# Statistical physics of a “ball on a hill”

There's something basic about the application of statistical physics which I do not seem to understand. Let's say we have some potential, U(x), which looks like a mountain with valleys: (França, H. M., Gomes, G. G., & Parra, R. L.. (2007). Tunneling and the Vacuum Zero-Point Radiation. Brazilian Journal of Physics, 37(1a), 13-16. https://dx.doi.org/10.1590/S0103-97332007000100006)

Let's say I put my "system" at $x_a$. This is a "metastable" state. In case of a temperature higher than 0, we expect that at some point the system ends up in the stable state at $x_c$. Now my problem arises when I think of this in a "statistical physics" kind of way.

The relative probability to find the particle at state a or state b is given by the Boltzmann probability:

$exp(\frac{-\Delta U}{kT})$

Here $\Delta U$ is the difference in energies of the states, which happens to be $-U_0$. Plugging this into the formula gives, for all T>0, a probability larger than one. Extending this result, if I were to use the Metropolis algorithm to simulate this potential / Hamiltonian, I would always find my particle at the lowest energy state ($x_c). My questions are: 1. What am I doing wrong in this analysis? How do I include the "hill energy"? 2. How to simulate this system using a statistical algorithm (Metropolis?) As a follow-up question: How do I include "friction" in this model? (How does statistical physics account for losing energy? Do I need a reservoir?) This might be better as a separate question, though. • Two things, first you are missing the normalization factor in your Boltzmann distribution (sum of all probabilities should be 1). The other thing is that Boltzmann distributions is statistically valid for large number of systems in equilibrium. When you start in a metastable state, you are far from equilibrium. To reach the Boltzmann distribution, you can start with many systems at random locations and wait a long long time. – Ali May 23 '17 at 18:11 • What you refer to "relative probability" is infact the ratio of probabilities, and it may be greater than$1$: This does not imply the probability is greater than or equal to$1$. Also, your system is not statistical in nature. You are talking about Boltzmann distribution but there is no entropy or entropy-related data in the question. – Ranc May 23 '17 at 18:13 • This is indeed a mistake - I meant relative probability of course. – JBrouwer May 23 '17 at 19:10 • @Ali make it an answer! +1 – innisfree May 23 '17 at 19:58 ## 1 Answer Let's proceed by parts. More than statistical mechanics, I think the best is to use the stochastic processes formalism. I will try to be very simple and not going into the subtle mathematical details. First of all, I think your problem is a particle that can move with some kind of randomness over a potential. If the movement is overdamped, then you can neglect the second order derivative, and then you have the following Langevin equation for the movement: $$\frac{dx}{dt} = -\frac{dV(x)}{dx}+\sqrt{2D}\xi(t)$$ where$\xi(t)$is a white noise. If you are not familiar with this, it is basically a random number every single time. The noise has a constant intensity$D$. Disclaimer: I know white noise is really more complicated than that, but I said I am not very interested in the details. The way to analyze this kinds of systems is usually to write down the Fokker-Planck equation. This is an equation which will say you how the probability to find the particle in position$x$changes with time. If your Langevin is$\dot{x}=f(x)+g(x)\xi(t)$, then Fokker-Planck is given by: $$\frac{\partial p(x,t)}{\partial t}=-\frac{\partial J}{\partial x}=\frac{\partial }{\partial x} \left [ -f(x)p(x,t) + \frac{g(x)}{2} \frac{\partial }{\partial x}[g(x)p(x,t)] \right]$$ This equation is usually very complicated to solve, but at least you are able to get rid of the noise. However, you are interested in which the probability to find the particle at$t\rightarrow +\infty$at certain point, so you want the stationary solutions:$\partial _t p(x,t)=0$. Setting the current of probability$J=0$(some cases can have a current different from zero, but I don't want to go into deep details) you get a ODE for the probability: $$-f(x)p(x) + \frac{g(x)}{2} \frac{d }{dx}[g(x)p(x)] = 0$$ In our case, when you put the explicit form for$f(x)$and$g(x)$, you find $$p(x) =\frac{1}{Z} \exp(-V(x)/D)$$ where$Z$is a normalization factor. This is a function that has maxima in the minima of your potential and viceversa, so the most probable place to find the particle is at the minima. You can check that it has a similar shape to your Boltzmann distribution since the noise intensity$D$is proportional to the temperature. Note that increasing or decreasing$D$will change the width of maxima and minima, as you expect -high temperature, less probability to be in the metastable state. To solve your doubt number (1), you obtain a probability which is greater than one because you forgot to divide by a normalization factor$Z\$. However, for this kind of systems you ask I think it is worth to point out the existence of Fokker-Planck equations and so on...

...because to simulate this behaviour, you can use, for example, a Milstein algorithm to solve the Langevin equation (the first one I have written). This is very easy to code (like an Euler but with a bit of random number generation), and you can see the noisy movement of your particle inside the potential.

• While I agree that this would be an approach to this problem, I'm looking specifically into the statistical mechanics of this problem, as there lies my problem of understanding. – JBrouwer May 23 '17 at 23:05