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The Quantum Mechanics of
Systems of Two Nucleonic Clusters:
The General Case

One of the major enigmas of nuclear physics is the relatively high value of the binding energy of the Helium 4 nuclide, the alpha particle, compared to those of the smaller nuclides such as the Hydrogen 2 nuclide (deuteron) and the Hydrogen 3 (triteron). The binding energy of the deuteron is about 2.225 million electron volts (MeV), that of the triteron 8.48 MeV and the alpha particle an enormous 28.29 MeV. To a degree the differences could be due to the number of nucleon-nucleon interactions (bonds) of the different nuclides. The deuteron has only one bond; the triteron has three and the alpha particle has six. The number of bonds however, is at best only a partial explanation of the differences.

A previous study accounted for the binding energy of the alpha particle to within a few percent by treating the alpha particle as two deuterons spinning about their center of mass. A search for a model that would give a more accurate explanation of the binding energy was not successful so this is a further investigation and generalization of nuclides as systems of clusters.

This approach treats the clusters as particles and analyzes the tractible two-body problem when in reality such a system is an intractible n-body problem. This is based up the adage

It is better to be approximately right
than precisely undecided.

What came out of a previous investigation in which the clusters were identical and each contained q nucleons is that the energy levels of such systems are roughly proportional to the fourth and fifth powers of q. This is the investigation of the more general case in which the number of nucleons in the clusters can be different.

The Model

Consider a system of two clusters in which one cluster has q1 nucleons and a mass of M1 and the other q2 nucleons and a mass of M2. This is a deuteron of clusters. The alpha particle might be a deuteron of deuterons. On the other hand it might not be. It could be a neutron pair and a proton pair revolving about their center of mass. Or, it could be something entirely different. An accurate explanation of the binding energy of the alpha particle would have to take into account the moderation of the strong force attraction by the electrostatic repulsion of its two protons. What is not taken into account here is the electrostatic repulsion of the protons.

The Reduced Mass of the System

The reduced mass μ is such that

1/μ = 1/M1 + 1/M2
which is equivalent to
μ = M1M2/(M1+M2)

The Orbit Radii

Let r1 and r2 be the distances of the centers of the two clusters from the center of mass of the system. Then

M1r1 = M2r2
and thus
r2/r1 = M1/M2

The separation distance s for the centers of the two clusters is then

s = r1 + r2
which is then equal to
s = (1 + M1/M2)r1
and hence
r1 = s/(1 + M1/M2)
and likewise
r2 = s/(1 + M2/M1)

The above expressions for r1 and r2 can be rewritten as

r1 = [M2/(M1+M2)]s
r2 = [M1/(M1+M2)]s

Angular Momentum

Let ω be the rate of rotation of the system. The angular momentum of one cluster is M1ωr1² and that of the other is M2ωr2². The total angular momentum of the system, pθ can then be expressed as

pθ = [M1ωr1]ωr1 + [M2ωr2]ωr2

But both [M1r1] and [M2 r2] are equal to [M1M2/(M1+M2)]s which is equal to μs. Therefore

pθ = μωs(r1+r2) = μωs²

The system angular momenum is quantized so

pθ = μωs² = nh

where n is an integer, called the principal quantum number, and h is Planck's constant divided by 2π. Thus

ω = nh/(μs²)
and, for later use,
ω² = n²h²/(μ²s4)

Dynamic Balance

Without any loss of generality the nuclear force between clusters of size q1 and q2 may be expressed as

F = Hq1q2f(s)/s²

where H is a constant and f(s) is a function such that f(0)=1. Later it will be assumed that f(s)=exp(-s/s0), but for now the form of the force formula will be kept general.

The attractive nuclear force on each cluster must balance the centrifugal force on that cluster. The centrifugal force on the first cluster is equal to

M1ω²r1 = [M1M2/(M1+M2)]ω²s.
which reduces to


μω²s = Hq1q2f(s)/s²
and therefore
ω² = Hq1q2f(s)/(μs³)

Equating the two expressions for ω² gives

Hq1q2f(s)/(μs³) = n²h²/(μ²s4)
which reduces to
sf(s) = n²h²/(μHq1q2)

This is the quantization condition for the separation distance of the centers of the two clusters. Note that if M1=mq1 and M2=mq2 then μ is equal to mq1q2/(q1+q2). This would mean that

sf(s) = n²h²(q1+q2)/(mHq1²q2²)

In effect, this would make sf(s) inversely proportional to the third power of the average cluster size.

