# QM Compton scattering energy loss; check of derivation

1. Oct 4, 2004

### Ingwe

Okay, so the question is an electron of energy 100 MeV collides with a photon of wavelength 3x10^(-3) m (ie: the CMB). What is the maximum energy loss of the electron?
After doing a few derivations for formulae, I came up with this one at work...could someone please let me know if there is anything is wrong with it?

Momentum Conservation:

Pe + Pγ = Pe' + Pγ'
Pe + hν/c = Pe' + hν'/c
c[Pe' - Pe] = h(ν-v')
c²Pe'² + c²Pe² - 2c²PePe'[cosθ] = (hv-hv')²
c²Pe'² = (hv-hv')² - c²Pe² + 2c²Pe'Pe[cosθ]
Where cosθ would be the angle between Pe and Pe'

Energy Conservation:

Ee + Eγ = Ee' + Eγ'
Ee + hv = [(mc²)² + (Pe'c)²]^½ + hv'
[Ee + (hv-hv')]² = m²c^4 + Pe'²c²
Ee² + 2(hv-hv') + (hv-hv')² = m²c^4 + (hv-hv')² - c²Pe² + 2c²Pe'Pe[cosθ]
Ee² + 2cPe' - 2cPe = m²c^4 - Pe² + 2c²Pe'Pe[cosθ]
Pe'[2c - 2c²Pe[cosθ]] = m²c^4 - c²Pe² - Ee² + 2cPe
Pe' = [m²c^4 - c²Pe² - Ee² + 2cPe]/2c[1 - cPe[cosθ]]

So since the energy loss is given as
ΔEe = Ee - Ee' = Ee - [(mc²)² + (Pe'c)²]^½, to maximize it we need to make Pe' as small as we can, which occurs when cosθ = (-1) ==> θ = 180

Does this make sense?

Last edited: Oct 4, 2004