Derivation of Ee in Inverse Compton Scattering

  • Thread starter Alkass
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  • #3
BvU
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Sorry if that seems a bit coarse. After posting I noticed this was your first post in the homework area. Didn't mean to shoo you away. Anyway, the homework area has some guidelines with mandatory parts. Showing an effort is the main one. To prevent us from being used as workhorses instead of helpers. :wink:
 
  • #4
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Sorry if that seems a bit coarse. After posting I noticed this was your first post in the homework area. Didn't mean to shoo you away. Anyway, the homework area has some guidelines with mandatory parts. Showing an effort is the main one. To prevent us from being used as workhorses instead of helpers. :wink:
Sorry, I just wanted to avoid redundancy as I already posted the link.

What I have already in the reference frame of the e- for the photon

[itex] E_0 = \gamma E_0^* (1- \frac{u}{c}cos\theta_0^*) \\

E_1 = \frac{E_0}{1+ \frac{E_0}{(m_e*c^2)}(1-cos\phi)} \\

E_1^*= \gamma E_1(1- \frac{u}{c}cos\theta_0^*)[/itex]


where [itex]*[/itex] denotes the lab reference frame

[itex] \theta_0^* [/itex] the angle between the direction of the e- and the incoming photon in the lab frame

[itex] \phi [/itex] the angle between the incoming and outgoing photon in the rest frame of the e-..

Now, I cannot figure out how to get the energy of the electron in the lab frame...
 
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  • #5
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Hi

Any hint for this ?

Thanks again
 
  • #6
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@Orodruin : can you help this guy out ? I don't get much further than googling to see what limits can be taken safely. (Sorry Alk...)
 
  • #7
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@Orodruin : can you help this guy out ? I don't get much further than googling to see what limits can be taken safely. (Sorry Alk...)
I think what I am not sure about, is how to derive the Ee given the fact that the the photon energy has to be corrected with the Doppler term and has been "boosted" by the gamma...
 
  • #8
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Hi

@Orodruin @BvU sorry for keep bugging you again, but any help would be highly appreciated
 

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