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.

This question already has an answer here:

If the Higgs Boson is supposed to be the particle responsible for other particles having mass. How can it itself have a mass?

Is it not then a 'who came first, the chicken or the egg' situation?

share|improve this question

marked as duplicate by Danu, John Rennie, Ross Millikan, Alfred Centauri, BMS Jul 6 at 16:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1  
Possible duplicates: physics.stackexchange.com/q/30732/2451 and links therein. –  Qmechanic Jul 6 at 15:41

2 Answers 2

No.

The mass of most particles is not a problem. But for the force carriers, i.e. gauge bosons like the gluon or the W and Z bosons, it is a theorem that they must be (naively) massless. But we find that the W and Z bosons act as if they have a mass! The mechanism by which this mass arises is the Higgs mechanism, but we can have masses without it - just not on gauge bosons. Since the Higgs is no gauge boson, it can have mass without its own mechanism (fortunately).

share|improve this answer
1  
Also the fermions in the SM need a Higgs field with a non-vanishing vev to get their masses since the SM is a chiral theory. No mass term is allowed by the gauge invariance which thus need to be broken spontaneously to generate the fermion masses. –  TwoBs Jul 6 at 16:12

It is not the Higgs boson, but rather the Higgs field, that gives mass to the elementary particles, Higgs boson included. In fact, even in Higgsless theories, e.g. such as a technicolor, the W and Z get mass but there is no Higgs boson (although there is a composite Higgs field made of techniquarks).

Said this, the Higgs boson has a finite mass because its quartic selfcoupling $\lambda$, in the SM, is not vanishing, $V=\lambda(|H|^2-\frac{v^2}{2})^2$, which implies $m_h^2=2\lambda v^2$.

share|improve this answer

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