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Differential cross-section divergence

  1. Jul 24, 2008 #1
    Hello, I was wondering if anyone could explain the troubling divergence here of the differential cross-section for rutherford scattering for [tex]\theta = 0[/tex]. I know it must have something to do with the fact that the em force extends to infinity, which makes sense to me for the [tex]total[/tex] cross section.. Why would the differential be 0 for [tex]\theta = 0[/tex] and not over all angles?

    [tex]\frac{d\sigma}{d\Omega} \theta = -\frac{b}{\sin \theta}\frac{db}{d\theta}[/tex]

    Thanks..
    Anna
     
  2. jcsd
  3. Jul 24, 2008 #2
    Hi, and welcome on PF,

    the formula
    [tex]\frac{d\sigma}{d\Omega} = \left|\frac{b}{\sin \theta}\left(\frac{db}{d\theta}\right)\right|[/tex]
    is not the Rutherford formula. It is hard sphere scattering, and is merely geometrical.

    Rutherford formula is
    [tex]\frac{d\sigma}{d\Omega} = \left(\frac{q_1q_2}{4E\sin^2 (\theta/2)}\right)^2[/tex]

    I have no clue whether you are refering to the optical theorem which relates the total cross section to the forward amplitude.
     
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