Fermions

Fermion integrations

(Slowly moving the documentation from fermion_rel and fermion_deriv_rel to this location.)

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 $ . In the case where $ \nu $ includes the rest mass (still denoted $ m $), the fermionic distribution function is

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

where the left expression is used when the chemical potential includes the rest mass and the energy density includes the rest mass energy density, (o2scl::part::inc_rest_mass is true) and the right expression is used when the rest mass is not included (o2scl::part::inc_rest_mass is false). For convenience, we define $ E^{*} \equiv \sqrt{k^2+m^{* 2}} $ .

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

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

Evaluation of the 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 direct) 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

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

2) 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

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

3) 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

\[ \left(\frac{d s}{d \mu}\right)_T = \left(\frac{d n}{d T}\right)_{\mu} \]

4) 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 $

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

5) 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{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

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

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{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*}

The entropy terms are

\begin{eqnarray*} s_k &=& \frac{2 P_k}{t} - \frac{k (x+1) P_k}{t^2} + \frac{(-1)^{k+1}}{2k} e^{k x/t} e^{k/t} \left[ K_1(k/t) + K_3(k/t)\right] \nonumber \\ \bar{s}_k &=& \frac{2 \bar{P}_k}{t} + \frac{k \bar{P}_k}{t^2} - \frac{k (x+1) \bar{P}_k \mathrm{tanh}[k(x+1)/t]}{t^2} + \frac{(-1)^{k+1}}{k} e^{-k/t} \mathrm{cosh}[k(x+1)/t] e^{k/t} \left[ K_1(k/t) + K_3(k/t)\right] \end{eqnarray*}

if antiparticles are included. For the derivatives

\begin{eqnarray*} \left(\frac{dn}{d\mu}\right)_k &=& \frac{k^2}{t^2}{P_k} \\ \left(\frac{d\bar{n}}{d\mu}\right)_k &=& \frac{k^2}{t^2}{\bar{P}_k} \\ \left(\frac{dn}{dT}\right)_k &=& \frac{k}{t} s_k - \frac{k}{t^2} P_k \\ \left(\frac{d\bar{n}}{dT}\right)_k &=& \left( \frac{k}{t} \bar{s}_k - \frac{k}{t^2} \bar{P}_k \right) \mathrm{tanh}\left[k(x+1)/t\right] - \frac{k^2(x+1)}{t^2} \bar{P}_k \mathrm{sech}^2 \left[k(x+1)/t\right] \\ \left(\frac{ds}{dT}\right)_k &=& \left[ \frac{2 k (x+1)-2 t}{t^3}\right] P_k + \left[ \frac{2 t-k (x+1)}{t^2}\right] s_k \\ && - \frac{(x+1)(-1)^{k+1}}{2 t^2} e^{k x/t} e^{k/t} \left[ K_1(k/t)+K_3(k/t) \right] - \frac{(-1)^{k+1}}{4 k} e^{k x/t} e^{k/t} \left[ K_0 (k/t) + 2 K_2 (k/t) + K_4 (k/t)\right] \\ \left(\frac{d\bar{s}}{dT}\right)_k &=& \left[ \frac{-2(t+k)}{t^3}\right] \bar{P}_k + \left[ \frac{2 t+k}{t^2}\right] \bar{s}_k - \frac{k (x+1) \bar{s}_k \mathrm{tanh}[k(x+1)/t]}{t^2} + \frac{k^2 (x+1)^2 \bar{P}_k \mathrm{sech}^2[k(x+1)/t]}{t^4} \\ && - \frac{(x+1)(-1)^{k+1}}{2 t^2} e^{-k/t} \mathrm{sinh}\left[k(x+1)/t\right]e^{k/t} \left[ K_1(k/t)+K_3(k/t) \right] - \frac{(-1)^{k+1}}{2 k} e^{-k/t} \mathrm{cosh}\left[k(x+1)/t\right] e^{k/t} \left[ K_0 (k/t) + 2 K_2 (k/t) + K_4 (k/t)\right] \end{eqnarray*}

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

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