91影视

The equations for the distributions will be needed;

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

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

$N_{FD}\left(E\right)=\frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{e^{\frac{E}{k_{B}T}}鈭�1}+1}$ 鈥︹��. (3)

Here,

$E鈫�$Energy

$k_B鈫�$Boltzmann's constant

$T鈫�$Temperature

$E=\frac{N{\mathrm{鈩�}}_{{蝇}_0}}{2s+1}$ 鈥︹��. (4)

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 of equation (2) is checked first, using that $E$will be 0:

$\begin{array}{c}{N}_{\text{Boltz}}\left(E\right)=\frac{E}{k_{B}T}\frac{1}{e^{E{k}_{B}T}}\\ N_{\text{Boltz}}\left(0\right)=\frac{E}{k_{B}T}\frac{1}{e^{(0/V_{B}T}}\\ =\frac{E}{k_{B}T}\frac{1}{e^0}\\ =\frac{E}{k_{B}T}\end{array}$

So, the number of oscillators expected in the ground state if they were classically distinguishable would be .

$N_{Boltz}\left(0\right)=\frac{E}{k_{B}T}$

The Bose-Einstein distribution of equation (3) is checked with

being 0.

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

At E=0,

$\begin{array}{c}{N}_{BE}\left(0\right)=\frac{1}{\frac{e^{[0/K_{B}T}}{1鈭�e^{鈭�\frac{E}{k_{B}T}}}鈭�1}\\ =\frac{1}{\frac{e^0}{1鈭�e^{鈭�E/K_{B}T}}鈭�1}\\ =\frac{1}{\frac{e^0}{1鈭�e^{鈭�E//{/}_{B}T}}鈭�1}\\ =\frac{1}{\frac{1}{1鈭�e^{鈭�E/k_{B}T}}鈭�1}\end{array}$

$\begin{array}{c}{N}_{BE}\left(0\right)=\frac{1}{\frac{1}{1鈭�e^{鈭�\frac{E}{k_{B}T}}}鈭�\frac{1鈭�e^{\frac{E}{k_{B}T}}}{1鈭�e^{\frac{E}{k_{B}T}}}}\\ =\frac{1}{\frac{1鈭�(1鈭�e^{鈭�\frac{E}{k_{B}T}})}{1鈭�e^{鈭�\frac{E}{k_{B}T}}}}\\ =\frac{1鈭�e^{鈭�\frac{E}{k_{B}T}}}{e^{鈭�\frac{E}{k_{B}T}}}\end{array}$

$\begin{array}{c}{N}_{BE}\left(0\right)=\frac{1鈭�e^{鈭�\frac{E}{k_{B}T}}}{e^{鈭�\frac{E}{k_{B}T}}}\\ =\frac{1}{e^{鈭�\frac{E}{k_{B}T}}}鈭�\frac{e^{鈭�\frac{E}{k_{B}T}}}{e^{鈭�\frac{E}{k_{B}T}}}\\ =e^{\frac{E}{k_{B}T}}鈭�1\end{array}$

Since $k_{B}T$is very small, that$尉/k_{B}T$means that will be very large, and consequently, so$e^{E/k_{B}T}$. Thus subtracting 1 from it won't significantly affect it, and can be ignored:

$\begin{array}{l}{N}_{BE}\left(0\right)=e^{E/k_{B}T}鈭�1\\ N_{BE}\left(0\right)鈮�e^{E/k_{B}T}\end{array}$

So, the number of oscillators expected in the ground state if they were bosons would be$N_{BE}\left(0\right)鈮�e^{E/k_{B}T}$ and matches the expected value for the Bose-Einstein distribution, thus verifying it.

The Fermi-Dirac distribution of equation (4) is checked for an $E$of 0.

$\begin{array}{c}{N}_{FD}\left(E\right)=\frac{1}{\frac{E^{\frac{E}{k_{B}T}}}{e^{{}^{\frac{E}{k_{B}T}}}鈭�1}+1}\\ N_{FD}\left(0\right)=\frac{1}{\frac{e^{(0)/k_{B}T}}{e^{E/k_{B}T}}+1}\\ =\frac{1}{\frac{c^0}{e^{E/k_{B}T}}+1}\\ =\frac{1}{\frac{1}{e^{E/k_{B}T}}+1}\\ =\frac{1}{\frac{1}{e^{E{k}_{B}T}}+1}\end{array}$

$\begin{array}{c}{N}_{FD}\left(0\right)=\frac{1}{\frac{1}{e^{{}^{\frac{E}{k_{B}T}}}}+\frac{e^{{}^{\frac{E}{k_{B}T}}}}{e^{{}^{\frac{E}{k_{B}T}}}}}\\ =\frac{1}{\left[\frac{1+e^{{}^{\frac{E}{k_{B}T}}}}{e^{{}^{\frac{E}{k_{B}T}}}}\right]}\\ =\frac{e^{{}^{\frac{E}{k_{B}T}}}}{1+e^{{}^{\frac{E}{k_{B}T}}}}\end{array}$

Since $k_{B}T$is very small, that means that $尉/k_{B}T$will be very large, and consequently, so will $e^{E/k_{B}T}$. Thus adding 1 to it won't significantly affect it, and can be ignored

$\begin{array}{c}{N}_{FD}\left(0\right)=\frac{e^{\frac{E}{k_{B}T}}}{1+e^{\frac{E}{k_{B}T}}}\\ 鈮�\frac{e^{\frac{E}{k_{B}T}}}{e^{\frac{E}{k_{B}T}}}\\ 鈮�1\end{array}$

So, the number of oscillators expected in the ground state if they were fermions would be$N_{FD}\left(0\right)鈮�1$ and matches the expected value from equation (1) for the Fermi-Dirac distribution thus verifying it.

For the three distributions, the results are as expected in terms of their relationship to the Pauli Exclusion Principle. Because fermions resist being clumped together in the same state (because Pauli stated that no two fermions can share the same state), a very low temperature is required to view an entire oscillator in the ground state. The bosons, on the other hand, have no such constraints and can pack a large number of oscillators into the ground state. And, ignoring spin, the Boltzmann conclusion is simply another way of saying that the average number of oscillators is equal to the total energy of the oscillators divided by the energy of each oscillator.