# Fermion Details¶

O2scl

## Interacting and non-interacting fermions¶

In many cases, the non-interacting expressions for fermion thermodynamics can be used in interacting systems as long as one replaces the mass with an effective mass, $$m^{*}$$ and the chemical potential with an effective chemical potential, $$\nu$$ . When $$\nu$$ includes the rest mass (denoted $$m$$), o2scl::part_tl::inc_rest_mass should be true, and the fermionic distribution function is

$f = \frac{1}{1+e^{(\sqrt{k^2+m^{* 2}}-\nu)/T}}$

Then the energy density will also include the rest mass energy density, $$n m$$. When $$\nu$$ does not include the rest mass, the fermionic distribution function is

$f = \frac{1}{1+e^{(\sqrt{k^2+m^{* 2}}-\nu-m)/T}}$

For convenience, we often define $$E^{*} \equiv \sqrt{k^2+m^{* 2}}$$. In the documentation below, expressions to the left of the semicolon apply when inc_rest_mass is true, and expressions to the right of the semicolon apply when inc_rest_mass is false.

## Relativistic versus non-relativistic fermions¶

There are a few distinctions between how relativistic and nonrelativistic fermions are handled in o2scl which are worth noting. For an interacting relativistic fermion, the effective mass, $$m^{*}$$, and the effective chemical potential, $$\nu$$ are defined so that the energy density is

$\begin{split}\begin{eqnarray} {\varepsilon}_{\mathrm{R}} &=& \frac{g}{2 \pi^2} \int dk~\frac{k^2 \sqrt{k^2+m^{* 2}}} { 1+\exp\left[\left(\sqrt{k^2+m^{*2}}- \nu_{\mathrm{R}}\right)/T\right]} \\ &=& \frac{g}{2 \pi^2} \int dk~\frac{k^2 \sqrt{k^2+m^{* 2}} } {1+\exp\left[\left(\sqrt{k^2+m^{*2}}- \bar{\nu}_{\mathrm{R}}-m\right)/T\right]} \end{eqnarray}\end{split}$

for a relativistic fermion, where we define the chemical potential without the rest mass with $$\bar{\nu}_{\mathrm{R}} \equiv \nu_{\mathrm{R}}-m$$. When o2scl::part_tl::inc_rest_mass is true, o2scl::part_tl::nu is equal to $$\nu_{\mathrm{R}}$$ and when o2scl::part_tl::inc_rest_mass is false, o2scl::part_tl::nu is equal to $$\bar{\nu}_{\mathrm{R}}$$ . If we define $$\psi_{\mathrm{R}} = (\nu_{\mathrm{R}}-m^{*})/T$$, $$\phi = m^{*}/T$$, $$u \equiv k/T$$ then the energy density is

${\varepsilon}_{\mathrm{R}} = \frac{g T^4}{2 \pi^2} \int du~\frac{u^2 \sqrt{u^2+\phi^2}} { 1+\exp\left[\sqrt{u^2+\phi^2} - \psi_{\mathrm{R}} - \phi \right]}$

This expression is used for o2scl::part_calibrate_class_tl::part_calibrate() because $$\varepsilon_{\mathrm{R}}/(g T^4)$$ depends only on $$\psi_{\mathrm{R}}$$ and $$\phi$$. For a nonrelativistic fermion,

$\begin{split}\begin{eqnarray} \bar{\varepsilon}_{\mathrm{NR}} &=& \frac{g}{2 \pi^2} \int dk~ \frac{k^4}{2 m^{*}} \left\{ 1+\exp\left[\left(\frac{k^2}{2 m^{*}}- \bar{\nu}_{\mathrm{NR}}\right)/T\right] \right\}^{-1} \\ &=& \frac{g}{2 \pi^2} \int dk~ \frac{k^4}{2 m^{*}} \left\{ 1+\exp\left[\left(\frac{k^2}{2 m^{*}}- \nu_{\mathrm{NR}}+m\right)/T\right] \right\}^{-1} \end{eqnarray}\end{split}$

