the 2 lagrangians
Nov4-06, 03:37 PM
I just found Max's question on the 2 Lagrangians giving the same EOM's
(8 Mar 06).
[ Mod. note: The poster is probably referring to the message
Please quote when replying. -ik ]
As I didn't see a reaction , here is one .....
Using the identity (x+y)^2 = x^2 + y^2 - 2 x y
the 2 Lagrangians are connected by
F = m/2 (dx - dy)^2 -a/2 (x - y) ^2
where dX stands for x_dot.
It is easy to see that F is the difference between the energy
(hamiltonian) of the two formulations.
F = H - H' , hence F = d [ (H - H') t] and the text-book
property is satisfied.
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