I The Higgs Field and Fermion Generations: Stability and Mass

kodama
Messages
1,074
Reaction score
144
if there were no higgs field, would the second and third generation of fermions, such as the top quark, be exactly the same mass as first generation?

is the coupling between the top quark and the higgs field the sole reason the top quark is heaviest SM particle?

is there a reason the top quark, and second and third generation fermions couple to the higgs field more strongly than the first generation?

if there were no higgs field, would second and third generation fermions be stable and long-lived as first generation?
 
Physics news on Phys.org
Could you elaborate about what alternative mass mechanism are you thinking about?
 
arivero said:
Could you elaborate about what alternative mass mechanism are you thinking about?
why do third generation fermions interact more strongly with the higgs field than second, and second first. why does the top quark interact with the higgs field more strongly than an electron?
 
kodama said:
why do third generation fermions interact more strongly with the higgs field than second, and second first. why does the top quark interact with the higgs field more strongly than an electron?

The couplings of the Higgs field are free parameters in the Standard Model and directly related to the masses of the quarks. Therefore, a quark with a larger mass will interact more strongly with the Higgs field. In effect, it is the interaction with the Higgs field that provides the mass, implying that a quark that interacts more strongly with the Higgs field will have a larger mass.
 
  • Like
Likes vanhees71
i understand it is a free parameter, but are there any physics explanations why second and third generation fermions interact more strongly with the higgs, such as a degree of freedom

is there any reason why second and third generation fermions couldn't be lighter and couple less strongly than first generation to the higgs, since it is a free parameter and can take any value
 
kodama said:
i understand it is a free parameter, but are there any physics explanations why second and third generation fermions interact more strongly with the higgs, such as a degree of freedom

is there any reason why second and third generation fermions couldn't be lighter and couple less strongly than first generation to the higgs, since it is a free parameter and can take any value
The lighter quarks are what we define as the first generation.
 
  • Like
Likes ohwilleke
Orodruin said:
The lighter quarks are what we define as the first generation.
since higgs coupling is a free parameter isn't it certainly conceivable that all 3 generations have the same coupling to the higgs and therefore same masses
 
kodama said:
since higgs coupling is a free parameter isn't it certainly conceivable that all 3 generations have the same coupling to the higgs and therefore same masses
Yes, but this would be a rather strange situation where we would probably suspect that some symmetry was in play because there is no a priori reason for this to be the case.
 
  • #10
Orodruin said:
Yes, but this would be a rather strange situation where we would probably suspect that some symmetry was in play because there is no a priori reason for this to be the case.

In fact all the masses are almost zero compared to the top quark, so we could suspect a symmetry where all masses are null except the top quark, but this scenario is not asked/developed frequently, neither in literature nor even here in forums.

EDIT: AFAIK, of course
 
Last edited:
  • Like
Likes ohwilleke
  • #11
Given that the W bosons are the means by which quarks and charged leptons, at least, change from one flavor to another at frequencies that show a some crude relationship to the mass differences involved, surely the Higgs boson couplings and the CKM matrix that governs W boson quark flavor transformations, have some deeper connection to which we are not yet privy.

Also, it isn't obvious that it is possible for particles to be distinct from each other in some means other than their propensity to change from one type to another (which would have no observable consequences if all other properties were the same) if they do not have some different properties, so it stands to reason that given that they are identical in everything else, that they ought to have different masses, and are assignment of those masses to particular generations in order of mass is something that can be done without loss of generality in every case where there are distinct masses.
 
  • #12
kodama said:
since higgs coupling is a free parameter isn't it certainly conceivable that all 3 generations have the same coupling to the higgs and therefore same masses

Not if the Higgs coupling is the source of masses in the universe. If all the couplings were the same the Higgs boson and field would be much simpler mathematically.
 
  • #13
ohwilleke said:
Also, it isn't obvious that it is possible for particles to be distinct from each other in some means other than their propensity to change from one type to another (which would have no observable consequences if all other properties were the same)

Pauli exclusion principle effects should be observable: in the "Higgsless Universe", electrons and muons would look the same, but you'd notice that sometimes you can cram two of them into the same state, and sometimes you can't.
 
  • #14
Hmm if all the three masses are equal, does the CKM matrix still has the same physical components? Or does some component become unphysical?
 
