Why Neutron Does Not Decay

Why does a neutron not decay in some nuclei? Why does it decay in other nuclei? Is there some mechanism suppressing the d quark -> u quark + W- boson channel? I'm looking for a fundamental QFT / Standard Model explanation.

Thanks!
 

Meir Achuz

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Whether or not a neutron can decay in any given nucleus is determined only by conservation of energy. No deeper explanation is necessary.
The decay A-->A' + e +\nu can only happen if M_A>M_A' + M_e.
 
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Meir Achuz said:
Whether or not a neutron can decay in any given nucleus is determined only by conservation of energy.
I'm not sure if this is an explanation. Why would a neutron get lighter when bound to a nuclei, if this were really the case?
 
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Can I add a question to this - something related that I've never really understood.

Why are free Neutrons unstable? What stops atoms from existing made only (or mainly) of neutrons? The strong nuclear force would bind them together, and there would be no electrostatic repulsion forces... Surely a two neutron pair would be stable?
 
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lonelyphysicist said:
I'm not sure if this is an explanation. Why would a neutron get lighter when bound to a nuclei, if this were really the case?
Meir is very correct in his explanation.

You are forgetting that mass is equivalent to energy and if for example binding energy is negative then mass will reduce in value

Well the case of the nucleus mass being smaller then the constituent nuclei is indeed due to the negative binding energy. This is shown by the semi-empirical mass formula. The nucleon-nucleon potential becomes repulsive at very short distances.

Inside the nucleon things are totally different though :

The sum of the constituent quarkmasses is much smaller then the mass of the hadron. The extra mass comes from the potential and kinetic energy of the quarks and also from dynamical quarks.

For example the proton contains three valence quarks of three different colours (red, green and blue), but it also contains dynamical (sea) quarks. These are quark-antiquark pairs that appear and disappear through energy fluctuations in the vacuum.

These dynamical quarkpairs will generate mass. The mass of a hadron is bigger then the sum of the masses of the constituent quarks (the three quarks of the proton). But the dynamical quarks also generate mass (via symmetry breaking) , so in the end the mass of a proton is BIGGER then the sum of the three quark masses.

Keep in mind that the three quarks are confined, yielding a rise in their linear potential (dominant in the long range). Once a certain distance is exceeded there is enough energy to create a quark antiquark pair


marlon
 
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Adrian Baker said:
Can I add a question to this - something related that I've never really understood.

Why are free Neutrons unstable? What stops atoms from existing made only (or mainly) of neutrons? The strong nuclear force would bind them together, and there would be no electrostatic repulsion forces... Surely a two neutron pair would be stable?
well you are forgetting about energy conservation not being respected and what are you gonna do with Medelejev's table ? How are you gonna incorporate charge conservation and chemical properties based upon valence electrons ? It goes on and on

Besides it is NOT the strong force that binds together several protons. This is done by the residual strong force which is quite different in nature. The exchange particle itself is not elementary, ie the pion. This pion is lightest meson (ie lowest potential energy or 'most easy to fabricate') and is a composition of a quark and anti-quark pair. Here is another thing : the interaction that you mention is explained in terms of quarks yet you only wanna work with neutrons? You need to be careful when engaging in spectacular effective field theory-endeavours :wink:

regards
marlon
 
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Meir Achuz said:
Whether or not a neutron can decay in any given nucleus is determined only by conservation of energy. No deeper explanation is necessary.
The decay A-->A' + e +\nu can only happen if M_A>M_A' + M_e.
Can the energy difference be absorbed from an external source, in order to enable the "decay"?
 
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That's annoying. I was just asking a serious question.
 
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Meir Achuz said:
Whether or not a neutron can decay in any given nucleus is determined only by conservation of energy. No deeper explanation is necessary.
The decay A-->A' + e +\nu can only happen if M_A>M_A' + M_e.
Hi. Maybe he was thinking of reaction like:

[tex]p_{\mathrm{bound}} \to n + e^+ + \nu_e[/tex]

which is possible only within a nucleus (if, for example, produced neutron is more strongly bind - then it would compensate for the greater mass on the right side, I guess :) ). The simple answer to the original question

lonelyphysicist said:
Why does a neutron not decay in some nuclei? Why does it decay in other nuclei? Is there some mechanism suppressing the d quark -> u quark + W- boson channel? I'm looking for a fundamental QFT / Standard Model explanation.
would be: sometimes it decays because the final state is lower in energy and more stable. But it's surely not what you're looking for...

Adrian Baker said:
Why are free Neutrons unstable? What stops atoms from existing made only (or mainly) of neutrons? The strong nuclear force would bind them together, and there would be no electrostatic repulsion forces... Surely a two neutron pair would be stable?
A bare two neutron pair is not stable due to isospin (proton and neutron have different isospins) dependence of nuclear force (mentioned earlier by residual strong force). I think it's based on experimental evidence that no one has seen neutron-neutron or proton-proton bound states (only proton-neutron). I guess deeper answer would be obtained by studying many-body interactions between 6 quarks involved.
 
marlon said:
You are forgetting that mass is equivalent to energy and if for example binding energy is negative then mass will reduce in value

Well the case of the nucleus mass being smaller then the constituent nuclei is indeed due to the negative binding energy. This is shown by the semi-empirical mass formula. The nucleon-nucleon potential becomes repulsive at very short distances.
I'm seeking what the underlying mechanism is that's preventing the decay. If it's kinematical, as claimed, I'm trying to see if there's an easy way to understand why the mass of the neutron would be decreased sufficiently when bound inside a nucleus to disallow its decay.
 
