By Goldstein Herbert

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8 Prove directly that the transformation Q1 = q 1 , Q2 = p 2 , P1 P2 = p1 − 2p2 = −2q1 − q2 is canonical and find a generating function. After a little hacking I came up with the generating function F13 (p1 , Q1 , q2 , Q2 ) = −(p1 − 2Q2 )Q1 + q2 Q2 9 Homer Reid’s Solutions to Goldstein Problems: Chapter 9 which is of mixed F3 , F1 type. This is Legendre-transformed into a function of the F1 type according to F1 (q1 , Q1 , q2 , Q2 ) = F13 + p1 q1 . The least action principle then says p1 q˙1 + p2 q˙2 − H(qi , pi ) = P1 Q˙ 1 + P2 Q˙ 2 − K(Qi , Pi ) + + ∂F13 ∂F13 ˙ p˙ 1 + Q1 ∂p1 ∂Q1 ∂F13 ∂F13 ˙ q˙2 + Q2 + p1 q˙1 + q1 p˙ 1 ∂q2 ∂Q2 whence clearly ∂F13 = Q1 ∂p1 ∂F13 P1 = − = −p1 − 2Q2 ∂Q1 = −p1 − 2p2 ∂F13 = Q2 p2 = ∂q2 ∂F13 P2 = − = −2Q1 − q2 ∂Q2 q1 = − = −2q1 − q2 .

Hence the minimum energy situation is that in which the K and Λ both travel in the direction of the original pion’s motion. (This is equivalent to Goldstein’s conclusion that, just at threshold, the produced particles are at 7 Homer Reid’s Solutions to Goldstein Problems: Chapter 7 rest in the COM system). Then the momentum conservation relation becomes simply pπ = p K + p λ (8) and the energy conservation relation is (with c = 1) (m2π + p2π )1/2 + mn = (m2K + p2K )1/2 + (m2Λ + p2Λ )1/2 . (9) The problem is to find the minimum value of pπ that satisfies (9) subject to the constraint (8).

Find m3 and v3 in terms of m1 , m2 , v1 , and v2 . Would it be possible for the resultant particle to be a photon, that is m3 = 0, if neither m1 nor m2 are zero? Equating the 3rd and 4th components of the initial and final 4-momentum of the system yields γ 1 m1 v 1 − γ 2 m2 v 2 = γ 3 m3 v 3 γ 1 m1 c + γ 2 m2 c = γ 3 m3 c Solving the second for m3 yields m3 = γ1 γ2 m1 + m2 γ3 γ3 (5) and plugging this into the first yields v3 in terms of the properties of particles 1 and 2: γ 1 m1 v 1 − γ 2 m2 v 2 v3 = γ 1 m1 + γ 2 m2 Then γ 1 m 1 β1 − γ 2 m 2 β2 v3 = c γ 1 m1 + γ 2 m2 2 2 γ m + 2γ1 γ2 m1 m2 + γ22 m22 − [γ12 m21 β12 + γ22 m22 β22 − 2γ1 γ2 m1 m2 β1 β2 ] 1 − β32 = 1 1 (γ1 m1 + γ2 m2 )2 2 2 2 2 2 γ m (1 − β1 ) + γ2 m2 (1 − β22 ) + 2γ1 γ2 m1 m2 (1 − β1 β2 ) = 1 1 (γ1 m1 + γ2 m2 )2 2 2 m + m2 + 2γ1 γ2 m1 m2 (1 − β1 β2 ) = 1 (γ1 m1 + γ2 m2 )2 β3 = and hence γ32 = (γ1 m1 + γ2 m2 )2 1 = 2 .

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