Why don't neutrons bind into large out of control masses?

In summary, neutrons do not tend to form large neutral nuclei due to several reasons. First, the strong nuclear force is repulsive at short distances, making the di-neutron just slightly unbound. Second, free neutrons have a mean lifetime of 880 seconds, limiting their ability to self-aggregate. Third, the spin-dependent component of the nuclear force favors particles with aligned spins, making it difficult for neutrons to bind. Lastly, the Pauli repulsion between same quantum states further prevents neutrons from forming larger bound states.
  • #1
FrankJ777
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I'm trying to understand why neutrons don't just continually bind into large masses. As I understand it proton binding in a nucleus is governed mostly by the strong nuclear force which attracts at close distance and electromagnet force that repels. So for protons to bind, they must have enough energy to get close enough together to overcome the Coulomb barrier. Neutrons on the other hand are subject to the strong nuclear force, but with a neutral charge are not subject to Coulomb forces. So why don't neutrons tend to drift together and form large neutral nuclei?
 
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  • #2
One reason is that free neutrons undergo spontaneous decay with a mean lifetime of 880 seconds.
 
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  • #3
Another reason is that the strong force is repulsive at short diatances. The di-neutron is just slightly unbound.
 
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  • #5
Vanadium 50 said:
Another reason is that the strong force is repulsive at short diatances. The di-neutron is just slightly unbound.
Why then can a proton and neutron bind to form a deuteron? Is it just the electromagnetic interaction between the individual quarks?
 
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  • #6
First, most nucleii are bound at the order of 8000 MeV per nucleon. The deuteron is bound by 1100, and the dineutron is unbound by 20. I don't think the tetraneutron is known well, but probably less than that. Since these are edge cases, you need to consider every feature of the strong interaction to understand why +20 and not -1 (out of 8000).

That said, the fact that the proton and neutron have different isospin helps their binding.
 
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  • #7
vela said:
Why then can a proton and neutron bind to form a deuteron? Is it just the electromagnetic interaction between the individual quarks?
https://en.wikipedia.org/wiki/Nuclear_force#/media/File:ReidForce2.jpg
https://en.wikipedia.org/wiki/Nuclear_force#/media/File:ReidPotential.jpg

https://en.wikipedia.org/wiki/Nuclear_force
The nuclear force has a spin-dependent component. The force is stronger for particles with their spins aligned than for those with their spins anti-aligned. If two particles are the same, such as two neutrons or two protons, the force is not enough to bind the particles, since the spin vectors of two particles of the same type must point in opposite directions when the particles are near each other and are (save for spin) in the same quantum state
 
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  • #8
While neutrons are not repelled by Coulomb repulsion like protons are, both neutrons and protons are subject to Fermi repulsion. Even for neutrons which suffer from Fermi repulsion alone, it seems to be enough to make nuclei unbound if there are too few protons (or, presumably, other baryons) to strengthen the strong force. Which means that neutrons suffer from neutron dripline.
For some oddities, see isotopes of hydrogen:
https://en.wikipedia.org/wiki/Isotopes_of_hydrogen#List_of_isotopes
All known isotopes past 3 are unbound (naturally... first 2 neutrons fill the 1s states, but L shell is too energetic to be bound). But now look at the lifetimes/widths. After H-5 (86 ys), the lifetimes increase to H-7. Isotopes beyond (H-8 and beyond) are unknown, so their widths are unknown... do they keep increasing?
 
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  • #9
snorkack said:
Even for neutrons which suffer from Fermi repulsion alone
Do you mean that the spin-dependent part of the Nucleon-Nucleon potential is solely due to Fermi repulsion?
 
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  • #10
Hyperfine said:
One reason is that free neutrons undergo spontaneous decay with a mean lifetime of 880 seconds.
The spontaneous lifetime is irrelevant for bound neutrons. There are many stable nuclei with many neutrons that never decay.
 
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  • #11
Put "neutron star" into Google.
 
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  • #12
Meir Achuz said:
The spontaneous lifetime is irrelevant for bound neutrons. There are many stable nuclei with many neutrons that never decay.
FrankJ777 said:
So why don't neutrons tend to drift together and form large neutral nuclei?
The question was not about bound neutrons or stable nuclei containing many neutrons. The question was about neutrons, presumably free neutrons, self-aggregating to form larger neutral entities.
Meir Achuz said:
Put "neutron star" into Google.
I fail to see the relevance of this comment given the question posed. Neutron stars, which certainly exist, are not formed by self-aggregation of neutrons. They are formed by collapse of already gravitationally bound systems composed of normal nuclei.
 
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  • #13
Vanadium 50 said:
First, most nucleii are bound at the order of 8000 MeV per nucleon. The deuteron is bound by 1100, and the dineutron is unbound by 20. I don't think the tetraneutron is known well, but probably less than that. Since these are edge cases, you need to consider every feature of the strong interaction to understand why +20 and not -1 (out of 8000).
That should be keV each.
Meir Achuz said:
The spontaneous lifetime is irrelevant for bound neutrons. There are many stable nuclei with many neutrons that never decay.
Yes, they all have protons already. Without protons they could all do beta- decay.
 
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  • #14
mfb said:
That should be keV each.
Yes it should Thanks.
 
