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Unit 3
Ideal Bose Gas
Ideal Bose Gas: Introduction
Properties of Ideal Bose Gas
- An ideal Bose gas is a quantum-mechanical phase of matter, analogous to a classical ideal gas.
- It has integral spins.
- Its constituents are called bosons.
- Its examples are photons, which travel at the speed of light and are quanta of electromagnetic radiation.
- Bosons obey Bose-Einstein statistics.
- Bosons are neutral.
- 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:
- The rest mass of a photon is zero, whereas a gas molecule has a finite mass.
- Photons in a photon gas are indistinguishable, while molecules in a classical ideal gas are distinguishable.
- A photon gas obeys Bose-Einstein statistics, whereas a classical ideal gas obeys Maxwell-Boltzmann statistics.
- Photons have integral spin, while molecules are spinless.
- Photons in a photon gas travel at the speed of light, but gas molecules travel with speeds ranging from 0 to infinity.
- Photons are not conserved during collisions, whereas gas molecules are conserved during collisions.
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:
$$
\begin{aligned}
S & =k \ln W_{\max } \\
\Rightarrow \frac{S}{k} & =\ln W_{\max }\\
&=\sum \ln \left(\frac{\left(g_{i}+n_{i}-1\right)!}{n_{i}!\left(g_{i}-1\right)!}\right) \approx \sum \ln \left(\frac{\left(g_{i}+n_{i}\right)!}{n_{i}!\left(g_{i}\right)!}\right) \because g_{i}+n_{i} \gg 1 \& g_{i} \gg 1\\
& =\sum\left[\left(g_{i}+n_{i}\right) \ln \left(g_{i}+n_{i}\right)-g_{i}-n_{i}-n_{i} \ln n_{i}+n_{i}-g_{i} \ln g_{i}+g_{i}\right] \\
& =\sum\left[g_{i} \ln \left(g_{i}+n_{i}\right)+n_{i} \ln \left(g_{i}+n_{i}\right)-g_{i}-n_{i}-n_{i} \ln n_{i}+n_{i}-g_{i} \ln g_{i}+g_{i}\right] \\
& =\sum\left[g_{i} \ln \left(g_{i}+n_{i}\right)-g_{i} \ln g_{i}+n_{i} \ln \left(g_{i}+n_{i}\right)-n_{i} \ln n_{i}\right] \\
& =\sum\left[n_{i} \ln \left(\frac{g_{i}+n_{i}}{n_{i}}\right)+g_{i} \ln \left(\frac{g_{i}+n_{i}}{g_{i}}\right)\right] \\
& =\sum\left[n_{i} \ln \left(\frac{g_{i}}{n_{i}}+1\right)+g_{i} \ln \left(1+\frac{n_{i}}{g_{i}}\right)\right] \\
& =\sum\left[n_{i} \ln \left(e^{\alpha+\beta E_{i}}-1+1\right)+g_{i} \ln \left(1+\frac{1}{e^{\alpha+\beta E_{i}}-1}\right)\right] \\
& =\sum\left[n_{i}\left(\alpha+\beta E_{i}\right)+g_{i} \ln \left(1-\frac{1}{e^{\alpha+\beta E_{i}}-1}\right)\right] \\
& =\sum\left[n_{i}\left(\beta E_{i}-\ln z\right)+g_{i} \ln \left(\frac{e^{\alpha+\beta E_{i}}}{e^{\alpha+\beta E_{i}}-1}\right)\right] \because e^{\alpha}=z^{-1}, \alpha=-\ln z \\
