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I am trying to derive Rutherford's scattering formula, with the coordinate system and polar coordinates chosen as in the picture below.

enter image description here

Angular momentum conservation yields $mvb = mr^2 \dot{\varphi}$. I then tried to make use of this in Newton's equation along the $x$-direction \begin{equation} m \frac{\mathrm{d}v_x}{\mathrm{d}t} = F_x = \frac{2Ze^2}{4\pi\epsilon_0} \frac{\sin(\varphi) \dot{\varphi}}{bv} \end{equation} from the starting point on the left to the endpoint on the right-hand side with scattering angle $\theta$. Integrating over time gives \begin{equation} \int_{t_{\text{start}}}^{t_{\text{end}}} \frac{\mathrm{d}v_x}{\mathrm{d}t} \mathrm{d}t = \frac{2Ze^2}{4\pi\epsilon_0 mvb} \int_{t_{\text{start}}}^{t_{\text{end}}} \sin(\varphi)\dot{\varphi}\, \mathrm{d}t. \end{equation} Introducing substitutions results in \begin{align} \int_{0}^{v\sin(\theta)}\, \mathrm{d}v_x = \frac{2Ze^2}{4\pi\epsilon_0 mvb} \int_{\pi}^{\theta} \sin(\varphi)\,\mathrm{d}\varphi,\\ v\sin(\theta) = -\frac{2Ze^2}{4\pi\epsilon_0 mvb} \left(1+\cos(\theta)\right). \end{align} At this point, I do not see where I made the mistake which led to the minus sign on the right-hand side. Taking a look at Rutherford's formula, it should not appear there. Does anybody see it? Any help is appreciated.

My assumption: With $\varphi$ chosen as in the picture, $\dot{\varphi}(t)$ is negative for all $t$. Then, the equation for the angular momentum conservation might also be written with an additional minus sign, as $mvb$ is positive, and both would cancel at the end. However, a formal derivation of $mr^2\dot{\varphi}$ does not lead to such an additional minus sign.

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  • $\begingroup$ Which textbook are you following? Also, why are you assuming v=0 at t_start? $\endgroup$
    – daydreamer
    Oct 14, 2020 at 12:45
  • $\begingroup$ I am not particularly following a textbook as my definition of $\varphi$ is different from the standard literature (usually, $\varphi$ is measured to the negative $z$-axis in my picture and then integration is performed from 0 to $\pi-\theta$). However, my solution is geared to "Atom- und Quantenphysik" by Haken and Wolf (a German book). $\endgroup$ Oct 14, 2020 at 12:49
  • $\begingroup$ Note that I am assuming the $x$-component of $v$ to be zero at $t_{\text{start}}$. $\endgroup$ Oct 14, 2020 at 12:50
  • $\begingroup$ Notice that if you make theta -> -theta you recover the usual equation. Does that make any sense as to how you defined the other variables? $\endgroup$
    – daydreamer
    Oct 14, 2020 at 12:50
  • $\begingroup$ But $\theta$ should be positive when defining $\varphi$ like in the picture, right? $\endgroup$ Oct 14, 2020 at 12:52

1 Answer 1

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I found the mistake. Indeed, my assumption gave the correct hint: The term $mvb$ needs to have an additional minus sign. We have \begin{equation} \vec{L}_{\text{start}} = \vec{r}_{\text{start}}\times \vec{p}_{\text{start}} = (b\,\hat{e_x} - |z|\,\hat{e_z})\times mv\,\hat{e_z} = -mvb\,\hat{e_y}. \end{equation} All in all, angular momentum conservation reads \begin{equation} -mvb = L_y = mr^2\dot{\varphi} \end{equation} and the two minus signs cancel after the integration.

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