## 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
g(u,y)=f(x,y)-ux
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

K=H+∂F1(q,Q,t)/∂t
(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

K=H+∂F2(qj,Pj,t)/∂t

(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

K=H+∂F3(pj,Qj,t)/∂t
(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
K=H+∂F4(pj,Pj,t)/∂t