Neutron matter is matter comprised entirely of neutrons, as it exists in neutron stars.

Most optical phenomena encountered in everyday life, such as light reflection and spectral absorption (i.e. color appearance) are the result of mechanisms involving electrons.

My simple question: How would a macroscopic sample of matter consisting entirely of neutrons (without electrons) appear to the naked eye? Assume the matter is degenerate and stable.

Let me add here that I'm not specifically asking about the appearance of a neutron star, as Wikipedia states that it would radiate so much that it appears white.

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    $\begingroup$ Note that even in neutron stars the equilibrium between neutrons and protons/electrons still consists of plenty of the latter, especially when considering the absolute density. $\endgroup$
    – user10851
    Dec 7, 2013 at 8:35
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    $\begingroup$ Related: Are Neutron stars transparent? $\endgroup$ Dec 8, 2013 at 4:03
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    $\begingroup$ Possible duplicate of physics.stackexchange.com/questions/22722/… $\endgroup$ Dec 8, 2013 at 4:05
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    $\begingroup$ If the specific heat of neutronium per gram is anything comparable to the specific per gram of ordinary matter, it is going to take an enormous amount of time for neutron stars to cool down. So this question can't be answered by observation, and has to be answered by theory, which I suspect is not up to the task. $\endgroup$ Dec 12, 2013 at 22:14
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    $\begingroup$ Pure neutron matter cannot be "degenerate and stable". To be stable it has to have a population of degenerate protons and electrons. This question is way too hypothetical. It has to be a question about the appearance of neutron stars because that is actually the equilibrium state of a big ball of neutrons. A much "smaller" neutron star (below about 0.2 solar masses) cannot be stable either. Any attempt to "make" macroscopic degenerate neutron matter would result in atom bomb type explosions. $\endgroup$
    – ProfRob
    Feb 14, 2015 at 14:46

5 Answers 5


This is mostly conjecture, based on physics and common sense.

We know that photons couple to other (charged) particles via the electromagnetic force. Whilst neutrons themselves have zero charge, they are comprised of bound (u,d,d) quarks, which are charged, and with which, the photons could interact.

The density of pure neutron matter would be extremely high, so even a small amount of it would contain a lot of neutrons, and thus many opportunities for photons to interact with quarks. Photons will either scatter directly off the neutrons or briefly induce an excited state, which would decay of the order of $10^{-24}$s, emitting a photon of equivalent energy.

As a simple model, this is not too conceptually different from why clouds appear the way they do (white when thin, black when dense - if the light source is behind the cloud). Therefore, I would conjecture that a lump of neutronic matter would appear black if in front of a light source and white if behind it.


Leaving aside the very hypothetical nature of your question - neutronium or pure neutron matter cannot exist in a stable form because of beta decay - the answer is that it would depend whether you were viewing it in the light it emitted or you were illuminating it.

However, assuming it is cold, and therefore not emitting light, and that you illuminate it with white light, then it will appear transparent. Neutrons have no electric dipole moment, but they do have a magnetic dipole moment. This means they do interact with photons and can scatter them in a similar way to protons or electrons, but the cross-section is much smaller.

The cross-section, when the photon energy is much less than the neutron rest mass energy is approximately $3\times 10^{-81}\nu^2$ m$^2$, where $\nu$ is the photon frequency (Gould 1993).

Assuming visible light with $\nu \sim 5\times 10^{14}$ Hz and a neutron number density of $n\sim 10^{44}$ m$^{-3}$, then the photon mean free path in the material is $\sim 10^7$ m. Thus a lab-sized lump will be transparent to visible photons.

In fact the mean free path would be longer than this because the calculation of the cross-section assumes that scattering states are available for the neutron. If the neutrons are degenerate, then the scattering rate would be reduced by another factor of $k_BT/E_F$, where $E_F$ was the Fermi energy of the degenerate neutrons, since only neutrons at the top of the "Fermi sea" could participate. On the other hand, the cross-section goes as $\nu^2$, so higher energy X-ray or Gamma ray photons are much more likely to be scattered.

Real neutron star matter, at densities of a few $10^{17}$ kg/m$^3$ would consist of neutrons, protons and electrons in beta equilibrium. If you do the sums you find that by equating the Fermi energies and demanding baryon conservation and charge neutrality, that there are of order 10-100 times (depends on the exact density) more neutrons than protons or electrons.

