﻿ A Testing of the Conventional Model of Nuclear Structure Which Involves the So-Called Strong Force
San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

A Testing of the Conventional Model
of Nuclear Structure Which Involves
the So-Called Strong Force

## Background

The conventional model of nuclear structure asserts that the neutrons and protons of a nucleus are held together by their attraction due to a strong force. This theory goes back to Werner Heisenberg in the early 1930's. The supposed strong force is conjectured to drop off with distance more rapidly than the electrostatic repulsion between protons. Thus, according to the model, there is an attraction between two protons at small separation distances that becomes a repulsion at larger separation distances. This is all very plausible but, as will be shown below, empirically invalid.

## Nature of a Force and Change in Potential Energy

For a potential field V(s) the force on a particle is given by

#### F(s) = − ∂V/∂s

Thus if ∂V/∂s>0 then the force is an attraction and if ∂V/∂s<0 it is a repulsion.

Likewise

#### ΔV = −∫abF(s)ds for a<b

Therefore, if ΔV is known over some interval one can ascertain the nature of the force over that interval.

## Potential Energy and Binding Energy

Conceptually binding energy is entirely different from potential energy, but increases in binding energy occur where there would be a greater loss in potential energy.

The mass of a nucleus is generally less than the sum of the masses of its constituent nucleons. The difference is called the mass deficit of the nucleus. The mass deficits of 2931 nuclides have been measured. Binding energy is the mass deficit of a nucleus expressed in energy units through the Einstein formula E=mc².

When a nucleus is formed there is a loss of potential energy due to the coming together of its constituent parts.

Conceptually binding energy is entirely different from potential energy, but increases in binding energy occur where there would be a greater loss in potential energy. There is reason to believe that the binding energy (BE) of a nucleus is proportional to the loss of potential energy which occurred when it was formed. See Binding Energy and Potential Energy. Thus in the analysis BE will be taken to be equivalent to the loss of potential energy.

## The Interactions of Nucleons

Let n and p be the numbers of neutrons and protons, respectively, in a nuclide. The number of neutron-neutron interactions is equal to n(n-1)/2. This will be denoted as nn. Likewise the number of proton-proton interactions is p(p-1)/2 and this will be denoted as pp. The number of neutron-proton interactions is np.

The binding energy due to these interactions is a function of the separation distances of the nucleons. Here no distinction is made for separation distances so the results will be for the average separation distance of the nucleon.

The regression equation expressing the attempt to predict the binding energy of a nuclide from the numbers of the interactions of its nucleons is

#### BE = cnnnn + cnpnp + cpppp

There is no constant term because if nn=np=pp=0 the BE must be zero.

## The Conventional Model of Nuclear Structure

The conventional model of nuclear structure is then expressed as

#### cnn = cnp > 0 0 < cpp < cnn

The coefficient for proton-proton interactions would be less than that for neutron-neutron interaction because of the electrostatic repulsion between protons.

## Regression Results

Here are the results of the regression analysis for the 2931 nuclides.

#### BE = −0.69377nn + 0.89685np −0.68818pp [-5.8]          [5.4]        [-2.9]

The figures in the square brackets below a coefficient is its t-ratio, the ratio of the coefficient to its standard deviation. The t-ratios indicate that the coefficients are statistically significantly different from zero.

The assertions of the conventional model of nuclear structure are not born out. Two of the three coefficients are negative. The negative values for cnn and cpp indicate that the force between two like nucleons is a repulsion. The positive value for cnp indicates the force between two unlike nucleons is an attraction.

The value of cpp is not numerically less than that of cnn; it is numerically larger. This cannot be, because the the electrostatic force between two protons is known to be a repulsion.

The coefficient of determination (R²) for the above regression equation is 0.924. But given that almost all of the regression coefficients are wrong in terms of sign or relative magnitude a higher value of R² is evidence against the conventional model rather than for it.

The above regression coefficient values can be explained by allowing for a neutron to have a different nucleonic charge than a proton. But more importantly the conventional model leaves out the effects of the spin pairing of nucleons. A model taking spin pairing into account explains 99.995 percent of the variation in the binding energies of the 2931 nuclides. See What holds a nucleus together.

## Conclusions

The conventional model of nuclear structure is decidedly not empirically valid. Almost all of its predictions concerning the regression coefficients are found to be wrong. The conventional nuclear strong force is an inappropriate conflation of nucleon spin pairing and nucleonic interactions.

Dedicated to Lydia,
my friend of many years
for her many abilities
and for the many wonderful
conversations we shared
over the years.