Different approximations of Compton scattering equation

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The discussion focuses on deriving the change in frequency of low-energy photons scattered by ultrarelativistic electrons using the Compton scattering equation. Participants clarify the simplification of the equation under specific conditions, noting that for low-energy photons, the relationship reduces significantly. The importance of correctly evaluating the vector difference (Ω'-Ω).β is emphasized, with confirmation that it can be expressed as Ω'.β - Ω.β. A participant successfully simplifies the expression by combining fractions, leading to the desired result. The exchange highlights the mathematical intricacies involved in understanding Compton scattering.
ck99
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Homework Statement



Show that, for low energy photons scattered by ultrarelativistic electrons, the cange in frequency of the photon is given by

(v'-v) / v = [(Ω'-Ω).β] / [1-Ω'.β]


Homework Equations



The full/general form of Compton scattering is given by

v'/v = (1-Ω.β) / [(1-Ω'β) + hv/(γmc2) (1 - Ω.Ω') ]

where v is photon frequency
m is electron mass
β is electron velocity divided by c
c is speed of light
γ is Lorentz factor
Ω is unit vector of propagation of the photon

and primed quantities are those quantities after scattering

The Attempt at a Solution



I have attempted the following. For low energy photons, hv << mc2 so that reduces the equation to

v'/v = (1-Ω.β) / (1-Ω'β)

or (v'-v)/v = [ (1-Ω.β) / [(1-Ω'β) ] - 1

For ultra-relativistic electrons, velocity is almost c, so β = 1 but looking at the target answer it is not helpful to remove β from the equation.

I think maybe I am missing something to do with vectors. How do I properly evaluate (Ω'-Ω).β ?

Is it just (Ω'-Ω).β = Ω'.β - Ω.β ?
 
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ck99 said:
How do I properly evaluate (Ω'-Ω).β ?

Is it just (Ω'-Ω).β = Ω'.β - Ω.β ?
Right. The scalar product and vector addition are distributive.

(v'-v)/v = [ (1-Ω.β) / [(1-Ω'β) ] - 1 is identical to the given formula, just written in a different way.
 
Hi mfb and thanks very much for your response. Just to clarify, when you say "Right." do you mean

1) Right, you have not expanded the brackets correctly

2) Right, you have expanded the brackets correctly

If I was any sort of mathematician I am sure I would be able to tell which you mean, but I'm not, and I can't!

If I have expanded the brackets correctly, I can't see how the two versions of the expression are compatible. I have three or four pages of algebra here, trying to make it work, but I must be missing something. If I am incorrect in the expansion, could you elaborate on how it should b done properly?
 
Right, is it just (Ω'-Ω).β = Ω'.β - Ω.β

Start with [ (1-Ωβ) / [(1-Ω'β) ] - 1
write 1 as (1-Ω'β)/(1-Ω'β) and combine the fractions:
(1-Ωβ-(1-Ω'β)) / (1-Ω'β)
Simplify, using (Ω'-Ω).β = Ω'.β - Ω.β:
((Ω'-Ω)β) / (1-Ω'β)
Done.
 
Ah, thanks! I as trying Taylor expansions and all sorts of things. I think it was the second step I was missing.

Cheers!
 
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