# I Weak force - only free neutrons decay?

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1. Sep 10, 2016

### arlesterc

Reading this article - http://hyperphysics.phy-astr.gsu.edu/hbase/particles/proton.html#c4 there is a statement -

"A free neutron will decay with a half-life of about 10.3 minutes but it is stable if combined into a nucleus."

So is it only free neutrons that decay into protons?

In conjunction with this question, does a neutron have to be accelerating at a specific speed to decay? So if could fix a neutron theoretically to as close to a single point as possible versus accelerating it as close as I can to the speed of light would there be a difference in decay 'strength' - the faster moving neutron would decay more quickly, the decay particles emerging it from it having more energy?

2. Sep 10, 2016

### Staff: Mentor

No, beta decay of nuclei is possible, too. Neutrons in nuclei can be stable - it depends on the nucleus.

No, the motion of neutrons does not matter at all. Also note that acceleration and speed are different things.

If a neutron is moving fast in a given reference frame, its decay products have (on average) more energy, but this is nothing special about the decay, you would get the same result in classical mechanics and with explosives.

3. Sep 12, 2016

### arlesterc

Thanks for the responses. The reason I am asking is I am trying to have a basic level of understanding of the weak force. The other forces seem to be a 'force' between particles - two electrons repelling each other, two bodies attracting each other via gravity, quarks held together by gluons. However it seems that the weak force doesn't involve two particles. It's as if there were something inside a quark/combination of quarks in the neutron that almost inevitably causes an explosion and a W-boson emerges. So if I had a time lapse camera pointed at a free neutron I would initially see two down quarks and an up quark. Approximately 10 minutes later I would see two up quarks and a down quark and a W-boson and then shortly thereafter an electron and an electron antineutrino. The W-Boson has been born and has died in a very short time but its birth to me is immaculate in comparison with my understanding of the other forces and how they arise.
For instance, electron/electron, proton/electron, proton/proton interactions produce photons - a star is born so to speak. Is there something analogous in the weak force? Two particles/fields that when they interact cause a W boson to appear? I can understand that the Boson can be felt by an electron or other particles once they are around. (I believe that to be true anyway.) But that's a different interaction from the birth of photons - photons being 'born' when charged particles interact versus impacting particles once they are alive - their presence being felt by other particles just like the W boson once it shows up affecting other particles. It's the coming into existence mechanism that is unclear to me. How is a W boson born? Thanks in advance for any enlightenment.

4. Sep 12, 2016

### Staff: Mentor

There is a weak force between particles that doesn't change their type (mediated by the Z boson), but there is a reason it is usually called weak interaction: if it is relevant, then it typically changes the particle content. The electromagnetic interaction can do that as well (e. g. electrons and positrons can annihilate, or be produced as pair), and so does the strong interaction (producing or annihilating pairs of quarks), but not as much as the weak interaction.
You won't see a W boson. For a beta decay, it is a tool in calculations - it does not exist as actual particle in the decay.

Not without changing a particle type (charge has to be conserved!), but this is possible with a Z boson. As example, if a Higgs boson is produced at the LHC, there is a chance that it emits a Z boson.

5. Sep 15, 2016

### ChrisVer

explosion?
Well the free neutron is heavy enough to be able to decay into a proton+electron+ electron neutrino.

as already been said: you won't see the W... if you would see the W it would have to be a real particle (on-shell) and that would violate energy conservation (a neutron has less mass than 1GeV, while W's have almost 90GeV)... W's in neutron decay are off-shell or virtual and are not observable.

Yes... for example when two protons collide you may have the one up-quark of the proton1 to "interact" with the down-quark of the proton2... then a W can be produced in s-channel and then decay (to neutrino+lepton or to hadrons)... give the protons enough energy so that the two quarks, when interacting, have a center of mass energy of ~90GeV and you will produce a W-boson.

The W and Z bosons are pretty heavy and decay fast...

6. Sep 15, 2016

### nikkkom

There are two steps. (1) A photon is emitted by one interacting particle. (2) Then it is absorbed by another particle. both particles change their energy and momenta - hence, "interacted".

Same happens with gluon exchange between quarks.

The emitted photon can also be "absorbed" by an electron which "comes from the future" (when drawn as Feynmann diagram). In more normal view, such photon is not absorbed by anything, it creates (or you may see it as "decays into") two new particles, electron and positron. (The positron is "electron which comes from the future").

