A Custodial Symmetry in the Standard Model

Click For Summary
Custodial symmetry in the Standard Model is a global symmetry extending SU(2)L to SU(2)LxSU(2)R, respected by SU(2) gauge interactions and the Higgs self-potential but not by U(1) hypercharge interactions or Yukawa terms. After electroweak symmetry breaking (EWSB), this symmetry reduces to a residual form, often referred to as "custodial" symmetry. The three SU(2)L gauge bosons (W±, W3) form a triplet under this symmetry, leading to equal masses in the approximation where the symmetry holds. However, deviations occur due to the influence of hypercharge and Yukawa terms. Thus, custodial symmetry is considered "approximate" because it is only valid when certain terms in the Lagrangian are neglected.
Luca_Mantani
Messages
33
Reaction score
1
Hi,
I am reading about this symmetry but I'm struggling to have a deep understanding of it. Would somebody please explain this symmetry to me from a conceptual point of view?

Thanks in advance,
Luca
 
Physics news on Phys.org
Luca_Mantani said:
Hi,
I am reading about this symmetry but I'm struggling to have a deep understanding of it. Would somebody please explain this symmetry to me from a conceptual point of view?

Thanks in advance,
Luca

The question seems to me to general. Due you have a more specific question?
A few basics:

Custodial symmetry is a symmetry that is respected by the SU(2) gauge interactions and the higgs self potential. It is not respected by the U(1) hypercharge interactions and yukawa terms for the fermions.
It extends the SU(2)L in the standard model to SU(2)LxSU(2)R (global symmetry only). After electroweak symmetry breaking(EWSB) ,
SU(2)LxSU(2)R\rightarrowSU(2)L+R. This residual symmetry after EWSB is usually referred to as the "custodial" symmetry.

The three SU(2)L Gauge bosons (W\pm,W3) form a triplet under this symmetry, and thus their masses are equal in the approximation that the symmetry is valid. The deviation from that is due to hypercharge and yukawa terms.
 
  • Like
Likes ohwilleke and vanhees71
ofirg said:
The question seems to me to general. Due you have a more specific question?
A few basics:

Custodial symmetry is a symmetry that is respected by the SU(2) gauge interactions and the higgs self potential. It is not respected by the U(1) hypercharge interactions and yukawa terms for the fermions.
It extends the SU(2)L in the standard model to SU(2)LxSU(2)R (global symmetry only). After electroweak symmetry breaking(EWSB) ,
SU(2)LxSU(2)R\rightarrowSU(2)L+R. This residual symmetry after EWSB is usually referred to as the "custodial" symmetry.

The three SU(2)L Gauge bosons (W\pm,W3) form a triplet under this symmetry, and thus their masses are equal in the approximation that the symmetry is valid. The deviation from that is due to hypercharge and yukawa terms.
In which sense this is a "approximate" symmetry? How can a symmetry be approximate?
 
Luca_Mantani said:
In which sense this is a "approximate" symmetry? How can a symmetry be approximate?

In the sense that it is not respected by some terms in the Lagrangian. The symmetry is only present when those terms are neglected.
 
  • Like
Likes ohwilleke and Luca_Mantani
Thread 'Some confusion with the Binding Energy graph of atoms'

Thread 'Some confusion with the Binding Energy graph of atoms'

My question is about the following graph: I keep on reading that fusing atoms up until Fe-56 doesn’t cost energy and only releases binding energy. However, I understood that fusing atoms also require energy to overcome the positive charges of the protons. Where does that energy go after fusion? Does it go into the mass of the newly fused atom, escape as heat or is the released binding energy shown in the graph actually the net energy after subtracting the required fusion energy? I...
View full post »

Thread 'Calculate the energy-loss straggling of particles using LISE++'

Hello everyone, I am trying to calculate the energy loss and straggling of alpha particles with same energy, I used LISE++ to obtain the energy loss in every layer of the materials using Spectrometer Design of LISE++, but I can only calculate the energy-loss straggling layer by layer. Does anyone know the way to obtain the energy-loss straggling caused by every layer? Any help would be appreciated. J.
View full post »

Similar threads

I Are SM B & L conservation violations through sphalerons possible?
Replies
1
Views
2K
B Can a hypothetical atom be made out of positrons and electrons ?
Replies
9
Views
3K
I Spontaneous symmetry breaking from a QM perspective
Replies
34
Views
4K
I Does the New Muon g-2 Measurement Indicate Physics Beyond the Standard Model?
Replies
2
Views
2K
A Supplementary material for standard model
Replies
11
Views
2K
A Mass of Higgs, W and Z bosons going to zero at high energies
Replies
9
Views
3K
I The standard model from loop quantum gravity via spinors octonions
Replies
26
Views
5K
A What does gauge theory explain?
Replies
3
Views
3K
I Understanding crossing symmetry: inverse beta decay
Replies
4
Views
2K
I Why does a Lagrangian matter for the standard model?
Replies
3
Views
2K