For the (standard) derivation from Lagrange's equations see http://en.wikipedia.org/wiki/Hamilton%27s_principle.
I don't know a direct derivation from Hamilton's equations (would such a thing be possible? Hamilton's principle is in terms of the action which involves the Lagrangian).
It should be possible,,Its my exercise.I will study more about it.......
In my opinion, the problem is not so simple, it is not only an "exercise". It is known that the Euler-Lagrange equations and the canonical system are equivalent, via Legendre transform (at least, for regular applications). As regards the corresponding variational principles (the Hamilton principle for Lagrange function, resp. the Least Action principle for Hamilton function), they are not equivalent, acting on different admissible curves. The second implies the first one. A detailed discussion on the subject may be found in well known monograph of V. Arnold "Mathematical Methods of Classical Mechanics", in chapters 3 and 9.
Hi!
the derivation from the Lagrange equation of motion should not be too hard, you just need the definition of the physical action:
S = integral(L) between to abitrary times t1 and t2 (L is the Lagrangian and it holds that L = l (q(t), q`(t), t)), where q' is the time derivative of your generalised coordinate q.
Now you have to prove that every variation of S with respect to q vanishes, which means that: DS/Dq(t) = 0 (D in this case is the variation of the action integral).
As you are dealing with a curve of motion which has a determined starting and end point you can, you can "fix" the start and end point. In mathematical terms that means that a small displacement d vanishes at the start and the end, so d(t1) = d(t2) = 0 (that are your boundary conditions.
In order to get DS, you have to make a variation of your Lagrangian: L(q, q', t) --> L(q+d, q'+d') (time is invariant in the non-relativistic limes and the following integral is not explicitly dependent on t, because the t-dependency drops out of you Lagrangian!)
No you make a Taylor-expansion of your new L and you get from the first definition:
DS = int(L(q+d, q'+d')-L(q,q'))dt between t1 and t2 (note that you still have a time dependence in your generlised variables q and q')
the rest is easy: you directly calculate DS/Dq and use the fact that integral and derivation can be interchanged:
DS/Dq = int(DL(q+d, q'+d')/Dq - DL(q,q')/Dq)dt between t1 and t2.
as the small variation d does not depend on q, you get immdiately:
DS/Dq = int (DL/Dq - DL/Dq)dt = int(0)dt = (some constant C) between t1 and t2 = C - C = 0 q.e.d
if you have this done, the second is also not a big problem anymore: you just have to know that the Hamiltionian is connected with the Lagrangian via Legendre-Transformation (like Olivian Simionescu-Panait already noted).
Or, H (q_j, p_j, t) = sum_i (q_i', p_i) - L(q, q', t)
so, you just have to solve this for L and put the whole thing again into the definition of S and redo the integral stuff with the help of the canonical equations of Hamilton - that should solve your problem.
Thanx to all of u,,for sharing ur views,,,its now enough clear to me..........
- Badges
- Science topic
- Similar topics
- Mathematics
- Applied Mathematics