Nuclear Strong Force is a Fiction

[ben m]

the proton and the neutron are both nucleons but they are different, they have different magnetic moments for example, completely neglected by Meissner and co.

You're flat wrong. Meissner (like all nuclear physics) use strong interactions---treating both nucleons the same---for the strong-interaction part of the calculation. Then they add in the electromagnetic interaction. I clicked through a couple of papers and this is stated clearly.

In Table II we show results for the low-lying excited states of 12C at leading order (LO), next-to-leading order (NLO), next-to-leading order with isospin-breaking and electromagnetic corrections (IB & EM), and next-to-next- to-leading order (NNLO).

In addition to isospin-symmetric interactions, we also include isospin-breaking (IB) and electromagnetic (EM) interactions.

Explain, please, why you made this accusation that Prof. Dr. Meissner doesn't include E&M correctly in his nuclear theory? According to his papers, he IS including E&M effects.

 

[ben m]

My approach works as you can see in my paper "Electromagnetic Theory of the Binding Energy of the Hydrogen Isotopes" http://www.springerlink.com/content/...tation/?MUD=MP

By the way, I found (more) errors in this paper.

1) You take a vague drawing of the "neutron" charge separation, and the proton charge, and compute an "electric quadrupole". You forgot to compute, also, the electric dipole moment---which is certainly nonzero, and very large (of order 10^-15 e*m) in your picture. This is important, because no nucleus has ever shown a nonzero electric dipole moment in experiments sensitive to ~10^-29 e*m. Your model can only get to this level if there's absolutely perfect matching between your neutron's internal +/- separation, and the external separation between the neutron's - and the proton's +. If this is to be consistent with data, these two distances would have to be identical to a precision of 0.000000000000001%. (One part in 10^14, in case I miscounted my zeros)

2) You recognize that the neutron has no large permanent electric dipole moment. You suggest, in the text, that the dipole might be induced by a nearby charge. Now, induced dipoles are an idea from real-world physics, but that's not the physics you used. All of your energy calculations are the calculations associated with a non-induced, large, static dipole. You did *not* include any polarization energy, nor indeed do you appear to be aware that this exists. You used a neutral object with a giant static dipole, an object which is obviously not a neutron.

3) You did force-calculations between the proton and the "-" half of the "neutron", which you put at a distance r_{np}. But then you put the neutron magnetic dipole also at r_{np}. Shouldn't the magnetic dipole be associated with both halves of the "neutron", rather than perfectly co-located with the "-" half? Since this dipole is, you say, repelled from the proton, it would be (if anything) pushed more towards "+" side---i.e., much, much farther from the "proton" than you assumed.

4) You extend your model to other isotopes, saying "the magnetic moments of the neutron and the proton are collinear in the deuteron and perpendicular in the other isotopes." You don't bother analyzing why, or how, the magnetic moments might be "perpendicular"---you just plunk the particles down and pretend that they stay where you put them. For example, your picture of tritium is unstable---it's not the ground state. The total energy will be lower if the central magnet is allowed to flip sideways. For another example, your "planar" drawing of 5H is nonsense---even assuming you know the spin directions and the force directions, which you don't, the ground-state configuration of this set of objects is a tetrahedron, not a flat square.

5) You have your E&M equilibria entirely wrong. You chose to guess all of the spin directions, then calculate linear forces, and find an "equilibrium". Unfortunately, you forgot about torques. The state you identify is never the ground state; the energy would be lowered (continuously) by rotating the proton spin, making the magnetic force attractive, removing the hard-core you point out and making the proton "fall into" the negative end of the neutron. Why is this obvious? It's an instance of Earnshaw's Theorem, the general proof that electromagnetic forces in 3D do not have stable static equilibria. Never. Look it up.

 

[fuelair]

Still wrong (a term actual scientists use to mean incorrect/completely not proven/laughable/incompetent..........). YMMV.

 

[bjschaeffer]

No matter how big or bold you make your text - you are the one who has proved that your model is wrong! Read your web page:
Binding energy of the hydrogen isotopes
The first standard yous fail are not presenting your claculations and using a graph to list these binding energies. But this is not a problem becauase the graph shows that you are wrong!



