## m Theory of the Nuclear Strong Force

OK many thanks, the computer gives the right result for the Woods Saxon (WS) force between neutrons and protons, and the right units of potential energy over meters. I had inadvertently included a superfluous factor of one. I suggest graphing the WS force for a given R, a sub N and U sub 0. Then graph the m force of eq. (4) for various m(r) and dm(r) / dr. Alternatively graph the WS potential as a function of r for a given U0, R and a sub N, then fit this function with a two variable least means squares fitting program, the two variables being m(r) and dm(r) / dr. These two variable least means squares programs are available in libraries of code. I used the NAG library in the early seventies to explain the far infra red. Maxima may have its own fitting routine.This procedure transforms the m force into the WS force. So we can interpret the m force as a WS force between protons and neutrons, which is the strong nuclear force, QED. Finally adjust r so that 2m(r) = r dm(r) / dr. At this point the strong force becomes infinite and the proton and Ni64 become one unstable entity. This breaks apart into copper63, mega electron volts, and other products. This theory can be used for any mixture. A third method is to eliminate r between the corrected equation (3) and (4) of Note 431(5), and set up a differential equation involving dm(r) / dr and m(r). This will find them in terms of the WS parameters U0, R and a sub N. I will see if I can develop this last method by hand to a certain point and then let the computer take over. I am receiving all e mails – apparently there was a communications problem caused by gmail giving out spurious messages. I have found that g mail often does this. So we have forged a new theory of the nuclear strong force by identifying it with the force of m space. This shows that the nuclear strong force is not a force of classical physics , neither is it a force of special relativity. It needs dm(r) / dr not equal to zero. We can gradually make the quark gluon model obsolete. We have already shown in notes for UFT431 that the elementary particle masses are given by m(r) and by Cartan geometry.

431(5): Development of the Woods Saxon Potential

It seems that the derivative of the Woods-Saxon potential (eq.3) is not correct. There is no need to introduce a constant a0, see attachment. I will work through the note and see if the rest is consistent.

Horst

Am 14.02.2019 um 14:47 schrieb Myron Evans:

431(5): Development of the Woods Saxon Potential

The Woods Saxon potential was developed for LENR in UFT227 ff, and in this note the attractive force of the potential, Eq. (3), is identified with the attractive force of the m space, Eq. (4) for a static proton and Eq. (11) for a moving proton. It is shown that the resonance condition 2m(r) = rdm(r)/dr is equivalent to the approach to zero of the normalized surface thickness of the Woods Saxon model. When the surface thickness disappears there is no barrier to Ni(64) merging with p. So m (r) and dm(r) / dr can be understood in terms of the well known parameters of the nuclear Woods Saxon potential. The total potential inside the nucleus is Eq. (10). Inside the nucleus (fused Ni64 and p) the repulsive potential is Eq. (8). Outside it is Eq.(9) (p approaching Ni64). Finally the proton wave equation is Eq. (18), the ECE wave equation. By wave particle dualism the proton approaching a Ni64 atom is both a wave and a particle. The nickel atom is also a wave as well as a particle and after the fused entity Ni64 and p breaks apart, wave particles of various mass are generated. In the quark gluon model the nucleus consists of quarks, and they can be related to the eigenvalues of the ECE wave equation (18), the wave equation of m space. At this point I will pause to write up Sections 1 and 2 of UFT431.

431(5).pdf