What is Gibbs' Paradox? How is it resolved by Gibbs?

Gibbs' Paradox arises from classical statistical mechanics, where the entropy appears to increase upon mixing two identica gases — violating the expected additive nature of entropy. This paradox occurs due to the \textbf{incorrect assumption of distinguishability} of identical particles in classical statistics.

Resolution

The paradox is resolved using \textit{quantum statistics}, which treats identical particles as \textbf{indistinguishable}. The meaningless overcounting in classical Boltzmann statistics is corrected by dividing the number of microstates by \( \mathbf{n!} \), where \( n \) is the number of molecules.

Explanation

Consider two systems A and B, each at the same temperature \( T \). Let the systems be characterized by: \[ (N, V, T, m) \quad \text{and} \quad (N, V, T, m) \] where:
\(N\) : Number of molecules,
\(V\) : Volume,
\(m\) : Mass of each molecule,
\(T\) : Temperature
Assume the partition separating systems A and B is both insulating and impermeable Then the entropy of the combined system (A + B) is: Now, if we remove the partition and allow the gases to mix (even though they are identical), then: Total number of molecules becomes \( 2n \), Total volume becomes \( 2V \). Then, the entropy becomes: Conclusion: The additional term \( 2nk \ln 2 \) implies an artificial increase in entropy upon mixing, which is paradoxical since the gases are identical. Quantum mechanics resolves this by acknowledging the indistinguishability of particles and dividing by \( n! \), thereby canceling the extra \( \ln 2 \) term. Hence it is obserbed that before and after removal of Partition the entropy differes by a value $2 n k \ln 2$ violating the additive nature of entropy.

Removal of Gibbs' Paradox

The partition function for one molecule is given by: \[ Z = \frac{V}{h^{3}} (2 \pi m k T)^{3/2} \] The partition function for \( n \) molecules (assuming distinguishable particles) is: \[ Z_n = \left( \frac{V}{h^{3}} (2 \pi m k T)^{3/2} \right)^n \] \[= \frac{V^n}{h^{3n}} (2 \pi m k T)^{3n/2} \] However, since the molecules are indistinguishable, the corrected partition function should be: \[ Z_n = \frac{1}{n!} \cdot \frac{V^n}{h^{3n}} (2 \pi m k T)^{3n/2} \] Now, the correct entropy is: Thus, the final expression for entropy is: This is the famous Sackur–Tetrode equation Now, using this equation, the entropy of the system (A + B) before removal of the partition is: \[ \begin{aligned} S_{A+B} &= S_A + S_B \\ &= 2 \left[ n k \ln \left( \frac{V}{n h^3} (2 \pi m k T)^{3/2} \right) + \frac{5}{2} n k \right] \end{aligned} \] And after removal of the partition, the entropy becomes: Conclusion: There is no change in entropy before and after mixing identical gases — which resolves the Gibbs' Paradox. The paradox vanishes when quantum indistinguishability is properly accounted for.Which proves both are equal and hence the additive nature of entropy. ⬆️ Back to Table of Contents

Classical Ideal Gas

A classical ideal gas is a theoretical gas composed of many identical point particles that do not interact with each other except through elastic collisions. It follows the laws of classical mechanics (Newtonian mechanics), and its statistical properties can be described using Maxwell-Boltzmann statistics.

Properties of a Classical Ideal Gas:

• Particles are indistinguishable and non-interacting.
• Each particle moves in straight lines unless it collides elastically with a wall or another particle.
• The gas obeys the ideal gas law: \( PV = NkT \).
• Energy is purely kinetic: \( E = \dfrac{1}{2}mv^2 \).
• The probability distribution of particle speeds is given by the Maxwell-Boltzmann distribution.

Density of States:

The number of quantum states available to particles in a given energy range is:

\[ g(E) = \frac{V \cdot 2\pi(2m)^{3/2} E^{1/2}}{h^3} \]

where \( g(E) \) is the density of states, \( V \) is the volume, \( m \) is the particle mass, and \( h \) is Planck's constant.

Average Energy:

\[ \langle E \rangle = \frac{\int_0^\infty E^{3/2} e^{-\beta E} dE}{\int_0^\infty E^{1/2} e^{-\beta E} dE} = \frac{3}{2\beta} = \frac{3}{2}kT \]

This result shows that each degree of freedom contributes \( \frac{1}{2}kT \) to the average energy of the system, as predicted by the equipartition theorem.

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What is Partition Function and write down its properties:

Ans: \begin{align*} n_{i}&=g_{i} e^{\alpha} e^{-E / k T}\\ &\Rightarrow \sum n_{i}=e^{\alpha} \sum g_{i} e^{-E / k T}\\ &\Rightarrow N=e^{\alpha} \sum g_{i} e^{-E / k T}\\ &\Rightarrow \dfrac{n_{i}}{N}=\dfrac{g_{i} e^{\alpha} e^{-E / k T}}{e^{\alpha} \sum g_{i} e^{-E / k T}}\\ & \Rightarrow \dfrac{n_{i}}{N}=\dfrac{g_{i} e^{-E / k T}}{\sum g_{i} e^{-E / k T}}\\ & \Rightarrow \dfrac{n_{i}}{N}=\dfrac{g_{i} e^{-E / k T}}{Z} \end{align*} Z is called as the partition function of the system. It represents how N particles are distributed among thier energy levels.

Features of Partition Function(Z)

1. It indicates the mode of distribution of energy among various energy levels.
2. It is a pure number, hence it is dimensionless.
3. It can never be zero.
4. Its lowest value is 1 at absolute zero temperature as molecules stay in the ground state.
5. It is much larger than 1 at higher temperature as fewer molecules stay in the ground state.
6. It is also a measure of the extent to which particles may escape from the ground state.

Partition Function for Canonical Ensembles with no degeneracy \[Z=\sum{{{g}_{i}}{{e}^{-\beta {{E}_{i}}}}}\] Partition Function for Canonical Ensembles with degeneracy \[Z=\sum{{{e}^{-\beta {{E}_{i}}}}}\] Partition Function for Grand Canonical Ensembles with \[ Z=\sum{{{n}_{i}}{{e}^{-\beta \left( {{E}_{i}}-\mu \right){{n}_{i}}}}}\]

Fluctuations in the Density of a Grand Canonical Ensemble

In a Grand Canonical Ensemble, the number of particles \(N\) and energy \(E\) both fluctuate. Since the density \(\rho = \dfrac{N}{V}\) and the volume \(V\) is fixed, fluctuations in density are equivalent to fluctuations in \(N\).

The average number of particles is:

\[ \left\langle N \right\rangle = \dfrac{\sum_N \sum_n N e^{-\dfrac{E_{nN} - \mu N}{kT}}}{\sum_N \sum_n e^{-\dfrac{E_{nN} - \mu N}{kT}}} \] \[\begin{aligned} &= \dfrac{1}{kT} \left( \left\langle N^2 \right\rangle - \left\langle N \right\rangle^2 \right) \\ &= \dfrac{(\Delta N)^2}{kT} \\ \Rightarrow \left\langle (\Delta N)^2 \right\rangle &= kT \left( \dfrac{\partial \left\langle N \right\rangle}{\partial \mu} \right)_T \end{aligned} \]

This gives the mean square fluctuation in the number of particles (or density).

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Classical Limit of Quantum Statistics

Quantum Statistics will reduce to Classical Statistics(more specifically MB statistics) if the mean separation between the particles exceeds much larger than the mean debroglie wavelength. In classical statistics the particles are identical , neutral and noninteracting but distinguishable. In Quantum statistics the Particles are so closer that their wave functions overlap ,Hence the particles become indistnguishable. Their mean separation is comparable to the debroglie wavelength. Moreover the system that adopts Quantum statistics is generally at a very low temperature so that the randomness of particles decreases to such a level that the particles are closer enough. In addition to it the density of the system is also very high.

Derivation of Mathematical Condition:( VALIDITY OF MB STATISTICS)

From Heisenberg Uncertainity Principle $$\Delta x\Delta p>>\hbar$$ $$\Rightarrow \Delta x>>\lambda $$ $$\Rightarrow{{\left( \Delta x \right)}^{3}}\propto \frac{V}{N} $$ If the system contains N particles enclosed in volume V , Then Volume per particle is\( N/V \) Assuming the system in a Cube, the length of the cube is given by \(\Delta x \)where : \[\begin{align} & \Delta x={{\left( \frac{V}{N} \right)}^{\frac{1}{3}}} \\ & {{\left( \frac{V}{N} \right)}^{\frac{1}{3}}}>>\lambda \\ & {{\left( \frac{M}{\rho } \right)}^{\frac{1}{3}}}<<\lambda \\ & {{\left( \frac{1}{\rho } \right)}^{1/3}}<<\lambda \\ & \rho {{\lambda }^{3}}>>1 \\ \end{align}\] \( \rho\) is density of the system. Other Conditions So other conditions for classical approximation is :
1. $N/V$ ( concentration of molecules) must be very small.
2. The temperature T msut be very high
3. The mass of the molecules is not to be too small. ⬆️ Back to Table of Contents

IDEAL FERMI GAS :

Properties of FERMI GAS and it's Comparison with Classical Ideal Gas

1. Fermions are negatively charged while classically ideal gas is neutral.
2. Fermions travel with finite speed while classical particles may be treated differently.
3. Fermions have odd spins while molecules have no spins.
4. Fermions obey FERMI-DIRAC statistical distribution while classical gas obeys Maxwell-Boltzmann statistics.
5. Fermions are indistinguishable in FERMI GAS while molecules in classical gas are distinguishable.
6. Fermion density in FERMI GAS is much greater than the molecular density in classical ideal gas.
7. Fermions obey Pauli's Exclusion Principle while molecules do not.

