Partition function and energy fluctuation of a canonical ensemble

The basic requirement or hypothesis of statistical mechanics is : the only dynamical feature on which the probability of a state may depend on is the energy of the state. The probability P_i of the ith N-particle quantum state is a function of E_i only:

$$P_i~=~P_i(E_i~\mbox{only})~~.$$ (11)

In the long run, the justification of the above equation rests in the continuing agreement between experimental results and statistical mechanical predictions.

Determination of the functional dependence of P_i on E_i proceeds by recognizing that the energy of the N-particle system in quantum state i can be interpreted meaningfully as the sum of small energy contributions. These may be, for example, the kinetic energies of atoms and molecules, potential energies, internal energies, and many others. This fact of summation could be written for independent particles as

$$E_i~=~\epsilon_1~+~\epsilon_2~+~\epsilon_3~+...=\sum_j ~\epsilon_j~~.$$ (12)

Combining equations (11) and (12) yields

$$P_i~=~P_i(~\epsilon_1~+~\epsilon_2~+~\epsilon_3~+...)~~.$$ (13)

We could now factorize the above probability as

$$P_i(~\epsilon_1~+~\epsilon_2~+~\epsilon_3~+...)~=~p_1(\epsilon_1)p_2(\epsilon_2)p_3(\epsilon_3)...~~.$$ (14)

The microcanonical distribution does not factor, because $E_{\mbox{system}}=\epsilon_1+\epsilon_2+\epsilon_3+...$. Thus if the second particle had a large amount of energy, there will be less energy available for the first particle. But there is no such requirement on the total energy when the system is in equilibrium with a heat bath, as is the case in canonical ensembles, rather than being isolated as in the microcanonical ensembles. Taking the logarithm of both sides of the above equation

$$\ln P_i(\epsilon_1+\epsilon_2+\epsilon_3+...)=\ln p_1(\epsilon_1)+\ln p_2(\epsilon_2)+\ln p_3(\epsilon_3)+...$$ (15)

The left hand side can therefore be expanded in a power series as f(x)=a₀+xa₁+x²a₂+x³a₃+..., which in this case x is the sum of $\epsilon$'s:

$$\ln P_i(\epsilon_1+\epsilon_2+\epsilon_3+...)=a_0+(\epsilon_1... ...ilon_3+...)a_1+(\epsilon_1+\epsilon_2+\epsilon_3+...)^2a_2+...$$ (16)

where the a are arbitrary constants. Now, however, the physical content of equation (16) is the same as that for (13). $\ln P_i$ must equal a sum of functions of each separate $\epsilon$, with no cross terms that couple two or more different $\epsilon$. That can be the case only if a₂ and all higher a are zero. So we get

$$\ln P_i(\epsilon_1+\epsilon_2+\epsilon_3+...)=a_0+(\epsilon_1+\epsilon_2+\epsilon_3+...)a_1$$

$$P_i~=~e^{a_0+(\epsilon_1+\epsilon_2+\epsilon_3+...)a_1}~=~e^{a_0}~e^{a_1E_i}$$

and

$$P_i~=~a~e^{-\beta E_i}~~.$$ (17)

In equation (17) we have simply defined two new constants, a=e^a₀ and $\beta =-a_1$: We find the parameter a by normalizing equation (17)

$$\sum_iP_i=1=a \sum_i e^{-\beta E_i}$$

$$a=\frac{1}{\sum_ie^{-\beta E_i}}=\frac{1}{Z}$$

$$P_i~=~\frac{e^{-\beta E_i}}{\sum_ie^{-\beta E_i}}~=~\frac{e^{-\beta E_i}}{Z}~~.$$ (18)

In equation (18) the sum over all states of $e^{-\beta E_{\mbox{\scriptsize {state}}}}$ is given by the symbol Z which is the partition function. The symbol was taken as Z because Planck has called the function Zustandssumme.

