# The Microcanonical Ensemble

[ "article:topic", "Author tag:Tuckerman", "showtoc:no" ]

The microcanonical ensemble is built upon the so called postulate of equal a priori probabilities:

Postulate of equal a priori probabilities: For an isolated macroscopic system in equilibrium, all microscopic states corresponding to the same set of macroscopic observables are equally probable.

### Basic definitions and thermodynamics

Consider a thought experiment in which N particles are placed in a container of volume V and allowed to evolve according to Hamilton's equations of motion. The total energy $$E = H (x)$$ is conserved. Moreover, the number of particles N and volume V are considered to be fixed. This constitutes a set of three thermodynamic variables N,V,E that characterize the ensemble and can be varied to alter the conditions of the experiment.

The evolution of this system in time generates a trajectory that samples the constant energy hypersurface $$H (x) = E$$. All points on this surface correspond to the same set of macroscopic observables. Thus, by the postulate of equal a priori probabilities, the corresponding ensemble, called the microcanonical ensemble, should have a distribution function $$F (H (x) )$$ that reflects the fact that all points on the constant energy hypersurface are equally probable. Such a distribution function need only reflect the fact that energy is conserved and can be written as

$F ( H (x) ) = \delta ( H (x) - E )$

where $$\delta (x)$$ is the Dirac delta function. The delta function has the property that

$\int \limits _{-\infty}^{\infty} \delta (x - a) f (x) dx = f (a)$

for any function f(x).

Averaging over the microcanonical distribution function is equivalent to computing the time average in our thought experiment. The microcanonical partition function $$\Omega (N, V, E )$$ is given by

$\Omega (N, V, E ) = C_N \int dx \delta ( H (x) - E )$

In Cartesian coordinates, this is equivalent to

$\Omega (N, V, E) = C_N \int d^m p \int \limits _{D(V)} d^N r \delta (H (p, r) - E )$

where $$C_N$$ is a constant of proportionality. It is given by

$C_N = \frac {E_0}{N! h^{3N}}$

Here h is a constant with units Energy Time, and $$E_0$$ is a constant having units of energy. The extra factor of $$E_0$$ is needed because the $$\delta$$ function has units of inverse energy. Such a constant has no effect at all on any properties). Thus, $$\Omega (N, V, E )$$ is dimensionless. The origin of $$C_N$$ is quantum mechanical in nature (h turns out to be Planck's constant) and must be put into the classical expression by hand. Later, we will explore the effects of this constant on thermodynamic properties of the ideal gas.

The microcanonical partition function function measures the number of microstates available to a system which evolves on the constant energy hypersurface. Boltzmann identified this quantity as the entropy, S of the system, which, for the microcanonical ensemble is a natural function of N, V and E:

$S = S (N, V, E )$

Thus, Boltzmann's relation between $$\Omega (N, V, E )$$, the number of microstates and S(N,V,E) is

$S (N, V, E ) = k \ln \Omega (N, V, E )$

where k is Boltzmann's constant 1/k= 315773.218 Kelvin/Hartree. The importance of Boltzmann's relation is that it establishes a connection between the thermodynamic properties of a system and its microscopic details.

Recall the standard thermodynamic definition of entropy:

$S = \int \frac {dQ}{T}$

where an amount of heat dQ is assumed to be absorbed reversibly, i.e., along a thermodynamic path, by the system. The first law of thermodynamics states that the energy, E of the system is given by the sum of the heat absorbed by the system and the work done on the system in a thermodynamic process:

$E = Q + W$

If the thermodynamic transformation of the system is carried reversibly, i.e., along a thermodynamic path, then the first law will be valid for the differential change in energy, dE due to absorption of a differential amount of heat, $$dQ_{rev}$$ and a differential amount of work, dW done on the system:

$dE = dQ + dW$

The work done on the system can be in the form of compression/expansion work at constant pressure, P, leading to a change, dV in the volume and/or the insertion/deletion of particles from the system at constant chemical potential,$$\mu$$, leading to a change dN in the particle number. Thus, in general

$dW = - PdV + \mu dN$

(The above relation for the work is true only for a one-component system. If there are M types of particles present, then the second term must be generalized according to $$\sum _{k=1}^M \mu _k d N_k$$). Then, using the fact that $$dQ = TdS$$, we have

$dE = T dS - P dV + \mu dN$

or

$dS = \frac {dE}{T} + \frac {P}{T} dV - \frac {\mu}{T} dN$

But since S=S(N,V,E) is a natural function of N, V, and E, the differential, dS is also given by

$dS = \left ( \frac {\partial S}{\partial E} \right ) _{N, V} dE + \left ( \frac {\partial S}{\partial V} \right ) _{N, E} dV + \left ( \frac {\partial S}{\partial N} \right ) _{E, V} dN$

Comparing these two expressions, we see that

$\left ( \frac {\partial S}{\partial E} \right ) _{N, V} = \frac {1}{T}$

$\left ( \frac {\partial S}{\partial V} \right ) _{N, E} = \frac {P}{T}$

$- \left ( \frac {\partial S}{\partial N} \right ) _{E, V} = \frac {\mu}{T}$

Finally, using Boltzmann's relation between the entropy S and the partition function $$\Omega$$, we obtain a prescription for obtaining the thermodynamic properties of the system starting from a microscopic, particle-based description of the system:

$\frac {1}{T} = k \left ( \frac {\partial \ln \Omega }{\partial E} \right ) _{N, V}$

$\frac {P}{T} = k \left ( \frac {\partial \ln \Omega }{\partial V} \right ) _{N, E}$

$\frac {\mu}{T} = -k \left ( \frac {\partial \ln \Omega }{\partial N} \right ) _{V, E}$

Of course, the ultimate test of Boltzmann's relation between entropy and the partition function is that the above relations correctly generate the known thermodynamic properties of a given system, e.g. the equation of state. We will soon see several examples in which this is, indeed, the case.