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marlon said:
well you are forgetting about energy conservation not being respected....
No, I just don't have the knowledge to understand how this works here. As I said, I've never understood this, and I'm clearly not a particle Physicist!


marlon said:
and what are you gonna do with Medelejev's table ? How are you gonna incorporate charge conservation and chemical properties based upon valence electrons ? It goes on and on
Sorry, I didn't mean atoms as such. I meant neutron based particles, such as a two neutron pair. I didn't mean atoms with electrons. Perhaps nuclei would have been a better term?

marlon said:
Besides it is NOT the strong force that binds together several protons. This is done by the residual strong force which is quite different in nature. The exchange particle itself is not elementary, ie the pion. This pion is lightest meson (ie lowest potential energy or 'most easy to fabricate') and is a composition of a quark and anti-quark pair.
Hmm, I certainly didn't know this. I haven't heard of the 'residual strong force' before. I have some research to do... However, why does this residual strong force act on groups of protons, but not on groups of neutrons?


marlon said:
Here is another thing : the interaction that you mention is explained in terms of quarks yet you only wanna work with neutrons? You need to be careful when engaging in spectacular effective field theory-endeavours :wink:
I don't understand what you mean here at all, sorry. Surely quark interactions are relevant with neutrons? If not why not?

I am self-taught in particle Physics and have never studied it at a higher level. I have a good understanding of the basics (VERY basic!) but one of the problems with being self-taught is that you can't ask questions of books when you want to. I still have no real idea why neutron pair particles can't exist from what you have said above.

Igor_S said:
A bare two neutron pair is not stable due to isospin (proton and neutron have different isospins)
Thanks Igor. Something else I need to expand my knowledge on obviously. :rolleyes:
 
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lonelyphysicist said:
Why does a neutron not decay in some nuclei? Why does it decay in other nuclei? Is there some mechanism suppressing the d quark -> u quark + W- boson channel? I'm looking for a fundamental QFT / Standard Model explanation.

Thanks!
I think the explanation lies with the Pauli principle. It's a lot easier to explain in a nuclear shell model than going down to quark level.

Basically protons and neutrons each obey the rule that no two of them can exist in the same state. So energy levels get filled one by one with their maximum number of protons and neutrons. The only neutrons that can decay to protons are those for which there exists an available (i.e. empty) proton slot at the same (approximately) or lower energy.
 
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mikeyork said:
I think the explanation lies with the Pauli principle. It's a lot easier to explain in a nuclear shell model than going down to quark level.

Basically protons and neutrons each obey the rule that no two of them can exist in the same state. So energy levels get filled one by one with their maximum number of protons and neutrons. The only neutrons that can decay to protons are those for which there exists an available (i.e. empty) proton slot at the same (approximately) or lower energy.
I'm not sure this is the reason. The first closed shell in shell model is with 2 nucleons * (either protons or neutrons) :smile: , with both of them having orbital angular momentum zero, but different orientations of spins.


* There's a recent short article involving shell model, that you may find useful: http://www.physicsweb.org/articles/news/9/6/9/1 .

P.S. About isospin, a somewhat simple explanation is given at http://en.wikipedia.org/wiki/Isospin , in a first paragraph.
 

Astronuc

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jackle said:
Can the energy difference be absorbed from an external source, in order to enable the "decay"?
That is usually not the case. As far as we know from studying populations of a given radionuclide, external energy does not influence a particular decay mode (under normal circumstances). On the other hand, perhaps with a sufficient amount of energy (as from an acceleraterd particle), one could possibly change the decay mode.

There are so-called photo-neutron reactions, but that is different than neutron decay by beta emission.

As for Adrian asked, as far as we know, neutrons do not pair, at least not in conditions found on earth. In a neutron star, that is a different matter altogether.

I suppose it comes down to how the energy is distributed among the nucleons in the nucleus. At some point, a neutron receives sufficient energy to break 'confinement' (I am appropriating this term).

Also see discussion of proton, neutron and their decay at http://hyperphysics.phy-astr.gsu.edu/hbase/particles/proton.html
 

ZapperZ

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And talk about perfect timing, there's an excellent review article of the nuclear shell model in the current online issue of Rev. Mod. Phys.:

Abstract
For almost half a century, shell-model theory has served as a reference tool for the interpretation of nuclear spectroscopy. However, its impact was restricted at first by a limited knowledge of the effective Hamiltonian on the one hand and by the magnitude of the associated computations on the other hand. In the last decade, an improvement in our understanding of nuclear dynamics and the availability of ever growing computing power have radically changed the status of shell-model theory and have pushed it to the forefront among nuclear theoretical models. This article, which provides an introduction to the modern literature on the subject, offers many examples of its achievements, including an enhanced understanding of nuclei very far from stability.
E. Caurier et al., Rev. Mod. Phys. 77, 427 (2005).

