# The different lifetimes of W boson and Pion?

1. Mar 3, 2008

### isospin

The positive W boson decays to a anti-muon and a muon neutrino, it should be a weak interaction.
And the positive Pion can decays to a anti-muon and a muon neutrino, too.
But the lifetimes of them are totally different, so why?
I know the positive Pion is composed of two quarks, but could you please give me a quantitative analysis?

Last edited: Mar 3, 2008
2. Mar 3, 2008

Staff Emeritus
There is no theorem that states that particles with the same decay products have the same lifetime.

The pion lifetime is proportional to m_W^4/m_pi^5.

The W lifetime is proportional to m_W.

3. Mar 3, 2008

### arivero

To note: while the charged pion lifetime is proportional to m^pi^-5, the lifetime of the neutral pion (whose decay is electromagnetic) is proportional to m_pi^-3

And now, if you consider the product of lifetime times m^3 for all the electromagnetic decays of neutral particles, you will we surprised they keep all in the same order of magnitude, from pi0 up to Z0. And thus also the charged W boson keeps near of this line. In fact the fit is better than order of magnitude, I have told about it elsewhere.

As you say, there is not a theorem. But it is a general and almost exhaustive empirical rule (it only fails, afaik, for the Upsilon).

And of course, for weak decays there is "almost a theorem", or a couple of them, and there was an empirical rule already in 1930, sometimes named "Sargent's rule", from a paper of an obscure physicist having this name.

4. Mar 3, 2008

### arivero

Hmm I am prety sure https://www.physicsforums.com/search.php?searchid=1028755 [Broken] having some part of the kind of quantitative analysis you are interested on....

Ah, here! But it only works out how the weak decay rate of a quark (or a charged lepton, too?) does depend of its mass. See figure 2 of post #13.

Perhaps someone would like to expand the thread adding a calculation of charged pion decay as a function of its mass. I have never seen such beast, but it could exist in some obscure paper.

Last edited by a moderator: May 3, 2017
5. Mar 3, 2008

Staff Emeritus
Yes, but that's a statement that the lifetimes are different (and go as m^5), not that they are the same.

One can calculate the lifetime of the charged pion, but one will end up needing to include a factor f_pi, which is a measure of the wavefunction overlap between the two quarks. Typically, one gets f_pi from the measured lifetime of the pion, although I suppose one could use the lattice instead.

6. Mar 4, 2008

### isospin

Thanks for all your replies.
I just thought that the lifetime of W boson, with the weak interaction decay, would be much larger, as the characteristic constant of weak interaction is much smaller that that of Strong interaction.
But the fact is just opposite. What is the right relationship of the lifetime and interacton constant?
Excuse me, I am not expert on it......

7. Mar 4, 2008

Staff Emeritus
The pion decays by the weak interaction.

8. Mar 4, 2008

### isospin

Really?
In my opinion, the poin decays to a quark, an anti-quark and a W boson first. Then the W boson decay into leptons via the weak interaction. But the annihilation of the left q and anti-q should be a strong interaction precess.
Is it true?

9. Mar 4, 2008

Staff Emeritus
Do you think I am making this stuff up?

This isn't an opinion poll. What happens happens.

10. Mar 4, 2008

### arivero

Indeed. A 99.98% of the time.

http://pdglive.lbl.gov/Rsummary.brl?nodein=S008&sub=Yr&return=MXXX005 [Broken]

Now, while calling for strong process is out of reach because we do not know what to do with the resulting gluon, I wonder if it is possible to think of the process
μ ν γ [c] ( 2.00 ± 0.25 ) × 10 − 4
in the lines of Isospin's suggestion. Ie with the gamma coming from anhiquilation of quark and antiquark.

On the other hand, the process claimed by Isospin exists somehow when the W decays to electrons instead of muons:

e+ ν π0 ( 1.036 ± 0.006 ) × 10 − 8

it is just that the extant quarks take some time to decay, they do live as a neutral pion. An then the neutral pi0 decay is not really "strong". The QCD aspects of the interaction are hidden in the "pion decay constant", as V50 told before somewhere. The products of the decay are a pair of photons.

Last edited by a moderator: May 3, 2017
11. Mar 4, 2008

### arivero

just a point, isospin. In modern terms, the W boson *IS* the weak interaction. The W comes from "Weak". Try to keep focused in post- QCD models.

Of course in the 1950s the terms weak and strong, both, were standing for different objects that nowadays.

So if you tell that something decays to a W boson, and its mass is less than the mass of the W, it is the same that telling that it decays via the Weak interaction.

Last edited: Mar 4, 2008
12. Mar 4, 2008

### isospin

I am really sorry because I didn't make the question clear and said some silly words. To be honest, I was confused by those conceptions.

Ok, I just want to know why the positive W boson is so short-lived with a lifetime of about 3 × 10−25 s, which is almost the typical tme scale for a strong reaction.
But in fact the decay of W+ should be a weak reaction. Is it just because the heavy mass of it?

Last edited: Mar 4, 2008
13. Mar 4, 2008

### arivero

Point is, the weak decay is weak only when (or because) the mass of the decaying particle is smaller than the mass of the weak intermediate boson, W. The detailed calculation for the simplest case is in the thread I referred to, and you can follow the integrals.

At higher scales, weak decay is not weak.

Plus, all the interactions have an scaling with the some power of the mass, so you can not compare straighforwardly the decay rates at 1GeV and 100GeV.

14. Mar 4, 2008