Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Universe w/o Higgs

  1. Jul 12, 2012 #1

    tom.stoer

    User Avatar
    Science Advisor

    How would the universe with standard model w/o Higgs look like?

    I guess confinement would still cause hadronization. Due to zero rest mass pions would be stable. Nucleons as lightest baryons would be stable b/c there is not lighter state into wich they could decay (even so the weak force is now a long-range force). The typical fusion processes of light nuclei would produce at least alpha particles.

    Therefore there would be a dilute gas of massless pions, nucleons, some alpha particles, massless leptons and photons as well as massless W- and Z-bosons.

    Am I correct?
     
  2. jcsd
  3. Jul 12, 2012 #2
    Just thinking about pions to start things off, surely charged ones would osscilate freely into (eg) muon and neutrino pairs, with some sort of equilibrium being reached. Neutral ones could still transmute into pairs of photons, but could equally easily be formed from photon collisions.

    So wouldn't it actually be more like a plasma or "soup" than a gas?
     
  4. Jul 12, 2012 #3

    tom.stoer

    User Avatar
    Science Advisor

    My idea is the following:

    The pion-lepton decay vanishes in the chiral limit; the pion-photon decay may still be non-zero due to the axial anomaly, therefore neutral pions would indeed oscillate into photon pairs.

    But we do not have two but six flavors and therefore only the U(6) singulet state (= the eta-prime with other quark contributions c, b, t) would oscillate into photon pairs; the pion - or whatever it may be with quark mixing of c, b, t - would have vanishing el.-mag. decay constant and would be stable; therefore all U(6) states but the singulet are stable with zero coupling constant.
     
    Last edited: Jul 12, 2012
  5. Jul 12, 2012 #4
    Are there any interesting consequences of the fact that SU(2)xU(1) is unbroken? I think that there are enough fermions around that the SU(2) is not confining. But there should be a new long-range force mediated by the massless SU(2) gauge bosons, yes? What does a long-range nonabelian gauge field look like?

    In addition we have the unbroken U(1) hypercharge which I guess looks like electromagnetism with different charges and a different coupling strength.

    The parity-violating "weak" force is no longer weak or short-ranged, so presumably parity violation becomes a lot more obvious?

    Do the massless pions effectively mediate another new long-range force?

    It seems like the U(6) equivalents of the nucleons are the only stable massive particles around. Are there atom-like bound states of multiple nuclei, held together by one or more of the new long-range forces?

    Wouldn't the U(6) singlet be massive because of the anomaly, and thus simply decay into U(1) gauge bosons?
     
    Last edited: Jul 12, 2012
  6. Jul 12, 2012 #5

    tom.stoer

    User Avatar
    Science Advisor

    Interesting ideas!

    This is something I forgot! The beta-function for QCD with 6 flavors reads

    [tex]\beta(g) = -\left(\frac{11}{3}N_C - \frac{2}{3}N_f\right)\frac{g^3}{16\pi^2} = -\left(11 - 4\right)\frac{g^3}{16\pi^2} \lt 0 [/tex]

    I do not know how to count the chiral fermions for the SU(2), including both leptons and quarks, so I can't say whether this SU(2) is confining or not. If yes, the 'SU(2) hadrons' would have to be SU(2) singulets just like SU(3) hadrons. Then all what I said above was nonsense and the world would look totally different.

    I don't know how a non-confining i.e. long-range non-abelian gauge theory would look like, but certainly very different from what we know and again all what I said above was nonsense.

    Yes, something like that; the other mesons remain massless Goldstone bosons.
     
    Last edited: Jul 12, 2012
  7. Jul 12, 2012 #6

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Confinement seems like an important issue to me, and I don't think there are enough massless fermions to keep the SU(2) deconfined. For example, SU(3) confinement should still render all the fermions carrying weak charge due to quarks massive. That just leaves 3 generations of chirally coupled doublets.

    If the SU(2) did go to weak coupling in the IR, then similar to N=4 SYM, I would think that an IR free gauge theory would still have something like the coulomb law (perhaps with a minor logarithmic modification). There would of course also be lots of non-abelian radiation.

