Containing Free Neutrons: Could it be Done?

[Kostik]

I understand that neutron stars arise because of gravitational forces in a collapsing star. But consider an isolated free neutron moving at slow speed. Leaving aside its decay (12m half life), can they be "contained"? If two of them came together, would the strong nuclear force bind them? Is there any reason why you could not contain a small "cloud" of free neutrons? You cannot do so electromagnetically, but what if you built a container from atoms that do not interact with free neutrons?

If you contained a cloud of free neutrons, could you cool it to a liquid or a solid? Unlike a neutron star, which was already extremely massive and compressed before the protons and electrons were squeezed together, could a solid chunk of free neutrons be contained at normal pressures?

 

[Bystander]

from atoms that do not interact with free neutrons?

What might those be?

 

[Kostik]

What might those be?

The container material isn't important. Let them collect in empty space, or let them orbit a small massive body. I supposed maybe a container made from very stable nuclei, such as a cold, solidified noble gas might be notionally used.

 

[e.bar.goum]

The container material isn't important. Let them collect in empty space, or let them orbit a small massive body. I supposed maybe a container made from very stable nuclei, such as a cold, solidified noble gas might be notionally used.

The chemical stability of nuclei doesn't matter. Neutrons interact with atoms via the strong interaction - that is, they interact with nuclei. And it's hard to trap neutrons. I'd go further: it's hard to manipulate neutrons, full stop.

This group trapped neutrons using a magnetic trapped filled with ultracold helium. The trapped density of ultra cold neutrons was 2 per cubic cm. Extremely impressive experimental work! http://www.nature.com/nature/journal/v403/n6765/full/403062a0.html

This group used a highly neutron reflective chemical: Fomblin Oil (F3CCF2OCF2CF5)n which has a UCN reflection loss rate of
(2-3)*10^-5/bounce at 20 degrees C, which is pretty impressive.
http://journals.aps.org/prl/pdf/10.1103/PhysRevLett.63.593[/URL]

Even so, the trick with the above measurements is to slow down and quantify the losses rather than stop them.

 

[Kostik]

Assume for the moment that they ARE already together in a high concentration cloud, moving at slow speed. What would this "gas" be like? Would it behave like an ideal gas? Could it be liquified or solidified? Would two or more free neutrons bind together under the nuclear force and form a "nucleus" of neutrons?

This is what I'm interested in, rather than the question of containment and concentration.

 

[Drakkith]

So you're pretty much asking how low energy neutrons interact with each other?

 

[Kostik]

Right!

 

[Kostik]

can anyone shed any light on this?

 

[snorkack]

Dineutron is unbound, for the same reasons as singlet deuteron.
Yet the existence of low-lying virtual state of singlet deuteron is the reason proton has such a huge cross-section for moderation.
What is the cross-section for neutron-neutron collision?
Furthermore, 2 neutrons are unbound, but so are 2 helium 3 atoms. Does not stop He-3 from liquefying... provided there are more than about 30 atoms.
Now about containment: the only stable isotope which does not absorb neutrons is He-4.
Liquid He-4 has neutron chemical potential of about 10 neV.
Does a neutron make a bubble in liquid helium 4?
How do neutrons in liquid helium 4 interact with one another?

 

[mfb]

Free neutrons behave like an ideal gas to a very good approximation - and follow the Fermi-Dirac statistics (which is notable at low temperature) because neutrons are fermions.
They cannot bind to each other so there are no nuclear reactions going on if we can neglect neutron decays.

 

[snorkack]

Free neutrons behave like an ideal gas to a very good approximation - and follow the Fermi-Dirac statistics (which is notable at low temperature) because neutrons are fermions.

Fermi statistics is itself a nonideal behaviour?

They cannot bind to each other so there are no nuclear reactions going on if we can neglect neutron decays.

How does the elastic scattering cross-section of a neutron from an opposite spin neutron compare against the elastic scattering cross-section of the neutron from an opposite spin proton at the same energy?
Also, as pointed out: 2 He-3 atoms cannot bind to each other, but multiple He-3 atoms can. So how about multiple neutrons?

 

[mfb]

Fermi statistics is itself a nonideal behaviour?

Depends on the definition, and I won't argue about definitions.

How does the elastic scattering cross-section of a neutron from an opposite spin neutron compare against the elastic scattering cross-section of the neutron from an opposite spin proton at the same energy?

I would expect differences, but it does not really matter. It just changes thermalization time.

Also, as pointed out: 2 He-3 atoms cannot bind to each other, but multiple He-3 atoms can. So how about multiple neutrons?

Those Efimov states exist for bosons only, neutrons are fermions.

 

[snorkack]

neutrons are fermions.

So are He-3 atoms.

 

[mfb]

Oh, He-3. Do you have a reference for its production? I only see predictions.

That's still a chemical reaction. A third neutron would need a higher energy level which makes things even worse.

 

[snorkack]

Oh, He-3. Do you have a reference for its production? I only see predictions.

Um. Bulk liquid He-3 is a well known phenomen with a lot of properties.

That's still a chemical reaction. A third neutron would need a higher energy level which makes things even worse.

Yes, and so does a third He-3 atom. Which is why it is not bound... but three millionth He-3 atom IS bound despite needing a higher energy level again.

 

[mfb]

Um. Bulk liquid He-3 is a well known phenomen with a lot of properties.

How is that related to the structure of a helium molecule with 3 atoms?

 

[snorkack]

How is that related to the structure of a helium molecule with 3 atoms?

It shows that the impossibility of a bound state for 2 or 3 He-3 atoms, or neutrons, does not rule out existence of bound states for bulk liquid He-3.

 

[mfb]

You get such a state in neutron stars, but I don't see the relevance.
Every collection of multiple neutrons would be instable against beta decay (this is not the same as the free neutron decay), even if everything else fails - and I don't see an argument why a collection of multiple neutrons should be a bound state at all (neglecting gravitationally bound neutron stars).