Equation of state for Ideal fermi Gas

Prove the following using Summerfield's Expansion

Proof :

The Summerfield Expansion for the fermi function is :

Prove that Fermi Energy is

: $$E_{F}=\left(\frac{3 N}{4 \pi V}\right)^{2 / 3}\left(\frac{h^{2}}{2 m}\right)$$

and the chemical Potential is:

$$\mu=E_{F}\left(1-\frac{\pi^{2}}{12} \frac{1}{(\ln z)^{2}}+\ldots.\right)$$\\

Proof:

$$ \operatorname{At} \mathbf{T}=\mathbf{0 K}, \mu=E_{F} $$ $$ \Rightarrow E_{F}=\left(\frac{3 N}{4 \pi V}\right)^{2 / 3}\left(\frac{h^{2}}{2 m}\right) $$

At any Temperature $T$

Prove that

Proof:

Prove that Pressure of the Fermi Gas is :

$$P=\frac{2}{5} n E_{F}\left(1+\frac{5 \pi^{2}}{12}\left(\frac{k T}{E_{F}}\right)^{2}\right)$$ Proof $$ \begin{aligned} P&=\frac{2}{3} \frac{U}{V}\\ &=\frac{2}{3} \frac{U / N}{V / N} \\ & =\frac{2}{3} \frac{\frac{3}{5} E_{F}\left(1+\frac{5 \pi^{2}}{12}\left(\frac{k T}{E_{F}}\right)^{2}\right)}{V / N}\\ &=\frac{2}{5} n E_{F}\left(1+\frac{5 \pi^{2}}{12}\left(\frac{k T}{E_{F}}\right)^{2}\right) \end{aligned} $$

Prove that Specific heat at low temperature is

. $$\dfrac{C_{V}}{N k}=\dfrac{\pi^{2} k T}{2 E_{F}}$$

Proof :

Prove that for Ideal Fermi Gas

Proof:

Prove that the equation of state for an Ideal Fermi gas is given by:

Prove that

$$ \frac{P}{k T}=\frac{f_{5 / 2}(z)}{\lambda^{3}}$$

Proof:

Prove that the number of Particles in an Ideal Fermi Gas is giiven by :

$$ N=\frac{V f_{3 / 2}(z)}{\lambda^{3}} $$

Proof:

Prove that Internal energy of the Fermi gas is given by :

$$ U=\frac{3}{2} N k T \frac{f_{5 / 2}(z)}{f_{3 / 2}(z)} $$

Proof:

Prove that

IDEAL BOSE GAS

1. An ideal Bose gas is a quantum-mechanical phase of matter, analogous to a classical ideal gas.
2. It has integral spins.
3. Its constituents are called bosons.
4. Its examples are photons, which travel at the speed of light and are quanta of electromagnetic radiation.
5. Bosons obey Bose-Einstein statistics.
6. Bosons are neutral.
7. The theory of Bose gas can explain many phenomena that were classically inexplicable, such as:

Black body radiation

Heat capacity variation

Superconductivity and so on

Properties of PHOTON GAS and it's Comparision with Classical Ideal Gas:

1. The rest mass of a photon is zero, whereas a gas molecule has a finite mass.
2. Photons in a photon gas are indistinguishable, while molecules in a classical ideal gas are distinguishable.
3. A photon gas obeys Bose-Einstein statistics, whereas a classical ideal gas obeys Maxwell-Boltzmann statistics.
4. Photons have integral spin, while molecules are spinless.
5. Photons in a photon gas travel at the speed of light, but gas molecules travel with speeds ranging from 0 to infinity.
6. Photons are not conserved during collisions, whereas gas molecules are conserved during collisions.