    Assuming confinement, neutrinos, etc. would no longer be around, but probably there are composite fields with quantum numbers similar to the fields we observe. I guess electromagnetism would now be pure hypercharge?
     
  8. Jul 12, 2012 #7

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Right - the proton would be a little lighter, but it would still be there.

    Part 1: Bosons

    There wouldn't be a pion.

    You're right that the pion becomes a true massless Goldstone boson. But then it gets "eaten" to become the longitudinal component of the W and Z by something we call "the Higgs mechanism". So the W and Z gain a mass, to the tune of 20-30 MeV. Interestingly, the lifetime of these particles becomes close to that of the charged pions.
     
  9. Jul 12, 2012 #8

    tom.stoer

    User Avatar
    Science Advisor

    which field would cause this Higgs mechanism?
     
  10. Jul 12, 2012 #9

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Part 2: Nuclei

    I'm not 100% sure what happens to nuclei. The N-N force still has a repulsive core, but the attractive piece is now mediated by the W and Z. These are vectors, so that means that the p-p and n-n forces are repulsive, and only p-n is attractive. The range of this force is longer, but the strength - the coupling - is less. I suspect you still have nuclei, but they are larger, more weakly bound, and probably more prone to alpha-decay.

    Part 3: Leptons

    The electron gains a mass from QCD at loop level. It's very small - maybe meV or less. So atoms are very large: microns, possibly millimeters. The muon and tau gain mass through the same mechanism, so have the same mass. So flavor becomes a quantum number much like color today. We wouldn't perceive them as distinct particles, but our atomic shells could hold 3 times as many electrons: carbon would be a noble gas.
     
  11. Jul 12, 2012 #10

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    I think the point is that the chiral condensate [itex] \langle q \bar{q} \rangle [/itex] carries SU(2) charge and hence leads to another Higgs another effect.

    Is it possible that the weak force reaches strong coupling before the scale of chiral symmetry breaking, so that we should think in the opposite order?
     
  12. Jul 12, 2012 #11

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    I believe it's a quark bilinear term - you get a vev of order Lambda_QCD. But the key is that the pion has the same quantum numbers as the longitudinal piece of the W, so once it becomes a Goldstone, it's eaten. Actually, it's eaten today, and contributes a tiny piece to the W mass, but since the Higgs vev is so much larger, we don't care about it.

    This is how Technicolor works. You add pions with ~4000x the mass to the theory, and now the W gets a mass from them comparable to the observed mass.
     
  13. Jul 12, 2012 #12
    If we have six flavors of massless quarks, aren't there 6^2 - 1 = 35 pions? Then if 3 get eaten we should still have 32 massless pions.

    Can you explain how this works?
     
  14. Jul 13, 2012 #13

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    That's a good point - there would be 30 nucleons, 15 charged and 15 neutral. (I'm assuming N-type is lighter than Lambda-type)
     
  15. Jul 13, 2012 #14

    tom.stoer

    User Avatar
    Science Advisor

    Regarding the beta-function and confinement for an unbroken SU(2): the weak isospin would replace the color, i.e. NC=3 → 2. But how do you count the 'flavors'? The flavor in QCD are 'identical copies' of quarks which are not distinguished by the color-gauge interaction. Now what about weak force? what are these new flavours? how do you count them? You have again 3 generations i.e. 6 flavors (u,d,s,c,b,t), I guess you have another 6 leptonic flavors (electron, electron-neutrino, ...), but what about an additional factor of three for the quarks? the weak SU(2) does not distinguish between different colors. So do we have 6+6 = 12? Or do we have 3*6+6 = 24?
     
  16. Jul 13, 2012 #15
    But [itex]c\bar{s}[/itex] and [itex]t\bar{b}[/itex] have the same quantum numbers as [itex]u\bar{d}[/itex], so wouldn't they get eaten by the [itex]W^+[/itex] too?

    Could the latter swallow all of these together, or would some form of "quantum indigestion" occur? :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook