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Simplifying scattering integral, goldstein derivation 3.4

  1. Sep 12, 2006 #1
    Hi everyone, could anyone give me a hint on Goldstein derivation 3.4? Starting from

    [tex] \theta = \pi - 2 \int_{r_{m}}^{\infty} \frac{s / r^{2} dr}{\sqrt{1 - V(r)/E - s^{2}/r^{2}}}
    [/tex]

    they do a change of variables to get

    [tex] \theta = \pi - 4 s \int_{0}^{1} \frac{\rho d\rho}{\sqrt{r_{m}^{2} (1 - V(r)/E)^{2} - s^{2} (1-\rho^{2})}}
    [/tex]

    where

    [tex] 1 - V(r_{m})/E - s^{2}/r^{2} = 0
    [/tex]

    Naturally I want the mystery function [tex] \rho(r) [/tex]. I have gotten to the expression

    [tex]
    \theta = \pi - 2 \int_{0}^{1} \frac{s du}{\sqrt{r_{m}^{2}(1 - V(u)/E) - s^{2}u^{2}}}
    [/tex]

    by making the transformation [tex]u = r_{m} / r[/tex], but no further. I haven't been able to find this transformed integral in the literature either. Note that this is from the third edition, 6th printing of Goldstein; earlier versions had an error where a square exponent was omitted.

    Thanks!
     
    Last edited: Sep 12, 2006
  2. jcsd
  3. Sep 13, 2006 #2
    problem solved, typo in Goldstein

    Many thanks to Ravinder Abrol for solving this problem. There is a typo in the most current printing of Goldstein, the correct equation should be

    [tex] \theta = \pi - 4 s \int_{0}^{1} \frac{\rho d\rho}{\sqrt{r_{m}^{2} (1 - V(r)/E) - s^{2} (1-\rho^{2})^{2}}}
    [/tex]

    with the mystery function

    [tex]
    1 - \rho(r)^{2} = r_{m}/r
    [/tex]

    Interestingly, the online errata for the earlier printings

    http://astro.physics.sc.edu/goldstein/4-5To6.html

    stated that the (1 - ...) term should be squared, but in the most recent printing they squared the wrong term.

    I have confirmed with Mathematica that the transformation does remove the singularity, and that the integral value is equal.
     
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