I'm working through the problems in Chapter 3 of the 3rd edition of Goldstein's Classical Mechanics and I'm stuck on Derivation 4. This problem asks the reader to rewrite the scattering angle

\begin{equation*} \Theta(s) = \pi - 2\int_{r_{m}}^{\infty}{\frac{s\mathrm{d}r}{\displaystyle{r\sqrt{r^{2}\left[1-\frac{V(r)}{E}\right] - s^{2}}}}} \end{equation*}


\begin{equation*} \Theta(s) = \pi - 4s\int_{0}^{1}{\frac{\rho\mathrm{d}\rho}{\displaystyle{\sqrt{r_{m}^{2}\left[1-\frac{V(r)}{E}\right]^{2} - s^{2}(1-\rho^{2})}}}} \end{equation*}

(where $r_{m}$, the distance of nearest approach, is such that $V(r_{m})/E = 1$) via a change of variables $r\to \rho(r)$. To attack this problem, I tried writing

\begin{equation*} \frac{\mathrm{d}r}{\displaystyle{r\sqrt{r^{2}\left[1-\frac{V(r)}{E}\right] - s^{2}}}} = \frac{\displaystyle{\frac{r_{m}}{r^{2}}\sqrt{1-\frac{V}{E}}\mathrm{d}r}}{\displaystyle{\sqrt{r_{m}^{2}\left(1-\frac{V}{r}\right)^{2} - \frac{r_{m}^{2}s^{2}}{r^{2}}\left(1-\frac{V}{E}\right)}}} \end{equation*}

which would seem to hint at $\rho$ of a form such that

\begin{equation*} s^{2}(1-\rho^{2}) = \frac{r_{m}^{2}s^{2}}{r^{2}}\left(1-\frac{V}{E}\right) \end{equation*}

or, solving for $\rho^{2}$,

\begin{equation*} \rho^{2} = 1-\frac{r_{m}^{2}}{r^{2}}\left(1-\frac{V}{E}\right). \end{equation*}

Unfortunately, this form of $\rho$ doesn't yield the correct bounds of integration, and I haven't even incorporated the Jacobian term

\begin{equation*} \mathrm{d}\rho = \frac{\mathrm{d}\rho}{\mathrm{d}r}\mathrm{d}r. \end{equation*}

Should I keep working the problem this way, or have I missed something obvious? The physical insight offered by this problem hardly seems worth the effort necessary to solve it (the author suggests its benefit is numerical stability), so I can't help feeling there's an easier way. All the same, I'd love to know what $\rho(r)$ is supposed to be.


1 Answer 1


$$ \begin{align} \tag1\text{Let }r&=\frac{r_m}{1-\rho^2} \quad\text{so that}\quad \frac{\mathrm dr}{\mathrm d\rho}=\frac{2r_m\rho}{(1-\rho^2)^2}=\frac{2r^2\rho}{r_m}\\ \tag{3.96}\Theta(s)&=\pi-2\int_{r_m}^\infty \frac{s\,\mathrm dr}{r\sqrt{r^2\left[1-\frac{V(r)}E\right]-s^2}}\\ \tag2\therefore\qquad\Theta(s)&=\pi-4s\int_0^1\frac{\rho\,\mathrm d\rho}{\sqrt{r_m^2\left[1-\frac{V(r)}E\right]-s^2(1-\rho^2)^2}} \end {align} $$ That is, the question had a typo at this point by missing the square on the $(1-\rho^2)^2$ in the square root. If it had it correct, then it would connect to the next form of the integral because $(1-\rho^2)^2=1-2\rho^2+\rho^4$ is needed, with the fact that $\frac{s^2E}{r_m^2}=\frac{\ell^2}{2mr_m^2}=E-V(r_m)$

  • $\begingroup$ Thank you so terribly much! In the errata, the authors mention that in this derivation, " (1-...) in deominator is squared" (sic), and it was unclear that they meant $(1-\rho^{2})$ instead of $[1-V(r)/E]$ (which, already being squared, might seem to have already been corrected in my printing of the book). $\endgroup$
    – kandb
    Mar 6 at 11:56
  • 1
    $\begingroup$ You might be amused that it took me a while to figure out which particular choice of integral would get this correct. It is definitely not trivial. $\endgroup$ Mar 6 at 16:19
  • $\begingroup$ In that case, thanks again! $\endgroup$
    – kandb
    Mar 7 at 7:31
  • $\begingroup$ Incidentally, in the next part of the problem, is it true that the $(2-\rho^{2})$ is correct in the book (i.e. does not need to be squared)? $\endgroup$
    – kandb
    Mar 7 at 9:28
  • $\begingroup$ I specifically wrote why it is correct. That is my last statement up there. $\endgroup$ Mar 7 at 12:06

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