Random Diffusion - Variance of the distribution

In most of the textbooks, I have read, for a random diffusion, it is given that the random displacements are chosen from a normal distribution with zero mean, and variance $$\sigma^2 = \frac{2k_BT\Delta t}{m\gamma} \, .$$ However, I have not seen the derivation for this variance, and the books haven't mentioned the reason behind choosing this variance or for choosing zero mean. Where does this equation for the variance come from, and why is the mean zero?

• What are the variables? Jun 19, 2021 at 18:49
• Variables are Boltzmann constant, Temperature, Time step, mass and friction coefficient. Jun 19, 2021 at 19:07
• Try Kubo's Nonequilibrium statistical mechanics vol 2, Gardiner's stochastic processes in physics, Gillespie's Markov processes in the physical sciences, or the youtube lectures on the Langevin equation by Balakrishnan Jun 20, 2021 at 0:45
• Thanks, I checked those resources, but still could not find the reason behind choosing that particular variance, and 0 as mean. If you could show me your showing, I would be very grateful. Thanks. Jun 20, 2021 at 12:43
• Perhaps, my memory is rusty... but when do $\Delta t$ and $\gamma$ appear in the same formula? Without damping the diffusion is as $\langle(\delta x)^2\rangle=2D\Delta t$, whereas with damping we replace $\Delta t$ by $\gamma$. Jun 22, 2021 at 13:36

The distribution of random diffusion: $$n(x,t) = \frac{1}{\sqrt{2\pi D t}} \exp\left\{-\frac{x^2}{2Dt}\right\}.$$

The deviation of this distribution $$\sigma^2 = 2 D t.\tag{1}$$

Then using the Einstien relation for Brownian motion for the diffusion constant: $$D = \mu K_B T, \tag{2}$$ where $$\mu$$ is the mobility, defined as $$v_d = \mu F$$ as the proportional parameter between drift velocity $$v_d$$ and force $$F$$. This renders the deviation: $$\sigma^2 = 2 \mu K_B T t. \tag{3}$$

Now, we need a model for mobillity $$\mu$$. This problem seems that the mobility was derived from the stochastic damping equation: $$m\frac{d^2x}{dt} + m\gamma \frac{dx}{dt} = F_{drive} + F_{random}.$$ where $$m\gamma$$ stands for the damping coefficient. The $$m$$ added to the damping coefficient is typically to make the equation simpler in expression after divided by $$m$$.

The average terminal velocity (under condition: $$\langle a \rangle =\left\langle\frac{d^2x}{dt^2}\right\rangle = 0$$). The bra-ket is for time average.): \begin{align} m\left\langle\frac{d^2x}{dt}\right\rangle &+ m\gamma \left\langle\frac{dx}{dt}\right\rangle = \langle F_{drive}\rangle + \langle F_{random} \rangle.\\ 0 &+ m\gamma v_d = \langle F_{drive}\rangle + 0\\ v_d &= \frac{\langle F_{drive} \rangle}{ m\gamma} = \mu \langle F_{drive}\rangle. \end{align}

The model renders the mobility $$\mu = \frac{1}{ m\gamma}.$$

Using this mobility, the deviation of Eq.(3) leads to the answer: $$\sigma^2 = \frac{2 K_B T t}{m\gamma}.$$

• If one calculates $n(x,t)$ from your eq. (4), its time dependence will be different from your eq. (0). Jun 23, 2021 at 5:44
• Yes. Eq (4) added a driven force in order to derived the mobility. Eq. (0) is distribution without driven force. It is similiar in deriving Einstein relation (Eq.(2) here), one have to impose an external potential to derive the relation.
– ytlu
Jun 23, 2021 at 6:12
• For example, I want to draw a harmonic oscillation, but didn't know its natural frequency. To get this information, I apply an external driven force and find the resonace.
– ytlu
Jun 23, 2021 at 6:21
• It is not very consistent, in my opinion... but this is how it is done in textbooks. +1 anyhow. Jun 23, 2021 at 7:11
• @RogerVadim I don't see it in that way. The equation for $<x^2(t)>$ with $F_drive=0$ reads $\frac{d^2<x^2>}{dt^2}=\gamma\frac{d<x^2>}{dt}-3k_BT$. The solution of this equation is $<x^2(t)>=Cte(1-e^{-\gamma t})+\frac{3k_BT}{\gamma}t$. So $<x^2(t)>$ (and therefore, $D(t)$) is proportional to t in the stationary limit.
– Javi
Jun 23, 2021 at 12:53