Let me count the ways that you are wrong:

1. Deuterium binding energy measured to be 2,224.52±0.20 keV.
   You have ~1.0 MeV.
2. Tritium binding energy measured to be 8,481.821± 0.004 keV
   You have ~2.8 MeV.
3. 4H has a half-life of (1.39 ± 0.10) × 10−22 seconds and so a negative binding energy.
   You have a positive binding energy in MeV.
4. 5H has a half-life of ~9.1 × 10−22 seconds and so a negative binding energy.
   You have a positive binding energy in MeV.
5. 6H has a half-life of 2.90×10−22 seconds and so a a negative binding energy.
   You have a positive binding energy.
6. 7H has a half-life of 2.3(6)×10−23 seconds and so a negative binding energy.
   You have a positive binding energy in MeV.
7. Helium-4 has a binding energy of 7.718 MeV.
   You have ~2.5 MeV.
8. Helium-4 has a binding energy of 28300.7 keV.
   You have ~7.0 MeV.
9. Helium-5 has a half-life of 700(30)×10−24 seconds and so a negative binding energy.
   You have a positive binding energy in MeV.
10. Helium-6 has a half-life of 806.7(15) ms and so a (probably!) negative binding energy.
    You have a positive binding energy in MeV.
11. Helium-7 has a half-life of 2.9(5)×10−21 seconds and so a negative binding energy.
    You have a positive binding energy in MeV.
12. Helium-8 has a half-life of 119.0(15) ms and so a (probably!) negative binding energy.
    You have a positive binding energy in MeV.
13. Helium-9 has a half-life of 7(4)×10−21 seconds and so a negative binding energy.
    You have a positive binding energy in MeV.
14. Helium-10 has a half-life of 2.7(18)×10−21 seconds and so a negative binding energy.
    You have a positive binding energy in MeV.

N.B. A positive binding energy in MeV means that a isotope is stable.
Generally the lower the half-life of an unstable isotope, the lower the binding energy and when the half-life is small enough the binding energy is negative. My impression is that a half-life less than a second implies a negative binding energy (a real nuclear physicist will probably correct me!).



One good thing about your web site, bjschaeffer, is that you are honest about the rejecttions of your paper by several journals:

• Nuclear Physics A
• European Physical Journal A
• Few-Body Systems
• Physics Letters B
• Physical Review C
• Europhysics Letters

The last comment sums it up quite well:
"This article has to be rejected with no further consideration. The author seems to be unaware that there are in nature strong nuclear forces that cannot be reduced to electromagnetic ones."

The binding energies are always negative even for unstable nuclides. When I drop the minus sign it is only for convenience.

You make the confusion between the total binding energy and the binding energy per nucleon. For 2H they are 2.2 and 1.1 MeV respectively.

I have published one paper on this subject and 40 on other subjects
http://link.springer.com/article/10.1007%2Fs10894-010-9365-0
and a second one is under way.

The fundamental laws of nuclear physics are still unknown by the mainstream nuclear physics. The "strong force"and its companion the "weak force" are myths as well as the "color" electric and magnetic interactions. In fact they are ordinary electric and magnetic interactions: the nucleus is not like an atom.

 

[bjschaeffer]

By the way, I found (more) errors in this paper.

1) You take a vague drawing of the "neutron" charge separation, and the proton charge, and compute an "electric quadrupole". You forgot to compute, also, the electric dipole moment---which is certainly nonzero, and very large (of order 10^-15 e*m) in your picture. This is important, because no nucleus has ever shown a nonzero electric dipole moment in experiments sensitive to ~10^-29 e*m. Your model can only get to this level if there's absolutely perfect matching between your neutron's internal +/- separation, and the external separation between the neutron's - and the proton's +. If this is to be consistent with data, these two distances would have to be identical to a precision of 0.000000000000001%. (One part in 10^14, in case I miscounted my zeros)

2) You recognize that the neutron has no large permanent electric dipole moment. You suggest, in the text, that the dipole might be induced by a nearby charge. Now, induced dipoles are an idea from real-world physics, but that's not the physics you used. All of your energy calculations are the calculations associated with a non-induced, large, static dipole. You did *not* include any polarization energy, nor indeed do you appear to be aware that this exists. You used a neutral object with a giant static dipole, an object which is obviously not a neutron.

