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PS5-solutions - Physics 253a 1 Problem Set 5 Solutions...

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Unformatted text preview: Physics 253a 1 Problem Set 5 Solutions October 26, 2010 1. (a) We need a Weν e coupling and a Wμν μ coupling in order to get a decay of the form W- μ- ν μ e- ν e An appropriate Lagrangian is L = X f = e,μ,ν e ν μ- 1 2 φ f ( + m 2 f ) φ f- 1 2 W ( + m 2 W ) W + gWμν μ + gWeν e Here, we’ve ignored charge and spin. In real life, the W would be a complex vector, and the μ , e , and neutrinos would be spinors, so everything would look a little different. The only fermion mass that will matter in the problem is m μ , since m e is much smaller, and the m ν ’s are incredibly small. (b) The amplitude corresponding to the diagram above is i M = ( ig ) 2 i ( p μ- p ν μ ) 2- m 2 W ≈ ig 2 m 2 W when m W is large. Remember that we’re supposed to fudge the answer by replacing g → gE cm = gm μ , so the amplitude is M = g 2 m 2 μ m 2 W . (c) |M| 2 is dimensionless, but Γ should have dimensions of 1 time , or mass. After we compute M , we don’t know that it came from a W diagram anymore, so the only mass scale left in the problem is m μ (remember that m e and m ν are small enough in comparison that we can take them to vanish). Thus, to get the right units, we should toss in another m μ . We get Γ = m μ (2 π ) 3 g 4 m 4 μ m 4 W This kind of reasoning is a quick and dirty way to avoid doing the (fairly com- plicated) 3 body phase space integral in the definition of Γ. Knowing that the only relevant mass scale is m μ is a huge simplification. Physics 253a 2 (d) The muon decays almost exclusively to e- ν e ν μ , and its decay rate is (lifetime)- 1 = (2 . 2 × 10- 6 s)- 1 = 3 × 10- 16 MeV. Let’s say g = 1 2 , so that m W = 1 2 m μ m μ (2 π ) 3 Γ 1 4 ≈ 320GeV This is too big by about a factor of 4, stemming from our approximation to the phase space integral, our choice of g , and some other ways order-one constants can creep into the answer. (e) The τ lifetime is 3 × 10- 13 s , corresponding to Γ τ = 2 × 10- 9 MeV. Very naively, we should have m τ = (2 π ) 3 Γ τ m 4 W g 4 1 5 ≈ 2 . 5GeV (f) The formula we used for Γ τ actually comes from the single decay channel τ → eν τ ν e , so we should only use the contribution from that channel. Γ τ → Fraction( Γ i Γ τ )Γ τ = . 18Γ τ Our answer changes by ( . 18) 1 5 = . 7, so we predict m τ ≈ 1 . 7GeV, which is pretty frickin’ close. The reason is that our estimate for m W cancels. Just knowing that the rate for our three body decay is proportional to m 5 f ( f = τ,μ ), we can say m τ m μ = Γ τ → eν τ ν e Γ μ → eν μ ν e 1 5 = Fraction( Γ i Γ τ )Γ τ Γ μ ! 1 5 (g) The reason we only see the combination g 2 m 2 W is because we only worked to zeroth order in q 2 m 2 W , where q = p μ- p ν μ . (Recall our approximation in part ( b )). Let’s keep the next order term, i M = ( ig ) 2 i q 2- m 2 W = ig 2 m 2 W (1 + q 2 m 2 W + O q 4 m 4 W ) We see that we get a different combination of g and m W at this order, so if we could measure the decays precisely enough we could determine...
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This note was uploaded on 02/04/2012 for the course PHYS 253A at Harvard.

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PS5-solutions - Physics 253a 1 Problem Set 5 Solutions...

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