# Are neutrino stars theoretically possible?

Since neutrinos

1. have a small mass and
2. are affected by gravity,

wouldn't it be theoretically possible to have such a large quantity of them so close to each other, that they would form a kind of a stellar object, i.e. one that would keep itself from dissolving due to the large gravity.

If such objects were possible, how would they interact with rest of the world? Would they be invisible (dark matter?) because of neutrinos' lack of electromagnetic charge? Would this lack also make ordinary matter pass through them, or would the Pauli exclusion principle prevent this passing through due to the high density of neutrinos?

Instead of the massive compact objects which could serve as a 'replacement' for the supermassive black hole inside the galactic center (which are discussed in the Viollier and Tupper paper from Anna's answer) I would like to point another possibility: halos of degenerate neutrino gas around galactic clusters.

The order of magnitude calculations for the sphere of degenerate matter held together by gravity could be made using energy equipartition. Here we have calculations for the white dwarfs (that is electron-degenerate matter), but generalization for neutrino star is rather straightforward: there both the degeneracy pressure and gravitational pull is produced by neutrino, so we should simply substitute electron and proton masses to neutrino mass: $m_e \to m_\nu$ and $m_p \to m_\nu$.

Then, the two main equations would be

1. Chandrasekhar limit: $$M_\text{Ch} = C \cdot \left ( \frac{\hbar c}{G}\right )^{3/2}\frac{1}{m_\nu^2} \tag{1},$$ which is the maximum possible mass for the star at equilibrium. (C is a $O(1)$ constant). For a $m_\nu$ of 1eV we would have $M_\text{Ch}$ on the order of $10^{48} kg$ which is the $10^6$ mass of the Milky Way.

2. Relationship between mass of neutrino star and it's radius which in the nonrelativistic limit (which would be justified for 'stars' with mass less than Milky Way) is: $$R_{*} = C' \cdot \frac{\hbar^2}{G m_\nu ^{8/3}}M_{*}^{-1/3}\tag{2}$$ If we assume $M_*$ on the order of $4 \cdot 10^6 M_\odot$ (mass of Sagittarius A*) then for the $m_\nu$ of 1eV the $R_*$ would be hundreds of Mpc which is unrealistic.

Aside: For the $m_\nu$ of 17keV $R_*$ would be ~100 light hours, which is much more reasonable.

So we see the problem with the galactic center neutrino star: the experimental upper limit for the neutrino masses (~1eV) means that, unless there are sterile neutrino with masses substantially larger than those of active neutrinos, neutrino stars would be either too light to be noticeble or too large to be called 'stars'.

So we come to conclusion that if there are a lot of cold neutrinos in the universe, then they would form not small compact objects (stars) but rather halos around galaxies. Large scale objects of baryonic matter (galactic clusters) will have clouds of neutrino degenerate neutrino gas around them. In this case the gravitational pull is generated both by neutrinos and ordinary matter, so the eq. (1) and (2) do not really apply. That way neutrino could constitute a noticable portion of dark matter.

Turning to Google Scholar in support of such hypothesis we can find a recent paper which discusses this possibility:

Theo M. Nieuwenhuizen and Andrea Morandi. "Are observations of the galaxy cluster A1689 consistent with a neutrino dark matter scenario?" Mon. Not. R. Astron. Soc. 434 no. 3 (2013), pp. 2679-2683. arXiv:1307.6788.

The Abstract:

Recent weak and strong lensing data of the galaxy cluster A1689 are modelled by dark fermions that are quantum degenerate within some core. The gas density, deduced from X-ray observations up to 1 Mpc and obeying a cored power law, is taken as input, while the galaxy mass density is modelled. An additional dark matter tail may arise from cold or warm dark matter, axions or non-degenerate neutrinos. The fit yields that the fermions are degenerate within a 430-kpc radius. The fermion mass is a few eV and the best case involves three active plus three sterile neutrinos of equal mass, for which we deduce 1.51 ± 0.04 eV. The eV mass range will be tested in the KATRIN experiment.

So, at a glance the scenario seems plausible!

• so, why not assume these cold neutrinos and dark matter ARE THE SAME THING? Oct 11 '13 at 20:51
• @diffeomorphism Heavy sterile neutrinos are a candidate for the dark matter. People are working on detecting them if present, but you don't go assuming that you have the answer to a puzzle when you haven't even got any evidence for the thing you claim is the solution. Heavy sterile neutrinos are admitted but the current state of the data but there is no positive evidence for them: they just are not ruled out. Oct 11 '13 at 21:07
• you say sterile neutrinos, but what about normal neutrinos, have we ruled out that they can be dark matter? Isn't thermalization quicker for normal neutrinos rather than for steriles? (assuming steriles means that they only interact gravitationally) Oct 12 '13 at 2:10
• @diffeomorphism No. The known neutrino flavors are too light and insufficiently numerous to make up a significant fraction of the dark matter. There are vast numbers of relic neutrinos left over from the Big Bang (the neutrino equivalent of the CMB), but they are even colder that the CMB and so have essentially only their intrinsic mass which is exceedingly small. Jan 30 '14 at 2:33

It seems that theoretically neutrino stars have been postulated. A google search came up with

Supermassive neutrino stars and galactic nuclei. R.D. Viollier, D. Trautmann and G.B. Tupper. Phys. Lett. B 306 no. 1-2 (1993), pp. 79-85.