where $$\bar{\nu}_{\mathrm{NR}} = \nu_{\mathrm{NR}} - m$$ . Note that the rest mass energy density is $$\varepsilon_{\mathrm{rest}} = n m$$ (not $$n m^{*}$$) in both cases, but it is included in $$\varepsilon_{\mathrm{R}}$$ while it is not included in $$\bar{\varepsilon}_{\mathrm{NR}}$$ . Taking the nonrelativsitic limit of the relativistic energy density shows that $$\nu_{\mathrm{R}} - m^{*} = \bar{\nu}_{\mathrm{NR}}$$. Thus the class fermion_nonrel_tl uses the value stored in o2scl::part_tl::nu slightly differently than does fermion_rel_tl and fermion_eff . The Fermi momentum is also handled slightly differently, $$k_{F,\mathrm{R}} \equiv \sqrt{\nu_{\mathrm{R}}^2-m^{* 2}}$$ and $$k_{F,\mathrm{NR}} \equiv \sqrt{2 \bar{\nu}_{\mathrm{NR}} m^{*}}$$.

Now if we define $$u_{\mathrm{NR}} \equiv k^2/(2 m^{*} T)$$ and $$\psi_{\mathrm{NR}} \equiv (\nu_{\mathrm{NR}}-m^{*})/T$$ then the argument of the exponential is

$\frac{k^2}{2 m^{*} T } - \frac{\bar{\nu}_{\mathrm{NR}}}{T} = u_{\mathrm{NR}} - \psi_{\mathrm{NR}} + \frac{m}{T}- \phi$

which is inconvenient because then $$\varepsilon_{\mathrm{NR}}/(g T^4)$$ is no longer a function of $$\psi_{\mathrm{NR}}$$ and $$\phi$$ alone. Thus we define $$\psi_{\mathrm{NR}} \equiv \bar{\nu}_{\mathrm{NR}}/T$$ and then the energy density is

$\bar{\varepsilon}_{\mathrm{NR}} = \frac{g T^4}{2 \pi^2} \int du_{\mathrm{NR}}~\frac{\sqrt{2}~u_{\mathrm{NR}}^{3/2} \phi^{3/2}} { 1+\exp\left[u_{\mathrm{NR}} - \psi_{\mathrm{NR}} \right]}$

which is now a function of $$\psi_{\mathrm{NR}}$$ and $$\phialone. This is the form used to compute the energy density in :ref:fermion_nonrel_tl <fermion_nonrel_tl>$$ and the definition of $$\psi_{\mathrm{NR}}$$ used for nonrelativistic fermions in ref o2scl::part_calibrate_class_tl::part_calibrate().

## Upper limits¶

The fermionic integrands vanish when the argument of the exponential becomes large compared to a positive number $$\zeta$$. This condition is

$\sqrt{k^2+m^{* 2}}-\nu \gg \zeta T \quad ; \quad \sqrt{k^2+m^{* 2}}-\nu-m \gg \zeta T$

Thus solving for the momentum, an upper limit, $$k_{\mathrm{ul}}$$ is

$k_{\mathrm{ul}} = \sqrt{\left(\zeta T + \nu\right)^2-m^{* 2}} \quad ; \quad k_{\mathrm{ul}} = \sqrt{\left(\zeta T + m + \nu\right)^2-m^{* 2}}$

The entropy is only significant at the Fermi surface, thus in the degenerate case, the lower limit of the entropy integral can be given be determined by the value of $$k$$ which solves

$- \zeta = \frac{\sqrt{k^2+m^{* 2}}-\nu}{T} \quad ; \quad - \zeta = \frac{\sqrt{k^2+m^{* 2}}-\nu-m}{T}$

The solution is

$k_{\mathrm{ll}} = \sqrt{(-\zeta T+{\nu})^2-m^{*,2}} \quad ; \quad k_{\mathrm{ll}} = \sqrt{(-\zeta T + m +\nu)^2-m^{*,2}}$

which is a valid lower limit only if the argument under the square root is positive.

