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Nuclear force as residual color force

  1. Feb 27, 2010 #1
    Hi all,

    I'm can't wrap my head around this. I get how the strong interaction bonds quarks into baryons and I get the pion exchange model for the attraction between nucleons (well sort of, I still don't understand how exchanging massive particles can result in an attractive force, but OK).

    What I don't understand is how the latter is a residual effect of the former. The color interaction leaves flavor untouched, but the pion exchanges flavors between (the quarks of) nucleons. So how exactly does the pion form?

    There is a Feynman-Diagram in the Wikipedia article on nuclear force (http://en.wikipedia.org/wiki/Nuclear_force), but I don't understand that either. Not very good with Feynman diagrams.

    Can anyone explain this to me in preferably plain English? I'm a physics student, so jargon is OK, but I have almost no background in QCD, so go easy on the maths, if possible.

    Thanks in advance.
    /W
     
  2. jcsd
  3. Feb 27, 2010 #2

    bcrowell

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    OK, this may be an even more naive explanation than you're looking for, but here it is at the crayon level at which I understand it.

    Naively, you wouldn't expect two hydrogen atoms to bond to form a molecule. They're both electrically neutral, so the force between them should vanish. It fails to vanish only because of polarization effects, which produce an interaction that's weak and falls off more rapidly than the Coulomb force (say [itex]r^{-7}[/itex] (?), rather than [itex]r^{-2}[/itex]).

    Same thing with a neutron and a proton. They're both color-neutral, so naively they shouldn't bond to form a deuteron. The attraction fails to vanish only because of polarization effects, which produce an interaction that's weak and falls off more rapidly than the color force (say [itex]e^{-kr}[/itex], rather than [itex]r[/itex]).

    I'm not even sure you need anything explicitly quantum-mechanical in order to understand this. The electrical version can certainly be understood qualitatively in terms of classical E&M. So it may not be necessary to invoke exchange of quanta at all.
     
  4. Feb 27, 2010 #3
    Thanks bcrowell. I've heard that analogy before, and I'm quite OK with it (NB: Hydrogen bonds are covalent, not van der Waals-like). However, my question is specifically about the transition from the color force to the nuclear force, or, in terms of particles, how the gluon-based force becomes a meson-based force. (I can accept that you would use different models for different kinds of interactions, but since it is said that one results from the other, there should be an argument connecting both.)
     
  5. Feb 27, 2010 #4
    For two sources of the same charge, the exchange of spin zero particles produces an attractive force, the exchange of spin one particles produces a repulsive force, and the exchange of spin two particles results in an attractive force. Mass doesn't matter, just spin. The pattern is +-+-+- (where + is attractive, - repulsive)... Tony Zee's book "QFT in a Nutshell" explains this very well (and very early on).

    There's actually a name for the meson-exchange picture of QCD: chiral perturbation theory, which is an effective field theory that works for energies less than .2 GeV. It is based on flavor symmetry, whereas the color theory of quarks and gluons is based on color symmetry.

    I don't really have a problem with flavor of the quarks in the baryons being altered as long as the flavor is conserved (which it is, in the meson), because quarks can wander where they want, so I don't necessarily view it as a flavor-changing process, but a quark leaving process. This is just my interpretation though.
     
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