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Suppose we start with the standard Langevin equation for a free Brownian particle in an infinite medium:

\begin{equation} m \frac{dv}{dt} = -\gamma v + \sqrt{2 \gamma k_B T} \, \eta(t), \end{equation}

where $ \eta(t)$ is Gaussian white noise with zero mean and delta-correlated in time, and $T$ is the temperature of a heat bath.

We find that there exists a characteristic velocity relaxation time $ \tau_v = \frac{m}{\gamma}$, beyond which the particle equilibrates in velocity space. Specifically, the velocity distribution becomes Maxwellian, with

\begin{equation} \langle v^2 \rangle_{\text{eq}} = \frac{k_B T}{m} \end{equation}

So the system reaches equilibrium in velocity space.

Next, we evaluate the mean-square displacement (MSD) at long times and obtain:

\begin{equation} \langle x^2(t) \rangle = 2 D t, \quad \text{where } D = \frac{k_B T}{\gamma} \end{equation}

which implies that the position distribution is time-dependent and continues to spread indefinitely.

Now, this is the source of my confusion. I thought equilibrium meant that the probability distribution becomes time independent and local detailed balance is satisfied. The velocity distribution is Maxwellian (time-independent), whereas the distribution in position is time-dependent.

If the Brownian particle is said to be ``in equilibrium'' with the thermal bath, how can this be the case when its position distribution does not reach a stationary state and keeps spreading? There is a non-zero diffusion current in position space, so the position variable is not in equilibrium? Moreover, if it is not in equilibrium, how can one write $D =k_{B}T/\gamma$? I know that it has to do something with local and global equilibrium. Could someone please clarify??

If there were some harmonic confinement potential, then things would be obvious, what it means to be in equilibrium.

A better way to phrase this might be:

What does it mean for a Brownian particle to be in equilibrium? Is equilibrium only achieved in velocity space? Or is there a deeper sense in which the full phase space (position and velocity) is in equilibrium, even though the marginal distribution over position is time-dependent?

Mayank
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1 Answers1

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When we think about a distribution of many particles in a container, then a similar thing happens: their velocities evolved towards a common distribution, but the particles continue to bounce back and forth between the container walls. However, when averaging over a reasonable measurement time and/or length, the concentration of particles becomes homogeneous. (See discussion for a toy model of 1D billiard in this answer to Proof of pressure of ideal gas from first principles.) This is actually a restatement of the ergodicity assumption, where a single trajectory in a phase space is replaced by a cloud of points in this space, obeying the Liouville theorem.

Now, if we consider a gas without a container it will not be in equilibrium, but rather will spread over space. This is what corresponds to the model of Brownian motion described in the Q. In fact, by solving a Fokker-Planck equation we can obtain the position distribution, something like $$ w(x,t)=\frac{1}{\sqrt{4\pi Dt}}e^{-\frac{x^2}{4Dt}}. $$ If we were solving FPE with boundary conditions, such as reflecting walls, the distribution would evolve to and equilibrium one (an amusing way to do it is through the method of images.)

What accounts for the fact that velocity comes to equilibrium distribution rather quickly, while the position evolves to equilibrium but very slowly (or never reaching it) is that Brownian motion is different from that of gas molecules - it applies to large particles acted upon by many small forces (collisions with myriads of small particles), rather than via collisions with equal sized particles (which becomes rarer and rarer as the gas expands.)

Roger V.
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