## Integrands¶

The energy density is

$\varepsilon = \frac{g}{2 \pi^2} \int_0^{\infty} k^2~dk~\sqrt{k^2+m^{* 2}} f \quad ; \quad \varepsilon = \frac{g}{2 \pi^2} \int_0^{\infty} k^2~dk~\left(\sqrt{k^2+m^{* 2}}-m\right) f \, ,$

the number density is

$n = \frac{g}{2 \pi^2} \int_0^{\infty} k^2~dk~f \, ,$

and the entropy density is

$s = \frac{g}{2 \pi^2} \int_0^{\infty} dk~(-k^2 {\cal S})$

where

${\cal S}\equiv f \ln f +(1-f) \ln (1-f) \quad ; \quad \frac{\partial {\cal S}}{\partial f} = \ln \left(\frac{f}{1-f}\right) \, .$

The derivative can also be written

$\frac{\partial {\cal S}}{\partial f} = \left(\frac{\nu-E^{*}}{T}\right) \quad ; \quad \frac{\partial {\cal S}}{\partial f} = \left(\frac{\nu-E^{*}+m}{T}\right)$

In the degenerate regime, $${\cal S}$$, can lose precision when $$(E^{*} - \nu)/T$$ is negative and sufficiently large in absolute magnitude. Thus when $$(E^{*} - \nu)/T < \xi$$ (for $$\xi \rightarrow - \infty$$ ) an alternative expression

${\cal S} \approx e^{(E^{*}-\nu)/T} \left( \frac{E^{*} -\nu-T}{T} \right) \quad ; \quad {\cal S} \approx e^{(E^{*}-\nu-m)/T} \left( \frac{E^{*} -\nu-m-T}{T} \right) \,$

can be used.

## Non-degenerate integrands¶

The integrands in the non-degenerate regime are written in a dimensionless form, by defining $$u=(E^{*}-m^{*})/T$$ (this choice ensures $$k=0$$ corresponds to $$u=0$$), $$y \equiv \nu/ T$$ (or $$y = (\nu+m)/T$$ if the chemical potential does not include the mass), and $$\eta \equiv m^{*}/T$$. Then $$k/T = \sqrt{u^2+2 u \eta}$$, $$(1/T) dk = E^{*}/k du = (u+\eta)/\sqrt{u^2+2 u \eta}~du$$, and $$f = 1/(1+e^{u+\eta-y})$$ . The density is

$n = \frac{g T^3}{2 \pi^2} \int_0^{\infty}~du~ \sqrt{u^2+2 u \eta} (u+\eta) \left(1+e^{u+\eta-y}\right)^{-1}$

the energy density is

$\varepsilon = \frac{g T^4}{2 \pi^2} \int_0^{\infty}~du~ \sqrt{u^2+2 u \eta} (u+\eta)^2 \left(1+e^{u+\eta-y}\right)^{-1}$

and the entropy density is

$s = -\frac{g T^3}{2 \pi^2} \int_0^{\infty}~du~ \sqrt{u^2+2 u \eta} (u+\eta) {\cal S}$

## Distribution function derivatives¶

The relevant derivatives of the distribution function are

$\frac{\partial f}{\partial T}= f(1-f)\frac{E^{*}-\nu}{T^2} \quad ; \quad \frac{\partial f}{\partial T}= f(1-f)\frac{E^{*}-m-\nu}{T^2}$
$\frac{\partial f}{\partial \nu}= f(1-f)\frac{1}{T}$
$\frac{\partial f}{\partial k}= -f(1-f)\frac{k}{E^{*} T}$
$\frac{\partial f}{\partial m^{*}}= -f(1-f)\frac{m^{*}}{E^{*} T}$

The derivatives can be integrated directly or they may be converted to integrals over the distribution function through an integration by parts

$\int_a^b f(k) \frac{d g(k)}{dk} dk = \left.f(k) g(k)\right|_{k=a}^{k=b} - \int_a^b g(k) \frac{d f(k)}{dk} dk$

using the distribution function for $$f(k)$$ and 0 and $$\infty$$ as the limits, we have

$\frac{g}{2 \pi^2} \int_0^{\infty} \frac{d g(k)}{dk} f dk = \frac{g}{2 \pi^2} \int_0^{\infty} g(k) f (1-f) \frac{k}{E^{*} T} dk$

as long as $$g(k)$$ vanishes at $$k=0$$ . Rewriting using $$g(k) = h(k) E^{*} T/k$$

$\frac{g}{2 \pi^2} \int_0^{\infty} h(k) f (1-f) dk = \frac{g}{2 \pi^2} \int_0^{\infty} f \frac{T}{k} \left[ h^{\prime} E^{*}-\frac{h E^{*}}{k}+\frac{h k}{E^{*}} \right] dk$

as long as $$h(k)/k$$ vanishes at $$k=0$$ .

