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Using the Properties of the Deuteron
to Estimate the Parameters of the Nuclear Force

The deuteron, the nucleus of heavy hydrogen, is a very interesting, important special case. Its two nucleons, the proton and neutron, are 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 would be of the form


where d is the distance between the nucleons, 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 the deuteron.

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. This means the formula for the force experienced by each nucleon as a function of the radius r of their orbits with respect to their center of mass is

H*e−λ*2r/(2r)² = (H*/4)e−(2λ*)r/r²
= He−λr/r²

where H=H*/4 and λ=2λ*. It is conventient to express λ* as 1/d0; i.e., d0 = 1/λ*. This means that r0 = 1/(2λ*) = ½d0

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 1.54 fermi (1.54×10−15 meters). Thus r0 is about 0.77 fermi.

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. 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


where β is the velocity relative to the speed of light, v/c.

Circular Orbits

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

mv²/r = He−r/r0/r²
and hence
mv² = He−r/r0/r

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

v = [He−r/r0/(mr)]1/2

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

pθ = [Hmre−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

Hmre−r/r0 = l²h²
or, dividing by r0 and solving
(r/r0)e−r/r0 = l²h²/(Hmr0)

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/(Hmr0) multiplied by the square of an integer l. The value of this coefficient can be calculated. The value of h² in SI units is 1.1121×10−68. The value of m is 1.67×10−27 kg. As indicated above the value of r0 is 7.7×10−16 m.

It has been known since 1935 that if a deuteron is hit by a sufficiently high energy photon the deuteron disintegrates into a free proton and a free neutron. The energy level required for this photodisintergration of the deuteron is 2.22457 Mev. This is the binding energy of the deuteron. This energy is 3.5638×10−13 joules. This is reasonably identified as the potential energy of each of the l=1 level nucleons. A photon that knocks one nucleon out of its orbit disassociates the deuteron.

Let r1 be the radius of the nucleons in the deuteron. The value of r1/r0=2 is a significant figure. The accepted diameter of the deuteron is 4.2 fermis. The diameters of the proton and neutron are both about 1 fermi. Deducting 1 fermi, for the combined radii of the proton and neutron, from 4.2 fermis gives 3.2 fermis as the distance between the centers of mass of the two nucleons. This represents orbit radii of 1.6 fermis for both of the two nucleons. This figure divided by r0=0.77 fermi is 2.078.

Before estimating the value of H it is advisable to consider the relativistic adjustment which should be made.

The Relativistic Case

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) = ½mv2 + (3/8)mv4/c2 + ...

Circular Orbits

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

mv²/r = He−r/r0/r²
m = m0/(1-β²)1/2
m0v²/(r(1-β²)1/2)) = He−r/r0/r²

This can be reduced to

β4/(1-β²)) = [He−r/r0/(m0c²r)]²

Multiplying by the denominator on the left produces a quadratic equation in β²; i.e.,

β4 + β2ζ2 − ζ2 = 0

where ζ² = [He−r/r0/(m0c²r)]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

mv = He−r/r0/(vr)


pθ = [He−r/r0/(vr)]r
pθ = He−r/r0/v
pθ = He−r/r0/(cβ)

Since pθ = lh it follows that

β = γe−r/r0/l

where γ=H/(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 = [He−r/r0/(m0c²r0)], which is the same as γ(hc/(m0c²r0).

In principle 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[He−r/r0/(m0c²r)] = He−r/r0/(hcl)
which reduces to
[½[(1 + 4/ζ2)½−1]]½ = m0cr/(hl)

After squaring and rearranging the above equation reduces to

(1 + 4/ζ2)½ = 1 + 2m0²c²r²/(h²l²)
or, after again squaring,
(1 + 4/ζ2) = 1 + 4m0²c²r²/(h²l²) + 4[m0²c²r²/(h²l²)]²
and after elimination of the 1's
and division by 4
1/ζ2 = m0²c²r²/(h²l²) + [m0²c²r²/(h²l²)]²

Since 1/ζ=m0c²r/(He−r/r0) the previous equation is equivalent to

[m0c²r/(He−r/r0)]² =
m0²c²r²/(h²l²) + [m0²c²r²/(h²l²)]²
which upon division by (m0c²r)² gives
[er/r0/H]² = 1/(c²h²l²) + m0²r²/(hl)4)
or, equivalently,
e2r/r0/H² = 1/(chl)² + (m0c²)²r²/(chl)4

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

e2r/r0 = γ²/l² + γ4(m0c²/H)²r²/l4

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

ez = γ²/l² + γ4(m0c²r0/(2H))²z²/l4

To simplify matters still more for computation let (γ/l)² be denoted as ε and (m0c²r0/(2H)) as μ. The equation to be solved for z is then

ez = ε + μ²ε²z²

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 might be physically meaningless.

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 = (mc²r0/(2hcl))² ≥ 1
which further reduces to
l ≤ mc²r0/(2hc)
or when numerical values
are substituted into this expression
l ≤ 1.778

Thus l can only have the value 1 and hence there is only one angular momentum and energy state for the deuteron.

