| firstname.lastname@example.org (email@example.com) writes:|
> I wonder why pure small neutral nuclei (e.g. only 4 neutrons instead
> of 2 protons and 2 neutrons like in He) seemed not to exist.
Because the "strong force" is only strongly attractive between nucleons
with the same spin and opposite isospin, i.e., between spin-aligned protons
and neutrons. By contrast, the strong-force interaction between same isospin,
opposite spin nucleons (i.e., two protons or two neutrons) is only mildly
attractive, and not strong enough to form a "di-nucleon" --- and all other
combinations are repulsive. Since every combination of particles in a
"tetraneutron" is at best weakly attractive and at worst repulsive,
and since it is impossible to put all four neutrons into the ground state
"shell" without violating the pauli exclusion principle, a "tetraneutron"
is not a "bound state," i.e., not a lower energy configuration than four
free neutrons --- and certainly not a stable state!
(NOTE: There was a Russian group that _claimed_ to have observed a short-
lived "tetraneutron" about two years ago, but it has not been replicated,
and so far no one seems to be taking their claim very seriously, since
such a "tetraneutron" state would strongly conflict with our understanding
of the strong force and with constraints imposed by the pauli exclusion
> I know that giant stars can transform in so called "neutron stars",
> that consist only of neutrons.
So-called "neutron stars" are only _mostly_ neutrons. They have a crust
of "normal" matter (if one can call a coulomb crystal of nuclei in a sea
of degenerate electrons "normal" matter!) that gradually become more
neutron-rich as one goes deep; a "mantle" of superfluid neutrons in
weak-interaction equilibrium with a minority population of protons
and electrons, and possibly a "core" consisting of a hyperon or
> These kind of single neutral nuclei normally should be more likely
> and even more stable, because there is no electromagnetical repulsive
> forces between each other like in normal nuclei.
There is also no strong attraction between opposite isospin nucleons, nor a
natural satisfaction of the pauli exclusion principle, as in normal nuclei ---
only mild attraction, strong repulsion, and no way to put all four nucleons
into the ground state without violating the pauli exclusion principle.
> It is clear that single neutrons are not stable and decay within
> about 10 Min., but bounded neutrons should exist.
No, they should not: Theory shows that "dineutrons" and "tetraneutrons"
are not bound states, and with the exception of the single non-replicated
Russian claim, so does experiment.
> In standardmodel of particle physics all quarks are equal to the strength
> of the strong interaction. The strong force do not distinguish between
> the flavor, so there is no sensible reason why neutral nuclei should not
It is inaccurate to say that the strong force "[does] not distinguish"
between flavor," since the QCD interaction also depends on both spin and
"weak isospin" (`up' quarks are "weak isospin up," and `down' quarks are
"weak isospin down). The spin and weak-isospin dependence of the QCD force
plus the need to satisfy the pauli exclusion principle again ensures that
a "tetraneutron bag" is not bound.
-- Gordon D. Pusch
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