Degenerate Fermi Gases

The most familiar degenerate gas is the conduction electrons inside a chunk of metal. At room temperature an electron’s quantum volume is \(v_Q \approx (4.3\ \mathrm{nm})^3\), while a typical metal has about one conduction electron per \((0.2\ \mathrm{nm})^3\). So \(V/N \ll v_Q\): the opposite of the Boltzmann limit, even though room temperature isn’t especially cold.

At \(T = 0\) the Fermi–Dirac distribution becomes a step function. Every state below \(\mu\) is occupied and every state above it is empty, and the Fermi energy is defined as \(\varepsilon_F \equiv \mu(T=0)\). Build the gas by adding one electron at a time into the lowest available state. For a box of side \(L\), the allowed energies are

so the energy grows with the squared distance from the origin in “\(n\)-space.” Each lattice point takes two electrons (spin up and down). In three dimensions the filled region is an eighth of a sphere. Drag the view to turn it, or switch to the flat two-dimensional version, a quarter of a disk.

Counting the lattice points inside an eighth-sphere of radius \(n_{\max}\), with two spin orientations each,

This is an intensive quantity: it depends only on the number density \(N/V\). Integrating \(\varepsilon(\vec n)\) over the eighth-sphere in spherical coordinates gives the total energy and the average energy per electron:

Because \(U\) depends on \(V\) through \(\varepsilon_F \propto V^{-2/3}\), the electrons push outward. This degeneracy pressure has nothing to do with electrostatic repulsion between the electrons; it exists purely by virtue of the exclusion principle. Compressing the gas shortens every wavelength, which raises every energy:

The pressure itself is largely canceled by the electrostatic forces that hold the electrons inside the metal, but the bulk modulus is not, and the free-electron formula agrees with experiment within a factor of 3 or so for most metals. Pick a metal to check.

Problems 7.23 and 7.24. A white dwarf is essentially a degenerate electron gas, with nuclei mixed in to balance the charge and provide the gravitational attraction. Treat it as a uniform sphere with one proton and one neutron per electron, so \(N = M/2m_p\). The gravitational energy (from assembling the sphere shell by shell) and the kinetic energy of the electrons are

The total, \(A/R^2 - B/R\), has a minimum at \(R_{\text{eq}} = 2A/B \propto M^{-1/3}\): heavier white dwarfs are smaller. A neutron star is the same calculation with \(N = M/m_n\) and the neutron mass in place of \(m_e\).

Problem 7.22. If the particles are highly relativistic, \(\varepsilon = pc = (hc/2L)\sqrt{n_x^2+n_y^2+n_z^2}\), and the same counting gives

Now both \(U_{\text{kinetic}}\) and \(U_{\text{grav}}\) scale as \(1/R\), so \(U_{\text{total}} = C/R\) has no stable equilibrium radius, whatever the sign of \(C\). Since \(\varepsilon_F \propto M^{4/3}\), the average kinetic energy reaches \(mc^2\) at about 3 solar masses for a white dwarf and about 12 for a neutron star in this simple model.

At small but nonzero \(T\), electrons deep in the sea can’t absorb thermal energy: every nearby state is already full. Only those within about \(k_{\mathrm{B}}T\) of the Fermi energy can, and there are about \(N k_{\mathrm{B}}T/\varepsilon_F\) of them, each gaining about \(k_{\mathrm{B}}T\). So the extra energy goes as \(N(k_{\mathrm{B}}T)^2/\varepsilon_F\), and the heat capacity is proportional to \(T\), vanishing as \(T \to 0\) in agreement with the third law and with experiments on metals at low temperatures.

To be quantitative, rewrite the energy integral in terms of \(\varepsilon\) to find the density of states, the number of single-particle states per unit energy:

At any temperature, multiply by the probability that each state is occupied:

Because \(\bar{n}_{\mathrm{FD}}\) is symmetric about \(\mu\) but \(g(\varepsilon)\) increases with \(\varepsilon\), keeping the area (that is, \(N\)) fixed forces \(\mu\) to drop below \(\varepsilon_F\) as the gas warms. The demo solves these integrals numerically and compares them with the Sommerfeld expansion.

Integrate by parts so that \(d\bar{n}_{\mathrm{FD}}/d\varepsilon\), sharply peaked at \(\mu\), appears in the integrand. With \(x = (\varepsilon-\mu)/k_{\mathrm{B}}T\), extend the lower limit to \(-\infty\) and expand \(\varepsilon^{3/2}\) about \(\mu\). Three integrals are needed:

They give

That is a lot of work for a factor of \(\pi^2/4\) already guessed by dimensional analysis, but it is typical of how physicists make controlled approximations.

Problems 7.28 and 7.31. For fermions confined to an area \(A = L^2\), a quarter-disk in \(n\)-space gives

The density of states is a constant, so the number integral can be done exactly:

For \(k_{\mathrm{B}}T \ll \varepsilon_F\) this is \(\mu \approx \varepsilon_F\), and for \(k_{\mathrm{B}}T \gg \varepsilon_F\) it becomes \(\mu \approx k_{\mathrm{B}}T\ln(\varepsilon_F/k_{\mathrm{B}}T) = -k_{\mathrm{B}}T\ln(2A/NA_Q)\), the ordinary ideal gas result with the quantum area \(A_Q = h^2/2\pi m k_{\mathrm{B}}T\) and \(Z_{\text{int}} = 2\) for spin. The energy integral still needs the Sommerfeld approach, giving \(C_V = \pi^2 N k_{\mathrm{B}}^2 T/3\varepsilon_F\) at low temperature and \(C_V = Nk_{\mathrm{B}}\), the classical equipartition value for two degrees of freedom, at high temperature. Switch the demo to 2D gas to compare.