section*{Prove that for an Ideal Bose gas $\frac{S}{k}=\sum g_{i}\left(\frac{\beta E_{i}-\log z}{z^{-1} e^{\beta E_{i}}-1}-\log \left(1-z e^{-\beta E_{i}}\right)\right)$} Proof: $$ \begin{aligned} S & =k \ln W_{\max } \\ \Rightarrow \frac{S}{k} & =\ln W_{\max }=\sum \ln \left(\frac{\left(g_{i}+n_{i}-1\right)!}{n_{i}!\left(g_{i}-1\right)!}\right) \approx \sum \ln \left(\frac{\left(g_{i}+n_{i}\right)!}{n_{i}!\left(g_{i}\right)!}\right) \because g_{i}+n_{i} \gg 1 \& g_{i} \gg 1 \end{aligned} $$ $$ \begin{aligned} & =\sum\left[\left(g_{i}+n_{i}\right) \ln \left(g_{i}+n_{i}\right)-g_{i}-n_{i}-n_{i} \ln n_{i}+n_{i}-g_{i} \ln g_{i}+g_{i}\right] \\ & =\sum\left[g_{i} \ln \left(g_{i}+n_{i}\right)+n_{i} \ln \left(g_{i}+n_{i}\right)-g_{i}-n_{i}-n_{i} \ln n_{i}+n_{i}-g_{i} \ln g_{i}+g_{i}\right] \\ & =\sum\left[g_{i} \ln \left(g_{i}+n_{i}\right)-g_{i} \ln g_{i}+n_{i} \ln \left(g_{i}+n_{i}\right)-n_{i} \ln n_{i}\right] \\ & =\sum\left[n_{i} \ln \left(\frac{g_{i}+n_{i}}{n_{i}}\right)+g_{i} \ln \left(\frac{g_{i}+n_{i}}{g_{i}}\right)\right] \\ & =\sum\left[n_{i} \ln \left(\frac{g_{i}}{n_{i}}+1\right)+g_{i} \ln \left(1+\frac{n_{i}}{g_{i}}\right)\right] \\ & =\sum\left[n_{i} \ln \left(e^{\alpha+\beta E_{i}}-1+1\right)+g_{i} \ln \left(1+\frac{1}{e^{\alpha+\beta E_{i}}-1}\right)\right] \\ & =\sum\left[n_{i}\left(\alpha+\beta E_{i}\right)+g_{i} \ln \left(1-\frac{1}{e^{\alpha+\beta E_{i}}-1}\right)\right] \\ & =\sum\left[n_{i}\left(\beta E_{i}-\ln z\right)+g_{i} \ln \left(\frac{e^{\alpha+\beta E_{i}}}{e^{\alpha+\beta E_{i}}-1}\right)\right] \because e^{\alpha}=z^{-1}, \alpha=-\ln z \\ & =\sum\left[n_{i}\left(\beta E_{i}-\ln z\right)+g_{i} \ln \left(\frac{1}{1-e^{-\alpha-\beta E_{i}}}\right)\right] \\ & =\sum\left[n_{i}\left(\beta E_{i}-\ln z\right)-g_{i} \ln \left(1-e^{-\alpha-\beta E_{i}}\right)\right] \\ & =\sum\left[\frac{g_{i}}{e^{\alpha+\beta E_{i}}-1}\left(\beta E_{i}-\ln z\right)-g_{i} \ln \left(1-z e^{-\beta E_{i}}\right)\right] \\ & =\sum g_{i}\left[\frac{\left(\beta E_{i}-\ln z\right)}{z^{-1} e^{\beta E_{i}}-1}-\ln \left(1-z e^{-\beta E_{i}}\right)\right] \end{aligned} $$ \section*{Prove that for an Ideal Bose gas the equation of State is given by :} \begin{align} \frac{P V}{k T} &= \log Z_{BE} = -\sum \log \left(1 - z e^{-\beta E_i}\right) \\ &= \frac{V}{\lambda^{3}} \, g_{5/2}(z) \end{align} where \[ g_r(z) = z + \dfrac{z^2}{2^r} + \dfrac{z^3}{3^r} + \dfrac{z^4}{4^r} + \ldots = -\sum_{n=1}^{\infty} (-1)^{2n+1} \dfrac{z^n}{n^r}\] \section*{Proof:} Since $\dfrac{S}{k}=\sum g_{i}\left[\dfrac{\left(\beta E_{i}-\ln z\right)}{z^{-1} e^{\beta E_{i}}-1}-\ln \left(1-z e^{-\beta E_{i}}\right)\right]$ \begin{align} \frac{S}{k} - \sum g_i \left( \frac{\beta E_i - \ln z}{z^{-1} e^{\beta E_i} - 1} \right) &= -\sum g_i \ln \left( 1 - z e^{-\beta E_i} \right) \\ \frac{S}{k} - \sum n_i \left( z^{-1} e^{\beta E_i} - 1 \right) \frac{\beta E_i - \ln \left( e^{\frac{\mu}{kT}} \right)}{z^{-1} e^{\beta E_i} - 1} &= -\sum g_i \ln \left( 1 - z e^{-\beta E_i} \right) \quad \because n_i = \frac{g_i}{z^{-1} e^{\beta E_i} - 1} \\ \frac{S}{k} - \sum \left( \frac{n_i E_i}{kT} - \frac{n_i \mu}{kT} \right) &= -\sum g_i \ln \left( 1 - z e^{-\beta E_i} \right) \\ \frac{S}{k} - \left( \frac{E - G}{kT} \right) &= -\sum g_i \ln \left( 1 - z e^{-\beta E_i} \right) \\ \frac{TS - E + G}{kT} &= -\sum g_i \ln \left( 1 - z e^{-\beta E_i} \right) \\ \frac{G - (E - TS)}{kT} &= -\sum g_i \ln \left( 1 - z e^{-\beta E_i} \right) \\ \frac{H - TS - U + TS}{kT} &= -\sum g_i \ln \left( 1 - z e^{-\beta E_i} \right) \\ \frac{PV}{kT} &= -\sum g_i \ln \left( 1 - z e^{-\beta E_i} \right) \end{align} $$ \begin{aligned} & \dfrac{S}{k}-\sum g_{i}\left(\frac{\beta E_{i}-\ln z}{z^{-1} e^{\beta E_{i}}-1}\right)=-\sum g_{i} \ln \left(1-z e^{-\beta E_{i}}\right)\\ & \frac{S}{k}-\sum n_{i}\left(z^{-1} e^{\beta E_{i}}-1\right) \frac{\beta E_{i}-\ln \left(e^{\frac{\mu}{k T}}\right)}{z^{-1} e^{\beta E_{i}}-1}=-\sum g_{i} \ln \left(1-z e^{-\beta E_{i}}\right)\because n_{i}=\frac{g_{i}}{z^{-1} e^{\beta E_{i}}-1} \\ & \frac{S}{k}-\sum\left(\frac{n_{i} E_{i}}{k T}-\frac{n_{i} \mu}{k T}\right)=-\sum g_{i} \ln \left(1-z e^{-\beta E_{i}}\right) \\ & \frac{S}{k}-\left(\frac{E-G}{k T}\right)=-\sum g_{i} \ln \left(1-z e^{-\beta E_{i}}\right) \\ & \frac{T S-E+G}{k T}=-\sum g_{i} \ln \left(1-z e^{-\beta E_{i}}\right) \\ & \frac{G-(E-T S)}{k T}=-\sum g_{i} \ln \left(1-z e^{-\beta E_{i}}\right) \\ & \frac{H-T S-U+T S}{k T}=-\sum g_{i} \ln \left(1-z e^{-\beta E_{i}}\right) \\ & \frac{P V}{k T}=-\sum g_{i} \ln \left(1-z e^{-\beta E_{i}}\right) \end{aligned} $$ \section*{To Prove $\frac{P}{k T}=\frac{g_{5 / 2}(z)}{\lambda^{3}}$}\\ Proof: $$ \begin{aligned} & \frac{P V}{k T}=-\sum g_{i} \ln \left(1-z e^{-\beta E_{i}}\right) \\ & \frac{P V}{k T}=-\int_{0}^{\infty} \ln \left(1-z e^{-\beta E_{i}}\right) \frac{V}{h^{3}} 4 \pi p^{2} d p \\ & =-\frac{4 \pi V}{h^{3}} \int_{0}^{\infty} \ln \left(1-z e^{-\beta E_{i}}\right) p^{2} d p \\ & =-\frac{4 \pi V}{h^{3}} \int_{0}^{\infty} \sum_{1}^{n}(-1)^{n+1} \frac{\left(-z e^{-\beta E_{i}}\right)^{n}}{n} p^{2} d p \because \ln (1+y)=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{y^{n}}{n} \\ & =-\frac{4 \pi V}{h^{3}} \int_{0}^{\infty} \sum_{1}^{n}(-1)^{2 n+1} \frac{z^{n}}{n} e^{-n \beta E} p^{2} d p \\ & =-\frac{4 \pi V}{h^{3}} \sum_{1}^{n}(-1)^{2 n+1} \frac{z^{n}}{n} \int_{0}^{\infty} e^{-n p^{2} / 2 m k T} p^{2} d p \\ & =-\frac{4 \pi V}{h^{3}} \sum_{1}^{n}(-1)^{2 n+1} \frac{z^{n}}{n} \int_{0}^{\infty} e^{-x} p^{2} d p \end{aligned} $$ Let $n p^{2} / 2 m k T=x \Rightarrow p^{2}=\frac{2 m k T x}{n}, d p=\frac{\sqrt{2 m k T} d x}{2 \sqrt{n} \sqrt{x}}$ $$ \begin{aligned} & =-\frac{4 \pi V}{h^{3}} \sum_{1}^{n}(-1)^{2 n+1} \frac{z^{n}}{n} \int_{0}^{\infty} e^{-x}\left(\frac{2 m k T x}{n}\right)\left(\frac{\sqrt{2 m k T} d x}{2 \sqrt{n} \sqrt{x}}\right) \\ & =-\frac{4 \pi V}{h^{3}} \sum_{1}^{n}(-1)^{2 n+1} \frac{z^{n}}{n} \int_{0}^{\infty} e^{-x} x^{1 / 2} d x\left(\frac{2 m k T}{n}\right)^{3 / 2} \frac{1}{2} \\ & =-\frac{4 \pi V}{h^{3}} \sum_{1}^{n}(-1)^{2 n+1} \frac{z^{n}}{n} \Gamma(3 / 2)\left(\frac{2 m k T}{n}\right)^{3 / 2} \frac{1}{2} \\ & =-\frac{4 \pi V}{h^{3}} \sum_{1}^{n}(-1)^{2 n+1} \frac{z^{n}}{n} \frac{\sqrt{\pi}}{2}\left(\frac{2 m k T}{n}\right)^{3 / 2} \frac{1}{2} \\ & =\frac{V}{h^{3}} \sum_{1}^{n}-(-1)^{2 n+1} \frac{z^{n}}{n}\left(\frac{2 m \pi k T}{n}\right)^{3 / 2} \\ & =\frac{V}{h^{3}}(2 m \pi k T)^{3 / 2} \sum_{1}^{n}-(-1)^{2 n+1} \frac{z^{n}}{n^{5 / 2}} \\ & =\frac{V}{h^{3}}(2 m \pi k T)^{3 / 2} g_{5 / 2}(z) \\ & =\frac{V}{\lambda^{3}} g_{5 / 2}(z) \\ & \frac{P V}{k T}=\frac{V}{\lambda^{3}} g_{5 / 2}(z) \end{aligned} $$ For Ideal Bose gas the Grand parttion function is given by : $$ \begin{aligned} Z& = z_{1} z_{2} z_{3} \ldots \ldots\\ &=\sum_{0}^{\infty}\left(z e^{-\beta E_{0}}\right)^{n} \\ & =1+\left(z e^{-\beta E_{0}}\right)+\left(z e^{-\beta E_{0}}\right)^{2}+\left(z e^{-\beta E_{0}}\right)^{3} \ldots\left(z e^{-\beta E_{0}}\right)^{\infty} \\ & =\frac{1}{1-\left(z e^{-\beta E_{0}}\right)} \end{aligned} $$ Where $$ z=e^{-\alpha}=e^{\frac{\mu}{k T}} \quad \text { and } \beta=1 / k T \quad \mu=\text { Chemical Potential } $$ \subsubsection*{The average number of particles in an ideal Bose Gas in terms of partiton function is:} $$ \begin{aligned} n_{i}& =\frac{g_{i}}{e^{\alpha} e^{\beta E}-1}\\ \Rightarrow \frac{n_{i}}{g_{i}}&=\frac{e^{-\alpha} e^{-\beta E_{i}}}{1-e^{-\alpha} e^{-\beta E_{i}}} \\ &=\frac{z e^{-\beta E_{i}}}{1-z e^{-\beta E_{i}}} \\ \Rightarrow N&=\sum \frac{z e^{-\beta E_{i}}}{1-z e^{-\beta E_{i}}} \\ &=-z \frac{\partial}{\partial z} \sum \log \left(1-z e^{-\beta E_{i}}\right) \end{aligned} $$ \section*{Prove that average Number of Particles in an Ideal Bose Gas system is :} $$ N=\frac{V g_{3 / 2}(z)}{\lambda^{3}} $$ Proof: $$ $$ \begin{aligned} N &=\sum n_{i}\\ &=\sum \frac{g_{i}}{e^{\alpha+\beta E_{i}}-1}\\ & =\sum \frac{g_{i} e^{-\alpha-\beta E_{i}}}{1-e^{-\alpha-\beta E_{i}}} \\ & =\sum \frac{g_{i} z e^{-\beta E_{i}}}{1-z e^{-\beta E_{i}}}=z \sum \frac{\partial}{\partial z}\left(-g_{i} \ln \left(1-z e^{-\beta E_{i}}\right)\right) \\ & =z \sum \frac{\partial}{\partial z} g_{i} \ln \left(\frac{1}{1-z e^{-\beta E_{i}}}\right) \\ & =z \sum \frac{\partial}{\partial z} g_{i} \ln Z \quad \mathrm{Z}=\text { Gramd Partition function of Bose Gas } \\ & =z \sum \frac{\partial}{\partial z}\left(\frac{P V}{k T}\right) \\ & =z V \sum \frac{\partial}{\partial z}\left(\frac{g_{5 / 2}(z)}{\lambda^{3}}\right) \\ & =\frac{V}{\lambda^{3}} z \sum \frac{\partial}{\partial z}\left(g_{5 / 2}(z)\right)\\ &=\frac{V}{\lambda^{3}} z \frac{\partial}{\partial z} \sum \frac{z^{n}}{n^{5 / 2}} \\ & =\frac{V}{\lambda^{3}} z \sum \frac{n z^{n-1}}{n^{5 / 2}}\\ &=\frac{V}{\lambda^{3}} \sum \frac{z^{n}}{n^{3 / 2}} \\ & =\frac{V}{\lambda^{3}} g_{3 / 2}(z) \\ & \Rightarrow \frac{1}{v}=\frac{g_{3 / 2}(z)}{\lambda^{3}} \end{aligned} $$ \section*{Prove that $z \frac{\partial}{\partial z} g_{r}(z)=g_{r-1}(z)$} Hence $$ \begin{aligned} z \frac{\partial}{\partial z} g_{5 / 2}(z)&=g_{3 / 2}(z) \\ z \frac{\partial}{\partial z} g_{3 / 2}(z)&=g_{1 / 2}(z) \end{aligned} $$ \section*{Proof: Method 1} $$ \begin{aligned} z \frac{\partial}{\partial z} g_{r}(z) &=z \frac{\partial}{\partial z}\left(z+\frac{z^{2}}{2^{r}}+\frac{z^{3}}{3^{r}}+\frac{z^{4}}{4^{r}}+\ldots .