# I Pion+/- decay - what *Exactly* is happening?

1. Aug 15, 2017

### Asgrrr

I have seen vague explanations and (to me) misleading diagrams, and I am still not completely sure what is supposed to be happening in pion+/- decay. This Feynman diagram is a starting point.
https://en.wikipedia.org/wiki/Pion#/media/File:PiPlus_muon_decay.svg
What is happening? What happens to the quarks? I'm not asking for the unknowable, just a reasonable description of events and the fate of particles.

Thank you.

2. Aug 15, 2017

### Staff: Mentor

A pion decays to a muon plus a neutrino. The coupling is mediated by the W field.
The quarks stop existing.

Don't take Feynman diagrams too literally.

3. Aug 16, 2017

### Orodruin

Staff Emeritus
To build on that, Feynman diagrams are really mathematical objects that represent different terms in a series expansion. It is easy to be misled into thinking that there are particles running around like that and physicists will often use a language that suggests that.

4. Aug 16, 2017

### Asgrrr

I'm grateful for the replies I'm sure. However, I'm working on teaching materials for high school/college. I am hoping for some way to explain pion decay in the terms of already established weak interactions involving the W particle. Help with that would be much appreciated.

5. Aug 16, 2017

### Orodruin

Staff Emeritus
I am not sure you can explain Feynman diagrams to high school students without a significant amount of "lying to children". You need a large amount of background to just understand what a particle physicist means when (s)he says "particle", let alone what the lines in a Feynman diagram represent.

6. Aug 16, 2017

### ChrisVer

Since it's for teaching matterial:
The level of the class should be considered. As a result if you, as a a teacher, can't interpret well the lines of a Feynman Diagram, you should rethink of introducing the decay through it... If Ws are already introduced (wow), then you can just tell that the udbar quarks annihilate via a virtual W boson which couples to a lepton+neutrino.
How would you answer the simpler question of "what happens to the muons" when a muon and antimuon annihilate to electron+positron?
In fact, outside the context of a quantum theory, the only way you can introduce Feynman diagrams is by giving the students a list of conservation laws (eg charge conservation) and rules (eg flavor changing happens only via Ws), and making them play with the possible Feynman Diagrams (since they are made such that all conservations laws are satisfied in every vertex). They can't be thought as what "really" happens as others mentioned, or as you can understand by the fact that you integrate over all the spacetime the vertices of the internal lines.

Last edited: Aug 16, 2017
7. Aug 16, 2017

### Asgrrr

To clarify, it was never my intention to use Feynman diagrams as the basis of the treatment. I might show one or two for curiosity. What I am after in this thread is connecting pion decay to specific, established weak interaction(s) involving quarks, that's all.

Last edited by a moderator: Aug 16, 2017
8. Aug 16, 2017

### ChrisVer

as an up-quark can be transformed to a down-quark by emitting a W (eg in beta processes), an up-quark and an anti-down-quark can "decay" into a W (virtual or real depending on the available energy, for the pion decay that's virtual)... The W decays to quarks (giving birth to mesons/hadrons) or leptons (e/mu/tau+neutrino), however because the pion is the lightest meson, the only energy-conserving decays are those to leptons (and in particular to electron or muon).

9. Aug 16, 2017

### Asgrrr

PS: for some reason a quote from "Fear and Loathing in Las Vegas" keeps coming to mind when looking at this thread:

"Finish the f-ing story! What happened? What about the glands?"

For "glands" insert "quarks".

Don't take it the wrong way please.

10. Aug 16, 2017

### Staff: Mentor

See above: They stop existing.

11. Aug 16, 2017

### ChrisVer

what about them? they're are no more (before you had a pion, then you have the leptons)... if they didn't disappear we wouldn't call it a decay but radiation (or something).

12. Aug 18, 2017

### Asgrrr

Sorry I didn't see your reply earlier, ChrisVer, my silly quote was not a response to you. I feel we are getting somewhere now.

I'll express antiquarks as ~x.

We start with a let's say u~d (Pion). That decays into W, disappearing both u and ~d.

That doesn't feel right to me as the end of the story. I have a problem with two particles decaying at once in this manner.

Can we say that the established decay of u -> d + W leaves us with d and ~d which immediately annihilate each other in the established way? Photons are not produced because they are not needed. The W carries away the energy and momentum.

That sounds much better to me.

13. Aug 18, 2017

### ChrisVer

what's that problem? They don't decay at once (it's not like a double beta decay), they annihilate each other.

