Compton Scattering

A photon bounces off a particle, and you treat it just like an elastic collision of billiard balls or something, except that if a photon loses energy it declines in frequency and gains in wavelength. In what follows, I will use m_o for mass measured at rest. This is for emphasis; most people use m.

Before doing the conventional Compton equation, let us look at a head-on collision: A particle has velocity v (+ if to the right), and a photon of wavelength l is moving to the right. After collision, the particle has velocity v' and the photon is going to the left. Now it has wavelength l '.
Recall g = (1- v²/c²)^-1/2 The momentum and energy equations are

h/l + g m_ov = -h/l ' + g ' m_ov' and

hc/l + g m_oc² = hc/l + g 'm_oc².

Do some algebra on this and show that
g ' = (A² + 1)/(2A), where A is a dimensionless quantity that depends on the initial conditions:

A = 2h/(l m_oc) + g v/c + g .

Then show that 1/l ' = 1/l +m_oc(g - g ')/h. This is not Compton’s equation.

We can check this result against the Compton equation, which is for v = 0, so g = 1. Insert these, do more algebra, and show that l ' = l + 2h/m_oc. See below for the Compton equation, and you will see that it agrees with this when you plug in the photon scatter angle of 180^o. (If you want to use Compton's equation when v ¹ 0, you will need to use the system in which v=0 and modify l and l ' by the Doppler effect. Yecchhh.)

Compton did the case of the particle initially at rest, and the collision can be a glancing blow, so the photon scatters at angle q and the particle at angle f with respect to the initial photon direction, which we take to be the x direction, as shown below. After collision the particle has momentum p = g m_ov. Then we have the two momentum equations (x and y components) and the energy equation:

h/l = h/l 'cosq + pcosf

0 = h/l 'sinq - psinf

hc/l + m_oc² = hc/l ' + g m_oc²

Then with a with a whole lot of algebra and gnashing teeth, cursing, etc., you can show that

D l = (h/m_oc)(1 - cosq )

Now that you know it involves cosq and not f , here is an easier derivation: Draw the vector diagram for the conservation of momentum, and write the law of cosines equation involving cosq .

h/l’

p

q f

p² = (h/l )² +(h/l ')² - 2[(h)²/(l l ')]cosq h/l

Then get the g m_oc² term by itself in the energy equation, square the equation, and replace
(g m_oc²)² with (pc)² + (m_oc²)² (a handy identity) to rid yourself of g . Then divide by c² and combine the two equations to get rid of p. Multiply by l l ' and simplify and you are finished.

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