Is the weak interaction asymptotically free?

In summary, the sign of the beta function for SU(N) (unbroken) with nf fermions is determined by the expression ##\dfrac{-11N}{6} + \dfrac{n_f}{3}##, where N is the number of fermions in the fundamental representation of SU(N). For N = 2, this means that ##n_f < 11## and the SU(2) gauge theory is asymptotically free. However, this would result in a strange theory with massless gauge-bosons and is not the weak interaction as we know it. The weak force weakens at higher energy scales and there is some debate in QCD phenomenology about whether the strong force coupling constant has a low
  • #1
HomogenousCow
737
213
Is the ##SU(2)_L## part of the SM asymptotically free like a typical non-Abelian gauge theory? I've been made to understand that confinement does not occur for ##SU(2)_L## because the spontaneous symmetry breaking scale is above the confinement scale, however I can't find any information on whether this affects asymptotic freedom.
 
Physics news on Phys.org
  • #2
The sign of the beta function for SU(N) (unbroken) with nf fermions is determined of the expression ##\dfrac{-11N}{6} + \dfrac{n_f}{3}##, note that this is for fermions in the fundamental representation of SU(N). With N = 2 we get that ##n_f < 11## then SU(2) gauge theory is asymptotical free

But this is would be a strange theory, the gauge-bosons would be massless, it would not be the weak-interaction we know of.

Anyway, back to the standard model.
1657522852073.png

The weak force indeed becomes weaker at shorter distances (higher energy scales) [note - the vertical axis is 1/strenght]
 
Last edited by a moderator:
  • Like
Likes ohwilleke, Orodruin and vanhees71
  • #3
drmalawi said:
The sign of the beta function for SU(N) (unbroken) with nf fermions is determined of the expression ##\dfrac{-11N}{6} + \dfrac{n_f}{3}##, note that this is for fermions in the fundamental representation of SU(N). With N = 2 we get that ##n_f < 11## then SU(2) gauge theory is asymptotical free

But this is would be a strange theory, the gauge-bosons would be massless, it would not be the weak-interaction we know of.

Anyway, back to the standard model.
View attachment 303989
The weak force indeed becomes weaker at shorter distances (higher energy scales) [note - the vertical axis is 1/strenght]
It is worth noting that the strength of the strong force isn't a straight line as the chart depicts in the very low infrared. Instead, the red line should surge up on the far left from close to 1 or maybe a bit more than 1 to the top left of the chart.

Instead gets stronger down to a peak energy scale, below which it gets weaker to zero or near zero in the limit of zero energy, which is why we say that the strong force becomes asymptotically free at some point within hadrons as shown here:

1657555979675.png


From:

I.L. Bogolubsky, et al., "Lattice gluodynamics computation of Landau-gauge Green's functions in the deep infrared" 676 Phys.Lett.B 69-73 (2009). https://doi.org/10.1016/j.physletb.2009.04.076 (conformed free preprint at https://arxiv.org/abs/0901.0736).

There is some debate in QCD phenomenology over whether the strong force coupling constant has a low energy limit of zero or of a small finite value more than zero, which has significant theoretical implications.

Thus, the strong force coupling constant is asymptotically zero both at zero and at infinite energy scales with a peak value at q2 = ca. 300 MeV.

Of course, we have no good reason to think that the weak force behaves in this manner, and the electromagnetic force certainly does not.
 
  • #4
ohwilleke said:
It is worth noting that the strength of the strong force isn't a straight line as the chart depicts in the very low infrared. Instead, the red line should surge up on the far left from close to 1 or maybe a bit more than 1 to the top left of the chart.
Logarithmic scale?
ohwilleke said:
There is some debate in QCD phenomenology over whether the strong force coupling constant has a low energy limit of zero or of a small finite value more than zero, which has significant theoretical implications.
indeed, this is beyond perturbation theory calculation
 
  • #5
drmalawi said:
indeed, this is beyond perturbation theory calculation
If the question is "Is the weak interaction asymptotically free?" it is necessary to consider all cases and not just those that are subject to perturbative theory calculation, although this is essentially a null set in the case of the weak force.

[Not trying to shout. Can't figure out how to unbold the text.]
[Fixed it for you (berkeman)] :wink:
 
Last edited by a moderator:
  • Wow
Likes malawi_glenn
  • #6
ohwilleke said:
If the question is "Is the weak interaction asymptotically free?" it is necessary to consider all cases and not just those that are subject to perturbative theory calculation, although this is essentially a null set in the case of the weak force.
Well, you started the QCD-spinn off...

If you want to discuss the validation of the running of the SU(2)L Gauge-coupling calculations for low energy regimes, that is another story.
 
  • #7
If you don't have confinement, what can asymptotic freedom possibly mean except a beta function such that the coupling constant goes down with increasing energy?

Further, at high energy, "the weak force" is a poor description - it is better to discuss hypercharge and weak isospin, and one increases with energy and the other decreases.
 
  • Like
Likes malawi_glenn

FAQ: Is the weak interaction asymptotically free?

What is the weak interaction?

The weak interaction is one of the four fundamental forces of nature, along with gravity, electromagnetism, and the strong interaction. It is responsible for radioactive decay and plays a crucial role in the structure and behavior of subatomic particles.

What does it mean for a force to be asymptotically free?

Asymptotic freedom refers to the behavior of a force at very high energies. In the case of the weak interaction, it means that at extremely high energies, the force becomes weaker and eventually disappears. This is in contrast to the strong interaction, which becomes stronger at high energies.

How does the weak interaction become asymptotically free?

The weak interaction becomes asymptotically free due to the effects of quantum chromodynamics (QCD). QCD is a theory that describes the strong interaction and predicts that at high energies, the force between particles becomes weaker, allowing particles to move more freely.

What evidence is there for the weak interaction being asymptotically free?

Experimental evidence for the asymptotic freedom of the weak interaction comes from particle accelerator experiments, such as those conducted at CERN in Switzerland. These experiments have shown that at high energies, the weak interaction becomes weaker, supporting the predictions of QCD.

What are the implications of the weak interaction being asymptotically free?

The asymptotic freedom of the weak interaction has important implications for our understanding of the behavior of subatomic particles and the structure of matter. It also plays a key role in the development of theories that aim to unify all four fundamental forces of nature.

Similar threads

Replies
2
Views
2K
Replies
6
Views
2K
Replies
9
Views
3K
Replies
27
Views
5K
Replies
6
Views
4K
Replies
6
Views
4K
Replies
26
Views
6K
Replies
10
Views
4K
Replies
7
Views
2K
Back
Top