## Explicit forms¶

1. The derivative of the density wrt the chemical potential

$\left(\frac{d n}{d \mu}\right)_T = \frac{g}{2 \pi^2} \int_0^{\infty} \frac{k^2}{T} f (1-f) dk$

Using $$h(k)=k^2/T$$ we get

$\left(\frac{d n}{d \mu}\right)_T = \frac{g}{2 \pi^2} \int_0^{\infty} \left(\frac{k^2+E^{*2}}{E^{*}}\right) f dk$
1. The derivative of the density wrt the temperature

$\left(\frac{d n}{d T}\right)_{\mu} = \frac{g}{2 \pi^2} \int_0^{\infty} \frac{k^2(E^{*}-\nu)}{T^2} f (1-f) dk \quad ; \quad \left(\frac{d n}{d T}\right)_{\mu} = \frac{g}{2 \pi^2} \int_0^{\infty} \frac{k^2(E^{*}-m-\nu)}{T^2} f (1-f) dk$

Using $$h(k)=k^2(E^{*}-\nu)/T^2$$ we get

$\left(\frac{d n}{d T}\right)_{\mu} = \frac{g}{2 \pi^2} \int_0^{\infty} \frac{f}{T} \left[2 k^2+E^{*2}-E^{*} \nu - k^2 \left(\frac{\nu}{E^{*}}\right)\right] dk$

when the rest mass is included in the chemical potential and

$\left(\frac{d n}{d T}\right)_{\mu} = \frac{g}{2 \pi^2} \int_0^{\infty} \frac{f}{T} \left[2 k^2+E^{*2}-E^{*}\left(\nu+m\right)- k^2 \left(\frac{\nu+m}{E^{*}}\right)\right] dk$

when the rest mass is not included in the chemical potential.

1. The derivative of the entropy wrt the chemical potential

$\left(\frac{d s}{d \mu}\right)_T = \frac{g}{2 \pi^2} \int_0^{\infty} k^2 f (1-f) \frac{(E^{*}-\nu)}{T^2} dk \quad ; \quad \left(\frac{d s}{d \mu}\right)_T = \frac{g}{2 \pi^2} \int_0^{\infty} k^2 f (1-f) \frac{(E^{*}-m-\nu)}{T^2} dk$

This verifies the Maxwell relation

$\left(\frac{d s}{d \mu}\right)_T = \left(\frac{d n}{d T}\right)_{\mu}$
1. The derivative of the entropy wrt the temperature

$\left(\frac{d s}{d T}\right)_{\mu} = \frac{g}{2 \pi^2} \int_0^{\infty} k^2 f (1-f) \frac{(E^{*}-\nu)^2}{T^3} dk \quad ; \quad \left(\frac{d s}{d T}\right)_{\mu} = \frac{g}{2 \pi^2} \int_0^{\infty} k^2 f (1-f) \frac{(E^{*}-m-\nu)^2}{T^3} dk$

Using $$h(k)=k^2 (E^{*}-\nu)^2/T^3$$

$\left(\frac{d s}{d T}\right)_{\mu} = \frac{g}{2 \pi^2} \int_0^{\infty} \frac{f(E^{*}-\nu)}{E^{*}T^2} \left[E^{* 3}+3 E^{*} k^2- (E^{* 2}+k^2)\nu\right] d k$

when the rest mass is included in the chemical potential and

$\left(\frac{d s}{d T}\right)_{\mu} = \frac{g}{2 \pi^2} \int_0^{\infty} \frac{f(E^{*}-m-\nu)}{E^{*}T^2} \left[E^{* 3}+3 E^{*} k^2- (E^{* 2}+k^2)(\nu+m)\right] d k$

when the rest mass is not included in the chemical potential.

