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Characteristics in a Central Force Field, Including Their Rest Masses |
Previous work derived the quantization scheme that exists for particles in a central force field when the orbital angular momentum is quantized. It included relativistic conditions as well as nonrelativistic conditions. That analysis started a single particle of mass m traveling in a circular orbit of radius r. The central force F on a particle at a distance r from the center of the force field is given by the formula
where k is the constant for the force field, q is the charge generating the force field. The charge might be electrostatic charge or mass. The charge would be the number of electrostatic charges or it could be the number of nucleons or the number of particles. The charge of the central force field is presumed to be quantized. The charge of the single particle under consideration is presumed to be unity.
The quantity f(r) is a dimensionless quantity. For forces carried by particles that do not decay such a the electrostatic force or gravity the quantity f(r)=1. For a force carried by a decaying particle such as a pi meson the quantity f(r) is of the form e^{-λr}. The force may be a combination of nuclear force, electrostatic force and gravity. The force may be an attraction or a repulsion, depending upon the sign of k.
For a circular orbit the attractive force must balance the centrifugal force; i.e.,
.
Angular momentum is equal to mvr so its quantization means
where n is an integer and h is Planck's constant divided by 2π.
Expressing v as βc this quantization condition along with the condition for a circular orbit are:
The division of the first equation by the second results in
so that if the quantization of r is determined then the quantization of velocity is given by
the above expression. The quantity (k/(hc)) is a dimensionless constant that may be
denoted as γ, a characteristic of the force field. Thus the general relationship,
which applies
in both the relativistic and nonrelativistic case, is
For the electrostatic force γ is the fine structure constant 1/137.036. An estimate of γ for the nuclear force is 1.5217246. For more on this quantity γ see Force Constants.
Under relativistic conditions the mass of the particle is given by
where m_{0} is the rest mass of the particle. For the immediate analysis m_{0} is taken to be a constant. However later analysis will allow for the possibility that m_{0} is quantized.
The angular momentum condition then takes the form
For convenience let nh/(m_{0}cr) be denoted as D.
The previous equation may be solved for β² as
On the other hand from the general expression previously found for β
Equating the two expressions for β² gives
Replacing D by its definition and carrying out some simple algebraic manipulation gives
The quantity hc/(m_{0}c²) has special significance but it has the
dimensions of length. Let r_{0} be some significant unit of length. The variable r then can
be expressed as r=zr_{0}.
The quantization condition then takes the form
Now let (hc/(r_{0}m_{0}c²)=μ.
The quantization condition then takes the form
The function f(r) could be a constant or it could be e^{−λr}, in which case r_{0}=1/λ. In that case f(zr_{0})=e^{−z}. Let f(zr_{0}) be denoted as φ(z). Then the quantization condition takes the form
This is the condition sought for the quantization of z and r.
The constant hc is equal 3.1616×10^{-26} kg m³/s².
For a proton m_{0}c² = 1.503353×10^{-10} kg m²/s² (joules).
Then for r_{0}=(0.5)1.522×10^{-15} m, the nondimensional constant μ = 27.635116 and μ² = 763.7.
As stated previously an estimate of γ for the nuclear force is 1.5217246 and hence γ²=2.31564583.
For a given value of n the quantization condition may have one solution, no solution or more than one solution.
Let N be the set of values of n for which the quantization condition has solutions and R the set
{r_{n}, n∈N}. Then the set of allowable β's B={β_{n}, n∈N} is given
by β_{n}=f(r_{n})/(nhc). These values depend upon the dimensionless
parameter μ which in turn depends upon the rest mass m_{0}.
The orbital angular momentum condition implies that
The value of (hc/r_{0}) is 4.155285×10^{-11} kg m²/s² (joules). The value of this constant
relative to the rest mass energy of a free proton is 0.2764. Relative to the rest mass energy of a neutron
it is 0.275355.
Since the quantized values of β and z depend upon m_{0} through the parameter μ there has to be a simultaneous determination of m_{0}, β and z as functions of the quantum number n. Quite likely the solution could be found through an iteration scheme.
Since μ was defined to be hc/(r_{0}m_{0}c²) the previous
equation can be expressed as
The general condition for β can be expressed as
Therefore
The quantization condition found previously was
The above two equations for given values of n and γ then might determine the quantized values of z_{n} (and thus r_{n}) and μ_{n} (and thus m_{0}_{n}). Once values of r_{n} and m_{0}_{n} are found then the values of β_{n} and m_{n}c² could be found.
The kinetic energies κ_{n} are given by
Thus kinetic energies are known only if the rest masses are known.
The potential energies can be expressed as
However the potential energy is really a function of the separation distance of the particles rather than orbit radius. In a deuteron the separation distance s is approximately twice the orbit radius r; i.e., s_{n}=2r_{n}.
When a transition in n results in a lower potential energy some of that energy loss goes into an change in mass energy mc² but some of it goes into the energy of an emitted photon. . The change in mc² may involve a decrease in rest mass energy (due to a decrease in rest mass) and an increase in kinetic energy (due to an increase in velocity). Thus the enigma of mass deficits of nuclei may be explanable as a quantization phenomenon.
Although it was not obvious the equation for determining z and the equation for μ² are not independent. Effectively they are the same equation. Thus a quantization of the rest mass could not be determined. Nevertheless some of the computational results may be of interest.
The case of the deuteron is a good starting point. For n=1, q=1, γ²=2.31564583 and μ²=763.7 the solution for z is 0.4199598. This means the orbit radius is about (0.42)(0.5)(1.522 fm)=0.32 fm. For this value of r, the value of β is 0.999884519. The potential energy of a neutron at a distance of 2(0.32 fm)=0.64 fm from a proton is
(To definitely be continued.)
The problem reduced to its mathematical basics is that there are three unknowns, r, β and m_{0}, and that there are two equations to be satisfied; i.e.,
Because of the structure of these two equations it was simple to eliminate m_{0} and get one equation in two unknowns, r and β. But it is a mathematical impossibility to determine unique values for all three variables. Of course if the value of m_{0} is given then values may be determined for r and β. From the values of r and β the values of all derived quantities such as potential and kinetic energy may be determined.
A third independent equation is needed to complete the system. Such a third equation might be a separate quantization. Or, it might be
(To definitely be continued.)
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