Maxwell Boltzman Statistics

Prove that the expression for occupation index (Number of particles per phase space cell) of Maxwell-Boltzmann statistical distribution is:

\[ W = N! \cdot \dfrac{g_{1}^{n_{1}} g_{2}^{n_{2}} g_{3}^{n_{3}} \ldots g_{l}^{n_{l}}}{n_{1}! n_{2}! n_{3}! \ldots n_{l}!} = N! \cdot \prod_{i=1}^{l} \dfrac{g_{i}^{n_{i}}}{n_{i}!} \]

Proof: Consider a system of $N$ identical, indistinguishable, and weakly interacting particles contained in a vessel of volume $V$ at an equilibrium temperature $T$ and total energy $U$. Potential energy is zero; the energy is purely kinetic. This is a monoatomic ideal gas. Since the mean separation between molecules is larger than the thermal wavelength, they are distinguishable and any number of particles can occupy any energy level.

Let there be $l$ states with energies $E_1, E_2, E_3, \ldots, E_l$ such that:

$n_1$ particles with energy $E_1$
$n_2$ particles with energy $E_2$
$n_3$ particles with energy $E_3$
$n_l$ particles with energy $E_l$

with degeneracies:

$g_1$ levels in energy $E_1$
$g_2$ levels in energy $E_2$
$g_3$ levels in energy $E_3$

Subject to the constraints:

The number of ways to distribute the particles is the product of the number of ways of selecting and arranging them.

The selection ways: This way the total selection will be :

Each group is arranged in their respective $g_i$ levels:

Total ways:

Summary of Distribution Functions

\[ n_{1}+n_{2}+n_{3}+\ldots n_{l}=\sum_{1}^{l} n_{i}=N \] \[ \& \quad n_{1} E_{1}+n_{2} E_{2}+n_{3} E_{3}+\ldots n_{l} E_{l}=\sum_{1}^{l} n_{i} E_{i}=U \]

which means the total number of particles is constant and total energy of the system is constant. The particles can be distributed in a number of ways subject to the constraints and liberty that any number of particles can occupy any number of energy states.

The number of ways is thus the product of number of ways of particle's selection (considering distinguishability) and the number of arrangements among various energy states.
$n_{1}$ particles can be selected from $N$ particles in $^{N}C_{n_{1}}$ ways.
The next $n_{2}$ particles will be selected from remaining $N-n_{1}$ particles in $^{N-n_{1}}C_{n_{2}}$ ways.
This way the total selection will be:

\[ \begin{align*} W_{\text{selection}} &= W_1 W_2 W_3 \ldots W_l \\ &= {}^N C_{n_1} \times {}^{N-n_1} C_{n_2} \times {}^{N-n_1-n_2} C_{n_3} \ldots {}^{N-n_1-n_2 \ldots - n_{l-1}} C_{n_l} \end{align*} \]

The arrangement can be done as follows. The first particle from $n_{1}$ particles can occupy $g_{1}$ levels in $g_{1}$ ways. Similarly, the second particle from $n_{1}$ particles in $g_{1}$ ways, the third particle in $g_{1}$ ways, and so on, as there is no restriction. Hence $n_{1}$ particles can be arranged in $g_{1}$ energy levels in $g_{1}^{n_{1}}$ ways. Similarly, $n_{2}$ particles in $g_{2}$ energy levels in $g_{2}^{n_{2}}$ ways and so on.

Thus, the number of arrangements for $N$ particles in $g_{1}, g_{2}, \ldots, g_{l}$ energy levels is: \[ W_{\text{arrangements}} = {g}_{1}^{n_{1}} {g}_{2}^{n_{2}} {g}_{3}^{n_{3}} \cdots {g}_{l}^{n_{l}} \]

