Nuclear Strong Force is a Fiction

[bjschaeffer]

Let me re-re-reexplain. I'll stick to the easy part.

There are three charges in your setup. Right? The proton charge (p+), your half-neutron (n-), and your half-neutron (n+). In doing your calculation, you took into account the electrostatic potential between p+ and n-. You call that distance r_{np}, so you plugged in binding energy -1/r_{np}.

My first calculation was by neglecting the positive charge of the neutron it gave 1.6 MeV see #525

My second calculation was with the simple dipole formula, see #525 giving a minimum at 2.3 MeV

It is not a high precision but better than nothing… It will be always possible to enhance the precision by specialists of numerical calculation.

My third calculation was without approximation but with a flat spot instead of a minimum, giving a binding energy of 2.2 MeV (to be published in a Chinese journal). This comes from Coulomb singularity but it doesn't change much the result.

You *ignored* the separation (d) between n- and n+ inside the "neutron". But that's is indeed a charge-separation, and incurs a binding energy -1/d. You thought you could ignore this term and call it an "approximation", but the term you ignored is much, much more important than the term you included.

The separation distance between n- and n+ is 2a or d if you prefer, the dipole moment distance.

(You also ignored the repulsive p+ n+ potential, at distance (d+r_{np}). At least you *knew* you were ignoring this one, and since you guessed that d was large it was reasonable to ignore it. But d isn't large, as we will see.)

There is no repulsive force between a charge and a dipole, only an attractive force because the positive charge is farther away from the proton than the negative charge (see Feyman).

Go ahead, BJschaeffer, try this: calculate the potential energy of a state with r_np = 1 fm and d=1. That's pretty close to the thing *you* think is the bound state. (Include the magnetic repulsion or not, I don't care, it doesn't help.) Tell me the total energy.

The potential is infinite, it is the Coulomb singularity. As I showed above it is not a real problem, the numerical results being not very different. A so-called cut-off or a "Taylor expansion" are not necessary. More over it is a problem because it needs adjasted parameters…

Try this: calculate the potential energy of a state with r_np = 1 and d=0.01---a case where the neutron *does not* "polarize", but stays compact . Tell me the total energy. It's lower. How is it that you "minimized the energy" but I can find a lower-energy state? Because your approximation was missing an important term, and mine is not.

The graphs shown have been obtained with Excel. I don't understand your problem.

In fact, the binding energy is arbitrarily large for d=0. What does this tell us? Well, it tells *me* the same thing that it told Niels Bohr: charge bound states are constrained by quantum mechanics, so your whole electrostatic model, which ignores quantum mechanics, is gibberish. What should it tell *you*? It should tell you that your electrostatic model was a "nice try", and now that you've worked it out better, you see that it does NOT have a deuteron-like bound state.

It is not necessary to know the value at when r=0, only the minimum of the potential. There are no "states" in my theory, only classical electromagnetism.

 

[ben m]

It is not necessary to know the value at when r=0, only the minimum of the potential. There are no "states" in my theory, only classical electromagnetism.

You are jumping to the end and guessing that you have the right answer, without doing the work that will show otherwise.

Please go through my post sentence-by-sentence and work out the energies I suggested calculating.

Sorry, you screwed up the quoting in the above---you have your text mixed in with mine. Fix?

 

[ben m]

After trying to parse the above text: you're still missing the point. Start with a free "Schaeffer-neutron". Its two charges are unseparated, at d=0. Allow the Schaeffer-neutron to approach a proton. You claim that its charges separate to d= ~1fm, maybe more. But it requires energy to separate charges like this, which you ignore.

This energy is very, very large, and you are wrong to ignore it. The true energy "minimum" for the system of a Schaeffer-neutron near a proton is at r_np = infinity and d=0; the potential is actually purely repulsive.

Again, Mr. Schaeffer, I repeat that I am using your model exactly, and simply removing your mistaken "approximation". I'm calculating the *three* Coulomb potentials (n+ to p+, n- to p+, and the n- to n+ potential you forgot) and one magnetic dipole-dipole potential (n to p). I'm minimizing the energy by varying both r_np and d.

