Derivation of equation$\mathbf{(}\mathbf{9}\mathbf{-}\mathbf{40}\mathbf{)}$: Our model for calculating$\overline{\mathbf{E}}$is equation (9-26), whose denominator is the total number of particles$\mathit{N}$and whose numerator is the total energy of the system, which we here call${\mathit{U}}_{\mathbf{\text{total}}}$. State with the denominator:

$\mathit{N}\mathbf{=}{\int }_0^{\infty }N\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{\cap }\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{d}\mathit{E}$

Insert the quantum gas density of states and an expression for the distribution. using$\mathbf{\pm }$to distinguish the Bose-Einstein from the Fermi-Dirac. Then change variables:$\mathit{E}\mathbf{=}{\mathit{y}}^{\mathbf{2}}$, and factor$\mathit{B}{\mathit{e}}^{\mathbf{+}{\mathbf{r}}^{\mathbf{2}}\mathbf{/}{\mathbf{k}}_{\mathbf{U}}\mathbf{T}}$out of the denominator. In the integrand will be a factor

${\mathbf{(}\mathbf{1}\mathbf{\mp }\frac{\mathbf{1}}{\mathbf{B}}{\mathbf{e}}^{\mathbf{-}{\mathbf{y}}^{\mathbf{2}}\mathbf{/}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}\mathbf{)}}^{\mathbf{-}\mathbf{1}}$

Using,${\mathbf{(}\mathbf{l}\mathbf{\mp }\mathbf{\epsilon }\mathbf{)}}^{\mathbf{-}\mathbf{1}}\mathbf{\cong }\mathbf{1}\mathbf{\pm }\mathit{\epsilon }$ a sum of two integrals results, each of Gaussian form. The integral thus becomes two terms in powers of$\mathbf{1}\mathbf{/}\mathit{B}$. Repeat the process. but instead find an expression for${\mathit{U}}_{\mathbf{\text{total}}}$in terms of$\mathbf{1}\mathbf{/}\mathit{B}$, using

${\mathit{U}}_{\mathbf{\text{|ntal}}}\mathbf{=}{\int }_0^{\infty }E\mathit{N}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{D}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{d}\mathit{E}$

Divide your expression for${\mathit{U}}_{\mathbf{\text{total}}}$by that for$\mathit{N}$. both in terms of$\mathbf{1}\mathbf{/}\mathit{B}$. Now$\mathbf{1}\mathbf{/}\mathit{B}$can safely be eliminated by using the lowest-order expression for$\mathit{N}$in terms of$\mathbf{1}\mathbf{/}\mathit{B}$.

The expression of total number of particles is given by,

$\mathit{N}\mathbf{=}{\int }_0^{\infty }N\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{D}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{d}\mathit{E}$

The expression of total energy is given by,

${\mathit{U}}_{\mathbf{\text{total}}}\mathbf{=}{\int }_0^{\infty }E\mathit{N}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{D}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{d}\mathit{E}$

The expression for average energy is given by,

$\overline{\mathbf{E}}\mathbf{=}\frac{{\mathbf{U}}_{\mathbf{\text{total}}}}{\mathbf{N}}$

The total number of particles is $N$.

The total energy of system is $U_{\text{total}}$Calculation:

The expression of total number of particles is given by,

$\begin{array}{c}N={\int }_0^{\infty }N\left(E\right)D\left(E\right)dE\\ ={\int }_0^{\infty }\left[\frac{(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}{E}^{\frac{1}{2}}\right]\left[\frac{1}{B{e}^{\frac{E}{k_{\text{B}}T}}\mp 1}\right]dE\\ =\frac{(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}{\int }_0^{\infty }\frac{E^{\frac{1}{2}}}{B{e}^{\frac{E}{{\text{k}}^{2}T}}\mp 1}dE\end{array}$

Let$E=y^2$ then$dE=2ydy$,

Substitute$2ydy$for$dE$and$y^2$for $E$in equation (1).

$\begin{array}{c}dN=\frac{(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}{\int }_0^{\infty }\frac{{\left(y^2\right)}^{\frac{1}{2}}}{B{e}^{\frac{E}{k_{B}T}}\mp 1}\left(2ydy\right)\\ =\frac{2(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}{\int }_0^{\infty }\frac{y^2}{B{e}^{\frac{E}{k_{\text{B}}T}}\mp 1}dy\\ =\frac{2(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}{\int }_0^{\infty }\frac{y^2}{B{e}^{\frac{y^2}{k_{B}T}}}(1\mp e^{-\frac{y^2}{k_{\text{B}}T}})dy\\ =\frac{2(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}[{\int }_0^{\infty }\frac{y^2}{B{e}^{\frac{y^2}{k_{\text{B}}T}}}\pm \frac{y^2{e}^{-\frac{y^2}{k_{\text{B}}T}}}{e^{\frac{y^2}{k_{\text{B}}T}}}dy]\end{array}$