  • #15
ohwilleke said:
surely the Higgs boson couplings and the CKM matrix that governs W boson quark flavor transformations, have some deeper connection to which we are not yet privy.
The mixing in the quark sector is directly given by the mismatch between the left-handed transformations that diagonalise the up and down type Yukawa couplings.

arivero said:
Hmm if all the three masses are equal, does the CKM matrix still has the same physical components? Or does some component become unphysical?
If all masses are equal the entire CKM matrix is unphysical. (In fact, it is sufficient that either the up or down type quarks have the same masses.)
 
  • #16
Orodruin said:
Yes, but this would be a rather strange situation where we would probably suspect that some symmetry was in play because there is no a priori reason for this to be the case.

what would be a reason a priori they would be different? how can you tell apart a massless muon from a massless electron or tau
 
  • #17
ohwilleke said:
Not if the Higgs coupling is the source of masses in the universe. If all the couplings were the same the Higgs boson and field would be much simpler mathematically.

in a higgless universe is it possible that there is NO difference between an electron tau or muon? they are all the same particle, the only difference is that sometimes the higgs is more strongly attracted to muon or tau, or that a muon or tau are quantized excited versions of electron
 
  • #18
nikkkom said:
electrons and muons would look the same, but you'd notice that sometimes you can cram two of them into the same state, and sometimes you can't.

Not exactly. What would be observed is you can get 3 electrons into the same state. Eventually it would be explained as a hidden quantum number called "infracolor" or "ultracolor" or something like that, in analogy with QCD color.

Orodruin said:
If all masses are equal the entire CKM matrix is unphysical.

I wouldn't say "unphysical". "Trivial" and "unnecessary" seem to be more descriptive.
 
  • #19
Vanadium 50 said:
Not exactly. What would be observed is you can get 3 electrons into the same state. Eventually it would be explained as a hidden quantum number called "infracolor" or "ultracolor" or something like that, in analogy with QCD color.
It is not gauge, so flavor is still a good name.
I wouldn't say "unphysical". "Trivial" and "unnecessary" seem to be more descriptive.
My vote is with "Trivial".

Is it right to think that while we need three different masses to have a full non trivial CKM matrix, we do not need six?
 
  • #20
arivero said:
It is not gauge

I am not sure that people would immediately conclude that. I suspect they would start putting together "ultraweak" theories to try and explain flavor. We sort of do this today with topcolor-style theories.

arivero said:
Is it right to think that while we need three different masses to have a full non trivial CKM matrix, we do not need six?

Has to be the right three. All the +2/3 or all the -1/3. That allows you to change the basis so the CKM matrix is the identity. If you make e.g. two of the +2/3's degenerate with one of the -1/3's, you still have a nontrivial matrix. However, that matrix will be real, so there will be no CP violation.
 
  • #21
Vanadium 50 said:
I wouldn't say "unphysical". "Trivial" and "unnecessary" seem to be more descriptive.
I think unphysical nails it. The precedent here being that we call the Majorana phases unphysical as they can be reabsorbed in the definitions of the fields, the entire CKM matrix would be unphysical as you could reabsorb it in the field definitions or chose to work in a basis with an arbitrary CKM matrix and it would make no observable difference.

Vanadium 50 said:
Has to be the right three. All the +2/3 or all the -1/3. That allows you to change the basis so the CKM matrix is the identity. If you make e.g. two of the +2/3's degenerate with one of the -1/3's, you still have a nontrivial matrix. However, that matrix will be real, so there will be no CP violation.
There is no requirement that a -1/3 has to be degenerate with the degenerate 2/3 pair. It is sufficient that a single 2D rotation in flavor space commutes with the up type Yukawas.
 
  • #22
The question was 3 mass degeneracies. There are two ways to do this: 3 and 0 and 2 and 1 - i.e. all of the +2/3s, all of the -1/3's or one degeneracy within a charge and one across charges. In the 3-0 case you can adjust your basis so that the CKM matric is diagonal. In the 2-1 case you can't make it diagonal, but it is always real. That's all I am saying.
 
  • #23
Vanadium 50 said:
In the 2-1 case you can't make it diagonal, but it is always real.
To be specific: It does not need to be real, but you can absorb any of the complex phases in field redefinitions and even if you do not it does not the phases are unphysical and therefore do not lead to any CP violation
 

Similar threads

Replies
2
Views
2K
Replies
8
Views
2K
Replies
13
Views
4K
Replies
11
Views
3K
Replies
5
Views
2K
Replies
13
Views
3K
Back
Top