3) You did force-calculations between the proton and the "-" half of the "neutron", which you put at a distance r_{np}. But then you put the neutron magnetic dipole also at r_{np}. Shouldn't the magnetic dipole be associated with both halves of the "neutron", rather than perfectly co-located with the "-" half? Since this dipole is, you say, repelled from the proton, it would be (if anything) pushed more towards "+" side---i.e., much, much farther from the "proton" than you assumed.

4) You extend your model to other isotopes, saying "the magnetic moments of the neutron and the proton are collinear in the deuteron and perpendicular in the other isotopes." You don't bother analyzing why, or how, the magnetic moments might be "perpendicular"---you just plunk the particles down and pretend that they stay where you put them. For example, your picture of tritium is unstable---it's not the ground state. The total energy will be lower if the central magnet is allowed to flip sideways. For another example, your "planar" drawing of 5H is nonsense---even assuming you know the spin directions and the force directions, which you don't, the ground-state configuration of this set of objects is a tetrahedron, not a flat square.

5) You have your E&M equilibria entirely wrong. You chose to guess all of the spin directions, then calculate linear forces, and find an "equilibrium". Unfortunately, you forgot about torques. The state you identify is never the ground state; the energy would be lowered (continuously) by rotating the proton spin, making the magnetic force attractive, removing the hard-core you point out and making the proton "fall into" the negative end of the neutron. Why is this obvious? It's an instance of Earnshaw's Theorem, the general proof that electromagnetic forces in 3D do not have stable static equilibria. Never. Look it up.

1) The drawing is schematic.
I neglected it in a first approximation. I have also calculated with a dipole giving a better result of course but not yet published. With the dipole is that there is no analytical formula.

The dipole moment you think about is a permanent dipole so small that it is in fact imaginary. The dipole I use is induced by the proton in 2H and therefore is zero in an isolated neutron.

2) The induced dipole is the product of polarization by the proton. Unfortunately the polarizability cannot be used here because the electric field is not uniform and would give wrong results. It is more correct to use directly Coulomb's laws.

3) your reasoning is purely qualitative. The following figure shows the comparison of what is obtained with and without the positive charge:



4) Usually the spins are assumed to be like that ↑↓ but there is a couple producing a rotation, physically absurd. I assume that the spins or more exactly the magnetic moments are collinear and opposite because the magnetic moments of the deuteron are known to be opposite and different. Same thing for 4He where the proton's magnetic moments are opposite and equal. Same thing for the neutrons. The proton's magnetic moments are perpendicular to the neutrons' by reason of symmetry.

My planar drawing of 5H is not nonsense, it is an approximation: the difference with 3D is not critical for a first calculation. Now I have an even simpler approximation by assuming a zero binding energy between the excess neutrons and the proton, giving similar results. With this approximation for the He isotopes I get a better result than those obtained with a super computer… see my post #414

5) It would be a catastrophy if there were torques!
I agree that the reason for equilibrium may not be clear. I suppose that the equilibrium is possible because of the spins are like a spinning top like the Levitron where Earnshaw's Theorem seems to be invalid.

 

[smartcooky]

You, however, have accused Ulf Meißner of not knowing the difference between a proton and a neutron. While making that accusation, you revealed that you yourself did not know that "nucleon" is a technical term that includes both protons and neutrons. Scientifically literate readers will notice such glaring mistakes, and will conclude that you are far from expert in the field in which you pretend to have superior expertise.

Now I am hardly what you would call "scientifically literate" when it comes to nuclear physics (prybar/asprin) and even I know this.

 

[bjschaeffer]

You're flat wrong. Meissner (like all nuclear physics) use strong interactions---treating both nucleons the same---for the strong-interaction part of the calculation. Then they add in the electromagnetic interaction. I clicked through a couple of papers and this is stated clearly.





Explain, please, why you made this accusation that Prof. Dr. Meissner doesn't include E&M correctly in his nuclear theory? According to his papers, he IS including E&M effects.

The strong interaction is a myth: nobody knows its fundamental laws and constants. It was invented because of the belief that that the nucleus is almost like a Bohr atom and that the nucleons are orbiting, but around what? That is the question. This hypothesis needs to equilibrate the centrifugal force which is more then ten times larger than the electric forces in the nucleus.

Without the hypothesis of the orbiting nucleons the electromagnetic interaction works fine. They didn't add completely the electromagnetic interaction, only the long range Coulomb repulsion between the protons. No attraction between neutron and proton and no magnetic interaction are taken into account. Meissner considers that the force between nucleons is the same for neutrons and protons called NN (2 nucleons force) or NNN (3 nucleons force) force. He may only take into account the long range repulsion between the protons but I am not sure.

