Why does the strong nuclear force become repulsive at small distances?

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SUMMARY

The strong nuclear force becomes repulsive at small distances due to quantum mechanics and the Heisenberg uncertainty principle, which prevents quarks from collapsing into a nucleon. This repulsion is essential for maintaining nuclear stability, similar to how electrons do not fall into the nucleus of an atom. The concept of a hard repulsive core in nucleon-nucleon interactions is supported by the Pauli exclusion principle, which restricts the proximity of quarks. Understanding these principles is crucial for grasping the behavior of nuclear forces.

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Hootenanny
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This isn't a traditional homework question but here goes. I asked my physics tutor (A-Level) how the strong nuclear force becomes repulsive at small distances and he said he didn't know. I know that it is a nessecity that it is repulsive to prevent the nucleus collapsing, but I would like to know howit happens, are there any law's or mathematical derivations to prove this?

Any help / explanations would be great!
 
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Hootenanny said:
This isn't a traditional homework question but here goes. I asked my physics tutor (A-Level) how the strong nuclear force becomes repulsive at small distances and he said he didn't know. I know that it is a nessecity that it is repulsive to prevent the nucleus collapsing, but I would like to know howit happens, are there any law's or mathematical derivations to prove this?

Any help / explanations would be great!
I'll take a stab at it. But you really should be talking to ZapperZ.

I would also suggest this be moved to the High energy, Nuclear, Particle physics section.

What makes you think that the strong force becomes repulsive at small distances? The reason the nucleus does not collapse is similar to the reason an atom does not collapse (negative electrons do not fall into the positive nucleus). It has to do with quantum mechanics. It does not require the concept of a force to keep the quarks apart or the electrons and protons apart.

If the electron fell into the nucleus, we would know its position to within a very small \Delta x. So, the Heisenberg uncertainty principle says that its momentum range would be enormous. But this would require a huge amount of energy. Hence the probability that the electron will fall to the nucleus on its own is practically 0. Similarly, the energy required to collapse the three quarks in a nucleon is really enormous so the probability of nucleon collapse on its own is virtually 0.

But as I say, talk to ZapperZ.

AM
 
Thank's the reason I think that it becomes negative is because of the graphs shown in my textbook, something similar to this;
http://webs.mn.catholic.edu.au/physics/emery/assets/9_5_op21.gif
I understand what you mean about the uncertainty principle though, has it got something to do with zero point motion?
 
It is generally believed that the N-N potential has a hard repulsive core.
There have been attempts to get this with vector meson exchanges, but it is more easily seen as the Pauli principle for the quarks.
You can't put six quarks that close together.
The hard core has little to do with nuclear stability.
That is no different than hydrogen atom stability.
They both come from the lowest quantum state having a finite size.
 

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