Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

This is a manual from a Monte Carlo physics manual:

Computing the occurrence of a process

Can anyone tell me why the number of mean free paths of a particle is independent of the material which it traverses? It does not make sense to me.

share|improve this question

1 Answer 1

The mean free path denotes the average length the particle travels between interactions. But, while the actual length of the mean free path, depends on the material in question, the number of mean free paths does not. This is because all the material dependent factors are already in the length of the mean free path.

It is the same as in radioactive decay: The amount left after three half-lives passed is always $\frac{1}{8}$ or the original amount, no matter which radioactive material you look at. But the actual time that passed is different for each material i.e. $3\tau_{\frac{1}{2}}$ where $\tau_{\frac{1}{2}}$ contains all the material dependent information.

So in the context of MC simulation of material interactions, you roll dice in the beginning to decide how many mean free paths the particle travels, and then step through the detector until you reach the point where the interaction occurs. The actual physical length travelled of course depends on the material. In some sense it cancels out, in the equation of your link it says

$n_\lambda = \int \frac{dx}{\lambda(x)}$

both $dx$, the physical distance travelled, and $\lambda(x)$, the mean free path, depend in the same way on the material, so the end-result does not.

share|improve this answer
Thank you for that, I had presumed that dx referred to an actual physical distance. For example I thought that the text implied that an electron would have the same number of mean free paths in 1cm of Lead as in 1cm in air. I did have a feeling though that dx had some kind of material dependency. Thanks for clarification –  bateman Mar 5 '13 at 14:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.