There are hundreds of imaginary forces called or not "strong" force. Now they are so many that they use symbols like NN, 3N, Paris, Bonn, Argonne alias AV13… The last one I discovered is the "color force".

 

[bjschaeffer]

Now I am hardly what you would call "scientifically literate" when it comes to nuclear physics (prybar/asprin) and even I know this.

Yes see preceding post

 

[catsmate]

Meißner rejected my paper :

I didn't obtain this standards, presumably inexistent.
<snip>

Did you bother to actually look? Or do you expect people to publish whatever crap you submit? I found the journals guidelines in about thirty seconds.

<snip>

<snip>
Gee, big fonts. You must to correct.

 

[Dancing David]

The binding energies are always negative even for unstable nuclides. When I drop the minus sign it is only for convenience.

You make the confusion between the total binding energy and the binding energy per nucleon. For 2H they are 2.2 and 1.1 MeV respectively.

I have published one paper on this subject and 40 on other subjects
http://link.springer.com/article/10.1007%2Fs10894-010-9365-0
and a second one is under way.

The fundamental laws of nuclear physics are still unknown by the mainstream nuclear physics. The "strong force"and its companion the "weak force" are myths as well as the "color" electric and magnetic interactions. In fact they are ordinary electric and magnetic interactions: the nucleus is not like an atom.

Then why the weak force chirality?

How does the weak force work as an EM field? How do you get the predicted masses of the
W and Z bosons?

 

[ben m]

1) The drawing is schematic.
I neglected it in a first approximation. I have also calculated with a dipole giving a better result of course but not yet published. With the dipole is that there is no analytical formula.

So you *guessed* that the distance ratio was ~1. You have to *hope* that it's 1.0 to first approximation, and 1.00 to 2nd approximation, and 1.000 to 3rd approximation, and 1.000 to 4th ... that's a lot of approximations that have to "magically" work out your way in order for the answer to be 1.0000000000000 and agree with data.

2) The induced dipole is the product of polarization by the proton. Unfortunately the polarizability cannot be used here because the electric field is not uniform and would give wrong results. It is more correct to use directly Coulomb's laws.

I'm not asking you to use a linear polarizability, I'm asking you to include any form of polarization energy whatsoever---even just electromagnetic. You get all excited about the -1.6 MeV potential gained by the "n-" getting close to the "p+", and at the same time you allow the "n-" and "n+" to wander from arbitrarily-close to arbitrarily-far and you don't even look at the energy required to do this.

3) your reasoning is purely qualitative. The following figure shows the comparison of what is obtained with and without the positive charge:

I think you misunderstand the objection. Where, within the neutron, is the effective location of the magnetic dipole? Your equation 3 assumes that the magnetic dipole is at the *same* location as the "n+" charge---that's why the attraction term is 1/r_{np}, and the repulsion term is 1/r_{np}^3. If you hadn't colocated them, you'd have two distances to deal with---1/r_np and, e.g., (1/r_m)^3. The difference between this and your approximation is, to say the least, enormous.

The proton's magnetic moments are perpendicular to the neutrons' by reason of symmetry.

I know that's why you *put* them that way. What force do you think *keeps* them that way?

With this approximation for the He isotopes I get a better result than those obtained with a super computer… see my post #414

I remind you that I find your results to be, not merely "approximate", but dramatically and qualitatively different than reality. The only features you have "right" are (a) the order of magnitude, and (b) some sort of inflection point associated with 2H and 4He. (This is unsurprising, perhaps, given that you hand-engineered a special spin states for 2H.)

5) It would be a catastrophy if there were torques!
I agree that the reason for equilibrium may not be clear. I suppose that the equilibrium is possible because of the spins are like a spinning top like the Levitron where Earnshaw's Theorem seems to be invalid.

"If there were torques"? Of course there are torques.

Levitron? This is a very strange thing to assume. It's rather like announcing your foolproof method for jumping out of a plane, then saying "I suppose you might happen to land on a ski-jump at the correct angle." The Levitron's dynamic stability is very unusual, and if you think about it it's completely impossible for your system (where *both* magnets are free to move---compare to the Levitron, with one moving magnet and a mechanically-fixed base) because the "deuteron"'s long axis can rotate without torquing the spins themselves.

