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Nuclear force between neutrons

  1. May 31, 2010 #1
    Why is it that neutrons don't stick together? As far as I know the nuclear force is also acting between 2 neutrons and not just between neutrons and protons.
    So why are there no ultra dense neutron objects consisting of a large number of neutrons, except of course for neutron stars where gravity pulls the particles together?
     
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  3. May 31, 2010 #2

    Meir Achuz

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    The neutron-neutron force is a bit two weak to form pure bound neutron states.
     
  4. May 31, 2010 #3

    tom.stoer

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    Afaik it's due to combination the Pauli principle with the effective nucleon-nucleon interaction. Neutrons are fermions, that means their spins must be antiparallel. But the strong force is repulsive for antiparallel spins. By the same principle diprotons are forbidden.
     
  5. May 31, 2010 #4

    Meir Achuz

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    No. The N-N force is not repulsive for anti-parallel spins.
    It is still attractive, but a little bit weaker than for parallel spins.
     
  6. May 31, 2010 #5
    In 3d, a sufficiently shallow potential well has no bound states.
     
  7. Jun 1, 2010 #6

    tom.stoer

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    Then please explain the absence of a di-neutron and a di-proton state.
     
  8. Jun 1, 2010 #7
    di proton = electric repulsion
     
  9. Jun 1, 2010 #8

    tom.stoer

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    That doesn't matter; there is no di-neutron, either; electromagnetic forces are mostly irrelevant.
     
  10. Jun 1, 2010 #9

    jimgraber

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    A quick, naive look at the chart of nuclides would seem to say that both the effective proton-proton force and the effective neutron-neutron force are repulsive but that the effective neutron-proton force is attractive. (regardless of spin?. or overwhelming spin?) what does chromodynamics say?
    yours naively,
    Jim Graber
     
  11. Jun 1, 2010 #10

    tom.stoer

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    Looking a phenomenologically successful nucleon-nucleon potentials derived from one-pion exchange one finds

    [tex]V_C(r) \sim \frac{e^{-m_\pi r}}{r}[/tex]

    [tex]V_S(r,S) \sim V_S(r) (\vec{s}_1\vec{s}_2)[/tex]

    plus further terms, e.g. tensor and LS.

    For the spin-spin coupling one has

    [tex]\langle \vec{s}_1 \vec{s}_2\rangle = \frac{1}{2}\left[S(S+1) - s_1(s_1+1) -s_2(s_2+1)\right][/tex]

    Therfeore

    [tex]\langle \vec{s}_1 \vec{s}_2\rangle_{S=0} = -\frac{3}{4}[/tex]

    [tex]\langle \vec{s}_1 \vec{s}_2\rangle_{S=1} = +\frac{1}{4}[/tex]

    As the diproton or dineutron is symmetric in isospin it must by antisymmetric in spin which corresponds to S=0. So the term

    [tex]V_S(r,S=0) \sim -\frac{3}{4} V_S(r) [/tex]

    has - minus sign whereas

    [tex]V_S(r,S=1) \sim +\frac{1}{4} V_S(r) [/tex]

    has a + sign.

    I do not know if for S=0 this results in a repulsive potential [tex]V_C(r) + V_S(r,S=0)[/tex]. I have found my old notes indicating that at very low energies the scattering phases for NN scattering indicate an attractive potential for both S=0 and S=1, but that attraction is not strong enough to form a bound state.

    This goes into the direction of Dickfore and Meir Achuz.
     
  12. Jun 1, 2010 #11

    Meir Achuz

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    Dickfore:
    "Re: nuclear force between neutrons
    In 3d, a sufficiently shallow potential well has no bound states."
     
  13. Jun 1, 2010 #12
    binding energy per nucleon...


    Because of the Pauli exclusion principle.

    According to the semi-empirical mass formula (SEMF) or liquid drop model:

    Diprotium binding energy per nucleon:
    [tex]E_b = -26.5625 \; \text{MeV}[/tex]

    Dineutronium binding energy per nucleon:
    [tex]E_b = -26.7005 \; \text{MeV}[/tex]

    Deuterium binding energy per nucleon:
    [tex]E_b = -2.90318 \; \text{MeV}[/tex]

    According to quantum mechanics:

    Deuterium binding energy per nucleon: spin 0 'singlet' - virtual
    [tex]E_b = -30 \; \text{keV}[/tex]

    The negative signs indicates that the nuclei are repulsive and the model experiences spontaneous fission.

    According to mass defect:

    Deuterium binding energy per nucleon: spin +1 'triplet'
    [tex]E_b = 1.11226 \; \text{MeV}[/tex]

    In nuclear terms, the Deuterium nucleus is just barely stable and susceptible to fracture inside stellar cores. Although Diprotium and Dineutronium cannot form bound states due to the Pauli exclusion principle, they can experience a nuclear reaction via the weak nuclear force resulting in a bound state Deuterium.