It is in fact *the same*, on the fundamental level, for the weak force. Two particles can exchange a W or Z boson (although that happens far, far, far more rarely than with photons), or one particle can emit a W or Z boson which then decays into two particles.

7. Sep 15, 2016

### ChrisVer

future/past can be confusing for interpreting the Feynman diagrams [which don't correspond to physical processes]... the reason the future/past can be used is because of the spinor usage (under the interpretation of spinors moving backward in time for their negative energy eigenstates).
Afterall an interaction is taken by integrating over the propagator... and so over all the positions a W (or Z) can be interchanged from. In the end you measure only the final states and not the intermediate ones... that's the reason you cannot detect a W or Z even if they are created on-shell, because they are unstable and so they cannot be final states.

8. Sep 15, 2016

### Staff: Mentor

Just a detail:
One of the quarks has to be an antiquark.

9. Sep 16, 2016

### snorkack

Why?
Could you have a reaction like
p+p+their kinetic energy (but no antiquarks)->p+n+W+?

10. Sep 16, 2016

### ChrisVer

that's the beta+ decay... also you shouldn't write a W/Z in the final states.
not what I was talking about...
when you talk about 90GeV parton CoM energies, you are already considering "quarks","antiquarks" and "gluons" and not protons.

11. Sep 16, 2016

### Staff: Mentor

At the energies necessary to produce a W, the colliding components are the partons.
You can have such a reaction, but if you want it to be s-channel ("things => W"), it has to happen with a sea antiquark.

Please open a new thread if you want to discuss this in more detail, I think it is beyond the scope of this thread.

12. Sep 27, 2016

### arlesterc

Thanks. There should be a search and replace function on the Internet as hardly any article I came across mentioned that it was virtual Bosons that were created.

13. Sep 27, 2016

### arlesterc

Thanks for the further points. I am still a little stuck on the word 'decay'. Would it be fair to say that 'decay' is the interaction you describe above
"you may have the one up-quark of the proton1 to "interact" with the down-quark of the proton2? " So the W boson is the 'exchange' particle of that interaction?

Following from that then what is the interaction when there is only a free neutron? What are the two partners in the exchange of the W particle? Interaction with vacuum field? Or is decay of a free neutron some kind of quantum probability event - so the free neutron/its components can exist in several different states/shift through different states because of the probabalistic nature of quantum entities and at some point in the life of a free neutron - usually within - 10 - 15 minutes - the lower energy option is chosen by the nature of quantum events. So three quarks in a free neutron can in theory for some interval of time - maybe incredibly small - exist as one up, two down or two up, one down, three up, etc. and 'two up one down' being lower energy than 'one up and two down' when that possibility is explored as part of the nature of quantum activity it becomes pretty well irreversible as it would take energy to bump the up back into a down. So please excuse the corny analogy but it's as if Humpty Dumpty at some point out of inevitable curiosity peeks down the wall he is sitting on and the peeking leads to his fall. Once down, Humpty will be in no position to look up/get back up without a tremendous amount of assistance.

14. Sep 27, 2016

### arlesterc

Thanks for the response. Is there any reason why one electron emits and the other absorbs? Or is it just that two electrons when they come close to each other produce a field which is seen as a photon the force of which they both create/feel simultaneously so to speak?

15. Sep 27, 2016

### nikkkom

https://en.wikipedia.org/wiki/Feynman_diagram

16. Sep 27, 2016

### ohwilleke

* This has more to do with the inadequacies of the English language which naturally try to fit events into a cause and effect sequence than any difference that is relevant to how the interaction works. The more neutral and precise language would be to say that each of the two electrons "couple" to the photon.

The Standard Model physical constant called the electromagnetic "coupling constant" alpha which is approximately equal to 1/137 describe the probability of a photon coupling to a particle with an electromagnetic charge of magnitude one (such as an electron), which refers to the technical and precise terminology "coupling" in this sense. The weak force has an analogous coupling constant that governs couplings between particles that interact via the weak force and weak force bosons (W and Z). And the strong force, in turn, has an analogous coupling constant that governs the couplings between gluons and either quarks or other gluons. (All three of these coupling constants are "dimensionless" numbers in contrast to the physical constant known as Newton's constant that governs the strength of the gravitational force which is not dimensionless in the conventional formulation of General Relativity - one more barrier to a theory of quantum gravity.)

* Why would someone want to use a technical word like "coupling" instead of plain old vernacular English?