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  • #15
malawi_glenn said:
Do you mean that the spin-dependent part of the Nucleon-Nucleon potential is solely due to Fermi repulsion?
Maybe the concept I meant is Pauli repulsion?
If you look at the isobars, like dineutron, para- and ortodeuteron and diproton (of which only ortodeuteron is bound) then the spin-dependent part favours ortodeuteron. But it is the Pauli repulsion that repels both dineutron and diproton from the more strongly bound orto-state.
Dineutron and diproton are both unbound because both have too much Pauli repulsion; but diproton is higher in energy because of Coulomb repulsion. Looking at bound isobars, you can look at the (unbound) trineutron and triproton along with the (bound) triton and 3-He. Triton has the bigger binding energy (and 3-He is a good neutron absorber) because of the Coulomb repulsion which triton lacks.
 
  • #16
Pauli repulsion is for same quantum state.
 
  • #17
malawi_glenn said:
Pauli repulsion is for same quantum state.
Yes. And even for dinucleon, same quantum state (two nucleons sharing spin, too) turns out to be the ground state... and the only bound state. It is Pauli repulsion that stops even neutrons from going to the lowest state... and turns out that for neutrons alone, the lowest available state is unbound for any number of them.
 
  • #18
Sharing spin?
 
  • #19
malawi_glenn said:
Sharing spin?
Yes. The ground (and the only bound) state of deuteron has the proton and neutron having same 1s orbit and also same spin orientation. Paradeuteron (proton and neutron having 1s orbit but opposite spins) turns out to be higher and unbound.
 
  • #20
snorkack said:
Yes. The ground (and the only bound) state of deuteron has the proton and neutron having same 1s orbit and also same spin orientation. Paradeuteron (proton and neutron having 1s orbit but opposite spins) turns out to be higher and unbound.
You mean that they have the same spin. Sharing is something else.
 
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  • #21
vela said:
Why then can a proton and neutron bind to form a deuteron? Is it just the electromagnetic interaction between the individual quarks?
Another important point in physics is to first figure out the "right degrees of freedom" to describe a given problem. It's of course doomed to fail to understand the nuclear-binding problem using quarks and gluons. Fortunately that's not necessary, because the binding of nucleons (protons and neutrons) into atomic nuclei is a very-low energy phenomenon, and you can forget about the quark-gluon substructure of the nucleons.

The nucleon-nucleon interaction can be theoretically described by using effective hadronic models. Nowadays one uses chiral effective theory.

https://s3.cern.ch/inspire-prod-files-9/906bf5d9f9c565f7cec533011f40090f
 
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  • #22
vanhees71 said:
It's of course doomed to fail to understand the nuclear-binding problem using quarks and gluons
While I agree with your point, this may not be the best example. In grad school, one problem was to work out the deuteron in terms of quarks. It was a horrible exercise in 9j, 12j etc. symbols and I still have nightmares, but it was not "doomed".

IIRC, you find:
1. The ground state is T=0, J=1,
2, The proton and neutron at some level keep their identities, in that there is a (uud) and (udd) combination that is alwayus J = 1/2.
3. The "proton" and "neutron" may not always be in color singlets. So in that sense, they don't retain their identities.

So I wouldn't say "doomed". Maybe "needlessly painful and more trouble than its worth".
 
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  • #23
Interesting! I guess, that wasn't an "ab-initio calculation" of QCD. How was it set up? A kind of "MIT bag model"? Do you have a reference, where this "nightmarish calculation" is presented?
 
  • #24
This is all very interesting, but have we not strayed rather far afield for a category B question? Note that we have lost @FrankJ777.
 
  • #25
Hyperfine said:
The question was not about bound neutrons or stable nuclei containing many neutrons. The question was about neutrons, presumably free neutrons, self-aggregating to form larger neutral entities.
There's a pretty recent hint concerning "tetra neutrons":

https://www.nature.com/articles/s41586-022-04827-6
 
  • #26
vanhees71 said:
How was it set up? A kind of "MIT bag model"?
No dynamics at all. Just dealing with spin and color indices.

The trick was to look at your trial wavefunction and minimize/eliminate (qqq) with J-3/2 in a color singlet. Those are Δ's and are so much heavier than the nucleons that they can't possibly be the ground state, But it's been a long time, the details are fuzzy, and I'm sure not going to repeat the calculation!
 
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  • #27
Both n and 3-He atom have spin 1/2.
Both n and 3-He atom have zero charge and electric dipole moment, but nonzero magnetic dipole moment, and nonzero polarizability. I suspect, however, that the polarizability of a neutron (only tightly bound quarks) is somewhat smaller than the polarizability of 3-He atom (two loosely bound electrons).
Neutron also is unstable, while 3-He atom is stable.
It is an experimental observation that 2, 3 or 4 neutrons do not form a bound system. Neither do 2, 3 or 4 3-He atoms.
However, it is an experimental observation that a large number of 3-He atoms do form a bound system.

How do you estimate from first principles the strength of van der Waals forces between neutrons? Are sufficient numbers of neutrons bound by van der Waals forces, or is it provable that they are not?
 
  • #28
snorkack said:
van der Waals forces between neutrons
You mean some electromagnetic interaction due to the quark composition of the neutron?
 
  • #29
It's rather the analogue of the van der Waals interaction concerning the strong interactions. It's the interaction between hadrons, i.e., color-neutral bound states of quarks and gluons, which can be described by effective field theories, which make use of the approximate chiral symmetry of the light-quark sector of QCD.
 

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