Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The central importance of Higgs boson would be that the Higgs mechanism gives rest mass to fundamental particles. It seems like a very natural argument that fundamental particles need to be given rest mass by a field interaction because as something fundamental (or a perturbation in a field as its often described), it can't have any properties beyond what a QFT can explain through interactions with a field.

What I don't understand is the role of matter-energy in this story. If the interaction of a particle with a field gave that particle rest mass, it seems logical that the field in the vicinity of the particle is in an excited state, which, integrated over the area of influence, is exactly equal to the rest mass of the particle. Why wouldn't this happen with interactions with other fields? Why does the (baseline) interaction of fundamental particles with the Higgs field create mass while interactions with the electric field, for instance, won't? In fact, it seems like we have lots of fields that could account for particle mass. Are those all just zero net-energy fields? What makes the Higgs capable of this while the others are not?

On this point, if the electric field is mediated by the photon force carrier, then how can the electric field from a charged particle have anything less than infinite energy? I believe it has zero energy, but I lack the tools to argue this. Are the photons mediating an electrostatic interaction countable? Or do they exist in a less real sense? The magnetic field, on the other hand, obviously has energy as evidenced by the existence of inductors. The gravitational field is the strangest of all since gravitational flux is proportional to mass itself. Rest mass is proportional to interaction with the Higgs field, and mass interacts proportionally with the gravitational field, but yet the Higgs field is not the gravitational field. Can't mass come from any of the fields? Surely, if negative charge orbits closely to a positive charge, the system's mass is less due to a phenomenon associated with the electromagnetic field. Is this all the correct perspective, or close?

share|improve this question
    
How does your rest mass translate into total mass? –  Argus Jul 6 '12 at 1:19
    
@Argus Well, I was speaking of rest mass of fundamental particles. I need to be careful of that, because other people frequently get corrected when speaking of the proton mass, for instance. Total mass has contributions from the fundamental particle masses as well as kinetic and (I think) field potential energy. I think the idea would be that fundamental particle rest mass would be embodied in a higher energy state of the Higgs field, and I'm speculating here about the simultaneous presence of mass from other field excitations in configurations more complicated than fundamental particles. –  AlanSE Jul 6 '12 at 1:36
add comment

1 Answer 1

up vote 1 down vote accepted

I guess this was asked a long time ago but maybe this will still help.

The essential difference between the Higgs field and other fields (for our purpose) is that only the Higgs has a nonzero VEV (vacuum expectation value). I'll try to show why this matters for your question.

A QFT is described by it's Lagrangian. An explicit mass term for a (scalar) field $\phi$ with mass $m$ is

$\frac{1}{2}m^2\phi^2 $.

Interaction terms look something like

$g H^2 \phi^2$,

where here $g$ is some coupling constant and $H$ is another field. Now usually the lowest energy state - the vacuum - will have $H=0$ (along with $\phi=0$, by the way, but right now we are focusing on $H$ as the coefficient of the field of interest, $\phi$). Then interactions occur via excitations in the field $H$ around this minimum at $H=0$. Such is the case if we imagine $\phi$ to be an electron and $H$ to be the electromagnetic field, for instance (I am totally glossing over the differences between spin-0, 1/2, and 1 fields but that is irrelevant for now) - the EM field has a VEV of zero and an excitation can occur only if some energy/momentum is given to it from another field - say, the electron field in the case of an electron radiating a photon.

However, what makes the Higgs field different is that it does not have a minimum of its energy at $H=0$ but rather at about $H=246$ GeV, call this $H=v$. Then fluctuations and excitations in the $H$ field occur not around zero but around $v$. It is useful to separate out these two parts and define $H=v+h$, where $h$ is the fluctuation around $v$. Substituting this into the equation above obviously gives rise to a term

$gv^2\phi^2$,

where $v$ is just our constant VEV. But this is exactly the same form as a mass term, with $gv^2$ playing the role of $\frac{1}{2}m^2$. Thus the nonzero VEV of the field $H$ makes it "look like" the particle associated with the field $\phi$ has a mass $m=v\sqrt{2g}$.

There is of course also a term

$gh^2\phi^2$

and I suppose a $2ghv\phi^2$ term in this case. But the point is: you are right that there is some energy in excitations of whatever fields; however, in the case of the Higgs field these are excitations around a nonzero VEV while in every other field they are around zero. Really, any term with greater than two powers of fields is an interaction and the only "true" mass term is one like the first thing I wrote. Everything else is an interaction and not a mass. It's just that matter fields are always interacting with the Higgs field since it always has a nonzero value (this is why people sometimes describe the Higgs as like molasses a marble has to move through or something like that). Because these interactions are always happening, it becomes nearly indistinguishable from the case where the particle was given an explicit mass in the Lagrangian.

I hope I addressed your question.

share|improve this answer
    
Oh, and also, regarding the infinite energy bit- check out the topic of renormalization. There are infinities all over the place in QFT. –  gn0m0n Dec 4 '12 at 5:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.