Comment: 431(5): Classical wave equation with m(r)

Comment: 431(5): Classical wave equation with m(r)

This is very interesting! The wave function is the wave in the wave particle interpretation of the particle of mass m, this can by Ni64 or the proton or any particle or combination. LENR is the interaction of two particles or of two waves, or a wave with a particle as in Compton scattering. The wave particle dualism is almost always applied to the electron, and in electron diffraction the wave nature can be observed as you know. However there can be proton diffraction or Ni64 diffraction. Any particle is also a wave. Compton’s initial experiments were met with great derision, because it was thought that a particle cannot be a wave. Human nature never changes. I do not use the word "scepticism" because that is rational doubt. Derision is based on mindless ignorance.

I found out that the wave equation (13) of the note can be solved analytically with a slightly simplified exponential m(r) function. Then psi may be interpreted as an electric field for example, or what do you think? The result is a "wave function" concentrated around r=0 in a spherically symmetric symmetry. Will send over the details later. m(r)=const will probably give sinusoidal waves or Bessel functions.

Horst

Am 15.02.2019 um 11:15 schrieb Myron Evans:

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 beteen 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.

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Graph of Woods-Saxon force

I can confirm that our calculations of the Woods Saxon force are correct. The force is given in Georgia J. Sci.,p. 5, Eq. (37) and also Fig. (2), which is the same as your figure. This is found by googling "Woods Saxon force and nuclear strong force". I would suggest broadening the Woods Saxon force by adjusting R, a sub N and U sub 0 , it is the strongly attractive nuclear force that keeps the nucleus together. By googling "Woods Saxon potential" it is seen that it is used universally in nuclear physics. It is know known that it is due to m space. Your graph is correct , the force is very short ranged. The nucleus is a small entity, so the regions in which the force in your graph is negative are the regions inside the nucleus.

Graph of Woods-Saxon force

The Woods-Saxon potential and force are graphed in the plot. There is a
negative force peak at the nuclear radius of Ni. There is no force in
the inner region. This means that nucleons are drawn to the inner at the
boundary but are force-free when moving within the nucleus. The only
possibility I see to increase the Woods-Saxon force is to minimize the
"surface thickness" a_N. then the force peak becomes sharper.

Horst

431(6): The m Theory of the Nuclear Strong Force and Conditions for LENR

431(6): The m Theory of the Nuclear Strong Force and Conditions for LENR

Many thanks! Agreed with all these ideas, the U. S. and other Governments have taken out many patents on LENR, and have heavily funded it. So have Mitsubishi heavy industry in Japan. I agree with all your ideas and a demonstration of E Cat by Rossi took place in January. I wonder whether Steve Bannister has any news about this demonstration. The E Cat industrial unit is on sale already, so if its a fraud or a scam and so on therefore are going to be a lot of very angry customers. However I don’t think for one minute that it is a fraud or a scam. If so all the U. S. Government patents would be worthless. The video you saw was the replication of LENR by leading experimentalists in 2017.

431(6): The m Theory of the Nuclear Strong Force and Conditions for LENR

This is excellent Myron. Exactly what has been missing from the cold fusion debate. Intuitively it makes good sense and is now supported by a rigorous and meticulously checked m-theory. As you say, this does away with the need for some more obscure and unproven ideas in the standard approach (that could not explain cold fusion anyway).

It may well be that Ni64 is a great candidate for optimised cold fusion but there may also be others – including other metals, Rossi mentions palladium, compounds and mixtures (Horst’s group have suggested nickel hydride that is used in rechargeable batteries etc). It would also be interesting to know if external electric fields, for example, could have any influence on the cold fusion process (Ni64, for example, neat or as a compound, dissolved in a conducting aqueous solution with an external electric field applied). It will be surprising if cold fusion does not take place in other situations based on the ideas that are currently emerging.

This will end up in a vast research field once these new energy sources are successfully used widely in living environments.

Sent from my Samsung Galaxy smartphone.

