# Electron density in the Sun

I'm currently working on solar neutrino and in order to make a numerical simulation, I need the potential felt by electron-neutrino : $$$$V_e(r) = \sqrt{2} G_F N_e(r)$$$$ where $$N_e(r)$$ is the electron density perceived by the neutrino and $$G_F$$ the Fermi coupling constant associated to the weak interaction. Do you know any website able to provide me with data or formulas? (I just know that $$N_e$$ is roughly exponential).

• $N_e(r)$ is the electron density perceived by the neutrino and $G_F$ the Fermi coupling constant associated to the weak interaction. Apr 21, 2018 at 15:11
• You should edit the information about symbols into the body of the question. It's not a good idea to expect people to look for additional info like this in comments. Apr 21, 2018 at 16:30
• See this question for density of the sun as a function of radius. Apr 24, 2018 at 11:54

Recently I am also looking for the electron density of sun, and luckily I find it, you can get everything you want in this website, http://www.sns.ias.edu/~jnb/SNdata/solarmodels.html. But it just provide the mass density and the fraction of main elements, so you just need to consider the Molar mass of them.

It depends how accurate you need the answer to be...

What you can do is use the mass density of the Sun as a function of radius and combine that with what we know about $$\mu_e$$, the number of mass units per electron in the solar interior.

The Sun will be reasonably modelled by two zones. The convective core where nuclear fusion is occurring is about 36% H, 62% He and 2% heavier elements (mainly O, Fe, N). The mass-weighted number of mass units per electron is therefore $$\mu_e= (0.36\times 1 + 0.62 \times 2 + 0.02\times \sim 2) = 1.64$$, where I've assumed that both He and heavy elements have 2 mass units (a neutron and a proton) for every electron.

The core is well mixed out to about $$0.25R_{\odot}$$, but beyond that, into the radiative and convective zones, the composition more closely reflects the initial composition of the Sun and is reasonably uniform at 75% H, 23% He and 2% heavier elements. Thus $$\mu_e \simeq 1.25$$.

With this two-zone $$\mu_e$$, you can take the mass-density profile $$\rho(r)$$ and then obtain the number density of electrons as $$\rho(r)/\mu_e m_u$$.

To be more accurate you can simply do a similar, row-by-row calculation on a standard solar model (which has the mass fraction of H, He and heavier elements, as well as the density).

Initially the number of electrons was equal that of protons, i.e. the ratio of numbers e-/p=1. Nuclear reactions convert protons to neutrons, then to heavier nuclei but the e-/nucleon ratio remains the same. The Sun is not a mixture of neutral gases in a vessel that a helium atom be counted 2 electrons. There are however factors decreasing the electron density, namely annihilation with positrons which appear in certain nuclear reactions (pp, 8Be) as well as through electrons entering directly into reactions like pep and Be-7 (small). The Wolfenstein potential in the centre of the Sun is about 0.012 neV if one takes the ratio 1:1.

• The e/nucleon ratio does not remain constant. Feb 25, 2022 at 7:26