This is the beginning of an extensive study the nuclear magnetic moments of nuclides as a means of measuring the rates of rotation of nuclei. On a microscopic level the magnetism of a particle or nucleus is better represented as an electromagnet which results from an electrical current traveling in a circular orbit.

For a particle with a net charge of Q that is spinning at a rate of ω (radians per second) or ν (turns per second) the effective current is i=Qν=Qω/(2π). The area of the loop which the current surrounds is πr². Thus the magnetic moment μ is given by

μ = iA = Qνπr²
or, equivalently
μ = Q(ω/(2π))πr² = (Q/2)ωr² = (Q/2)vr

where v is the tangential velocity of the charge. This is analogous to the angular momentum of a particle. The angular momentum involves the mass of the particle rather than the term (Q/2).

When the charge is not concentrated at a specific radius but has a radial distribution ρ(r) then the calculation of the magnetic moment is more complicated. The details are given in Nuclear Magnetic Moments

Nuclear Dipole Moments

The magnetic dipole moment of a proton, measured in magneton units, is +2.79285. That of a neutron is −1.9130. The ratio of these two numbers is −0.685, intriguingly close to −2/3.

The first nuclide to be considered is a deuteron, which is a proton and a neutron spinning about their center of mass. They are held together by the formation of a spin pair, by the force of nucleon interactions, and by the elctrostatic force. In such a system only the proton would generate a dipole moment. The sum of the dipole moments of a neutron and proton is 0.87980 magnetons, intriguingly close to the magnetic dipole moment of a deuteron, which is 0.8574. The difference of 0.02237 magnetons, which is only 2.6 percent of that value. The small value due to the deuteron rotation could come from that rotation being at a much slower rate than the spins of the nucleons. The difference may be due to a rotation of the deuteron about its center of mass, as depicted diagrammatically below.

The value of the nuclear magnetron μ_N in SI units is

μ_N = eh/(2m_p) = 5.051×10⁻²⁷ Joules per second

where e and m_p are the charge and mass of a proton, respectively.

This means that the magnetic moment due to the rotation of the deuteron is 1.1466×10⁻²⁸ Joules per second.

Since μ is equal to (e/2)ωr²

ω = 2μ/(er²) = 2(0.02237)eh/(2m_p)/[(e/2)r²]
which reduces to
ω = 2(0.02237)h/[m_pr²] = (0.04474)(1.055×10⁻³⁴)/[1.6727×10⁻²⁷(2.252×10⁻¹⁵)²
or, equivalently
ω = 5.56×10²⁰ radians/sec
and hence
ν = 8.84849×10¹⁹ rotations/sec

or roughly 88.5 million trillion rotations per second, or equivalently, 88.5 billion billion rotations per second.

In the Guide to the Nuclear Science Wall Chart there is this statement

[…] scientists can create nuclei which have very high angular momentum. Nuclei respond to this rotation, which can be as fast as a hundred billion billion revolutions per second[…]

The figure cited is roughly the same as the figure computed.

The second nuclide to consider is the triteron, a proton with two neutrons. The two neutrons form a spin pair of opposite spin so the neutrons' magnetic moments cancel out. Seemingly only the proton's magnetic moment and that due to rotation of the triteron contribute to the magnetic moment of a triteron. The magnetic moment of a triteron is 2.97896244 magnetons. Since the moment of a proton is 2.79285 magnetons the rotation of the triteron accounts for 0.18612 magnetons.

The third nuclide to consider is the helium isotope containing two protons and one neutron. The two protons form a pair of opposite spin so the protons' magnetic moments cancel out. Again seemingly only the neutron's magnetic moment and that due to rotation of the He3 nucleus contribute to the magnetic moment of the He3. The magnetic moment of a He3 nucleus is −2.12749772 magnetons. Since the moment of a neutron is −1.9130427 magnetons the rotation of the He3 seemingly accounts for −0.21445502 magnetons. But there is a problem. The triteron has only a single positive charge rotating whereas a He3 has two. Furthermore the deuteron with one positive charge rotating about its center of mass has only 0.02237 magnetons due to rotation. Note that the difference in the magnitudes of nuclear rotation of a triteron and a He3 is 0.02833502 magnetons. The ratio of this figure to the rotational momement of a deuteron is 1.26665. This is a far cry from a value of 2.0.

Apparently there is something that accounts nuclear moments besides the net spins of the nucleons and the rotation of the nucleus. And furthermore the effect of the net charge's rotation is not simple. The magnetic moment due to a rotationg charge depends not only upon the amount of the charge but also the distance from the axis of rotation. The near-equality of the magnitudes of the magnetic moments of the H3 and He3 nuclei indicates that the axis of roration for an He3 passes through one of its protons, or at least nearly so.

The nuclear magnetic moments given below are from the compilation prepared by N.F. Stone. The magnetic moments due to nucleonic spin are computed by the simple scheme:

Spin pairs of like nucleons account for zero moment
because nucleons pair with other nucleons of opposite spins.

The moment due to a singular proton is 2.79285 magnetrons
and −1.9130 magnetrons for a singular neutron.

It is also found that an alpha particle has no magnetic moment despite having two prositive charges that could be rotating. This suggests that its axis of rotation passes through its two protons.

In Stone's compilation a magnetic moment is given a sign only if it can be established experimentally. In the tabulation the ambiguous signs for three cases were resolved to their obvious values.

There is one glaring anomally in the table. The moment for the carbon isotope C13 with its 6 protons and 7 neutrons should be in the neighborhood of −1.7 magnetrons but instead the accepted value for it is +0.702 magnetrons. In spite of this one anomally the graph of measured moments versus the values predicted from the nucleonic spins shows a reaonably close relationship.

The regression equation with the C13 anomaly left out is

MMM = 0.37510 + 0.94986MMNS
      [4.1]      [20.1]

where MMM is the measured magnetic moment and MMNS is the magnetic moment due nucleonic spins.

The figures in the square brackets [ ] are the t-ratios for the coefficient above. In order for a variable to be statistically significant in explaining the variation in the dependent variable at the 95 percent level of confidence the magnitude of the t-ratio must be 2 or greater.

The coefficient of determination (R²) for this equation is 0.95. The regression coefficient for MMNS is statistically not significantly different different from 1.0 at the 95 percent level of confidence.

Nuclear Rotation

The magnetic moment due to the rotation of the nucleus is just the difference between the measured moment and that due the spins of its nucleons. The values of these moments due to nuclear rotation are:

The graph of this data shows there is no dependence of the moment due to nuclear rotation and the numbers of protons or neutrons in the nuclide.

When the anomaly of C13 is ignored there is no relationship. To verify this the magnetic moment due to nuclear rotation was regressed on the numbers of protons and neutrons the following result was obtained.

MMDR = −0.060136 + 0.14134p −0.018804n
   [−0.17]            [1.4]     [−0.35]

Neither the coefficient for the number of protons nor the number of neutrons is statistically different from zero at the 95 percent level of confidence.

(To be continued.)