Microcanonical ensemble probability density distribution

Question

In microcanonical ensemble the probability density function is postulated as $\rho(q,p)=const.\times\delta(E-E_0)$ so the probability of an ensemble being in an element of phase space $\mathrm{d} q \mathrm{d} p$ is $\mathrm{d} P = \rho(p,q) \mathrm{d} p \mathrm{d} q$. But since $\rho(p,q)$ is constant for a given energy $E_0$ of the ensemble, and we know that for example all gas particles being in one half of a container is highly unlikely but still allowed, does this mean that phase space $\mathrm{d} p \mathrm{d} q$ belonging to a state describing the aforementioned example is much smaller than the phase space belonging to the equilibrium state? Is my conclusion correct? And if it is, can I conclude it in a more rigorous way than pointing out the example?

Thanks for answering

Answer 1

There are a few concepts that should be better focused, to formulate this question precisely.

An ensemble of Classical Statistical Mechanics is the set of all possible configurations in phase space, each configuration being characterized by the set of its Hamiltonian coordinates $q=(q_1,q_2,\dots,q_N)$ and $p=(p_1,p_2,\dots,p_N)$. Therefore, there is nothing like the probability of an ensemble being in an element of phase space. Instead, we can safely speak about the probability of a system of the ensemble being in a volume of the phase space. When such a volume is so small that the variations of the probability density over the volume are negligible, we can say that the probability of that microscopic state is ${\mathrm dP}=\rho(q,p){\mathrm dq}{\mathrm dp}$.

If $\rho(q,p)$ is a constant over the hypersurface $H(q,p)=E$ (where $H$ is the Hamiltonian, and $E$ a possible value of the energy), all the subsets of the phase space on such a hypersurface with the same volume ${\mathrm dq}{\mathrm dp}$ have the same probability.

This fact implies that each microstate is as probable as any other. However, a set of microstates may be overwhelming more probable than others. In particular, the collection of microstates such that all the particles occupy only half of the volume has a negligible probability compared to the set where there is almost the same number of particles in the two half-volumes.

Answer 2

The volume of phase space when particles are in one half of the box is smaller –much, much smaller– than the phase space at equilibrium, i.e., when particles occupy the entire volume of the box. This is the reason why the spontaneous concentration of particles in one half of the box is so unlikely.

Here is a simple demonstration with a lattice system: Take a lattice with $L$ sites and $N$ particles. The number of microstates is

$$\Omega(L,N) = \frac{L!}{N!(L-N)!}$$

If we restrict the particles to be in one half of the lattice the number of microstates is

$$\Omega(L/2,N) = \frac{(L/2)!}{N!(L/2-N)!}$$

The probability to observe the spontaneous segregation of particles in one half of the lattice is

$$ p = \frac{\Omega(L,N)}{\Omega(L/2,N)}$$

With $L=100$, $N=10$ this probability is 0.00059342. With $L=1000$, $N=100$ it becomes $3.19762\times10^{-33}$. As we see, this probability drops very quickly as the system grows in size.

Answer 3

We could start with the microcanonical probability density and calculate the probability of all the gas particles being in a particular sub-volume of the reservoir. Let us use the ideal gas for simplicity: $$ H(\mathbf{p},\mathbf{q})=\sum_{i=1}^{3N}\frac{\mathbf{p}^2}{2m},\\ \rho(\mathbf{p},\mathbf{q})=C\delta\left(H(\mathbf{p},\mathbf{q})-E_0\right) $$ The normalization constant is obtained by integrating over all the momenta and the volume of the reservoir: $$ \int d^{3N}\mathbf{p}\int_Vd^{3N}\mathbf{q}\rho(\mathbf{p},\mathbf{q})= CV^N\int d^{3N}\mathbf{p}\delta\left(H(\mathbf{p},\mathbf{q})-E_0\right)=1\\\Rightarrow C=\frac{1}{V^N\int d^{3N}\mathbf{p}\delta\left(H(\mathbf{p},\mathbf{q})-E_0\right)} $$ Let us now calculate the probability of finding the system in a microstate, where all the particles have assembled in a sub-volume $v$: $$ \int d^{3N}\mathbf{p}\int_vd^{3N}\mathbf{q}\rho(\mathbf{p},\mathbf{q})= Cv^N\int d^{3N}\mathbf{p}\delta\left(H(\mathbf{p},\mathbf{q})-E_0\right)=\left(\frac{v}{V}\right)^N $$ Thus, if the sub-volume is a half-volume of the reservoir, i.e. $v=V/2$, this probability is $$\frac{1}{2^N},$$ where $N$ is of the order of the Avogadro number, $N_A\sim 10^{24}$.

One can indeed see this as a restriction of the phase space, since the position variables are integrated only over a part of the full phase space (volume $v$ instead of $V$.)