8.10. Ideal Fermi Gas

We start with a grand potential for fermions and use a Thomas-Fermi approximation (that allows us to change the discrete sum below into a continuous integral):

\Omega(\beta, V, \mu) = -\sum_i {1\over\beta} \log\left(\sum_{N=0}^1 e^{-\beta\left(N\epsilon_i - N\mu\right)}\right) = = -\sum_i {1\over\beta} \log\left(1 + e^{-\beta\left(\epsilon_i - \mu\right)}\right) = = -{1\over\beta} \int \int {2\d^3 x \d^3 p \over (2\pi)^3} \log\left(1 + e^{-\beta\left({p^2\over 2} - \mu\right)}\right) = = -{2\over\beta} \int \d^3 x \int_0^\infty{ 4\pi p^2 \d p \over (2\pi)^3} \log\left(1 + e^{-\beta\left({p^2\over 2} - \mu\right)}\right) = = -{1\over \pi^2 \beta} \int \d^3 x \int_0^\infty p^2 \log\left(1 + e^{-\beta\left({p^2\over 2} - \mu\right)}\right) \d p = = -{2\sqrt2 \over 3 \pi^2 \beta^{5\over2}} \int \d^3 x \int_0^\infty {u^{3\over2} \over 1 + e^{u-\beta\mu}} \d u = = -{2\sqrt2 \over 3 \pi^2 \beta^{5\over2}} \int I_{3\over2}\left(\beta\mu\right) \,\d^3 x = = -{2\sqrt2 V \over 3 \pi^2 \beta^{5\over2}} I_{3\over2}\left(\beta\mu\right) \,.

Note: to write this thermodynamic potential in the canonical form \Omega=\Omega(T, V, \mu), we simply use the relation \beta = {1 \over k_B T} and get:

\Omega(T, V, \mu) = -{2\sqrt2 V (k_B T)^{5\over2} \over 3 \pi^2} I_{3\over2}\left(\mu\over k_B T\right) \,.

Let us compute the particle density:

n_e = - \left({\partial^2 \Omega(\beta, V, \mu) \over \partial V \partial \mu}\right)_\beta = {2\sqrt2 \over 3 \pi^2 \beta^{5\over2}} {\partial \over \partial \mu} I_{3\over2}\left(\beta\mu\right) = {2\sqrt2 \over 3 \pi^2 \beta^{5\over2}} \beta {3\over 2} I_{1\over2} \left(\beta\mu\right) = {\sqrt2 \over \pi^2 \beta^{3\over2}} I_{1\over2} \left(\beta\mu\right)

and express the chemical potential \mu=\mu(n_e) as a function of the particle density n_e:

(8.10.1)\mu = {1\over\beta} I_{1\over2}^{-1}\left( {\pi^2 \beta^{3\over2} \over \sqrt 2} n_e \right)

We write the grand potential using n_e as follows:

(8.10.2)\Omega(\beta, V, n_e) = -{2\sqrt2 V \over 3 \pi^2 \beta^{5\over2}} I_{3\over2}\left( I_{1\over2}^{-1}\left( {\pi^2 \beta^{3\over2} \over \sqrt 2} n_e \right) \right)\,.

Now we can calculate the free energy:

F_e(\beta, V, n_e) = \Omega(\beta, V, n_e) + \mu N = = -{2\sqrt2 V \over 3 \pi^2 \beta^{5\over2}} I_{3\over2}\left(\beta\mu \right) + \mu n_e V = = -{2\sqrt2 V \over 3 \pi^2 \beta^{5\over2}} I_{3\over2}\left( I_{1\over2}^{-1}\left( {\pi^2 \beta^{3\over2} \over \sqrt 2} n_e \right) \right) + {1\over\beta} I_{1\over2}^{-1}\left( {\pi^2 \beta^{3\over2} \over \sqrt 2} n_e \right) n_e V

where we used (8.10.2), (8.10.1) and the fact that n_e = N / V. Note: we can express the free energy in canonical form F = F(T, V, N) using \beta = {1 \over k_B T} and n_e = N / V:

F_e(T, V, N) = -{2\sqrt2 V (k_B T)^{5\over2} \over 3 \pi^2 } I_{3\over2}\left( I_{1\over2}^{-1}\left( {\pi^2 N \over \sqrt 2 (k_B T)^{3\over2} V} \right) \right) + k_B T I_{1\over2}^{-1}\left( {\pi^2 N \over \sqrt 2 (k_B T)^{3\over2} V} \right) N \,.

