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Everything we learn about colour in relation to matter is based on "normal" matter that has electrons around it. Absorption and emission of electromagnetic radiation is explained in terms of electrons transitioning between quantum levels with different colours being caused by the energy difference between transitions.

In this thought experiment I have a piece of neutronium at room temperature. As there are no electrons to interact with light of any wavelength, what colour would it be? (You need to do the measurement quickly in the few attoseconds before it explodes).

I can only make 3 guesses but I can't think of any way of deciding which one is the least unlikely.

1 - Transparent.

2 - Perfect mirror

3 - Perfectly black.

My limited physics suggests #1 as being plausible as there are no electrons, no orbitals and therefore no interactions with light. However, a totally non-scientific gut feeling says that transparent is ridiculous. How can something with such insane density have no interaction with light and look like it's essentially invisible? Surely it has to either reflect light perfectly, or absorb it perfectly?

Then again, perhaps there would be the neutron matter equivalent of an absorption spectrum. That at relatively low photon energies (visible light), neutronium would be transparent, but at stupidly high energies (cosmic rays from matter falling into black holes,) it would absorb photons.

There's no great reason for asking, just intellectual curiosity. A mental itch that needs scratching.

Barry Stone
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2 Answers2

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Neutrons are composed of quarks and quarks do have electric charge and so clearly photons would interact with neutrons. Light interacts with all charged particles and not just electrons. Because of its nature, neutronuim would behave like a black body and therefore would emit light in the form of black-body radiation. By definition, a black-body is "black" and so you would probably be right with answer "3. perfectly black".

joseph h
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  • I understand that quarks have a charge of +/- 1/3 e. But overall, neutrons do not have any electric charge, that's why they are neutrons not protons, so I don't understand how light could "see" the neutrons. I'm used to the idea that you need large (gamma ray) photon energies to interact directly with the quarks that make up neutrons. There are no electromagnetic forces that I am aware of holding the neutrons together, just gravity, so once again, I don't understand how there can be any classical interaction at such low photon energies, hence my comment about an absorption spectrum. – Barry Stone Aug 23 '20 at 00:56
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    First of all, photons can be scattered by neutrons, $n + \gamma \rightarrow n + \gamma'$ And even though the neutron as a whole is electrically neutral, it is not a point particle. The neutron has a multipole moment (electric and magnetic). A photon can interact with a particle if it has a multipole moment, even if its net charge is zero. – joseph h Aug 23 '20 at 01:14
  • The neutron has a magnetic dipole moment, but no electric dipole or quadrupole. The cross-section for scattering is very, very small at visible wavelengths. – ProfRob Aug 23 '20 at 13:48
  • What's the mechanism behind the emission of light in the form of black-body radiation? – Deschele Schilder Aug 23 '20 at 17:44
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Hi there and welcome to the family!
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. I can see no reason why the neutrons don't have associated 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.

Did this scratch your back?

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. After reading a comment I'm not so sure about this anymore. Inside a neutron, the quarks experience a quite small mutual attraction (here is a relation with quark confinement). If the photon has certain energy it's quite possible that the neutron will absorb and re-emit the photon (the photon gets scattered). So, in that case, neutronium has color if the scattered photons have a frequency falling within the range of visible light. They are transparent though (or black) if the photon can't be promoted to a higher energy state.

Also, one can read in the first citation:

It is not a bound particle but had been proposed as an extremely short-lived resonance state produced by nuclear reactions involving tritium.

So the particle is not a bound state of six quarks. It's a resonance, caused by the nuclear force. If this resonance can be excited (the resonance resonates(?)) in a very short time span by a photon, depends on how strongly the resonance is held together. If that force is stronger than a gamma photon can deliver to the resonance, no absorption will occur but if not then the photon that has enough energy to excite a quark in the short-lived resonance will be absorbed.
Draw your conclusion. This problem is hitting me in the face now too! But in a kind way...