Thus, the number of ways of distributing $N$ particles in $l$ states is: \[ W = W_{\text{selection}} \times W_{\text{arrangements}} \]

\[ \begin{align*} &= {}^N C_{n_1} \times {}^{N-n_1} C_{n_2} \times {}^{N-n_1-n_2} C_{n_3} \ldots {}^{N-n_1-n_2 \ldots - n_l} C_{n_l} {g}_{1}^{n_1} {g}_{2}^{n_2} {g}_{3}^{n_3} \ldots {g}_{l}^{n_l} \\ =\ & \dfrac{N!}{n_1!(N-n_1)!} \cdot \dfrac{(N-n_1)!}{n_2!(N-n_1-n_2)!} \cdot \dfrac{(N-n_1-n_2)!}{n_3!(N-n_1-n_2-n_3)!} \cdots \\ & \cdot \dfrac{(N-n_1-n_2 \ldots -n_{l-1})!}{n_l!(N-n_1-n_2-\ldots-n_l)!} \cdot g_1^{n_1} g_2^{n_2} \ldots g_l^{n_l} \\ =\ & \dfrac{N!}{n_1!n_2!n_3!\ldots n_l!} g_1^{n_1} g_2^{n_2} \ldots g_l^{n_l} \\ =\ & N!\dfrac{g_1^{n_1} g_2^{n_2} \ldots g_l^{n_l}}{n_1!n_2!n_3!\ldots n_l!} \\ =\ & N! \prod_{1}^{l} \dfrac{g_i^{n_i}}{n_i!} \end{align*} \]

Thus the thermodynamic probability of MB distribution is:

\begin{align} W= N! \prod_{1}^{l} \dfrac{g_i^{n_i}}{n_i!} \end{align}

Derivation of Maxwell-Boltzmann Distribution Function

Taking Logarithm of the thermodynamic probability we have:

\[ \begin{aligned} \ln W= & (N \ln N-N)+\sum n_{i} \ln g_{i}-\sum n_{i} \ln n_{i}+\sum_{1}^{l} n_{i} \\ & =N \ln N+\sum n_{i} \ln g_{i}-\sum n_{i} \ln n_{i} \\ \Rightarrow d(\ln W) & =0+\sum d n_{i} \ln g_{i}-\sum d n_{i} \ln n_{i}-\sum n_{i} \dfrac{1}{n_{i}} d n_{i} \\ & =\sum \ln g_{i} d n_{i}-\sum \ln n_{i} d n_{i}-\sum d n_{i}=\sum \ln g_{i} d n_{i}-\sum \ln n_{i} d n_{i}-0\\ &=\sum \ln \dfrac{g_{i}}{n_{i}} d n_{i} \end{aligned}\]

For most probable state $W$ is maximum .

\[ \begin{aligned} & \Rightarrow d(\ln W)=0 \\ & \Rightarrow d\left(\alpha \sum_{1}^{l} n_{i}\right)=\alpha \sum_{1}^{l} d n_{i}=0 \\ & \Rightarrow d\left(\beta \sum_{1}^{l} n_{i} E_{i}\right)=\beta \sum_{1}^{l} E_{i} d n_{i}=0 \end{aligned} \]

Combining all these we get

\[ \begin{aligned} & d(\ln W)=\alpha \sum_{1}^{l} d n_{i}+\beta \sum_{1}^{l} E_{i} d n_{i} \\ & \sum_{1}^{l} \ln \dfrac{g_{i}}{n_{i}} d n_{i}=\sum_{1}^{l}\left(\alpha+\beta E_{i}\right) \\ & \Rightarrow \ln \dfrac{g_{i}}{n_{i}}=\left(\alpha+\beta E_{i}\right) \\ & \Rightarrow \dfrac{g_{i}}{n_{i}}=e^{\alpha+\beta E_{i}} \\ & \Rightarrow n_{i}=\dfrac{g_{i}}{e^{\alpha+\beta E_{i}}} \\ & \Rightarrow \dfrac{n_{i}}{g_{i}}=\dfrac{1}{e^{\alpha+\beta E_{i}}} \\ & f\left(E_{i}\right)=\dfrac{1}{e^{\alpha+\beta E_{i}}} \end{aligned} \]

. Evaluate $\alpha$ and $\beta$ appearing in the three types of distribution functions as follows :

$$ \begin{array}{ll} n_{i}=\dfrac{g_{i}}{e^{\left(\alpha+\beta E_{i}\right)}} & \text { ( MB statistics ) } \\ n_{i}=\dfrac{g_{i}}{e^{\left(\alpha+\beta E_{i}\right)}-1} & \text { ( BE statistics ) } \\ n_{i}=\dfrac{g_{i}}{e^{\left(\alpha+\beta E_{i}\right)}}+1 & \text { ( FD statistics ) } \end{array} $$