 

[sol invictus]

It is not necessary to know the value at when r=0, only the minimum of the potential.

If the energy at d=0 is less than the energy at your "minimum" with d non-zero, then it's not a minimum.

 

[bjschaeffer]

After trying to parse the above text: you're still missing the point. Start with a free "Schaeffer-neutron". Its two charges are unseparated, at d=0. Allow the Schaeffer-neutron to approach a proton. You claim that its charges separate to d= ~1fm, maybe more. But it requires energy to separate charges like this, which you ignore.

The two charges of the neutron are separated only when the proton is nearby the neutron.
The energy needed to separate charges is the dipole energy. The total energy is the sum of the dipole energy and the energy between the proton and each charge of the neutron. Thus the total energy is twice the exact dipole energy.

This energy is very, very large, and you are wrong to ignore it. The true energy "minimum" for the system of a Schaeffer-neutron near a proton is at r_np = infinity and d=0; the potential is actually purely repulsive.

How do you know that the energy is very, very large?
This is only an assumption, there is no experimental proof of it. The potential between a positive charge and a dipole is attractive. It seems that you ignore electrostatics. Feynman says:
“When you bring a positive charge up to a conducting sphere, the positive charge attracts negative charges to the side closer to itself and leaves positive charges on the surface of the far side”

Again, Mr. Schaeffer, I repeat that I am using your model exactly, and simply removing your mistaken "approximation". I'm calculating the *three* Coulomb potentials (n+ to p+, n- to p+, and the n- to n+ potential you forgot) and one magnetic dipole-dipole potential (n to p). I'm minimizing the energy by varying both r_np and d.

I don't understaind this paragraph. It seems that you made a confusion between my three approximations:
1) n+ neglected, only p+, n- potential
2) n+p+, - n-p+ interaction between proton and neutron charges
plus approximated dipole n+ n- potential gives 3 potentials
3) The total exact potential is twice the exact dipole potential (to be published).

All these three methods give energy values near the experimental result.

You may repeat what you want : I believe only what is proved. If you use my model, please show me the graph of the potential you obtain

 

[Reality Check]

Originally Posted by ben m
Start with a free "Schaeffer-neutron". Its two charges are unseparated, at d=0. Allow the Schaeffer-neutron to approach a proton. You claim that its charges separate to d= ~1fm, maybe more. But it requires energy to separate charges like this, which you ignore.

This is really easy to understand, bjschaeffer.
Your "n+" and "n-" only exist because the proton induced the dipole.
No proton = no dipole = no "n+" or "n-".
It costs energy to separate charges.
You ignore this energy.
Your model is thus wrong.

Anything there that you cannot understand bjschaeffer?

 

[ben m]

I don't understaind this paragraph. It seems that you made a confusion between my three approximations:
1) n+ neglected, only p+, n- potential
2) n+p+, - n-p+ interaction between proton and neutron charges
plus approximated dipole n+ n- potential gives 3 potentials

Why the heck would you "approximate" the n+ n- potential? Just state d and use Coulomb's Law. Nothing approximate about it. Except the correct, exact calculation---the one I did---does NOT give a deuteron-like energy minimum. It gives an energy minimum with d=0 and the neutron repelled from the proton.

 

[ben m]

3) The total exact potential is twice the exact dipole potential (to be published).

No it's not. The exact potential is

U(r,d) = - 1/r - 1/d + 1/(r+d) + 1/r^3

which has a global minimum at d=0, r=infinity no matter what the charges and moments are. It has no other minima, not even local ones. I have no idea what mistake you made to find a minimum at nonzero d or r.

If you displace the magnet from the charges, to a distance of (say) a, such that r < a < r+d, then d=0 r=infinity remains the global minimum, but there's a wacky local alternative minimum at r=0 and a nonzero d whose exact value depends on how a is determined.