Because, a Feynman diagram of an interaction is equally valid and happens with the same probability regardless of how it is rotated in four dimensions space-time, even though in the English language the way that the interaction depicted by the Feynman diagram would be described is very different in different rotations of the diagram and in different interpretations of a given Feynman diagram with the same rotation.

For example, one way to describe a particular Feynman diagram is that two photons suddenly transforms into an electron and a positron. An equivalent description is that the positron is moving back in time, emits a photon backward in time, and then reverses its direction emitting another photon to become an electron moving forward in time. Rotate that Feynman diagram 180 degrees and a common way to describe the same interaction (from a mathematics of quantum electrodynamics perspective) is that an electron and positron collide and annihilate into two photons. But, that rotated diagram could also equivalently be described as an electron moving forward in time, emitting a photon forward in time and emitting another photon as it is reversing its direction to go backwards in time. All four descriptions describe equivalent Feynman diagrams with the same probabilities of occurring and count as a single possible type of interaction when you add up the probability of all possible ways that something can happen to evaluate the "path integral" the tells you what the probabilities of a particular change in the status quo is.

(I would draw pictures of the Feynman diagram in question and its 180 degree rotated equivalent if I knew how to do that with this forum interface, but I am not so talented. Maybe someone else could drop in a couple of images into a post.)

* Another question you asked which wasn't really answered squarely is what we mean when we say that a particle (composite or fundamental) "decays".

Basically, a decay is a transformation into other particles that is not prohibited by energy conservation even in the absence of any mass-energy other than the rest mass of the particle. In other words, a decay is an interaction that produces different particles and excess kinetic energy in addition to the decay products.

(I'm not sure if a transformation of one particle to another that was kinetic energy neutral would count as a decay or not, perhaps because I can't think of any good examples at the moment.)

Interactions that convert one or more particles plus additional energy into different particles without the excess energy (i.e. the inverse of a decay interaction) are also possible (because all interactions except weak force interactions are perfectly reversible, and weak force interactions are reversible too subject to a small adjustment for CP violation in a matrix called the CKM matrix that contains a bunch of physical constants pertinent to weak force interactions). But, so far as I know, there is no catchy exact antonym to the word "decay" as used in the technical physics sense that I have just described in the English language. (If I am wrong and simply limited in my vocabulary, I will take no offense if a reader or forum participant supplies the word I lack in my vocabulary and makes us all wiser.)

* Anyway, and finally, somewhere along the thread your original question - how can a bound neutron be stable when a free one is not? - seems to have been lost along the way, but you deserve an answer.

A nice understandable presentation answering that question can be found at Matt Strassler's blog called Of Particular Significance: https://profmattstrassler.com/artic...-together/neutron-stability-in-atomic-nuclei/

In a nutshell, this is true in some atoms due to conservation of mass-energy. Sometimes a bound nucleus which includes a neutron has less mass as a single unit than the free neutron and the rest of the nucleus would have if broken into parts. In those cases, the bound neutron is stable.

It is possible for the whole to have less mass than the sum of its parts, because the nuclear binding energy of the nuclei of atoms less heavy than iron is a negative number (which is why energy is emitted in nuclear fusion involving these elements).

(If I were more nimble I would insert a chart of the binding energy per nucleon of the various chemical elements here, but I'm not so you'll just have to imagine a U-ish shape with hydrogen at one end, iron at the bottom of the U, and a right hand high end that extends up into numbered elements with no name.)

Quantum mechanics can seem to cheat mass-energy conservation in interactions like quantum tunneling and interactions that involve virtual or "off shell" particles, but these little short term "energy lending" transaction (to horribly bastardize the correct technical description of these processes in the interest of a heuristic description) don't change the iron law that the mass-energy present in an initial state must always be exactly the same in an end state. (Indeed, this iron law of mass-energy conservation is the foundational principle of the two main ways that the equations of particle physics can be written -- Hamiltonians and Lagrangians - both of which can also be used in classical mechanics as well.)

On the other hand, if you have a big nucleus (bigger than iron), nuclear binding energy is positive so the parts have less mass than the whole and splitting the atom creates energy. In those atoms, beta decay can be possible and the bound neutrons are not entirely stable, even though they may still be much more stable than free neutrons. So, the statement that a bound neutron is stable is only partially true. Neutrons bound in stable isotypes of light elements really are stable, while other bound neutrons are merely "metastable" with a stability that is a function of the particular isotype in question.

