When is the force of strong nuclear interaction repulsive?

[user-000]

TL;DR Summary

In what cases the force of the strong interaction is the force of repulsion?

In what cases the force of the strong nuclear interaction (between nucleons or between quarks) is the force of repulsion?

Thank you.

 

[Vanadium 50]

The strong force depends on flavor (proton vs. neutron), spin, and distance. There is no simple relationship between these and its direction.

 

[user-000]

The strong force depends on flavor (proton vs. neutron), spin, and distance. There is no simple relationship between these and its direction.

Thank you

Can you, please, tell when exactly (which effects or reactions) the strong interaction-repulsion takes place or send something where I can find some examples of such case?

 

[Vanadium 50]

When two protons get too close together the force switches from attraction to repulsion.

 

[FusionJim]

When two protons get too close together the force switches from attraction to repulsion

How does then the p-p chain in stars take place when you need to overcome not just electrostatic repulsion but also the very close range (less than 2 fm?) strong force repulsion?

This is not a problem for fusion of ions with more than one proton, correct? IIRC there the protons/neutrons never have to get as close as two protons in the p-p chain?

 

[Vanadium 50]

The first step of the pp chain goes by the weak force, not the strong force. Good thing too, as you want the sun to last a good long time.

 

[Bandersnatch]

You don't need to overcome the strong force to fuse the two protons in a p-p reaction. It's attractive on that scale. It's what snaps the two nucleons together once the electrostatic potential barrier is overcome, to form a deuterium nucleus (following a beta decay).

The repulsive aspect of the strong force is what prevents the nucleus from collapsing further.

 

[FusionJim]

The first step of the pp chain goes by the weak force, not the strong force. Good thing too, as you want the sun to last a good long time.

Ok, so there is no stable bound state with 2 protons in a nucleus, is this solely due to electrostatics or is also the strong force involved in this, due to it becoming repulsive at close range? As for the weak force, what is the order in which this happens?

I would guess that first two protons get close enough and because they can exist in a "bound state" only very briefly before they again split apart then sometimes in that brief moment one of them undergoes beta decay which turns one of them into a neutron and releases a positron and electron neutrino both with some kinetic energy.

I read that beta plus decay normally happens in existing nucleus and not for single protons, for energy conservation reasons I assume.

So does this mean that the reason the proton sometimes decays to neutron in the p-p fusion is because of the energy involved in that brief moment when two protons manage to get close enough during pp fusion?

So what fundamentally determines the low rate of weak force proton decay in such pp fusion events, is it dependent on both quantum tunneling for the number of protons that get close enough together in the first place and then on top of that the probabilistic nature of weak force in terms of how many of those tunneled protons then undergo decay to form deuterium?

 

[Vanadium 50]

There is no pp bound state period.

The idea of force vs. distance is inherently classical, and this is a quantum system. You are stretching the concept past its degree of validity.

The distances we are talking about are less than the size of a proton, sometimes much less. See above.

 

[ohwilleke]

This answer is a little more technically detailed and precise than the previous answers, which are trying to keep things simple for this "basic" rated question.

But I am doing so, because there are statements from FusionJim that indicate some really basic misunderstandings about the issue and make incorrect statements that need to be cleared up. And, it is pretty much impossible to clear up these misunderstandings in a satisfactory way, without being more technical and precise. I've done my best, however, to still make this answer as accessible and understandable as possible.

Ok, so there is no stable bound state with 2 protons in a nucleus, is this solely due to electrostatics or is also the strong force involved in this, due to it becoming repulsive at close range? As for the weak force, what is the order in which this happens?

Keep in mind that the true strong force is a force operating "within" protons and neutrons. The actual strong force itself confines quarks and gluons within strong force bound composite particles known as "hadrons" (a category that includes protons, neutrons, and many other short-lived composite particles).

The force binding protons and neutrons to each other within an atomic nucleus is the non-fundamental "residual strong force" (a.k.a. "nuclear force" or "nuclear binding force") carried mostly by neutral pions (and to a much smaller second order degree by neutral kaons), both of which are composite particles made primarily out of quarks bound by the strong force, rather than directly by the strong force, which is carried by fundamental particles which are somewhat analogous to photons for electromagnetism, called gluons.