 

[bjschaeffer]

Why the heck would you "approximate" the n+ n- potential? Just state d and use Coulomb's Law. Nothing approximate about it. Except the correct, exact calculation---the one I did---does NOT give a deuteron-like energy minimum. It gives an energy minimum with d=0 and the neutron repelled from the proton.

What is your calculation?
I cannot reply if I don't see it.

 

[bjschaeffer]

No it's not. The exact potential is

U(r,d) = - 1/r - 1/d + 1/(r+d) + 1/r^3

which has a global minimum at d=0, r=infinity no matter what the charges and moments are. It has no other minima, not even local ones. I have no idea what mistake you made to find a minimum at nonzero d or r.

If you displace the magnet from the charges, to a distance of (say) a, such that r < a < r+d, then d=0 r=infinity remains the global minimum, but there's a wacky local alternative minimum at r=0 and a nonzero d whose exact value depends on how a is determined.

Your formula is strange.
The real formula is
U(r,d) = - 2/(r - d) + 2/(r+d) + 1/r^3
or, with simplified dipole
U(r,d) = - 1/(r - d) + 1/(r+d) - 2d/r^2 + 1/r^3

 

[ben m]

Your formula is strange.
The real formula is
U(r,d) = - 2/(r - d) + 2/(r+d) + 1/r^3
or, with simplified dipole
U(r,d) = - 1/(r - d) + 1/(r+d) - 2d/r^2 + 1/r^3

First off, you've changed notation. In your JFE paper, you used "r_np" to mean the distance from the p+ to the "n-". (That's why you had terms like 1/r_np and 1/r_np^3, right?). Now you're using r to mean the distance from the proton to the *center* of the neutron, or something. In any case, you've done something wrong, because there's no "d/r^2" potential under any definition of r that I can think of.

My notation uses these simple definitions:

        r            d   
p+  <---->  n-  <----->  n+

There are three terms in the Coulomb potential (-1/r, -1/d, and +1/(r+d) ) and one magnetic-dipole-repulsion term (+1/something^3, where the "something" depends where the neutron's magnetic dipole is located---maybe 1/r^3, maybe 1/(r+d/2)^3, whatever).


ETA: OH WAIT, I think I know where you got a "d/r^2" term. With the effective neutron electric dipole moment proportional to d, a dipole oriented in a field of strength E has an potential energy d/E, thus d/r^2. Very funny, and very inappropriate. Why? It's redundant with the Coulomb terms you've already included. Since you have already written down the two proton-based Coulomb terms ( - 1/(r - d) + 1/(r+d) ), you're done. The d/r^2 term is the term you'd use if you were trying to replace the Coulomb terms with their Taylor expansion---not a term you can add to them. (i.e., 0 + d/r^2 + d^2/r^4 ... is the same thing as 1/(r+d) - 1/(r-d). And only if d<<r, which is not even true here.)

And you're still missing the interesting term. The n+ and the n- are a distance d apart (maybe 2d in your notation). Use Coulomb's Law, exactly---you have the charges and the distance, that's all you need, no more dipole mistakes---to tell me how they got that far apart, and how much energy it took to get them there.

(Maybe you were under the impression that the d/r^2 term took care of this? You're wrong. That's the energy required to reorient (i.e. rotate) a dipole whose d is fixed. It is not the energy required to change d.)

 

[bjschaeffer]

First off, you've changed notation. In your JFE paper, you used "r_np" to mean the distance from the p+ to the "n-".

My first approximation is the only one published in a journal and the only one giving an analytical formula for the binding energy. The two other approximations take into account the dipole.

(That's why you had terms like 1/r_np and 1/r_np^3, right?). Yes


Now you're using r to mean the distance from the proton to the *center* of the neutron, or something. In any case, you've done something wrong, because there's no "d/r^2" potential under any definition of r that I can think of.



