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The Trajectory of a Nucleon in a Nuclear Force Field

This material is to show the trajectory of a particle subject to a potential of the form

V(r) = −H∫_{r}^{∞(}e^{-λs}/s²)ds

This is the potential for a force carried by particles which decay; as do the π mesons,
the carriers of the nuclear force.

A particle moving in a central force field characterized by a potential function V(r) conserves
total energy and angular momentum. The conservation of angular momentum q with respect to the center of the
force field requires that

q = mr²(dθ/dt)

be constant. This may be solved for (dθ/dt) and the result substituted into the energy equation to
obtain

The ± comes from taking the square root. The equation for the orbit has an
upper and lower branch which meet in tangency at the upper and lower limits for r.

The limits on r are established by finding the roots of

E − V(r) − q²/(2mr²) = 0

For the nuclear force potential V(r) = − H∫_{r}^{∞}(e^{-λs}/s²)ds
this reduces to

E + H∫_{r}^{∞}(e^{-λs}/s²)ds
− q²/(2mr²) = 0

It is worthwhile to change the variable of integration from s to z=λs. Thus the equations
becomes

E + Hλ∫_{λr}^{∞}(e^{-z}/z²)dz
− q²/(2mr²) = 0

Now it is desirable to divide through the equation by Hλ to make it nondimensional; i.e.,

The angular momentum q is quantized; i.e., q=nh, where n is an integer and
h is Planck's constant divided by 2π. The value of h
is 1.05457×10^{-34} kg m²/s. Thus if q=h then
q²=1.1212×10^{-68} kg^{2} m^{4}/s^{2}. The mass
of the neutron is 1.6749×10^{-27} kg so q²/m is
6.64×10^{-42} kg m^{4}/s^{2}.
The value of λ is 1/(1.522×10^{-15} m^{-1}
and thus q²λ/m is 4.36268×10^{-27} kg m^{3}/s^{2}

An estimate of H is
1.92570×10^{-25} kg*m^{3}/s^{2}.
Therefore q²λ/(Hm) is 2.2655×10^{-2}=1/44.14, a dimensionless number.