As $\alpha$ and $\beta$ do not depend on types of distribution we can take MB distribution function for easy evaluation. Then $$ \Rightarrow n_{i}=g_{i} e^{-\alpha} e^{-\beta E_{i}} $$ $\Rightarrow n(E)=g(E) e^{-\alpha} e^{-\beta E}$ when energy levels are continuos we can replace $E_{i}$ by $E$ and $g_{i}$ by $g(E)$ and $n_{i}$ by $n(E)$ $\Rightarrow n(E)=g(E) e^{-\alpha} e^{-\beta E}$

Let us express $g(E)$ in terms of energy.

\[ \begin{align*} g(E) &= \dfrac{\int dx\, dy\, dz\, dp_x\, dp_y\, dp_z}{h^3} \\ &= \dfrac{\int dx\, dy\, dz \int dp_x\, dp_y\, dp_z}{h^3} \\ &= \dfrac{V \int dp_x\, dp_y\, dp_z}{h^3} \\ &= \dfrac{V \cdot \dfrac{4}{3} \pi \left(p^3(E + dE) - p(E)^3\right)}{h^3} \\ &= \dfrac{V \cdot \dfrac{4}{3} \pi \left((2m(E + dE))^{3/2} - (2mE)^{3/2}\right)}{h^3} \\ &= \dfrac{V \cdot \dfrac{4}{3} \pi (2mE)^{3/2} \left((1 + \dfrac{dE}{E})^{3/2} - 1\right)}{h^3} \\ &= \dfrac{V \cdot \dfrac{4}{3} \pi (2mE)^{3/2} \left(1 + \dfrac{3}{2}\dfrac{dE}{E} + \dfrac{1}{4}\left(\dfrac{dE}{E}\right)^2 + \ldots - 1\right)}{h^3} \\[-1ex] &\text{(applying Binomial expansion)} \\ &= \dfrac{V \cdot \dfrac{4}{3} \pi (2mE)^{3/2} \cdot \dfrac{3}{2}\dfrac{dE}{E}}{h^3} \\ &= \dfrac{V \cdot 2\pi (2mE)^{3/2} dE}{Eh^3} \\ &= \dfrac{V \cdot 2\pi (2m)^{3/2} E^{1/2} dE}{h^3} \end{align*} \]

Thus, $g(E) = V \cdot 2\pi\left(\dfrac{2m}{h^2}\right)^{3/2} E^{1/2} dE$
(For particles with no spin. For particles with spin $\pm 1/2$, we multiply by 2)

Average Energy and Evaluation of $\beta $

\[ \begin{align*} \langle E \rangle &= \dfrac{E_{\text{total}}}{N} \\ &= \dfrac{\sum\limits_{i=0}^{\infty} n(E) E}{\sum\limits_{i=0}^{\infty} n(E)} \\ &= \dfrac{\int g(E) e^{-\alpha} e^{-\beta E} E\, dE}{\int g(E) e^{-\alpha} e^{-\beta E} \, dE} \\ &= \dfrac{\int g(E) e^{-\beta E} E\, dE}{\int g(E) e^{-\beta E} \, dE} \\ &= \dfrac{\int_0^\infty V \cdot 2\pi\left(\dfrac{2m}{h^2}\right)^{3/2} E^{3/2} e^{-\beta E} \, dE}{\int_0^\infty V \cdot 2\pi\left(\dfrac{2m}{h^2}\right)^{3/2} E^{1/2} e^{-\beta E} \, dE} \\ &= \dfrac{\int_0^\infty E^{3/2} e^{-\beta E} \, dE}{\int_0^\infty E^{1/2} e^{-\beta E} \, dE} \\ &= \dfrac{\dfrac{3\sqrt{\pi}}{4\beta^{5/2}}}{\dfrac{\sqrt{\pi}}{2\beta^{3/2}}} \\[1ex] &\text{(Using } \int_0^\infty x^{n-1} e^{-ax} \, dx = \dfrac{\Gamma(n)}{a^n}\text{)} \end{align*} \]

Using values:

$\Gamma(1/2) = \sqrt{\pi},\quad \Gamma(3/2) = \dfrac{\sqrt{\pi}}{2},\quad \Gamma(5/2) = \dfrac{3\sqrt{\pi}}{4}$

\[ \begin{align*} \Rightarrow \langle E \rangle &= \dfrac{3}{2\beta} \\ \Rightarrow \dfrac{3}{2}kT &= \dfrac{3}{2\beta} \\ \Rightarrow \beta &= \dfrac{1}{kT} \end{align*} \]

Evaluation of $\alpha$

As we have number of partcles occupying energy $\mathrm{E}_{\mathrm{i}}$ in $\mathrm{g}_{\mathrm{i}}$ number of cells, Then $n_{i}=g_{i} e^{-\alpha} e^{-E_{i} / k T}$

Thus Total number of particles is

$$ N=\int_{0}^{\infty} n(E) d E $$ ( where $n(E)$ represents number of particles in the energy range $E$ and $E+d E$ )

$$ \begin{aligned} \Rightarrow N&=\int_{0}^{\infty} g(E) e^{-\alpha} e^{-E / k T} d E\\ & =e^{-\alpha} \int_{0}^{\infty} 2 \pi V\left(\dfrac{2 m}{h^{2}}\right)^{3 / 2} E^{1 / 2} e^{-E / k T} d E \\ & =e^{-\alpha} 2 \pi V\left(\dfrac{2 m}{h^{2}}\right)^{3 / 2} \int_{0} E^{1 / 2} e^{-E / k T} d E \\ & =e^{-\alpha} 2 \pi V\left(\dfrac{2 m}{h^{2}}\right)^{3 / 2}\left(\dfrac{\sqrt{\pi}}{2(1 / k T)^{3 / 2}}\right) \\ & =e^{-\alpha} 2 \pi V\left(\dfrac{2 m}{h^{2}}\right)^{3 / 2}\left(\dfrac{(k T)^{3 / 2} \sqrt{\pi}}{2}\right) \\ & =e^{-\alpha} \pi^{3 / 2} V\left(\dfrac{2 m}{h^{2}}\right)^{3 / 2}(k T)^{3 / 2} \\ & =e^{-\alpha} V\left(\dfrac{2 m \pi k T}{h^{2}}\right)^{3 / 2} \\ & \Rightarrow e^{-\alpha}=\dfrac{N}{V}\left(\dfrac{h^{2}}{2 m \pi k T}\right)^{3 / 2} \end{aligned} $$

Relation of $e^{-\alpha}$ with Partition function :

$$ \begin{aligned} & n_{i}=g_{i} e^{-\alpha} e^{-E_{i} / k T} \\ \Rightarrow & \sum n_{i}=\sum g_{i} e^{-\alpha} e^{-E_{i} / k T} \\ \Rightarrow & N=e^{-\alpha} \sum g_{i} e^{-E_{i} / k T} \\ \Rightarrow & N=e^{-\alpha} Z \\ \Rightarrow & e^{-\alpha}=\dfrac{N}{Z} \end{aligned} $$

Prove that the number of particles in the energy range $E$ and $E+dE$ is :

$n(E) d E=\dfrac{2 N}{\sqrt{\pi}}\left(\dfrac{1}{k T}\right)^{3 / 2} \sqrt{E} e^{-E / k T} d E$\\

Proof :

$$ \begin{aligned} n(p) d p&=4 \pi N\left(\dfrac{1}{2 \pi m k T}\right)^{3 / 2} p^{2} e^{-p^{2} / 2 m k T} d p\\ & \Rightarrow n(E) d E=4 \pi N\left(\dfrac{1}{2 \pi m k T}\right)^{3 / 2}(2 m E) e^{-E / k T} d(\sqrt{2 m E}) \\ & \Rightarrow n(E) d E=4 \pi N\left(\dfrac{1}{2 \pi m k T}\right)^{3 / 2}(2 m E) e^{-E / k T}(\sqrt{2 m}) \dfrac{1}{2 \sqrt{E}} d E \\ & \Rightarrow n(E) d E=4 \pi N\left(\dfrac{1}{2 \pi m k T}\right)^{3 / 2}(2 m)^{3 / 2} E^{1 / 2} e^{-E / k T} \dfrac{1}{2 \sqrt{E}} d E \\ & \Rightarrow n(E) d E=\dfrac{2 N}{\sqrt{\pi}}\left(\dfrac{1}{k T}\right)^{3 / 2} E^{1 / 2} e^{-E / k T} d E \end{aligned} $$

