This might be a stupid question, but I just don't get it. In Hamiltonian mechanics when examining conditions for a (q,p)→(Q,P) transformation to be canonical one starts with ˙qipi−H(q,p,t)=˙QiPi−ˉH(Q,P,t)+ddtW(q,Q,t)
Answer
These are very good questions.
Let us start with the old phase space variables (qk,pℓ). The Hamiltonian action is SH = ∫tftidt LH,LH := ˙qjpj−H(q,p,t).
Its infinitesimal variation reads δSH = bulk-terms + boundary-terms,where bulk-terms = ∫tftidt(δSHδqjδqj+δSHδpjδpj)yield Hamilton's equations, and where boundary-terms = [pjδqj⏟=0]t=tft=ti = 0vanish as they should because of, say1, essential/Dirichlet boundary conditions (BCs) qj(ti) = 0andqj(tf) = 0.Notice that the momenta2 pj are unconstrained at the boundary.Next let us consider new phase space variables (Qk,Pℓ). The action of type 1 reads3 S1 := ∫tftidt L1 = SK+[F1(q,Q,t)]t=tft=ti,SK := ∫tftidt LK,
L1 := LK+dF1(q,Q,t)dt,LK := ˙QjPj−K(Q,P,t),where the old positions qj=qj(Q,P,t) are implicit functions of the new phase space variables (Qk,Pℓ). Its infinitesimal variation reads δS1 = bulk-terms + boundary-terms,where bulk-terms = ∫tftidt(δS1δQjδQj+δS1δPjδPj)yield Kamilton's equations, and where boundary-terms = [(Pj+∂F1∂Qj)⏟=0δQj+∂F1∂qiδqj⏟=0]t=tft=ti = 0vanish as they should.
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1 Alternatively, one could impose natural BCs, or perhaps some mixture thereof.
2 Note that in QM it would conflict with the HUP to simultaneously impose BCs on a canonical conjugate pair.
3 Notation conventions: Kamiltonian K≡ˉH and type 1 generating function F1≡G1≡W.
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