where r is the separation distance between the two particles. The negative sign indicates that it is an attractive force. The parameter λ will depend upon the time rate of decay of the force-carrying particles and their velocity. It is convenient to express λ as the reciprocal of a scale distance r₀. The purpose of this page is obtain an estimate of the constant H* using the properties of the deuteron.

An estimate of the value of r₀ is available from the mass of the π mesons. Hideki Yukawa preposed a relationship between the mass of the force-carrying particle and the scale distance of the force. From Yukawa's relation the scale distance for the nuclear force should be 1.522 fermi; i.e., 1.522×10⁻¹⁵ meters.

The potential energy function V(r) is defined to be such that (∂V/∂r)=−F(r), where F is the force function.

For the above formula for the nuclear force the potential energy for one nucleon separated from another by a distance of r is

V(r) = ∫_r^∞H*(exp(−z/r₀)/z²)dz = H*∫_r^∞(exp(−z/r₀)/z²)dz

This may be simplified by the substitution z/r₀=s and thus z=r₀s and dz=r₀ds; i.e.,

V(r) = −(H*/d₀)∫_r/r0^∞((exp(−s)/s²)ds

Consider the integral W(x) = ∫_x^∞((exp(−s)/s²)ds and the integration-by-parts formula of ∫GdF=GF−∫FdG, with G=exp(-s) and F=−1/s. Thus dF=1/s² and dG=−exp(−s). Therefore, GdF=exp(−s)/s², FG=−exp(−s)/s and FdG=+exp(−s)/s. Thus,

W(x) = [−exp(−s)/s]_x^∞ − (∫_x^∞(−exp(−s)/s)ds)
or, equivalently
W(x) = exp(−x)/x + (∫_x^∞(exp(−s)/s)ds)

The integral on the right is closely related to the function called the exponential integral and its values have been tabulated.

The value of the ratio H/d₀ is energy. The disassociation and binding energy of the deuteron is 2.22457 MeV. The accepted diameter of the deuteron is 4.2 fermi. The nucleons themselves are believed to have diameters of 1 fermi. Thus the diameter of the deuteron would include the two half diameters of the nucleons so the separation distance of the centers of the two nucleons is 3.2 fermi. The ratio of this distance to r₀=1.522 fermi is 2.1. The value of W(2.1) is 0.100927. Thus the ratio (H/r₀) has the value of 20.83 MeV. In joules this is 3.337×10⁻¹². This figure multiplied by r₀=1.522×10⁻¹⁵ gives H*=5.079×10⁻²⁷ kg m³/s².

A value of 1 fermi for the diameters of the nucleons is conceptually and numerically uncertain. Therefore it is not advisable to rely too strongly on this figure. If the diameters of the nucleons is ignored and the separation distance of the nucleons is taken to be 4.2 fermi then the estimate of H* is 1.3333×10⁻²⁶. This suggests that H* has a value on the order of 1×10⁻²⁶ kg m³/s². A value of 51.84 MeV for the ratio (H*/r₀) and a separation distance of 4.13 fermi gives a potential of 2.22457 MeV for the deuteron.

(To be continued.)