Prove that the number of particles in the range $v$ and $v+d v$ is :

$n(v) d v=4 \pi N\left(\dfrac{m}{2 \pi k T}\right)^{3 / 2} e^{-m v^{2} / 2 k T} v^{2} d v$

Proof:

$$ \begin{aligned} & n(p) d p=4 \pi N\left(\dfrac{1}{2 \pi m k T}\right)^{3 / 2} p^{2} e^{-p^{2} / 2 m k T} d p \\ & \Rightarrow n(v) d v=4 \pi N\left(\dfrac{1}{2 \pi m k T}\right)^{3 / 2} m^{2} v^{2} e^{-m v^{2} / 2 k T} m d v \\ & \Rightarrow n(v) d v=4 \pi N\left(\dfrac{m}{2 \pi m k T}\right)^{3 / 2} e^{-m v^{2} / 2 k T} v^{2} d v \end{aligned} $$

Prove that the most probable speed of gas molecules is $v_{m p}=\sqrt{\dfrac{2 k T}{m}}$

Proof:

Since most probable speed is the speed that is possessed by maximum number of molecules particles then $\dfrac{d}{d v}(P(v))=0$,
Where $P$ is the probability that particles have speed between $ v $ and $v+dv$

$$ \begin{aligned} &\Rightarrow \dfrac{d}{d v}\left(e^{-m v^{2} / 2 k T} v^{2}\right)=0, \quad P(v)=n(v) / N \\ &\Rightarrow e^{-m v^{2} / 2 k T}\left(-\dfrac{m}{2 k T} 2 v\right) v^{2}+e^{-m v^{2} / 2 k T}(2 v)=0\\ & \Rightarrow-\dfrac{m}{k T} v^{3}+2 v=0 \\ & \Rightarrow-\dfrac{m}{k T} v^{2}+2=0 \\ & \Rightarrow v^{2}=\dfrac{2 k T}{m} \\ & \Rightarrow v_{m p}=\sqrt{\dfrac{2 k T}{m}} \end{aligned} $$

Prove that the average ( mean ) speed of the gas molecules is: $v_{a v g}=\sqrt{\dfrac{3 k T}{m}}$

\begin{aligned} v_{\text {avg }}&=\dfrac{n_{1} v_{1}+n_{2} v_{2}+\ldots}{n_{1}+n_{2}+\ldots .}\\ &=\dfrac{\sum_{0}^{\infty} n v}{\sum_{0}^{\infty} n} \\ &=\dfrac{\int_{0}^{\infty} n(v) v}{\int_{0}^{\infty} n(v)}\\ &=\dfrac{\int_{0}^{\infty}\left(4 \pi N\left(\dfrac{m}{2 \pi k T}\right)^{3 / 2} e^{-m v^{2} / 2 k T} v^{2} d v\right) v}{\int_{0}^{\infty} 4 \pi N\left(\dfrac{m}{2 \pi k T}\right)^{3 / 2} e^{-m v^{2} / 2 k T} v^{2} d v}\\ &=\dfrac{\int_{0}^{\infty}\left(e^{-m v^{2} / 2 k T} v^{3} d v\right)}{\int_{0}^{\infty} e^{-m v^{2} / 2 k T} v^{2} d v}\\ &=\dfrac{\int_{0}^{\infty} x^{3} e^{-a x^{2}} d x}{\int_{0}^{\infty} x^{2} e^{-a x^{2}} d x} \quad(\text { Taking } m / 2 k T=a)\\ &=\dfrac{\dfrac{1}{2 a^{2}}}{\dfrac{1}{4 a} \sqrt{\dfrac{\pi}{a}}}\\ &=\dfrac{2}{\sqrt{\pi a}}\\ &=\dfrac{2 \sqrt{2 k T}}{\sqrt{\pi m}}\\ &=\sqrt{\dfrac{8 k T}{\pi m}} \end{aligned}

