Why are metastable crossover times equal to the exponential of the free energy barrier?

Question

Many different models in stastical mechanics (for example, Curie-Weiss models) share the property that the (expected value of the) time it takes the system to move from one local minimum of the free energy to another equals the exponential of the free energy barrier. I am wondering whether there is a general heuristic explanation. Or is this just a feature that is remarkably shared by different models?

Answer 1

Let's start from an overdamped langevin equation for some variable $A$:

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

$A$ could represent the size of a nucleus in classical nucleation theory, a particle in an energy landscape $F$... Anything really. $F$ is akin to an energy for the variable $A$. If $A$ is some coarse grained variable, then $F$ is a variational free energy that can be obtained from:

$$e^{-F(A)/T} = \int\dots\int d^Np d^Nq \delta(A-a(p, q))e^{-E(q, p)/T}\tag 2$$

For example, if $A$ is supposed to be the size of a nucleus in a metastable fluid; $F(A)$ represents the effective energy landscape in which the nucleus evolve which has a maximum at $A^*$ ($A^*$ corresponds to the critical size of the nucleus above which the system fully changes its phase). Anyway, more generally the situation you are interested in corresponds to a particle that evolves in this kind of potential:

Where $A_0$ is a metastable state and the stable state (in the thermodynamic sense) is on the right of $A^*$, let's call it $A^\infty$. Using a saddle point approximation, far from any metastable state, the thermodynamic free energy of the system is well approximated by $F(\text{argmin}[ F(A)])=F(A^{\infty})$. Hence, in order to reach the thermodynamic stable state, the system has to overcome a potential barrier at $A^*$.

In any case, Eq. (1) is really a relaxation dynamics for the variable $A$ that makes you explore the configuration space and, eventually lets you reach $A^\infty$. Please note that: this is very general*, and hence is why Kramer's rule/Arhenius law is so ubiquitous.

The question you are asking for is:

What's the average waiting time corresponding to a transition $A^0\to A^*$?

This is really not easy to compute in fact (hence my reluctance to do more in the first post :) ). Kramer's original derivation, uses the Fokker-Plank representation of the Langevin equation (1). And, I believe, some kind of Taylor expansion of the potential/variational free energy. We will do it using the path integral representation of (1), which is perhaps not the simplest approach, but the more straightforward one. These kinds of approach are colloquially known now as Transition State Theory. They are more or less exactly equivalent to some WKB approximations that you would perform in quantum mechanics except that quantum noise is replaced by thermal noise and $\hbar\to 0$ is replaced by $T\to 0$. So, this already gives you an heuristic to why Arhenius law is exponential, it's equivalent to asking the waiting time for a particle tunnelling in a potential. The equivalence holds because of the formal similarity between a diffusion equation (Or equivalent a Fokker-Plank equation) and the Schrodinger equation.

Note also the limitations of the derivation I will give:

1. $T\to 0$ limit: Very important, everything is approximately valid as long as the effective energy/free energy the particle has to overcome is significantly larger than the thermal energy of the system ($\Delta F\simeq 10^1-10^2 T$ is a good range, below that, you have additional contributions to the purely exponential rate that start to play some role)
2. Overdamped system: Not too important as an assumption. In the long time limit, an underdamped system would behave as an overdamped one, and since we assume $T\to 0$, we also assume a long time to cross the barrier.
3. Markovian: We assume that the system haas no memory. If the memory is short, it's alright. If the memory if decaying as a power law or is roughly equal to the average crossing time you predict for an similar system without memory, you are in trouble.
4. Equilibrium: We assume that the system is in equilibrium. Hence, that, forces are 'potential' and that the noises respect the Fluctuation dissipation theorem. This last condition is not too important, the first one is.

Mathematical derivation of Arhenius law in (1).

We will want to find a distribution probability of the path, such that we can count all the paths that go from $A^0$ to $A^\infty$. Such probability distribution can be found using the so called Onsager-Machlup Formalism which is straightforward to define but leads to complicated computations. A more convoluted approach, which proves to be useful is the so called Martin-Siggia-Rose-Janssen-deDominicis (MSRJD) formalism.

We start from the definition of an average of some function of $A$: $f$:

$$\langle f\rangle \propto \int \mathcal D\eta e^{-\frac{1}{2}\int \eta^2(t) dt}\int \mathcal D A f(A)\delta(\dot A + F'(A) - \sqrt{2T}\eta)).$$

This might seem daunting but it's quite straightforward. The first path integral is just the probability distribution of a given realisation of the noise. Since the noise is gaussian at every time and uncorrelated, its probability distribution is just proportional to $e^{-\sum_t \eta(t)^2/2 }$, which in the continuum limit becomes an integral. The second path integral enforces Eq. (1).

