Magnetic moment and binding energy

[x_engineer]

I ran across the fact that the magnetic moment of the neutron is approx 9.66e-27 J/T. When I plug it into the formula for attractive/repulsive force between 2 magnetic dipoles within 1 femtometer of each other (nuclear dimensions) I get almost 9000N, and the energy to separate them (if attractive) would be like MeV, about the nuclear binding energy.

Of course neutrons for some reason like to orient anti-parallel to the local magnetic field so they will likely repel each other.

Has anybody proposed this as the origin of the binding force?

 

[daschaich]

The strong nuclear force overwhelms electromagnetic effects.

 

[Creator]

I ran across the fact that the magnetic moment of the neutron is approx 9.66e-27 J/T. When I plug it into the formula for attractive/repulsive force between 2 magnetic dipoles within 1 femtometer of each other (nuclear dimensions) I get almost 9000N, and the energy to separate them (if attractive) would be like MeV, about the nuclear binding energy.

?

Which formula are you using may I ask?

Creator

 

[Bill_K]

It's easy to estimate things like this, provided you remember a few constants - and don't try to use SI units! The constants to remember are just the fine structure constant e²/hc = 1/137 (forgive me, I'm writing h instead of h-bar) and Planck's constant, hc = 200 MeV-f. (Actually it's 197.5, but I'm rounding off.)

First the Coulomb interaction energy between two protons. It is V = e²/r = (e²/hc) (hc/r) = (1/137)(200 MeV-f/1 f) = about 1.5 MeV. Note, the Coulomb energy is *not* overwhelmed by the strong force, which is about 8 MeV/nucleon. Yes it's smaller, but not by much.

Second, the dipole-dipole interaction. The magnetic dipole moment M of the neutron is about -2 nuclear magnetons. A nuclear magneton is the unit of magnetic moment used in nuclear physics and defined as eh/2m, where e is the charge of the proton and m is the mass of the proton, which in turn is about 1 GeV. So M = -2 (eh/2m) = -eh/m. The dipole-dipole interaction energy is similar to the Coulomb energy, only it goes like 1/r³. (Never mind all the vectors and factors of 3 and stuff. We're looking for an order of magnitude.)

So V = M²(1/c²)(1/r³) = (eh/mc)²(1/r³) = (h/mc)²(e²/r³). Well we don't even have to work this out, because it's just like the Coulomb energy except for the extra factors (h/mc)²(1/r²). Here r = 1 fermi, and h/mc = the Compton wavelength for the proton, which is hc/mc² = 200 MeV-f/1GeV = 1/5 f. So the extra factors amount to (1/5)². Voila, the dipole-dipole interaction energy is about 1/25 of the proton-proton Coulomb energy.

 

[daschaich]

Terminology aside, the physics content of my previous post is the simple observation that the strong nuclear force can bind multiple protons into nuclei, despite their electromagnetic repulsion. I think this is an even easier way to answer the original question.

 

[bcrowell]

I ran across the fact that the magnetic moment of the neutron is approx 9.66e-27 J/T. When I plug it into the formula for attractive/repulsive force between 2 magnetic dipoles within 1 femtometer of each other (nuclear dimensions) I get almost 9000N, and the energy to separate them (if attractive) would be like MeV, about the nuclear binding energy.

I don't think this can be right, since the straight Coulomb repulsion would be about a couple of hundred Newtons, and the magnetic force is going to be less than the electrical one.

 

[x_engineer]

The formula I used is here:-

http://en.wikipedia.org/wiki/Magnetic_moment

Making suitable assumptions about the angles

F(r) = 3 * mu * (M^2)/ (4 pi R^4)

I made a mistake in my calculation, the force comes out to 27.99N at 1 femtometre. Because of the R^4 dependence it does not take much of a change to get to 9000 N - that is what it would be at .24 femtometre.

The Coulomb force is about 231 N at 1 femtometre, and about 4000 N at .24 femtometre.


The reason I bring this up is because at small scales (of the order of the nuclear radius) the magnetic moment can completely dominate force. Under .5 femtometer it becomes larger than the Coulomb force and grows much more rapidly than the Coulomb force. It is kind of like the binding force in that respect. It would take only a small assumption about the structure of a nuclear particle behaving like a ferromagnet/paramagnet in those enormous magnetic fields to explain why a nucleus is around femtometres in radius.

 

[Creator]

The formula I used is here:-

http://en.wikipedia.org/wiki/Magnetic_moment

Making suitable assumptions about the angles

F(r) = 3 * mu * (M^2)/ (4 pi R^4)

I made a mistake in my calculation, the force comes out to 27.99N at 1 femtometre. Because of the R^4 dependence it does not take much of a change to get to 9000 N - that is what it would be at .24 femtometre.

The Coulomb force is about 231 N at 1 femtometre, and about 4000 N at .24 femtometre.

Thanks for providing the formula, X Engineer; I figured that's the one you were using , and that's why I asked... the 1/ r^4 dependence for Force...

The reason I bring this up is because at small scales (of the order of the nuclear radius) the magnetic moment can completely dominate force. Under .5 femtometer it becomes larger than the Coulomb force and grows much more rapidly than the Coulomb force.

Using that formula you make a good point.
However...
You should realize this is a classical formula and that force equation was derived from the magnetic dipole field equation (also given in the link you supplied) ...and is an approximation of the field strength based upon approximation for distances where r is LARGER than the dipole length, which at these tiny distance may be quite an unreliable assumption.

Furthermore, when protons are present, and even though its dipole magnetic moment is comparable to that of the Neutron, its electric field (from its charge) can be so great at such close radii that there are radiative corrections; IOW, the high fields (both electric and magnetic) in effect may change the vacuum permeabilty & permittivity 'constants' upon which the above 'classical' dipole magnetic field formula is based.

Now you know why QED is normally used.
However, it is an interesting point that I think bears attention.

Creator

 

[x_engineer]

Wow! so there is a mechanism for correction!

I can see how a high field can increase the permittivity of free space in the presence of vacuum fluctuations. I guess the same could happen with the permeability in which case the range of the magnetic moment would increase... As for the approximation condition, how big is the dipole actually, and maybe that explains why the nucleus does not just disappear into a point (at least not with measurable probability - we currently do believe in black holes)

I am basically an electrical engineer, so QED is way out of my league. But I have found there are so many solutions to the electromagnetic field equations; so that just because you find one solution does not mean the system will not behave totally differently, because there might be another solution with different properties. With things happening so fast, you need a boundary on the range of possible solutions to say something meaningful in our timeframe, and QED probably is that.

 

[x_engineer]

I should have said --- with things happening so fast and almost anything possible, you need a probability distribution on the range of possible solutions to say anything meaningful about the system in timeframes at our scale ---

 

[daschaich]

... As for the approximation condition, how big is the dipole actually, and maybe that explains why the nucleus does not just disappear into a point (at least not with measurable probability - we currently do believe in black holes)

To calculate electromagnetic form factors of nucleons, one typically uses lattice quantum chromodynamics (some observational information should be obtainable from deep inelastic scattering experiments)... the approximation of pretending we don't know about QCD is not going to be reliable.