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The Internal Dynamics
of Two Particle Systems

In both the astronomic and the atomic realm the analysis of the dynamics of the interaction of two particles is dominated by the case in which one particle is so massive compared with the other that effectively the massive particle remains fixed the lighter particle revolves around it. This material examines the general case in which the particles can be of arbitrary relative masses. It is also general in the sense that the central force field the particles generate and which holds them together can be of arbitrary functional form. The purpose of the analysis is to examine the implications of quantization in nuclei.

For electrons in atoms the quantization of angular momentum results in the quantization of orbit radius, orbit velocity, kinetic energy and potential energy. The quantization is such that when an electron moves from a higher level to a lower level the kinetic energy and the potential energy undergo a quantum change, but the decrease in potential energy is greater than the increase in kinetic energy and the difference in these energies goes into the creation of a photon of radiation. It happens in the case of electrons that the decrease in potential energy goes in equal portions into the increase in kinetic energy and the energy of the emitted photon.

On a nuclear level something analogous but more complex happens. When a proton and a neutron come together to form a deuteron there is a gamma ray emitted. There is also a decrease in mass. The mass of the deuteron is less than the sum of the masses of its component proton and neutron. This is an unexplained phenomenon. One purpose of the analysis here is to investigate whether the mass deficit is a result of the quantization of the energy characteristics of the particles in system.

The mass of the neutron is not directly measurable; instead it is estimated from the measured masses of the deuteron and the proton and the energy of the gamma ray involved in the formation or disassociation of a deuteron. The traditional analysis derives the mass of a neutron by presuming that the energy of the gamma ray is exactly equal to the mass deficit of the deuteron. If the conversion of mass goes partially into the energy of the gamma ray and the rest goes into an increase in the kinetic energies of the proton and neutron then the accepted estimate of the mass of the neutron in error by being less than the actual mass. The accepted estimate of the mass of the neutron has been used to compute the mass deficits (usually referred to as the binding energies) of about three thousand nuclides. The binding energies are positive for all but one nuclide, the berylium isotope with one neutron (and four protons), Be5. The binding energy of BE5 is approximately a negative 0.75 million electron volts (MeV). This could be the result of an underestimate of the mass of the neutron.

The Analysis of a Two Particle System Subject to a Central Force

Let m1 and m2 be the masses of two particles. Let r1 and r2 be their distances from their center of mass. These quantitites have to be such that

m1r1 = m2r2
and hence
r2/r1 = m1/m2

Let the attractive force F that holds the particles in the system be of the form

F = −kq1q2f(s)/s²

where s is the separation distance between the particles. In this case s=r1+r2. The quantities q1 and q2 are the charges of the particles for the force field. They would be the masses of the particles if the force is gravitational. If the force is electrostatic they are the electrostatic (coulombic) charges. If the force is nuclear they would be the number of nucleons. The constant k is the constant for the force field. The functional form f(s)/s² indicates that the force is carried by particles which spread out over a sphere of area s². If the force is carried by non-decaying particles the function f(s) is equal to unity. This is the case for the photons of the electrostatic field and the gravitons of the gravitational field. The pi mesons carrying the nuclear force decay and hence f(s)=exp(−s/s0), where s0 is a constant.

To simplify the analysis let kq1q2 be denoted as H.

Classical Analysis

For circular orbits the centrifugal force must be balanced by the attractive force. Thus

m1r1ω² = Hf(s)/s²
m2r2ω² = Hf(s)/s²

where ω is the rate of rotation of the particle system and s=r1+r2.

The kinetic energy K of the particles is:

K = ½m1r1²ω² + ½m2r2²ω²
which reduces to
K = ½(m1r1²+m2r2²)ω²

The orbit radii can be expressed in terms of the separation distance s; i.e.,

s = r1+r2 = (1+r2/r1)r1
r2/r1 = m1/m2
r1 = s/(1+m1/m2) = s[m2/(m1+m2)]
and likewise
r2 = s/(1+m2/m1) = s[m1/(m1+m2)]

The expressions on the right in the last two equations are the mass shares; i.e., the shares of total mass in particle 1 and in particle 2. For convenience let these relationships between orbit radii and separation distance be expressed as

r1 = a1s
r2 = a2s
a1 + a 2 = 1

Therefore the kinetic energy K can be expressed as

K = Js²ω²

where J is complex algebraic expression of the particle masses that need not be displayed here.

