Energy Fluctuations in the Canonical Assembly

We must now perform the averaging in Eq. (1.4). We note that there are two sorts of terms first, those for which i = j secondly, those for which i j. We shall now show that the terms of the second sort average to zero. The reason is the statistical independence of two molecules i and j in the Boltzmann distribution. To find the average of such a term, we multiply by the fraction of all systems of the assembly in which the ith and jth molecules have the particular energies and ej, where k% and k3 are indices referring to particular cells in phase space, and sum over all states of the assembly. From Eq. (1.2) of Chap. IV, giving the fraction of all systems of the assembly in which each particular molecule, as the ith, is in a particular state, as the th, we see that this average is 

Having eliminated the terms of Eq. (1.4) for which i j, we have left only 

That is, the mean square deviation of the energy from its mean equals the sum of the mean square deviation of the energies of the separate molecules from their means. Each molecule on the average is like every other, so that the terms in the summation (1.6) are all equal, and we may 

We can understand Eq. (1.7) better by putting it in a slightly different form. We divide the equation by U2, so that it represents the fractional deviation of the energy from the mean, squared, and averaged. In computing this, we use Eq. (1.2) but note that the mean energy of each molecule is equal, so that Eq. (1.2) becomes 

To evaluate the fluctuations of Eq. (1.9) exactly, we must find the fluctuations of energy of a single molecule. We have 

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