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Stable Two-Body Systems
with Zero Angular Momentum

In the quantum mechanics of atoms one quantum number represents the angular momentum of the system. Zero is an allowable value and this is interpreted as an enigma from the viewpoint of classical mechanics. However there is a puzzle only if the subatomic particles are taken to be solid units enclosing their centers of mass and charge. However if either of the particles has an empty center the particles can pass through each other without a collision. In the case of the hydrogen atom, suppose the electron is a point particle and the proton is a triadic structure of quarks such that the center is empty. Then the electron can pass back and forth through the proton without collision and the system would have zero angular momentum.

There are also macroscopic examples of such systems. When two vortex rings (smoke rings) approach each other one enlarges in radius and the other diminishes in radius so that it can pass through the other one. In mathematics it was discovered the solitons (solutions to particular nonlinear partial differential equations) can merge and then the originals re-emerge unaffected by the collision. However below only classical models of two particle systems such as deuterons will be considered.

Such zero momentum systems are one dimensional; i.e., have only one degree of freedom. As such they are relatively easy to analyze mathematically. Let f(x) is the attractive force between the particles where x is the separation distance. The dynamics of the particle is given by the equation

m(d²x/dt²) = f(x)

A conservative force is given by the negative of the gradient of a potential function V(x); i.e.,

f(x) = −(dV(x)/dx)
and thus integrating down
from +∞ to s>0 gives
s(−f(x))dx = V(+∞) − V(s)

The asymptotic value of the potential function may be taken to be zero and hence

s(−f(x))dx = − V(s)
or, equivalently
V(s) = −∫s(−f(x))dx
or V(s) = ∫sf(x)dx

Now consider two particles with masses m1 and m2. Let the center of mass be at the origin of the coordinate system and the particles located at distances r1 and r2 from that center. This means that

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

Thus

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

The dynamics equations for the two particles with respect to the conservative force are

m1(d²r1 /dt²) = f(s)
and
m2(d²r2 /dt²) = f(s)
which, when the expressions for r1 and r2
in terms of s are substituted into the equations,
(m1m2/(m1+m2))(d²s/dt²) = f(s)
(m2m1/(m1+m2))(d²s/dt²) = f(s)

These last two equations are the same.

The expression (m1m2/(m1+m2)) is just the reduced mass μ since

1/μ = 1/m1 + 1/m2
and thus
μ = 1/[1/m1 + 1/m2]
which upon multiplication of
the numerator and denominator
by m1m2 gives
μ = m1m2/(m1+m2)

Thus the dynamics of the system is given by the single equation

μ(d²s/dt²) = f(s) = −dV/ds

When both sides of this equation are multiplied by (ds/dt) the result is

μ(ds/dt)(d²s/dt²) = −(dV/ds)(ds/dt)

Note now that

dV(s)/dt = (dV(s)/ds)(ds/dt)

This means that the previous equation can be represented as

μ(ds/dt)(d²/dt²) = − dV/dt
or, letting (ds/dt) be denoted as v
μv(dv/dt) = −(dV/dt)
or, equivalently
μ(d(½v²)/dt) = −(dV/dt)

When this equation is integrated (indefinitely) with respect to t the result is

½μv² = E − V(s(t))

where E is a constant.

Rewritten, the above equation is the familiar conservation of energy equation

½μv² + V(s(t)) = E

In words, the kinetic energy, ½μv², plus the potential energy V is equal to a constant at all times. However there might be a problem in that the kinetic energy of the two particles might perhaps not be equal to ½μ(ds/dt)². This must be determined.

The kinetic energy of the two particles is given by

½m1(dr1/dt)² + ½m2(dr2/dt)²
which expressed in terms of (ds/dt) is
½[(m1m2² + m1²m2)/(m1+m2)²](ds/dt)²
which may be factored to
½[(m1m2)(m1+m2)/(m1+m2)²](ds/dt)²
which reduces to
½[(m1m2)/(m1+m2)](ds/dt)²
which is none other than
½μ(ds/dt)²

The equation ½μv² + V(s(t)) = E can be solved for v=(ds/dt) to give

v = ds/dt = [(2/μ)(E−V(s))]½
and hence
ds/[1−V(s)/E)]½ = (2E/μ)½dt

The quadrature (integration) of this last equation gives the trajectory of s as a function of time t. Once s(t) is found then the radial momentum μ(ds/dt) can be determined for application of the Wilson-Sommerfeld quantization condition.

