# On a point of logic in the Higgs vs strong force origin of mass

Tags: force, higgs, logic, mass, origin, point, strong
 P: 428 Recently, in the Scientific American blog http://blogs.scientificamerican.com/...ysics_20121130 the author criticizes Time Magazine for its science reporting, which of course is nothing new, but one of the criticisms is that Time's author wrote that without Higgs there wouldn't be any mass around (Sentence 1), and the blog counters by pointing out that most mass in the universe is due to the strong force, not Higgs. I am not sure that the blog's counter, although correct, actually destroys Time's contention, in that without Higgs, the elementary particles would not have mass, including the gluons and quarks upon which the strong force acts -- and therefore the question is whether the strong force could act on massless particles. If not, then there would be nothing around (without Higgs) for the strong force to act on, and hence nothing would have mass. (Whereupon Time's contention would be correct, even if it does give the wrong impression.) I would be happy to be corrected on this; I would not want to be seen as a defender of Time magazine.
 Sci Advisor P: 3,447 The Higgs is an essential part of nature, and so asking what nature would be like without it is a bit meaningless. However regarding the strong force, its action doesn't depend on quarks and gluons having mass. Gluons, in fact, ARE massless. And the two most common quarks, the up and down, nearly so.
 P: 925 just a question:why not photon interact with higgs field to get a mass?
P: 3,447

## On a point of logic in the Higgs vs strong force origin of mass

In a single sentence, because the Higgs field is electrically neutral.

In more detail... electroweak theory has the symmetry group SU(2) x U(1) (Isospin T x Hypercharge Y) with four generators, and corresponding to each generator is a gauge boson (the W's, the Z and the photon).

The Higgs field breaks this symmetry by acquiring a vacuum expectation value. Each symmetry transformation that is broken results in a mass for the corresponding boson. But if the vacuum is still left invariant by some subgroup of gauge transformations, the gauge bosons associated with that subgroup will remain massless.

The generator corresponding to electric charge is Q = T3 + Y/2. So we make the simplest choice and assume the Higgs field to be an isospin doublet, T = 1/2, with hypercharge Y = 1, and assume that its vacuum expectation value φ0 = (0 v) has only a T3 = -1/2 component.

Then Qφ0 = (T3 + Y/2)φ0 = 0, the vacuum is invariant under electromagnetic gauge transformations ("electrically neutral"), and the photon remains massless!
 P: 428 Thanks very much, Bill K. That helped me understand these issues quite a lot. (If you ever quit whatever profession you are in, you should go into popular science writing. Most explanations about Higgs etc. are either oversimplified or too technical, with nothing in between for those with a moderate physics background.)
P: 925
 Quote by Bill_K In a single sentence, because the Higgs field is electrically neutral. In more detail... electroweak theory has the symmetry group SU(2) x U(1) (Isospin T x Hypercharge Y) with four generators, and corresponding to each generator is a gauge boson (the W's, the Z and the photon). The Higgs field breaks this symmetry by acquiring a vacuum expectation value. Each symmetry transformation that is broken results in a mass for the corresponding boson. But if the vacuum is still left invariant by some subgroup of gauge transformations, the gauge bosons associated with that subgroup will remain massless. The generator corresponding to electric charge is Q = T3 + Y/2. So we make the simplest choice and assume the Higgs field to be an isospin doublet, T = 1/2, with hypercharge Y = 1, and assume that its vacuum expectation value φ0 = (0 v) has only a T3 = -1/2 component. Then Qφ0 = (T3 + Y/2)φ0 = 0, the vacuum is invariant under electromagnetic gauge transformations ("electrically neutral"), and the photon remains massless!
with neutrino?
P: 925
 in a single sentence, because the Higgs field is electrically neutral.
what has this to do with mass?Even U(1) gauge symmetry is not broken,how can one conclude that photon is massless?
 Sci Advisor P: 3,447 The gauge symmetry is a local symmetry, a position-dependent phase change, ψ → eiα(x)ψ, which we want to be a symmetry of the Lagrangian. But the Lagrangian contains derivative terms, so we must prevent derivatives of α(x) from appearing by using a modified derivative: Dμ = ∂μ - ieAμ where Aμ is the gauge field and transforms as Aμ → Aμ + (1/e)∂μα. Then the gauge field must be massless because a mass term m2AμAμ would not be gauge invariant.
P: 925
 Quote by Bill_K The gauge symmetry is a local symmetry, a position-dependent phase change, ψ → eiα(x)ψ, which we want to be a symmetry of the Lagrangian. But the Lagrangian contains derivative terms, so we must prevent derivatives of α(x) from appearing by using a modified derivative: Dμ = ∂μ - ieAμ where Aμ is the gauge field and transforms as Aμ → Aμ + (1/e)∂μα. Then the gauge field must be massless because a mass term m2AμAμ would not be gauge invariant.
I already know it, defining covariant derivative gives the required form,also used in non abelian gauge theory.I just want to know why spontaneous symmetry braking does not give mass to photons.(or only something like goldstone bosons appear?)
 Sci Advisor P: 744 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 $$g_{Y} = g_{L} \sin \theta$$ Sam
 P: 925 oh nice way,can I get some reference for it.thank you, sam.
P: 136
 Quote by samalkhaiat ...This happens for $A_{\mu}$ provided that we CHOOSE the couplings such that $$g_{Y} = g_{L} \sin \theta$$
Surely the couplings have already already determined by nature/more fundamental physics we haven't yet discovered. Isn't the real point that, for the field configuration of the GWS theory, there exists a θ for which the Aμ remains massless?

