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Neutron-only nuclei

  1. Jan 21, 2009 #1
    Since both neutrons and protons feel the attractive strong interaction, and only protons feel the repulsive coulomb interaction, shouldn't 'nuclei' made of neutrons and no protons, be more stable than nuclei with roughly equal numbers of protons and neutrons?
    Or am I missing something here? (does the strong force affect neutrons and protons differently?)
     
  2. jcsd
  3. Jan 21, 2009 #2
    Is it to do with protons being more stable (weak force)?
     
  4. Jan 21, 2009 #3

    vanesch

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    The nuclear force is not purely attractive, because it is spin-dependent. But "attractive" isn't the right language. One needs to have a bound state as ground state. A neutron-neutron bound state requires the neutrons to have opposite spins (Fermi exclusion principle) and then there is simply no bound state for this case.
     
  5. Jan 21, 2009 #4

    turin

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    How did you come to this conclusion?
     
  6. Jan 21, 2009 #5

    vanesch

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    I didn't come to this conclusion from first principles! It just turns out for NN to be so.
    The potential determines if a bound state exists or not, and in this case, it seems not to.
    The trick is that the potential is spin-dependent. So for deuterium (spin +1), we have the proton and the neutron with parallel spins, and then a bound state exists (deuterium) ; but for the NN case, the spins cannot be parallel (fermi) and then it turns out (it could have been different) that for *this* potential, there is no bound state.
     
  7. Jan 22, 2009 #6

    Vanadium 50

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    It's also true that the di-neutron is almost bound. But as they say, close only counts in horseshoes, dancing and hand grenades.
     
  8. Jan 22, 2009 #7
    nuclei with too many neutrons tend to decay by beta decay whereby a neutron converts to a proton and an electron (and a neutrino).
     
  9. Jan 22, 2009 #8

    turin

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    However, doesn't this raise the question of how a nucleas with any neutrons at all can be stable against beta decay? For example, why does an isolated neutron beta decay with a half-life of 10 minutes whereas a neutron bound to a proton (i.e. deuterium) is quite stable. I suspect that this notion of an identifiable neutron and proton as two hadrons in a bound state is implausible. Apparently there is a formalism called chiral perturbation theory that allows the calculation of the properties of bound states of hadrons, but I am quite unfamiliar with this formalism, and I don't understand how it incorporates the weak force. It almost seems as if certain configurations of hadrons can allow the strong force to block the weak force.

    There seem to be two very different decay mechanisms begin discussed. For the di-neutron, it has been claimed that there can be no bound state, and I am interpretting the reason as neutron repulsion via the strong force. This is the explanation for why a nucleas of two neutrons and no protons cannot exist, and it apparently has nothing to do with beta decay. For a large nucleas of many more neutrons than protons, the instability seems to be due to the completely different mechanism of the weak force, and has apparently nothing to do with the strong force repulsion of hadrons.
     
    Last edited: Jan 22, 2009
  10. Jan 24, 2009 #9
    There is one exotic case, in the words of my lecturer: "Do neutron stars count as whop-off nuclei?"
     
  11. Jan 24, 2009 #10

    turin

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    Supposedly, neutron stars have complex composition, and they are not simply a lump of neutrons as is typically naively assumed.
     
  12. Jan 24, 2009 #11

    malawi_glenn

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    and also, neutron stars are held togheter by Gravitation, not the strong force.
     
  13. Jan 24, 2009 #12
    Yes, I have always found it rather strange that people would make such an analogy. I guess, the reason is that nuclear model predictions nowadays are tested not against atomic nuclei data, but also against other model predictions wrt neutron stars, for so called nuclear matter "equation of state".
     
  14. Jan 25, 2009 #13

    turin

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    The density of a neutron star is even greater than the density in an atomic nucleas, suggesting that the nucleons in a neutron star are even closer together than the nucleons in an atomic nucleas. Therefore, the strong force between neighboring nucleons in a neutron star is no more negligible than it is between neighboring nucleons in an atomic nucleas. So, if it is the strong force that protons exert on neutrons that is responsible for inhibitting beta decay of the neutrons, then this effect should also be important in neutron stars, and perhaps provides a consistent reason why the neutrons in a neutron star are stable against beta decay.
     
  15. Jan 25, 2009 #14
    You are perfectly right in your description. Does this structure which you seem to know qualifies to be called "nucleus" in your opinion ?
     
  16. Jan 28, 2009 #15

    turin

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    I would emphasize the different mechanisms at work in the case of atomic nuclei vs. neutrons stars. As a thought experiment, I would think that a neutron star would simply drift apart into microscopic fragments if gravity were suddenly turned off, since there is no other attractive force that would hold together fragments on the order of kilometer separation. So, I don't consider a neutron star as anything like an atomic nucleas, because, even though I believe the strong force remains important, I don't believe that gravity can be neglected as it can be for an atomic nucleas.
     
  17. Jan 28, 2009 #16
    I heard that something like 3000 neutrons would make a stable nucleus.
     
  18. Jan 29, 2009 #17

    Vanadium 50

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    From whom? Is this published anywhere?
     
  19. Jan 29, 2009 #18
    I herd it from the grapevine.
     
  20. Jan 29, 2009 #19

    malawi_glenn

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    and that is?
     
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