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Thursday, December 16, 2010

Hamiltonian formulism of mechanics (part 2)

-Following four forms of generator are possible : F1(q,Q,t) , F2(q,P,t) , F3(p,Q,t) , F4(p,P,t)

  Legendre transforms are
    where u=(∂ F/∂ x)
 (i) F1=F1(q,Q,t) putting it in equation 1 and finding derivative of F w.r.t. we get
          pj = ∂F1(q,Q,t)/∂qj

          Pj = - ∂F1(q,Q,t)/∂Qj

(ii) putting u = -Pj , x = Qj , y = qj , g = F2(q,P,t) , f=F1 and using Legender transforms we get
          F2(qj,Pj,t) = F1(qj,Qj,t)+ΣPjQj
         Putting this in 1 and solving we get
          pj = ∂F2(qj,Pj,t)/∂qj

          Qj =  ∂F2(qj,Pj,t)/∂Pj


 (iii) F3 = F3(Q,p,t) is the third form
            Connect it to the first form using legender transforms
            Since pj = ∂F1(qj,Pj,t)/∂qj and u=∂f/∂x
             this implies that u=pj , x=qj , y=Qj , g=F3 , f=F1
              Thus,  F3 = F1(Q,q,t)-Σpjqj
              or, F1(Q,q,t)=F3+Σpjqj
              putting these in equation 1
            qj = -∂F3(pj,Qj,t)/∂pj
            Pj = -∂F3(pj,Qj,t)/∂Qj

(iv)    Fourth form is F4(p,P,t) , connecting F4 by F1 through legender transforms and solving we get
         qj = -∂F4(pj,Pj,t)/∂pj

         Qj = ∂F4(pj,Pj,t)/∂Pj

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