What is the relation between amount of electrons and protons in Universe? I expect that the Universe should not be charged, so the estimation is 1:1.

But then, why there should be the same amount of leptons and quarks/3 ? (if I generalize a bit the question) I remember, that lepton and baryon numbers are conserved, which sounds to me like we speak about two completely different species. That have the same number of members by chance...


The Universe is indeed electrically neutral at the cosmological length scales which means that the total charge of the positively charged particles is equal to (minus) the total charge of the negatively charged particles.

However, one must be more careful what these particles are. Electrons and protons are two dominant charged particle species. However, the Universe also contains other charged particles including antiprotons, positrons, and, less importantly, some unstable particles.

But if one ignored all charged particles except for protons and electrons, $N_e=N_p$ would really arise from the neutrality of the Universe, and it is approximately obeyed by the Universe around us, anyway. (Most of electrons and protons in the Universe exist in the form of hydrogen atoms, anyway.)

If one talks about the number of quarks, the counting is different. A proton contains 3 (valence) quarks so the number of (valence) quarks is $N_q=3N_p+\dots $. However, protons aren't the only particles that contain quarks. There are lots of neutrons, so neglecting all other hadrons, $N_q=3N_p+3N_n\gt 3N_p\approx 3N_e$.

However, I have mentioned that most of the electrons and protons in the Universe come in the form of hydrogen-1 which has no neutrons, so the number of neutrons in the Universe is much smaller than the number of the protons, and this term may be approximately neglected. However, heavy elements – like those on Earth – actually contain a greater number of neutrons than protons. And neutron stars are full of neutrons. One must be careful about this counting.

| cite | improve this answer | |
  • $\begingroup$ My curiosity was about the equivalence of # of protons and # of electrons (as other cases are either symetric or just small). But I forgot neutrons - yes, if there existed only neutrons at time=0, then we get #electrons = #protons from a weak decay (lepton # conserved with a-neutrino). However, I am missing something here... all were neutrons, electrons born from nothing after? $\endgroup$ – jaromrax Apr 15 '15 at 18:04
  • $\begingroup$ Quantum field theory shows that particles may be born out of nothing as long as a few general conservation laws - energy, momentum, electric charge, angular momentum - are conserved. So yes, they did arise from "nothing" corpuscular (at the end of inflation). $\endgroup$ – Luboš Motl Apr 16 '15 at 5:09
  • $\begingroup$ Ok, so I summarize on my question: the total Universe charge is zero (to fulfill the conservation conditions) and the electrons (leptons) were actually created the way they leave the same number of protons. Approximately. And the lepton number is this way just correlated to the charge of all electrons... $\endgroup$ – jaromrax Apr 17 '15 at 8:35
  • 1
    $\begingroup$ I'm not sure this really answers the question as posed? If you assume the creation of the baryons and leptons each is done in a charge-neutral way (by particle/antiparticle creation) at the end of inflation, the question could be why, after the baryo/lepto-genesis matter/antimatter annihilation imbalanced process, the resulting numbers of matter baryons and leptons match in charge? $\endgroup$ – BjornW Jul 2 '17 at 15:43
  • $\begingroup$ I d0n't think that the author of the question was asking anything about inflation. At any rate, the post-annihilation matter only contains matter - like electrons, protons, neutrons - and no antimatter because antimatter has annihilated with the (slightly) prevailing matter. The equal number of electrons and protons in this final state follows from the vanishing zero charge of the Universe which holds at all times, with inflation or without, doesn't it? I don't need to discuss the the amount of lepto- and bary0genesis separately to prove what I want to prove. $\endgroup$ – Luboš Motl Jul 3 '17 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.