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What quanta mediates the nuclear forces?

  1. Mar 27, 2008 #1
    What particle mediates the nuclear forces amongst the protons and neutrons inside nuclei? Yukawa thought it was the pions that mediated the nuclear force, but I am yet to see a good theory (or even a model) that can explain nuclear binding quantitatively.:confused:

    Before you answer, a clarification is in order: I am not asking about the strong forces that confine the quarks and make hadrons. That's QCD physics with quarks and gluons. However, inside nuclei the protons and neutrons do not dissolve into a quark and gluon soup. They remain as protons and neutrons. Specifically, they remain as color-singlet objects. This is not unexpected, since the nuclear binding energy (per nucleon) is at least two orders of magnitude smaller than the QCD binding energy, or the mass of the nucleons. (The highest nuclear binding energy per nucleon is only 8.9 Mev for some Iron isotopes.)

    So what is the modern prevailing view on the mediator of nuclear forces? Pions still? Or other mesons? Or even multiple gluon exchanges? It would be very difficult to get flavor (isospin) dependent forces by gluon exchanges, which nuclear forces seem to be. (The attraction in between two neutrons (or two protons) is a lot weaker than the attraction between a proton and neutron.)
     
  2. jcsd
  3. Mar 27, 2008 #2
    Pions!
     
  4. Mar 27, 2008 #3
  5. Mar 27, 2008 #4

    malawi_glenn

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    yeah, also etas and omegas are exhange particles in the "meson exhange model of the strong nuclear force" :-)
     
  6. Mar 27, 2008 #5

    blechman

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    If you like, you can think of it as the nucleons are exchanging "gluons" but since (as you so correctly pointed out) you're in the far IR as far as QCD is concerned, there is an exchange of an "infinite number of gluons" (perturbation theory has broken down) that then have an "effective description" in terms of mesonic modes such as pions, etc.
     
  7. Mar 28, 2008 #6
    I thank Humanino for the two papers he suggested. I downloaded them immediately and I have been reading them since yesterday. Both papers are very interesting and they both are related to my question. However, neither paper addresses the question directly, especially the second one. Still, reading these papers was helpful for me.

    I like Blechman's response very much, because that's exactly how I would like to think about the nuclei. The problem with that is that while it is a good (and probably the correct) answer qualitatively, it is not of much use in quantitative computations. I am wondering if there may be models for computing nuclear binding energies for many hundreds of nuclear isotopes with reasonable accuracy (say 10% or better.) Any model that lends itself to computation can tune its parameters on a few isotopes, and make predictions on the other 500. That's the sort of information I am after, if indeed it exists.
     
  8. Mar 28, 2008 #7

    blechman

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    You're right: it's completely useless quantitatively!

    Actually computing binding energies is a very nontrivial problem, and to my knowledge it has not been solved. In fact, it's quite a serious problem: the binding energies of nuclei are of order a few MeV, and the only physical energy scale we have to describe such things is the QCD scale, of order a few HUNDRED MeV. I do not believe we have a satisfactory answer for why these scales are so disparate! There are plenty of models, each one as contrived as the others, but no first principles. Nuclear theorist colleagues of mine sometimes call this the "deuteron problem" (since that's pretty much the simplest nontrivial nucleus around, they start with that: the binding energy is 2.2 MeV - about a factor of 100 smaller than theoretical estimates from QCD).
     
  9. Mar 28, 2008 #8

    Vanadium 50

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    I don't understand this argument at all. You could make the same argument about the hydrogen atom - the only energy scale is the (reduced) mass of the electron, but the actual binding energy is only 1/35000 of this.
     
  10. Mar 28, 2008 #9

    Haelfix

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    The hyrdogen atom as you know, is a ridiculously special case (and a historically unfortunate one since it probably delayed QFT from developing for a good ten years).

    Of course, I think Blechman is also referring to infrared slavery here, so that further complicates matters.
     
  11. Mar 29, 2008 #10

    Vanadium 50

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    Why is it a "ridiculously special case"?

    I would have expected a 1940's physicist to think along the following lines: the binding energy of the hydrogen atom is alpha^2/4pi * the reduced mass of the electron. If the same expression is true for the deuteron - it's bound by alpha_nuclear^2/4p * (mp/2), that means alpha_nuclear is about 1/4. (Of course, the force isn't inverse square, so this is really more an order of magnitude estimate than a calculation)

    This makes sense - nucleons interact quite strongly, and we know the interaction strength has to be greater than that of electromagnetism to bind all this positive charge in a small space. It also makes a prediction - a large alpha_nuclear means that spin effects are important, and they are: the only (n,p) bound state has spin 1.
     
