4
$\begingroup$

There has been extensive discussion in the literature and on this forum regarding the question of "whether a given system has a Lagrangian" (e.g. post1, post2, post3, and paper1, paper2).

The fundamental approach involves framing this question as the "inverse problem of the calculus of variations", wherein, given a set of second-order differential equations, one seeks the necessary and sufficient conditions for transforming them into the Euler-Lagrange equations. The results of this approach can be found on various platforms, including wiki.

However, I still harbor doubts about the underlying premise of this approach. Our understanding of a system may not necessarily begin with its differential equations; rather, we might only be aware of the components of the system (particles, interactions, etc., similar to the positive problem). From this perspective, there seems to be a gap between the question of "whether a given system has a Lagrangian" and the "inverse problem of the calculus of variations."

How can we address this gap?

$\endgroup$
1

1 Answer 1

5
$\begingroup$

Briefly speaking OP's question seems to be: Given a physical system, how to model it and find the equations of motion? That's a good but broad question whose answer presumably depends on the circumstances. However, ideally speaking$^1$, once the equations of motion are known, it then becomes the inverse problem for Lagrangian mechanics.


$^1$It should perhaps be mentioned that model building in modern theoretical physics (say e.g. beyond the standard model) often takes the opposite route: Start with an action principle respecting certain symmetries and properties, and then work out the equations of motion.

$\endgroup$
2
  • $\begingroup$ 1. Does the term "opposite route" here already imply the assumption that the system has a Lagrangian (and thus an action can be formulated)? 2. So, if we want to determine whether a system has a Lagrangian, we either need to start from its equations of motion by solving the inverse problem, or directly formulate the Lagrangian based on specific circumstances. Is this statement accurate? $\endgroup$
    – Luessiaw
    Feb 12 at 8:54
  • 1
    $\begingroup$ 1. Yes. 2. It could be a mix/hybrid/combination thereof. $\endgroup$
    – Qmechanic
    Feb 12 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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