Microcanonical Ensemble

As seen in the onboarding example, consider one point particle with constant total energy \(E\) somewhere in a container of constant volume \(V\). When it reaches a container wall, the particle will elastically collide, also known as specular reflection. We know the kinetic energy of this particle, but we are ignorant to where it sits and in which direction it is moving.

There are many possible states–sets of position and momentum data–this particle can be in that agree with our observations. In classical dynamics, we say that these states–constrained by our observables–constitute a surface in the space of all possible states, called phase space. In statistical mechanics, we call these states microstates and the collection of observables the macroscopic state, or macrostate.

If we put a probabilistic spin on this problem by assuming that the particle is equally likely to be in any of these accessible (phase space surface) microstates, we can begin to answer probabilistic questions such as “how likely is it that the particle is in the left half of the container?” This is a silly question because the statement includes the answer, but more difficult questions are tackled using this same principle.

As a brief aside, this assumption is called the ‘‘postulate of a priori equal probabilities’’. It defines the probability distribution at a point \(\Gamma\) in phase space as \[p_{NVE}(\Gamma) = \frac{\delta(H(\Gamma) - E)}{\int \delta(H(\Gamma) - E)\ d\Gamma}\]

where the denominator is often called \(\Omega\). Whether or not this postulate is appropriate in our classical system is discussed in the classical mechanics pages and in the onboarding example. The NVE label on the distribution is a reminder that the phase space surface we are constrained to actually also depends on the number of particles \(N\) and the volume \(V\) being held constant. This specific distribution accounts for all realizations of the system consistent with the \((N,V,E)\) macrostate, and the collection of all of these realizations is called the microcanonical ensemble.

In Bayesian statistics, we would say this postulate corresponds to the construction of a flat prior distribution, and the specific phrase “a priori” typically means deducible without experience, often from system symmetries. For one possible justification of this prior: the time needed to conduct a macroscopic measurement on an observable, \(f(\cdot)\), of a system is far longer than typical microscopic timescales. If long-time averages on a specific orbit of the ensemble (parameterized by the initial condition \(\Gamma_0 = \gamma(0)\)), \(\gamma(t; \Gamma_0)\), is equal to an ensemble average, \[\lim_{T\rightarrow\infty} \frac{1}{T}\int_0^T f(\gamma(t; \Gamma_0)) dt = \int_{\Gamma_0 \in \Gamma} f(\Gamma_0) p_{NVE}(\Gamma_0)d\Gamma_0\]

we can take \(f\) to be an indicator, or step, function which turns on in a region of interest \(R\) and off everywhere else. Then the equality becomes \[\tau_R = \int_R p_{NVE}(\Gamma_0)d\Gamma_0\]

for \(\tau_R\) the proportion of time our propagating orbit spent in \(R\). The microcanonical ensemble is valid when the time a system spends in a region of phase space scales with the volume of that region. \[p_{NVE}(\Gamma = R) = \operatorname{const.} \implies \tau_R = \operatorname{const.}\int_{\Gamma = R} d\Gamma\]

Dynamical systems theorists define this as an ergodic system and put a great deal of effort into proving when single trajectories can sample all of accessible phase space uniformly. Ergodicity also provides another key insight: simulation of a single copy of the system over long times will equal the ensemble average.

Finding the escape rate in the onboarding question means determining \(k_{\operatorname{esc}}(t)\) either analytically or numerically in the expression \[\frac{dS(t)}{dt} = - k_{\operatorname{esc}}(t)S(t).\]

The initial value, \(S(0)\) for this initial value problem is 1, as there is no chance the particle has escaped at the very instant the hole has been punched out. If we are to count trajectories analytically, we must normalize the count by the total volume of phase space, \[\Omega = \int \delta(H(\Gamma) - E) d\Gamma = 8\pi ml^3E.\]

Then, our survival probability can be written as \[S(t) = 1 - \int \int_0^t \delta(\gamma(t'; \Gamma_0) - B) p_{NVE}(\Gamma_0) dt' d\Gamma_0\]

for \(B\) the opening in the container in phase space coordinates. As used earlier on this page, \(\gamma(t'; \Gamma_0)\) is the pair \((\vec r(t'),\vec p(t'))\) plus the initial condition where the trajectories are periodic and piecewise functions while the boundary \(B\) can be written as \[B(x,y,z) = \theta\left(R^2 - \left(y-\frac{l}{2}\right)^2 - \left(z-\frac{l}{2}\right)^2 \right)\delta(x-l)\]

for the circular punch-out centered on the cube wall at \(x = l\). This integral is formidable, and attempting to integrate our functions by discretizing each of the six phase space variables poses two immediate issues: (1) For an exceedingly poor grid spacing of 10 points along \(x,y,z\) and \(p_x,p_y,p_z\), we must evaluate this integral at \(10^6\) points, and (2) we know from our energy condition that the momentum vector is constrained to lie on the surface of a sphere, which would mean even the “clever” choice to discretize the momentum integrals in spherical coordinates still creates a nonuniform density biased towards the sphere’s poles. These numerical hiccups point to the use of Monte Carlo simulation.

Now, consider a gas of particles with no interatom forces but constant total energy. The particles have different momenta but the sum of all kinetic energies equals the prescribed total. In the single particle case, uniform sampling of the particle momentum requires choosing points uniformly on a sphere. These questions will explain how to efficiently do this sampling for our ideal gas.

(1) For this new system of N particles, what object’s surface do we need to uniformly sample? If we attempt to use the same method as before, what happens to the efficiency? Give an estimate for the probability of accepting a trial when there are just 4 particles in the container.

(2) We can investigate the momentum distribution for the ith particle by writing \(E' = E - p_i^2/2m\) which equals the sum of kinetic energies for all other particles. Note that by studying one particle out of many, you are analyzing a subsystem that no longer has constant energy, but is connected to a thermal reservoir (all the other particles). In other words, you have made a choice to move from the microcanonical ensemble to the canonical ensemble. If the momenta are random variables (no matter what kind), then what is the distribution for \(E'\)? This important result shows why for our system in the thermodynamic limit, the microcanonical and canonical ensembles are equivalent. You can read more about the canonical ensemble in your favorite stat mech book, but if you want to know the distribution of momenta for this single particle, one way that Maxwell originally thought about it is by noting that the distribution must be rotationally symmetric and isotropic, so \(f(p^2) = f(p_x^2)f(p_y^2)f(p_z^2)\). Solving this functional equation will give you the appropriate distribution up to normalization.

(3) As a quick example, return to the single particle. By drawing random numbers from the distribution determined in the previous question for each momentum component \(p_x,p_y,p_z\), create a figure to show that the final distribution of points satisfies the energy constraint. We call this a rejection-free sampling technique as we no longer need to discard points outside of the sphere.

Starting a system at the appropriate total energy is an important step in any molecular dynamics (MD) simulation, and rarely, the Maxwell-Boltzmann distribution is even an appropriate estimate for the system behavior. If the particles interact, you can choose the kinetic energy of every particle to be equal and you will have a very similar waiting time to come to equilibrium.