# Strong nuclear force's repulsive aspect

1. Aug 29, 2010

### hkyriazi

I was curious about the evidence for the above. I found an old discussion here:

But, I wasn't sure better answers, involving experimental evidence, weren't available. It seems to me that high speed collisions between heavy nuclei might, by not "collapsing" the nucleons, suggest minimums on the repulsive forces involved (if such exist).

2. Aug 29, 2010

### mathman

Within the nucleus, the strong force has a weird behavior as a function of separation. Specifically, the closer together things are, the weaker the force. There is no need for any assumption about repulsive force.

3. Aug 29, 2010

Staff Emeritus
If there were no repulsive force between nucleons, why aren't nuclei infinitely small? There's your experimental evidence right there.

4. Aug 29, 2010

### humanino

The quark-quark interaction vanishes at very high energies, or very close distances. This is asymptotic freedom, the running of the strong coupling constant.

The nucleon-nucleon interaction is an effective interaction at longer distances resulting from this strong interaction. At very long distances, or very low energy, nucleons exchange pions. Pions are (pseudo)scalars so the long distance nucleon-nucleon interaction is attractive. At not-so-long distances, heavier mesons start contributing, in particular say the rho or the omega. Those are vectors, and with vector exchanges like repel each other. So at distances of the order of the nucleon size, nucleon repel each other (at least, this is a reasonable expectation even without going into full details) At even shorter distances, it makes no more sense to use nucleons, and we are back to the quark quark interaction.

To describe how this is extracted from the data would take quite a bit of discussion. The nucleon-nucleon cross-sections are decomposed into partial waves and phase shifts tell us wether the interaction is attractive or repulsive.

5. Aug 29, 2010

### bcrowell

Staff Emeritus
It would be nice to have a model-independent answer, such as one directly from experiment. Unfortunately there is no model-independent way of extracting a nucleon-nucleon interaction from experimental data. Ultimately this is because a nucleon isn't even a fundamental particle. If you want to probe very short distance scales, which is where some models have a repulsive core, you have to do high-energy scattering experiments, but these are exactly the experiments in which you're most likely to see effects arising from the composite nature of nucleons.

No, this is incorrect.

FAQ: Why do nuclei have the sizes they do?

The strong nuclear force has a range of about 1 fm, so if the separation between a neutron or proton and its nearest neighbors was much more than that, there would be no binding. Since the strong nuclear force is generally attractive, we could ask why they don't pack down closer than this. The reason is the Heisenberg uncertainty principle. The smaller the space you pack the nucleons into, the faster they have to move, and the greater their kinetic energy. The nucleus has a size that minimizes the sum of its kinetic and potential energies.

Some models of the strong nuclear force include a repulsive core. This feature is not necessary in order to explain the sizes of nuclei. Models without it can reproduce the observed sizes of nuclei.

6. Aug 29, 2010

Staff Emeritus
Well, I was trying to get the idea across without resorting to partial-wave analysis of scattering data, but I think that the finite size does tell you something about the strength of the interaction.

The nuclear radius will be of order

$$r = \frac{\hbar}{m_N c \alpha}$$

So you're right - QM makes this finite so long as alpha is as well. However, you can take this one step further: you can infer the size you expect nuclei to be based on the strength of the p-n interaction, and you discover they are substantially larger than that. Of course, a good deal of this is due to Fermi repulsion rather than rho exchange, but I would argue repulsion is repulsion.

7. Aug 29, 2010

### bcrowell

Staff Emeritus
No, nucleon-nucleon interactions are models with lots of adjustable parameters that are fitted to data. One of those pieces of data is the sizes of nuclei. So if you calculate the size you expect nuclei to be based on the strength of some model of the nucleon-nucleon interaction, you will find that they are exactly right, on the average, because the interaction's adjustable parameters were carefully adjusted to make them come out exactly right, on the average.

8. Aug 29, 2010

Staff Emeritus
Exactly. If you take the long distance p-n attraction, and model the nuclear force by that (essentially, the OPE model) you get nuclei that are too small. You need to put in a repulsive vector term in to get the sizes right (and to get the partial waves right as well).

9. Aug 29, 2010

### bcrowell

Staff Emeritus
Well, of course OPEP isn't going to give realistic results for nuclear structure. Your reasoning in #3 is still incorrect. You said, "If there were no repulsive force between nucleons, why aren't nuclei infinitely small? There's your experimental evidence right there." Purely attractive models of the nucleon-nucleon interaction exist that get the sizes of nuclei right. Therefore the finite size of nuclei is not evidence for repulsive interactions between nucleons.

10. Aug 29, 2010

Staff Emeritus
Like I said, I had hoped to get away with an oversimplification to avoid going into partial-wave analyses. So "infinitely small" was overambitious - but the fact that nuclei are a few fm across is evidence that the forces do have have a repulsive component, otherwise they would be smaller.

11. Aug 29, 2010

### bcrowell

Staff Emeritus
No, that's incorrect. You can take a purely attractive force and tune its strength to make nuclei any size you like. You could make it so weak that nuclei would be a meter in diameter.