The Wilson-Sommerfeld Quantization Condition

According to the Wilson-Sommerfeld quantization condition

(psds) = nsh

where the integration is over a complete cycle, ns is an integer, h is Planck's constant and ps is the momentum associated with s, μ(ds/dt).

Since for a system in which energy is conserved

½μ(ds/dt)² + V(s) = E
and hence
(ds/dt) = [(2/μ)(E−V(s))]½
and thus the momentum is
ps = μ(ds/dt) = [(2μ(E−V(s))]½

The quantization condition is then

[(2μ(E−V(s))]½ds = nsh

This condition can be put into a nondimensional form. First both sides of the equation are divided by h and then h is taken into the square root expression to give

[(2μE/h² − 2μV(s)/h²)]½ds = ns

Note that both E and V are negative with V≤E. Let V(s) be expressed as −CW(s) where C is force formula constant with dimensions [ML³/T²] and W(s) is a positive function with dimension [1/L].

Thus

[(2μE/h² + (2μC/h²W(s))]½ds = ns

Planck's constant, h, has dimensions [ML²/T] so h²/(2μC) has dimension [L]. It is a scale length. For the hydrogen atom and using h instead of h it is the Bohr radius of the ground state electron orbit. Let h²/(2μC) be denoted as σ. Since a force constant such as k has the same dimensions as hc or hc, C may be expressed as ζhc, where ζ, a pure number, is called a structure constant. Thus

σ = h²/(2μC) = h²/(2μζhc) = h/(2μζc) = hc/(2ζμc²)

where μc² is related to the rest mass energy of the particles in the system.

The quantization now has the form

[(2μE/h² + W(s)/σ)]½ds = ns
which can be more
conveniently expressed as
[((2μE/h²)(σ²/σ²) + σW(s))/σ²]½ds = ns
which, upon factoring out
a (1/σ) term from the
square root term,
becomes
[((2μE/h²)σ² + σW(s))]½(ds/σ) = ns

The term

(2μE/h²)σ²
reduces to
2Eh²/(μC²)
and further to
(2E/C)σ
and still further to
(2E/(ζhc))(hc/(2μc²) = (E/(μc²)/ζ

This is a negative dimensionless constant. Let it be denoted as −ψ and let z denote s/σ. Further let σW(s) be denoted as the dimensionless function U(z). Thus the quantization condition is

[(U(z) − ψ]½dz = ns

Let S be the maximum separation. The cycle may be considered to consist of four phases: From +S to 0, from 0 to −S, from −S to 0 and from 0 to +S. These phases are essentially identical so the quantization condition may be reduced to

4∫0S([(2μ(E−V(s))]½ds) = nsh
or, in terms of the
previous formulation,
4∫0S/σ([U(z)−ψ]½dz) = ns

The maximum separation distance S is defined by the condition that V(S)=E; i.e., the velocity is zero at the maximum value of s and hence the kinetic energy is also equal to zero. Empirically its value should be equal or proportional to the measured diameter of the particle system. Thus the above condition would establish the quantization of ψ and thereby the quantization of the energy E.

The Average Values of the Quantities Which Vary

Although total energy may be quantized and constant the kinetic and potential energies vary throughout the cycle. The time the system spends at separation distance s is inversely proportional to |(ds/dt)|. By integrating 1/(ds/dt) over a cycle one can construct a probability distribution function q(s); i.e.,

q(s) = (1/|ds/dt|/τ
where
τ = |1/ds/dt|ds

The quantity τ appears to be simply the cycle time. From this probability distribution it is possible to compute the average values of the kinetic energy K; i.e.,

K = q(s)K(s)ds

Since K(s)=½μ(ds/dt)² and q(s) is proportional to 1/|ds/dt| the evaluation of K reduces to evaluating the integral |μ(ds/dt)|ds which by the Wilson-Sommerfeld condition is quantized. Since K+V=E this means the average potential energy V is also quantized.