\right) \\ & =z\left(1+\frac{2 z}{2^{r}}+\frac{3 z^{2}}{3^{r}}+\frac{4 z^{3}}{4^{r}}+\ldots . .\right) \end{aligned} $$ $$ \begin{aligned} & =\left(z+\frac{z^{2}}{2^{r-1}}+\frac{z^{3}}{3^{r-1}}+\frac{z^{4}}{4^{r-1}}+\ldots . .\right) \\ & =g_{r-1}(z) \end{aligned} $$ \section*{Method-2} $$ \begin{aligned} & z \frac{\partial}{\partial z} g_{r}(z)=z \frac{\partial}{\partial z} \sum \frac{z^{n}}{n^{r}} \\ & =z \sum \frac{n z^{n-1}}{n^{r}} \\ & =z \sum \frac{z^{n-1}}{n^{r-1}}=\sum \frac{z^{n}}{n^{r-1}} \\ & =g_{r-1}(z) \end{aligned} $$ \section*{Prove that $z \dfrac{\partial}{\partial z} f_{r}(z)=f_{r-1}(z)$} \section*{Proof: Method -1} $$ \begin{aligned} z \frac{\partial}{\partial z} f_{r}(z)&=z \frac{\partial}{\partial z}\left(z-\frac{z^{2}}{2^{r}}+\frac{z^{3}}{3^{r}}-\frac{z^{4}}{4^{r}}+\ldots .\right) \\ & =z\left(1-\frac{2 z}{2^{r}}+\frac{3 z^{2}}{3^{r}}-\frac{4 z^{3}}{4^{r}}+\ldots . .\right) \\ & =\left(z-\frac{z^{2}}{2^{r-1}}+\frac{z^{3}}{3^{r-1}}-\frac{z^{4}}{4^{r-1}}+\ldots . .\right) \\ & =f_{r-1}(z) \end{aligned} $$ Method 2: $$ \begin{aligned} z \frac{\partial}{\partial z} f_{r}(z)&=z \frac{\partial}{\partial z} \sum(-1)^{n+1} \frac{z^{n}}{n^{r}} \\ & =z \sum(-1)^{n+1} \frac{n z^{n-1}}{n^{r}} \\ & =z \sum(-1)^{n+1} \frac{z^{n-1}}{n^{r-1}} \\ & =\sum(-1)^{n+1} \frac{z^{n}}{n^{r-1}} \\ & =f_{r-1}(z) \end{aligned} $$ hence : $$z \dfrac{\partial}{\partial z} f_{5 / 2}(z)=f_{3 / 2}(z) \& z \dfrac{\partial}{\partial z} f_{3 / 2}(z)=f_{1 / 2}(z)$$ \section*{ Prove that $\left[\frac{\partial}{\partial T} g_{3 / 2}(z)\right]_{V}=-\frac{3}{2 T} g_{3 / 2}(z)$} Proof: $$ \begin{gathered} {\left[\frac{\partial}{\partial T} g_{3 / 2}(z)\right]_{V}=\frac{\partial}{\partial T} \frac{N \lambda^{3}}{V}} \\ =\frac{N}{V} \frac{\partial}{\partial T} \lambda^{3} \\ =\frac{N}{V} 3 \lambda^{2} \frac{d \lambda}{d T} \\ =\frac{N}{V} 3 \lambda^{2} \frac{d}{d T}\left(\frac{h}{\sqrt{2 \pi m k T}}\right) \\ =\frac{N}{V} 3 \lambda^{2} \frac{h}{\sqrt{2 \pi m k}}\left(-\frac{1}{2} T^{-3 / 2}\right) \\ =-\frac{3 N}{2 V} \lambda^{2} \frac{h}{\sqrt{2 \pi m k T}} \frac{1}{T}=-\frac{3}{2 T} \frac{N \lambda^{3}}{V} \\ =-\frac{3}{2 T} g_{3 / 2}(z) \end{gathered} $$ \section*{ Prove that : $\dfrac{1}{z}\left(\dfrac{\partial z}{\partial T}\right)_{V}=-\dfrac{3}{2 T} \dfrac{g_{3 / 2}(z)}{g_{1 / 2}(z)}$} Proof : Already we have proved $$ \begin{aligned} \frac{\partial}{\partial T} g_{3 / 2}(z)&=-\frac{3}{2 T} g_{3 / 2}(z) \\ \Rightarrow\left(\frac{\partial}{\partial z} \frac{\partial z}{\partial T}\right)_{V} g_{3 / 2}(z) \frac{\partial z}{\partial T}&=-\frac{3}{2 T} g_{3 / 2}(z) \\ \Rightarrow\left(\frac{\partial}{\partial z} g_{3 / 2}(z)\right)_{V}\left(\frac{\partial z}{\partial T}\right)_{V}&=-\frac{3}{2 T} g_{3 / 2}(z) \\ \Rightarrow z\left(\frac{\partial}{\partial z} g_{3 / 2}(z)\right)_{V} \frac{1}{z}\left(\frac{\partial z}{\partial T}\right)_{V}&=-\frac{3}{2 T} g_{3 / 2}(z) \\ \Rightarrow g_{1 / 2}(z) \frac{1}{z}\left(\frac{\partial z}{\partial T}\right)_{V}&=-\frac{3}{2 T} g_{3 / 2}(z) \\ \Rightarrow \frac{1}{z}\left(\frac{\partial z}{\partial T}\right)_{V}&=-\frac{3}{2 T} \frac{g_{3 / 2}(z)}{g_{1 / 2}(z)} \end{aligned} $$ \section*{Prove that $\dfrac{1}{z}\left(\dfrac{\partial z}{\partial T}\right)_{P}=-\dfrac{5}{2 T} \dfrac{g_{5 / 2}(z)}{g_{3 / 2}(z)}$}\\ Proof : $$ \begin{aligned} \left[\frac{\partial}{\partial T} g_{5 / 2}(z)\right]_{P}&=\frac{\partial}{\partial T}\left(\frac{P}{k T} \lambda^{3}\right)\\ & =\frac{P}{k} \frac{\partial}{\partial T}\left(\frac{1}{T}\left(\frac{h}{\sqrt{2 \pi m k T}}\right)^{3}\right)\\ &=\frac{P}{k} \frac{h}{\sqrt{2 \pi m k}}\left(\frac{-5}{2}\right) T^{-7 / 2} \\ &= -\frac{5 P}{2 k}\left(\frac{h}{\sqrt{2 \pi m k T}}\right)^{3} \frac{1}{T^{2}} \\ &= -\frac{5 P}{2 k T}\left(\frac{h}{\sqrt{2 \pi m k T}}\right)^{3} \frac{1}{T} \\ &= -\frac{5 P}{2 k T} \lambda^{3} \frac{1}{T} \\ & = -\frac{5}{2}\left(g_{5 / 2}(z)\right) \frac{1}{T}\\ \because \frac{P}{k T}&=\frac{g_{5 / 2}(z)}{\lambda^{3}} \\ &= -\frac{5}{2 T} g_{5 / 2}(z) \end{aligned} $$ Thus $$\left[\frac{\partial}{\partial T} g_{5 / 2}(z)\right]_{P}=-\frac{5}{2 T} g_{5 / 2}(z)$$ \\ Now modifying the left hand side term we will have\\ \begin{align} \left[ \frac{\partial}{\partial T} g_{5/2}(z) \right]_P &= \left[ \frac{\partial g_{5/2}(z)}{\partial z} \cdot \frac{\partial z}{\partial T} \right]_P \\ & = -\frac{5}{2T} g_{5/2}(z) \\ \Rightarrow \left[ z \frac{\partial g_{5/2}(z)}{\partial z} \cdot \frac{1}{z} \cdot \frac{\partial z}{\partial T} \right]_P &= -\frac{5}{2T} g_{5/2}(z) \\ \Rightarrow g_{3/2}(z) \cdot \frac{1}{z} \left[ \frac{\partial z}{\partial T} \right]_P &= -\frac{5}{2T} g_{5/2}(z) \\ \Rightarrow \frac{1}{z} \left[ \frac{\partial z}{\partial T} \right]_P &= -\frac{5}{2T} \cdot \frac{g_{5/2}(z)}{g_{3/2}(z)} \end{align} Prove that specific heat of the Ideal Bose Gas is : $$ \frac{C_{V}}{N k}=\frac{15}{4} \frac{g_{5 / 2}(z)}{g_{3 / 2}(z)}-\frac{9}{4} \frac{g_{3 / 2}(z)}{g_{1 / 2}(z)} $$ Proof : $$ \begin{aligned} \frac{C_{V}}{N k}=\frac{1}{N k} \frac{\partial U}{\partial T}&=\dfrac{1}{N k} \frac{\partial}{\partial T}\left(\frac{3}{2} P V\right) \\ & =\frac{3}{2} \frac{\partial}{\partial T}\left(\frac{P}{k} / N / V\right)=\frac{3}{2} \frac{\partial}{\partial T}\left(T \frac{P}{k T} / N / V\right)\\ &=\frac{3}{2} \frac{\partial}{\partial T}\left(\frac{T g_{5 / 2}(z)}{\lambda^{3}} / \frac{g_{3 / 2}(z)}{\lambda^{3}}\right) \\ &=\frac{3}{2} \frac{\partial}{\partial T}\left(\frac{T g_{5 / 2}(z)}{g_{3 / 2}(z)}\right) \\ &=\frac{3}{2}\left(\frac{g_{5 / 2}(z)}{g_{3 / 2}(z)}+T \frac{\partial}{\partial T}\left(\frac{g_{5 / 2}(z)}{g_{3 / 2}(z)}\right)\right) \\ &=\frac{3}{2}\left(\frac{g_{5 / 2}(z)}{g_{3 / 2}(z)}+T \frac{g_{3 / 2}(z) \frac{\partial}{\partial T} g_{5 / 2}(z)-g_{5 / 2}(z) \frac{\partial}{\partial T} g_{3 / 2}(z)}{\left(g_{3 / 2}(z)\right)^{2}}\right) \\ &=\frac{3}{2}\left(\frac{g_{5 / 2}(z)}{g_{3 / 2}(z)}+T \frac{\frac{\partial}{\partial T} g_{5 / 2}(z)}{g_{3 / 2}(z)}-T \frac{g_{5 / 2}(z) \frac{\partial}{\partial T} g_{3 / 2}(z)}{\left(g_{3 / 2}(z)\right)^{2}}\right) \\ &=\frac{3}{2}\left(\frac{5}{2 g_{5 / 2}(z)}+T\left(\frac{1}{2} \frac{\partial z}{\partial T}\right) g_{3 / 2}(z)\right. \\ &=\frac{3}{2}\left(\frac{5 g_{5 / 2}(z)}{2 g_{3 / 2}(z)}+T \frac{g_{5 / 2}(z)}{g_{3 / 2}(z)}+T \frac{\frac{\partial}{\partial T} g_{5 / 2}(z) \frac{1}{z} \frac{\partial z}{\partial T}}{g_{3 / 2}(z)}-T \frac{g_{5 / 2}(z)\left(-\frac{3}{2 T} g_{3 / 2}(z)\right)}{\left(g_{3 / 2}(z)\right)^{2}}\right) \\ \left.\\ &=\frac{3}{2}\left(\frac{5 g_{5 / 2}(z)}{2 g_{3 / 2}(z)}+T \frac{g_{5 / 2}(z)}{g_{3 / 2}(z)}+T \frac{\frac{\partial}{\partial T} g_{5 / 2}(z)}{g_{3 / 2}(z)}+\frac{3}{2} \frac{g_{5 / 2}(z)}{g_{3 / 2}(z)}\right) \\ &=\frac{3}{2}\left(\frac{5 g_{5 / 2}(z)}{2 g_{3 / 2}(z)}+T \frac{\frac{\partial}{\partial T} g_{5 / 2}(z)}{g_{3 / 2}(z)}\right) \\ g_{3 / 2}(z) z & =\frac{3}{2}\left(\frac{5 g_{5 / 2}(z)}{2 g_{3 / 2}(z)}+T\left(-\frac{3}{2 T} \frac{g_{3 / 2}(z)}{g_{1 / 2}(z)}\right)\right) \\ & =\frac{3}{2}\left(\frac{5 g_{5 / 2}(z)}{2 g_{3 / 2}(z)}-\frac{3}{2} \frac{g_{3 / 2}(z)}{g_{1 / 2}(z)}\right) \\ & =\frac{15 g_{5 / 2}(z)}{4 g_{3 / 2}(z)}-\frac{9}{4} \frac{g_{3 / 2}(z)}{g_{1 / 2}(z)} \end{aligned} $$ \textbf{Special Case:} \\ \begin{align*} \text{1. At high temperature:} \quad \frac{C_V}{N k} &= \left( \frac{15}{4} - \frac{9}{4} \right) = \frac{3}{2} \Rightarrow C_V = \frac{3R}{2} \\[1ex] \text{2. At low temperature:} \quad \frac{C_V}{N k} &= \left( \frac{15}{4} \cdot 0 - \frac{9}{4} \cdot 0 \right) = 0 \Rightarrow C_V = 0 \end{align*}

