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Deriving Rutherford's Formula

  1. Apr 27, 2010 #1
    1. The problem statement, all variables and given/known data
    2. Relevant equations
    This is in my mechanics book. It gives the final formula, only saying "carrying out the elementary integration..."
    Derive Rutherford's Formula:
    [tex] \phi_0=cos^{-1}\frac{a}{\sqrt{1+a^2}} [/tex]
    where
    [tex] a=\frac{\alpha}{mv_\infty^2 \rho} [/tex]

    From the equation
    [tex] \phi_0 = \int_{r_{min}}^\infty \frac{\frac{\rho}{r^2}dr}{\sqrt{1-\frac{\rho^2}{r^2}-\frac{2U}{mv_\infty^2}}} [/tex]

    Using [tex] U=\frac{\alpha}{r} [/tex]


    3. The attempt at a solution
    This question comes up regularly on exams, and the hint that is given is that

    [tex] \int_{x_+}^\infty \frac{1}{x\sqrt{(x-x_+)(x-x_-}} = arccos \frac{a}{\sqrt{1+a^2}}[/tex]

    where [tex] x_+=a+\sqrt{1+a^2} [/tex] and [tex] x_-=a-\sqrt{1+a^2} [/tex]

    I don't see how this helps, since the integral is from r min, which corresponds to [tex] x_- [/tex] I think. Also, I have [tex] \rho [/tex] instead of 1 in the expressions for [tex] x_-[/tex] and [tex] x_+ [/tex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
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