Pion Lifetime. A. George January 18, 2012

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1 Pion Lifetime A. George January 18, 01 Abstract We derive the expected lifetime of the pion, assuming only the Feynman Rules, Fermi s Golden Rule, the Dirac Equation and its corollary, the completeness relations for Dirac Spinors), and the definition of the γ matrices anti-commutator). Negligible masses or momenta are dropped. The decay process is π µ + ν µ. We ll label the pion s momentum with q, the neutrino s with k, and the muon s with p. Let s construct M using the Feynman Rules. 1. i, since the presented Feynman rules are for im.. The W propagator: Since M W is so large, we can simplify to: ig µν p µ p ν ) p M W ig µν M W 3. Vertex 1. This is a bound state, so it does not have the usual vertex factor. It must be a Lorentz Invariant. Since the pion has no spin, the only four vector available is q µ. We will put in a form factor, which can be a function of a scalar. However, the only scalar available is q, which is a constant. We ll put in the same constants as for the non-bound vertex. Hence, the form factor will also be constant, and the vertex factor is: igf π qµ 4. Vertex. This time we use the normal vertex factor, with vk) for the antimuon neutrino leaving the vertex, and up) for the muon leaving the vertex. i g up)γ ν 1 1 γ5 )vk) 5. The external lines contribute nothing beyond the spinors already included. Multiplying all these together, we are left with: M = i ig µν igf π MW qµ up) i) g γ ν 1 1 γ5 )vk) Pulling out the constants, we have: Using the metric: Then q µ = p µ + k µ. Then, M = f πg g µν q µ up)γ ν 1 γ 5 )vk) M = f πg q µ up)γ µ 1 γ 5 )vk) M = f πg up) p+ k)1 γ 5 )vk) 1

2 which implies: Then: Using the Dirac Equations: Now we are ready to take the Hermitian Conjugate: Distributing the dagger, we find: M = f πg up) p1 γ 5 )vk) + f πg up) k1 γ 5 )vk) M = f πg up) p1 γ 5 )vk) + f πg up) k1 γ 5 )vk) M = f πg m µ up)1 γ 5 )vk) M = f πg m µ 8M W up)1 γ 5 )vk) ) M = f πg m µ vk) 1 γ 5 ) up) 1 and γ 5 are equal to 1 and γ 5 respectively. Further, we can use the identity u = u γ 0 to find that u = γ 0 u and also that u = uγ 0. Then, M = f πg m µ vk)γ 0 1 γ 5 )γ 0 up) Commuting the left γ 0 through the 1 γ 5 ), we switch the negative to a positive. Then, the two γ 0 s cancel, leaving: Putting these together, we have: M = f πg m µ vk)1 + γ 5 )up) Using the definition of G, this simplifies to: M = f πg 4 m µ 64MW 4 vk)1 + γ 5 )up)up)1 γ 5 )vk) M = f πg m µ vk)1 + γ 5 )up)up)1 γ 5 )vk) This time there is only one initial spin that of the pion), but we must still sum over all the final states. Hence, M = f πg m µ vk)1 + γ 5 )up)up)1 γ 5 )vk) spins With the up) we can use the completeness relation to find: M = f πg m µ vk)1 + γ 5 ) p + m)1 γ 5 )vk) spin We ll use the usual trick of writing the last two terms in index notation, bringing the final term to the front, and then dropping the index notation to write a trace, leaving us with: M = f πg m µ Using the completeness relation again, we find: T r vk)vk)1 + γ 5 ) p + m µ )1 γ 5 ) ) spin M = f πg m µ T r k1 + γ 5 ) p + m µ )1 γ 5 ) )