Number of cells with momentum in the range $p$ and $p+d p$ as per BE statistics :

$$g(p) d p=2 \times \frac{V 4 \pi p^{2} d p}{h^{3}}$$

Number of cells with frequency in the range $f$ and $f+d f$ as per BE statistics :

$$ g(f) d f=\frac{8 \pi V}{h^{3}}\left(\frac{h f}{c}\right)^{2} \frac{h}{c} d f$$ $$=\frac{8 \pi V f^{2}}{c^{3}} d f $$

Number of Photons/particles in the momentum range $p$ and $p+d p$ as per BE statistics :

$$ \begin{aligned} & n(p) d p=\frac{g(p) d p}{e^{\beta E}-1}\\ &=\frac{8 \pi p^{2} V d p}{h^{3}\left(e^{\beta E}-1\right)} \\ \Rightarrow n(p) d p &= h^{3}\left(e^{\frac{p^{2}}{2 m k T}}-1\right) \end{aligned} $$

Number of Photons/particles in the momentum range $p$ and $p + dp $ as per BE statistics :

$$n(f) d f=\frac{g(f) d f}{\left(e^{\frac{n f}{k T}}-1\right)}$$ $$\Rightarrow n(f) d f=\frac{8 \pi V f^{2} d f}{c^{3}\left(e^{\frac{h f}{k T}}-1\right)}$$

Energy of Photons in the range $f$ and $f+d f$

\begin{align*} \Rightarrow E(f) d f &=h f n(f) df \\ &=h f \frac{8 \pi V f^{2} d f}{c^{3}\left(e^{\frac{h f}{h t}}-1\right)} \end{align*}

Energy density in the range $f$ and $f+df$

\begin{align*} u(f) d f&=\frac{E(f) d f}{V}\\ &=\frac{8 \pi h f^{3} d f}{c^{3}\left(e^{\frac{H f}{e r}}-1\right)} \end{align*}

Number of Photons/Particles in the wavelength range $\lambda$ and $\lambda+d \lambda $ as per BE statistics :

$$n(\lambda) d \lambda=\frac{8 \pi d \lambda}{\lambda^{4}\left(e^{\dfrac{h c}{\lambda k T}}-1\right)}$$

Energy density in the wavelength range $\lambda$ and $\lambda+d \lambda$

> $$ u(\lambda) d \lambda=\frac{8 \pi h c d \lambda}{\lambda^{5}\left(e^{\frac{h c}{\lambda k t}}-1\right)} $$

Specialc case (1) when $\lambda$ is very long

$$ u(\lambda) d \lambda=\frac{8 \pi h c d \lambda}{\lambda^{5}\left(1+\frac{h c}{\lambda k T}-1\right)}$$ $$=\frac{8 \pi k T}{\lambda^{4}} d \lambda $$

Specialc case (2) when $\lambda$ is very short

\[ \begin{align} u(\lambda) \, d\lambda &= \frac{8 \pi h c \, d\lambda}{\lambda^{5} \left(e^{\frac{h c}{k k T}} - 1\right)} \\ &= \frac{8 \pi h c \, d\lambda}{\lambda^{5} \left(e^{\frac{h c}{k k T}}\right)} \\ &= \frac{8 \pi h c}{\lambda^{4}} \, e^{- \left( \frac{h c}{k c t} \right)} \, d\lambda \end{align} \]

Wein's Displacement Law :

It states that as temperature of black body is raised , the maximu intensity of the radiation emmitted by it is displaced towards shorter side such that the product of wavelength at which maximum intensity is radiated and corresponding temperature is constant . Mathematically $\lambda_{m} T=b$\\ Where $b$ is wien's constant $=b=2 \cdot 898 \times 10^{-5} \mathrm{~m}-K$.

