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Lecture 13 Muon and Tau Decay 1 Introduction 2 Lepton Number

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Nuclear and Particle Physics - Lecture 13 Muon and tau decay 1 Introduction We have seen the Feynman diagram vertices for weak charged current interactions and now we want to apply them. There are extra complications with quarks which we will discuss later, so first we will look at purely leptonic interactions and in particular, muon and tau decay. In the absence of charged current weak interactions, these leptons would be stable, so their decays provide a good test of the weak theory. 2 Lepton number conservation We saw the vertices have the W coupling the e to the νe. We can add further generations for µ/νµ and τ/ντ but these will be independent; the W does not couple e to νµ, for example. Each vertex has the same generation of lepton on both fermion lines so we can introduce the concept of \lepton numbers" and get conserved quantities. We need three separate lepton number for each of e, µ and τ and their neutrinos. Lepton Le Lµ Lτ − e , νe +1 0 0 + e , νe −1 0 0 − µ , νµ 0 +1 0 + µ , νµ 0 −1 0 − τ , ντ 0 0 +1 + τ , ντ 0 0 −1 Table 1: Lepton number values This number is the equivalent of particle type (or flavour) in the electromagnetic and strong interactions, as it is never changed by those forces. 3 Muon decay Because the type of particle can change in a weak interaction, the heavier fundamental fermions can decay. The first example of this which we shall look at is the muon, mass 0.1057 GeV, which has only one significant decay mode − − µ ! νµe νe This couples µ to νµ and e to νe only and, as stated above, therefore conserves each of the lepton numbers. Explicitly, the initial values are Lµ = +1 and Le = 0. The final values are Lµ = +1 and Le = (+1) + (−1) = 0 again. The Feynman diagram is below. This is structurally very similar to the ones we have seen before. However, the final state is much more complicated because we have a three-body final state. In two-body final states, the overall energy and momentum conservation completely fixes both outgoing momentum magnitudes and only the overall θ and φ directions are left as free variables. Thus, the decay width is only a function of solid angle, dΓ=dΩ and the total width is given by integrating over the two solid angle variables. 1 νµ g - µ- W e W gW ν e It is straightforward to count the number of variables; each final state particle has three momentum components (from which the energy can be calculated using E2 = p2c2 + m2c4). Hence, for a two-body final state, there are 2×3 = 6 variables needed to describe the final state. Total energy and momentum conservation give four constraints (one for energy and one for each momentum component) so the number of remaining free variables is 6 − 4 = 2. However, for a three-body decay, there are now 3 × 3 = 9 variables needed to describe the final state. The total energy and momentum conservation gives the same four constraints, so the number of free variables is 9 − 4 = 5 free variables, rather than 2. The decay rate will, in general, be a function of all five. Hence we need to integrate over five variables to get the total width. In particular, this tells us that the energy of the electrons from muon decay is not constrained to a single value (as would be the case for a two-body decay) but instead come out with a spread of energies, i.e. a spectrum ν e e ν ν ν In fact, as the neutrinos are almost always unobserved in this decay, then often the only possible observables of the five are the three which give the electron energy and direction (θ and φ). Sometimes, e.g. when the muon is unpolarised, the only interesting variable is the electron energy. Hence, the usual calculation is to integrate over the other four and just express the width as a function of the electron energy. This is a laborious calculation, even ignoring the electron mass, but the result is 2 dΓ 1 g 2 2m2 c4E2 4E G2 2m2 c4E2 4E = 2 W µ e 1 − e = F µ e 1 − e dE 8 M c2 (2π)3 3m c2 (¯hc)6 (2π)3 3m c2 e " W # µ ! µ ! Note, this is in the low energy (i.e. four-fermion vertex) limit as we use GF , which is perfectly valid here as all energies are ∼ 100 MeV or less. This spectrum is called the Michel distribution 2 and it peaks at the maximum allowed value of Ee = mµc =2. 