Recent content by 13Treize

Dang, and here I thought I was being all careful and I would get the final result right. Well, I've done it, and I wouldn't have without help from you guys. Thank you very much to both of you.

I don't see where I'm wrong: \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0 L_1 = \frac{A}{2}\dot{q}^2 + \frac{d}{dt} \left(\frac{AB}{2}q^2e^{-\alpha t}} \right) - \frac{kq^2}{2} Using L_1, I get: A\ddot{q} -...

That's a good hint. Obviously, integrating the (d/dt) term results in a constant which doesn't affect optimization. So, the time-independent Lagrangian is "equivalent" to the time-dependent Lagrangian minus the total time derivative part? That seems a bit strange to me, because inserting the...

Err, ehm... I don't recall any property in particular. It does seem a bit peculiar (which is why I made it explicit), but I don't know what to do with it.

Thank you for your answer, gabbagabbahey. It was very helpful. Indeed, intuition can often lead to error, as it did in my problem above. :smile: I've used Hamilton's equation as you suggested, and I've made some progress. Starting with the original Hamiltonian and using \dot{q} =...

Homework Statement A system with only one degree of freedom is described by the following Hamiltonian: H = \frac{p^2}{2A} + Bqpe^{-\alpha t} + \frac{AB}{2}q^2 e^{-\alpha t}(\alpha + Be^{-\alpha t}) + \frac{kq^2}{2} with A, B, alpha and k constants. a) Find a Lagrangian...