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Attractive nature of nuclear forces

  1. Feb 14, 2012 #1
    "At a distance of 1.5 fermi, the nuclear force becomes maximum and remains attractive in nature. However, at a distance of 0.5 fermi, the nuclear force suddenly becomes repulsive. "

    Revered Members,
    Nuclear force is a short range force, so naturally when the distance between the nucleons decrease, there should be more attractive force. I can't understand why repulsive force becomes predominant when the distance between the nucleons is of the order of 0.5 fermi.
  2. jcsd
  3. Feb 14, 2012 #2


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    To really answer this question is quite difficult, since nucleons are made of quarks and gluons, and a full explanation must be sought in lattice QCD calculations. However you can understand it qualitatively the old-fashioned way, in terms of meson exchange. The original idea by Yukawa was that the nuclear force was caused by exchange of an as yet unknown particle. The range of the force would be inversely related to the particle's mass. Sure enough the pion was discovered, with a mass of 140 MeV, implying a range of 197.5/140 = 1.4f.

    But is the exchange force attractive or repulsive? That depends on the properties of the particle, namely its behavior under charge conjugation. Pion exchange, it turns out, is attractive. The repulsive part has been sometimes explained by exchange of another particle, the ω meson with spin one and a mass of 780 MeV. The range this time is 197.5/780 = 0.25 f.

    There are more elaborate explanations floating around. My main point is that particle exchange does not have to imply an attractive potential, it can be either.
  4. Feb 16, 2012 #3


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    Could you please elaborate more on this?
    I also have no real understanding as how to read of from the characteristics of the particle exchanged whether it will lead to attraction or repulsion. I remember Zee discussing it in his QFT book and linking it to the spin of the particle although I found his argument still quite involved.
  5. Feb 16, 2012 #4

    'why' questions ultimately have very few good answers. 'what' happens we are better at describing because we have observational evidence.

    As easy way to think about the replusion is Pauli exclusion...but of course that doesn't
    really explain 'why'.....

    Check this out for a description of observed characteristics,

    but again exactly 'why' all this occurs is still a bit beyond our reach.
  6. Feb 16, 2012 #5
    ok, it IS a result of 'quantum degeneracy pressure'.....I've only read about the electron and neutron degeneracy in stars
    version, and wikie doesn't have an article on the more general topic.


    edit: Found this additional tidbit:
    "Neutron Degeneracy Pressure: Quantum mechanics restricts the number of neutrons that can have low energy. Each neutron must occupy its own energy state. When neutrons are packed together, as they are in a neutron star, the number of available low energy states is too small and many neutrons are forced into high energy states. These high energy neutrons make up the entire pressure supporting the neutron star. Because the pressure arises from this quantum mechanical effect, it is insensitive to temperature, i.e., the pressure doesn't go down as the star cools. Similar to electron degeneracy pressure but, because the neutron is much more massive than the electron, neutron degeneracy pressure is much larger and can support stars more massive than the Chandrasekhar mass limit..."

  7. Feb 16, 2012 #6


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    Naty1, Pauli degeneracy cannot be the explanation, since the same repulsive core exists in the interaction between a proton and a neutron. Also between two protons, the exclusion principle needs to be taken into account in the triplet state (spins the same) but does not affect the singlet state.
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