Knowing P_i we can calculate the ensemble average of the energy $\bar E$ of the N-particle system using the equation $\bar E=\sum_iP_iE_i$

$$\bar E~=~\frac{\sum_ie^{-\beta E_i}E_i}{\sum_ie^{-\beta E_i}}$$ (19)

or

$$\bar E~=~-~\partial_\beta ^{V,N} \ln Z~~.$$ (20)

The proof that the latter equation is equivalent to the former is as follows

$$-~\partial_\beta ^{V,N} \ln Z~=-\frac{1}{Z}{\partial_\beta Z}... ... E_i}}=-\frac{1}{\sum_ie^{-\beta E_i}}\sum_i-E_ie^{-\beta E_i}$$

which is the same as equation (19) (for notation see [9]).

In thermodynamics, whether a system is isolated or in thermal equilibrium with a heat bath is immaterial to the final equilibrium relations. If statistical mechanics is to agree with thermodynamics (macroscopic observations), one must be able to predict that the energy of the system will be $\bar E$ and that significant fluctuations about $\bar E$ are unlikely. To show this, let us find the mean square deviation of the energy.

From equation (8) we define

$$\overline{\delta E^2}~:=~\overline{E^2}~-~\bar E^2~~.$$ (21)

Differentiating either equation (19) or (20) with respect to $\beta$ yields

$$\partial_\beta \bar{E}~=~\frac{-\sum_ie^{-\beta E_i}E_i^2}{\s... ...{-\beta E_i}E_i \big{)}}{\big{(}\sum_ie^{-\beta E_i}\big{)}^2}$$

where we have used the formula

$$\partial_\beta(\frac{a}{b})=\frac{1}{b}\partial_\beta a-\frac{a}{b^2}\partial_\beta b ~~.$$

Comparing this equation with equation (21), we see that the first term on the right-hand side is recognized from equation (5) as $-\overline{E^2}$. The second term is $\bar E^2$, and we have proved that

$$\overline{\delta E^2}~=~-\partial_\beta ^{V,N} \bar{E}~~.$$ (22)

In the next section we prove that $\beta=\frac{1}{T}$ see [9], one always uses the same units on both sides of a relation between real observables, and the equation of state of an ideal gas could be written as PV=T, where T is measured in energy units). We use the above value of $\beta$ here to find the magnitude of $\overline{\delta E^2}$:

$$\overline{\delta E^2}~=~-\partial_\beta ^{V,N} \bar{E}=-\partial_T ^{V,N} \bar{E}{\partial_\beta T}=T^2C_v$$ (23)

where we have used equation (42) for the last step [4].

One of the conclusions from the above equation is that when the energy fluctuations become large, C_v becomes large also. This sort of thing happens near critical points when we have phase transitions [5].

If we take as an example an ideal gas consisting of N monatomic molecules, and so we have from the kinetic theory [4] 

$$\bar E=\frac{3}{2}NT\quad\mbox{and}\quad C_v=\frac{3}{2}N ~~$$

and we get

$$\frac{\overline{(E-\bar E)^2}}{\bar E^2}=\frac{T^2C_v}{(3/2 NT)^2}=\frac{2}{3N} ~~.$$

This means that the fluctuations in energy become very small relative to the magnitude of the energy itself. In the themodynamic limit ( $N\rightarrow 0$), the canonical ensemble becomes equivalent to the microcanonical ensemble [7]. If we follow the same above steps to derive a formula for the grand canonical ensemble, we will find at the end that the fluctuation in the number of particles in some chosen volume is :

$$\overline{\delta N^2}=\overline{N^2}-\bar N^2=N ~~.$$

If we take the density $\rho=N/V$, the same conclusion holds for density fluctuations: when density is measured by looking in a volume containing an average of N particles, the ratio of $\sqrt{\overline{ \delta\rho^{2}}} / \bar\rho$ is N^-1/2. Thus relative fluctuations in density can be made as small as desired by looking in larger and larger volumes.

If we take air as our grand canonical ensemble we will reach a conclusion that due to the density fluctuations we have a blue sky [8]. If there were no fluctuations in density, sunlight would not be appreciably scattered by the atmosphere; it would go in straight lines -as it does, for example, in the vacuum over the surface of the moon- and the sky would be nearly black. A significant density fluctuation must occur over a region whose dimensions are of the order of the wavelength of the light before there is appreciable scattering. Since blue light has shorter wavelength than red, and since significant fluctuations are more likely in small regions than large, blue light is preferentially scattered over red.