One can define a reduced cluster size ν as

1/ν = 1/q1 + 1/q2
which can be expressed as
ν = q1q2/(q1+q2)

This means that for cluster masses proportional to cluster size μ=mν. It also means that the previous equation for sf(s) can be expressed as

sf(s) = n²h²/(mHν²(q1+q2))

Quantization of the Kinetic Energy

A quantization condition for ω may be obtained from

ω² = Hq1q2f(s)/(μs³) = Hq1q2sf(s)/(μs4)
which, by replacing sf(s) with n²h²/(μHq1q2)
can be reduced to
ω² = n²h²/(μs4)

Since s is a function of n, the above equation quantizes ω.

The kinetic energy K of the system is given by

K = ½M1ω²r1² + ½M2ω²r2²
which reduces to
K = ½ω²[M1r1² + M2r2²]
and further to
K = ½ω²μs²

Replacing ω² by n²h²/(μs4) gives

K = ½n²h²/s²

This quantizes kinetic energy but the dependence of K on n and q is obscure. A simple approximation helps reveal that dependence.

At least over some range the function sf(s) can be approximated by γs, where γ is a constant. This follows from sf(s) being zero at s=0. Thus

γs = n²h²/(μHq1q2)
and hence
s = n²h²/(γμHq1q2)
and therefore
s² = n4h4/(γ²μ²H²q1²q2²)

Since K = ½n²h²/s²

K = ½(γ²μ²H²q1²q2²/h²)/n²)

Note again that if M1=mq1 and M2=mq2 then μ is equal to mq1q2/(q1+q2). This would mean that

s² = n4h4(q1+q2)²/(γ²m²H²q14q24)
and hence
K = ½(γ²m²H²q14q24/((q1+q2)²(h²)/n²)

Thus for q1=2 and q2=2, a deuteron of deutrons, would have kinetic energy equal to 2424/42=16 times the kinetic energy of a deuteron for the same principal quantum number.

Likewise if q1=2 and q2=1 then kinetic energy is equal to 24/32=16/9=1.78 times the energy of a deuteron.

The Quantization of Potential Energy

The potential energy of the system is a function only of the separation distance of the centers of the clusters is given by

V(s) = ∫s+∞[−Hq1q2f(p)/p²]dp
which reduces to
V(s) = −Hq1q2s+∞(f(p)/p²)dp

Over some range the integral ∫s+∞(f(p)/p²)dp can be approximated by α/sζ, where ζ≥1. Let q denote the geometric mean as an average cluster size. It was previously noted that if M1=mq1 and M2=mq2 it would mean that sf(s) and hence s would be inversely proportional to the third power of the average cluster size q. Then given the inverse dependence of s on q³ this means that the above integral and hence potential energy is proportional to q. From the above expression for V(s) this means that the potential energy and the hence the binding energy of a cluster deuteron has potential energy is proportional to q3ζ−2.

Some Interesting Rough Computations

For f(s)=1, ζ is equal to 1 and hence V(s) is proportional to q. Thus a deuteron of deuterons when the nuclear strong force is simply an inverse distance squared force would have two times the potential energy of a deuteron and its binding energy would be only twice as large.

For f(s)=exp(-s/s0) the value of ζ depends upon the value of s/s0. For s<s0 the value of ζ is about 2. In that case V(s) would be proportional to q4. Thus the potential energy and hence the binding energy of a deuteron of deuterons would be about 16 times that of a deuteron. With the binding energy of the deuteron of 2.225 MeV this would make the binding energy of a deuteron of deuterons 35 MeV.

Consider a triteron as deuteron and neutron revolving about their center of mass. The geometric mean q of the cluster sizes of 1 and 2 is √2. With the potential energy and binding energy proportions to q4 this would make the binding energy of the triteron equal to (√2)4=4 times that of the deuteron. With binding energy of the deuteron being 2.225 MeV this would make the binding energy of the triteron equal to 8.9 MeV whereas the conventional estimate is 8.5 MeV.

More Detailed Analysis

Here it is assumed that ∫s+∞(f(p)/p²)dp is can be adequately approximated by K/sζ over some appropriate range where K and ζ are constants. From previous analysis when the masses of the clusters are proportional to cluster size

s = n²h²/(γμHq1q2)
μ = mq1q2/(q1+q2)
and thus
s = n²h²(q1+q2)/(γmH(q1²q2²)
which for convenience can
be expressed as
s = φ(q1+q2)/(q1q2

where φ=n²h²/(γmH)


V = −Hq1q2(K/φζ)(q1q2)/(q1+q2)ζ
which reduces to
V = −H(K/φζ)(q1q2)2ζ+1/(q1+q2)ζ


It is not surprising that an alpha particle has a binding energy of 28.29 MeV compared to a binding energy of 2.225 MeV for a deuteron, or that the binding energy of the triteron is 8.48 MeV. It may be just a matter of scale; i.e., cluster size, given the nature of the dependence of the nuclear force on distance.

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