# In particle physics what is the derivation of the mean free path length: $\ell=\frac{1}{n \sigma}$?

From my lecture notes (ICL, dept. of Physics) it is written: Consider a thin piece of material with thickness $$d$$ containing target particles with number density $$n$$, as illustrated in Figure $$\bf{2.3}$$. Each target particle presents an area, the cross-section $$\sigma$$, for the reaction. For a front surface area $$A$$, there are $$Adn$$ targets, which therefore have a total target area $$Adn \sigma$$. Hence, the probability of an incoming particle hitting one of the targets is $$Adn \sigma/A = dn\sigma$$. Therefore, in this ‘thin target approximation’ the reaction rate per unit area of target is simply: $$R=\phi d n \sigma$$ where $$\phi$$ is the incident particle flux (particles per unit area per unit time); the point is that $$d$$ and $$n$$ can be different for different experiments, but $$\sigma$$ is the physical property we compare. Note that this only works for thin materials, meaning $$d$$ is small enough that the probability of a reaction $$dn \sigma \ll 1$$. Otherwise, every incoming particle will see several targets and have multiple interactions. In this situation, a more appropriate measure is the mean free path, $$\ell$$, of the particle in the material. Clearly, on average the particle will move this distance before interaction ($$\color{red}{\text{corresponding to a probability of one}}$$), so the mean free path is given by $$\color{red}{\ell n \sigma=1}$$, or $$\ell=\frac{1}{n \sigma}\tag{1}$$

I don't understand the parts in red. How does it correspond to a probability of one? I looked on page 34 of W.S.C Williams "Nuclear and Particle Physics. Published by Oxford science publications 1991. But the formula, $$(1)$$, is simply stated without proof. But Williams does mention the kinetic theory of gases, so I searched the internet and found this which looks nothing like $$(1)$$. The closest to a success I have so far is Wikipedia (which even uses the same setup and diagram as ICL). But still, there is no clear derivation that $$\ell=\frac{1}{n \sigma}$$. Does anyone know of (rigorous) proof of $$(1)$$?

As it moves along covering some total travel distance $$L$$, the particle's motion carves out a tube of length $$L$$ and cross sectional area $$\sigma$$, making a volume $$L \sigma$$. If $$n$$ is the number density of other particles that it might hit, then there are $$N = n L \sigma$$ such particles in the tube, so that is the number of collisions it will have in the distance $$L$$. Clearly then, the mean travel distance per collision is $$L/N$$, giving $$\lambda = \frac{L}{N} = \frac{1}{n \sigma}$$ This model is precise when one particle is moving and the others are sitting still.

When the other particles are moving then the answer depends both on their velocity distribution and on the velocity of the particle under consideration. When its velocity is high compared to all the others, then the above result applies. At the other extreme, when its velocity is low but other particles are moving, then in the limit $$v \rightarrow 0$$ its mean free path tends to zero, since it will not move anywhere before another particle comes along and hits it.

In elementary kinetic theory of gases, we wave our hands and average over all this, and a reasonable estimate for the mean free path averaged over both time and velocity is $$\lambda \simeq \frac{1}{\sqrt{2} \sigma n}$$ The $$\sqrt{2}$$ factor here is approximate.

• What happens if the others are moving as well, like a gas? Dec 4, 2020 at 20:57
• @OfekGillon I added some material to cover that. Dec 4, 2020 at 21:29
• @OfekGillon beams are fast. A fixed target experiment at SLAC (40 GeV electrons) on a 30 cm liquid hydrogen target, the beam traverses it in 1 ns. Throw in relativity, and it takes 13 fs: the target is stationary as far as the beam is concerned.
– JEB
Dec 4, 2020 at 22:51
• @JEB yeah but this question bothered me as someone who just deals with gasses from time to time (for optical uses) and I wondered about things like that. Dec 4, 2020 at 23:31
• @AndrewSteane Nice work, thank you. Dec 10, 2020 at 4:23

Suppose the beam comes from the $$-x$$ direction, with number density:

$$N(x) = N_0\ \ \ \ \ (x <0)$$

(where $$N_0$$ is large), and then interacts with the target, which extends from $$x=0$$ to positive $$x$$.

For a small region in the target, of length $$dx$$, at $$x$$, the flux coming in from the left is:

$$N_+(x) = N(x)$$

while the flux leaving the region is:

$$N_-(x+dx) = N(x) - N(x)n\sigma dx$$

Thus:

$$dN(x) \equiv N_+(x) - N_-(x+dx) = -n\sigma N(x)dx$$

$$\frac{dN(x)}{dx} = - n\sigma N(x)$$

and that is solved by:

$$N(x) = Ae^{-\sigma/\lambda}$$

with:

$$\lambda \equiv \frac 1 {n\sigma}$$

With the boundary condition $$N(0)=N_0$$ fixing $$A$$, the beam density in the target is:

$$N(x) = N_0e^{-x/\lambda}$$

The parameter lambda is traditionally called the mean-free path.

If you expand the exponential to 1st order, which is valid if $$x \ll \lambda$$:

$$N(x) \approx N_0(1 - x/\lambda)$$

you that probability of scattering is $$x / \lambda$$, hence if $$x=\lambda$$, one may say that probability of scattering is "1", but it's not because you need to correct for the fact that you are double counting probability, the fix is $$+\frac 1 2 x^2/\lambda^2$$...but that over corrected, so you need to subtract $$\frac 1 6 x^3/\lambda^3$$, and that goes on for infinity until you are back to $$\exp{-x/\lambda}$$.

A probability of one means the particle will definitely travel that distance in the given amount of time using the defined path (through the thin target surface), instead of some other path (say, if its path is curved instead of straight).