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An Analysis of a Two-particle System
Using Methods Derived from
the Old Quantum Physics
of Niels Bohr

The components of a two-particle system of the nucleus would be held together by the nuclear force alone. This nuclear force is carried by the π meson. As with gravitation or the Coulomb force there is an inverse-square-of-the-distance dependence because force-carrying particles are spread over a spherical surface whose area is proportional to the square of the distance from the source.. However there is an essential difference for the nuclear force as compared to the electrostatic Coulomb force in that the particles carrying the electrostatic force, photons, do not decay whereas the π mesons of the nuclear force, a.k.a the strong force, do decay quite rapidly with distance. The proportion of the mesons which survive is an exponential function of distance. Therefore the formula for the nuclear force between a subsystem of n1 nucleons and a subsystem of n2 would be of the form


where d is the distance between the nucleon subsystems. H* and λ* are constants with dimensions of kg·m³/s² and m−1, respectively. The material which follows is an attempt to work out the implications of this force formula for the quantum level phenomena of such a two-subsystem subsystem. The subsystems hereafter will be referred to as particles and their locations are just their centers of mass. The analysis applies only for the case in which the separation of the components of the the subsystems is insignificant compared to the separation between the subsystems. This analysis is a necessary preliminary to the analysis for subsystems with significant internal separation.

At this point it is convenient to switch notation. Let r stand for the radius of the orbit of a particle with respect to the center of mass of the particle pair. For the electron in an proton-electron pair r is essentially equal to the distance d between the particles. For the proton-neutron pair of a deuteron r is essentially equal to one half of the distance between the particles.

To keep things simple it will be presumed that all nucleons have the same mass, mp. The proportional difference between the masses of the proton and neutron is only 0.1 of 1%. The relationship between the particle orbit radii, r1 and r2, and the distance of their separation is derived as follows.

Let the center of mass be at the origin with particle one at −r1 and particle two at +r2. Since the masses of the particles are mpr1 and mpr2 it must hold that

−mpn1r1 + mpn2r2 = 0
and thus
n1r1 = n2r2
r1/r2 = n2/n1
and hence
r1 = (n2/n1)r2
r2 = (n1/n2)r1

Since d=r1+r2 this means that

d = (1 + n1/n2)r1
d = (1 + n2/n1)r2
and consequently
r1 = (n2/(n1+n2))d
r2 = (n1/(n1+n2))d

This means the formula for the force experienced by Particle 1 as a function of the radius r1 of its orbit with respect to the center of mass of the system is

H*e−λ*(1 + n1/n2)r1/((1 + n1/n2)r1
= (H*/(1 + n1/n2)²)e−(((1 + n1/n2))λ*)r1/r1²
= He−λr1/r1²

where H=H*/(1 + n1/n2)² and λ=(1 + n1/n2)λ*. It is conventient to express λ* as 1/d0; i.e., d0 = 1/λ*. This means that

r0 = d0/(1 + n1/n2) = [n2/(n1+n2)]d

Hideki Yukawa established that the mass of the force-carrying particles is related to the λ parameter. From the mass of the π mesons it is known that d0 is about 1.5 fermi (1.5×10−15 meters).

Potential Functions

In mechanics the force is conveniently represented as the negative of the derivative of a potential function V(r). This means that the potential function is given by the integration of the force function with with respect to distance. The potential at an infinite distance is taken to be zero.

For the Coulomb force of −α/r² this gives a potential function of V(r)=−α/r. Yukawa hypothesized a potential function which is just this potential function multiplied by an exponential factor; i.e., V(r)=(−α/r)e−λr. However it is the force function that is multiplied by the exponential factor due to the decay of the force-carrying mesons; i.e.,

F = −H*e−λd/d²

where d is the distance between the two nucleons and H* and λ* are constants characteristic of the nuclear force. Therefore the potential function for the nuclear force is really of the form

V(d) = −∫d(H*e−λ*s/s²)ds

Quantum Mechanics

The standard procedure in quantum mechanics is to solve Schroedinger's equation for the particular potential function which applies. This works beautifully for the Coulombic potential and about a half dozen other special cases but gives no analytic results for the cases that deviate from those special cases. The Schroedinger equation cannot be solved even for the Yukawa potential function. There is no way to solve the Schroedinger equation analytically for even more complicated potential functions. Physicists then rely upon purely numerical methods of solution based upon perturbation theory.

What does work analytically is the analysis of Niels Bohr's Old Quantum Theory. Using Bohr's analysis it can be established that the angular momentum is quantized for any potential function in exactly the same way that it is for the Coulombic force. This means that angular momentum must be equal to an integral multiple of h, Planck's constant divided by 2π. This is also true whether the kinetic energy is of the Newtonian form ½mv², where m is the mass of the particle and v is its velocity or the relativistic form m0c²[1/(1−β²)½ − 1 ] where β is the velocity relative to the speed of light, v/c and m0 is the rest mass of the particle.

