Invariance of Poisson bracket under canonical transformation Let u and v be two functions such that u = u (q i , p i , t) and v = v (q i , p i , t) Let a canonical transformation is from (q i , p i , t) → (Q i , P i , t) Here q = q (Q, P, t) and p = p (Q, P, t) Corresponding to it the transformation in u and v are u (q i , p i , t) → u′ (Q i , P i , t) and v (q i , p i , t) → v′ (Q i , P i , t) Now we have to prove that if (q, p, t) → (Q, P, t) is canonical then [u, v] p, q = [u′, v′] P, Q It means the Poisson bracket are invariant under a canonical transformation. Proof If F 1 and F 2 are generating function, then the transformation relation for the variables are F 2 = F 1 + PQ Thus the Poisson brackets are invariant under a canonical transformation . To know more about Invariance of Poisson bracket under canonical transformation click on the link for English and click on the link for Hindi
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