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Simple variable elimination

  1. Oct 30, 2009 #1
    1. The problem statement, all variables and given/known data
    From the equations:

    [tex]h\nu - h\nu' = T= m_o c^2 (\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1)[/tex]
    [tex]\frac{h\nu}{c}=\frac{h\nu'}{c}cos\theta+\frac{m_o v}{\sqrt{1-\frac{v^2}{c^2}}}cos\phi[/tex]
    [tex]\frac{h\nu'}{c}sin\theta = \frac{m_o v}{\sqrt{1-\frac{v^2}{c^2}}}sin\phi[/tex]
    [tex]\alpha=\frac{h\nu}{m_oc^2}[/tex]

    (representing the energy and momentum conservation for Compton scattering, theta = photon scatter angle and phi = electron scatter angle)

    Eliminate v and [tex]\theta[/tex] to obtain:

    [tex]T=m_oc^2 \frac{2\alpha^2cos^2\phi}{1+2\alpha+\alpha^2sin^2\phi}[/tex]

    2. Relevant equations
    See above


    3. The attempt at a solution
    This problem has been driving me insane. I know the physics of Compton scattering but when I try to eliminate the v and theta from these 3 equations I just get hopelessly lost in the math... I have about 5 pages of scribbles leading to nothing.

    Desperate at this point, some math help would be appreciated.
     
  2. jcsd
  3. Oct 31, 2009 #2

    gabbagabbahey

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    HInt: [itex]\sin^2\theta+\cos^2\theta=1[/itex]:wink:
     
  4. Nov 2, 2009 #3
    I know that.... the problem is when I try to make theta disappear, I have to bring in a v. If I want to make v disappear, I bring in a theta. The only way that it seems to "work" gives me the following mess:

    [tex]\sqrt{1-\frac{v^2}{c^2}}= \left( \frac{T}{mc^2}+1 \right) ^{-1}[/tex]
    [tex]v=c\sqrt{\left( 1-\left(\frac{T}{mc^2}+1 \right)^{-2} \right)}[/tex]
    [tex]cos\theta = \sqrt{1-\left(\frac{c}{h\nu'} \right)^2 \frac{(mv)^2}{1-\frac{v^2}{c^2}} sin^2\phi}[/tex]

    As you can imagine, taking v and substituting it into the cos equation makes for a disgusting mess that doesn't lead to anything very productive.
     
  5. Nov 2, 2009 #4

    gabbagabbahey

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    I don't see why you say this... just solve the equation

    [tex]\frac{h\nu}{c}=\frac{h\nu'}{c}cos\theta+\frac{m_o v}{\sqrt{1-\frac{v^2}{c^2}}}cos\phi[/tex]

    for [itex]\frac{h\nu'}{c}\cos\theta[/itex], square both sides, and then square both sides of the equation

    [tex]\frac{h\nu'}{c}sin\theta = \frac{m_o v}{\sqrt{1-\frac{v^2}{c^2}}}sin\phi[/tex]

    and add the two resulting equations together...
     
  6. Nov 2, 2009 #5
    Ok so that gives me:

    [tex]\left(\frac{h\nu'}{c}\right)^2=p_e^2-2p_ecos\phi\frac{h\nu}{c}+\left(\frac{h\nu}{c}\right)^2[/tex]

    where [tex]p_e=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}[/tex] (for simplicity)

    Not sure how to proceed... if I solve for p I have to use a quadratic eqn giving another big mess?
     
  7. Nov 2, 2009 #6

    gabbagabbahey

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    The next step would be to eliminate [itex]\nu'[/itex] in favor of [itex]T[/itex] by using your first equation...
     
  8. Nov 2, 2009 #7
    So:

    [tex]\frac{(h\nu)^2-2Th\nu+T^2}{c^2}=p_e^2-2p_ecos\phi\left(\frac{h\nu}{c}\right)+\left(\frac{h\nu}{c}\right)^2[/tex]

    My instinct says to solve for [tex](cp_e)^2[/tex] and then substitute that into energy conservation equation...

    [tex]h\nu+mc^2=h\nu'+\sqrt{(mc^2)^2+(cp)^2}[/tex]

    ...but I'll still have a [tex]p_e[/tex] in there. Not sure if that's the right course of action...
     
  9. Nov 2, 2009 #8

    gabbagabbahey

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    Why not solve the energy conservation equation

    [tex]T+m_0c^2=\sqrt{(m_0c^2)^2+p_e^2c^2}[/tex]

    for [itex]p_e[/itex] and then substitute that into

    [tex]\frac{(h\nu)^2-2Th\nu+T^2}{c^2}=p_e^2-2p_ecos\phi\left(\frac{h\nu}{c}\right)+\left(\frac {h\nu}{c}\right)^2[/tex]
     
  10. Nov 2, 2009 #9
    so

    [tex]c^2p_e^2=T^2+2Tmc^2[/tex]

    After substituting...

    [tex](h\nu)^2-2Th\nu=2Tmc^2-2h\nu\sqrt{T^2+2Tmc^2}cos\phi+\frac{(h\nu)^2}{c}[/tex]

    And getting rid of the root gives me a massive equation with 12 terms on the right :S
     
  11. Nov 2, 2009 #10

    gabbagabbahey

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    Why do you still have a factor of [itex]1/c[/itex] in the last term?
     
  12. Nov 2, 2009 #11
    My bad, the last one doesn't have 1/c.

    Subbing in I get:

    [tex]2h\nu cos\phi \sqrt{T^2+2Tmc^2}=2Tmc^2+2Th\nu[/tex]

    I'll try to solve for T...
     
  13. Nov 2, 2009 #12
    Awesome! FINALLY got it.

    Thanks a million for your help!
     
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