Derivation of Planck's Law :

When intensity of radiation is maximum $$ \begin{aligned} & \frac{d}{d \lambda}\left(E_{\lambda}\right)=\frac{d}{d \lambda}\left(\frac{8 \pi h c \lambda^{-5}}{e^{\frac{h c}{\lambda k T}}-1}\right)=0 \\ & \Rightarrow-5 \lambda^{-6}\left(e^{\frac{h c}{\lambda k T}}-1\right)^{-1}+\lambda^{-5}\left(e^{\frac{h c}{\lambda k T}}-1\right)^{-2} e^{\frac{h c}{\lambda k T}} \frac{h c}{k T} \lambda^{-2}=0 \\ & \Rightarrow-\frac{5}{\lambda^{6}} \frac{1}{e^{\frac{h c}{\lambda k T}}-1}+\frac{1}{\lambda^{7}} \frac{h c}{k T\left(e^{\frac{h c}{\lambda k T}}-1\right)^{2}}=0 \\ & \Rightarrow \frac{5}{\lambda^{6}}=\frac{1}{e^{\frac{h c}{\lambda k T}}-1}\left(\frac{h c}{k T}\right) \frac{1}{\lambda^{7}} \\ & \Rightarrow e^{\frac{h c}{\lambda k T}}-1=\frac{e^{\frac{h c}{\lambda k T}} h c}{5 \lambda k T} \\ & \Rightarrow e^{\frac{h c}{\lambda k T}}\left(1-\frac{h c}{5 \lambda k T}\right)=1 \\ & \Rightarrow e^{x}\left(1-\frac{x}{5}\right)=1 \\ & \Rightarrow x=4.965 \\ & \Rightarrow \frac{h c}{\lambda k T}=4.965 \\ & \Rightarrow \lambda_{m} T=\frac{4.965 k}{h c}=b=2 \cdot 898 \times 10^{-3} m-K \\ & \therefore \lambda_{m} \alpha \frac{1}{T} \end{aligned} $$

Stefan's Law :

It states that the total energy radiated per unit area per unit time by a black body at any temperature T is directly proportional to the fourth power of it's absolute temperature . Mathematically: $E \propto T^{4}$ $$ \begin{aligned} & \text { or } \\ & E / A=\sigma T^{4} \quad\left(\sigma=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2}-K^{4}\right. \text { called as Stefan's Constant.) } \end{aligned} $$

Proof :

Energy density in the frequency range $f$ and $f+d f$ $$ \begin{aligned} \Rightarrow u(f) d f&=\frac{E(f) d f}{V}\\ &=\frac{8 \pi h f^{3} d f}{c^{3}\left(e^{\frac{h f}{k T}}-1\right)} \\ & \left.=8 \pi h\left(\frac{k T}{h}\right)^{3} \frac{k T}{h} \int_{0}^{\infty} \frac{x^{3} d x}{c^{3}\left(e^{x}-1\right)} \quad \text { Putting } \frac{h f}{k T}=x \Rightarrow f=x k T / h, \Rightarrow d f=d x k T / h\right) \\ & =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} \frac{x^{3} d x}{\left(e^{x}-1\right)} \\ & =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} \frac{x^{3} e^{-x} d x}{e^{-x}\left(e^{x}-1\right)} \\ & =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} \frac{x^{3} e^{-x} d x}{\left(1-e^{-x}\right)} \\ & =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} x^{3} e^{-x}\left(1-e^{-x}\right)^{-1} d x \\ & =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty} x^{3} e^{-x}\left(1+e^{-x}+e^{-2 x}+e^{-3 x}+\ldots\right) d x \\ & =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}} \int_{0}^{\infty}\left(x^{3} e^{-x}+x^{3} e^{-2 x}+x^{3} e^{-3 x}+x^{3} e^{-4 x}+\ldots\right) d x \\ & =\frac{8 \pi k^{4} T^{4}}{c^{3} h^{3}}\left(6\left(1+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\frac{1}{4^{4}}+\ldots\right)\right) \text { using } \int_{0}^{\infty} x^{3} e^{-a x} d x=\frac{6}{a^{4}} \\ & =\left(\frac{8 \pi^{5} k^{4}}{15 c^{3} h^{3}} T^{4}\right. \\ & \therefore u_{f} \alpha T^{4} \end{aligned} $$

Determination of Stefan's Constant :

$$ \begin{aligned} & \Rightarrow \frac{1}{4} c u_{f}=\sigma T^{4} \\ & \Rightarrow \frac{1}{4} c \frac{8 \pi^{5} k^{4}}{15 c^{3} h^{3}} T^{4}=\sigma T^{4} \\ & \Rightarrow \sigma=\frac{2 \pi^{5} k^{4}}{15 c^{2} h^{3}}=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2}-K \end{aligned} $$

Number of Photons in Volume $\mathbf{V}$ at any temperature T is :

$$ \begin{aligned} N&=\int_{0}^{\infty} n(f) d f \\ & =\int_{0}^{\infty} \frac{8 \pi V f^{2} d f}{c^{3}\left(e^{\frac{b f}{k T}}-1\right)} \\ & =\frac{8 \pi V}{c^{3}} \int_{0}^{\infty} \frac{f^{2} d f}{\left(e^{\frac{H f}{h_{T}}}-1\right)} \\ & =\frac{8 \pi}{c^{3}}\left(\frac{k T}{h}\right)^{2 \infty} \frac{x^{2} d x}{\left(e^{x}-1\right)}\left(\frac{k T}{h}\right) \quad \text { Substituiting } \frac{h f}{k T}=x \\ & =8 \pi\left(\frac{k T}{h c}\right) \int_{0}^{3 \infty} \frac{x^{2} d x}{\left(e^{x}-1\right)} \\ & =8 \pi\left(\frac{k T}{h c}\right) \int_{0}^{3 \infty} \frac{x^{2} e^{-x} d x}{e^{-x}\left(e^{x}-1\right)} \\ & =8 \pi\left(\frac{k T}{h c}\right)^{3 \infty} \int_{0}^{2} e^{-x}\left(1-e^{-x}\right)^{-1} d x \\ & =8 \pi\left(\frac{k T}{h c}\right)^{3 \infty} \int_{0}^{2} x^{2} e^{-x}\left(1+e^{-x}+e^{-2 x}+e^{-3 x}+e^{-4 x}+\ldots\right) d x \\ & =8 \pi\left(\frac{k T}{h c}\right)^{3 \infty} \int_{0}^{2}\left(x^{2} e^{-x}+x^{2} e^{-2 x}+x^{2} e^{-3 x}+x^{2} e^{-4 x}+x^{2} e^{-5 x}+\ldots\right) d x \\ & =8 \pi\left(\frac{k T}{h c}\right)^{3} \times 2 \times \sum \frac{1}{n^{3}} \quad \text { using } \int_{0}^{\infty} x^{m} e^{-a x} d x=\frac{m!}{n^{m+1}} \end{aligned} $$

WHAT IS BOSE-EINSTEIN CONDENSATION ? DERIVE BOSE TEMPERATURE

Ans : It is the fifth state of matter where all bosons are compressed share a common minimum energy state.\\ When BOSONS are cooled to extemely low temperature they fall or condednse into a lowest accessible quantum quantum state. This state is called BE condensate.\\ This Phenomenon can explain Superfluidity of liquid Helim.

EXPLANATION :

\[ \begin{align} f(E) &= \frac{1}{e^{\left(\frac{\alpha + \beta E}{kT}\right)} - 1} \\ &= \frac{1}{e^{\alpha} e^{\frac{\beta E}{kT}} - 1} \\ &= \frac{1}{A \, e^{\frac{\beta E}{kT}} - 1} \end{align} \]

Thus, \[ f(E) = \frac{1}{A \, e^{\frac{\beta E}{kT}}} \] for a Maxwell-Boltzmann gas at \( A \gg 1 \), where \[ A = e^{\alpha} = \frac{V}{N} \left( \frac{2\pi m k T}{h^2} \right)^{3/2} \]

For a Bose-Einstein Gas:

• \( A \geq 1 \), so the chemical potential is negative.
• \( A > 1 \) when \( T > T_B \)   and   \( A = 1 \) when \( T \leq T_B \).

\[ \begin{align} N_{\text{Total}} &= N_{\text{Ground}} + N_{\text{Excited}} \\ N_{\text{Total}} &= N_{\text{Excited}} \quad \text{for} \quad T > T_B \\ N_{\text{Total}} &= N_{\text{Ground}} \quad \text{for} \quad T < T_B \\ N_{\text{Total}} &\approx N_{\text{Excited}} \quad \text{for} \quad T = T_B \end{align} \]

\[ \Rightarrow N_{\text{Ground}} = N_{\text{Total}} - N_{\text{Excited}} = N_{\text{Total}} \left( 1 - \left( \frac{T}{T_B} \right)^{3/2} \right) \]

\[ \begin{align} A &= 1 = \frac{V}{N} \left( \frac{2\pi m k T}{h^2} \right)^{3/2} \\ \Rightarrow N &= V \left( \frac{2\pi m k T}{h^2} \right)^{3/2} \\ \Rightarrow \frac{N_{\text{Excited}}}{N_{\text{Total}}} &= \left( \frac{T}{T_B} \right)^{3/2} \\ \Rightarrow N_{\text{Ground}} &= N_{\text{Total}} \left( 1 - \left( \frac{T}{T_B} \right)^{3/2} \right) \end{align} \]

This shows that when \( T < T_B \), a large number of particles accumulate in the ground state. At \( T = 0\,K \), all particles are in the ground state.

