Fermions

The pressure for non-interacting fermions with degeneracy $ g $ is

\[ P = \frac{g T}{2 \pi^2} \int_0^{\infty} k^2~dk~\ln \left[ 1 + e^{-(E-\mu)/T}\right] = \frac{g}{2 \pi^2} \int_0^{\infty} k^2\left(\frac{k^2}{3 E}\right)~dk~ \frac{1}{1 + e^{-(E-\mu)/T}} \, , \]

where the second form is obtained with an integration by parts. We use units where $\hbar=c=1$. The variable substitutions from Johns96 are $ \ell = k/m $, $\psi = (\mu-m)/T$, and $t=T/m$. (Presumably this choice of variables gives better results for non-relativistic fermions because the mass is separated from the chemical potential in the definition of $\psi$, but I haven't checked this). These replacements give

\[ P = \frac{g m^4}{2 \pi^2} \int_0^{\infty} d\ell~\frac{\ell^4}{3 \sqrt{\ell^2+1}} \left( \frac{1}{1 + e^{z/t-\psi}} \right) \]

where $ z = \sqrt{\ell^2+1}-1$ . Re-expressing in terms of $z$, one obtains

\[ \frac{\ell^4}{3 \sqrt{\ell^2+1}} = \frac{z^2(2+z)^2} {3 (1+z)} \quad\mathrm{and}\quad \frac{d \ell}{d z} = \frac{1+z}{\sqrt{z(2+z)}} \, . \]

The pressure is

\[ P = \frac{g m^4}{2 \pi^2} \int_0^{\infty} dz~\frac{1}{3}[z(2+z)]^{3/2} \left[ \frac{1}{1 + e^{(z-x)/t}} \right] \, . \]

where $x = \psi t = (\mu-m)/m$.

Degenerate expansion

The Sommerfeld expansion for $t \rightarrow 0$ is

\begin{eqnarray} \int_0^{\infty} dz~\frac{f(z)}{1 + e^{(z-x)/t}} &=& \int_0^{x} f(z) + \frac{\pi^2 t^2}{6} f^{\prime}(x) + \frac{7 \pi^4 t^4}{360} f^{(3)}(x) + \frac{31 \pi^6 t^6}{15120} f^{(5)}(x) + \ldots \nonumber \\ &=& \int_0^{x} f(z) + \sum_{n=1}^{\infty} \pi^{2n}t^{2n} \left[f^{(2n -1)}(x) \right] \left[ \frac{2 (-1)^{1+n}(2^{2n-1}-1)B_{2n}}{(2n)!} \right] \nonumber \end{eqnarray}

This is an asymptotic expansion, and must thus be used with care. Define $\tilde{P}(x,t) \equiv 2 \pi^2 P/(g m^4)$. The first term in the Sommerfeld expansion for $\tilde{P}$ depends only on $x$ alone:

\[ P_0 \equiv \frac{1}{24} (1+x)\sqrt{x(2+x)} \left[ -3 + 2 x(2+x)\right] + \frac{1}{4} \log \left[ \frac{ \sqrt{x}+\sqrt{2+x}}{\sqrt{2}} \right] \]

where $ x = \psi t$ . This expression cannot be used when $x$ is small, but a Taylor series expansion can be used instead. A few terms are

\begin{eqnarray} \frac{2 \pi^2 P}{g m^4} = P_0 + \frac{\pi^2 t^2}{6} \sqrt{x(2+x)}(1 + x) + \frac{7 \pi^4 t^4}{360} \left\{\frac{(1+x)(2 x^2+4x-1)}{[x(2+x)]^{3/2}} \right\} -\frac{31\pi^6 t^6}{1008} \frac{(1+x)\sqrt{x(2+x)}}{x^4 (2+x)^4} + \ldots \nonumber \end{eqnarray}

The number density is

\[ n = \frac{dP}{d \mu} = \frac{d P}{d x} \frac{d x}{d \mu} = \frac{1}{m} \left(\frac{d P}{d x}\right)_t \]

Note that because the density is a derivative, it is possible that the terms in the density fail before the terms in the pressure, thus we should use one less term for the density when using the expansion. The entropy is

\[ s = \frac{dP}{d T} = \frac{d P}{d t} \frac{d t}{d T} = \frac{1}{m} \left(\frac{d P}{d t}\right)_x \]