Circular Orbits

In a circular orbit the attractive force balances the centrifugal force; i.e.,

mv²/r = Hn1n2e−r/r0/r²
and hence
mv² = Hn1n2e−r/r0/r
and, since m=n1mp,
mpv² = Hn2e−r/r0/r

This means that orbital velocity v is a function of orbit radius r; i.e.,

v = [Hn2e−r/r0/(mpr)]1/2

Angular momentum is defined as pθ=mvr, which on the basis of the relation between v and r means

pθ = n1[Hn2mpre−r/r0]1/2

But previous analysis found angular momentum to be quantized. That is to say,

pθ = lh

where l is an integer and h is Planck's constant divided by 2π.

This means that

n1²Hn2mpre−r/r0 = l²h²
or, dividing by r0 and solving
(r/r0)e−r/r0 = l²h²/[n1²(Hn2mpr0)]

Now let r/r0 be denoted as z.

The function ze−z rises from 0 at z=0 to a maximum of e−1 at z=1 and falls asymptotically to 0 as z goes to +∞, as shown below.

For sufficiently small values of angular momentum there will be two values of r but there is a maximum angular momentum for which there are any solutions for r.

The levels shown in green above correspond to the quantity h²/[n1²(Hn2mpr1)] multiplied by the square of an integer l. The value of this coefficient can be calculated. The value of h² in SI units is 1.11×10−68. The value of mp is 1.67×10−27 kg.

The potential energy function is

V(r) = −∫r(He−s/r0/s²)ds

The first step in its evaluation is to express the integral in terms of the dimensionless variable q=s/r0. This substitution gives

V(r) = −(H/r0)∫p(e−q/q²)dq

where p=r/r0.

The integral can be, by integration by parts, put into the form

V(r) = (H/r0)[e-p/p −∫p(e−q/q)dq]

The integral in the above expression is the E1 function which is closely related to the exponential integral function, Ei(z). It has been evaluated and is available in standard tables. In particular E1(1) = 0.2193839344.

Relativistic Considerations

Under the Special Theory of Relativity the kinetic energy is given by the expression

K(β) = m0c²[(1−β²)−½−1].

where β=v/c and m0 is the rest mass of the particle.

The first two terms of this kinetic energy function are

K(v) = ½m0v2 + (3/8)m0v4/c2 + ...

Circular Orbits

The balance of the attractive nuclear strong force with the centrifugal force under Special Relativity is

mpv²/(r(1-β²)1/2)) = Hn2e−r/r0/r²

This can be reduced to

β4/(1-β²)) = [Hn2e−r/r0/(mpc²r)]²

Multiplying by the denominator of the left-hand side produces a quadratic equation in β²; i.e.,

β4 + β2ζ2 − ζ2 = 0

where ζ² = [Hn2e−r/r0/(mpc²r)]². The solutions are

β² = ½[−ζ² ± (ζ4 + 4ζ2)½]


The negative solution must be discarded. Thus

β² = ½[(ζ4 + 4ζ2)½−ζ²]
or, equivalently
β² = ½[(ζ2(1 + 4/ζ2)½−ζ²]

For large values of ζ the solution is approximately β=1.


It is shown elsewhere that even in the relativistic case angular momentum pθ=mvr is quantized in increments of h so

pθ = lh

where l is an integer.

For a circular orbit

mpv = Hn2e−r/r0/(vr)


pθ = n1[Hn2e−r/r0/(vr)]r
pθ = n1Hn2e−r/r0/v
pθ = n1Hn2e−r/r0/(cβ)

Since pθ = lh it follows that

β = γe−r/r0/l

where γ=Hn1n2/(hc). This is not an entirely satisfactory expression of quantization because r on the RHS is quantized in some as yet undetermined manner, yet the similarity with the non-relativistic case makes it of interest.

From the analysis of the previous section it is known that

β² = ½[(ζ2(1 + 4/ζ2)½−ζ²]
or, equivalently
β² = ½[(1 + 4/ζ2)½−1]ζ2
and hence
β = [½[(1 + 4/ζ2)½−1]]½ζ

where ζ = [Hn2e−r/r0/(mpc²r)], which is the same as γ(hc/(n1mpc²r).

The two expressions for β can be equated the result solved for r as a function of l. This provides the quantization of r and subsequently that of β and the rest of the characteristics of the system.