(This was first achieved in a laboratory in 1995 by Eric Cornell and Carl Wieman at the University of Colorado at Boulder (NIST–JILA lab), for which they were awarded the Nobel Prize in 2001.)

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WHAT IS BOSE-EINSTEIN CONDENSATION?

Bose-Einstein Condensation and Bose Temperature

Definition:

Bose-Einstein Condensation (BEC) is a quantum phenomenon in which a gas of bosons at low density, when cooled to a sufficiently low temperature, locks together to share a common minimum energy state.\\ This state of matter is called the \textbf{Bose-Einstein Condensate} — often referred to as the \textbf{fifth state of matter}.\\ In BEC, the bosons condense in momentum space, with their wavefunctions overlapping and becoming indistinguishable from each other. The theoretical prediction was given by \textbf{Satyendra Nath Bose} in 1924 and later extended by \textbf{Albert Einstein}. The phenomenon was experimentally observed in laboratories in \textbf{1995} (Rubidium-87) and in \textbf{2001} (Sodium-23).\\ This phenomenon explains key quantum behaviors like:

1. Superfluidity in liquid Helium
2. Superconductivity in materials

Derivation of Bose Temperature \( T_B \)

The number of bosons in the excited states is given by: \begin{align*} N_e &= \int_0^\infty \frac{g(E)}{e^{\beta(E - \mu)} - 1} \, dE \end{align*} At the \textbf{critical temperature} (Bose temperature), the chemical potential \( \mu = 0 \). So: \begin{align*} N &= N_e = \int_0^\infty \frac{g(E)}{e^{\beta E} - 1} \, dE \end{align*} The density of states in 3D is: \begin{align*} g(E) &= 2\pi V \left( \frac{2m}{h^2} \right)^{3/2} E^{1/2} \end{align*} So: \begin{align*} N &= \int_0^\infty \frac{2\pi V \left( \frac{2m}{h^2} \right)^{3/2} E^{1/2}}{e^{\beta E} - 1} \, dE \\ &= V \left( \frac{2\pi m k T}{h^2} \right)^{3/2} \int_0^\infty \frac{u^{1/2}}{e^u - 1} \, du \quad (\text{Substitute } u = \beta E) \\ &= V \left( \frac{2\pi m k T}{h^2} \right)^{3/2} \Gamma(3/2) \zeta(3/2) \end{align*} Since \( \Gamma(3/2) = \frac{\sqrt{\pi}}{2} \), and \( \zeta(3/2) \approx 2.612 \), we have: \begin{align*} N &= V \left( \frac{2\pi m k T}{h^2} \right)^{3/2} \zeta(3/2) \end{align*} Solving for \( T \) gives the \textbf{Bose temperature} \( T_B \): \begin{align*} T_B &= \frac{h^2}{2\pi m k} \left( \frac{N}{V \zeta(3/2)} \right)^{2/3} \end{align*}

Proof of BEC Below \( T_B \)

When \( T < T_B \), the number of particles in excited states becomes: \begin{align*} N_e &= N \left( \frac{T}{T_B} \right)^{3/2} \end{align*} Then, the number of particles in the ground state is: \begin{align*} N_0 &= N - N_e = N \left[ 1 - \left( \frac{T}{T_B} \right)^{3/2} \right] \end{align*}

Conclusion:

As \( T \to 0 \), almost all bosons fall into the ground state: \[ \lim_{T \to 0} N_0 = N \] This marks the formation of a Bose-Einstein Condensate — a macroscopic occupation of the lowest energy state, a purely quantum phenomenon.\\

Derive Bose Temperature and Prove that below Bose temperature The number of bosons will increase in the ground state to show the Bose-Einstein Condensation

ANS :\\ It is a quantum phenomenon in which a gas of bosons of low density when cooled to a suffiencently low temperature locked together to share a common minimum energy state. This state of matter is called the Bose-Einstein Condensate. It is called as the 5th state of matter. The bosons condense in the momentum space with their wavefunctions overlapping with each other. The theory for the possibility of such a state of matter was given by SN bose in 1924 but was made practically possible in the laboratory after 71 years in 1995 and 2001. This Phenomenon can explain Superfluidity of liquid Helim \& the phenomenon of superconductivity

BOSE EINSTEIN GAS AT ANY TEMPERATURE T

$$ \begin{aligned} N&=\int_{0}^{\infty} f(E) g(E) d E \\ & =\int_{0}^{\infty} \frac{1}{e^{(\alpha+\beta E)}-1} g(E) d E \\ & =\int_{0}^{\infty} \frac{1}{e^{\alpha} e^{\beta E}-1}\left(2 \pi V\left(\frac{2 m}{h^{2}}\right)^{3 / 2} E^{1 / 2}\right) d E \\ & =\int_{0}^{\infty} \frac{1}{z^{-1} e^{\beta E}-1}\left(2 \pi V\left(\frac{2 m}{h^{2}}\right)^{3 / 2} E^{1 / 2}\right) d E \\ & =2 \pi V\left(\frac{2 m}{h^{2}}\right)^{3 / 2}(k T)^{1 / 2}(k T) \int_{0}^{\infty} \frac{u^{1 / 2}}{z^{-1} e^{u}-1} d u \\ & (\beta E=u \Rightarrow E=u k T) \end{aligned} $$ $$ \begin{aligned} & =2 \pi V\left(\frac{2 m k T}{h^{2}}\right)^{3 / 2} \frac{\sqrt{\pi}}{2} \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{u^{3 / 2-1}}{z^{-1} e^{u}-1} d u \\ & =V\left(\frac{2 m \pi k T}{h^{2}}\right)^{3 / 2} \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{u^{3 / 2-1}}{z^{-1} e^{u}-1} d u \\ & =V\left(\frac{2 m \pi k T}{h^{2}}\right)^{3 / 2} \frac{1}{\Gamma(3 / 2)} \int_{0}^{\infty} \int_{z^{-1} e^{u}-1}^{u^{3 / 2-1}} d u \\ & =V\left(\frac{2 m \pi k T}{h^{2}}\right)^{3 / 2} F_{3 / 2}(z) \\ & =\frac{V}{\left(\frac{h^{2}}{2 m \pi k T}\right)^{3 / 2}} F_{3 / 2}(z) \\ & =\frac{V}{\lambda^{3}} F_{3 / 2}(z) \end{aligned} $$ Hence $$F_{n}(z)=\dfrac{1}{\Gamma(n)} \int_{0}^{\infty} \dfrac{u^{n-1}}{z^{-1} e^{u}-1} d u$$

SIMPLIFICATION OF THE FUNCTION IN TERMS OF REIMAN ZETA FUNCTION

$$ \begin{aligned} & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} \frac{u^{n-1}}{z^{-1} e^{u}\left(1-\frac{1}{z^{-1} e^{u}}\right)} d u \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} \frac{u^{n-1}}{z^{-1} e^{u}}\left(1-\frac{1}{z^{-1} e^{u}}\right)^{-1} d u \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} z \frac{u^{n-1}}{e^{u}}\left(1+\frac{z}{e^{u}}+\frac{z^{2}}{e^{2 u}}+\ldots\right) d u \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} z \frac{u^{n-1}}{e^{u}} \sum_{r=1}^{r=\infty}\left(\frac{z}{e^{u}}\right)^{r-1} d u \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} \sum_{r=1}^{r=\infty}\left(\frac{z}{e^{u}}\right)^{r} u^{n-1} d u \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} \sum_{r=1}^{r=\infty} z^{r} e^{-r u} u^{n-1} d u \quad \text { Taking } r u=x \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} \sum_{r=1}^{r=\infty} z^{r} e^{-x}\left(\frac{x}{r}\right)^{n-1} \frac{d x}{r}\\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} \sum_{r=1}^{r=\infty} z^{r} e^{-x} x^{n-1} \frac{d x}{r^{n}} \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} e^{-x} x^{n-1} d x \sum_{r=1}^{r=\infty} \frac{z^{r}}{r^{n}} \\ & =\sum_{r=1}^{r=\infty} \frac{z^{r}}{r^{n}} \end{aligned} $$ Hence, \begin{align*} F_{n}(z)&=\sum_{r=1}^{r=\infty} \frac{z^{r}}{r^{n}}\\ \Rightarrow N&=\frac{V}{\lambda^{3}} F_{3 / 2}(z) \& \end{align*} Since\begin{align*} U &= \int_{0}^{\infty} f(E) g(E) E \, dE \\ &= \int_{0}^{\infty} \frac{1}{z^{-1} e^{\beta E}-1} \cdot 2 \pi V \left( \frac{2m}{h^2} \right)^{3/2} E \cdot \frac{1}{2\sqrt{E}} \cdot E \, dE \\ &= \pi V \left( \frac{2m}{h^2} \right)^{3/2} \int_{0}^{\infty} \frac{E^{3/2}}{z^{-1} e^{\beta E} - 1} \, dE \\ &= \pi V \left( \frac{2m}{h^2} \right)^{3/2} \frac{3\sqrt{\pi}}{2} \left( \frac{1}{\frac{3\sqrt{\pi}}{2}} \int_{0}^{\infty} \frac{E^{3/2}}{z^{-1} e^{\beta E} - 1} \, dE \right) \\ &= V \left( \frac{2\pi m}{h^2} \right)^{3/2} \frac{3}{2} \left( \frac{1}{\frac{3\sqrt{\pi}}{2}} \int_{0}^{\infty} \frac{u^{3/2}}{z^{-1} e^{u} - 1} \, du \cdot (kT)^{3/2} \cdot kT \right) \\ &= V \left( \frac{2\pi m kT}{h^2} \right)^{3/2} kT \cdot \frac{3}{2} \left( \frac{1}{\frac{3\sqrt{\pi}}{2}} \int_{0}^{\infty} \frac{u^{3/2}}{z^{-1} e^{u} - 1} \, du \right) \\ &= \frac{V}{\left( \frac{h^2}{2\pi m kT} \right)^{3/2}} kT \cdot \frac{3}{2} \left( \frac{1}{\frac{3\sqrt{\pi}}{2}} \int_{0}^{\infty} \frac{u^{3/2}}{z^{-1} e^{u} - 1} \, du \right) \\ &= \frac{V}{\lambda^3} \cdot \frac{3kT}{2} \cdot \frac{1}{\Gamma(5/2)} \int_{0}^{\infty} \frac{u^{5/2 - 1}}{z^{-1} e^{u} - 1} \, du \\ &= \frac{V}{\lambda^3} \cdot \frac{3kT}{2} \cdot F_{5/2}(z) \\ U &= \frac{3kT V}{2 \lambda^3} F_{5/2}(z) \end{align*}