& =\sum\left[n_{i}\left(\beta E_{i}-\ln z\right)+g_{i} \ln \left(\frac{1}{1-e^{-\alpha-\beta E_{i}}}\right)\right] \\
& =\sum\left[n_{i}\left(\beta E_{i}-\ln z\right)-g_{i} \ln \left(1-e^{-\alpha-\beta E_{i}}\right)\right] \\
& =\sum\left[\frac{g_{i}}{e^{\alpha+\beta E_{i}}-1}\left(\beta E_{i}-\ln z\right)-g_{i} \ln \left(1-z e^{-\beta E_{i}}\right)\right] \\
& =\sum g_{i}\left[\frac{\left(\beta E_{i}-\ln z\right)}{z^{-1} e^{\beta E_{i}}-1}-\ln \left(1-z e^{-\beta E_{i}}\right)\right]
\end{aligned}
$$
Prove that for an Ideal Bose gas the equation of State is given by :
\begin{align}
\dfrac{P V}{k T} &= \log Z_{BE} \\
&= -\sum \log \left(1 - z e^{-\beta E_i}\right) \\
&= \dfrac{V}{\lambda^{3}} \, g_{5/2}(z)
\end{align}
where
\[ g_r(z) = z + \dfrac{z^2}{2^r} + \dfrac{z^3}{3^r} + \dfrac{z^4}{4^r} + \ldots
= -\sum_{n=1}^{\infty} (-1)^{2n+1} \dfrac{z^n}{n^r}\]
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{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}
To Prove $ \dfrac{P}{kT}=\dfrac{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}}$
$$
\begin{aligned}
& =-\frac{4 \pi V}{h^{3}} \sum_{1}^{n}(-1)^{2 n+1} \frac{z^{n}}{n} \int_{0}^{\infty} e^{-x}\left(\frac{2 m k T x}{n}\right)\left(\frac{\sqrt{2 m k T} d x}{2 \sqrt{n} \sqrt{x}}\right) \\
& =-\frac{4 \pi V}{h^{3}} \sum_{1}^{n}(-1)^{2 n+1} \frac{z^{n}}{n} \int_{0}^{\infty} e^{-x} x^{1 / 2} d x\left(\frac{2 m k T}{n}\right)^{3 / 2} \frac{1}{2} \\
& =-\frac{4 \pi V}{h^{3}} \sum_{1}^{n}(-1)^{2 n+1} \frac{z^{n}}{n} \Gamma(3 / 2)\left(\frac{2 m k T}{n}\right)^{3 / 2} \frac{1}{2} \\
& =-\frac{4 \pi V}{h^{3}} \sum_{1}^{n}(-1)^{2 n+1} \frac{z^{n}}{n} \frac{\sqrt{\pi}}{2}\left(\frac{2 m k T}{n}\right)^{3 / 2} \frac{1}{2} \\
& =\frac{V}{h^{3}} \sum_{1}^{n}-(-1)^{2 n+1} \frac{z^{n}}{n}\left(\frac{2 m \pi k T}{n}\right)^{3 / 2} \\
& =\frac{V}{h^{3}}(2 m \pi k T)^{3 / 2} \sum_{1}^{n}-(-1)^{2 n+1} \frac{z^{n}}{n^{5 / 2}} \\
& =\frac{V}{h^{3}}(2 m \pi k T)^{3 / 2} g_{5 / 2}(z) \\
& =\frac{V}{\lambda^{3}} g_{5 / 2}(z) \\
\frac{P }{k T} &=\frac{1}{\lambda^{3}} g_{5 / 2}(z)
\end{aligned}
$$
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 $$ z=e^{-\alpha}=e^{\frac{\mu}{k T}} \quad \text { and } \beta=1 / k T \quad \mu=\text { Chemical Potential } $$
The average number of particles in an ideal Bose Gas in terms of partiton function is:
$$
\begin{aligned}
n_{i}& =\frac{g_{i}}{e^{\alpha} e^{\beta E}-1}\\
