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Empirical tests of quark charges, any info?

  1. Jul 17, 2012 #1
    Hi, I just remembered a question I have wondered about for countless of years without finding any answer to:

    Have (any of) the electric charges of quarks been empirically measured/tested/verified?

    (Yes, I know about the Standard Model and what the charges are supposed to be, that's not what I'm after. I've searched for it on this forum and on the net, but failed to find any info about empirical tests).

    If anybody has good info about this and/or good link(s) I'll be very interested to know about it. My regards!
     
  2. jcsd
  3. Jul 17, 2012 #2

    mfb

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    Well, you cannot use individual quarks and let them fly around. But there are indirect methods:

    - note that a proton is "up up down" and a neutron is "up down down". Measure proton and neutron charge, solve the linear equations. Do the same with other hadrons. This does not work with the top-quark.
    - measure the cross-section [itex]e^- e^+ \to \gamma \to q \overline{q}[/itex], which depends on the (absolute) quark charges. This would work with the top-quark, but there is no collider with the required energy.
    - measure some other processes where photons couple to quarks
     
  4. Jul 17, 2012 #3

    Bill_K

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    Since you can't have one quark in isolation there's no way to directly measure its charge. You have to work with combinations of them and test constraints on their total charge. Besides the obvious ones, like the total charge of quarks in a proton or neutron is +1 or 0, there are tests which involve quadratic combinations. For example a ratio of certain scattering cross-sections,

    R = σ(e+e- → hadrons)/σ(e+e- → μ+μ-)

    The standard model predicts R = 3[(2/3)2 + (1/3)2 + (1/3)2] = 2

    and sure enough, that's what is observed. At energies > 10 GeV where c-quarks and b-quarks can be produced,

    R = 3[(2/3)2 + (1/3)2 + (1/3)2 + (2/3)2 + (1/3)2] = 11/3

    and again that agrees with experiment also.
     
  5. Jul 17, 2012 #4
    Thanks a lot, both of you! Yes, I understood you would have to work on combinations of them and do some cross-section "magic". I will digest the info and also search a little. And if anyone comes to think of more and knows any links to experiment papers/reports, I'm happy. If they're old and/or technical doesn't matter, I'm interested anyway.

    I now also remember why this issue started troubling me once, now a long time ago; it's those 1/3 and 2/3 charges which seem to fit so perfectly into the scheme, conveniently enough. Or, more specifically, I think my original thoughts concerned why the denominator would turn out to be 3. For this, I believe there's no particularly good answer (?) nor theory (?) (I don't know). That is, beside the Standard Model scheme itself?
     
    Last edited: Jul 17, 2012
  6. Jul 17, 2012 #5
    That's due to anomaly cancellation where the anomalies are certain diagrams that cannot be evaluated self-consistently. The triangle anomaly features a fermion loop with three gauge particles coming out of it. It cannot be evaluated consistently if the gauge-fermion interactions violate parity. The only way to get rid of them is to make them cancel each other out, and that supplies constraints on the fields.

    Anomalies and the Standard Model

    For gauge particles a, b, and c, with operators on the fermions Ta, Tb, Tc, operators like electric charge.

    Anomaly ~ Tr(Ta.{Tb,Tc})left - Tr(Ta.{Tb,Tc})right

    If all the interactions conserve parity, then the triangle anomaly will cancel out. So the interesting case is when they don't, like the electroweak interactions.

    Electroweak multiplets have two quantum numbers, weak isospin and weak hypercharge
    Weak isospin (WIS) is like 3D angular momentum. Members have (WIS)3 form - (WIS) to + (WIS) in integer steps
    Weak hypercharge (WHC) is like electric charge
    Electric charge Q = (WIS)3 + (WHC)

    Elementary fermions come in left-handed doublets and right-handed singlets. The right-handed singlets must have hypercharge values that enable the electromagnetic interaction to conserve parity. A left-handed multiplet with hypercharge Y has charges
    up: Y + 1/2
    dn: Y - 1/2
    Right-handed:
    up: WHC = Y + 1/2
    dn: WHC = Y - 1/2
    Right-handed neutrinos are electroweak and Standard-Model singlets, so their presence or absence does not affect this argument.


    Let's see if we can get constraints on possible hypercharge values.

    QCD-QCD-QCD - cancels, from # left-handed degrees of freedom equaling # right-handed ones
    QCD-QCD-WIS - cancels, from Tr(WIS operator) = 0
    QCD-QCD-WHC - cancels, from left-handed hypercharges 2Y equaling right-handed ones (Y+1/2)+(Y-1/2) = 2Y
    QCD-(WIS/WHC)-(WIS/WHC) - cancels, from Tr(QCD operator) = 0
    WIS-WIS-WIS - cancels, since it's zero for (WIS) = 0 and 1/2
    WIS-WIS-WHC - gives 2Y
    WIS-WHC-WHC - cancels, since Tr(WIS operator) = 0
    WHC-WHC-WHC - gives 2Y3 - (Y+1/2)3 - (Y-1/2)3 = -(3/2)Y

    So the sum of the hypercharges of the left-handed multiplets must equal zero, with the sums being weighted by the multiplets' multiplicities.

    Let's now look at lepton and quark triangle anomalies.

    Leptons first. Electrons have charge -1 and neutrinos 0, giving their left-handed multiplets a WHC of -1/2
    2*that = -1 -- anomaly!

    Now quarks. Up-like ones have charge 2/3 and down-like ones charge -1/3, giving their left-handed multiplets a WHC of +1/6
    Being sure to count colors,
    3*2*that = +1 -- anomaly also!

    But the lepton and quark ones cancel out.

