91影视

The quantum distributions for some energy $E$are the Bose-Einstein distribution $N_{BE}\left(E\right)$and the Fermi-Dirac distribution$N_{FD}\left(E\right)$

$N_{BE}\left(E\right)=\frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{1鈭�e^{\frac{鈭�E}{k_{B}T}}}鈭�1}$ 鈥︹��. (1)

$N_{FD}\left(E\right)=\frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{e^{\frac{E}{k_{B}T}}鈭�1}+1}$ 鈥︹��. (2)

Here,$E鈫�$Energy

$k_B鈫�$Boltzmann's constant

$T鈫�$Temperature

$E=\frac{N\mathrm{鈩�}{蝇}_0}{2s+1}$ 鈥︹��. (3)

Here,

$\text{N}鈫�$Number of oscillators

$\mathrm{鈩�}鈫�$Planck's reduced constant

${蝇}_0鈫�$Fundamental angular frequency

$s鈫�$Spin

The Boltzmann distribution$N_{\text{Boltz}}\left(E\right)$is given by

$N_{\text{Boltz}}\left(E\right)=\frac{E}{k_{B}T}\frac{1}{e^{\frac{E}{k_{B}T}}}$ 鈥︹��. (4)

An approximation for$e^z$when $z$is very small will be helpful:

$e^z鈮�1+z$ 鈥︹��. (5)

$N_{BE}\left(E\right)=\frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{1鈭�e^{鈭�\frac{E}{k_{B}T}}}鈭�1}$

If $k_{B}T>>E$then the$E/k_{B}T$in the exponent of the denominator will be very small, meaning that the approximation of equation (5) can be used with$z$being .$E/k_{B}T$

$\begin{array}{c}{N}_{BE}\left(E\right)鈮�\frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{1鈭�(1鈭�\frac{E}{k_{B}T}}鈭�1}\\ 鈮�\frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{1鈭�1+\frac{E}{k_{B}T}}鈭�1}\\ 鈮�\frac{1}{\frac{k_{B}T}{E}{e}^{\frac{E}{k_{B}T}}鈭�1}\end{array}$

\\

Further solving,

$\begin{array}{c}{N}_{BE}\left(E\right)鈮�\frac{1}{\frac{k_{B}T}{E}{e}^{\frac{E}{k_{B}T}}鈭�1}\\ 鈮�\frac{1}{\frac{k_{B}T}{E}{e}^{\frac{E}{k_{B}T}}}\\ 鈮�\frac{E}{k_{B}T}\frac{1}{e^{\frac{E}{k_{B}T}}}\end{array}$

So, the Bose Einstein distribution approaches $N_{BE}\left(E\right)鈮�\frac{E}{k_{B}T}\frac{1}{e^{\frac{E}{k_{B}T}}}$and matches the Boltzmann distribution of equation (4), thus showing their correlation at high temperature.

To show that the Fermi-Dirac distribution of equation (2).

$N_{FD}\left(E\right)=\frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{e^{\frac{E}{k_{B}T}}鈭�1}+1}$

It approaches the Boltzmann distribution when$k_{B}T>>E$, the approximation of equation (5) is used, with $z$being $E/k_{B}T$(since that ratio is very small for the given condition).

$\begin{array}{c}{N}_{FD}\left(E\right)鈮�\frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{{(1+\frac{E}{k_{B}T})}^{鈭�1}}+1}\\ 鈮�\frac{1}{\left(\frac{e^{\frac{E}{k_{B}T}}}{\frac{E}{k_{B}T}}\right)+1}\\ 鈮�\frac{1}{\frac{k_{B}T}{E}{e}^{\frac{E}{k_{B}T}}+1}\end{array}$

Since$k_{B}T$is much greater than$E$then $k_{B}T/E$will be very large, making$(k_{B}T/E)e^{E/k_{B}T}$very large as well. Given that, adding 1 to it won't significantly affect it; so, it can be ignored.

$\begin{array}{c}{N}_{FD}\left(E\right)鈮�\frac{1}{\frac{k_{B}T}{E}{e}^{\frac{E}{k_{B}T}}+1}\\ 鈮�\frac{1}{\frac{k_{B}T}{E}{e}^{\frac{E}{k_{B}T}}}\\ 鈮�\frac{E}{k_{B}T}\frac{1}{e^{\frac{E}{k_{B}T}}}\end{array}$

So, the Fermi-Dirac distribution approaches to $N_{FB}\left(E\right)鈮�\frac{E}{k_{B}T}\frac{1}{e^{\frac{E}{k_{B}T}}}$and matches the Boltzmann distribution of equation (4) thus showing their correlation at high temperature