My notation uses these simple definitions:

        r            d   
p+  <---->  n-  <----->  n+

There are three terms in the Coulomb potential (-1/r, -1/d, and +1/(r+d) ) and one magnetic-dipole-repulsion term (+1/something^3, where the "something" depends where the neutron's magnetic dipole is located---maybe 1/r^3, maybe 1/(r+d/2)^3, whatever).

The simple first approximation is not (-1/r, -1/d, and +1/(r+d) ) but
(-1/r + 1/r^3 )
electric+magnetic with one + charge only plus two interacting magnetic moments.
The plus charge attracts the negative charge of the neutron and the magnetic moments of the proton and the neutron repulse themselves.

The second approximation is
-1/(r-d/2)+1/(r+d/2) + d/r^2 + 1/r^3

The third approximation is
-2/(r-d/2)+2/(r+d/2) + 1/r^3



ETA: OH WAIT, I think I know where you got a "d/r^2" term. With the effective neutron electric dipole moment proportional to d, a dipole oriented in a field of strength E has an potential energy d/E, thus d/r^2. Very funny, and very inappropriate. Why? It's redundant with the Coulomb terms you've already included. Since you have already written down the two proton-based Coulomb terms ( - 1/(r - d) + 1/(r+d) ), you're done. The d/r^2 term is the term you'd use if you were trying to replace the Coulomb terms with their Taylor expansion---not a term you can add to them. (i.e., 0 + d/r^2 + d^2/r^4 ... is the same thing as 1/(r+d) - 1/(r-d). And only if d<<r, which is not even true here.)

And you're still missing the interesting term. The n+ and the n- are a distance d apart (maybe 2d in your notation). Use Coulomb's Law, exactly---you have the charges and the distance, that's all you need, no more dipole mistakes---to tell me how they got that far apart, and how much energy it took to get them there.

(Maybe you were under the impression that the d/r^2 term took care of this? You're wrong. That's the energy required to reorient (i.e. rotate) a dipole whose d is fixed. It is not the energy required to change d.)

d=2a is the distance between the two neutron charges
The only Coulomb exact formula for the dipole is:
- 1/(r - a) + 1/(r+a) (≈ 2a/r^2)
The dipole in the neutron is not permanent : it is induced by the proton. The interaction between the proton and the dipole is given by the same exact formula.
Therefore the exact interaction (using Coulomb's laws) is twice
- 1/(r - a) + 1/(r+a) (≈ 2a/r^2)
That is
- 2/(r - a) + 2/(r+a) (≈ 4a/r^2)

The reason of the 2 coefficient is that energy is needed to create the induced dipole. Unfortunately I did'nt find this in any book. They all use the approximate formula d/r^2. For induced electric dipole, they use the polarizability, which is an empirical constant.

It seems that it is the same as you say, but with other words.

 

[ben m]

The reason of the 2 coefficient is that energy is needed to create the induced dipole. Unfortunately I did'nt find this in any book. They all use the approximate formula d/r^2. For induced electric dipole, they use the polarizability, which is an empirical constant.

Sorry, bjschaeffer, you have made a mistake. The coefficient of 2 that you have invented is a mistake, it has no relationship to the energy needed to create the dipole.

Here is what you tried to find in a book: you tried to find information about polarization in atoms and solid materials. Let's talk about why those systems are treated "empirically", and why I am able to treat your system exactly.

In real-world systems, at zero applied field you have a complex, spatially-extended charge distribution determined by quantum mechanics. If you apply a field, the charge distribution shifts, to a different value also determined by quantum mechanics. In these cases, you need to know quantum mechanics to figure out the induced dipoles. (An engineer won't bother doing this quantum mechanically, they'd just look up the "polarizability" in a table.) But there's nothing magical about it. If you know the applied fields, you just solve for the energy minimum of the system. The applied field wants to pull the charges apart; the internal fields want to pull them back together; and the system will find a point where the forces balance.

