# Difference between $W^-$ and $\pi^-$

Maybe it's a very naif question, but what is the difference between a $$W^-$$ and a $$\pi^-$$? I mean they both change a $$d$$ into a $$u$$ right?

$$d \rightarrow u W^- \quad \text{and} \quad d \rightarrow u \pi^-$$

Of course, the $$W^-$$ also interacts with leptons, but at the level of quarks, what is their difference?

The pion is the lightest hadron, a bound state of $$\bar u$$ and d, and a Goldstone boson of an axial flavor current with its isospin and parity quantum numbers. It is a pseudoscalar, so it is unlike the W - which is a vector. The $$\rho^-$$, also a hadron, couples to $$\bar u ~d$$ and is a vector, so it has the spin of W.

The W is over 500 times heavier than the π and couples to all quarks and all leptons. (Except in an extremely technical/fanciful Higgs-related way, it has no $$\bar u ~d$$ content.) The scale at which the W couples to the quarks is close to the mass of it (too technical), around 80GeV, and that mass enters by square in the denominator of the effective weak coupling constant (Fermi's coupling $$G_F$$) which is then substantially smaller than the strong-scale coupling the pion couples to these quarks with, called $$f_\pi$$, about 90 MeV.

At very low energies, the Goldstone coupling of the pion is crucially implicated in the chiral-symmetry breaking strong-interaction phenomenon that "converts" the quarks with mass 5-10MeV to "constituent quarks" of mass about 300MeV.

When the pion decays, it first converts to the pair of quarks you mention, and the pair of quarks further couple to the W which couples to μν, the decay products. It has to go to these, since there is no lighter hadron for it to decay to, strongly. So the so called "weak decay" (intermediated by a virtual W) is the only thing that can happen. And it happens proportionately to both $$f_\pi$$ and $$G_F$$ in the amplitude, so squared in the rate. This is where the name of $$f_\pi$$ comes from: it is measured in this decay and is characteristic of the pion, but part of the mysterious workings of nonperturbative QCD, and not really computable.

Lots of very different particles may couple to the same particles, with very different couplings indeed, and you hit on the one that virtually takes the prize in this sort of thing.

• You've introduced the $\rho$ without really saying how it relates to the question. Or did you mean $\pi$ in the last sentence of the first paragraph? – dmckee Feb 16 at 23:38
• Well,I indirectly reassured the OP that the spin of the pion is not an essential part of the contrast, as there are other hadrons, like the ρ with the spin of the W, and vector couplings of which the W has one too, among others (the axial). If the language is confusing, I could improve, but am not sure how. – Cosmas Zachos Feb 16 at 23:42

The most fundamental difference is that a $$W^-$$ is an elementary particle not composed of anything else, while a $$\pi^-$$ is a bound state of two quarks.

Another difference is that the $$W^-$$ is one of three particles that transmit the weak nuclear force, while the $$\pi^-$$ is unrelated to the weak nuclear force.

• The other big difference is the mass. 140 MeV compared to 80 GeV – RogerJBarlow Feb 16 at 23:22
• @RogerJBarlow Good point. Thank you for mentioning that. – G. Smith Feb 17 at 0:13

The W interacts with quarks in a nucleon by transforming a u to d or vice versa.

(From Wikipedia)

This is different than exchanging a pion. When a nucleon exchanges a pion, that pion carries away e.g. the original d quark, leaving behind a new/different u quark. The pion does this by containing that original d, plus an anti-u.

So the W is a fundamentally transforming interaction. The pion is a way of reshuffling things bound together by the strong force.

• Curious about the downvotes. What’s objectionable? – Bob Jacobsen Feb 18 at 4:57
• I downvoted because what you are describing is somewhat vague and it's not even presented in most standard references as an intuitive picture of what is going on (afaik). It has no concrete mathematical content --- nor physical, in my opinion. A good answer to the question would need to contain the actual differences between $W_\mu$ and $\pi$, such as, their spin, their mass, the fact that one of them is composite, their couplings, their interactions with other fundamental forces etc... – MannyC Feb 21 at 4:23
• Thanks for clarifying. I added diagrams to make the distinction clear. As to whether this is a better way to answer a naive question about transformation roles, as opposed to introducing Goldstone bosons, reasonable people can differ. – Bob Jacobsen Feb 21 at 19:38