1. The derivative of the density wrt the effective mass

$\left(\frac{d n}{d m^{*}}\right)_{T,\mu} = -\frac{g}{2 \pi^2} \int_0^{\infty} \frac{k^2 m^{*}}{E^{*} T} f (1-f) dk$

Using $$h(k)=-(k^2 m^{*})/(E^{*} T)$$ we get

$\left(\frac{d n}{d m^{*}}\right)_{T,\mu} = -\frac{g}{2 \pi^2} \int_0^{\infty} m^{*} f dk$

## Expansions for fermions¶

Presuming the chemical potential includes the rest mass, and $$E=\sqrt{k^2+m^2}$$, 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

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

This is an asymptotic expansion, and must thus be used with care. In the case where $$f(z)=z^n$$,

$\int_0^{\infty} dz~\frac{f(z)}{1 + e^{(z-x)/t}} = \int_0^{x} f(z) + \sum_{n=1}^{\infty} \pi^{2n}t^{2n} (2n-1)! z^{2n-1} \left[ \frac{2 (-1)^{1+n}(2^{2n-1}-1)B_{2n}}{(2n)!} \right]$

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

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

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.

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

These expansions are used in o2scl::fermion_thermo_tl::calc_mu_deg() and o2scl::fermion_deriv_thermo_tl::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

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

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

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

where the scaled Bessel function has been separated out. Similarly defining

$n = \frac{g m^3}{2 \pi^2} \sum_{k=1}^{\infty} n_k \, ,$

the terms in the expansion for the density (without and with antiparticles) are

$\begin{split}\begin{eqnarray} n_k &=& \frac{k}{t}{P_k} \nonumber \\ \bar{n}_k &=& \frac{k}{t}{\bar{P}_k} \mathrm{tanh} \left[k (x+1)/t\right] \end{eqnarray}\end{split}$

The entropy terms (with and without antiparticles) are

$\begin{split}\begin{eqnarray} s_k &=& \left( \frac{4t-kx-k}{kt}\right) n_k + \frac{(-1)^{k+1}}{k} e^{k x/t} \left[ e^{k/t} K_1(k/t) \right] \nonumber \\ \bar{s}_k &=& -\frac{(1+x)\bar{n}_k}{t} + \frac{2(-1)^{k+1}}{k} e^{-k/t} \mathrm{cosh}[k(x+1)/t] \left[ e^{k/t} K_3(k/t) \right] \end{eqnarray}\end{split}$

included. To obtain these expressions, the recurrence relation for the modified Bessel function of the second kind has been used

$K_{\nu+1}(x) = K_{\nu-1}(x) + \frac{2 \nu}{x} K_{\nu}(x)$

For the derivatives, no additional Bessel functions are required.

$\begin{split}\begin{eqnarray} \left(\frac{dn}{d\mu}\right)_k &=& \frac{k}{t}{n_k} \\ \left(\frac{d\bar{n}}{d\mu}\right)_k &=& \frac{k}{t}{\bar{n}_k} \\ \left(\frac{dn}{dT}\right)_k &=& \frac{k}{t} s_k - \frac{1}{t} n_k \end{eqnarray}\end{split}$
$\begin{split}\begin{eqnarray} \left(\frac{d\bar{n}}{dT}\right)_k &=& \frac{k}{t} \bar{s}_k \mathrm{tanh}\left[k(x+1)/t\right] - \left\{ t+2 k (1+x) \mathrm{csch}\left[k(x+1)/t\right] \right\} \frac{\bar{n}_k}{t^2} \\ \left(\frac{ds}{dT}\right)_k &=& \left[ \frac{3t -2k x -2 k}{t^2}\right] s_k + \left[ \frac{5 k t - 2 k^2 x +5 k t x - k^2 x^2}{k t^3}\right] n_k \\ \left(\frac{d\bar{s}}{dT}\right)_k &=& \left\{2 k (1+x) \mathrm{tanh}\left[ k(1+x)/t\right] - 3 t\right\} \frac{\bar{s}_k}{t^2} + \left\{2 k^2 (1+x)^2 \mathrm{tanh}\left[ k(1+x)/t\right] - \right. \nonumber \\ && \left. k^2 (2 + 2 x + x^2) \mathrm{coth}\left[ k(1+x)/t\right] - 5 k(1+x) t \right\} \frac{\bar{n}_k}{k t^3} \end{eqnarray}\end{split}$

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