# Graviton vs Higgs boson

by teeninventor
Tags: boson, gravitation, graviton, higgs
 P: 2 How can we propose the existence of a massless graviton when the recent discovery of the higgs boson means that there is a Higgs field that endows everything with mass? Doesn't the higgs field disprove the graviton? In that case, we would now have no idea where gravity fundamentally comes from, right?
Mentor
P: 10,523
 when the recent discovery of the higgs boson means that there is a Higgs field that endows everything with mass?
No, that is not an implication of the Higgs mechanism. It gives a mass to quarks, leptons, the Z and W boson and the Higgs boson, but not to gluons, photons or (hypothetical) gravitons.
P: 2,194
 Quote by teeninventor In that case, we would now have no idea where gravity fundamentally comes from, right?
Well, we still have no idea about that...

P: 2

## Graviton vs Higgs boson

 Quote by mfb It gives a mass to quarks, leptons, the Z and W boson and the Higgs boson, but not to gluons, photons or (hypothetical) gravitons.
Any idea why photons and gluons are unaffected by the field? I know photons and gluons are considered massless, but by definition doesn't the field endow particles with mass? Is it the photons and gluons' extremely small size that allows it to pass through the field or between it's energy? That part is confusing to me
P: 3,719
 by definition doesn't the field endow particles with mass?
No, it does not. Please disabuse yourself of that notion.

The role played by the Higgs field is to break electroweak symmetry. As a side effect, it allows those particles that participate in the weak interaction to have mass. Gluons do not feel the weak force, and are for that reason unaffected. Photons remain massless basically because the Higgs field is neutral, and so the vacuum remains invariant under the electromagnetic gauge,

This topic has been discussed here many, many times before. If you look down at the bottom of this page you will find links to some of the earlier discussions.
P: 955
here is a post by Sam which clarifies the idea of photon not getting mass in electroweak symmetry breaking
 Quote by samalkhaiat In $SU(2)_{L}\times U(1)_{Y}$ theory, we have two coupling constants $g_{L}$ and $g_{Y}$, and four MASSLESS gauge fields: $W_{\mu}^{a}, \ a = 1,2,3$ and $B_{\mu}$. We can redefine these fields by introducing two electrically charged fields $$W^{\pm}_{\mu} = \frac{1}{\sqrt{2}}( W^{1}_{\mu} \mp i W^{2}_{\mu}),$$ and two neutral fields $$Z_{\mu} = W^{3}_{\mu}\cos \theta - B_{\mu} \sin \theta$$ $$A_{\mu} = W^{3}_{\mu} \sin \theta + B_{\mu} \cos \theta$$ We still have no photon in here, because the gauge group is not $U(1)_{em}$, all fields are still massless and (more important) the two couplings $g_{L}$ and $g_{Y}$ are unrelated. To break $SU(2)_{L}\times U(1)_{Y}$ down to $U(1)_{em}$, we need to introduce a set of scalar fields $\Phi$ which has $U(1)_{em}$ invariant non-zero vacuum expectation value $< \Phi > = v$, i.e. it vanishes under the action of the $U(1)_{em}$ generator (the electric charge) $$Q_{em}< \Phi > = 0. \ \ \ \ (1)$$ Next, we introduce a small perturbation $H(x)/ \sqrt{2}$ around the VEV of the scalar field $< \Phi >$. This will provides masses to ALL four gauge fields $W^{\pm}_{\mu}, Z_{\mu}$ and $A_{\mu}$. So, in order to satisfy eq(1) one of the neutral fields must remain massless, so that it can be identified with the gauge field of the (unbroken) $U(1)_{em}$ group, i.e. the photon. This happens for $A_{\mu}$ provided that we CHOOSE the couplings such that $$g_{Y} = g_{L} \tan \theta .$$ Sam