The recent observation of a tetra-neutron resonance suggests that clusters of 6 or 8 neutrons stand a strong chance of being bound and stable. The shell configuration of a tetra-neutron should be approximately that of the neutrons in beryllium-8 (itself an unstable resonance). Two of the neutrons should occupy the $s_{1/2}$ orbital leaving the remaining two neutrons in the incompletely occupied $p_{3/2}$ orbital. Adding two more neutrons would complete the $p_{3/2}$ shell yielding the neutron configuration of carbon-12 and adding a further two neutrons would complete the $p_{1/2}$ shell yielding the neutron configuration of oxygen-16 (a doubly magic nucleus). The enhances stability of magic nuclei (those having completely filled shells) is well known in nuclear physics.

The experiment used to create the tetra-neutron employed the exotic isotope helium-8 which upon bombardment with a proton yielded helium-4 and the tetra-neutron resonance. It is probably unlikely that any similar experiment could be used to create and observe the 6 or 8 neutron cluster states, but my question is whether nature may already have created these states in abundance and could they be the dark matter that has puzzled astrophysicists for such a long time?

Aside from atomic nuclei the only other environment where neutrons exist in such close proximity involve stellar collapse leading to neutron stars and/or black holes. Might these small clusters of neutrons be copiously shed during stellar collapse with sufficient energy to escape the gravitational pull of the collapsing core to form the galactic gas we know as dark matter?

  • $\begingroup$ The problem is that a pn pair is more strongly bound than an nn pair so half the neutrons would beta decay to produce a regular nucleus. See Why doesn't a nucleus-like body made up of just neutrons exist? for details. $\endgroup$ Commented Jun 28, 2022 at 4:25
  • $\begingroup$ Related: physics.stackexchange.com/q/176006/2451 , physics.stackexchange.com/q/1008/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Jun 28, 2022 at 7:44
  • 1
    $\begingroup$ "to form the galactic gas we know as dark matter" - We don't - "dark matter" has not been directly observed, but is just a speculation intended to "explain" why General Relativity doesn't work at galactic distances. $\endgroup$
    – safesphere
    Commented Jun 28, 2022 at 8:41
  • 1
    $\begingroup$ Isn't this a duplicate of physics.stackexchange.com/q/176006 ? Say why not. $\endgroup$
    – ProfRob
    Commented Jun 28, 2022 at 9:52
  • 2
    $\begingroup$ There's a real hypothesis with some resemblance to what you propose -- look up "sexaquark dark matter". $\endgroup$
    – knzhou
    Commented Jun 28, 2022 at 14:50

3 Answers 3


Let’s grant you the existence of one or more stable isotopes of “neutronium,” with atomic mass six or eight, $\rm^6_0n$ or $\rm^8_0n$. Now we need to re-do our calculations for big-bang nucleosynthesis with these new stable states available. From the Particle Data Group we get this summary plot of BBN calculations:

enter image description here

Figure 24.1: The primordial abundances of 4He, D, 3He, and 7Li as predicted by the standard model of Big-Bang nucleosynthesis — the bands show the 95% CL range [50]. Boxes indicate the observed light element abundances. The narrow vertical band indicates the CMB measure of the cosmic baryon density, while the wider band indicates the BBN D+4He concordance range (both at 95% CL).

The major feature of this plot is that more massive nuclei were produced at the Big Bang in smaller numbers than less-massive nuclei. If we added a stable $\rm^6_0n$ or $\rm^8_0n$, we would expect it in the part-per-billion or part-per-trillion range of primordial mass fractions. However, dark matter makes up more of the mass of the universe than ordinary matter, by a factor of about six.

This is separate from the analysis of CMB brightness fluctuations which also indicate that the total non-baryonic matter is about six times more massive than the total baryonic matter.

Furthermore, based on the Milky Way’s dark matter distribution, we expect the local density of dark matter to be about $0.3\,\mathrm{GeV}\,c^{-2}\,\mathrm{cm}^{-3}$. That is a huge density of sexaneutrons or octoneutrons —— much higher than the density of mononeutrons from cosmic-ray spallation, which have observable effects like the continuous replenishment of carbon-14 from atmospheric nitrogen. A stable neutron cluster might be invisible, but it would certainly be strongly-interacting. Thermalized neutronium would induce lots of nuclear reactions as it passed through Earth’s atmosphere and crust.

