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I Euler-Lagrange equation even if the action isn't stationary?

  1. Mar 2, 2016 #1
    It seems like I could get the Euler-Lagrange equation for any function that allows symmetry of second derivatives even when the action is not stationary.

    Suppose ##L=L(q_1, q_2, ... , q_n, \dot{q_1}, \dot{q_2}, ... , \dot{q_n}, t)##, where all the ##q_i##'s and ##\dot{q_i}##'s are functions of ##t##. ##(\dot{x}## represents ##\frac{dx}{dt}##.##)##

    ##\frac{d}{dt}\frac{\partial L}{\partial\dot{q}}=\Sigma_i\Big(\frac{dq_i}{dt}\frac{\partial}{\partial q_i}\frac{\partial L}{\partial\dot{q}}+\frac{d\dot{q_i}}{dt}\frac{\partial}{\partial\dot{q_i}}\frac{\partial L}{\partial\dot{q}}\Big)+\frac{\partial}{\partial t}\frac{\partial L}{\partial\dot{q}}##

    ##=\Sigma_i\Big(\frac{dq_i}{dt}\frac{\partial}{\partial\dot{q}}\frac{\partial L}{\partial q_i}+\frac{d\dot{q_i}}{dt}\frac{\partial}{\partial\dot{q}}\frac{\partial L}{\partial\dot{q_i}}\Big)+\frac{\partial}{\partial\dot{q}}\frac{\partial L}{\partial t}##

    ##=\frac{\partial}{\partial\dot{q}}\frac{dL}{dt}##

    ##=\frac{\partial\dot{L}}{\partial\dot{q}}##

    ##=\frac{\partial L}{\partial q}##

    Therefore, ##\frac{d}{dt}\frac{\partial L}{\partial\dot{q}}-\frac{\partial L}{\partial q}=0##.

    What's wrong?

    EDIT: I think I found the mistake. ##\frac{\partial\dot{L}}{\partial\dot{q}}=\frac{\partial L}{\partial q}## when ##L=L(q, t)## but not so when ##L=L(q, \dot{q}, t)##.
     
    Last edited: Mar 2, 2016
  2. jcsd
  3. Mar 7, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Mar 8, 2016 #3

    vanhees71

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    No, your derivation was correct. The reason is that indeed concerning the partial derivatives used here you consider the ##q##, the ##\dot{q}##, and ##t## as independent variables, and the Euler-Lagrange equation are valid thus also if ##L## is explicitly time dependent.
     
  5. Mar 8, 2016 #4
    Consider ##L=q\dot{q}##.

    ##\frac{\partial L}{\partial q}=\dot{q}##

    whereas ##\dot{L}=q\ddot{q}+\dot{q}^2##

    ##\frac{\partial \dot{L}}{\partial \dot{q}}=2\dot{q}\neq\frac{\partial L}{\partial q}##
     
  6. Mar 8, 2016 #5

    vanhees71

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    Well, maybe I didn't follow your derivation carfully enough, but nevertheless the Euler-Lagrange equations are also correct for explicitly time-dependent Lagrangians. The reason is simply that in the Hamilton principle ##\delta t=0## by definition, i.e., you define the equations of motion as the stationary points of the action functional
    $$A[q]=\int \mathrm{d} t L(q,\dot{q},t) \; \Rightarrow \; \delta A[q]=\int \mathrm{d} t \left (\delta q \frac{\partial L}{\partial q} + \delta \dot{q} \frac{\partial L}{\partial \dot{q}} \right ).$$
    Now since ##\delta t=0## by definition you have
    $$\delta \dot{q}=\frac{\mathrm{d}}{\mathrm{d} t} \delta q$$
    and thus via integration by parts
    $$\delta A[q]=\int \mathrm{d} t \delta q \left (\frac{\partial L}{\partial q} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}} \right ).$$
    This must vanish for all ##\delta q## in order to make ##\delta A## stationary, which leads to the Euler-Lagrange equations
    $$\frac{\partial L}{\partial q} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}}=0.$$
    I didn't need the total time derivative of ##L## at all.

    Sorry for the confusion.
     
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