0
$\begingroup$

I am a math major who recently started to study thermodynamics seriously. I have some confusing points while studying it, so I'd appreciate it if you'd correct me and give me some answers.

(1) As far as I know, a small particle immersed in a fluid is in Brownian motion, which takes place in thermal equilibrium (if the temperature is maintained constant).

(2) The special Fokker-Plank equation describes this Brownian motion. (For example, one dimensional Fokker-Plank equation is $\frac{\partial W}{\partial t} = \gamma \frac{\partial (vW)}{\partial v} + \gamma \frac{kT}{m}\frac{\partial^2 W}{\partial v^2}$, where $W(v,t)$ is the distribution for the particle.)

However, the solution of the (general) Fokker-Planck equation $W(v,t)$ is not in equilibrium? If so, what sense of equilibrium is this?

(3) I've seen the phrase "convergence to equilibrium" in the literature about the Fokker-Planck equation. Here, even though $W(v,t)$ is not in equilibrium, it converges to equilibrium?

I am confused about the terminology 'equilibrium' here. Please help me!

$\endgroup$
0
$\begingroup$

The stationary solution of your Fokker-Planck equation is an equilibrium distribution. Here, assuming that $v$ is the variable for the velocity, that mean a Maxwell-Boltzmann distribution.

So if your initial condition for the distribution $W(v,t)$ is not an equilibrium distribution, your system will not be at equilibrium. However the evolution of your distribution via the Fokker-Planck equation will brings it towards an equilibrium distribution, hence the convergence towards the equilibrium.

Fokker-Planck equation that do not contain non-conservative forces are also called equilibrium Fokker-Planck equation, as theirs stationary solution are equilibrium distribution, and the evolution is a relaxation towards an equilibrium distribution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.