\Rightarrow \frac{n_{i}}{g_{i}}&=\frac{e^{-\alpha} e^{-\beta E_{i}}}{1-e^{-\alpha} e^{-\beta E_{i}}} \\
&=\frac{z e^{-\beta E_{i}}}{1-z e^{-\beta E_{i}}} \\
\Rightarrow N&=\sum \frac{z e^{-\beta E_{i}}}{1-z e^{-\beta E_{i}}} \\
&=-z \frac{\partial}{\partial z} \sum \log \left(1-z e^{-\beta E_{i}}\right)
\end{aligned}
$$
Prove that average Number of Particles in an Ideal Bose Gas system is :$$
N=\frac{V g_{3 / 2}(z)}{\lambda^{3}}
$$
Proof:
$$
\begin{aligned}
N &=\sum n_{i}\\
&=\sum \frac{g_{i}}{e^{\alpha+\beta E_{i}}-1}\\
& =\sum \frac{g_{i} e^{-\alpha-\beta E_{i}}}{1-e^{-\alpha-\beta E_{i}}} \\
& =\sum \frac{g_{i} z e^{-\beta E_{i}}}{1-z e^{-\beta E_{i}}}=z \sum \frac{\partial}{\partial z}\left(-g_{i} \ln \left(1-z e^{-\beta E_{i}}\right)\right) \\
& =z \sum \frac{\partial}{\partial z} g_{i} \ln \left(\frac{1}{1-z e^{-\beta E_{i}}}\right) \\
& =z \sum \frac{\partial}{\partial z} g_{i} \ln Z \quad \mathrm{Z}=\text { Grand Partition function of Bose Gas } \\
& =z \sum \frac{\partial}{\partial z}\left(\frac{P V}{k T}\right) \\
& =z V \sum \frac{\partial}{\partial z}\left(\frac{g_{5 / 2}(z)}{\lambda^{3}}\right) \\
& =\frac{V}{\lambda^{3}} z \sum \frac{\partial}{\partial z}\left(g_{5 / 2}(z)\right)\\
&=\frac{V}{\lambda^{3}} z \frac{\partial}{\partial z} \sum \frac{z^{n}}{n^{5 / 2}} \\
& =\frac{V}{\lambda^{3}} z \sum \frac{n z^{n-1}}{n^{5 / 2}}\\
&=\frac{V}{\lambda^{3}} \sum \frac{z^{n}}{n^{3 / 2}} \\
& =\frac{V}{\lambda^{3}} g_{3 / 2}(z) \\
& \Rightarrow \frac{N}{V}=\frac{g_{3 / 2}(z)}{\lambda^{3}}
\end{aligned}
$$
Prove that $z \frac{\partial}{\partial z} g_{r}(z)=g_{r-1}(z)$
Hence
$$
\begin{aligned}
z \frac{\partial}{\partial z} g_{5 / 2}(z)&=g_{3 / 2}(z) \\
z \frac{\partial}{\partial z} g_{3 / 2}(z)&=g_{1 / 2}(z)
\end{aligned}
$$
Proof: Method 1
$$
\begin{aligned}
z \frac{\partial}{\partial z} g_{r}(z) &=z \frac{\partial}{\partial z}\left(z+\frac{z^{2}}{2^{r}}+\frac{z^{3}}{3^{r}}+\frac{z^{4}}{4^{r}}+\ldots .\right) \\
& =z\left(1+\frac{2 z}{2^{r}}+\frac{3 z^{2}}{3^{r}}+\frac{4 z^{3}}{4^{r}}+\ldots . .\right)
\end{aligned}
$$
$$
\begin{aligned}
& =\left(z+\frac{z^{2}}{2^{r-1}}+\frac{z^{3}}{3^{r-1}}+\frac{z^{4}}{4^{r-1}}+\ldots . .\right) \\
& =g_{r-1}(z)
\end{aligned}
$$
Method-2
$$
\begin{aligned}
& z \frac{\partial}{\partial z} g_{r}(z)=z \frac{\partial}{\partial z} \sum \frac{z^{n}}{n^{r}} \\
& =z \sum \frac{n z^{n-1}}{n^{r}} \\
& =z \sum \frac{z^{n-1}}{n^{r-1}}=\sum \frac{z^{n}}{n^{r-1}} \\
& =g_{r-1}(z)
\end{aligned}
$$