 

[bjschaeffer]

So you *guessed* that the distance ratio was ~1. You have to *hope* that it's 1.0 to first approximation, and 1.00 to 2nd approximation, and 1.000 to 3rd approximation, and 1.000 to 4th ... that's a lot of approximations that have to "magically" work out your way in order for the answer to be 1.0000000000000 and agree with data.

I guess nothing I apply Coulomb's laws with some approximations. I have tried different approximations, for example with or without the positive charge of the neutron. The calculated values differ of course by I don't pretend to obtain a high precision like people who look at the figures after the comma but don't know those before the comma. For example they calculate the binding energies of the helium isotopes with a super computer without being able to calculate the simplest nucleus beyond the proton, the deuteron, even with a bad precision.

I'm not asking you to use a linear polarizability, I'm asking you to include any form of polarization energy whatsoever---even just electromagnetic. You get all excited about the -1.6 MeV potential gained by the "n-" getting close to the "p+", and at the same time you allow the "n-" and "n+" to wander from arbitrarily-close to arbitrarily-far and you don't even look at the energy required to do this.

I prefer to use directly the Coulomb's laws than an empirical polarizability, valid only in a uniform electric field which is not the case here.



I think you misunderstand the objection. Where, within the neutron, is the effective location of the magnetic dipole? Your equation 3 assumes that the magnetic dipole is at the *same* location as the "n+" charge---that's why the attraction term is 1/r_{np}, and the repulsion term is 1/r_{np}^3. If you hadn't colocated them, you'd have two distances to deal with---1/r_np and, e.g., (1/r_m)^3. The difference between this and your approximation is, to say the least, enormous.

This approximation is not enormous quantitavely as proved by my calculation with the dipole. Now I put the magnetic moment at the center of the neutron but this calculation is no more analytical.

I know that's why you *put* them that way. What force do you think *keeps* them that way?



I remind you that I find your results to be, not merely "approximate", but dramatically and qualitatively different than reality. The only features you have "right" are (a) the order of magnitude, and (b) some sort of inflection point associated with 2H and 4He. (This is unsurprising, perhaps, given that you hand-engineered a special spin states for 2H.)

All experimental and calculated results are approximate. It is better to have a rough approximation than nothing. At the present time nobody is able to calculate the binding energy of the heavy hydrogen nucleus as Bohr did for the hydrogen atom.

"If there were torques"? Of course there are torques.

What kind of torques?

Levitron? This is a very strange thing to assume. It's rather like announcing your foolproof method for jumping out of a plane, then saying "I suppose you might happen to land on a ski-jump at the correct angle." The Levitron's dynamic stability is very unusual, and if you think about it it's completely impossible for your system (where *both* magnets are free to move---compare to the Levitron, with one moving magnet and a mechanically-fixed base) because the "deuteron"'s long axis can rotate without torquing the spins themselves.

I don't understand what is your strange thing.
For me, the deuteron may rotate around the neutron and proton common axis, also the axis of the magnetic moments. It's perhaps impossible but I am the only one able to calculate the hydrogen and helium isotopes from the fundamental laws of electricity.

 

[bjschaeffer]

Then why the weak force chirality?

How does the weak force work as an EM field? How do you get the predicted masses of the
W and Z bosons?

I din't study the weak force but Iknow that it is now called electroweak, that is electromagnetic.

I have no time to solve everything. For the time being I am solving the problem of the binding energy of the nuclides.

 

[ben m]

More generally, your answers above reveal the huge amounts of guesswork, approximation, and ignore-X-and-hope-it's-not-important behavior that's actually present in your model. I'd like to contrast this with your attitude in the rest of the thread, in which you state (repeatedly) that you "proved" that E&M forces "work" better than QCD.

"Proved" is the problem work here.

You did the first 5% of the work towards a calculation that you hope may turn out to work. That tiny amount of work, which got you to the (correct order of magnitude) deuteron binding = 1.6 MeV, might be said to be encouraging. You might have said, "A first approximation, using just E&M, got surprisingly close to the right answer, and I have high hopes for further success". But no, you insisted over and over that you had "proven" something.

