# Is it possible that dark-matter is composed of a large number of neutrinos from the big bang? [duplicate]

They seem to have all the properties of dark-matter (massive, with no electromagnetic interaction). Could it be that many of the neutrinos produced since the big bang have formed massive neutrino clusters that act as dark-matter?

Short answer: dark matter appears to be cold, but for any reasonable neutrino mass and dark matter temperature, neutrinos would be relativistic.

• Thanks for the quick reply.Could you please explain what " neutrinos would be relativistic" means.I'm in high school so please forgive my lack of knowledge :) – mkmkmk4 Feb 19 '16 at 19:06
• relativitic, the speed is comparable to the speed of light. – Mikey Mike Feb 19 '16 at 19:29
• What is important is that they were relativistic at the epoch when the horizon contained a typical mass for the cosmic structures one is trying to explain. – Rob Jeffries Feb 19 '16 at 19:44
• The most massive neutrino must be more than 0.04 eV. The sum of all three has been estimated as between 0.3 and $<2$ eV. Current neutrino temperature 1.95K. So $kT/m_{\nu}c^2 \sim 10^{-3}$, so not highly relativistic. – Rob Jeffries Feb 20 '16 at 8:44

There are a couple of reasons why the three types of known neutrino (and their antiparticles) cannot be the missing dark matter.

We now know roughly what the masses of neutrinos are. The sum for all 3 is about 0.3eV and the heaviest one must be bigger than about 0.04eV.

If you work out how many are produced in the big bang, and you assume they are non-relativistic (see below) it turns out that their density today (about 56 cm$^{-3}$ for each of the six species) is about a tenth that of the luminous matter in the universe and about 0.3% of the density required to produce a flat universe and nowhere near the 25% required of any dark matter candidate.

The neutrinos decoupled from the rest of the matter shortly after the big bang when the universe was at temperatures of $10^9$ K, and they have now cooled to just below 2 K. Thus current background neutrinos have $kT/mc^2\sim 10^{-3}$ and are not highly relativistic, but still to fast to be trapped by individual galaxies. However, what is important is that when large scale structures were beginning to form, when the universe was at $10^6$ K, the neutrinos were very relativistic (hot dark matter). Such particles would act to erase structure and the universe would not be the way we see it today.

Thus even if you found a way to make neutrinos clump together (to overcome the second objection), there are not enough of them to overcome the first objection.

• So when we look back the radiation increases with a^-4, normal and dark matter with a^-3, curvature with a^-2, and dark energy with a^0. But what would be the equation of density and pressure for those previously hot and now cold neutrinos? – Yukterez Feb 19 '16 at 20:55
• @Симон Тыран it would change from $1/a^4$ to $1/a^3$ I suppose. – Rob Jeffries Feb 19 '16 at 20:58
• How do we know they roughly sum to 0.3 eV? I believe you but I'm curious how that was measured / calculated. – Brandon Enright Feb 19 '16 at 21:11
• @BrandonEnright I followed the references on the wikipedia page on neutrinos, which suggest that the CMB gives that constraint. There are other measurements suggesting a limit below 2eV. – Rob Jeffries Feb 19 '16 at 21:18

Not really, there are many models in which dark matter is assumed to be relativistic, and hence warm. For example here there are discussed some features in the cases where dark matter is composed from sterile neutrinos or light gravitinos. Also, dark matter can be a multi component field. One answer with high positive probability would be that dark matter, if it exist and the General Relativity in the shape as we know today is valid, would not be cold, but non-relativistic, i.e. warm.

Neutrinos were not just created in the Big Bang but ceased interacting with baryonic matter after one second when the temperature of the universe was approximately 10 billion kelvins, or 1 MeV. Because neutrinos have mass the force of gravity restricts them to galaxies. All main sequence stars produce neutrinos from the fusion process and if they don't interact with baryonic matter there is a good chance most of them are still around. Lisa Randall hypothesised a disc of 'dark matter' running right through the Milky Way. A 180-trillion-mile 'disc of death' was how the Sunday Times reported it in a review of her book on Dark Matter and the Dinosaurs. It is theorised that the Earth gets a gravitational 'nudge' every 35 million years as it passes through an alignment with this disc of dark matter. This parallels thinking on a dilemma posed by Jupiter's moon Io. Because Io is so far from the sun it should have cooled down by now but there appear to be volcanoes on the surface. This has been linked to the fact that Io is regularly stretched 100 metres by gravity alignment with Jupiter's other moons. Friction is thought to be the cause of the volcanoes on Io, which effectively means they may be fuelled by gravity alone as a power source. Lisa Randall's disc of dark matter is probably all the neutrinos in the galaxy obeying the speed limit. Because they cannot acquire energy to reverse time they are stuck in a galaxy until the black hole at its centre does something about it.

• "Because neutrinos have mass the force of gravity restricts them to galaxies." Nonsense. All neutrinos are ultra-relativistic in the rest frame of their creation reaction which means that with the possible exception of the CNB all neutrinos have far more than the paltry velocity needed to escape a galaxy. – dmckee --- ex-moderator kitten Feb 22 '16 at 0:13
• I thought it was accepted that dark matter formed a spherical "halo" about the galaxy, not a disk. Wouldn't that put the kybosh on any effect from dark matter on the solar system oscillating backwards and forwards through the galactic plane? – Errol Hunt Nov 17 '16 at 0:25