Mean square speed from mean square displacement?

Consider a distribution of particles in space with a mean square displacement defined in each spatial direction $$q$$ as $$\langle d_q^2 \rangle = \langle |q(t) - q_0|^2 \rangle$$, where $$q_0$$ is the initial position along the given axis. If the displacement occurs over a time $$\Delta t^2$$, does it make sense to say that the mean square speed in each direction is $$\langle v_{q}^2 \rangle= \langle d_q^2 \rangle/\Delta t^2$$? More precisely, is such a definition of the mean square speed equivalent to the usual one derived via the Maxwell-Boltzmann distribution?

• What is the definition of $d_x$? Commented Aug 23, 2022 at 15:13
• You actually didn't define what you mean by "mean square displacement", just provided the symbols. Could you define what you mean by $\langle d^2_n\rangle$?
– Javi
Commented Aug 23, 2022 at 15:15
• @GiorgioP I’ve now edited the question with a definition for the mean square displacement. Commented Aug 23, 2022 at 15:48
• Ok, but I do not see why you indicate the time as $\Delta t^2$. According to your formula for the mean squared displacement, it should be $t$. Shouldn't? Commented Aug 23, 2022 at 15:59
• @GiorgioP Why would it be $t$? $\langle d^2\rangle$ has units of length square... Commented Aug 23, 2022 at 22:02

Consider a particle moving according to a Langevin equation, $$m\frac{\mathrm d\mathbf{v}}{\mathrm dt} =\frac{\mathbf{v}}{\mu}+\boldsymbol{\eta}(t)$$ where $$\boldsymbol{\eta(t)}$$ represents the stochastic white noise. The general solution for this is, $$\mathbf{v}(t)=\mathbf{v}(0)\mathrm{e}^{-t/\tau}+\frac{1}{m}\int_0^t\boldsymbol{\eta}(t')\mathrm{e}^{-(t-t')/\tau}\,\mathrm{d}t' \tag{1}$$ where $$\tau=m\mu$$ is the relaxation time. The mean square of Eq (1) is found to be, $$\boxed{\langle v^2(t)\rangle=\langle v^2(0)\rangle\mathrm{e}^{-2t/\tau}+\frac{kT}{m}\left(1-\mathrm{e}^{-2t/\tau}\right)}\tag{2}$$ where the $$kT$$ comes from the equipartition theorem and the autocorrelation of the noise term, $$\langle\eta(t)\eta(t')\rangle=2kT\delta(t-t')$$.

We can integrate Eq (1) again to find the displacement from the original position, $$\mathbf{d}(t)=\mathbf{v}(0)\tau\left(1-\mathrm{e}^{-t/\tau}\right)+\mu\int_0^t\boldsymbol{\eta}(t')\left(1-\mathrm{e}^{-(t-t')/\tau}\right)\,\mathrm{d}t'$$ From which the average distance is equal to the first term on the right. If you want the mean square displacement, then you need to square this and after some work, you'd find, $$\boxed{\langle d^2(t) \rangle = v^2(0) \tau^2 \left(1 - e^{-t/\tau}\right)^2 - \frac{3kT}{m}\tau^2f\left(t,\,\tau\right)+\frac{6kT}{m}\tau t}\tag{3}$$ where $$f(t,\tau)$$ is some function of $$\exp(-t/\tau)$$ that isn't exactly important right now. The $$kT$$ terms also arise due to the same reasons in Eq (2).

Direct comparison of Eq (2) and Eq (3) clearly show that it is not generally true that $$\langle d^2\rangle=\langle v^2\rangle\,t^2$$. However, we can look at two limiting cases to see if it could be the case.

Short time scales

For $$t\ll\tau$$, Eq (2) reduces to, $$\langle v^2(t\ll\tau)\rangle\approx\langle v^2(0)\rangle$$ while Eq (3) reduces to, $$\langle d^2(t\ll\tau)\rangle\simeq \langle v^2(0)\rangle t^2,\tag{4}$$ This is close to your wanted equation (if we integrated from $$t_1$$ to $$t_2$$ and defined $$\Delta t=t_2-t_1$$, then I think it'd be an exact relation), but as Eq (4) is valid only for short timescales, then the wanted relation doesn't hold for particularly long.

Long time scales

The long time solution to the Langevin equation should replicate the Boltzmann distribution. Hence, in this limit Eq (2) reduces to, $$\lim_{t\to\infty}\langle v^2(t)\rangle=\frac{kT}{m},$$ as it should. Eq (3) in this limit, the last term dominates, $$\lim_{t\to\infty}\langle d^2(t)\rangle=\frac{6kT}{m}t\tau =6\langle v^2(t)\rangle\,t\tau$$ which also isn't quite the relationship you were expecting, due to the relation being linear with time $$t$$ (with $$\tau$$ providing the other unit of time to make the units match).

Your derivation ends with an incorrect assertion:1 that $$\langle d^2\rangle=\langle v^2\rangle/\Delta t^2$$ is valid for (arbitrary) finite time changes due to the "product property of kernels". While this is correct for kernels, it is not correct for the definition of the velocity, $$v\overset{\text{def}}{=}\lim_{\Delta t\to0}\frac{x(t+\Delta t)-x(t)}{\Delta t}\equiv\frac{\mathrm{d}x}{\mathrm{d}t}.$$ This relation is valid only for infinitesimal increments, it is incorrect for the finite durations (which is why calculus prefers using $$\mathrm{d}$$ to denote the infinitesimal and $$\Delta$$ for the finite, to avoid such confusions). Since your derivation required the use of infinitesimal durations to replace the velocity with $$\Delta x/\Delta t$$ in the MB distribution, then your are still bound by this condition.
1. Master Drifter has since deleted their answer, but the solution they proposed was based on matching the heat kernel to the Maxwell-Boltzmann distribution by using a Wick rotation and assuming that the change in time was sufficiently small that $$v=\mathrm{d}x/\mathrm{d}t\equiv\Delta x/\Delta t$$ to use a change in variables in the Boltzmann distribution. This allowed them to show that $$\langle d^2\rangle\sim\langle v^2\rangle\Delta t$$, but extended this small-time scale comparison to arbitrary $$\Delta t$$ due to the product property of the heat kernel, forgetting that the match to the MB distribution required small $$\Delta t$$.