I am trying to understand Lagrangian mechanics. I am having trouble capturing all of the nuances in one gulp. I can see the equations, but not necessarily the semantics behind such equations. I would like to develop the skill to identify a system which is likely well modeled using Lagrangian mechanics without having to actually develop the equations (think of it as a screening step before trying to do the number crunching).

I see there is a path-independence rule, which makes sense due to the conservation laws which are so entwined with Lagrangian systems. Thus, if a system exhibits path-dependent behavior, it is not Lagrangian. The first half of my question is: is it necessary for a system to exhibit only path-independent behavior for it to be well modeled using Lagrangian mechanics.

The second half is what else is sufficient for a system to be well modeled using Lagrangian mechanics? Is path independence sufficient? Or is there another constraint that I am missing when I look at the actual equations.

My goal is to not only identify good candidates for systems worth exploring using Lagrangian mechanics, but also to identify subtleties which would make such modeling ineffective. Lagrangian mechanics appear so deeply entwined with physics, and my imagination is so used to thinking of things physically, that I don't want to mis-model something because I was oblivious to a necessary detail to make these mechanics work. I understand that, if we ignore waste-heat, systems cease to be well defined using Lagrangian mechanics, but beyond that I am less clear.


The only condition I think is important is that you can quantify the energy in different forms of the system-- this includes a clear definition of its coordinate system ($x$) and their derivetives ($\dot{x}$) in some maybe abstract basis.

Edit: replaced momentum with $\dot{x}$ as discussed.

| cite | improve this answer | |
  • $\begingroup$ You don't even need a clear definition of momentum; that gets defined for you by the Euler-Lagrange equations as $\frac{\partial L}{\partial\dot x}.$ This is one of the huge advantages of Lagrangians over the Hamiltonian approach to classical dynamics, which expects you to just magically know what momentum is conjugate to your generalized coordinate. $\endgroup$ – CR Drost Oct 6 '15 at 0:33
  • $\begingroup$ That's right. I really should replace momentum with $\dot{x}$ in my answer. $\endgroup$ – Xiaodong Qi Oct 6 '15 at 1:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.