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Compton Scattering and Conservation of Momentum

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data

    photon scatter angle is theta, electron recoil angle is phi

    prove

    tan(phi) = (1+hf/mc^2)^-1 cot(theta/2)




    2. Relevant equations



    3. The attempt at a solution

    Conservation of energy is the same as in here:

    http://en.wikipedia.org/wiki/Compton_scattering


    Conservation of Momentum

    In x :

    p_gamma = p_gamma' cos(theta) + p_e' cos(phi)

    In y:

    0 = p_gamma' sin(theta) - p_e' sin(phi)

    p_gamma'=p_e' sin(phi)/sin(theta)


    Let's solve for p_e'^2 since we can directly substitute it in the energy term.

    p_e'^2 = p_gamma^2 sin^2(theta)/ [sin^2(phi)cos^2(theta) + 2 sin(phi)cos(theta)sin(theta)cos(phi) + sin^2(theta)cos^2(phi)]

    I'm not sure where the cot(theta/2) comes from which we want to show.
     
  2. jcsd
  3. Nov 19, 2009 #2
    Is the problem at least set up correctly?
     
  4. Nov 20, 2009 #3

    Redbelly98

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  5. Nov 25, 2009 #4
    If I'm to prove


    tan(phi) = (1+hf/mc^2)^-1 cot(theta/2)

    Would you have any idea where the hf' disappears if you look at the energy equations in the wikipedia link?
     
  6. Nov 25, 2009 #5
    In a frams S, two identical particles with electric charge q move abreast along lines parallel to the x-axis, a distance r apart and with velocity v. Determine the force in S that each exerts on the other, by use of the force law for a uniformly moving charge.



    I would need a lorentz forc e law in 4 vector notation,

    what is it? Then where do I proceed?
     
  7. Nov 26, 2009 #6

    Redbelly98

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    Well, you have 3 equations (x-momentum, y-momentum, energy). The idea is to use those 3 equations to eliminate 2 of the variables -- hf' and p_e'. That will leave you with just hf, θ, and φ in a single equation. The problem is in working out the algebra to get there.

    I recommend starting a new thread for this problem.
     
  8. Dec 3, 2009 #7
    [tex]
    E_{\gamma} = hf
    [/tex]
    [tex]
    E_{\gamma'} = hf'
    [/tex]
    [tex]

    E_e = m_e c^2

    [/tex]
    [tex]

    E_e'= \sqrt{ (p_e' c)^2 + (m_e c^2)^2 }
    [/tex]
    [tex]

    hf + m_e c^2 = hf' + \sqrt{ (p_e' c^2)^2 + (m_e c^2)^2 }

    [/tex]
    [tex]

    p_e'^2 c^ = (hf +m_e c^2 - hf')^2 - m_e^2 c^4
    [/tex]


    Conservation of Momentum

    In x :

    [tex] p_\gamma = p_\gamma' cos(\theta) + p_e' cos(\phi)
    [/tex]
    In y:

    [tex] 0 = p_{\gamma'} sin(\theta) - p_{e'} sin(\phi)
    [/tex]
    [tex]
    p_{\gamma'}=p_{e'} sin(\phi)/sin(\theta)
    [/tex]

    Let's solve for p_e'^2 since we can directly substitute it in the energy term.
    [tex]
    p_e'^2 = p_{\gamma}^2 sin^2(\theta)/ [sin^2(\phi)cos^2(\theta) + 2 sin(\phi)cos(\theta)sin(\theta)cos(\phi) + sin^2(\theta)cos^2(\phi)]
    [/tex]
    I'm not sure where the cot(theta/2) comes from which we want to show.
     
  9. Dec 3, 2009 #8

    Redbelly98

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    You still haven't eliminated pe from the equations yet.

    I suggest changing the cot(θ/2) expression using the half-angle formulas (see my link in Post #3). Then the expression you are trying to prove will be in terms of θ, and you can stop wondering about the θ/2. Note, there are several different, equivalent expressions for cot(θ/2)
     
  10. Dec 3, 2009 #9
    Where do you see to eliminate hf'?
     
  11. Dec 4, 2009 #10

    Redbelly98

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    Disclaimer: I have not actually solved this problem myself.
    You can use the p_x and p_y equations to eliminate pe and get an expression for hf'. That expression can be substituted into the energy equation, eliminating hf'. HOWEVER -- the expression you get is very messy, and it's not obvious to me how to simplify it.

