Is the weak nuclear force attractive or repulsive or both?

[abdossamad2003]

Hi
Of the 4 fundamental forces, I did not understand the weak nuclear force. Is the weak nuclear force attractive or repulsive or both? It works between two particles, that is, it is the interaction of two particles?

 

[Drakkith]

It is neither. The weak interaction isn't a force in the sense of Newton's laws where a force causes an acceleration. That's why it's better to call the fundamental forces 'interactions' instead of forces. The weak force, or weak interaction, is responsible for certain types of particle decays. It does not accelerate objects.

 

[abdossamad2003]

Does this force act as repulsive or attractive? How does this force act in decay? Is it large and directional?

 

[PeroK]

Does this force act as repulsive or attractive? How does this force act in decay? Is it large and directional?

https://physics.stackexchange.com/questions/77196/weak-force-attractive-or-repulsive

 

[abdossamad2003]

https://physics.stackexchange.com/questions/77196/weak-force-attractive-or-repulsive

this is not useful

 

[weirdoguy]

Does this force act as repulsive or attractive?

You've already got an answer, right before you posted that question the second time:

It is neither. The weak interaction isn't a force in the sense of Newton's laws where a force causes an acceleration.

 

[abdossamad2003]

" weak interaction, is responsible for certain types of particle decays"
That is, if this force does not exist, the particles will not decay.

Pay attention to this sentence
The weak force between two neighboring protons is about 10^-7 of the strong force between them, and the range of the weak force is smaller than 1 fm.
repulsive or attractive?

 

[weirdoguy]

repulsive or attractive?

Asking the same question over and over won't change the answer.

 

[vanhees71]

The answer has been given in the above quoted stackexchange discussions in #4.

 

[phyzguy]

this is not useful

Why do you not find it useful? The answer seems very clear. Whether it is attractive or repulsive depends on the particle values of weak isospin and weak hypercharge, just like whether the electric force is attractive or repulsive depends on the particle values of electric charge.

 

[haushofer]

Does this force act as repulsive or attractive? How does this force act in decay? Is it large and directional?

See post #2. Welcome in the loop.

 

[PeterDonis]

You've already got an answer

Actually he's been given two contradictory answers, since post #2 says it's not a force at all but the Stack Exchange thread says it is.

 

[PeterDonis]

Is the weak nuclear force attractive or repulsive or both? It works between two particles, that is, it is the interaction of two particles?

As the Stack Exchange thread that was linked to says, the weak interaction is mediated by spin 1 bosons, like the electromagnetic force, and that means like charges repel and opposite charges attract. The charges for the weak force are weak isospin and weak hypercharge; on the "spin 1 boson" view these charges are actually charges for two different kinds of "weak interaction", one mediated by the $W$ bosons and the other (the "weak neutral current") mediated by the $Z$ boson.

However, because this interaction is so weak, and because it is never present by itself--it is only present in situations like atomic nuclei and radioactive decay where the strong and electromagnetic interactions are also present--it is never observed as an actual "force" that causes attraction or repulsion because its attraction or repulsion is too small compared to the strong and electromagnetic interactions; it is only observed by the particular radioactive decays that it causes. That is what @Drakkith was describing in post #2.

 

[Bandersnatch]

Ah, yes. The perennial problem with the weak force...

 

[fresh_42]

It would be nicer - for symmetry reasons - if the last one would read: ... It turns neutrons into protons.

 

[vanhees71]

However, because this interaction is so weak, and because it is never present by itself--it is only present in situations like atomic nuclei and radioactive decay where the strong and electromagnetic interactions are also present--it is never observed as an actual "force" that causes attraction or repulsion because its attraction or repulsion is too small compared to the strong and electromagnetic interactions; it is only observed by the particular radioactive decays that it causes. That is what @Drakkith was describing in post #2.

I'd say the fact that the "weak interaction" (I'd abandon the word "force" from all relativistic physics right from the very beginning ;-)) "is never present by itself", which I guess means that you don't "feel" it in our macroscopic world, is due to the fact that it's of very short range, which is due to the large masses of the corresponding gauge fields, whose particle-like excitations are the $W$ and $Z$ bosons. That's of course due to the Higgs mechanism at work in the description of the weak interaction. Note that there's also the photon within this theory ("quantum flavor dynamics" aka "Glashow-Salam-Weinberg model"), which is "un-Higgsed" and thus massless. The corresponding gauge field is of course the electromagnetic field, which is observable in our macroscopic world, even in its "free states", i.e., electromagnetic waves of a wide range of wave lenghts (note that these "macroscopic fields" are no photon states but rather coherent states or thermal (Planck) radiation, etc.).

To make the discussion complete, there's also the strong interaction, which holds the quarks together to form hadrons (among others protons and neutrons, making up the nuclei forming the macroscopic matter around us). It is described by quantum chromodynamics, QCD, which is also a gauge theory, but it's completely "un-Higgsed". Nevertheless, we also don't "feel" it although it's by far the strongest interaction in Nature. Here the reason is much more complicated: It's called "confinement".