The potential energy V is

V(s) = −H∫s(f(z)/z²)dz

The total energy of the system E is then

E = Js²ω² + V(s)

In order for s and ω to be determined the angular momentum L of the system must be given. This is given by

L = m1r1²ω + m2r2²ω
which, since r1 and r2 are
proportional to s, reduces to
L = Is²ω

where I is an algebraic expression of the masses.

The equilibrium state of the system is then found by

ω = (L/I)/s²
and hence
E = Js²(L/I)²/s4 + V(s)
which reduces to
E = [J(L/I)²]/s² + V(s)
or, equivalently
E = (J/I²)L²/s² + V(s)

In principle, this equation can be solved for s as a function of E and L. For example, if the force is carried by a non-decaying particle then V(s)=−H/s and the equation to be solved as simply a quadratic equation in (1/s); i.e.,

AL²(1/s)² − H(1/s) - E = 0

where A=J/I². Thus

1/s = [H ± (H² + 4EAL²)½]/(2AL²)

From s the rotation rate ω can be determined as ω=(L/I)²(1/s)² and the orbit radii as r1=s/(1+m1/m2) and r2=s/(1+m2/m1).

For the general case the equilibrium s cannot be determined analytically but the effect of changes in total energy E or angular momentum L on s can be determined. If the total energy equation is differentiated with respect to E the result is

1 = AL²(−2/s³)(∂s/∂/E) + (Hf(s)/s²)(∂s/∂/E)
and hence
(∂s/∂/E) = 1/[−2AL²/s³ + Hf(s)/s²]
which reduces to
(∂s/∂/E) = s²/[−2AL²/s + Hf(s)]

Likewise, differentiation with respect to L gives

0 = 2AL/s² + AL²(−2/s³)(∂s/∂/L) + Hf(s)/s²(∂s/∂/L)
and thus
(∂s/∂/L) = 2AL²/[AL²(2/s³) − Hf(s)/s²]
which reduces to
(∂s/∂/L) = 2AL²s²/[AL²(2/s) − Hf(s)]

The Partitioning of Angular Momentum and Kinetic Energy

The components of angular momentum are m1r1²ω and m2r2²ω. Let these be denoted as L1 and L2, respectively. Therefore the particle shares of angular momentum are

L1/L = m1a1²/[m1a1²+m2a2²]
L2/L = m2a2²/[m1a1²+m2a2²]

The components of kinetic energy are ½m1r1²ω² and ½m2r2²ω². Let these be denoted as K1 and K2, respectively. Note that K1=½L1ω and K2=½L2ω. Therefore the particle shares of kinetic energy are the same as those for anglular momentum.

If m1=m2 then the particle shares of angular momentum and kinetic energy are equal to one half. If m1/m2=2000, as is approximately the case of the proton and electron, then a1=1/2001 and a2=2000/2001 and the particle shares of angular momentum and kinetic energy are 0.0005 for the heavy particle and 0.99995 for the light particle.

The Quantization of Angular Momentum

Suppose that L1 and L2 must be integral multiples of h, Planck's constant divided by 2π. Let the multiples be n1 and n2. Then

L1/L = m1a1²/[m1a1²+m2a2²] = n1/(n1+n2)
L2/L = m2a2²/[m1a1²+m2a2²] = n2/(n1+n2)

The left sides of the above equations are entirely functions of the masses. If the right sides are given then the masses cannot be arbitrary values. The solution(s) can be obtained by first dividing the first equation by the second. This gives

(m1/m2)(a1/a2)² = n1/n2
which reduces to
(m1/m2)(m2/m1)² = n1/n2
and consequently to

m2/m1 = n1/n2

This would mean that particle mass ratios must be rational numbers. This would be the case if all particles had masses which were integral multiples of some unit of mass, say the mass of the electron. However that does not have to be the case for the condition of the mass ratios being rational numbers to hold. Consider some of the ratios.


If m1>m2 and Δm=m1-m2 then the condition for the rationality of the masses can be expressed as

Δm/m1 = (n2-n1)/n2

This means that for the neutron-proton comparison one should look for a rational number expression for 0.00137841918. The ratio of 2 to 1451 comes close; it being 0.001378359752. Thus the ratio of neutron mass to proton mass is then approximately 1453/1451. Of course, one can approximate any real number as close as desired with rational numbers and any terminating or repeating decimal number is automatically a rational number.

For the proton/electron mass ratio a rational number which closely approximates it is 240,536/131. The decimal version of this number is 1836.152671756. This duplicates the empirical value to nine places and deviates from it by slightly less than 4×10-10.

