San José State University

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 Quantization of Particle EnergyCharacteristics 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

#### F = −kqf(r)/r²

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.,

#### mv²/r = kqf(r)/r² or, upon expressing the velocity v as βc where c is the speed of light mc²β²/r = kqf(r)/r² which reduces to mc²β² = kqf(r)/r

.

Angular momentum is equal to mvr so its quantization means

#### mvr = nh

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:

#### mc²β² = kqf(r)/r mcβ = nh/r

The division of the first equation by the second results in

#### β = (kq/(hc))f(r)/n = (k/(hc))qf(r)/n

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

###### β = γqf(r)/n

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.

## The Quantization of Orbit Radius

Under relativistic conditions the mass of the particle is given by

#### m = m0/(1−β²)½

where m0 is the rest mass of the particle. For the immediate analysis m0 is taken to be a constant. However later analysis will allow for the possibility that m0 is quantized.

The angular momentum condition then takes the form

#### m0βcr/(1−β²)½ = nhor, equivalently β/(1−β²)½ = nh/(m0cr) or, upon squaring both sides β²/(1−β²) = [nh/(m0cr)]²

For convenience let nh/(m0cr) be denoted as D.

The previous equation may be solved for β² as

#### β² = D²/(1+D²) = 1/(1/D² +1)

On the other hand from the general expression previously found for β

#### β² = [γqf(r)/n]²

Equating the two expressions for β² gives

#### [γqf(r)/n]² = 1/(1/D²+1) so [γqf(r)/n]²(1/D²+1) = 1

Replacing D by its definition and carrying out some simple algebraic manipulation gives

#### γ²q²f²(r)[r² + n²(hc/(m0c²))²] = n4(hc/(m0c²))²

The quantity hc/(m0c²) has special significance but it has the dimensions of length. Let r0 be some significant unit of length. The variable r then can be expressed as r=zr0.

The quantization condition then takes the form

#### γ²q²f²(zr0)[z²r0² + n²(hc/(m0c²))²] = n4(hc/(m0c²))² which, upon division by r0², becomes γ²q²f²(zr0)[z² + n²(hc/r0(m0c²))²] = n4(hc/(r0m0c²))²

Now let (hc/(r0m0c²)=μ.

The quantization condition then takes the form

#### γ²q²f²(zr0)[z² + n²μ²] = n4μ²

The function f(r) could be a constant or it could be e−λr, in which case r0=1/λ. In that case f(zr0)=e−z. Let f(zr0) be denoted as φ(z). Then the quantization condition takes the form

###### γ²q²φ²(z)[z² + n²μ²] = n4μ²

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 m0c² = 1.503353×10-10 kg m²/s² (joules). Then for r0=(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 {rn, n∈N}. Then the set of allowable β's B={βn, n∈N} is given by βn=f(rn)/(nhc). These values depend upon the dimensionless parameter μ which in turn depends upon the rest mass m0.

## Quantization of the rest mass m0

The orbital angular momentum condition implies that

#### m0 = nh(1−β²)½/(βczr0) which reduces to m0 = (h/(cr0))n(1−β²)½/(βz) or, in terms of the rest mass energy m0c² = (hc/r0)n(1−β²)½/(βz)

The value of (hc/r0) 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 m0 through the parameter μ there has to be a simultaneous determination of m0, 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/(r0m0c²) the previous equation can be expressed as

#### μ = βz/(n(1-β²)½) and thus μ² = β²z²/(n²(1-β²))

The general condition for β can be expressed as

Therefore

#### μ² = γ²q²z²φ²(z)/(n4(1−γ²q²φ²(z)/n²))

The quantization condition found previously was

#### γ²q²φ²(z)[z² + n²μ²] = n4μ²

The above two equations for given values of n and γ then might determine the quantized values of zn (and thus rn) and μn (and thus m0n). Once values of rn and m0n are found then the values of βn and mnc² could be found.

The kinetic energies κn are given by

#### κn = m0nc²[1/(1−βn²]½−1]

Thus kinetic energies are known only if the rest masses are known.

The potential energies can be expressed as

#### Vn = ∫rn∞(kqf(r)/r²)dr

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., sn=2rn.

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.

## Computational Results

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 m0, and that there are two equations to be satisfied; i.e.,

#### m0c²β²/(1-β²)½ = kqf(r)/r and m0cβ/(1-β²)½ = nh/r

Because of the structure of these two equations it was simple to eliminate m0 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 m0 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.)