In your system ... well, it's just Coulomb's Law holding the neutron together. And it's just Coulomb's Law that resists the formation of the dipole. The n+ is attracted to the n-, and you know this attraction. The n+ is close to the n-, and you know the distance. That's why I know exactly how much energy to create the dipole---it's just Coulomb's Law.

Let me switch to your notation and write the correct, complete potential energy of the p+ "n+/n-" system, including all of the physics you're attempting to use. The proton and the Schaffer-neutron center are a distance r apart. The "n+" and "n-" are displaced by a in either direction. I'll leave the magnetic moment in the center of the neutron.

 <-------r------>
            <-a-><-a-> 
p+         n-        n+

U = -1/(r-a) + 1/(r+a) - 1/(2a) + 1/r^3

That's the exact solution, bjschaeffer. We do not need any dipole-expansions, additional dipole energies, or empirical polarization terms---we just apply Coulomb's Law to each of the charges (including within the dipole). The ground state of this system looks like this:

 <---------------------------r goes to infinity------>
                                                   <aa> 
p+                                                 n-n+ 
                                              a goes to zero

 

[bjschaeffer]

Sorry, bjschaeffer, you have made a mistake. The coefficient of 2 that you have invented is a mistake, it has no relationship to the energy needed to create the dipole.

Here is what you tried to find in a book: you tried to find information about polarization in atoms and solid materials. Let's talk about why those systems are treated "empirically", and why I am able to treat your system exactly.

In real-world systems, at zero applied field you have a complex, spatially-extended charge distribution determined by quantum mechanics. If you apply a field, the charge distribution shifts, to a different value also determined by quantum mechanics. In these cases, you need to know quantum mechanics to figure out the induced dipoles. (An engineer won't bother doing this quantum mechanically, they'd just look up the "polarizability" in a table.) But there's nothing magical about it. If you know the applied fields, you just solve for the energy minimum of the system. The applied field wants to pull the charges apart; the internal fields want to pull them back together; and the system will find a point where the forces balance.

In your system ... well, it's just Coulomb's Law holding the neutron together. And it's just Coulomb's Law that resists the formation of the dipole. The n+ is attracted to the n-, and you know this attraction. The n+ is close to the n-, and you know the distance. That's why I know exactly how much energy to create the dipole---it's just Coulomb's Law.

Let me switch to your notation and write the correct, complete potential energy of the p+ "n+/n-" system, including all of the physics you're attempting to use. The proton and the Schaffer-neutron center are a distance r apart. The "n+" and "n-" are displaced by a in either direction. I'll leave the magnetic moment in the center of the neutron.

 <-------r------>
            <-a-><-a-> 
p+         n-        n+

U = -1/(r-a) + 1/(r+a) - 1/(2a) + 1/r^3

That's the exact solution, bjschaeffer. We do not need any dipole-expansions, additional dipole energies, or empirical polarization terms---we just apply Coulomb's Law to each of the charges (including within the dipole). The ground state of this system looks like this:

 <---------------------------r goes to infinity------>
                                                   <aa> 
p+                                                 n-n+ 
                                              a goes to zero

I had tried this formula also but the result is bad:

 

[ben m]

When you try the best calculation your model can do, and it gives the wrong answer, that's evidence against the model. That's exactly what the scientific method is built to do-- to efficiently throw away wrong models, allowing people to focus on the right one.

Throw away your "electromagnetic nucleus" model, bjschaeffer, it's wrong. You did not obtain "1.6 MeV" by applying the physics of electromagnetism to the nucleus. You obtained "1.6 MeV" by applying *throwing away* bits of the physics, selectively, until you got the answer you wanted.

Also: I still don't think that you have jointly minimized r and a properly. maybe you guessed a and only minimized r? The ground state really is at a=0. If you can't see that mathematically, graph it.

 

[bjschaeffer]

When you try the best calculation your model can do, and it gives the wrong answer, that's evidence against the model. That's exactly what the scientific method is built to do-- to efficiently throw away wrong models, allowing people to focus on the right one.

