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In order to show that the total energy $U_{\text{total}}$of an $\mathit{N}$number of$\mathbf{spin}\mathbf{1}\mathbf{/}\mathbf{2}$fermions at$\mathbf{0}\mathbf{.}\mathbf{\text{K}}$is:

${\mathit{U}}_{\mathbf{\text{local}}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{10}}{\mathbf{\left(}\frac{\mathbf{3}{\mathbf{蟺}}^{\mathbf{2}}{\mathbf{鈩�}}^{\mathbf{3}}}{{\mathbf{m}}^{\mathbf{3}\mathbf{/}\mathbf{2}\mathbf{V}}}\mathbf{\right)}}^{\mathbf{2}\mathbf{/}\mathbf{3}}{\mathit{N}}^{\mathbf{5}\mathbf{/}\mathbf{3}}$ 鈥︹�� (1)

Here V is their total volume.$\mathit{h}$is Planck's reduced constant, and$\mathit{m}$is the particle mass).

The equation for the total energy of a group of particles is used:

${\mathit{U}}_{\mathbf{l}\mathbf{o}\mathbf{c}\mathbf{a}\mathbf{l}}\mathbf{=}鈭�E\mathit{N}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{D}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathit{d}\mathit{E}$ 鈥︹�� (2)

Here$\mathit{E}$is the energy$\mathit{N}\mathbf{\left(}\mathit{E}\mathbf{\right)}$is the number of particles per state as a function of energy, $\mathit{D}\mathbf{\left(}\mathit{E}\mathbf{\right)}$is the density of states as a function of energy, and$\mathit{d}\mathit{E}$is the differential of energy.

The density of states$\mathit{D}\mathbf{\left(}\mathit{E}\mathbf{\right)}$for a gas in a well $\mathbf{3}\mathbf{鈭�}\mathit{D}$of volumelocalid="1658398051531" $\mathit{V}$is also needed:

$\mathit{D}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{(}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{)}{\mathbf{m}}^{\mathbf{3}\mathbf{/}\mathbf{2}}\mathbf{V}}{{\mathbf{蟺}}^{\mathbf{2}}{\mathbf{鈩�}}^{\mathbf{3}}\sqrt{\mathbf{2}}}{\mathit{E}}^{\mathbf{1}\mathbf{/}\mathbf{2}}$ 鈥︹�� (3)

With s being the spin of the particles, $\mathit{m}$as their mass,$\mathit{鈩�}$as Planck's reduced constant, and$\mathit{E}$as the energy.

The expression for the Fermi energy$$$E_F$at $0\text{K}$will also be useful:

$E_F=\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3}{(2s+1)蟺\sqrt{2}V}\frac{N}{V}\right]}^{2/3}$ 鈥︹�� (4)

With the variables having the same meanings as in the other equations.

To find the total energy of the fermions, equation (2) is used, with equation (3) being used for $D\left(E\right)$If the energy is below the Fermi energy, then the number of particles per state $N\left(E\right)$will just be 1, with the limits of integration thus being from 0 to the Fermi energy :$E_F$

$\begin{array}{c}{U}_{\text{local}}=鈭�EN\left(E\right)D\left(E\right)dE{U}_{\text{local}}\\ ={鈭�}_{E=0}^{E=E_F}E\left[1\right]\left[\frac{(2s+1)m^{3/2}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{E}^{1/2}\right]dE\\ ={鈭�}_{E=0}^{E=E_F}\frac{(2s+1)m^{3/2}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{E}^{3/2}dE\end{array}$

Since its fermions that are being used, this will be$\frac{1}{2}$ (given that fermions are spin$\frac{1}{2}$ particles), and the 2's combined with rationalizing the denominator:

$\begin{array}{c}{U}_{\text{total}}={鈭�}_{E=0}^{E=E_F}\frac{(2s+1)m^{3/2}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{E}^{3/2}dE\\ ={鈭�}_{E=0}^{E=E_f}\frac{(2\left[\frac{1}{2}\right]+1)m^{3/2}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{E}^{3/2}dE\\ ={鈭�}_{E=0}^{E=E_f}\frac{2m^{3/2}V}{{蟺}^2{\mathrm{鈩�}}^3\sqrt{2}}{E}^{3/2}dE\\ ={鈭�}_{E=E_f}\frac{m^{3/2}V\sqrt{2}}{{蟺}^2{\mathrm{鈩�}}^3}{E}^{3/2}dE\end{array}$

So then the constants can be pulled out, and then just integrate with respect to $E$:

$\begin{array}{c}{U}_{\text{total}}={鈭�}_{E=0}^{E=E_f}\frac{m^{3/2}V\sqrt{2}}{{蟺}^2{\mathrm{鈩�}}^3}{E}^{3/2}dE\\ =\frac{m^{3/2}V\sqrt{2}}{{蟺}^2{\mathrm{鈩�}}^3}{鈭�}_{E=0}^{E=E_f}{E}^{3/2}dE\text{鈥�}\\ =\frac{m^{3/2}V\sqrt{2}}{{蟺}^2{\mathrm{鈩�}}^3}{\left[\frac{2}{5}{E}^{3/2}\right]}_0^{E_F}\\ =\frac{2\sqrt{2}{m}^{3/2}V}{5{蟺}^2{\mathrm{鈩�}}^3}{E}_F^{3/2}\end{array}$ 鈥︹�� (5)

The Fermi energy can be inserted, but it's helpful to simplify its expression a little first, with using$\frac{1}{2}$ for the :$S$

