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Why does the top quark interact so much more strongly with Higgs field than the other quarks?

  1. Oct 23, 2014 #1
    The title says it all.
     
  2. jcsd
  3. Oct 23, 2014 #2

    Orodruin

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    The vacuum expectation value of the Higgs field gives a mass to the quarks which is proportional to how strongly the field couples to it. Since the top quark is the heaviest fermion, it interacts more strongly with the Higgs field. As to why it interacts more strongly (and hence obtains more mass) is not known, the coupling strength is a parameter of the standard model.
     
  4. Oct 23, 2014 #3

    arivero

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    Your question can be tranlated to other twho:

    1) Why all the other 21 particles almost massless, respective to the scale of the Higgs Field? Is there some hidden symmetry that should appear when the masses of such particles are all of them (except, as we said, the top quarks) exactly zero.

    2) Why is the yukawa coupling of the top quark compatible with 1.00? Note the two zero digits.
     
  5. Oct 24, 2014 #4

    arivero

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    Thinking about this, two subquestions:

    1a) Is there some elementary group with a 21 irreducible representation that could guide on the restoration of this symmetry? SO(7) and SU(6) look fine, but their decomposition towards SU(3) singlets and multiplets, which could be a guide to separate leptons and quarks, is not convincing. Sp6 is also tempting

    1b) Has some role the restauration of, ahem, exact chiral symmetry breaking?. This is, if all the quarks (and leptons, but this is irrelevant) except the top are massless, then all the pseudoscalar mesons (K, D, B, ...) become true goldstone bosons, massless.

    1b') Are diquarks massless in this limit?
     
  6. Oct 25, 2014 #5
    [hep-ph/9309293] Infrared Fixed Point Solution for the Top Quark Mass and Unification of Couplings in the MSSM
    Infrared fixed point of the top Yukawa coupling in split supersymmetry
    are some of several papers that discuss the infrared-fixed-point hypothesis for the top quark's mass. It's essentially that its mass is a result of its Higgs coupling getting fixed to a function of other couplings. In simple form, we can write its renormalization equation as
    $$ \frac{dy}{dt} = 2y (y^2 - g^2) $$
    where t = log( (interaction momentum) / (reference value) ), y = Higgs coupling, and g = combination of other couplings, like gauge coupling. I'm leaving out a lot of multiplicative factors. That can be turned into
    $$ \frac{d}{dt} \left( \frac{1}{y^2} \right) = \frac{g^2}{y^2} - 1 $$
    with solution
    $$ \frac{1}{y^2} = e^G \left(c - \int e^{-G} dt \right) $$
    where
    $$ G = \int g^2 dt $$
    and c is a constant of integration. Changing the limits of integration,
    $$ G'(t) = \int_t^{GUT} g(t')^2 dt' $$
    $$ \frac{1}{y(t)^2} = \frac{1}{y_{GUT}^2} e^{-G'(t)} + \int_t^{GUT} e^{G'(t') - G'(t)} dt' $$
    In the limit of constant g, this becomes
    $$ \frac{1}{y(t)^2} = \left( \frac{1}{y_{GUT}^2} - \frac{1}{g^2} \right) e^{-G} + \frac{1}{g^2} $$
    where
    $$ G = g^2 (t_{GUT} - t) $$

    So in the limit of low t, y -> g.
     
  7. Oct 25, 2014 #6
    It's fairly easy to get the above result in more qualitative, hand-waving fashion. If y is less than g, then as t increases, y gets closer and closer to g. As it does so, it slows down and it gets dragged along by g.

    We can get some improved bounds if g increases as t decreases, as it does in the (MS)SM.
    $$ \frac{1}{y(t)^2} = \frac{1}{y_{GUT}^2} e^{-G'(t)} + \int_t^{GUT} e^{G'(t') - G'(t)} dt' $$
    we can change the integration variable in the second integral to G'(t'), giving
    $$ \frac{1}{y(t)^2} = \frac{1}{y_{GUT}^2} e^{-G'(t)} + \int_0^{G'(t)} \frac{1}{g(t')^2} e^{G'(t') - G'(t)} d(G'(t')) $$
    So y(t) <= g(t') averaged over some t' values greater than t. This makes y(t) <= g(t).
     
  8. Oct 25, 2014 #7
    There is a further problem. In the unbroken Standard Model, the quarks and leptons are treated as members of three ur-generations, with the familiar generations being mass eigenstates of them. For the quarks, these states are not orthogonal to each other, producing cross-generation decays, and their mixing matrix contains complex terms, making CP violation. Thus, the Higgs-coupling matrices must be different 3*3 non-diagonal complex matrices in ur-generation space. There has been a lot of effort to try to explain these matrices using simple theoretical principles and presumed forms like several components being zero, but none has been very successful. It's very unlike the case of the gauge-symmetry multiplets, where Grand Unified Theories offer elegant unifications of the elementary fermions and Higgs particles.
     
  9. Oct 25, 2014 #8

    arivero

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    What about the other yukawas? Can the RG be used to argue that they will go to zero, or at least [itex]{y_x\over y_t} << 1[/itex]??
     
  10. Oct 25, 2014 #9
    I don't recall anyone claiming that for any (MS)SM RGE solutions. That's likely because these particles' gauge couplings are usually larger than their Yukawa ones, and when gauge couplings dominate, you get behavior like y ~ tc.
     
  11. Oct 25, 2014 #10

    arivero

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    I think to remember now that the article of Cecilia Jarlskow with Georgi was also a RG argument.
     
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