Monday, April 09, 2007

Why di-neutron does not exist?

As previous postings (see this and this) should make clear, nuclear string model works amazingly well. There is however an objection against the model. This is the experimental absence of stable n-n bound state analogous to deuteron favored by lacking Coulomb repulsion and attractive electromagnetic spin-spin interaction in spin 1 state. Same applies to tri-neutron states and possibly also tetra-neutron state. There has been however speculation about the existence of di-neutron and poly-neutron states. One can consider a simple explanation for the absence of genuine poly-neutrons.
  1. The formation of negatively charged bonds with neutrons replaced by protons would minimize both nuclear mass and Coulomb energy although binding energy per nucleon would be reduced and the increase of neutron number in heavy nuclei would be only apparent. As found, this could also explain why heavy nuclei become unstable.

  2. The strongest hypothesis is that mass minimization forces protons and negatively charged color bonds to serve as the basic building bricks of all nuclei. If this were the case, deuteron would be a di-proton having negatively charged color bond. The total binding energy would be only 2.222 -1.293=.9290 MeV. Di-neutron would be impossible for this option since only one color bond can be present in this state.

  3. The small mass difference m(3He)-m(3H)=.018 MeV would have a natural interpretation as Coulomb interaction energy. Tri-neutron would be allowed. Alpha particle would consist of four protons and two negatively charged color bonds and the actual binding energy per nucleon would be by mn-mp/2 smaller than believed. Tetra-neutron would also consist of four-protons and binding energy per nucleon would be smaller by mn-mp than what obtains from the previous estimate. Beta decays would be basically beta decays of exotic quarks associated with color bonds.

Addition. I performed the calculations for the binding energies by assuming that ordinary nuclei have protons and neutral and negatively charged color bonds as building bricks.
  1. The resulting picture is not satisfactory. The model with ordinary neutrons and protons and color bonds works excellently if one assumes that standard isospin dependent strong interaction is present at nuclear space-time sheets besides the color interaction mediated by much longer color magnetic flux tubes. This fits nicely with the visualization of nucleus as a kind of plant such that nuclear space-time sheet serves as a "seed" from which the long color flux tubes emanate from nucleons and return back.

  2. For pn states, which are singlets with respect to strong isospin, this contribution to energy turns out to be surprisingly small, of order .1 MeV: this explains why the fit without this contribution was so good. One can obtain a complete fit for A≤4 nuclei by simple fractal scaling arguments from that for A>4 nuclei by adding this contribution.

  3. If isospin dependent strong contribution is much larger in non-singlet states (expressible in terms of isospin Casimirs) one can understand the experimental absence of poly-neutrons in standard sense of the word. Since color bonds can carry em charges (0,1,-1), exotic nuclear states are however predicted. For instance, 3H with 3 color bonds in principle extends to a multiplet with charges running from +4 to -2. This seems to be an unavoidable prediction of TGD.

For more details see the chapter TGD and Nuclear Physics and the new chapter Nuclear String Hypothesis of "p-Adic Length Scale Hypothesis and Dark Matter Hierarchy".


At 1:47 AM, Anonymous Anonymous said...

search for tri-neutrons :


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