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Interaction of particle with Higgs field provides mass to the particles, then why still some particles are massless?

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They don't interact with the Higgs field. –  jhobbie Aug 6 at 14:57
what is the criteria for to interact and to not interact? –  user56751 Aug 6 at 15:28

2 Answers 2

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John Rennie's answer is good, but I'll try to explain intuitively why the symmetry breaking leaves some symmetry unbroken.

Start with a sphere. You can rotate a sphere in three independent ways—around the x axis, around the y axis, and around the z axis, if you like. All of these are symmetries of the sphere, i.e., they leave the sphere unchanged. These rotations are called $SU(2)$ [almost—there's a technicality that I'll ignore], and saying that an $SU(2)$ gauge theory has three gauge bosons (which it does) is the same as saying that a sphere can be rotated in three independent ways.

Draw a dot somewhere on the sphere. All three of the above rotations will probably move the dot, which means they are not symmetries of the sphere plus dot. But there is still a rotational direction that leaves the dot in place, and there was never any reason to prefer those other axes that don't. So forget about those axes and instead pick the axis of symmetry (an axis going through the dot and its antipodal point), and two other axes perpendicular to that one and each other. The axis of symmetry is the "photon", and I suppose you could call the other two the W+ and W. This analogy is inaccurate in that the real electroweak gauge group is $SU(2)\times U(1)$, not $SU(2)$, and has four bosons, not three. But a sphere is much easier to visualize, and the basic principle is the same. Before the symmetry is broken (by drawing the dot), you can't be sure that any particular boson (rotational axis) will remain a symmetry (remain massless), but there will always be some combination of those bosons (some other axis) that remains a symmetry, and we call that the photon.

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thank you very much for this valuable information –  user56751 Aug 7 at 4:30

Elementary particles are conveniently divided into the fermions and the gauge bosons. The fermions are what we think of as matter, e.g. protons (i.e. quarks) and electrons, while the gauge bosons provide the forces that act between the particles of matter.

Fermions get their mass from an interaction with the Higgs field called a Yukawa coupling, and the strength of this interaction determines the mass of the fermion. We do not have a theory that predicts the coupling with the Higgs field, so we can't say why the fermions have the masses they do. What we can say is that for all fermions the coupling to the Higgs field is non-zero i.e. no fermions are massless. There is a discussion of this on Matt Strassler's blog, and Luboš covers some of the ground in his answer to Why some particles interact with the Higgs field and others don't?. Both of these are rather technical I'm afraid.

A quick aside: the neutrinos are fermions, and it is currently unknown if they are all massive since one of the three types of neutrino could be massless. We also don't know if neutrinos get their mass by interacting with the Higgs or from some other mechanism such as the seesaw mechanism.

Anyhow back to the Higgs. The standard model is based on a type of symmetry called gauge symmetry, and this requires that the gauge bosons be massless. The gauge bosons in question are the electroweak gauge bosons, the W$^+$, W$^-$, Z and photon, and the strong force gauge boson called the gluon. However there is a get-out clause that allows some of the gauge bosons to acquire a mass by a process called symmetry breaking. Electroweak symmetry breaking allows three out of the four gauge bosons to acquire a mass, so the two Ws and the Z are massive while the photon loses out and has to stay massless.

Although the electroweak bosons can gain mass by symmetry breaking there is no symmetry breaking of the strong force, so there is no way for the gluon to acquire a mass.

So to summarise:

  • There are only two particles known to be massless: the photon and the gluon

  • the photon is massless because the electroweak symmetry breaking can give mass to only three out of the four electroweak bosons and the photon was the unlucky one

  • the gluon is massless because there is no symmetry breaking mechanism that could give it a mass

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thank you very much, i have learned much from your posted answer. –  user56751 Aug 6 at 16:12
Maybe you should make clear that the Higgs field interacts with the weak interaction only, that is why the gluons escape, there is no weak vertex to connect them to the all pervading Higgs field. –  anna v Aug 6 at 17:46

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