    Another observation: if you take the energy equation and divide by mc2, the left-hand side becomes

    1 + hf/(mc2),​

    a term that appears in the result you are trying to prove. Maybe (I repeat: maybe) working with the equation in that form would help.

    Have you recast the equation-to-be-proved in terms of θ, to get rid of the θ/2?
     
  12. Dec 4, 2009 #11

    kuruman

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    The momentum conservation equations can be written as
    [tex]p \: sin\phi=\frac{\epsilon '}{c}sin\theta[/tex]
    [tex]p \: cos\phi=\frac{\epsilon}{c}-\frac{\epsilon '}{c}cos\theta[/tex]

    from which

    [tex]tan\phi=\frac{\epsilon 'sin\theta}{\epsilon-\epsilon ' cos\theta}[/tex]

    where ε and ε' are the photon energies before and after the scattering process. Replace the wavelengths with energies in the Compton equation to get

    [tex]\frac{1}{\epsilon '}-\frac{1}{\epsilon}=\frac{1-cos\theta}{mc^2}=\frac{2sin^2(\theta /2)}{mc^2}[/tex]

    Solve this last equation for ε' and plug back in the tangent equation. You should get the answer without too much algebra. Don't forget to use the half-angle identities for theta.
     
  13. Dec 7, 2009 #12
    I've only gotten as close as:

    [tex]

    tan(\phi) = \frac{1}{mc^2 (\frac{sin(\theta)}{2\epsilon + cos^2(\theta)/sin^2(\theta))})}

    [/tex]
     
  14. Dec 7, 2009 #13

    Redbelly98

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    Something is wrong with that equation, which is evident by examining the units. In this expression:

    2ε + cos2θ/sin2θ​

    a term with units of energy is being added to a unitless quantity, which is not allowed. That indicates an error has been made somewhere.
     
  15. Dec 7, 2009 #14

    kuruman

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    What is your expression for ε' when you solved for it?
     
  16. Dec 7, 2009 #15
    [tex]
    \epsilon ' = (\frac{2sin^2(\theta /2)}{mc^2} + \frac{1}{\epsilon})^{-1}
    [/tex]

    ah I see now, let me fix it

    [tex]

    tan(\phi) = \frac{sin(\theta)}{\frac{\epsilon}{\frac{2sin^{2}(\theta/2)}{mc^2}} +\frac{1}{\epsilon}}

    [/tex]
     
    Last edited: Dec 7, 2009
  17. Dec 7, 2009 #16

    kuruman

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    It still doesn't work dimensionally. It seems you have trouble adding fractions. Look

    [tex]
    \frac{1}{\epsilon'}=\frac{1}{\epsilon}+\frac{2sin^2(\theta /2)}{mc^2}
    [/tex]

    [tex]
    \frac{1}{\epsilon'}=\frac{mc^2+2 \epsilon \: sin^2(\theta /2)}{\epsilon \: mc^2}
    [/tex]

    [tex]\epsilon'=\frac{\epsilon \: mc^2}{mc^2+2 \epsilon \: sin^2(\theta /2)}[/tex]

    Stick this in the expression for the tangent and see what happens. Be sure to add fractions and simplify correctly. If in doubt, do dimensional analysis on both sides.
     
  18. Dec 7, 2009 #17
    I got it thanks.
     
    Last edited: Dec 7, 2009
  19. Nov 16, 2010 #18
    thanks for such an impressive discussion.. would you mind to post the whole final equation on here please ? i got some misunderstandings i think..
     
  20. Nov 16, 2010 #19

    Redbelly98

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    I don't know if I understand your question. The whole final equation is already given in the problem statement:
    In other words

    [tex]\tan \phi = \frac{\cot(\theta/2)}{1 + \frac{hf}{mc^2}}[/tex]​

    If you're trying to work this out yourself, show us how far you have gotten and where you're stuck. Then we'll know how to help.
     
    Last edited: Nov 16, 2010
  21. Nov 16, 2010 #20
    I just wanna see the whole solution step by step with explanation in one post . so i can translate it to my own language and use in my lectures. if this is possible it would be great but if not it's ok. thank you again.
     
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