Deschele Schilder
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  • If the neutronium is dark, how is the energy absorbed? If it can't emit photons, then by symmetry, it can't absorb photons. – Barry Stone Aug 23 '20 at 01:31
  • It can absorb gluons. – Deschele Schilder Aug 23 '20 at 01:33
  • I'm not arguing about the gluons. At this level of particle physics, I have no intuitive understanding of the nature of the interactions. I'm happy to accept your assertion that neutronium can absorb and emit gluons. But I can't see gluons, only photons. If the strong force is too strong for gamma photons to overcome, would the photons be absorbed or reflected? I take it that what happens to gamma photons would also happen to visible radiation? – Barry Stone Aug 23 '20 at 01:52
  • The gamma photon just goes on very happily. It can't interact with the quarks, so nothing changes in its situation. There is some math behind this, but believe me. So there is no reflection, no absorption, absolutely no interaction going on. The gamma photon just travels along. Now if we would be standing on the side where the gamma photon is heading and if we could see the gamma photon, in that case, the neutronium would be tranparent. – Deschele Schilder Aug 23 '20 at 02:21
  • I have Anthony Dwayne Chapelle suggesting neutronium can be any colour, Dr jh saying it is black and you saying it is transparent. This is a fun discussion! It also tells me that this is a seriously hard problem. I thought that I would be less confused at the end of this than at the start, but sadly not. It's been an interesting journey though. – Barry Stone Aug 23 '20 at 07:01
  • All right then! Well, it's black in the sense that it doesn't emit photons (black), while at the same time it can let photons go right through (transparent). – Deschele Schilder Aug 23 '20 at 07:11
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    Which means that it can't change temperature (no photon absorption or emission). It's a black body radiator with an emissivity of zero? My brain hurts! It's been a long time since I thought about any science problem in depth like this. Thanks to everyone for taking part but to understand this any further I would need to dive in to maths that is way beyond my ability. – Barry Stone Aug 23 '20 at 07:22
  • The approximation that nucleons within nuclei move in orbitals is used in the shell model and in mean-field theories, but it's only a good approximation in large-ish nuclei. Small nuclei tend to have only unbound excitations that decay by particle emission. In the mass-2 system, the deuteron nucleus is transparent to photons below 2 MeV, but higher-energy photons dissociate it. A hypothetical stable dineutron would have to have $L\neq 0$ in its first exited state, due to exchange symmetry; it's hard to imagine such an excitation would be bound. – rob Aug 23 '20 at 13:19
  • So the difference in energy between, say the two lowest energy levels in the deuteron (caused by the nuclear force) can be overcome by a photon hitting a quark? Isn't the color force between two quarks stronger than the electromagnetic force? Or do the quarks move almost freely inside a neutron (quark confinement)? – Deschele Schilder Aug 23 '20 at 13:26
  • (1) The deuteron does not have two energy levels. There is one bound state, and the continuum. (2) The quark degrees of freedom are not relevant in deuteron photodissociation. The parsimonious model is the nucleon-nucleon strong interaction, mostly mediated by pion exchange. The behavior of the quarks inside the nucleons is strongly collective; you don't get to talk about "a quark." (3) If you do want to talk about striking quarks, the first thing that happens is that you make excited baryons. The first is the delta; the phenomenon is pion photoproduction. – rob Aug 23 '20 at 16:41
  • I am talking about the excitation of the three quarks in either neutron. Though the neutrons are intimately connected, neutronium is not a single particle consisting of six quarks. One thus can speak of an excited quark (though it is not known which) inside the neutron, even if it's part of the neutronium. So seen separately, it depends on how strong force, mediated by pions, is compared to the e.m. force hitting a charged quark. And when one does consider the neutronium as a bound state of six quarks, why shouldn't it be possible to excite this state? – Deschele Schilder Aug 23 '20 at 17:40
  • Some good reading (though not up to date): Some good reading: https://www.jstor.org/stable/97540?read-now=1&seq=1#page_scan_tab_contents – Deschele Schilder Aug 23 '20 at 17:40
  • The first such excitation would be $\rm nn\to n\Delta$; your out-of-date reading predates the discovery of the $\Delta$. – rob Aug 23 '20 at 19:15