The <chemical potential> is the change in <internal energy> when one particle is added at fixed <entropy> and <volume>:
$$
\mu=\left(\frac{\partial E}{\partial N}\right)_{S,V}.
$$
Equivalently, $\mu=(\partial F/\partial N)_{T,V}$ for the <Helmholtz free energy>.

Let a microstate $s$ have <energy> $E_s$ and particle number $N_s$. Maximizing the <Gibbs entropy> subject to normalization and fixed mean values of $E$ and $N$ is the <maximum-entropy derivation of equilibrium ensembles>. With <Lagrange multipliers> $\alpha$, $\beta$, and $-\beta\mu$, vary
$$
-\sum_s p_s\log p_s
-\alpha\left(\sum_s p_s-1\right)
-\beta\sum_s p_s(E_s-\mu N_s).
$$
The stationarity equation is $-\log p_s-1-\alpha-\beta(E_s-\mu N_s)=0$. Normalization therefore gives the <grand canonical ensemble>
$$
\boxed{
p_s=\frac{e^{-\beta(E_s-\mu N_s)}}{\mathcal Z},
\qquad
\mathcal Z=\sum_s e^{-\beta(E_s-\mu N_s)},
\qquad
\beta=\frac1{k_BT}.}
$$
Here $\mathcal Z$ is the <grand canonical partition function> and $k_B$ is the <Boltzmann constant>.

For one fermionic <quantum state> of energy $\varepsilon$, the <Pauli exclusion principle> permits occupation numbers only $n=0,1$. Its two grand-canonical weights are $1$ and $e^{-\beta(\varepsilon-\mu)}$, so its mean occupation is the <Fermi-Dirac distribution>
$$
\boxed{
n_F(\varepsilon)
=\frac{e^{-\beta(\varepsilon-\mu)}}{1+e^{-\beta(\varepsilon-\mu)}}
=\frac1{e^{\beta(\varepsilon-\mu)}+1}.}
$$

For a free nonrelativistic particle, $\varepsilon=\hbar^2k^2/(2m)$. In a region of area $A$, the number of wave-vector states in the annulus from $k$ to $k+dk$, including the two <spin angular momentum> states, is
$$
2\frac{A}{(2\pi)^2}2\pi k\,dk.
$$
Since $k\,dk=(m/\hbar^2)d\varepsilon$, the <two-dimensional free-electron density of states> is constant:
$$
\boxed{g(\varepsilon)=\frac{Am}{\pi\hbar^2}.}
$$

We now use units in which $k_B=1$, as in the question. At zero <temperature>, the <Fermi-Dirac distribution> is a <step function>, and hence
$$
N=g\int_0^{\varepsilon_F}d\varepsilon=g\varepsilon_F.
$$
At positive temperature the same fixed particle number satisfies the exact relation
$$
N=g\int_0^\infty\frac{d\varepsilon}{e^{(\varepsilon-\mu)/T}+1}
=gT\log(1+e^{\mu/T}).
$$
Thus
$$
\mu=T\log(e^{\varepsilon_F/T}-1)
=\varepsilon_F+T\log(1-e^{-\varepsilon_F/T})
=\varepsilon_F+O(Te^{-\varepsilon_F/T}).
$$
This is the <low-temperature particle-number cancellation for constant density of states>. Therefore, for $T\ll\varepsilon_F$, the mean number of particles in the energy interval $[\varepsilon,\varepsilon+d\varepsilon]$ is
$$
\boxed{
g(\varepsilon)n_F(\varepsilon)d\varepsilon
\simeq
\frac{N}{\varepsilon_F}
\frac{d\varepsilon}{e^{(\varepsilon-\varepsilon_F)/T}+1}.}
$$

Finally, compare the <internal energy> with its zero-temperature value. The thermally excited particles above $\varepsilon_F$ and holes below $\varepsilon_F$ have equal leading particle numbers, so their terms proportional to $\varepsilon_F$ cancel. With $x=|\varepsilon-\varepsilon_F|/T$,
$$
U(T)-U(0)
=2gT^2\int_0^\infty\frac{x\,dx}{e^x+1}
=\frac{\pi^2}{6}gT^2,
$$
where the integral follows from the <fermion-to-boson thermal integral ratio>. Differentiation gives the <linear low-temperature heat capacity of a two-dimensional Fermi gas>:
$$
\boxed{
C_A=\left(\frac{\partial U}{\partial T}\right)_{N,A}
=\frac{\pi^2}{3}gT
=\frac{\pi^2N}{3\varepsilon_F}T.}
$$
The requested <power law> is therefore \b[linear in $T$], with exponent one.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 35D / Solution

For a two-dimensional <Fermi gas> of $N$ spin-one-half particles with <Fermi energy> $E_F$, the constant <density of states> makes the low-temperature <internal energy> correction and <heat capacity at constant area>
$$
U(T)-U(0)=\frac{\pi^2N}{6E_F}(k_BT)^2+o(T^2),
\qquad
C_A=\frac{\pi^2Nk_B^2}{3E_F}T+o(T).
$$
Thus its low-temperature heat capacity obeys a <power law> with exponent one.

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 Incoming links: Linear low-temperature heat capacity of a two-dimensional Fermi gas