Empirical Estimates of the Model Parameters

There are several relationships involving H and the radius of the nucleon orbits r1; i.e.,

(r1/r0)e−r0/r0 = l²h²/(Hmr0)
r1/r0 = z/2
where z is the solution to
z = ln(ε) + ln(1+εμ²z²)
ε = (H/(hcl
μ = m0c²r0/(2H)

The value of H which is consistent with all relationships must be found by iteration. An iteration scheme which can achieve this is as follows:

For illustration, take the trial value of r1 to be 1.54×10−15 so ρ=2.0. Then H is computed to be 3.24×10−26

The values of the parameters of the transcental equation are:

Thus ε=1.051.
The value of μ is then

The square of μ is then 3.01766 and εμ²=3.3386. The equation to be solved is then

z = 0.05 + ln(1+3.3386z²)

Its solution is 4.09025 which, since z=2r/r0, means that r1=(4.09025/2)×0.77=1.5756 fermis. This is not far different from the value of r=1.54 fermi upon which the value of H was based. Their potential energy is

V(r1) = (H/r0)(∫2(e−s/s²)ds)
V(r1) = [(3.24×10−26/(7.7×10−16)](0.0188)
V(r1) = 7.91065×10−13 joules = 4.9374 Mev

Although this is off by a factor of 2.2 from the empirical value of 2.22457 Mev found for the photodisintergration of deuterons, it is clear the trial value of H is in the ballpark.

If r1/r0=2.5 then H has to be equal to 4.326×10−26. The computed value of V(r1) is then

V(r1) = (H/r0)(∫2.5(e−s/s²)ds)
V(r1) = [4.326×10−26/(7.7×10−16)](0.00792)
V(r1) = 4.45×10−13 joules = 2.778 Mev

If r1/r0=2.7` then H has to be equal to 4.7661×10−26. The computed value of V(r1) is then

V(r1) = (H/r0)(∫2.7(e−s/s²)ds)
V(r1) = [4.7661×10−26/(7.7×10−16)](0.005709)
V(r1) = 3.5338×10−13 joules = 2.206 Mev

If r1/r0=2.69 then H has to be equal to 4.7320×10−26. The computed value of V(r1) is then

V(r1) = (H/r0)(∫2.69(e−s/s²)ds)
V(r1) = [4.732×10−26/(7.7×10−16)](0.005802)
V(r1) = 3.5656×10−13 joules = 2.2257 Mev

If r1/r0=2.6905 then H has to be equal to 2.7333×10−26. The computed value of V(r1) is then

V(r1) = (H/r0)(∫2.691(e−s/s²)ds)
V(r1) = [4.7333×10−26/(7.7×10−16)](0.00579735)
V(r1) = 3.5637×10−12 joules = 2.22454 Mev

For H=4.7333×10−26, γ=1.4961, ε=2.2383 and μ=1.2225. For these values of the parameters the solution of the transcendental equation is z=5.39404 and thus ρ=2.697. This is pretty close to the trial value of ρ=2.6904, differing from it by only about 0.25 of 1%. The value of r1 from this computed value of ρ is thus (2.697)(0.77)=2.0767 fermis as compared with the value of (2.6904)(0.77)=2.072 fermis from the trial value. The tangential velocity of the nucleons may be obtained from the expression for β; i.e., β = γe−r/r0/l. Since γ/l=1.4961 and ρ=2.69, β=(1.4961)(0.0679)=0.102. The results indicate that for the deuteron that the velocity of the nucleons in the lowest energy state is 10 percent of the speed of light. Their individual kinetic energies as a ratio of their rest mass energy are given by [(1−β²)½−1]. This means that in the deuteron the kinetic energy of the nucleons about 0.5 of 1 percent of their rest mass energy.

Since H=4.7333×10−26 then H* is four times that value, 1.8933×10−25. The constant for the electrostatic force is 2.31×10−27. This means that the nuclear force is greater than the electrostatic force up to the point where eρ=1.8933/0.23102 or ρ= ln(8.1954)=2.10. This corresponds to a distance 2.1d0 or 3.24 fermi. The ratio H*/(hc), which corresponds to the fine structure constant, is equal to 5.98. Since this value is greater than unity the analysis of the nuclear force cannot be achieved by a series approximation in the way that the electrostatic force can be where the fine structure constant is 1/137.036. This does not mean that the nuclear force cannot be analyzed; it just means that it has to be done by a different scheme than for the electrostatic force.


The results imply that the formula for nuclear force expressed with respect to particle separation d is:

F = −H*e−d/d0/d²
H* = 1.8933×10−25 kg·m³/s²
d0 = 1.54×10−15 m.

The radius of the single feasible orbit for the nucleons is (2.6904)(0.77)=2.072 fermis. The separation distance d of the nucleons is then 4.143 fermi. The potential energy for a nucleon in the deuteron is then 2.225 Mev. The tangential velocity of the component particles of the deuteron relative to the speed of light is 0.28. The angular velocity is then 2.04×1023 radians per second. This corresponds to a frequency of 3.245×1022 per second.

The value of ε is 2.2383 and μ equals 1.2225. For these values the solution to the transcendental equation is 5.39404 which is 2r1/r0 so r1/r0=2.697, very close to the value derived from the photodisintegration energy of the deuteron. The two values are within 0.25 of 1% of each other.

Thus the model of the nuclear force presented here is logically consistent and empirically compatible with the data for the deuteron.

The above formula is for the force between two single nucleons. The general formula for the force between one collection of n1 nucleons and another of n2 nucleons is;

F = −H*(e−d/d0)n1n2/d²
H* = 1.8933×10−25 kg·m³/s²
d0 = 1.54×10−15 m.

(To be continued.)

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