The "residual strong force" that binds protons and neutrons into a nucleus is not nearly as strong the strong force itself, even though the residual strong force is so strong that it, and not the direct strong force itself, gives rise to atomic bombs and nuclear power.

In general, we usually only directly observe the direct strong force at work in particle colliders so powerful that they can blow protons and neutrons to bits, something that doesn't happen in nuclear fusion, and doesn't happen (except via weak force beta decay) in nuclear fission.

In graph form, via this Wikipedia link, the direction and magnitude of the residual strong force that binds atomic nuclei is as follows:

The range of the residual strong force is a function of the mass of the primary particle (the neutral pion) that carries it between protons and neutrons in a nucleus.

I would guess that first two protons get close enough and because they can exist in a "bound state" only very briefly before they again split apart then sometimes in that brief moment one of them undergoes beta decay which turns one of them into a neutron and releases a positron and electron neutrino both with some kinetic energy.

I read that beta plus decay normally happens in existing nucleus and not for single protons, for energy conservation reasons I assume.

So does this mean that the reason the proton sometimes decays to neutron in the p-p fusion is because of the energy involved in that brief moment when two protons manage to get close enough during pp fusion?

So what fundamentally determines the low rate of weak force proton decay in such pp fusion events, is it dependent on both quantum tunneling for the number of protons that get close enough together in the first place and then on top of that the probabilistic nature of weak force in terms of how many of those tunneled protons then undergo decay to form deuterium?

Protons never experience beta decay, or any other kind of decay. They are stable.

Protons never decay into neutrons, an interaction which is prohibited by mass-energy conservation because a neutron has 1.29 MeV/c² more mass-energy than a proton. There is no such thing as weak force proton decay.

Proton decay is impossible in the Standard Model of Particle Physics (due to a combination of mass-energy conservation and a less widely known law of physics known as baryon number conservation).

Experimental measurements have shown that if proton decay does exist, that the mean lifetime of a proton is, at a minimum, vastly longer the the current age of the universe. The experimental bound on the mean lifetime of a proton which was at least 1.29*10³⁴ years (as of the year 2014), which implies that it would happen in not more than one in 10²⁴ protons over the entire current 1.38*10¹⁰ year lifetime of the universe.

A proton only transforms into a neutron or other composite particle when it interacts with outside particles that provide the necessary electromagnetic charge reduction and mass-energy. Interactions that require a new particle not emitted by the original particle to occur are not decays. It you smash protons into other protons with enough energy, however, this can produce hundreds of different kinds of composite particles (called "hadrons"), all of which have mean lifetimes of microseconds or less (sometimes as much as eighteen orders of magnitude less), except for protons, bound neutrons, electrons, neutrinos, and their respective anti-particles, which are all stable.

Free neutrons experience beta decay via the weak force (as one of its valence down quarks transforms into a valence up quark when the down quark emits a fundamental W- boson of the weak force, one of the massive carrier bosons of the weak force which is the weak force's version of the photon, which in turn decays into an electron and an electron anti-neutrino) on average, after a period of about fourteen minutes and 42 seconds ± 1.5 seconds (the imprecision of this measurement is somewhat shocking compared to the precision with which so many other properties of protons and neutrons are known and is plagued by the inherent systemic errors sources present when one tries to make "inclusive" v. "exclusive" measurements of this mean lifetime).

About 99.9% of the time, beta decay produces a proton, an electron, and an anti-neutrino, each of which has kinetic energy, because the mass-energy of a neutron is greater than the mass-energy of a proton, an electron, and an anti-neutrino. The kinetic energy and the photon energy combined are equal to the difference in rest mass between the neutron and combined rest mass of the proton, electron and electron anti-neutrino times the speed of light squared, which turns out to be 0.782343 MeV/c².

About 0.1% of the time, beta decay produces a proton, an electron, and an electron anti-neutrino, each of which has kinetic energy (but less combined kinetic energy than in the usual case), and a photon. This happens because the extremely short lived electromagnetically charged W- boson emitted by the down quark in the neutron when it changed into an up quark (and thus turns the neutron in to a proton) sometimes has enough time to emit a photon in the way that all electromagnetically charged particles do, before the W- boson decays into an electron and an electron anti-neutrino.