    Endothermic weak nuclear reaction:
    [tex]^1_1p + ^1_1p \rightarrow ^2_1D + e^+ + \nu_e + 0.42 \; \text{MeV}[/tex]

    Exothermic weak nuclear reaction:
    [tex]^1_0n + ^1_0n \rightarrow ^2_1D + e^- + \overline{\nu}_e + 3.00691 \; \text{MeV}[/tex]


    Reference:
    http://en.wikipedia.org/wiki/Pauli_exclusion_principle" [Broken]
    http://en.wikipedia.org/wiki/Semi-empirical_mass_formula" [Broken]
    http://en.wikipedia.org/wiki/Deuterium#Quantum_properties"
     
    Last edited by a moderator: May 4, 2017
  14. Jun 2, 2010 #13

    tom.stoer

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    Re: binding energy per nucleon...

    No quite; look at our discussion above and your own arguments below: it's a it's a bit more complicated than that. The Pauli exclusion principle only tells you that neither diproton nor dineutron can exist in a spin triplet state. Only via further calculations from a specific NN-potential it follows that the singlet state is not a bound state.

    This is not a derivation, but only a "collection of facts". A derivation should start from an NN-potential.

    According to quantum mechanics:

    Which NN-potential is used?
     
  15. Jun 2, 2010 #14

    Vanadium 50

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    I'm afraid Wikiscraping isn't sufficient here. This takes some actual understanding.

    As pointed out, the n-n potential is attractive. It's just not attractive enough to bind.

    More interesting is the case of the tetraneutron (4n). Calculations have oscillated back and forth over the years on whether it's expected to be bound or not. Experimental evidence for one is not very good - most searches are negative. It's absolutely certain that if it is bound, it's loosely bound (less than 0.7 MeV per nucleon, less than a tenth of typical nuclear binding).

    I don't believe it exists. I find the lack of a signal in DCX reactions compelling evidence for its absence. Others have different opinions.
     
  16. Jun 2, 2010 #15

    tom.stoer

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    I found this overview http://www.physics.umd.edu/courses/Phys741/xji/chapter6.pdf which presents a first introduction.

    It is explained that at low energies S-wave scattering is dominant. A few remarks regarding scattering phases and scattering length follow. A virtual bound state for T=1 is mentioned, too. The conclusion is - again: "the potential is attractive but not attractive enough".

    For further investigation I think the "Nijmegen group" and the work around the "CD Bonn potential" is interesting:

    http://arxiv.org/abs/nucl-th/0410042v1
    Partial-Wave Analyses of all Proton-Proton and Neutron-Proton Data Below 500 MeV
    M.C.M. Rentmeester (1), R.G.E. Timmermans (2), J.J. de Swart (1) ((1) Radboud University Nijmegen, (2) KVI, Groningen)
    (Submitted on 9 Oct 2004)
    Abstract: In 1993 the Nijmegen group published the results of energy-dependent partial-wave analyses (PWAs) of the nucleon-nucleon (NN) scattering data for laboratory kinetic energies below Tlab=350 MeV (PWA93). In this talk some general aspects, but also the newest developments on the Nijmegen NN PWAs are reported. We have almost finished a new energy-dependent PWA and will discuss some typical aspects of this new PWA; where it differs from PWA93, but also what future developments might be, or should be.

    http://arxiv.org/abs/nucl-th/0006014
    The high-precision, charge-dependent Bonn nucleon-nucleon potential (CD-Bonn)
    R. Machleidt (University of Idaho)
    (Submitted on 9 Jun 2000)
    Abstract: We present a charge-dependent nucleon-nucleon (NN) potential that fits the world proton-proton data below 350 MeV available in the year of 2000 with a chi^2 per datum of 1.01 for 2932 data and the corresponding neutron-proton data with chi^2/datum = 1.02 for 3058 data. This reproduction of the NN data is more accurate than by any phase-shift analysis and any other NN potential. The charge-dependence of the present potential (that has been dubbed `CD-Bonn') is based upon the predictions by the Bonn Full Model for charge-symmetry and charge-independence breaking in all partial waves with J <= 4. The potential is represented in terms of the covariant Feynman amplitudes for one-boson exchange which are nonlocal. Therefore, the off-shell behavior of the CD-Bonn potential differs in a characteristic and well-founded way from commonly used local potentials and leads to larger binding energies in nuclear few- and many-body systems, where underbinding is a persistent problem.
     
  17. Jun 3, 2010 #16
    tetraneutronium...


    The model is based upon a 3 dimensional infinitely repulsive spherical finite potential well of nuclear radius [tex]r_0[/tex].

