# Higgs - pls help with feynman diagram

1. Feb 3, 2010

### Dmitry67

I need help - Higgs for idiots:

So, how a particle which have no mass without higgs (so it travels at c, and on feynman diagram we would draw its worldline at 45deg) can slow down or even be at rest when Higgs is switched on?

Could anyone draw a feynman diagram?

2. Feb 3, 2010

### ansgar

feynman diagram are not spacetime diagrams thus the lines are NOT worldlines.

3. Feb 4, 2010

### naima

I think that massification by interacting with the higgs field is a not ending process. If you could switch the fields electrons would loose their mass.
You can find a diagram on page 6 of http://arxiv.org/PS_cache/hep-ph/pdf/9803/9803257v2.pdf" [Broken]

Last edited by a moderator: May 4, 2017
4. Feb 4, 2010

### turin

What do you mean by this phrase? Are you talking about cosmology or something?

5. Feb 5, 2010

### Dmitry67

yes, when temperature is low enough so some particles gain mass

6. Feb 5, 2010

### turin

Ah, OK. Think of fields rather than particles. Then there really isn't a notion of "slowing down".

A very simplistic way of looking at it is that, say some matter field is coupled to the Higgs field. At "high temperature", while there are effects on the matter field due to the interaction, these effects are in random SU(2)W directions. Once the "temperature" lowers to some point, the Higgs potential pushes the Higgs field away from |φ|=0 where there is a local maximum, since the energy becomes less sufficient to push the Higgs field "over the hump". As the energy reduces far below this level, a particular SU(2)W direction is chosen (because there is not enough energy to supply the derivative terms for suffiicient variations in the SU(2)W direction). When the Higgs is pushed in this direction, it "drags" the coupled matter field with it. Since the coupling is Yukawa, this "dragging" appears in the Lagrangian as a product of the squared matter field with some constant value of the Higgs that occurs at the minimum of the Higgs potential. Such terms have the same form as mass terms.

So, the "acquisition of mass" is not manifested as "slowing down", but rather as the appearance of a "mass term".

7. Feb 6, 2010

### Dmitry67

Thank you

Another question: Mass of Higgs boson itself - is it also given by Higgs mechanism?

8. Feb 8, 2010

### turin

Sort of.

Yes, in the sense that it is NOT given by the |φ|2 in the Lagrangian, but rather is derived from the shifted potential.

No, in the sense that "quantum corrections" can be huge. In other words, even after considering the Higgs mechanism at a naive classical level, the derived mass can be quite different from the actual mass.

9. Feb 10, 2010

### Frame Dragger

If the mass of the Higgs boson isn't a result of the boson, but the mechanism... this is starting to sound endlessly reducible.

I love and hate the idea of scalar fields, and I respect and believe that the Higgs Mechanism is probably a good description of how other particles are imparted with mass. When it comes to self-interaction of the field however, I start to find it spooky. Of course that doesn't change matters, but still... hrm.

10. Feb 10, 2010

### Dmitry67

How it is different of 'dressed' particles where their 'dress' consists of all sorts of virtual particles, including themselves, all other particles, including the ones we don't know yet?

11. Feb 10, 2010

### Frame Dragger

It isn't. They're both weirder than hell. I'm not saying they're not potentially useful or true... just weird.

12. Feb 10, 2010

### turin

Your confusion probably arises from an imprecise notion of mass and "dressing". I will assume that, by "dressing", you are referring to the higher order Feynman diagrams as compared to a given "tree-level". In particular, a tree-level propagator is "dressed" with loops, but there remains only one input line and one output line. If this is what you are talking about, then the relevant notion of mass is the pole mass of the Fourier transform of the 2-point correlation function when the input time is taken to -∞ and the output time is taken to +∞; i.e. the pole of the propagator. This is actually one of the three ways of defining the mass of the particle.

