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.

Suppose you reset the parameters of the standard model so that the Higgs field average value is zero in the vacuum, what would happen to standard matter?

If the fundamental fermions go from a finite to a zero rest mass, I'm pretty sure that the electrons would fly away from nuclei at the speed of light, leaving positively charged nuclei trying to get away from each other. Looking at the solution for the Hydrogen atom, I don't see how it would be possible to have atoms with zero rest mass electrons.

What happens to protons and neutrons? Since only a very small part of the mass of protons and neutrons is the rest mass of the quarks, and since they're flying around in there at relativistic speeds already, and since the nuclear force is so much stronger than the electrical force with an incredible aversion to naked color, would protons and neutrons remain bound assemblages of quarks and virtual gluons? Would they get a little larger? A little less massive?

What would happen to nuclei? Would they stay together? If the protons and neutrons hold together and their properties change only some, then I might expect the same of nuclei. Different stable isotopes, different sizes, and different masses, but I would expect there would still be nuclei.

Also the W and Z particles go to zero rest mass. What does that do to the electroweak interactions? Does that affect normal stable matter (outside of nuclear decay modes)? Is the weak force no longer weak? What happens to the forces overall?

share|improve this question
add comment

2 Answers 2

up vote 10 down vote accepted

The analysis of the phase structure of gauge theories is a whole field. Some major breakthroughs were the t'Hooft anomaly matching conditions, the Banks-Zaks theories, Seiberg duality, and Seiberg Witten theory. There is a lot of controversy here, because we don't have experiment or simulation data for most of the space, and there is much more unknown than known.

The first thing to note is that when the Higgs field vacuum expectation value is zero, the Higgs doesn't touch the low energy physics. You can ignore the Higgs at energy scales lower than it's mass, and if this mass is much greater than the proton mass, the result is indistinguishable qualitatively from the Higgsless standard model. So I'll describe the Higgsless standard model.

Higgsless standard model

Even without the Higgs, electroweak symmetry is broken anyway by QCD condensates. When the Higgs VEV is zero, the W and Z do not become completely massless, although they become much much lighter.

The reason is that QCD has a nontrivial vacuum, where quarks antiquark pairs form a q-qbar scalar fluid that breaks the chiral symmetry of the quark fields spontaneously. This phenomenon is robust to the number of light quark flavors, assuming that there aren't so many that you deconfine QCD. QCD is still asymptotically free with 6 flavors, and it should be confining even with 6 flavors of quarks. So I have no compunctions about assuming the confinement mechanism still works with 6 flavors, and all 6 are now like the up and down quark. Assuming the qualitative vacuum structure is analogous to QCD is plausible and consistent with the anomaly conditions, but if someone were to say "no, the vacuum structure of QCD with 6 light quarks is radically different from the vacuum structure of QCD", I wouldn't know that this is wrong with certainty, although it would be strange.

Anyway, assuming that QCD with 6 light quarks produces the same sorts of condensates as QCD with 3 light quarks (actually 2 light quarks and a semi-light strange quark), the vacuum will be filled with a fluid which breaks SU(6)xSU(6) chiral rotations of quark fields into the diagonal SU(6) subgroup. The SU(6) is exact in the strong interactions and mass terms, it is only broken by electroweak interactions.

The electroweak interactions are entirely symmetric between the 3 families, so there is a completely exact SU(3) unbroken to all orders. The SU(6)xSU(6) breaking makes a collection of massless Goldstone bosons, massless pions. The number of massless pions is the number of generators of SU(6), which is 35. Of these, 8 are exactly massless, while the rest get small masses from electroweak interactions (but 3 of the remaining 27 go away into W's and Z's by Higgs mechanism, see below). The 8 massless scalars give long-range nuclear forces, which are an attractive inverse square force between nuclei, in addition to gravity.

The hadrons are all nearly exactly symmetric under flavor SU(6) isospin, and exactly symmetric under the SU(3) subgroup. All the strongly interacting particles fall into representation of SU(6) now, and the mass-breaking is by terms which are classified by the embedding of SU(3) into SU(6) defined by rotating pairs of coordinates together into each other.

The pions and the nucleons are stable, the pion stability is ensured by being massless, the nucleon stability by approximate baryon number conservation. At least the lowest energy SU(3) multiplet

The condensate order-parameter involved in breaking the chiral SU(6) symmetry of the quarks is $\sum_i \bar{q}_i q_i$ for $q_i$ an indexed list of the quark fields u,d,c,s,t,b. The order parameter is just like a mass term for the quarks, and I have already diagonalized this order parameter to find the mass states. The important thing about this condensate is that the SU(2) gauge group acts only on the left-handed part of the quark fields, and the left-handed and right handed parts have different U(1) charge. So the condensate breaks the SU(2)xU(1) gauge symmetry.

The breaking preserves a certain unbroken U(1) subgroup, which you find by acting the SU(2) and U(1) generators. The left handed quark field has charge 1/6 and makes a doublet, so for the combination $I_3+Y/2$ where I is the SU(2) generator and Y is the U(1) generator, you get a transformation of 2/3 and 1/3 on the top and bottom component, which is exactly the same as $I_z + Y/2$ on the singlets (since they have no I). So this combination isn't chiral, and preserves the vacuum. So the QCD vacuum preserves the ordinary electromagnetic subgroup, which means it makes a Higgs, just like the real Higgs, which breaks the SU(2)xU(1) down to U(1) electromagnetic, with W and Z bosons just like in the standard model.

This is not really as much of a coincidence as it appears to be--- a large part of this is due to the fact that QCD condensates in our universe are not charged, so that they don't break electromagnetism, because u-bar and u have opposite electromagnetic charge transformation. This means that a u-bar u condensation leaves electromagnetism unbroken, and it isn't a surprise that it doesn't leave any of the rest of SU(2) and U(1) unbroken, because it's a chiral condensate, and these are chiral gauge transformations.

The major difference is that there are 3 separate Higgs-like condensates, one for each family, each with an identical VEV, all completely symmetric with each other under the global exact SU(3) family symmetry.

The W's and Z's get a mass from an arbitrary one of these 3, leaving 2 dynamical Higgs-like condensates. The main difference is that these scalar condensates don't necessarily have a simple distinguishable higgs-boson-like oscillation, unlike a fundamental scalar Higgs. The result of this is that the W's and Z's acquire QCD-scale masses, so around 100 MeV for the W's and Z's, as opposed to approximately 100 GeV in the real world. The ratio of the W and Z mass is exactly as in the standard model.

Behavior of analogs of ordinary objects

The low energy spectrum of QCD is modified drastically, due to the large quark number. The 8 massless pions and 24 nearly massless pions (three of the pions are eaten by the W's and Z's to become part of the massive vectors) include all the diquark degrees of freedom that we call the pions,kaons and certain heavy quark mesons. There will still be a single instanton heavy eta-prime from the instanton violated chiral U(1) part of U(6)xU(6). There should be 35 rho particles splitting into 8 and 27 and 35 A particles splitting into 8 and 27 effectively gauging the flavor symmetry.

The 6 quarks could be thought of as getting a mass from their strong interaction with the Higgs-like condensates, of order some meVs, but since the mass of a quark is defined at short distances, from the propagator, it might be more correct to say the quarks are massless. Some of the particles you see in the data-book, the sigma(660), the f0(980) should disappear (as these are weird--- they might be the product of pion interactions making some extremely unstable bound states, something which wouldn't work with massless pions)

The electron and neutrino will be massless except for nonrenormalizable quark-lepton direct coupling, which would couple the electron to the Higgslike chiral quark condensate. This effect is dimension 6, so the compton wavelength of the electron will be comparable to the current radius of the visible universe. The neutrino mass will be even more strongly suppressed, so it might as well be exactly massless.

The massless electron will lead the electromagnetic coupling (the unHiggsed U(1) left over below the QCD scale) to logarithmically go to zero at large distances, from the log-running of QED screening. So electromagnetism, although it will be the same subgroup of SU(2) and U(1) as in the Higgsed standard model, will be much weaker at macroscopic distances than it is in our universe.

Nuclei should form as usual at short distances, although Isospin is now a nearly exact SU(6) symmetry broken only by electromagnetism, and not by quark mass, and with an exact SU(3) subgroup. So all nuclei come in SU(6) multiplets slightly split into SU(3) multiplets. The strong force will be longer ranged, and without the log-falloff of the electromagnetic force, because the pions quickly run to a free-field theory, since the pion self-interactions are of a sigma-model type. The pion interactions will look similar to gravity in a Newtonian approximation, but scalar mediated, so not obeying the equivalence principle, and disappearing in scattering at velocities comparable to the speed of light.

