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Why no neutral quarks?

  1. Jan 14, 2012 #1
    For leptons we have charged leptons electron, muon, tau and uncharged leptons electron neutrino, muon nutrino, tau neutrino. For quarks we have charged quarks pairs down, up and strange, charm and truth, beauty. But we have no neutral quarks. Why?

    The force carrying particle for the charged particles in each case is massless. The force carrying particle for the neutral leptons is massive. By analogy why is there no massive boson triplet to couple with the neutral quarks?
    Last edited: Jan 14, 2012
  2. jcsd
  3. Jan 14, 2012 #2

    Simon Bridge

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    If you like you could work out what the particle possibilities are if there were neutral quarks and see what their properties would have to be. The basic reason there are no neutral quarks in the standard model is that there are no particles that require them. You understand why we have a quark model to start with right?

    Though I think you may need to distinguish electric charge from color charge.
    (JIC: the third quark pair you are looking for is possibly top/bottom rather than truth/beauty)

    The only uncharged leptons are neutrinos.
    All leptons feel the weak nuclear interaction.
    Presumably any hypothetical neutral quarks would have the same characteristic interactions as the other quarks. You are thinking that gluons are to quarks what photons are to leptons?
  4. Jan 15, 2012 #3


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    Nobody knows.

    The gauge symmetry group of the SM is SU(3)color * [U(1)*SU(2)]el.-weak where color is described by the 1st factor and electro-weak charges (especially the electric charge) is described by the 2nd factor. Nobody knows whether there is a relation between these two gauge groups, why they are there, why exactly this group structure, why three generations etc.

    These are question to be answered by GUTs, a ToE or something like that
  5. Jan 15, 2012 #4
    Simon, yes I am thinking gluons are to quarks as photons are to charged leptons.

    When I say neutral quark I mean both color neutral and electric charge neutral.
  6. Jan 15, 2012 #5

    Simon Bridge

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    Like tom said - "we don't know" is the bottom line here.
    We don't have neutral quarks in our model because we don't need them to describe anything that we have seen yet.

    A quark that is both color and electric neutral would not be a quark. Imagine a lepton that feels neither weak nor EM interactions ... it wouldn't be a lepton right?

    You are thinking of something with a quark-like rest-mass which feels only a force mediated by an hitherto unsuspected heavy baryon (and gravity?). Well, that's pretty wide open then... you may as well ask why there are the particles that there are.

    (Hmmm ... would the Higgs boson fit that description I wonder?)
  7. Jan 15, 2012 #6
    A spin-1/2 uncharged both for electric and color interaction would very much behave like a neutrino. Why would you call it "quark" if it does not feel the color interaction ?

    There are within the standard model itself constraint between the electric charges of all fermions. They are not completely arbitrary. Those constraints are called "anomaly cancellations".
  8. Jan 15, 2012 #7


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  9. Jan 15, 2012 #8
    Tom thanks for the reference.
  10. Jan 16, 2012 #9
    Also at http://www.nikhef.nl/pub/theory/academiclectures/ is Beyond the Standard Model lecture 1, PDF pages 77 - 81, especially page 80.

    Anomalies are (sum over left-handed elementary fermions) - (same for right-handed ones) of Tr(Ta.{Tb,Tc})
    where the T's are gauge-group generators.

    The Standard Model's symmetry is SU(3) * SU(2) * U(1) -- QCD * (weak isospin) * (weak hypercharge)

    Weak hypercharge Y: average electric charge of a multiplet:
    each member's charge Q = (weak isospin)3 + Y

    Some of the anomalies cancel as a result of gauge-group structure: SU(3)2*SU(2), SU(3)*SU(2)2, SU(2)3, SU(3)*SU(2)*U(1), SU(3)*U(1)2, SU(2)*U(1)2, (gravity)2*SU(3), (gravity)2*SU(2), and apparently (gravity)*(nongravitational)(2) and (gravity)3

    This leaves SU(3)3, SU(3)2*U(1), SU(2)2*U(1), U(1)3, and (gravity)2*U(1). They mean:

    All these must add up to zero, taking (sum of left-handed multiplet members) - (likewise for right-handed ones):
    1. Quarks 1, leptons 0
    2. Quark (hyper)charge
    3. Weak-isospin doublet (hyper)charge
    4. (hypercharge)3 value
    5. (hyper)charge
    The first one is (number of left-handed quarks) = (number of right-handed quarks)

    Let's work it out for the Standard Model. It has a left-handed weak-isospin doublet of quarks and one for leptons, with all the right-handed ones being weak-isospin singlets.

    Anomaly #1: 1 quark doublet, 2 quark singlets. Check.
    Anomaly #2: 6Y(QL) = sum of 3Y(QR's)
    Anomaly #3: 6Y(QL) + 2Y(LL) = 0
    Anomaly #4: 6Y(QL)3 + 2Y(LL)3 = sum of 3Y(QR's)3 + sum of Y(LR's)3
    Anomaly #5: 6Y(QL) + 2Y(LL) = sum of 3Y(QR's) + sum of Y(LR's)

    Let's see how much we can constrain the hypercharge values. Let Y(LL) = -1/2 for definiteness, as in the Standard Model. Then Y(QL) = 1/6, check, though Y(QR's) and the Y(LR's) are more difficult.

    From #2, Y(QR1) + Y(QR2) = 1/3
    From #2 and #5, sum of Y(LR's) = -1
    From #4, -2/9 = 3Y(QR1)3 + 3Y(QR2)3 + sum of Y(LR's)3

    This is still a little ambiguous, so let us suppose that only one right-handed lepton has a nonzero hypercharge. Then Y(LR) = -1. Check.

    Y(QR1) + Y(QR2) = 1/3
    Y(QR1)3 + Y(QR2)3 = 7/27
    with solution Y(QR1) = 2/3 and Y(QR2) = -1/3. Check.

    So anomaly cancellation forces Y(QL) = 1/6, Y(LL) = -1/3, Y(QR1) = 2/3, Y(QR2) = -1/3, Y(LR) = -1 to within a multiplicative factor.

    Electrically-neutral quarks? They'd have to be added on to the Standard Model.
  11. Jan 16, 2012 #10

    Simon Bridge

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    Though all this basically rewrites the question to "why does the Universe show this kind of symmetry?" ... come to think of it, this is the sort of thing someone has probably pondered along the lines of "What kinds of symmetries could produce Universes with physicists to ponder what kind of...(repeats)?"

    A bit like how come we see 3+1 dimensions.
  12. Jan 16, 2012 #11


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    Yep, it smeels a little bit as "lepton number as the fourth color". Moreover, one can separate the electric charge coming brom B-L and then neutrinos, (and leptons and quarks) are not so different.
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