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Euler-Lagrangian Equations

  1. Sep 24, 2006 #1
    Euler-Lagrangian Equations

    Let [tex]L=L(q_i(t), \dot{q}_i (t)[/tex] be a Lagrangian of a mechanical system, where [tex]q_i(t) [/tex] and [tex] \dot{q}_i (t) [/tex] are the short hand notations for [tex] q_1(t), q_2(t), . . . q_N(t)[/tex] and [tex] \dot{q}_1(t), \dot{q}_2(t), . . . \dot{q}_N(t)[/tex], respectively.

    I need to prove that if [tex] L ' =L+\frac{d \phi}{dt} [/tex], then [tex]L=L(q_i(t), \dot{q}_i (t)[/tex] and [tex]L ' =L(q_i(t), \dot{q}_i (t)[/tex] give the same equations of motion (Euler-Lagrangian equations).

    The Euler-Lagrangian equations is [tex] \frac{\partial L}{\partial x_i} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}_i }= 0[/tex]

    I really have no idea where to begin. Any suggestions would be much appreciated.
     
    Last edited: Sep 24, 2006
  2. jcsd
  3. Sep 24, 2006 #2

    quasar987

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    More generally, the equations of motions are the same up to an added time dependant function in the lagrangian. The proof is as direct as can be... just write the E-L equations for L and L' (and perform the derivatives on L').
     
  4. Sep 24, 2006 #3
    [tex] \frac{\partial L}{\partial x_i} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}_i }= 0[/tex]

    [tex] L ' =L+\frac{d \phi}{dt} [/tex]

    [tex] \frac{\partial L}{\partial x_i} + \frac{\partial }{\partial x_i} \frac{d \phi }{dt}- \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}_i } + \frac{\partial }{\partial \dot{x}_i } \frac{d \phi }{dt} \right)= 0[/tex]

    [tex]\frac{\partial }{\partial \dot{x}_i } \frac{d \phi }{dt} = 0 [/tex]

    [tex] \frac{\partial L}{\partial x_i} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}_i }= 0[/tex]

    something like that?
     
  5. Sep 25, 2006 #4

    dextercioby

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    Does [itex] \phi [/itex] depend on the depend on the generalized coodinates & velocities...?

    Daniel.
     
  6. Sep 25, 2006 #5
    I am told that [tex]\frac{d \phi }{dt}[/tex] is a "total time derivative"
     
  7. Sep 25, 2006 #6

    dextercioby

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    As i know it, the trick is that [itex] \phi=\phi\left(t,q^{i}\right) [/itex].

    Therefore

    [tex]\frac{d\phi}{dt} =\frac{\partial \phi}{\partial t}+\frac{\partial \phi}{\partial q^{i}} \dot{q}^{i} [/tex].

    Daniel.
     
  8. Sep 25, 2006 #7

    quasar987

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    Dexter is probably right on this, because I checked the definition of "total time derivative" on wiki and it says "a derivative which takes indirect dependencies into account"*

    This property of the equ. of motion you're trying to prove is mentioned in the "Landau & Lif****z" on mechanics in the first few pages of the book.


    *http://en.wikipedia.org/wiki/Total_time_derivative
     
  9. Sep 25, 2006 #8
    frankly, i need some real help. My class has an awful professor and using a very mathematical book that I cant seem to grasp. Before I try to attack this problem, could someone explain to me what exactly the Euler-Lagrangian does? What does it solve for?

    Also the notation [tex]L ' =L(q_i(t), \dot{q}_i (t)[/tex]

    Why is there a comma between q and q dot? What does the comma symbolize?

    Also, could someone please explain [tex]\delta[/tex] notation?

    What does it mean when [tex] \frac{dJ}{d \alpha} d \alpha = \delta J[/tex] ?

    My whole class seems confused as to what the lectures are about, so I have taken it upon myself to learn the material by myself.

    In class today, we were told that [tex] \phi=\phi\left(q_i,\dot{q}_i (t),t\right) [/tex]

    What is the significance of this statement dextercioby?

    I am sorry for my ignorance, but I am willing to learn.
     
  10. Sep 25, 2006 #9
    The Euler-Lagrange equations are the equations of motion for a system which motion is described by the generalized coordinates [itex] q_i [/itex]. They are, so to say, for generalized coordinates what the Newtonian equations are for cartesian coordinates.

    [itex] L(q_i(t), \dot q_i(t) ) [/itex] simply means that L ist a function of both q_i and the total time-derivatives [itex] \dot q_i [/itex]- of generalized velocities, so to say.

    edit: The Euler-Lagrangian equations can be derived from the "priciple of least action", which you can take as an axiom instead of Newton's equations. From the formulas I think you did this in your course. For an introduction what it's all about I suggest Feynman's Lectures on physics, Volume 2, Chapter 19, after this f.e. Goldstein for a more detailed explanation.
     
    Last edited: Sep 25, 2006
  11. Sep 25, 2006 #10
    what about [tex]\delta[/tex] notation? and what dextercioby wrote?
     
  12. Sep 25, 2006 #11
    If [itex] \Phi(q_i (t), \dot q_i(t),t ) [/itex] the total time derivative becomes
    [itex] \frac{d}{dt} \Phi = \sum \limits_i \frac{\partial \Phi}{\partial q_i} \dot q_i + \sum \limits_i \frac{\partial \Phi}{\partial \dot q_i} \ddot q_i + \frac{\partial \Phi}{\partial t}[/itex] (dexter didn't write the sum out but used Einstein convention - sum is taken over i, since i is double).
    [itex] \delta [/itex] means the "variation".

    For example, [itex] \delta \int \limits L(q_i, \dot q_i, t) dt =0[/itex] means that the integral doesn't change when you change the "pathfunction" q_i over which is integrated a little (just as a function has a maximum or minimum and doesn't change "nearby" when the derivative is zero). This is the "Principle of least action", from which the Euler-Lagrange-equations can be derived.
     
    Last edited: Sep 25, 2006
  13. Sep 26, 2006 #12
    was what I did here okay?
     
  14. Sep 28, 2006 #13
    Hi....

    I don't understand where you got [tex] \left( \frac{\partial L}{\partial \dot{x}_i } + \frac{\partial }{\partial \dot{x}_i } \frac{d \phi }{dt} \right)= 0[/tex]. It is supposed to be equal to L right? How is it equal to L?
     
  15. Sep 28, 2006 #14
    sorry, I just meant
    [tex] \left( \frac{\partial L}{\partial \dot{x}_i } + \frac{\partial }{\partial \dot{x}_i } \frac{d \phi }{dt} \right)[/tex],
    not the equal to zero part.
     
    Last edited: Sep 28, 2006
  16. Sep 29, 2006 #15

    dextercioby

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    It doesn't work if [itex] \phi [/itex] depends on the generalized velocities.

    Daniel.
     
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