Neutrons bound in atomic nuclei are basically stable, but can experience beta decay when they are in an atomic nucleus with many protons and neutrons, via the same weak force mechanism. The rather technical and complicated reasons that bound neutrons sometimes experience beta decay, and sometimes don't, is explored in a Physics Stack Exchange post here, if you'd like to read further about this reality.

Empirically, the neutron to proton ratio needed to make an atomic nucleus stable is between 1 and 1.537 (gradually increasing with larger atomic numbers), except in the degenerate cases of hydrogen (a bare proton without a neutron), and helium-3 (two protons and a neutron). As Wikipedia explains:

Neutron-proton ratio (N/Z ratio or nuclear ratio) is the ratio of the number of neutrons to protons in an atomic nucleus. The ratio generally increases with increasing atomic numbers due to increasing nuclear charge due to repulsive forces of protons. Light elements, up to calcium (Z = 20), have stable isotopes with N/Z ratio of one except for beryllium (N/Z ratio=1.25), and every element with odd proton numbers from fluorine to potassium. Hydrogen-1 (N/Z ratio=0) and helium-3 (N/Z ratio=0.5) are the only stable isotopes with neutron–proton ratio under one. Uranium-238 and plutonium-244 have the highest N/Z ratios of any primordial nuclide at 1.587 and 1.596, respectively, while lead-208 has the highest N/Z ratio of any known stable isotope at 1.537.

One can imagine an atomic nucleus which has zero protons and two neutrons, which is called a "bound dineutron". An atomic nucleus with no protons and an arbitrary number of neutrons is known, in general, as neutronium. A bound dineutron particle, like a neutron, would have no electrons and would therefore be chemically inert similar to noble gases like Helium and Neon. It would have a rest mass of about 1.879 GeV/c² before adjusting for the mass due to the nuclear binding energy of the two neutrons. But, in reality, there is a modest shortfall of nuclear binding force between two two neutrons. The minimum nuclear binding force necessary to create a stable bound atomic nucleus is greater than the amount of nuclear binding force between two neutrons in a bineutron state. So, bound bineutrons are not a thing (strictly speaking bineutrons can be created, or at least strongly correlated with each other, but are extremely short lived).

A biproton state (a.k.a. Helium-2) would have essentially the same insufficient amount of nuclear binding force as a hypothetical dineutron particle (the shortfall is actually slightly smaller, since protons are a little more than 1% less massive than neutrons), but would also, unlike a dineutron particle (involving electromagnetically neutral neutrons), have to overcome the electromagnetic repulsion between two protons (which more than offsets the reduced nuclear binding energy required due to the 1% smaller mass of the proton). So stable Helium-2 is also not a thing (strictly speaking Helium-2 can be created, but is so unstable that it has a undetermined mean lifetime that is, at a minimum, less than a billionth of a second).

The relative strengths of the residual strong force and the electromagnetic force (a.k.a. the Coulomb force) between two protons is show below (from here):

The average distance between protons is the distance at which the attractive pull of the residual strong force balances the repulsive force of electromagnetism, which happens at a little less than one femtometer.

Footnote Regarding The Strong Force Itself

The direct strong force between quarks of different strong force color charge within a proton, neutron, or other hadron, is always attractive (although the strong force is repulsive between quarks of the same strong force color charge). The strong force color charge is the strong force analog to electromagnetic charge.

The direct strong force gets weaker (but not below zero) at closer distances and in higher energy scale interactions, which gives rise to a phenomena called asymptotic freedom. This is illustrated by the following chart where the Y axis is the strength of the strong force (shown on the solid line) and the X axis shows distance and energy scale (with the left side representing longer distances and lower energy scales and the right side representing shorter distances and higher energy scales). The dotted line illustrates electromagnetic forces for comparison's sake.

In contrast, at low energies and long distances, the strong force gets stronger and stronger.

This is why quarks and gluons are never observed in a free state outside a composite particle bound by the strong force called a hadron (such as a proton or neutron). Instead, quarks and gluons are "confined" within a hadron, except for (1) top quarks which decay via the weak force so fast that they decay before they can be bound into a composite particle, and (2) at extremely high temperatures (hotter than inside any star) that are only found on Earth inside the Large Hadron Collider, and even there only momentarily, where they briefly transform into a state known as "quark gluon plasma".