Determination of the Two Particle System Characteristics

There are four unknowns; s, ω, m1 and m2. There appears to be four equations to be satisfied; two for the balances of the nuclear force and the centrifugal force and two for angular momentum quantification. But the force balance equations for the two particles are the same equation; i.e.,

[m1m2/(m1+m2)]sω² = Hf(s)/s²

This equation can be put into the form

ω² = [Hf(s)/s³][1/m1 + 1/m2]

The angular momentum equations are

L1 = m1[m2/(m1 +m2)]²s²ω = n1h
L2 = m2[m2/(m1 +m2)]²s²ω = n2h

The masses and rotation rate can be determined as a function of the separation distance s.

The previous equations can be put into the forms

(1/m1)ω = n1h[1/m1 + 1/m2]²/s²
(1/m2)ω = n2h[1/m1 + 1/m2]²/s²

If 1/mi is defined as bi for i=1,2 and Hf(s)/s³ as g(s) then the equations to be solved are

ω² = g(s)(b1+b2)
b1ω = n1h[b1+b2]²/s²
b2ω = n2h[b1+b2]²/s²

The last two equations imply that

(b1+b2)ω = (n1+n2)h[b1+b2]²/s²
which reduces to
ω = (n1+n2)h[b1+b2]/s²
and hence
ω² = (n1+n2h²[b1+b2]²/s4

When this last equation is combined with

ω² = g(s)(b1+b2)
the result is
(n1+n2h²(b1+b2)/s4 = g(s)
or, equivalently
(b1+b2) = g(s)s4/[(n1+n2h²]

This last equation implies that

ω² = [g(s)s²/((n1+n2)h)]²
and hence
ω = ± g(s)s²/((n1+n2)h)
and note that
ωs is proportional to g(s)s³.

Ignoring the negative sign (which only indicates that two directions of rotation are possible), then

(b1+b2)/ω = s²/((n1+n2)h)

There b1 and b2 may be solved for as

b1 = n1hg(s)s²/[((n1+n2)h)³]
b2 = n2hg(s)s²/[((n1+n2)h)³]

This means that

m1 = (n1+n2h²/(n1g(s)s²)
m2 = (n1+n2h²/(n2g(s)s²)

For the nuclear force g(s)s³=H*exp(−s/s0) and s0=1.522 fm, (where 1 fermi = 10-15 meter). An estimate of H is 1.92570×10-25 kg·m³/sec². The dimensions of g(s)s³ are kg m³/sec² and its value for s=1.0 fm is 1.92570×10-25. The value of h is 1.05457×10-34 kg m²/sec and thus h² is 1.11212×10-68. This means that for s=1 fm and n1=n2=1 the value of m1 would be 4.62×10-43 kg. The mass of a proton is 1.6726×10-27 kg so the values of n1 and n2 would have to be quite large to bring the value of m1 to the order of magnitude of the mass of the proton.

Thus according to the analysis the smaller the separation distance the smaller the mass (and the larger the potential energy). The functional dependence of mass on separation distance is just the reciprocal of g(s)s² which is proportional to exp(s/s0)s. This is a very strong dependence and incompatible with the empirical observation of the mass deficits being relatively small.

The masses m1 and m2 considered above are the rest masses. The masses with the relativistic enhancements are more relevant. The tangential orbital velocities of the particles are ωri and thus ωais. The relativistic adjustment factors for the masses are then

1/(1−β²)½ = 1/(1−(ai/c)²ω²s²)½

Since ωs is just equal to Hf(s)/s² the factors become larger and larger the smaller s becomes. Thus the measured mass of a system would be proportional to exp(s/s0)s/(1−γ/s²)½, where γ is a constant. The factor in the denominator at least moderates the diminution of mass as separation distance decreases.

For s large the expression exp(s/s0)s/(1−γ/s²)½ is increasing rapidly with s. However as s approaches √γ the expression goes to infinity. Thus there is some value of s greater than √γ where the expression reaches a minimum and near that minimum changes in s have near zero effect on the masses of the particles.

(To be continued.)


A model of a two particle system in which the particles revolve around their center of mass and their angular momenta are separately quantized implies that the ratio of masses must be a rational number. In such a system the rest masses of the particles are dependent upon the separation distance between the particles. The rate of rotation increases as the separation distance decreases. The tangential velocities of the particles thus increases as separation distance of the particles decreases. Therefore the relativistic enhancement of mass increases as separation distance decreases. Perhaps over some range of separation distances the decrease in the rest mass and the relativistic enhancement essentially offset one another as separation distance decreases.

(To be continued.)

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