  12. Mar 29, 2008 #11

    blechman

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    hydrogen is NOT a counterexample; in fact it works perfectly because you also have to count powers of the fine structure constant! Ryberg's energy is [itex]\frac{1}{2}mc^2\alpha^2[/itex], exactly what power counting would have you believe (well, up to the 1/2, but big deal!). QCD doesn't work that way: the "fine-structure constant of QCD" is order 1 (or even larger!) so you don't get the same kind of suppression you get for QED.
     
  13. Mar 29, 2008 #12

    blechman

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    You're argument about trying to reproduce the Rydberg formula in QCD leads me to estimate the binding energy as [itex]\frac{1}{2}\Lambda c^2\alpha_s^2[/itex] where [itex]\Lambda\sim 200[/itex] MeV is the QCD scale (roughly of order mp/2, so I won't begrudge you that) - how do you get this to reproduce 2.2 MeV when [itex]\alpha_s[/itex] is experimentally large? I don't see where the [itex]4\pi[/itex]'s in your formula came from (I'm using [itex]\alpha\equiv\frac{e^2}{4\pi\hbar c}=\frac{1}{137}[/itex]).
     
  14. Mar 29, 2008 #13

    Vanadium 50

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    Remember, alpha_nuclear is not the same as alpha_s, even though they both arise from the strong interaction. Alpha_s is the strength of the quark-gluon coupling (ignoring terms of order 1 from the contraction of the T-matrices). Alpha_nuclear is more like the strength of the proton-pion coupling (off the Delta resonance), which is a smaller number - the proton and the pion are both colorless objects.

    An EM analogy would be that van der Waals-type forces exist between neutral objects, but are weaker than the forces between charged objects.
     
  15. Mar 31, 2008 #14
    Can we really go through this analogy between nuclear forces and Van der Waals as we often hear ?
    I mean I understand there is a charge cancellation in both cases and so we are only sensitive to higher order effects. But, as QCD is highly non-perturbative in this energy domain contrary to QED, how far can we go with the comparison ?
     
  16. Mar 31, 2008 #15

    Haelfix

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    Vanadium, the reason Hydrogen is a special case is that in the QFT treatment of bound states (already a notoriously difficult approximative scheme, where you often treat things like they were scattering rather than continously interacting) it is the one example where it fits to a few percent with experiment. The reason for this, is b/c in the dynamics there are unusually small matrix elements that make certain approximations tractable. This is related to the fact that radiative corrections are small, and things like the Dirac equation work pretty well.

    Not so with QCD, where you really have to work hard to understand the scales in question, and why the bound state for the Deuteron yields rather poor results. Not only is the interaction much stronger, as you end up having to deal with the full dynamics of a strongly coupled regime, but the further complication of adding even more bodies to the problem (making things like treating the proton as an external potential hard).

    Even taking the nonrelativistic limit doesn't help the calculation much (assuming you are starting with field theory ad initio).

    So it is true to say that the full treatment of the bound state problem is at this time, limited to straight quantum mechanics and treating things much like those 1940 theorists did (eg pion exchange)
     
  17. Mar 31, 2008 #16
    The question is rather involved. The reason for this can be traced back to the computation of the Yang-Mills propagator in the infrared and what the quasi-particles of the theory are in this regime. I should expect that these particles are just glueballs that are bound states of Yang-Mills theory in the infrared.

    The computation of the proper gluon propagator in the infrared is currently hot debated and lattice results give evidence for a finite value of the propagator at zero momentum. Once the propagator is known, QCD reduces to a Nambu-Jona-Lasinio model if indeed the propagator is finite at zero momentum. You can see all this in

    http://arxiv.org/abs/0803.0319

    Jon
     
  18. Mar 31, 2008 #17
    Hello Jon,

    thanks for the link. Looking a bit at the article, I have a naive question :
    what does the assumption [tex]\lambda^{a}D^{ab}\lambda^{b} \rightarrow \lambda^{a}G(x-y)\lambda^{a}[/tex] come from ? Is it this line which means that we have colorless objects as a first assumption ?
     
  19. Apr 1, 2008 #18
    Hi,

    the reason for this is a standard result of QCD that the gluon propagator is diagonal in the generator indeces and then can always be written as

    [tex]D^{ab}(x-y)=\delta^{ab}G(x-y)[/tex].

    Jon
     
    Last edited: Apr 1, 2008
  20. Apr 1, 2008 #19
    Ok thanks... Looking back to my old courses, indeed I found something like [tex]\frac{\delta^{ab}}{k^{2} - i\epsilon}(g_{\mu\nu} - (1-\varsigma)(\frac{k^{\mu}k^{\nu}}{k^{2}}))[/tex]
     
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