BOSE-EINSTEIN GAS AT $T \geq T_{B}$

At sufficeinttly high temperature

1. All the particles are in the exicited state
2. The number of paricles in the ground state with energy $E=0$ is neglibly small
3. $A$ is quite larger than 1

So neglecting the higher terms and considering first two terms.\\ \begin{align*} \Rightarrow \frac{U}{N} &= \frac{3kT}{2} \cdot \frac{f_{5/2}(z)}{f_{3/2}(z)} \\ &= \frac{3kT}{2} \cdot \frac{z + \frac{z^2}{2^{5/2}} + \cdots}{z + \frac{z^2}{2^{3/2}} + \cdots} \\ &= \frac{3kT}{2} \cdot \left( \frac{1 + \frac{z}{4\sqrt{2}}}{1 + \frac{z}{2\sqrt{2}}} \right) \\ &= \frac{3kT}{2} \cdot \left(1 + \frac{z}{4\sqrt{2}} \right) \left(1 + \frac{z}{2\sqrt{2}} \right)^{-1} \\ &= \frac{3kT}{2} \cdot \left(1 - \frac{z}{4\sqrt{2}} \right) \\ &= \frac{3kT}{2} \left(1 - \frac{1}{4\sqrt{2}} \cdot \frac{N}{V} \left( \frac{h^2}{2\pi m kT} \right)^{3/2} \right) \\ \Rightarrow U &= \frac{3NkT}{2} \left(1 - \frac{1}{4\sqrt{2}} \cdot \frac{N}{V} \left( \frac{h^2}{2\pi m kT} \right)^{3/2} \right) \\ &= \frac{3Nk}{2} \left(T - \frac{1}{4\sqrt{2}} \cdot \frac{N}{V \sqrt{T}} \left( \frac{h^2}{2\pi m k} \right)^{3/2} \right) \end{align*}

A. Specific heat :

The first term is for the classical ideal gas . Second term is for the first order correction due to the deviation from the classical behaviour. When the temperature will be sufficiently large the correction terms will be very small.\\

B. Pressure

3. BOSE-EINSTEIN GAS AT TEMPERATURE $ T

Behavious of Chemical Potential;

$$ \begin{aligned} & NHence,=\sum n_{i}=n_{0}+n_{1}+n_{2}+n_{3}+\ldots \\ & =\frac{g_{0}}{A e^{\beta E_{0}}-1}+\frac{g_{1}}{A e^{\beta E_{1}}-1}+\frac{g_{2}}{A e^{\beta E_{2}}-1}+\ldots \quad \text { Where } \mathbf{g}=\text { degeneracy of the level. } \\ & =\frac{g_{0}}{A-1}+\frac{g_{1}}{A e^{\beta E_{1}}-1}+\frac{g_{2}}{A e^{\beta E_{2}}-1}+\ldots \end{aligned} $$

The first term is for the number of particles in the ground state. The above expression is valid for all temperatures.Since number of particles in a state at any temperature can not be negative, $$ A \geq 1 \Rightarrow e^{-\mu / k T} \geq 1 \Rightarrow-\mu / k T \geq 0 \Rightarrow \mu \leq 0 $$ since $ A=e^{\alpha}=e^{-\mu / k T}\right)$ at sufficiently low temperature $\mu \approx-10^{-38} \mathrm{erg}$ which is very small and can be taken to be zero.Thus the chemical potential can be negative or zero for the bosons.\\ it is zero for photons. It can be positive and negative for Fermions and as well as for Maxwell gas. The chemical potential at temperatute greater than the Bose temperature is negative. So $$ \text { A>1, when } T>T_{B} $$ But when the temperature is lowered below $T_{B}$, the chemical potential becomes 0 . So A=1, when $TCases

\textbf{Case 1:} When \( T > T_B \)\\ The system is entirely in the gaseous phase. All the particles occupy excited states only.\\ \textbf{Case 2:} When \( T = T_B \)\\ \begin{align*} N_{\text{ground}} &= N_{\text{total}} \left( 1 - \left( \frac{T_B}{T_B} \right)^{3/2} \right) \\ &= N_{\text{total}} (1 - 1) = 0 \end{align*} So, no particles occupy the ground state. All particles are in the excited states.\\

Case 3:

When \( T < T_B \)\\ The system may be regarded as a mixture of two phases:
1. A Bose-Einstein condensate (ground state occupation).
2. A thermal gas (excited states).
a gaseous phase in the exicited states with number of particles as $N_{e}=N_{\text {Total }}\left(\frac{T}{T_{B}}\right)^{3 / 2}$
a condensed phase with number of particles in the ground state $N_{0}=N_{\text {total }}-N_{e} \ll N_{\text {total }}$
A large number of particles are in the ground state and some are in the excites state too.

Case 4

: When $\mathrm{T}=0\left(T \rightarrow 0, N_{0} \rightarrow N_{\text {total }} \rightarrow N\right)$\\ Since $N_{\text {ground }}=N_{\text {Total }}\left(1-\left(\frac{0}{T_{B}}\right)^{3 / 2}\right)=N_{\text {total }}$\\ All the particles occupy the ground state. The condesation of the particles takes place in the momentum space not like the condensation of gas in real space. All the wavefunctons of particles overlap with each other due to increasing wavlength. The state is called Bose-Einstein Condensate and the process as Bose-Einstein Condensation\\ ( 1924 theoritical predication and practical realisation in laboratory in 1995 and 2001)

Graphical Explanation

Fermi-Dirac Distribution and Chemical Potential

The Fermi-Dirac distribution is given by:

\( f(\varepsilon) = \frac{1}{e^{(\varepsilon - \mu)/k_B T} + 1} \)

where:

• \( \mu \): Chemical potential
• \( k_B \): Boltzmann's constant
• \( T \): Temperature

Total Number of Fermions

The total number of particles is given by:

\( N = \int_0^\infty g(\varepsilon) f(\varepsilon) \, d\varepsilon \)

For a 3D free electron gas, the density of states is:

\( g(\varepsilon) = \frac{V}{2\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt{\varepsilon} \)

So:

\( N = \int_0^\infty \frac{g(\varepsilon)}{e^{(\varepsilon - \mu)/k_B T} + 1} \, d\varepsilon \)

At T = 0 K (Zero Temperature)

\( f(\varepsilon) = \begin{cases} 1, & \varepsilon < \mu \\ 0, & \varepsilon > \mu \end{cases} \)

So the number of particles becomes:

\( N = \int_0^{\mu} g(\varepsilon) \, d\varepsilon \)

Substitute \( g(\varepsilon) \):

\( N = \int_0^{\mu} \frac{V}{2\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt{\varepsilon} \, d\varepsilon \)

After integration:

\( N = \frac{V}{2\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \cdot \frac{2}{3} \mu^{3/2} \)

Solving for \( \mu \):

\( \mu = \left( \frac{3N}{V} \cdot \frac{\pi^2 \hbar^3}{(2m)^{3/2}} \right)^{2/3} \)

This value at \( T = 0 \) is called the Fermi energy:

\( \mu(T = 0) = E_F \)

At Low Temperatures

Using Sommerfeld expansion, for low \( T \):

\( \mu(T) \approx E_F \left[ 1 - \frac{\pi^2}{12} \left( \frac{k_B T}{E_F} \right)^2 \right] \)