Compton Scattering

## **Problem**

Consider a photon of wavelength

λ

{\displaystyle \lambda}

 and frequency

ν

{\displaystyle \nu}

 moving towards a stationary electron of mass

m
0

{\displaystyle m\_0}

 along the x-axis. The energy and momentum of the photon are

# E
h
ν

{\displaystyle E = h \nu}

 and

# p

h
ν

c

=

h
λ

{\displaystyle p = \frac{h\nu}{c} = \frac{h}{\lambda}}

 respectively. After the photon collides with the electron, the photon is scattered at an angle

θ

{\displaystyle \theta}

 and its frequency shifts. Show that the change in the photon's wavelength is

Δ
# λ

h

m
0

c

(
1
−
cos
⁡
θ
)

{\displaystyle \Delta \lambda = \frac{h}{m\_0 c} (1-\cos \theta)}

.

## **Solution**

The incident photon with wavelength

λ

{\displaystyle \lambda}

 and frequency

ν

{\displaystyle \nu}

 collides with a stationary electron of mass

m
0

{\displaystyle m\_0}

. The collision causes the electron to recoil, and a new photon with wavelength

λ
′

{\displaystyle \lambda'}

 and frequency

ν
′

{\displaystyle \nu'}

 emerges at angle

θ

{\displaystyle \theta}

 from x-axis.

Applying the conservation of energy, we get

E
γ

E

e

=

E

γ
′

E

e
′

{\displaystyle E\_\gamma + E\_{e} = E\_{\gamma'} + E\_{e'}}

h
ν

m
0

c
2

=
h

ν
′

E

e
′

{\displaystyle h \nu + m\_0 c^2 = h \nu' + E\_{e'}}

Applying the conservation of momentum, we get

p

γ

# 0

p

γ
′

p

e
′

{\displaystyle \mathbf{p}\_\gamma + 0 = \mathbf{p}\_{\gamma'} + \mathbf{p}\_{e'}}

h
ν

c

# 0

h

ν
′

c

p

e
′

{\displaystyle \frac{h\nu}{c}+ 0 = \frac{h\nu'}{c} + {p}\_{e'}}

Applying the law of cosines to the triangle in the vector diagram (Figure 2, left), we get

p

e
′

2

=

p

γ

2

p

γ
′

2

−
2

p

γ

p

γ
′

cos
⁡
θ

{\displaystyle
p\_{e'}^{2} = p\_{\gamma}^{2} + p\_{\gamma'}^{2} - 2 p\_{\gamma} p\_{\gamma'} \cos\theta}

p

e
′

2

=

h
2

ν
2

c
2

h
2

ν

′

2

c
2

−
2

h
2

ν

ν
′

c
2

cos
⁡
θ

{\displaystyle
{p}\_{e'}^2 = \frac{h^2 \nu^2}{c^2} + \frac{h^2 \nu'^{\; 2}}{c^2} - 2 \frac{h^2 \nu \nu'}{c^2} \cos\theta}

p

e
′

2

c
2

=

h
2

ν
2

h
2

ν

′

2

−
2

h
2

ν

ν
′

cos
⁡
θ

{\displaystyle
{p}\_{e'}^2 c^2 = h^2 \nu^2 + h^2 \nu'^{\; 2} - 2 h^2 \nu \nu' \cos\theta}

After scattering, the possibility that the electron might be accelerated to a significant fraction of the speed of light, requires that its total energy be represented using the relativistic energy-momentum relation

E

e
′

=

(

p

e
′

c

)
2

(

m
0

c
2

)
2

.

{\displaystyle E\_{e'} = \sqrt{(p\_{e'} c)^2 + (m\_0 c^2)^2}.}

Substituting these quantities into the expression for the conservation of energy gives

h
ν

m
0

c
2

=
h

ν
′

(

p

e
′

c

)
2

(

m
0

c
2

)
2

.

{\displaystyle h\nu + m\_0 c^2 = h\nu' + \sqrt{(p\_{e'}c)^2 + (m\_0 c^2)^2}.}

Thus,

p

e
′

2

c
2

=

h
2

ν
2

h
2

ν

′

2

−
2

h
2

ν

ν
′

cos
⁡
θ

{\displaystyle
{p}\_{e'}^2 c^2 = h^2 \nu^2 + h^2 \nu'^{\; 2} - 2 h^2 \nu \nu' \cos\theta}

 can be expressed as

(
h
ν
−
h

ν
′

m
0

c
2

)
2

−
(

m
0

c
2

)
2

=

(

h
ν

)

2

(

h

ν
′

)

2

−
2

h
2

ν

ν
′

cos
⁡

θ

.

{\displaystyle (h\nu - h\nu' + m\_0 c^2)^2 - (m\_0c^2)^2 = \left(h \nu\right)^2 + \left(h \nu'\right)^2 - 2h^2 \nu\nu'\cos{\theta}.}

After evaluating the square and rearranging terms, we get

2
h
ν

m
0

c
2

−
2
h

ν
′

m
0

c
2

=
2

h
2

ν

ν
′

(

1
−
cos
⁡
θ

)

.

{\displaystyle 2 h \nu m\_0 c^2-2 h \nu' m\_0 c^2 = 2 h^2 \nu \nu' \left( 1 - \cos \theta \right). }

Dividing both sides by

2
h
ν

ν
′

m
0

c

{\displaystyle 2h\nu\nu' m\_0 c}

 gives

c

ν
′

−

c
ν

=

h

m
0

c

(

1
−
cos
⁡
θ

)

.

{\displaystyle \frac{c}{\nu'} - \frac{c}{\nu} = \frac{h}{m\_0 c}\left(1-\cos \theta \right). }

Since

λ
# ν
c

{\displaystyle \lambda \nu = c}

 and

λ
′

ν
′

=
c

{\displaystyle \lambda' \nu' = c}

, we get

λ
′

−
# λ

h

m
0

c

(

1
−
cos
⁡
θ

)

{\displaystyle \lambda' - \lambda = \frac{h}{m\_0 c}\left(1-\cos \theta \right)}

Δ
# λ

h

m
0

c

(

1
−
cos
⁡
θ

)

.

{\displaystyle \Delta \lambda = \frac{h}{m\_0 c}\left(1-\cos \theta \right). }