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How to determine correct Lagrangian?

  1. Sep 3, 2015 #1
    First, let me take as the definition of a Lagrangian the quantity that when put into the Euler Lagrange equations, it gives the correct equation of motion.

    It sounds like we need to know the equations of motion first. For example. the Lagrangian for a particle subject to a constant magnetic field. It is not your standard L=T-U.

    1. With this in mind, when I write down T-U for a system, how do I know if it is also the Lagrangian of a system?
    2. Also, this seems somewhat circular as to get the equations of motion we use the Lagrangian, but the Lagrangian is defined by the correct equations of motion. Can someone clarify this for me?
     
  2. jcsd
  3. Sep 3, 2015 #2

    Orodruin

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    This is a matter of how you write down models. Writing down a Lagrangian is the Lagrange mechanics equivalent of writing down the inertia and force relations in Newtonian mechanics. You can do this however you like (you could introduce a gravitational force proportional to the distance instead of the inverse square law), but ultimately you must test the model against experiments.

    What you are talking about here is just the proof of equivalence between Newtonian and Lagrange mechanics. You are showing that you can get the equations of motion from the variation of the action and that you can get the Lagrangian from the equations of motion. In itself, Lagrange mechanics does not require your Lagrangian to be of a particular form. The Lagrangian defines your model.
     
  4. Sep 3, 2015 #3
    So what the Lagrangian does is that it gives us an ability to create and test models for systems in a way that is easier than using Newton's law?
     
  5. Sep 4, 2015 #4

    Orodruin

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    If you are using Newton's law you need to model the forces. The Lagrangian approach has some advantages and the Newtonian (and also the Hamiltonian approach) has some. What is better suited really depends on what you want to do. Things such as symmetries and constants of motion are more apparent in the Lagrangian approach.
     
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