Higgs field

  • Thread starter kurious
  • Start date
  • #1
641
0
If the Higgs theory is correct shouldn't it be able to predict the rest masses of all particles?
 

Answers and Replies

  • #2
selfAdjoint
Staff Emeritus
Gold Member
Dearly Missed
6,786
9
I don't think they can predict the values of the masses from the Higgs mechanism. Just that the symmetry is broken and mass is acquired.

But if I'm wrong about this, I welcome correction.
 
  • #3
ahrkron
Staff Emeritus
Gold Member
736
1
In a shell, it goes like this:

While developing what we now call the "standard model", people got to a point where everything worked great, except for the inconvenient fact that all particles were predicted to be massless. That is of course not a good thing, since particle's masses were already known to be different from zero.

The way out was what we now call the "Higgs mechanism", by which particles acquire their masses via their interaction with the Higgs field. If that is the case, then each particle's mass tells you how strongly the said particle couples to the Higgs field, and you get a prediction: since the Higgs field is also a QM field, there should be possible to detect a corresponding particle.

If the current model of particle physics is correct, there should be a Higgs particle. Which has not been detected yet.
 
  • #4
vanesch
Staff Emeritus
Science Advisor
Gold Member
5,028
17
kurious said:
If the Higgs theory is correct shouldn't it be able to predict the rest masses of all particles?

Predict, no. In the standard model, the coupling of the higgs field to each particle predicts the mass of that particle... but each of these couplings is of unknown strength! So it just displaces the problem from knowing the masses to knowing the coupling constants. The reason people worked out the higgs mechanism is not the prediction of the mass values, but the respect of the symmetries thought to be necessary, because the normal way of putting in masses (just writing a mass term m^2 phi^2) didn't allow the right symmetries.

cheers,
patrick
 
  • #5
641
0
Coupling constants are usually associated with probabilities.
Is it right to say a particle has a probability of having a certain mass?
 
  • #6
2,425
7
kurious said:
Coupling constants are usually associated with probabilities.
What do you mean ? The coupling constants are parameters of the theory used to calculate probabilites (amplitudes). The coupling might evolve with the energy scale. But they are not probabilities :confused:
 
  • #7
641
0
Since the rest mass of one particle can be equivalent in magnitude to the mass of another faster moving particle which has a smaller rest mass,
can't the coupling constants be related to one particle moving at different speeds?
 
  • #8
2,425
7
I never heard of this idea, and I have a hard time to actually understand your proposal. It is interesting. Althoug, one should have to deal not only with mass, but with all quantum numbers in order to calculate a coupling.
 
  • #9
3,763
9
kurious said:
Since the rest mass of one particle can be equivalent in magnitude to the mass of another faster moving particle which has a smaller rest mass,
can't the coupling constants be related to one particle moving at different speeds?


yes for example the coupling constant for the strong force depends on the speed at which the quarks move, or the distance scales at which we look at them. It is the fact that when the energies become higher, the coupling constant get's weaker. Slow moving quarks are more stricktly bound to eachother then fast moving quarks. This is called the asymptotic freedom.

regards
marlon
 
  • #10
2,425
7
Are you absolutely positive on that Marlon ?
If you look at the evolution of parton densities in the nucleon (DGLAP & ERBL) you should notive that at higher [tex]Q^2[/tex] the densities increase at small [tex]x_{Bjorken} = \frac{Q^2}{2 M \nu} [/tex].

The asymptotic freedom says that the coupling goes to zero with [tex]Q^2[/tex] increasing, not the speed. The coupling constant goes to zero when the quarks are close to each other, or at higher energies. But I am not certain (I have not heard about) relative speed interpretation.

At higher [tex]Q^2[/tex], one probes the structure at smaller distances at observes more and more parton pair fluctuations, which are located at smaller and smaller [tex]x_{Bjorken}[/tex] which is the fraction of momentum carried by the parton. As far as I understand, if the fraction of momentum is smaller, the speed is smaller too. I am not too sure about this. The problem is, their is no satisfying way of defining the mass of the quarks which should depend on [tex]Q^2[/tex] too. So talking about relative speed, I am not too sure.
 
