Markov Chain Derivation of the Mean Free Path

1. Memoryless (Markov) Property

A fundamental assumption in deriving the mean free path is the memoryless property. Specifically, let $X$ be the random variable for the distance traveled by a particle before it collides. The Markov (or memoryless) property states:

$P\bigl(X > x + \Delta x \,\big\vert\, X > x\bigr) \;=\; P\bigl(X > \Delta x\bigr), \quad \forall\, x, \,\Delta x \,\ge 0.$

In words, “the probability of traveling an additional distance $\Delta x$ without collision, given that the particle has already traveled $x$ without collision, does not depend on $x$ .” This memoryless trait is the defining feature of the exponential distribution.

2. Collision Rate and Exponential Distribution

Suppose the medium has:

Number density of scatterers: $n$ (scatterers per unit volume),
Effective cross-section: $\sigma$ .

Then, in a small distance $\mathrm{d}x$ , the probability of collision can be approximated as:

$\text{Prob}(\text{collision in } \mathrm{d}x) \;=\; n\,\sigma \,\mathrm{d}x,$

provided $n,\sigma,\mathrm{d}x \ll 1$ . Consequently, the probability of not colliding in that interval is $1 - n,\sigma,\mathrm{d}x$ .

Let $S(x) = P(X > x)$ be the survival function, i.e., the probability that the particle has traveled farther than $x$ without colliding. Because collisions are assumed memoryless,

$S(x + \mathrm{d}x) \;=\; S(x)\,\bigl[1 – n\,\sigma \,\mathrm{d}x\bigr].$

Taking the limit $\mathrm{d}x \to 0$ , we get the differential equation:

$\frac{\mathrm{d}S}{\mathrm{d}x} \;=\; -\,n\,\sigma \,S.$

Its solution is the familiar exponential function:

$S(x) \;=\; \exp\!\bigl(-\,n\,\sigma\,x\bigr).$

Thus, the probability density function (PDF) for $X$ is:

$f_X(x) \;=\; \frac{\mathrm{d}}{\mathrm{d}x}\bigl[1 – S(x)\bigr] \;=\; n\,\sigma \,\exp\!\bigl(-\,n\,\sigma\,x\bigr), \quad x \ge 0.$

3. Mean Free Path

From the properties of the exponential distribution, the expected value of $X$ is the reciprocal of the rate:

$\langle X \rangle \;=\; \int_0^\infty x \, \bigl(n\,\sigma \,e^{-\,n\,\sigma\,x}\bigr)\, \mathrm{d}x \;=\; \frac{1}{n\,\sigma}.$

We call this value the mean free path $\lambda$ :

$\boxed{\lambda \;=\; \frac{1}{n\,\sigma}.}$

4. Markov Chain Interpretation

In a discrete Markov chain, the system moves from one state to another with a transition probability depending only on the current state.
Here, in a continuous sense, the position of the particle changes incrementally, but the chance of collision in the next infinitesimal interval is constant—independent of how far it has already traveled.

Thus, the distance to collision is a waiting-time-like variable with the memoryless property, which is exactly what defines the exponential distribution. Hence, viewing collision events as a Markov process directly yields:

$\lambda \;=\; \frac{1}{n\,\sigma}.$

5. Summary

Memoryless assumption $\rightarrow$ exponential free-path distribution.
Rate parameter $n,\sigma$ $\rightarrow$ mean free path $\lambda = \frac{1}{n,\sigma}$ .
Markov chain viewpoint $\rightarrow$ collisions occur in a Poisson-like manner with no dependence on prior “collision-free” distance.

This completes the Markov chain derivation of the mean free path in a uniform medium.