We can calculate the entropy S=-\left(\partial\Omega\over\partial T\right)_{V,\mu} as follows:

TS =-T \left(\partial\Omega\over\partial T\right)_{V,\mu} = =\beta \left(\partial\Omega\over\partial \beta\right)_{V,\mu} = =\beta {\partial\over\partial \beta}\left( -{2\sqrt2 V \over 3 \pi^2 \beta^{5\over2}} I_{3\over2}\left(\beta\mu\right) \right) = =\beta \left( {5\over2}{2\sqrt2 V \over 3 \pi^2 \beta^{7\over2}} I_{3\over2}(\beta\mu) -{2\sqrt2 V \over 3 \pi^2 \beta^{5\over2}} {3\over2} I_{1\over2}(\beta\mu) \mu \right) = = {5\over3}{\sqrt2 V \over \pi^2 \beta^{5\over2}} I_{3\over2}(\beta\mu) -{\sqrt2\over \pi^2 \beta^{3\over2}} I_{1\over2}(\beta\mu) \mu V = = {5\over3}{\sqrt2 V \over \pi^2 \beta^{5\over2}} I_{3\over2}(\beta\mu) -n_e \mu V = = {5\over3}{\sqrt2 V \over \pi^2 \beta^{5\over2}} I_{3\over2}(\beta\mu) -\mu N \,.

The total energy U is then equal to:

U = \Omega + \mu N + TS = = -{2\sqrt2 V \over 3 \pi^2 \beta^{5\over2}} I_{3\over2}\left(\beta\mu\right) + \mu N + {5\over3}{\sqrt2 V \over \pi^2 \beta^{5\over2}} I_{3\over2}(\beta\mu) -\mu N = = {\sqrt2 V \over \pi^2 \beta^{5\over2}} I_{3\over2}\left(\beta\mu\right) \,.

Note: the kinetic energy E_{kin} = U is equal to the total energy, as the gas is non-interacting.

The pressure p can be calculated from:

p = - \left(\partial\Omega\over\partial V\right)_{\mu,\beta} = {2\sqrt2 \over 3 \pi^2 \beta^{5\over2}} I_{3\over2}\left(\beta\mu\right) = = {2\sqrt2 \over 3 \pi^2 \beta^{5\over2}} I_{3\over2}\left( I_{1\over2}^{-1}\left( {\pi^2 \beta^{3\over2} \over \sqrt 2} n_e \right) \right) \,.

Note that we got p = {2 U \over 3 V}, \Omega=-{2\over3} U, F=-{2\over3} U + \mu N and TS = {5\over 3} U -\mu N.

8.10.1. Low Temperature Limit

At low temperature (T\to0) we have \beta \to \infty, I_{1\over2}(x) \to {2\over3} x^{3\over 2} (for x\to\infty) and we obtain:

n_e = {\sqrt2 \over \pi^2 \beta^{3\over2}} I_{1\over2} \left(\beta\mu\right) \to {2\sqrt 2\over 3\pi^2 \beta^{3\over2}} (\beta\mu)^{3\over2} ={(2\mu)^{3\over2} \over 3\pi^2}

Identical with the zero temperature Thomas-Fermi equation where the chemical potential \mu = E_f = {p_f^2\over 2} becomes the Fermi energy in the limit T \to 0. We now express \mu in terms of n_e at T=0:

\mu \to \half (3\pi^2 n_e)^{2\over 3}

and compute pressure at T=0 using I_{3\over2}(x) \to {2\over5} x^{5\over 2} for x\to\infty:

p = {2\sqrt2 \over 3 \pi^2 \beta^{5\over2}} I_{3\over2}\left(\beta\mu\right) \to {2\sqrt2 \over 3 \pi^2 \beta^{5\over2}} I_{3\over2}\left(\beta \half (3\pi^2 n_e)^{2\over 3} \right) \to {2\sqrt2 \over 3 \pi^2 \beta^{5\over2}} {2\over 5} \left(\beta \half (3\pi^2 n_e)^{2\over 3} \right)^{5\over 2} = {3^{2\over 3} \pi^{4\over 3} \over 5} n_e^{5\over 3}

8.10.2. High Temperature Limit

At high temperature (T\to\infty) we have \beta \to 0, I_{1\over2}(x) \to {\sqrt\pi\over2} e^x (for x\to0) and we obtain:

n_e = {\sqrt2 \over \pi^2 \beta^{3\over2}} I_{1\over2} \left(\beta\mu\right) \to {\sqrt 2\over\pi^2 \beta^{3\over2}} {\sqrt\pi\over2} e^{\beta\mu}

We now express \mu in terms of n_e at T\to\infty:

\mu \to {1\over\beta}\log\left( n_e {2\pi^2 \beta^{3\over2} \over \sqrt 2 \sqrt\pi} \right)

In the limit \beta\to0 we get \beta\mu\to-\infty. Let us compute pressure at T\to\infty using I_{3\over2}(x) \to {3\sqrt\pi\over4} e^x for x\to-\infty:

p = {2\sqrt2 \over 3 \pi^2 \beta^{5\over2}} I_{3\over2}\left(\beta\mu\right) \to {2\sqrt2 \over 3 \pi^2 \beta^{5\over2}} I_{3\over2}\left( \log\left( n_e \over {\sqrt 2\over\pi^2 \beta^{3\over2}} {\sqrt\pi\over2} \right) \right) \to \to {2\sqrt2 \over 3 \pi^2 \beta^{5\over2}} {3\sqrt\pi\over 4} \exp\log\left( n_e \over {\sqrt 2\over\pi^2 \beta^{3\over2}} {\sqrt\pi\over2} \right) = = {n_e \over \beta} = n_e k_B T = {N k_B T \over V}\,.

We obtained the ideal gas equation pV = N k_B T.