Phenomenologically it means that we never observe free quarks (the fundamental/elementary spin-1/2 particles carrying color charge and thus participate in the strong interaction) nor free gluons (the quanta of the gauge field) or the corresponding free field. The reason is that in a non-Abelian gauge theory the gauge bosons themselves carry charge.

The formal argument, why we don't observe anything carrying color charge is gauge invariance: We can't observe anything which is not gauge invariant and we can't form color-charged gauge-invariant observables. That's why all we can observe are color-neutral objects, the socalled hadrons, which are very complicated bound states of quark and gluon fields. Usually they come as mesons (bound-states of a quark and an antiquark) or baryons (bound-states of three quarks). There are also hints of more "exotic" varieties like tetra quarks (consisting of two quarks and two antiquarks) as well as glue balls (formed purely by the gauge field/gluons). All we can observe of the strong interaction in a sense of "force" (if you want to use this word at all ;-)) is the residual interaction between color-neutral hadrons.

From a theoretical point of view in our macroscopic world, quarks and gluons are not the appropriate degrees of freedom to describe the strong interaction but rather effective theories of hadrons, which use the (residual) symmetries of QCD to constrain the corresponding models. Most important in this context is the approximate chiral symmetry of QCD in the light-quark sector (u- and d-quarks; to lesser extent also s-quarks), from which the pretty complicated interactions, e.g., between nucleons (protons and neutrons) can be understood, which describes then both the scattering between the nucleons as well as the formation of atomic nuclei as bound states of protons and neutrons. These interactions go also beyond simple two-body interactions, and an understanding of nuclear structure from these "first-principle approaches" is a (already quite successful) topic of ongoing research.

The understaning of the formation of the hadrons as bound states of quarks and gluons, however, is much more complicated. The most successful technique is lattice-gauge theory, where QCD is numerically evaluated approximating its action on a discrete space-time lattice using Monte-Carlo techniques. There are two varieties: "vacuum lattice-QCD". Here one of the greatest successes is the calculation of the hadronic mass spectrum, including the prediction of higher-mass states not yet observed. The other variety is "finite-temperature lattice-QCD", which studies the equation of state of strongly interacting matter and the corresponding phase diagram. In Nature this is relevant for (a) neutron stars and neutron-star mergers, which nowadays can be observed with "multi-messenger astronomy", i.e., via electromagnetic waves in a large range of wave lengths and, since 2015, with gravitational waves (as well as with neutrinos, at least in principle), and (b) in collisions of heavy nuclei (relativsitic heavy-ion collisions) as done at the LHC and SPS (CERN), RHIC (BNL), GSI (Darmstadt) and in upcoming new facilities like FAIR (Darmstadt), and NICA (Dubna).

 

[snorkack]

It is neither. The weak interaction isn't a force in the sense of Newton's laws where a force causes an acceleration. That's why it's better to call the fundamental forces 'interactions' instead of forces. The weak force, or weak interaction, is responsible for certain types of particle decays. It does not accelerate objects.

When a neutrino undergoes elastic collision with an electron, there is no decay. Electron and neutrino before, electron and neutrino after. Weak force must have accelerated the electron after all.
But this does not yet give you whether the force was attractive or repulsive, because of the wave nature and Heisenberg uncertainty of the position of electron (and neutrino). Compare electromagnetic interaction. If you watch an electron collide with a nucleus and bounce to the right, then because of wave uncertainty of electron position, you do not know whether the electron passed to the right of nucleus and was repelled, or to the left of nucleus and was attracted. Finding the sign of force is another and more complex exercise.

But weak interaction has the property of being sensitive to helicity... and the right helicity particles don´t just undergo opposite weak interaction but no weak interaction at all. Whereas electromagnetic and strong interaction do not care about helicity, except through interaction with other spins.
Are there any bound states where energy level splitting can be observed which is specifically due to weak interaction (states that would be degenerate under electromagnetic and strong interaction) and which indicate the sign of weak force?

 

[vanhees71]

One example is the CP violation of the weak interaction, which was discovered by observing kaon oscillations, i.e., the oscillation between neutral kaons and antikaons:

https://en.wikipedia.org/wiki/Neutral_particle_oscillation

 

[snorkack]

One example is the CP violation of the weak interaction, which was discovered by observing kaon oscillations, i.e., the oscillation between neutral kaons and antikaons:

https://en.wikipedia.org/wiki/Neutral_particle_oscillation

Yes, but the oscillations are not interactions between two separate particles, right? They are two forms of same particle.
So, where are the systems where weak interaction is evidenced by energy level shift between two levels which would be degenerate by strong and electromagnetic forces alone, one of which is unaffected by weak interaction because of right helicity, and the other affected because of left helicity?