2
$\begingroup$

I have what is probably a bitter misconception about the strong force which I would like to clarify.

Here's my (probably flawed) reasoning.

  • The strong forces holds protons and neutrons together at close range, overcoming the repulsion between protons.

  • The strong forces cannot identify or choose which numbers of configurations of protons and neutrons it should hold together.

  • Hence, the strong force would hold together any number of protons and neutrons if they were brought close enough.

But this doesn't sound right, since not all "possible nuclei" are nuclei of actual elements, and when fusion happens the products are very specific nuclei instead of a much more diverse salad.

So what's telling the strong force "when" to hold stuff together?

$\endgroup$
  • 1
    $\begingroup$ What you are referring to is the "nuclear force", not the strong force. The nuclear force is not considered a fundamental interaction, it is a residual interaction of the actual strong force, which acts between quarks. Similarly to how the van der Waals force relates to the EM interaction. While the strong force is mediated by gluons, the nuclear force is mediated by mesons, which themselves contain a quark and an antiquark. $\endgroup$ – Bence Racskó Jul 17 '16 at 14:56
  • $\begingroup$ @Uldreth forgive me if this is ridiculous, but the wiki entry says the strong force acts at two scales, the larger of which includes gluing of protons and neutrons. At any rate, what you say does not change my question. Why is it that some configurations of protons and neutrons stay together while others do not? $\endgroup$ – Arrow Jul 17 '16 at 14:59
  • $\begingroup$ physics.stackexchange.com/q/267909 .This link might help. The strong force cannot hold an unlimited protons and neutrons together otherwise we would have elements beyond uranium, with 92 protons, and we don't. $\endgroup$ – user108787 Jul 17 '16 at 15:01
  • $\begingroup$ @count_to_10 that doesn't explain why not all possible smaller configurations of protons and neutrons are nuclei of some atoms. $\endgroup$ – Arrow Jul 17 '16 at 15:05
  • $\begingroup$ The residual force that acts at the nuclear scale is sometimes called the "nuclear force", "strong nuclear force", "effective strong force" and of course "residual strong force". Nor does there seem to be any real standardization between authors. $\endgroup$ – dmckee Jul 17 '16 at 15:58
4
$\begingroup$

The reason why not any nuclei (for e.g. one which has only protons) can exist in nature is due to the fact that they will have a very low Binding Energy, compared to a nuclei which has the same number of nucleons, but with a more stable proton/neutron ratio.

A simple model which allows us to get fairly accurate estimated of Binding Energies is the Liquid Drop model. Based on this model, the binding energy is given by the Semi Empirical Mass Formula :https://en.wikipedia.org/wiki/Semi-empirical_mass_formula

It is true that the strong force does not distinguish between protons and neutrons. However, we know that nature favours some combinations of protons and neutrons better than others. For example, a nuclei with a nucleon number of 40 (Calcium) is most stable if it has 20 protons and 20 neutrons.

Another possibility could be 30 protons and 10 neutrons, but having 30 protons would cause the nuclei to be more unstable because of the increased electrostatic repulsion.

What about having 10 protons and 30 neutrons? This would have lower electrostatic repulsion. But now the asymmetry term in the semi empirical mass formula will cause the nuclei to be less stable. This is because protons and neutrons stack up in energy independently.

$\endgroup$
2
$\begingroup$

The limit on the strong force holding things together is given by the combination of a range and the degeneracy of the nuclear matter.

The residual strong force is well described mathematically by a Yukawa potential $$ V(r) = - \frac{g^2}{4 \pi c^2} \frac{e^{-mr}}{r} \,,$$ where the mass $m$ that appears in there is roughly the pion mass and $g$ is an effective coupling constant. This form causes the strength of the force to die off strongly above a couple of femptometers.

At the same time, the nucleons in an large nucleus form a degenerate gas, and can not be compressed further without a large increase in energy. Their equation of state is fairly stiff. As a consequence adding additional nucleons necessarily increases the size of the nucleus. This isn't just conjecture, it is supported by the measured sizes which goes roughly as the cube-root of the nucleon number $$r_\text{nucleus} \propto A^{1/3} \,,$$ implying a constant density.

The only place where the energy for further comp[action can be found is in very dense astronomical objects.


Walecka's book has a nice treatment of this stuff in the first few chapters.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.