What colour would neutronium be?

Question

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.

Answer 1

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".

Answer 2

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...