Can photon interact with neutron?

[skeleton]

I want to know if a photon can interact with a neutron?

I understand that a photon will mediate the electromagnetic force between electrically charged particles (such as electron and proton). This is even manifested in the photoelectric effect. So, I presume when we "see" matter through visible light, it is the photons shone onto the matter are interacting with the electrons bound to those atoms.

Since the neutron is electrically neutral, I am presuming that a photon might not be able to interact with it.

So, this leads to my question. If a beam of neutrons was sprayed with photons, would their be any interaction? Would the photons reveal the presents of the passing neutrons?

(Curious ...)

[jtbell]

Since the neutron is electrically neutral,

Neutrons are made up of quarks which are electrically charged.

[malawi_glenn]

So yes, photoproduction of hadrons from neutron target are possible.

[hamster143]

Can anyone give an order-of-magnitude calculation of photon-neutron scattering cross section, in the limit of low-energy photons (visible light, ~1 eV)? Or the ratio of that cross section to photon-hydrogen? IOW how much more "transparent" would a pound of neutrons be, compared with a pound of hydrogen atoms?

[malawi_glenn]

why not search in papers yourself?

[Vanadium 50]

Also, for visible light, you aren't interacting with the quarks. You're interacting with the whole neutron, and there the interaction is predominantly magnetic.

[hamster143]

Also, for visible light, you aren't interacting with the quarks. You're interacting with the whole neutron, and there the interaction is predominantly magnetic.

That's what I thought, too, but apparently the dominant term is driven by electric polarizability of the neutron: T \approx \alpha E^2, \alpha \approx 1.2 * 10^{-3} fm^3.

[Vanadium 50]

Indeed.

[skeleton]

Thanks for your insight.

So, let's see if I understand this right ...

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Neutron composition
1) Up-Quark Charge = +2/3
2) Down-Quark Charge = -1/3
3) Down-Quark Charge = -1/3

A) Electric charge
==============
Sum of quark charges = 0; thus electrically neutral. Good.

B) Electric dipole moment
===================
Where both down-quarks are bounded together as a geometric pair (in space),
then this pair can couple with the up-quark to form an electric dipole.

The dipole equation:
e = 1.6E-19 C
q = 2/3*e (1*Up or 2*Down quarks)

p < 2.9E-26 e*cm (Source: http://en.wikipedia.org/wiki/Neutron)
p = q*r

r < p/q = 4.2E-26 cm = 4.2E-28 m

C) Size of neutron
=============
d = 1E-15 m (Source: http://wiki.answers.com/Q/What_is_the_size_of_a_proton_neutron_and_electron)

D) Compare
=========
Diameter of neutron / Radius of quark separation:
ratio = d/r = 2.3E+10
But, shouldn't this ratio = 1 ?

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It looks like my conclusion (ratio d/r) is wrong. What is my mistake, if so?

Also, my calculation presupposes that the three quarks will be spatially oriented, so as to create an electric dipole. Further, since a neutron can manifest electric polarizability, I believe this implies that the spatial orientation can be maintained over time. (This is all surprising to me since I would have thought the three quarks would all occupy concurrently the same spatial region of the neutron.) Are these suppositions correct?

(Thanks again in advance.)

[jtbell]

TB) Electric dipole moment
===================
Where both down-quarks are bounded together as a geometric pair (in space),
then this pair can couple with the up-quark to form an electric dipole.

The dipole equation:
e = 1.6E-19 C
q = 2/3*e (1*Up or 2*Down quarks)

p < 2.9E-26 e*cm (Source: http://en.wikipedia.org/wiki/Neutron)


No one has actually detected a neutron dipole moment yet, as far as I know. This is an experimental upper limit, i.e. the maximum sensitivity of experiments done to date. It means that the neutron could have a dipole moment less than this value but we would not have been able to detect it.

As experiments improve, the upper limit on p will decrease, until we actually detect a neutron dipole moment (if we ever do).