Equilibrium in isolated systems

We consider an isolated system made of $N$ identical particles in a volume $V$, that cannot exchange energy with the environment and so the value of the energy $E$ is fixed. The particles could also interact amongst themselves. The totality of the microstates associated with such a system forms the microcanonical ensemble, which has $\Omega(E,V,N)$ elements, and the system has probability $1/\Omega(E,V,N)$ to be found in any of these states. We have tried to make this equiprobability intuitive and plausible with the discussion of a simplified system in Sec. , but for a general system this is not necessarily obvious and therefore we postulate it here, as mentioned at the beginning of the chapter. It is clear that $\Omega$ is a function of state, i.e. it is completely defined by the physical state of the system and by its constraints. The dependence of $\Omega$ on $N$ and $E$ is clear from the discussions in the previous sections, the one on $V$ comes from the spatial degrees of freedom, which are also quantised. More precisely, we will see that each state occupies a finite volume in phase space (the combined space of positions and momenta), and so changing the volume can change the number of available states. The single particle energies also depend on the volume.

If we imagine the volume $V$ divided into small cells, labelled $v_1, \dots, v_s$, $\dots, v_r, \dots, v_M$, we see that the number of volume cells is proportional to the volume of the system, and so the number of possible states for each particle must also be proportional to the volume of the system. The total number of states is the product of the number of possible states for each particle, and therefore a factor $V^N$ will be present in the statistical weight.

If the particles of gas are all identical and have equal access to the whole volume $V$, it doesn't matter if particle $i$ occupies cell $v_s$ and particle $j$ cell $v_r$ or viceversa, we would not be able to distinguish the two configurations. Therefore, any permutation of the $N$ particles will result in a configuration that cannot be distinguished from the others. It follows that the statistical weight must include a factor ($1/N!$), and will have the general form:

$$\Omega(E,V,N) = k(E,V)^N\frac{V^N}{N!},$$ (3.4)

where $k(E,V)$ is some appropriate function of the energy and possibly of the volume. It is easy to persuade ourselves that $k(E,V)$ must be an intensive function, i.e. not proportional to the size of the system and must enter $\Omega$ as an $N^{th}$ power, as shown in Eq. . Let us imagine to partition our system into two halves, each having $N/2$ particles, volume $V/2$ and energy $E/2$. The partitioning of the volume and the number of particles does not cause any issue, but that of the energy into two equal halves requires the discussion of one point. The separation in two subsystems removes the cross interactions of the particles between the two systems and so in principle the total energy would be modified. However, if the range of the interactions is short compared to the size of the sub-systems, neglecting these surface interactions is usually a good approximation. Under these circumstances the partitioning does not affect the total number of available states and the two partitions are independent. The total number of states, therefore, must be the product of the statistical weights in the two subsystems and since they are identical they also have the same statistical weight, $\Omega(E/2,V/2,N/2)$, which means that we must have:

$$\Omega(E,V,N) = \left[\Omega\left(\frac{E}{2},\frac{V}{2},\frac{N}{2}\right)\right]^2.$$ (3.5)

Using form  for $\Omega$ we can express the r.h.s. of Eq.  as:

$$\left[\Omega\left(\frac{E}{2},\frac{V}{2},\frac{N}{2}\right)\righ... ...rac{N}{2} \frac{\left(\frac{V}{2}\right)^\frac{N}{2}}{\frac{N}{2}!} \right]^2 =$$

$$k\left(\frac{E}{2},\frac{V}{2}\right)^N V^N \left [\frac{2^{-\frac{N}{2}}}{\frac{N}{2}!}\right ]^2.$$ (3.6)

and using the Stirling approximation for the factorial: ^3.3

$$N! \simeq e^{-N} N^N,$$ (3.7)

we obtain

$$\left [\frac{2^{-\frac{N}{2}}}{\frac{N}{2}!}\right ]^2 \simeq \le... ...N}{2}\right)^\frac{N}{2}}\right ]^2 = \frac{1}{e^{-N} N^N} \simeq \frac{1}{N!},$$ (3.8)

which together with Eq.  gives $k(E/2,V/2) = k(E,V)$, proving that $k$ is an intensive quantity and can be written as $k(E/N,V/N)$. Using the Stirling formula  we rewrite Eq.  as:

$$\Omega(E,V,N) = \left [k\left(\frac{E}{N},\frac{V}{N}\right) \frac{eV}{N} \right ]^N,$$ (3.9)

which shows that $\Omega$ is the $N^{th}$ power of an intensity quantity. This suggests that the logarithm of this quantity may be an interesting physical property, because it is proportional to the size of the system.