Prove that the rms speed of gas molecules is $v_{r m s}=\sqrt{\dfrac{3 k T}{m}}$

Proof:

$$ \begin{aligned} v_{r m s} & =\sqrt{\dfrac{n_{1} v_{1}{ }^{2}+n_{2} v_{2}{ }^{2}+n_{3} v_{3}{ }^{2}+\ldots}{n_{1}+n_{2}+n_{3}+\ldots}}\\ &=\sqrt{\dfrac{\int_{0}^{\infty} n(v) v^{2}}{\int_{0}^{\infty} n(v)}}=\sqrt{\dfrac{\int_{0}^{\infty}\left(e^{-m v^{2} / 2 k T} v^{4} d v\right)}{\int_{0}^{\infty} e^{-m v^{2} / 2 k T} v^{2} d v}} \\ & =\sqrt{\dfrac{\int_{0}^{\infty} x^{4} e^{-a x^{2}} d x}{\int_{0}^{\infty} x^{2} e^{-a x^{2}} d x}}(\text { Taking } m / 2 k T=a) \\ & =\sqrt{\dfrac{\dfrac{3}{\dfrac{8 a^{2}}{\sqrt{\dfrac{\pi}{a}}}}}{\dfrac{1}{4 a} \sqrt{\dfrac{\pi}{a}}}}\\ &=\sqrt{\dfrac{3}{2 a}}\\ &=\sqrt{\dfrac{6 k T}{2 m}} \\ & =\sqrt{\dfrac{3 k T}{m}} \end{aligned} $$

Prove that the mean Kinetic energy of gas molecules is $\langle E\rangle=\dfrac{3}{2} k T$

Proof :

$$ \begin{aligned} & \langle E\rangle=\dfrac{n_{1} E_{1}+n_{2} E_{2}+n_{3} E_{3}+\ldots \ldots . .}{n_{1}+n_{2}+n_{3}+\ldots \ldots .}\\ &=\dfrac{\int_{0}^{\infty} n(E) E d E}{\int_{0}^{\infty} n(E) d E} \\ & =\dfrac{\dfrac{2 N}{\sqrt{\pi}}\left(\dfrac{1}{k T}\right)^{3 / 2} \int_{0}^{\infty} E^{3 / 2} e^{-E / k T} d E}{\dfrac{2 N}{\sqrt{\pi}}\left(\dfrac{1}{k T}\right)^{3 / 2} \int_{0}^{\infty} E^{1 / 2} e^{-E / k T} d E} \\ & =\dfrac{\int_{0}^{\infty} E^{3 / 2} e^{-E / k T} d E}{\int_{0}^{\infty} E^{1 / 2} e^{-E / k T} d E} \\ & =\dfrac{\int_{0}^{\infty} x^{5 / 2-1} e^{-a x} d x}{\int_{0}^{\infty} x^{3 / 2-1} e^{-a x} d x}(\text { Taking } a=1 / k T, x=E) \\ & =\dfrac{\Gamma(5 / 2) / a^{5 / 2+1}}{\Gamma(3 / 2) / a^{3 / 2+1}}\\ &=\dfrac{\dfrac{3 \sqrt{\pi}}{4} \dfrac{1}{a^{5 / 2}}}{\frac{\sqrt{\pi}}{2} \frac{1}{a^{3 / 2}}}\\ &=\frac{3}{2} \frac{1}{a} \\ &=\frac{3}{2} k T \end{aligned} $$

Comparison of MB, FD, and BE Statistics

Property	Maxwell-Boltzmann (MB)	Fermi-Dirac (FD)	Bose-Einstein (BE)
Particle Type	Classical, distinguishable	Fermions (half-integer spin)	Bosons (integer spin)
Occupation Rule	No restriction	At most 1 particle per state	Multiple particles per state allowed
Distribution Function	$ f(E) = A e^{-E / kT} $	$ f(E) = \frac{1}{e^{(E - \mu)/kT} + 1} $	$ f(E) = \frac{1}{e^{(E - \mu)/kT} - 1} $
Applicable To	High T, low density gases	Electrons, protons, neutrons	Photons, Helium-4 atoms
Quantum Effects	Negligible	Important at low T	Important at low T
Obeys Pauli Exclusion	No	Yes	No
Behavior at T → 0	Particles fall to lowest energy	Fills up states up to $ E_F $	Bose-Einstein Condensation possible