Meanwhile, you took QCD's calculation problems as a complete failure. QCD can't yet calculate binding energies from pure fundamental constants, and BJSchaeffer can't calculate whether or not two charged magnets can hover close to one another. But you labeled the former as a failure, and you ignored the latter while pretending to have a "proof".

 

[ben m]

I don't understand what is your strange thing.
For me, the deuteron may rotate around the neutron and proton common axis, also the axis of the magnetic moments. It's perhaps impossible but I am the only one able to calculate the hydrogen and helium isotopes from the fundamental laws of electricity.

Correction:

You did not calculate the isotopes. You calculated the behavior of an extremely non-neutron-like object next to a proton-like object, and the result of the calculation is an object with (unless you get an unprecedented lucky cancellation) a huge CP-violating electric dipole moment.

You did not use "fundamental laws of electricity". Earnshaw's Theorem, which you ignore and hope will go away, is a fundamental law. "The neutron spontaneously charge-separates with no internal energy cost" is not a fundamental law. "the neutron magnetic moment exerts a 1/r^3 force centered at the neutron's negative end" is not a fundamental law.

 

[bjschaeffer]

More generally, your answers above reveal the huge amounts of guesswork, approximation, and ignore-X-and-hope-it's-not-important behavior that's actually present in your model. I'd like to contrast this with your attitude in the rest of the thread, in which you state (repeatedly) that you "proved" that E&M forces "work" better than QCD.

I never could find the fundamental basis of QCD, physical laws and fundamental constants. It seems also that you don't know either because you say "QCD" without knowing what are its fundamental constants and laws. An approximation with a proved theory and fundamental constants is always better than fitted results.

"Proved" is the problem work here.

You did the first 5% of the work towards a calculation that you hope may turn out to work. That tiny amount of work, which got you to the (correct order of magnitude) deuteron binding = 1.6 MeV, might be said to be encouraging. You might have said, "A first approximation, using just E&M, got surprisingly close to the right answer, and I have high hopes for further success". But no, you insisted over and over that you had "proven" something.

It is easy to check my calculations : the formula is on the graph on post #493.

Meanwhile, you took QCD's calculation problems as a complete failure. QCD can't yet calculate binding energies from pure fundamental constants, and BJSchaeffer can't calculate whether or not two charged magnets can hover close to one another. But you labeled the former as a failure, and you ignored the latter while pretending to have a "proof".

Nobody except me has ever calculated the simplest nucleus from fundamental laws, you included.

 

[Reality Check]

The binding energies are always negative even for unstable nuclides. When I drop the minus sign it is only for convenience.

Wrong: You calculated the binding energies. That results in a number. That number is either positive or negative. You then plot that number (divied by the nuclen nucleon). Negative values appear below the axis !
The convention is that binding energies are written as positive numbers for bound systems and negative for not bound systems.
For example: Helium-4

Binding energy 28300.7 keV

You are right - you plot binding energy per nucleon.

Let me count the ways that you are wrong (from that graph):

1. Deuterium binding energy measured to be 2,224.52±0.20 keV.
   You have ~2.0 MeV.
2. Tritium binding energy measured to be 8,481.821± 0.004 keV
   You have ~8.4 MeV (close but still wrong).
3. 4H has a half-life of (1.39 ± 0.10) × 10−22 seconds and so a negative binding energy.
   You have a positive binding energy in MeV.
4. 5H has a half-life of ~9.1 × 10−22 seconds and so a negative binding energy.
   You have a positive binding energy in MeV.
5. 6H has a half-life of 2.90×10−22 seconds and so a a negative binding energy.
   You have a positive binding energy.
6. 7H has a half-life of 2.3(6)×10−23 seconds and so a negative binding energy.
   You have a positive binding energy in MeV.
7. Helium-3 has a binding energy of 7.718 MeV.
   You have ~7.5 MeV.
8. Helium-4 has a binding energy of 28300.7 keV.
   You have ~28.0 MeV.
9. Helium-5 has a half-life of 700(30)×10−24 seconds and so a negative binding energy.
   You have a positive binding energy in MeV.
10. Helium-6 has a half-life of 806.7(15) ms and so a (probably!) negative binding energy.
    You have a positive binding energy in MeV.
11. Helium-7 has a half-life of 2.9(5)×10−21 seconds and so a negative binding energy.
    You have a positive binding energy in MeV.
12. Helium-8 has a half-life of 119.0(15) ms and so a (probably!) negative binding energy.
    You have a positive binding energy in MeV.
13. Helium-9 has a half-life of 7(4)×10−21 seconds and so a negative binding energy.
    You have a positive binding energy in MeV.
14. Helium-10 has a half-life of 2.7(18)×10−21 seconds and so a negative binding energy.
    You have a positive binding energy in MeV.