    Nuclear potential well boundary conditions:
    [tex]V(r) = \left\{ \begin{array}{ll} V_{0} &, \ 0 \le r \le r_{0} \\ 0 &, r_{0} < r \right. \end{array}[/tex]

    Matter wave wavenumber: (ref. 1)
    [tex]k = \frac{\sqrt{2m E}}{\hbar}[/tex]

    Matter wavenumbers: (ref. 2)
    [tex]K' = \frac{\sqrt{m(V_0 + E)}}{\hbar}[/tex] and [tex]k = \frac{\sqrt{mE}}{\hbar}[/tex]

    Deuterium spin 0 'singlet' virtual state energy:
    [tex]\boxed{E_1 = -\frac{\pi \hbar^2 E_b}{3 \pi \hbar^2 - m_p \sigma_p E_b}}[/tex]

    [tex]E_b[/tex] - total binding energy
    [tex]\sigma_p[/tex] - proton cross section

    Deuterium binding energy per nucleon: spin 0 'singlet' virtual state
    [tex]E_1 = -34.9685 \; \text{keV}[/tex]

    Deuterium binding energy per nucleon: spin +1 'triplet'
    [tex]E_b = 1.11226 \; \text{MeV}[/tex]

    According to the semi-empirical mass formula (SEMF) or liquid drop model:

    Tetraneutronium binding energy per nucleon:
    [tex]E_b = -17.9688 \; \text{MeV}[/tex]

    Substituting in neutron mass and Helium binding energy:

    Tetraneutronium spin 0 'singlet' virtual state energy:
    [tex]\boxed{E_1 = -\frac{\pi \hbar^2 E_b}{3 \pi \hbar^2 - m_n \sigma_p E_b}}[/tex]

    Tetraneutronium binding energy per nucleon: spin 0 'singlet' virtual state
    [tex]E_1 = -16.05825 \; \text{keV}[/tex]

    The negative signs indicates that the nuclei are repulsive and the model experiences spontaneous fission and is nearly as unstable as Diprotonium and Dineutronium and singlet Deuterium.

    Tetraneutronium cannot form a stable bound state, however it can experience a nuclear reaction via the weak nuclear force resulting in a bound state Helium nucleus.

    Exothermic weak nuclear reaction:
    [tex]4 ^1_0n \rightarrow ^4_2He + 2 e^- + 2 \overline{\nu}_e + 29.8615 \; \text{MeV}[/tex]

    Reference:
    http://en.wikipedia.org/wiki/Scattering_length" [Broken]
    http://mightylib.mit.edu/Course%20Materials/22.101/Fall%202004/Notes/Part3.pdf" [Broken]
    http://en.wikipedia.org/wiki/Semi-empirical_mass_formula" [Broken]
    http://en.wikipedia.org/wiki/Wavenumber" [Broken]
     
    Last edited by a moderator: May 4, 2017
  18. Jun 3, 2010 #17
    can you write down the NN potential or not dear Orion1???
     
  19. Jun 3, 2010 #18
    NN potentials...

    Negative, the reference does not state which NN potential is used in the differential scattering model and the NN potential for Quantum Chromodynamics (QCD) is unknown to me.

    According to Wikipedia the most prevalent models are based on Yukawa, Paris, Argonne AV18, CD-Bonn and Nijmegen potentials.

    My preference is the Yukawa potential:
    [tex]V(r)= -g^2 \; \frac{e^{-m_{\pi} r}}{r}[/tex]

    [tex]V_0 = 18 \; \text{MeV}[/tex] - Deuterium spin = 0 'singlet'
    [tex]V_0 = 36 \; \text{MeV}[/tex] - Deuterium spin = +1 'triplet'

    Reference:
    http://en.wikipedia.org/wiki/Nuclear_force#Nucleon-nucleon_potentials"
    http://en.wikipedia.org/wiki/Yukawa_potential" [Broken]
     
    Last edited by a moderator: May 4, 2017
  20. Jun 3, 2010 #19
    Re: tetraneutronium...

    Actually, it was stated in the first statement of his previous post:
    This would mean:

    [tex]
    V(r) = \left\{ \begin{array}{ll}
    V_{0} &, \ 0 \le r \le r_{0} \\

    0 &, r_{0} < r
    \right. \end{array}
    [/tex]

    EDIT:

    A repulsive force is assumed here, hence no bound state is possible.
     
  21. Jun 3, 2010 #20

    tom.stoer

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    Re: NN potentials...

    That does not really help. Yukawa is simple but far from realistic. In our context here it fails to explain the difference between S=0 and S=1 (Yukawa is explicitly spin independent).

    Unfortunately you have to use the more complicated potentials like CD-Bonn etc.
     
    Last edited: Jun 3, 2010
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