Do Neutrons in Stable Helium-4 Undergo Continuous Transformation?

[Rade]

I have a question about current experimental findings on the status of the neutron N while contained within nuclear radius of a stable atom, say Helium-4.

It is well known that the N will undergo beta(-) decay when it is free from a nucleus (takes ~ 14 minutes). But...

My question is--do the two N in stable Helium-4 maintain a stable identity with no beta (-) decay or, do they exist as a continuous back-forth transformation of N <----> P mediated by mesons ?

Thanks for any help. If you can point me to a peered reviewed citation where this question has been addressed that would be appreciated.

 

[mormonator_rm]

I would say that virtual pion exchange may account for that. Virtual pi- emission from a neutron would allow the neutron to become a proton, and the absorbing proton could become a neutron, with both of the nucleons remaining in the same spin state, and thus not changing their status with respect to the Pauli exclusion principle. This mechanism would result in the same long-range attractive force experienced by nucleons when pi0 exchange occurs.

Anyone else have a comment or correction?

 

[Meir Achuz]

My question is--do the two N in stable Helium-4 maintain a stable identity with no beta (-) decay or, do they exist as a continuous back-forth transformation of N <----> P mediated by mesons ?

A neutron in Helium cannot beta decay because any (pppn) state has higher mass than He4.
The two n and two p in He are in an isospin eigenstate (I=0).
As an easier example in the deuteron, \psi=(pn+np )/sqrt{2}.
Thus particle 1 is a mixture of p and n. This need not (although it happens to be) be mediated by pions. It is not "a continuous back-forth transformation of N <----> P".

 

[Rade]

A neutron in Helium cannot beta decay because any (pppn) state has higher mass than He4.
The two n and two p in He are in an isospin eigenstate (I=0).
As an easier example in the deuteron, \psi=(pn+np )/sqrt{2}.
Thus particle 1 is a mixture of p and n. This need not (although it happens to be) be mediated by pions. It is not "a continuous back-forth transformation of N <----> P".

Thank you, but I am not sure I made myself clear.

So, for your deuteron example, I am not asking if the N and P within deuteron have a continuous back-forth N <-----> P transformation.

What I am asking is if the [N] in the deuteron is itself undergoing a transformation independent of the nearby [P]. Thus the picture of interactions would be:

{ [N] <----> [P] } <-----> [P]

If so, then we can say the [N] is "unstable" within a "stable" deuteron.

The other option is that the [N] is "stable" within a "stable" deuteron, and the picture of interactions would then be this:

[N] <------> [P]

Hope this makes sense.

 

[mormonator_rm]

A neutron in Helium cannot beta decay because any (pppn) state has higher mass than He4.
The two n and two p in He are in an isospin eigenstate (I=0).
As an easier example in the deuteron, \psi=(pn+np )/sqrt{2}.
Thus particle 1 is a mixture of p and n. This need not (although it happens to be) be mediated by pions. It is not "a continuous back-forth transformation of N <----> P".

I had to think about that, but that does make more sense than what I said at first. I'm with you on that one.

 

[mormonator_rm]

Thank you, but I am not sure I made myself clear.

So, for your deuteron example, I am not asking if the N and P within deuteron have a continuous back-forth N <-----> P transformation.

What I am asking is if the [N] in the deuteron is itself undergoing a transformation independent of the nearby [P]. Thus the picture of interactions would be:

{ [N] <----> [P] } <-----> [P]

If so, then we can say the [N] is "unstable" within a "stable" deuteron.

The other option is that the [N] is "stable" within a "stable" deuteron, and the picture of interactions would then be this:

[N] <------> [P]

Hope this makes sense.

Actually, Meir is right. The reason it is stable is because the binding energy of the nucleus makes it the lightest available nuclear state, and hence a beta decay would cause it to increase in mass rather than decrease. Because these consequenses are incompatible, the helium-4 nucleus is literally forced to remain in its stable state. I believe this is what Meir was getting at.

 

[Rade]

Actually, Meir is right. The reason it is stable is because the binding energy of the nucleus makes it the lightest available nuclear state, and hence a beta decay would cause it to increase in mass rather than decrease. Because these consequences are incompatible, the helium-4 nucleus is literally forced to remain in its stable state. I believe this is what Meir was getting at.

Thank you, but in beta (-) decay (decay of N to P) after the transformation we have a decrease in atomic mass (not increase as you say) because the P (1007825.03207 mass units) is lighter mass than the N (1008664.9157 mass units). So, if we "start" the motion with beta (-) decay [ N -----> P], then quickly (say at same speed (17 trillion times/sec) recently documented for transformation of matter & antimatter quarks in Bs-meson--see this link http://www.photonics.com/content/news/2006/April/5/82000.aspx ) the reverse motion of beta (+) decay [ P ------> N], you see, the net mass must remain constant if we observe at any moment of time.

Recall my OP question, I am asking if the 2 neutrons in He-4 can undergo this type of transformation independent of the 2 protons.

But as Meir suggests, let us consider the more simple case of deuteron [NP]. Where is the experimental evidence that we do not have the [N] as unstable with quick (say many trillion times/sec) back-forth beta (-) <-----> beta (+) decay while the [P] remains unchanged ?

Of course, we really must look to the dynamics at the microscopic level of the quarks. Now the [N] has quark structure (ddu) and the [P] has (uud). So my OP question, now for deuteron [NP] at level of quarks, becomes this question, is it possible that we have this type of transformation within deuteron ?:

{ (ddu) <-- many trillion times/sec --> (uud)} bonded to {(uud)}

ps/ Does anyone know the "speed" of the weak force d ----> u transformation or the reverse u -----> d, is it at speed of light ?

I hope I am making myself clear.

in edit: Recall that while [P] is very, very stable outside nucleus, inside nucleus the [P] will undergo beta (+) decay--this is the source of the positron in PET scans used in medical research:

...When a nucleus decays by positron emission, a proton in the nucleus converts into a neutron, and a positron and a neutrino are ejected. The neutrino leaves the scene without a trace, while the positron rapidly annihilates with a nearby electron... see here: http://physicsweb.org/articles/world/15/6/7