1

I think that my question is more general, but here is a brief description of the system I am looking at:

I am studying the motion of particles on a lattice and the system can viewed as a simple canonical ensemble with two parameters $L$ (the number of lattice sites) and $N$ (the particle number). Let $F(L,N)$ be the free energy. Roughly speaking, it turns out that there is a physical quantity $M$ (the maximal occupation number) and a function $F(L,N,M)$ such that $$F(L,N)=\min_{M}F(L,N,M)$$ and $$M(L,N)=\underset{M}{\operatorname{argmin}} F(L,N,M)$$ is the value of the maximum in equilibrium. This suggests to call $F(L,N,M)$ the free energy and to view this as a case of free energy minimzation.

That being said, it was suggested to me that I should assume that the system started in the state $(L,N,M_0)$ most likely moves in the direction of the gradient of the free energy $M\mapsto F(L,N,M)$, meaning that on short time scales it typically ends up in a local minimum instead of the global minimum.

  • Is there any justification for this assumption?$^1$

  • Are there are any other systems where this assumption is known to hold true?

$^1$ Note that in my particular situation the equilibrium distribution of the maximum can be expressed in terms of the free energy and the principle of free energy minimization is equivalent to the intuitive maximization of the probabiltiy density. Hence this preceding question of mine.

Filippo
  • 2,000

2 Answers2

1

Non-equilibrium dynamics isn't quite encoded in free energy alone. The coupling with a heat bath allows relaxation, but just looking at $F(L,N,M)$ we have no slightest idea how the relaxation occurs. Not to mention that, on the very microscopic scale, the system gets excited from time to time due to the coupling; it is not a one way street, even though on average you tend to move one way.

I think it is just an empirical thing at the end. It make sense that a piece of real-life ferromagnet doesn't flip all magnetic domain over at once, right? The reason is the microscopic dynamics and beyond the simple stat mech treatment. But from there it seems reasonable to assume that the system can only explore its immediate neighborhood in the phase space. Then it is reasonable to write done an empirical relaxation law where the rate is proportional to $\nabla F$.

Note that the story is very different from your previous question. All because now there is implicit relaxation.

T.P. Ho
  • 777
1

Just to go on with T.P. Ho answer. The relaxation dynamics of a system coupled to an equilibrium bath is indeed complicated and not universal. Here are some thought.

The relaxation dynamics of a Brownian particle can be written as:

$$\frac{\partial X}{\partial t} = -\frac{\partial U(X)}{\partial X} + \sqrt{2T}\eta(t)\tag 1$$

where $\eta(t)$ is a white Gaussian noise with unit variance and zero mean. This is the typical equation for equilibrium Langevin dynamics, we set $\gamma=1$ (the damping/coupling constant associated to the bath) and $k_B = 1$ for simplicity. Using the associated Fokker-Planck equation you can prove that at long time, the stationary distribution probability of $X$ is: $P_{S}(X)\sim e^{-U(X)/T}$. Which means that this equation samples correctly stationnary distribution because you obtain the results of Gibbs theory of statistical mechanics. It turns out that in most cases the relaxation dynamics of an experimental system is well predicted by this equation too. But of course, it will depend on: How close the dynamics in Eq. (1) is from the realistic dynamics. Notably you could have memory, in which case you would have a memory kernel and a noise with an autocorrelation (this is called a generalized langevin equation). However, the stationary state would be the same. But the dynamics would not! You could for example add inertia..

The take home message is that: You know for sure that your equation of relaxation must predict a Gibbs weight as a stationary state. But there are an infinite number of equations that can do that.

Now, going to your equation. Equilibrium statistical mechanics tells us that:

$$e^{-F_{thermo}(L, N)/T}=\int dM e^{-F(L, N, M)/T}\simeq \alpha e^{-F(L, N, M_{min})/T}\tag 3$$

It is indeed reasonable to say that, similarly to the Brownian case, you perform an overdamped gradient descent on $F$:

$$\frac{\partial M}{\partial t}=- \frac{\partial F(L, N, M)}{\partial M} + \sqrt{2T}\eta(t)\tag 4$$

But there is no way from thermodynamics alone to know if this is the right equation or not! If you get rid of the noise, you find indeed that you reach a local minimum. With noise, you sample the configuration space with proba $P_{s}\sim e^{-F/T}$

Additional note

In the field of non-equilibrium relaxation phase transition interested in questions such as: what's the power law associated to relaxation of the critical Ising model? Of course the answer will vary according to the dynamics of the experimental system. Is the magnetization conserved (Glauber dynamics/model B) or is it not (model A)? Nice reviews (explaining the nomenclature model A, model B) can be found here (Theory of dynamic critical phenomena) or here. In their case they are not too much concerned by memory because of the divergence of time scale and the universality classes. But still, they have to include conservation law for example.

For example, let's say that we want to study the relaxation dynamics of a conserved quantity $\rho$. We could possibly write:

$$\frac{\partial \rho(x, t)}{\partial t}=\frac{\partial^2}{\partial x^2}\frac{\delta F}{\delta \rho(x, t)}+\sqrt{2T}\frac{\partial \eta(x, t)} {\partial x}\tag 5$$

which is different from the simple model that we could come up with if $\rho(x, t)$ was not conserved:

$$\frac{\partial \rho(x, t)}{\partial t}=-\frac{\delta F}{\delta \rho(x, t)}+\sqrt{2T}\eta(x, t)\tag 6$$

(5) and (6) have the same steady state probability distribution of $\rho$!

The take home message being here: The Ising model can relax in a bunch of different way! Even if the stationary distribution of each relaxational model is the same..

Syrocco
  • 2,660