1
$\begingroup$

I have an MD simulation that I'm pretty confident it works correctly. One of my tests was to check if the velocity distribution follows a Maxwell Boltzmann distribution.

I printed out the velocity of just one atom to file, and when I histogrammed it, I obtained the following plot. The solid curve is the Maxwell Boltzmann distribution with $kT=0.5$. enter image description here

As indicated, the simulated data match the theoretical curve. However, there is something I do not understand.

As I recall from my studies of statistical mechanics, the Maxwell Boltzmann distribution is a probability distribution for the speed of all particles in a system that has reached equilibrium.

In my case, I'm considering the velocities of only one particle since the beginning of simulation to its end. That is, I'm not looking at the speeds of all the particles, and I'm not waiting for the system to reach equilibrium. Yet, the data fits theory almost perfectly. What could be the explanation for my results? (The Wikipedia page defines the distribution as giving the probability of finding the particle near a speed $v$. Could it be that the distribution actually concerns one particle and I'm remembering it wrong?)

Improve this question
$\endgroup$
  • 1
    $\begingroup$ One would need more details about the simulation to give a precise answer, however (let alone very specific cases) you can assume most of the time that each atom, over long times, follows the same statistics of the ensemble. Think of it this way: why would this given atom be special? Given long enough time one can't be able to distinguish it from the other ones. As for the equilibrium question, either you have been at equilibrium long enough to "hide" the small error due to the transient time before reaching it, or the simulation, for some reason, starts at equilibrium. $\endgroup$ – JalfredP Jan 25 '18 at 20:29
  • $\begingroup$ I think your first point regarding the equilibrium case is reasonable. I do run the simulation for a very long time, so perhaps non-equilibrium effects are hidden. I doubt it starts from equilibrium as the simulation starts with all atoms with $v=0$ sitting on a lattice. I understand that there's nothing unique about one atom, but if the MB distribution is for a collection of atoms, why should it also hold for just one atom? $\endgroup$ – Ptheguy Jan 25 '18 at 20:34
  • 2
    $\begingroup$ This implies the ergodic hypothesis. That a configurational average is equivalent to a average over an infinite-time horizon. Since a collision randomly changes the velocity of a particle, it takes on a new state (configuration). If we assume molecular chaos, then each new state is IID. Averaging over particle states is then equivalent to the statistics of a single particle over an infinite time-horizon. $\endgroup$ – user18764 Jan 25 '18 at 21:21
1
$\begingroup$

As I recall from my studies of statistical mechanics, the Maxwell Boltzmann distribution is a probability distribution for the speed of all particles in a system [...]

The Maxwell-Boltzmann (MB) distribution $p(v)$ gives you the probability that a given particle has speed $v$. To be more precise, the integral $\int_{u}^{u'} p(v)dv$ gives you the probability that a given particle has speed between $u$ and $u'$. In other words, it tells you if you measure the velocity of a random particle in the whole system, how likely it is that you find it around the value $v$.

So, no contradiction here, and Wikipedia is right. You are not looking at some probability density $p(\mathbf v_1 \dots \mathbf v_N)$ for the whole system, but really at something that concerns single particles.

The results are well fitted by an equilibrium distribution probably because you are averaging over a time interval much larger than the time needed for the system to reach equilibrium. Take smaller and smaller time intervals (all of them starting at $t=0$), and you will start to see deviations from the MB. However, if you want to do this I suggest that you average over all the particles, because reducing the time window means that you are reducing your statistical sample.

It could actually be very instructive to plot $p(v,t)$ for some values of $t$ starting at $t=0$, because you will be able to see the system evolving from some non-equilibrium speed distribution to a MB, and you will also know exactly how long it takes to do so.

Improve this answer
$\endgroup$
  • $\begingroup$ Great, then there would be no contradiction as you pointed out. Regarding my simulation, I am starting from $t=0$, ending at $t=1000$ with time step $dt=0.02$. Effectively, I take my system of 50 atoms through many steps. I believe it is enough to reach equilibrium. Do you think I'm somehow biasing equilibrium results? $\endgroup$ – Ptheguy Jan 25 '18 at 20:53
  • 1
    $\begingroup$ @Ptheguy As I suggested, try to plot $p(v,t)$ (averaged over all the atoms) for many values of $t$ starting from $t=0$: when you see that the curves start to superimpose to the MB distribution, you will know that equilibrium has been reached, and the time $t_e$ at which this happens will give you an estimate of the time needed by your system to reach equilibrium with the initial state you chose. Note that size of the system (number of particles), initial configuration and thermodynamic parameters (temperature, density) can affect greatly the value of $t_e$. $\endgroup$ – valerio Jan 25 '18 at 21:15

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.