Nuclear Strong Force is a Fiction

The question was "How do they separate again" not 'When do they separate again'.

How about 'when an electron approaches'? Do your Neutronic charges separate then as well?

It's evident, the electric field of the proton attracts the negative charge(s) and repulses the positive charge(s). They are thus again separated in an electric dipole.
 
When a proton approaches

Wrong. The binding energy is infinite in classical electromagnetism when two charges coincide (and a=0). That's Ben's -1/(2 a) term. A proton approaching will have no effect whatsoever.

This is one reason why classical electromagnetism is a total failure at describing atoms, let alone nuclei.
 
Wrong. The binding energy is infinite in classical electromagnetism when two charges coincide (and a=0). That's Ben's -1/(2 a) term. A proton approaching will have no effect whatsoever.

This is one reason why classical electromagnetism is a total failure at describing atoms, let alone nuclei.

How do you know?
 
It's evident, the electric field of the proton attracts the negative charge(s) and repulses the positive charge(s). They are thus again separated in an electric dipole.

Put numbers on this. The proton, currently far away, attracts the "-" very gently. It repels the "+' very gently. Meanwhile, the + and - are attracting each other as strongly as possible, because they're starting at a=0.

The proton's field tries to pull them into a dipole, but it will not succeed. I've just shown you the math on this six times. Do it yourself and stop guessing the answer.
 
When a is small the classical formula is 2a/r^2 = 0 when a=0

No. I've already explained this. A "2a/r^2" term is an energy associated with reorienting a dipole of fixed a. (It's the magnitude of a d*cos(theta)/E term.) It is nothing but the first term in a Taylor-expansion of 1/(r+a) - 1/(r-a), which you've already included.

It is not the polarization energy. The polarization energy is 1/2a.
 
No. I've already explained this. A "2a/r^2" term is an energy associated with reorienting a dipole of fixed a. (It's the magnitude of a d*cos(theta)/E term.) It is nothing but the first term in a Taylor-expansion of 1/(r+a) - 1/(r-a), which you've already included.

It is not the polarization energy. The polarization energy is 1/2a.

What do you mean by polarization energy?
 
Put numbers on this. The proton, currently far away, attracts the "-" very gently. It repels the "+' very gently. Meanwhile, the + and - are attracting each other as strongly as possible, because they're starting at a=0.

This is qualitative reasoning

The proton's field tries to pull them into a dipole, but it will not succeed. I've just shown you the math on this six times. Do it yourself and stop guessing the answer.

I don't see your numbers
 
Last edited:
Please learn how to use the "quote" tags. If you're trying to quote me, all of my text should be between a "quote" and a "/quote". None of your text should be between such tags. Use the "preview" button if you're unsure.

This is qualitative reasoning

I don't see your numbers

When you had the concepts/qualitative reasoning wrong, we gave you numbers and you rejected them. Now that we're talking through the concepts, you're rejecting it by demanding numbers.

If you want numbers, scroll back one page, look at the equation I wrote for you---which, indeed, it looks like you tried and rejected yourself---and minimize the energy with respect to both a and r. You'll get a=0, r=infinity, as several people have told you. If you don't understand why the minimum is at a=0, well, we're trying to help you with that using qualitative explanations.
 
It's evident, the electric field of the proton attracts the negative charge(s) and repulses the positive charge(s). They are thus again separated in an electric dipole.

How much energy does it take to so 'evidently' separate those charges?

Again how about an approaching Electron, what is equally and oppositely 'evident' from its equal and opposite electric field?
 
When a is small the classical formula is 2a/r^2 = 0 when a=0

No, it is not. The classical formula for the interaction energy between two opposite charges a distance 2a apart is given by Coulomb's law, and it's proportional to -1/(2a), which goes to -infinity when a goes to zero. The presence of the proton adds a finite correction.
 
Last edited:
It's evident, the electric field of the proton attracts the negative charge(s) and repulses the positive charge(s). They are thus again separated in an electric dipole.
Oh dear - you do not know how charges attract or repulse :eye-poppi. Otherwise your answer to the question would be:
It's evident, the electric field of the electon attracts the positive charge(s) and repulses the negative charge(s). They are thus again separated in an electric dipole
Thus your model predicts the existence of a electron + neutron atomic nucleus :p.

As the other posters have told you: Classical binding energy is always infinite when 2 charges coincide. This is qute basic physics.
 
Oh dear - you do not know how charges attract or repulse :eye-poppi. Otherwise your answer to the question would be:
It's evident, the electric field of the electon attracts the positive charge(s) and repulses the negative charge(s). They are thus again separated in an electric dipole
Thus your model predicts the existence of a electron + neutron atomic nucleus :p.

As the other posters have told you: Classical binding energy is always infinite when 2 charges coincide. This is qute basic physics.

I don’t think bjschaeffer was responding to the question of an approaching Electron, Reality Check. Seems to be a question he simply does not want to address. Just pretending to answer the question of “How” his Neutronic charges get separated (polarized) by simply claiming (again) it (his Neutron) gets, well, polarized.
 

Back
Top Bottom