The derivative of the number density with respect to the chemical potential is

\[ \frac{d n}{d \mu} = \frac{d^2P}{d \mu^2} = \frac{d}{d \mu} \left(\frac{d P}{d x} \frac{d x}{d \mu}\right) = \frac{d^2 P}{d x^2} \left(\frac{d x}{d \mu}\right)^2 + \frac{d P}{d x} \frac{d^2 x}{d \mu^2} = \frac{1}{m^2} \left(\frac{d^2 P}{d x^2}\right)_t \, . \]

The derivative of the number density with respect to the temperature is

\[ \frac{d n}{d T} = \frac{d^2P}{d \mu dT} = \frac{1}{m^2} \frac{d^2 P}{d x d t} \, , \]

and the derivative of the entropy density with respect to the temperature is

\[ \frac{d s}{d T} = \frac{d^2P}{d T^2} = \frac{1}{m^2} \left(\frac{d^2 P}{d t^2}\right)_x \, . \]

Finally, the derivative of the number density with respect to the mass is more involved because of the mass-dependent prefactor.

\begin{eqnarray} \frac{d n}{d m} &=& \frac{4 n}{m}+ \left(\frac{g m^4}{2 \pi^2}\right) \frac{d}{d m} \left(\frac{1}{m}\frac{d \tilde{P}}{d x} \right) = \frac{4 n}{m} + \left(\frac{g m^4}{2 \pi^2}\right) \left[\frac{1}{m}\left(\frac{d^2\tilde{P}}{dx^2}\frac{dx}{dm}+ \frac{d^2\tilde{P}}{dt dx}\frac{dt}{dm}\right)- \frac{1}{m^2}\frac{d \tilde{P}}{d x}\right] \nonumber \\ &=& \frac{4 n}{m} - \left(\frac{g m^2}{2 \pi^2}\right) \left( \frac{d\tilde{P}}{dx} +\frac{\mu}{m} \frac{d^2\tilde{P}}{dx^2} +\frac{T}{m} \frac{d^2\tilde{P}}{dt dx} \right) = \frac{3n}{m} -\left[(x+1) \left(\frac{dn}{d\mu}\right) + t \left(\frac{dn}{dT}\right) \right] \nonumber \end{eqnarray}

These expansions are used in o2scl::fermion_eval_thermo::calc_mu_deg() and o2scl::fermion_deriv_thermo::calc_mu_deg() .

Nondegenerate Expansion

There is a useful identity (Chandrasekhar10 and Tooper69)

\[ \int_0^{\infty} \frac{x^4 \left(x^2+z^2\right)^{-1/2}~dx} {1+e^{\sqrt{x^2+z^2}-\phi}} = 3 z^2 \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} e^{n \phi} K_2(n z) \]

which works well when $\phi-z < -1$. This result directly gives the sum in Johns96

\[ P = \frac{g m^4}{2 \pi^2} \sum_{k=1}^{\infty} P_k \equiv \frac{g m^4}{2 \pi^2} \left[ \sum_{k=1}^{\infty} \frac{t^2 (-1)^{k+1}}{k^2} e^{k \psi} e^{k/t} K_2\left(\frac{k}{t}\right) \right] \]

The function $ e^{x} K_2(x) $ is implemented in GSL as gsl_sf_bessel_Kn_scaled. In the case that one wants to include antiparticles, the result is similar

\[ P = \frac{g m^4}{2 \pi^2} \sum_{k=1}^{\infty} Q_k \equiv \frac{g m^4}{2 \pi^2} \left\{ \sum_{k=1}^{\infty} \frac{2 t^2 (-1)^{k+1}}{k^2} e^{-k/t} \mathrm{cosh} \left[k \mu/(t m)\right] \left[ e^{k/t} K_2\left(\frac{k}{t}\right) \right] \right\} \]

where the scaled Bessel function has been separated out. The density is

\[ n = \frac{g m^4}{2 \pi^2} \sum_{k=1}^{\infty} \frac{k}{T}{P_k} \]

and

\[ n = \frac{g m^4}{2 \pi^2} \sum_{k=1}^{\infty} \frac{k}{T}{Q_k} \mathrm{tanh} \left[k \mu/(t m)\right] \]

if antiparticles are included. The entropy density can also be written down similarly.

These expansions are used in o2scl::fermion_eval_thermo::calc_mu_ndeg() .

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