The combination of a long-ranged attractive nuclear force and a log-running screened electromagnetic force might give you nuclear bound galaxies, held at fixed densities by the residual slowly screened electrostatic repulsion. These galaxies will be penetrated by a cloud of massless electrons and positrons constantly pair-producing from the vacuum.

share|improve this answer
1  
@MarkAdler: The pion is not a simple bound state of quarks, this is just wrong (but commonly misrepresented in the literature, and sold to the public). The pion is a Goldstone boson, it's like a sound mode of the quark fluid. The case where the quarks are massless, the symmetry becomes exact, and the pions exactly massless (except for electroweak interactions. And, oops, I just realized that I didn't subtract out the massless pions doing the Higgs mechanism out from the pions, there are 8 massless and 24 nearly massless pions, and 3 pions eaten to make W's and Z's at the QCD scale). –  Ron Maimon Jul 6 '12 at 16:00
1  
The massless pion is like saying a phonon in a solid is massless (meaning linear dispersion) even though the atoms are not massless. I don't see you asking "is a phonon are the atoms making the phonon squished really small?" The pion is a state where a coherent superposition of the condensed quarks are rotated in opposite directions chirally. BTW, I fixed the eaten pions, and yes, that's what I meant by nuclear bound galaxies. –  Ron Maimon Jul 6 '12 at 16:12
2  
@MarkAdler: You can't see the "little suckers" in a pion, that's just false. The pion is produced by acting "q-bar \gamma5 q" on the vacuum--- this is an operator which makes two quarks in the microscopic theory, and has a matrix element to the pion in the low-energy theory. The link between a two quark state and a pion state is real. The charge radius of a pion is how the electromagnetic field couples to a chiral shaking of the condensate, this can be sort of local, it doesn't contradict the goldstone property. The idea that the pion is made of two quarks is just an outright embarassment. –  Ron Maimon Jul 6 '12 at 17:49
2  
I should point out that if you translate a single atom, you have a matrix element with phonon states. This does not mean that a phonon is made from a single atom. Similarly, if you use a pair-operator on electrons, you make a BCS condensate excitation, but the BCS-pair is not exactly composed of two isolated electron excitations, it is a collective excitation. There is a training divide between condensed matter and high energy which made these types of things alien to high energy folks, and led them to dismiss vacuum structure, even though the evidence since Weinberg was overwhelming. –  Ron Maimon Jul 6 '12 at 17:57
2  
@MarkAdler: You are repeating wrong stuff people say: the deep inelastic regime is high energy, much larger than the mass of the pion, at least some GeV's. In this regime, you can't say you have a single pion, you make more. The deep inelastic probe (if done on a pion) will not see two point quarks, it will see a distribution of quarks, with a large unpredictable number of 10-100 MeV range quarks and gluons (wee partons) which you ignore. What you see is a high energy quark every once in a while carrying most of the energy/momentum of the pion. This is no information on how quarks make a pion. –  Ron Maimon Jul 8 '12 at 6:52
show 8 more comments

The Higgs field has a non-zero average value. And because it does, many particles have mass, including the electron, quarks, and the W and Z particles of the weak interactions. If the Higgs field was zero, those particles would be massless or very light. That would be a disaster; atoms and atomic nuclei wouldn't exist. Nothing like human beings, or the earth we live on, could exist without the Higgs field having a non-zero average value.

But, the particles in universe would be more organized.

Let's put it under microscope..

  • Instead of the electromagnetic and weak nuclear force present in our world, with its non-zero-Higgs-field, a zero-Higgs-field world has these forces scrambled and rearranged. The rearranged forces are called hypercharge and isospin (for historical reasons; the names are just that, names, without other significance.)

  • As part of this scrambling, the force carrier particles are changed; there are 3 W particles and an X particle, and the Z° and photon are missing. And the W and X particles are all massless now.

  • The force carriers are now simpler in another sense. The photon affects the W and W particles directly. But the X particle does not directly affect any of the three W particles. The gluons affect themselves as before; the W’s affect themselves too; but the X particle affects no force carriers at all.

Read more at paper by Matt Strassler

share|improve this answer
    
Thanks for the link! That answers a lot of questions (and raises many more ...). –  Mark Adler Jul 6 '12 at 4:44
    
This is repeating the wrong information in the linked article. There is a nontrivial vacuum even without the Higgs, it is the pion vacuum in our universe. Pion style condensation breaks SU(2)xU(1) to U(1) too, but in a little different way than the Higgs does. –  Ron Maimon Jul 6 '12 at 5:08
    
The article does not answer part of my question. Will there be protons and neutrons (or more likely many flavors of protons and neutrons since all the quark flavors have zero rest mass)? From what I can tell, the strong nuclear force in this model is pretty much independent of the Higgs shenanigans, and it would still require extremely large energies to separate quark colors. So in a relatively cool universe, there should be these composite particles even with massless quarks. Correct? –  Mark Adler Jul 6 '12 at 5:13
    
@MarkAdler: Correct, although there are lots of different neutrons and protons of exactly the same mass, and split electromagnetically from each other. Further, you still get W's and Z's massive. –  Ron Maimon Jul 6 '12 at 7:22
add comment

protected by Qmechanic Dec 27 '13 at 20:31

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

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