The equating of the two expressions for β gives

[½[(1 + 4/ζ2)½−1]]½[Hn2e−r/r0/(mpc²r)] = Hn1n2e−r/r0/(hcl)
which reduces to
[½[(1 + 4/ζ2)½−1]]½ = n1mpcr/(hl)

After squaring and rearranging the above equation reduces to

(1 + 4/ζ2)½ = 1 + 2n1²mp²c²r²/(h²l²)
or, after again squaring,
(1 + 4/ζ2) = 1 + 4n1²mp²c²r²/(h²l²) + 4[n1²mp²c²r²/(h²l²)]²
and after elimination of the 1's
and division by 4
1/ζ2 = n1²mp²c²r²/(h²l²) + [n1²mp²c²r²/(h²l²)]²

Since 1/ζ=mpc²r/(Hn2e−r/r0) the previous equation is equivalent to

[mpc²r/(Hn2e−r/r0)]² =
mp²c²r²/(h²l²) + [mp²c²r²/(h²l²)]²
which upon division by (mpc²r)² gives
[er/r0/Hn2]² = 1/(c²h²l²) + mp²r²/(hl)4)
or, equivalently,
e2r/r0/(Hn2)² = 1/(chl)² + (mpc²)²r²/(chl)4

Previously the expression Hn1n2/(hc) was defined as γ. Utilizing this term the previous equation can be put into the form

e2r/r0 = (γ/n1)²/l² + (γ/n1)4(mpc²/Hn2)²r²/l4

It is mathematically convenient to deal with z=2r/r0 as a variable so the above equation takes the form

ez = (γ/n)²/l² + [(γ/n1)4/l4](mpc²r0/(2Hn2))²z²

To simplify matters still more for computation let (γ/(n1l))² be denoted as ε and (mpc²r0/(2Hn2)) as μ. The equation to be solved for z is then

ez = ε + μ²ε²z²
ε = [Hn2/(hcl)]²
μ = (mpc²r0/(2Hn2))

Although the above equation is transcendental and cannot be solved analytically, approximations to any degree of accuracy can be readily be obtained numerically, provided of course the equation has a solution at all. Since ε and μ² are positive there will always be a negative solution, but a negative solution is physically meaningless.

The Potential Energy

The potential envergy V(r) at r=r1 is given by

V(r1) = [Hn1n2/r0]∫r1/r0e−s(ds/s²)

The Implication of the Analysis
for the Number of States

The equation ez=ε+μ²ε²z² has solutions for some values of ε and μ and not for others. At the dividing point between the range for solutions and the range for no solution there would be a tangency solution. In other words there would be a value of z such that not only are the LHS and RHS equal but also the derivative of the two sides are equal. This means that

ez = ε+μ²ε²z²
ez = 2μ²ε²z

Equating the expressions for ez gives a quadratic equation in z which has a solution that can be put into the form

z = 1 ± (1 − 1/(μ²ε))½

Thus there will be a real valued solution for z if μ²ε≥1. The values of ε and μ reduces this condition to

μ²ε = (mpc²r0/(2hcl))² ≥ 1
which further reduces to
l ≤ mpc²r0/(2hc)

Since r0=d0/(1+n1/n2) the above reduces to the condition

l ≤ [mpc²/(2hc)]d0/(1+n1/n2)

When numerical values are substituted into this expression

l ≤ 3.556/(1+n1/n2)

Thus when n1/n2=1, l ≤ 1.778 so l can only have the value 1 and hence there is only one angular momentum and energy state.

If n1/n2=1/2, l ≤ (2/3)(3.556)=2.3 so l can have the values 1 and 2 and hence there are two angular momentum and energy states.

Subsystems with Significant Internal Separation

Consider a proton-neutron pair interacting with a single neutron. There are three distance quantities:

The distance r is the radius of the orbit of the single neutron about the center of mass of the system. The angle θ is half of the angle spanned by the proton-neutron pair from the viewpoint of the single neutron. The relations which prevail among these distances are

d = s cos(θ)
s = d/cos(θ)
r = (2/3)d
d = (3/2)r

The force on the single neutron is

F = 2H*cos(θ)e−s/s0)/s²
or, in terms of d
F = 2H*cos(θ)e−d/(cos(θ)s0))/(d/cos(θ))²
and finally in terms of r
F = 2H*cos(θ)e−r/((2/3)cos(θ)s0)/((3/2)r/cos(θ))²
which reduces to
F = 2H*(2/3)²cos³(θ)e−r/((2/3)cos(θ)s0)/r²

Thus the force can be presented as

F = He−r/r0/r²
with H = 2(2/3)²cos³(θ)
r0 = (2/3)cos(θ)s0

This means the previous analysis may be applied using these values for H and r0.


(To be continued.)

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