Prove that $z \dfrac{\partial}{\partial z} f_{r}(z)=f_{r-1}(z)$
Proof: Method-1
$$
\begin{aligned}
z \frac{\partial}{\partial z} f_{r}(z)&=z \frac{\partial}{\partial z}\left(z-\frac{z^{2}}{2^{r}}+\frac{z^{3}}{3^{r}}-\frac{z^{4}}{4^{r}}+\ldots .\right) \\
& =z\left(1-\frac{2 z}{2^{r}}+\frac{3 z^{2}}{3^{r}}-\frac{4 z^{3}}{4^{r}}+\ldots . .\right) \\
& =\left(z-\frac{z^{2}}{2^{r-1}}+\frac{z^{3}}{3^{r-1}}-\frac{z^{4}}{4^{r-1}}+\ldots . .\right) \\
& =f_{r-1}(z)
\end{aligned}
$$
Proof: 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 :
$$z \dfrac{\partial}{\partial z} f_{5 / 2}(z)=f_{3 / 2}(z) \& z \dfrac{\partial}{\partial z} f_{3 / 2}(z)=f_{1 / 2}(z)$$
Prove that $\left[\frac{\partial}{\partial T} g_{3 / 2}(z)\right]_{V}=-\frac{3}{2 T} g_{3 / 2}(z)$
Proof:
$$
\begin{gathered}
{\left[\frac{\partial}{\partial T} g_{3 / 2}(z)\right]_{V}=\frac{\partial}{\partial T} \frac{N \lambda^{3}}{V}} \\
=\frac{N}{V} \frac{\partial}{\partial T} \lambda^{3} \\
=\frac{N}{V} 3 \lambda^{2} \frac{d \lambda}{d T} \\
=\frac{N}{V} 3 \lambda^{2} \frac{d}{d T}\left(\frac{h}{\sqrt{2 \pi m k T}}\right) \\
=\frac{N}{V} 3 \lambda^{2} \frac{h}{\sqrt{2 \pi m k}}\left(-\frac{1}{2} T^{-3 / 2}\right) \\
=-\frac{3 N}{2 V} \lambda^{2} \frac{h}{\sqrt{2 \pi m k T}} \frac{1}{T}=-\frac{3}{2 T} \frac{N \lambda^{3}}{V} \\
=-\frac{3}{2 T} g_{3 / 2}(z)
\end{gathered}
$$
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
$$
\begin{aligned}
\frac{\partial}{\partial T} g_{3 / 2}(z)&=-\frac{3}{2 T} g_{3 / 2}(z) \\
\Rightarrow\left(\frac{\partial}{\partial z} \frac{\partial z}{\partial T}\right)_{V} g_{3 / 2}(z) \frac{\partial z}{\partial T}&=-\frac{3}{2 T} g_{3 / 2}(z) \\
\Rightarrow\left(\frac{\partial}{\partial z} g_{3 / 2}(z)\right)_{V}\left(\frac{\partial z}{\partial T}\right)_{V}&=-\frac{3}{2 T} g_{3 / 2}(z) \\
\Rightarrow z\left(\frac{\partial}{\partial z} g_{3 / 2}(z)\right)_{V} \frac{1}{z}\left(\frac{\partial z}{\partial T}\right)_{V}&=-\frac{3}{2 T} g_{3 / 2}(z) \\
\Rightarrow g_{1 / 2}(z) \frac{1}{z}\left(\frac{\partial z}{\partial T}\right)_{V}&=-\frac{3}{2 T} g_{3 / 2}(z) \\
\Rightarrow \frac{1}{z}\left(\frac{\partial z}{\partial T}\right)_{V}&=-\frac{3}{2 T} \frac{g_{3 / 2}(z)}{g_{1 / 2}(z)}
\end{aligned}
$$
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) \\
&=\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}
$$
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 :
$$n(\lambda) d \lambda=\frac{8 \pi d \lambda}{\lambda^{4}\left(e^{\dfrac{h c}{\lambda k T}}-1\right)}$$