I will use the photon as a clean example. You can speak of the tree-level photon propagator, and it has a naive pole at E2-p2=0. This means that the "undressed", or "bare", photon is massless. However, quantum theory tells us that we must add every possible contribution to a given process, and for the photon propagator that means we have to add the loop diagrams (the vacuum polarization etc.). In general, loop diagrams can "shift" mass poles away from their tree-level locations. However, the Standard Model imposes electromagnetic gauge invariance and Lorentz invariance. These two requirements restrict the form of the full propagator such that the pole is not allowed to shift. (See, for example, Peskin & Schroeder, Chapter 7, Section 5, regarding "Renormalization of the Electric Charge".) We say that the masslessness of the photon is "protected by (electromagnetic gauge) symmetry". (I don't know if anyone even bothers to mention the Lorentz symmetry, even though it is essential for the protection of the zero mass pole. I think that particle physicist either actively try to test for violations of Lorentz symmetry, or they just take it for granted.)

Of course, you could argue the restrictions of invariance under the various groups themselves (and physics isn't science unless we always do so). Asking these kinds of questions is rather contextual. One unavoidable fact remains: the photon is massless (or at least very very very ... very light). So, if you argue away the symmetry that elegantly protects this fact, then you have quite an undertaking. The photon is special. It enjoys the above-mentioned protection and also remains free of (direct) self-interactions or interactions with the Higgs field (well, sort of.). This nature of the photon is the big clue that leads to the notion of gauge symmetry in the first place, not the other way around.

The other gauge bosons do not provide such a clean demonstration of how the symmetry protects their mass. Furthermore, most other particles, including some of the gauge bosons, are given some coupling to the Higgs field in the Standard Model, and this is so because they have experimentally nonzero masses. (Neutrinos are another can of worms. Gluons are supposedly massless, but I don't know how seriously I should take gluons.)

However, if you take the symmetry principle seriously, then you can make the same basic argument regarding the other particles that acquire their mass from the Higgs mechanism: their masslessness is protected by symmetry, and they acquire their mass through the Higgs mechanism that breaks one of the symmetries. In fact, the Higgs mechanism breaks the SU(2)W symmetry directly, and this is why it "gives" mass to the weak bosons. Then, there is a clever interplay between the two chiralities of fermions in the Yukawa interactions, so that the Higgs mechanism breaks the chiral symmetry of the fermions in a sort of indirect way.

It should be noted that the Higgs boson is not included in this list of protected particles. First of all, the Higgs field is set up to be strange from the beginning, having a |φ|2 term in the Lagrangian with a positive coefficient! You may be tempted to interpret that as an indication of imaginary mass! I don't know a good way to interpret it "at tree-level". At any rate, I wouldn't say that the bare Higgs boson is massless, so it is hard to say whether or not the Higgs mechanism is responsible for the Higgs mass. The two concepts are inextricable. I suppose that you could talk about a dynamical Higgs potential such that the coefficient of the |φ|2 can become negative (or could have been negative a long time ago). Then, you could naively think of that as the mass term. However, the "quantum corrections" will destroy this notion regardless of what the Higgs mechanism does, because the mass is not protected by a symmetry.

If you want to consider such things, then all bets are off.

Last edited: Feb 10, 2010
13. Feb 10, 2010

### Dmitry67

turin, at first, thank you for so detailed answer!
From my layman perspective you have very deep understanding of that stuff... so I wanted to ask another related question.

How do you interpret Scharnhorst effect?

1. Light moves FTL between plates.

2. c in Minkowsky metrics is slightly > than pysical spped of light. Light in vacuum is slightly slowed down because of the vacuum interactions, like in water or glass.

both interpretations lead to some problems.

14. Feb 11, 2010

### turin

I have never heard of this effect; I'll have to look into it. Thank you for bringing it to my attention.

There are some members on this forum who like to get onto me when I relay my interpretation of such things. I will just point out that this claim is ambiguous. Not the most exciting comment, but the safest one for me to make here.

The only vacuum interaction of light that comes to my mind is the so-called vacuum polarization, and as I tried to explain above, this does not "slow down photons". The effect of vacuum polarization is to slightly reduce effective charge as distance increases. For two points in space with macroscopic separation, d, the time delay of a light pulse is given by t=d/c as far as I know. Of course, light passing through matter is an entirely different ball game.