Weak interaction left handed v. right handed

[kashiark]

(correct me if i state anything wrong) the weak interaction only works on left handed particles and right handed antiparticles; so if there was a right handed muon what would it decay into? and is it possible for a left handed particle to become a right handed particle? (not just a change of perspective) also are the Z bosons just used in electroweak interactions?

 

[malawi_glenn]

a muon does not have a definite handedness since it is massive. You can go to a frame where the spin flips direction vs. its momentum.

Z bosons are only in electroweak yes.

 

[Meir Achuz]

The word "handedness" refers to the weak interaction of the particle, not its actual helicity.
A left handed particle like an electron has predominantly negative helicity, but as MG says the helicity can be changed by a boost. This does not affect its "handedness".
If you postulate a right handed muon, it is up to you to also postulate its weak interaction, and probably other off-handed particles for it to decay to.

 

[malawi_glenn]

the muon (and an electron) will have one left-handed part and one right handed part, which is proportional to its speed.

That is why pion decay to electron and neutrino is heavily supressed in comparison with pion decay to muon and neutrino.

 

[blechman]

A right-handed particle cannot decay directly, since as you say it is neutral under the weak force (it does not couple to the W bosons). However, thanks to the fact that it's massive (or for the nit-pickers, it has a coupling to the Higgs boson), the right-handed muon can shift to a left-handed muon, which can then decay the usual way. This is all built into the calculation when you sum over helicities and include the appropriate factors of mass and spin-projection operators.

The W,Z bosons, together with the photon, are the gauge bosons of the electroweak interaction. So I suppose the answer to your last question is "yes". But Z bosons are not responsible for decays - only the W boson can decay a particle.

clem is right: you have to be very careful not to confuse "helicity" (the component of angular momentum along an axis) vs "chirality" (the electroweak quantum number) - these only coincide for massless particles. But I'm a little confused by your statement "postulate a right-handed muon". You don't have to postulate anything - the right-handed muon is there in the standard model, and its quantum numbers are dictated by things like anomaly cancellation and observation.

 

[whynothis]

Blechman, I am confused on what you said about Z not be responsible for decays. Why is that? I mean couldn't you have, for instance, higgs ---> lepton anti-lepton via a Z?

 

[hamster143]

Higgs does not couple to a single Z.

It does couple to two Z's, which is the only loophole to blechman's categorical statement. Most other Z's couplings are to pairs of identical particles. For example, muon can decay into muon neutrino by emitting a virtual W. If it emits a virtual Z, it stays a muon.

 

[genneth]

The Higgs couples to fermions via Yukawa type couplings, i.e. direct H ---> lepton+antilepton. The Z and photons are diagonal in the fermions, so they do not cause one fermion to turn into anything other than the same fermion plus one gauge boson.

It's impossible to emphasise enough the need to keep clear about chirality and helicity, even though both are often referred to (in the same context) as being left- or right-handed. The weak interaction couples to the left chiral fermions. Whatever effect this has on helicity depends on the kinematics.

 

[kashiark]

ok so the weak interaction only affects left chiral fermions. i don't really understand chirality. is the chirality the same for all particles of the same type? (btw ty guys )

 

[whynothis]

Agreed.

 

[blechman]

genneth and hamster143 have answered whynothis's question. i was thinking about the fermions, not about the higgs. fermions can ONLY decay through a W boson (to be TECHNICALLY correct, fermions decay though the charged Higgs, which is the goldstone boson eaten by the W boson, but this is a technicality). But never a Z boson, which couples to flavor-diagonal currents.

 

[kashiark]

(to be TECHNICALLY correct, fermions decay though the charged Higgs, which is the goldstone boson eaten by the W boson, but this is a technicality).

so the charge of the W boson is given to it by the Higgs boson it "eats?" are Z and W bosons the same and have just "eaten" different Higgs bosons?

 

[blechman]

so the charge of the W boson is given to it by the Higgs boson it "eats?" are Z and W bosons the same and have just "eaten" different Higgs bosons?

no, it's more complicated than that. The W^+ boson is charged because it is the gauge boson paired with the SU(2) generator T^+ which transforms under the unbroken U(1)_{EM} gauge group as an object of charge +1. The fact that the charged W ate the charged Higgs is just a statement that EM is not broken!

 

[Vanadium 50]

fermions can ONLY decay through a W boson (to be TECHNICALLY correct, fermions decay though the charged Higgs, which is the goldstone boson eaten by the W boson, but this is a technicality).

Are you sure about this? Why can't you have a decay like \mu \rightarrow w_1 + \nu followed by w_1 \rightarrow e + \nu \overline?

But never a Z boson, which couples to flavor-diagonal currents.

I'm not sure about this either - see Hou, Willey and Soni, where they calculate b \rightarrow s + \mu^+ \mu^- which has an important Z contribution.

 

[blechman]

V-50: I'm quite positive!

The Z only couples to flavor diagonal currents! It can't do any flavor violation. Both of your examples involve the W boson in a key way. Just write down the Feynman diagrams that generate your examples. [i'm removing a totally wrong statement i made, see my next post]; the second is a Penguin diagram (the Z contribution is what makes the muons).

Of course, this is ALL assuming there's no other physics out there (SUSY, extra dimensions, etc). Then all bets are off. But strictly within the Standard Model, only the W boson can be implicated in a flavor changing process.

Flavor-Changing Charged Currents involve the W boson directly (as your first example).

Flavor-Changing Neutral Currents must involve the W boson but in LOOPS (as your second example). That's why FCNC are suppressed in the Standard Model.

 

[blechman]

AH, sorry, I misinterpreted your first example (I thought you were going through a hadronic resonance, that was stupid on my part).

No, muon decay goes through a W+ boson, that is correct. Can a muon decay through a purely transverse W boson? I might have been a little too quick with my "technically" comment.

But I do stand by my statement that Z does not decay anything.

 

[Vanadium 50]

My first question was addressing the charged Higgs part. I think I provided a case where the fermions coupled to the triplet W and not the doublet phi. As far as the second part, you have W's, but you also have Z's. In the loop, the Z comes off a fermion. You also have a W, of course.

 

[blechman]

i acknowledge your point on the first example, and have removed the offending statements from my earlier post. as to the second example: I'm not saying the Z boson doesn't have anything to do with physics! I'm just saying that it has nothing to do with flavor violation! the flavor-violating part of the diagram is the part with the W's. The Z boson is just window dressing.

 

[kashiark]

ah ok and perhaps with this next question I'm going beyond our understanding but why do the w and z bosons "eat" higgs bosons when photons and gluons and gravitons(if they exist) dont?

 

[turin]

blechman, are you really trying to say that there is no such vertex that will couple, say, s,d,Z? When one speaks of particle decay, is one not speaking of the mass states rather than the flavor states? Doesn't the CKM mixing provide some component of the s mass state that projects onto the d flavor state?

 

[Vanadium 50]

blechman, are you really trying to say that there is no such vertex that will couple, say, s,d,Z?

No vertex. You can get these interaction through loops - often represented as "blobs" where the vertex would be.

Doesn't the CKM mixing provide some component of the s mass state that projects onto the d flavor state?

No. The CKM matrix rotates d-type quarks with respect to u-type quarks. So it appears when you have one of each. The Z couples to either a pair of u's or a pair of d's, so the CKM doesn't enter into it. (Strictly speaking, can appears twice, in opposite directions, but the effect is the same - it cancels)

 

[turin]

(Strictly speaking, can appears twice, in opposite directions, but the effect is the same - it cancels)

Aha! That's what I was forgetting about. Thank you.