the The Separation Distance In Spin Pairs of Like-Nucleons
San José State University

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Thayer Watkins
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The Separation Distance In
Spin Pairs of Like-Nucleons

Consider two nucleons of a like nature each with a charge q of some type and rate of rotation ω. Because they are of a like nature the are repelled from each other and the repulsive force is given by

F = Hq²f(s)/s²

where s is the separation distance of the centers of the nucleons, H is a constant and f(s) is a monotonically non-increasing function. This function f(s) is f(s)=1 for electrical charge and may be exp(−s/σ) for nucleonic charge.

Special Relativity requires that associated with a moving charge of any type there is a magnetism-like field. The spinning nucleons then generate a magnetism-like field with poles which aligns the spins and attracts the nucleons together. The magnetic field intensity B for a spinning sphere of electrical charge is given by the formulas

B(r, φ, θ) = ∇×A(r, φ, θ)
where the vector potential A is
A = (μ0qω/6π)(R/r)²sin(φ)θ^

where μ0 is the permittivity of space, R is the radius of the spherical charge and θ^ is the unit vector is the direction of increasing θ. The important thing is that the magnetic field intensity due to one spinning sphere of charge is proportional the total charge q of the sphere, say

B(r, 0, 0) = K(R/r)²qω

where K is a constant. The force FM of attraction between two spinning spheres each with charge q is proportion to the product of the field intensities.

FM = L(R/r)4q²ω²

The two nucleons would assume a configuration in which the repusion of their charges is balanced by the attraction due the magnetism-like field due to their spinning; i.e.,

F = FM
Hq²f(s)/s² = L(R/r)4q²ω²

The distance r is equal to s/2. Both sides of the above equation are proportional to q² and this term may be cancelled out. Thus the above equation reduces to

Hf(s)/s² = 16L(R/s)4ω²
or, further to
Hf(s)s² = 16LR4ω²

This equation has a solution for separation distance that is independent of charge. For electrical charge for which f(s)=1 the solution is

s = 4(L/H)½R²ω

Spin pairs of nucleons with different charges but the same radius and spin rate would have the same separation distance. Thus spin pairs of neutrons and of protons would have the same separation distance.

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