Further simplify the above,

$\begin{array}{c}N=\frac{2(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}[\frac{1}{4B}\sqrt{\pi {\left(k_{\text{B}}T\right)}^3}\pm \frac{1}{4B^2}\sqrt{\pi {\left(\frac{k_{\text{B}}T}{2}\right)}^3}]\\ =\frac{(2s+1)V}{B}{\left(\frac{m{k}_{\text{B}}T}{2\pi {\hslash }^2}\right)}^{\frac{3}{2}}[1\pm \frac{1}{2B\sqrt{2}}]\end{array}$

The expression of total energy is given by,

$\begin{array}{c}U\left({}_{\text{total}}\right)={\int }_0^{\infty }EN\left(E\right)D\left(E\right)dE\\ ={\int }_0^{\infty }E\left[\frac{(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}{E}^{\frac{1}{2}}\right]\left[\frac{1}{B{e}^{\frac{E}{k_{\text{B}}T}}\mp 1}\right]dE\\ =\frac{(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}{\int }_0^{\infty }\frac{E^{\frac{3}{2}}}{B{e}^{\frac{E}{k_{B}T}}\mp 1}dE\end{array}$

Let $E=y^2$then$dE=2ydy$,

Substitute $2ydy$for$dE$ and $y^2$for$E$ in equation (2).

$\begin{array}{c}{U}_{\text{total}}=\frac{(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}{\int }_0^{\infty }\frac{{\left(y^2\right)}^{\frac{3}{2}}}{B{e}^{\frac{E}{k_{\text{B}}T}}\mp 1}\left(2ydy\right)\\ =\frac{2(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}{\int }_0^{\infty }\frac{y^4}{B{e}^{\frac{E}{k_{B}T}}\mp 1}dy\\ =\frac{2(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}{\int }_0^{\infty }\frac{y^4}{B{e}^{\frac{y^2}{k_{\text{B}}T}}}(1\mp e^{-\frac{y^2}{k_{\text{B}}T}})dy\\ =\frac{2(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}[{\int }_0^{\infty }\frac{y^4}{B{e}^{\frac{y^2}{k^2}}}\pm \frac{y^4{e}^{-\frac{y^2}{k_{\text{B}}T}}}{e^{\frac{y^2}{k_{\text{B}}T}}}dy]\end{array}$

Further simplify the above,

$\begin{array}{c}{U}_{\text{total}}=\frac{2(2s+1)m^{\frac{3}{2}}V}{{\pi }^2{\hslash }^3\sqrt{2}}[\frac{3}{8B}\sqrt{\pi {\left(k_{\text{B}}T\right)}^5}\pm \frac{3}{8B^2}\sqrt{\pi {\left(\frac{k_{\text{B}}T}{2}\right)}^5}]\\ =\frac{3(2s+1)V{k}_{\text{B}}T}{2B}{\left(\frac{m{k}_{\text{B}}T}{2\pi {\hslash }^2}\right)}^{\frac{3}{2}}[1\pm \frac{1}{4B\sqrt{2}}]\end{array}$

The expression for average energy is calculated as,

$\begin{array}{c}\overline{E}=\frac{U_{\text{total}}}{N}\\ =\frac{\frac{3(2s+1)V{k}_{\text{B}}T}{2B}{\left(\frac{m{k}_{\text{B}}T}{2\pi {\hslash }^2}\right)}^{\frac{3}{2}}[1\pm \frac{1}{4B\sqrt{2}}]}{\frac{(2s+1)V}{B}{\left(\frac{m{k}_{\text{B}}T}{2\pi {\hslash }^2}\right)}^{\frac{3}{2}}[1\pm \frac{1}{2B\sqrt{2}}]}\\ \cong \frac{3}{2}{k}_{\text{B}}T[1\pm \frac{1}{4B\sqrt{2}}+\dots ]\\ \cong \frac{3}{2}{k}_{\text{B}}T[1\pm \frac{1}{4\sqrt{2}}\left\{\frac{1}{(2s+1)}{\left(\frac{2\pi {\hslash }^2}{m{k}_{\text{B}}T}\right)}^{\frac{3}{2}}\left(\frac{N}{V}\right)\right\}+\dots ]\end{array}$

Further simplify the above,

$\begin{array}{c}\overline{E}\cong \frac{3}{2}{k}_{\text{B}}T[1\pm \frac{2\sqrt{2}{\pi }^3{\hslash }^3}{2\left(2\right)\frac{3}{2}(2s+1)}{\left(\frac{1}{\pi m{k}_{\text{B}}T}\right)}^{\frac{3}{2}}\left(\frac{N}{V}\right)+\dots ]\\ \cong \frac{3}{2}{k}_{\text{B}}T[1\pm \frac{\sqrt{2}{\pi }^3{\hslash }^3}{(2s+1)}{\left(\frac{1}{2\pi m{k}_{\text{B}}T}\right)}^{\frac{3}{2}}\left(\frac{N}{V}\right)+\dots ]\end{array}$