Are electrically neutral particles invisible?

[pelmel92]

As the thread title suggests, I was wondering if electrically neutral particles like neutrons can absorb/emit/reflect photons. I don't really have any sort of physics background so I'm not sure this question even makes much sense... but I vaguely remember hearing that photons don't necessarily interact with all particles, which got me to wondering about which particles they do/don't affect. Since photons are, as I understand it, electromagnetic in nature, it made sense to me that maybe they wouldn't be able to interact with electrically neutral particles... which I imagine would render them invisible.

Anyone willing to alleviate my breathtaking ignorance would be greatly appreciated :)

Cheers

 

[Delta2]

As wikipedia suggests here a neutron does not have NET electric charge. But it consists of up and down quarks which have electric charge (+2e/3 for the up quark, -1e/3 for the down quark, e the charge of electron) so most likely interracts with photons . (Common matter has no net electric charge too but still interacts with photons ).

 

[lpetrich]

Neutrons also have a magnetic-dipole moment, this also due to their quark content. However, it interacts much more weakly than an electric charge. I estimate for the photon scattering cross section:

CS(M1) ~ (w/m)² CS(E0)
M1 = magnetic dipole
E0 = electric charge

CS(E0) = Thomson scattering (scattering off of free charged particles)
CS(E0, neutron) = (m_electron/m_neutron)² CS(E0, electron)

w = angular frequency of photon, m = mass of neutron

CS(M1)/CS(E0,electron) is about 10^-24 for visible light.

Neutrons' electric-dipole moments are too small to measure. Experimental limits are about 10^-13 e*fermi (neutron size), while the Standard Model predicts a value 10^-6 times smaller than that. A neutron's electric-dipole moment would have to be about 1 e*fermi to produce an amount of scattering comparable to what a neutron's magnetic-dipole moment produces.

But there's another electrical property we can use: polarizability. That's even worse:

CS(pol) ~ (w/m)⁴ CS(E0)

So CS(pol)/CS(E0,electron) ~ 10^-42 for visible light.