There are easily enough electrons to scatter visible photons in any non-microscopic piece of neutron star material (the fact that the electrons are degenerate, makes no odds here, because typical neutron star interiors are at millions, if not billions of degrees, compared to tens of MeV electron Fermi energies and $\sim$eV photons). It will therefore be completely opaque.

As for its appearance I could not say; I am sure it would be determined by things going on at the surface. At a guess I would say it behaves like a perfect scatterer and that if it has a smooth surface it would look like a mirror. Entirely hypothetical though, since I don't see how you can prevent the neutron star matter reverting to its low density equilibrium composition at the surface - i.e. changing back into iron peak nuclei. From this non-hypothetical view I suspect that your lump of neutron star material would be a pretty good approximation to a black body at whatever temperature you had it and would absorb almost all light incident upon it.


Although Neutronium is used in SF a lot, I got a different take on it from a video of a researcher who put his iPhone in a neutron beam to see what happens (but that's another story).

He pointed out that a marble sized ball of neutronium would not only need a SF "force field" to keep it under pressure, but also be in time stasis because it would give off as much energy as some rediculous number of atomic weapons per second.

A time stasis field would block anything inside it from view (in Larry Nivin's stories they are perfect mirrors) and can't be effectivly transparent because the stopped time inside would stop the light from coming or going.

If it were contained and stabilized in some manner that did not cloak it, what might it look like? Bulk objects reflect light selectively to give color, due to the electrons blending into a net of sorts that have many ways to stash energy so "take" light of any frequency. A pure element gas would show spectral lines, coming or going as the case may be, and ignores light of different frequencies because there are only specific energy levels present. So, for example, it would be perfectly transparent except for a narrow bit of yellow taken out. Hey, sounds like air, nice and clear to our range of frequencies.

So are the electrons in our sample a mesh like a mineral grain, finiky like a gas, or liquidy like a metal? Uh... what electrons? Actually any charged particle will do. Nope, none of any kind, just neutrons.

So, it would be perfectly transparent. Not considering gravity lensing, and the effects of whatever is stabilizing it. Or maybe not... if it's really "like a neutron star" you would have other stuff in it at equilibrium, plus a crust of other materials covers it so you can't see it anyway so let's suppose pure contained neutrons somehow. But, in a story setting that would give you licence to make it clear, black, white, or mirror, "depending".

Actually, neutrons have a concentric separation of charges, so from the outside the concentric charges cancel out, but might that give some degrees of freedom for light to interact with? If the skin's charges can somehow blend together and work like a mineral grain, I think it would not react to anything less than hard x-rays, so still invisible to our eyes.

Effects of a very small electric dipole moment (and I do mean very) beyond the Standard Model would be correspondingly small, so don't expect anything visible from unknown effects yet to be discovered.

Does magnetism affect light, if very strong? That's a question for another topic. If you could get all the neutrons to line up their fields...

  • $\begingroup$ Why should neutron star matter "give off as much energy as some rediculous number of atomic weapons per second"? Clearly the matter of actual neutron stars is very hot, and I can imagine that when neutron fluid is formed there would be lots of high energy nuclear reactions (exothermic? no idea.) but why would this be continuous? $\endgroup$
    – ikrase
    Sep 9, 2019 at 7:07
  • $\begingroup$ Because neutrons decay. $\endgroup$
    – JDługosz
    Oct 2, 2019 at 18:14
  • $\begingroup$ Doesn't the pressure holding it in degeneracy prevent that from happening (or hold it in equilibrium)? $\endgroup$
    – ikrase
    Oct 3, 2019 at 5:20
  • $\begingroup$ If it’s a star, held together by gravity. If you had a marble-sized piece, there is nothing stopping it. $\endgroup$
    – JDługosz
    Oct 3, 2019 at 14:46
  • $\begingroup$ But if it's held in an imaginary force-field, that will hold it in degeneracy pressure with no need for time stasis. $\endgroup$
    – ikrase
    Jan 9, 2020 at 5:46

From Wikipedia:

Dineutron: The dineutron, containing two neutrons, was unambiguously observed in 2012 in the decay of beryllium-16. It is not a bound particle but had been proposed as an extremely short-lived resonance state produced by nuclear reactions involving tritium. It has been suggested to have a transitory existence in nuclear reactions produced by helions (helium 3 nuclei, completely ionized) that result in the formation of a proton and a nucleus having the same atomic number as the target nucleus but a mass number two units greater. The dineutron hypothesis had been used in nuclear reactions with exotic nuclei for a long time. Several applications of the dineutron in nuclear reactions can be found in review papers. Its existence has been proven to be relevant to the nuclear structure of exotic nuclei. A system made up of only two neutrons is not bound, though the attraction between them is very nearly enough to make them so. This has some consequences on nucleosynthesis and the abundance of the chemical elements. Trineutron: A trineutron state consisting of three bound neutrons has not been detected, and is not expected to exist[citation needed] even for a short time. Tetraneutron: A tetraneutron is a hypothetical particle consisting of four bound neutrons. Reports of its existence have not been replicated.