    Thus, quarks' fractional charges are due to anomaly cancellation and their coming in 3 colors.
     
    Last edited: Jul 18, 2012
  7. Jul 18, 2012 #6

    tom.stoer

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    That's a nice summary on anomaly cancellation and consistency; if you use one generation (e, ν, u, d) as input and if you fix some of the charges like q(e) = -1, q(ν) = 0, ... you can derive consistency conditions for the remaining ones; but wouldn't it be possible to get similar cancellations e.g. for different multiplets or different charges to start with? I think the anomaly cancellation allowes for the SM generations, but is does not rule out other structures (unfortunately).
     
  8. Jul 18, 2012 #7
    That's right. The number of anomaly constraints stays fixed independent of how many multiplets there are and what their parameters are.

    There's one mentioned in my link that I didn't mention earlier: U(1) interactions + 2 gravity vertices:
    WHC-grav-grav - cancels, since 2Y = (Y+1/2) + (Y-1/2) for each multiplet


    Interestingly, the MSSM Higgs particles also have anomaly cancellation, though strictly speaking, it's for the higgsinos. The MSSM has two Higgs doublets, one up-like with Y = 1/2 and the other down-like with Y = -1/2. It's easy to verify anomaly cancellation: sum of Y's = sum of Y3's = 0.
     
  9. Jul 18, 2012 #8
    Many, many thanks, lpetrich! You have definitely opened up a door that's previously been closed for me, so to say. Very interesting! I will take time to thoroughly go through your description & pdf. I'm a little rusty on the terminology but I certainly recognize it (except anomaly cancellation); it was a long time ago I dived into the quark world, and when I did, it was self-studies only, and never quite this deep. Thanks again, I really appreciate it!
     
  10. Jul 18, 2012 #9
    You're welcome.

    Parent directory of the document I linked to: PHY 522 Topics in Particle Physics and Cosmology - Anomalies in Quantum Field Theory


    Anomaly cancellation is likely a pleasant side effect of the ultimate origin of the Standard Model's particle multiplets: some Grand Unified Theory. I won't go into detail unless anyone wants to see it, but I'll mention [hep-ph/0606206] Anomaly for Model Building.

    Seems like a good subject for the PF Library. Should I contribute something to it?

    As tom.stoer noted, if one adds enough particles and adjusts their properties appropriately, one can always get anomaly cancellations.

    The simplest GUT that unites the gauge particles in one multiplet is the Georgi-Glashow model, with symmetry group SU(5). The unbroken Standard Model has this symmetry group:
    SU(3)QCD * SU(2)WIS * U(1)WHC

    SU(5) still has triangle anomalies, but they cancel in GG for both the elementary fermions and the Higgs particles. The elementary fermions are in 2 multiplets per generation, or 3 if you count right-handed neutrinos, and the Higgs particles in 2 multiplets.

    The next step up is SO(10), and it has the nice property that all possible triangle anomalies will cancel. The elementary fermions are in 1 multiplet per generation, and the Higgs particles in 1 multiplet.


    The only additional elementary-fermion flavor is the right-handed neutrino, and that's only necessary in SO(10). It's a Standard-Model and Georgi-Glashow singlet, however. Likewise, the only additional Higgs particle is a "Higgs triplet" with the quantum numbers of the down quark, and that one is in both SU(5) and SO(10).

    So SU(5) and SO(10) make only a few extra relatives of the elementary fermions and the Higgs particles.
     
  11. Jul 18, 2012 #10

    tom.stoer

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    what about E(6) ?
     
  12. Jul 18, 2012 #11
    Triangle anomalies cancel in E6 also.

    Harish-Chandra isomorphism - Wikipedia lists the "degrees of fundamental invariants" for all the semisimple Lie algebras. Here is what I think that they are.

    From the generators Ti of a Lie algebra, construct commutator f: [Ti,Tj] = fijkTk

    metric g: gij = fiab fjba and its inverse gij

    Construct Cij = fiaj gab Tb

    Create invariants I(p) = Tr(matrix product of p C's)

    I(p) will be polynomials with degree p in the highest-weight components of the generators' representation. Since the algebra has rank n, the I's will be constructed from n polynomials in the highest weight, what I will call basis polynomials.

    I(0) = (dimension of representation)
    I(1) = 0 (for semisimple algebras; may be nonzero if U(1)'s are present)
    I(2) ~ Casimir invariant

    Basis polynomials' degrees for the simple algebras:
    A(n), SU(n+1): 2, 3, 4, ..., n+1
    B(n), SO(2n+1) / C(n), Sp(2n): 2, 4, 6, ..., 2n
    D(n), SO(2n): 2, 4, 6, ... 2n-2, n (the missing 2n contains the square of a degree-n basis polynomial)
    G2: 2, 6
    F4: 2, 6, 8, 12
    E6: 2, 5, 6, 8, 9, 12
    E7: 2, 6, 8, 10, 12, 14, 18
    E8: 2, 8, 12, 14, 18, 20, 24, 30

    The triangle anomaly will be present only if I(3) is nonzero, and that requires that a basis polynomial with degree 1 or 3 be nonzero. I'll now list some Standard-Model and GUT gauge-symmetry algebras and bold those that can have triangle anomalies.

    U(1) - weak hypercharge, electromagnetism
    SU(2) - weak isospin
    SU(3) - QCD "color"
    SU(3) * SU(2) * SU(1) - unbroken Standard Model
    SU(5) - Georgi-Glashow
    SU(4) * SU(2) * SU(2) / SO(6) * SO(4) - Pati-Salam
    SU(6)
    SU(3) * SU(3) * SU(3) - trinification
    SO(10)
    E6
     
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