Generally, when you are given the action
S=∫t2t1dt(p˙q−H)
the boundary conditions are
q(t1)=q1andq(t2)=q2.
This is useful because to calculate δS we do an integration by parts with boundary term
[pδq]t2t1=0.
But suppose I give you different boundary conditions for the action, namely q(t1)=q1andp(t2)=p2.
Then solving δS=0 should still give you Hamilton's equations, I think, but I'm having trouble showing this, as I get annoying boundary terms since δq(t2)≠0.
Answer
I) In general, for a given choice of boundary conditions, it is important to adjust the action with compatible boundary terms/total divergence terms in order to ensure the existence of the variational/functional derivative. As OP observes, the problem is (when deriving the Euler-Lagrange expression) that the usual integration by parts argument fails if the boundary conditions (BCs) and the boundary terms (BTs) are not compatible.
II) Concretely, for the mixed BCs
q(ti) = qiandp(tf) = pf,
which OP considers, we need to prepare the standard Hamiltonian action
S0[p,q] = ∫tftidt {p˙q−H}
with a total divergence term −ddt(pfq). The new action becomes
S[p,q] = ∫tftidt {(p−pf)˙q−H},
or what amounts to the same,
S[p,q] = ∫tftidt {˙p(qi−q)−H}.
It is straightforward to use the BCs (1) to show that the actions (3) and (4) are equal.
III) Now when we vary the action (3)
δS[p,q] = [(p−pf)δq]tfti+∫tftidt {(˙q−∂H∂p)δp−(˙p+∂H∂q)δq},
the BCs (1) cancels the total derivative term
[(p−pf)δq]tfti (1)= 0,
so that the variation (5) only contains bulk terms. The corresponding Euler-Lagrange equations become Hamilton's equations.
IV) The above example can be generalized to other BCs. We leave it to the reader to work out the compatible BTs.
No comments:
Post a Comment