This question already has an answer here:

This is more of a request for clarification of understanding and intuition rather than a question, but I hope people can help me with it.

I have learned calculus of variations and have subsequently used Lagrangian mechanics at a mathematical level, however, I'm unsure whether I've grasped the physical intuition behind the formulation correctly. In the following, I have written down how I "see" it and hope that people will be able to give me feedback on whether I have understood it correctly at all, or not.

Instead of determining all the forces that are acting on a particular system and then solving Newton's 2nd law to determine the physical path of the system, we instead take a different approach by considering the possible paths that a system could take between two different configurations and employ variational techniques to determine the actual physical path the system takes between these two points. This is advantageous over the Newtonian approach since we don't have to worry about all the different forces acting on the system and avoid the awkwardness of changing between Euclidean coordinates and other curvilinear coordinate systems encountered in solving the equations of motion in Newtonian mechanics, since such a variational approach is coordinate independent and so we can judiciously choose a set of "generalised" coordinates, $\lbrace q_{i}\rbrace$ that enable one to solve the problem as efficiently as possible.

To use such an approach we first need a function that characterises the dynamics of a physical system for every possible configuration that it could assume. Empirically, we know that the physical state of a system, at a given instant in time, is fully specified through knowledge of the positions $\lbrace q_{i}\rbrace$ and velocities, $\lbrace \dot{q}_{i}\rbrace$ of all the constituent components of the system at that instant. Thus, such a function, which we call the Lagrangian of the system, must depend on the state of the system at each point in its so-called configuration space. A priori, before considering any particular path of the system through configuration space, the positions and velocities defining the state of the system at a given instant can be chosen independently. This implies that the Lagrangian should be a function of both position and velocity, i.e. $\mathscr{L}=\mathscr{L}(q_{i},\dot{q}_{i})$.

Given this, we now wish to distinguish each path that the system can take through configuration space. We do so by assigning a number to each path. This is achieved by defining a functional, $S$, the action, which maps each given path, $q(t)=\left(q_{1},q_{2},\ldots ,q_{n}\right)$ that the system can take, to a real number. Since the Lagrangian evaluated along a given path characterises the dynamics of the system at each instant in time as the system moves along the path, we define the action in terms of this, i.e. $$S\left[q(t)\right]=\int_{t_{i}}^{t_{f}}\mathscr{L}(q(t),\dot{q}(t))dt$$ where $t_{i}$ and $t_{f}$ are the initial and final instants in time, enabling us to quantify the end points of the section of the path we are considering. Note also that we have now evaluated the Lagrangian along a particular path such that $\mathscr{L}(q(t),\dot{q}(t))$, and $q$ and $\dot{q}$ are no longer independent, but related by $\dot{q}(t)=\frac{d}{dt}q(t)$.

With this initial formalism in place, to find the true physical path of the system between two configurations (at two instants in time $t_{i}$ and $t_{f}$) we invoke a variational principle. This is the so-called principle of stationary action, motivated by empirical observations, it states that the physical path taken by a given system (through configuration space) is the one that extremises the action, $S$ of the system. Thus, we take a putative curve $\bar{q}(t)$ with fixed end points at $\bar{q}(t_{i})$ and $\bar{q}(t_{f})$, and make variations of the path in the neighbourhood of this curve between these two end points. This induces a variation in the action, and we require that this variation vanishes at first-order. We thus find that for $\bar{q}(t)$ to be the physical path taken by the system (i.e. an extremal path of $S$), it must satisfy the Euler-Lagrange equation $$\frac{\partial\mathscr{L}}{\partial q}-\frac{d}{dt}\left(\frac{\partial\mathscr{L}}{\partial \dot{q}}\right)=0$$ which is the equation of motion for the system.

Sorry this is long-winded, but I really want to check that I understand the concept correctly (at least at an intuitive level), so I have put down my thoughts on the subject.


marked as duplicate by Qmechanic classical-mechanics Mar 2 at 12:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @Qmechanic Thanks for the links. I was hoping that someone would be able to verify whether I have explained the intuitive "set-up" correctly or not, though?! $\endgroup$ – user35305 Apr 24 '16 at 11:45
  • $\begingroup$ Just a remark: The variational principle determines the equation of motion not the solutions of these equations. This is because when you fix the endpoints of the varied paths, generally speaking you can have none or many extremal paths joining those endpoints. Under some hypotheses on the structure of the Lagrangian, the boundary condition problem is well posed if the endpoints are sufficiently close to each other. $\endgroup$ – Valter Moretti Apr 24 '16 at 15:25
  • $\begingroup$ @Moretti Yes, I understand that, but one usual derives the Euler-Langrange equation by considering a putative extremal curve and then making arbitrary (up to fixed end points) variations around this curve. Then the condition is found that such a curve must satisfy the equations of motion if it is to be an extremal curve, right?! Is what I put on the whole correct though? $\endgroup$ – user35305 Apr 24 '16 at 15:55
  • $\begingroup$ Yes, it is correct as you wrote. $\endgroup$ – Valter Moretti Apr 24 '16 at 17:04