We continue by getting rid of the dirac delta by using its integral representation, $\delta(a)\propto \int d\bar a e^{-i\bar a a}$:

$$\langle f\rangle \propto \int \mathcal D\eta e^{-\frac{1}{2}\int \eta^2(t) dt}\int \mathcal D A \mathcal D \bar A f(A)e^{ -i\int\bar A(\dot A + F'(A) - \sqrt{2T}\eta)dt}).$$

Now, we can perform the gaussian integral on $\eta$:

$$\langle f\rangle \propto \int \mathcal D A \mathcal D \bar A f(A) e^{ i\int\bar A(\dot A + F'(A))dt}\underbrace{\int \mathcal D\eta e^{-\int(\frac{1}{2} \eta^2(t) - i\bar A\sqrt{2T}\eta(t))dt})}_{\displaystyle{\propto e^{-\int \bar A^2 Tdt}}}.$$

You see now that some stochastic process $A(t)$ has a probability distribution (it's the probability distribution of the full process at everytime) proportional to (with the right change of variable to get rid of the imaginary constant:

$$P[A, \bar A]\propto e^{ -\int\left[\bar A(\dot A + F'(A))-\bar A^2T\right]dt}.$$

At this point you might wonder about this $\bar A$, its physical interpretation. Well there is nothing much to say about it, except that it is an auxiliary field, that plays the same role as the momentum in the Hamiltonian version of the quantum mechanical path integral. If you integral over $\bar A$, you obtain a path integral only in term of $A$ exactly as in quantum mechanics, you can get rid of the momentum to only have an expression in term of the field in the Lagrangian. $\bar A$ is also called the response field because functional differentiation of the functional integral with respect to it, leads to linear response functions.

Now, to the interesting part! How to obtain the the probability/waiting time/rate of jumping from $A^0$ to $A^*$. Well, you can just sum the probability of all the paths that satisfy this condition:

$$\text{rate of }A^0\to A^*\propto\sum_{A^0\to A^*} P[A, \bar A] \propto \int_{A^0\to A^*} \mathcal D A \mathcal D \bar A e^{ -\int\left[\bar A(\dot A + F'(A))-\bar A^2T\right]dt}$$

Which seems pretty complicated :). First, we surely will want to do this integral using a saddle point approximation. So we can first try to obtain this saddle point (which with a change of variable, can be shown to be exact in the limit $T\to 0$), which is given by the minimum of the exponential which verify:

$$\frac{\delta S[A, \bar A]}{\delta A}=0~~~\frac{\delta S[A, \bar A]}{\delta \bar A}=0$$

Where $S=-\int\left[\bar A(\dot A + F'(A))-\bar A^2T\right]dt$. This leads to:

\begin{split} \dot{\bar A}&= \bar A F''(A)\\ \dot{A}&= 2T\bar A - F'(A) \end{split}

As foreshadowed before, this has the form of an hamiltonian equation:

\begin{split} \dot{\bar A}&= -\frac{\partial H}{\partial A}\\ \dot{A}&= \frac{\partial H}{\partial \bar A} \end{split}

with $H(A, \bar A) = -\bar A F'(A) + T\bar A^2$, using the above equations, you can prove that: $H(A, \bar A) = \frac{\dot{A}^2-F'(A)^2}{4T}$. The usefulness of this method is that, we know that from the Hamiltonian structure of the saddle point method, $H$ is conserved along a saddle path. Moreover, we will ask for transitions that barely manage to escape $A^*$ such that the velocity $\dot A_{A^*}$ of the process is 0. This results again, from the fact that we ask only for the most likely trajectory. The one that manage to escape with a velocity larger than $0$ are exponentially less likely, hence they are not taken into account in the saddle point approximation. Moreover, by definition of a maximum $F'(A^*)= 0$. Hence, we find that at the end of the trajectory $H(A^*, \bar A)=0$. Since it is conserved all along the saddle point path, we can assume $H(A, \bar A)=0$. This shaky explanation is better explained (more formally in these lecture notes, specifically, the ones on large deviation theory.

Anyway, we go on, and use the saddle point approximation:

$$\text{rate of }A^0\to A^* \propto\int_{A^0\to A^*} \mathcal D A \mathcal D \bar A e^{ -\int\left[\bar A(\dot A + F'(A))-\bar A^2T\right]dt}\simeq e^{ -\int\left[\bar A_{saddle}\dot A_{saddle} - H_{saddle}\right]dt} $$

The hamiltonian of the saddle point approximation is 0. Moreover, since we have $H=\frac{\dot{A}^2-F'(A)^2}{4T}$, we see that the saddle point path respect: $\dot A = \pm F'(A)$. The minus equation describe a solution of $A^*\to A^0$ while the positive one is the one we want since it describes $A^0\to A^*$. Now, we use the Hamiltonian equation for $\dot A$, from which we obtain together with $\dot A=F'(A)$: $\bar A = F'(A)/T$. Which finally gets us the thing we wanted:

$$\text{rate of }A^0\to A^* \propto e^{ -\int \bar A_{saddle}\dot A_{saddle} dt} = e^{ -\dfrac{1}{T}\int \frac{\partial F}{\partial A}\frac{\partial A}{\partial t} dt}=e^{ -\dfrac{1}{T}\int \frac{\partial F}{\partial A}dA}=e^{-\dfrac{F(A^*)-F(A^0)}{T}}$$

Note that, if we included the additional terms of the Saddle point approximation, we would obtain terms proportional to $F''(A)$ (the curvature). This result, where curvature is included, is called the Eyring–Kramers law.

Old Lazy Post:

This is more or less "exact" for markovian overdamped systems with small noise and relatively well defined minimum and maximum concerning the energy landscape altough you may have subdleading terms proportional to the curvature of the energy (this is called the Eyring–Kramers law when you add additional contributions). The exponential form of the mean waiting time is usually called Kramer's rule when derived from a langevin equation and more generally Arhenius law in chemistry. This is also used in classical nucleation theory where I believe it is also called Arhenius law. Mathematical treatments of these kinds of problems in dimension higher than 1, are usually grouped under the term transition path theory or state transition theory (they are usually studied from a path integral approach).

With all this terminology you should be good. Tell me if you wish something a bit more quantitative.