A definition of the chemical potential that has always served me well is

$$\mu_i = \Big(\frac{\partial U}{\partial N_i}\Big)_{S,V},$$

that is, the amount of energy one would have to add to a system in order to counteract the change in entropy caused by adding one particle of species $i$.

In my cosmology course, the lecturer has said that in the context of finding the relative abundances of neutrons and protons, we may neglect the chemical potentials of electrons and neutrinos.

I'm looking for some intuition as to why the chemical potentials of these particles are negligible in comparison to protons and neutrons. It seems too simplistic to just say 'they are smaller, therefore they change the entropy less' or something along those lines.

Any ideas are appreciated.


1 Answer 1


Ok, lets look at how we determine $\mu$ in a cosmological setting.

In order to determine $\mu_i$, we can use the fact that, in equilibrium, $\mu$ is conserved in all reactions. This means that if we have a scattering process $i + j \rightarrow a+b$, then we know that $\mu_i + \mu_j = \mu_a + \mu_b$.

Fermions in equilibrium, like electrons and neutrinos in the early universe, follow the Fermi-Dirac distribution $$f_i(p) = \frac{1}{e^{\frac{E_i(p) - \mu_i}{T}} + 1}.$$

We also know that, since photon number is not conserved, the chemical potential of photons is zero. This means that for any species in equilibrium with photons, the chemical potential of the anti-particles are negative those of the particles. This means that, for particles that have an antiparticle, a non-zero chemical potential signifies an asymmetry between the number of particles and the number of anti-particles. In the relativistic limit, the difference in number densities is given by \begin{equation} n_i - \bar{n_i} = \frac{g_i}{6} T_i^3 \left[\frac{\mu_i}{T_i} + \frac{1}{\pi^2}\left(\frac{\mu_i}{T_i}\right)^3\right]. \end{equation} When the universe cools down to temperatures below the rest mass of a given species, the particles and anti-particles start to annihilate with eachother leaving just this small excess.

To quantify exactly how small this exess was, we can use the charge neutrality of the universe to infer

$$ \frac{\mu_e}{T} \sim \frac{n_e - \bar{n_e}}{n_\gamma} = \frac{n_p}{n_\gamma} \sim 10^{-10}.$$

So $\mu_e\ll (m_p - m_n) \sim $ MeV around $T \sim $ MeV, when the ratio of neutrons to protons get decided.

You can make a similar argument for neutrinos, so we expect $\mu_\nu$ to be very small as well, but since we cannot observe the neutrino bacground, this is only an assumption. I am not an expert, but I guess if $\mu_\nu$ was too large it could seriously screw up BBN.

  • $\begingroup$ Thanks for your answer! I'm happy with this for the neutrinos, but not sure though that we can consider the electrons to be relativistic at the time of recombination/BBN (when we have protons and neutrons) though, because electrons/positrons have decoupled by this point and become non-relativistic. Is there another possible argument for this? $\endgroup$
    – Orca
    May 7, 2016 at 11:26
  • 2
    $\begingroup$ See edit for answer. $\endgroup$
    – Ihle
    May 7, 2016 at 11:58

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