Show that the Energy Transfer is given by

AI Thread Summary
The discussion revolves around deriving the energy transfer formula for a photon undergoing Inverse Compton Scattering when colliding with a relativistic charged particle. Participants are using conservation laws of energy and linear momentum to set up equations but are facing challenges with the variable u' and its resolution. There is a suggestion to express u' in terms of E' to simplify the equations, which participants find tedious but necessary for the proof. The conversation highlights the importance of careful algebraic manipulation to achieve the desired result. Overall, the thread emphasizes the complexity of the problem and the collaborative effort to find a solution.
vmr101
Gold Member
Messages
25
Reaction score
1

Homework Statement


A Photon has undergone Inverse Compton Scattering, a charged particle of rest mass m0 has relativistic energy E >> m0, collides head on with a photon of frequency v, where hv << m0. Assume the complete process takes place in one spatial dimension, say x.
Using the conservation laws of rel. energy and rel. linear momentum, show the energy transfer to the photon is given by:
hv&#039; = \frac{hvE(1+u)} {2hv+E(1-u)}

Homework Equations


where the rel. momentum of the charged particle before the collision is px = -Eu

The Attempt at a Solution


I have E+hv = E&#039;+hv&#039; (cons. of energy) and -Eu+hv = E&#039;u&#039; - hv&#039; (cons. of linear momentum)
I sub one into the other, yet am stuck with u' which I can not resolve. I believe this can be solved with the required information, but I am unsure how to proceed. Any help would be appreciated. Thank you.
 
Physics news on Phys.org
There is a fixed relation between E' and u', which is a third equation for your three unknowns.
 
Do you mean E' = M(u') = gamma (u') m0
which seems to still keep the factor of U' in there
 
Forget the concept of a relativistic mass, it is not used in physics.

Yes, it has u' in it. That is great, because it allows to express u' in terms of E', which you can plug into the equation you got before.
 
There was a ' in the question, but looked like typo as it was different format to the other dash '
 
Last edited:
That would surprise me, as the one-dimensional case has a clear solution, there is no need to introduce unknown quantities in the equation.
 
Now I am more confused :(

I added eq 1 and 2, and got 2hv+E(1-u) = E'(1+u') but solving this for u' seems to get complicated. Is there a simpler way to show this proof?
 

Attachments

  • q2.png
    q2.png
    998 bytes · Views: 500
Last edited:
That looks like a comma, not like a dash. And it would not make sense to add a dash to a bracket.
vmr101 said:
Is there a simpler way to show this proof?
There is no way to tell if you don't show your steps.
 
mfb said:
That looks like a comma, not like a dash. And it would not make sense to add a dash to a bracket.
That is what I first thought.

So solve for u' then sub that back into one of the equations and rearranging should produce the proof. The algebra seems to be tedious, ill keep trying to make it work. Thanks for the help mfb
 
  • #10
Hey vmr101, have you solved the problem? I'm stuck on the same part as you are, a little help will be appreciated!
 
  • #11
I added Eq 1 to 2 to remove the hv', and rearranged to have u' in terms of all other parameters, giving 2hv + E(1-u) = E'(1+u'), then subbed E' [where E'=m_0 / sqrt(1-u')] back in, and then have (2hv+E(1-u))/m_0 (set this equal to x) = (1+u') / sqrt(1-u'^(2)), then solved for u' in terms of x.
Then subbed back in the terms, and put all this back into eq 1.
I haven't pushed through with the working as the algebra is tedious, but that should show the proof.
 
  • #12
deleted
 

Similar threads

Back
Top