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A Attractive strong force, isospin and hypercharges

  1. Aug 12, 2017 #1
    In the electromagnetic interaction, opposite electric charges q attract each other.

    In the strong nuclear force,
    • the proton p(uud) is attracted to p(uud) and the neutron n(udd), and
    • n(udd) is attracted to p(uud) and n(udd).
    Both neutrons and protons have
    • a hypercharge Y=+1, and
    • 3rd component of the isospin I3=-1/2 and + 1/2 respectively.
    There are many more heavy baryons (Σ,Δ,Λ) with different I3 (3rd component of isospin) and Y (hypercharge) , positive as well as negative.
    • Are all the baryons attracted to each other, regardless of their 2 quantum numbers Y and I3 (assuming strong interaction acts within their lifetime)?
    • How the sign of the 3rd component of the isospin (I3) and the sign of the hypercharge (Y) are related to this attraction between quarks and between baryons ?
     
  2. jcsd
  3. Aug 12, 2017 #2

    Orodruin

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    Isospin and hypercharge have nothing to do with the strong interactions.
     
  4. Aug 12, 2017 #3

    mfb

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    Yes. A better electromagnetic equivalent would be uncharged molecules interacting via the van der Waals force.
     
  5. Aug 13, 2017 #4

    vanhees71

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    This is a bit misleading. What you probably have in mind is weak isospin and hypercharge.

    On the other hand there's a very important approximate "accidental symmetry" of QCD, the socalled chiral symmetry in the light-quark sector. The reason is that the light quarks have pretty small masses (small compared to the typical hadronic scale of around ##1 \;\text{GeV}##). Thus, neglecting the u- and d-quark masses, leads to a chiral symmetry of the strong interaction with symmetry group ##\mathrm{SU}(2)_{\text{L}} \times \mathrm{SU}(2)_{\text{R}}##.

    This symmetry is, however, spontaneously broken due to strong attraction in the quark-antiquark channel, leading to the formation of a quark condensate, i.e., ##\langle \bar{\psi} \psi \rangle \neq 0##, which makes the chiral symmetry broken to the vector part, ##\mathrm{SU}(2)_{\text{V}}##, leading to three massless pseudoscalar Goldstone bosons, identified with the pions.

    Now the pions are not massless, because also the quarks are not strictly massless, and that's why chiral symmetry is also explicitly broken, but that explicit symmetry breaking is small and can be treated as a perturbation, leading to chiral perturbation theory, which is the most important way to build effective low-energy hadronic models based on fundamental symmetries of QCD.

    The ##\mathrm{SU}_{\text{V}}## symmetry stays intact even if the quarks are massive, but u- and d-quarks should have the same mass then, which is not the case in nature. Thus you have isospin symmetry on the same level of accuracy as chiral symmetry, and that's why isospin symmetry of the strong interactions has been discovered very early by Heisenberg, who grouped proton and neutron to an isospin doublet.

    You can also extent the idea of chiral symmetry to strange quarks, which are however a bit heavier than the u and d quarks. This leads to a modern understanding of Gell-Mann's and Zweig's "eightfold way", which has been discovered as a mathematical pattern to bring order into the zoo of light+strange hadrons and lead to the discovery of quarks and finally QCD as the modern description of the strong interaction.

    For some more details on chiral symmetry on a pretty elementary level, see my transparencies from a recent Lecture Week (Lecture I):

    http://th.physik.uni-frankfurt.de/~hees/hgs-hire-lectweek17/

    A very nice introduction (on which also my transparencies are mostly based) can be found here:

    https://arxiv.org/abs/nucl-th/9706075
     
  6. Aug 13, 2017 #5

    Vanadium 50

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    I'm not sure this is an A level thread, because

    is not entirely right. The n-p force is negligible at large distances, attractive at short distances, and repulsive at even shorter distances. So there isn't a single "sign of the force". (And isospin does play a role here, although weak isospin does not)
     
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