1. Feb 5, 2016

### Garlic

Hello everyone,
Why does the residual strong force arise?
Hadrons carry no net color charge, so why are they experiencing the strong force?

2. Feb 5, 2016

3. Feb 5, 2016

### Garlic

"This same force is much weaker between neutrons and protons, because it is mostly neutralized within them, in the same way that electromagnetic forces between neutral atoms (van der Waals forces) are much weaker than the electromagnetic forces that hold the atoms internally together."

Thank you, this helps a lot.

4. Feb 6, 2016

### kiwaho

The average strong force between a neutron and a proton is about 46kg.

5. Feb 6, 2016

### Garlic

I'm sorry, I don't understand what you mean. How can you show the force strength with a mass unit?

6. Feb 6, 2016

### kiwaho

OK, the said strong force should be 456N.

Last edited: Feb 6, 2016
7. Feb 6, 2016

### Staff: Mentor

Or ~3 MeV/fm. What does "average" mean? Averaged over what? The order of magnitude is certainly right, but 456 N looks very specific.

8. Feb 6, 2016

### kiwaho

The 456N is calculated from classical kinematic equation Work = Average_Forece*Distance.
I use the data of deuterium. The binding energy is 2.22456MeV, and given 1MeV= 1.602*10^(-13)J.
The Distance = 3 - 2.22 = 0.78fm, because after 3fm, the strong force is 0, and the 2.22fm is the minimum distance between neutron and proton in deuterium. When 2 particles are kissing together, their distance equals to the sum of their radii, and both radii of proton and neutron are officially given.
Now we have: the Work= 2.22456* 1.602*10^(-13)J = 0.78*10^(-15) *Average_Force.
So the Average_Force= 2.22456* 1.602*10^(-13)J/0.78*10^(-15) = 456N
The max force occurs while 2 hadrons kissing together, of course, it will be far greater than the Average_Force 456N.
If the strong force is reduced to 0 at 3fm linearly, then the max force = 2* Average_Force = 912N, but probably the attenuation is not by linear but by exponential, so the max force may be a couple of thousands Newtonian level.

9. Feb 6, 2016

### Staff: Mentor

This is not true.
And this is completely wrong.
Proton and neutron are not billard balls. Their wave functions overlap. And they don't have a well-defined boundary either. The radius value is usually the root mean square of something (charge, mass, or whatever).

10. Feb 6, 2016

### kiwaho

The specific weight of neutron is very very high, so it is the hardest matter in universe! do you think they can overlap concentrically?
Even the minimum distance is not the sum of their radii, the split distance may be doubled, that means the average force will be half of 456N, still in the same order of magnitude.

11. Feb 6, 2016

### Staff: Mentor

This is just nonsense.
I don't just "think" so, I work with the consequences of it (parton collisions at high energy) every day.
This does not make sense.

As Orodruin and ZapperZ already told you here: quantum mechanics is not classical mechanics on a smaller scale, it is completely different.
Feel free to ask questions, but don't claim things that are simply wrong.

12. Feb 6, 2016

### kiwaho

I believe quantum mechanics QM is right. But for estimation, why not to hug classical theory occasionally? You know, QM only talks about potential or energy, but never quantitizes force although QM often talks about force.
Bohr calculated the radius of hydrogen atom by method of classical mechanics, even the controversial "classic radius of electron" 2.8fm has the same base.

13. Feb 6, 2016

Staff Emeritus
Classical mechanics is inappropriate to use at nuclear scales.

Yes, and he got the wrong answer by a factor of 3/2.

Exactly.

14. Feb 7, 2016

### Staff: Mentor

Then I suggest you learn quantum mechanics.
You can calculate forces in quantum mechanics, but they are rarely single values, and rarely useful.
Not purely classical mechanics, he used something between classical mechanics and quantum theory. And he got it wrong, as V50 noted already.
And we know today this is not the electron radius. If it has a radius at all, it has to be much smaller.