By  L. Kaliambos (Λευτέρης Καλιαμπός) Τ.Ε. Institute of Larissa Greece.( June 2014)


NCSR "Demokritos" (2002)

It is indeed unfortunate that the discovery of the assumed uncharged neutron along with the invalid relativity led to the abandonment of the well-established laws of electromagnetism. Thus the various  nuclear theories and structure models  could not lead to the nuclear structure . The problem became more complicated when Heisenberg assumed that the proton-proton and  neutron-neutron repulsions create fallacious attractive nuclear forces.

 For example in the “Evidence for Alpha Particle Substructures in nuclei-San Jose State University ” one reads:The nuclear force was believed to exist equally between protons and neutrons and between neutrons as well as between protons.  This led Werner Heisenberg to speculate that a proton and a neutron are merely different forms of a single particle which he called the nucleon. This nucleon hypothesis came to be widely accepted even though there is considerable evidence against it. It was just a convenient assumption that made theorizing simpler. If Heisenberg's hypothesis were really true there would exist He2 nuclei, two protons bound together by the overwhelming nuclear force. Such a nuclide does not exist. There would also be bound neutron complexes and the emission of gamma rays from neutron collections when such complexes form.”

Under this physics crisis in my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003) I revived the natural laws which led to  my discovery of extra 9 charged  quarks in proton and 12 ones in neutron responsible for the nuclear structure and binding. Note that such extra charged quarks ate a part of  288 quarks in nucleons.

 Today it is well-known that the structures and binding energies of nuclei are based not on invalid nuclear theories but on the well-established laws of electromagnetism. You can see my DEUTERON STRUCTURE AND BINDING  for  the simplest explanation of the deuteron structure and binding able to tell us how the charges of two spinning nucleons interact electromagnetically with  parallel spin ( S = 1) for giving the nuclear binding and force in the simplest nuclear structure. Also you can see my paper SRUCTURE AND BINDING  OF H3 AND He3 in my FUNDAMENTAL PHYSICS CONCEPTS . To compare the structure of He-4 with  the structures of Be-8, O-16, and Pb-208 see the following figures


structures of He-4, Be-8, O-16, and Pb-208

Also in the following diagrams of He4 and He6 you see that the spinning nucleons in He4 form a simple rectangle with S=0  which explains the stability of alpha particles.



                n2(-1/2)..p2 (- 1/2)




               n2 (-1/2)..p2(- 1/2)..n4(-1/2)                   



Here you see that the simplest p1n1 and p2n2 systems  ( deuterons ) are characterized by the following weak binding energy

 B(p1n1) = B(p2n2)  = -2.2246 MeV

According to electromagnetic laws they are coupled along the radial direction or along the x axis  giving the total parallel spin along the spin axis or along the z axis as

 S = +1/2 +1/2  =  1

which invalidates the so-called Pauli Principle. Then from the structure of the mirror nuclei H3 and He3  it was  possible to find the repulsive energies

U(p1p2) =  0.863 MeV  and U(n1n2 )  = 0.099 MeV

Then  from the strong binding energy of He4

B(He4) = - 28.29 MeV

one concludes that the two deuterons are coupled along the spin axis or along the z axis with very strong binding energies

B(p1n2) = B(n1p2)  =  -12.4 MeV

Note that this very strong nuclear binding energy is derived after a large number of integral equations when z = 0.41 Rp where Rp is the proton radius (Rp  =  0.84 /1015 m). This short separation justifies the oblate spheroid of nucleons.

Under this condition we write

B(He4) = B(p1n1) + B(p2n2) + B(p1n2) + B(n1p2) + U(p1p2) + U(n1n2)

Therefore the above equation in terms of MeV can be written as

-28.29  = -2.2246 -2.2246 -12.4 -12.4 + 0.863 + 0.099



 In the He6 the greater number of bonds and repulsions complicates the calculations of energies. However the shape as shown in the second diagram explains very well the total spin S=0 and the decay of He6 since the repulsive forces of the nn systems contribute to the reduction of the p1n3 bond and the p2n4 bond. Note that in the absence of nn repulsions the binding energy of them should be equal to the binding energy of deuteron.

  For example the repulsions  n3n2 and  n3n1, (the repulsion n3n4 is negligible) contribute to the reduction of the p1n3 bond. In the same way the repulsions n4n1 and n4n2  contribute to the reduction of the p2n4 bond.

Although the distances n1n3 and n2n4 are greater than the distances n2n3 and n1n4  we assume that all the repulsions have the same value because in the systems n3n1 and n4n2 the  neutrons have parallel spin with electric and magnetic repulsions.


  U(nn) = U(n3n2) +U(n3n1)  + U(n4n1) + U(n4n2)  = 4U(n1n2) = 4(0.099) = 0.396 MeV  

Then from the binding energy of He6

B(He6) = -29.27 MeV

one concludes  that the weak bonds of p1n3 and p2n4  have a binding energy weaker than the binding energy of -2.2246 MeV (deuteron). Such weak bonds along with the repulsions of the nn systems give a total energy

B(p1n3) + B((p2n4) + U(nn ) = B(He6)- B(He4) =  (-29.27) – (-28.29) = -0.98 MeV

Because of  symmetry we have B(p1n3) = B(p2n4).  Thus  we may write

2B(p1n3) +U(nn)) = -0.98  or 2B(p1n3) + 0.396  = - 0.98 MeV

That is  B(p1n3) = Bp2n4) =  - 0.344 MeV.  Since this value is weaker than the - 1.29 MeV it leads to the decay. Note that a free neutron or a single np bond with a binding energy B(np) <  1.29 MeV leads to the decay, because the difference in energies between a down quark (3.69 MeV) and an up quark (2.4 MeV) is 1.29 MeV. For example for the decay of a neutron we discovered that the neutron with 288 quarks turns into the proton with 288 quarks as

n = p +e +ν    or    [92(dud) +4u +8d] = [(92(dud) +dud +4 +5d ] +e + ν   οr  d = u +e +ν.