PHY 396 K/L. Solutions for problem set #. Problem : Note the correct muon decay amplitude M(µ e ν µ ν e = G F ū(νµ ( γ 5 γ α u(µ ū(e ( γ 5 γ α v( ν e. ( The complex conjugate of this amplitude M = G F ū(µ γ β ( + γ 5 u(ν µ v( ν e γ β ( + γ 5 u(e (S. has the opposite sign of the γ 5 because γ 5 γ 0 (γ 5 γ 0 = γ 5. Consequently, M = G F ū(ν µ ( γ 5 γ α u(µ ū(µ γ β ( + γ 5 u(ν µ (S. ū(e ( γ 5 γ α v( ν e v( ν e γ β ( + γ 5 u(e and hence M = 4 G F (( tr γ 5 γ α ( p µ + M µ γ β ( + γ 5 ( p νµ + m νµ all spins ( tr ( γ 5 γ α ( p νe m νe γ β ( + γ 5 ( p e + m e ( 4 G F tr ( γ 5 γ α ( p µ + M µ γ β ( + γ 5 p ν tr (( γ 5 γ α p ν γ β ( + γ 5 p e (S.3 where the approximation is m e 0; we also make use of m νe = m νµ = 0 (which may be exactly true or just a very good approximation, future data will tell and simplify notations: p ν p νµ and p ν p νe. Please note that here and henceforth the indices µ, ν, ν, e denote the particles to which respective momenta belong and have nothing to do with the Lorentz indices of those momenta. For the Lorentz indices, I use here α, β and later also γ, δ, σ and ρ. Thus, p µα is the α s component of the muon s 4 momentum, etc., etc.
Having derived eq. (S.3, we now need to evaluate the traces. For the first trace, we have tr (( γ 5 γ α ( p µ + M µ γ β ( + γ 5 p ν ( = tr ( γ 5 γ α ( p µ + M µ γ β p ν ( γ 5 = tr (( γ 5 γ α ( p µ + M µ γ β p ν = tr (( γ 5 γ α ( p µ + M µ γ β p ν using tr(γ α γ β p ν = tr(γ 5 γ α γ β p ν = 0 = tr (γ α p µ γ β p ν tr (γ 5 γ α p µ γ β p ν = 8 p α µp β ν + p β µp α ν g αβ (p µ p ν (S.4 + 8iɛ αγβδ p µγ p νδ. Similarly, the second trace evaluates to tr ( ( γ 5 γ α p e γ β ( + γ 5 p ν = 8 (peα p νβ + p eβ p να g αβ (p e p ν + 8iɛ αρβσ p ρ νp σ e. (S.5 It remains to substitute the trace formulæ (S.4 and (S.5 back into eq. (S.3 and contract the Lorentz indices. Thus, M all spins ( = 6G F p α µp β ν + p β µp α ν g αβ (p µ p ν + iɛ αγβδ p µγ p νδ ( p eα p νβ + p eβ p να g αβ (p e p ν + iɛ αρβσ p ρ νp σ e using symmetry/antisymmetry of factors under α β ( = 6G F p α µp β ν + p β µp α ν g αβ (p µ p ν p eα p νβ + p eβ p να g αβ (p e p ν ɛ αγβδ p µγ p νδ ɛ αρβσ p ρ νp σ e ( = 6G F (p µ p e (p ν p ν + (p µ p ν(p ν p e (p µ p ν (p e p ν (p µ p ν (p e p ν + 4(p µ p ν (p e p ν + (p µ p ν (p ν p e (p µ p e (p ν p ν Q.E.D. = 64G F (p µ p ν (p ν p e. (S.6
Lemma: Consider a decay or a scattering process producing three final state particles. For any such process, the phase space of the final states is d 3 p (π 3 (E d 3 p (π 3 (E d 3 p 3 (π 3 (E 3 (π3 δ (3 (p + p + p 3 (πδ(e + E + E 3 E tot (S.7 in the center-of-mass frame. In this lemma, we shall see that regardless of the three particles masses, the phase space (S.7 can be written as d3 Ω 56π 5 de de de 3 δ(e + E + E 3 E tot (S.8 where d 3 Ω refers to three angular variables parametrizing the directions of the three particles relative to some external frame but not affecting the angles between the three momenta. For example, one may use two angles to describe the orientation of the decay plane (the three momenta are coplanar, p + p + p 3 = 0 and one more angle to fix the direction of e.g., p in tha plane. Altogether, d 3 Ω = 4π π = 8π. We begin the proof with the obvious step of eliminating the momentum conservation deltafunction, thus d3 p d 3 p δ(e + E + E 3 E tot 56π 5 E E E 3 Next, we use spherical coordinates for the two remaining momenta, p3= (p +p. (S.9 d 3 p = p dp d Ω, d 3 p = p dp d Ω, (S.0 and then replace the d Ω describing the direction of the second particle s momentum relative to the fixed external frame with d Ω ( = dθ sin θ dφ ( describing the same direction of p relative to the frame centered on the p. Consequently, d Ω d Ω = d Ω d Ω ( = d Ω dφ ( dθ sin θ d 3 Ω d(cos θ (S. 3
and hence d3 Ω 56π 5 p p dp dp d(cos θ δ(e + E + E 3 E tot E E E 3 p3= (p +p. (S. Next, we use the cosine theorem p 3 = (p + p = p + p + p p cos θ which gives (for fixed p, p and therefore d(cos θ = p 3 dp 3 p p d3 Ω 56π 5 p p p 3 E E E 3 dp dp dp 3 δ(e + E + E 3 E tot. (S.3 Finally, we notice that for a relativistic particle of any mass, pdp = EdE and hence (S.8. It remains to determine the limits of kinematically allowed ways to distribute the net energy E tot of the process among the three final particles. Such limits follow from the triangle inequalities for the three momenta, p p + p 3, p p + p 3, p 3 p + p, (S.4 which look simple but produce rather complicated inequalities for the energies. However, when all three final particles are massless, the kinematic restrictions become simplify to E, E, E 3 E tot, E + E + E 3 = E tot. (S.5 4
Problem : We are now ready to calculate the muon decay rate and the electons spectrum. According to the general rules of decay dγ = M M dp, (S.6 thus in light of eqs. ( and (S.8, dγ(µ e ν µ ν e = G F 8π 5 M µ (p µ p ν (p e p ν de e de ν de ν d 3 Ω δ(e e +E ν +E ν M µ. (S.7 In the muon s frame, (p µ p ν = M µ E ν (S.8 while (p e p e = E e E ν p e p ν cos θ eν = E e E ν + p e + p ν p ν; (S.9 neglecting the electron and neutrino s masses, we may rewrite (S.9 as (E e + E ν E ν = M µ(m µ E ν. (S.0 Consequently, dγ(µ e ν µ ν e = G F 6π 5 M µe ν (M µ E ν de e de ν de ν d 3 Ω δ(e e +E ν +E ν M µ. (S. At this point we are ready to integrate over the final-state variables. In light of d 3 Ω = 8π and the kinematic limits (S.5, we immediately obtain Γ(µ e ν µ ν e = G F M µ π 3 = G F M µ π 3 Mµ de e de ν de ν E ν (M µ E ν δ(e e + E ν + E ν M µ 0 Mµ 0 de e Mµ de ν E ν(m µ E ν Mµ Ee (S. = G F M µ π 3 Mµ 0 de e E e ( M µ 3 E e. In other words, the partial muon decay rate with respect to the final electron s energy is given 5
by dγ de e = G F M µ π 3 E e (3M µ 4E e (S.3 or rather Graphically, dγ de e { G F π 3 M µ E e (3M µ 4E e for E e < M µ, 0 for E e > M µ. (S.4 dγ/de e... M µ.. E e Note how this curve smoothly reaches its maximim at E e = M µ and then abruptly falls down to zero. It remains to calculate the total decay rate of the muon by integrating the partial rate (S.4 over the electron s energy. The result is Γ tot (µ eν ν = G F M 5 µ 9π 3. (S.5 6