91Ó°ÊÓ

We have been given that the density of states and the chemical potential for a two-dimensional Fermi gas which we have found earlier in Problem $7.28$is$g\left(\mathrm{Î\mu }\right)=\frac{N}{{\mathrm{Î\mu }}_F}$.

We need to find the heat capacity of this gas in the limit $kT≪{\mathrm{Î\mu }}_F$and show that the heat capacity has the expected behavior when $kT≫{\mathrm{Î\mu }}_F$. Also we need to sketch the heat capacity as a function of temperature.

The total energy is given by the following integral:

$U={∫}_0^{∞}\mathrm{Î\mu }g\left(\mathrm{Î\mu }\right){\stackrel{¯}{n}}_{FD}\left(\mathrm{Î\mu }\right)d\mathrm{Î\mu }$ (Let this equation be (localid="1650103209259" $1$))

We have found the energy density earlier in Problem $7.28$such as

localid="1650102270353" $g\left(\mathrm{Î\mu }\right)=\frac{N}{{\mathrm{Î\mu }}_F}$

Substitute the value of $g\left(\mathrm{Î\mu }\right)$in equation (localid="1650103517684" $1$), we get:

$U=\frac{N}{{\mathrm{Î\mu }}_F}{∫}_0^{∞}\mathrm{Î\mu }{\stackrel{¯}{n}}_{FD}\left(\mathrm{Î\mu }\right)d\mathrm{Î\mu }$

Integrating by parts, we get:

$U=\frac{N}{{\mathrm{Î\mu }}_F}\frac{{\mathrm{Î\mu }}^2}{2}{\stackrel{¯}{n}}_{FD}{|}_0^{∞}-\frac{N}{{\mathrm{Î\mu }}_F}{∫}_0^{∞}\frac{{\mathrm{Î\mu }}^2}{2}\frac{∂{\stackrel{¯}{n}}_{FD}}{∂\mathrm{Î\mu }}d\mathrm{Î\mu }$ (Let this equation be ($2$))

If we substitute $\mathrm{Î\mu }=0$, the integral will become zero due to the dependence of the term on ${\mathrm{Î\mu }}^2$and the first term vanishes.

The equation ($2$) reduces to :

$U=-\frac{N}{{\mathrm{Î\mu }}_F}{∫}_0^{∞}\frac{{\mathrm{Î\mu }}^2}{2}\frac{∂{\stackrel{¯}{n}}_{FD}}{∂\mathrm{Î\mu }}d\mathrm{Î\mu }$ (Let this equation be ($3$))

We know that ${\stackrel{¯}{n}}_{FD}=\frac{1}{e^{{\displaystyle \raisebox{1ex}{$(e-\mathrm{μ})$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}$

Taking derivative with respect to $\mathrm{Î\mu }$, we get:

localid="1650103912932" $\frac{∂{\stackrel{¯}{n}}_{FD}}{∂\mathrm{Î\mu }}=\frac{∂\left[{\displaystyle \frac{1}{e^{{\displaystyle \raisebox{1ex}{$(\mathrm{Î\mu }-\mathrm{μ})$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}}\right]}{∂\mathrm{Î\mu }}$

$\frac{∂{\stackrel{¯}{n}}_{FD}}{∂\mathrm{Î\mu }}=-\frac{1}{kT}\frac{e^{{\displaystyle \raisebox{1ex}{$(\mathrm{Î\mu }-\mathrm{μ})$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}}{{(e^{{\displaystyle \raisebox{1ex}{$(\mathrm{Î\mu }-\mathrm{μ})$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1)}^2}$

Let $\raisebox{1ex}{$(\mathrm{Î\mu }-\mathrm{μ})$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.=x$, then we can write as:

$dx=\frac{d\mathrm{Î\mu }}{kT}$

We also need to change the boundaries of integration, so:

$\mathrm{Î\mu }→∞$ $x→∞$

$\mathrm{Î\mu }→0$ $x→-\frac{\mathrm{μ}}{kT}$

As $kT≪\mathrm{μ}$, we can put $-∞$as the lower limit of integration.

The integral in equation ($3$) will become:

$U=\frac{N}{2{\mathrm{Î\mu }}_F}{∫}_{-∞}^{∞}{\mathrm{Î\mu }}^2\frac{e^x}{{(e^x+1)}^2}dx$ (Let this equation be (localid="1650114310341" $4$))

Now expanding ${\mathrm{Î\mu }}^2$about $\mathrm{μ}$using Taylor series, we get:

$$\begin{gathered} {\mathrm{Î\mu }}^2={\mathrm{μ}}^2+2\mathrm{μ}(\mathrm{Î\mu }-\mathrm{μ})x+(\mathrm{Î\mu }-\mathrm{μ})x^2 \\ {\mathrm{Î\mu }}^2={\mathrm{μ}}^2+2\mathrm{μ}\left(kTx\right)+{\left(kTx\right)}^2 \end{gathered}$$

By substituting the value of in equation ($4$), we get three integrals say $I_1,I_2,I_3$

The equation ($4$) will become:

where,

$$\begin{gathered} I_1={\mathrm{μ}}^2{∫}_{-∞}^{∞}\frac{e^x}{{(e^x+1)}^2}dx \\ {I}_2=2kT\mathrm{μ}{∫}_{-∞}^{∞}\frac{x{e}^x}{{(e^x+1)}^2}dx \\ {I}_3={\left(kT\right)}^2{∫}_{-∞}^{∞}\frac{x^2{e}^x}{{(e^x+1)}^2}dx \end{gathered}$$

Simplifying the integrals $I_1,I_2,I_3$, we get:

$I_1={\mathrm{μ}}^2$

The second integral is odd integral and the limits of integration are from to , so this integral is simply zero.

$I_2=0$

$I_3={\left(kT\right)}^2\left[\frac{\mathrm{Ï€}2}{3}\right]=\frac{{\left(kT\mathrm{Ï€}\right)}^2}{3}$

Substitute the values of in equation ($5$), we get:

$U=\frac{N}{2{\mathrm{Î\mu }}_F}\left[{\mathrm{μ}}^2+\frac{{\left(kT\mathrm{Ï€}\right)}^2}{3}\right]$

Let $\mathrm{μ}={\mathrm{Î\mu }}_F$, we get:

$$\begin{gathered} U=\frac{N}{2{\mathrm{Î\mu }}_F}\left[{\mathrm{Î\mu }}_F^2+\frac{{\left(kT\mathrm{Ï€}\right)}^2}{3}\right] \\ U=\frac{N{\mathrm{Î\mu }}_F}{2}+\frac{N{\left(kT\mathrm{Ï€}\right)}^2}{6{\mathrm{Î\mu }}_F} \end{gathered}$$

The heat capacity is the partial derivative of with respect to the temperature, that is:

$$\begin{gathered} C_V=\frac{∂U}{∂T} \\ {C}_V=\frac{∂\left[{\displaystyle \frac{N{\mathrm{Î\mu }}_F}{2}}+{\displaystyle \frac{N{\left(kT\mathrm{Ï€}\right)}^2}{6{\mathrm{Î\mu }}_F}}\right]}{∂T} \\ {C}_V=\frac{N{\left(k\mathrm{Ï€}\right)}^{2}T}{3{\mathrm{Î\mu }}_F} \end{gathered}$$

The graph of the heat capacity as a function of temperature is linear, where the slope of the line is $Slope=\frac{N{\left(k\mathrm{Ï€}\right)}^2}{3{\mathrm{Î\mu }}_F}$.