  • 3
    $\begingroup$ Your last point is definitive. Neutrons from cosmic ray spallation are easily detectable with the right benchtop setup. Forget interactions in the atmosphere and crust: any sensitive radiation detector would be pounded by these things. But we don't see that. $\endgroup$
    – John Doty
    Commented Jun 28, 2022 at 12:38
  • 4
    $\begingroup$ To the contrary, if there was a local density of many octoneutrons per cubic meter, I think it would be impossible to live long enough to invent a “sensitive radiation detector.” $\endgroup$
    – rob
    Commented Jun 28, 2022 at 13:50
  • 1
    $\begingroup$ True. Forget interactions in the atmosphere and crust, consider the radiation biology... $\endgroup$
    – John Doty
    Commented Jun 28, 2022 at 13:55

Such neutron clusters are baryonic matter. However, we know from measurements of the primordial abundances of helium and deuterium and within the context of big-bang nucleosynthesis models, that most of the matter in the universe must be non-baryonic - the primordial abundances show that baryonic matter makes up ony 1/6 of the matter required to explain the dynamics of galaxies and clusters (e.g. Pettini & Cooke 2012). Independently, fitting the angular fluctuations in the otherwise isotropic cosmic microwave background to the same model also shows that baryonic matter makes up only $18.3 \pm 0.5$% of the gravitational matter (Peebles 2015).

Thus dark matter must be predominantly non-baryonic in the context of the big-bang model.

A further property of dark matter is that it should only interact gravitationally. I am sure that would not be the case for nuclei made of neutrons, which could interact with other nuclei and particles and with photons.

See also:

Why can't dark matter be baryonic?

Why isn't dark matter just ordinary matter?

Can Dark Matter just be clumps of Neutrons

  • 1
    $\begingroup$ Careful. We have models of primordial nucleosynthesis that are incompatible with baryonic dark matter given the data. However, the models involve a drastic backward extrapolation of physical conditions from the portion of the universe accessible to observation. Astrophysics is full of surprises. Few theorists thought hypernovae possible until observations of long/soft gamma-ray bursts could not be explained any other way. The hypothesis in the question seems unlikely to be true, but I wouldn't want to declare this opinion to be settled science. $\endgroup$
    – John Doty
    Commented Jun 28, 2022 at 12:28
  • $\begingroup$ @JohnDoty I suppose I agree and will change my wording. But I am unaware of ANY other model that provides a simultaneous explanation of the helium and deuterium abundances, let alone also explain the angular power spectrum of the CMB. This is spectacular agreement. $\endgroup$
    – ProfRob
    Commented Jun 28, 2022 at 14:44
  • $\begingroup$ @ProfRob For me, the agreement of the angular power spectrum with theoretical prediction was jaw-dropping. Still, overall we have only a few lines of evidence and a theory that requires a great deal of extrapolation. And then there's the disagreement on the Hubble parameter, modest by historical standards, but troubling for a heavily extrapolated theory. $\endgroup$
    – John Doty
    Commented Jun 28, 2022 at 15:40
  • $\begingroup$ @JohnDoty the angular spectrum has free parameters - so you can get agreement by messing with those. The nucleosynthesis model, which is quite independent physics, has some of the same free parameters and it requires almost exactly the same $\Omega_b h^2$ to fit both. Whatever the value of $h$, the same $\Omega_b$ is needed for both. In the context of this question that is all thatis needed. If you don't believe $\Lambda$CDM cosmology then you possibly don't need any dark matter at all. $\endgroup$
    – ProfRob
    Commented Jun 28, 2022 at 15:45
  • $\begingroup$ @ProfRob If the problem with h is measurement error, you have a good point. But the problem with h may be unknown physics. $\endgroup$
    – John Doty
    Commented Jun 28, 2022 at 15:52

Might these small clusters of neutrons be copiously shed during stellar collapse with sufficient energy to escape the gravitational pull of the collapsing core to form the galactic gas we know as dark matter?

As far as our observations of the universe and the current model go, dark matter is stable. You are talking of resonances, and resonances decay to their components, so they cannot make up dark matter. See the answers here and here ,that discuss the problem of bound states with neutrons.

  • 1
    $\begingroup$ No, I'm assuming that the extra binding from shell closures will render the 6 and 8 neutron clusters stable. $\endgroup$ Commented Jun 28, 2022 at 12:11
  • $\begingroup$ in the inks I give there are the arguments why this assumption is wrong, $\endgroup$
    – anna v
    Commented Jun 28, 2022 at 12:52
  • $\begingroup$ I don't trust the arguments in your links, but this is a question that can only be settled by an experiment. Perhaps an alpha bombardment of Ca-48 analyzed in the same way that was used to find the quartaneutron resonance? $\endgroup$ Commented Jun 30, 2022 at 2:22
  • $\begingroup$ Sorry, I meant tetraneutron resonance. $\endgroup$ Commented Jun 30, 2022 at 13:30
  • $\begingroup$ @LewisMiller I do not think it is a possible experimental detection scenario. If instead of a resonance whose decay products are seen in the experiment one has a bound non decaying state of four neutrons it would leave no trace except in the missing mass, a very difficult to measure reliably in the LHC environment of jets and multiple tracks. $\endgroup$
    – anna v
    Commented Jun 30, 2022 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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