hν/c = p₁ cos a + p₁ sin a cos b/sin b = p₁ sin(a + b)/sin b
Thus:
p₁ = (hν/c) sin b/ sin( a+b) (8)
And:
p₂ = (hν/c) sin a/ sin ( a+b) (9)

The relation (4) becomes then:
hν = (p₁²c ² + m₀² c⁴)^1/2 + ( p₂²c ² +m₀² c⁴)^1/2

Substituing (8) and (9) in the latter equation, we can write it:
(((sin b)/sin (a+b))²+ (m_oc²/hν)²)^1/2 + (((sin a)/sin (a+b))²+ (m_oc²/hν)²)^1/2 = 1

We have used the available 3 equations and we have 4 unknown values (p₁, p₂, a and b) We get a relionship between a and b.

If we assume that p₁= p₂ , then a = b and E₁= E₂ we will get from ( 8) :
p₁ = p₂ = (hν/c)/ 2 cos a (10)
The inverse reaction to pair production is the annihilation reaction.