I was struggling to understand what a scattering cross section is (in a classical setting), and I found this document:


If I understand correctly from page 3 if we use a "particle current density" $\vec J (\vec x,t)$ (analogously to electromagnetism) then the differential cross section is given by:

$$\frac{d\sigma}{d \Omega}=\frac{R^2 \vec J_{\infty} \cdot \hat r}{\vec J_0 \cdot \hat n}$$

Where $\vec J_{\infty}$ is the current when the interaction is over and $\vec J_0$ is the current before the interaction happens. This resembles some expressions I've seen regarding scattering of electromagnetic waves, so I thought it might at least make some sense. When we plug in $\vec J$ though problems arise. If we write $\vec J$ for $N$ point particles:

$$\vec J (\vec x, t)=\sum_{i=1}^N \vec v_i \delta (\vec x-\vec x_i (t))$$

Then we have delta functions all over the place and I do not know how to give an interpretation to that thing anymore.

My question is: am I completely off target about this? Is there a way to give meaning to what I wrote and use it to actually calculate a "toy" cross section (for example, Rutherford scattering)?

Thank you in advance for your answers!

  • 1
    $\begingroup$ I'm a little irritated that Rutherford scattering would be considered a 'toy' problem, particularly since it works for all $1/r^{2}$ potentials. $\endgroup$
    – Jon Custer
    Commented Jan 8, 2016 at 20:12
  • 1
    $\begingroup$ It's 'toy' in the sense that it makes for a good example. It's not derogatory. $\endgroup$ Commented Jan 8, 2016 at 20:15
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    $\begingroup$ I believe that the equation you started with implicitly assumes a continuum approximation, either by averaging on a time/length scale such that the particle discreteness doesn't matter or going straight to the quantum probability current and not using a classical-point-particle concept at all. Even in a classical model, to make sense of the first equation in the large-R limit, you have to make some smoothing approximation and/or a probabilistic interpretation of J. Basically I think you're trying to use a discrete and continuum viewpoint at the same time and it's getting messy. $\endgroup$
    – elifino
    Commented Jan 8, 2016 at 22:54


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