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Rutherford Scattering in Landau book

  1. Mar 28, 2009 #1
    I'm reading Landau and Lifgarbagez's Mechanics book and am having a hard time proving the following:

    On page 53, they present theta_0 = arccos( ... ). As described on page 48 eqn 18.2 the integral should produce this theta_0. However, I am not quite sure what r_min is? On page 48, they say 'It should be recalled that r_min is a zero of the radicand. I *think* that means rmin is a turning point, yes?

    On page 36, they integrate eqn 18.2 (I assume with the same limits: rmin to inf), however notice there is an extra term. (Also they integrated it for U(r) = - alpha/r, but that is just a sign change).

    Furthermore, eqn 15.14 on page 38 (evaluated for a repulsive potential U = +a/r) shows:
    p/r = -1 + e cos(phi). However, if I am not mistaken, for Rutherford's problem, cos(phi) = -1/e, where does the p/r term go?

    Here's a summary of what I am asking if it didn't make sense:
    1) I am trying to find phi_0 for a repulsive potential U(r) = a/r
    2) Using Kepler's problem with the signs changed, phi_0 on page 53 and phi on the top of page 36 don't seem to match (different limits of integration?) I did of course convert to E = 1/2 m vinf^2 and M = mv_inf p, but that still doesn't appear to yield the same result.
  2. jcsd
  3. Mar 28, 2009 #2
    (I don't have a copy of L & L) The variable r min is usually used to connotate the distance of closest approach of the alpha particle in Rutherford scattering. The scattering, being elastic, has a well defined differential cross ection in a central Coulomb field (point source). If r-min is too small, the alpha particle hits the nucleus, and the differential cross section is modified.
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