Specifically |μ(ds/dt)|ds is quantized as nh so

K = ½nhτ

Along the same line of analysis the average radial momentum pr is given by

ps = ps(s)g(s)ds = μ(ds/dt>[|ds/dt|/τ]ds
which reduces to
ps = (μ/τ)ds = (μ/τ)L

where L is the path length of the trajectory. Since L is equal to 4S the pertinent relationship is

ps = 4μS/τ

Since by the Wilson-Sommerfeld condition img src="circint.gif">ps(s)ds is quantized as nh there is a relationship of the nature

ps(4S) = nh
which implies that
μ/τ = nh
and hence the
quantization condition
1/τ = n(h/μ)

The Electrostatic Force Function

The formula for the electrostatic force is k/s² where k is a constant incorporating unit charges. The constants of force functions have the dimensions of Planck's constant h times the speed of light c, or hc, in which h is Planck's constant divided by 2π. Thus k may be expressed as ζhc where ζ is the fine structure constant 1/137.036. For the electrostatic force

V(s) = −ζhc/s

The nondimensional version of the quantization condition is

4∫0S/σ[1/z −ψ]½dz = ns

The integral in the above equation has an analytic evaluation, but it is instructive to compute the value numerically in preparation for the evaluating the corresponding integral for the nuclear force which does not have an analytic solution.

The problem in the numeric evaluation is how to handle the singularity at z=0.

The first task is to evaluate the parameter σ. Its value is 1.0430952l×10-9 m, or about one million fermi.

The value of the maximum separation is such that

|E| = k/S
and thus
S = k/|E|
and hence
S/σ = (k/σ)/|E|
which reduces to
S/σ = (2ζμc²/hc)/|E| = 2ζ/ψ

Now the quantization condition takes the form

4∫02ζ/ψ[1/z −ψ]½dz = ns

The left-hand side of the above equation is to be evaluated as a function of ψ with ζ being 1/137.036.

If φ is defined to be (2ζ)/ψ and w to be (2ζ)z then the quantization condition takes the form

(4/(2ζ)½)∫0φ[1/w −1/φ]½dw = ns

The value of the coefficient of the integral is 33.11.

The integral is to be evaluated for a number of different values of φ. For a specific value of φ the range from 0 to φ is divided up into intervals of width Δz=φ/N. Because of the singularity at z=0 a value of δz will be used for the first point. Ultimately the behavior of the estimate must be investigated both for N→∞ and δz→0.

Here is the table of values involved in estimating the integral ∫0φ(1/w−1/φ)½dz for φ=0.1 and δz=0.0001.

w 1/w 1/w-1/φ (1/w-1/φ)½
0.00001 100000 99990 316.21195
0.005 200 190 13.78405
0.01 100 90 9.486832981
0.015 66.66667 56.66667 7.52773
0.02 50 40 6.32456
0.025 40 30 5.47723
0.03 33.33333 23.33333 4.83046
0.035 28.57143 18.57143 4.30946
0.04 25 15 3.87298
0.045 22.22222 12.22222 3.49603
0.05 20 10 3.16228
0.055 18.18182 8.18182 2.86039
0.06 16.66667 6.66667 2.58199
0.065 15.38462 5.38462 2.32048
0.07 14.28572 4.28572 2.07020
0.075 13.33333 3.33333 1.82574
0.08 12.5 2.5 1.58114
0.085 11.76471 1.764706 1.32842
0.09 11.11111 1.11111 1.05409
0.095 10.52632 0.52632 0.72548
0.1 10 0 0

The trapezoidal rule for the estimation of J(φ)=∫abf(z)dz leads to the formula 0.5f(a)Δz+Σf(zi)Δz+0.5f(b)Δz. Because of the singularity at z=a this formula has to be modified to 0.5f(a+δz)(Δz-δz)+(Δz-δz)f(a+2Δz)+Σf(zi)Δz+0.5f(b)Δz .

For the data in the above table this leads to

J(φ) = 0.5(99.94998749)(0.00499)
+0.5(13.78404875)(0.00999)
+64.83547042(0.005)
+0.5(0.72547625)(0.005)
= 14.28599702

When this value is multiplied by the coefficient 33.11 the result is 473.0128. Because this is not exactly an integer it means that φ=0.1 is not a quantum level for φ, but it is close to a quantum level.

The relationship between J and φ is displayed below in terms of the common logarithms of the two quantities.

The slope of the relationship is 0.8 and thus J(φ) is approximately proportional to φ4/5.

(To be continued.)

A Nuclear Force Function

Now consider a force of the form

f(x) = −H*e−λ*x/x²

(To be continued.)


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