  • #11
3,763
9
humanino said:
Are you absolutely positive on that Marlon ?
If you look at the evolution of parton densities in the nucleon (DGLAP & ERBL) you should notive that at higher [tex]Q^2[/tex] the densities increase at small [tex]x_{Bjorken} = \frac{Q^2}{2 M \nu} [/tex].

The asymptotic freedom says that the coupling goes to zero with [tex]Q^2[/tex] increasing, not the speed. The coupling constant goes to zero when the quarks are close to each other, or at higher energies. But I am not certain (I have not heard about) relative speed interpretation.

At higher [tex]Q^2[/tex], one probes the structure at smaller distances at observes more and more parton pair fluctuations, which are located at smaller and smaller [tex]x_{Bjorken}[/tex] which is the fraction of momentum carried by the parton. As far as I understand, if the fraction of momentum is smaller, the speed is smaller too. I am not too sure about this. The problem is, their is no satisfying way of defining the mass of the quarks which should depend on [tex]Q^2[/tex] too. So talking about relative speed, I am not too sure.

Humanino, the example you give is entirely true, yet it IS also a fact that quarks in accelerators are more "loosly" bound to pther quarks then when they are in the QCD-groundstate. Asymptotic freedom states that the coupling constant get's stronger at lower energies or at bigger distances (Heisenberg uncertainty). Part of that energy can be gained or lost through a change in the kinetic energy.

regards
marlon
ps : i have posted a problem of resistances in the homework-department. I don't seem to find it , Humanino, perhaps you wanna give it a go...Thanks
 
  • #12
2,425
7
Yes, now I understand. You are right Marlon, thanks ! I am not too familiar with colliders, I am working on lepton scattering on fixed nucleon target.
 
  • #13
Haelfix
Science Advisor
1,955
222
Thats more or less correct, the Higgs mechanism was invoked to preserve guage symetry not to predict masses.

Otoh, it does place natural orders of magnitudes on what masses particles should have. Usually, there is finetuning constants that are experimentally determined in order to have things work out. However, when the ratio of those constants turns out to be very large (see for instance the Neutrino mass vs the expected electroweak symmetry breaking scale) then its a hint that there is something else involved.

To get models that are more predictive, you have to go to something like SUSY. There the masses are usually constrained much more severly (for example, if you know such and such a mass for particle x and y.. that implies particle z is fixed).. In stringy physics, there are additional constraints (though far more degrees of freedom elsewhere)

But thats the state of the art.. Unfortunately we don't have a theory that has identically zero quantities that need to be found via semiempirical measurements. But then again, thats the fact of life even in classical theories (why does the mass of the electron have the value it has).
 
  • #14
641
0
Haelfix:
Why does the electron have the mass it does?

Kurious:
Electron and proton have same charge and spin but different masses.
This can only mean that the electron and proton have another property new to science (as colour once was) that has a different magnitude in each.
This is why I don't believe Higgs theory can be complete.
 
  • #15
2,425
7
yes, they do have another property, but it is not new : it is about color ! The electron is a trivial (white) singlet under color transformation, because it has no component link to color. You can ever exchange blue and green and red the way you like, the electron won't care.
On the other hand, the proton is a non trivial singlet, because the quarks that constitute it are in an antisymmetric state invariant under color (non trivial singlet).

This has nothing to do with Higgs. This very much explains to me why the proton is so heavy (take a look at mass gap story : https://www.physicsforums.com/showthread.php?t=38964

Higgs theory seems to be fine to me. It has succesfully been applied in condensed matter, but its relevance in fundamental interactions is not yet established.
 

Related Threads on Higgs field

  • Last Post
Replies
9
Views
978
  • Last Post
Replies
8
Views
3K
Replies
16
Views
779
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
5
Views
776
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
689
  • Last Post
Replies
1
Views
2K
Top