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Deriving Rutherford Scattering Angle

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data

    We want to fin an expression for [itex]\theta[/itex], the Rutherford scattering angle.

    I have an expression that I derived for a hyperbolic trajectory, of the form [itex]\frac{(x+x_0)^2}{A^2}-\frac{y^2}{B^2} = 1[/itex]. We wanted it in terms of α, ε, where α is a constant ([itex] \alpha = \frac{L^2}{mk}[/itex]) and [itex] \frac{\alpha}{r} = 1 + \epsilon \cos \theta[/itex].

    So what I got was:

    [tex] \left( x - \frac{\alpha \epsilon}{(\epsilon^2-1)} \right)\frac{(\epsilon^2-1)}{\alpha^2} - \frac{y^2(\epsilon^2-1)}{\alpha^2} =1 [/tex].

    Well an good, I thought. Anyhow, the next bit is deriving an equation for the scattering angle, knowing that the asymptotes are [itex]y = \pm \tan \left(\frac{\pi - \theta}{2}\right)(x+x_0)[/itex].

    I have [itex](x+x_0)[/itex], that's [itex] \left( x - \frac{\alpha \epsilon}{(\epsilon^2-1} \right)[/itex].

    so if I want θ i first solved for [itex]\tan \left(\frac{\pi - \theta}{2}\right)[/itex].

    I got:

    [tex]\tan\left(\frac{\pi - \theta}{2}\right) = \frac{y}{x+x_0} = \frac{y}{\left( x - \frac{\alpha \epsilon}{\epsilon^2-1}\right)} = \frac{y(\epsilon^2-1)}{x(\epsilon^2-1)+\alpha \epsilon}[/tex]

    From there it seems to me I need to solve for θ. The simplest thing seemed to be to take an inverse tangent of the rather messy expression on the right. But I suspect something else is wanted here. That's what I am trying to figure out.

    Otherwise Id have tried treating [itex]\tan ((\theta - \pi)/2)[/itex] as [itex]\frac{-\sin(\theta)}{1-\cos \theta}[/itex] on the left and substituting in the conversion to cartesian coordinates for cos θ, and then take an inverse sine of the (again) messy stuff on the right.

    Anyhow, if anyone can tell me if I am doing something wrong that would be much appreciated, and I thank you.
     
  2. jcsd
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