# Understanding the need for the Higgs Field

1. ### rpthomps

44
Hello,

I am a layman who just read Sean Carroll's The Particle at the End of the Universe and I am trying to understand what I think is the crux of the book, the need for the Higg's boson according to the Standard model. This is my thought process:

1. The weak interaction only interacts with left-handed fermions
2. The above implies the weak bosons are massless because only bosons which are massless can travel at light speed and therefore observe only one type of helicity
3. Experimentally, however the weak bosons are shown to have mass due to the their limited range of interaction.
4. Therefore, mass can't be inherent to the W and Z bosons but rather acquired through the interaction of another field. (i.e. the Higg's Field)

Is my thought process even remotely close?

Thanks

Ryan

2. ### fzero

2,601
This is true.

The fact that the weak interaction only involves left-handed fermions does not itself place a restriction on the mass or helicity of the gauge bosons. Rather, the restriction on the mass comes from something called renormalizability of the gauge theory. Renormalizability, in one sense, refers to the fact that observables of a quantum field theory often involve certain divergences related to momentum integrals. These integrals can diverge because we integrate over all momenta with magnitudes from zero to infinity. The standard methods for making sense of these divergences to extract finite, physical observables comes under the name renormalization.

However, it can be shown that renormalization of a gauge theory requires that the gauge symmetry be a good symmetry at high energies (which the source of the infinities). A mass term for a gauge boson explicitly breaks the gauge symmetry, so the quantum field theory of a massive gauge boson is not renormalizable. The electroweak theory evades this restriction because the gauge bosons are only massive below the scale where the gauge symmetry is broken. At high energies, the gauge bosons are massless and the full gauge symmetry is restored, allowing for renormalization.

This is true.

Again, it is the requirement that the theory be renormalizable that leads to the conclusion that the W and Z bosons can't have an explicit mass, but can have a mass generated at low energy by the Higgs mechanism.

3. ### Bill_K

4,160
The W boson only interacts with left-handed fermions. The Z boson interacts to a certain extent with right-handed fermions as well.

4. ### rpthomps

44
Thanks a lot for your response. I well need to digest your response a little. I appreciate it all the same

5. ### kurros

383
But, to elaborate, this is because it isn't a purely "weak" object, in that it is a mixture of both SU(2) and U(1)_Y gauge bosons. This is partially just the semantics of what one calls the weak force though.

6. ### Bill_K

4,160
Of course it is! The Z is a weak boson, and deserves the name every bit as much as the W does. It interacts via the weak force, with weak coupling constants g and g'. And the symmetry group it gauges is weak isospin X weak hypercharge.

(But feel free to substitute "electroweak" everywhere you see "weak".)

7. ### kurros

383
Haha, but if you try to make this argument then the photon is also a "weak" gauge boson. So it seems to me the substitution 'electroweak' is indeed necessary. But anyway, if we want to say things like "the weak force only interacts with left-handed fermions" then it seems like we are talking about SU(2)_L, and to me it seems more intuitive to then explain the Z's interaction with right handed fermions as coming from the "not weak" part of the SM lagrangrian. But it's just words, so whatever. I don't know what is less confusing for people who don't already know what is really meant.

Last edited: Nov 23, 2013

### Staff: Mentor

That will fail at "with weak coupling constants g and g'". You can use the photon with "electroweak", but without that unification the coupling constant is the fine-structure constant and the coupled charge is the electric charge.