12. Aug 30, 2010

Staff Emeritus
Which is why I said, "If you take the long distance p-n attraction, and model the nuclear force by that (essentially, the OPE model) you get nuclei that are too small."

13. Aug 30, 2010

### hkyriazi

Thanks, guys. Regarding the repulsion, I was going on figures such as this (which was posted in that old PF discussion I'd linked to): http://webs.mn.catholic.edu.au/physics/emery/assets/9_5_op21.gif

The red curve shows the increasing repulsion at small distances. I like Vanadium 50's response that, without such (apparent) repulsion, nuclei could be infinitely small, which they obviously aren't. Bcrowell, I don't see how a purely attractive force, however one configures it, could avoid the possibility of infinitely small nuclei, barring the application of Heisenberg's Uncertainty Principle to quarks (if I've even phrased that properly - I haven't studied quantum chromodynamics).

Last edited: Aug 30, 2010
14. Aug 30, 2010

### bcrowell

Staff Emeritus
That is one model of the nuclear interaction. Other models do not have any such hard core. For example, the radial variation of the one-pion exchange potential is $1+3/mr+3/(mr)^2$. This never changes sign as a function of r, so there is no repulsive core.

The Heisenberg uncertainty principle certainly applies to quarks. Actually in low-energy nuclear structure it's usually simpler to think in terms of nucleons (neutrons and protons) as the fundamental objects -- but in any case the H.U.P. applies to them as well. Vanadium 50's statement that without repulsion nuclei would have zero size is completely incorrect. This is a very common beginner's misconception about nuclear forces, and that's why I wrote it up as a FAQ.

15. Aug 30, 2010

### humanino

He already mentioned that it was a voluntary oversimplification, so I do not feel there is any need to continue a polemic. I too did not want to go into the details, partly because they are quite technical, and partly because they are not fully successful in all details. It remains that one should not stretch the parameters out of reasonable ranges either. We can expect in-medium modifications of the various parameters, but they should remain moderate enough that we can recognize the states. Otherwise those approaches sort of loose relevance.

16. Aug 30, 2010

### hkyriazi

Thanks again, folks. That seems to be about as complete an answer as I can expect, given our current state of knowledge.

17. Dec 20, 2010

### Leperous

I apologize for the thread necro, but seeing as this page seems to be (relatively) important, I feel it should be pointed out that the above is itself incorrect. I should know as I'm writing a PhD on neutron stars and this error is in my correction list - I'm told by several people that this is a very common misconception.

Pauli Exclusion and the residual strong force are not the same thing, and it is the repulsive force generated by the latter at short distances which maintains nuclear size (and also prevents NS collapse into a black hole).

18. Dec 20, 2010

### hkyriazi

Thanks, Leperous. It seems to me that this question lies at the border between what I think of as "physical" models, involving tangible repulsive forces, and "spiritual," quantum mechanical models, involving mysterious rules, such as Heisenberg's Uncertainly Principle and the Pauli Exclusion Principle. You seem to fall on the side of physical models. Fair characterization?

19. Dec 20, 2010

### bcrowell

Staff Emeritus
Expanding a little more on what I wrote above, note that we can have three effects going on:

1. the Pauli exclusion principle
2. the Heisenberg uncertainty principle
3. a repulsive core that is present in some models of the nucleon-nucleon interaction

Note that in the material I wrote I only referred to 2 and 3, but Leperous's reply referred to 1. Leperous, maybe you misread what I wrote and thought I was referring to 1?

Clearly 1 is not necessary in order to explain all nuclear sizes. For example, 4He has a certain nonzero size, but 1 is irrelevant.

Leperous, if you think about it for a minute, I think you'll also see that the answer can't possibly be that 3 is the only relevant effect. If 1 were irrelevant, then it wouldn't be relevant that nuclei and neutron stars are composed of fermions -- but obviously that is highly relevant in determining many aspects of their structures, including their sizes. In fact, 3 is the only one that is optional, because one can make models of nuclei with purely attractive nucleon-nucleon interactions, and by tuning up the parameters of the model one can get the sizes of nuclei right.

Last edited: Dec 20, 2010
20. Dec 20, 2010

### Norman

browell and Leperous-

can either of you reference some peer-reviewed literature to back your claims? I am genuinely interested and curious about this, as I was always under the impression that it was the HUP which maintained the nuclear size.

21. Dec 20, 2010

### bcrowell

Staff Emeritus
Hi, Norman -- I spent some time looking at papers. There are certainly lots of examples of successful Hartree-Fock calculations that do not involve anything like a repulsive core in the potential, e.g.:
Chamel and Pearson, "The Skyrme-Hartree-Fock-Bogoliubov method: its application to finite nuclei and neutron-star crusts," http://arxiv.org/abs/1001.5377
Stone and Reinhard, "The Skyrme Interaction in finite nuclei and nuclear matter," http://arxiv.org/abs/nucl-th/0607002
The Skyrme interaction has zero range, although it incorporates momentum-dependent terms that partially serve to mock up a finite range. However they also have density-dependent terms that are repulsive, so they aren't a perfect example of a purely attractive interaction.
-Ben