2 It is now straightforward to get the total width by integrating the above expression 2 2 4 2 2 4 mµc =2 2 mµc =2 3 2 3 4 GF 2mµc 2 4Ee GF mµc Ee Ee Γ = 6 3 Ee − 2 dEe = 6 3 − 2 (¯hc) (2π) 0 3m c (¯hc) 4π 3 3m c Z µ ! " µ #0 2 2 4 3 6 3 6 2 2 4 3 6 GF mµc mµc mµc GF mµc mµc = 6 3 − = 6 3 (¯hc) 12π " 8 16 # (¯hc) 12π 16 2 5 10 2 2 5 G m c G (m c ) − = F µ = F µ = 3:0 × 10 19 GeV (¯hc)6 192π3 (¯hc)6 192π3 −6 which corresponds to a lifetime of τµ = 2:2 × 10 s, in good agreement with the experimental value of (2:19703 0:0004) × 10−6 s. We have seen a picture of e+e− ! µ+µ− at Aleph, where the muons leave tracks in the detector. How come this is observed if they decay? The average time a muon lives for in its rest frame is ht0i = τµ so in a boosted frame, because of time dilation, this is lengthened to hti = γτµ. In this time, it will go an average distance of hdi = vhti = γβcτµ. This is usually easiest to evaluate by noting p = γβmc, so hdi = (p=mc)cτµ. The combination cτ for any particle sets the scale of the decay lengths and for the muon, it is cτµ = 660 m. For the muons at Aleph, p ∼ 45 GeV so hdi ∼ 280 km, which takes them all the way through the atmosphere, so they normally decay well after leaving the detector. 4 Mass dependence and the tau lifetime 5 The dependence on the muon mass is striking; Γ / mµ. Where does this come from? The only 2 masses in the problem are the muon and W masses and we know the amplitude / 1=MW so the 4 width / 1=MW Therefore, on dimensional grounds, the width (which has dimensions of energy) must be 4 mµ 2 5 Γ / 4 mµc / mµ MW We can use this to look at the tau. One decay of the tau is − − τ ! ντ e νe 3 which is exactly the same as for the muon, except the muon mass is replaced by the tau mass mτ = 1:777 GeV, i.e. mτ =mµ = 16:8. Hence, we would calculate the partial width for this decay mode to be 5 − − mτ − − −13 Γ(τ ! ντ e νe) = 5 Γ(µ ! ντ e νe) = 4:0 × 10 GeV mµ This is not the only decay mode for the tau, as the extra mass makes other modes accessible. There is a muon decay mode − − τ ! ντ µ νµ which, ignoring the muon mass relative to the tau mass, will have the same rate as the electron decay. Since the tau mass is more than pion mass, then there are also decays to hadrons, which obviously go via quarks − τ ! ντ du ντ τ- u W d If we assume asymptotic freedom holds, then from universality, the rate might be expected to be e : µ : hadrons = 1 : 1 : 1. However, this ignors colour, and just as the e+e− cross section gets enhanced by a factor of three, so does the decay rate here. Hence, we might expect e : µ : hadrons = 1 : 1 : 3, i.e. 20% e, 20% µ and 60% hadrons. In fact, the measured rates are 17.8%, 17.4% and 64.8% so this is not a bad approximation. The total width is the sum of the partial widths and in this approximation would be five times the electron partial width − − −12 Γ = 5Γ(τ ! ντ e νe) = 2:0 × 10 GeV −13 This gives ττ = 3:2 × 10 s in reasonable agreement with the experimental value of (2:906 0:011) × 10−13 s. How far does a tau go before decaying? Applying the above formula gives cττ = 87 µm and for 45 GeV momentum at Aleph, then the average distance is 2.2 mm. Hence, all taus decay quite close to the beam collision point, well before they enter the tracking detectors; we don't usually see charged tracks due to taus themselves, only their decay products. So how can we measure the tau lifetime? With accurate tracking using a silicon vertex detector, the actual tau decay positions can be reconstructed and the lifetime distribution unfolded. 4 10 3 10 2 Entries / 0.05 cm 10 1 −1 0 1 2 3 Decay length (cm) 5.
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