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Show that the Energy Transfer is given by

  1. May 15, 2015 #1

    vmr101

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    1. The problem statement, all variables and given/known data
    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:
    [tex] hv' = \frac{hvE(1+u)} {2hv+E(1-u)} [/tex]

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

    3. The attempt at a solution
    I have [tex]E+hv = E'+hv'[/tex] (cons. of energy) and [tex]-Eu+hv = E'u' - hv'[/tex] (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.
     
  2. jcsd
  3. May 15, 2015 #2

    mfb

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    There is a fixed relation between E' and u', which is a third equation for your three unknowns.
     
  4. May 15, 2015 #3

    vmr101

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    Do you mean E' = M(u') = gamma (u') m0
    which seems to still keep the factor of U' in there
     
  5. May 16, 2015 #4

    mfb

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    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.
     
  6. May 16, 2015 #5

    vmr101

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    There was a ' in the question, but looked like typo as it was different format to the other dash '
     
    Last edited: May 16, 2015
  7. May 16, 2015 #6

    mfb

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    That would surprise me, as the one-dimensional case has a clear solution, there is no need to introduce unknown quantities in the equation.
     
  8. May 16, 2015 #7

    vmr101

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    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?
     

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    Last edited: May 16, 2015
  9. May 16, 2015 #8

    mfb

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    That looks like a comma, not like a dash. And it would not make sense to add a dash to a bracket.
    There is no way to tell if you don't show your steps.
     
  10. May 16, 2015 #9

    vmr101

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    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
     
  11. May 19, 2015 #10
    Hey vmr101, have you solved the problem? I'm stuck on the same part as you are, a little help will be appreciated!
     
  12. May 19, 2015 #11

    vmr101

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    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.
     
  13. May 19, 2015 #12
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