Neutrons have no charge so they would not, I think, interact with photons. Would a neutron star be transparent?
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1$\begingroup$ Related, perhaps an effective duplicate: physics.stackexchange.com/questions/22722/… . Opinions? $\endgroup$– dmckee --- ex-moderator kittenCommented Mar 22, 2013 at 0:04
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1$\begingroup$ @dmckee Pretty close duplicate. $\endgroup$– MichaelCommented Mar 22, 2013 at 0:26
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$\begingroup$ related physics.stackexchange.com/q/130459/44176 $\endgroup$– Nikos M.Commented Aug 11, 2014 at 16:46
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$\begingroup$ While a neutron star wouldn't be, it's an interesting question whether in theory, pure neutrons would be, not that there would be much use for a transparent few miles across, super dense object that would bend light around it and you couldn't get close to anyway, but it's a fun question. Likely such an object, made of pure Neutrons, the outer layers would degenerate into protons anyway, so, transparency wouldn't be possible. $\endgroup$– userLTKCommented Apr 8, 2015 at 0:25
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4 Answers
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Neutron stars are one of the least transparent objects in the universe. There are still lots of protons and free electrons, and most importantly the density of these charged species is extremely high. Also, the quarks, which constitute neutrons, are charged---so due to their density, I would guess (hopefully an expert will chime in) that that is also plenty to keep them optically thick. Definitely once you get to gamma-ray energies, neutrons will interact with photons.
answered Mar 21, 2013 at 21:18
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1$\begingroup$ +1 for pointing out that neutron stars don't just consist of neutrons. $\endgroup$– ProfRobCommented Jan 14, 2015 at 18:09
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$\begingroup$ Concerning quarks: If quarks are presented they are most probably in a superconducting state. Some of the most popular candidates, the CFL phase is actually nearly transparent due to the formation pattern of the Cooper pairs $\endgroup$– NoldigCommented Feb 19, 2019 at 19:03
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$\begingroup$ I think it even absorbs a substantial fraction of neutrinos. $\endgroup$ Commented Sep 19, 2019 at 2:03
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1$\begingroup$ However, pure neutrons are almost transparent. The magnetic dipole scattering cross-section for a neutrons with visible photons is extremely small (neutrino-like). physics.stackexchange.com/questions/89330/… $\endgroup$– ProfRobCommented Aug 23, 2020 at 11:29
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Neutron stars are not charged, but they definitely have magnetic dipole moments, and these DO interact with the electromagnetic field.
answered Mar 22, 2013 at 2:19
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$\begingroup$ Hi Jerry, I was wondering about this while formulating my response. Would you need to worry about the magnetic field before it's energy density became comparable to that of the radiation field? Or is that the relevant regime? $\endgroup$ Commented Mar 22, 2013 at 4:03
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1$\begingroup$ In principle the gamma could, indeed, couple pretty well to the neutron EDM, but in practice the vast majority of those neutrons have no accessible states to scatter into---you have to get them above the Fermi surface, right? I think that Ron and John's analyses in the proposed duplicate are the way to go. $\endgroup$ Commented Mar 22, 2013 at 14:03
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$\begingroup$ @dmckee: I'm sure you could work out what the scattering depth of a solid substance with a constant magnetic polarization. And I'm pretty sure that it would be pretty shallow for something like a neutron star, which is basically a ferromagnet. $\endgroup$ Commented Mar 23, 2013 at 0:34
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Light with wavelength $\lambda\gg10\,\mathrm{km}$ can pass around a neutron star thanks to diffraction, even if the star is made of a perfectly absorbing material. That's exceptionally low-energy radio waves, though, and pretty different from what you probably had in mind.
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You can think about it in more practical terms. It is known that one metre thick lead can stop very energetic $\gamma$-photons. Then, given that a neutron star is billions of times denser than lead and several kilometres across, it is not very likely that a neutron star will be transparent.
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3$\begingroup$ Note that lead is a good photon absorber because it has a high electron number density, so while the facts here are correct the reasoning is suspect. $\endgroup$ Commented Mar 22, 2013 at 13:53
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$\begingroup$ @dmckee If you take into account the fact that there are some $10^{44}$ neutrons per m$^3$(only a rough estimate,) about 16 orders of magnitude higher than the number of electrons in lead in the same volume, and that each neutron has three charged particles (quarks,) then perhaps the answer maybe not such suspect. If the question is talking about transparency to neutrinos, then the answer could be different, as it could be possible for neutrinos to travel through the neutron star. $\endgroup$– JKLCommented Mar 22, 2013 at 23:29
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$\begingroup$ A visible photon can not interact with the bulk of those neutrons because they are far, far below the Fermi surface. The bulk material is presumably opaque anyway simply because the conversion of proton to neutrons only proceeds until the electron Fermi surface is in agreement with that of the neutrons, so there are free protons in the mix. $\endgroup$ Commented Mar 22, 2013 at 23:39
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$\begingroup$ @dmckee if you are still around, could you explain the "Fermi surfaces in agreement"? $\endgroup$– JDługoszCommented Jan 14, 2015 at 11:21
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1$\begingroup$ @jdlugosz By "in agreement" I mean at the same energy. Of course, that is only helpful if you understand what "Fermi surface" implies. The reason for this is that converting a proton requires energy to get the neutron above the neutron Fermi surface, but removes an electron from the electron Fermi surface. $\endgroup$ Commented Jan 14, 2015 at 14:16