where J is the moment of inertia of the nucleus, ω is its rate of rotation in radians per second, h is Planck's constant, divided by 2π and I is an integer.

When this equation is applied to computing the rates of rotation of a deuteron and an alpha particle the results are several billion trillion rotations per second. See Deuteron and Alpha Particle. This is an unbelievably large rate. It is so unbelievable that hardly any physicist publicizes such values even when they are working in the field of nuclear rotation. See Nuclear Rotation.

This raises the issue of finding an empirical determination of the rate of rotation of some nuclide. It might be that the constant in the Bohr-Mottelson is not necessarily the conventional reduced Planck's constant. To find out what that constant would be requires only the determination of the rate of rotation for one nuclide. Such a method of determination exists in the matter of nuclear magnetic moments.

A classical and macroscopic definition of magnetic dipole moment, such as for a bar magnet, would be the product of the pole strength and their separation distance, μ=ms.

The torque exerted on a bar magnet by an external magnetic field would be proportional to this magnetic moment and also to the strength of the external magnetic field B.

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 magnetic moment generated by the motion of the electron in its trajectory is the product of the effective current times the area surrounded by the path of the particle. The area of the loop which the current surrounds is πr². Thus

μ = iA = Qνπr²
or, equivalently
μ = iA = 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).

It is worthwhile to establish the relationship between the magnetic moment and angular momentum. Let angular momentum be denoted by L. The if L is divided by the mass of the particle and the result multiplied by (Q/2) the result is the magnetic moment μ.; i.e.,

μ = QL/(2m)

When the charge is not concentrated at a specific radius but has a radial distribution ρ(r) then the moment is given by

μ = ∫dQνπr² = ∫₀^∞(2πrρ(r))(ω/(2π))πr²dr = ωπ∫r³ρ(r)dr
with the total charge being given by
∫₀^∞(2πrρ(r))dr = Q

The radial distributions of charge for a proton and a neutron has been found to be

Although the net charge of the neutron is zero, because the negative charge is located at a greater distance from the center of rotation than the positive charge, the neutron has a dipole moment and it is negative.

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.

A deuteron is a proton and rotating about theircenter of mass. The magnetic moment of a deuteron is the sum of the magnetic moments of the proton and neutron plus the magnetic moment due to the rotation of the charge of the proton rotating about the center of mass of the deuteron.

Any difference in magnetic moments of the measured and the computed value may be due to a rotation of the deuteron about its center of mass, as depicted diagrammatically below.

In such a system only the proton would generate a dipole moment. A smaller value due to the deuteron rotation could come from that rotation being at a much slower rate than the spins of the nucleons.

Suppose that difference of 0.00502 magnetons. The actual value is larger than this, but it is worthwhile displaying the implications of this value. The reason for this tactic is the implications are so clear and uncomplicated. The analysis of the actual value is carried out below and elsewhere.

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 would be 2.5255×10⁻²⁸ Joules per second.

Since μ is equal to (e/2)ωr²

ω = 2μ/(er²) = 2(0.00502)eh/(2m_p)/[(e/2)r²]
which reduces to
ω = 2(0.00502)h/[m_pr²] = (0.01)(1.055×10⁻³⁴)/[1.6727×10⁻²⁷(2.252×10⁻¹⁵)²
or, equivalently
ω = 1.24365×10²⁰ radians/sec
and hence
ν = 1.98×10¹⁹ rotations/sec

or roughly 20 million trillion rotations per second or, equivalently, 20 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[…]

So this unbelievably high rate is not inconsistent with an empirical measurement.

Now consider the Bohr-Mottelson equation for nuclear rotations, but suppose the constant in the equation is not necessarily the conventional reduced Planck's constant h but instead another value that applies to the force of interactions between nucleons, call it q.

Jν = q(I(I+1))^½
with I=1

where J is the moment of inertia.

What would the value of q be? From the previous computations it would be

q = Jν/√2 = (m_pr²)ν/√2
and thus
q = 0.01004h/√2 = 0.00707h = (1/141)h

The value 141 is close to the reciprocal of the fine structure constant, the coupling constant for the electromagnetic force. The ratio of q to h should be inversely proportional to the coupling constants of the two fields; i.e.,

q/h = α_EM/α_N

If the ratio is about 1/137 then the coupling constant α_N for the nuclear force would be about 1. A quick check online reveals that accepted estimate of the coupling constant for the nuclear force is 1. My estimate of the coupling constant for the nuclear force is given in Coupling Constants.

The actual value of the difference between the magnetic moment of a deuteron and the sum of the magnetic moments a proton and a neutron is 0.02237 magnetrons, instead of the value of 0.00501 magnetrons used in the above. The ratio of 0.02237 to 0.00501 is 4.46507. This should be equal to (I(I+1))^½ where I is the quantum number for the deuteron. Note that

I(I+1) = (4.46507)² = 19.93685

Obviously a quantum number of 4 along with a quantum constant q=h/137 explains the data for a deuteron. Note that the minimum quantum number for a structure is not necessarily unity. That minimum depends upon the number of separate modes of vibration and rotation. This is also called the degrees of freedom of the system. In the case of a ring structure there are four modes: 1. As a vortex ring, 2. Spinning like a wheel, 3. Flipping like a coin about a horizontal axis, 4. Spinning like a coin about a vertical axis.

Conclusion

The above computed constant q is smaller than the reduced Planck's constant h by a factor of about 1/141. Theoretically it probably should be h/137. Thus for the rate of rotation of a deuteron based upon the empirical measurement of the magnetic moment of a deuteron the analysis provides a theoretical derivation of the quantum constant for nuclear force phenomena. The implication of this is however quite profound; i.e., that Planck's constant applies only in electromagnetic phenomena and at the atomic scale and above whereas at the nuclear level concerning nuclear forces a significantly smaller constant applies.

(To be continued.)