Q. 7.31 - An Introduction to Thermal Physics

Question

In Problem $7.28$ you found the density of states and the chemical potential for a two-dimensional Fermi gas. Calculate the heat capacity of this gas in the limit $kT \ll {\epsilon }_F$· Also show that the heat capacity has the expected behavior when $kT \gg {\epsilon }_F$. Sketch the heat capacity as a function of temperature.

Step-by-Step Solution

Answer

The heat capacity of the gas in the limit  is $C_V = \frac{N{\left(k\pi \right)}^{2}T}{3{\epsilon }_F}$.

The graph of heat capacity as a function of temperature is

Step 1: Given information

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$g\left(\epsilon \right) = \frac{N}{{\epsilon }_F}$.

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

Step 2: Simplify

The total energy is given by the following integral:

$U = {\int }_0^{\infty }\epsilon g\left(\epsilon \right){\overline{n}}_{FD}\left(\epsilon \right)d\epsilon$                                                       (Let this equation be ($1$))

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

$g\left(\epsilon \right) = \frac{N}{{\epsilon }_F}$

Substitute the value of $g\left(\epsilon \right)$ in equation ($1$), we get:

$U = \frac{N}{{\epsilon }_F}{\int }_0^{\infty }\epsilon {\overline{n}}_{FD}\left(\epsilon \right)d\epsilon$

Integrating by parts, we get:

$U = \frac{N}{{\epsilon }_F}\frac{{\epsilon }^2}{2}{\overline{n}}_{FD}{|}_0^{\infty } - \frac{N}{{\epsilon }_F}{\int }_0^{\infty }\frac{{\epsilon }^2}{2}\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }d\epsilon$                          (Let this equation be ($2$))

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

The equation ($2$) reduces to :

$U = -\frac{N}{{\epsilon }_F}{\int }_0^{\infty }\frac{{\epsilon }^2}{2}\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }d\epsilon$                                                    (Let this equation be ($3$))  

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

Taking derivative with respect to $\epsilon$, we get:

$\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }=\frac{\partial \left[{\displaystyle \frac{1}{e^{{\displaystyle \raisebox{1ex}{$(\epsilon -\mu )$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}} + 1}}\right]}{\partial \epsilon }$

$\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }=-\frac{1}{kT}\frac{e^{{\displaystyle \raisebox{1ex}{$(\epsilon -\mu )$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}}{{(e^{{\displaystyle \raisebox{1ex}{$(\epsilon -\mu )$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}} + 1)}^2}$

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

$dx = \frac{d\epsilon }{kT}$

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

$\epsilon \to \infty$            $x\to \infty$

$\epsilon \to 0$               $x\to -\frac{\mu }{kT}$ 

As $kT \ll \mu$, we can put $-\infty$ as the lower limit of integration.

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

$U = \frac{N}{2{\epsilon }_F}{\int }_{-\infty }^{\infty }{\epsilon }^2\frac{e^x}{{(e^x + 1)}^2}dx$                                              (Let this equation be ($4$))

Step 3: finding the values of integrals

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

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

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

The equation ($4$) will become:

$U = \frac{N}{2{\epsilon }_F}[I_1 + I_2 + I_3]$                                                                   (Let this equation be ($5$))

where,

 $$\begin{gathered} I_1 = {\mu }^2 {\int }_{-\infty }^{\infty }\frac{e^x}{{(e^x + 1)}^2}dx \\ {I}_2 = 2kT\mu {\int }_{-\infty }^{\infty }\frac{x{e}^x}{{(e^x + 1)}^2}dx \\ {I}_3 = {\left(kT\right)}^2 {\int }_{-\infty }^{\infty }\frac{x^2{e}^x}{{(e^x + 1)}^2}dx \end{gathered}$$

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

$I_1 = {\mu }^2$

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

$I_2 = 0$

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

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

$U = \frac{N}{2{\epsilon }_F}\left[{\mu }^2 + \frac{{\left(kT\pi \right)}^2}{3}\right]$

Let $\mu = {\epsilon }_F$, we get:

$$\begin{gathered} U = \frac{N}{2{\epsilon }_F}\left[{\epsilon }_F^2 + \frac{{\left(kT\pi \right)}^2}{3}\right] \\ U = \frac{N{\epsilon }_F}{2} + \frac{N{\left(kT\pi \right)}^2}{6{\epsilon }_F} \end{gathered}$$

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

$$\begin{gathered} C_V = \frac{\partial U}{\partial T} \\ {C}_V = \frac{\partial \left[{\displaystyle \frac{N{\epsilon }_F}{2}}+{\displaystyle \frac{N{\left(kT\pi \right)}^2}{6{\epsilon }_F}}\right]}{\partial T} \\ {C}_V = \frac{N{\left(k\pi \right)}^{2}T}{3{\epsilon }_F} \end{gathered}$$

Step 4: Graph

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