Determining free Fermi/Bose gas chemical potentials, given temperature, number density, and baryon density

Question

I'm attempting to recreate some plots of individual particle chemical potentials from S. Reddy et. al.'s paper on neutrino interactions in hot and dense matter (of the same title). The specifics aren't, I think, terribly important. In particular, I'd like to recreate the plot in Figure 2.

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Consider an absorption reaction $\nu_e + B_2 \to e^- + B_4$ (where $B_2$ and $B_4$ are baryons — for instance, $\nu_e + n \to e^- + p$) in thermal equilibrium (hence $\mu_n + \mu_p = \mu_e + \mu_{\nu_e}$).

The paper plots $\mu_e, \mu_{\nu_e}$, and uses the values $\mu_{B_2}$, $\mu_{B_4}$, as a function of $u = n_B/n_0$ (where $n_0$ is the empirical nuclear equilibrium density, and $n_B$ the baryon density) for particular values of $Y_L$ (where $Y_i = n_i/n_B$ for $i=n,p,e^-,\nu_e$, and where $Y_L = Y_{\nu_e} + Y_{e^-}$).

I'd like to determine numerical values for $\mu_{e^-}, \mu_{\nu_e}, \mu_{B_2}$ and $\mu_{B_4}$ for use in further calculations.Here's what I've tried:

I know that the distribution functions for leptons and baryons can both be written as:

$$f_\pm(p) = \frac{1}{e^{(E(p)-\mu)/T} \pm 1}$$

In some notes from a statistical mechanics class (unfortunately, I cannot find an more authoritative reference) I find that:

$$\mu_i = T\textrm{ln}\bigg(\frac{n_i}{n_{max,i}}\bigg)$$

We have that $u = n_B/n_0$, hence $n_B = u \times n_0$. We have also that $Y_i = n_i/n_B$, hence $n_i = Y_i \times n_B = Y_i \times (u \times n_0)$. Consequently:

$$\mu_i = T\textrm{ln}\bigg(\frac{Y_i\times u \times n_0}{n_{max,i}}\bigg)$$

It isn't clear to me, however, how I might go about determining $n_{max,i}$. Moreover, if I simply plug in the values of $T,Y_i$ and $n_0$ used in the paper, it seems as though I cannot find $n_{max,i}$ such that the plot of this function fits that given in the paper.

How do I find numerical values for the chemical potentials, as a function of potential energy $u$ and for given $Y_i$?

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Edit: A related paper ("Neutrino Opacity in High Density Nuclear Matter", S. Santos et. al) gives:

The chemical potential for particle $i$ is given by:

$$\mu_i^* = \mu_i - \chi_{\omega i}\frac{g_\omega^2}{m_\omega^2}\rho_B -\chi_{\rho i}I_{3,i}\frac{g_\rho^2}{m_\rho^2}\rho_3$$ ---where $\rho_e,\rho_\nu$, and $\rho_B$ are the electron, neutron, and baryon densities, respectively, and the lepton fraction is defined as the ratio $Y_L = (\rho_e + \rho_\nu)/\rho_B$.

Unfortunately, it does not define $I_{3,i}$, $\chi_{\omega i}$, $\chi_{\rho i}$, $g_\omega$, or $g_\rho$ (I think the latter are degeneracies?).

Answer 1

Your expression for the chemical potential as a function of density is the usual text book result for a non-interacting classical gas (although $n_{max}$ is just terrible notation). I have been teaching a course based on Sethna's text book "Statistical Mechanics", and this result is given in (6.55) $$ \mu = k_BT\log\left( n\lambda^3\right) $$ where $k_B$ is Boltzmann's constant, $T$ is the temperature, $n=N/V$ is the density, and $\lambda=h/\sqrt{2\pi mk_B T}$ ($h$ is Planck's constant, and $m$ is the mass of the particle). So you can see that $n_{max}\sim \lambda^{-3}$ is the "degeneracy density", the density at which quantum effects (like Bose/Fermi statistics) become important.

The result for free Fermions/Bosons is a little more complicated (but also standard text book material). The pressure of a free Fermi gas is $$ P(\mu,T)=dT\lambda^{-3}{\it Li}_{5/2}(-\zeta^{-1}) $$ where $d$ is the degeneracy factor, ${\it Li}_p$ is the order $p$ Polylog function, and $\zeta=\exp(-\mu/T)$ is the inverse fugacity. You can now solve for the density $n=(\partial P)/(\partial \mu)$, which is also a Polylog, but (I think) there is no closed form expression for the inverse function $\mu(n,T)$ (except in certain limits). However, the result is easily obtained numerically.

The "related paper" appears to be based on a specific model, relativistic mean field theory, and requires fitted values for various couplings (the coefficients $g_\rho$ and g$_\omega$ are coupling constants, not degeneracies).