Maxwell-Boltzmann Distribution Function:

Let us consider a system of N distinguishable particles having total energy E which is divided into several levels. The \(i^{th}\) level has \(g_{i}\) states containing \(n_i\) particles each of the energy \(\epsilon_i\) subject to the conditions that both N and E are constants i.e.

\(N=\sum_1^{i=n}n_i = constant \tag{1}\)

and

\(E=\sum_1^{i=n}n_i\epsilon_i = constant \tag{2}\)

The number of the ways in which the N particles can be placed in different levels so that \(i^{th}\) level contains \(n_i\) particles is

\(\displaystyle{\Omega_1=\frac{N!}{n_1! \cdot n_2! \cdot n_3! \cdots n_n!}=\frac{N!}{\Pi_{i=1}^n n_i!}}\)

here the symbol \(\Pi\) stands for the product of the terms.

Now each \(n_i\) particles in the \(i^{th}\) level can be placed in \(g_i\) ways. Hence \(n_i\) particles can be placed in \(g_i\) states in the \(i^{th}\) level in \({(g_i)}^{n_i}\) ways, because a state can contain any number of particles and particles are distinguishable. Considering all the levels, the total number of arrangement is

\(\displaystyle{\Omega_2={(g_1)}^n_1\times{(g_2)}^n_2\times{(g_3)}^n_3\times\cdots\times{(g_n)}^n_n=\Pi_{i=1}^n{(g_i)}^{n_i}}\)

Thus the total number of the distribution obeying the conditions (1) and (2) is

\(\displaystyle{\Omega=\Omega_1\times\Omega_2=\frac{N!}{\Pi_{i=1}^n n_i!}\times\Pi_{i=1}^n{(g_i)}^{n_i}=N!\ \Pi_i\frac{{(g_i)}^{n_i}}{n_i!}}\tag{3}\)

The quantity \(\Omega\) is the thermodynamic probability of the system according to Maxwell-Boltzmann Statistics.

Most Probable Distribution: Maxwell-Boltzmann Distribution Function:

For most probable distribution \(\Omega\) is maximum i.e. \(\ln\Omega\) is also maximum. Thus \(d(\ln\Omega)=0\).

Now from equation (3) we get

\(ln\Omega=\ln{N!}+\displaystyle{\sum_i}(n_i\ln{g!}-\ln{n_i}!) \\ =N\ln{N}-N+\displaystyle{\sum_i}(n_i\ln{g!}-n_i\ln{n!}+n_i)\)

\(=N\ln{N}-N+\displaystyle{\sum_i}(n_i\ln{g!}-n_i\ln{n!})\) [applying Stirling’s formula \(\ln{n!}=n\ln{n}-n\) ]

[since, \(\displaystyle{\sum_i=N}\)]

For most probable distribution

\(\delta(\ln{\Omega})=0\)

or, \(\displaystyle{\sum_i(\delta{n_i}\ \ln{g_i}-\delta{n_i}\ \ln{n_i}-n_i\frac{1}{n_i}\delta{n_i})=0}\)

Since, \(g_i\) = constant

or, \(\displaystyle{\sum_{i}\delta{n_i}\ \ln\frac{g_i}{n_i}=0}\tag{4}\)

Since, \(\sum\delta{n_i}=\delta\displaystyle{\sum_{i}}n_i=\delta{N}=0\)

Subject to the condition that \(\displaystyle{N=\sum_i=constant}\) and \(\displaystyle{E=\sum_i{n_i}{epsilon_i}=constant}\) , we have two other consitions,

\(\displaystyle{\delta{N}=\sum_{i}\delta{n_i}=0}\tag{5}\) \(\displaystyle{\delta{E}=\sum_{i}\epsilon_{i}\delta{n_i}=0}\tag{6}\)

Using method of undetermined multipliers, multiplying (5) and (6) by \(\alpha\) and \(\beta\) and subtracting from (4) we get,

\(\displaystyle{\sum_{i}(ln\frac{g_i}{n_i})-\alpha-\beta\epsilon_{i})\delta{n_i}=0}\)

Since all the \delta{n_i} are arbitrary, we must have

\(ln\frac{g_i}{n_i}-\alpha-\beta\epsilon_{i}=0\)

or, \(ln\frac{g_i}{n_i}=\alpha+\beta\epsilon_{i}\tag{7}\)

or, \(\displaystyle{\frac{g_i}{n_i}=e^{\alpha+\beta\epsilon_{i}}}\)

or, \(\displaystyle{n_i=\frac{g_i}{e^{\alpha+\beta\epsilon_{i}}}}\)

This is Maxwell-Boltzmann distributions law.

Evaluation of \(\alpha\) and \(\beta\) :

We have the entropy of the system is given by,

\(S=k\ \ln{\Omega}\), where \(k\) is the Boltzmann constant.

\(\displaystyle{\frac{S}{k}=\ln{\Omega}}\)

\(or,\ \displaystyle{\delta\frac{S}{k}=\delta\ln{\Omega}}\\=\displaystyle{\sum_{i}\ln\frac{g_i}{n_i}\cdot\delta{n_i}},\ using\ (4)\\=\displaystyle{\sum_{i}(\alpha+\beta\epsilon_{i}\delta{n_i})},\ using\ (7)\)

\(or,\ \displaystyle{\frac{\delta{S}}{k}=\alpha\sum_{i}\delta{n_i}+\beta\sum_{i}\epsilon_i\delta{n_i}}\) \(or,\ \delta{S}=k\alpha\ \delta{N}+k\beta\ \delta{E}\tag{9}\)

Again we have Gibb’s free energy,

\(G=\mu{N}=E+pV-TS\)

where, \(\mu\) is the chemical potential of the system.

\(TS=E+pV-\mu{N}\)

\(T\delta{S}=\delta{E}-\mu{N}\)

\(\displaystyle{\delta{S}=\frac{\delta{E}}{T}-\mu\frac{\delta{S}}{T}}\tag{10}\)

Comparing (9) and (10) we get

\(\displaystyle{\alpha=-\frac{\mu}{kT}}\tag{11}\)

and, \(\displaystyle{\beta=\frac{1}{kT}}\tag{12}\)

Therefore, \(\displaystyle{n_i=\frac{g_i}{e^{\alpha+\frac{\epsilon_i}{kT}}}}\)

and the Maxwell-Boltzmann distribution function will be,

\(\displaystyle{f(\epsilon_{i})=\frac{n_i}{g_i}=\frac{1}{e^{\alpha+\frac{\epsilon_i}{kT}}}}\)