$\begin{array}{c}{E}_F=\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3}{(2s+1)蟺\sqrt{2}}\frac{N}{V}\right]}^{2/3}\\ =\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3}{(2\left\{\frac{1}{2}\right\}+1)蟺\sqrt{2}}\frac{N}{V}\right]}^{2/3}\\ =\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3N}{2\sqrt{2蟺V}}\right]}^{2/3}\end{array}$

That can be simplified a little more by combining the 2's, then factoring them out, and factoring the$鈰�$ is:

$\begin{array}{c}{E}_V=\frac{{蟺}^2{\mathrm{鈩�}}^2}{m}{\left[\frac{3N}{2\sqrt{2蟺V}}\right]}^{2/3}\\ =\frac{{\mathrm{鈩�}}^2}{m}{\left[\frac{3{蟺}^{3}N}{2^{3/2}蟺V}\right]}^{2/3}\\ =\frac{{\mathrm{鈩�}}^2}{2m}{\left(\frac{3{蟺}^{3}N}{V}\right)}^{2/3}\end{array}$

So then replace the$E_F$in equation (5) with that, and apply the exponent:

$\begin{array}{c}{U}_{\text{total}}=\frac{2\sqrt{2m^{3/2}V}}{5{蟺}^2{\mathrm{鈩�}}^3}{E}_F^{5/2}\\ =\frac{2^{3/2}{m}^{3/2}V}{5{蟺}^2{\mathrm{鈩�}}^3}{\left[\frac{{\mathrm{鈩�}}^2}{2m}{\left(\frac{3{蟺}^{2}N}{V}\right)}^{2/3}\right]}^{5/2}\\ =\frac{{\left(2m\right)}^{3/2}V}{5{蟺}^2{\mathrm{鈩�}}^3}\left[\frac{{\mathrm{鈩�}}^5}{{\left(2m\right)}^{5/2}}{\left(\frac{3{蟺}^{2}N}{V}\right)}^{5/3}\right]\end{array}$

The terms outside the parentheses can combine with the term outside the brackets:

$\begin{array}{c}{U}_{\text{total}}=\frac{{\left(2m\right)}^{3/2}V}{5{蟺}^2{\mathrm{鈩�}}^3}\left[\frac{{\mathrm{鈩�}}^5}{{\left(2m\right)}^{5/2}}{\left(\frac{3{蟺}^{2}N}{V}\right)}^{5/3}\right]\\ =\frac{{\mathrm{鈩�}}^{2}V}{5{蟺}^2\left(2m\right)}{\left(\frac{3{蟺}^{2}N}{V}\right)}^{5/3}\\ =\frac{{\mathrm{鈩�}}^{2}V}{10m{蟺}^2}{\left(\frac{3{蟺}^{2}N}{V}\right)}^{5/3}\end{array}$

Then bringthe $V$and $鈰�$inside the parentheses to combine them with the others:

$\begin{array}{c}{U}_{\text{total}}=\frac{{\mathrm{鈩�}}^{2}V}{10m{蟺}^2}{\left(\frac{3{蟺}^{2}N}{V}\right)}^{5/3}\\ =\frac{{\mathrm{鈩�}}^2}{10m}{\left(\frac{3{蟺}^{2}N{V}^{3/5}}{{蟺}^{6/5}V}\right)}^{5/3}\\ =\frac{{\mathrm{鈩�}}^2}{10m}{\left(\frac{3{蟺}^{4/5}N}{V^{2/5}}\right)}^{5/3}\end{array}$

With the variables combined now, the $5/3$can be applied to the parentheses, and then the 3 rewritten slightly:

$\begin{array}{c}{U}_{\text{total}}=\frac{{\mathrm{鈩�}}^2}{10m}{\left(\frac{3{蟺}^{4/5}N}{V^{2/5}}\right)}^{5/3}\\ =\frac{{\mathrm{鈩�}}^2}{10m}\frac{{\left(3\right)}^{5/3}{蟺}^{4/3}{N}^{5/3}}{V^{2/3}}\\ =\frac{{\mathrm{鈩�}}^2}{10m}\frac{3{\left(3\right)}^{2/3}{蟺}^{4/3}{N}^{5/3}}{V^{2/3}}\end{array}$

Given that some of the exponents are multiples of $2/3$, they can be grouped together:

$\begin{array}{c}{U}_{\text{total}}=\frac{{\mathrm{鈩�}}^2}{10m}\frac{3{\left(3\right)}^{2/3}{蟺}^{4/3}{N}^{5/3}}{V^{2/3}}\\ =\frac{3{\mathrm{鈩�}}^2}{10m}{\left(\frac{3{蟺}^2}{V}\right)}^{2/3}{N}^{5/3}\end{array}$

Some of the other variables can be brought inside the parentheses.

$\begin{array}{c}{U}_{\text{total}}=\frac{3{\mathrm{鈩�}}^2}{10m}{\left(\frac{3{蟺}^2}{V}\right)}^{2/3}{N}^{5/3}\\ =\frac{3}{10}{\left(\frac{3{蟺}^2{\mathrm{鈩�}}^3}{m^{3/2}V}\right)}^{2/3}{N}^{5/3}\end{array}$

$\begin{array}{c}{U}_{\text{total}}=\frac{3{\mathrm{鈩�}}^2}{10m}{\left(\frac{3{蟺}^2}{V}\right)}^{2/3}{N}^{5/3}\\ =\frac{3}{10}{\left(\frac{3{蟺}^2{\mathrm{鈩�}}^3}{m^{3/2}V}\right)}^{2/3}{N}^{5/3}\end{array}$

The total energy is

$U_{\text{total}}=\frac{3}{10}{\left(\frac{3{蟺}^2{\mathrm{鈩�}}^3}{m^{3/2V}}\right)}^{2/3}{N}^{5/3}$