* Let's take the question one step further than Strassler does, because negative energy sounds like a pretty mysterious thing and really, in this circumstance it isn't, because the fact that binding energy is negative is simply a product of our arbitrary (but natural, convenient, and intuitively helpful) choice of where to put zero in the coordinate system by which we measure nuclear binding energy.

This negative binding energy is possible because the force that binds protons and neutrons together in an atomic nucleus is just a second order effect of the strong force that binds the quarks in individual protons and neutrons together, and the strong force binding the quarks together is the source of most of the mass of a proton or neutron.

When arranged in the right kind of light atomic nucleus, the amount of strong force energy per quark necessary to hold all of the quarks in the entire nucleus together in the right configuration is slightly less than the amount of strong force energy per quark necessary to do so in free proton or neutron (heuristically, it is helpful to think of it as similar to the way you need less material to build to buildings that share a party wall between them than to build two free standing buildings).

If we measured the total energy in a nucleus created by the strong force, by starting with the total mass of the nucleus and then subtracting out the mass of the quarks in the nucleus and the Mass-energy contribution for the electromagnetic fields in the nucleus, the gluon field mass-energy of the system would be positive in both the bound state and the divided state and the difference in the gluon field mass-energy would be equal to the difference in binding energy. The physics would be the same and the description is equivalent, but the coordinate system we used to measure the change in energy would have a true zero baseline, rather than the arbitrary baseline that is used for the nuclear binding energy.

The normal zero for nuclear binding energy is analogous to measuring temperature in degrees Celsius, while the approach taken above in the previous paragraph is analogous to measuring temperature in degrees Kelvin. So, even though nuclear binding energy is negative in light nuclei that doesn't really mean that there is negative energy in the sense that we talk about the hypothetical and probably unphysical concept of "true negative energy" in the context of the theoretical physics of general relativity.

Last edited: Sep 28, 2016
17. Sep 28, 2016

### nikkkom

I don't think it's right.
Any bound system has negative binding energy. A system with positive binding energy is not bound, like two electrons pushed together: they will fly apart at once.

No. Binding energy of e.g. uranium nucleus is still negative. It is just less negative than sum of binding energies of its decay products (be it alpha decay, or fission).

No. Negative binding energy is possible because it's just a mathematical construct: it's a difference between energy of a bound system and energy of this same system "disassembled" into its parts and each part pulled away far enough that they don't interact: Ebound - Eparts. Both E's are positive. Only the difference is negative. No mystery here.

Wrong. If you pull quarks apart, their energy will not be small. In fact, our current best guess is that their energy will tend to infinity as you pull them farther and farther away. (Yes, I know if you actually try to do it in an experiment, this would just create hadron showers).

In other words, it's logical to see proton's binding energy as negative, because energy of "free three u,u,d quarks" is bigger than the energy of their bound system (proton).

18. Sep 28, 2016

### Staff: Mentor

It is the same reaction for both a free and a bound neutron: neutron->proton+electron+antineutrino. No other nucleons are involved (at least not in a way you could describe via Feynman diagrams). If you draw the Feynman diagram of this decay there is a virtual W boson in it.

19. Sep 28, 2016

### ChrisVer

it depends how the proton quarks-antiquarks interact...
but yes W plays the role of the propagator.

yes it is, but in general we call the interactions occuring on a single particle that decays a "decay"...

Obviously the decay of a neutron is governed by probabilities... Feynman diagrams (not quarks) represent the amplitudes and not the actual physical processes (which cannot be represented in any way). In explicitly I'm saying that Feynman diagrams correspond to mathematical expressions and not to something that you can say "ah a quark moves and suddenly it turns into something + something + something etc" without knowing the limitations of that saying... what you measure (and use the Feynman diagrams as a tool to calculate) are cross sections....
What happens depends on how you wanna see it, for neutron decays the Fermi theory of point interaction can give qualitive results because the Ws are very off-shell (due to low mass difference between the neutron and the proton).
In the quark level the down quark of a neutron undergoes a trassition to up quark and there you use the W propagator.

No. The three valance quarks of the neutron are there (udd)... however d can turn into a u as explained by the weak force, which as a result explain the neutron decay.

20. Sep 29, 2016

### nikkkom

I think you are overdoing it with "it's not what happens", followed with jargon-laden explanation. Think about it. How likely is it that a reader of your post would understand what does "very off-shell" mean?

The picture "d quark emits a W- boson and turns into u quark, W- boson then decays into electron and neutrino" is plenty enough for the "layman-level" understanding of neutron decay. Yes, the deep, fundamental picture is different, but it's much more difficult to understand.