Mean Free Path — Question 8

Question 8

Random flights accumulate path length faster than net displacement

In a deliberately simplified three-dimensional random walk, each flight has fixed length $\ell=100$ nm, speed $v=500$ m/s and an independent isotropic direction. Use $\langle\mathbf r_i\cdot\mathbf r_j\rangle=0$ for distinct flights. This fixed-length model differs from exponential free paths.

1. Derive rms displacement after $N$ flights and find the number needed for an rms displacement of 1.00 mm.

2. Find the elapsed time and total path length. Compare with a straight 1.00 mm flight at the same speed.

Question 8 - Solution

Random flights accumulate path length faster than net displacement

(a) Method and calculation

$$\langle R^2\rangle=\left\langle\sum_{i,j}\mathbf r_i\cdot\mathbf r_j\right\rangle=N\ell^2\ \Rightarrow\ R_{\mathrm{rms}}=\ell\sqrt N.$$$$N=(10^{-3}/10^{-7})^2=10^8.$$

(b) Interpretation and further calculation

$$t=N\ell/v=10^8(10^{-7})/500=.0200\,\mathrm s,\quad L_{\mathrm{path}}=N\ell=10.0\,\mathrm m.$$$$t_{\mathrm{straight}}=10^{-3}/500=2.00\times10^{-6}\,\mathrm s,\quad t/t_{\mathrm{straight}}=10^4.$$An rms displacement target is not a first-passage-time condition.

Physical check

The cancellation of differently directed steps separates diffusion from ballistic travel.