By Goldstein Herbert
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Extra resources for Classical Mechanics solution manual
6 MeV) stationary in the laboratory system to produce a K + meson (rest mass 494 MeV) and a Λ hyperon (rest mass 1115 MeV). What is the threshold energy for this reaction in the laboratory system? We’ll put c = 1 for this problem. The four-momenta of the pion and neutron before the collision are pµ,π = (pπ , γπ mπ ), pµ,n = (0, mn ) and the squared magnitude of the initial four-momentum is thus pµ,T pµT = −|pπ |2 + (γπ mπ + mn )2 = −|pπ |2 + γπ2 m2π + m2n + 2γπ mπ mn = m2π + m2n + 2γπ mπ mn = (mπ + mn )2 + 2(γπ − 1)mπ mn (7) The threshold energy is the energy needed to produce the K and Λ particles at rest in the COM system.
Then the energy conservation relation becomes πc2 = (µ2 c4 + p2 c2 )1/2 + pc −→ (πc − p)2 = µ2 c2 + p2 −→ p= π 2 − µ2 c. 2π Then the total energy of the muon is Eµ = (µ2 c4 + p2 c2 )1/2 = c 2 µ2 + (π 2 − µ2 )2 4π 2 1/2 c2 4π 2 µ2 + (π 2 − µ2 )2 2π c2 2 = (π + µ2 ) 2π = 1/2 Then subtracting out the rest energy to get the kinetic energy, we obtain c2 2 (π + µ2 ) − µc2 2π c2 2 = (π + µ2 − 2πµ) 2π c2 = (π − µ)2 2π K = Eµ − µc2 = as advertised. 6 MeV) stationary in the laboratory system to produce a K + meson (rest mass 494 MeV) and a Λ hyperon (rest mass 1115 MeV).
The Jacobian of the transformation is M= = ∂Q ∂q ∂P ∂q ∂Q ∂p ∂P ∂p − q1 cot p cot p −q csc2 p . 6 Homer Reid’s Solutions to Goldstein Problems: Chapter 9 Hence − q1 cot p cot p −q csc2 p ˜ MJM = cot p − 1q cot p −q csc2 p = cot p − 1q cot p −q csc2 p 0 1 −1 0 cot p −q csc2 p 1 − cot p q 0 csc2 p − cot2 p 2 cot p − csc p 0 = 2 = =J 0 1 −1 0 so the symplectic condition is satisfied. 5 Show directly for a system of one degree of freedom that the transformation Q = arctan αq , p P = αq 2 2 1+ p2 α2 q 2 is canonical, where α is an arbitrary constant of suitable dimensions.