431(6): The m Theory of the Nuclear Strong Force and Conditions for LENR

431(6): The m Theory of the Nuclear Strong Force and Conditions for LENR

This note derives the differential equation Eq. (5) which can be solved by computer for the m(r) of the nuclear strong force. The correct Woods Saxon force is given in Eq. (1), checked by computer algebra and it is equated with the m force to give Eq. (5), a differential equation for the m(r) of the nuclear strong force. The resonance condition (7) of m theory is shown to correspond to an infinitely thin surface layer of the nucleus, which is made up of a hard outer surface layer of neutrons and an interior of neutrons and protons. At the surface of the nucleus, r = R, the differential equation (12) may be used to find m (r ) at the nuclear surface. The 1954 Woods Saxon potential is a mean field theory based on the shell model of the nucleus. The shell model can be used to give magic numbers which indicate which nuclei are stable. Wigner et al were awarded a Nobel prize for the nuclear shell model, which can be extended to describe spin and spin orbit interaction. The shell model could be developed with m theory. Google "Surface thickness of a nucleus" to find that the surface thickness a sub N in lead is of the order of ten power minus sixteen metres according to the latest experiments. So it makes perfect sense to deduce that the force of attraction between Ni64 and p is maximized when the shell thickness is minimized, reduced to zero by the resonance condition of the m theory. The shell thickness is the "protective armour" of the nucleus, an outer layer of neutrons. Inside is a mixture of neutrons and protons. Ni64 is the isotope of nickel which has the most number of neutrons. So it is overloaded with neutrons but it is stable. Once a proton enters by m space resonance, it becomes unstable, and transmutes to Cu63, megabytes of energy and other particles. The challenge is therefore to solve Eq. (5) by computer. The result would be dm(r) / dr and m(r) for the nuclear strong force.

a431stpapernotes6.pdf

431(5): Development of the Woods Saxon Potential

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 beteen 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

Resonance condition

Resonance condition

Thanks again! The resonance condition is the most important feature of the m theory of LENR, so the comparison with the Woods Saxon (WS) potential can be carried out in various ways, giving expressions for m(r) and dm(r)/dr in terms of the parameters of the WS potential. This goes beyond the models used so far for m(r) and dm(r) / dr. It is also necessary to be careful about the differentiation of the WS potential, Eq. (1) of Note 431(5), to give the WS force Eq. (3), because the parameter a0 must appear in the force in order to keep dimensional units right. If r =0 in the WS potential it becomes a constant so the WS force vanishes. Clearly, this is incompatible with an infinite m force at r = 0. So a more general solution of 2m(r) = rdm(r) / dr is needed, because the solution r = 0 means no WS force. The comparison of Eqs. (3) and (4) can be carried out in various ways with computer algebra, so m(r) and dm(r) / dr can be expressed in tems of R, U sub 0, a sub N and a0. Firstly my hand calculation of Eq. (3) should be run through Maxima to check it. Maxima will give a result without a sub 0 being present, but that is dimensionally incorrect. So a sub 0 has to be devised to keep the dimensions correct.

Resonance condition

The condition 2m(r) –> r*dm(r)/dr is only reached for r–>0 in the
models we have investigated so far. I am not sure if this has an impact
on the comparison with the Woods-Saxon potential.

Horst

Graphics of Eqs. (1) and (2) of Note 431(4)

The answer is given in UFT247 and UFT248, which develop collision and transmutation processes in general, for any number of nuclear reaction products. The total force is calculated from the total energy of p and Ni(64). However Ni(64) is much heavier and generates a much greater force due to m space. In nuclear theory the Coulomb repulsion is treated differently for separate p and Ni(64) and for the combined p and Ni(64) (see latest note for UFT431).

Graphics of Eqs. (1) and (2) of Note 431(4)

Does the total energy E (eq.2) relate to the proton or the Ni nucleus? The nuclear mass m and relativistic momentum p has to be chosen appropriately. It seems that p may be negleced for v<<c.

Horst

Am 13.02.2019 um 11:40 schrieb Myron Evans:

Graphics of Eqs. (1) and (2) of Note 431(4)

It would be very interesting to plot Eqs. (1) and (2) as a function of m(r), dm(r) / dr and p, the linear relativistic momentum of the proton. Quantization of Eqs. (1) and (2) should produce a wealth of new information, also the extension of UFT246 UFT248 with m theory should give important results. I will look in to this extension next.

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