Throw away your "electromagnetic nucleus" model, bjschaeffer, it's wrong. You did not obtain "1.6 MeV" by applying the physics of electromagnetism to the nucleus. You obtained "1.6 MeV" by applying *throwing away* bits of the physics, selectively, until you got the answer you wanted.

Also: I still don't think that you have jointly minimized r and a properly. maybe you guessed a and only minimized r? The ground state really is at a=0. If you can't see that mathematically, graph it.

a=0 is not possible, the potential being infinite.
For the first approximation giving 1.6 MeV, there was no "a". Therefore it gives an analytical formula for the binding energy. All my calculations with different approximations give similar results proving that the electromagnetic approach is correct.

There are more than hundred theories of the nuclear interaction. None is able to calculate the deuteron binding energy from fundamental laws and constants, even with supercomputers and without fitting, of course.

My formula giving the order of magnitude of the nuclear energy is αm_pc² the nuclear equivalent of the Bohr formula for the atom αm²_ec². The physical origin of the nuclear energy is electromagnetic all other self called "modern" theories will disappear as many old theories.

 

[ben m]

For the first approximation giving 1.6 MeV, there was no "a". Therefore it gives an analytical formula for the binding energy.

It give the analytical formula for something, but not the binding energy of a deuteron. You calculated the binding energy of a "+" charge with a proton-like magnetic moment, attracted to a "-" charge with a neutron-like magnetic moment. Your analytic formula is for an negatively-charged "neutron", which you stick to a proton to make a neutral "deuteron". Nobody cares if you have an analytic formula for the binding energy of a "neutral deuteron".

a=0 is not possible, the potential being infinite.

Let me put it this way: what do you think the value of "a" is? Well, the forces in this system are trying to make "a" smaller than that.

Sorry. Another way to think about it is: do your usual "approximation" in which you ignore the "n+". Go ahead with your weird approximation, and draw your "neutral deuteron". Got it? Now, place the other half of the neutron nearby (the "n+") and find the force on it due to the "p+/n-", "neutral deuteron" dipole. You find that the "n+" is attracted to the "n-", right?

How are you going to stop it? Why would it do anything other than obeying this attractive force?

 

[bjschaeffer]

It give the analytical formula for something, but not the binding energy of a deuteron. You calculated the binding energy of a "+" charge with a proton-like magnetic moment, attracted to a "-" charge with a neutron-like magnetic moment. Your analytic formula is for an negatively-charged "neutron", which you stick to a proton to make a neutral "deuteron". Nobody cares if you have an analytic formula for the binding energy of a "neutral deuteron".



Let me put it this way: what do you think the value of "a" is? Well, the forces in this system are trying to make "a" smaller than that.

Sorry. Another way to think about it is: do your usual "approximation" in which you ignore the "n+". Go ahead with your weird approximation, and draw your "neutral deuteron". Got it? Now, place the other half of the neutron nearby (the "n+") and find the force on it due to the "p+/n-", "neutral deuteron" dipole. You find that the "n+" is attracted to the "n-", right?

How are you going to stop it? Why would it do anything other than obeying this attractive force?

The calculation is such that there is equilibrium, the potential being at a minimum.

 

[sol invictus]

The calculation is such that there is equilibrium, the potential being at a minimum.

Your calculation is wrong, because you left out the important terms. If the energy is lower with a very small then it is with a at the value you think it takes (which it is), then the "minimum" you found is not a minimum.

You're clearly no good at math, so forget it for a moment. Think about the physics. Question: what force prevents the two halves of the neutron from falling together?

 

[bjschaeffer]

Your calculation is wrong, because you left out the important terms. If the energy is lower with a very small then it is with a at the value you think it takes (which it is), then the "minimum" you found is not a minimum.

I agree that it is not always a real minimum, only a flat. Do you have a solution?

You're clearly no good at math, so forget it for a moment. Think about the physics. Question: what force prevents the two halves of the neutron from falling together?

They fall together. It is the same as electric charges in a cloud