We measure its value experimentally.
P: 744
 Quote by AdrianTheRock Surely the couplings have already already determined by nature/more fundamental physics we haven't yet discovered. Isn't the real point that, for the field configuration of the GWS theory, there exists a θ for which the Aμ remains massless? We measure its value experimentally.
Yes, that is true. In a self-consistence theory, that does not change the fact that the photon was introduced by that particular choice of the $SO(2)$ parameter.

Sam
P: 744
 Quote by andrien oh nice way,can I get some reference for it.thank you, sam.
All textbooks on Weinberg-Salam theory go through the details. After introducing the perturbation $H(x)/ \sqrt{2}$ and performing local gauge transformation on all the fields in the theory, the rest is just an algebra. The vector mesons masses are contained in the $|D_{\mu}\Phi |^{2}$ part of the Lagrangian:
$$D_{\mu}\Phi = \left( \partial_{\mu} + i \frac{g_{L}}{2}\tau_{a} W^{a}_{\mu} - i \frac{g_{Y}}{2}B_{\mu} \right) \left( \begin{array}{c} 0 \\ (v + H/ \sqrt{2}) \end{array} \right)$$
$$V( \Phi ) = - \mu^{2} H^{2}(x) + \mathcal{O}^{3}(H) + \mathcal{O}^{4}(H) + \mbox{const.} \ \ \ (1)$$
So,
$$( D_{\mu}\Phi )^{\dagger}( D^{\mu} \Phi ) = \frac{1}{2} ( \partial_{\mu}H )^{2} + \frac{g_{L}^{2}v^{2}}{4} \left[ ( W_{\mu}^{1})^{2} + (W_{\mu}^{2})^{2} + ( W_{\mu}^{3} - \frac{g_{Y}}{g_{L}} B_{\mu})^{2}\right] + \mbox{interaction terms} \ (2)$$
From (1) and (2) we read off the mass of the Higgs field: $M_{H}= \sqrt{2 \mu^{2}}$.
Now, we write eq(2) in terms of the fields $W^{\pm}_{\mu}, Z_{\mu}$ and $A_{\mu}$ given in my previous post. The relevant terms become (I):
$$\frac{g_{L}^{2}v^{2}}{4}[ ( W_{\mu}^{1})^{2} + ( W_{\mu}^{2})^{2}] = \frac{g_{L}^{2}v^{2}}{2} W_{\mu}^{+} W^{\mu -}$$
This gives the mass term for the $W^{\pm}$ bosons: $M_{W}= \frac{g_{L}v}{\sqrt{2}}$
And (II):
$$\frac{g_{L}^{2}v^{2}}{4}( W_{\mu}^{3} - \frac{g_{Y}}{g_{L}}B_{\mu} )^{2} = \frac{g_{L}^{2}v^{2}}{4}[( \sin \theta - \frac{g_{Y}}{g_{L}}\cos \theta ) A_{\mu} + ( \cos \theta + \frac{g_{Y}}{g_{L}} \sin \theta ) Z_{\mu}]^{2}$$
Now, if we make the choice
$$g_{Y} = g_{L} \tan \theta ,$$
the above reduces to
$$(0) A_{\mu}^{2} + (0) A_{\mu}Z^{\mu} + \frac{g_{L}^{2}v^{2}}{2 \cos^{2}\theta}Z_{\mu}^{2}.$$
This means that the $A_{\mu}$ is massless; there is no e-m interaction between the A and the Z fields and the mass of the neutral boson Z is given by $$M_{Z}= \frac{vg_{L}}{\sqrt{2}\cos \theta} = \frac{M_{W}}{\cos \theta}.$$

Sam
P: 744
 Quote by samalkhaiat This happens for $A_{\mu}$ provided that we CHOOSE the couplings such that $$g_{Y} = g_{L} \sin \theta$$
CORRECTION: The correct form is
$$g_{Y} = g_{L} \tan \theta .$$
 Quote by samalkhaiat CORRECTION: The correct form is $$g_{Y} = g_{L} \tan \theta .$$
No, you are not missing anything. This is what I said in post #13. The point is this: in the unbroken $SU(2)_{L}\times U(1)_{Y}$ that choice of $\theta$ does not give you photon. Only when $<\Phi >:SU(2)_{L}\times U(1)_{Y}\rightarrow U(1)_{em}$, we identify the A-field with the massless photon.