So let's consider the dineutron only first. I can see no reason why the neutrons don't have associated ( and correlated, in the case of a neutron star) orbitals, caused by the strong force tough (this is the most important). Suppose the neutrons are in an excited state. When they fall back to the ground state, no photons will be produced because the force holding the neutrons together is the strong nuclear force. Then, what does the system emit? Non-virtual gluons. And certainly no photons, so neutronium is dark.

Gluons were first conclusively proven to exist in 1979, though the theory of strong interactions (known as QCD) had predicted their existence earlier. Gluons were detected by the jets of hadronic particles they produce in a particle detector soon after they are first created.

So, although neutronium has no color it can be "seen" (without a color tough) by particle detectors.

One more thing. Gamma photons can't interact with the charged quarks because the strong force that holds the quarks together is too strong even for a gamma photon to overcome.

Therefore in neutron stars (considered as a multi-neutronium system), packed with neutrons, it will be transparent to every kind of photon. So the star doesn't absorb and neither emits photons., so it can't be seen.


The hypothetical matter you are referring to is called neutronium. It is basically a huge nucleus of pure neutrons. The problem is that current physics predicts that it would instantly decay via radioactive decay and explode much like an atomic bomb. thats how it would look like.

Edit: My answer concerns the only qualitatively different case from the neutron star case, because the only force that could keep your nucleus together is gravity.

the decay of individual neutrons can be neglected here.

Edit: And now i have the cooling down neutron star:

After several billion years they are still at many thousands of degrees, and it would take a large multiple of the age of the universe to get to room temperature (exact values depend on unknowns about the neutron star; for example, it matters how much energy they have lost in neutrinos). The surface of such a star would probably be made of hydrogen in long chains of atoms. At room temperature, the emission would be almost all in the infrared, so the star would appear completely dark. If you were to shine a light on the neutron star, its appearance would depend on whether it had maintained its strong magnetic field. If so, the light would not be absorbed until fairly deep in the atmosphere, and when it was radiated it would again be in the infrared, so it would appear black. If the magnetic field had decayed away, the star might have a metallic glint; I'm not sure, because it's difficut to project the optical look of such an object.


note that you most likely cannot get a qualitatively different phenomenon even if you replace the gravity with a fictional force of your choice.

  • $\begingroup$ I have edited the question to assume that the matter is degenerate and therefore stable. The question is not whether neutronium can exist under normal conditions, but rather, assuming it exists, how would it interact with light. $\endgroup$ Dec 13, 2013 at 6:29
  • $\begingroup$ Is this so? Could you explain some more: yhis is not my field, but neutrons decay into protons, electrons and antineutrinos and the process would only be "explosive" if one decay provokes more than one other, which is not clear to me. Moreover, there has to be a way whereby gravity forestalls this decay when you've got enough of this stuff in one place to form a neutron star, which is stable. $\endgroup$ Dec 13, 2013 at 6:31
  • $\begingroup$ Yes, it would be a spectacular explosion, because anything it would decay into would also be extremely unstable already. unlike regular atomic bombs where you would need high energy neutrons to induce nuclear fission of the relatively stable nucleons. $\endgroup$ Dec 13, 2013 at 6:41
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    $\begingroup$ @Wet Actually, all gravity does is shift the equilibrium concentration. Neutrons still decay in neutron stars, but they are being reformed by combining protons and electrons. Because both reactions release high-energy neutrinos, neutron stars cool by maintaining equilibrium, paradoxically enough. $\endgroup$
    – user10851
    Dec 13, 2013 at 7:47
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    $\begingroup$ @WetSavannaAnimalakaRodVance The explosion occurs because of the kinetic energy density of getting-on-for relativistically degenerate neutrons. See physics.stackexchange.com/questions/10052/… Nothing to do with beta decay on a timescale of 10 minutes! $\endgroup$
    – ProfRob
    Jan 14, 2015 at 17:31

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