# Does differential cross section assume no multiple collisions in a material?

The differential cross section definition is often expressed as $$\frac{d\sigma}{d\Omega} = \frac{n(\theta,\phi)}{N_{target} j_A}=\frac{n(\theta,\phi)}{p_N \Delta x \space j_A}$$

My question concerns how we simply divide by $$N_{target}$$. I understand the logic of the incoming particle having 'more opportnities to scatter. However does this only work if we assume that once a particle has scattered of one target, it goes straight to a detector without any more scatters? This question is only relevant if we assume the target is thick so we use the $$\Delta x$$ thickness notation (so multiple successive scatters are possible).

I would think that in cases where a back scatter is incredibly unlikely (say negligible) but a 'sideways' scatter is quite likely, if we account for a particle undergoing multiple scatters then having more particles would increase the liklihood of 'two successive sideways scatters' and therefore skew the final distribution to have back scatters no longer neglible.

Therefore am I correct in my assumption that we only include $$N_{target}$$ as 'more opportunities to scatter one' and not 'more opportunities to also scatter multiple times'?

• Yes, it generally is not applied to multiple scattering. Jan 7, 2021 at 20:12

I like the following description of the microscopic cross section. Let $$\Sigma$$ be the mean free path for a specific interaction. The microscopic cross section $$\sigma$$ is defined as $$\Sigma/N$$ where $$N$$ is the density of atoms/nuclei in the target.