PS5_253A_solutions.pdf - Problem Set 5 Solutions Physics...

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Problem Set 5 Solutions Physics 253A Fall 2018 1(a) The decay process μ - e - ¯ ν e ν μ proceeds through an intermediate W - . From the charge of each particle involved, it follows that this process must be μ - ν μ W - followed by W - e ¯ ν e . Since they are charged, let us represent μ - , W - , and e - with complex scalars. We will use real scalars for the neutral neutrinos. The Lagrangian that allows this decay process is then L = - φ * μ φ μ - φ * W φ W - φ * e φ e - 1 2 φ ν μ φ ν μ - 1 2 φ ν e φ ν e - m 2 μ φ * μ φ μ - m 2 W φ * W φ W (1) + g φ ν μ ( φ μ φ * W + h.c.) + g φ ¯ ν e ( φ W φ * e + h.c.) . (2) 1(b) The tree level diagram for the decay process is μ - ν μ W - e - ¯ ν e (3) which corresponds to the amplitude i M = ig i ( p μ - p ν μ ) 2 - m 2 W ig . (4) Squaring this leads to |M| 2 = g 4 [( p μ - p ν μ ) 2 - m 2 W ] 2 (5) = g 4 m 4 W + O ( m 2 μ /m 2 W ) . (6) 1(c) The muon decay rate depends on |M| 2 in the following way: Γ μ = 1 2 m μ Z d Π LIPS |M| 2 , (7) which shows that Γ μ g 4 m 4 W . (8) In the Lagrangian (1) the coupling g has mass dimension 1. However, the coupling constant of the real weak interaction is dimensionless, just like the coupling constant for electromagnetism. So we will let g in equation (8) be dimensionless. The decay rate Γ μ has mass dimension 1, so there should be 5 factors of m μ to make this work: Γ μ g 4 m 5 μ m 4 W . (9) These factors of m μ come from momentum factors in the Feynman rules for the weak interactions, and from the integral over final-state phase space. To a first approximation (in the limit of large m W and negligible m e ) the muon’s 1
mass is the only dimensionful quantity controlling the momentum flowing through the legs of the diagram as well as the size of final-state phase space. This justifies the factor of m 5 μ in equation (9). The integral over 3-body phase space also leads to a numerical factor of 1 / 192 π 3 . This leads to Γ μ = g 4 m 5 μ 192 π 3 m 4 W . (10) We arrived at this form by a direct calculation in Problem Set 3. 1(d) The muon has a mass of m μ = 105 . 7 MeV and lifetime of 2 . 197 microseconds. If we guess that g 1 / 137 as in electromagnetism, this leads to g 0 . 3. Working to a single significant figure, since we are hoping for an order-of- magnitude estimate, we have ~ 2 × 10 - 6 sec (0 . 3) 4 (100 MeV) 5 192 π 3 m 4 W , (11) which implies m W 80 GeV . (12) 1(e) By applying the arguments above to the τ - e - ¯ ν e ν τ decay, we can write Γ τ e ¯ ν e ν τ = g 4 m 5 τ 192 π 3 m 4 W . (13) If this were the only decay channel for the tau, we would have Γ τ = m 5 τ m 5 μ Γ μ . (14) This would allow us to estimate m τ in terms of only Γ τ , Γ μ , and m μ ; 1 3 × 10 - 13 sec m 5 τ (100 MeV) 5 1 2 × 10 - 6 sec , (15) which would imply m τ 2 GeV . (16) 1(f) In reality, the tau decays to e - ¯ ν e ν τ only 17.82% of the time, meaning that Γ τ e ¯ ν e ν τ = 0 . 1782 Γ τ . (17) The muon virtually always decays through this channel. Using this, we can get an accurate estimate of m τ from 0 . 1782 2 . 903 × 10 - 13 sec = m 5 τ (105 . 7 MeV) 5 1 2 . 197 × 10 - 6 sec . (18) We find m τ = 1 . 778 GeV , (19) which is very close to the actual value.

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