3 Rearranging the trace, we have: M = f πg m µ T r p + m µ )1 γ 5 ) k1 + γ 5 ) ) If we do the distribution, the terms involving m µ involve either 1, 5, or 9 gamma matrices remember that γ 5 contains four gamma matrices. The trace of the product of and odd number of gamma matrices is zero, so: M = f πg m µ T r p1 γ 5 ) k1 + γ 5 ) ) This trace can be evaluated simply. First, do the remaining distribution: M = f πg m µ [ T r p k) + T r p kγ 5 ) T r pγ 5 k ) T r pγ 5 kγ 5)] These correspond to simple trace identities. 1 The middle two vanish; for the last, the γ 5 anticommutes with the k, giving a minus sign and a term that is identical to the first term. Then, M = f πg m µt r p k) This is also a trace identity. Hence, M = 4f πg m µp k) This is two body decay, so in the π rest frame, k = p, and p k = E k k p = E k + k = E k + E k ). Then, M = 4f πg m µe k + E k ) The decay rate Γ is given by the integral of M over the Lorentz Invariant Phase Space. Hence, dγ = 4fπG m 1 d 3 p d 3 k µe k + E k ) m π π) 3 π) 3 π) 4 δ 4 q p k) E k Cleaning this up, we have: dγ = f πg m µ 8π m π + E k ) d 3 pd 3 kδ 4 q p k) It remains to do six integrals. The spatial part of the delta function fixes q = p + k, but none of these enter into the calculation, so doing those integrals are trivial. dγ = f πg m µ 8π m π + E k ) d 3 kδe q E k ) Three integrals left. We re working in the pion s rest frame, so E q = m π. So, dγ = f πg m µ 8π m π + E k ) d 3 kδm π E k ) Let s switch to polar coordinates, noting that the radial component k ) equals the energy E k since the neutrino s mass is neglected). Hence, dγ = f πg m µ + E k )Ek 8π de k sinθ)dθdφδm π E k ) m π The angular integrals are trivial, so: dγ = f πg m µ πm π + E k )E k de k δm π E k ) 1 These identities are so simple that I won t prove them here. If necessary, see section 1 of my derivation of the muon lifetime; all these identities are proven there. 3

4 One integral left. We ll simply integrate over E k : Γ = f πg m µ πm π Ep + E k )E k de k δm π E k ) This is a bit complicated, because and E k are related. To see how, we ll consider the muon: E µ = m µ + p µ We ve already noted that the muon s momentum is equal and opposite that of the neutrino, so: Eµ = m µ + p ν The neutrino has no mass, so: Eµ = m µ + Eν Switching to the momentum convention in the integral, Ep = m µ + Ek Then, = m µ + E k Now we can explicitly see that our delta function is a function of only one variable E k ). We ll simplify by using the delta function identity. δ[fω)] = δω ω 0) f ω ω=ω 0 which holds whenever there is a root at ω 0. As a totally random guess, let s try ω 0 = m π m µ m π The argument of our delta function evaluated at this point is: m π m π + m µ m µ m π + m π 4 + m4 µ 4m = m π m π π + m µ m π m π m µ = 0 m π So our guess was good, ω 0 is indeed a root of this function. Then, we can evaluate the derivative of the argument and plug into the identity above, yielding: δm π E k ) = δe k ω 0 ) 1 + ω0 Hence, Rewriting a bit: Evaluating the integral, we find: Hence, Γ = f πg m µ πm π Γ = f πg m µ πm π Ep + E k )Ek δe k ω 0 ) de k 1 + ω0 1 + E k )E δe k ω 0 ) E kde k p 1 + ω0 Γ = f πg m µ 1 + ω 0 )ω 1 0 πm π 1 + ω0 Γ = f πg m µω 0 πm π To first order this can be thought of as a Taylor Series. It turns out that it s not an approximation, but it holds exactly. I won t go through it here, but it is proved directly by evaluating performing algebraic substitution let u = fx)) on the integral gx)δ[fx)]dx, then comparing the answer with the result of performing the same procedure on the integral gx) δx x 0) dx. f x) 4

5 Plugging in for ω 0 : The lifetime is therefore Γ = f πg m µm π 8π 1 m µ m π 8π m τ = π fπg m µm π m π m µ ) ) Without a numerical value for f π, we are not able to compute the lifetime directly but we could use this compute the ratio of decay modes. 5