# Compton effect

1. Dec 29, 2015

### simon96c

Hello everyone, I have a question regarding a formula which can be derived from conservation of momentum and energy in the Compton effect.
From conservation of momentum and energy during the collision of a photon with an electron, it is possible to get two expressions for the final energy E of the electron:

E^2=(Q_0-Q)^2+2(Q_0-Q)m_0*c^2+(m_0*c^2)^2
and
(cP)^2=Q_0^2-2*QQ_0cosΘ+Q^2

where Q and Q_0 are the final and initial energies of the photon and Θ the angle between the momentum vector of the photon before hitting the electron and after.

The problem arises when, to derivate the final formula, the two equations are subtracted and one term seems to go missing, as the book says it is:

2QQ_0(1-cosΘ)-2(Q_0-Q)m_0c^2=0

whereas I have

2QQ_0(1-cosΘ)-(2Q_0-2Q-m_0c^2)m_0c^2=0, so I basically have a "m_0c^2" which should not be there.
Could some explain what I'm missing there?

I'm sure it's something really silly, so thanks in advance for the answers! :)

2. Dec 29, 2015

### Orodruin

Staff Emeritus
E and cP are not equal. They satisfy the dispersion relation for a particle of mass m_0 ...

3. Dec 29, 2015

### simon96c

I see. In fact, the book does not mention it and I'm realising why.
So, since to get the formula I'm subtracting the two equations and I now that the RHS is zero, does this mean that E^2-(m_0*c^2)^2 = (cP)^2? This would explain why I could not get the correct final formula, but I'm still not completely sure about why E^2-(m_0*c^2)^2 = (cP)^2.

4. Dec 29, 2015

### Orodruin

Staff Emeritus
This is the fundamental dispersion relation for a particle of mass m_0. In fact, it is how we define mass in relativity.

5. Dec 29, 2015

### simon96c

I've looked it up and now I think I have understood. Thank you for the answer!