Zz.
 
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Igor

I am well aware of the fact that protons and neutrons are both members of the same nucleon isospin multiplet and this can tell us the relative mixture of neutrons and protons in a closed shell -- and in any case, the origin of the isospin rules in the shell model is, again, the Pauli rule. But this symmetry is broken and so isospin rules are approximate. Since neutron decay is a weak interaction, I chose to describe neutrons and protons separately.

As regards the first closed shell, I don't understand what you are trying to say here. Could you explain what about this contradicts what I wrote? Are you claiming that the Pauli rule is not the reason why some neutrons don't decay?

In the first closed shell, the Pauli rule requires I=0 (since L=S=0). This is why you get 1n and 1p and not 2p or 2n. If the n were to decay, the new p would have to form an I=1 state with the other p, this would require a higher energy level and so it cannot happen. In other words, there is no available proton slot at the same (approximately) or lower energy. and so deuterium is stable. Tritium, on the other hand is not.
 
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mikeyork said:
Igor

I am well aware of the fact that protons and neutrons are both members of the same nucleon isospin multiplet and this can tell us the relative mixture of neutrons and protons in a closed shell -- and in any case, the origin of the isospin rules in the shell model is, again, the Pauli rule. But this symmetry is broken and so isospin rules are approximate. Since neutron decay is a weak interaction, I chose to describe neutrons and protons separately.
Breaking of isospin symmetry is very small in 2 nucleon system, so the approximation (or introduction of isospin) is very good. The error is only due to different masses. I mentioned isospin only to explain why 2 neutrons don't form a bound state, unrelated to the shell model.

mikeyork said:
As regards the first closed shell, I don't understand what you are trying to say here. Could you explain what about this contradicts what I wrote? Are you claiming that the Pauli rule is not the reason why some neutrons don't decay?

In the first closed shell, the Pauli rule requires I=0 (since L=S=0). This is why you get 1n and 1p and not 2p or 2n. If the n were to decay, the new p would have to form an I=1 state with the other p, this would require a higher energy level and so it cannot happen. In other words, there is no available proton slot at the same (approximately) or lower energy. and so deuterium is stable. Tritium, on the other hand is not.
I didn't involve isospin when I spoke about shell model. Sorry, in my course in nuclear physics (I'm an undergrad student) we didn't mention any isospin rules in shell model. We used harmonic approximation so there was no energy dependence on isospin.

And we treated both p and n separately, so I figured there are actually 2 sets of shells ? (and you fill them independent of each other)

Anyway, I didn't figured out why you wrote S=0 (total spin?), because if total spin is zero, then p and n already differ in spins and there is no need to make their isospins different.
 
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Igor_S said:
Anyway, I didn't figured out why you wrote S=0 (total spin?), because if total spin is zero, then p and n already differ in spins and there is no need to make their isospins different.
You have to consider the symmetry of the two-particle wave-function under permutation. If you treat both proton and neutron as different states of the same particle (nucleon) then you must also consider the effect of the combined isospin multiplet in the symmetry.
 
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Hmm, but I thought deuteron has S=1. You are correct that two-particle wave function has to be antisymmetric, but that would be total wave function which is space part x spin part x isospin part ... The - (minus) comes from the isospin part (T=0; antisymmetric state). If S would be 0, total wave function would be symmetric (you would get +).

Btw, since deuteron state is not spherical in shape, it's potential is also non-central and therefore assumption of central effective potential (which is used in shell model) may not hold...
 
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arivero

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Hmm my guess is that there is some detail lacking in the understanding of the role of the electroweak sector in the nucleus; it seems to be implicated in the generation of spin-orbit couplings too (thus in the shell corrections).

See
http://arxiv.org/abs/hep-ph/0405076 esp. figure 5
http://dftuz.unizar.es/~rivero/research/LS9530.pdf [Broken]
http://arxiv.org/abs/nucl-th/0312003 figs 9
http://dftuz.unizar.es/~rivero/research/uno.gif [Broken]
 
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Igor_S said:
Hmm, but I thought deuteron has S=1.
Yes you are correct. I was thrown your previous comment of 2 nucleons with opposite spins and assumed (incorrectly) you were saying that the first closed shell has S=0.

I was also in error in stating that L=S=0 implied I=0. The correct exclusion rule is that L+S+I-2i must be even in the CM frame. (At 58 I should stop relying on my memory!) So, if two nucleons have L=0, S=1, then one gets I=0 (which requires 1n and 1p). The S=0 states (which require I=1 if L=0) are actually at higher energy (I don't remember why, exactly, but it is presumably something to do with magnetic coupling) and are all unstable.

Whatever, the details of the energy levels of individual shells, I think you'll get back to my original suggestion that it is primarily the generalized exclusion rule that prevents some neutrons from decaying into protons -- there have to be vacant proton positions at lower energy.
 
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Yeah, ok... if you do need to satisfy that exclusion rule, this (Pauli principle) could be a good explanation of the original question.
 

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