The calculations have been done for you :) if you have access to a library:

### Abstract

The characteristics of supermassive ‘stars’ consisting of self-gravitating degenerate neutrino (or neutralino) matter are studied with particular emphasis on fermion masses around 17 keV/c2. Such compact dark objects could be as massive as 10^9.5 to 10^6.5 solar masses, with radii of about one to ten light days; they might thus mimic phenomena that are expected around the supermassive black holes recently purported at the centres of some galaxies and quasi-stellar objects.

I suppose as neutrinos have masses, this means that they can be at rest with each other and allow gravitational forces to dominate.

• Should just mention that $17\ \mathrm{keV}$ is a good six or seven orders of magnitude above typical theoretical neutrino masses these days. Oct 11 '13 at 15:53
• @MichaelBrown is right about the masses of the known neutrino mass states (1,2,3), but there is still some interest in various heavy sterile neutrino states which are not yet ruled out. Indeed some meta studies suggest that 2 or 3 sterile states may be favored over just one. Oct 11 '13 at 16:42
• @MichaelBrown the experimental limits are of order of ev. I do not know how that would translate in the size in the paper's calculations. Oct 11 '13 at 19:07

The problem is that because neutrinos have a very small mass they are generally moving at speeds near the speed of light. See the hot dark matter Wikipedia entry for more details. The high speed means their velocity is greater than any escape velocities found in the universe, so they aren't gravitationally bound to anything.

I don't think it would even be possible to make slow neutrinos in the lab. As far as I know, all the ways to produce neutrinos result in neutrinos with relativistic speeds. The problem is that with a very small mass even low kinetic energies correspond to very high velocities.

• I was guessing that the typically relativistic speeds of neutrinos would pose a problem. However, neutrinos still can't escape black holes. If there were enough of them close enough to each other, wouldn't that be sufficient to cause a gravity which the neutrinos couldn't escape? Or, on the other hand, would that just cause a black hole? I was wondering, whether there could exist a state that wouldn't collapse into a black hole and would neither dissolve into space due to high velocities. Oct 11 '13 at 14:51
• The escape velocity of a black hole is of course the speed of light, so I think it's hard to see how any particle travelling at near the speed of light could be bound by an object that wouldn't collapse to form a black hole. I'm sure it's possible, but I'm equally sure that it's highly implausible. Oct 11 '13 at 14:57
• John & @miikkas I was about to answer with this but then I checked the cosmic neutrino background temperature is actually less than the expected neutrino mass scale. So cold neutrinos actually can in principle exist in the current universe. That said the density fluctuations of primordial neutrinos are unlikely to trigger gravitational collapse (Jeans instability) and so couldn't account for natural dark matter.. But perhaps a neutrino star could be constructed. Whether such a bound state could exist in principle is an interesting question... Oct 11 '13 at 15:44
• My dimensional analysis fu tells me that degeneracy pressure can support an absolute zero temp neutrino "star" larger than the Hubble radius before it undergoes runaway collapse to a black hole. Such a beast has more mass than the observable universe (says wolframalpha). I have not included electroweak pressure, nor the effect of background radiation heating and disrupting the star. Both of these would be important compared to gravity. (EDIT: Just saw anna's answer go up.) Oct 11 '13 at 15:48
• @MichaelBrown: I'd be interested to see your calculation. Can you post it as an answer? I'll track down the paper Anna mentions, though your comment suggests it's out of date. Oct 11 '13 at 16:10

Look at the "bar" region of, e.g., NGC1300 and wonder whether you are seeing a neutrino star. If so it could be of greater mass than the visible galaxy so accounting for the linear g:r relationship in the bar region. If it is a Fermi condensate it could have a "liquid" surface at the tip of the bar with a lesser density tapering off in the spiral arms region. The small rest mass of the neutrino would explain how a "star" of such immense radius could exist. The neutrinos at the liquid limit could be far from cold due to the accumulated quantum number.

Paul E G Cope

• This sounds much like the proposal that annav's answer links to, but without a relevant computation to back it up. Nov 7 '15 at 16:02