I don't understand what you mean here, is ~d a $\bar{d}$ ? And why would you need a ddbar (or where did you find it)? I also don't understand the "photons are not needed"...

14. Aug 18, 2017

### Asgrrr

Yes, a ~d is as you print it, anti-d. I'm not that good with symbols here.

My problem is, what is it that gets u and ~d to annihilate each other? What allows that? They are not each other's antiparticle, and it doesn't happen inside protons for example (or does it?). This doesn't feel like a fundamental interaction, which leads me to think there's more to the story.

What I was trying to say, is that if we start out with a pion, we have a u~d. That's a pion. Now, if the u decays to d via

u → d +W

which is an established, basic decay mode, then what we are left with is d~d, or a particle and its antiparticle. THAT can reasonably decay in an established fashion (pair annihilation). Usually, pair annihilation would produce photons to carry away the energy and momentum of the destroyed particles, but when the W is forming at just about the same time, the W can carry away the energy/momentum, so there is no need for photons to be emitted. This is a much better explanation of pion decay because it only appeals to already established decay modes.

PS: scratch the proton example, I was mistaken there.

15. Aug 18, 2017

### ChrisVer

Weak interactions do (via the charged W bosons). W bosons allow transitions/couplings between up-type quarks ($u,c,t$) and down-type quarks ($d,s,b$) and generally violate the flavour conservation (it's the only Standard Model interaction which does that)...
The vertices will then have any combination depending on what is incoming/outgoing (so for the ud, you can have any vertex with $ud$, $\bar{d}\bar{u}$, $\bar{u}d$ or $u\bar{d}$ )

in your $u\rightarrow d W$ where is the $\bar{d}$ ? This process has the following outcomes:
$u \rightarrow d W \rightarrow d \ell \nu$
$u \rightarrow d W \rightarrow d q q'$

Pair annihilation between quark-antiquarks will preferably happen via gluons, rather than photons, because of the stronger strong interactions. The only exception would be the neutral pion because again of energy conservation (at least for its decay).
The photons can't transform an up-quark to a down-quark (this would even violate charge conservation), so there can be no $ud$, $\bar{u}d$ or $u\bar{d}$ vertices with a photon coming out of them.

16. Aug 18, 2017

### Asgrrr

Thank you ChrisVer.

I think you have perhaps misunderstood some things I've said, but I'm satisfied with your explanations.

17. Aug 18, 2017

### Orodruin

Staff Emeritus
This is incorrect. What is established is the vertex between the u, d, and W with a particular fermion and boson charge flow. It certainly is no "basic decay mode" as the W is much heavier than both quarks. What is established from the theory are the Feynman rules and later you have to figure out what this means for the interactions among the asymptotic states.

18. Aug 18, 2017

### ChrisVer

OK I think I see what you mean now: you are saying that a charged pion coul decay to a neutral pion and a lepton(and neutrino) like a semileptonic decay? however that's very unlikely and it's a very rare way for the charged pion to decay (because a lot of energy will be kept by the neutral pion, and the W would have only energy to transfer equal to the charged minus neutral pions' masses difference (which makes it even more off-shell W).
If that's what you meant, then the neutral pion will indeed decay to 2 photons (main decay channel for the neutral pions)...

19. Aug 18, 2017

### Staff: Mentor

That is the process that happens. Nature doesn't care if you have a problem with its processes.

20. Aug 19, 2017

### Orodruin

Staff Emeritus
There is only one decaying particle, the pion. If you look at the quark level it is a binary interaction between the bond state constituents.

21. Aug 19, 2017

### Asgrrr

That is almost what I mean but not quite. A neutral pion is (u~u - d~d)/root2 or something like that. But after u decays into d in the charged pion we would have a pure d~d. I realize this is a state that would probably be impossible to detect, I'm looking for a rationalization.

22. Aug 19, 2017

### vanhees71

23. Aug 19, 2017

### ChrisVer

uubar or ddbar can be neutral pions... the expression you show doesn't make sense in terms of quark content (what does MINUS stand for?)...
Nevertheless, the $\pi^\pm \rightarrow \pi^0 e^\pm \nu$ exists, however the dominant decay is the fully leptonic one to muons (mostly) and electrons (less)

24. Aug 19, 2017

### Orodruin

Staff Emeritus
Uhmmmm, it is the linear combination of states that makes a neutral pion. The other linear combination is not a pion.

25. Aug 19, 2017

### Asgrrr

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

See table at bottom of article.