Hi everyone, could anyone give me a hint on Goldstein derivation 3.4? Starting from(adsbygoogle = window.adsbygoogle || []).push({});

[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!

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# Homework Help: Simplifying scattering integral, goldstein derivation 3.4

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