Of course your entire web page (and the paper you published) is wrong

The strong force exists and you do not include it. Thus all of your calculations are wrong.
​

bjschaeffer, one sign that a paper is a crank paper is when it is published in an inappropriate journal. That suggests that the author has touted the paper around journals who have knowledgeable peer reviewers who rejected it. Your paper is on nuclear physics and has no mention of fusion. It was rejected by

• Nuclear Physics A
• European Physical Journal A
• Few-Body Systems
• Physics Letters B
• Physical Review C
• Europhysics Letters

It was accepted by the Journal of Fusion Energy

The Journal of Fusion Energy features original research contributions and review papers examining the development of thermonuclear fusion as a useful power source

The fundamental laws of nuclear physics are still unknown by the mainstream nuclear physics.
[/quoute]
You continuously repeating your ignorance of nuclear physics does not mean that the strong force will magically vanish !

Gibberish about imaginary

"color" electric and magnetic interactions

does not impress anyone because it demonstrates your lack of knowledge of nuclear physics.

the nucleus is not like an atom.

At last you get something right: No one with any knowledge of nuclear physics thinks that the nucleus is like an atom!

The nucleons in a nucleus follow vastly more complex laws than the electrons outside of the nucleus. The same language is used, e.g. nucleons have orbitals. One source of complexity is that the nucleons create the potential in which they move. It is simpler for electrons because they just move in the potential well of the nucleus.

An anology from classical physics:
Planets are bound to move in the gravitational well from a star.
Galaxies in galaxy clusters are also bound. They move in the gravitational well caused by the other galaxies.

 

[Reality Check]

Nobody except me has ever calculated the simplest nucleus from fundamental laws, you included.

bjschaeffer: you are lying.

• The two-nucleon system at next-to-next-to-next-to-leading order
  E. Epelbaum, W. Glöckle, Ulf-G. Meißner
• Chiral effective field theory and nuclear forces, R. Machleidt, D. R. Entem (Phys.Rept.503:1-75,2011).

Of course the "simplest nucleus" is 1H .
ETA
You have not "calculated the simplest nucleus from fundamental laws" because your calculations violate a fundamental law as ben m pointed out: Earnshaw's Theorem

Earnshaw's theorem states that a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges. This was first proven by British mathematician Samuel Earnshaw in 1842.

 

[ben m]

Nobody except me has ever calculated the simplest nucleus from fundamental laws, you included.

I agree that you did some sort of calculation. Your calculation describes something which is quite different than an actual nucleus. Your calculation attempts to use fundamental laws, but does so incorrectly or nonsensically.

I do not find this to be worthwhile or encouraging.

Let me again point out your more serious problem/hypocracy:

Nobody except QCD theorists have ever calculated the simplest scattering cross sections from fundamental laws, you included.

And the big difference is that, while QCD is merely untestable (due to computer limitations as of 2012) in low-energy systems, your theory appears to be already falsified at high energies, and (I argue above) equally false---or at best untestable because you haven't solved enough details---at low energies.

 

[bjschaeffer]

I agree that you did some sort of calculation. Your calculation describes something which is quite different than an actual nucleus. Your calculation attempts to use fundamental laws, but does so incorrectly or nonsensically.
How is your "actual nucleus?"

I do not find this to be worthwhile or encouraging.

Let me again point out your more serious problem/hypocracy:



And the big difference is that, while QCD is merely untestable (due to computer limitations as of 2012) in low-energy systems, your theory appears to be already falsified at high energies, and (I argue above) equally false---or at best untestable because you haven't solved enough details---at low energies.

It is not the details (the decimal figures) that are important it is first the order of magnitude.

The computers are a bad excuse. Garbage in a computer remains garbage out. Perhaps you believe that a computer can replace the human intelligence. In fact it is the problem of the QCD which is missing the target and therefore will never be able to calculate the binding energy of a single nucleus because it is high fantasy.

The nuclear interaction is electric and not a mysterious "strong" force accompanied with an equally imaginary "weak" force.