Starting from the Langevin equation for a 1D Brownian particle, the mean square displacement (MSD) is calculated: $$\langle r^2 \rangle = \frac{2k_BT}{ζ}[t+τ_B(e^{-t/τ_B}-1)]$$ When $t\rightarrow \infty$, this expression shows that the MSD increases linearly with time. On the other hand, taking the limit for $t\rightarrow 0$ shows a quadratic dependence of MSD on time. This is a standard result from many textbooks and it is being referred to as the "ballistic" regime, where the particles are still under the effect of their initial speed/inertia.

This does not make much sense to me, since we have already made use of the equipartition theorem in order to extract this result, thus implying that the system is at equilibrium. Therefore, it is conceptually wrong to compute the limit for $t\rightarrow 0$ in the first place. Of course, since this is a standard result, it is probably me who is wrong. Can somebody explain in a more specific way what textbook writers mean when they refer to the "ballistic" regime of Brownian particles?


1 Answer 1


The ballistic regime can be derived without considering the equipartition theorem, though it does require that the autocorrelation of the noise term is finite, $$\langle\eta(t)\eta(t')\rangle=\alpha\delta(t-t').$$ In this case, the Langevin equation leads to a mean square speed of, $$\langle v(t)^2\rangle=\langle v(0)^2\rangle\mathrm{e}^{-2t/\tau}+\frac{\alpha}{m}\left(1+\mathrm{e}^{-2t/\tau}\right)\tag{1}$$ where $\tau=m\mu$ is the relation time. The mean square distance is then, $$\langle d(t)^2\rangle=\langle v(0)^2\rangle\tau^2\left(1-\mathrm{e}^{-t/\tau}\right)^2-\frac{\alpha}{m}\tau^2f(t,\,\tau)\tag{2}$$ where $f(t,\,\tau)$ is some function involving $\exp(-t/\tau)$. Then in the limit $t\to0$, you compare Equations (1) and (2) to find that, $$\lim_{t\to0}\langle d(t)^2\rangle=\langle v(0)^2\rangle\cdot t^2,\tag{3}$$ which, as you point out, indicates that the distance the particle travels is largely determined by its initial momenta/speed.

Since we derived the above only by asserting a finite autocorrelation, it should be clear that the equipartition theorem isn't needed for the analysis of the short (ballistic) timescale; that part would be required for the longer timescales when the system does thermalize--in which case Eq (3) is false (cf. this answer of mine).

Now as far as I can tell, there is no physical process in the system that could change the constant $\alpha$ in the autocorrelation. So if you can determine it by studying the short-term or long-term evolution of the system, it should hold for the entire evolution, hence $\alpha\sim kT$ should also be valid for the ballistic regime.

  • $\begingroup$ This answer seems reasonable. However, many textbooks (e.g. Nonequilibrium Statistical Mechanics by Robert Zwanzig) seem to present this argument after applying the equipartition theorem and without the initial speed appearing in any way. Would you say that their explanation is conceptually incorrect but they were lucky because the limit of the expression I wrote in my question also gives a quadratic dependence on time? $\endgroup$
    – Christos
    Nov 15, 2022 at 14:29
  • $\begingroup$ No, I would not agree with that. $\alpha$ is constant for all time $t$, so solving it using $t\to\infty$ via equipartition theorem is conceptually correct. Add to that the fact that one obtains $d^2\sim v^2\cdot t^2$ in the ballistic regime indicates the model is correct. $\endgroup$
    – Kyle Kanos
    Nov 15, 2022 at 22:51
  • $\begingroup$ I totally agree with your point when $t \rightarrow \infty$. However, most textbooks seem to use the same expression (one that does not contain the initial speed) for $t \rightarrow 0$. $\endgroup$
    – Christos
    Nov 16, 2022 at 9:22

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